{ "catalog_hash": "5269b9f2521e3bf14db293fea34895f6e44a387c2479c164407b4d16dd9a2247", "items": [ { "assumptions": [ "v is an arbitrary real sequence; only its first k+1 entries are constrained", "derivatives are ordinary real polynomial derivatives" ], "context_hash": "44f87c243bc4e27604d9c9483b9ad14d0a84b7313d562f865ed6eb6227fbb938", "conventions": [ "the center a is fixed and the polynomial is expressed in the shifted basis (x-a)" ], "core_id": "qcb-core-sp-101", "difficulty": { "level": "intermediate", "rationale": "Requires linking the ordinary iterated derivative to Taylor coefficients and tracking the factorial normalization, not just quoting a derivative rule." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-jet-interpolation", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-43433f837de106948066", "key": "qcb-core-sp-101-corrected-v2", "primary_sources": [], "question": "Fix a real point a and a natural number k, and prescribe target values v(0), v(1), ..., v(k). Define a degree-at-most-k real polynomial P by taking the local Taylor polynomial centred at a whose coefficient of (x-a)^j is v(j)/j!. Prove that for every j with 0 <= j <= k the ordinary j-th derivative of P evaluated at a equals exactly v(j). You may use the fact that the j-th derivative of a polynomial at a point equals j! times the degree-j Taylor coefficient of the polynomial about that point.", "reference_answer": "Write P = sum over i in {0,...,k} of (v(i)/i!) * (x-a)^i. The key bridge is that for any polynomial Q, (d^j/dx^j Q)(a) = j! * c_j, where c_j is the coefficient of (x-a)^j in the Taylor expansion of Q about a; equivalently c_j is the degree-j coefficient of the shifted polynomial Q(x+a). Step 1: the Taylor coefficients of P about a are read off directly: the coefficient of (x-a)^j is v(j)/j! for j <= k and 0 for j > k, because P is already written in the (x-a) basis. Step 2: apply the bridge at order j <= k: (d^j P/dx^j)(a) = j! * (coefficient of (x-a)^j) = j! * (v(j)/j!). Step 3: since j! is a nonzero natural number cast into the reals, j! * (v(j)/j!) = v(j) exactly. Hence every requested jet value is reproduced exactly. Each summand has degree at most k, so the finite sum has degree at most k. This is only an upper bound on the degree of the polynomial for the prescribed data; for example all-zero data gives the zero polynomial. No minimality claim or numerical approximation is needed.", "rubric": [ { "criterion": "Writes P in the (x-a) monomial basis with coefficient v(j)/j! and identifies these as the Taylor coefficients about a.", "id": "r1", "weight": 1 }, { "criterion": "Invokes the derivative/Taylor-coefficient bridge: j-th derivative at a equals j! times the degree-j Taylor coefficient.", "id": "r2", "weight": 1 }, { "criterion": "Cancels j! against the 1/j! normalization to conclude the j-th derivative at a equals v(j) for all j <= k.", "id": "r3", "weight": 1 }, { "criterion": "Notes the degree bound k and that the identities are exact (no numerical approximation).", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The j-th derivative at a of any polynomial equals j! times its degree-j Taylor coefficient about a.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.eval_iterate_derivative" ] }, { "claim": "The polynomial written in the (x-a) basis has degree-j coefficient v(j)/j! for j<=k.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.jetPolynomial" ] }, { "claim": "Multiplying the coefficient by j! cancels the normalization, giving exactly v(j).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.jetPolynomial_jet" ] } ], "source_item_digest": "b8d06ee21a9000b00a45e7a559e8ec84e1a3821bc80cf948e134426bf5687741", "source_refs": [ { "anchor": "lines 27-49: def jetPolynomial (a : R) (k : N) (v : N -> R) := taylor (-a) (sum_{j in range (k+1)} monomial j (v j / j!)); theorem eval_iterate_derivative (p) (a) (j) : (derivative^[j] p).eval a = j! * (taylor a p).coeff j; theorem jetPolynomial_jet (a k j v) (hj : j <= k) : (derivative^[j] (jetPolynomial a k v)).eval a = v j", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermitePolynomial.lean", "sha256": "f8a3e1d5f29aed1228b6ede3577a9af1e76301559172ee67a266fcb0d56fee1e" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.jetPolynomial", "declaration:QuantumBlockEncoding.HermitePolynomial.jetPolynomial_jet", "declaration:QuantumBlockEncoding.HermitePolynomial.eval_iterate_derivative" ], "task_hash": "331d0222ebb11ab28e804f84093e5393a8c762fa8d161bec37c86238cc40cbc7", "task_kind": "proof", "topic": "structured_preparation" }, { "assumptions": [ "The eight per-counter polynomial budgets are as stated", "Totalling means summing the eight counters" ], "context_hash": "780ff88d57dab968e9613fe2ee265df7bc0f8a33baa4aeeafe497d13eef609bb", "conventions": [ "Exact-real word-RAM operation model with no bit-level charge; not a finite-bit runtime", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-205", "difficulty": { "level": "intermediate", "rationale": "Careful polynomial summation of eight counters with an explicit model caveat." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-205", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-dee10ebc75a67678466d", "key": "qcb-core-la-205-corrected-v2", "primary_sources": [], "question": "A full stored rectangular compilation of an N-by-M matrix produces a reduced residual, a chronological list of rotation steps, and a fully accumulated N-by-N transform. In the eight-counter exact-real model its per-counter cost has been bounded by an explicit polynomial budget whose values on the eight counters are: field = N*M*(6*M+6*N+9); square root = N*M; inverse-cosine = N*M; other trigonometric = 4*N*M; comparison = 2*N*M + M + N*N; read = N*M*(10*M+14*N+10) + 2*N*N + 2*N + 1; write = N*M*(6*M+10*N+1) + 2*N*N + 2*N; emitted records = N*M. Prove that the sum of all eight counters is bounded by N*M*(22*M+30*N+29) + 5*N*N + 4*N + M + 1, and explain why this single scalar is an operation count in an extended exact-real model and not a finite-bit running time.", "reference_answer": "Add the eight per-counter polynomials. Group the N*M*(...) coefficients: from field 6*M+6*N+9, from read 10*M+14*N+10, from write 6*M+10*N+1, giving 22*M+30*N+20 inside the N*M factor; the square-root, inverse-cosine, emitted and part of the comparison counters each contribute an extra N*M (that is 1+1+1 from sqrt/angle/emit, 4 from the other trig, and 2 from the comparison), adding 9 more inside the same N*M factor and yielding N*M*(22*M+30*N+29). The remaining non-N*M terms are: comparison M + N*N, read 2*N*N + 2*N + 1, write 2*N*N + 2*N; summing the N*N terms gives 5*N*N, the linear-in-N terms give 4*N, the M term stays M, and the constants give 1. Hence the grand total is N*M*(22*M+30*N+29) + 5*N*N + 4*N + M + 1. The quantity counts exact-real primitive operations (field arithmetic, exact square roots and trigonometric evaluations, exact comparisons) and stored-word traffic in a word-RAM style model; it charges no bit-level cost for representing or rounding reals, so it is explicitly not a finite-bit running-time bound.", "rubric": [ { "criterion": "Correctly collect the N*M coefficient to 22*M+30*N+29 across all eight counters", "id": "c1", "weight": 1 }, { "criterion": "Sum the lower-order terms to 5*N*N + 4*N + M + 1", "id": "c2", "weight": 1 }, { "criterion": "Explain the exact-real word-model reading and that it is not finite-bit complexity", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The N*M-scaled parts of all eight counters collect to 22*M+30*N+29.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.polynomialBudget" ] }, { "claim": "The remaining quadratic, linear and constant terms sum to 5*N*N+4*N+M+1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.total" ] }, { "claim": "The scalar total is an exact-real operation count, not finite-bit complexity.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.compile_total_cost_le" ] } ], "source_item_digest": "3e52d9ef18efc62d9cff3b7087e30bb8b30494040dba6f8b2292182c2274badc", "source_refs": [ { "anchor": "polynomialBudget213-221; total236-238; compile_total_cost_le240-253", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredRectangularGivens.lean", "sha256": "f8638a6d287b3b984fa530b13eeb18e3992904543236bd0756eda41fa6490d98" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.polynomialBudget", "declaration:QuantumBlockEncoding.StoredRectangularGivens.total", "declaration:QuantumBlockEncoding.StoredRectangularGivens.compile" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.compile_total_cost_le" ], "task_hash": "9b25a6f4e4f97a7f62960e57b2731bfca9d7afb61ec3c7918d165e9899ce1de7", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "A is diagonal with ||A||_max <= 1", "O_A acts as the single-index amplitude oracle", "Positive subnormalization scales only; the scaling construction A/r is used only for r=||A||>0, with A=0 handled separately." ], "context_hash": "035e87b808cb66b3300bf5e517d65e33bc69fae0e82e1dc2fc1adbeeee44b645", "conventions": [ "BE_{alpha,m}(A) denotes an (alpha,m)-block-encoding of A", "The ancilla success flag is the first qubit returning |0>" ], "core_id": "qcb-core-plan-210", "difficulty": { "level": "foundational", "rationale": "A short exact calculation plus two conceptual observations (permutation triviality and the carefully-bounded sense in which alpha = 1 is optimal); requires not overclaiming global optimality." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-sparse-block-encoding", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-3f568dad9e98559c03dc", "key": "qcb-core-plan-210-diagonal-one-sparse-block-encoding-identity-corrected-v2", "primary_sources": [ { "anchors": "Example 9.17; Eq. (9.72)-(9.73); U_A in BE_{1,1}(A)", "edition": "29 April 2026", "inspection": "full bounded primary text read", "pages": "155", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text or external relicensing", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Let A be a diagonal N x N matrix (a 1-sparse matrix whose only nonzero entries are A_ii), with ||A||_max <= 1. Using the simplified amplitude oracle O_A |0>|i> = (A_ii |0> + sqrt(1 - |A_ii|^2) |1>) |i>, set U_A := O_A. Prove by direct evaluation of the matrix element <0||j> that U_A is a (1,1)-block-encoding of A, i.e. <0||j> = A_ii delta_ij, so THIS construction has subnormalization factor alpha = 1 with a single ancilla qubit. Explain why no index-permutation oracle is needed here, in contrast to the general 1-sparse case. Then discuss precisely in what sense alpha = 1 is or is not optimal: state the general lower bound any block encoding must obey, and say whether alpha = 1 is a global optimum for every diagonal A with ||A||_max <= 1 or only in a specific case.", "reference_answer": "Apply U_A = O_A to |0>|j>: U_A |0>|j> = (A_jj |0> + sqrt(1-|A_jj|^2)|1>)|j>. Now take the overlap with <0||j> = A_jj <0|0> + sqrt(1-|A_jj|^2) <0|1> . The second term vanishes because <0|1> = 0. The first term is A_jj = A_jj delta_ij. When i = j this equals A_jj = A_ii; when i != j it is 0. Hence <0||j> = A_ii delta_ij, which is precisely the (i,j) entry of the diagonal matrix A. By the definition of block encoding, (<0| tensor I) U_A (|0> tensor I) = A exactly, with one ancilla qubit (m = 1) and subnormalization alpha = 1, so U_A is in BE_{1,1}(A). No permutation oracle is needed because for a DIAGONAL matrix the row index equals the column index: the single nonzero entry in column j sits at row c(j) = j, i.e. the sparsity permutation c is the identity. In the general 1-sparse case the unique nonzero entry of column j lives at row c(j) != j, so one must apply an index permutation O_c |j> = |c(j)> to move the amplitude-carrying basis state from |j> to |c(j)>; for the diagonal case c = id, so O_c = I and U_A = O_A alone suffices. Optimality -- stated carefully. The general lower bound is that any (alpha,m)-block-encoding of A must satisfy alpha >= ||A|| (operator norm), because the encoded top-left block equals A/alpha and a sub-block of a unitary has operator norm at most 1, forcing ||A||/alpha <= 1. For a diagonal A, ||A|| = max_i |A_ii|. The present construction FIXES alpha = 1 regardless of the entries; this value saturates the lower bound (hence is globally optimal) ONLY in the case ||A|| = max_i |A_ii| = 1, e.g. when some |A_ii| = 1. If instead 0 < max_i |A_ii| = r < 1 strictly, then alpha = 1 is a VALID but NOT optimal subnormalization: the lower bound only demands alpha >= r, and a smaller alpha = r is mathematically achievable -- for example by encoding A/r (whose max entry has modulus 1) with this very construction to obtain an (r,1)-block-encoding of A. Thus alpha = 1 is a property of THIS fixed construction, globally optimal exactly when ||A|| = 1, and for ||A|| < 1 merely a valid upper bound rather than the minimum. The honest statement: the construction yields alpha = 1; it meets the alpha >= ||A|| floor iff ||A|| = 1, and is not subnormalization-optimal when ||A|| < 1. At A=0, every strictly positive scale is feasible (a zero clean block), but scale0 is not an admissible positive normalization; the infimum0 is not an attained positive minimum, and division by the zero norm is never used.", "rubric": [ { "criterion": "Applies U_A = O_A to |0>|j> and expands the overlap <0||j>", "id": "r1", "weight": 1 }, { "criterion": "Uses <0|1> = 0 to kill the second term and = delta_ij to obtain A_ii delta_ij, concluding U_A in BE_{1,1}(A) (this construction's alpha = 1)", "id": "r2", "weight": 1 }, { "criterion": "Explains the sparsity permutation is the identity for a diagonal matrix (c(j)=j), so no O_c is needed, unlike the general 1-sparse case", "id": "r3", "weight": 1 }, { "criterion": "States alpha >= ||A|| and correctly limits optimality: alpha = 1 is globally optimal ONLY when ||A|| = max_i|A_ii| = 1; for 0 < ||A|| < 1 it is valid but NOT optimal (a smaller alpha = ||A|| when ||A||>0 is achievable, e.g. by encoding A/||A||)", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "U_A|0>|j> = (A_jj|0> + sqrt(1-|A_jj|^2)|1>)|j>.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "<0||j> = A_jj <0|0> + sqrt(1-|A_jj|^2)<0|1>, and <0|1>=0 kills the second term.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": " = delta_ij gives <0||j> = A_ii delta_ij, so U_A in BE_{1,1}(A) with this construction's alpha = 1.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "For diagonal A the sparsity permutation is c(j)=j, so O_c = I and no permutation oracle is required.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "Any block encoding needs alpha >= ||A||; alpha = 1 saturates this iff ||A|| = 1, and for 0 < ||A|| < 1 a smaller alpha = ||A|| for nonzero A is achievable, so alpha = 1 is not globally optimal there.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "1f65e62ee73cdfa4f023076fc698fed90d81a1dbaf24524784aa24219df9cfe4", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:block-encoding", "concept:matrix-analysis", "concept:operator-theory" ], "task_hash": "baef5b2da813b58de9a27ced74535ef26cc3fff1aa2f0d2e19f00a55698bf1b8", "task_kind": "proof", "topic": "planned_scientific_algorithms" }, { "assumptions": [ "Finite-dimensional real matrices; two selected rows and one column", "The split angle reproduces the pair up to the nonnegative length r" ], "context_hash": "cd592959332ac9cb259a06affebdeb8dac7d00065cbc911ed94cd224114acfda", "conventions": [ "RY half-angle convention: a plane rotation uses cos(t/2) and sin(t/2)", "Elimination angle is the negative of the exact preparation angle; zero pivots get angle 0", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-101", "difficulty": { "level": "intermediate", "rationale": "Multi-step exact trigonometric cancellation plus a totality boundary case." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-givens-pair-elimination", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8bdfbacf5d982d096ac2", "key": "qcb-core-la-101", "primary_sources": [], "question": "A planar rotation acts on two rows by the half-angle convention: the first row becomes cos(t/2)*x - sin(t/2)*y and the second becomes sin(t/2)*x + cos(t/2)*y, where (x,y) are the two entries in some column. Choosing the elimination angle to be the negative of the exact angle whose cosine and sine (times the pair length r=sqrt(x^2+y^2)) reproduce x and y, prove that this rotation sends the pair (x,y) to (r,0). Then explain why the special pair x=y=0 needs no separate division-by-zero case.", "reference_answer": "By construction the exact split angle a satisfies cos(a/2)*r = x and sin(a/2)*r = y. The elimination angle is -a, so cos(-a/2)=cos(a/2) and sin(-a/2)=-sin(a/2). The transformed first entry is cos(a/2)*x + sin(a/2)*y = cos(a/2)*(cos(a/2)*r) + sin(a/2)*(sin(a/2)*r) = (cos^2(a/2)+sin^2(a/2))*r = r by the Pythagorean identity. The transformed second entry is -sin(a/2)*x + cos(a/2)*y = -sin(a/2)*cos(a/2)*r + cos(a/2)*sin(a/2)*r = 0. Hence the pair maps to (r,0) exactly, with no approximation. For x=y=0 the length r is 0 and the chosen elimination angle is defined to be 0, so the rotation is the identity and the already-zero pair stays (0,0); no reciprocal of a vanishing norm is ever formed, so the construction is total. This is an exact-real elimination identity, not a floating-point or arithmetic-cost claim.", "rubric": [ { "criterion": "Use cos(a/2)*r=x, sin(a/2)*r=y and the Pythagorean identity to get first entry r", "id": "c1", "weight": 1 }, { "criterion": "Show the second entry cancels to exactly 0", "id": "c2", "weight": 1 }, { "criterion": "Explain the zero pair is handled by angle 0 with no division", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Record the defining identities cos(a/2)*r=x and sin(a/2)*r=y of the split angle.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.splitAngle_real_firstColumn" ] }, { "claim": "Substitute into the first transformed entry and collapse via cos^2+sin^2=1 to get r.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.eliminate_pair" ] }, { "claim": "Substitute into the second transformed entry and cancel to 0.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.eliminate_pair" ] }, { "claim": "Handle the zero pair by the defined angle 0, giving the identity with no division.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.eliminationAngle_zero" ] } ], "source_item_digest": "5d349b058b56a55a91b036f2d61d63d6c9cdc6be9ba99988d6e0d0ecea65486e", "source_refs": [ { "anchor": "splitAngle_real_firstColumn84-97; eliminationAngle99-99; eliminationAngle_zero101-102; eliminate_pair104-119", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.eliminationAngle", "declaration:QuantumBlockEncoding.AdjacentGivens.eliminationAngle_zero", "declaration:QuantumBlockEncoding.AdjacentGivens.splitAngle_real_firstColumn" ], "target_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.eliminate_pair" ], "task_hash": "f3142eb714ed991a640119ded4d070769103d75dc361ede1fd526b085f65fb14", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite-dimensional real square matrix; two distinct selected rows", "Determinant is multilinear and alternating in rows" ], "context_hash": "e9ae52c1a7af88ee1dde0c7721ab7ce001f1a6ba441f7056050f5cf59508a4f7", "conventions": [ "RY half-angle convention for the rotation coefficients", "A single row transposition contributes sign -1", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-102", "difficulty": { "level": "intermediate", "rationale": "Multilinear determinant expansion with a sign bookkeeping and a trigonometric specialization." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-givens-determinant", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-77d2592075419412d878", "key": "qcb-core-la-102", "primary_sources": [], "question": "Replacing two distinct rows i and j of a square matrix by the linear mixtures (a*row_i + b*row_j) and (c*row_i + d*row_j) multiplies the determinant by (a*d - b*c). Use this to prove that a plane rotation acting on rows i and j by the half-angle convention leaves the determinant unchanged.", "reference_answer": "Expanding the determinant linearly in the two updated rows produces four terms. The two terms that reuse the same original row twice (coefficients a*c on row_i in both slots, and b*d on row_j in both slots) are determinants with two equal rows, hence 0. The surviving cross terms are a*d times the original determinant and b*c times the determinant with rows i and j swapped; the swap introduces one transposition, so its sign is -1 and that determinant equals -det. Collecting gives (a*d)*det + (b*c)*(-det) = (a*d - b*c)*det. For a plane rotation the mixing coefficients are a=cos(t/2), b=-sin(t/2), c=sin(t/2), d=cos(t/2), so a*d - b*c = cos^2(t/2) + sin^2(t/2) = 1. Therefore the rotation preserves the determinant exactly. This is an exact algebraic identity about the determinant, not an arithmetic evaluation cost.", "rubric": [ { "criterion": "Kill the repeated-row terms via equal-row determinant 0", "id": "c1", "weight": 1 }, { "criterion": "Identify the swap term sign as -1 and assemble a*d - b*c", "id": "c2", "weight": 1 }, { "criterion": "Specialize rotation coefficients to get factor cos^2+sin^2=1", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Expand the determinant multilinearly in both replaced rows into four terms.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.det_two_row_mix" ] }, { "claim": "Discard the two equal-row terms as zero determinants.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.det_two_row_mix" ] }, { "claim": "Evaluate the swap term with sign -1 and collect the factor a*d - b*c.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.det_two_row_mix" ] }, { "claim": "Substitute the rotation coefficients to obtain the factor cos^2+sin^2=1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.rotateRows_det" ] } ], "source_item_digest": "470ce4707a5d7cebab5a056b624cf70eb6448b2fdee43c594b57cf19d273943f", "source_refs": [ { "anchor": "det_two_row_mix341-365; rotateRows_det367-378; planeMatrix_det381-383", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.planeMatrix_det" ], "target_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.det_two_row_mix", "declaration:QuantumBlockEncoding.AdjacentGivens.rotateRows_det" ], "task_hash": "138320c5e1dfcc843ef7e537a688fc9b48e63e9950f9083058097aa58ca8dda6", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite-dimensional real matrices; each step acts on a distinct ordered row pair", "Chronological application: the first listed rotation acts first" ], "context_hash": "90927953c61f5a59855ba30f5030f48ec17de7c5934647358dd76cc2be1d1ef4", "conventions": [ "RY half-angle convention; the inverse rotation negates the angle", "Undo list is the reversed list with every angle negated", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-103", "difficulty": { "level": "intermediate", "rationale": "Structural list induction resting on an exact two-row inverse identity." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-rotation-reversibility", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-37d9f0eff8e7a9de552b", "key": "qcb-core-la-103", "primary_sources": [], "question": "A sequence of plane rotations is applied to a matrix in chronological order (the first listed rotation acts first). Prove that reversing the list and negating every rotation angle undoes the whole sequence exactly, returning the original matrix.", "reference_answer": "First, a single plane rotation by angle t followed by the plane rotation on the same ordered pair by -t is the identity: on the chosen two rows the composed linear map has matrix [[cos,-sin],[sin,cos]] times [[cos,sin],[-sin,cos]], which equals the identity because cos^2+sin^2=1 and the cross terms cancel; all other rows are untouched. This gives the single-step inverse law. Now argue by induction on the applied list. For the empty list there is nothing to undo. For a list (step :: rest), applying it means first acting by step then by rest; the proposed undo is (reverse rest ++ [step^{-1}]) with every angle negated, i.e. first undo rest then undo step. By the induction hypothesis undoing rest returns the matrix to the state right after step, and then the single-step inverse law removes step, recovering the original matrix. Because each step is distinct on an adjacent ordered pair, the inverse law applies at every stage. The result is an exact recovery statement, not an assertion about numerical conditioning.", "rubric": [ { "criterion": "Prove the single-step inverse law plane(t) then plane(-t) is identity", "id": "c1", "weight": 1 }, { "criterion": "Set up induction on the list with correct reversed/negated undo order", "id": "c2", "weight": 1 }, { "criterion": "Combine induction hypothesis and single-step law to recover the matrix", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Establish the two-row inverse law plane(-t)*plane(t)=1 via cos^2+sin^2=1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.planeMatrix_inverse" ] }, { "claim": "Rewrite the reversed-negated undo of (step::rest) as undo-rest then undo-step.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps_append" ] }, { "claim": "Apply the induction hypothesis to cancel rest.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps_reverse_inverse" ] }, { "claim": "Apply the single-step inverse law to cancel the leading step and recover the matrix.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.Step.inverse" ] } ], "source_item_digest": "169363b29832bc0b484a2c0a455b9b07a8fce233631ec20ffbd7868be52abb66", "source_refs": [ { "anchor": "planeMatrix_inverse57-60; Step.inverse197-198; applySteps_append200-205; applySteps_reverse_inverse207-220", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.planeMatrix_inverse", "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps_append", "declaration:QuantumBlockEncoding.AdjacentGivens.Step.inverse" ], "target_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps_reverse_inverse" ], "task_hash": "edb43405ae7aa9da95bc424017a4863e4513af2696648cb4cb4a542ca80f4902", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite-dimensional real matrix; the swept window stays within the row range", "Each elimination touches only one adjacent ordered row pair" ], "context_hash": "64a4faefeefc97a202d7ed1b277ffffdf2b87a84a2c63a15561ac56cb5551ba6", "conventions": [ "Bottom-up sweep: lower rows are folded into the next higher row", "RY half-angle convention; a zero pair produces an identity step", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-104", "difficulty": { "level": "intermediate", "rationale": "Nested induction coordinating a zeroing invariant with a window-outside invariant." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-sweep-zeroing-invariant", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-9a8db20121cb23001ada", "key": "qcb-core-la-104", "primary_sources": [], "question": "A column is cleared by a bottom-up sweep: starting from a low index, adjacent rows are rotated upward one pair at a time so that each lower entry is folded into the one above it. Prove that after sweeping the window of rows from lo to lo+count, every entry of that column strictly below the top row lo is zero, while any row lying outside the window (above lo or below lo+count) is left exactly as it was.", "reference_answer": "Argue by induction on the number of adjacent eliminations. The outside-invariance is proved first: a single adjacent elimination on the pair (r,r+1) changes only rows r and r+1, so any row whose index is below lo or above lo+count is never one of these pairs across the whole sweep, and its value is unchanged; this is a direct per-step bookkeeping over the recursion. For the zeroing claim, consider a row index v with lo < v <= lo+count. In the inductive step the sweep performs the elimination that folds row v into v-1 (clearing the entry at v) and then recurses on the remaining upper eliminations, which only touch rows at or above v-1 and hence, by outside-invariance for the smaller window, do not disturb the freshly created zero at v. The base pivot pair uses the exact elimination identity, which places 0 in the lower slot. Thus every strictly-below entry ends at 0 while the frame outside the window is untouched. This describes exact-real elimination semantics, not an iteration-count cost.", "rubric": [ { "criterion": "Prove rows outside the window are unchanged by every step", "id": "c1", "weight": 1 }, { "criterion": "Use the pair elimination to zero the swept entry, by induction over the window", "id": "c2", "weight": 1 }, { "criterion": "Argue later eliminations do not disturb an already-created zero", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Show one adjacent elimination fixes every row other than the touched pair.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.eliminateEntry_unchanged" ] }, { "claim": "Propagate this to full outside-invariance of the sweep by induction.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_outside" ] }, { "claim": "Zero the current swept entry using the exact pair-elimination identity.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.eliminateEntry_pivot" ] }, { "claim": "Use outside-invariance on the smaller window to protect the created zero.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_zeroed" ] } ], "source_item_digest": "d116f4f01eb9ee69ff710e3d88155094fba848cfbe1475bc166d29434e5c98ea", "source_refs": [ { "anchor": "eliminateEntry_pivot145-149; eliminateEntry_unchanged151-154; columnSweep_outside265-281; columnSweep_zeroed283-303", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_outside", "declaration:QuantumBlockEncoding.AdjacentGivens.eliminateEntry_pivot", "declaration:QuantumBlockEncoding.AdjacentGivens.eliminateEntry_unchanged" ], "target_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_zeroed" ], "task_hash": "259766fa07dfb4753018c891df58ee503e55446ceb0600666a53a686e44d619d", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite-dimensional real matrix with B^T B = I", "The named column is supported only on its diagonal row" ], "context_hash": "99f747a1446fe482b76bb953b290233f28d93a1ee86a0ed6f55a1b855717f58b", "conventions": [ "Pivots produced by elimination equal a nonnegative pair length", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-105", "difficulty": { "level": "intermediate", "rationale": "Gram-entry collapse followed by a sign argument tying square-one to a nonnegativity bound." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-orthonormal-pivot-sign", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-7d91bebff0bb25ec6f4f", "key": "qcb-core-la-105", "primary_sources": [], "question": "Let B have orthonormal columns (B^T B = I) and suppose one column c of B is supported only on its diagonal row, i.e. every entry of column c off that row is zero. Prove that the diagonal entry B[c,c] squares to 1. If in addition the sweep guarantees that this pivot is nonnegative, prove that it equals exactly +1.", "reference_answer": "The (c,c) entry of B^T B is the inner product of column c with itself, which equals the sum over all rows of B[row,c]^2. By the support hypothesis only the diagonal row survives, so this sum collapses to B[c,c]^2. Orthonormality gives the (c,c) entry of B^T B equal to 1, hence B[c,c]^2 = 1. A real number whose square is 1 is +1 or -1; combined with the sweep guarantee 0 <= B[c,c], the only possibility is +1. The nonnegativity itself comes from the elimination placing the pivot equal to a pair length sqrt(x^2+y^2) >= 0, propagated through the recursion. This settles the pivot sign without any normalization or square-root reciprocal, purely from orthonormality and a sign bound.", "rubric": [ { "criterion": "Collapse the diagonal Gram entry to a single squared term using support", "id": "c1", "weight": 1 }, { "criterion": "Conclude the square equals 1 from orthonormality", "id": "c2", "weight": 1 }, { "criterion": "Use the nonnegative-pivot guarantee to pin the value to +1", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Write the diagonal Gram entry as the sum of squared column entries.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.orthogonal_supported_column_sq" ] }, { "claim": "Collapse the sum to the single diagonal term via the support hypothesis.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.orthogonal_supported_column_sq" ] }, { "claim": "Deduce the pivot squares to 1 from orthonormality.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.orthogonal_supported_column_sq" ] }, { "claim": "Combine the square-one condition with the nonnegativity bound to force +1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_pivot_nonneg" ] } ], "source_item_digest": "059fbedeca9a54a2029f3769ccac4ed3d567bda7d0899acd6cb3cc9cc3fb0c2d", "source_refs": [ { "anchor": "columnSweep_pivot_nonneg422-439; orthogonal_supported_column_sq441-455; columnSweep_prefix_succ457-483", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_pivot_nonneg", "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_prefix_succ" ], "target_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.orthogonal_supported_column_sq" ], "task_hash": "8e8e9e056cc93392963e1e601721cf41bea5d5e4d6d8964b47c5830111c386d9", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite-dimensional real matrix with A^T A = I and det A = 1", "First N-1 columns are already the standard identity prefix" ], "context_hash": "29b62afebf9c3f917275ecca6f8612f5c3a3b0647f91e2f9e8292558c89b7191", "conventions": [ "The determinant of a diagonal matrix is the product of its diagonal entries", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-106", "difficulty": { "level": "advanced", "rationale": "Combines an orthogonality off-diagonal argument with a determinant orientation lock and an explicit alternative." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-so-identity-termination", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-88cbd0db23c2f46f217d", "key": "qcb-core-la-106", "primary_sources": [], "question": "Suppose a real NxN matrix A is orthogonal (A^T A = I), has determinant 1, and its first N-1 columns have already been reduced to the standard identity prefix (column c equals the c-th coordinate vector for every c < N-1 index, where N-1 denotes the last index k with k+1=N). Prove that A must equal the identity matrix. Explain precisely where the determinant hypothesis is used.", "reference_answer": "First show A is diagonal. For any column, if it is one of the fixed early columns it is already a coordinate vector, so its off-diagonal entries vanish. For the single last column, orthogonality against each fixed early column forces its entries in the early rows to be zero: the inner product of the last column with a fixed coordinate column equals that one surviving entry, which must be 0. The remaining possible nonzero off-diagonal entries lie in the last column at the last row region, but the symmetric orthogonality argument eliminates those too, leaving A diagonal. On the diagonal, every early entry is 1 by the prefix hypothesis, so A = diag(1,...,1,A[last,last]). Taking determinants of a diagonal matrix gives the product of diagonal entries, which here is exactly A[last,last]. The determinant hypothesis det A = 1 therefore forces A[last,last] = 1. Without det = 1 the last diagonal could be -1 (a reflection), which is still orthogonal with an identity prefix but is not the identity; the determinant is precisely what selects the +1 orientation. Hence A = I.", "rubric": [ { "criterion": "Use orthogonality with fixed coordinate columns to force off-diagonal zeros", "id": "c1", "weight": 1 }, { "criterion": "Reduce A to a diagonal with 1s except possibly the last entry", "id": "c2", "weight": 1 }, { "criterion": "Use det = 1 to pin the last diagonal to +1 and note -1 is the alternative", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Zero every off-diagonal entry by orthogonality with the fixed coordinate columns.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.orthogonal_zero_of_fixed_column" ] }, { "claim": "Conclude A is diagonal with 1s in the first N-1 positions.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.matrix_eq_one_of_last_prefix" ] }, { "claim": "Compute the determinant as the single remaining diagonal entry.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.matrix_eq_one_of_last_prefix" ] }, { "claim": "Apply det = 1 to force the last diagonal to +1, excluding the -1 reflection.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.matrix_eq_one_of_last_prefix" ] } ], "source_item_digest": "2e7a6b4d161e7e91748a4d476a0fc00e1f24695c727136037c0478ed59488ad3", "source_refs": [ { "anchor": "orthogonal_zero_of_fixed_column403-408; matrix_eq_one_of_full_prefix485-488; matrix_eq_one_of_last_prefix490-525", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.orthogonal_zero_of_fixed_column", "declaration:QuantumBlockEncoding.AdjacentGivens.matrix_eq_one_of_full_prefix" ], "target_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.matrix_eq_one_of_last_prefix" ], "task_hash": "c4e5e16089be23e21f996a9cf9f4d2c28167f6916f999f8a8ca259e8f2b48a6f", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Little-endian primitive basis strings of a fixed width", "Gray recursion: head of 2*q+b is b xor (q mod 2) over the Gray label of q" ], "context_hash": "686fc0ddde0bf196de71a4095485a3d3e334ede931f0497491be5dd8b777c654", "conventions": [ "A single X action flips exactly one chosen wire of the basis string", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-107", "difficulty": { "level": "advanced", "rationale": "Width induction over a reflected Gray recursion with parity-dependent head/tail casework." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-gray-single-wire-adjacency", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-ac3378e59656015e092e", "key": "qcb-core-la-107", "primary_sources": [], "question": "A width-n binary-reflected Gray labeling assigns to each index in 0..2^n-1 a computational-basis string, little-endian, with the recursion that index 2*q+b carries head bit (b xor (q mod 2)) over the Gray label of q. Prove that two numerically consecutive indices (second = first + 1) carry labels that differ in exactly one physical bit, i.e. the second label is obtained from the first by a single X action on one wire.", "reference_answer": "Induct on the width n. For n=0 there are no two distinct indices, so the claim is vacuous. For width n+1 write first = 2*q + b. There are two cases by the parity of first. If first is even (b=0), then first+1 = 2*q+1 shares the same tail q, so the Gray tails coincide and only the head bit can change; the head recursion gives head(first+1) = flip(head(first)) because the tail q is unchanged and the stored head toggles, so exactly the head wire flips. If first is odd (b=1), then first+1 = 2*(q+1), the head bit is governed by the parity of the quotient and stays equal while the tail advances from q to q+1; applying the induction hypothesis to the tail yields a single flipped wire there, and lifting that wire through the tail embedding gives one physical X action on the full string. In both cases exactly one wire changes, and the explicit head-parity lemmas certify which. This is a statement about which single basis relabeling realizes the step, not a gate-cost estimate.", "rubric": [ { "criterion": "Set up induction on width with the correct even/odd split on the first index", "id": "c1", "weight": 1 }, { "criterion": "Even case: tail fixed, head toggles via the head-parity lemma", "id": "c2", "weight": 1 }, { "criterion": "Odd case: head fixed, tail advances and the induction hypothesis gives one flipped wire", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Reduce to width induction and split on the parity of the first index.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayBasis.adjacent" ] }, { "claim": "Even first index: tail unchanged, head bit toggles.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayBasis.headBit_even" ] }, { "claim": "Odd first index: head bit unchanged, tail quotient advances.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayBasis.headBit_odd" ] }, { "claim": "Lift the inductive single-wire flip of the tail to the full string.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayBasis.equiv" ] } ], "source_item_digest": "87706ea5c66f54bf0bffd4b1d4fcaab38075224230201e30e5d530af8ae448e2", "source_refs": [ { "anchor": "equiv15-22; headBit_even37-42; headBit_odd44-52; adjacent54-94", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/GrayBasis.lean", "sha256": "658e01d906fc1ed4dd12983ae62c1a45f5639a9d0cbff35c825db9b78445d3db" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayBasis.headBit_even", "declaration:QuantumBlockEncoding.GrayBasis.headBit_odd", "declaration:QuantumBlockEncoding.GrayBasis.equiv" ], "target_nodes": [ "declaration:QuantumBlockEncoding.GrayBasis.adjacent" ], "task_hash": "5c47b48347ea770ce40a5d1a0acef281acd0e678d5ae7c98b2bfb0ed774cc2da", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite-dimensional real rectangular matrix; the transform is the accumulated rotation product", "Each plane rotation is orthogonal with determinant 1" ], "context_hash": "6246e4091dbd8d0e087b833dcb415fd3828e930090a7b4c98ac65963620e8977", "conventions": [ "Chronological product: the transform left-multiplies the matrix to the residual", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-108", "difficulty": { "level": "intermediate", "rationale": "Short orthogonality-based recovery argument with an explicit determinant remark." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-rectangular-exact-recovery", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-36f3daa7645c49ea7423", "key": "qcb-core-la-108", "primary_sources": [], "question": "A rectangular NxM real matrix A is reduced to an upper-trapezoidal residual by accumulating plane rotations into a single orthogonal transform matrix T, so that T*A equals the residual. Prove that T^T times the residual recovers A exactly, and state why T^T is a valid left inverse of T here.", "reference_answer": "The accumulated transform T is a product of plane rotations, each of which is orthogonal; a product of orthogonal matrices is orthogonal, so T^T T = I. The construction guarantees T*A = residual by chaining the per-step actions (each plane rotation left-multiplies the current matrix, matching the chronological product). Multiplying this identity on the left by T^T gives T^T*(T*A) = T^T*residual. The left side regroups by associativity as (T^T T)*A = I*A = A, using orthogonality. Hence T^T*residual = A, an exact reconstruction. T^T is the valid left inverse precisely because T is orthogonal: no numerical pseudo-inverse or least-squares step is involved; the inverse is the transpose by the orthogonality identity. Furthermore T has determinant 1 since each plane rotation has determinant 1, so T is a genuine rotation, not merely an orthogonal reflection. These are exact matrix identities, not an evaluation-cost claim.", "rubric": [ { "criterion": "Establish T^T T = I from a product of orthogonal plane rotations", "id": "c1", "weight": 1 }, { "criterion": "Left-multiply T*A = residual by T^T and regroup via orthogonality", "id": "c2", "weight": 1 }, { "criterion": "Note T has determinant 1 so the transpose is the exact inverse", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Show the accumulated transform is orthogonal (T^T T = I).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RectangularGivens.transform_orthogonal" ] }, { "claim": "Record the reduction identity T*A = residual.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RectangularGivens.transform_mul" ] }, { "claim": "Left-multiply by T^T and collapse T^T T to the identity.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RectangularGivens.exact_recovery" ] }, { "claim": "Note determinant 1 confirms a genuine rotation and exact transpose inverse.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RectangularGivens.transform_det" ] } ], "source_item_digest": "b49acfec30465319f4637b8b0bc8fb1655146b68a99b8b9d6688fff48ec32fa7", "source_refs": [ { "anchor": "transform_mul130-133; transform_orthogonal144-145; transform_det147-148; exact_recovery150-153", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RectangularGivens.lean", "sha256": "8066d811985b60d5f30eeb6cef50c516a00d9f8b95182a2d541c7c9eb35fa133" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RectangularGivens.transform_orthogonal", "declaration:QuantumBlockEncoding.RectangularGivens.transform_mul", "declaration:QuantumBlockEncoding.RectangularGivens.transform_det" ], "target_nodes": [ "declaration:QuantumBlockEncoding.RectangularGivens.exact_recovery" ], "task_hash": "c9f60dbf1fa94375ebe18f80b5dc9924038bc1dd131b66837f3f01083cd4e884", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Fixed-width real cores; left boundary vector of length D; terminal bond 1", "Amplitudes are finite chronological matrix contractions" ], "context_hash": "904ccbd1b4ef03a15dd63795868cade37ddde8402832dad1cf4d2d0d206ca540", "conventions": [ "Closing the left boundary absorbs the vector into the first core only", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-109", "difficulty": { "level": "intermediate", "rationale": "Contraction identity resting on a finite double-sum exchange." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-mpc-left-boundary", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-13449011c051c0807334", "key": "qcb-core-la-109", "primary_sources": [], "question": "A finite matrix-product chain of width-D cores produces an amplitude for a bit string by chronological matrix contraction, with a fixed left boundary vector of length D. Contracting that left boundary into the first core yields a chain with left bond 1. Prove that the single scalar amplitude of this closed chain equals the sum over a of left[a] times the open chain's amplitude indexed by a.", "reference_answer": "Write the first core as a function A[a,out] of an incoming bond a and an output index out, and let the rest of the chain contract to a vector indexed by its own incoming bond. Closing the left boundary replaces the first core by the single-row core whose entry at out is sum over a of left[a]*A[a,out]. The amplitude of the closed chain for a string x is then sum over out of (sum over a of left[a]*A[a,out]) * (tail amplitude at bond out). Expanding the inner definition this is a double sum over a and out of left[a]*A[a,out]*(tail amplitude at out). The open chain's amplitude indexed by a is sum over out of A[a,out]*(tail amplitude at out). So the target expression sum over a of left[a]*(open amplitude at a) is sum over a of left[a]* sum over out of A[a,out]*(tail at out), which is the same double sum with the two finite summations exchanged. Exchanging the order of a finite double sum is valid, so the two expressions are equal. This is an exact contraction identity about absorbing a boundary vector, not a claim about contraction runtime.", "rubric": [ { "criterion": "Express the closed-chain amplitude as a double sum over the first bond and output", "id": "c1", "weight": 1 }, { "criterion": "Express the target as left-weighted open amplitudes, also a double sum", "id": "c2", "weight": 1 }, { "criterion": "Equate the two by exchanging the finite summation order", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Expand the closed-chain amplitude using the folded first core.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.closeLeft" ] }, { "claim": "Rewrite it as a double sum over the first bond and the output index.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.closeLeft_contract" ] }, { "claim": "Express the left-weighted open amplitudes as the same double sum.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.readout" ] }, { "claim": "Conclude equality by exchanging the finite summation order.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.closeLeft_contract" ] } ], "source_item_digest": "1fd3017d71601c3fa4b81db41b1b04a8d4a9fd00e609a2ba2b87f5082ddbb14b", "source_refs": [ { "anchor": "readout14-17; closeLeft38-40; closeLeft_contract42-49", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/MatrixProductChain.lean", "sha256": "d916236ffc04fb1d09c9737d77ad83c965ea12716d1f22d6e7c0ec7b1a34e974" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.closeLeft", "declaration:QuantumBlockEncoding.MatrixProductChain.readout" ], "target_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.closeLeft_contract" ], "task_hash": "13c736ff5325c1173f61bf5c5a5e06e0519feb2102682653a2ab0374481f16a1", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Fixed-width real cores; right boundary vector of length D", "Readout is the chronological matrix-vector product from the right boundary" ], "context_hash": "84159849f20d5e013c6b91d43f1b3d03a117265b1063ba08f2fe488a3eda6ae7", "conventions": [ "Absorbing the right boundary makes the terminal bond 1", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-110", "difficulty": { "level": "intermediate", "rationale": "Structural induction identifying a built chain's contraction with a matrix-vector readout." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-mpc-terminal-absorption", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-daf9e6cc892e74694277", "key": "qcb-core-la-110", "primary_sources": [], "question": "In a matrix-product chain with width-D cores and a fixed right boundary vector of length D, define the readout of a bit string as the chronological matrix-vector product that starts from the right boundary and applies each core's selected matrix in turn. Prove that building the chain by absorbing the right boundary into the last core (so the terminal bond is 1) reproduces exactly this readout at every incoming bond.", "reference_answer": "Induct on the number of cores beyond the last. The base case is a single core: absorbing the right vector makes the last core's entry at incoming bond a equal to sum over out-dependent action of (selected matrix times right)[a], which is by definition the readout of a length-one word at bond a; the terminal bond-1 contraction is just this scalar. For the inductive step with one more core on the left, the contraction of the built chain at bond a is sum over b of (selected matrix of the new core)[a,b] times the contraction of the shorter built chain at bond b. By the induction hypothesis the shorter contraction equals the readout starting one step later, so the whole expression is sum over b of M[a,b] * readout(rest)[b], which is exactly the matrix-vector product defining readout at the current step. Hence the built chain contracts to the readout for every word and every incoming bond. This identifies two definitions of the same amplitude; it makes no contraction-cost assertion and the terminal rank is one by construction.", "rubric": [ { "criterion": "Base case: single absorbed core equals the length-one readout", "id": "c1", "weight": 1 }, { "criterion": "Inductive step expands the contraction as a matrix-vector product over the next bond", "id": "c2", "weight": 1 }, { "criterion": "Apply the induction hypothesis to match the readout recursion", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Set up induction on the number of cores left of the last.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.tailChain" ] }, { "claim": "Base case: the single absorbed core equals the length-one readout.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.tailChain_contract" ] }, { "claim": "Inductive step: expand the contraction as a sum over the next bond.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.tailChain_contract" ] }, { "claim": "Match the readout recursion by the induction hypothesis.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.readout" ] } ], "source_item_digest": "fad513662c7653c200c59df3ebf0190dd261233cbc497793dff6c0821c892001", "source_refs": [ { "anchor": "readout14-17; tailChain20-24; tailChain_contract26-35", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/MatrixProductChain.lean", "sha256": "d916236ffc04fb1d09c9737d77ad83c965ea12716d1f22d6e7c0ec7b1a34e974" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.tailChain", "declaration:QuantumBlockEncoding.MatrixProductChain.readout" ], "target_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.tailChain_contract" ], "task_hash": "7672204c291c6a13069f7cf5f6089e8ac606e795f5c57a701a06c4ac2090d4f9", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Fixed-width real cores; every bond dimension is at most D", "Stored-scalar count sums 2*l*m over all cores" ], "context_hash": "2febea361fb7db57f1cf1c658628f891346502518ba6991322e7b9d19f8fe338", "conventions": [ "Stored scalars count physically laid-out entries, not arithmetic operations", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-111", "difficulty": { "level": "intermediate", "rationale": "Clean storage induction with an explicit boundary between an address count and a runtime claim." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-mpc-storage-bound", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-0e36f7cd1e8b72129198", "key": "qcb-core-la-111", "primary_sources": [], "question": "A matrix-product chain stores dense local cores; the number of stored scalars sums 2*l*m across cores, where l and m are the left and right bond dimensions of each core. Given that every bond of the chain is at most D and the chain has n cores, prove the stored-scalar count is at most 2*n*D^2. Then explain why this counts stored addresses only, and does not by itself bound arithmetic runtime.", "reference_answer": "Induct on the chain. The empty chain stores 0 scalars, which is at most 2*0*D^2. For a chain whose first core has bonds l (left) and m (right) followed by a tail, the stored count is 2*l*m + stored(tail). The bond bound gives l <= D and m <= D, so 2*l*m <= 2*D*D = 2*D^2. By the induction hypothesis stored(tail) <= 2*(n-1)*D^2. Adding, the total is at most 2*D^2 + 2*(n-1)*D^2 = 2*n*D^2. The required m <= D for the first core is read from the bound on the tail's leading left bond (or directly for a terminal core), so the inequality closes at every step. This counts the number of real entries physically laid out in the cores: it is a storage/address measure. It says nothing about the number of multiplications or additions used to contract the chain, nor about numerical precision; those are separate quantities. A per-bit-string amplitude still requires its own contraction work, which this bound does not estimate.", "rubric": [ { "criterion": "Per-core bound 2*l*m <= 2*D^2 from both bond bounds", "id": "c1", "weight": 1 }, { "criterion": "Induction summing to 2*n*D^2 with correct bond extraction", "id": "c2", "weight": 1 }, { "criterion": "Explain this is a stored-address count, not a runtime bound", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Bound a single core's stored entries by 2*D^2 using both bond bounds.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.storedScalars" ] }, { "claim": "Extract the tail's leading bond to justify m <= D at each step.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.storedScalars_le" ] }, { "claim": "Sum over cores by induction to reach 2*n*D^2.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.storedScalars_le" ] }, { "claim": "Distinguish stored addresses from contraction arithmetic cost.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.ofKernel_storedScalars" ] } ], "source_item_digest": "913e7d71276fa5074d8dde978f1e7bb69b6655f8aaa0087808b89035aad8a38e", "source_refs": [ { "anchor": "storedScalars75-77; storedScalars_le79-93; ofKernel_storedScalars95-98", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/MatrixProductChain.lean", "sha256": "d916236ffc04fb1d09c9737d77ad83c965ea12716d1f22d6e7c0ec7b1a34e974" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.storedScalars", "declaration:QuantumBlockEncoding.MatrixProductChain.ofKernel_storedScalars" ], "target_nodes": [ "declaration:QuantumBlockEncoding.MatrixProductChain.storedScalars_le" ], "task_hash": "8f80fc079a05269fc955b7d15d15622e5a7472bf1ca03c58b7c78bc269df4169", "task_kind": "boundary", "topic": "matrix_constructions" }, { "assumptions": [ "e is an injection from r original indices into N physical positions", "The extension is built by greedy transpositions starting from the identity" ], "context_hash": "ac404e6ce2dcc0698991f378d424ad69c43fed4eb86cc7e863572b5403a06f3f", "conventions": [ "A transposition fixes every point other than its two endpoints", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-114", "difficulty": { "level": "advanced", "rationale": "Strengthened induction with injectivity-driven case analysis on transposition fixed points." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-greedy-injection-extension", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-3337b848207352e516fd", "key": "qcb-core-la-114", "primary_sources": [], "question": "A greedy swap process extends a finite injection e from r original indices into N physical positions: starting from the identity permutation, at stage k it swaps the current image of index k with the target e(k). Prove that after r stages the resulting permutation p agrees with e on every original index, i.e. p(castLE a) = e(a) for all a < r.", "reference_answer": "Induct on the stage count k, proving the stronger statement: after k stages the permutation agrees with e on every index a with a.val < k. At k=0 the claim is vacuous. At stage k+1 the permutation is p_k followed by the transposition swapping p_k(k) with e(k). Take any index a with a.val < k+1. If a.val = k, then the fresh swap sends p_k(k) to e(k); since the composition first applies p_k (landing on p_k(k)) and then the swap (sending it to e(k)), index k maps to e(k) as required. If a.val < k, the induction hypothesis gives p_k(a) = e(a); it remains to see the new transposition fixes e(a). It does because e(a) differs from both swap endpoints: e(a) != p_k(k) since p_k is injective and a != k (otherwise p_k(a)=p_k(k)=e(a) would force, via injectivity of e, a=k), and e(a) != e(k) since e is injective and a != k. A transposition fixes any point distinct from its two endpoints, so p_{k+1}(a) = p_k(a) = e(a). Taking k=r gives agreement on all original indices. This is an exact placement guarantee, not a sorting-cost estimate.", "rubric": [ { "criterion": "Strengthen to agreement on indices below k and induct", "id": "c1", "weight": 1 }, { "criterion": "Stage k: the fresh swap sends index k to e(k)", "id": "c2", "weight": 1 }, { "criterion": "Earlier indices preserved because e(a) avoids both swap endpoints (injectivity)", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "State the stronger below-k agreement invariant and set up induction.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.extendPrefix" ] }, { "claim": "Stage k case: the fresh transposition carries index k to e(k).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.extendPrefix_agrees" ] }, { "claim": "Earlier index case: apply the induction hypothesis.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.extendPrefix_agrees" ] }, { "claim": "Show e(a) differs from both swap endpoints using injectivity of p_k and e.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.extendPrefix_agrees" ] } ], "source_item_digest": "d786b4f7dd920203358d420044561a976ab5979789857cbfc7dc3a3c73a6aa96", "source_refs": [ { "anchor": "extendPrefix167-172; extendPrefix_agrees174-198", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ConstructiveIsometryCompletion.lean", "sha256": "1b61cbc9ad587cc2ae2a7cdc7bacd3b7d39fb05274833a3fea5dbb4dff69678c" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.extendPrefix" ], "target_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.extendPrefix_agrees" ], "task_hash": "3b91fa211b782d5140a5c43f50812f56e4bc2d1258e2fe66339421756c3c66ae", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite real orthogonal matrix with determinant 1 before reorientation", "The sign flip acts on a single designated unused column" ], "context_hash": "4a92a13a9ca331d654ab71b1d3f0a186a4a9e39bccd38abf8a3a6bc0c7d002c2", "conventions": [ "Branch selection uses the permutation parity, a discrete value in {+1,-1}", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-115", "difficulty": { "level": "intermediate", "rationale": "Two-branch determinant/orthogonality bookkeeping driven by discrete permutation parity." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-parity-orientation-fix", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-3831b2e8492a5f18f1bd", "key": "qcb-core-la-115", "primary_sources": [], "question": "An orthogonal matrix U is reoriented to have determinant 1 by multiplying it with a sign-flip on a single designated unused column exactly when U already has the wrong sign; the choice is made by the parity (sign) of an accompanying column permutation, not by evaluating the determinant. Prove the reoriented matrix is orthogonal with determinant 1 regardless of which branch is taken.", "reference_answer": "Let P be the column permutation and consider two branches. If sign(P) = +1, the output is the permuted matrix permuteColumns(U,P) alone; a column permutation of an orthogonal matrix is orthogonal (its Gram matrix is the identity permuted on both sides, still the identity), and its determinant equals sign(P) times det(U) = (+1)*(1) = 1 when U has determinant 1. If sign(P) = -1, the output is permuteColumns(U,P) multiplied by a single-column sign flip S. Orthogonality: S is orthogonal (it is a diagonal of plus/minus ones), and a product of orthogonal matrices is orthogonal, with S^T (permuteColumns Gram) S = S^T I S = I. Determinant: det(permuteColumns(U,P)*S) = det(permuteColumns(U,P)) * det(S) = (sign(P)*det U) * (-1) = (-1)*(1)*(-1) = 1. Because the orthogonal group over the reals has only signs plus or minus one for a permutation, these two branches are exhaustive, and both yield orthogonal determinant-one matrices. Crucially the branch is selected by the discrete permutation parity, so no real determinant comparison or thresholding is performed. This is an exact orientation-fix construction, not a numerical sign test.", "rubric": [ { "criterion": "Positive-parity branch: permuted matrix orthogonal with det 1", "id": "c1", "weight": 1 }, { "criterion": "Negative-parity branch: single sign-flip restores det 1 and keeps orthogonality", "id": "c2", "weight": 1 }, { "criterion": "Note the branch is chosen by discrete parity, not a determinant evaluation", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Positive-parity branch: permuted matrix is orthogonal with determinant sign(P)*det U = 1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.permuteColumns_det" ] }, { "claim": "Negative-parity branch: compose with a single-column sign flip to multiply the determinant by -1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.orientColumns_det" ] }, { "claim": "Show orthogonality survives the sign flip since it is itself orthogonal.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.orientColumns_orthogonal" ] }, { "claim": "Conclude both exhaustive parity branches give orthogonal determinant-one matrices.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.permuteColumns_orthogonal" ] } ], "source_item_digest": "c1f7bebe1314ec75cc91e72658ca0457c6bc62518acffb2a8b53db63dd6bb50d", "source_refs": [ { "anchor": "permuteColumns_orthogonal223-229; permuteColumns_det231-234; orientColumns237-240; orientColumns_orthogonal242-254; orientColumns_det256-267", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ConstructiveIsometryCompletion.lean", "sha256": "1b61cbc9ad587cc2ae2a7cdc7bacd3b7d39fb05274833a3fea5dbb4dff69678c" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.orientColumns_orthogonal", "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.permuteColumns_det", "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.permuteColumns_orthogonal" ], "target_nodes": [ "declaration:QuantumBlockEncoding.ConstructiveIsometryCompletion.orientColumns_det" ], "task_hash": "b8c6dc0cbf6c9223be288fdbffc4586b2acb2faa73d228c94ac3a8fa58c0c9f9", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite real matrix indexed by width-(q+1) labels; a fixed label/index bijection", "The same bijection relabels rows and columns" ], "context_hash": "a7a55b80adb3bee422f705a0ec94756b1da5b8648c7ffcc310eec66fa8e88820", "conventions": [ "Symmetric reindexing is an algebra equivalence on matrices preserving multiplication and the identity", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-116", "difficulty": { "level": "intermediate", "rationale": "Transport of orthogonality and determinant through a symmetric reindexing with an explicit single-bijection necessity argument." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-gray-coordinate-transport", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-65376e82073d6e193504", "key": "qcb-core-la-116", "primary_sources": [], "question": "A real matrix indexed by width-(q+1) computational-basis labels is converted to Gray coordinates by relabeling both its rows and its columns through a fixed bijection between labels and the numerical indices 0..2^(q+1)-1 (the same bijection on each side). Prove that this relabeling preserves orthogonality, i.e. if A^T A = I then the Gray-coordinate matrix is again orthogonal, and that it preserves the determinant exactly. Identify precisely why using the same bijection on rows and columns is essential.", "reference_answer": "The symmetric relabeling is the reindexing map that sends a matrix A to the matrix whose (i,j) entry is A at the labels corresponding to i and j under the fixed bijection. Because the same bijection is applied to rows and to columns, this map is multiplicative: the reindexing of a product equals the product of the reindexings, since the internal summation index of matrix multiplication is itself relabeled consistently and a bijective reindexing of a finite sum does not change its value. It also sends the identity matrix to the identity, because the identity's entries depend only on whether the two labels coincide, a property preserved by any bijection applied equally to both sides. Orthogonality: apply the reindexing to the hypothesis A^T A = I. Transpose commutes with symmetric relabeling (swapping row and column roles is unaffected by using one bijection on both), so reindex(A)^T reindex(A) = reindex(A^T A) = reindex(I) = I. Hence the Gray-coordinate matrix is orthogonal. Determinant: a symmetric reindexing is a similarity-type conjugation by a permutation, and the determinant is invariant under relabeling both axes by the same permutation, so det is unchanged exactly. The same-bijection requirement is essential: if rows and columns were relabeled by different bijections, the map would not respect multiplication (the shared summation index would be split), transpose would not commute with it, and the determinant could pick up a relative permutation sign. Thus using one bijection on both axes is exactly what makes relabeling an algebra isomorphism preserving these invariants. This is an exact transport fact, not a numerical change of basis cost.", "rubric": [ { "criterion": "State that symmetric relabeling is multiplicative and fixes the identity", "id": "c1", "weight": 1 }, { "criterion": "Derive orthogonality by applying the relabeling to A^T A = I, using transpose compatibility", "id": "c2", "weight": 1 }, { "criterion": "Argue determinant invariance and explain why one shared bijection is essential", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Describe symmetric relabeling as a multiplicative, identity-fixing reindexing map.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.grayCoordinates" ] }, { "claim": "Apply the map to A^T A = I and use transpose compatibility to get orthogonality.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.grayCoordinates_orthogonal" ] }, { "claim": "Conclude the determinant is unchanged under the symmetric relabeling.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.grayCoordinates_det" ] }, { "claim": "Explain that one shared bijection is required for multiplicativity and sign-free determinant.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.grayCoordinates" ] } ], "source_item_digest": "e13c0d1bd06158bd627632029146044c6323a71aaaf22ded1d505d62b6f8e2e2", "source_refs": [ { "anchor": "grayCoordinates159-162; grayCoordinates_orthogonal169-175; grayCoordinates_det177-180", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/GrayGivensCompiler.lean", "sha256": "21bb92938d2d0d81341700123e66195fdebb12e72db7faf066a897479fe942ce" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.grayCoordinates", "declaration:QuantumBlockEncoding.GrayGivensCompiler.grayCoordinates_det" ], "target_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.grayCoordinates_orthogonal" ], "task_hash": "4fd8a6c0cf4b9c687f571f0519678f28f1983555ea92b65fc816080f7976397a", "task_kind": "boundary", "topic": "matrix_constructions" }, { "assumptions": [ "A real SO matrix on 2^(q+1) indices; the adjacent pair differs by one Gray wire flip", "Every non-target wire is a control; the target is the Gray-flip wire" ], "context_hash": "62fc01f88c96c98a914d3d9a2fc1b4d9803ac3231e5cf9a591b7579637b96e19", "conventions": [ "Plane orders the pair numerically; the selected plane orders by target bit 0 then 1", "Reversed ordering negates the RY half-angle", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-117", "difficulty": { "level": "advanced", "rationale": "Basis-relabeling correspondence with careful control-context matching and an orientation-dependent sign." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-gray-plane-transport", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-87c93d40a67d1573befc", "key": "qcb-core-la-117", "primary_sources": [], "question": "A real SO matrix on 2^(q+1) Gray-ordered indices is compiled to a circuit by first decomposing it into adjacent plane rotations, then realizing each plane as a selected RY rotation whose target wire is the single Gray-flip wire of that adjacent pair, with every other wire a control. Prove that reindexing the adjacent plane matrix by the Gray labeling equals the selected two-level plane on that target wire, and explain how the half-angle sign depends on the target bit of the first index.", "reference_answer": "Fix an adjacent pair (first, second) with second = X_target(first) in Gray labels. Reindexing the plane matrix by the Gray equivalence sends row/column named states back to their numerical indices. The plane rotates only the two indices first and second and fixes all others; under Gray labeling these two indices are exactly the two computational-basis states that share the same control context (all non-target bits equal) and differ only on the target wire, which is precisely the pair acted on by the selected RY plane with that chosen control context. For a named state outside this control context, both the reindexed plane and the selected plane act as the identity, matching entrywise. Inside the context, the 2x2 action is the real RY block; here the orientation matters: the plane's convention lists the pair by numerical order (first, then second), whereas the selected plane orders the pair by the target bit value 0 then 1. If the first index already has target bit 0 the orders agree and the angle is theta; if the first index has target bit 1 the two orderings are swapped, which conjugates the RY block by the bit-swap and is exactly the RY block at angle -theta. Hence the reindexed plane equals the selected plane at angle theta when the first target bit is 0 and at -theta otherwise. This matches a basis relabeling to a controlled rotation; it is a semantic correspondence, not a gate-count claim.", "rubric": [ { "criterion": "Identify the Gray pair as the shared-control-context two-level pair", "id": "c1", "weight": 1 }, { "criterion": "Match identity action for states outside the control context", "id": "c2", "weight": 1 }, { "criterion": "Explain the half-angle sign flip from first index target bit via pair ordering", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Characterize the two reindexed-plane indices as the shared-context target-wire pair.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.same_context_iff" ] }, { "claim": "Match identity action off the control context entrywise.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.selectedRyPlaneMatrix" ] }, { "claim": "On the context, reduce both sides to the real RY 2x2 block.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.realRyPlaneBlock" ] }, { "claim": "Track the pair ordering to get angle theta or -theta from the first index target bit.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.edge_plane_transport" ] } ], "source_item_digest": "c187f71e1150c9fa98992c1dc15dbe5e22c6e35ff1be3cc2a8acccb8244daf96", "source_refs": [ { "anchor": "same_context_iff21-37; edge_plane_transport54-86", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/GrayGivensCompiler.lean", "sha256": "21bb92938d2d0d81341700123e66195fdebb12e72db7faf066a897479fe942ce" }, { "anchor": "realRyPlaneBlock42-45; selectedRyPlaneMatrix58-66", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SelectedRyPlane.lean", "sha256": "0098a9ef2bfc1af9eeaa44357b0818628084be35f88854232496e14d715c2f40" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.same_context_iff", "declaration:QuantumBlockEncoding.selectedRyPlaneMatrix", "declaration:QuantumBlockEncoding.realRyPlaneBlock" ], "target_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.edge_plane_transport" ], "task_hash": "e89afac279a9451712ccae636734137d14f7a125cf3402564f27162d863dc9be", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Basis states are bit strings; the control context is the non-target restriction", "A single X action flips only the target wire" ], "context_hash": "4ddf646f8c946262427ed8d7d4ee832899948d7c7151804983fbd1f3b251595f", "conventions": [ "The selected plane fixes states outside the chosen context", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-118", "difficulty": { "level": "intermediate", "rationale": "Iff characterization by target-bit casework with an explicit converse." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-control-context-pair", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-2e5a12c0cc09c8924830", "key": "qcb-core-la-118", "primary_sources": [], "question": "A selected RY plane fixes every computational-basis vector whose control context differs from the chosen one, and acts as a 2x2 real rotation on the single pair sharing the chosen context and differing on the target bit. Prove that two basis states lie in the same control context if and only if they are equal or related by a single X action on the target wire.", "reference_answer": "The control context of a basis state is the restriction of its bit string to all wires other than the target. Two states share this context exactly when they agree on every non-target wire. Consider their target bits. If the two target bits are equal, then agreeing on all other wires plus agreeing on the target wire means the two full strings are identical, so the states are equal. If the two target bits differ, then they agree everywhere except the target wire, which is precisely the relation that one is obtained from the other by flipping the target bit, i.e. a single X action on the target wire; the non-target context is manifestly unchanged by flipping only the target bit. Conversely, equal states trivially share context, and a single target-wire X action changes only the target bit and leaves every other wire fixed, so it preserves the control context. Hence same context holds if and only if the states are equal or target-wire X-related. This pins down exactly the two-dimensional invariant subspace a selected rotation can touch, and is a statement about which basis pair is affected, not about implementation cost.", "rubric": [ { "criterion": "Define control context as the non-target restriction and split on target-bit equality", "id": "c1", "weight": 1 }, { "criterion": "Equal target bits force equal states; differing bits give a single X relation", "id": "c2", "weight": 1 }, { "criterion": "Converse: equality and target X both preserve the context", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Reduce same-context to agreement on all non-target wires and split by target-bit equality.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.bit_eq_or_flip" ] }, { "claim": "Equal target bits give equal states; differing bits give a target-wire X relation.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.same_context_iff" ] }, { "claim": "Show a target X action preserves the non-target context (converse direction).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.x_context" ] }, { "claim": "Relate this invariant pair to the fixed-column behavior of the selected plane.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.selectedRyPlaneMatrix_fixed_column" ] } ], "source_item_digest": "0d08607d4fa98a7c4e6c2f3cef267a9b7e35329c0894b2c6272a11c823fe4ffd", "source_refs": [ { "anchor": "bit_eq_or_flip12-13; x_context15-19; same_context_iff21-37", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/GrayGivensCompiler.lean", "sha256": "21bb92938d2d0d81341700123e66195fdebb12e72db7faf066a897479fe942ce" }, { "anchor": "selectedRyPlaneMatrix_fixed_column122-129", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SelectedRyPlane.lean", "sha256": "0098a9ef2bfc1af9eeaa44357b0818628084be35f88854232496e14d715c2f40" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.x_context", "declaration:QuantumBlockEncoding.GrayGivensCompiler.bit_eq_or_flip", "declaration:QuantumBlockEncoding.selectedRyPlaneMatrix_fixed_column" ], "target_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.same_context_iff" ], "task_hash": "e068d6b6f0961aba472c2390dc70ef0417dc6c94b74386ec1765a800fd941592", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Finite real matrices; each list entry is a plane rotation with its own matrix", "Chronological convention: the first listed rotation acts first" ], "context_hash": "09d1bd2afce3c3ec5a08d7c64d2de5ab293ccbc1f6e0491a556bb9552c9b91c3", "conventions": [ "The accumulated matrix multiplies the list in reverse order (last entry left-most)", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-120", "difficulty": { "level": "intermediate", "rationale": "List induction reconciling an iterated action with a single reverse-order matrix product via associativity and a convention argument." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "matrix-constructions-step-action-matrix-equivalence", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-5588e9ee86a96021a408", "key": "qcb-core-la-120", "primary_sources": [], "question": "A list of plane rotations can be applied to a matrix in two ways. The step-by-step action takes the list and the matrix and applies the first listed rotation to the matrix, then the next to the result, and so on. Alternatively one forms a single accumulated matrix from the list (built so that the last listed rotation multiplies on the left-most, i.e. the product is taken in reverse-list order) and multiplies it once against the matrix. Prove that these two procedures agree for every list and every matrix, and explain why the accumulated product must be taken in reverse list order.", "reference_answer": "Induct on the list. For the empty list, the step-by-step action returns the matrix unchanged and the accumulated matrix is the identity, so both give the matrix. For a list (step :: rest) applied to M, the step-by-step action first forms step.matrix * M (applying step) and then applies rest to that, so by the induction hypothesis it equals accMatrix(rest) * (step.matrix * M). Associativity of matrix multiplication rewrites this as (accMatrix(rest) * step.matrix) * M. The accumulated matrix of the whole list is defined as accMatrix(rest) * step.matrix precisely so that the leading step sits on the right of the product; hence (accumulated matrix of the full list) * M, matching the claim. The reverse-list product order is forced by the chronological convention: the first listed rotation acts first on M, so it must be the right-most factor (closest to M), which means later list entries accumulate to its left. Taking the product in forward order would apply the rotations in the wrong temporal sequence and break the agreement. Therefore the two procedures coincide exactly for all lists, with the reverse order being the unique convention consistent with first-listed-acts-first. This is an exact identity about two equal formulations, not a claim about multiplication cost.", "rubric": [ { "criterion": "Base case: empty list gives the matrix via identity accumulation", "id": "c1", "weight": 1 }, { "criterion": "Inductive step uses associativity to peel the leading step onto the right", "id": "c2", "weight": 1 }, { "criterion": "Explain reverse product order is forced by first-listed-acts-first", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Base case: empty list gives the matrix, matching identity accumulation.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps" ] }, { "claim": "Expand the step-by-step action of a cons list as leading step then rest.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.Step.matrix" ] }, { "claim": "Apply the induction hypothesis and reassociate to isolate the full accumulated matrix.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps_eq_matrix_mul" ] }, { "claim": "Explain the reverse product order is the unique convention matching first-listed-acts-first.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.stepsMatrix" ] } ], "source_item_digest": "378ddb50eda58d0fe217c02313ba7c13f7550d190ac84862faa25f8e3a1cb315", "source_refs": [ { "anchor": "Step.matrix188-189; applySteps192-195; stepsMatrix633-635; applySteps_eq_matrix_mul637-642", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps", "declaration:QuantumBlockEncoding.AdjacentGivens.stepsMatrix", "declaration:QuantumBlockEncoding.AdjacentGivens.Step.matrix" ], "target_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.applySteps_eq_matrix_mul" ], "task_hash": "cd6ac5be7d47e516fb257ff8b4041237e28670f2ef7de5509a8cb8c8c841adc6", "task_kind": "proof", "topic": "matrix_constructions" }, { "assumptions": [ "Slot values range over {0,...,7}, so the outer mod 8 never fires", "The loader is XOR with the fixed mask 3" ], "context_hash": "12458a8b8546d7051e53f9c341a7af0fe86b8f934718d5037c8905f524bf72af", "conventions": [ "XOR with a constant mask toggles exactly the bits set in the mask", "The loader is realized by two X gates on the two low wires" ], "core_id": "qcb-core-be-101", "difficulty": { "level": "intermediate", "rationale": "Requires recognizing XOR-by-constant as an involution, proving the self-inverse law, and computing the explicit permutation table while connecting it to the gate realization." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-xor-loader-involution", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-815b7e8e6e923b22e193", "key": "qcb-core-be-101", "primary_sources": [], "question": "A reversible source-loader on three little-endian bits is defined by the rule that maps each slot value s in {0,...,7} to (s XOR 3) mod 8, i.e. it flips the two low-order bits of s. Prove that this loader is its own inverse (an involution) as a permutation of {0,...,7}, and give the explicit image list obtained by applying it to 0,1,2,3,4,5,6,7 in order. Explain why the involution property is exactly what lets the two bit-flip gates that realize it also undo it when applied a second time.", "reference_answer": "The loader is L(s) = (s XOR 3) mod 8. Since every s in {0,...,7} is below 8, the outer mod 8 is inert and L(s) = s XOR 3. Bitwise XOR with the fixed mask 3 (binary 011) toggles bit 0 and bit 1 and leaves bit 2 untouched. XOR with a fixed mask is self-inverse because XOR is associative and commutative with x XOR x = 0: L(L(s)) = (s XOR 3) XOR 3 = s XOR (3 XOR 3) = s XOR 0 = s, so L is an involution and in particular a bijection of the 8-element set. Applying L to 0..7 in order gives [3, 2, 1, 0, 7, 6, 5, 4]: 0->3, 1->2, 2->1, 3->0, 4->7, 5->6, 6->5, 7->4. Operationally the loader is implemented by two X (bit-flip) gates on wires 0 and 1; because each X is self-inverse and the two flips commute, running the same two gates again applies 3 XOR 3 = 0, returning every slot to its original value, which is the circuit-level statement of the involution.", "rubric": [ { "criterion": "State L(s)=s XOR 3 (mod 8 inert for s<8) and that it toggles the two low bits, mask 3 = binary 011", "id": "c1", "weight": 1 }, { "criterion": "Prove the involution L(L(s))=s using x XOR x=0 and associativity/commutativity of XOR, concluding L is a bijection", "id": "c2", "weight": 1 }, { "criterion": "Give the exact image list [3,2,1,0,7,6,5,4] and connect the involution to the two X gates undoing themselves on a second pass", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Because every slot value is below 8, (s XOR 3) mod 8 equals s XOR 3.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveOffset3" ] }, { "claim": "XOR with the fixed mask 3 toggles bit 0 and bit 1 and fixes bit 2.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveOffset3" ] }, { "claim": "Self-inverse: (s XOR 3) XOR 3 = s XOR (3 XOR 3) = s, so L is an involution and a bijection.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveOffset3" ] }, { "claim": "Evaluating L on 0..7 yields the ordered image list [3,2,1,0,7,6,5,4].", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveOffset3_table" ] }, { "claim": "The two X gates implementing the mask are each self-inverse and commute, so a second pass applies 3 XOR 3 = 0 and restores every slot.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3ReversibleProgram" ] } ], "source_item_digest": "c121d5a0d54af38e9206a89b7e5c8acbab4eca77fe5c2e3ef5dad9fc0267a2a1", "source_refs": [ { "anchor": "def primitiveOffset3 (slot : Fin 8) : Fin 8 := ⟨(slot.val ^^^ 3) % 8, Nat.mod_lt _ (by decide)⟩ (lines 29-30); @[simp] theorem primitiveOffset3_table : List.ofFn primitiveOffset3 = [3, 2, 1, 0, 7, 6, 5, 4] := by native_decide (lines 32-34)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/BandedSparseAccessPrimitive.lean", "sha256": "e4f35bfdd690445de12d5a7e6692b9ab86d1a096b9ed3f4d1a41e07b99181ca7" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3ReversibleProgram" ], "target_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveOffset3", "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveOffset3_table" ], "task_hash": "e8083994a4b578fd6ef1f4034f65bd98136d6988955023f1a549a61452b14c91", "task_kind": "proof", "topic": "oracle_composition" }, { "assumptions": [ "c and s are real scalars embedded in the complex field", "Unitarity means the conjugate transpose times the matrix equals the identity" ], "context_hash": "7a130562529c5b4e7415c19c7d5696426f71f8b78d1e26fbc28bab895c66d7f3", "conventions": [ "The matrix has top row (c,-s) and bottom row (s,c)", "Conjugation is the identity on real entries" ], "core_id": "qcb-core-be-105", "difficulty": { "level": "basic", "rationale": "Requires an explicit 2x2 Gram-matrix computation and the observation that conjugation is trivial on real entries, reducing unitarity to the Pythagorean normalization." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-real-rotation-unitary-iff", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b7863163dc4db0132f8f", "key": "qcb-core-be-105", "primary_sources": [], "question": "A real planar rotation embedded over the complex numbers is the 2x2 matrix with top row (c, -s) and bottom row (s, c) for real scalars c and s. Prove that this matrix is unitary exactly when c^2 + s^2 = 1, by computing all four entries of the product of its conjugate transpose with itself, and explain why no further hypothesis on c and s is needed and why complex conjugation plays no role here.", "reference_answer": "Call the matrix M with M = [[c, -s], [s, c]], entries real but viewed in the complex field. Its conjugate transpose M* has entries that are complex conjugates of the transpose; since c and s are real, conjugation is the identity on them, so M* = [[c, s], [-s, c]] (the ordinary real transpose). Compute M* M entrywise using Fin-2 sums. (1,1) entry: c*c + s*s = c^2 + s^2. (1,2) entry: c*(-s) + s*c = -cs + cs = 0. (2,1) entry: (-s)*c + c*s = -sc + cs = 0. (2,2) entry: (-s)*(-s) + c*c = s^2 + c^2 = c^2 + s^2. So M* M = [[c^2+s^2, 0],[0, c^2+s^2]], which equals the 2x2 identity if and only if c^2 + s^2 = 1. Hence M is unitary precisely under the normalization c^2 + s^2 = 1. No further hypothesis is required: the off-diagonal entries cancel identically for every real c, s (they are structural cancellations -cs+cs), so the only constraint comes from the diagonal. Conjugation plays no role because the entries are real: for a genuinely complex rotation one would need |entries|^2, but here conjugation acts trivially and the unitarity condition collapses to the real Pythagorean identity c^2 + s^2 = 1. In particular, taking c = cos t and s = sin t makes the identity automatic, so every angle-parameterized real rotation is unitary.", "rubric": [ { "criterion": "Compute M* as the real transpose [[c,s],[-s,c]], noting conjugation is trivial on real entries", "id": "c1", "weight": 1 }, { "criterion": "Compute all four entries of M* M and show the off-diagonals cancel identically while the diagonal equals c^2+s^2", "id": "c2", "weight": 1 }, { "criterion": "Conclude unitarity iff c^2+s^2=1, with no extra hypothesis, and note cos/sin make it automatic", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Since c,s are real, the conjugate transpose M* equals the real transpose [[c,s],[-s,c]].", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realOrthogonalRotation" ] }, { "claim": "The diagonal entries of M* M are both c^2 + s^2.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realOrthogonalRotation_unitary" ] }, { "claim": "The off-diagonal entries cancel identically to 0 as -cs+cs for every real c,s.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realOrthogonalRotation_unitary" ] }, { "claim": "Therefore M* M = (c^2+s^2) I, equal to the identity iff c^2+s^2 = 1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realOrthogonalRotation_unitary" ] }, { "claim": "With c=cos t, s=sin t the Pythagorean identity makes unitarity automatic for every angle.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realRotation_unitary" ] } ], "source_item_digest": "15381b174d3517cf5ade51fd5a8bfdf0eee977640793ea43e93c0627b2eaede1", "source_refs": [ { "anchor": "noncomputable def realOrthogonalRotation (cosine sine : Real) : _root_.Matrix (Fin 2) (Fin 2) ℂ := fun row column => match row.val, column.val with | 0, 0 => (cosine : ℂ) | 0, _ => -(sine : ℂ) | _, 0 => (sine : ℂ) | _, _ => (cosine : ℂ) (lines 23-29); theorem realOrthogonalRotation_unitary (cosine sine : Real) (normalization : cosine * cosine + sine * sine = 1) : realOrthogonalRotation cosine sine ∈ _root_.Matrix.unitaryGroup (Fin 2) ℂ (lines 32-57)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realOrthogonalRotation", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realRotation_unitary" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realOrthogonalRotation_unitary" ], "task_hash": "6da356ccfd1d2eab75c8fc797b19af7bb6737c8a198b3864b7c5fa2181fb1f29", "task_kind": "proof", "topic": "oracle_composition" }, { "assumptions": [ "The angle t(x) satisfies cos t(x) = x on -1 <= x <= 1 and is extended in some unspecified way elsewhere", "The clean entry is read as the top-left matrix element" ], "context_hash": "155da9f9b71a08a88720c258bf9adcd8db519b62f572f5cb88c113d0ccbc340e", "conventions": [ "The amplitude gate is the planar rotation through an inverse-cosine branch of x", "The cosine of any real angle lies in the closed interval [-1, 1]" ], "core_id": "qcb-core-be-106", "difficulty": { "level": "intermediate", "rationale": "Requires separating a structural unitarity guarantee from a domain-restricted semantic readout and proving failure for |x|>1 purely from the bounded range of cosine, with no appeal to a specific extension value." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "be-amplitude-unitary-vs-readout", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-46f6d73e40d459e27ce4", "key": "qcb-core-be-106", "primary_sources": [], "question": "An amplitude-loading gate is built from a planar rotation through an angle t(x), where t is any fixed real-valued function of the real parameter x chosen so that on the interval -1 <= x <= 1 the cosine of t(x) equals x (an inverse-cosine branch), and extended in some unspecified way to the rest of the real line. Two claims are on the table: (A) the gate is unitary for every real x, with no restriction and independent of how t is extended; (B) the top-left entry of the gate equals x. Decide for which values of x each claim holds. Prove the unconditional claim, prove that claim (B) holds for -1 <= x <= 1, and prove that claim (B) FAILS whenever |x| > 1 without using any particular value of the extension. Explain the asymmetry: why unitarity is free but the clean-entry readout needs -1 <= x <= 1.", "reference_answer": "The gate is the planar rotation with top row (cos t(x), -sin t(x)) and bottom row (sin t(x), cos t(x)). Claim (A) holds unconditionally. For every real angle, cos^2 + sin^2 = 1, so this matrix is a planar rotation and hence unitary (its two entries lie on the unit circle); this is true whatever real number t(x) happens to be, so unitarity holds for all real x and does not depend on how t is extended outside [-1,1]. Claim (B): the top-left entry equals cos t(x). On the interval -1 <= x <= 1 the angle was chosen so that cos t(x) = x, so the top-left entry equals x exactly and the amplitude is loaded correctly. For |x| > 1 the claim fails, and this can be proved without knowing the extension value: the cosine of any real angle lies in the closed interval [-1, 1], so cos t(x) lies in [-1, 1] for every possible choice of t(x); but |x| > 1 means x lies outside [-1, 1], so cos t(x) cannot equal x. Thus the readout is wrong for every |x| > 1 regardless of how t is extended, and no guess about the extension is needed. The asymmetry is the key point: unitarity is a structural property of the rotation shape (two unit-circle entries for any angle), so it survives even a meaningless input; but the clean-entry readout is a semantic promise that the stored amplitude is exactly x, and a cosine can only ever equal numbers in [-1, 1]. A construction can therefore be perfectly unitary yet load the wrong number: a unitarity certificate must not be mistaken for a certificate that the intended coefficient was loaded.", "rubric": [ { "criterion": "Prove unitarity holds for all real x because every planar rotation satisfies cos^2+sin^2=1 regardless of the angle, independent of how t is extended", "id": "c1", "weight": 1 }, { "criterion": "Prove the top-left entry cos t(x) equals x on -1 <= x <= 1 by the defining choice of the angle", "id": "c2", "weight": 1 }, { "criterion": "Prove claim (B) fails for every |x|>1 because the cosine of any angle lies in [-1,1] while x does not, with no appeal to a particular extension value", "id": "c3", "weight": 1 }, { "criterion": "Explain the asymmetry: unitarity is structural and free, but the clean-entry readout is a semantic promise a cosine can satisfy only for values in [-1,1]", "id": "c4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "The amplitude gate is a planar rotation through the angle t(x) with top row (cos t(x), -sin t(x)).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation" ] }, { "claim": "Every planar rotation is unitary since cos^2+sin^2=1 for any angle, so the gate is unitary for all real x independent of the extension of t.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation_unitary" ] }, { "claim": "On -1 <= x <= 1 the angle is chosen so cos t(x) = x, so the top-left entry equals x.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation_cleanEntry" ] }, { "claim": "For |x|>1 the cosine of any angle lies in [-1,1] while x lies outside it, so cos t(x) cannot equal x for any extension value.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation_cleanEntry" ] }, { "claim": "Unitarity is structural and hypothesis-free, but the clean-entry readout is a semantic promise valid only where x is a possible cosine, so unitarity does not certify the right amplitude.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation_unitary" ] } ], "source_item_digest": "a01f8f5245eb84999d3a73584d8019263b296e87e308ca3a6b391d676bbcc2ea", "source_refs": [ { "anchor": "theorem amplitudeRotation_unitary (coefficient : Real) : amplitudeRotation coefficient ∈ _root_.Matrix.unitaryGroup (Fin 2) ℂ := realRotation_unitary _ (lines 93-96); theorem amplitudeRotation_cleanEntry (coefficient : Real) (lower : -1 ≤ coefficient) (upper : coefficient ≤ 1) : amplitudeRotation coefficient 0 0 = (coefficient : ℂ) := by simp [amplitudeRotation, Real.cos_arccos lower upper] (lines 99-102)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.realRotation_unitary" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation_unitary", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeRotation_cleanEntry" ], "task_hash": "c18c7f56cef06abc1e2f41f11e7b138a09af8723a451ce63248484dfda4667aa", "task_kind": "boundary", "topic": "oracle_composition" }, { "assumptions": [ "Each diagonal block B_k is unitary on the shared target space", "The full operator is block-diagonal in the control index" ], "context_hash": "b96740d4bd40f02165dd3d3d9a88c47387aa2acc454c500e5a57f913822e2f5a", "conventions": [ "Off-diagonal blocks of a block-diagonal matrix are zero", "Unitarity means the conjugate transpose times the operator is the identity" ], "core_id": "qcb-core-be-107", "difficulty": { "level": "intermediate", "rationale": "Requires knowing how conjugate transpose and multiplication interact with block-diagonal structure and assembling per-block unitarity into a global identity." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-blockdiagonal-unitary", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-4a974ac3a9b28b706e9d", "key": "qcb-core-be-107", "primary_sources": [], "question": "A controlled family of operators is placed block-diagonally: for each control value k there is a block B_k acting on the same target space, and the full operator is the block-diagonal matrix whose k-th diagonal block is B_k. Prove that if every block B_k is unitary then the whole block-diagonal operator is unitary, by showing how the conjugate transpose interacts with the block-diagonal structure and that the product collapses block-by-block to the identity.", "reference_answer": "Let D = blockDiagonal(B) be the operator on the product index (target x control) whose (k,k) block is B_k and whose off-diagonal blocks are 0. First, the conjugate transpose commutes with the block-diagonal construction: star(D) is the block-diagonal operator whose k-th block is star(B_k). This is checked entrywise: for two compound indices with control parts kR and kC, if kR != kC the entry is 0 on both sides (a block-diagonal matrix has zero off-diagonal blocks, and so does its conjugate transpose), and if kR = kC the entry is the conjugate-transpose entry of the single block B_k. Hence star(D) = blockDiagonal(k -> star(B_k)). Next, the product of two block-diagonal matrices is block-diagonal with the blocks multiplied pointwise: star(D) * D = blockDiagonal(k -> star(B_k) * B_k). Now use the hypothesis: each B_k is unitary, so star(B_k) * B_k = I for every k. Therefore the family k -> star(B_k) * B_k is the constant identity family, and blockDiagonal of the all-identity family is the identity matrix on the whole product space. Thus star(D) * D = I, which is exactly unitarity of D. The proof uses only (1) that conjugate transpose respects the block structure, (2) that block-diagonal multiplication is pointwise in the control index, and (3) the per-block unitarity hypothesis; no relation between different control values is needed because distinct blocks never interact.", "rubric": [ { "criterion": "Show the conjugate transpose of a block-diagonal matrix is block-diagonal with each block conjugate-transposed (off-diagonal blocks zero on both sides)", "id": "c1", "weight": 1 }, { "criterion": "Show the product of block-diagonal matrices multiplies blocks pointwise in the control index, giving blockDiagonal(k -> star(B_k)*B_k)", "id": "c2", "weight": 1 }, { "criterion": "Apply per-block unitarity star(B_k)*B_k = I to conclude the product is the identity, hence D is unitary; note distinct blocks never interact", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "The conjugate transpose of a block-diagonal matrix is block-diagonal with each block replaced by its conjugate transpose.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.blockDiagonal_unitary" ] }, { "claim": "Off-diagonal compound entries vanish on both sides because distinct control values index different blocks.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.blockDiagonal_unitary" ] }, { "claim": "The product of two block-diagonal matrices multiplies blocks pointwise in the control index.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.blockDiagonal_unitary" ] }, { "claim": "Per-block unitarity gives star(B_k)*B_k = I for every k, so the block family is the constant identity.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.blockDiagonal_unitary" ] }, { "claim": "Block-diagonal of the all-identity family is the identity, so star(D)*D = I and D is unitary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeLift_unitary" ] } ], "source_item_digest": "1b36266ef6c6f4eed5fc5fd43aada972b13c343a97ddce568581c553671be494", "source_refs": [ { "anchor": "theorem blockDiagonal_unitary {ι κ : Type*} [Fintype ι] [DecidableEq ι] [Fintype κ] [DecidableEq κ] (blocks : κ → _root_.Matrix ι ι ℂ) (unitary : ∀ k, blocks k ∈ _root_.Matrix.unitaryGroup ι ℂ) : _root_.Matrix.blockDiagonal blocks ∈ _root_.Matrix.unitaryGroup (ι × κ) ℂ (lines 122-145)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeLift_unitary" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.blockDiagonal_unitary" ], "task_hash": "8114a0a389962dc50c6a7bfca7eb6274bf529650c8adcaeb666fb67f72a96b71", "task_kind": "proof", "topic": "oracle_composition" }, { "assumptions": [ "PREPARE is unitary on the selector register alone", "The lift tensors identities on the coefficient and system registers" ], "context_hash": "9b66aa4a12e9112047a6121278eba262a97fc498902de9e33038d988ce50137c", "conventions": [ "The identity matrix is a unitary on any register", "star of a Kronecker product is the Kronecker product of the stars" ], "core_id": "qcb-core-be-109", "difficulty": { "level": "intermediate", "rationale": "Requires applying the Kronecker-product unitarity law twice through a nested tensor structure and interpreting identity factors as inert registers." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-tensor-lift-unitary", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-924a045577dd96107568", "key": "qcb-core-be-109", "primary_sources": [], "question": "A state-preparation unitary acts only on a selector register, but inside a composite machine it is applied on the full product register coefficient x (selector x system) as identity-on-coefficient tensor (PREPARE tensor identity-on-system). Prove that if PREPARE is unitary on the selector register alone then this tensor-lifted operator is unitary on the whole product register, and explain the physical reading: tensoring a gate with 'do nothing' on untouched registers never spoils unitarity.", "reference_answer": "The lifted operator is L = I_c ⊗ (PREPARE ⊗ I_s), a Kronecker product acting on the compound index coefficient x (selector x system). The governing fact is that the Kronecker product of two unitaries is unitary: if M and N are unitary on their respective factors then M ⊗ N is unitary on the product, because star(M ⊗ N) = star(M) ⊗ star(N) and (star(M) ⊗ star(N))(M ⊗ N) = (star(M) M) ⊗ (star(N) N) = I ⊗ I = I. Apply this twice. Inner step: PREPARE is unitary by hypothesis and I_s (the identity on the system) is unitary (its conjugate transpose is itself and the product is the identity), so PREPARE ⊗ I_s is unitary on selector x system. Outer step: I_c (identity on coefficient) is unitary, and PREPARE ⊗ I_s is unitary from the inner step, so I_c ⊗ (PREPARE ⊗ I_s) is unitary on the whole register. The physical reading: 'do nothing' on a register is represented by the identity, which is a trivial unitary; tensoring any unitary with identities on registers it does not touch leaves it unitary because the Kronecker factorization of the unitarity check splits into independent per-factor identities. A unitary gate that physically touches only one subsystem therefore remains a legitimate reversible operation on the enlarged machine, with the untouched registers carried along inertly.", "rubric": [ { "criterion": "State the lift as I_c ⊗ (PREPARE ⊗ I_s) and invoke that the Kronecker product of unitaries is unitary via star(M⊗N)=star(M)⊗star(N)", "id": "c1", "weight": 1 }, { "criterion": "Apply the Kronecker fact twice: PREPARE ⊗ I_s unitary (inner), then I_c ⊗ (that) unitary (outer), using that identities are unitary", "id": "c2", "weight": 1 }, { "criterion": "Give the physical reading: tensoring a gate with identity on untouched registers preserves unitarity because the check factorizes per subsystem", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "The lifted operator is the Kronecker product I_c ⊗ (PREPARE ⊗ I_s) on coefficient x (selector x system).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectorLift" ] }, { "claim": "The Kronecker product of two unitaries is unitary because star(M⊗N)=star(M)⊗star(N) and (star(M)⊗star(N))(M⊗N)=I⊗I.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectorLift_unitary" ] }, { "claim": "Inner step: PREPARE is unitary and I_s is unitary, so PREPARE ⊗ I_s is unitary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectorLift_unitary" ] }, { "claim": "Outer step: I_c is unitary, so I_c ⊗ (PREPARE ⊗ I_s) is unitary on the whole register.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectorLift_unitary" ] }, { "claim": "Physically, identities represent inert untouched registers, and the unitarity check factorizes per subsystem, so tensoring with them never spoils unitarity.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectorLift_unitary" ] } ], "source_item_digest": "76cc51c2e49f3f7062d8732b280e44f07723d2cfe3a72ceb5bf003e71965a5ff", "source_refs": [ { "anchor": "def selectorLift (prepare : _root_.Matrix selector selector ℂ) : ... := (1 : _root_.Matrix coefficient coefficient ℂ) ⊗ₖ (prepare ⊗ₖ (1 : _root_.Matrix system system ℂ)) (lines 254-264); theorem selectorLift_unitary ... := by apply _root_.Matrix.kronecker_mem_unitary ... (lines 267-281)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectorLift" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectorLift_unitary" ], "task_hash": "1faf9eaa4fca8f545dc8f1cfcf536a424b97f631bf0143299dcb84b43bc89b9b", "task_kind": "proof", "topic": "oracle_composition" }, { "assumptions": [ "Each selector value s names a permutation p_s of the system register", "SELECT preserves the coefficient and selector registers" ], "context_hash": "76f86abe3d4cc775ee052257ec6b09a550944f32b37bcbb829741cb3be406a1a", "conventions": [ "The forward map carries the selector through unchanged", "A basis permutation's matrix is a 0/1 permutation matrix" ], "core_id": "qcb-core-be-110", "difficulty": { "level": "intermediate", "rationale": "Requires constructing a two-sided inverse of a controlled permutation and reasoning about why the control register must be preserved for coherent inversion." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-controlled-select-bijection", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b49873b0ff2b1a2695f8", "key": "qcb-core-be-110", "primary_sources": [], "question": "A coherently controlled SELECT applies, for each selector value s, a permutation p_s of the system register, while leaving the coefficient and selector registers fixed. Prove that the map sending (coeff, (s, x)) to (coeff, (s, p_s(x))) is a genuine bijection of the whole product register by exhibiting its two-sided inverse, and explain why its matrix is automatically unitary. Why is it essential that the inverse applies the inverse permutation (p_s)^{-1} indexed by the SAME selector value s that is read on input?", "reference_answer": "Define F(coeff, (s, x)) = (coeff, (s, p_s(x))). Candidate inverse: G(coeff, (s, y)) = (coeff, (s, (p_s)^{-1}(y))). Check both composites. G(F(coeff,(s,x))) = G(coeff,(s,p_s(x))) = (coeff,(s,(p_s)^{-1}(p_s(x)))) = (coeff,(s,x)) since (p_s)^{-1} undoes p_s. F(G(coeff,(s,y))) = F(coeff,(s,(p_s)^{-1}(y))) = (coeff,(s,p_s((p_s)^{-1}(y)))) = (coeff,(s,y)). Both composites are the identity, so F is a bijection with inverse G. Its matrix is the permutation matrix of F (a 0/1 matrix with a single 1 per row and column), and every permutation matrix over C is unitary: its conjugate transpose is the permutation matrix of F^{-1}, and the product of a permutation matrix with its inverse is the identity. Hence SELECT is unitary purely because it is a basis permutation, with no amplitude arithmetic required. It is essential that G applies (p_s)^{-1} indexed by the SAME selector s: the selector is preserved by F, so on input the controlling value is s and the system permutation applied is p_s; to invert we must read that same preserved selector s and undo exactly p_s. If the inverse used a different selector's permutation p_{s'} the composite would apply p_s then (p_{s'})^{-1}, which is not the identity unless the permutations happen to coincide. Coherent control means the control register both selects which permutation acts and is itself carried through unchanged so the inverse can re-read it.", "rubric": [ { "criterion": "Exhibit the candidate inverse G applying (p_s)^{-1} on the system while preserving coefficient and selector", "id": "c1", "weight": 1 }, { "criterion": "Verify both composites G∘F and F∘G are the identity, concluding F is a bijection", "id": "c2", "weight": 1 }, { "criterion": "Conclude unitarity because the matrix is a basis permutation, and explain that the inverse must use the same preserved selector s so coherent control re-reads the correct permutation", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Define F(coeff,(s,x)) = (coeff,(s,p_s(x))) and the candidate inverse G applying (p_s)^{-1} on the system.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.controlledSystemEquiv" ] }, { "claim": "G∘F = identity because (p_s)^{-1} undoes p_s under the preserved selector s.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.controlledSystemEquiv" ] }, { "claim": "F∘G = identity because p_s undoes (p_s)^{-1}, so F is a bijection.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.controlledSystemEquiv" ] }, { "claim": "The matrix of F is a basis permutation matrix, which is unitary over C.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectLift_unitary" ] }, { "claim": "The inverse must apply (p_s)^{-1} for the same preserved selector s; coherent control carries the selector through so the inverse re-reads the correct permutation.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.controlledSystemEquiv" ] } ], "source_item_digest": "0cc5f51f216a0d5dbe1cc2aa09034b4fe15c07d1d632632b930ae11ec0505a4d", "source_refs": [ { "anchor": "def controlledSystemEquiv (permutation : selector → system ≃ system) : LCUIndex coefficient selector system ≃ LCUIndex coefficient selector system where toFun index := (index.1, (index.2.1, permutation index.2.1 index.2.2)) invFun index := (index.1, (index.2.1, (permutation index.2.1).symm index.2.2)) left_inv index := by simp right_inv index := by simp (lines 315-327); theorem selectLift_unitary ... := equivPermutationMatrix_unitary _ (lines 340-349)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectLift", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_unitary" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.controlledSystemEquiv", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectLift_unitary" ], "task_hash": "499ff5ab00014a4936cd6e54f83756f612f7c9858c56297b1e96fb0823ebe95f", "task_kind": "proof", "topic": "oracle_composition" }, { "assumptions": [ "Each permutation matrix has exactly one 1 per row and per column", "Matrices act on column vectors by left multiplication" ], "context_hash": "6e1a46205af169b134fbc6ddc81efb973b59d76765e2ee659ecc78fade3737db", "conventions": [ "P(e)[r,c] = 1 iff r = e(c)", "The composite 'first then second' sends c to second(first(c))" ], "core_id": "qcb-core-be-111", "difficulty": { "level": "intermediate", "rationale": "Requires proving the single-surviving-term rule for permutation-matrix multiplication and reasoning carefully about the forced order reversal under left action." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-permutation-composition-order", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-225bf1c006db10d58b18", "key": "qcb-core-be-111", "primary_sources": [], "question": "Finite basis permutations are represented by 0/1 matrices: for a bijection e of the index set, its matrix has a 1 at (row, column) exactly when row = e(column). Prove that composing two such permutation matrices multiplies them in reverse order: the matrix of the composite that does 'first then second' equals (matrix of second) times (matrix of first). Make explicit how matrix multiplication selects a single surviving term and why the order reversal is forced rather than a convention.", "reference_answer": "Let P(e) denote the permutation matrix of e, with P(e)[r,c] = 1 if r = e(c) else 0. Multiplying P(e) on the left of any operator applies e^{-1} to the row index: (P(e) * X)[r,c] = X[e^{-1}(r), c], because the matrix product sums over an intermediate index k the term P(e)[r,k] X[k,c], and P(e)[r,k] is nonzero (equal to 1) only at the single index k = e^{-1}(r). Now take X = P(first) and e = second: (P(second) * P(first))[r,c] = P(first)[second^{-1}(r), c] = 1 iff second^{-1}(r) = first(c), i.e. iff r = second(first(c)). That is exactly the condition defining P(first then second), the composite that applies 'first' and then 'second'. Hence P(second) * P(first) = P(first then second). The single surviving term is the heart of the computation: because each column of a permutation matrix has exactly one 1, the convolution sum over the contracted index collapses to one summand, turning matrix multiplication into function composition. The order reversal is forced, not conventional: a matrix acts on column vectors from the left, so applying 'first' then 'second' to a state means second*(first*v), and reading off the induced map on basis labels gives the composite in the order r = second(first(c)) with the matrices written right-to-left. Any other ordering would contradict how left multiplication selects the inverse-image row.", "rubric": [ { "criterion": "Establish that left multiplication by P(e) sends row index r to the single term at e^{-1}(r), because each column has exactly one 1", "id": "c1", "weight": 1 }, { "criterion": "Compute (P(second)*P(first))[r,c] = 1 iff r = second(first(c)), matching P(first then second)", "id": "c2", "weight": 1 }, { "criterion": "Explain the order reversal is forced by left action on column vectors (second*(first*v)), not a free convention", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Left multiplication by P(e) picks the single term at the inverse image: (P(e)*X)[r,c] = X[e^{-1}(r),c].", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_mul_apply" ] }, { "claim": "Setting X = P(first), e = second gives (P(second)*P(first))[r,c] = P(first)[second^{-1}(r),c].", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_mul" ] }, { "claim": "This entry is 1 iff second^{-1}(r) = first(c), equivalently r = second(first(c)).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_mul" ] }, { "claim": "That is the matrix of the composite 'first then second', so P(second)*P(first) = P(first then second).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_mul" ] }, { "claim": "The order reversal is forced because left action means second*(first*v); any other order contradicts the inverse-image selection.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_mul_apply" ] } ], "source_item_digest": "9b7764d055cff46ba40805e22a51db043464ec809fb8b49c72ab4ffd7bed969a", "source_refs": [ { "anchor": "theorem equivPermutationMatrix_mul_apply (equiv : ι ≃ ι) (operator : _root_.Matrix ι κ ℂ) (row : ι) (column : κ) : (equivPermutationMatrix equiv * operator) row column = operator (equiv.symm row) column (lines 167-187); theorem equivPermutationMatrix_mul (first second : ι ≃ ι) : equivPermutationMatrix second * equivPermutationMatrix first = equivPermutationMatrix (first.trans second) (lines 206-223)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_mul_apply", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_mul" ], "task_hash": "675d4553fe6286cc20b12350272df173682da9e9f7494a092d6a952e44a13cb8", "task_kind": "proof", "topic": "oracle_composition" }, { "assumptions": [ "The dagger is stored as the same matrix as the oracle", "The oracle's integer image map is asserted self-inverse" ], "context_hash": "b712c194e863de6607940ae481fb66832a8cf6e7bbf4432075a818a1174ef8e1", "conventions": [ "A permutation matrix P is unitary with P-dagger = P^{-1}", "An uncompute restores the register iff forward then inverse is the identity" ], "core_id": "qcb-core-be-120", "difficulty": { "level": "advanced", "rationale": "Requires distinguishing involutivity from mere permutation unitarity and seeing that the self-inverse image hypothesis is exactly what licenses reusing the same oracle as its own uncompute." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "be-indicator-involution-uncompute", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-bd9d24f365d0ad4463fd", "key": "qcb-core-be-120", "primary_sources": [], "question": "A fixed indicator oracle and its stated dagger are certified for a warm Robin instance by four facts bundled together: the dagger's matrix equals the oracle's own matrix; the integer image map of the oracle is self-inverse; and the oracle matrix is a permutation (each row has a unique column with a 1, and each column a unique row). From these, explain the precise sense in which this indicator oracle is self-inverse, why 'dagger equals itself' plus 'image map self-inverse' together certify a clean uncompute, and what would break the uncompute if only the permutation property held without the self-inverse image.", "reference_answer": "The bundle certifies that the indicator oracle U is an involutive permutation and that its dagger is realized by the SAME matrix. Reading the four facts: (a) the dagger's stored matrix equals the oracle's matrix, so applying 'U-dagger' is operationally identical to applying U again; (b) the oracle's integer image map j |-> image(j) satisfies image(image(j)) = j, i.e. it is self-inverse as a function on labels; (c) and (d) the matrix is a genuine permutation matrix (unique 1 per row and per column). Combining (b),(c),(d): U is the permutation matrix of a self-inverse label map, so U is an involution, U*U = I, hence U-dagger = U already follows abstractly, and (a) confirms the stored dagger matches. The precise sense of self-inverse is this involution U^2 = I. Why (a) plus (b) certify a clean uncompute: an uncompute step must restore the register, and it does so iff running the 'inverse' block after the forward block returns to the start. Here the inverse block is stored as U-dagger = U (fact a), and because the image map is self-inverse (fact b) applying U twice returns every basis label to itself, so the composite forward-then-uncompute is exactly the identity on labels. That is a clean, exact uncompute with no residual. What breaks if only the permutation property (c),(d) held WITHOUT self-inverse image: a general permutation P is unitary with P-dagger = P^{-1}, but P^{-1} need not equal P. Then applying the SAME stored matrix twice gives P^2, not the identity, so re-running the forward block does not uncompute; the register ends in the permuted-twice state and the ancilla is not restored. The self-inverse image hypothesis is exactly what guarantees that the physically convenient choice 'use the same oracle to uncompute' is correct rather than an error; permutation-ness alone gives unitarity but not involutivity.", "rubric": [ { "criterion": "Interpret the bundle as: U is a permutation matrix of a self-inverse label map, hence an involution U^2 = I with U-dagger = U", "id": "c1", "weight": 1 }, { "criterion": "Explain that dagger-equals-itself (fact a) plus self-inverse image (fact b) make forward-then-same-oracle an exact clean uncompute with no residual", "id": "c2", "weight": 1 }, { "criterion": "Show that permutation-ness alone gives P-dagger = P^{-1} but not P^2 = I, so reusing the same matrix would give P^2 and fail to uncompute without the self-inverse image hypothesis", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The oracle matrix is a permutation of a self-inverse label map, hence an involution U^2 = I with U-dagger = U.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RobinEvolution.warmRobinIndicatorCertificate" ] }, { "claim": "Fact (a) stores the dagger as the same matrix, and fact (b) self-inverse image makes applying U twice the identity on labels.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RobinEvolution.warmRobinIndicatorCertificate" ] }, { "claim": "Thus forward-then-same-oracle is an exact clean uncompute with no residual on the register.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RobinEvolution.warmRobinIndicatorCertificate" ] }, { "claim": "With only permutation-ness, P-dagger = P^{-1} need not equal P, so reusing the same matrix gives P^2, not the identity, and the uncompute fails.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RobinEvolution.warmRobinIndicatorCertificate" ] } ], "source_item_digest": "e5fce40032b6c3cf22a772b67ac145bb32118f946cb8534d087744848f82c38b", "source_refs": [ { "anchor": "theorem warmRobinIndicatorCertificate : (GHL2025.oneTermRobinGate_U_indic_dagger warmRobinParameters).matrix = (GHL2025.oneTermRobinGate_U_indic warmRobinParameters).matrix ∧ (∀ j : Nat, GHL2025.indicatorOracleImage warmRobinParameters (GHL2025.indicatorOracleImage warmRobinParameters j) = j) ∧ (... indicatorOracleMatrix is a permutation ...) (lines 180-210)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RobinEvolution.lean", "sha256": "8673572458ff260cf35d6cfce60ffcc52b5b2be721671b3171919e32d0b73a37" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RobinEvolution.warmRobinParameters" ], "target_nodes": [ "declaration:QuantumBlockEncoding.RobinEvolution.warmRobinIndicatorCertificate" ], "task_hash": "e0a3005a91cab5b8ca784f6d0d96e9da609421af7a992239cb3ac61258ed590d", "task_kind": "boundary", "topic": "oracle_composition" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-101", "difficulty": { "level": "basic", "rationale": "A short structured proof from group axioms; requires naming the uniqueness/cancellation steps rather than quoting the result, but no deep machinery." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-representation-foundations", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-91bad03f60703a92a44c", "key": "qcb-core-plan-101-homomorphism-preserves-identity-inverse", "primary_sources": [ { "anchors": "Definition 3.5 (group homomorphism) and Eq. (3.23)-(3.24) proving phi(e_G)=e_H and phi(g^{-1})=phi(g)^{-1}; Definition 3.6 (representation) and the remark that a representation satisfies phi(e)=1_V and phi(g^{-1})=phi(g)^{-1}", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "17", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let (G, .) and (H, *) be groups with identities e_G and e_H, and let phi: G -> H be a group homomorphism, meaning phi(x.y) = phi(x)*phi(y) for all x, y in G. Prove from the homomorphism property alone (without assuming phi is injective, surjective, or that G or H is finite) that phi(e_G) = e_H and that phi(g^{-1}) = phi(g)^{-1} for every g in G. State explicitly which group axioms (uniqueness of identity, uniqueness of inverses, cancellation) you invoke at each step, and explain why these two facts justify the standing claims that every representation phi satisfies phi(e) = 1_V and phi(g^{-1}) = phi(g)^{-1}.", "reference_answer": "For the identity: pick any g in G and write g = e_G . g, so phi(g) = phi(e_G . g) = phi(e_G) * phi(g). Thus phi(e_G) * phi(g) = phi(g) = e_H * phi(g). Right-multiplying both sides by phi(g)^{-1} in H (which exists because H is a group) and cancelling gives phi(e_G) = e_H; equivalently, phi(e_G) acts as a left identity on phi(g), and since the identity of H is unique, phi(e_G) = e_H. For the inverse: fix g in G. Then e_H = phi(e_G) = phi(g . g^{-1}) = phi(g) * phi(g^{-1}), and likewise e_H = phi(g^{-1} . g) = phi(g^{-1}) * phi(g). So phi(g^{-1}) is a two-sided inverse of phi(g) in H; because inverses in a group are unique, phi(g^{-1}) = phi(g)^{-1}. The only axioms used are the homomorphism identity, the group axioms e_G . g = g and g . g^{-1} = g^{-1} . g = e_G in G, existence of inverses in H (to cancel), and uniqueness of the identity and of inverses in H. Applied to a representation phi: G -> GL(V) with H = GL(V) and identity 1_V, these say precisely phi(e) = 1_V and phi(g^{-1}) = phi(g)^{-1}; in particular every phi(g) is automatically invertible with the stated inverse, so invertibility need not be postulated separately.", "rubric": [ { "criterion": "Derives phi(e_G) = e_H by applying the homomorphism identity to g = e_G . g and cancelling / invoking uniqueness of the identity in H", "id": "r1", "weight": 1 }, { "criterion": "Derives phi(g^{-1}) = phi(g)^{-1} by showing phi(g^{-1}) is a two-sided inverse of phi(g) and invoking uniqueness of inverses in H", "id": "r2", "weight": 1 }, { "criterion": "Identifies the specific group axioms used (homomorphism property, identity/inverse axioms in G, cancellation/uniqueness in H) rather than asserting the results", "id": "r3", "weight": 1 }, { "criterion": "Connects the two facts to the representation statements phi(e) = 1_V and phi(g^{-1}) = phi(g)^{-1}, noting invertibility of phi(g) is a consequence not a hypothesis", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Applying phi to e_G . g = g gives phi(e_G)*phi(g) = phi(g).", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Cancellation / uniqueness of identity in H forces phi(e_G) = e_H.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Applying phi to g.g^{-1} = e_G = g^{-1}.g gives phi(g)*phi(g^{-1}) = e_H = phi(g^{-1})*phi(g).", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Uniqueness of inverses in H forces phi(g^{-1}) = phi(g)^{-1}.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Specializing H = GL(V) yields phi(e) = 1_V and phi(g^{-1}) = phi(g)^{-1}, so each phi(g) is invertible automatically.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "39cf1c51e2f1204e64df33d8a8210e06de5ea44fdc2525809858621ecd97be1a", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "2402a26f91ab8dd36aab5ed55c7e1d4fb465da101143f204e5445e6268a05ab6", "task_kind": "proof", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-102", "difficulty": { "level": "intermediate", "rationale": "Requires correctly handling the averaged inner product, the reindexing bijection, and the adjoint-equals-inverse step; a clean multi-step argument beyond definition recall." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-representation-foundations", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b87b2fed369724fb0cfe", "key": "qcb-core-plan-102-weyl-unitarity-trick-invariant-complement", "primary_sources": [ { "anchors": "Second (alternative) proof of Proposition 3.14 via Weyl's unitarity trick: the averaged inner product _G, G-invariance of W^perp, and (phi,V) becoming unitary with phi(g^{-1})=phi(g)^dagger", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "19-20", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let (phi, V) be a finite-dimensional complex representation of a finite group G, and let <.|.> be any inner product on V. Define the averaged form _G := (1/|G|) sum_{g in G} . Prove that (a) <.|.>_G is again an inner product on V; (b) every phi(g) is unitary with respect to <.|.>_G, i.e. _G = _G; and (c) consequently, for any G-invariant subspace W, its orthogonal complement W^perp taken with respect to <.|.>_G is also G-invariant, so that V = W (+) W^perp as representations. Explain exactly where the ability to average over G is used.", "reference_answer": "(a) Each summand is sesquilinear in (v,w) because phi(g) is linear and <.|.> is sesquilinear, and a finite average of sesquilinear forms is sesquilinear; Hermitian symmetry is inherited termwise. For positivity, _G = (1/|G|) sum_g >= 0 as an average of nonnegative terms, and it vanishes only if every term is 0; the g = e term gives = 0, hence v = 0 since <.|.> is definite. So <.|.>_G is a genuine inner product. (b) Fix h in G. Then _G = (1/|G|) sum_g = (1/|G|) sum_g by the homomorphism property. As g ranges over G so does g' = gh (left translation is a bijection of G), so the sum equals _G. Thus phi(h) preserves <.|.>_G and is unitary for it. (c) Let W be G-invariant and u in W^perp. For any g and w in W, _G = _G = _G, using that the adjoint of the <.|.>_G-unitary phi(g) is phi(g)^{-1} = phi(g^{-1}). Invariance gives phi(g^{-1}) w in W, so the pairing is 0; hence phi(g)u in W^perp and W^perp is invariant. Then V = W (+) W^perp (nondegeneracy) with both summands invariant, a decomposition of representations. The averaging over G is used exactly twice: forming the normalized sum (1/|G|) sum_g requires it to be finite (compact groups replace it by Haar integration), and the reindexing g -> gh being a bijection is what makes the form G-invariant.", "rubric": [ { "criterion": "Shows <.|.>_G is sesquilinear, Hermitian, and positive-definite, using the g=e term for definiteness", "id": "r1", "weight": 1 }, { "criterion": "Proves G-invariance of <.|.>_G via the homomorphism property and the reindexing g -> gh being a bijection of G", "id": "r2", "weight": 1 }, { "criterion": "Deduces each phi(g) is unitary for <.|.>_G and uses phi(g)^* = phi(g^{-1}) to show W^perp is invariant when W is", "id": "r3", "weight": 1 }, { "criterion": "Concludes V = W (+) W^perp as representations and pinpoints the normalized average and reindexing as the essential ingredient", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Each term is sesquilinear/Hermitian; the g=e term gives definiteness, so <.|.>_G is an inner product.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "_G = (1/|G|) sum_g by the homomorphism property.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Reindexing g'=gh (a bijection of G) shows the average is unchanged, so phi(h) is unitary for <.|.>_G.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Unitarity gives phi(g)^* = phi(g^{-1}); for u in W^perp, w in W, _G = _G = 0.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Hence W^perp is G-invariant and V = W (+) W^perp as representations.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "4aa8a582922a72a46c4a3189cc1902dd7961a753460a2c5189793a0372a0dff8", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "3f1ec006682acdbb536e22d7543b1189b81f3d92028b90f918bfe316bed58c2e", "task_kind": "proof", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-103", "difficulty": { "level": "intermediate", "rationale": "A multi-step operator computation (equivariance, image, idempotency) plus recognizing the role of the characteristic hypothesis in the normalization." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-representation-foundations", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-a2935ca908a46361c8c8", "key": "qcb-core-plan-103-invariant-complement-averaged-projection", "primary_sources": [ { "anchors": "Proof sketch of Proposition 3.14: the averaged projection Q_W = (1/|G|) sum_g phi(g) P_W phi(g)^{-1} (Eq. 3.26) with im Q_W = W, phi(g)Q_W = Q_W phi(g), and W' = ker Q_W the invariant complement; characteristic-not-dividing-|G| hypothesis", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "19-20", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let (phi, V) be a representation of a finite group G over a field whose characteristic does not divide |G|, and let W be a G-invariant subspace with some (not necessarily invariant) projection P_W onto W (P_W^2 = P_W, im P_W = W). Define Q_W := (1/|G|) sum_{g in G} phi(g) P_W phi(g)^{-1}. Prove that Q_W commutes with every phi(g), that im Q_W = W and Q_W is a projection, and conclude that ker Q_W is a G-invariant complement, so V = W (+) ker Q_W as representations. Indicate where the hypothesis on the characteristic is used.", "reference_answer": "Equivariance: for h in G, phi(h) Q_W phi(h)^{-1} = (1/|G|) sum_g phi(h)phi(g) P_W phi(g)^{-1} phi(h)^{-1} = (1/|G|) sum_g phi(hg) P_W phi(hg)^{-1}. Reindexing g' = hg (a bijection of G) shows this equals Q_W, so phi(h) Q_W = Q_W phi(h) for all h. Image in W: invariance means phi(g)^{-1} maps V into V, P_W maps into W, and phi(g) maps W into W, so each term phi(g) P_W phi(g)^{-1} has image in W; the average does too, giving im Q_W subset W. Q_W fixes W pointwise: for w in W, phi(g)^{-1} w in W, P_W fixes it, and phi(g) phi(g)^{-1} w = w, so each term returns w; the normalized average (1/|G|) sum_g w = w. This normalization is exactly where the characteristic hypothesis enters: 1/|G| must exist in the field, i.e. |G| must be invertible, which holds iff the characteristic does not divide |G|. Hence Q_W|_W = id_W, so Q_W maps everything into W and acts as the identity there, giving Q_W^2 = Q_W: a projection onto W. Finally, since Q_W commutes with all phi(g), both im Q_W = W and ker Q_W are G-invariant, and V = im Q_W (+) ker Q_W = W (+) ker Q_W. Thus W has a G-invariant complement and the decomposition is of representations.", "rubric": [ { "criterion": "Proves phi(h) Q_W phi(h)^{-1} = Q_W via the reindexing g -> hg, i.e. Q_W commutes with phi", "id": "r1", "weight": 1 }, { "criterion": "Shows im Q_W subset W (each averaged term lands in W by invariance) and Q_W fixes W pointwise, hence im Q_W = W and Q_W^2 = Q_W", "id": "r2", "weight": 1 }, { "criterion": "Identifies that the normalization 1/|G| requires |G| invertible, i.e. char does not divide |G|", "id": "r3", "weight": 1 }, { "criterion": "Concludes ker Q_W is a G-invariant complement giving V = W (+) ker Q_W as representations", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "phi(h)Q_W phi(h)^{-1} = (1/|G|) sum_g phi(hg)P_W phi(hg)^{-1}; reindexing g->hg shows this equals Q_W.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Hence Q_W commutes with every phi(h).", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Invariance of W makes each phi(g)P_W phi(g)^{-1} have image in W, so im Q_W subset W.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "For w in W each term returns w; the normalized average (needing |G| invertible, i.e. char not dividing |G|) gives Q_W w = w.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Thus im Q_W = W, Q_W^2 = Q_W, and commuting with phi makes ker Q_W invariant: V = W (+) ker Q_W.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "da4e2877f4284ace0995f01785a6c7c6c5c209746ca8a2c28b52e5252370b82f", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "4b017d9b5abb1a8bc7943761bbadd084a862c7d9c3aaf3cf8b7bcffab77790a4", "task_kind": "proof", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-109", "difficulty": { "level": "intermediate", "rationale": "A multi-step argument chaining additivity, orthonormality, an integrality observation, and an iff for isomorphism; genuinely uses the orthogonality relations." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-character-theory", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-062ced27e7c425e06b4d", "key": "qcb-core-plan-109-multiplicity-character-inner-product", "primary_sources": [ { "anchors": "Proposition 3.30 (orthonormality of irreducible characters, inner product Eq. 3.85) and Proposition 3.31(i)-(iii): multiplicity n_i = (chi_V, chi_{W_i}), irreducibility iff (chi_V, chi_V)=1, isomorphism iff equal characters", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "29-31", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let G be a finite group and let the characters of its pairwise inequivalent complex irreducible representations W_1, ..., W_k be chi_1, ..., chi_k, orthonormal for the inner product (x, y) = (1/|G|) sum_{g} conj(x(g)) y(g); so (chi_i, chi_j) = delta_{ij}. Let (phi, V) be an arbitrary finite-dimensional complex representation with isotypical decomposition V ~ (+)_i W_i^{(+) n_i} (n_i the multiplicity of W_i). Prove that n_i = (chi_i, chi_V), where chi_V is the character of V. Then prove the two consequences: V is irreducible iff (chi_V, chi_V) = 1, and two representations are isomorphic iff they have equal characters.", "reference_answer": "By additivity of the character under direct sums (chi_{A (+) B} = chi_A + chi_B, which follows from tr being additive on block-diagonal operators), the isotypical decomposition V ~ (+)_j W_j^{(+) n_j} gives chi_V = sum_j n_j chi_j. Taking the inner product with chi_i and using bilinearity (conjugate-linearity in the first slot but the n_j are real nonnegative integers) together with orthonormality, (chi_i, chi_V) = (chi_i, sum_j n_j chi_j) = sum_j n_j (chi_i, chi_j) = sum_j n_j delta_{ij} = n_i. This proves the multiplicity formula. Consequence 1: compute (chi_V, chi_V) = (sum_i n_i chi_i, sum_j n_j chi_j) = sum_{i,j} n_i n_j delta_{ij} = sum_i n_i^2. Since the n_i are nonnegative integers, sum_i n_i^2 = 1 holds iff exactly one n_i equals 1 and all others are 0, i.e. iff V is (isomorphic to) a single irreducible W_i; thus V is irreducible iff (chi_V, chi_V) = 1. Consequence 2: if V ~ V' then they have the same character (characters are isomorphism invariants, since conjugate representation matrices have equal trace). Conversely, the multiplicities n_i = (chi_i, chi_V) are determined entirely by chi_V; so if chi_V = chi_{V'} then n_i(V) = n_i(V') for all i, meaning V and V' have the same isotypical decomposition and hence V ~ V'. Therefore two representations are isomorphic iff their characters coincide.", "rubric": [ { "criterion": "Expands chi_V = sum_j n_j chi_j from additivity of characters over the isotypical decomposition", "id": "r1", "weight": 1 }, { "criterion": "Computes (chi_i, chi_V) = sum_j n_j delta_{ij} = n_i using orthonormality of the irreducible characters", "id": "r2", "weight": 1 }, { "criterion": "Derives (chi_V, chi_V) = sum_i n_i^2 and argues it equals 1 (n_i nonnegative integers) iff V is irreducible", "id": "r3", "weight": 1 }, { "criterion": "Proves V ~ V' iff chi_V = chi_{V'} by noting the multiplicities, hence the whole decomposition, are determined by the character", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Additivity of characters gives chi_V = sum_j n_j chi_j from V ~ (+)_j W_j^{(+)n_j}.", "supporting_nodes": [ "concept:harmonic-analysis" ] }, { "claim": "Orthonormality (chi_i, chi_j) = delta_{ij} yields (chi_i, chi_V) = n_i.", "supporting_nodes": [ "concept:harmonic-analysis" ] }, { "claim": "(chi_V, chi_V) = sum_i n_i^2, which equals 1 iff a single n_i = 1: V irreducible.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Equal characters force equal multiplicities n_i = (chi_i, chi_V), hence the same decomposition and V ~ V'.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "940d170e2bda6414a28189165ce2bc37170ee0653d2cf788e54f04b40a60ecd5", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:harmonic-analysis", "concept:matrix-analysis" ], "task_hash": "ea113d36c58925e6934f6ca5d497fc01dc25549c636995883d751af4bddcfb4f", "task_kind": "proof", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-110", "difficulty": { "level": "advanced", "rationale": "Requires combining the additivity of Hom spaces with both parts of Schur and tracking inequivalence; a conceptual proof underlying the realization of multiplicity spaces as intertwiner spaces." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-intertwiners-multiplicity", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-c78c1706e82383b2075f", "key": "qcb-core-plan-110-intertwiner-space-dimension-multiplicity", "primary_sources": [ { "anchors": "Proposition 3.33(i): k = dim hom_G(U, V); Eq. (3.89)-(3.91) additivity of Hom and the Schur evaluations hom_G(U,F)=0, dim hom_G(U,U)=1; Corollary 3.34 isotypical decomposition V ~ (+)_i W_i (x) hom_G(W_i, V)", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "31-32", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let (phi, V) be a finite-dimensional complex representation of a finite group G, and let ( phi_U, U) be a complex irreducible representation of G. Recall that Hom_G(U, V) is the space of intertwiners f: U -> V, i.e. linear maps satisfying f . phi_U(g) = phi(g) . f for all g, and that Hom_G is additive in each argument: Hom_G(X, Y (+) Z) ~ Hom_G(X, Y) (+) Hom_G(X, Z). Prove that dim Hom_G(U, V) equals the multiplicity k of U in the isotypical decomposition of V. You should use Schur's Lemma to evaluate Hom_G(U, U) and Hom_G(U, F) where F collects the irreducible summands of V inequivalent to U.", "reference_answer": "Decompose V ~ V_U (+) F, where V_U ~ U^{(+)k} is the isotypical component of U (U appearing with multiplicity k) and F is the direct sum of all irreducible summands of V inequivalent to U. By additivity of Hom_G in the second argument, Hom_G(U, V) ~ Hom_G(U, U^{(+)k}) (+) Hom_G(U, F) ~ ((+)_{i=1}^{k} Hom_G(U, U)) (+) Hom_G(U, F). Now evaluate the pieces with Schur's Lemma. First, Hom_G(U, U) is the space of G-equivariant endomorphisms of the irreducible U over C; by Schur (ii) every such endomorphism is a scalar multiple of 1_U, so Hom_G(U, U) = C . 1_U has dimension 1. Second, for any irreducible summand W of F, W is inequivalent to U, so by Schur (i) any equivariant map U -> W is either zero or an isomorphism; an isomorphism would make U ~ W, contradicting inequivalence, so the only equivariant map is 0. Hence Hom_G(U, W) = 0 for each such W, and by additivity Hom_G(U, F) = 0. Combining, dim Hom_G(U, V) = sum_{i=1}^{k} dim Hom_G(U, U) + dim Hom_G(U, F) = k . 1 + 0 = k. Therefore the dimension of the intertwiner space Hom_G(U, V) equals the multiplicity of U in V, which realizes the multiplicity space of U as the space of intertwiners from U into V.", "rubric": [ { "criterion": "Decomposes V ~ U^{(+)k} (+) F and uses additivity of Hom_G to split Hom_G(U, V) accordingly", "id": "r1", "weight": 1 }, { "criterion": "Uses Schur (ii) to show Hom_G(U, U) = C 1_U is one-dimensional", "id": "r2", "weight": 1 }, { "criterion": "Uses Schur (i) and inequivalence to show Hom_G(U, W) = 0 for every irreducible summand W of F, hence Hom_G(U, F) = 0", "id": "r3", "weight": 1 }, { "criterion": "Concludes dim Hom_G(U, V) = k, the multiplicity of U in V (multiplicity space realized as intertwiners)", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "V ~ U^{(+)k} (+) F and additivity give Hom_G(U, V) ~ ((+)_{i=1}^k Hom_G(U,U)) (+) Hom_G(U, F).", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Schur (ii) over C: Hom_G(U, U) = C 1_U, dimension 1.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Schur (i) plus inequivalence: Hom_G(U, W) = 0 for irreducible W !~ U, so Hom_G(U, F) = 0.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Summing, dim Hom_G(U, V) = k.1 + 0 = k, the multiplicity of U in V.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "9b9f633d427e365f4b941d9ba0ac65a7198e50c1ed20cb2f793d3398242285ab", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "8f166953014899b3cfc29f716a7967dd2c1d25a63af12f3847ee563dff86e16c", "task_kind": "proof", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-111", "difficulty": { "level": "intermediate", "rationale": "Requires the equivariance computation plus the correct block interpretation and both parts of Schur; a focused instance of the symmetrization theorem." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-schur-lemma", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-1731d8dbf2a88cbaddab", "key": "qcb-core-plan-111-symmetrization-block-diagonal-isotypical", "primary_sources": [ { "anchors": "Proposition 3.23 (symmetrization block form, Eq. 3.32) and the worked multiplicity-free S_2 / S_3 computations in Section 3.4.1 showing off-diagonal blocks vanish (Schur i) and diagonal blocks are scalar (Schur ii)", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "21-24", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let (phi, V) be a finite-dimensional complex representation of a finite group G, and for X in L(V) define the symmetrized operator Xbar := (1/|G|) sum_{g in G} phi(g) X phi(g)^{-1}. (a) Prove that Xbar is G-invariant, i.e. phi(h) Xbar = Xbar phi(h) for all h in G. (b) Using Schur's Lemma, prove that if V = W_1 (+) W_2 is a multiplicity-free decomposition into two inequivalent irreducibles, then in a basis adapted to W_1, W_2 the operator Xbar is block diagonal with each diagonal block a scalar multiple of the identity, and the off-diagonal blocks vanish. Explain which part of Schur kills the off-diagonal blocks and which part scalarizes the diagonal blocks.", "reference_answer": "(a) For h in G, phi(h) Xbar phi(h)^{-1} = (1/|G|) sum_g phi(h)phi(g) X phi(g)^{-1}phi(h)^{-1} = (1/|G|) sum_g phi(hg) X phi(hg)^{-1}. Reindexing g' = hg (a bijection of G) leaves the average unchanged, so phi(h) Xbar phi(h)^{-1} = Xbar, i.e. phi(h) Xbar = Xbar phi(h). Thus Xbar commutes with the whole representation; it is G-equivariant. (b) Write Xbar in block form with respect to V = W_1 (+) W_2: Xbar = [[A_{11}, A_{12}], [A_{21}, A_{22}]], where A_{ij}: W_j -> W_i. Equivariance phi(h) Xbar = Xbar phi(h) restricted to blocks says A_{ij} phi_j(h) = phi_i(h) A_{ij} for all h, i.e. each A_{ij} is an intertwiner from the irreducible W_j to the irreducible W_i. For the off-diagonal blocks i != j: W_1 and W_2 are inequivalent irreducibles, so by Schur part (i) any intertwiner between them is zero; hence A_{12} = 0 and A_{21} = 0. For the diagonal blocks i = j: A_{ii} is an equivariant endomorphism of the irreducible W_i over C, so by Schur part (ii) A_{ii} = lambda_i . 1_{W_i} for some scalar lambda_i. Therefore Xbar = lambda_1 1_{W_1} (+) lambda_2 1_{W_2}, block diagonal with scalar blocks. Part (i) of Schur kills the off-diagonal blocks (maps between inequivalent irreps), and part (ii) scalarizes the diagonal blocks (equivariant endomorphisms of an irrep). This is exactly the small multiplicity-free instance of the general symmetrization result; with multiplicities the diagonal blocks would instead be 1_{W_k} tensored with an arbitrary operator on the multiplicity space.", "rubric": [ { "criterion": "Proves phi(h) Xbar phi(h)^{-1} = Xbar via the reindexing g -> hg, so Xbar is G-equivariant", "id": "r1", "weight": 1 }, { "criterion": "Translates the block A_{ij} into an intertwiner W_j -> W_i using equivariance", "id": "r2", "weight": 1 }, { "criterion": "Uses Schur (i) and inequivalence of W_1, W_2 to force the off-diagonal blocks A_{12} = A_{21} = 0", "id": "r3", "weight": 1 }, { "criterion": "Uses Schur (ii) over C to make each diagonal block A_{ii} = lambda_i 1_{W_i}, giving the scalar block-diagonal form and naming which part of Schur does which", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Reindexing g -> hg gives phi(h) Xbar phi(h)^{-1} = Xbar, so Xbar commutes with the representation.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Block A_{ij} satisfies A_{ij} phi_j(h) = phi_i(h) A_{ij}: an intertwiner W_j -> W_i.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Schur (i): off-diagonal intertwiners between inequivalent irreps vanish, A_{12}=A_{21}=0.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Schur (ii) over C: each A_{ii} = lambda_i 1_{W_i}, so Xbar = lambda_1 1_{W_1} (+) lambda_2 1_{W_2}.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "5c86b4af86f4716aeec0f5399496dacdd9f853ece3b24ed6d0c0810ec154546e", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "cd506ddfe5d47a38676df65ef4a8662220d1bfe3626c4bfc4bca8d66bf8b1402", "task_kind": "proof", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-113", "difficulty": { "level": "intermediate", "rationale": "Combines the averaging/reindexing argument with the image characterization and the trace-equals-dimension fact; ties cleanly to the general character projector." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-character-theory", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-179a913dcd0b76b7062e", "key": "qcb-core-plan-113-character-projector-isotypical-component", "primary_sources": [ { "anchors": "Proposition 3.32 (character projector P_W = (dim W/|G|) sum_g chi_W(g) phi(g), Eq. 3.87) and the special case P = (1/|G|) sum_g phi(g) projecting onto V^G", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "30-31", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let (phi, V) be a finite-dimensional complex representation of a finite group G. For the trivial representation, define P := (1/|G|) sum_{g in G} phi(g). Prove that (a) P is idempotent (P^2 = P), (b) im P = V^G := {|v> : phi(g)|v> = |v> for all g}, and (c) P commutes with every phi(h). Then explain, by taking the trace, why tr P equals the multiplicity of the trivial representation in V (the number of independent G-invariant vectors), connecting to the general character-projection formula P_W = (dim W / |G|) sum_g conj(chi_W(g)) phi(g).", "reference_answer": "(a) and (c) together: for any h in G, phi(h) P = (1/|G|) sum_g phi(h)phi(g) = (1/|G|) sum_g phi(hg) = (1/|G|) sum_{g'} phi(g') = P, reindexing g' = hg (a bijection of G); likewise P phi(h) = P by reindexing on the right. So phi(h) P = P = P phi(h) for all h, which gives both that P commutes with every phi(h) and that P is left/right absorbing for the group action. Idempotency follows: P^2 = (1/|G|) sum_g phi(g) P = (1/|G|) sum_g P = P, using phi(g) P = P from the previous line. (b) If |v> in im P, write |v> = P|w>; then phi(h)|v> = phi(h) P|w> = P|w> = |v> for all h, so |v> in V^G, giving im P subset V^G. Conversely if |v> in V^G then phi(g)|v> = |v> for all g, so P|v> = (1/|G|) sum_g phi(g)|v> = (1/|G|) sum_g |v> = |v>, hence |v> in im P; thus V^G subset im P. Therefore im P = V^G. (b continued, trace) Since P is idempotent and (being an average of the unitaries phi(g) in a suitable invariant inner product) an orthogonal projection, its trace equals the dimension of its image: tr P = dim im P = dim V^G, which is precisely the number of independent G-invariant vectors, i.e. the multiplicity of the trivial representation in V. This is the chi_W = chi_triv = 1 case of the general formula P_W = (dim W/|G|) sum_g conj(chi_W(g)) phi(g): the trivial representation has dimension 1 and character identically 1, so P_triv = (1/|G|) sum_g phi(g) = P, and its trace recovers the multiplicity, consistent with the character inner-product formula for multiplicities.", "rubric": [ { "criterion": "Proves phi(h) P = P = P phi(h) via reindexing, giving both commutation with phi(h) and absorption", "id": "r1", "weight": 1 }, { "criterion": "Deduces idempotency P^2 = P from phi(g) P = P", "id": "r2", "weight": 1 }, { "criterion": "Proves im P = V^G by double inclusion (image vectors are invariant; invariant vectors are fixed by P)", "id": "r3", "weight": 1 }, { "criterion": "Explains tr P = dim V^G = multiplicity of the trivial representation, as the chi_triv case of P_W = (dim W/|G|) sum_g conj(chi_W(g)) phi(g)", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Reindexing g -> hg gives phi(h) P = P (and P phi(h) = P), so P commutes with the representation.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "P^2 = (1/|G|) sum_g phi(g) P = P, so P is idempotent.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "im P subset V^G (image vectors fixed) and V^G subset im P (invariant vectors satisfy P|v>=|v>), so im P = V^G.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "As an orthogonal projection tr P = dim V^G = multiplicity of the trivial irrep, the chi_triv case of the character projector.", "supporting_nodes": [ "concept:harmonic-analysis" ] } ], "source_item_digest": "86e1d04d033535a8c92558d1206d86b9032daab481cb0b41dfbdc72641fd4800", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:matrix-analysis" ], "target_nodes": [ "concept:harmonic-analysis", "concept:operator-theory" ], "task_hash": "9efb104cea1d42f81e2940887d22f66bc4a27c33b947de88fb7e77e5fcdad674", "task_kind": "proof", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-114", "difficulty": { "level": "advanced", "rationale": "Requires recognizing the subtle false half of the claim and constructing an explicit multiplicity-driven counterexample with a verified computation; tests the limitation of Schur." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-schur-lemma", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-c9acfbfe820076580eed", "key": "qcb-core-plan-114-multiplicity-breaks-schur-scalar-counterexample", "primary_sources": [ { "anchors": "Section 3.4.2 (Limitations of Schur's Lemma: Multiplicities): the S_2 action on (C^2)^{(x)2}, isotypical decomposition (C^2)^{(x)2} = Vtriv (+) Vsgn with Vtriv ~ W_triv (x) C^3 of multiplicity 3, and the symmetrized operator having form 1_{W_triv} (x) X_triv (Eq. 3.69-3.72) which is generally not scalar", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "26-27", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "A student claims: for any representation (phi, V) of a finite group G and any X in L(V), the symmetrized operator Xbar = (1/|G|) sum_g phi(g) X phi(g)^{-1} is always block diagonal with respect to the isotypical decomposition AND each diagonal block is a scalar multiple of the identity. Decide whether the second half of this claim is true in general. Give an explicit counterexample using the action of S_2 = {e, (12)} on (C^2)^{(x)2} by permuting the two tensor factors (so phi((12)) = F, the swap), and explain precisely what feature of the decomposition makes the diagonal block fail to be scalar.", "reference_answer": "The block-diagonal part of the claim is correct, but the assertion that every diagonal block is a scalar multiple of the identity is FALSE in general; it holds only when each isotypical component is a single copy of an irrep (multiplicity one). Counterexample: let S_2 act on (C^2)^{(x)2} by phi(e) = 1 and phi((12)) = F (swap). The isotypical decomposition is (C^2)^{(x)2} = Vtriv (+) Vsgn, where Vtriv is the 3-dimensional +1 eigenspace of F (symmetric subspace, spanned by |00>, |11>, (|01>+|10>)/sqrt 2) carrying three copies of the trivial irrep, and Vsgn is the 1-dimensional -1 eigenspace (spanned by (|01>-|10>)/sqrt 2) carrying the sign irrep. Now take X = |00><11| (a rank-one operator). Symmetrizing: Xbar = (1/2)(X + F X F^{-1}) = (1/2)(|00><11| + F|00><11|F) = (1/2)(|00><11| + |00><11|) = |00><11|, because F|00> = |00> and F|11> = |11> (both symmetric), so F X F = X. The operator Xbar = |00><11| maps entirely within the symmetric (trivial-isotypical) subspace Vtriv, so it is block diagonal (its Vsgn block and the off-diagonal blocks are zero, consistent with Schur part (i)). However, restricted to the 3-dimensional Vtriv it is the rank-one operator |00><11|, which is NOT a scalar multiple of the 3x3 identity (it is not even diagonal in the symmetric basis and has rank 1, trace 0). The reason the diagonal block is not scalar is multiplicity: Vtriv contains the trivial irrep three times, so Hom_G(triv, Vtriv) is 3-dimensional and the symmetrized operator restricted to Vtriv is of the form 1_{W_triv} (x) X_triv with X_triv an arbitrary 3x3 operator on the multiplicity space C^3 - only this multiplicity-space factor is scalar when the multiplicity is 1. Hence Schur's Lemma forces scalar diagonal blocks only in the multiplicity-free case; with multiplicity greater than 1 the diagonal block is merely of the form (identity on the irrep) (x) (arbitrary operator on the multiplicity space).", "rubric": [ { "criterion": "Correctly judges the claim: block-diagonal part holds but diagonal-block-scalar part is false in general", "id": "r1", "weight": 1 }, { "criterion": "Gives the explicit S_2 action on (C^2)^{(x)2}, the isotypical decomposition Vtriv (3 copies of trivial) (+) Vsgn, and a concrete X whose symmetrization stays non-scalar on Vtriv", "id": "r2", "weight": 1 }, { "criterion": "Verifies the symmetrization computation (e.g. Xbar = |00><11| via F|00>=|00>, F|11>=|11>) showing a rank-one, non-scalar diagonal block", "id": "r3", "weight": 1 }, { "criterion": "Explains that multiplicity > 1 is the cause: the diagonal block is 1_W (x) X_k with X_k arbitrary on the multiplicity space, scalar only when multiplicity is 1", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Schur forces scalar diagonal blocks only when each isotypical component is a single irrep copy (multiplicity one).", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "For S_2 on (C^2)^{(x)2}, Vtriv is the 3-dim symmetric subspace (trivial irrep, multiplicity 3) and Vsgn the 1-dim sign irrep.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "With X = |00><11|, F X F = X so Xbar = |00><11|, a rank-one non-scalar operator supported on Vtriv.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Multiplicity 3 makes the Vtriv block 1_{W_triv} (x) X_triv with X_triv arbitrary on C^3, hence not scalar.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "be8cfff02ab1e6793f2ec533ebd7da483867295b6376480ae2ff6811b33f7663", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "8f25d699b1394213fd163f484bf17adb5dfb96c04f932235d8eae45b0ca2b797", "task_kind": "counterexample", "topic": "planned_representation_qit" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-116", "difficulty": { "level": "intermediate", "rationale": "Requires distinguishing isotypical uniqueness from irreducible-splitting non-uniqueness and producing two explicit distinct splittings; a precise conceptual boundary of the uniqueness theorem." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-isotypical-decomposition", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-ab9e69a77679b622b412", "key": "qcb-core-plan-116-nonunique-irrep-splitting-isotypical-counterexample", "primary_sources": [ { "anchors": "Proposition 3.22 (uniqueness of isotypical components and multiplicities) contrasted with Section 3.4.2, where Vtriv on (C^2)^{(x)2} has multiplicity 3 and the three trivial copies depend on a basis choice (the symmetric subspace is unique but its splitting into irreps is not)", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "21,26-27", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Proposition 3.22 says the decomposition of a representation into isotypical components is unique. A student over-reads this as: 'the decomposition of a representation into irreducible subrepresentations is unique.' Show by counterexample that the finer decomposition into individual irreducible subspaces is NOT unique, even though the isotypical components are. Use the trivial-isotypical component of the S_2 action on (C^2)^{(x)2} (equivalently, any space carrying the trivial representation with multiplicity >= 2) and exhibit two genuinely different splittings into irreducible lines.", "reference_answer": "The student's reading is wrong: it is the isotypical components that are unique, not the splitting of a multiplicity-greater-than-one component into individual irreducible subspaces. Take the S_2 action on (C^2)^{(x)2} with phi((12)) = F. The symmetric subspace Vtriv = Sym^2(C^2) is the +1 eigenspace of F, a 3-dimensional space on which S_2 acts trivially; it is the trivial-isotypical component and is uniquely determined (it equals the +1 eigenspace of F, basis-independent). But because S_2 acts as the identity on all of Vtriv, EVERY one-dimensional subspace of Vtriv is an irreducible (trivial) subrepresentation, and Vtriv can be written as a direct sum of three trivial lines in infinitely many ways. Concretely, one splitting is Vtriv = span(|00>) (+) span(|11>) (+) span((|01>+|10>)/sqrt 2); a genuinely different splitting is Vtriv = span((|00>+|11>)/sqrt 2) (+) span((|00>-|11>)/sqrt 2) (+) span((|01>+|10>)/sqrt 2). Both are decompositions of Vtriv into three irreducible (trivial) subrepresentations, but the first line span(|00>) of the first splitting is not among the lines of the second splitting, so the two decompositions into irreducibles are different. This does not contradict Proposition 3.22: the isotypical component Vtriv itself is identical in both (it is the full +1 eigenspace of F), and only its internal basis / line-by-line splitting changed. The source of the non-uniqueness is multiplicity: when an irrep occurs with multiplicity n >= 2, the choice of the n individual copies corresponds to a choice of basis of the n-dimensional multiplicity space, and any invertible change of that basis gives an equally valid finer decomposition. Uniqueness is recovered precisely in the multiplicity-free case.", "rubric": [ { "criterion": "Diagnoses the over-reading: isotypical components are unique, but the splitting of a multiplicity >= 2 component into irreducible subspaces is not", "id": "r1", "weight": 1 }, { "criterion": "Identifies Vtriv = Sym^2(C^2) as the unique trivial-isotypical component (the +1 eigenspace of F), 3-dimensional with trivial action", "id": "r2", "weight": 1 }, { "criterion": "Exhibits two genuinely different direct-sum splittings of Vtriv into trivial lines and notes a line in one is absent from the other", "id": "r3", "weight": 1 }, { "criterion": "Attributes the non-uniqueness to multiplicity (choice of basis of the multiplicity space) and notes uniqueness holds in the multiplicity-free case, consistent with Proposition 3.22", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Proposition 3.22 fixes the isotypical components, not the internal irreducible splitting of a multiplicity > 1 component.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Vtriv = Sym^2(C^2) (the +1 eigenspace of F) is the unique 3-dim trivial-isotypical component with trivial S_2 action.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Since S_2 acts as identity on Vtriv, every line is a trivial subrep; two different line-splittings of Vtriv both valid.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Non-uniqueness comes from the freedom to change basis of the multiplicity space; multiplicity-free decompositions are unique.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "ab9df40fd93fe05c0101d8c2d0d0b36cb55443c64c0780a092a57db7377c5e67", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "945d58f54d5d0936ba2f5cd3fbc4e68bb83df419865f9b10f5145c304d97c9c1", "task_kind": "counterexample", "topic": "planned_representation_qit" }, { "assumptions": [ "Each U_i is a (beta,a)-block-encoding of sqrt(P_i) with common a and common beta", "The output register is prepared in the exact uniform superposition over 2^m labels", "|psi> is a normalized pure system state" ], "context_hash": "c35213b8aa20330189dcd4e8ef35a4ff65ffbfacf5225e5744826035be337904", "conventions": [ "sqrt(P_i) denotes the unique positive semidefinite square root of the PSD effect P_i", "Postselection on 0^a means projecting the ancilla onto |0^a> and renormalizing" ], "core_id": "qcb-core-plan-201", "difficulty": { "level": "intermediate", "rationale": "A multi-step derivation combining block-encoding action, postselection, orthonormality bookkeeping and two distinct uses of the POVM axioms; more than recall but a clean structured argument." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-generalized-measurement", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-f10f96112b8c111b9f55", "key": "qcb-core-plan-201-povm-lcu-success-probability-completeness", "primary_sources": [ { "anchors": "Lemma 9.15; Eq. (9.55)-(9.62); W = sum_i |i> 0, so that U_i |0^a> |psi> = (1/beta) |0^a> sqrt(P_i) |psi> + |perp_i>, where (<0^a| tensor I)|perp_i> = 0. Define the controlled unitary W = sum_i |i> = (1/sqrt(2^m)) sum_i |i>, an ancilla register |0^a>, and the system register |psi>. After applying W one measures the ancilla register and postselects on the outcome 0^a. Starting from |u> |0^a> |psi>, derive in full the probability P(succ) of obtaining 0^a on the ancilla, show that it equals exactly 1/(beta^2 2^m) independent of the input state |psi>, and identify precisely where the POVM completeness relation sum_i P_i = I and the mutual orthogonality of the output-register basis states are used.", "reference_answer": "Apply W to the uniform-superposition input. Because W is block-diagonal in the output register, W |u> |0^a> |psi> = (1/sqrt(2^m)) sum_i |i> (U_i |0^a> |psi>) = (1/(beta sqrt(2^m))) sum_i |i> |0^a> sqrt(P_i)|psi> + |perp>, where |perp> = (1/sqrt(2^m)) sum_i |i> |perp_i> satisfies (I tensor <0^a| tensor I)|perp> = 0 because each (<0^a| tensor I)|perp_i> = 0. Postselecting the ancilla on 0^a projects out |perp> entirely, since |perp> has zero overlap with the subspace spanned by |0^a> on the ancilla. The unnormalized postselected vector is therefore (1/(beta sqrt(2^m))) sum_i |i> |0^a> sqrt(P_i)|psi>. Its squared norm is the success probability. The output-register states |i> are mutually orthonormal, so the cross terms vanish and the norm squared factorizes outcome by outcome: P(succ) = (1/(beta^2 2^m)) sum_i || sqrt(P_i)|psi> ||^2 = (1/(beta^2 2^m)) sum_i = (1/(beta^2 2^m)) sum_i . Here we used sqrt(P_i)^dagger sqrt(P_i) = P_i because each P_i is positive semidefinite and sqrt(P_i) is its unique PSD square root. Finally, by POVM completeness sum_i P_i = I, so sum_i = = 1 for any normalized |psi>. Hence P(succ) = 1/(beta^2 2^m), which is manifestly independent of |psi>. Completeness is what collapses the state-dependent sum to the constant 1; the orthogonality of the |i> is what prevents interference between different outcomes so the sum of norms (not a norm of a sum) appears.", "rubric": [ { "criterion": "Applies W to |u>|0^a>|psi>, obtains the branch (1/(beta sqrt(2^m))) sum_i |i>|0^a> sqrt(P_i)|psi> plus an orthogonal remainder annihilated by <0^a|", "id": "r1", "weight": 1 }, { "criterion": "Shows postselection on 0^a removes the perpendicular part and computes P(succ) as the squared norm of the surviving vector", "id": "r2", "weight": 1 }, { "criterion": "Uses orthonormality of the |i> to get sum_i ||sqrt(P_i)|psi>||^2 = sum_i (PSD square root), and completeness sum_i P_i = I to reduce it to 1", "id": "r3", "weight": 1 }, { "criterion": "Concludes P(succ) = 1/(beta^2 2^m) and states it is independent of |psi>, correctly attributing the two uses (completeness -> constant; orthogonality -> no interference)", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "W acts block-diagonally, so W|u>|0^a>|psi> = (1/sqrt(2^m)) sum_i |i> U_i|0^a>|psi>.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Substituting the block-encoding action gives a clean branch (1/(beta sqrt(2^m))) sum_i |i>|0^a> sqrt(P_i)|psi> plus |perp> with zero 0^a-overlap.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "Postselection on 0^a annihilates |perp> because each (<0^a| tensor I)|perp_i> = 0.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "Orthonormality of the |i> makes the squared norm a sum of per-outcome squared norms, with no cross terms.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "sum_i ||sqrt(P_i)|psi>||^2 = sum_i using sqrt(P_i)^dagger sqrt(P_i)=P_i for PSD P_i.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Completeness sum_i P_i = I collapses the sum to = 1, giving P(succ)=1/(beta^2 2^m) independent of |psi>.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "5432c2b2b9dea37c3425626c4e2f81de0bc5efe924d1f09178943b56ade40d74", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:block-encoding", "concept:probability", "concept:operator-theory" ], "task_hash": "19349f95b6d1eb044233b2b5ab0452f983a5327f4dd85cd0fae1a0930283e574", "task_kind": "proof", "topic": "planned_scientific_algorithms" }, { "assumptions": [ "Postselection on 0^a has already succeeded", "The |i> form an orthonormal computational basis of the output register", "P_i are PSD effects summing to the identity" ], "context_hash": "f99fb79bdfb7572f5dbd7e2ca61851f3b5778660b1d971f2e5396e80dfac50b8", "conventions": [ "sqrt(P_i) is the unique PSD square root", "Mixed-state results are obtained by convex-linear extension of the pure-state map" ], "core_id": "qcb-core-plan-202", "difficulty": { "level": "intermediate", "rationale": "Requires chaining normalization, conditional measurement statistics, the pure-to-mixed lift by linearity, and the composition of success and conditional probabilities." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-generalized-measurement", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-e65658d05b17c01a0444", "key": "qcb-core-plan-202-povm-conditional-output-state-update-rule", "primary_sources": [ { "anchors": "Lemma 9.15; Eq. (9.63)-(9.66); update rule Eq. (9.51)", "edition": "29 April 2026", "inspection": "full bounded primary text read", "pages": "153-154", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text or external relicensing", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Continue the POVM-by-block-encoding construction of Lemma 9.15: after applying W = sum_i |i> |0^a> |psi> and successfully postselecting the block-encoding ancilla on 0^a, the (renormalized) joint state of the output register and the system register is proportional to sum_i |i> tensor sqrt(P_i)|psi>. Prove that this joint state is correctly normalized, that measuring the output register in the computational basis yields outcome i with conditional probability P(i | succ) = , and that the resulting post-measurement system state is sqrt(P_i)|psi> / sqrt(). Then explain how the mixed-state update rule rho -> sqrt(P_i) rho sqrt(P_i) / Tr(P_i rho) follows from the pure-state result, and why the overall outcome probability for i (not conditioned on success) is still proportional to .", "reference_answer": "Normalization: the squared norm of sum_i |i> sqrt(P_i)|psi> is, using orthonormality of the |i>, sum_i ||sqrt(P_i)|psi>||^2 = sum_i = = = 1 by completeness. So after postselection the normalized joint state is exactly |Phi> = sum_i |i> tensor sqrt(P_i)|psi>. Measuring the output register in the computational basis: the probability of reading label i is the squared norm of the component attached to |i>, namely ||sqrt(P_i)|psi>||^2 = = (PSD square root), so P(i|succ) = . Conditioned on reading i, the system register collapses to the normalized vector sqrt(P_i)|psi> / || sqrt(P_i)|psi> || = sqrt(P_i)|psi> / sqrt(), which is exactly the pure-state version of the generalized-measurement update. Mixed states: for rho = sum_k q_k |psi_k> sqrt(P_i)|psi> sqrt(P_i) rho sqrt(P_i) / Tr(P_i rho). Linearity is what lets the pure-state formula be lifted verbatim to density operators. Unconditional probability: P(outcome i) = P(succ) * P(i|succ) = (1/(beta^2 2^m)) , i.e. proportional to with the i-independent proportionality constant 1/(beta^2 2^m); amplitude amplification later removes the constant penalty but never changes the relative weights, so the correct POVM statistics Tr(P_i rho) are reproduced.", "rubric": [ { "criterion": "Proves normalization of sum_i |i> sqrt(P_i)|psi> via orthonormality + completeness sum_i P_i = I", "id": "r1", "weight": 1 }, { "criterion": "Derives P(i|succ) = as the squared norm of the |i> component, using the PSD square-root identity", "id": "r2", "weight": 1 }, { "criterion": "Identifies the collapsed system state sqrt(P_i)|psi>/sqrt() and lifts it to the mixed-state rule by linearity in rho", "id": "r3", "weight": 1 }, { "criterion": "Explains the unconditional probability is (1/(beta^2 2^m)) , i.e. proportional to with an outcome-independent constant", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Orthonormality + completeness give ||sum_i |i> sqrt(P_i)|psi>||^2 = sum_i = 1, so the joint state is normalized.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "The weight on label |i> is ||sqrt(P_i)|psi>||^2 = , so P(i|succ) = .", "supporting_nodes": [ "concept:probability" ] }, { "claim": "Collapsing on i leaves the normalized system vector sqrt(P_i)|psi>/sqrt().", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The map |psi> sqrt(P_i)|psi> sqrt(P_i) rho sqrt(P_i)/Tr(P_i rho).", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Chaining with P(succ) gives unconditional P(i) = (1/(beta^2 2^m)) , proportional to the target statistic.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "4d0389f52f14f202bd366a400d21ee868cd70f95d739c66384d54c616d48666f", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:block-encoding", "concept:probability", "concept:operator-theory" ], "task_hash": "11b0bc7e058f3512393acaff01fa70778e93a9d9a5ec6af77e0e324ba32daf5d", "task_kind": "proof", "topic": "planned_scientific_algorithms" }, { "assumptions": [ "|c| < 1 and both states normalized", " = c*, = c" ], "context_hash": "cb129e7ef7666c40c2455c4827479bd2b507c08eecac2b90c8b9d3c60463de47", "conventions": [ "|psi_perp>, |phi_perp> are the unit complements from the text", "Positivity of P_? is checked on span{|psi>,|phi>} since it acts as identity off that plane" ], "core_id": "qcb-core-plan-205", "difficulty": { "level": "advanced", "rationale": "Requires an explicit 2x2 diagonalization of a non-orthogonal rank-one sum to certify PSD of P_?, beyond quoting the POVM axioms." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-state-discrimination", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-c2665c41e7fc3381116a", "key": "qcb-core-plan-205-state-discrimination-povm-validity-unambiguity", "primary_sources": [ { "anchors": "Example 9.14; Eq. (9.53)-(9.54); checks P_psi,P_phi,P_? >= 0 and sum = I, unambiguity", "edition": "29 April 2026", "inspection": "full bounded primary text read", "pages": "152-153", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text or external relicensing", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "For the unambiguous-discrimination POVM with P_psi = (1/(1+|c|)) |phi_perp>, |c| < 1, and |psi_perp>, |phi_perp> are defined as in the text, prove the three properties that make it a legitimate unambiguous POVM for state discrimination: (i) P_psi, P_phi, P_? are all positive semidefinite; (ii) P_psi + P_phi + P_? = I; (iii) the conclusive outcomes are unambiguous, i.e. = 0 and = 0. For (i) you must actually verify that P_? >= 0, not merely assert it, by diagonalizing I - P_psi - P_phi on the two-dimensional space span{|psi>,|phi>}.", "reference_answer": "Property (ii) is immediate from the definition P_? := I - P_psi - P_phi, so P_psi + P_phi + P_? = I exactly, with no assumption on c. Property (iii): compute . Since |phi_perp> = (|psi> - c*|phi>)/sqrt(1-|c|^2), we get = ( - c*)/sqrt(1-|c|^2) = (c* - c*)/sqrt(1-|c|^2) = 0 (using = c* and =1). Hence = (1/(1+|c|))||^2 = 0. Symmetrically, = ( - c)/sqrt(1-|c|^2) = (c - c)/sqrt(1-|c|^2) = 0, so = 0. The conclusive outcomes never fire on the wrong state. Property (i): P_psi, P_phi are nonnegative multiples of rank-one projectors |phi_perp>,|phi>} (off this plane I - P_psi - P_phi acts as the identity, which is PSD). Both |phi_perp> and |psi_perp> are unit vectors; a direct computation of their overlap gives || = |c|. In the (non-orthogonal) spanning pair {|phi_perp>, |psi_perp>}, the operator S := P_psi + P_phi = (1/(1+|c|))(|phi_perp>= 0. Therefore P_? = I - S has eigenvalues 1 - 1 = 0 and 1 - (1-|c|)/(1+|c|) = 2|c|/(1+|c|), both >= 0 for |c| in [0,1). So P_? >= 0 (with a zero eigenvalue, showing the weight 1/(1+|c|) is the largest that keeps P_? PSD). All three properties hold, so the family is a valid unambiguous-discrimination POVM.", "rubric": [ { "criterion": "States (ii) P_psi+P_phi+P_? = I holds by definition of P_?", "id": "r1", "weight": 1 }, { "criterion": "Proves (iii) by computing = 0 and = 0 explicitly, hence = = 0", "id": "r2", "weight": 1 }, { "criterion": "Argues P_psi, P_phi >= 0 as nonnegative scalings of rank-one projectors", "id": "r3", "weight": 1 }, { "criterion": "Verifies P_? >= 0 by diagonalizing S = P_psi+P_phi on the 2D span, finding eigenvalues {1, (1-|c|)/(1+|c|)} so I - S has eigenvalues {0, 2|c|/(1+|c|)} >= 0", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "P_psi+P_phi+P_? = I holds by definition of P_?.", "supporting_nodes": [ "concept:probability" ] }, { "claim": " = (c* - c*)/sqrt(1-|c|^2) = 0 and = 0, giving unambiguity = = 0.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "P_psi, P_phi are nonnegative scalings of rank-one projectors, hence PSD.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "In basis {|phi_perp>,|psi_perp>} with Gram overlap |c|, S = P_psi+P_phi has eigenvalues {1, (1-|c|)/(1+|c|)}.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Thus ||S|| = 1 and I - S = P_? has eigenvalues 0 and 2|c|/(1+|c|), both >= 0 for |c|<1.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "ac1b3b3d96f318f7409e0ff115c76d8ac4d471ea0f8b8acf2b79810e14205eb7", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis", "concept:probability" ], "task_hash": "ad368185fb19df8737cfe936f1893bc6157404dff08516ed089144fe3df00efd", "task_kind": "proof", "topic": "planned_scientific_algorithms" }, { "assumptions": [ "Pi, Pi' are orthogonal projectors of ranks N, N' with the given orthonormal bases", "Xi, Xi' are unitaries whose first N, N' columns are those bases", "N' <= MN" ], "context_hash": "8ea4118fee9a5db6183fe6cc856791f6377241620121c9d2d8b203505390eb34", "conventions": [ "Xi_N, (Xi')_{N'} are the first-column blocks; A~ = ((Xi')_{N'})^dagger U_A Xi_N", "Operator-norm error ||.|| defines the approximation" ], "core_id": "qcb-core-plan-216", "difficulty": { "level": "intermediate", "rationale": "A linear-algebra derivation identifying a compression with a basis-expressed induced block; conceptually clean but requires careful projector/column bookkeeping." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-generalized-block-encoding", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-63f0c8db36edc34fd9b7", "key": "qcb-core-plan-216-block-encoding-beyond-computational-basis-def925", "primary_sources": [ { "anchors": "Section 9.9; Eq. (9.99)-(9.105); Definition 9.25 generalized block encoding", "edition": "29 April 2026", "inspection": "full bounded primary text read", "pages": "161-162", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text or external relicensing", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "The block-encoding notion generalizes beyond computational-basis subspaces. Fix rank-N and rank-N' orthogonal projectors Pi, Pi' on an ambient space C^{MN} (M = 2^m), with orthonormal bases {|phi_j>}_{j in [N]} for range(Pi) and {|psi_i>}_{i in [N']} for range(Pi'). Collect these as the first columns of unitaries Xi, Xi' and let Xi_N, (Xi')_{N'} be those column blocks. For a unitary U_A on C^{MN}, define the induced block A~ := ((Xi')_{N'})^dagger U_A Xi_N in C^{N' x N}. Prove that the (i,j) entry of A~ equals , that A~ coincides with the compression Pi' U_A Pi expressed in the chosen bases, and hence that the exact relation Pi' U_A Pi = sum_{i,j} |psi_i> A_ij <0^m| tensor I and N' = N.", "reference_answer": "Entry formula: Xi_N has columns |phi_0>,...,|phi_{N-1}>, so Xi_N |e_j> = |phi_j> for the standard basis vector |e_j> of C^N; similarly (Xi')_{N'} |e_i> = |psi_i>, hence ((Xi')_{N'})^dagger |x> has i-th component . Therefore the (i,j) entry of A~ = ((Xi')_{N'})^dagger U_A Xi_N is A~_{ij} = = (((Xi')_{N'})|e_i>)^dagger U_A (Xi_N |e_j>) = . Compression identity: the orthogonal projectors can be written Pi = sum_j |phi_j> A~_{ij} }, {|phi_j>}, has exactly the matrix of coefficients A~. Consequently the exact encoding requirement Pi' U_A Pi = sum_{i,j} |psi_i> A_ij <0^m| tensor I and N' = N. Then range(Pi) = range(Pi') is the computational ancilla-0^m subspace; choosing Xi = Xi' = I (identity, whose first N columns are |0^m>|j>) gives Xi_N |e_j> = |0^m>|j> and (Xi')_{N'} |e_i> = |0^m>|i>, so A~_{ij} = <0^m||j>, which is exactly the standard top-left block (<0^m| tensor I) U_A (|0^m> tensor I). Thus ||A - alpha A~|| <= eps becomes the ordinary (alpha,m,eps)-block-encoding condition. The generalization is purely a change of input/output subspaces (via Pi, Pi' and the basis unitaries) from the fixed computational ancilla-flag subspace to arbitrary ones.", "rubric": [ { "criterion": "Shows A~_{ij} = using that Xi_N, (Xi')_{N'} map standard basis vectors to the chosen orthonormal basis vectors", "id": "r1", "weight": 1 }, { "criterion": "Writes Pi = sum_j |phi_j> A~_{ij} A_ij <0^m| tensor I, N' = N (A~_{ij} = <0^m||j>)", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Xi_N|e_j> = |phi_j>, (Xi')_{N'}|e_i> = |psi_i>, so A~_{ij} = .", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "With Pi = sum_j |phi_j> A~_{ij} A_ij <0^m| tensor I, N'=N, A~_{ij} = <0^m||j>, the standard block.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "So the generalized condition reduces to the ordinary (alpha,m,eps)-block-encoding definition.", "supporting_nodes": [ "concept:block-encoding" ] } ], "source_item_digest": "25c3a802898fe52a55571fa3631a3925ffbb6c11f24473f29e40723e7f634ad2", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:block-encoding", "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "afb656d87a470be889e81399de157ae85d25c650fa3023f617133653a2d93b1e", "task_kind": "proof", "topic": "planned_scientific_algorithms" }, { "assumptions": [ "U_A in HBE_{1,m}(A) is Hermitian with ||A|| <= 1", "A = sum_i lambda_i |v_i><0^m| tensor I; Z_Pi = 2Pi - I" ], "context_hash": "732b2449205f9d1936d39c6a286fbd5b2833745d8b16bf19de6bbc4651a97ce4", "conventions": [ "lambda_i = cos theta_i with 0 < theta_i < pi", "T_k(cos theta) = cos(k theta), U_{k-1}(cos theta) = sin(k theta)/sin theta" ], "core_id": "qcb-core-plan-220", "difficulty": { "level": "advanced", "rationale": "The central qubitization theorem: it requires the Hermitian two-step subspace closure, the reflection-times-reflection = rotation identity, and the Chebyshev power computation summed over the spectrum." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-qubitization", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-eb0e575d5a33ae0e37ea", "key": "qcb-core-plan-220-qubitization-iterate-chebyshev-block-encoding", "primary_sources": [ { "anchors": "Section 10.2.1; Eq. (10.25)-(10.41); iterate O = U_A Z_Pi; O^k in BE_{1,m,0}(T_k(A))", "edition": "29 April 2026", "inspection": "full bounded primary text read", "pages": "167-168", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text or external relicensing", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Let U_A in HBE_{1,m}(A) be a Hermitian block encoding of a Hermitian A with ||A|| <= 1, spectral decomposition A = sum_i lambda_i |v_i>|v_i> = lambda_i |0^m>|v_i> + sqrt(1-lambda_i^2)|perp_i> with Pi|perp_i> = 0, Pi = |0^m><0^m| tensor I. For -1 < lambda_i < 1, prove that H_i = span{|0^m>|v_i>, |perp_i>} is a 2D invariant subspace of U_A on which U_A is a reflection [[lambda_i, sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2), -lambda_i]], that the iterate O = U_A Z_Pi (with Z_Pi = 2 Pi - I) acts on H_i as the rotation [[lambda_i, -sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2), lambda_i]], and hence that O^k restricted to H_i has top-left entry T_k(lambda_i) (Chebyshev polynomial of the first kind), so O^k = (U_A Z_Pi)^k is a (1,m,0)-block-encoding of T_k(A). State where Hermiticity of U_A is used.", "reference_answer": "Invariant subspace and reflection form: by the HBE relation, U_A|0^m>|v_i> = lambda_i|0^m>|v_i> + sqrt(1-lambda_i^2)|perp_i>, with |perp_i> normalized and orthogonal to the Pi-subspace. Since U_A is Hermitian (U_A = U_A^dagger) and U_A is unitary, U_A^2 = I. Apply U_A again to the identity above: U_A(U_A|0^m>|v_i>) = |0^m>|v_i>, i.e. lambda_i U_A|0^m>|v_i> + sqrt(1-lambda_i^2) U_A|perp_i> = |0^m>|v_i>. Substituting U_A|0^m>|v_i> and solving gives U_A|perp_i> = sqrt(1-lambda_i^2)|0^m>|v_i> - lambda_i|perp_i>. Thus both U_A|0^m>|v_i> and U_A|perp_i> lie in H_i = span{|0^m>|v_i>, |perp_i>}, so H_i is invariant, and in the ordered basis B_i = {|0^m>|v_i>, |perp_i>} the matrix of U_A is [[lambda_i, sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2), -lambda_i]], a (symmetric, trace-0, determinant -1) reflection. Hermiticity is used precisely here: applying U_A = U_A^dagger a SECOND time (via U_A^2 = I) is what produces the closed expression for U_A|perp_i> and hence the 2D invariance; without Hermiticity U_A|perp_i> need not stay in H_i. Iterate as rotation: H_i is invariant for Pi too, with [Pi]_{B_i} = [[1,0],[0,0]] (since |0^m>|v_i> is in range(Pi) and |perp_i> is orthogonal to it). Hence Z_Pi = 2Pi - I has [Z_Pi]_{B_i} = [[1,0],[0,-1]], a reflection. The iterate O = U_A Z_Pi has matrix [[lambda_i, sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2), -lambda_i]] [[1,0],[0,-1]] = [[lambda_i, -sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2), lambda_i]], which with lambda_i = cos theta_i is exactly the rotation [[cos theta_i, -sin theta_i],[sin theta_i, cos theta_i]] (product of two reflections = rotation). Chebyshev powers: therefore [O^k]_{B_i} = [[cos k theta_i, -sin k theta_i],[sin k theta_i, cos k theta_i]] = [[T_k(lambda_i), -sqrt(1-lambda_i^2) U_{k-1}(lambda_i)],[sqrt(1-lambda_i^2) U_{k-1}(lambda_i), T_k(lambda_i)]], using T_k(cos theta) = cos(k theta) and U_{k-1}(cos theta) = sin(k theta)/sin theta. The top-left entry is T_k(lambda_i). Block encoding of T_k(A): the vectors {|0^m>|v_i>} span range(Pi), and on each H_i the (0^m,0^m) component of O^k equals T_k(lambda_i). Summing over i, Pi O^k Pi = sum_i T_k(lambda_i) |0^m>|v_i><0^m| tensor T_k(A). Hence (<0^m| tensor I) O^k (|0^m> tensor I) = T_k(A), i.e. O^k = (U_A Z_Pi)^k in BE_{1,m,0}(T_k(A)). (For lambda_i = pm 1 the subspace is already 1D-invariant and T_k(pm1) = (pm1)^k is reproduced trivially.)", "rubric": [ { "criterion": "Uses U_A Hermitian (U_A^2 = I) to apply U_A twice and derive U_A|perp_i> = sqrt(1-lambda_i^2)|0^m>|v_i> - lambda_i|perp_i>, proving H_i is 2D-invariant with the reflection matrix", "id": "r1", "weight": 1 }, { "criterion": "Explicitly identifies where Hermiticity is used (the second application / U_A^2 = I closing the subspace)", "id": "r2", "weight": 1 }, { "criterion": "Computes [Z_Pi]_{B_i} = diag(1,-1) and the product O = U_A Z_Pi as the rotation [[lambda_i,-sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2),lambda_i]]", "id": "r3", "weight": 1 }, { "criterion": "Raises to the k-th power to get top-left entry T_k(lambda_i) and sums over i to conclude O^k in BE_{1,m,0}(T_k(A))", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "U_A Hermitian unitary gives U_A^2 = I; applying U_A twice yields U_A|perp_i> = sqrt(1-lambda_i^2)|0^m>|v_i> - lambda_i|perp_i>, so H_i is 2D invariant.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "In B_i, [U_A]_{B_i} = [[lambda_i, sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2), -lambda_i]], a reflection.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "[Z_Pi]_{B_i} = diag(1,-1), so O = U_A Z_Pi has [O]_{B_i} = [[lambda_i,-sqrt(1-lambda_i^2)],[sqrt(1-lambda_i^2),lambda_i]], a rotation.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "[O^k]_{B_i} has top-left entry cos(k theta_i) = T_k(lambda_i).", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Summing over i gives Pi O^k Pi = |0^m><0^m| tensor T_k(A), so O^k in BE_{1,m,0}(T_k(A)).", "supporting_nodes": [ "concept:block-encoding" ] } ], "source_item_digest": "17fa9371009c4d46bf734d23da434d92f20544b991214dd9f1936d2ae4633bcc", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:block-encoding", "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "93f4c3a8bdd9c533c8ae158597d33c7a309ce8922cc13034d4fecb1f1f761a19", "task_kind": "proof", "topic": "planned_scientific_algorithms" }, { "assumptions": [ "Finite-dimensional real matrices stored as nested vectors; all scalars are exact reals", "Eight independent primitive counters; each primitive call costs one unit of its own counter" ], "context_hash": "8801dc8303e843d317119d144d4009ea578e1578b5b4a3238f92e73f9b7f3bdb", "conventions": [ "Proof checking, counter bookkeeping and loop-index arithmetic are free; a materialized pass of length M charges two reads and two writes per index", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-201", "difficulty": { "level": "intermediate", "rationale": "Several-step exact operation count requiring correct per-pass read/write accounting." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-201", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-f8c828d6288ae2b58400", "key": "qcb-core-la-201", "primary_sources": [], "question": "Work in an exact-real word-operation model that keeps eight separate running counters: field operations, square roots, inverse-cosine calls, other trigonometric calls, exact comparisons, stored-word reads, stored-word writes, and emitted records. Each primitive call adds one unit to its own counter; proof checking, counter bookkeeping and loop-index arithmetic are free, and stored data are nested vectors. A plane rotation acts on two stored length-M real rows u and v with coefficients c and s, returning rows whose j-th entries are c*u_j - s*v_j and s*u_j + c*v_j. The implementation first materializes a stored vector of M computed entry-pairs, then materializes the first output row and the second output row as two further stored passes. Every materialized pass over M indices additionally charges two reads and two writes per index to write each intermediate value and read it back for projection, and each single entry-pair is computed with exactly the six field operations c*u_j, s*v_j, s*u_j, c*v_j, their difference, and their sum. Prove that rotating the two rows costs exactly 6*M field operations, 10*M reads and 6*M writes, and account for the origin of every read and write.", "reference_answer": "Count the three materialized passes separately. Pass 1 computes the entry-pairs: for each index j it reads u_j and v_j (2 reads) and performs the six field operations, so its own work is 6 field + 2 reads per index; the materialization surcharge adds 2 reads + 2 writes per index, giving 6*M field + 4*M reads + 2*M writes. Pass 2 materializes the first output row by reading the stored pair at each index (1 read) and projecting its first component (free), with surcharge 2 reads + 2 writes, giving 3*M reads + 2*M writes. Pass 3 does the same for the second component, another 3*M reads + 2*M writes. Summing: field = 6*M; reads = 4*M + 3*M + 3*M = 10*M; writes = 2*M + 2*M + 2*M = 6*M. The field count is purely the arithmetic of the rotation; the reads split into 2*M genuine input reads, 2*M pair reads for the two projections, and 6*M materialization read-backs; the writes are the three stored passes at 2*M each. This is an exact operation count in the extended exact-real model, not a finite-bit running-time bound, and it does not assert that any of these operations is physically free.", "rubric": [ { "criterion": "Each entry-pair uses exactly six field operations and two input reads", "id": "c1", "weight": 1 }, { "criterion": "Correctly charge the two-read/two-write materialization surcharge on all three passes", "id": "c2", "weight": 1 }, { "criterion": "Derive the exact totals 6*M field, 10*M read, 6*M write with per-pass accounting", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A single entry-pair costs six field operations plus two input reads.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.entryPair_cost" ] }, { "claim": "Each materialized pass of length M adds two reads and two writes per index on top of its entry work.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.collect_cost" ] }, { "claim": "Summing the three passes yields exactly 6*M field, 10*M read and 6*M write.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.rowPair_cost" ] } ], "source_item_digest": "4b2411a0ba33a8286691461446ef4686f5dd20ab90d92061bc04fe1fbd74eb65", "source_refs": [ { "anchor": "entryPair123-130; entryPair_cost135-138; collect60-64; collect_cost83-85; rowPair141-153; rowPair_cost272-277", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredGivens.lean", "sha256": "87a1f69d5b138f23c96a6f77a438aa81edef27427cb2581c4defe019079d2ee4" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.entryPair_cost", "declaration:QuantumBlockEncoding.StoredGivens.collect_cost", "declaration:QuantumBlockEncoding.StoredGivens.rowPair" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.rowPair_cost" ], "task_hash": "67d0d8df3dbda464c93aa56d53720ec9c46f299b5b12ba566bbf3a0cdc70ae5a", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "Finite-dimensional real matrices stored as nested vectors of N rows", "Replacing a stored row is charged as a full persistent copy of N reads and N writes" ], "context_hash": "d8339a5488e377183373e63874afa2d011fefad014539d8679ee591f843b194a", "conventions": [ "No in-place-update or copy-on-write optimization is assumed; the bound is an exact-real operation count, not a finite-bit runtime", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-202", "difficulty": { "level": "intermediate", "rationale": "Compositional exact count plus a model-level justification of the persistent-copy charge." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-202", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8b5706a83a1150200f95", "key": "qcb-core-la-202", "primary_sources": [], "question": "Continue in the eight-counter exact-real operation model where stored matrices are vectors of N stored rows and replacing one stored row is charged as a full persistent copy costing N reads and N writes, never an in-place update. A rotation step on an N-by-M stored matrix reads the two rows i and j to be combined, computes the two rotated rows (whose production costs 6*M field, 10*M read and 6*M write), and then stores the two new rows with two separate persistent row-replacements. Prove that one such rotation step costs 6*M field operations, 10*M + 2*N + 2 reads, and 6*M + 2*N writes, and explain why the step cannot be made cheaper by assuming in-place row updates.", "reference_answer": "Add the charged subparts. Reading rows i and j from the stored matrix contributes 2 reads. Producing the two rotated rows contributes 6*M field, 10*M reads and 6*M writes. Each persistent row-replacement copies the whole length-N spine of rows, charging N reads and N writes; two replacements add 2*N reads and 2*N writes. Totals: field = 6*M; reads = 10*M + 2 + 2*N; writes = 6*M + 2*N. The 2*N read/write term is a deliberate feature of the cost contract: the model charges a full persistent row-replacement as N reads and N writes of the row-index spine. This is a stipulated charge, not a theorem that no sharing is possible; a functional persistent vector may in fact share the unaffected rows, but the model declines to credit any such copy-on-write or structural-sharing optimization and bills the full spine copy regardless. Charging it this way makes the bound robust: it holds for any implementation of row replacement without relying on aliasing analysis, and it is an exact-real word-operation count rather than a finite-bit machine running time. The 2*N term would only vanish under a model that explicitly credits in-place or shared updates, which this cost contract does not.", "rubric": [ { "criterion": "Account the two input row reads and the row-production cost", "id": "c1", "weight": 1 }, { "criterion": "Charge each persistent replacement as a full N-read, N-write spine copy", "id": "c2", "weight": 1 }, { "criterion": "Explain the 2*N term is a stipulated spine-copy charge in the cost contract, not a proof that sharing is impossible", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Reading the two combined rows and producing the rotated rows contributes 2 reads plus 6*M field, 10*M read, 6*M write.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.rowPair_cost" ] }, { "claim": "Each persistent row-replacement copies the N-row spine for N reads and N writes.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.replace" ] }, { "claim": "Summation gives 6*M field, 10*M+2*N+2 read, 6*M+2*N write, and the 2*N term is the contract's stipulated spine-copy charge.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.rotate_cost" ] } ], "source_item_digest": "cffaec368aa74c7a29a98568529ed0821b8c45328ba670593e9b00c3d43c19f3", "source_refs": [ { "anchor": "replace67-68; rotate161-167; rowPair_cost272-277; rotate_cost279-285", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredGivens.lean", "sha256": "87a1f69d5b138f23c96a6f77a438aa81edef27427cb2581c4defe019079d2ee4" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.rowPair_cost", "declaration:QuantumBlockEncoding.StoredGivens.replace", "declaration:QuantumBlockEncoding.StoredGivens.rotate" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.rotate_cost" ], "task_hash": "f45ab52aa0ae2187f3251598a846a9c08025a7fb4d15ab7f9cec21cf0f0058f7", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "Finite-dimensional real stored matrices; eight independent primitive counters", "The per-step budget is independent of the angle-computation branch" ], "context_hash": "bfd56a3228867191e6bccf2c01343f2b3c61811c6e557441c1ebe2fcb60b61cc", "conventions": [ "A zero-angle rotation still emits one record; the budget is an upper bound on arithmetic counters only", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-203", "difficulty": { "level": "intermediate", "rationale": "Induction proving an upper bound and an exact count simultaneously with a branch-independence subtlety." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-203", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-16789e0cd201c61f5633", "key": "qcb-core-la-203", "primary_sources": [], "question": "In the eight-counter exact-real model, a column sweep eliminates entries below a pivot by performing a chronological chain of elimination steps. Each step first computes an elimination angle and the two rotation coefficients (bounded by a fixed per-step budget B that is independent of which branch of the angle computation is taken), applies a rotation to the stored matrix, and then emits exactly one rotation record (one unit on the emitted-record counter). A sweep performing k steps is defined by recursion: zero steps emit nothing and do no work; a sweep of k+1 steps performs one elimination step and then recurses on the resulting matrix, emitting one record for the step. Prove by induction on k that (a) the total cost of a k-step sweep is at most k*(B + one emit) on every counter, and (b) the emitted-record counter equals exactly k, even though some steps may rotate by a harmless zero angle.", "reference_answer": "Argue by induction on the number of steps k. Base case k=0: the sweep returns the input unchanged with zero cost on every counter, so both the bound 0*(B+emit) and the exact emit count 0 hold. Inductive step k+1: by the recursive definition the sweep cost equals the cost of one elimination step, plus the cost of the k-step sweep on the updated matrix, plus one emitted record. For (a), the single step is bounded by B on every counter (the per-step budget already absorbs the branch-independent angle budget, both rotation coefficients, the rotation, and the stored reads), the recursive part is bounded by k*(B+emit) by the induction hypothesis, and the emit contributes one unit; adding gives at most (k+1)*(B+emit) on every counter. For (b), the single step emits exactly one record and the recursive sweep emits exactly k by hypothesis, with no other operation touching the emit counter, so the total is exactly k+1. The exact emit equality holds regardless of the actual angles because a zero-angle rotation still produces and emits its record; the step budget B is only an upper bound on the arithmetic counters, so the two claims are of different strengths: an exact count for emissions and an inequality for the rest.", "rubric": [ { "criterion": "Set up induction on step count with correct base and recursive decomposition", "id": "c1", "weight": 1 }, { "criterion": "Prove the k*(B+emit) upper bound using the branch-independent step budget", "id": "c2", "weight": 1 }, { "criterion": "Prove the emit counter equals exactly k including zero-angle steps, distinguishing exact count from upper bound", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The recursion splits a (k+1)-step sweep into one step, the k-step tail, and one emission.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.columnSweep" ] }, { "claim": "The branch-independent step budget bounds each step, giving the k*(B+emit) upper bound by induction.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.columnSweep_cost_le" ] }, { "claim": "Each step emits exactly one record, so the emit counter equals k exactly, even for zero-angle steps.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.columnSweep_emit" ] } ], "source_item_digest": "d3c7ffab655e801f0c3275b31d71aff77610f9d16830e0dfc9b47683fbb647f7", "source_refs": [ { "anchor": "columnSweep219-229; stepBudget300; columnSweep_cost_le313-326; columnSweep_emit343-353", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredGivens.lean", "sha256": "87a1f69d5b138f23c96a6f77a438aa81edef27427cb2581c4defe019079d2ee4" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.stepBudget", "declaration:QuantumBlockEncoding.StoredGivens.columnSweep" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.columnSweep_cost_le", "declaration:QuantumBlockEncoding.StoredGivens.columnSweep_emit" ], "task_hash": "9654b480868274c42c9972e64e62e848de27b78a9d1599a421be0879c3a72a57", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "Recorded steps carry an exact angle and a pair of adjacent rows", "Eight independent primitive counters in the exact-real model" ], "context_hash": "8fd51f96ad53c362a6e07c2de9226f0d5e674b85c7458325733188ddbcf0a5c4", "conventions": [ "The accumulated transform is replayed, never formed as a dense matrix product; costs are not finite-bit runtimes", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-204", "difficulty": { "level": "advanced", "rationale": "Joint refinement-and-cost induction over a recorded step list with a non-dense accumulation strategy." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-204", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-ab016d966c7415d35386", "key": "qcb-core-la-204", "primary_sources": [], "question": "A rectangular elimination records a chronological list of two-row rotation steps. Rather than forming the accumulated orthogonal transform by multiplying out dense plane-rotation matrices, a stored replay starts from a materialized identity and re-applies each recorded two-row rotation to the stored matrix in order; the empty list charges one read of the end marker, and each recorded step charges one read of the record, one field operation for halving its angle, two trigonometric calls for the pair of coefficients, and one stored two-row rotation. On a k-by-N working matrix a single such rotation costs the fixed amount described earlier (field, read and write terms linear in N and in the matrix dimension). Prove two things about replaying a list of L steps: first, that the result denotes exactly the product of the recorded plane rotations applied to the starting matrix, and second, that the total cost is L times a fixed per-step budget plus one extra read, with the extra read coming from the empty-list base case.", "reference_answer": "Both claims go by induction on the step list. Correctness: for the empty list, replay returns the starting matrix, which equals the empty product applied to it. For a cons step followed by the rest, replay rotates the current matrix by the step's recorded angle and recurses; by the induction hypothesis the recursion denotes the remaining product, and one rotation realizes exactly the plane-rotation matrix of this step, so composing gives the full product applied to the start. This shows the stored replay refines the dense matrix-product specification without ever evaluating that product. Cost: for the empty list the only charge is the single end-marker read, matching 0*budget + one read. For a cons step, the step charges one read of the record, one field halving, two trigonometric calls and one two-row rotation; summing with the inductive cost L_tail*budget + one read gives (L_tail+1)*budget + one read, where the per-step budget collects exactly the field, two trig, rotation field/read/write and the three reads (two from the rotation, one from the record). Hence an L-step replay costs L*budget + one read, and the two reads inside the rotation plus the record read explain the three-read-per-step term, while the single trailing read is the base case.", "rubric": [ { "criterion": "Prove by list induction that replay denotes the product of recorded rotations without forming the dense product", "id": "c1", "weight": 1 }, { "criterion": "Identify the per-step budget: one field, two trig, one rotation, and three reads", "id": "c2", "weight": 1 }, { "criterion": "Derive total cost L*budget + one read with the trailing read from the empty-list base case", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The empty list returns the start matrix and the cons case rotates once then recurses.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.replay" ] }, { "claim": "List induction shows the replay denotes the composed product of recorded plane rotations.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.replay_value" ] }, { "claim": "Each step charges one field, two trig, a rotation and three reads, giving total L*budget + one trailing read.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.replay_cost" ] } ], "source_item_digest": "705a52ec7b52b60d5ea01c2f2bf97d622094dcf936d909b80d18cfd481591425", "source_refs": [ { "anchor": "replay110-118; replay_value120-128; replayStepBudget130-133; replay_cost135-143", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredRectangularGivens.lean", "sha256": "f8638a6d287b3b984fa530b13eeb18e3992904543236bd0756eda41fa6490d98" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.replay", "declaration:QuantumBlockEncoding.StoredGivens.rotate_cost" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.replay_value", "declaration:QuantumBlockEncoding.StoredRectangularGivens.replay_cost" ], "task_hash": "d91d06db12c3a3286f6741f3b47ed84dba0691a13cac895fe5d3668d76728135", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "Reading a stored core entry is a row read followed by an entry read (two reads)", "Eight independent primitive counters; pure zero carries no charge" ], "context_hash": "ed5b2a7690f25a5193e2def62df2229bc4a0867df1dcc12ef82f9b1ee1231912", "conventions": [ "The accumulation addition is one field operation per summand; counts are exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-207", "difficulty": { "level": "intermediate", "rationale": "Cost induction over a monadic sum with a specialization that isolates the accumulation field." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-207", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-c089baece659f2d59a86", "key": "qcb-core-la-207", "primary_sources": [], "question": "In the eight-counter exact-real model, a bounded sum over an index set of size k is evaluated by a monadic fold: starting from the pure value zero (no charge), each of the k summands is produced and then added to the running accumulator with one field addition. Suppose every summand has the same per-counter cost B. Prove that the whole sum costs exactly k*(B + one field operation). Then specialize to a single contraction entry over a bond of size D, where each summand reads one entry of a stored core (two reads) and one entry of a stored vector (one read) and multiplies them (one field), so B is three reads plus one field, and conclude that the entry costs exactly D*(three reads + two field operations).", "reference_answer": "Induct on k. Base case k=0: the empty sum is the pure zero with no charge, matching 0*(B+field). Inductive step: a sum over k+1 indices produces the first summand (cost B), binds an addition of it to the recursively evaluated tail (that addition is one field operation), and the tail is a sum over k indices. By the induction hypothesis the tail costs k*(B+field); adding the first summand's B and the one field for its combination gives B + field + k*(B+field) = (k+1)*(B+field). So the sum costs exactly k*(B+field) on every counter. For the contraction entry: each summand reads a stored core entry, which is a row read followed by an entry read (two reads), reads one stored vector entry (one read), and multiplies the two values (one field), so B = three reads + one field. Substituting into the law gives D*((three reads + one field) + one field) = D*(three reads + two field operations). The second field per index is exactly the accumulation addition, which is why the entry has two field operations per bond index rather than one. This is an exact count in the extended exact-real model, not a finite-bit cost.", "rubric": [ { "criterion": "Prove the sum cost law k*(B+field) by induction with correct base and accumulation step", "id": "c1", "weight": 1 }, { "criterion": "Identify the summand cost B as three reads plus one field for a contraction entry", "id": "c2", "weight": 1 }, { "criterion": "Conclude D*(three reads + two field), attributing the extra field to accumulation", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "An empty bounded sum has zero cost and a (k+1)-sum charges B plus one accumulation field over the k-tail.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.terminalEntry" ] }, { "claim": "Induction yields the exact sum cost k*(B+field).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.terminalEntry_cost" ] }, { "claim": "With B = three reads + one field per summand, the contraction entry costs D*(three reads + two field).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredThinLQ.entry_cost" ] } ], "source_item_digest": "172127f1825454b15f930b84064179096752d4fe8288d27eba49940a6a8d600e", "source_refs": [ { "anchor": "terminalEntry11-16; sumEntries_cost_eq125-133; terminalEntry_cost135-143", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredMatrixProductChain.lean", "sha256": "7a697d064a5fddd909bba048d6adc36cb645421bd565e2357588fef1ff0449f2" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.terminalEntry", "declaration:QuantumBlockEncoding.StoredThinLQ.entry_cost" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.terminalEntry_cost" ], "task_hash": "51d2992bc1b340eaef534bc1b0cf40e4055797fc0d19dbeb3c5a0106d28c8653", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "All cores stored and bond-bounded by D; reference copying is charged per remaining core", "Eight independent primitive counters; input cores already materialized" ], "context_hash": "a802a3c217fe35c67f3b627d5a49d4265d65bbb38bbb816ca1741526f424960e", "conventions": [ "No kernel callback is evaluated in production; counts are exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-208", "difficulty": { "level": "advanced", "rationale": "Structural cost decoupling with an arithmetic-series argument explaining the quadratic length dependence." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-208", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-d2319217120afeeafa0b", "key": "qcb-core-la-208", "primary_sources": [], "question": "A stored matrix-product chain over n+1 cores, each of bond size D, is closed into a scalar-boundary chain in two conceptual halves. First the tail is processed recursively: at the last position the terminal core is contracted against a right boundary vector, and at every earlier position the current core is kept while the remaining table is copied by reference and the tail is recursed; a fixed small per-node record is charged at each position. Second the leftmost core is contracted against a left boundary vector. The returned chain exactly refines the kernel chain contracted against the two boundary vectors. Prove that the summed cost of the whole closure is bounded by a polynomial that is quadratic in the chain length n and cubic-free but quadratic in the bond D, and explain why the tail recursion contributes an n-squared term while the single left closure contributes only a bond-dependent term with no dependence on n.", "reference_answer": "Split the accounting as the construction does. The tail recursion visits n+1 positions. The terminal contraction at the last position contracts a D-by-D core against a length-D vector, costing on the order of D-squared field and read work plus boundary materialization, a fixed function of D. Each earlier position copies the remaining table of references (cost linear in the number of remaining cores, i.e. linear in the position index) and charges a fixed node record; summing the reference-copy cost over the n earlier positions gives an arithmetic series that is quadratic in n, which is the source of the n-squared term, while the fixed node records contribute a term linear in n. Hence the tail total is bounded by a constant times D-squared plus a constant plus terms 5*n-squared + 7*n + constant. The left closure performs exactly one initial contraction of a D-by-D core against a length-D left vector plus one node record; this is a single bond-dependent cost with no sum over positions, hence no n dependence. Adding the two halves gives the closure bound 20*D-squared + 30*D + 5*n-squared + 7*n + constant. The decoupling is structural: the tail walks the whole chain so its copying accumulates with length, whereas the left boundary touches only the one leftmost core. All of this is counted in the extended exact-real word model, not as finite-bit running time, and the referenced cores are already materialized so no kernel callback is evaluated.", "rubric": [ { "criterion": "Separate the n+1-position tail recursion from the single left closure", "id": "c1", "weight": 1 }, { "criterion": "Explain the n-squared term as the arithmetic series of reference-copy costs over positions", "id": "c2", "weight": 1 }, { "criterion": "Explain why the left closure is bond-dependent only, with no n dependence, and note exact-real counting", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The tail recursion over n+1 positions copies references whose per-position cost grows with the index.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.tailChain_total_cost_le" ] }, { "claim": "The accumulated reference-copy series is quadratic in n, giving the 5*n-squared+7*n term.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.tailChain_total_cost_le" ] }, { "claim": "The single left closure is bond-dependent only, so the total is 20*D-squared+30*D+5*n-squared+7*n+constant.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.closeLeft_tail_total_cost_le", "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_total_cost_le" ] } ], "source_item_digest": "2af4654334ec83645d32e3ac7541aa9ca67e8c78afa43aa5ed50f9d40b1ea312", "source_refs": [ { "anchor": "ofTable55-59; tailChain_total_cost_le202-214; closeLeft_tail_total_cost_le216-226; ofTable_total_cost_le230-237", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredMatrixProductChain.lean", "sha256": "7a697d064a5fddd909bba048d6adc36cb645421bd565e2357588fef1ff0449f2" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.tailChain_total_cost_le", "declaration:QuantumBlockEncoding.StoredMatrixProductChain.closeLeft_tail_total_cost_le", "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_total_cost_le" ], "task_hash": "f6dff3feb169b3675b7651b6963959841395d22329c0a3f5c5964711e64af180", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "Every bond along the chain is at most D; input cores are stored tables", "Eight independent primitive counters plus a fixed six-word node record" ], "context_hash": "5120e371586197bda6d763c26e3bb3cb9750618e176b06005366ae9d7c6f0d28", "conventions": [ "The recursive environment is read, not recomputed; counts are exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-209", "difficulty": { "level": "advanced", "rationale": "Chain induction with a cubic-in-bond per-node bound and a cached-environment subtlety." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-209", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-0049b6643cf6b76ff522", "key": "qcb-core-la-209", "primary_sources": [], "question": "A stored tensor-train norm is computed by streaming local Gram environments from right to left. At the empty tail the environment is a materialized identity of bond size r. At a non-empty node carrying a core of left bond l and right bond m, the recursively computed right environment E (an m-by-m stored matrix) is updated: for each of the two physical bits the core slice is contracted with E (a first materialized pass) and the result is contracted with the transposed slice (a second materialized pass), and the two per-bit matrices are added; a fixed six-word node record is charged. Each per-node update therefore costs on the order of l*m*m + l*l*m matrix-contraction work plus additive and materialization terms. Assume every bond along a length-n chain is at most D. Prove by induction on the chain that the total summed cost of the Gram computation is bounded by n*(24*D-cubed + 25*D-squared + 20*D + 6) + 5*D-squared + 4*D + 6.", "reference_answer": "Induct on the chain. Base case (empty tail): the environment is the materialized identity of bond r at most D, whose summed cost is 5*r-squared + 4*r at most 5*D-squared + 4*D, plus the fixed node record 6; this matches the n=0 instance of the bound. Inductive step at a node with left bond l, right bond m and tail chain C: the recursive Gram of C is bounded by the induction hypothesis (its bonds are at most D). The one-node update contracts the stored core against the already-stored environment E for both bits and adds the results; bounding l and m by D, the update's summed cost 12*l*m*m + 12*l*l*m + 8*l*m + 17*l*l + 20*l is at most 24*D-cubed + 25*D-squared + 20*D, and the node record adds 6, so one node contributes at most 24*D-cubed + 25*D-squared + 20*D + 6. Adding this to the inductive bound for the tail increments the n-coefficient by one, giving (n+1)*(24*D-cubed + 25*D-squared + 20*D + 6) + 5*D-squared + 4*D + 6. The cubic-in-D term comes from the two matrix contractions, each costing on the order of a bond cubed when all three dimensions reach D. These are exact-real word-operation counts including every materialization pass, not finite-bit or whole-machine runtimes; the two uses of E are stored reads of the same cached environment, not a recomputation.", "rubric": [ { "criterion": "Base case bounds the identity environment plus node record by 5*D-squared+4*D+6", "id": "c1", "weight": 1 }, { "criterion": "Bound one update by 24*D-cubed+25*D-squared+20*D using l,m at most D, attributing the cubic term to the contractions", "id": "c2", "weight": 1 }, { "criterion": "Complete the induction to the per-node coefficient and note exact-real counting with a cached environment", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The empty-tail identity environment plus its node record is bounded by 5*D-squared+4*D+6.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.gram" ] }, { "claim": "One node update is bounded by 24*D-cubed+25*D-squared+20*D with the cubic term from the two contractions.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.update_total_cost_le" ] }, { "claim": "Induction over the chain accumulates the per-node coefficient n times.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.gram_total_cost_le" ] } ], "source_item_digest": "6e688a66b2ad5ce4760787a4e1ca2cbadba1fb9f536048ff3512ea316b3b0157", "source_refs": [ { "anchor": "update80-86; gram103-108; update_total_cost_le234-242; gram_total_cost_le256-284", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredTensorTrainNorm.lean", "sha256": "60fa71fc821f7598f5df739b231f5b25cf1927824d2b028ef723f9ee04074011" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.update_total_cost_le", "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.gram", "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.update" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.gram_total_cost_le" ], "task_hash": "b4358b9c337147c7a945c846a49c21d3c3e81d3a00398cbb2711f6ae2cc0ba1c", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "Scalar-boundary chain; the Gram environment is a 1-by-1 stored matrix", "Reading a stored matrix entry is one row read plus one entry read" ], "context_hash": "e462bd30ff72572f513c1d53bc7843445eb3773fc2fa0ff22385049ec8afd7b1", "conventions": [ "The environment is already materialized before the boundary step; counts are exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-210", "difficulty": { "level": "intermediate", "rationale": "Exact marginal-cost accounting combined with a precise statement of the computed quantity." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-210", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b8f44402027ba18e2e26", "key": "qcb-core-la-210", "primary_sources": [], "question": "Given the stored right-to-left Gram-environment computation of a scalar-boundary tensor-train chain (which returns a 1-by-1 stored environment), the stored norm is obtained by reading the single scalar entry of that environment and taking its exact square root. Prove that the norm computation's cost equals the Gram computation's cost plus exactly two reads and one square root on the appropriate counters, and hence that its summed total exceeds the Gram total by exactly three. Then state precisely what mathematical quantity the returned scalar equals and why charging only two reads and one square root is the correct marginal cost of the scalar boundary.", "reference_answer": "The norm is defined as: run the Gram computation to get the 1-by-1 environment E, read its entry E(0,0), and return the exact square root of that entry. Reading the entry of a stored matrix is a row read followed by an entry read, i.e. two reads; the square root is one square-root call. No other counter is touched. Therefore, counter by counter, the norm cost equals the Gram cost plus two on the read counter and plus one on the square-root counter, and is unchanged on the other six counters. Summing all eight counters, the norm total is the Gram total plus exactly 2 + 1 = 3. The returned scalar equals the square root of the sum over all computational-basis words of the squared contracted amplitude of the chain, i.e. the Euclidean norm of the vector the tensor train represents; equivalently it is the square root of the single top-left Gram entry, since the full Gram environment at the scalar boundary stores exactly that total squared mass. Charging only two reads and one square root is correct because the environment has already been fully materialized by the Gram computation, so the only additional work at the boundary is to look up the stored scalar and apply one exact-real square root; this is an exact-real operation count, not a finite-bit cost, and the heavy contraction work is all accounted inside the Gram stage.", "rubric": [ { "criterion": "Identify the marginal operations as two reads (row then entry) and one square root", "id": "c1", "weight": 1 }, { "criterion": "Conclude the summed total exceeds the Gram total by exactly three", "id": "c2", "weight": 1 }, { "criterion": "State the scalar equals the Euclidean norm (square root of total squared mass) and justify the small marginal charge", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The norm reads the single environment entry (two reads) and applies one square root.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm" ] }, { "claim": "Per counter the norm cost is the Gram cost plus two reads and one square root, so the total grows by exactly three.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm_cost", "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm_total_cost_le" ] }, { "claim": "The returned scalar is the Euclidean norm, the square root of total squared contracted mass.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm_eq_sum" ] } ], "source_item_digest": "f7b3e31bd2ea7ed164adf59e8ce49b69f7a86494a8f7b0de784805f53bf9efcb", "source_refs": [ { "anchor": "norm118-121; norm_eq_sum129-131; norm_cost286-290; norm_total_cost_le293-302", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredTensorTrainNorm.lean", "sha256": "60fa71fc821f7598f5df739b231f5b25cf1927824d2b028ef723f9ee04074011" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm", "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm_eq_sum" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm_cost", "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.norm_total_cost_le" ], "task_hash": "3b55f69ba3a6423468b1f8776b1be61a47673f700e049a7c1038e4325e9e0b4c", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "The operation tags form a finite type with exactly eight inhabitants", "Cost functions map tags to natural numbers" ], "context_hash": "ba9d7994ed0204438b9c3937bc26712e62145f4042615bff076dc916aec64d5b", "conventions": [ "Totals are operation counts in the exact-real model, not finite-bit runtimes", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-211", "difficulty": { "level": "basic", "rationale": "Elementary finite-sum algebra, but with a precise account of why it enables compositional cost proofs." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-211", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-bf4cebe7a3787551e5ab", "key": "qcb-core-la-211", "primary_sources": [], "question": "Two notions of total cost appear in the stored exact-real model. One, call it the ordinary total, is the sum of a cost function over the finite type of all operation tags. The other, call it the enumerated total, is the explicit eight-term sum field + square root + inverse-cosine + other-trig + comparison + read + write + emitted-record. Prove that these two totals are equal as functions of the cost, that the total of a sum of two costs is the sum of their totals, that the total of the zero cost is zero, and that the total of a single unit charge on one tag is one. Explain why these simple algebraic facts are exactly what licenses replacing a sum-over-tags by the fixed eight-term expression in later cost proofs.", "reference_answer": "The operation tags form a finite type with exactly the eight inhabitants field, square root, inverse-cosine, other-trig, comparison, read, write, emitted-record. The ordinary total sums the cost over the universal finite set of tags. Enumerating that finite set as the eight listed inhabitants, the sum over the set equals field + sqrt + angle + trig + compare + read + write + emit, which is precisely the enumerated total; hence the two totals coincide for every cost. Additivity: the sum over tags of (a+b) applied pointwise is the sum over tags of a plus the sum over tags of b, because finite summation distributes over pointwise addition; so the total of a sum is the sum of the totals. Zero: the zero cost assigns zero to every tag, and a finite sum of zeros is zero. Unit charge: a single unit charge on one tag assigns one to that tag and zero to the others, so its total over all tags is exactly one. These facts matter because cost expressions are built compositionally by adding sub-costs and charging unit ticks; additivity lets a total be computed term by term, the zero and unit laws evaluate the leaves, and the enumeration identity lets a proof pass freely between the abstract sum-over-tags and the concrete eight-term polynomial used in the final bound. All of this concerns operation counts in the exact-real model and makes no finite-bit runtime claim.", "rubric": [ { "criterion": "Prove ordinary total equals the enumerated eight-term total via finite enumeration of tags", "id": "c1", "weight": 1 }, { "criterion": "Prove additivity, the zero law, and the unit-charge law for totals", "id": "c2", "weight": 1 }, { "criterion": "Explain how these laws license term-by-term compositional cost computation", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Enumerating the eight tags identifies the ordinary total with the eight-term enumerated total.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_eq_total" ] }, { "claim": "Finite summation gives additivity, the zero law and the unit-charge law.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_add", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_zero", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_tick" ] }, { "claim": "These laws justify computing a total term by term and switching between abstract and enumerated forms.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_eq_total" ] } ], "source_item_digest": "8a26d0b580fa303806b022d75965d1d39401d78114de1ccc9363c4c6c4f582c6", "source_refs": [ { "anchor": "ordinary21; ordinary_eq_total24-29; ordinary_add31-32; ordinary_zero34; ordinary_tick36", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredHermiteRawCost.lean", "sha256": "315e9f210c44c15e41742ef673ac93b191dfbbe48bb4ba84cd9abdd35f6adf62" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_add", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_zero", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_tick" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.ordinary_eq_total" ], "task_hash": "4d20fe61475c51994c7b2fd6be69371e88a709e54129ad20658baa770252f45c", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "A single run returns the chain and its full ledger of ordinary and named integer counters", "The four integer counters take the stated closed forms" ], "context_hash": "452e0ce89dca39d2f22da78f76959868350db0bd9c2f4707945865943f5c7c90", "conventions": [ "The named integer counters cover only those operations; no finite-bit or whole-machine runtime, and the source is unnormalized", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-212", "difficulty": { "level": "advanced", "rationale": "Closed-form summation plus a careful delineation of what an exact-real same-run certificate does and does not cover." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-212", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8c3f378ad16b740e1d84", "key": "qcb-core-la-212", "primary_sources": [], "question": "A stored Hermite raw-source assembly builds a scalar-boundary tensor-train chain for a state-preparation amplitude. One single run of the assembly returns the chain together with a work ledger carrying both the eight ordinary exact-real counters and four separately named integer counters (quotient calls, remainder calls, integer doublings, integer additions), plus a count of exponential-function calls. For parameters k and n it is proved on the same run that the chain denotes the explicit raw-source kernel chain, that the ordinary total is bounded by a stated polynomial in k and n, that the exponential calls are at most 3*(n+1)+1, and that the four integer counters equal (n+1)*(n+2), (n+1)-squared, n, and 4*(n+1) respectively. Prove that the four integer counters sum to 2*n-squared + 10*n + 7, and explain carefully which costs this certificate does and does not account for, especially regarding finite-bit arithmetic and normalization.", "reference_answer": "Substitute the four certified integer-counter values and add: quotient (n+1)*(n+2) = n-squared + 3*n + 2; remainder (n+1)-squared = n-squared + 2*n + 1; integer doublings n; integer additions 4*(n+1) = 4*n + 4. Summing: (n-squared + 3*n + 2) + (n-squared + 2*n + 1) + n + (4*n + 4) = 2*n-squared + 10*n + 7. Scope: the certificate is a same-run guarantee, so the identical run that is proved to denote the raw-source kernel chain is the one whose costs are bounded; it does not build a second chain or substitute a hypothetical dimension-only budget. The ordinary total counts exact-real field, square-root, trigonometric, comparison and stored-word operations, including every materialization pass and the per-node records. The four integer counters count only those four named integer operations arising in the cache construction; they explicitly do not count every index manipulation, arbitrary-precision bit-level work, proof checking or counter bookkeeping. The exponential-call cap counts calls to the exponential primitive and is at most 3*(n+1)+1 because disabled guards contribute zero injections. Crucially the certificate makes no finite-bit running-time claim and no normalized state-preparation-compiler claim: the returned chain is an unnormalized source, so one may not read the bound as a complete machine runtime or as a normalized amplitude loader.", "rubric": [ { "criterion": "Correctly expand and sum the four integer counters to 2*n-squared+10*n+7", "id": "c1", "weight": 1 }, { "criterion": "State the same-run scope: the certified run is the one that denotes the kernel chain", "id": "c2", "weight": 1 }, { "criterion": "Explain the exclusions: not all integer/bit work, no finite-bit runtime, unnormalized source not a normalized loader", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The four certified integer-counter values expand and sum to 2*n-squared+10*n+7.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_quotientCalls", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_remainderCalls", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_integerAdditions", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_selected_integer_total" ] }, { "claim": "The certificate is a same-run guarantee tying the denoted chain to its costs.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_certified" ] }, { "claim": "The counters exclude general integer/bit work and make no finite-bit or normalized-loader claim.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_certified" ] } ], "source_item_digest": "3b06beca17d084d5e8dfc81822d969d9e1271b530e70525a1fb4f0e7713a7c24", "source_refs": [ { "anchor": "raw_quotientCalls193-195; raw_remainderCalls197-199; raw_integerDoublings201-203; raw_integerAdditions205-207; raw_selected_integer_total210-214; raw_certified218-227", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredHermiteRawCost.lean", "sha256": "315e9f210c44c15e41742ef673ac93b191dfbbe48bb4ba84cd9abdd35f6adf62" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_certified", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_quotientCalls", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_remainderCalls", "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_integerAdditions" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredHermiteRawCost.raw_selected_integer_total" ], "task_hash": "61b6799a77e1b16e9c1077b5bedc1b583d3c564f8bb73a77dfc57045a1e9d9a2", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "Depth is computed by the per-wire scheduling fold, not an assigned serial surrogate", "Gate count is the list length; touched wires determine the pushed layer" ], "context_hash": "1338d8882c4725aad5ef5f382883c6cab66581368420716e28e56fa0194ce1e2", "conventions": [ "The reported resource depth equals the scheduling depth; this is a layer-count bound, not a runtime claim", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-213", "difficulty": { "level": "intermediate", "rationale": "List induction driven by a per-step monotone inequality over a scheduling fold." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-213", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b9901d5e1d311c3aacc3", "key": "qcb-core-la-213", "primary_sources": [], "question": "A primitive quantum circuit is a chronological list of gates, each gate acting on a set of touched wires. A per-wire scheduling fold tracks, for every wire, the earliest layer at which its next gate can be placed: processing a gate pushes every touched wire to one more than the maximum current depth over the gate's touched wires, and leaves untouched wires unchanged. The circuit depth is the maximum over all wires of the final per-wire depths; the gate count is the length of the list. Prove that the circuit depth is at most the gate count, i.e. that scheduling in parallel layers never makes the reported depth exceed the number of gates. Identify the key per-step inequality that drives the induction.", "reference_answer": "The key per-step fact is: if every wire currently has depth at most some bound, then after processing one gate every wire has depth at most that bound plus one. For a touched wire the new depth is one plus the maximum current depth over the touched wires; since each of those is at most the bound, their maximum is at most the bound, so the new depth is at most bound plus one. For an untouched wire the depth is unchanged, hence at most the bound and so at most bound plus one. Now induct on the gate list to bound the folded profile: starting from the all-zero profile (every wire at most 0), each of the L gates raises the uniform upper bound by at most one, so after the whole circuit every wire has depth at most 0 + L = L, the gate count. Taking the maximum over wires, the circuit depth is at most L. Finally the reported resource depth is defined to equal this scheduling depth, so it too is at most the gate count. This bounds a genuine parallel-scheduling depth, computed by the per-wire fold rather than assigned as a serial surrogate; equality can fail because parallelizable gates on disjoint wires share a layer, making the true depth strictly smaller than the gate count.", "rubric": [ { "criterion": "State and prove the per-step inequality: one gate raises the uniform depth bound by at most one, covering touched and untouched wires", "id": "c1", "weight": 1 }, { "criterion": "Induct from the all-zero profile over the L gates to bound every wire depth by L", "id": "c2", "weight": 1 }, { "criterion": "Conclude depth and reported resource depth are at most gate count, noting parallel layers can make it strictly smaller", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "One gate raises every wire's depth bound by at most one, for touched and untouched wires alike.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveCircuit.nextWireDepth_le" ] }, { "claim": "Folding over L gates from the zero profile bounds every wire depth by L.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveCircuit.foldl_nextWireDepth_le" ] }, { "claim": "The maximum over wires, and the equal reported resource depth, are at most the gate count.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveCircuit.depth_le_gateCount", "declaration:QuantumBlockEncoding.PrimitiveCircuit.resource_depth_le_gateCount" ] } ], "source_item_digest": "e8d6aac4380e1be63a7428f60ed9f11f343696644445cfb3f1cae3cfe7f2ba36", "source_refs": [ { "anchor": "nextWireDepth_le12-19; foldl_nextWireDepth_le23-33; depth_le_gateCount40-42; resource_depth_le_gateCount45-47", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/PrimitiveDepthBound.lean", "sha256": "fd31cfd12a5ed193608b0f666ee1e990981f36b45182fc34dd0fc2db1cf62b16" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveCircuit.nextWireDepth_le", "declaration:QuantumBlockEncoding.PrimitiveCircuit.foldl_nextWireDepth_le", "declaration:QuantumBlockEncoding.PrimitiveCircuit.resource_depth_le_gateCount" ], "target_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveCircuit.depth_le_gateCount" ], "task_hash": "95dba6b4d5534fed046180a718df92a9bfd1485180fae20640c1585f7402ce3b", "task_kind": "proof", "topic": "stored_cost_refinement" }, { "assumptions": [ "The identity is materialized by an index-equality comparison per entry", "Materialization adds the two-read-two-write surcharge per entry and per row" ], "context_hash": "4c3cabc648b540934ae8317d9d7db948cf096d3f5b9e11c4f0f96cfcb2e61cca", "conventions": [ "Stored constants are not free primitives; counts are exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-216", "difficulty": { "level": "basic", "rationale": "A boundary analysis correcting the false intuition that a stored constant is free." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-216", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8e2dea76f95072f459b0", "key": "qcb-core-la-216", "primary_sources": [], "question": "In the stored exact-real model the identity matrix of size N is not a primitive constant but is materialized: each of the N*N entries is produced by an exact comparison deciding whether its row and column indices are equal (returning one or zero), and the materialization adds the standard two-read-two-write surcharge per entry and per row. The proved identity cost is N*N comparisons plus (N*N + N) times the two-read-two-write surcharge. A colleague treats the stored identity as a zero-cost input when bounding a larger algorithm. Explain why this is wrong in general, work out the exact summed cost, and determine the boundary size N at which the identity genuinely costs nothing.", "reference_answer": "Treating the stored identity as free is wrong because it is materialized entry by entry: producing it charges N*N exact comparisons (one per entry to test index equality) and the materialization surcharge of two reads and two writes on each of the N*N entries and each of the N rows. Summing all eight counters: the comparison counter contributes N*N, and the read and write counters each contribute 2*(N*N + N), so the summed total is N*N + 4*(N*N + N) = 5*N*N + 4*N. This is strictly positive for every N at least 1, so the identity is not free at any positive size, and in a composed algorithm this term must be added, not dropped. The unique boundary at which the stored identity costs nothing is N = 0: the empty matrix has no entries and no rows, so every counter is zero (5*0 + 0 = 0). Thus the only free identity is the degenerate zero-dimensional one; for all N at least 1 the quadratic comparison and storage work is real exact-real operation cost, carrying no finite-bit runtime claim but still contributing to the overall operation count.", "rubric": [ { "criterion": "Explain the identity is materialized with one comparison per entry plus the storage surcharge", "id": "c1", "weight": 1 }, { "criterion": "Compute the exact summed cost 5*N*N + 4*N", "id": "c2", "weight": 1 }, { "criterion": "Identify N=0 as the unique boundary where the cost vanishes, so it is nonzero for all N at least 1", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The identity is materialized with one index-equality comparison per entry.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.identity" ] }, { "claim": "The summed cost is N*N comparisons plus 2*(N*N+N) reads and writes, i.e. 5*N*N+4*N.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.identity_cost", "declaration:QuantumBlockEncoding.StoredRectangularGivens.identityBudget" ] }, { "claim": "Only N=0 makes every counter vanish, so the identity is nonzero at any positive size.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.identity_cost" ] } ], "source_item_digest": "06ed62cc08c6f5635662d9347275c6820fe449819d095626d4b19289a2acaf7f", "source_refs": [ { "anchor": "identity92-93; identityBudget100-101; identity_cost103-106", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredRectangularGivens.lean", "sha256": "f8638a6d287b3b984fa530b13eeb18e3992904543236bd0756eda41fa6490d98" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.identity", "declaration:QuantumBlockEncoding.StoredRectangularGivens.identityBudget" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.identity_cost" ], "task_hash": "0d9be32a7721524eb7be83a6f303f51dc61c8b227f194eb5e9c14e6202bb9745", "task_kind": "boundary", "topic": "stored_cost_refinement" }, { "assumptions": [ "Persistent append copies only the left spine and shares the right list", "Append is charged the left length on the read and write counters" ], "context_hash": "37125adaa0fae89fe7496aab7f2b72ad6c93ddf5cc6ae1fb6677d1b09fd8372f", "conventions": [ "Cost is asymmetric in the two operands; exact-real operation count, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-217", "difficulty": { "level": "basic", "rationale": "A two-sided counterexample pinning the asymmetric cost of a persistent append." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-217", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-52addc85f92446f231ab", "key": "qcb-core-la-217", "primary_sources": [], "question": "When a stored sweep concatenates its chronological step logs, it uses a persistent list append that copies the spine of the left list while sharing the right list unchanged; the append is charged exactly the length of the left list in reads and the same in writes, with the right list contributing nothing. A colleague asserts two things: first that appending two logs is a constant-cost pointer operation, and second that the cost depends on the total number of records in both logs. Show that both assertions are wrong by exhibiting append costs for concrete left/right lengths, and state the exact rule the cost actually follows.", "reference_answer": "Both assertions are wrong. Against the constant-cost claim: appending a left list of length p costs p reads and p writes, which grows without bound as p grows, so it is not constant; for example a left log of 100 records costs 100 reads and 100 writes regardless of pointer sharing. Against the total-length claim: the right list is shared, not copied, so it contributes zero to the cost; appending a left list of length 2 to a right list of length 1000 costs only 2 reads and 2 writes, not 1002. Concretely, left length 2 and right length 1000 cost (2,2); left length 1000 and right length 2 cost (1000,1000); same total length, vastly different cost, which also disproves any dependence on the sum. The exact rule is: the persistent append cost equals the left list's length on both the read and write counters and is independent of the right list's length, because only the left spine must be rebuilt to point into the shared right tail. This is an operation count in the exact-real model with no finite-bit content; the asymmetry is the whole point of charging prefix copying rather than a fictitious constant or total-length cost.", "rubric": [ { "criterion": "Refute constant cost by showing cost grows with left length", "id": "c1", "weight": 1 }, { "criterion": "Refute total-length dependence by showing the shared right list contributes zero, with a concrete asymmetric pair", "id": "c2", "weight": 1 }, { "criterion": "State the exact rule: cost equals left length in reads and writes, independent of right length", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Append charges the left list length in both reads and writes.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.append" ] }, { "claim": "The shared right list contributes zero, so equal total lengths can have very different costs.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.append" ] }, { "claim": "The exact rule depends only on the left length, as used in the sweep cost bound.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.sweep_cost_le" ] } ], "source_item_digest": "cf1248c62cc4a15751ecfbedf58495f047548317dc6baee60b7e784b4f6e4d99", "source_refs": [ { "anchor": "append21-22; sweep_cost_le67-89", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredRectangularGivens.lean", "sha256": "f8638a6d287b3b984fa530b13eeb18e3992904543236bd0756eda41fa6490d98" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.sweep_cost_le" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredRectangularGivens.append" ], "task_hash": "197a301537485f405e0e659bb8345d08d5cb8559bf3ca8a4f4fafe5b55a5c76f", "task_kind": "counterexample", "topic": "stored_cost_refinement" }, { "assumptions": [ "The dispatcher charges one comparison and branches on m at most n", "The thin inner dimension is the minimum of m and n" ], "context_hash": "3654a41c7de8fae300570a6785aebf7c796a3ff7da36ffdd45a237d680fdeb2c", "conventions": [ "The tall branch runs no elimination; counts are exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-218", "difficulty": { "level": "intermediate", "rationale": "A dispatch boundary analysis contrasting a trivial branch with the full-cost branch." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-218", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-df57f3046c2c05017fc1", "key": "qcb-core-la-218", "primary_sources": [], "question": "A stored thin-LQ dispatcher takes an m-by-n stored matrix and charges one exact comparison to decide the shape. When m is at most n (the wide case) it transposes, runs the full stored rectangular compilation, and extracts both factors. When m exceeds n (the tall case) it does none of that: it returns the input matrix as the triangular factor together with a materialized identity of size n as the orthonormal factor. Explain what the tall branch returns and why it is correct, and compare the cost of the tall branch against the wide branch. In particular, refute the claim that a thin-LQ factorization always performs a sweep of eliminations.", "reference_answer": "In the tall branch (m greater than n) the dispatcher performs no elimination sweep at all. It returns the original matrix unchanged in the triangular-factor slot and a freshly materialized n-by-n identity in the orthonormal-factor slot, after paying only the one dimension comparison plus the identity materialization. This is correct for the thin shape because the thin inner dimension is the minimum of m and n, which equals n here; pairing the matrix with the identity trivially reproduces the matrix as a product with an orthonormal (identity) factor of the right size, so the factorization identity and the orthonormality condition both hold by construction. The cost contrast is stark: the wide branch pays the transpose, the cubic-order rectangular compilation and two extractions, whereas the tall branch pays only one comparison plus the identity cost 5*n*n + 4*n, with no field, square-root, inverse-cosine, trigonometric or emit work from any elimination. This refutes the claim that thin-LQ always runs a sweep: on tall inputs the admissible route is the trivial dispatch, and the only elimination-bearing path is the wide branch. The boundary between the branches is the exact comparison m at most n; everything is counted in the exact-real model with no finite-bit runtime claim.", "rubric": [ { "criterion": "Describe the tall branch returning the input and an identity with only a comparison plus identity cost", "id": "c1", "weight": 1 }, { "criterion": "Justify correctness via the thin inner dimension being the minimum of m and n", "id": "c2", "weight": 1 }, { "criterion": "Refute 'thin-LQ always sweeps' by contrasting tall (no sweep) and wide (full compilation) costs", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The tall branch returns the input matrix and a materialized identity, charging only a comparison plus the identity cost.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredThinLQ.compileBody", "declaration:QuantumBlockEncoding.StoredThinLQ.compileBudget" ] }, { "claim": "It is correct because the thin dimension equals the minimum of m and n, giving a trivial orthonormal factor.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredThinLQ.compile_correct" ] }, { "claim": "Only the wide branch performs the compilation, refuting the always-sweep claim.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredThinLQ.compile_cost_le" ] } ], "source_item_digest": "2b9c9d2449bb0b0da1ab69a254c046c2da52c231a4d71d5ef29f1a828508ee72", "source_refs": [ { "anchor": "compileBody111-119; compile122-125; compile_correct162-166; compileBudget174-175; compile_cost_le177-187", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredThinLQ.lean", "sha256": "ed08ef278a93268f7d9620cb2ddafc2e3468fcc4f2543923251a3ce4e057177a" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredThinLQ.compileBody", "declaration:QuantumBlockEncoding.StoredThinLQ.compile_correct", "declaration:QuantumBlockEncoding.StoredThinLQ.compileBudget" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredThinLQ.compile_cost_le" ], "task_hash": "e16efef3b9c471a562bce8d736a5649f62f9f6b049b4faeaee842630899e45c1", "task_kind": "boundary", "topic": "stored_cost_refinement" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-102", "difficulty": { "level": "intermediate", "rationale": "Needs the order-dropping behaviour of differentiation on high-multiplicity zeros plus a sharpness witness, beyond rote derivative computation." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-jet-interpolation", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8f0cd5c551b9e1ec5882", "key": "qcb-core-sp-102", "primary_sources": [], "question": "Let p and q be real polynomials and a a real point. Suppose (x-a)^{k+1} divides the difference p-q. Prove that for every order j with 0 <= j <= k, the j-th derivatives of p and q agree at a: (d^j p/dx^j)(a) = (d^j q/dx^j)(a). Explain precisely why the hypothesis forces agreement of all jets strictly below the multiplicity k+1, and why it says nothing about the j = k+1 derivative in general.", "reference_answer": "Set r = p - q, so (x-a)^{k+1} | r. Step 1 (differentiation lowers the zero order by at most j): differentiating a function with a zero of order m at a exactly j times leaves a zero of order at least m-j; formally, (x-a)^{m-j} divides the j-th derivative of r whenever (x-a)^m | r. With m = k+1 and j <= k we get m-j = k+1-j >= 1. Step 2: therefore (x-a)^{k+1-j} divides (d^j r/dx^j); since k+1-j >= 1, in particular (x-a)^1 = (x-a) divides the j-th derivative of r. Step 3: a polynomial divisible by (x-a) vanishes at a; evaluating (d^j r/dx^j) at a gives 0. Step 4: derivatives are linear, so (d^j r/dx^j)(a) = (d^j p/dx^j)(a) - (d^j q/dx^j)(a) = 0, i.e. the jets agree. Why only below the multiplicity: at j = k+1 we have m-j = 0, so the argument only yields (x-a)^0 = 1 dividing the derivative, a vacuous statement; the (k+1)-th derivatives need not agree. Concretely p - q = (x-a)^{k+1} has k-th derivative 0 at a but nonzero (k+1)-th derivative (k+1)!, witnessing that the bound is sharp.", "rubric": [ { "criterion": "States that dividing by (x-a)^{k+1} and differentiating j<=k times leaves divisibility by (x-a)^{k+1-j} with exponent >= 1.", "id": "r1", "weight": 1 }, { "criterion": "Concludes (x-a) divides the j-th derivative of p-q, hence it vanishes at a.", "id": "r2", "weight": 1 }, { "criterion": "Uses linearity of the derivative to transfer vanishing of the difference's jet into equality of the jets of p and q.", "id": "r3", "weight": 1 }, { "criterion": "Explains sharpness: at order k+1 the exponent drops to 0, and p-q=(x-a)^{k+1} shows the (k+1)-th derivatives can differ.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A zero of order k+1 differentiated j<=k times retains a zero of order k+1-j >= 1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.jet_eq_of_pow_dvd_sub" ] }, { "claim": "Divisibility by (x-a) forces vanishing at a, so the j-th derivative of p-q is 0 at a.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.jet_eq_of_pow_dvd_sub" ] }, { "claim": "Linearity transfers this to equality of the individual jets of p and q for all j<=k, and the bound is sharp at k+1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.jet_eq_of_pow_dvd_sub" ] } ], "source_item_digest": "68c0514f0b9677c677caf2ac29f4872c3571042ec82a12426c71df57763e6c00", "source_refs": [ { "anchor": "lines 51-62: theorem jet_eq_of_pow_dvd_sub (p q : R[X]) (a : R) (k j : N) (hj : j <= k) (h : (X - C a)^(k+1) | p - q) : (derivative^[j] p).eval a = (derivative^[j] q).eval a; proof uses pow_sub_dvd_iterate_derivative_of_pow_dvd and X - C a | derivative^[j] (p-q)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermitePolynomial.lean", "sha256": "f8a3e1d5f29aed1228b6ede3577a9af1e76301559172ee67a266fcb0d56fee1e" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.jet_eq_of_pow_dvd_sub" ], "task_hash": "75288bc780ab7e196a476f3a4f58c22ad62eebccb099019ae7b46ceb4da0bdc5", "task_kind": "proof", "topic": "structured_preparation" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-103", "difficulty": { "level": "advanced", "rationale": "Combines a Cauchy-product coefficient identity, a nonnegativity-of-terms argument, a coordinate change, and a careful strictness case split." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-source-positivity", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-21934bffe092872bf6f1", "key": "qcb-core-sp-103", "primary_sources": [], "question": "Consider the generating power series obtained as the product of the exponential series exp(X) and the series 1/(1-X)^{k+1}. Let its degree-r coefficient be a_{k,r}. In the coordinate t, define A_k(t) as the truncation of this series to degrees 0..k, i.e. A_k(t) = sum_{r=0}^{k} a_{k,r} t^r. The source polynomial, in the shifted coordinate t = p+1, is a positive combination S(p) = exp(-1) * (1-t)^{k+1} * A_k(t) + t^{k+1} * A_k(1-t). Prove that A_k(t) > 0 for every t >= 0, and deduce that S(p) > 0 for every p in the closed interval [-1, 0].", "reference_answer": "Step 1 (coefficients are nonnegative): the coefficient a_{k,r} equals the Cauchy product sum_{m=0}^{r} (1/m!) * C(k+r-m, k), where C denotes a binomial coefficient coming from 1/(1-X)^{k+1} = sum C(k+i, k) X^i and 1/m! from exp. Each summand is a quotient of nonnegative reals, so a_{k,r} >= 0, and a_{k,0} = C(k,k)/0! = 1 > 0. Step 2 (positivity of A_k on t >= 0): A_k(t) = sum_{r=0}^{k} a_{k,r} t^r is a finite sum of nonnegative terms a_{k,r} t^r (products of nonnegative a_{k,r} and nonnegative t^r). The r=0 term is a_{k,0} = 1, which is strictly positive, so the whole sum is at least 1 > 0. Step 3 (positivity of S on [-1,0]): put t = p+1. For p in [-1,0] we have t in [0,1], so both t >= 0 and 1-t >= 0. Then A_k(t) > 0 and A_k(1-t) > 0 by Step 2. The prefactors exp(-1) > 0, (1-t)^{k+1} >= 0, t^{k+1} >= 0 are nonnegative. Step 4 (strictness): if t = 0 the second term vanishes but the first is exp(-1) * 1 * A_k(0) > 0; otherwise t > 0 makes t^{k+1} * A_k(1-t) > 0, so one term is always strictly positive while the other is nonnegative. Hence S(p) > 0 throughout [-1,0].", "rubric": [ { "criterion": "Computes a_{k,r} as the Cauchy product sum of 1/m! and binomial C(k+r-m,k), establishing nonnegativity and a_{k,0}=1.", "id": "r1", "weight": 1 }, { "criterion": "Concludes A_k(t) >= 1 > 0 for all t >= 0 because the degree-0 term is 1 and all other terms are nonnegative.", "id": "r2", "weight": 1 }, { "criterion": "Substitutes t=p+1, noting p in [-1,0] gives t and 1-t both in [0,1] so both A_k factors are positive.", "id": "r3", "weight": 1 }, { "criterion": "Handles strictness by a case split on t=0 vs t>0 so that at least one of the two nonnegative terms is strictly positive.", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The series coefficient a_{k,r} is a Cauchy product of 1/m! and binomials, hence nonnegative with a_{k,0}=1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff" ] }, { "claim": "A finite nonnegative sum whose constant term is 1 is at least 1, so A_k(t)>0 for t>=0.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientPolynomial_pos" ] }, { "claim": "With t=p+1 in [0,1], both A_k(t) and A_k(1-t) are positive and the prefactors nonnegative.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.sourceInterpolant_pos" ] }, { "claim": "A case split on t=0 vs t>0 keeps one strictly positive term, giving S(p)>0 on [-1,0].", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.sourceInterpolant_pos" ] } ], "source_item_digest": "cf9f012b3ae8f37e5d04817754ee8ca5785470f5f4faef3d82c859613ef20f54", "source_refs": [ { "anchor": "lines 168-205: def coefficientSeries (k) := PowerSeries.exp R * (invOneSubPow R (k+1)).val; theorem coefficientSeries_coeff : coeff r = sum_{m in range (r+1)} C(k+r-m,k)/m!; coefficientSeries_coeff_nonneg; coefficientSeries_coeff_zero = 1. lines 254-286: coefficientPolynomial_pos (k) (t) (ht: 0<=t) : 0 < eval t; sourceInterpolant_pos (k) (p) (hp : p in Icc (-1) 0) : 0 < (sourceInterpolant k).eval p", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermitePolynomial.lean", "sha256": "f8a3e1d5f29aed1228b6ede3577a9af1e76301559172ee67a266fcb0d56fee1e" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries", "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff", "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientPolynomial_pos", "declaration:QuantumBlockEncoding.HermitePolynomial.sourceInterpolant_pos" ], "task_hash": "7ef977c0771d5c3478263e39205e0c296f12c775eda106b7c8a116a93d7e80f2", "task_kind": "proof", "topic": "structured_preparation" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-104", "difficulty": { "level": "advanced", "rationale": "A genuine Chinese-remainder/degree-counting uniqueness argument requiring both the divisibility equivalence and coprime-product reasoning." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-interpolant-uniqueness", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-01f9c7efc192d885642b", "key": "qcb-core-sp-104", "primary_sources": [], "question": "Two real polynomials p and q both have degree at most 2k+1, and both satisfy the same two-point jet data: all derivatives through order k agree at the node -1, and all derivatives through order k agree at the node 0. Prove that p = q. Structure the argument around the two 'endpoint multiplicity' polynomials (x+1)^{k+1} and x^{k+1}, and explain where coprimality and the degree budget 2k+1 enter.", "reference_answer": "Step 1 (each endpoint gives a divisibility): since p and q share all jets through order k at -1, the difference p-q has a zero of order at least k+1 at -1, i.e. (x+1)^{k+1} divides p-q. The equivalence used is: agreement of all derivatives through order k at a point a is equivalent to (x-a)^{k+1} dividing the difference (one direction is the high-multiplicity-zero fact; the converse is proved by reading Taylor coefficients, all of which through degree k vanish). The same at 0 gives x^{k+1} divides p-q. Step 2 (coprimality combines the two): (x+1)^{k+1} and x^{k+1} are coprime because their bases x+1 and x differ by a unit (their difference is 1), and coprimality is preserved under taking powers. For coprime factors each dividing the same element, their product divides it; hence (x+1)^{k+1} * x^{k+1} divides p-q. Step 3 (degree accounting forces zero difference): the product has degree (k+1)+(k+1) = 2k+2. If p-q were nonzero, a nonzero multiple of a degree-(2k+2) polynomial would have degree at least 2k+2. But deg(p-q) <= max(deg p, deg q) <= 2k+1 < 2k+2, a contradiction. Therefore p-q = 0, i.e. p = q. The degree budget 2k+1 is exactly one below the combined multiplicity 2k+2, which is what makes the only common solution the zero difference; coprimality is what lets the two separate divisibilities multiply.", "rubric": [ { "criterion": "Translates jet agreement at each node into divisibility of p-q by (x+1)^{k+1} and by x^{k+1} respectively (both directions of the equivalence).", "id": "r1", "weight": 1 }, { "criterion": "Establishes coprimality of the two endpoint moduli (bases differ by a unit, preserved under powers) to multiply the divisibilities.", "id": "r2", "weight": 1 }, { "criterion": "Computes the product degree 2k+2 and compares with the degree budget 2k+1 on p-q.", "id": "r3", "weight": 1 }, { "criterion": "Concludes p-q=0 since a nonzero multiple of the degree-(2k+2) product cannot have degree <= 2k+1, and identifies 2k+1 as the sharp budget.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Jet agreement through order k at a node is equivalent to divisibility by (x-node)^{k+1}, applied at -1 and 0.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.pow_dvd_sub_of_jet_eq" ] }, { "claim": "The two endpoint moduli are coprime, so their product divides p-q.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.endpointModuli_coprime" ] }, { "claim": "The product has degree 2k+2 while deg(p-q) <= 2k+1, forcing p-q = 0.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.sourceInterpolant_unique" ] } ], "source_item_digest": "10130b96d45745bf1b25df4c137ef2168b7368cb7eef7fb384197c1df6db6fc7", "source_refs": [ { "anchor": "lines 119-127: leftModulus k = (X - C(-1))^(k+1); rightModulus k = (X - C 0)^(k+1); endpointModuli_coprime k : IsCoprime. lines 380-417: pow_dvd_sub_of_jet_eq (p q a k) (h : forall j<=k, jets equal) : (X-C a)^(k+1) | p - q; sourceInterpolant_unique (k p) (hp : natDegree <= 2k+1) (hl hr : endpoint jets) : p = sourceInterpolant k", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermitePolynomial.lean", "sha256": "f8a3e1d5f29aed1228b6ede3577a9af1e76301559172ee67a266fcb0d56fee1e" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.sourceInterpolant_unique", "declaration:QuantumBlockEncoding.HermitePolynomial.pow_dvd_sub_of_jet_eq", "declaration:QuantumBlockEncoding.HermitePolynomial.endpointModuli_coprime" ], "task_hash": "7c879d58b6a577ad86f9188d1c63bfedd8ef7f07aed48b1ae88b800691bf5aa1", "task_kind": "proof", "topic": "structured_preparation" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-106", "difficulty": { "level": "intermediate", "rationale": "A concrete Cauchy-product computation with a negative-binomial series plus a reindexing and nonnegativity argument." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-source-positivity", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8d5c81189bdb1bc8c34c", "key": "qcb-core-sp-106", "primary_sources": [], "question": "Let the generating series be C(X) = exp(X) * (1 - X)^{-(k+1)} as formal power series over the reals. Prove that its coefficient of X^r is exactly sum_{m=0}^{r} C(k + r - m, k) / m!, where C(.,.) is the binomial coefficient, and prove that this coefficient is nonnegative for every r, with the r=0 coefficient equal to 1. You may use that exp(X) has coefficient 1/m! in degree m and that (1-X)^{-(k+1)} has coefficient C(k+i, k) in degree i.", "reference_answer": "Step 1 (Cauchy product): the coefficient of X^r in a product of two power series F*G is the convolution sum_{m=0}^{r} [X^m]F * [X^{r-m}]G. Here [X^m] exp = 1/m! and [X^{i}] (1-X)^{-(k+1)} = C(k+i, k). Step 2 (substitute i = r-m): the convolution becomes sum_{m=0}^{r} (1/m!) * C(k + (r-m), k) = sum_{m=0}^{r} C(k+r-m, k)/m!, which is the asserted closed form. The index substitution is valid because as m ranges over 0..r the complementary index r-m ranges over r..0, covering the same antidiagonal. Step 3 (nonnegativity): every summand is a ratio of a nonnegative binomial coefficient and a positive factorial, hence nonnegative; a finite sum of nonnegative reals is nonnegative. Step 4 (value at r=0): the sum collapses to the single term m=0, giving C(k, k)/0! = 1/1 = 1. This closed form is what makes the truncation A_k the exact finite object underlying the positive source polynomial, and its nonnegativity is exactly the ingredient needed downstream for positivity.", "rubric": [ { "criterion": "Applies the Cauchy-product (convolution) formula for the coefficient of X^r in a product of two power series.", "id": "r1", "weight": 1 }, { "criterion": "Substitutes the known coefficients 1/m! (exponential) and C(k+i,k) (negative binomial) and reindexes to the closed form sum_{m} C(k+r-m,k)/m!.", "id": "r2", "weight": 1 }, { "criterion": "Shows each term is a nonnegative-binomial-over-positive-factorial ratio, so the coefficient is nonnegative.", "id": "r3", "weight": 1 }, { "criterion": "Evaluates the r=0 case as the single term C(k,k)/0!=1.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The coefficient of X^r in exp(X)*(1-X)^{-(k+1)} is the convolution of 1/m! and C(k+i,k).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff" ] }, { "claim": "Reindexing i=r-m yields the closed form sum_{m=0}^{r} C(k+r-m,k)/m!.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff" ] }, { "claim": "Each summand is nonnegative, so the coefficient is nonnegative; at r=0 it equals 1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff_nonneg" ] } ], "source_item_digest": "bfb8236c482b42681d8f649cd2b0f1c243cc6405d52b55cb9abb4ab337f16792", "source_refs": [ { "anchor": "lines 184-204: theorem coefficientSeries_coeff (k r) : coeff r (coefficientSeries k) = sum_{m in range (r+1)} C(k+r-m,k)/m!; uses invOneSubPow_val_succ_eq_mk_add_choose and coeff_exp; coefficientSeries_coeff_nonneg; coefficientSeries_coeff_zero : coeff 0 = 1", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermitePolynomial.lean", "sha256": "f8a3e1d5f29aed1228b6ede3577a9af1e76301559172ee67a266fcb0d56fee1e" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff", "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff_nonneg", "declaration:QuantumBlockEncoding.HermitePolynomial.coefficientSeries_coeff_zero" ], "task_hash": "5c4612dfdf8210f04b654cc325464bf49bf9bdca2b3a08e9fd57494637c1a4a1", "task_kind": "proof", "topic": "structured_preparation" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-109", "difficulty": { "level": "advanced", "rationale": "Requires an operator-algebra view of de Casteljau (linear step, semiconjugacy, iteration) rather than index bookkeeping, and translating back to Bernstein sums." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-bernstein-subdivision", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-263d965bb007c8b4eb97", "key": "qcb-core-sp-109", "primary_sources": [], "question": "A degree-d polynomial is held as a coefficient row c(0),...,c(d) in the Bernstein basis, so that its value at t is sum_{r=0}^{d} c(r) B_{d,r}(t). One de Casteljau step with parameter s replaces the row c by the row with entries (1-s) c(i) + s c(i+1). Let L_u denote the operation that reads the left edge of the de Casteljau triangle built with parameter u (its i-th output is the result of i de Casteljau steps at u applied to c, read at position 0). Prove that evaluating the L_u-transformed coefficients at parameter t reproduces the original polynomial at the rescaled parameter u*t: sum_r (L_u c)(r) B_{d,r}(t) = sum_r c(r) B_{d,r}(u t).", "reference_answer": "Model the Bernstein state as an infinite coefficient row acted on by the linear 'de Casteljau step' operator T_s := (1-s) * Id + s * Shift, where Shift sends c(i) to c(i+1). The i-th left-edge output is (T_u^i c)(0), so L_u c = (i -> (T_u^i c)(0)). Step 1 (parameter composition): a single step at a scaled parameter factors as T_{u t} = (1-t) * Id + t * T_u. This is checked entrywise: ((1-t)Id + t T_u) c (i) = (1-t) c(i) + t[(1-u)c(i) + u c(i+1)] = (1 - u t) c(i) + u t c(i+1) = T_{u t} c (i). Step 2 (semiconjugacy): consequently L_u intertwines the step at scaled parameter with the step at t: L_u (T_{u t} c) = T_t (L_u c). This is the operator identity that says applying one scaled step then taking the left edge equals taking the left edge then applying one unscaled step. Step 3 (iterate the semiconjugacy d times): because L_u is a semiconjugacy from T_{u t} to T_t, it intertwines their d-th iterates: L_u (T_{u t}^d c) = T_t^d (L_u c). Reading position 0 of both sides gives (T_t^d (L_u c))(0) = (T_{u t}^d c)(0). Step 4 (translate to Bernstein sums): the left-edge-at-0 of a d-fold de Casteljau triangle equals the Bernstein sum, i.e. (T_s^d c)(0) = sum_{r=0}^{d} c(r) B_{d,r}(s). Applying this with the row L_u c and parameter t on the left, and with c and parameter u t on the right, yields exactly sum_r (L_u c)(r) B_{d,r}(t) = sum_r c(r) B_{d,r}(u t). This is the exact affine-restriction law: left de Casteljau subdivision reparametrises the curve to [0,u].", "rubric": [ { "criterion": "Represents the de Casteljau step as the linear operator T_s=(1-s)Id+s Shift and identifies L_u as the left edge (T_u^i c)(0).", "id": "r1", "weight": 1 }, { "criterion": "Proves the parameter-composition identity T_{u t}=(1-t)Id + t T_u entrywise.", "id": "r2", "weight": 1 }, { "criterion": "Derives the semiconjugacy L_u T_{u t} = T_t L_u and iterates it d times, then reads position 0.", "id": "r3", "weight": 1 }, { "criterion": "Converts via (T_s^d c)(0)=sum_r c(r)B_{d,r}(s) to the Bernstein-sum identity, concluding the u*t reparametrisation.", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The de Casteljau step is a linear operator and the left restriction is its iterated left edge.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteBernstein.casteljauStep" ] }, { "claim": "The scaled step factors as T_{u t} = (1-t)Id + t T_u, verified entrywise.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteBernstein.casteljauStep_mul_parameter" ] }, { "claim": "Hence L_u is a semiconjugacy T_{u t} -> T_t, which iterates to the d-th powers.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteBernstein.leftRestriction_step", "declaration:QuantumBlockEncoding.HermiteBernstein.casteljau_leftRestriction" ] }, { "claim": "Reading the triangle apex gives the Bernstein-sum reparametrisation to u*t.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteBernstein.leftRestriction_eval" ] } ], "source_item_digest": "3a814cf53b0df4df2ad559be465c6234a73ffca971366dae638536dbc21981d0", "source_refs": [ { "anchor": "lines 219-282: casteljauStep t = t . coefficientShift + (1-t) . 1; casteljauStep_mul_parameter : casteljauStep (u*t) = (1-t).1 + t.casteljauStep u; leftRestriction u c i = casteljau i c u 0; leftRestriction_step : leftRestriction u (casteljauStep (u*t) c) = casteljauStep t (leftRestriction u c); casteljau_leftRestriction via Function.Semiconj.iterate_right; leftRestriction_eval : sum_r leftRestriction u c r * basis d r t = sum_r c r * basis d r (u*t)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermiteBernstein.lean", "sha256": "661618d9c10826491ee293c7c411ac2d55056426739f4bbfd3e7c410c05e4bff" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermiteBernstein.casteljauStep", "declaration:QuantumBlockEncoding.HermiteBernstein.casteljauStep_mul_parameter", "declaration:QuantumBlockEncoding.HermiteBernstein.leftRestriction_step", "declaration:QuantumBlockEncoding.HermiteBernstein.casteljau_leftRestriction", "declaration:QuantumBlockEncoding.HermiteBernstein.leftRestriction_eval" ], "task_hash": "9f60a897b2c80da54c7c08ba23c6d9623a83a357dc4f0c5ae4e2da9eff2fc134", "task_kind": "proof", "topic": "structured_preparation" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-110", "difficulty": { "level": "intermediate", "rationale": "Requires recognising the convex-hull property from partition of unity and nonnegativity, plus checking the derived subdivision parameter lies in the unit interval." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-bernstein-subdivision", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-118b1619a4aee1d7dd78", "key": "qcb-core-sp-110", "primary_sources": [], "question": "Let a degree-d polynomial be given by Bernstein coefficients c(0),...,c(d), all lying in a common interval [lo, hi]. Suppose we restrict the polynomial to a subinterval [u, v] with 0 <= u < v <= 1 by two successive de Casteljau subdivisions, producing new Bernstein coefficients c'(0),...,c'(d) for the restricted piece. Prove that every restricted coefficient c'(i) still lies in [lo, hi]. Identify the two structural facts about the Bernstein basis that make this true.", "reference_answer": "Two facts drive the proof. Fact A (partition of unity): the Bernstein basis functions of a fixed degree sum to 1 at every t, i.e. sum_{r=0}^{d} B_{d,r}(t) = 1. Fact B (nonnegativity on [0,1]): each B_{d,r}(t) >= 0 for t in [0,1]. Together they say that any Bernstein sum sum_r c(r) B_{d,r}(t) with t in [0,1] is a convex combination of the c(r), hence lies between min and max of the coefficients; in particular if all c(r) in [lo,hi] the value lies in [lo,hi]. Step 1 (one subdivision preserves bounds): each new coefficient produced by a single de Casteljau subdivision at a parameter in [0,1] is itself a Bernstein sum of a sub-row of the original coefficients evaluated at that parameter. By Facts A and B it is a convex combination of coefficients from [lo,hi], hence in [lo,hi]. This applies to the right-edge (upper) subdivision at u in [0,1]: all its outputs stay in [lo,hi]. Step 2 (second subdivision): the left-edge (lower) subdivision is applied at the parameter (v-u)/(1-u). Because 0 <= u < v <= 1 we have 0 <= (v-u)/(1-u) <= 1 (the numerator is nonnegative and at most the positive denominator 1-u). So the second subdivision is again evaluated at a parameter in [0,1], and by the same convex-combination argument its outputs — the restricted coefficients c'(i) — remain in [lo,hi]. Step 3 (conclusion): composing the two bound-preserving subdivisions keeps every c'(i) in [lo,hi]. This is the exact numerical-stability property of de Casteljau subdivision: restricting to a subinterval never amplifies coefficients beyond the original range, in contrast to monomial-basis restriction which can.", "rubric": [ { "criterion": "States the partition-of-unity and nonnegativity facts and concludes any Bernstein sum on [0,1] is a convex combination of its coefficients.", "id": "r1", "weight": 1 }, { "criterion": "Shows one de Casteljau subdivision at a parameter in [0,1] outputs convex combinations, hence stays in [lo,hi] (applied to the u subdivision).", "id": "r2", "weight": 1 }, { "criterion": "Verifies the second subdivision parameter (v-u)/(1-u) lies in [0,1] using 0<=u 0 the generic expression tends to exp(2 start) * count by L'Hopital/the limit of (q^count-1)/(q-1) at q=1, matching the step=0 value, so the piecewise definition is continuous across the boundary. A single formula without the case split would either divide by zero at step = 0 or silently return a wrong value, so the explicit boundary branch is required for a total, correct definition.", "rubric": [ { "criterion": "Rewrites exp(x_j)^2 = exp(2 start)*(exp(2 step))^j, exhibiting a geometric series with ratio exp(2 step).", "id": "r1", "weight": 1 }, { "criterion": "Gives the generic closed form and argues the ratio exp(2 step) != 1 (hence denominator nonzero) exactly when step != 0, by injectivity of exp.", "id": "r2", "weight": 1 }, { "criterion": "Handles step=0 as a separate branch E = count*exp(2 start), noting the generic formula is a genuine 0/0 there.", "id": "r3", "weight": 1 }, { "criterion": "Explains the two branches agree in the step->0 limit, so the piecewise definition is continuous and total.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Each squared exponential sample factors as exp(2 start) times (exp(2 step))^j, a geometric term.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteIntervalMass.exponentialMassClosed_eq" ] }, { "claim": "For step != 0 the ratio exp(2 step) != 1 by injectivity of exp, so the geometric-sum formula applies with nonzero denominator.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteIntervalMass.exponentialMassClosed_eq" ] }, { "claim": "For step = 0 the ratio is 1 and the sum is count*exp(2 start); the branches agree in the limit, giving a total continuous definition.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteIntervalMass.exponentialMassClosed" ] } ], "source_item_digest": "f461f617f938693f5dd6590f3941706aa04c057bf24d4f52eda14c0641aecfab", "source_refs": [ { "anchor": "lines 89-116: exponentialMassClosed start step count = if step = 0 then count * exp(2*start) else exp(2*start) * ((exp(2*step))^count - 1)/(exp(2*step)-1); exponentialMassClosed_eq : = sum_{j in range count} (exp(start+step*j))^2; proof needs exp(2*step) != 1 derived from step != 0 via exp_injective, then geom_sum_eq", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermiteIntervalMass.lean", "sha256": "ff87cc69d0527cb1a7001a7bae4b916cf0a2c2b8bd1e6ba017511c3be27bc69f" } ], "split": "dev", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermiteIntervalMass.exponentialMassClosed", "declaration:QuantumBlockEncoding.HermiteIntervalMass.exponentialMassClosed_eq" ], "task_hash": "46987e42bff116466072220be525f1b23ae1f61ddbdc98fe9d8607692167e149", "task_kind": "boundary", "topic": "structured_preparation" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-114", "difficulty": { "level": "intermediate", "rationale": "A clean separable-factorization proof hinging on the shifted-Taylor expansion of an additive argument and the rank-of-product bound, with attention to why only degree matters." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-cut-rank", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-937920e35f31bbc84aa2", "key": "qcb-core-sp-114", "primary_sources": [], "question": "Consider a bivariate real array F(x,y) = q(u(x) + v(y)), where q is a real polynomial of degree at most d and u, v are arbitrary real-valued functions of the row index x and column index y respectively. Prove that F factorizes as a matrix product A * B with inner dimension d+1, and conclude that the rank of F is at most d+1. Make explicit which polynomial identity provides the factorization and why only the degree of q — not the number of rows or columns — controls the inner dimension.", "reference_answer": "Step 1 (shifted-Taylor expansion of an additive argument): fix a row x and let a = u(x). Expand q about a: for any real s, q(a + s) = sum_{i=0}^{d} c_i(a) s^i, where c_i(a) is the degree-i coefficient of the Taylor expansion of q about a (the sum stops at d because deg q <= d, so the shifted polynomial also has degree <= d). Step 2 (separate the variables): apply this with a = u(x) and s = v(y): F(x,y) = q(u(x) + v(y)) = sum_{i=0}^{d} c_i(u(x)) * v(y)^i. Each summand is a product of a function of x alone, A(x,i) := c_i(u(x)), and a function of y alone, B(i,y) := v(y)^i. Step 3 (read off the matrix product): define A as the (number of rows) by (d+1) matrix with entries A(x,i) and B as the (d+1) by (number of columns) matrix with entries B(i,y). Then (A B)(x,y) = sum_{i=0}^{d} A(x,i) B(i,y) = F(x,y), i.e. F = A * B exactly, with inner dimension d+1. Step 4 (rank bound): the rank of a product A*B is at most the number of columns of A, which is d+1; equivalently the inner dimension bounds the rank. Hence rank F <= d+1. Why the inner dimension is degree-controlled: the only data that produce the finite factorization are the d+1 Taylor coefficients of q; the number of distinct values of u and v (i.e. the matrix dimensions) is irrelevant. The additive structure u(x)+v(y) is essential — it is what lets the single-variable Taylor expansion separate the two indices into a sum of d+1 rank-one pieces. A generic bivariate function without this additive polynomial structure would have no such bound.", "rubric": [ { "criterion": "Expands q about a=u(x) as a degree-<=d polynomial in the shift s, truncating at d because deg q <= d.", "id": "r1", "weight": 1 }, { "criterion": "Substitutes s=v(y) to write F(x,y)=sum_{i=0}^d c_i(u(x)) v(y)^i, separating each term into a function of x times a function of y.", "id": "r2", "weight": 1 }, { "criterion": "Assembles the matrices A (rows x (d+1)) and B ((d+1) x cols) so that A*B = F with inner dimension d+1.", "id": "r3", "weight": 1 }, { "criterion": "Concludes rank F <= d+1 via rank(A*B) <= inner dimension, and explains the bound depends only on deg q, not matrix size, with additivity essential.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Taylor-expanding q about u(x) gives q(u(x)+v(y)) = sum_{i<=d} c_i(u(x)) v(y)^i, a sum of d+1 separable terms.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteCutRank.polynomial_add_factorization" ] }, { "claim": "These terms assemble into a product A*B with inner dimension d+1 (the explicit finite-width factorization).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteCutRank.FactorsThrough" ] }, { "claim": "A matrix product has rank at most its inner dimension, so rank F <= d+1, controlled only by deg q.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteCutRank.FactorsThrough.rank_le", "declaration:QuantumBlockEncoding.HermiteCutRank.polynomial_add_rank_le" ] } ], "source_item_digest": "8692de94902e02e828e3bcd36f059d65ad388327e3f3824390ab974de1fa729b", "source_refs": [ { "anchor": "lines 23-50: FactorsThrough F := exists A : Matrix a i R, B : Matrix i b R, F = A*B; FactorsThrough.rank_le : F.rank <= Fintype.card i; polynomial_add_factorization (p) (d) (hd: p.natDegree <= d) (u v) : FactorsThrough (i := Fin (d+1)) (fun x y => p.eval (u x + v y)) via A x i = (taylor (u x) p).coeff i, B i y = v y^i; polynomial_add_rank_le : rank <= d+1", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermiteCutRank.lean", "sha256": "c0f038bd889c77163967c71df5f5987082f4b79ab7db2b7822efcdda7234aaf2" } ], "split": "dev", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermiteCutRank.FactorsThrough", "declaration:QuantumBlockEncoding.HermiteCutRank.polynomial_add_factorization", "declaration:QuantumBlockEncoding.HermiteCutRank.FactorsThrough.rank_le", "declaration:QuantumBlockEncoding.HermiteCutRank.polynomial_add_rank_le" ], "task_hash": "8b614e6706f44336d8bf2a029e216db595e59d3b13e231f89266a26cba617364", "task_kind": "proof", "topic": "structured_preparation" }, { "assumptions": [ "the suffix index satisfies v(y) < M so each threshold reduces to a two-state separation across the cut", "the bound is a rank/separation certificate and asserts no circuit compiler" ], "context_hash": "7570953e0d23a0cbeb553fda1718c5103e1b48ee07f771fdeceec493ff6bfbe2", "conventions": [], "core_id": "qcb-core-sp-116", "difficulty": { "level": "advanced", "rationale": "Requires assembling the full piecewise sample from indicator, exponential, and polynomial factorizations and bookkeeping the composed widths to the exact 8k+12, plus the two-state threshold reduction." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "hermite-cut-rank", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-caed3d27b15d8ccbb59b", "key": "qcb-core-sp-116", "primary_sources": [], "question": "The full target initial datum is a piecewise function: an exponential left tail for p < -1, a degree-<=2k+1 polynomial bridge for -1 <= p < 0, and an exponential right tail for p >= 0. Sample it on an integer-indexed grid and split the index as M*u(x) + v(y) across a binary cut, where v(y) < M. Prove that the resulting bivariate sample array has rank at most 8k+12, uniformly over the cut location, by exhibiting an explicit finite-width factorization. Account for every additive piece and the exact width it contributes.", "reference_answer": "Write the sampled datum as a sum of three terms selected by threshold indicators of the integer index n(x,y) = M u(x) + v(y). Let A, B be the integer cutoffs for the thresholds p < -1 and p < 0 respectively (obtained by a ceiling of an affine expression; across the cut each threshold indicator factors through width 2). Term 1 (left tail): 1[n < A] * exp(grid), width 2 * 1 = 2 (indicator times exponential). Term 2 (middle): (1[n < B] - 1[n < A]) * q(grid), where q is the source polynomial of degree <= 2k+1. The band indicator is a difference of two width-2 indicators, so by the sum/negation laws it factors through width 2 + 2 = 4; multiplied entrywise by the polynomial of width (2k+1)+1 = 2k+2 gives width 4 * (2k+2) = 8k+8. Term 3 (right tail): (1 - 1[n < B]) * exp(-grid), the complement indicator times an exponential; the complement factors through width 2 and the exponential through width 1, so width 2 * 1 = 2. Summing the three terms adds widths: 2 + (8k+8) + 2 = 8k+12. Each threshold indicator truly separates across the binary cut because 'M u(x) + v(y) < T' reduces, using v(y) < M, to a two-row condition (u(x) < T/M, or u(x) = T/M together with v(y) < T mod M) — exactly a two-state separation. The exponential separations use exp(a+b) = exp(a) exp(b) (width 1), and the polynomial separation uses the shifted-Taylor factorization (width deg+1). The combined explicit width is a finite index set of cardinality 8k+12, so the rank of the sample array is at most 8k+12, independent of the register width and of where the cut is placed. This is a separation certificate only; it does not by itself furnish a gate compiler or a finite-precision algorithm.", "rubric": [ { "criterion": "Decomposes the sampled datum into left-tail, middle-band, and right-tail terms selected by threshold indicators of the cut index.", "id": "r1", "weight": 1 }, { "criterion": "Assigns widths: left tail 2 (indicator x exp), right tail 2 (complement indicator x exp).", "id": "r2", "weight": 1 }, { "criterion": "Computes the middle width 8k+8 as band-indicator width 4 (difference of two width-2 indicators) times polynomial width 2k+2.", "id": "r3", "weight": 1 }, { "criterion": "Sums the widths 2+(8k+8)+2 = 8k+12, notes threshold separation uses the v(y)= c) is C^k. A student claims the jet-matching hypothesis is unnecessary — that merely requiring f and g to be C^k and to agree in value at c (f(c) = g(c)) already gives a C^k splice. Disprove this for k = 1 with an explicit example, and explain exactly which derivative order fails and why the full jet-matching hypothesis through order k is needed for a C^k conclusion.", "reference_answer": "Counterexample at k = 1, cut c = 0. Take f(x) = 0 and g(x) = x. Both are C^1 (indeed C^infinity), and they agree in value at 0: f(0) = 0 = g(0). The splice is h(x) = 0 for x < 0 and h(x) = x for x >= 0, which is exactly the ReLU function (max(x,0)). Step 1 (continuity holds): h is continuous because the one value-matching condition f(0) = g(0) is satisfied; so h is C^0. Step 2 (first derivative fails): the left derivative of h at 0 is f'(0) = 0, while the right derivative is g'(0) = 1. Since 0 != 1, h is not differentiable at 0, hence not C^1. The student's weakened hypothesis (value agreement only) is therefore insufficient: it controls order 0 but leaves the order-1 jets mismatched. Step 3 (what the real principle needs): to obtain C^1 we must also require f'(0) = g'(0); more generally for C^k we need iterated-derivative agreement at every order j = 0,1,...,k. The proof of the correct principle is by induction on k: the base case uses continuity from value matching; the inductive step differentiates the splice, which is itself a splice of f' and g' (using that the one-sided derivatives coincide precisely because f'(c)=g'(c)), and then applies the hypothesis one order lower. Each induction level consumes exactly one jet-matching equation, so all k+1 equations j = 0,...,k are genuinely used. Dropping any of them breaks the chain at that order, exactly as the ReLU example breaks at order 1. Hence the jet-matching hypothesis through order k is necessary, not decorative, and value agreement alone yields only continuity.", "rubric": [ { "criterion": "Gives an explicit k=1 example (f=0, g=x, i.e. ReLU) with f(0)=g(0) but mismatched first derivatives.", "id": "r1", "weight": 1 }, { "criterion": "Shows the splice is continuous (value match) but not differentiable at 0 because left derivative 0 != right derivative 1.", "id": "r2", "weight": 1 }, { "criterion": "States the correct hypothesis needs derivative agreement at every order j=0..k, not just the value.", "id": "r3", "weight": 1 }, { "criterion": "Explains the inductive proof consumes one jet-matching equation per order, so each of the k+1 equations is genuinely used.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "With only value agreement, the splice (ReLU) is continuous but has unequal one-sided first derivatives at the cut.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteSmoothness.hasDerivAt_splice" ] }, { "claim": "Hence it fails to be C^1, disproving the weakened claim; C^1 additionally requires f'(c)=g'(c).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteSmoothness.deriv_splice" ] }, { "claim": "The correct C^k gluing is proved by induction consuming one jet-matching equation per order, so all orders 0..k are needed.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteSmoothness.contDiff_splice" ] } ], "source_item_digest": "27638bfef87bf0b5f1fc109c2d14d973d255199199bfe386b9d7b6b53453ccdf", "source_refs": [ { "anchor": "lines 56-90: deriv_splice needs hv : f c = g c and hd : deriv f c = deriv g c; contDiff_splice (k c f g) (hf hg : ContDiff k) (hjet : forall j <= k, iteratedDeriv j f c = iteratedDeriv j g c) : ContDiff k (splice c f g); proof by induction on k, base uses value match for continuity, succ step differentiates and recurses with jets one order lower", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermiteSmoothness.lean", "sha256": "da47568584b8435a07431a7ebae29836796018f76d02b4d8b0c4169830ec68bc" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermiteSmoothness.contDiff_splice", "declaration:QuantumBlockEncoding.HermiteSmoothness.hasDerivAt_splice", "declaration:QuantumBlockEncoding.HermiteSmoothness.deriv_splice" ], "task_hash": "ad5b1531c748bbf99e004ffbac47e9511e7124505b135069e0ec96b85fd5050d", "task_kind": "counterexample", "topic": "structured_preparation" }, { "assumptions": [ "L > 0 so the grid step is positive and comparisons against -1 are well posed", "the cost model counts only real field operations and exact real comparisons; natural-index and bit-level costs are out of scope" ], "context_hash": "8e2c93cc7cca0006597ac42e4cf34e002a27c7af317fefdf67a12e3210d67cc8", "conventions": [], "core_id": "qcb-core-sp-120", "difficulty": { "level": "advanced", "rationale": "An invariant-based correctness proof of a recursive real-interval binary search with explicit boundary cases, plus an exact operation count under a stated restricted cost model." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "hermite-cutoff-search", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-67b95b72c0c86c5f45f5", "key": "qcb-core-sp-120", "primary_sources": [], "question": "To locate where the left exponential tail meets the polynomial bridge on a grid of 2^n cells, one performs an exact-comparison binary search carrying real interval endpoints: starting from the real interval [lower, lower+span] with span equal to the full width, each of n steps halves the span, forms the midpoint, compares it against the threshold -1, and recurses into the half that still contains the crossing. The search returns an index 'cutIndex'. Prove that this search returns exactly the true cutoff index for every width n and scale L > 0, including the boundary cases where the crossing is at the very first grid point or at the very last, and the single-cell (n = 0) base case. State the exact number of real comparisons and real field operations the search performs.", "reference_answer": "Model the search by an interval invariant: at a node with 'remaining' halvings left, 'start' already-decided low-order index, and current interval first point gridPoint(start) and width span = step * 2^remaining, assume the true cutoff c = cutIndex satisfies start <= c <= start + 2^remaining (it lies in the current interval). We prove the returned value equals c by induction on 'remaining'. Base case (remaining = 0, one cell): the interval has a single cell starting at 'start'. The search compares the grid point at 'start' against -1. If gridPoint(start) < -1, then 'start' is still below the threshold, so the crossing is at start+1 and the hypotheses force c = start+1; the search returns start+1. Otherwise 'start' is at or past the threshold, so c = start and the search returns start. In both branches the returned value equals c; this also covers the endpoint boundary cases (crossing at the first or last point) because the invariant pins c to one of the two interval ends when the cell is minimal. Inductive step (remaining+1): the search forms the midpoint, which equals gridPoint(start + 2^remaining) (the first point of the upper half), and compares it to -1. If it is below -1, the crossing lies in the upper half, so recurse with start' = start + 2^remaining and the same remaining; the lower bound updates to start' <= c (since the midpoint is still below threshold, c >= start + 2^remaining) and the width halves, preserving the invariant, so by induction the recursion returns c. If it is at or past -1, the crossing lies in the lower half; recurse with the same start and halved width, and the invariant's upper bound tightens to c <= start + 2^remaining, again returning c. The key geometric identities are span/2 = step * 2^remaining and gridPoint(start) + step*2^remaining = gridPoint(start+2^remaining), which make the carried real endpoints exact — no ceiling evaluation or grid enumeration occurs. Starting the top-level search with the full interval [-pi L, 0] and remaining = n satisfies the initial invariant (the cutoff lies between index 0 and the midpoint index 2^n), so the top-level call returns cutIndex exactly. Cost: each of the n internal halvings performs 2 real field operations (one division to halve the span, one addition to form the midpoint) and 1 real comparison; the base level performs 1 comparison. The top-level producer additionally does one multiplication (pi*L) and one subtraction (0 - width) to initialise the interval. Hence the search proper uses exactly 2n field operations and n+1 comparisons; including initialisation the producer uses 2n+2 field operations and n+1 comparisons. Natural-index arithmetic, powers of two, and counter bookkeeping are outside this real-arithmetic cost model; finite-bit comparison of arbitrary real inputs is not claimed.", "rubric": [ { "criterion": "Formulates the interval invariant start <= cutIndex <= start + 2^remaining and proves the base case returns start or start+1 by the single comparison, covering endpoint boundary cases.", "id": "r1", "weight": 1 }, { "criterion": "Proves the inductive step: the midpoint equals gridPoint(start+2^remaining), and each comparison recurses into the correct half while preserving the invariant.", "id": "r2", "weight": 1 }, { "criterion": "Uses the exact real identities span/2 = step*2^remaining and gridPoint(start)+step*2^remaining = gridPoint(start+2^remaining) so endpoints stay exact (no ceiling/enumeration).", "id": "r3", "weight": 1 }, { "criterion": "States the exact cost: 2n field operations and n+1 comparisons for the search (2n+2 field ops with the pi*L and subtraction initialisation), noting the real-arithmetic cost model's scope.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The interval invariant start <= cutIndex <= start+2^remaining with the base-case single comparison returns start or start+1, covering endpoint cases.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.search_value" ] }, { "claim": "The inductive step compares the exact midpoint gridPoint(start+2^remaining) and recurses into the correct half, preserving the invariant, so compute returns cutIndex.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.search", "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.compute_value" ] }, { "claim": "The search costs 2n field operations and n+1 comparisons (2n+2 field ops including the pi*L / subtraction initialisation).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.search_cost", "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.compute_cost" ] } ], "source_item_digest": "345c21e4f78ee80ed2cbaa7372d069aacbfb19d8c3c2f046e5b8df13c107196a", "source_refs": [ { "anchor": "lines 22-128: search (remaining start lower span) recursion; search_value (hlo : start <= cutIndex) (hhi : cutIndex <= start + 2^remaining) : (search ...).value = cutIndex via half_span and middle_point; compute n L = do width<-mul pi L; lower<-sub 0 width; search n 0 lower width; compute_value : = cutIndex; search_cost : cost op = 2*remaining*tick field op + (remaining+1)*tick compare op; compute_comparisons = n+1; compute_field_operations = 2n+2", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermiteBinaryCutoff.lean", "sha256": "3e236d6572ceba6d0418cb75da49512152bf03586cebc03b4d9cd8132029d098" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.search", "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.search_value", "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.compute_value", "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.search_cost", "declaration:QuantumBlockEncoding.HermiteBinaryCutoff.compute_cost" ], "task_hash": "0ae661d5bcfbd3f3f7820db0d44697a504cf2bf644caf423a2f841b95c06cb01", "task_kind": "boundary", "topic": "structured_preparation" }, { "assumptions": [ "d>=3 integer; k>=1 integer", "Werner alpha∈[-1,1]; isotropic x in its valid state range", "The supplied analytic criteria are exactly alpha>=(1-d)/k and x>=d/(d+1)*(1-1/k)" ], "context_hash": "c6771e45a4d4b702c2590163481cd496333ee75963f397290209efee14382645", "conventions": [], "core_id": "qcb-core-plan-402", "difficulty": { "level": "intermediate", "rationale": "A boundary analysis of the analytic extendibility formulas: solving for the maximal extension order and recovering separability as the infinite-k limit, using monogamy rather than any PPT argument." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "werner-isotropic-extendibility", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-3b9c0d7d556a8cb17f29", "key": "qcb-core-plan-402-werner-isotropic-k-extendibility-thresholds-corrected-v2", "primary_sources": [ { "anchors": "Example7.2 PDFpp59-60, Eq7.5 with fraction (1-d)/k and Eq7.6; printed fraction visually inspected, not OCR linearization.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "59-60", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Fix d>=3. A d-dimensional Werner state has parameter alpha in [-1,1] and is k-extendible exactly when alpha >= (1-d)/k. An isotropic state has valid parameter x>=0 and is k-extendible exactly when x >= [d/(d+1)](1-1/k), for positive integers k. Using these given criteria, prove every Werner state is2-extendible, determine the largest k for a fixed isotropic x below d/(d+1), and find the two parameter thresholds for extendibility at every k. Explain why the numerator grouping in (1-d)/k is essential.", "reference_answer": "For Werner states and k=2, (1-d)/2<=-1 because d>=3. Every allowed alpha>=-1 therefore satisfies the2-extendibility criterion. The thresholds (1-d)/k increase toward0 as k grows, since1-d<0. Thus satisfying every k is equivalent to alpha>=0: if alpha<0 a sufficiently large k violates the inequality. For isotropic states write c=d/(d+1)>0. The criterion x>=c(1-1/k) is equivalent to1/k>=1-x/c. For0<=x=c, every k satisfies the inequality. Hence all-k thresholds are alpha>=0 and x>=d/(d+1). This calculation uses (1-d)/k, not1-d/k. The latter tends to1 instead of0 and gives a different, false criterion for this parameterization. These conclusions follow from the stated analytic criteria, not from a new proof of the upstream extendibility theorem.", "rubric": [ { "criterion": "Use the exact grouped numerator and d>=3 to prove2extendibility", "id": "werner", "weight": 1 }, { "criterion": "Derive floor cutoff for0<=x=c", "id": "isotropic", "weight": 1 }, { "criterion": "Correct monotone thresholds and allk limits0,c", "id": "limits", "weight": 1 }, { "criterion": "Explain why (1-d)/k is not1-d/k", "id": "grouping", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "At k2 the Werner threshold(1-d)/2<=-1, so all allowed alpha satisfy it.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The Werner thresholds increase to0; satisfying every k is equivalent to alpha>=0.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Rearranging the isotropic criterion below c gives the exact integer cutoff; at or above c every k works.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "Changing fraction parentheses changes the supremum from0to1 and is not the same theorem.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "e44a4b02385956e555df305910fd11506ff3e11002d06ca0e3b57f07499a1820", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:matrix-analysis" ], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "9dae0547c3f5760f35e5660612b4b961e32402f269dda03065bb2a61285f0e4a", "task_kind": "boundary", "topic": "planned_quantum_information_apps" }, { "assumptions": [ "d>=1 finite; fixed common orthonormal basis defines transpose, conjugation and maximally entangled vector", "U is unitary; X arbitrary linear operator" ], "context_hash": "b3ed3047dd5b29b5469dd23e7ce6f21d3f84f22b179af4bca4dcb32720c9f742", "conventions": [], "core_id": "qcb-core-plan-411", "difficulty": { "level": "intermediate", "rationale": "A basis-level proof of the transpose trick and the U-conj(U) invariance of the maximally entangled state; this is the ring-ribbon identity underlying isotropic symmetry, kept distinct from any Werner/swap-commutant computation." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "maximally-entangled-intertwiner", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-f5668af8a7eb543e7a56", "key": "qcb-core-plan-411-transpose-trick-and-UU-invariance-corrected-v2", "primary_sources": [ { "anchors": "Exercise 6.1 on PDF page 58: (X_A tensor I_B)|Phi+> = (I_A tensor X^T_B)|Phi+> and invariance of |Phi+> under U tensor conj(U).", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "58", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let |Phi+> = (1/sqrt(d)) sum_{i=1}^d |i>_A |i>_B be the standard maximally entangled state on C^d tensor C^d. (a) Prove the transpose trick: for every operator X in L(C^d), (X_A tensor I_B)|Phi+> = (I_A tensor X^T_B)|Phi+>, where X^T is the transpose of X in the fixed computational basis. (b) Deduce that |Phi+> is invariant under every unitary of the form U tensor conj(U) (U_A tensor the complex conjugate of U on B). Carefully state the role of the computational basis in defining the transpose and conjugate.", "reference_answer": "Expand the left side: (X tensor I)|Phi>=(1/sqrt d)sum_{i,j}X(j,i)|j> tensor|i>. Expanding (I tensor X^T)|Phi> gives(1/sqrt d)sum_{i,j}X(i,j)|i> tensor|j>; interchanging dummy indices yields the same vector. For a unitary U, apply that identity with X=U and then apply I tensor bar U: (U tensor bar U)|Phi>=(I tensor bar U U^T)|Phi>. Since UU dagger=I, taking entrywise complex conjugates gives bar U U^T=I. The vector is therefore unchanged. Both transpose and bar refer to the same fixed orthonormal basis used to define|Phi>. In contrast U tensor U does not generally fix this vector: for d2 and U=diag(1,i), (|00>+|11>)/sqrt2 becomes(|00>-|11>)/sqrt2. Thus conjugation on the second factor is essential, not a removable notation convention.", "rubric": [ { "criterion": "Correct finite-index transpose calculation with no false intermediate identity", "id": "expansion", "weight": 1 }, { "criterion": "Correct bar U U^T=I from unitary identity", "id": "unitarity", "weight": 1 }, { "criterion": "Same fixed basis for transpose/conjugation and entangled vector", "id": "basis", "weight": 1 }, { "criterion": "Valid diag(1,i) counterexample to missing conjugation", "id": "counterexample", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Expand both tensor actions and relabel the finite double sum to prove the transpose trick.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Conjugating UU dagger=I yields bar U U^T=I.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Combine the two identities to prove U tensor bar U invariance.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "The diagonal phase example refutes U tensor U invariance without conjugation.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "05e0d2384e44696ebe45e9477ead16e724d11c08a081759a128d460e2e821847", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "2c3bd7c77e5078405d65f4eba49112b03858fbae596a9b14b5ab3c5905e8a44f", "task_kind": "proof", "topic": "planned_quantum_information_apps" }, { "assumptions": [ "d>=1; M>=N>=1; X positive semidefinite", "Haar pure-power projector identity as stated; fixed ancillary basis for Kraus insertions" ], "context_hash": "5fe18b5dd9398862eb513b0de02a99888e195158ed2f1508a1f647d68679982e", "conventions": [], "core_id": "qcb-core-plan-418", "difficulty": { "level": "intermediate", "rationale": "Verifies the structural (CP, trace-non-increasing) properties of the cloning map and characterizes exact trace preservation, complementing the fidelity-saturation item." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "approximate-cloning-optimal-map", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-94de1e94898503d72b45", "key": "qcb-core-plan-418-cloning-map-cp-trace-nonincreasing-corrected-v2", "primary_sources": [ { "anchors": "Lemma7.3 Eq7.9 Haar symmetric-projector identity; Eq8.25 cloning map and corresponding CP/trace exercise. Original trace-support proof supplied.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "60-61,67-69", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let H=C^d with d>=1, and integers M>=N>=1. Let Pi_L project onto the symmetric subspace of H tensor L and d_L its dimension. Define T(X)=(d_N/d_M)Pi_M(X tensor I^{tensor(M-N)})Pi_M. Prove T is completely positive and derive the exact trace formula tr T(X)=tr(X Pi_N) for every positive X. You may use Pi_L=d_L times the Haar average of|phi>v tensor e_j. Then sum_j K_j X K_j dagger=X tensor I, so this map is completely positive; generally it needs several Kraus maps, not one isometry. Compression Y->Pi_M Y Pi_M is also completely positive, and positive scaling preserves that property. Let Q be the partial trace of Pi_M over the last M-N factors. The given Haar identity gives Q=d_M integral |phi>=0 this is equivalent to support(X) contained in the symmetric subspace: the positive square-root identity tr[X(I-Pi_N)]=||(I-Pi_N)X^(1/2)||_HS^2 gives the equivalence. Every mixture of pure powers lies there, but not every symmetric state is such a mixture. For d2,N2, the entangled vector(|01>+|10>)/sqrt2 is symmetric, so its projector preserves trace under T, yet it is not a mixture of pure powers (a pure entangled state cannot be a convex combination of product pure states). Thus support in Sym^N, not separability into pure powers, is the correct equality condition.", "rubric": [ { "criterion": "Use a sum of insertion Kraus maps, not a single isometry for tensoring with identity", "id": "CP", "weight": 1 }, { "criterion": "Derive partial trace of Pi_M using Haar average with correct dimension ratio", "id": "partialtrace", "weight": 1 }, { "criterion": "Exact trace formula and positive support equality condition", "id": "trace", "weight": 1 }, { "criterion": "Valid symmetric entangled counterexample to pure-power-mixture equality characterization", "id": "boundary", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Ancillary identity extension has Kraus maps v->v tensor e_j; compression and positive scaling preserve complete positivity.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Taking a partial trace in the Haar projector identity gives Q=(d_M/d_N)Pi_N.", "supporting_nodes": [ "concept:tensor-networks", "concept:harmonic-analysis" ] }, { "claim": "Cyclicity and the partial-trace pairing give trT(X)=tr(XPi_N)<=trX.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The Hilbert-Schmidt square identifies equality exactly with symmetric support.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "A symmetric entangled pure state shows equality is broader than mixtures of pure powers.", "supporting_nodes": [ "concept:tensor-networks" ] } ], "source_item_digest": "759b3d5a55c7fb2b6fdecb9196f4d8a053109e944bb8a426cdefc98a5cc0a376", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:harmonic-analysis", "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "725a57d7569e1dd4e328632495035850f6fd70d234b83ef2f57049c7aaa2b02c", "task_kind": "proof", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "5c4b63ca58846e9c8e7055e0485fb73b8ff8129d5f36a24ad2c4990f2611cf83", "conventions": [ "T_k(cos p) = cos(k p) and U_{k-1}(cos p) = sin(k p)/sin p", "sqrt(1-t^2) is taken nonnegative, matching p in [0, pi]" ], "core_id": "qcb-core-plan-503", "difficulty": { "level": "intermediate", "rationale": "Standard rotation-power computation but requires correctly separating the finite second-kind value from its vanishing sqrt(1-t^2) prefactor at the endpoints." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-chebyshev-eigenvalue-transformation", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-33ef3d580b96537fb641", "key": "qcb-core-plan-503-second-kind-offdiagonal", "primary_sources": [ { "anchors": "Eq. (10.56) matrix power of the qubitization iterate with diagonal T_{2k} and off-diagonal -sqrt(1-lambda^2) U_{2k-1}, and the parallel singular-value forms Eqs. (10.72), (10.74) carrying the same second-kind off-diagonal structure", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "171-174", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "On the invariant plane attached to a value t in the closed interval from minus one to one, the qubitization iterate acts as the plane rotation M = [[t, -sqrt(1-t^2)],[sqrt(1-t^2), t]]. Prove that the k-th power of M has diagonal entries equal to the first-kind Chebyshev value T_k(t) and off-diagonal entries equal to plus or minus sqrt(1-t^2) times the second-kind Chebyshev value U_{k-1}(t). Make explicit the trigonometric definitions you use and verify the degenerate behavior at t = 1 and t = -1.", "reference_answer": "Write t = cos(p) with p = arccos(t) in [0, pi], so sqrt(1-t^2) = sin(p) (nonnegative on this range). Then M = [[cos p, -sin p],[sin p, cos p]] is exactly the rotation by angle p. Powers of a rotation add angles: M^k = [[cos(k p), -sin(k p)],[sin(k p), cos(k p)]]. By the definitions T_k(cos p) = cos(k p) and U_{k-1}(cos p) = sin(k p)/sin(p), the diagonal entries are cos(k p) = T_k(t) and the off-diagonal entries are minus sin(k p) = minus sin(p) [sin(k p)/sin(p)] = minus sqrt(1-t^2) U_{k-1}(t) (upper right) and plus sqrt(1-t^2) U_{k-1}(t) (lower left). Degenerate cases: at t = 1, p = 0, so M = identity and M^k = identity; indeed T_k(1) = 1 and sqrt(1-t^2) U_{k-1}(1) = 0 since the prefactor sin(p) vanishes even though U_{k-1}(1) = k is finite. At t = -1, p = pi, so M^k = diag(cos(k pi), cos(k pi)) = (-1)^k identity; consistently T_k(-1) = (-1)^k and the off-diagonal prefactor sqrt(1-t^2) = 0 kills the U term. Thus the off-diagonal genuinely carries the second-kind polynomial, but always multiplied by sqrt(1-t^2), which is why it vanishes exactly at the spectral endpoints.", "rubric": [ { "criterion": "Substitutes t = cos p and identifies M as a rotation by p with sin p = sqrt(1-t^2)", "id": "r1", "weight": 1 }, { "criterion": "Uses the angle-addition / rotation-power identity to get M^k with cos(kp), sin(kp)", "id": "r2", "weight": 1 }, { "criterion": "Matches diagonal to T_k(t) and off-diagonal to plus/minus sqrt(1-t^2) U_{k-1}(t) via the standard Chebyshev definitions", "id": "r3", "weight": 1 }, { "criterion": "Checks t = 1 and t = -1, noting the sqrt(1-t^2) prefactor annihilates the finite U_{k-1} at the endpoints", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "Set t = cos p so sqrt(1-t^2) = sin p and M is the rotation by p.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Rotation powers add angles: M^k rotates by k p.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Diagonal cos(k p) equals T_k(t).", "supporting_nodes": [ "concept:approximation", "concept:harmonic-analysis" ] }, { "claim": "Off-diagonal sin(k p) equals sin p times U_{k-1}(t), i.e. sqrt(1-t^2) U_{k-1}(t).", "supporting_nodes": [ "concept:approximation", "concept:harmonic-analysis" ] }, { "claim": "At t = 1 the prefactor sin p = 0 kills the finite U_{k-1}(1) = k, giving the identity.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "At t = -1 the power is (-1)^k identity, consistent with T_k(-1) = (-1)^k and zero off-diagonal.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "b6107ce9db2ef5b7186113a0124865a058dbf3c5319ce52791949fccae0d6d19", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:matrix-analysis", "concept:approximation", "concept:harmonic-analysis" ], "task_hash": "6ff7c709cd4f11815db5f15ad22db389bd2a2923edd6ba54675d4e611689c034", "task_kind": "boundary", "topic": "planned_polynomial_signals" }, { "assumptions": [], "context_hash": "7138e9c185f032cd5636250d0cbaf3cf4ee97f80523cf0f1becc76a19ea647c1", "conventions": [ "The flag/complement split is the grading into ancilla-zero and ancilla-nonzero sectors", "Z gives a diagonal reflection diag(+I,-I); X gives an off-diagonal swap in that grading" ], "core_id": "qcb-core-plan-507", "difficulty": { "level": "intermediate", "rationale": "A clean refutation hinging on the diagonal-versus-off-diagonal structure of the reflection relative to the grading; requires an explicit non-commuting product." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-cosine-sine-decomposition", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8bd8f3ad39a4012ca263", "key": "qcb-core-plan-507-z-reflection-commutes-not-xy", "primary_sources": [ { "anchors": "Theorem 10.10 and Eq. (10.96): Z_Pi decomposes as a direct sum of Z blocks, commuting with the outer unitaries W-tilde and V-tilde thanks to the block-diagonal structure, with the explicit remark that this would fail if Z were replaced by Pauli X or Y", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "176", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "In the cosine-sine view of qubitization the flag reflection is built from the single-qubit operator Z acting on the ancilla flag, and the large outer unitaries that diagonalize the construction are block-diagonal with respect to the split between the flag sector and its orthogonal complement. A student claims the construction would work identically if the flag reflection were instead built from the Pauli X operator, since X is also a self-inverse single-qubit unitary. Show that the student is wrong by proving that the Z-based reflection commutes with the block-diagonal outer unitaries whereas an X-based reflection generically does not, and explain why this commutation is essential to the clean cosine-sine factorization.", "reference_answer": "The flag reflection from Z is, in the flag-versus-complement split, the operator that is plus the identity on the flag sector and minus the identity on the complement; call it D = diag(+I, -I). The outer unitaries have the block-diagonal form diag(W, W') (and similarly diag(V, V')), where W acts entirely within the flag sector and W' entirely within the complement, because they come from singular value decompositions respecting that split. Two block-diagonal operators with respect to the SAME split commute when one of them is a scalar (plus or minus one) multiple of the identity on each block: D diag(W,W') = diag(+W, -W') = diag(W,W') D. Hence the Z-based reflection commutes with the outer unitaries, which is exactly why the reflection can be pulled through the factorization and its definition need not reference W, W', V, V'. Now replace Z by X. An X-based reflection acts within the ancilla as an off-diagonal swap, so in the flag-versus-complement split it has nonzero blocks COUPLING the flag sector to the complement: schematically it looks like [[0, I],[I, 0]] rather than diag(+I,-I). Multiplying this by diag(W,W') gives [[0, W'],[W, 0]], while diag(W,W') times it gives [[0, W],[W', 0]]; these are equal only if W = W', which fails generically. Therefore an X-based reflection does not commute with the outer unitaries, it entangles the two graded sectors, and the clean direct-sum-of-two-dimensional-rotations structure collapses. The essential point is that only a reflection that is diagonal in the grading (as Z provides) preserves the block-diagonal symmetry that makes qubitization factor into independent planar rotations.", "rubric": [ { "criterion": "Writes the Z-based reflection as diag(+I,-I) in the flag/complement grading and the outer unitaries as diag(W,W')", "id": "r1", "weight": 1 }, { "criterion": "Proves commutation diag(+I,-I) diag(W,W') = diag(W,W') diag(+I,-I)", "id": "r2", "weight": 1 }, { "criterion": "Shows an X-based reflection is off-diagonal [[0,I],[I,0]] and computes the two products disagreeing unless W = W'", "id": "r3", "weight": 1 }, { "criterion": "Explains commutation is what lets the reflection be pulled through the factorization, preserving independent planar rotations", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "The Z reflection is diag(+I,-I) in the flag/complement grading.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Outer unitaries are block-diagonal diag(W,W') respecting the same grading.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "A per-block scalar reflection commutes with any block-diagonal operator.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Commutation lets the reflection be pulled through without referencing W, W'.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "An X reflection is off-diagonal [[0,I],[I,0]], coupling the sectors.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Its products with diag(W,W') disagree unless W = W', so it fails to commute and breaks the factorization.", "supporting_nodes": [ "concept:matrix-analysis", "concept:operator-theory" ] } ], "source_item_digest": "8c10659977129d9beb6a9dc8a112cd04b982917267361a57dd4449e718f07091", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "5d33dbd730b9c3f9535af0e9a28d08bf79a1cbb5602669a18e3884eaec74ef7b", "task_kind": "counterexample", "topic": "planned_polynomial_signals" }, { "assumptions": [ "K is a power of two so the binary digit scheme is exact" ], "context_hash": "3f2a7e131dd70cda4725437f97b6eef4e3a579cee1c70cd746c90fd9dd61cac4", "conventions": [ "Query count means uses of the base block-encoding unitary; gate and state-preparation costs are not included", "The degree-2k transform uses exactly 2k base-unitary queries" ], "core_id": "qcb-core-plan-508", "difficulty": { "level": "intermediate", "rationale": "Requires both the quadratic naive bound and the geometric-series argument for the binary reuse, plus discipline to report only query counts." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-lcu-chebyshev-synthesis", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-086d0262eba86631e9eb", "key": "qcb-core-plan-508-lcu-select-query-count", "primary_sources": [ { "anchors": "Section 10.5 LCU and qubitization: the select oracle Eq. (10.100), the O(K^2) cost of the direct implementation because the iterates are not reused, and the efficient binary implementation Fig. 10.3 with total queries 2(1+2+...+2^{a-1}) = 2^{a+1}-2 = 2K-2 equal to one T_{2K-2} block encoding", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "177-178", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "To build a block encoding of a linear combination of even Chebyshev transforms, sum over k from 0 to K-1 of coefficients times the degree-2k Chebyshev transform, one needs a select operation that, conditioned on an index register holding k, applies the degree-2k transform. The degree-2k transform is itself built from 2k uses of the base block-encoding unitary. (a) Prove that a direct index-controlled implementation that applies each degree-2k transform independently uses a number of base-unitary queries that grows quadratically in K. (b) Prove that an implementation exploiting the binary digits of the index, which reuses a single squared iterate across the digits, uses only 2K-2 base-unitary queries when K is a power of two, and observe that this equals the query count of a single degree-(2K-2) transform. State the query counts without converting them into gate or state-preparation costs.", "reference_answer": "(a) In the direct implementation the branch for index k runs the degree-2k transform, costing 2k queries to the base unitary. Summing independently over k from 0 to K-1 gives total queries 2 times (0 + 1 + ... + (K-1)) = 2 times K(K-1)/2 = K(K-1), which is of order K squared. The inefficiency is that the squared base iterate used inside degree-2k is rebuilt from scratch for every k rather than reused. (b) Write K = 2^a and the index k in binary as digits k_{a-1} ... k_0. The degree-2k transform equals the squared base iterate raised to the power k. Using the binary expansion, k = sum over j of k_j 2^j, so raising to the power k is achieved by, for each digit position j, conditionally applying the squared iterate raised to the 2^j power exactly when digit k_j is one. The squared iterate to the 2^j power is applied once per position, contributing 2 times 2^j base-unitary queries (each squared iterate is two base-unitary uses). The total is 2 times (1 + 2 + 4 + ... + 2^{a-1}) = 2 times (2^a - 1) = 2^{a+1} - 2 = 2K - 2 queries. This matches the query count of a single degree-(2K-2) transform, which uses 2 times (K-1) = 2K-2 base-unitary queries, so the whole controlled sum costs no more base-unitary queries than one top-order term. These are counts of queries to the base unitary only; translating them into elementary-gate counts or into the cost of preparing the index-register superposition is a separate accounting that depends on additional oracle implementations and is not performed here.", "rubric": [ { "criterion": "Shows the direct implementation costs 2 times sum of k = K(K-1), order K squared, and identifies the non-reuse as the cause", "id": "r1", "weight": 1 }, { "criterion": "Explains the binary-digit scheme raises the squared iterate to the power k via per-digit controlled applications", "id": "r2", "weight": 1 }, { "criterion": "Computes the total as 2 times (2^a - 1) = 2K - 2 queries for K = 2^a", "id": "r3", "weight": 1 }, { "criterion": "Observes 2K-2 equals a single degree-(2K-2) transform's query count and keeps the accounting at query level without converting to gate or preparation cost", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "Each degree-2k branch costs 2k base-unitary queries.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "Independent summation over k gives K(K-1), order K squared.", "supporting_nodes": [ "concept:complexity" ] }, { "claim": "Binary expansion of k turns the power into per-digit controlled applications of the squared iterate to the 2^j power.", "supporting_nodes": [ "concept:complexity" ] }, { "claim": "Each position contributes 2 times 2^j queries.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "The geometric sum totals 2^{a+1}-2 = 2K-2 queries.", "supporting_nodes": [ "concept:complexity" ] }, { "claim": "This equals a single degree-(2K-2) transform; counts stay at query level only.", "supporting_nodes": [ "concept:complexity", "concept:block-encoding" ] } ], "source_item_digest": "99fb50087dd29dddc1503144dc867f4855efd2d53d33cf573ad8a8869fe5ed47", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:complexity", "concept:block-encoding" ], "task_hash": "5e2faa828a8b4adc864fe54585c9ed704958dc968fcbdc3c9f6838546b3cdbc7", "task_kind": "proof", "topic": "planned_polynomial_signals" }, { "assumptions": [ "A is accessed by a unitary with an input projector of rank N and an output projector of rank N'" ], "context_hash": "21bda55e5a762d3f13ddb066656b03625e03fff70a2d5351db9f84b36b8b8638", "conventions": [ "Each reflection is twice a projector minus the identity", "The iterate is adjoint(U) (output reflection) U (input reflection)" ], "core_id": "qcb-core-plan-511", "difficulty": { "level": "advanced", "rationale": "Requires understanding that qubitization beyond the computational basis is a two-reflection product and that the Jordan two-dimensional reduction is insensitive to rank equality." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-qubitization-beyond-basis", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-ae001ed77014e2db9df6", "key": "qcb-core-plan-511-subspace-blockencoding-reflections-reduce", "primary_sources": [ { "anchors": "Section 10.6 qubitization beyond the computational basis: reflections Z_Pi = 2 Pi - I and Z_{Pi'} = 2 Pi' - I (Eq. 10.106), the iterate U^dagger Z_{Pi'} U Z_Pi (Eq. 10.107), the remark that N' = N reduces to the usual iterate under a change of basis while N' != N remains well defined with the same CS-decomposition analysis, and the Notes-and-references statement of Jordan's lemma that a product of two reflections acts as a rotation on an invariant two-dimensional subspace", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "179-180", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Suppose an operator A maps an N-dimensional input space into an N'-dimensional output space and is accessed through a single unitary U on a larger space together with two projectors: an input projector of rank N and an output projector of rank N'. Define the two reflections as twice the respective projector minus the identity, and define the qubitization iterate as the product of the adjoint of U, the OUTPUT reflection, U, and the INPUT reflection. (a) Prove that when the input and output ranks are equal this iterate reduces, after a suitable change of basis, to the ordinary same-space qubitization iterate. (b) Explain precisely what feature of the construction continues to make sense when the ranks differ, and why the two-dimensional reduction underlying the cosine-sine analysis still applies.", "reference_answer": "(a) When N' = N the output projector has the same rank as the input projector, so there is a unitary change of basis identifying the output subspace with the input subspace; under this identification the output reflection becomes the input reflection and U together with its adjoint act between two copies of the same N-dimensional flagged subspace. Substituting, the iterate adjoint(U) (output reflection) U (input reflection) becomes adjoint(U) (input reflection) U (input reflection), which is exactly the standard same-space qubitization iterate adjoint(U) R U R with the single reflection R = 2(input projector) - I. Thus the general iterate specializes to the ordinary one up to the basis change. (b) When the ranks differ the construction still makes sense because the iterate is defined purely as a product of two reflections built from genuine projectors; nothing in that definition requires the two projector ranks to agree. The cosine-sine / two-dimensional reduction continues to apply because the product of two reflections always decomposes the ambient space into mutually orthogonal subspaces that are at most two-dimensional and invariant: on each such plane each reflection acts as a one-dimensional sign flip relative to the plane's intersection with the respective projector range, and the product of the two reflections acts as a rotation of that plane. The possibly different ranks only change how many planes are paired versus left as fixed vectors (the mismatch in dimension contributes extra invariant one-dimensional pieces), but every two-dimensional block still carries a planar rotation whose angle encodes a singular value of A. Hence the same singular-value transformation analysis via two-dimensional rotations survives when N' is not equal to N; this is the natural setting for a basis-changed singular value transformation.", "rubric": [ { "criterion": "Shows that equal ranks allow a basis change identifying output with input subspace, reducing the iterate to adjoint(U) R U R with one reflection", "id": "r1", "weight": 1 }, { "criterion": "States the general iterate is well defined for any ranks since it is just a product of two projector reflections", "id": "r2", "weight": 1 }, { "criterion": "Explains the product of two reflections always splits the space into at most two-dimensional invariant planes", "id": "r3", "weight": 1 }, { "criterion": "Notes rank mismatch only alters the count of paired planes versus fixed vectors, with each plane still a rotation encoding a singular value", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "Equal ranks give a unitary identifying the output subspace with the input subspace.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Under this identification the iterate becomes adjoint(U) R U R, the ordinary same-space iterate.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "For general ranks the iterate is still a well-defined product of two projector reflections.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "A product of two reflections splits the space into at most two-dimensional invariant planes.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Rank mismatch only changes the number of paired planes versus fixed one-dimensional pieces.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Each two-dimensional plane is a rotation whose angle encodes a singular value of A.", "supporting_nodes": [ "concept:matrix-analysis", "concept:block-encoding" ] } ], "source_item_digest": "216c79773b04ab1d4a02081409afd1d5bcb8b699bbc13e5e52d49665f0603746", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis", "concept:block-encoding" ], "task_hash": "9752fb0bb01b010d0f7d46108d00c2d2852fc81eabd3058763bc0060109bd576", "task_kind": "proof", "topic": "planned_polynomial_signals" }, { "assumptions": [ "theta lies strictly between zero and a right angle so cosine theta is a nondegenerate singular value" ], "context_hash": "013226851f8473ad6f61fd64f6b317fb36b1b005b6e64e7fce62223275bb513f", "conventions": [ "R(theta) is the planar rotation with cosine theta on the diagonal", "The input reflection restricts to Pauli Z on the first qubit" ], "core_id": "qcb-core-plan-512", "difficulty": { "level": "intermediate", "rationale": "A concrete worked instance of qubitization beyond the computational basis, requiring the overlap-matrix definition and the two-reflection-equals-double-rotation identity." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-qubitization-beyond-basis", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-06e37e8f96f59004bfe3", "key": "qcb-core-plan-512-rotation-reflections-overlap-example", "primary_sources": [ { "anchors": "Example 10.13: m = n = 1, U_A = I, Pi = |0><0| tensor I so Z_Pi = Z tensor I, R(theta) rotation (Eq. 10.109), Pi' = Xi' Pi adjoint(Xi'), overlap matrix A_ij = giving A = cos theta times I_2 (Eq. 10.110), and the iterate U^dagger Z_{Pi'} U Z_Pi = R(2 theta) tensor I with eigenphases plus/minus 2 theta determining the singular value cos theta (Eq. 10.111)", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "179-180", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Take the smallest nontrivial subspace-encoding example: a four-dimensional space, the identity as the accessing unitary, an input projector that keeps the first qubit in its zero state, and an output projector obtained by rotating the input projector by a real angle theta on the first qubit (via the planar rotation with cosine theta on the diagonal and the usual off-diagonal signs). With the encoded operator read off as the overlap between the rotated and unrotated flagged bases, prove that the encoded operator equals cosine theta times the two-dimensional identity, and that the qubitization iterate (product of the two reflections, since the unitary is the identity) equals the planar rotation by twice theta on the first qubit tensored with the identity. Deduce the eigenphases of the iterate and relate them to the encoded singular value.", "reference_answer": "The input projector keeps the first qubit at zero, so its range is spanned by the two states with first qubit zero; the unrotated flagged basis is these two states. The output projector is the input projector conjugated by the single-qubit rotation R(theta) on the first qubit, so its range is spanned by R(theta) applied to the same two states, giving the rotated flagged basis. The encoded operator is the matrix of overlaps between the rotated output basis vectors and the unrotated input basis vectors: each overlap is the inner product of a state with its R(theta)-rotation, which on the first qubit is cosine theta and on the second qubit is one; cross terms vanish by orthogonality of the second qubit. Hence the overlap matrix is cosine theta times the two-dimensional identity, so the encoded operator is cosine theta times the identity and its single distinct singular value is cosine theta. Because the accessing unitary is the identity, the qubitization iterate is just the product of the two reflections: (output reflection)(input reflection). On the first qubit the input reflection is the Pauli Z (plus one on zero, minus one on one), and the output reflection is R(theta) Z R(theta)-adjoint; their product R(theta) Z R(theta)-adjoint Z equals the planar rotation by twice theta. Tensoring with the untouched second qubit, the iterate is R(2 theta) on the first qubit tensored with the identity. A planar rotation by angle 2 theta has eigenvalues e to the plus and minus i 2 theta, each with multiplicity two here. These eigenphases plus and minus 2 theta determine the encoded singular value through cosine theta: the half of the eigenphase is theta, and its cosine is exactly the encoded singular value cosine theta. Thus the two-reflection product converts the geometric overlap cosine theta into a rotation whose eigenphase is twice the arccosine of that singular value.", "rubric": [ { "criterion": "Identifies the unrotated and rotated flagged bases and computes the overlap matrix as cosine theta times the identity", "id": "r1", "weight": 1 }, { "criterion": "Concludes the encoded operator is cosine theta times the identity with singular value cosine theta", "id": "r2", "weight": 1 }, { "criterion": "Shows the two-reflection product equals R(theta) Z R(theta)-adjoint Z = rotation by twice theta on the first qubit tensored with identity", "id": "r3", "weight": 1 }, { "criterion": "Gives eigenphases plus/minus 2 theta and relates them to the singular value via cosine theta = cosine of the half eigenphase", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "The input and output flagged bases are the first-qubit-zero states and their R(theta) rotations.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "Overlaps between the two bases give cosine theta on the diagonal, zero off-diagonal.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Hence the encoded operator is cosine theta times the identity, singular value cosine theta.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "With identity accessing unitary the iterate is the product of the two reflections.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "On the first qubit this product is R(theta) Z R(theta)-adjoint Z = R(2 theta).", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Eigenphases plus/minus 2 theta satisfy cosine of the half-phase equals the singular value cosine theta.", "supporting_nodes": [ "concept:matrix-analysis", "concept:operator-theory" ] } ], "source_item_digest": "f7c7bcf690a37698856d2c6713a064d8184919097eb397ba4fc74e1f628a723a", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis", "concept:block-encoding" ], "task_hash": "0886d4c092043d10f17c0a7d02ba6f8b42e0ac67add8d57cf3b47067f8ac38e4", "task_kind": "proof", "topic": "planned_polynomial_signals" }, { "assumptions": [ "Exactly one marked element; the candidate can be verified with one extra oracle query" ], "context_hash": "7ffb06caa2db6a02486b68095b2da41d2edd0f2a719f706de412a2db608cbd83", "conventions": [ "Half-angle = arcsine(1/sqrt(N)); failure probability = cosine squared of (2k+1) half-angles", "Counts are oracle queries, not elementary gates" ], "core_id": "qcb-core-plan-514", "difficulty": { "level": "intermediate", "rationale": "Requires the trigonometric failure bound plus the expected-query repetition argument, with care to keep the cost in oracle queries." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-grover-amplitude-amplification", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-78dfcf40c7801ba85af0", "key": "qcb-core-plan-514-grover-failure-bound", "primary_sources": [ { "anchors": "Theorem 11.2 and Eqs. (11.19)-(11.21): the choice k = floor(pi/(2 theta) - 1/2) giving (2k+1)theta/2 >= pi/2 - theta, the failure bound P(fail) = cos^2((2k+1)theta/2) <= sin^2(theta) = 4 sin^2(theta/2) cos^2(theta/2) <= 4/N, and the expected query complexity O(sqrt(N)/P(succ)) = O(sqrt(N)) with classical checking", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "185-186", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "For the single-marked search with N basis states, after choosing the number of iterations to be the nearest integer at or below a quarter of pi over the half-angle minus one half, prove that the failure probability (the probability of NOT measuring the marked state) is at most four over N. Then explain carefully why this residual failure does not spoil the overall square-root-of-N query scaling, by giving the expected-query argument. Keep the accounting in terms of oracle queries.", "reference_answer": "Let the half-angle be the arcsine of one over the square root of N, so the per-iteration rotation angle is twice the half-angle. After k iterations the marked-direction amplitude is the sine of (2k+1) times the half-angle, hence the failure probability is the cosine squared of (2k+1) times the half-angle. Choosing k as the floor of (a quarter of pi over the half-angle minus one half) guarantees that (2k+1) times the half-angle is at least a right angle minus the full rotation angle and at most a right angle. Therefore the cosine of (2k+1) half-angles is at most the cosine of (a right angle minus the full rotation angle) which equals the sine of the full rotation angle. So the failure probability is at most sine squared of the full angle = four times sine squared of the half-angle times cosine squared of the half-angle, which is at most four times sine squared of the half-angle = four over N (using sine of the half-angle = one over the square root of N and cosine squared at most one). This is the claimed bound. Why the scaling survives: a single run uses order square root of N queries to the marking oracle (k is of order square root of N) and succeeds with probability at least one minus four over N. If a run fails it can be detected by classically checking the returned candidate with one more oracle query, and the procedure is repeated; the number of runs until the first success is geometric with success probability at least one minus four over N, so the expected number of runs is one over that success probability, which is one plus a vanishing correction. The expected total queries are therefore order square root of N divided by (one minus four over N) = order square root of N. By a Markov-type argument the query count is also order square root of N with probability at least two thirds. Thus the residual failure of a single run only multiplies the query cost by a factor approaching one and does not change the square-root-of-N scaling. All counts here are oracle queries; translating to elementary gates would additionally require the cost of the diffusion and oracle circuits and is not performed.", "rubric": [ { "criterion": "Writes failure probability as cosine squared of (2k+1) half-angles and uses the k choice to bound the argument near a right angle", "id": "r1", "weight": 1 }, { "criterion": "Derives the bound failure at most sine squared of the full angle = 4 sine squared(half) cosine squared(half) at most 4/N", "id": "r2", "weight": 1 }, { "criterion": "Gives the expected-query argument: geometric repetition with success at least 1 - 4/N yields expected queries order sqrt(N)", "id": "r3", "weight": 1 }, { "criterion": "Keeps the accounting in oracle queries and does not silently convert to gate cost", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "Failure probability equals cosine squared of (2k+1) times the half-angle.", "supporting_nodes": [ "concept:approximation" ] }, { "claim": "The chosen k forces (2k+1) half-angles into a right-angle neighborhood.", "supporting_nodes": [ "concept:approximation" ] }, { "claim": "Bounding the cosine gives failure at most sine squared of the full angle.", "supporting_nodes": [ "concept:approximation" ] }, { "claim": "This is 4 sine squared(half) cosine squared(half) at most 4/N.", "supporting_nodes": [ "concept:approximation" ] }, { "claim": "Geometric repetition with success at least 1 - 4/N gives expected queries order sqrt(N).", "supporting_nodes": [ "concept:complexity" ] }, { "claim": "All counts are oracle queries; gate conversion is not performed.", "supporting_nodes": [ "concept:complexity", "concept:operator-theory" ] } ], "source_item_digest": "f8f0ce53cafa5cf44f72a178c5cf657a9a3d80928242702e3db089b8f6f7b21a", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:approximation", "concept:complexity", "concept:operator-theory" ], "task_hash": "d8a80df34fe72ffccddb2337a125a4c34943fb466ec58fa796ee4f53fc08b44c", "task_kind": "proof", "topic": "planned_polynomial_signals" }, { "assumptions": [ "Single marked element; a = 1/sqrt(N) in the interval from zero to one" ], "context_hash": "5ed931f463d2937cf7d7b2ceb75552694fed62362da8b83a850c9ef4707cb6c5", "conventions": [ "T_{2k+1}(a) = (-1)^k sine((2k+1) arcsine(a)) for a in [0,1]", "Global phases do not change measurement probabilities" ], "core_id": "qcb-core-plan-516", "difficulty": { "level": "advanced", "rationale": "Connects the amplitude-amplification picture to the Chebyshev singular value transformation via a special-argument polynomial identity and a careful global-phase accounting." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-grover-amplitude-amplification", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-54e14af65ea2b88a064d", "key": "qcb-core-plan-516-grover-odd-chebyshev-identity", "primary_sources": [ { "anchors": "Example 11.4 (Grover's algorithm as Chebyshev singular value transformation): R_{psi_0} as block encoding of the scalar a = 1/sqrt(N) = sin(theta/2) (Eq. 11.29), projectors Pi = |psi_0> - |1>)/sqrt(2)", "Query cost is counted in oracle calls; the reusable ancilla is prepared once" ], "core_id": "qcb-core-plan-519", "difficulty": { "level": "basic", "rationale": "A direct and self-contained computation of phase kickback with an explicit resource reading; conceptually central but technically short." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-grover-amplitude-amplification", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-43cf80978dcac35ab29a", "key": "qcb-core-plan-519-phase-kickback-bit-to-phase-oracle", "primary_sources": [ { "anchors": "Grover's algorithm phase-kickback step: V_f|x,-> = (1/sqrt(2))(|x,f(x)> - |x,1 + f(x)>) = (-1)^{f(x)}|x,-> (Eq. 11.4), the induced reflection R_{x_0} = I - 2|x_0> only changes up to a sign so it can be reused for both R_{psi_0} and R_{x_0} (around Eq. 11.22)", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "183-186", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "The marking information for search is first given as a bit oracle that adds the function value into an answer qubit modulo two. Prove that running this bit oracle with the answer qubit prepared in the minus state converts it into a phase oracle that multiplies each input basis state by minus one exactly when it is marked, leaving the answer qubit unchanged. Make explicit why this is a clean conversion (the answer qubit factorizes out and can be reused), and explain the resource consequence: that one phase-oracle reflection costs one bit-oracle query plus state preparation of the answer qubit, with no measurement or amplification required.", "reference_answer": "Prepare the answer qubit in the minus state, the equal superposition of zero and one with a relative minus sign, normalized by one over the square root of two. The bit oracle maps an input string x with answer bit y to the same x with answer bit y exclusive-or f(x). Apply it to x tensored with the minus state: by linearity it sends the minus state's two components, (x,0) with plus sign and (x,1) with minus sign, to (x, f(x)) with plus sign and (x, 1 exclusive-or f(x)) with minus sign. If f(x) = 0 this is (x,0) plus sign and (x,1) minus sign, i.e. x tensored with the minus state unchanged. If f(x) = 1 this is (x,1) with plus sign and (x,0) with minus sign, i.e. x tensored with (minus the minus state) = (minus one) times x tensored with the minus state. In both cases the result is (minus one) to the f(x) times x tensored with the minus state. Hence the overall action is x tensored with minus state mapping to (minus one) to the f(x) times the same, which is exactly a phase oracle applying the sign (minus one) to the f(x) on the input register while the answer qubit is returned unchanged in the minus state. The conversion is clean because the answer qubit comes out in the identical minus state regardless of x, so it factorizes completely out of the input register and is left unentangled; therefore it can be prepared once and reused across every application of the phase oracle (including both the marked-state reflection and, by the same device, the diffusion reflection). Resource consequence: one application of the phase-oracle reflection costs exactly one query to the bit oracle plus the one-time preparation of the answer qubit in the minus state (a single-qubit state preparation), and crucially it requires no measurement and no amplification, because phase kickback is a unitary, deterministic effect; the sign is imprinted coherently rather than read out. Thus converting bit access to phase access is free up to a reusable ancilla, and the search's query accounting can be stated directly in phase-oracle reflections.", "rubric": [ { "criterion": "Computes the bit oracle on x tensored with the minus state and obtains (minus one) to the f(x) times the same state", "id": "r1", "weight": 1 }, { "criterion": "Shows the answer qubit returns unchanged in the minus state for both f(x) = 0 and f(x) = 1", "id": "r2", "weight": 1 }, { "criterion": "Explains the answer qubit factorizes out unentangled and can be prepared once and reused", "id": "r3", "weight": 1 }, { "criterion": "States the resource consequence: one phase reflection = one bit-oracle query plus one-time minus-state preparation, with no measurement or amplification", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "Prepare the answer qubit in the minus state.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The bit oracle sends x tensored with minus state to (minus one) to the f(x) times the same.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "For both values of f(x) the answer qubit returns unchanged.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The answer qubit factorizes out unentangled and is reusable.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "One phase reflection costs one bit-oracle query plus a one-time minus-state preparation.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "No measurement or amplification is needed since phase kickback is a coherent unitary effect.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "afb80c476b2ea2e3b90eb6c6150ac9bab08f06123950b5cbfebb94a581d1d088", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:block-encoding" ], "task_hash": "e85be00402c7ccadd739f8f88c5357a11e16d0934ccd53400ce4e4a34f3e0558", "task_kind": "proof", "topic": "planned_polynomial_signals" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-403", "difficulty": { "level": "intermediate", "rationale": "A clean counterexample isolating the one-directional nature of extendibility monotonicity under channels, built from the depolarizing representation of isotropic states." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "extendibility-under-channels", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-1f02943f30b1537794e3", "key": "qcb-core-plan-403-extendibility-channel-converse-counterexample", "primary_sources": [ { "anchors": "Exercise 7.2 on PDF page 65 with the depolarizing representation Eq. (7.61) and isotropic condition Eq. (7.6) (page 60); pure-state extension obstruction from Exercise 7.3 and Eqs. (7.3)-(7.4) (page 59).", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "60, 65-66", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "It is a theorem that if rho_AB is k-extendible and N : B -> B' is any quantum channel, then (id_A tensor N)(rho_AB) is again k-extendible; applying a channel to one side can only preserve or improve extendibility. One might conjecture the converse: that if the output (id_A tensor N)(rho_AB) is k-extendible then the input rho_AB must already have been k-extendible. Disprove this converse. Build your counterexample from the isotropic family rho_AB(x) = (1 - x) Phi+ + (x/d^2) I, using that it is the image of the maximally entangled state under a depolarizing channel, rho_AB(x) = (id_A tensor D_x)(Phi+) with D_x(sigma) = (1 - x) sigma + x tr(sigma) I/d, together with the isotropic k-extendibility condition x >= (d/(d+1))(1 - 1/k). Exhibit an explicit input, channel, and k.", "reference_answer": "The converse fails. Take the input to be the maximally entangled state itself, rho_AB = Phi+ = rho_AB(x=0). This is a pure entangled state, hence NOT 2-extendible: a pure entangled bipartite state has no 2-extension, because any state whose B'-marginal equals a pure psi_AB must factorize as psi_AB tensor omega_{B'} (Exercise 7.3), and then its A B'-marginal is the uncorrelated psi_A tensor omega_{B'}, which cannot equal the correlated psi_AB. Now let the channel be the depolarizing channel N = D_x. The output is (id_A tensor D_x)(Phi+) = rho_AB(x), an isotropic state, which by the isotropic condition is k-extendible precisely when x >= (d/(d+1))(1 - 1/k). Concretely for k = 2 the condition is x >= (d/(d+1))(1/2) = d/(2(d+1)); pick any such x -- for instance x = d/(d+1) >= d/(2(d+1)), which even makes the output separable and hence infinitely extendible. Then the output rho_AB(x) is 2-extendible, yet the input Phi+ is a pure entangled state that is not even 2-extendible. This is a k-extendible output arising from a non-k-extendible input, disproving the converse. The mechanism: the depolarizing channel degrades entanglement by mixing the state toward the identity, so a channel can create extendibility that the input lacked -- extendibility is monotone under channels in one direction only.", "rubric": [ { "criterion": "Identifies that a pure entangled bipartite state (e.g. Phi+) has no 2-extension, with the Exercise-7.3 reason", "id": "r1", "weight": 1 }, { "criterion": "Uses rho_AB(x) = (id_A tensor D_x)(Phi+) to realize an isotropic output as a channel image of Phi+", "id": "r2", "weight": 1 }, { "criterion": "Chooses x satisfying the isotropic 2-extendibility window x >= d/(2(d+1)) so the output IS 2-extendible", "id": "r3", "weight": 1 }, { "criterion": "Concludes a k-extendible output from a non-k-extendible input, refuting the converse, and explains the one-way monotonicity", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A pure entangled state such as Phi+ is not 2-extendible.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The depolarizing channel maps Phi+ to the isotropic state rho(x).", "supporting_nodes": [ "concept:operator-theory", "concept:tensor-networks" ] }, { "claim": "Choosing x in the 2-extendible window gives an extendible output from a non-extendible input.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "e4a453dba730b81064c7eab905c12090da22e6fb67aa29a08ee0feccfc0ce3f0", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "6a0dc41199740a5d67ff86bc9bb9648d885d780516c130ab70bf3fe5f0af08f2", "task_kind": "counterexample", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-405", "difficulty": { "level": "intermediate", "rationale": "Sharpens the no-cloning theorem into an exact clonability boundary via the overlap fixed-point equation, with an explicit unclonable qubit pair." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "no-cloning-orthogonality", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-c88b583a5b6622106604", "key": "qcb-core-plan-405-no-cloning-orthogonal-or-equal-boundary", "primary_sources": [ { "anchors": "Theorem 8.1 proof on PDF pages 65-66: unitarity forces ^2 = , hence the overlap lies in {0,1}.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "65-66", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Suppose a fixed unitary U on A tensor B with reference state |0>_B is required to clone two particular states, U(|psi>_A |0>_B) = |psi>_A |psi>_B and U(|phi>_A |0>_B) = |phi>_A |phi>_B, where |psi>, |phi> are normalized. Determine exactly which pairs (|psi>, |phi>) can be simultaneously cloned by a single such unitary. Derive the constraint on the overlap from unitarity, solve it, and interpret the two surviving possibilities. Then explain why this makes exact cloning of an unknown state drawn from a non-orthogonal set impossible, and give a concrete two-state qubit example that cannot be cloned.", "reference_answer": "Unitarity preserves inner products: = ( tensor |0>) = ( tensor |phi>) = ^2. Writing z = , the constraint is z = z^2, i.e. z(z - 1) = 0, giving z = 0 or z = 1. Hence a single cloning unitary can serve the pair only if the two states are either orthogonal ( = 0) or identical (z = 1 forces |phi> = |psi> up to a phase). There is no middle ground: any pair with 0 < || < 1 is forbidden. Consequently, if a device must clone every state in a set containing two distinct non-orthogonal states, it cannot exist, because the fixed-point equation z = z^2 is violated for that pair. Concrete qubit example: |psi> = |0> and |phi> = |+> = (|0> + |1>)/sqrt(2) have overlap <0|+> = 1/sqrt(2), satisfying neither z = 0 nor z = 1; therefore no unitary can clone both |0> and |+>. (If one only ever needed to clone states from an orthonormal set, e.g. |0> and |1>, cloning is possible -- exactly the classical copyable case.) The boundary is sharp: clonability of a pair holds precisely on the set {orthogonal} union {equal}.", "rubric": [ { "criterion": "Derives = ^2 from unitarity of U", "id": "r1", "weight": 1 }, { "criterion": "Solves z = z^2 to get z = 0 or z = 1 and interprets them as orthogonal or equal states", "id": "r2", "weight": 1 }, { "criterion": "Concludes a non-orthogonal distinct pair cannot be cloned by any single unitary", "id": "r3", "weight": 1 }, { "criterion": "Gives a concrete qubit counterexample (|0> and |+>, overlap 1/sqrt(2)) that is unclonable", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Unitarity gives = ^2.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The only solutions are orthogonal or identical states.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "A non-orthogonal distinct pair such as |0>,|+> is unclonable.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "c77d7ed511391ab1864f7706a6fc6b3a476a578c5cbf169260939dff0ae77fef", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "257294cb9ee55d8c5a232a1204f8b5800e9d5fb588a3216241bafbd7cc5153b9", "task_kind": "boundary", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-406", "difficulty": { "level": "intermediate", "rationale": "Combines an asymptotic lower bound on the optimal cloning fidelity with the universal worst-case/average ordering and the collapse to equality for the input-independent optimal map." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "approximate-cloning-asymptotics", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-e1b935b05fa775c49d50", "key": "qcb-core-plan-406-cloning-fidelity-asymptotics-and-ordering", "primary_sources": [ { "anchors": "Eqs. (8.30)-(8.33) and Remark 8.3 on PDF page 68: F*(T) = d_N/d_M >= 1 - Kd/N and F(T) <= F_avg(T), with the optimal map achieving d_N/d_M.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "68", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "The optimal N -> M approximate cloner achieves worst-case fidelity F*(T) = d_N/d_M where d_L = dim Sym^L(H) = binom(d + L - 1, L) with d = dim H. It is stated that F*(T) >= 1 - Kd/N with K = M - N. (a) Prove the lower bound F*(T) >= 1 - Kd/N from the binomial ratio and conclude that for fixed K the optimal worst-case fidelity tends to 1 as N -> infinity. (b) The worst-case fidelity F(T) = inf_psi F(psi^{tensor M}, T(psi^{tensor N}))^2 and the Haar-average fidelity F_avg(T) satisfy a universal ordering for every map T; state and justify that ordering, and explain why nevertheless the same optimal map is simultaneously optimal for both criteria.", "reference_answer": "(a) With d_N = binom(d+N-1, N) and d_M = binom(d+M-1, M) and M = N + K, the ratio d_N/d_M after cancelling common factors is a quotient of two products of K consecutive integers: d_N/d_M = [(N+1)(N+2)...(N+K)] / [(N+d)(N+d+1)...(N+d+K-1)]. Each of the K factors has the form (N + j)/(N + d - 1 + j) = 1 - (d-1)/(N + d - 1 + j) >= 1 - (d-1)/N >= 1 - d/N. Multiplying K such factors and using the elementary inequality prod (1 - a_j) >= 1 - sum a_j for a_j in [0,1], we get d_N/d_M >= 1 - K(d-1)/N >= 1 - Kd/N. Hence F*(T) = d_N/d_M >= 1 - Kd/N. For fixed K and d, Kd/N -> 0 as N -> infinity, so F*(T) -> 1: cloning with a fixed number of extra copies becomes asymptotically perfect. (b) Universal ordering: F(T) <= F_avg(T) for every map T, because the infimum over pure inputs of a nonnegative quantity never exceeds its Haar average: inf_psi g(psi) <= integral g(psi) dpsi. So the worst-case criterion is the stronger (harder to maximize) one. Nevertheless the optimal symmetric cloner satisfies tr(phi^{tensor M} T(phi^{tensor N})) = d_N/d_M for EVERY pure input phi -- a constant independent of phi -- so its infimum and its average both equal d_N/d_M; and since d_N/d_M is also the universal upper bound for F_avg(T) over all maps, this map saturates both the worst-case and the average optima at once. The input-independence of its fidelity is exactly what collapses the ordering gap to equality for this particular map.", "rubric": [ { "criterion": "Proves F*(T) = d_N/d_M >= 1 - Kd/N via the K-consecutive-factor ratio and prod(1 - a_j) >= 1 - sum a_j", "id": "r1", "weight": 1 }, { "criterion": "Concludes F*(T) -> 1 as N -> infinity for fixed K", "id": "r2", "weight": 1 }, { "criterion": "States and justifies F(T) <= F_avg(T) for all maps because inf <= average", "id": "r3", "weight": 1 }, { "criterion": "Explains the optimal map has input-independent fidelity d_N/d_M, so its worst-case and average coincide and both are optimal", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The symmetric-subspace dimension ratio satisfies d_N/d_M >= 1 - Kd/N.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Fixed K gives asymptotically perfect cloning as N -> infinity.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "Worst-case fidelity never exceeds the average, with equality for the input-independent optimal map.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "8bf1b28a875f6ad1991b903518a8f3aba4204087b064b579c744440afa5614ca", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:matrix-analysis" ], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "b2cba577ab3ed65e085636a0c9806fcff3b2c681c31186f1cb902ef0b102a728", "task_kind": "boundary", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-408", "difficulty": { "level": "intermediate", "rationale": "A clean multi-step structural proof (explicit extension, marginal computation, monotone chain) with a monogamy interpretation; distinct from any PPT or symmetry-commutant argument." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "separable-infinite-extendibility", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-07021865621cb3f173c0", "key": "qcb-core-plan-408-separable-infinite-extendibility-and-hierarchy", "primary_sources": [ { "anchors": "Lemma 7.1 and the hierarchy (7.2) on PDF page 59, Exercise 7.1 on PDF page 65: separable states are infinitely extendible and k-extendibility is monotone in k.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "59, 65", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Call a bipartite state rho_AB k-extendible if there is a state rho_{A B_1 ... B_k} on A and k copies of B whose marginal on A B_i equals rho_AB for every i. Prove two structural facts and relate them. (a) Every separable state is k-extendible for every k (hence infinitely extendible). (b) Every k-extendible state is also k'-extendible for every k' <= k. Deduce the inclusion chain Ext_1 superset Ext_2 superset ... superset Ext_infinity and explain in one sentence why this exhibits extendibility as an obstruction to entanglement (monogamy). Give the explicit extension in (a) and the explicit marginal check.", "reference_answer": "(a) Let rho_AB = sum_i p_i sigma_A^{(i)} tensor sigma_B^{(i)} be separable, with p_i >= 0 summing to 1 and the sigma^{(i)} states. Define, for any k, the state rho_{A B_1 ... B_k} := sum_i p_i sigma_A^{(i)} tensor sigma_{B_1}^{(i)} tensor ... tensor sigma_{B_k}^{(i)}, i.e. put an independent copy of sigma_B^{(i)} on each B_j, all conditioned on the same classical index i. This is a valid state (a convex combination of product states, hence positive with unit trace). Its marginal on A B_j is obtained by tracing out all B_l with l != j: tr_{B_l, l != j} gives sum_i p_i sigma_A^{(i)} tensor sigma_{B_j}^{(i)}, since each traced copy contributes tr(sigma_B^{(i)}) = 1; this equals rho_AB for every j because the expression is symmetric in the B-labels. Thus rho_AB is k-extendible for every k, i.e. infinitely extendible. (b) If rho_AB is k-extendible with extension rho_{A B_1 ... B_k}, then for any k' <= k simply trace out the last k - k' systems B_{k'+1}, ..., B_k. The reduced state rho_{A B_1 ... B_{k'}} is a valid state, and for each j <= k' its marginal on A B_j is unchanged, because tracing out additional B-systems commutes with taking the A B_j marginal; hence it is a k'-extension. Therefore k-extendibility implies k'-extendibility. Combining (a) and (b), the sets Ext_k(A:B) satisfy Ext_1 superset Ext_2 superset ... superset Ext_infinity, with Ext_1 all states and Ext_infinity containing SEP by (a). Monogamy: because a k-extension forces A to be correlated in the same way with k copies of B, a strongly entangled state (in particular a pure entangled one) admits no large extension, so moving down the chain squeezes states toward separability -- entanglement cannot be shared with arbitrarily many parties.", "rubric": [ { "criterion": "Writes the explicit k-extension of a separable state by tensoring independent copies of sigma_B^{(i)} on each B_j under the shared index i", "id": "r1", "weight": 1 }, { "criterion": "Checks the A B_j marginal equals rho_AB using tr(sigma_B^{(i)}) = 1 and symmetry in the B-labels", "id": "r2", "weight": 1 }, { "criterion": "Proves k-extendible implies k'-extendible by discarding k - k' systems and noting the marginals are preserved", "id": "r3", "weight": 1 }, { "criterion": "Assembles the inclusion chain Ext_1 superset ... superset Ext_infinity containing SEP and interprets it as monogamy/obstruction to entanglement", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A separable state admits a k-extension by copying sigma_B on each B_j under a shared index.", "supporting_nodes": [ "concept:tensor-networks", "concept:operator-theory" ] }, { "claim": "The A B_j marginal recovers rho_AB via unit trace of the discarded factors.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Discarding systems reduces a k-extension to a k'-extension, giving the inclusion chain.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "682c79e81015ae2267b8ddef78c927dc7e1d68ec526c14ae7e30f002b5c310c8", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "a445a4b341becd397bb1bdd329fa90642de7611b3ae574aecbf194b1dcec0e91", "task_kind": "proof", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-409", "difficulty": { "level": "advanced", "rationale": "A Schmidt-rank rigidity argument: a marginal equal to a rank-one projector forces the whole purification to factorize, yielding the pure-state extension obstruction underlying monogamy." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "pure-state-extension", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-0f52a78a0b4180f6ebf3", "key": "qcb-core-plan-409-pure-bipartite-extension-forced-product", "primary_sources": [ { "anchors": "Exercise 7.3 on PDF page 65 (with the motivating remark around Eqs. (7.3)-(7.4) on page 59): any rho_{A B B'} with tr_{B'} rho = psi_AB equals psi_AB tensor omega_{B'}.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "59, 65", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let |psi>_AB be an arbitrary pure bipartite state. Prove that any state rho_{A B B'} with tr_{B'} rho_{A B B'} = psi_AB (the projector |psi>_{A B B' C} be a purification of rho_{A B B'}, with C a purifying system. Consider the bipartition (AB) versus (B' C) and write the Schmidt decomposition |chi> = sum_j sqrt(q_j) |u_j>_{AB} |w_j>_{B'C} with q_j > 0, {|u_j>} orthonormal in AB and {|w_j>} orthonormal in B'C. Tracing out B' and C gives the AB-marginal tr_{B'C} |chi> 0 is rank one only if there is a single term; hence the Schmidt rank across (AB : B'C) is 1, q_1 = 1, and |u_1>_{AB} = |psi>_{AB} up to a phase. The Schmidt decomposition collapses to |chi> = |psi>_{AB} tensor |w>_{B'C} for a single |w>_{B'C}. Therefore rho_{A B B'} = tr_C |chi>_AB is entangled, a 2-extension rho_{A B B'} (with tr_{B'} rho = psi_AB and tr_B rho = psi_{A B'} = psi_AB as a state on A and a B-copy) would, by the above, be forced to equal psi_AB tensor omega_{B'}; but then its A B'-marginal is psi_A tensor omega_{B'}, which is uncorrelated across A : B' and therefore cannot equal the correlated psi_AB. The contradiction shows a pure entangled state has no 2-extension.", "rubric": [ { "criterion": "Introduces a purification |chi>_{A B B' C} and Schmidt-decomposes across (AB : B'C)", "id": "r1", "weight": 1 }, { "criterion": "Computes the AB-marginal as sum_j q_j |u_j> = |psi>_{AB} |w>_{B'C}", "id": "r3", "weight": 1 }, { "criterion": "Concludes rho_{A B B'} = psi_AB tensor omega_{B'} and derives that a pure entangled state is not 2-extendible", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A purification Schmidt-decomposed across (AB : B'C) has AB-marginal sum_j q_j |u_j> = |psi>.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Hence rho_{A B B'} = psi_AB tensor omega_{B'}, so pure entangled states have no 2-extension.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "a64a1d8b27cf0fe3a7a66f8b4bf60d7ebf0a288ef911b5e8cdcc2fdaac26500a", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "190bd3f9babe57744d1ab9ad013c6fea459c3db08513922c2e84fd4f267a3f4f", "task_kind": "proof", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-410", "difficulty": { "level": "intermediate", "rationale": "A careful CPTP construction with an explicit marginal-reduction identity; shows precisely how complete positivity and trace preservation enter." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "extendibility-under-channels", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-1481015e6254781fe68b", "key": "qcb-core-plan-410-channel-preserves-k-extendibility", "primary_sources": [ { "anchors": "Exercise 7.2(i) on PDF page 65: (id_A tensor N)(rho_AB) is k-extendible whenever rho_AB is.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "65", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let rho_AB be k-extendible and let N : B -> B' be a quantum channel (a completely positive, trace-preserving linear map). Prove that sigma_{A B'} := (id_A tensor N)(rho_AB) is again k-extendible, by constructing an explicit k-extension of sigma from a k-extension of rho. Verify the required marginal conditions carefully, and state exactly where complete positivity and trace preservation are used.", "reference_answer": "Let rho_{A B_1 ... B_k} be a k-extension of rho_AB, so it is a valid state and tracing out all but B_i gives rho_{A B_i} = rho_AB for each i (identifying each B_i with B). Apply the channel N independently to each of the k B-systems: define sigma_{A B'_1 ... B'_k} := (id_A tensor N_{B_1 -> B'_1} tensor ... tensor N_{B_k -> B'_k})(rho_{A B_1 ... B_k}). This is a valid state because a tensor product of completely positive trace-preserving maps is itself completely positive and trace-preserving: complete positivity guarantees the output stays positive even in the presence of the untouched A-system and the other B-registers, and trace preservation guarantees unit trace. Now check the marginals. Fix i and trace out all B'_j with j != i. Because the maps on different systems act on independent tensor factors, the partial trace over B'_j commutes with the channels on the other systems, and for a trace-preserving map one has tr_{B'_j} N_{B_j}(X) = tr_{B_j} X. Applying this for every j != i gives tr_{B'_j, j != i} sigma_{A B'_1 ... B'_k} = (id_A tensor N_{B_i -> B'_i})( tr_{B_j, j != i} rho_{A B_1 ... B_k}) = (id_A tensor N)(rho_{A B_i}) = (id_A tensor N)(rho_AB) = sigma_{A B'}. Thus every marginal on A B'_i equals sigma_{A B'}, so sigma_{A B'_1 ... B'_k} is a k-extension of sigma_{A B'}, proving sigma is k-extendible. Complete positivity is used to ensure the constructed object is a genuine (positive) state; trace preservation is used both for its validity (unit trace) and, crucially, in the identity tr_{B'_j} N_{B_j}(X) = tr_{B_j} X that reduces the marginals on the discarded systems.", "rubric": [ { "criterion": "Starts from a k-extension of rho and applies N independently to each B_i, i.e. N^{tensor k} on the B-systems", "id": "r1", "weight": 1 }, { "criterion": "Argues the result is a valid state because tensor products of CPTP maps are CPTP (CP for positivity, TP for unit trace)", "id": "r2", "weight": 1 }, { "criterion": "Uses tr_{B'_j} N_{B_j}(X) = tr_{B_j} X (trace preservation) to commute partial trace past the channel on discarded systems", "id": "r3", "weight": 1 }, { "criterion": "Verifies each A B'_i marginal equals (id tensor N)(rho_AB) = sigma, establishing the k-extension and pinpointing where CP and TP are used", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Apply N independently to each B_i of a k-extension of rho.", "supporting_nodes": [ "concept:tensor-networks", "concept:operator-theory" ] }, { "claim": "Tensor products of CPTP maps are CPTP, so the output is a valid state.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Trace preservation lets partial trace commute past N on discarded systems, recovering sigma on each A B'_i.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "7f9d8bb4088fa19fc4b12e3fd2430d713e777a164111aa1fed72a0972f83e30e", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "d2a91669678d5a95bc1aa670a713c63ca1c29a1d6508e713516a5e2797849775", "task_kind": "proof", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-412", "difficulty": { "level": "intermediate", "rationale": "A representation-theoretic proof that the Haar average of pure n-fold states is the symmetric projector up to the Weyl dimension; it is the measure-theoretic input to de Finetti and is kept distinct from the Schur-Weyl commutant material handled elsewhere." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "symmetric-subspace-haar-average", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-51a2be653bf9e9306158", "key": "qcb-core-plan-412-haar-pure-state-average-equals-symmetric-projector", "primary_sources": [ { "anchors": "Lemma 7.3 with Eqs. (7.9)-(7.14) on PDF pages 60-61: Pi_sym = binom(n+d-1,n) integral dphi |phi>^{tensor n})(^{tensor n} in Sym^n(C^d); since such vectors |phi>^{tensor n} span Sym^n (every symmetric tensor is a combination of such product powers), every |phi> = V|phi_0> with V Haar on U_d; then U|phi> = (UV)|phi_0>, and by the left-invariance of the Haar measure the integral over V equals the integral over UV, so U^{tensor n} X (U^dagger)^{tensor n} = X. Thus X is invariant under conjugation by U^{tensor n} for all U; equivalently X commutes with the representation U^{tensor n} restricted to its support Sym^n. (iii) By a theorem (proved earlier in the course) the representation U -> U^{tensor n} restricted to Sym^n(C^d) is irreducible. An operator on an irreducible representation that commutes with the whole representation is, by Schur's lemma, a scalar multiple of the identity on that irreducible space; since X is supported on Sym^n and commutes there with U^{tensor n}, we get X = c Pi_sym for some constant c >= 0 (nonnegative because X is positive). (iv) Take traces: tr X = integral dphi tr(|phi>^{tensor n} span Sym^n", "id": "r1", "weight": 1 }, { "criterion": "Proves U^{tensor n}-invariance of X using the parametrization |phi> = V|phi_0> and left-invariance of the Haar measure", "id": "r2", "weight": 1 }, { "criterion": "Invokes irreducibility of U^{tensor n} on Sym^n and Schur's lemma to get X = c Pi_sym", "id": "r3", "weight": 1 }, { "criterion": "Fixes c = 1/d_sym by taking traces (tr X = 1, tr Pi_sym = d_sym), giving Pi_sym = d_sym integral dphi |phi> M cloning map T(X) = (d_N/d_M) Pi_M (X tensor I^{tensor (M-N)}) Pi_M, where Pi_M projects onto Sym^M(H), I is the identity on H, and d_L = dim Sym^L(H). Prove that this map achieves the worst-case fidelity bound with equality: for every pure input |phi>, tr(phi^{tensor M} T(phi^{tensor N})) = d_N/d_M, so that F(T) = d_N/d_M saturates the upper bound F(T) <= d_N/d_M. Justify each projector manipulation and the use of phi^{tensor M} being supported on Sym^M.", "reference_answer": "Fix any normalized |phi> in H and compute tr(phi^{tensor M} T(phi^{tensor N})). Insert the definition: tr(phi^{tensor M} T(phi^{tensor N})) = (d_N/d_M) tr( phi^{tensor M} Pi_M (phi^{tensor N} tensor I^{tensor (M-N)}) Pi_M ). Use cyclicity of the trace to move one Pi_M next to phi^{tensor M}: = (d_N/d_M) tr( Pi_M phi^{tensor M} Pi_M (phi^{tensor N} tensor I^{tensor (M-N)}) ). Now phi^{tensor M} = |phi>^{tensor M}>^{tensor M} is a symmetric vector; therefore Pi_M phi^{tensor M} Pi_M = phi^{tensor M} (the projector acts as the identity on it). This gives = (d_N/d_M) tr( phi^{tensor M} (phi^{tensor N} tensor I^{tensor (M-N)}) ). Factor the operators across the first N and last M - N tensor factors: phi^{tensor M} = phi^{tensor N} tensor phi^{tensor (M-N)}, and phi^{tensor N} tensor I^{tensor (M-N)} acts as phi^{tensor N} on the first N factors and I on the last M - N. Hence the product on the first N factors is phi^{tensor N} phi^{tensor N} = phi^{tensor N} (idempotent, since |phi>, so the infimum over inputs is also d_N/d_M, i.e. F(T) = d_N/d_M, saturating the upper bound F(T) <= d_N/d_M proved for every map. The two key facts used are: (1) phi^{tensor M} lives in Sym^M so Pi_M leaves it unchanged, and (2) |phi>, Q_pi |w> and |w> have the same component in Sym^n because Q_pi acts as the identity on Sym^n and permutes the orthogonal complement among itself (the decomposition into isotypic components is Q_pi-invariant, and the trivial/symmetric isotype is pointwise fixed). Concretely, Pi_n = (1/n!) sum_sigma Q_sigma (the projector onto the trivial isotype equals the group average), and Q_sigma Q_pi = Q_{sigma pi} ranges over all of S_n as sigma does, so Pi_n Q_pi = (1/n!) sum_sigma Q_{sigma pi} = (1/n!) sum_tau Q_tau = Pi_n; likewise Q_pi Pi_n = Pi_n and Pi_n is self-adjoint with Q_pi^dagger = Q_{pi^{-1}}, giving Pi_n Q_pi^dagger = Pi_n too. Now compute Pi_n rhobar Pi_n = Pi_n [ (1/n!) sum_pi Q_pi rho Q_pi^dagger ] Pi_n = (1/n!) sum_pi (Pi_n Q_pi) rho (Q_pi^dagger Pi_n). Using Pi_n Q_pi = Pi_n and Q_pi^dagger Pi_n = Pi_n for every pi, each summand becomes Pi_n rho Pi_n, independent of pi. Hence Pi_n rhobar Pi_n = (1/n!) sum_pi Pi_n rho Pi_n = (1/n!)(n!) Pi_n rho Pi_n = Pi_n rho Pi_n. The 1/n! average collapses precisely because the permutations are absorbed by the flanking symmetric projectors, leaving n! identical copies of Pi_n rho Pi_n. This shows that, as far as the symmetric-subspace compression is concerned, a state and its symmetrization are indistinguishable.", "rubric": [ { "criterion": "Establishes Pi_n Q_pi = Q_pi Pi_n = Pi_n using Pi_n = (1/n!) sum_sigma Q_sigma and the group reshuffle Q_sigma Q_pi = Q_{sigma pi}", "id": "r1", "weight": 1 }, { "criterion": "Writes Pi_n rhobar Pi_n = (1/n!) sum_pi (Pi_n Q_pi) rho (Q_pi^dagger Pi_n)", "id": "r2", "weight": 1 }, { "criterion": "Reduces each summand to Pi_n rho Pi_n via the absorption identity (including Q_pi^dagger Pi_n = Pi_n)", "id": "r3", "weight": 1 }, { "criterion": "Collapses the n! identical terms to conclude Pi_n rhobar Pi_n = Pi_n rho Pi_n", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Pi_n absorbs permutations: Pi_n Q_pi = Pi_n = Q_pi Pi_n.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Each averaged term becomes Pi_n rho Pi_n independent of pi.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The 1/n! average collapses to a single Pi_n rho Pi_n.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "e2e13bbb34865744a0e5a9f2ab454f606e78c8144fe0237f4ed4c708ddceebdc", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "eed9bb88cf62dda772a1440dae7ecf396552e70a5a0e91ae723ca9caed58a9f3", "task_kind": "proof", "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-419", "difficulty": { "level": "intermediate", "rationale": "Isolates and rigorously proves the telescoping fidelity inequality chain inside the de Finetti argument, naming each elementary step (monotone factors, Bernoulli, crude comparison)." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "fidelity-telescoping-bound", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b618c144189bf64b693b", "key": "qcb-core-plan-419-fidelity-telescoping-lower-bound", "primary_sources": [ { "anchors": "Eqs. (7.30)-(7.34) in the proof of Proposition 7.4 on PDF page 62: the average fidelity ratio is bounded below by ((m+1)/(k+m+1))^{d-1} >= 1 - kd/m.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "62", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "In the de Finetti proof one meets the average squared fidelity integral dphi p(phi) F(tilde-phi, phi^{tensor k})^2 = R, where R = [ (m+d-1)!/m! ] / [ (k+m+d-1)!/(k+m)! ] simplifies to the ratio of two products of (d-1) consecutive integers, R = [(m+1)(m+2)...(m+d-1)] / [(k+m+1)(k+m+2)...(k+m+d-1)]. Prove the chain of lower bounds R >= ((m+1)/(k+m+1))^{d-1} = (1 - k/(k+m+1))^{d-1} >= 1 - k(d-1)/(k+m+1) >= 1 - kd/m. State the elementary inequalities used and where each applies.", "reference_answer": "Write R as a product of d-1 factors, pairing the j-th numerator term (m+j) with the j-th denominator term (k+m+j) for j = 1, ..., d-1: R = prod_{j=1}^{d-1} (m+j)/(k+m+j). Each factor (m+j)/(k+m+j) is an increasing function of j (as j grows the ratio (m+j)/(k+m+j) increases toward 1), so the smallest factor is the one at j = 1, namely (m+1)/(k+m+1). Replacing every factor by this smallest one gives the first bound R >= ((m+1)/(k+m+1))^{d-1}; this uses only that each factor is at least the j=1 factor. Rewrite (m+1)/(k+m+1) = 1 - k/(k+m+1), which is a number in [0,1] since 0 <= k/(k+m+1) <= 1. Apply Bernoulli's inequality (1 - a)^p >= 1 - p a for a in [0,1] and p = d-1 >= 0: ((1 - k/(k+m+1))^{d-1} >= 1 - (d-1) k/(k+m+1). That is the second bound. Finally, since k+m+1 >= m and d-1 <= d, we have (d-1) k/(k+m+1) <= d k/m (replacing the denominator by the smaller m only increases the fraction, and d-1 by d only increases the numerator factor), hence 1 - (d-1)k/(k+m+1) >= 1 - kd/m. Chaining the three steps gives R >= 1 - kd/m. The elementary inequalities used are: (1) monotonicity of each factor in j to replace by the minimal factor; (2) Bernoulli's inequality (1-a)^p >= 1 - pa; (3) the crude comparison (d-1)/(k+m+1) <= d/m. This is exactly the fidelity lower bound 1 - kd/m that feeds, via Fuchs-van de Graaf and Jensen, into the de Finetti trace-distance bound sqrt(kd/(n-k)) with m = n - k.", "rubric": [ { "criterion": "Factorizes R = prod_{j=1}^{d-1} (m+j)/(k+m+j) and replaces each factor by the minimal j=1 factor to get R >= ((m+1)/(k+m+1))^{d-1}", "id": "r1", "weight": 1 }, { "criterion": "Rewrites (m+1)/(k+m+1) = 1 - k/(k+m+1) in [0,1]", "id": "r2", "weight": 1 }, { "criterion": "Applies Bernoulli (1-a)^{d-1} >= 1 - (d-1)a to get 1 - (d-1)k/(k+m+1)", "id": "r3", "weight": 1 }, { "criterion": "Uses (d-1)/(k+m+1) <= d/m to reach R >= 1 - kd/m and ties it to the de Finetti bound with m = n - k", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "R is a product of d-1 factors, minimized termwise by the j=1 factor.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Bernoulli's inequality lowers the power to a linear bound.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "A crude denominator/numerator comparison gives R >= 1 - kd/m.", "supporting_nodes": [ "concept:matrix-analysis", "concept:probability" ] } ], "source_item_digest": "59c6bcb17e48756c7d9d42d22d2a2834b87745b5f7064cefc99fb1b53aa8b270", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:matrix-analysis", "concept:probability" ], "task_hash": "050e1b64ad08f1dff084f70c6b3630e271fd00126612d44c49bab05f5a5af36c", "task_kind": "proof", "topic": "planned_quantum_information_apps" }, { "assumptions": [ "V and W are finite-dimensional complex vector spaces." ], "context_hash": "7b39191776c9d8a43eba370989b3aab8529a3ce1205dfa548824c47d87af586d", "conventions": [ "End(V tensor W) is identified with End(V) tensor End(W); 1 denotes the identity on the indicated factor." ], "core_id": "qcb-core-plan-301", "difficulty": { "level": "intermediate", "rationale": "A clean two-inclusion commutant computation requiring the block-matrix device and the fact that the center of End(V) is the scalars." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-tensor-commutant-lemma", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-48c6d2ff7917b2b74c2a", "key": "qcb-core-plan-301-tensor-commutant-lemma", "primary_sources": [ { "anchors": "Lemma 4.4 and its block-matrix proof around Eq. (4.2)-(4.3), p.39", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "39", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let V and W be finite-dimensional complex vector spaces and identify End(V tensor W) with End(V) tensor End(W). Consider the subalgebra A consisting of operators of the form a tensor 1 with a in End(V), where 1 is the identity on W. Prove that the commutant of A inside End(V tensor W), that is the set of all operators commuting with every element of A, is exactly the set of operators 1 tensor b with b in End(W).", "reference_answer": "Write n for the dimension of W and fix a basis of W, so every operator T on V tensor W becomes an n-by-n block matrix with blocks that are endomorphisms of V. In this basis a tensor 1 is block diagonal with the single block a repeated on the diagonal. First inclusion: any operator 1 tensor b has blocks that are scalar multiples of the identity on V, namely the entry of b times the identity, and these commute with the block-diagonal a tensor 1 because scalars commute with a; so every 1 tensor b lies in the commutant. Second inclusion: suppose a block operator T commutes with a tensor 1 for all a in End(V). Comparing blocks, the (i,j) block of (a tensor 1) T is a times the (i,j) block of T, while that of T (a tensor 1) is the (i,j) block of T times a, so block times a equals a times block for every a in End(V) and every pair i,j. An operator on V that commutes with all of End(V) must be a scalar multiple of the identity (apply the relation to rank-one operators, or use that the defining action of End(V) on V is irreducible). Hence each block equals a scalar times the identity on V, which means T = 1 tensor c where c in End(W) collects those scalars. Therefore the commutant equals the operators 1 tensor b, as claimed.", "rubric": [ { "criterion": "States both inclusions must be shown and proves the easy direction that every 1 tensor b commutes with every a tensor 1", "id": "r1", "weight": 1 }, { "criterion": "Introduces the block-matrix form in a basis of W and identifies a tensor 1 as block diagonal with repeated block a", "id": "r2", "weight": 1 }, { "criterion": "Derives from the commutation relation that each block commutes with every endomorphism of V", "id": "r3", "weight": 1 }, { "criterion": "Concludes each block is a scalar multiple of the identity because only scalars commute with all of End(V), giving T = 1 tensor c", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Operators 1 tensor b are block-scalar and commute with the block-diagonal a tensor 1.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "In a basis of W the commutation of T with every a tensor 1 forces each block to commute with every a.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Only scalar multiples of the identity commute with all of End(V), so each block is a scalar times the identity.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Collecting the scalar blocks shows T = 1 tensor c, proving the commutant is 1 tensor End(W).", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "b584bed80a1e10178eb2030fccf506dbf434129c43df6e2d8bfdb6f12531583e", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "d6d561066a400a89062e5d63b72a6a3372a01d1922b621e8e79dd376068c4e2a", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "The representation is of a finite group on a finite-dimensional complex space; the pieces V_alpha are pairwise inequivalent irreducibles." ], "context_hash": "b9f7e39f9f10bb017f7a8792200059804861b81cfabf5da25daed567877b6d3a", "conventions": [ "Direct sums are over the distinct irreducible types alpha; the identity is taken on the indicated factor." ], "core_id": "qcb-core-plan-302", "difficulty": { "level": "advanced", "rationale": "Builds on the commutant lemma and the isotypical decomposition; the student must track block structure and apply the lemma twice." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-double-commutant-bicommutant", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-11d11e3e95b8473301f5", "key": "qcb-core-plan-302-double-commutant-bicommutant", "primary_sources": [ { "anchors": "Proposition 4.5(i)-(iii) with Eq. (4.4)-(4.7); the final step is Exercise 4.2, p.39-40 and p.43", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "39-40, 43", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let a finite group act on a finite-dimensional space V with isotypical decomposition V = direct sum over alpha of (V_alpha tensor C^{n_alpha}) into pairwise inequivalent irreducible pieces V_alpha appearing with multiplicity n_alpha. Let A be the algebra of operators generated by the representation. Prove that the commutant A' equals the direct sum over alpha of (1 on V_alpha, tensored with End(C^{n_alpha})), and deduce that the double commutant equals A.", "reference_answer": "The isotypical projections onto the summands V_alpha tensor C^{n_alpha} lie in A, so any element b of the commutant A' commutes with each such projection and is therefore block diagonal: b is a direct sum of pieces b_alpha, each an endomorphism of V_alpha tensor C^{n_alpha}. On the alpha block the algebra A restricts to (End(V_alpha) tensored with the identity on the multiplicity space): the irreducible action on V_alpha with the multiplicity factor left untouched. By the commutant lemma for a tensor product, the commutant of (End(V_alpha) tensor identity) inside End(V_alpha tensor C^{n_alpha}) is (identity on V_alpha, tensored with End(C^{n_alpha})); hence each b_alpha has this form and A' is the stated direct sum. Conversely every such operator commutes with A, so this is an equality. Applying the same commutant lemma a second time: the commutant of (identity on V_alpha tensor End(C^{n_alpha})) inside each block is (End(V_alpha) tensor identity), and summing over alpha shows the double commutant equals the direct sum of (End(V_alpha) tensor identity), which is A. Thus the bicommutant recovers A.", "rubric": [ { "criterion": "Uses that the isotypical projections lie in A to show any element of A' is block diagonal across the alpha summands", "id": "r1", "weight": 1 }, { "criterion": "Identifies the restriction of A to the alpha block as endomorphisms of V_alpha with an untouched multiplicity factor", "id": "r2", "weight": 1 }, { "criterion": "Applies the tensor commutant lemma to get the multiplicity-space form of each block, hence A' is the stated direct sum", "id": "r3", "weight": 1 }, { "criterion": "Applies the commutant lemma a second time to the description of A' and sums over alpha to conclude the double commutant equals A", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Central isotypical projections lie in A, forcing elements of A' to be block diagonal.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "On each block A acts as endomorphisms of V_alpha with the multiplicity factor untouched.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The tensor commutant lemma gives A' as the direct sum of multiplicity-space endomorphism algebras.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Taking the commutant of A' blockwise returns End(V_alpha) tensor identity, so the double commutant equals A.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "eb90a4aa9096df7cd80102628985d2c7e4cf136eef91dcbf0eff10fed5c1451a", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "7165e61ccd6ba59df611fe528824c4e93b71f76c0135941580b06566f7704a8e", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "V is finite-dimensional, so finite-dimensional subspaces are closed." ], "context_hash": "f1f245ad97e4ac38d207f6984b8aa4de1a351e40e19d1918ebb00775eab8627b", "conventions": [ "A single-vector power repeats the same vector v in each of the n factors." ], "core_id": "qcb-core-plan-306", "difficulty": { "level": "advanced", "rationale": "The polarization identity and the closedness-of-finite-dimensional-subspace argument are the non-obvious ingredients." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-symmetric-power-spanning", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-d19bf4b3cd73cfdd88a0", "key": "qcb-core-plan-306-symmetric-power-spanning", "primary_sources": [ { "anchors": "Lemma 4.7 and the polarization identity Eq. (4.11)-(4.14), p.41", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "41", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let V be a finite-dimensional complex vector space and let the symmetric subspace of its n-fold tensor power be the set of tensors fixed by every permutation of the n factors. Prove that this symmetric subspace equals the span of the n-fold powers of single vectors v of V.", "reference_answer": "Call the span of the single-vector powers W. One inclusion is immediate: each n-fold power of a single vector is unchanged by any factor permutation, so W is contained in the symmetric subspace. For the reverse inclusion, start from the standard spanning set of the symmetric subspace obtained by symmetrizing basis tensors: for a basis of V the symmetrized basis tensors, each a sum over all permutations of a fixed arrangement of basis vectors, span the symmetric subspace. These symmetrized tensors are polarizations: each equals a mixed partial derivative, evaluated at the origin, of the single-vector power of a vector of the form (first basis vector plus t_2 times second plus ... plus t_n times last), differentiated once in each of t_2, ..., t_n. Each such derivative is a limit of difference quotients of elements that are themselves n-fold powers of single vectors, all of which lie in W. Because V is finite-dimensional, W is a finite-dimensional and hence closed subspace, so these limits remain in W. Therefore every symmetrizer, and thus the whole symmetric subspace, lies in W, giving equality.", "rubric": [ { "criterion": "Proves the easy inclusion that each single-vector power is permutation invariant, so the span W sits inside the symmetric subspace", "id": "r1", "weight": 1 }, { "criterion": "Recalls that symmetrized basis tensors span the symmetric subspace", "id": "r2", "weight": 1 }, { "criterion": "Expresses each symmetrized tensor as a mixed partial derivative (polarization) of a single-vector power", "id": "r3", "weight": 1 }, { "criterion": "Argues the derivative is a limit of difference quotients of single-vector powers in W, which is closed, so the symmetric subspace lies in W", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Single-vector powers are permutation invariant, so their span lies in the symmetric subspace.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Symmetrized basis tensors span the symmetric subspace.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Each symmetrizer is a polarization: a mixed derivative of a single-vector power.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Derivatives are limits of difference quotients in the closed finite-dimensional span, so the symmetric subspace lies in W.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "2c32e94daf140024914e71c5bb19e6cb754e24a522a54d1322c60d4271403a04", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:tensor-networks", "concept:operator-theory" ], "task_hash": "c42582416b0a80587350f8c210e7db701a3ed1cc653ee67ddfe5bb421a8c9659", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "The basis vectors form a fixed basis of C^d." ], "context_hash": "4c649f0153401385c1cab4d63cc95f80e10bdef51640d0cc2feb398ed0a51f5c", "conventions": [ "The totally antisymmetric subspace is the image of the signed average P." ], "core_id": "qcb-core-plan-308", "difficulty": { "level": "easy", "rationale": "A short boundary result: the antisymmetric power vanishes above the dimension, argued by distinct-index counting." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-antisymmetrizer-vanishing", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-7403da465cfb3c1c5676", "key": "qcb-core-plan-308-antisymmetrizer-vanishing", "primary_sources": [ { "anchors": "Exercise 5.2 on the antisymmetric projector Eq. (5.38), p.51", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "51", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let P be the antisymmetrizing projector on the n-fold tensor power of C^d, defined as the average over S_n of the sign of a permutation times the corresponding factor permutation. Prove that if d is strictly less than n then P annihilates the entire space, and interpret this as the statement that the totally antisymmetric subspace of the tensor power is zero whenever n exceeds d.", "reference_answer": "The image of P is the totally antisymmetric subspace, consisting of tensors that pick up the sign of any permutation when their factors are permuted. A spanning set for this image is obtained by antisymmetrizing basis tensors; the antisymmetrization of a basis tensor is, up to a scalar, the wedge of the chosen basis vectors. If two of the chosen indices coincide, this wedge is zero, because the transposition exchanging the two equal factors both fixes the tensor and multiplies it by minus one, forcing it to vanish. A nonzero antisymmetrized basis tensor therefore requires all n indices to be distinct, i.e. a strictly increasing choice of n labels drawn from the d-element alphabet. When n is greater than d there is no such choice: any n labels from a d-element set must repeat by the pigeonhole principle. Hence every antisymmetrized basis tensor is zero, the spanning set is trivial, the antisymmetric subspace is the zero subspace, and P is the zero operator. Equivalently its rank is the number of n-element subsets of a d-element set, which is zero for n greater than d.", "rubric": [ { "criterion": "Identifies the image of P as the totally antisymmetric subspace and gives its spanning set by antisymmetrized basis tensors", "id": "r1", "weight": 1 }, { "criterion": "Shows a repeated index makes the antisymmetrized tensor vanish via the fixing-but-sign-flipping transposition", "id": "r2", "weight": 1 }, { "criterion": "Observes a nonzero antisymmetrized tensor needs n distinct labels from a d-element set", "id": "r3", "weight": 1 }, { "criterion": "Concludes by pigeonhole that n greater than d forces every spanning tensor to zero, so P is zero and the dimension vanishes", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The image of P is spanned by wedges of basis vectors.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "A repeated index gives a vanishing wedge because the swap fixes the tensor yet flips its sign.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Nonzero wedges require n distinct labels from the d-element alphabet.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "For n greater than d pigeonhole forbids distinct labels, so the antisymmetric subspace is zero and P is zero.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "961a255fe1a9311fd2c896ffa61b8256cbd38f180ff5a734a2dfeb558b1ed39b", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "c844aba588b6c52d4bc23f4739bc6c8a5a56eee36f43de16b2495b7eace3e78f", "task_kind": "boundary", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "Diagrams with more than three rows are excluded for C^3." ], "context_hash": "18d615cdd82ff3f2da65038ed4f7dd7060a0249f628310bdeb5125c3a66fbf0a", "conventions": [ "Hook length of a box counts boxes to its right, boxes below it, and the box itself; the Weyl formula is the product over row pairs of the row-length differences adjusted by the index gap, divided by the index gap." ], "core_id": "qcb-core-plan-310", "difficulty": { "level": "intermediate", "rationale": "A concrete multi-part computation exercising both dimension formulas and the Schur-Weyl bookkeeping identity." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-schur-weyl-dimension-count", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-481f68f77ace606065ca", "key": "qcb-core-plan-310-schur-weyl-dimension-count", "primary_sources": [ { "anchors": "Schur-Weyl decomposition Eq. (5.8); dimension formulas Proposition 5.10 Eq. (5.9)-(5.11); Exercise 5.3(iii) and the sum rule Eq. (5.29), p.45-46 and p.50-51", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "45-46, 50-51", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "For the fourth tensor power of C^3, determine all Young diagrams that index irreducible pieces in the Schur-Weyl decomposition, compute for each the dimension of the symmetric-group piece using hook lengths and the dimension of the unitary-group piece using the Weyl dimension formula for dimension three, and verify that eighty-one equals the sum over diagrams of the product of these two dimensions.", "reference_answer": "The relevant diagrams are the partitions of 4 into at most 3 parts: (4), (3,1), (2,2), and (2,1,1). Compute each pair of dimensions. For (4): there is one standard tableau so the symmetric-group dimension is 1; the Weyl formula gives the unitary dimension 15 (the dimension of the degree-four symmetric power of C^3, which is six choose two). Product 15. For (3,1): hook lengths are four, two, one across the first row and one in the second, product eight, so the symmetric-group dimension is twenty-four over eight equal to 3; the Weyl formula with parts three, one, zero gives three times five halves times two equal to 15. Product 45. For (2,2): hook lengths are three and two across the first row and two and one across the second, product twelve, so the symmetric-group dimension is twenty-four over twelve equal to 2; the Weyl formula with parts two, two, zero gives one times two times three equal to 6. Product 12. For (2,1,1): hook lengths are four and one across the first row and two then one down the first column, product eight, so the symmetric-group dimension is twenty-four over eight equal to 3; the Weyl formula with parts two, one, one gives two times three halves times one equal to 3. Product 9. Summing the products fifteen, forty-five, twelve, nine gives eighty-one, which equals three to the fourth, confirming the identity.", "rubric": [ { "criterion": "Lists exactly the partitions of 4 into at most 3 parts: (4), (3,1), (2,2), (2,1,1)", "id": "r1", "weight": 1 }, { "criterion": "Computes each symmetric-group dimension correctly from hook lengths as one, three, two, three", "id": "r2", "weight": 1 }, { "criterion": "Computes each unitary dimension correctly from the Weyl formula as fifteen, fifteen, six, three", "id": "r3", "weight": 1 }, { "criterion": "Forms the products and sums them to eighty-one, verifying three to the fourth equals the sum of the products", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The contributing diagrams are the partitions of 4 with at most 3 rows.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Hook-length computations give symmetric-group dimensions one, three, two, three.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The Weyl dimension formula gives unitary dimensions fifteen, fifteen, six, three.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Summing the products gives fifteen plus forty-five plus twelve plus nine equal to eighty-one.", "supporting_nodes": [ "concept:tensor-networks" ] } ], "source_item_digest": "875a506ea7bb34fddaf430ef25b538ec70dd1707f5faa6172919f344a4a1faab", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:tensor-networks", "concept:matrix-analysis" ], "task_hash": "64788ad8a79009ae8116434c82be82e3441c38dff7e479689aee504e891588d4", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "lambda has exactly two parts with lambda_1 at least lambda_2 at least zero." ], "context_hash": "839aa520d919cc0427cb441965134614f4fbf50dfea7243f375b52834690b32e", "conventions": [ "The determinant representation sends a unitary to its determinant; the degree-m symmetric power of C^2 has dimension m plus one." ], "core_id": "qcb-core-plan-311", "difficulty": { "level": "intermediate", "rationale": "Two independent derivations of the same dimension, requiring both the tableau combinatorics and the determinant/symmetric-power structure." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-weyl-dim-u2", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-caaa71645efa2f1c8c05", "key": "qcb-core-plan-311-weyl-dim-u2", "primary_sources": [ { "anchors": "Proposition 5.10(ii) Eq. (5.11) and the U2 structure Eq. (5.26)-(5.28), p.46 and p.48", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "46, 48", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "For the two-dimensional unitary group, the irreducible pieces appearing in tensor powers of C^2 are indexed by partitions lambda = (lambda_1, lambda_2) with two parts. Prove that the dimension of the piece associated with such a partition equals lambda_1 minus lambda_2 plus one, and explain how this follows both from the semistandard-tableau count and from the explicit description of the piece as a determinant power times a symmetric power.", "reference_answer": "Semistandard count: the dimension equals the number of semistandard Young tableaux of shape lambda with entries from the two-letter alphabet {1,2}, that is fillings weakly increasing along rows and strictly increasing down columns. In the two-row shape the two cells of any column must be 1 above 2, which is forced; the only freedom is in the first row, which has lambda_1 minus lambda_2 cells beyond the part covered by the second row, and those cells must be weakly increasing in {1,2}, so they are some number of 1s followed by 2s. The number of such choices is lambda_1 minus lambda_2 plus one, giving the dimension. Explicit description: the piece is isomorphic to the determinant representation raised to the power lambda_2, tensored with the degree-m symmetric power of C^2, where m equals lambda_1 minus lambda_2. The determinant factor is one-dimensional, so it does not change the dimension; the symmetric power of degree m of a two-dimensional space has dimension m plus one. Hence the dimension is m plus one equal to lambda_1 minus lambda_2 plus one, matching the tableau count. Both routes agree because tensoring by the one-dimensional determinant shifts the highest weight without changing dimension, and the symmetric power is exactly the semistandard content of the first-row freedom.", "rubric": [ { "criterion": "States the dimension equals the number of two-letter semistandard tableaux of shape lambda", "id": "r1", "weight": 1 }, { "criterion": "Shows the columns are forced to be 1 over 2 and the only freedom is the lambda_1 minus lambda_2 extra first-row cells, giving lambda_1 minus lambda_2 plus one fillings", "id": "r2", "weight": 1 }, { "criterion": "Gives the determinant-power-times-symmetric-power description and notes the determinant factor is one-dimensional", "id": "r3", "weight": 1 }, { "criterion": "Concludes the symmetric power of degree lambda_1 minus lambda_2 has dimension lambda_1 minus lambda_2 plus one, matching the tableau count", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The dimension is the number of two-letter semistandard tableaux of shape lambda.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Columns are forced 1 over 2; only the lambda_1 minus lambda_2 extra first-row cells vary.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "The piece is a one-dimensional determinant power times a degree-m symmetric power.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The degree-m symmetric power of C^2 has dimension m plus one, agreeing with the count.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "a540284076b09a2f0fe5e2a88254a5329ad87d91586c481043740ab5db00e80a", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:tensor-networks", "concept:matrix-analysis" ], "task_hash": "aee26d0e7b77c22f2c3399b33b2f9fe1880ceb42fb1aeabf64ed9f39bc1e03f2", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "lambda = (4,2,1) is a partition of seven." ], "context_hash": "59cc38528b8d37bc45b0a72a4452e55d3b2d9d9103b3a56ec1aa50a28b949b11", "conventions": [ "The hook length of a box counts boxes to its right in its row, boxes below it in its column, and the box itself." ], "core_id": "qcb-core-plan-312", "difficulty": { "level": "easy", "rationale": "A direct application of the hook length formula with an explicit worked shape and two sanity-check extremes." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-hook-length-standard", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-46a15b0613a9a5a91d08", "key": "qcb-core-plan-312-hook-length-standard", "primary_sources": [ { "anchors": "Proposition 5.10(i) Eq. (5.9)-(5.10) and Example 5.11 Eq. (5.12)-(5.13), p.45-46", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "45-46", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "The number of standard Young tableaux of a shape lambda with n boxes equals n factorial divided by the product of all hook lengths of the diagram. Compute this number for the shape (4,2,1), displaying the hook length of every box, and separately compute the number for the single-row shape (n) and the single-column shape with n boxes to confirm the formula gives the expected extreme cases.", "reference_answer": "For the shape (4,2,1) with seven boxes, the hook lengths are as follows. First row, four boxes: the leftmost has three boxes to its right, two boxes below it (one in each of rows two and three in that column), plus itself, giving six; the next has two to the right, one below, plus itself, giving four; the next has one to the right, none below, plus itself, giving two; the last has none to the right, none below, plus itself, giving one. Second row, two boxes: the leftmost has one to the right, one below (row three in that column), plus itself, giving three; the next has none to the right, none below, plus itself, giving one. Third row, one box: hook length one. The product of hooks is six times four times two times one times three times one times one equal to one hundred forty-four. Then seven factorial is five thousand forty, and five thousand forty divided by one hundred forty-four equals thirty-five, so there are thirty-five standard tableaux of shape (4,2,1). Extremes: for the single row (n) every hook length is just the number of boxes to its right plus itself, giving n, n minus one, down to one, whose product is n factorial, so the count is n factorial over n factorial equal to one. For the single column with n boxes the hooks are likewise n, n minus one, down to one by symmetry, product n factorial, count one. Both extremes correctly give a single standard tableau.", "rubric": [ { "criterion": "Correctly assigns all seven hook lengths of (4,2,1) as six, four, two, one in the first row, three, one in the second, one in the third", "id": "r1", "weight": 1 }, { "criterion": "Computes the hook product as one hundred forty-four and seven factorial as five thousand forty", "id": "r2", "weight": 1 }, { "criterion": "Divides to obtain thirty-five standard tableaux for (4,2,1)", "id": "r3", "weight": 1 }, { "criterion": "Checks the single-row and single-column shapes both give hook product n factorial and hence exactly one standard tableau", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The hook lengths of (4,2,1) are six, four, two, one; three, one; one.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Their product is one hundred forty-four and seven factorial is five thousand forty.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The quotient is thirty-five standard tableaux.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Single-row and single-column shapes have hook product n factorial, giving count one.", "supporting_nodes": [ "concept:tensor-networks" ] } ], "source_item_digest": "c90a4685cd737246275c9615869fbb1cf529c906f12504a0b94d29215f79c7a9", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:matrix-analysis", "concept:tensor-networks" ], "task_hash": "a3fee229068938c51578772ab1b26b6b867cd687f92e32482864044627ddc9d8", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "Permutations act on a fixed finite set; every permutation has a disjoint cycle decomposition." ], "context_hash": "008554d364d3c44194734a0eb8c74328383261df43de669ef8d655e525adfbd1", "conventions": [ "Cycle type is the multiset of cycle lengths in non-increasing order." ], "core_id": "qcb-core-plan-313", "difficulty": { "level": "intermediate", "rationale": "Standard but complete proof of the cycle-type conjugacy criterion resting on the single-cycle conjugation identity." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-cycle-type-conjugacy", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-abfca0097d5005e6d61d", "key": "qcb-core-plan-313-cycle-type-conjugacy", "primary_sources": [ { "anchors": "Proposition 5.5(i)-(iii) with the conjugation identity Eq. (5.4) and Exercise 5.1 Eq. (5.37), p.44 and p.51", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "44, 51", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Two permutations in the symmetric group are conjugate if and only if they have the same cycle type. Prove the key computational identity that conjugating a single cycle by an arbitrary permutation sigma relabels its entries by sigma, and use it to establish both directions of the conjugacy criterion.", "reference_answer": "Key identity: let a cycle send each listed entry to the next and the last back to the first, and let sigma be any permutation. The conjugate sigma times cycle times sigma inverse acts on sigma applied to a listed entry by first undoing sigma, then applying the cycle to move to the next listed entry, then applying sigma; the net effect sends sigma of an entry to sigma of the next entry, and fixes everything outside the sigma-image of the listed entries. Thus the conjugate is again a cycle of the same length whose entries are the sigma-images of the original entries. Forward direction: write any permutation as a product of disjoint cycles. Conjugating the whole product by sigma conjugates each cycle, replacing its entries by their sigma-images while preserving all cycle lengths and keeping the cycles disjoint (sigma is a bijection). Hence the conjugate has the same multiset of cycle lengths, i.e. the same cycle type. Converse: given two permutations with the same cycle type, align their disjoint-cycle decompositions so that cycles of equal length are paired, and define sigma to map the entries of each cycle of the first permutation, in order, to the entries of the paired cycle of the second. This sigma is a well-defined permutation because the cycle decompositions partition the ground set, and by the key identity it conjugates the first permutation to the second. Therefore same cycle type is equivalent to conjugacy.", "rubric": [ { "criterion": "Proves the identity that conjugating a cycle by sigma yields the cycle with entries replaced by their sigma-images", "id": "r1", "weight": 1 }, { "criterion": "Uses disjoint cycle decomposition to show conjugation preserves the multiset of cycle lengths (forward direction)", "id": "r2", "weight": 1 }, { "criterion": "Constructs an explicit sigma matching cycle decompositions entrywise for two permutations of equal cycle type (converse)", "id": "r3", "weight": 1 }, { "criterion": "Verifies this sigma is a well-defined permutation and conjugates one to the other, completing the equivalence", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Conjugating a cycle by sigma produces the cycle on the sigma-images of its entries.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Conjugating a disjoint-cycle product preserves all cycle lengths, so cycle type is a conjugacy invariant.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "Matching equal-length cycles entrywise defines a permutation sigma relating two same-type permutations.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "This sigma conjugates one permutation to the other, giving the converse.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "2aa47cab42159b2c4d10c188b1f176c3bf6fa3799aff9b9f0e347853a9f92974", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "e19ca492a554395f0bbe3e4ca9118f07aef5905e198b241140a8a5b21d7cbb69", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "The invariant measure on the unitary group is normalized and left invariant." ], "context_hash": "633518706081d50e873ff25e67d71caeb59f43f520fd656aa4c8c020f31d2897", "conventions": [ "The swap operator exchanges the two subsystems; visibility is the trace of the state against the swap; twirling averages the unitary-tensor-unitary conjugation." ], "core_id": "qcb-core-plan-315", "difficulty": { "level": "intermediate", "rationale": "Combines invariance of the measure, idempotency of a projection-like average, and the swap-invariance trace trick." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-werner-twirl-visibility", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-bf931d2ce2392cdcad07", "key": "qcb-core-plan-315-werner-twirl-visibility", "primary_sources": [ { "anchors": "Proposition 6.3 properties of Werner states with the twirl Eq. (6.19)-(6.27), p.54-55", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "54-55", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Define the twirling map that averages an operator over all transformations of the form a unitary tensored with the same unitary, using the invariant measure on the unitary group. Prove that twirling any state produces a Werner state, that this Werner state is already invariant under further twirling, and that its visibility, defined as the trace of the state against the swap operator, equals the trace of the original state against the swap.", "reference_answer": "Twirling produces a Werner state: by left invariance of the invariant measure, applying a fixed unitary pair to the twirled operator and reaveraging merely reparametrizes the integration variable, leaving the integral unchanged; hence the twirled operator commutes with every unitary-tensor-unitary conjugation, which is the defining symmetry of a Werner state. Already invariant: if a state already has the Werner symmetry, then every term in the twirl average equals the state itself, so the average is the state; applying this to an already-twirled operator shows twirling is idempotent, i.e. the twirled state is a fixed point. Visibility preserved: the visibility is the trace of the state against the swap operator. Compute the visibility of the twirled state by moving the swap inside the average and using cyclicity of the trace to conjugate the swap by the unitary pair; the swap commutes with every unitary tensored with itself, so the conjugated swap equals the swap, and the unitary pair cancels against its adjoint. Each term of the average therefore reduces to the trace of the original state against the swap, and averaging a constant returns that constant. Hence the twirled state has the same visibility as the original.", "rubric": [ { "criterion": "Shows twirling yields the Werner symmetry using left invariance of the measure to reparametrize the average", "id": "r1", "weight": 1 }, { "criterion": "Proves twirling is idempotent: an already-symmetric state is a fixed point, so the twirled state is invariant under further twirling", "id": "r2", "weight": 1 }, { "criterion": "Uses cyclicity of the trace and the swap's invariance under unitary-tensor-unitary conjugation to evaluate the twirled visibility", "id": "r3", "weight": 1 }, { "criterion": "Concludes each averaged term equals the original trace against the swap, so visibility is preserved", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Left invariance of the measure makes the twirl commute with every unitary-tensor-unitary conjugation.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "An already-Werner state is unchanged by the twirl, so twirling is idempotent.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Cyclicity of the trace plus the swap's invariance let the unitary pair cancel inside the visibility integral.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "Averaging the constant original trace against the swap preserves the visibility.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "92a2b9973081ad5bf6b6ed502bb11e407263dfc26ea1de0bd92d9591e1674980", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "a58d00d28709efc01313a298b31e728bf07752e090aa9529aba30f41ef83f2e0", "task_kind": "proof", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "The local dimension d is fixed while n grows." ], "context_hash": "9a13b8df2dd195951ae56877aa0c12f5f58130b4ccc7cc4a6fe252487c4ae1f0", "conventions": [ "A permutation-invariant operator is a direct sum over diagrams of the identity on the symmetric-group piece times a free operator on the unitary-group piece." ], "core_id": "qcb-core-plan-318", "difficulty": { "level": "intermediate", "rationale": "A quantitative boundary result on parameter counts; requires combining the block structure with both polynomial estimates." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-permutation-invariant-poly", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-129c51974e84b7fa1408", "key": "qcb-core-plan-318-permutation-invariant-poly", "primary_sources": [ { "anchors": "Corollary 5.19 with the diagram-count bound Eq. (5.31) and unitary-dimension bound Eq. (5.33), p.50", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "50", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "A permutation-invariant operator on the n-fold tensor power of C^d, meaning one unchanged by conjugation by every factor permutation, is determined by only polynomially many parameters in n when d is held fixed. Explain precisely why, using the Schur-Weyl block decomposition together with the polynomial bounds on the number of diagrams and on the dimensions of the unitary-group pieces.", "reference_answer": "By Schur-Weyl duality the tensor power decomposes into blocks indexed by Young diagrams with at most d rows, each block a tensor product of a symmetric-group piece and a unitary-group piece. Permutation invariance, via the mutual-commutant structure, forces the operator to act as the identity on each symmetric-group factor and to be an arbitrary operator on each unitary-group factor; concretely the operator is a direct sum over diagrams of the identity on the symmetric-group piece tensored with a free operator on the unitary-group piece. The number of free parameters is therefore the sum over admissible diagrams of the squared dimension of the unitary-group piece. Two polynomial bounds control this. First, the number of admissible diagrams is at most the number of ways to fill d row lengths each between zero and n, bounded by n plus one to the power d, which is polynomial in n for fixed d. Second, the dimension of each unitary-group piece is bounded by a fixed polynomial in n of degree depending only on d, namely at most n plus one to the power d times d minus one over two. Multiplying a polynomial number of diagrams by a polynomial squared dimension gives a total parameter count polynomial in n with d fixed. Hence a permutation-invariant operator, though living in an exponentially large space, is pinned down by only polynomially many numbers.", "rubric": [ { "criterion": "Invokes the Schur-Weyl block decomposition into diagram-indexed blocks with at most d rows", "id": "r1", "weight": 1 }, { "criterion": "Uses permutation invariance to force the operator to be identity on each symmetric-group factor and free on each unitary-group factor", "id": "r2", "weight": 1 }, { "criterion": "States the parameter count is the sum over diagrams of the squared unitary-piece dimension", "id": "r3", "weight": 1 }, { "criterion": "Applies the polynomial bound on the number of diagrams and the polynomial bound on unitary-piece dimensions to conclude a polynomial total in n", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Schur-Weyl gives blocks indexed by diagrams with at most d rows.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Permutation invariance makes the operator identity on symmetric-group factors and free on unitary-group factors.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "The parameter count is the sum over diagrams of squared unitary-piece dimensions.", "supporting_nodes": [ "concept:complexity" ] }, { "claim": "Polynomially many diagrams times polynomially bounded dimensions give a polynomial total in n.", "supporting_nodes": [ "concept:complexity" ] } ], "source_item_digest": "73006cfae26652d2f1fde105a380b1da7f1b3994b6721ccaca20df28436a9ecc", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:complexity", "concept:tensor-networks" ], "task_hash": "af100cd6f7432655ab69b5b509fdb527e272c52b1e247bbc77e91b08dc272d05", "task_kind": "boundary", "topic": "planned_tensor_symmetry" }, { "assumptions": [ "Input and output pinned to the clean coefficient and clean selector", "rotation(j,systemColumn) is a coefficient-register unitary; only its clean diagonal entry survives" ], "context_hash": "769370beff1959279429c6dcf26434c8ad481568491ef275bea7bfa30afdeee6", "conventions": [ "SELECT uses the controlled system equivalence; matching condition is permutation(j)(systemColumn) = systemRow", "PREPARE amplitude is prepare(j, cleanSelector); UNPREPARE contributes its complex conjugate" ], "core_id": "qcb-core-be-006", "difficulty": { "level": "advanced", "rationale": "Full derivation of the exact clean LCU entry assembling four lifted-gate lemmas and interpreting the conjugate-squared selector weight; exact algebra, not an error bound." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-lcu-clean-entry", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-1d1ed8bfe8fc5407f079", "key": "qcb-core-be-006", "primary_sources": [], "question": "For the logical unprepare-select-amplitude-prepare LCU matrix on registers coefficient x (selector x system), derive the exact clean projected entry between clean input (clean coefficient, clean selector, systemColumn) and clean output (clean coefficient, clean selector, systemRow). Show that it is a finite sum over selector indices, that each surviving term carries the SELECT matching condition on the system permutation, and that each term's weight is the conjugate of a PREPARE amplitude times the controlled coefficient rotation diagonal entry times the PREPARE amplitude. Explain concretely why the selector weight appears as conjugate-times-itself and what this means physically.", "reference_answer": "Project the logical matrix onto the clean coefficient and clean selector on both sides, leaving system indices systemRow (output) and systemColumn (input). The clean entry equals the finite sum over all selector indices j of: [if permutation(j) applied to systemColumn equals systemRow then conj(prepare(j, cleanSelector)) * rotation(j, systemColumn)(cleanCoefficient, cleanCoefficient) * prepare(j, cleanSelector) else 0]. Derivation: UNPREPARE = PREPARE-dagger lifted on the selector contributes the clean bra, which turns the matrix product into a selector sum against conj(prepare(j, cleanSelector)) and forces the coefficient and system row labels; AMPLITUDE contributes the controlled coefficient-register entry rotation(j, systemColumn)(cleanCoefficient, cleanCoefficient) on the clean coefficient diagonal; SELECT contributes the indicator that the controlled system permutation for selector j maps the input systemColumn to the output systemRow; PREPARE contributes the ket amplitude prepare(j, cleanSelector). The selector weight is conj(prepare(j,cleanSelector)) * (...) * prepare(j,cleanSelector): the factor prepare(j,cleanSelector) comes from PREPARE loading the selector superposition from the clean selector, and the factor conj(prepare(j,cleanSelector)) comes from UNPREPARE = PREPARE-dagger projecting back onto the clean selector. Physically this is why PREPARE and UNPREPARE must be the SAME gate up to adjoint acting on the SAME selector register: the clean amplitude of each term is weighted by |prepare(j,cleanSelector)|^2 when the coefficient rotation diagonal is 1, i.e. by the probability of selecting term j, so the clean block is the probability-weighted (and coefficient-weighted) combination of the per-term system permutations. If prepare and unprepare were independent, the two factors would not form conj(x)*x and the clean weights would not be the intended squared selection amplitudes.", "rubric": [ { "criterion": "State the exact clean entry as a selector sum with the SELECT matching indicator permutation(j)(systemColumn)=systemRow and zero otherwise", "id": "c1", "weight": 1 }, { "criterion": "Identify the three nonzero factors per term: conj(prepare(j,cleanSelector)) from UNPREPARE, the clean coefficient rotation diagonal from AMPLITUDE, and prepare(j,cleanSelector) from PREPARE", "id": "c2", "weight": 1 }, { "criterion": "Explain the conjugate-times-itself weight as |prepare(j,cleanSelector)|^2 (selection probability) arising because UNPREPARE is PREPARE-dagger on the same selector register, and note it fails for independent gates", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The clean PREPARE-dagger bra turns the matrix product into a selector sum weighted by conj(prepare(j,cleanSelector)).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.star_selectorLift_mul_clean" ] }, { "claim": "SELECT after AMPLITUDE after PREPARE on a clean column yields the permutation indicator times the clean coefficient rotation diagonal times prepare(j,cleanSelector).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectLift_mul_amplitudeLift_mul_selectorLift_clean", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeLift_apply" ] }, { "claim": "Combining gives the exact clean entry with weight conj(prepare)*rotation-diagonal*prepare, i.e. |prepare|^2 when the diagonal is 1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.prepareAmplitudeSelectUnprepare_cleanEntry" ] } ], "source_item_digest": "5e2925c035693a4f8034296aa2d0448e56e059aac0db6f66de4b45cc440f9fb2", "source_refs": [ { "anchor": "theorem prepareAmplitudeSelectUnprepare_cleanEntry giving the selector sum 'if permutation selectorIndex systemColumn = systemRow then star (prepare ...) * rotation ... * prepare ... else 0' (lines 171-220); star_selectorLift_mul_clean (lines 147-169); selectLift_mul_amplitudeLift_mul_selectorLift_clean (lines 112-144)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCUProjection.lean", "sha256": "fa0b84ce5c819184f190b81b4d259e12f725e6196cd37147984fab76fe4ec340" }, { "anchor": "def prepareAmplitudeSelectUnprepare (lines 351-367)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.prepareAmplitudeSelectUnprepare", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.star_selectorLift_mul_clean", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.selectLift_mul_amplitudeLift_mul_selectorLift_clean", "declaration:QuantumBlockEncoding.Robin.ComplexLCU.amplitudeLift_apply" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.prepareAmplitudeSelectUnprepare_cleanEntry" ], "task_hash": "a9b3bd069a5b23dddc6b75dfef8f30d3c9d2d7790edaf659d831286417445a5b", "task_kind": "proof", "topic": "block_encoding" }, { "assumptions": [ "Normalizer N (normalizerA) is a complex scalar; the counterexample needs N != 0", "B is Hermitian (as in the PDE theorem) for the corrected block to equal S1" ], "context_hash": "8bf85ff5599197398b11c32c3392fe548ee050e09db65082298989f45eb70b29", "conventions": [ "Each term scaled by 1/2; filler is N*e^{i*phi} on the lower-right diagonal", "S1 is the Hermitian part of [[A,B],[0,0]] with lower-right block 0" ], "core_id": "qcb-core-be-011", "difficulty": { "level": "advanced", "rationale": "A concrete phase-cancellation counterexample with the exact -N lower-right entry, a disequality proof for nonzero normalizer, and the corrected phase-balanced block equal to S1; distinct from any additive/multiplicative LCU error core." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-lcu-phase-cancellation", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-6cb0ab49f8cefb8aa8af", "key": "qcb-core-be-011", "primary_sources": [], "question": "In an LCU-style assembly of a Hermitian piece, each selected term contributes a scaled controlled-phase source whose upper block carries the intended operator and whose lower-right filler block is normalizer times a phase e^{i*phi}. A source writes both filler phases as e^{i*pi} = -1 and claims the two filler blocks cancel so the clean block equals the intended Hermitian piece S1. Show by direct computation of the lower-right clean entry that the printed all-(-1)-phase choice does NOT cancel, give the exact value of that entry, prove the clean block therefore differs from S1 when the normalizer is nonzero, and exhibit the corrected phase assignment that does make the fillers cancel while preserving the intended operator blocks.", "reference_answer": "The assembly adds three scaled sources (each scaled by 1/2): a controlled-phase source for A with filler phase, one for A-dagger with filler phase, and a Pauli-X-tensor-B term that has zero lower-right block. The lower-right filler of a scaled controlled-phase source with normalizer N and phase e^{i*phi} is (1/2)*N*e^{i*phi} on the diagonal. With BOTH A and A-dagger terms using phase e^{i*pi} = -1, the lower-right diagonal clean entry is (1/2)*N*(-1) + (1/2)*N*(-1) + 0 = -N. So the printed clean lower-right block equals -N (the identity-filler blocks ADD rather than cancel). But the intended Hermitian piece S1 has lower-right block equal to 0 (S1 is the Hermitian part of the homogenized matrix [[A,B],[0,0]], whose lower-right block is 0). Therefore, if N != 0, the printed clean block has lower-right entry -N != 0 = S1's lower-right entry, so the printed-phase clean block is NOT equal to S1 - a genuine counterexample to the all-(-1) cancellation claim. Corrected assignment: give the two filler phases OPPOSITE signs, e.g. keep the A-dagger term at phase -1 but set the A term's filler phase to 0 (i.e. e^{i*0}=+1) while leaving its upper operator block intact. Then the lower-right clean entry becomes (1/2)*N*(+1) + (1/2)*N*(-1) + 0 = 0, matching S1's zero lower-right block, and the upper A, A-dagger and off-diagonal B blocks are unchanged, so the phase-balanced clean block equals S1 exactly (under the paper's Hermitian B). The lesson: filler phases are a real proof obligation - identical phases add, opposite phases cancel; only a phase-balanced choice recovers the intended block.", "rubric": [ { "criterion": "Compute the printed all-(-1)-phase lower-right clean entry as -N (fillers add, not cancel)", "id": "c1", "weight": 1 }, { "criterion": "Show S1's lower-right block is 0, hence for N!=0 the printed clean block differs from S1 (genuine counterexample)", "id": "c2", "weight": 1 }, { "criterion": "Give the corrected opposite/phase-balanced assignment making the lower-right entry 0 while preserving upper operator blocks, so the clean block equals S1", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The all-(-1) filler phases give lower-right clean entry (1/2)N(-1)+(1/2)N(-1) = -N.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.eq29PrintedClean_lowerRight", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.scaledControlledPhaseSource" ] }, { "claim": "S1 has lower-right block 0, so for N!=0 the printed clean block is not S1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.eq29PrintedClean_ne_S1", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.S1" ] }, { "claim": "Replacing one filler phase by its opposite (0 vs -1) cancels the lower-right fillers and preserves the upper blocks, giving S1 under Hermitian B.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.eq29PhaseBalancedClean_eq_S1" ] } ], "source_item_digest": "b64c8446192d113eddb6f829ef3cce766d74ac79527a75bec2bebc8674cce7fd", "source_refs": [ { "anchor": "def scaledControlledPhaseSource lower-right filler 'if i=j then normalizer*phase' (lines 191-198); def eq29PrintedClean (lines 221-229); theorem eq29PrintedClean_lowerRight = -normalizerA (lines 247-251); theorem eq29PrintedClean_ne_S1 (lines 254-262); theorem eq29PhaseBalancedClean_eq_S1 (lines 268-296)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/GHLHamiltonian.lean", "sha256": "2e453b72433b4031e29ba73e9e22c3be0f63e4640781b9c81219fdf908ee80c7" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.eq29PhaseBalancedClean_eq_S1", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.scaledControlledPhaseSource", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.S1" ], "target_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.eq29PrintedClean_lowerRight", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.eq29PrintedClean_ne_S1" ], "task_hash": "efc83fdaf4e8fed3071bca367e611bbc373a92490f4b44aa261c5593c84aa225", "task_kind": "counterexample", "topic": "block_encoding" }, { "assumptions": [ "Finite-dimensional complex Hilbert spaces unless real amplitudes are explicitly specified", "All basis/register permutations are fixed and applied to both input and output" ], "context_hash": "e833b486d4e9fcb2c28db31d0c060e32493023668cc2e193d14934756768c816", "conventions": [ "Column-vector action; Euclidean induced operator norm for norm claims", "Source inspection only; no fresh formal verification" ], "core_id": "qcb-core-la-014", "difficulty": { "level": "intermediate", "rationale": "A complete mathematical derivation with explicit boundary or convention, not interface recall." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-expansion-mechanism-014", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8413b0a4decea71b52c8", "key": "qcb-core-la-014", "primary_sources": [], "question": "A circuit uses only single-qubit rotations, bit flips and controlled bit flips. A bijection relabels all physical wires, preserving which wire controls which target. Which of total instruction count, rotation count and controlled-bit-flip count must stay unchanged? Does this alone prove equal nearest-neighbor routing cost on fixed hardware?", "reference_answer": "All three logical counts stay unchanged: map each instruction to the same kind with relabeled target/control, so there is a one-to-one position-preserving correspondence of lists. Counting all positions, rotation positions or controlled-bit-flip positions therefore yields equal counts. Hardware routing cost need not stay unchanged because the permutation can map an adjacent control-target pair to a distant pair while the hardware graph is fixed. Equal logical counts are a semantic placement invariant, not an assumption that connectivity or physical SWAP overhead is free.", "rubric": [ { "criterion": "All three logical counts invariant", "id": "c1", "weight": 1 }, { "criterion": "Preserve control-target roles", "id": "c2", "weight": 1 }, { "criterion": "No inferred physical routing equality", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Wire renaming maps each instruction to the same instruction kind.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveWireRename.gateCount" ] }, { "claim": "Count positions of each kind to preserve all logical counts.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveWireRename.gateCount", "declaration:QuantumBlockEncoding.PrimitiveWireRename.ryCount", "declaration:QuantumBlockEncoding.PrimitiveWireRename.cxCount" ] }, { "claim": "Fixed hardware distance is outside this semantic bijection.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveWireRename.gateCount" ] } ], "source_item_digest": "cc4258a0f2ab565d82b3abe29fb1f7cbd0112c16d5ad88e0b35d780ac7c0bdd1", "source_refs": [ { "anchor": "gate/circuit16-23; count theorems120-133 and module header", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/PrimitiveWireRename.lean", "sha256": "9f9b653ddda82a40bc9eafde31f6c932f59f631c03bfe24fbdc282acb8307727" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveWireRename.ryCount", "declaration:QuantumBlockEncoding.PrimitiveWireRename.cxCount" ], "target_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveWireRename.gateCount" ], "task_hash": "b881a2a341152237b49ee605317e92cae14c3fdf39f2586ddf424e4e7703fb7e", "task_kind": "concept", "topic": "wire_placement" }, { "assumptions": [ "Finite-dimensional complex Hilbert spaces unless real amplitudes are explicitly specified", "All basis/register permutations are fixed and applied to both input and output" ], "context_hash": "e833b486d4e9fcb2c28db31d0c060e32493023668cc2e193d14934756768c816", "conventions": [ "Column-vector action; Euclidean induced operator norm for norm claims", "Source inspection only; no fresh formal verification" ], "core_id": "qcb-core-la-015", "difficulty": { "level": "intermediate", "rationale": "A complete mathematical derivation with explicit boundary or convention, not interface recall." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-expansion-mechanism-015", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-5b5d7822642e661eb4fb", "key": "qcb-core-la-015", "primary_sources": [], "question": "Define R(theta)=[[cos(theta/2),-sin(theta/2)],[sin(theta/2),cos(theta/2)]]. Derive the exact operator distance between R(a) and R(b), and prove it is at most |a-b|/2. Why does adding any number of untouched spectator qubits not multiply the distance by a dimension factor?", "reference_answer": "Trigonometric sum-difference formulas factor R(a)-R(b)=2 sin((a-b)/4) R((a+b)/2+pi). The second factor is unitary, so its induced Euclidean operator norm is1, giving exactly 2|sin((a-b)/4)|. Since |sin x|<=|x|, the distance is <=|a-b|/2. Adding identity spectators, with any fixed wire placement, tensors the difference with an identity and conjugates by a permutation. The identity has norm1 and unitary conjugation preserves the norm; hence the same scalar-times-unitary factorization has no dimension multiplier. This is uniform operator error, not entrywise error or a finite-bit synthesis algorithm.", "rubric": [ { "criterion": "Exact quarter-angle sine formula", "id": "c1", "weight": 1 }, { "criterion": "Correct linear bound factor1/2", "id": "c2", "weight": 1 }, { "criterion": "Explain dimension independence in induced norm", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Use sum-difference identities to factor the difference by a unitary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.standardRy_difference_factor" ] }, { "claim": "Take scalar-times-unitary norm to obtain the exact sine expression.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance" ] }, { "claim": "Apply sine Lipschitz bound; spectator lift remains unitary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance", "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance_le" ] } ], "source_item_digest": "4ac0d74781d50ead023fd5830a7e8e5cda93b4580d32f2f4ef85299e32890567", "source_refs": [ { "anchor": "difference_factor28-37; liftedRy factor54-59; distance61-79", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/PrimitiveRyPerturbation.lean", "sha256": "57ad888af7a81ac5067f6655160fecf1f197a91b0414bc5e510786b9eed27e32" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.standardRy_difference_factor", "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance_le" ], "target_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance" ], "task_hash": "242ad29257e9eb9b7116b6087db3acec1a0f3c996f657866981eb925281dad66", "task_kind": "proof", "topic": "angle_approximation" }, { "assumptions": [ "Finite-dimensional complex Hilbert spaces unless real amplitudes are explicitly specified", "All basis/register permutations are fixed and applied to both input and output" ], "context_hash": "e833b486d4e9fcb2c28db31d0c060e32493023668cc2e193d14934756768c816", "conventions": [ "Column-vector action; Euclidean induced operator norm for norm claims", "Source inspection only; no fresh formal verification" ], "core_id": "qcb-core-la-016", "difficulty": { "level": "intermediate", "rationale": "A complete mathematical derivation with explicit boundary or convention, not interface recall." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-expansion-mechanism-016", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-49f19240811e3d6c12fa", "key": "qcb-core-la-016", "primary_sources": [], "question": "For the real rotation R(theta)=[[cos(theta/2),-sin(theta/2)],[sin(theta/2),cos(theta/2)]], compare R(theta+2pi) and R(theta). A colleague claims the two full operators have distance0 because every output probability agrees. Refute or prove that claim, and distinguish bare-system global-phase equivalence from coherent branch use.", "reference_answer": "Both sine and cosine change sign after adding pi to their arguments, so R(theta+2pi)=-R(theta). Their difference is -2R(theta), whose induced operator norm is2, not0. Bare-system output probabilities agree because the sign is a global phase. In a coherent controlled use, however, applying the minus sign only on the active control branch introduces a relative phase between control branches. Thus probabilities on one isolated output do not establish equality of operators; an error metric modulo global phase is different from the exact operator-norm contract. Equivalently the exact distance formula gives2|sin(2pi/4)|=2.", "rubric": [ { "criterion": "Distance2 not0", "id": "c1", "weight": 1 }, { "criterion": "Explain probability agreement without operator equality", "id": "c2", "weight": 1 }, { "criterion": "State exact-vs-phase-quotiented metric distinction", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A2pi parameter shift negates both matrix sine/cosine entries.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.standardRy_difference_factor" ] }, { "claim": "Unitary norm gives distance2 despite unchanged probabilities.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance" ] }, { "claim": "A controlled branch sign is relative rather than globally shared.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance" ] } ], "source_item_digest": "9c8a04ca1889fb0a56d3f452b52f5640741fb20e9f1d4e4962b89aef5642bc12", "source_refs": [ { "anchor": "exact difference factor and exact operator distance28-68", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/PrimitiveRyPerturbation.lean", "sha256": "57ad888af7a81ac5067f6655160fecf1f197a91b0414bc5e510786b9eed27e32" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.standardRy_difference_factor" ], "target_nodes": [ "declaration:QuantumBlockEncoding.PrimitiveRyPerturbation.liftedRy_distance" ], "task_hash": "5486b84e77ad663eb57a30b4d49c7de5cf3f19d6fd86e3afd6c081a666ab9d3e", "task_kind": "counterexample", "topic": "phase_semantics" }, { "assumptions": [ "Finite-dimensional complex Hilbert spaces unless real amplitudes are explicitly specified", "All basis/register permutations are fixed and applied to both input and output" ], "context_hash": "e833b486d4e9fcb2c28db31d0c060e32493023668cc2e193d14934756768c816", "conventions": [ "Column-vector action; Euclidean induced operator norm for norm claims", "Source inspection only; no fresh formal verification" ], "core_id": "qcb-core-la-017", "difficulty": { "level": "intermediate", "rationale": "A complete mathematical derivation with explicit boundary or convention, not interface recall." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-expansion-mechanism-017", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b8bdbeb7e17fc4a4eb4d", "key": "qcb-core-la-017", "primary_sources": [], "question": "A reversible gate flips a target bit exactly when two control bits are both1. Assume each control wire is distinct from the target, but allow the two control wires to coincide. Prove the gate is its own inverse and explain what it reduces to when the controls coincide.", "reference_answer": "If the control predicate is false, the gate leaves the whole bit string unchanged, so a second application also does nothing. If true, it flips only the target; each control remains1 because neither control is the target. Thus the predicate remains true for the second application, which flips the target back. All other bits stay fixed, proving involution. If the two controls coincide, the condition c=1 and c=1 is just c=1, so the action reduces to one controlled bit flip. Pairwise-distinct controls are required by some instruction encodings but not by this involution argument; control-target distinctness is the load-bearing condition.", "rubric": [ { "criterion": "Complete two-case inverse argument", "id": "c1", "weight": 1 }, { "criterion": "Correct minimal distinctness assumptions", "id": "c2", "weight": 1 }, { "criterion": "Identify coincident-control reduction", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Separate inactive and active control cases.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction" ] }, { "claim": "Active control values survive the target flip by wire distinctness.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction_involutive" ] }, { "claim": "Two target flips cancel; coincident controls collapse to a single predicate.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction_involutive", "declaration:QuantumBlockEncoding.ccxBasisEquiv" ] } ], "source_item_digest": "f717b346a68d9dddbefdc009acf698ff6aca04c0531629f56fab06200a6b0999", "source_refs": [ { "anchor": "ccxBasisAction24-28; involutive30-50; equivalence52-61", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ReversibleClassical.lean", "sha256": "01ea24f54500974985702a0f968e28d93b2bd04a1aefaf05da848c9b7fe64711" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction", "declaration:QuantumBlockEncoding.ccxBasisEquiv" ], "target_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction_involutive" ], "task_hash": "50b472475599286fe11b801718a3354581022be388574d11745bcc4808c44275", "task_kind": "proof", "topic": "reversible_classical" }, { "assumptions": [ "Finite-dimensional complex Hilbert spaces unless real amplitudes are explicitly specified", "All basis/register permutations are fixed and applied to both input and output" ], "context_hash": "e833b486d4e9fcb2c28db31d0c060e32493023668cc2e193d14934756768c816", "conventions": [ "Column-vector action; Euclidean induced operator norm for norm claims", "Source inspection only; no fresh formal verification" ], "core_id": "qcb-core-la-018", "difficulty": { "level": "intermediate", "rationale": "A complete mathematical derivation with explicit boundary or convention, not interface recall." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-expansion-mechanism-018", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-ff4c9f5ddef56430df40", "key": "qcb-core-la-018", "primary_sources": [], "question": "On a two-bit string(c,t), consider the update that flips c itself when c=t=1 and otherwise does nothing. Does the familiar self-inverse proof for a two-control target flip remain valid when one control is also the target? Give a concrete collision and explain the lost hypothesis.", "reference_answer": "No. The input(1,1) is sent to(0,1), while(0,1) already fails the activation predicate and is left at(0,1). Two distinct inputs therefore collide, so this map is not injective and cannot be reversible or unitary on basis states. Applying it twice to(1,1) does not return the original state: after the first flip the control predicate became false. The usual involution proof uses that flipping the target preserves all control bits, which fails exactly when a control coincides with the target. This is a mathematical counterexample, not a compiler rejection.", "rubric": [ { "criterion": "Explicit collision of two inputs", "id": "c1", "weight": 1 }, { "criterion": "Explain activation predicate changes", "id": "c2", "weight": 1 }, { "criterion": "Conclude noninjective not just unsupported API", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Evaluate active input(1,1) to(0,1).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction" ] }, { "claim": "Evaluate inactive input(0,1) to itself, exposing collision.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction" ] }, { "claim": "Identify failed preservation premise of the inverse proof.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction_involutive" ] } ], "source_item_digest": "c6d32bc846decf5ac3703d8ef0c1d92984d3bd2cacfb16960bdb04a891b21afd", "source_refs": [ { "anchor": "basis action24-28; control-target hypotheses and preserved controls30-49", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ReversibleClassical.lean", "sha256": "01ea24f54500974985702a0f968e28d93b2bd04a1aefaf05da848c9b7fe64711" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction_involutive" ], "target_nodes": [ "declaration:QuantumBlockEncoding.ccxBasisAction" ], "task_hash": "7c291c054fb57303c579fab08a1b5bafd71f6c36799a67ffd9671a04aeeee329", "task_kind": "counterexample", "topic": "reversible_classical" }, { "assumptions": [ "Finite-dimensional complex matrices; ordinary conjugate transpose", "All declared products have compatible dimensions" ], "context_hash": "124aacf20041b4b416e05378b83f926e88b9f98f9c4290b1e8968b51d2b2e8e6", "conventions": [ "Tensor factors ordered as written; stencil offsets are ordinary integers, not modular wrapping", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-019", "difficulty": { "level": "intermediate", "rationale": "Explicit several-step proof with a convention or missing-hypothesis counterexample." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-mathematical-expansion-019", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-1ce49b65237a2b88b7ce", "key": "qcb-core-la-019", "primary_sources": [], "question": "Let S and T be Hermitian matrices on one finite system and X a Hermitian matrix on a second system. Prove that H=S tensor X+T tensor I is Hermitian without assuming that S and T commute. Give a counterexample showing that the same conclusion can fail if X is not Hermitian.", "reference_answer": "The adjoint of a tensor product is the tensor product of the adjoints: entrywise, conjugating S(j,i)X(v,u) gives S(i,j)X(u,v) under Hermiticity of both factors. Hence (S tensor X) dagger=S tensor X. Similarly T tensor I is Hermitian because I dagger=I. Adjoint is additive, so H dagger=H. No product S T occurs anywhere in this argument, so S and T need not commute. The premise on X matters: take the first system one-dimensional with S=[1], T=[0], and X=[[0,1],[0,0]]. Then H=X, whereas X dagger=[[0,0],[1,0]] is different. This is an operator-algebra closure result, not a claim that implementing the factors or reading out a simulated evolution is free.", "rubric": [ { "criterion": "Correct tensor and additive adjoint calculation", "id": "c1", "weight": 1 }, { "criterion": "Explicitly do not assume S and T commute", "id": "c2", "weight": 1 }, { "criterion": "Valid non-Hermitian counterexample when X premise removed", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Check the tensor adjoint identity entry by entry and use both Hermitian premises.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.tensor_isHermitian", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.tensor" ] }, { "claim": "The identity factor and the sum preserve Hermiticity without a commutation assumption.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.add_isHermitian", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.oneDimHamiltonian_isHermitian" ] }, { "claim": "A one-dimensional outer factor reduces to a non-Hermitian X, exposing the missing hypothesis.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.tensor" ] } ], "source_item_digest": "2608e997670f03f1d37ccd000716a0a0478826035c40e22316b4639af5704267", "source_refs": [ { "anchor": "tensor367-369; tensor_isHermitian394-400; add_isHermitian402-408; oneDimHamiltonian_isHermitian410-417", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/GHLHamiltonian.lean", "sha256": "2e453b72433b4031e29ba73e9e22c3be0f63e4640781b9c81219fdf908ee80c7" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.tensor", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.oneDimHamiltonian_isHermitian" ], "target_nodes": [ "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.tensor_isHermitian", "declaration:QuantumBlockEncoding.GHL2025.Hamiltonian.add_isHermitian" ], "task_hash": "385a3ff7096cac772cca77ea20d294ca38ea56c3a278c20ef366f4673a858e19", "task_kind": "proof", "topic": "hamiltonian_composition" }, { "assumptions": [ "Finite-dimensional complex matrices; ordinary conjugate transpose", "All declared products have compatible dimensions" ], "context_hash": "124aacf20041b4b416e05378b83f926e88b9f98f9c4290b1e8968b51d2b2e8e6", "conventions": [ "Tensor factors ordered as written; stencil offsets are ordinary integers, not modular wrapping", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-020", "difficulty": { "level": "intermediate", "rationale": "Explicit several-step proof with a convention or missing-hypothesis counterexample." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-mathematical-expansion-020", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-5847f92e78677fb6dacd", "key": "qcb-core-la-020", "primary_sources": [], "question": "A finite row stencil is a list of offset/coefficient pairs. Its matrix entry at row i and column j is the sum of coefficients of every pair with i+offset=j; no modular wrapping is used. Prove that merging repeated offsets by summing their coefficients leaves every matrix entry unchanged. Use the pairs (1,2),(1,-2),(-1,3) to show why list width and the number of nonzero matrix entries need not agree.", "reference_answer": "Fix i,j. Every pair contributes its coefficient if its offset equals j-i and contributes zero otherwise. Group the finite sum by equal offsets; associativity and commutativity of scalar addition preserve the sum. Replacing each group by a single pair whose coefficient is that group sum therefore gives exactly the same entry for this arbitrary i,j, hence the same matrix. In the example the offset1 coefficient is2-2=0, while the offset-1 coefficient is3. Thus the entry at j=i+1 vanishes and the entry at j=i-1 equals3 if that column lies in the finite grid. The original list has three pairs, its merged list can have two pairs including a zero, and it has at most one nonzero matrix entry in a row; at a left boundary even that entry may disappear because j=i-1 is outside the grid. An access permutation or sparse-oracle cost cannot be inferred solely from the unmerged list length. This proof uses the stated additive matrix semantics, not a claim that arbitrary loaders with collided labels are automatically reversible.", "rubric": [ { "criterion": "Prove entrywise invariance including repeated offsets", "id": "c1", "weight": 1 }, { "criterion": "Compute cancellation and remaining coefficient3", "id": "c2", "weight": 1 }, { "criterion": "Distinguish list count, actual support and boundary truncation without claiming free access", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "For fixed i,j, only pairs with the single matching integer offset contribute.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.stencilRowCoeff" ] }, { "claim": "Regroup a finite scalar sum by equal offsets to preserve all entries.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.stencilRowCoeff", "declaration:QuantumBlockEncoding.buildRobinMatrix" ] }, { "claim": "Evaluate the cancellation example and separately check the finite-grid boundary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.stencilRowCoeff" ] } ], "source_item_digest": "af12f58eabab10496249efa91b5f940edea14c74f6121a7b860ae69e321d3fc6", "source_refs": [ { "anchor": "stencilRowCoeff15-26; buildRobinMatrix50-63", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RobinMatrix.lean", "sha256": "e957b6a5961ad2c22cfd154f17cbbec3e7668027b2f526d9f762798eb1c50038" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.buildRobinMatrix" ], "target_nodes": [ "declaration:QuantumBlockEncoding.stencilRowCoeff" ], "task_hash": "b8f02d87efb80643b127563aaaf0a9b4d2c530909cddacde1eb0297c57b1769f", "task_kind": "proof", "topic": "sparse_stencil_algebra" }, { "assumptions": [ "the core is a real matrix with composite column index (bit of 2 values, right bond of r values)", "a thin factorization with orthonormal rows exists in flat coordinates" ], "context_hash": "d6de3d6648c1c3654781aadf07988ec6e9f83c89a2f5c4dc5d21ccba0a62ec41", "conventions": [ "columns reshaped by a fixed finite-product bijection of size 2r", "orthonormal rows means Q*Q-transpose = identity" ], "core_id": "qcb-core-la-001", "difficulty": { "level": "intermediate", "rationale": "Requires understanding why a composite-to-flat column reshaping is a bijection and why matrix products and row-Gram sums are invariant under bijective reindexing." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "composite-index reshaping / bijective reindexing invariance", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-cfdd2b20b3b363c1b0f8", "key": "qcb-core-la-001", "primary_sources": [], "question": "A single core of a finite real tensor train is a real matrix whose rows are indexed by the left bond and whose columns are indexed by a composite label: an ordered pair of an emitted bit (two values) and a right-bond index (r values). To factorize this core, one reshapes that composite column label into a single flat column label and then performs a thin factorization in the flat coordinates. Explain (a) why the reshaping is a genuine bijection and what the flat label's size is, (b) why performing the factorization in flat coordinates and then relabeling the result back by the inverse reshaping produces valid factors of the original core, and (c) why the row-orthonormality of the orthonormal factor is unaffected by the relabeling. State the resulting inner (new left-bond) dimension.", "reference_answer": "(a) The composite column label ranges over all ordered pairs (bit, right-bond index) with the bit taking 2 values and the right-bond index taking r values. The set of such pairs has exactly 2 times r elements, and there is a fixed bijection between it and the flat label set {0,...,2r-1}; reshaping is simply applying this bijection, so no column is lost, duplicated, or merged. The flat label therefore has size 2r. (b) A thin factorization in flat coordinates writes the reshaped core as R times Q where Q has orthonormal rows. Relabeling the columns of Q back by the inverse bijection permutes/renames columns without changing any entry, so the matrix product R times (relabeled Q) reproduces the original core entry by entry: for each original column (bit, j) the entry equals the flat-coordinate entry at the image label, which the factorization already matches. Hence R and the relabeled Q are valid factors of the original core. (c) Row-orthonormality is the statement that Q times Q-transpose equals the identity, i.e. each pair of rows has the prescribed inner product summed over columns. Relabeling columns by a bijection only reorders the terms of that per-row-pair sum, and a finite sum is invariant under reindexing by a bijection, so the row-Gram matrix is unchanged; the relabeled Q still has orthonormal rows. The inner (new left-bond) dimension is the minimum of the original left bond and 2r, i.e. min(left bond, 2 * right bond).", "rubric": [ { "criterion": "States the composite (bit, right-bond) label is in bijection with a flat label of size 2r (2 bit-values times r right indices).", "id": "r1", "weight": 1 }, { "criterion": "Explains that relabeling columns by the inverse bijection leaves entries unchanged, so R times the relabeled factor reproduces the original core.", "id": "r2", "weight": 1 }, { "criterion": "Explains row-orthonormality survives because a finite column sum is invariant under bijective reindexing; states inner dimension min(left bond, 2*right bond).", "id": "r3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The composite (bit, right-bond) column label is bijective with a flat label of size 2r.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_core_lq" ] }, { "claim": "Factorizing in flat coordinates and relabeling back reproduces the original core because column relabeling preserves entries and products.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_core_lq" ] }, { "claim": "Row-orthonormality is preserved because the per-row-pair column sum is invariant under bijective reindexing; inner dimension is min(left bond, 2*right bond).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_core_lq" ] } ], "source_item_digest": "f871c32a18b71fb5ea483f94e2941d10f0b34a1ea600577f1e392fdb2d0412c3", "source_refs": [ { "anchor": "complete declaration exists_core_lq (lines 67-80) with Core (line 18) and slice (lines 20-21); the reshaping uses a finite-product bijection and sum-reindexing to recover A = R*Q and Q*Q-transpose = 1 with inner dimension min l (2*r)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/TensorTrainCanonical.lean", "sha256": "b8d1eb0651708ac2496be306e2d4b6f3025f7ea4d35a8c6907de02b89d1d981f" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.Core", "declaration:QuantumBlockEncoding.TensorTrainCanonical.slice" ], "target_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_core_lq" ], "task_hash": "2ffb4cbdbb0209bba0bfbd6abd58fd20ddec3b4586d34eb310f3c9d020c78650", "task_kind": "concept", "topic": "linear_algebra" }, { "assumptions": [ "A is a core emitting one bit of two values with a right-bond index", "R is a real matrix on the right-bond index" ], "context_hash": "d8775ba6a6aa8af4f4990ee542da3707c794f57450cf1cd54cc1769e4045b6d6", "conventions": [ "a bit-slice fixes the emitted bit and leaves an ordinary left-by-right matrix", "absorption acts only on the right-bond coordinate" ], "core_id": "qcb-core-la-002", "difficulty": { "level": "easy", "rationale": "Reading one definition and its slice identity; recognizing the bit coordinate is untouched." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "residual absorption / per-bit-slice right multiplication", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-991947ac5819309d2550", "key": "qcb-core-la-002", "primary_sources": [], "question": "During backward canonicalization of a finite real tensor train, a residual matrix R (left over from factorizing a later core) must be absorbed into the right bond of the preceding core A, where A emits one bit of two possible values and has a right-bond index. The absorption produces a new core whose entry at left index a and composite column (bit, j) is the sum over the old right-bond index b of A at (a, bit, b) times R at (b, j). Explain in matrix-slice terms what this absorption does to each of the two emitted-bit branches of A, and justify why it does not mix the two bit branches together. State the single slice identity that captures this.", "reference_answer": "View the core A, with the emitted bit fixed to a value, as an ordinary matrix called the bit-slice of A for that bit value: its rows are indexed by the left bond and its columns by the right bond. The absorption of R multiplies each such bit-slice on the right by R: the bit-value-0 slice of the absorbed core is the bit-value-0 slice of A times R, and the bit-value-1 slice of the absorbed core is the bit-value-1 slice of A times R. In other words absorption is just an independent right matrix multiplication performed within each emitted-bit branch. It cannot mix the two branches because the defining sum runs only over the old right-bond index b and carries the emitted bit through unchanged on both sides: the bit label of an output column equals the bit label fed in, and R acts only on the right-bond coordinate, never on the bit coordinate. The single identity that captures this is: the bit-b slice of (absorb A R) equals (the bit-b slice of A) times R, for each bit value b. This is why absorption changes bond dimensions on the right while preserving the bit structure the core emits.", "rubric": [ { "criterion": "Describes absorption as right-multiplying each fixed-bit slice of A by R.", "id": "r1", "weight": 1 }, { "criterion": "States the slice identity: slice(absorb A R, b) = slice(A, b) * R for each bit value b.", "id": "r2", "weight": 1 }, { "criterion": "Explains no bit-mixing: the sum is only over the right-bond index and the emitted bit is carried through unchanged.", "id": "r3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Absorption of R right-multiplies each fixed-bit slice of A by R.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.absorb" ] }, { "claim": "Formally, slice(absorb A R, bit) = slice(A, bit) * R for each bit value.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.absorb_slice" ] }, { "claim": "No bit branch is mixed because the contraction sum ranges only over the right-bond index and the emitted bit passes through unchanged.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.absorb" ] } ], "source_item_digest": "ca7a9fcd324216a1144307b9d48138420f5a9609c644b95f81d80c8ccc438408", "source_refs": [ { "anchor": "complete declarations absorb (lines 58-60) and absorb_slice (lines 62-64), with slice (lines 20-21) and Core (line 18)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/TensorTrainCanonical.lean", "sha256": "b8d1eb0651708ac2496be306e2d4b6f3025f7ea4d35a8c6907de02b89d1d981f" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.slice", "declaration:QuantumBlockEncoding.TensorTrainCanonical.Core" ], "target_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.absorb", "declaration:QuantumBlockEncoding.TensorTrainCanonical.absorb_slice" ], "task_hash": "8cd6beae636663fc110bf73e9725814d46830db87da13a2cc065fb2f83c9492a", "task_kind": "concept", "topic": "linear_algebra" }, { "assumptions": [ "system register has 2^n states", "a auxiliary qubits are used" ], "context_hash": "ea6e50c19d24efb4a23a5fffd94ab20672247d2495f924dc36b6050e27fb0d6c", "conventions": [ "signal/ancilla register is high-order, system register is low-order", "flat index = signal * systemDim + system" ], "core_id": "qcb-core-la-004", "difficulty": { "level": "intermediate", "rationale": "Combines a dimension count with a register-ordering convention and an equivalence of two indexing views." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "register tensor ordering / block indexing", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b17b7b36b8d49589df85", "key": "qcb-core-la-004", "primary_sources": [], "question": "A block-encoding unitary represents an operator that acts on the 2^n states of an n-qubit system register, using a auxiliary qubits. (a) What is the dimension of the square matrix representing the full unitary, and why? (b) The full index is a flattened combination of a signal/ancilla register and the system register; state which register is treated as the high-order (most significant) part of the index and which as the low-order part. (c) If one instead keeps the two registers explicit as an ordered pair (signal, system) and reads off the sub-matrix for one fixed value of the signal register, how does that sub-matrix relate to the sub-matrix obtained by projecting the single flattened unitary onto the same signal branch?", "reference_answer": "(a) The full unitary is a 2^(n+a) by 2^(n+a) matrix. The combined register has n + a qubits (n system qubits plus a auxiliary qubits), and a register of n+a qubits has 2^(n+a) basis states, so the unitary acting on all of them is a square matrix of that size. (b) The signal/ancilla register is the high-order (most significant) part of the flattened index and the system register is the low-order part; equivalently the flat index is signalIndex times (system dimension) plus systemIndex. (c) They are the same matrix. Taking the explicit product-register view of the flat unitary and then extracting the block for one fixed signal value yields exactly the same entries, position by position, as projecting the single flattened unitary onto that signal branch; the two descriptions are definitionally identical, so reasoning done in either view transfers to the other without any change of entries.", "rubric": [ { "criterion": "States the full dimension is 2^(n+a) and justifies it by the n+a combined qubits.", "id": "r1", "weight": 1 }, { "criterion": "Identifies signal/ancilla as high-order and system as low-order in the flattened index.", "id": "r2", "weight": 1 }, { "criterion": "States the explicit product-register branch equals the flattened block projection entry-for-entry.", "id": "r3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A system on n qubits with a auxiliary qubits yields a 2^(n+a)-dimensional unitary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.OperatorBlockEncodingCandidate" ] }, { "claim": "The flat index places the signal register high-order and the system register low-order.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.flatToProductRegister" ] }, { "claim": "The product-register branch and the flattened block projection are the identical matrix.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.productRegisterBlockProjection_flatToProductRegister" ] } ], "source_item_digest": "33cc06733514e0836a461f03019dbbcf4881b1b6963dfe5207cb05932deb4106", "source_refs": [ { "anchor": "declaration OperatorBlockEncodingCandidate with the dimension note (lines 97-116)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/BlockEncoding.lean", "sha256": "fe06f75a29822e28ccdbfedfb0b0e87f90872620fae7aed359085fad8fae64b4" }, { "anchor": "declarations flatToProductRegister (lines 190-198), productRegisterBlockProjection (lines 201-204), productRegisterBlockProjection_flatToProductRegister (lines 210-217)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ConcreteSemantics.lean", "sha256": "454c0c5e215ca197c5883bedf74882f392d259c7afeadc63912230437698ebe5" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.productRegisterBlockProjection" ], "target_nodes": [ "declaration:QuantumBlockEncoding.OperatorBlockEncodingCandidate", "declaration:QuantumBlockEncoding.ConcreteSemantics.flatToProductRegister", "declaration:QuantumBlockEncoding.ConcreteSemantics.productRegisterBlockProjection_flatToProductRegister" ], "task_hash": "1d31c3b53c14d764f0bb09400ae8f43f1e2294102aaea9e502028e34d642ac2b", "task_kind": "concept", "topic": "linear_algebra" }, { "assumptions": [ "V has orthonormal columns, i.e. V-transpose * V = I_r", "the placement of columns is an injection into the N positions" ], "context_hash": "f3d88c38e8dcd687556b74ed922bb68df0c0c19024fc5389ad6bc1c64e530e8a", "conventions": [ "real matrices", "orthogonal matrix means U-transpose * U = I", "inner-product space R^N has dimension N" ], "core_id": "qcb-core-la-005", "difficulty": { "level": "intermediate", "rationale": "Four-step existence argument hinging on the basis-extension theorem; no numerics but genuine structure." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "orthonormal completion / isometry extension", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-cce7fbbcaf7696206042", "key": "qcb-core-la-005", "primary_sources": [], "question": "You are given r real column vectors in R^N that are orthonormal (each has unit length and distinct ones are orthogonal), presented as the columns of an N-by-r matrix V satisfying V-transpose times V equals the r-by-r identity. You are also given an arbitrary assignment that places these r columns into r distinct column positions among the N columns of a target N-by-N matrix (the positions need not be the first r columns, and need not be contiguous). Prove that there exists an N-by-N real orthogonal matrix U (U-transpose times U equals the N-by-N identity) whose columns at exactly those r prescribed positions are the given vectors. Make explicit each ingredient your argument relies on.", "reference_answer": "First, the hypothesis V-transpose times V = identity says exactly that the r columns of V form an orthonormal family of vectors in R^N; in particular the family has r elements and since an orthonormal family is linearly independent, r <= N. Second, the prescribed placement is an injection from the r column-labels into the N positions, so it merely relabels the same r orthonormal vectors at distinct physical positions without changing their inner products; the placed family is therefore still orthonormal. Third, in the N-dimensional inner-product space R^N, any orthonormal family can be extended to an orthonormal basis of the whole space (basis-extension / Gram-Schmidt completion); this produces N - r additional unit vectors orthogonal to each other and to the prescribed ones, filling the N - r unused positions. Fourth, assemble U by taking as its columns these N orthonormal vectors, with the prescribed vectors sitting at their assigned positions and the completion vectors at the remaining positions. Because all N columns are orthonormal, U-transpose times U is the identity, i.e. U is orthogonal, and by construction its columns at the prescribed positions are exactly the given vectors. Hence such a U exists. The essential ingredients are: (i) the equivalence between the identity equation and column-orthonormality, (ii) invariance of orthonormality under an injective relabeling of positions, and (iii) the orthonormal-basis extension theorem in a finite-dimensional inner-product space whose dimension equals the number of positions.", "rubric": [ { "criterion": "Translates V-transpose V = I into orthonormality of the r columns and notes r <= N.", "id": "r1", "weight": 1 }, { "criterion": "Argues the injective placement preserves orthonormality (relabeling distinct positions keeps inner products).", "id": "r2", "weight": 1 }, { "criterion": "Invokes orthonormal-basis extension to fill the remaining N - r positions in R^N.", "id": "r3", "weight": 1 }, { "criterion": "Assembles U, concludes orthogonality from all-columns-orthonormal and that prescribed columns match.", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The identity equation is equivalent to the r columns being an orthonormal family of size r <= N.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ] }, { "claim": "An injective placement into distinct positions preserves the orthonormality of the family.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ] }, { "claim": "The family extends to an orthonormal basis of R^N because the space dimension equals the number of positions.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ] }, { "claim": "The basis assembled as columns gives an orthogonal U agreeing on all prescribed positions.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ] } ], "source_item_digest": "9de300e5af170f37dfbbb8e216ddc6a74157f274971779218e9477ade22a8db4", "source_refs": [ { "anchor": "complete declaration exists_orthogonal_completion (lines 20-59)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RealIsometryCompletion.lean", "sha256": "2a38bfa0dda457b563f5178e20dc2ccbb1281ff7d5f7357e4096a3c29fa2bb5d" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ], "task_hash": "5ad895b00de268977aa716f53882a6f23cf8a28bb323dc9fdcf740c4aae4e43a", "task_kind": "proof", "topic": "linear_algebra" }, { "assumptions": [ "an orthogonal completion with the prescribed columns exists", "r < N so at least one column is unused" ], "context_hash": "5fbb7ea2ea570330169c7cda2e76103a2c02795abacdac7086aa26d2aeab5f98", "conventions": [ "real matrices", "SO(N) = orthogonal matrices of determinant +1", "sign flip is a diagonal matrix with a single -1 entry" ], "core_id": "qcb-core-la-006", "difficulty": { "level": "intermediate", "rationale": "Multi-node composition (completion + three sign-flip facts) with a genuine role for the strict dimension bound." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "special-orthogonal completion / orientation repair", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-4781b1df2da2d22a6614", "key": "qcb-core-la-006", "primary_sources": [], "question": "Suppose r orthonormal real columns are placed at r distinct positions among the N columns of an orthogonal matrix, and suppose r < N so at least one column position is left free. The naive completion to an orthogonal matrix may land in the reflection family (determinant -1) rather than the rotation family (determinant +1). Prove that one can always repair the orientation so the final matrix has determinant exactly +1 while leaving every one of the r prescribed columns completely unchanged. Explain why the strict inequality r < N is exactly what makes the repair possible, and why the repair cannot in general be done when r = N.", "reference_answer": "Start from any orthogonal completion U whose prescribed columns are the given vectors. Because U is orthogonal, U-transpose U = I forces det(U)^2 = 1, so det(U) is either +1 or -1. If it is already +1 there is nothing to do. If it is -1, we fix it with a single-column sign flip at an unused position. Since r < N, the r prescribed positions do not exhaust the N columns, so there is at least one position, call it 'unused', that carries no prescribed column. Let S be the diagonal matrix that is the identity except for a -1 in the unused position. S is orthogonal (it is diagonal with entries +/-1, so S-transpose S = I) and det(S) = -1. Replace U by U*S. Then (U S)-transpose (U S) = S-transpose (U-transpose U) S = S-transpose S = I, so U*S is still orthogonal; and det(U S) = det(U) det(S) = (-1)(-1) = +1, so it is now a rotation. Crucially, right-multiplying by this diagonal S only rescales the single unused column by -1 and leaves every other column, in particular all r prescribed columns, exactly as it was. Hence the prescribed data is preserved and the orientation is corrected. The strict inequality r < N is essential because the repair needs a column that is NOT prescribed to absorb the sign flip; if r = N every column is prescribed, flipping any of them would change the required data, and indeed a fully prescribed orthonormal frame with determinant -1 simply cannot be reoriented to +1 without altering a prescribed column, so no determinant-+1 completion that fixes all N columns exists.", "rubric": [ { "criterion": "Derives det(U) in {+1,-1} from orthogonality (det squared = 1).", "id": "r1", "weight": 1 }, { "criterion": "Uses r < N to secure an unused column position and builds the diagonal sign-flip S there.", "id": "r2", "weight": 1 }, { "criterion": "Shows U*S stays orthogonal and has determinant +1 (product of determinants).", "id": "r3", "weight": 1 }, { "criterion": "Shows right-multiplication by S changes only the unused column, preserving all prescribed columns, and explains the r = N obstruction.", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Orthogonality forces the determinant to be +1 or -1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion_of_unused" ] }, { "claim": "r < N guarantees an unused column position to host the sign flip.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion" ] }, { "claim": "The single-column sign flip is orthogonal with determinant -1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.signFlip_orthogonal", "declaration:QuantumBlockEncoding.RealIsometryCompletion.signFlip_det" ] }, { "claim": "Right-multiplying by the sign flip corrects the determinant to +1 while preserving every prescribed column.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.mul_signFlip_preserves" ] } ], "source_item_digest": "4eae07a7fa7dc01f54e03e5d15700d11bec10c23f9368e6d01b5f44fc0d675f6", "source_refs": [ { "anchor": "complete declarations signFlip (lines 62-63), signFlip_orthogonal (lines 65-72), signFlip_det (lines 74-75), mul_signFlip_preserves (lines 77-79), exists_specialOrthogonal_completion_of_unused (lines 83-104), exists_specialOrthogonal_completion (lines 107-121)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RealIsometryCompletion.lean", "sha256": "2a38bfa0dda457b563f5178e20dc2ccbb1281ff7d5f7357e4096a3c29fa2bb5d" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion", "declaration:QuantumBlockEncoding.RealIsometryCompletion.signFlip", "declaration:QuantumBlockEncoding.RealIsometryCompletion.signFlip_orthogonal", "declaration:QuantumBlockEncoding.RealIsometryCompletion.signFlip_det", "declaration:QuantumBlockEncoding.RealIsometryCompletion.mul_signFlip_preserves" ], "target_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion", "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion_of_unused" ], "task_hash": "831b55b70b38e19adce962a75022f9c80cbf839fc8cd3e274e2793726a831b2b", "task_kind": "proof", "topic": "linear_algebra" }, { "assumptions": [ "finite complex matrices and vectors", "|left> and |right> are computational-basis states" ], "context_hash": "2aef7b285c53762e2773daad303b692003a8d89974eaf703e1bb19e9f97c0f57", "conventions": [ "basis ket is a Kronecker delta at its index", "matrix-vector action uses the standard finite sum" ], "core_id": "qcb-core-la-007", "difficulty": { "level": "intermediate", "rationale": "Short but genuine linearity derivation in both vector and scalar form plus its structural consequence." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "linearity of matrix-vector action / sparse-input collapse", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-ae29e06f605ae9485a6d", "key": "qcb-core-la-007", "primary_sources": [], "question": "In a finite complex matrix backend, acting with an operator on a computational-basis state (a vector with a single 1 at one position and 0 elsewhere) returns the corresponding column of the operator. Using only this fact and the linearity of matrix-vector multiplication, prove that acting with an operator A on the two-term superposition a*|left> + b*|right>, where |left> and |right> are two basis states, gives a*(column left of A) + b*(column right of A), and that the amplitude read at output row k equals a*A[k,left] + b*A[k,right]. Explain why this means a block-encoding or state-preparation witness never needs to reconstruct the whole matrix-vector product: only two named columns, or two named entries, are ever required.", "reference_answer": "Matrix-vector multiplication is linear in the vector argument: for scalars a, b and vectors u, w we have A(a*u + b*w) = a*(A u) + b*(A w), and A(c*u) = c*(A u). Apply this with u = |left> and w = |right>. By the basis-column fact, A|left> is exactly the 'left' column of A and A|right> is exactly the 'right' column of A. Therefore A(a*|left> + b*|right>) = a*(A|left>) + b*(A|right>) = a*(column left of A) + b*(column right of A). Reading the k-th coordinate of both sides and using that the k-th coordinate of the 'j' column of A is the entry A[k,j], the output amplitude at row k is a*A[k,left] + b*A[k,right]. Consequently, to certify what an operator does to any two-term superposition it suffices to inspect the two named columns (vector form) or, for a single output amplitude, just the two named matrix entries (scalar form). No sum over all basis states or full reconstruction of the matrix-vector product is needed; the sparsity of the input collapses the computation to exactly the terms whose input amplitude is nonzero. This is why finite preparation and block-encoding witnesses are stated and checked column-wise or entry-wise rather than by rebuilding a dense product.", "rubric": [ { "criterion": "States/uses linearity of matrix-vector action in the vector argument (additivity and scalar homogeneity).", "id": "r1", "weight": 1 }, { "criterion": "Identifies A on a basis ket with the corresponding column and derives the vector-form result a*col(left)+b*col(right).", "id": "r2", "weight": 1 }, { "criterion": "Reads coordinate k to get the scalar amplitude a*A[k,left]+b*A[k,right].", "id": "r3", "weight": 1 }, { "criterion": "Explains the practical consequence: only two columns/entries are needed, no full product reconstruction.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Acting on a single basis ket returns the corresponding matrix column.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec_basisKet" ] }, { "claim": "By linearity the action on a two-term superposition is the weighted sum of the two columns.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec_twoBasisSuperposition" ] }, { "claim": "Reading row k yields the scalar amplitude a*A[k,left] + b*A[k,right].", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec_twoBasisSuperposition_apply" ] } ], "source_item_digest": "f10f3a23733971a2aa0b150a90124936ed6d12d8b403eff4b77380607f85064c", "source_refs": [ { "anchor": "complete declarations applyVec_basisKet (lines 70-74), applyVec_twoBasisSuperposition (lines 79-88), applyVec_twoBasisSuperposition_apply (lines 93-100)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ConcreteSemantics.lean", "sha256": "454c0c5e215ca197c5883bedf74882f392d259c7afeadc63912230437698ebe5" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec_basisKet", "declaration:QuantumBlockEncoding.ConcreteSemantics.basisKet", "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec" ], "target_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec_twoBasisSuperposition", "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec_twoBasisSuperposition_apply" ], "task_hash": "86fb24816a0a438a3c334176690b6c6d4bb7b33db4d5b14e54219fc4412f8772", "task_kind": "proof", "topic": "linear_algebra" }, { "assumptions": [ "a fixed clean signal value is chosen", "finite matrices over a semiring backend" ], "context_hash": "335d16157ead17c0b5c2480f63cf2a561527e2908a5548e7b649d8cfca0a5594", "conventions": [ "signal-system flattening with signal high-order", "pointwise matrix equality quantifies over all entries" ], "core_id": "qcb-core-la-008", "difficulty": { "level": "intermediate", "rationale": "A biconditional resting on one bridging identity; requires handling both directions and the universal quantifier." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "block-encoding certification styles / projection-vs-action equivalence", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-55625631ad3256b66bc4", "key": "qcb-core-la-008", "primary_sources": [], "question": "Two proof styles are commonly used to certify that a flattened signal-system unitary block-encodes a target operator on the system register at a fixed 'clean' signal value. Style A proves a pointwise equality: the matrix block obtained by projecting the unitary onto that clean signal branch equals the target matrix entry by entry. Style B proves an action statement: for every clean-signal basis input and every clean-signal output row, the amplitude the unitary produces equals the corresponding target matrix entry. Prove that these two styles are logically equivalent, by first showing the single bridging identity that the clean-input/clean-output amplitude of the unitary equals exactly one entry of the projected block.", "reference_answer": "The bridge is: for a fixed clean signal value and any output and input system indices, the amplitude obtained by feeding the clean-signal basis input into the unitary and reading the clean-signal output row equals exactly the (output, input) entry of the projected block of the unitary at that signal value. This holds because acting on a basis ket selects a column (the basis-column fact), and selecting the clean-output coordinate of that column is by definition the same entry that the signal-branch projection reads off; so the amplitude and the projected-block entry are literally the same number for every output and input. Now prove the equivalence. (Style A implies Style B) Assume the projected block equals the target entrywise. Fix any output and input. By the bridge the clean-input/clean-output amplitude equals the projected-block (output,input) entry, which by assumption equals the target (output,input) entry; so the action statement holds for every output and input. (Style B implies Style A) Assume the action statement holds for all outputs and inputs. Fix any output and input; the projected-block (output,input) entry equals, by the bridge, the clean-input/clean-output amplitude, which by assumption equals the target (output,input) entry; so the projected block equals the target entrywise. Since each direction follows from the single bridging identity applied at an arbitrary (output, input) pair, the two certification styles are equivalent. (Linearity further extends the basis action to arbitrary system inputs, and any orthogonal failure branch is separate unitarity evidence rather than a competing block contract.)", "rubric": [ { "criterion": "States and justifies the bridge: clean-input/clean-output amplitude = projected-block entry (via basis-ket selects a column).", "id": "r1", "weight": 1 }, { "criterion": "Proves Style A => Style B by applying the bridge at an arbitrary (output,input).", "id": "r2", "weight": 1 }, { "criterion": "Proves Style B => Style A symmetrically.", "id": "r3", "weight": 1 }, { "criterion": "Notes the universal quantification over all (output,input) pairs is what delivers full equivalence (and linearity extends to arbitrary inputs).", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The clean-input/clean-output amplitude equals one entry of the signal-branch projected block.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.cleanBasisActionAmplitude_eq_signalSystemBlockProjection" ] }, { "claim": "Pointwise block equality implies the clean-basis action statement for all outputs/inputs.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.pointwiseProjection_iff_cleanBasisAction" ] }, { "claim": "The clean-basis action statement for all outputs/inputs implies pointwise block equality.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.pointwiseProjection_iff_cleanBasisAction" ] } ], "source_item_digest": "ffed957ea2517808e052738941799ab1478ba9bade87b7bf84f0987ffc74cd0d", "source_refs": [ { "anchor": "complete declarations cleanBasisActionAmplitude (lines 249-256), cleanBasisActionAmplitude_eq_signalSystemBlockProjection (lines 259-269), pointwiseProjection_iff_cleanBasisAction (lines 277-294)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ConcreteSemantics.lean", "sha256": "454c0c5e215ca197c5883bedf74882f392d259c7afeadc63912230437698ebe5" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.cleanBasisActionAmplitude", "declaration:QuantumBlockEncoding.ConcreteSemantics.applyVec_basisKet" ], "target_nodes": [ "declaration:QuantumBlockEncoding.ConcreteSemantics.pointwiseProjection_iff_cleanBasisAction", "declaration:QuantumBlockEncoding.ConcreteSemantics.cleanBasisActionAmplitude_eq_signalSystemBlockProjection" ], "task_hash": "938ce99e195dcde6306d2080eb9ceff6ac742884a4208ad9e67741dbeefcd0fb", "task_kind": "proof", "topic": "linear_algebra" }, { "assumptions": [ "V-transpose * V = I_r (orthonormal columns)", "the placement e is an injection from r labels into N positions", "part (b) additionally assumes r < N" ], "context_hash": "cb177fd8c4e42e119c0439365f3b5383250538542a80f62ab4b31d30ffd1b767", "conventions": [ "real matrices and the standard inner product on R^N", "orthonormal family is linearly independent; dimension bounds independent families" ], "core_id": "qcb-core-la-009", "difficulty": { "level": "intermediate", "rationale": "Two linked arguments: an independence/dimension bound and a finite-cardinality surjectivity contradiction, with the strict-regime consequence." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "isometry dimensional obstruction / finite-cardinality counting", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-447df43633bf6cb53175", "key": "qcb-core-la-009", "primary_sources": [], "question": "Let V be a real N-by-r matrix whose columns are orthonormal, expressed by the identity V-transpose times V equals the r-by-r identity. (a) Prove the dimensional obstruction r <= N. (b) Suppose further that the strict inequality r < N holds and that the r columns are placed at r distinct positions among the N available column positions via an injective assignment from the r column-labels into the N positions. Prove that this assignment cannot be surjective, i.e. at least one of the N positions receives no prescribed column. Make the counting argument explicit and explain how these two facts together guarantee a spare position exists exactly when r < N.", "reference_answer": "(a) The identity V-transpose times V = identity says that for every pair of column labels the inner product of the corresponding columns is 1 when the labels agree and 0 when they differ; that is, the r columns form an orthonormal family in R^N. An orthonormal family is linearly independent: if a linear combination of the columns were zero, taking the inner product with each column in turn (using orthonormality) forces every coefficient to be zero. A linearly independent family in an N-dimensional vector space has at most N members, because its size cannot exceed the dimension. Hence r, the number of columns, satisfies r <= N. This is the isometry dimensional obstruction: an inner-product-preserving (isometric) linear map from R^r into R^N can exist only when r <= N. (b) The placement is an injective function e from the r column-labels into the N positions. Suppose, for contradiction, that e were also surjective. A surjection from a finite set of size r onto a finite set of size N forces the size of the target to be at most the size of the source, i.e. N <= r (a surjection cannot increase cardinality). But we assumed r < N, i.e. N > r, contradicting N <= r. Therefore e is not surjective: there exists at least one position in the N positions that is not in the image of e, i.e. an unused position receiving no prescribed column. Together: part (a) shows orthonormal columns can only exist when r <= N, and part (b) shows that in the strict regime r < N the injective placement always leaves a spare position. When instead r = N, the injection is a bijection and no spare position exists, which is precisely the boundary that distinguishes the two regimes.", "rubric": [ { "criterion": "Translates V-transpose V = I into orthonormality and derives linear independence of the columns.", "id": "r1", "weight": 1 }, { "criterion": "Concludes r <= N because a linearly independent family has size at most the dimension (isometry obstruction).", "id": "r2", "weight": 1 }, { "criterion": "Gives the counting contradiction: a surjection Fin r -> Fin N would force N <= r, contradicting r < N, so e is non-surjective.", "id": "r3", "weight": 1 }, { "criterion": "Concludes a spare/unused position exists exactly in the strict regime r < N (and not when r = N).", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "V-transpose V = I makes the columns an orthonormal, hence linearly independent, family.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ] }, { "claim": "A linearly independent family in R^N has at most N members, so r <= N.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ] }, { "claim": "If the injection e were surjective then N <= r, contradicting r < N, so some position is unused.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion" ] } ], "source_item_digest": "b31607a7ea2a259a413bb5b7c11780ee9b302542f801bc1ce38ca72cfc3e1311", "source_refs": [ { "anchor": "complete declaration exists_specialOrthogonal_completion (lines 107-121), whose body derives non-surjectivity of e from r < N via a finite-cardinality (card_le_of_surjective) contradiction; orthonormality from V-transpose*V = 1 is set up in exists_orthogonal_completion (lines 20-59)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RealIsometryCompletion.lean", "sha256": "2a38bfa0dda457b563f5178e20dc2ccbb1281ff7d5f7357e4096a3c29fa2bb5d" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_orthogonal_completion" ], "target_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion" ], "task_hash": "acd79e90535b6b73651b347da8f15c1faa4fc21c0e6d861394efae9b28f0f4db", "task_kind": "proof", "topic": "linear_algebra" }, { "assumptions": [ "the original core is right-canonical (orthonormal rows before padding)", "padding appends zero outputs beyond the active rank" ], "context_hash": "7bda825f734504bfff26f5aed37f4adca5aab4a71072d2cfd04a093745d1927a", "conventions": [ "active clean-input columns are indexed by the original ranks", "column orthonormality is the conjugate-transpose-times-matrix Gram identity" ], "core_id": "qcb-core-la-010", "difficulty": { "level": "intermediate", "rationale": "A targeted counterexample distinguishing row- from column-orthonormality under zero padding, grounded in two complementary lemmas." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "zero-padding embedding / row-vs-column isometry under padding", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-87a232dc855788272f02", "key": "qcb-core-la-010", "primary_sources": [], "question": "When embedding a right-canonical tensor-train core into a larger fixed physical register by zero-padding its outputs, one claims the active clean-input columns remain orthonormal and that right-multiplication of the boundary stays mass-preserving. A reviewer argues: 'the padded core still satisfies core times core-transpose equals the identity on the physical register, so its rows are orthonormal and nothing changes'. Show by an explicit small example that the reviewer's blanket claim is false: padding generally destroys the ROW orthonormality relation even though it preserves the COLUMN orthonormality of the active clean-input columns. State precisely which relation survives padding and which does not, and why only the surviving one is used.", "reference_answer": "Take the simplest nontrivial case: a core whose matrix, after fixing the emitted bit, acts as the 1-by-1 identity (its single active row/column is the scalar 1), and pad its output into a physical register of dimension 2 by appending a zero output row. Writing the padded clean-input column as the 2-component vector (1, 0): the column relation (conjugate-transpose times itself) gives (1,0).(1,0) = 1, so the single active clean-input column is still a unit column orthonormal to the (absent) others. But the row relation (matrix times its own transpose on the physical register) now has rows (1,0) and (0,0): the second row is the zero vector, so the matrix-times-transpose is diag(1,0), NOT the 2-by-2 identity. Hence padding breaks 'rows orthonormal' (core*core-transpose = I fails) while keeping 'active clean-input columns orthonormal' (the column Gram matrix on active indices stays the identity). This is a general phenomenon: zero-padding adds all-zero output rows, and a matrix with a zero row can never satisfy matrix*matrix-transpose = I (that identity forces every row to have unit norm). What survives is the column isometry restricted to the active clean-input labels, because the extra padded output rows are identically zero and contribute nothing to the column inner products. The library therefore states the preserved property as orthonormality of the active clean-input columns after embedding, and uses only that; the naive 'rows stay orthonormal on the physical register' claim is false and is correctly avoided.", "rubric": [ { "criterion": "Gives a concrete small padded example (e.g. active block = [1], padded column (1,0) in a size-2 register).", "id": "r1", "weight": 1 }, { "criterion": "Shows the row relation fails: a zero padded row makes core*core-transpose = diag(1,0) != I.", "id": "r2", "weight": 1 }, { "criterion": "Shows the active-column relation survives: the clean-input column Gram matrix is still the identity.", "id": "r3", "weight": 1 }, { "criterion": "States the general reason (a zero row cannot satisfy row-orthonormality) and that only the active-column isometry is used.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Padding appends identically-zero output rows beyond the active rank.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.paddedCore_inactive_output" ] }, { "claim": "A matrix with a zero row cannot satisfy row-orthonormality (matrix*transpose = I).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.paddedCore_inactive_output" ] }, { "claim": "The active clean-input columns still form an orthonormal (identity Gram) family after padding.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.paddedCore_active_isometry" ] } ], "source_item_digest": "492aef4134070695970d651e98c5bf482a5c7d14719d34b8b92091a7a6178ff5", "source_refs": [ { "anchor": "complete declarations paddedCore (lines 255-258), paddedCore_inactive_output (lines 260-262), complexCore_isometry (lines 215-226), paddedCore_active_isometry (lines 295-315)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/TensorTrainCanonical.lean", "sha256": "b8d1eb0651708ac2496be306e2d4b6f3025f7ea4d35a8c6907de02b89d1d981f" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.paddedCore", "declaration:QuantumBlockEncoding.TensorTrainCanonical.complexCore_isometry", "declaration:QuantumBlockEncoding.TensorTrainCanonical.padVector" ], "target_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.paddedCore_active_isometry", "declaration:QuantumBlockEncoding.TensorTrainCanonical.paddedCore_inactive_output" ], "task_hash": "18a9866503da9a2e04a1c39bddaa84e502c5a12d874f8222c4a8db6ff9c236ce", "task_kind": "counterexample", "topic": "linear_algebra" }, { "assumptions": [ "prescribed columns are orthonormal", "the conjecture drops the strict inequality and allows prescribed-count = dimension" ], "context_hash": "c13f27011fff16f36526c63688f1ee6987df3d1eb1caa331edfe0f8a8790df41", "conventions": [ "real matrices", "determinant +1 means rotation (SO(N)), -1 means reflection" ], "core_id": "qcb-core-la-011", "difficulty": { "level": "intermediate", "rationale": "A sharp boundary counterexample pinpointing why a hypothesis is necessary, not merely convenient." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "necessity of the strict dimension bound / completion failure", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-62fef30d1298c65d628e", "key": "qcb-core-la-011", "primary_sources": [], "question": "The special-orthogonal completion result guarantees that prescribed orthonormal columns can be extended to a determinant-+1 orthogonal matrix WHENEVER the number of prescribed columns is strictly less than the ambient dimension. A student conjectures the strict inequality is just a technical convenience and that the result still holds when the prescribed columns already fill all the positions (number of prescribed columns equal to the ambient dimension). Give an explicit counterexample disproving the student's conjecture, and identify exactly which step of the strictly-fewer-columns argument has no analogue in the fully-prescribed case.", "reference_answer": "Work in dimension N = 2 with r = 2 prescribed columns, i.e. all columns prescribed. Let the two prescribed orthonormal columns be (1, 0) and (0, -1), placed at positions 1 and 2. They are orthonormal: each has unit length and they are orthogonal, so the matrix V = [[1,0],[0,-1]] satisfies V-transpose V = identity. The only orthogonal matrix whose columns at positions 1 and 2 are exactly these vectors is V itself, because every column is already dictated by the prescription; there is no free column to choose. But det(V) = (1)(-1) - 0 = -1, so V is a reflection, not a rotation, and it is the unique completion. Therefore no determinant-+1 orthogonal matrix has these prescribed columns, disproving the conjecture: the result genuinely fails when the prescribed columns fill all positions. The step with no analogue is the orientation-repair step. In the strictly-fewer-columns argument one selects an UNUSED column position and flips its sign to turn a determinant-(-1) completion into a determinant-(+1) one while leaving all prescribed columns untouched. When every position is prescribed there is no unused column to absorb the sign flip; any sign change or rotation would have to alter a prescribed column, which is forbidden. So the existence of a spare column, guaranteed only by the strict inequality (number of prescribed columns < ambient dimension), is exactly the ingredient that disappears, and with it the determinant-+1 guarantee.", "rubric": [ { "criterion": "Gives an explicit fully-prescribed orthonormal frame with determinant -1 (e.g. columns (1,0),(0,-1) in dimension 2).", "id": "r1", "weight": 1 }, { "criterion": "Argues the completion is forced to be V itself with determinant -1, so no determinant-+1 completion exists.", "id": "r2", "weight": 1 }, { "criterion": "Identifies the missing step as the sign-flip orientation repair, which needs an unused column available only when prescribed-count < dimension.", "id": "r3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A fully-prescribed orthonormal frame of determinant -1 exists in dimension 2.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.signFlip_det" ] }, { "claim": "With all columns prescribed the completion is unique and keeps determinant -1, so no SO completion exists.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion" ] }, { "claim": "The failing step is the sign-flip repair, which requires an unused column guaranteed only by prescribed-count < dimension.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion_of_unused" ] } ], "source_item_digest": "71e5ff52e00d5ec5c88b81ae1265a8094d961c43c77f466c3fb32c8b23f7196f", "source_refs": [ { "anchor": "complete declarations exists_specialOrthogonal_completion (lines 107-121) with its strict hypothesis hr : r < N, exists_specialOrthogonal_completion_of_unused (lines 83-104), signFlip_det (lines 74-75)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RealIsometryCompletion.lean", "sha256": "2a38bfa0dda457b563f5178e20dc2ccbb1281ff7d5f7357e4096a3c29fa2bb5d" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion_of_unused", "declaration:QuantumBlockEncoding.RealIsometryCompletion.signFlip_det" ], "target_nodes": [ "declaration:QuantumBlockEncoding.RealIsometryCompletion.exists_specialOrthogonal_completion" ], "task_hash": "ad2a2affbde35d46423697310aff24bfdc1b1ac95baed884f4f325d5a33478b8", "task_kind": "counterexample", "topic": "linear_algebra" }, { "assumptions": [ "canonicalization follows the backward min(l, 2*m') recurrence", "the terminal bond of a scalar-boundary chain is dimension one" ], "context_hash": "37fdac8eb39f556235b9b16034ed936e470cd0ca010c3ff80be35d0bf3978501", "conventions": [ "each core emits one bit of dimension 2", "bond dimensions are the dependent inner indices, not padded" ], "core_id": "qcb-core-la-012", "difficulty": { "level": "intermediate", "rationale": "Boundary reasoning from a min-recurrence to two simultaneous bounds and a concrete terminal specialization; genuine multi-node composition." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "bond-dimension recurrence / canonicalization boundary bounds", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-f99e973b80791f7531d2", "key": "qcb-core-la-012", "primary_sources": [], "question": "A finite real tensor train is right-canonicalized by sweeping from the terminal bond backward, replacing each core by one with orthonormal rows while absorbing a residual into the preceding core. The active left bond dimension produced at each core obeys the exact backward recurrence: the new left bond is the minimum of (the current left bond) and (twice the next active bond). Using only this recurrence and the fact that the sweep never enlarges any bond, prove two boundary facts: (a) after canonicalization every core's left bond is at most its original left bond AND at most twice its next active bond; and (b) for a scalar-boundary chain whose terminal bond is forced to dimension one, the active bond immediately before the terminal bond is at most two. Explain where the factor of two comes from.", "reference_answer": "The factor of two comes from the physical bit each core emits: a core carries an emitted bit of dimension 2 together with its right bond, so when it is reshaped and factorized the inner (new left) dimension cannot exceed the number of columns of the reshaped bit/right-bond block, which is 2 times the right (next active) bond. The factorization also cannot increase the row count, so the new left bond cannot exceed the original left bond either. Hence the new left bond is at most the minimum of the original left bond and twice the next active bond, which is exactly the stated recurrence. (a) Reading the recurrence directly: new left bond = min(original left bond, 2 * next active bond) <= original left bond, and <= 2 * next active bond. Both inequalities hold simultaneously because the minimum of two quantities is at most each of them. So every canonicalized core's left bond is bounded by both its original left bound and twice its next active bond. (b) Consider a chain of length one over a scalar (dimension-one) terminal boundary: the single core's right/next active bond is the terminal bond, which is forced to be dimension one. Apply the bound from (a) to this core: its left bond is at most 2 times its next active bond = 2 * 1 = 2. Therefore the active bond immediately preceding a dimension-one terminal bond is at most two. Intuitively, once the chain has been squeezed down to a scalar output, the preceding bond can retain at most the two values the final emitted bit can distinguish, so its dimension cannot exceed two.", "rubric": [ { "criterion": "Explains the factor of two as the emitted bit of dimension 2 multiplying the right/next-active bond (reshape column count).", "id": "r1", "weight": 1 }, { "criterion": "Derives (a): new left bond = min(original, 2*next) is at most each of the two quantities.", "id": "r2", "weight": 1 }, { "criterion": "Derives (b): for a length-one chain with terminal bond 1 the preceding bond is at most 2*1 = 2.", "id": "r3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The new left bond equals min(original left bond, twice the next active bond), from reshaping the emitted bit with the right bond and factorizing.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_core_lq" ] }, { "claim": "Therefore the new left bond is at most the original left bond and at most twice the next active bond simultaneously.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.RankReduced.head_bound" ] }, { "claim": "For a length-one scalar-boundary chain the bond before the dimension-one terminal bond is at most two.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.RankReduced.last_bond_le_two" ] } ], "source_item_digest": "b53e888a9d2bde591363f2eb202ee553d7d833c9cfc4f8c52e0ce21427338a09", "source_refs": [ { "anchor": "complete declarations RankReduced (lines 51-55), exists_core_lq (lines 67-80), RankReduced.head_bound (lines 104-110), RankReduced.last_bond_le_two (lines 113-117), RankReduced.maxBond_le (lines 203-207)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/TensorTrainCanonical.lean", "sha256": "b8d1eb0651708ac2496be306e2d4b6f3025f7ea4d35a8c6907de02b89d1d981f" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.RankReduced", "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_core_lq", "declaration:QuantumBlockEncoding.TensorTrainCanonical.RankReduced.maxBond_le" ], "target_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.RankReduced.head_bound", "declaration:QuantumBlockEncoding.TensorTrainCanonical.RankReduced.last_bond_le_two" ], "task_hash": "2dd97acc20e2b8f4de89801179f11bc0c1e99abe655f6a989548527791a19423", "task_kind": "boundary", "topic": "linear_algebra" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-005", "difficulty": { "level": "intermediate", "rationale": "Requires contrasting product, separable and pure-separable notions and producing an explicit correlated witness with a marginal/rank argument." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-composite-entanglement", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-e78c4883c82dc59a3e4a", "key": "qcb-core-plan-005-classical-correlation-vs-product-counterexample", "primary_sources": [ { "anchors": "Product-state factorization Eq. (2.16) tr[rho_AB(E_A tensor F_B)] = tr(omega_A E_A)tr(sigma_B F_B); separable-state definition as sum_i p_i omega_i tensor sigma_i encoding classical correlation; remark that a pure separable state is automatically a product state; partial-trace/marginal definition Eqs. (2.14)-(2.15)", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "5-6", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Call a bipartite state a product state if it factorizes as a single tensor product of local states, and separable if it is a probabilistic mixture of product states. (a) Show that in a product state the outcomes of any two local measurements are statistically independent (the joint expectation factorizes). (b) Exhibit a concrete separable two-qubit state that is NOT a product state and compute a pair of local measurements whose outcomes are correlated, demonstrating that separable states can carry classical correlation even though each term in the mixture is uncorrelated. (c) Reconcile this with the textbook remark that a PURE separable state is automatically a product state - i.e. explain why your correlated example must be mixed.", "reference_answer": "(a) For a product state omega tensor sigma and local measurement effects E on the first system and F on the second, the joint probability is tr[(omega tensor sigma)(E tensor F)] = tr(omega E) tr(sigma F), the product of the two marginal probabilities. Hence the two outcomes are independent and uncorrelated. (b) Take the classically correlated state rho = (1/2)(|0><0| tensor |0><0|) + (1/2)(|1><1| tensor |1><1|). It is separable (an equal mixture of the two product states |00><00| and |11><11|) but it is not a single product omega tensor sigma: its marginals are both the maximally mixed (1/2)I, whose product (1/4)I is full rank, while rho has rank 2, so rho != marginal tensor marginal. Measuring the first qubit in the computational basis (outcome a in {0,1}) and the second likewise (outcome b) gives joint probabilities Pr(0,0)=Pr(1,1)=1/2, Pr(0,1)=Pr(1,0)=0. The outcomes are perfectly correlated (a = b always) even though Pr(a=0)=Pr(b=0)=1/2, so Pr(a,b) != Pr(a)Pr(b); the correlation is exactly the shared classical index of the mixture. (c) A pure separable state is by definition an extreme point that is also a mixture of products; being extreme, the mixture must be trivial, so it equals one product term omega tensor sigma, and by part (a) it is uncorrelated. Therefore any separable state exhibiting correlation must be a nontrivial mixture, i.e. mixed - consistent with the example, whose rank is 2 and which is a genuine average over two distinct product states.", "rubric": [ { "criterion": "Proves factorization tr[(omega tensor sigma)(E tensor F)] = tr(omega E) tr(sigma F) and concludes local independence in a product state", "id": "r1", "weight": 1 }, { "criterion": "Gives an explicit separable non-product state (e.g. equal mixture of |00><00| and |11><11|) and argues it is not a single product via a marginal/rank comparison", "id": "r2", "weight": 1 }, { "criterion": "Computes local measurements with correlated outcomes (Pr(0,0)=Pr(1,1)=1/2) showing Pr(a,b) != Pr(a)Pr(b)", "id": "r3", "weight": 1 }, { "criterion": "Explains that a pure separable state is extreme hence a single product (uncorrelated), so a correlated separable state must be mixed", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "In a product state the joint expectation factorizes: tr[(omega tensor sigma)(E tensor F)] = tr(omega E) tr(sigma F).", "supporting_nodes": [ "concept:probability" ] }, { "claim": "The equal mixture of |00><00| and |11><11| is separable, not a product (rank 2 vs full-rank marginal product), and perfectly correlated.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "A pure separable state is extreme, hence a single product term and uncorrelated, so correlated separable states are mixed.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "75bdc7610d7bf3c2c4c4a2fb21a151bdd2966bb54443b4426e310f9b67f38bf5", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "59bcb9a36713b58d63cbd6e25ecaf9069cf4281831e663188428f9b827359772", "task_kind": "counterexample", "topic": "planned_quantum_information" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-010", "difficulty": { "level": "intermediate", "rationale": "Connects an operational entrywise verification recipe to the operator-norm accuracy definition and surfaces the standard entrywise-vs-spectral pitfall with a quantitative example." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-block-encoding-verification", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-498430b17f3a7e53e621", "key": "qcb-core-plan-010-entry-vs-operator-norm-verification", "primary_sources": [ { "anchors": "Eq. (9.11) definition ||A - alpha <0^m|UA|0^m>|| <= epsilon ((alpha,m,epsilon)-embedding, operator norm); Eq. (9.12) entrywise verification <0^m,i|UA|0^m,j> = A_{ij} for exact unit-factor embedding; Eq. (9.13) factorization <0^m,i|U_B U_C|0^m,j> = (U_B-dagger|0^m,i>)-dagger(U_C|0^m,j>) for composed unitaries and the circuit of Fig. 9.2", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "143-144", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "To certify that an (m+n)-qubit unitary U exactly embeds an n-qubit matrix A as its all-zeros block with unit factor, one checks the entrywise identities <0^m, i| U |0^m, j> = A_{ij} for all computational-basis indices i, j. (a) Explain operationally how each such entry can be read off by preparing the ancilla in all-zeros and the system in basis state j, applying U, and overlapping with the all-zeros-ancilla basis state i, and how a composed unitary U = U_B U_C lets one evaluate the two halves independently before overlapping. (b) Explain why checking all entries equal to A_{ij} is equivalent to the exact operator identity (all-zeros block of U equals A), but why for an APPROXIMATE embedding entrywise closeness of each A_{ij} is NOT the right accuracy notion: give the quantity that the definition of approximate embedding actually controls and explain how it can be large even when every entry error is tiny.", "reference_answer": "(a) Preparing |0^m>|j>, applying U and projecting onto <0^m, i| extracts exactly the number <0^m, i| U |0^m, j>, which by construction is the (i,j) entry of the all-zeros block; running over all i, j reads off the whole block. If U = U_B U_C, then <0^m, i| U_B U_C |0^m, j> = (U_B-dagger |0^m, i>)-dagger (U_C |0^m, j>), so one can prepare and evaluate the two states U_B-dagger|0^m,i> and U_C|0^m,j> independently and take their inner product; this is the standard way to verify a product construction entry by entry. (b) The all-zeros block is the operator B with matrix entries B_{ij} = <0^m,i|U|0^m,j>; the full set of equalities B_{ij} = A_{ij} for all i, j is literally the statement B = A as operators, so for EXACT embedding entrywise and operator identity coincide. For APPROXIMATE embedding, however, the definition controls the OPERATOR NORM of the difference, ||A - alpha B|| <= epsilon, not the individual entry errors. Operator norm is not controlled by entrywise size: a d-by-d matrix with every entry equal to a small delta has operator norm up to d*delta (e.g. the all-ones matrix times delta has norm d*delta), which grows with dimension even as each entry error delta stays tiny. Hence many small entry errors can accumulate coherently into a large operator-norm error, so entrywise closeness is neither the quantity the definition uses nor a sufficient accuracy guarantee; the correct certificate is a bound on the spectral/operator norm of A - alpha B.", "rubric": [ { "criterion": "Describes the state-preparation/overlap procedure that reads <0^m,i|U|0^m,j> and the factored evaluation for U = U_B U_C via (U_B-dagger|0^m,i>)-dagger(U_C|0^m,j>)", "id": "r1", "weight": 1 }, { "criterion": "States that the full set of entry equalities B_{ij} = A_{ij} is exactly the operator identity B = A for exact embedding", "id": "r2", "weight": 1 }, { "criterion": "Identifies that approximate embedding controls the operator norm ||A - alpha B|| <= epsilon, not per-entry errors", "id": "r3", "weight": 1 }, { "criterion": "Gives a dimension example (e.g. all-entries-delta matrix has norm up to d*delta) showing tiny entry errors can produce a large operator-norm error", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The entry <0^m,i|U|0^m,j> is read by prepare-apply-overlap, and factors as (U_B-dagger|0^m,i>)-dagger(U_C|0^m,j>) for U = U_B U_C.", "supporting_nodes": [ "concept:verification" ] }, { "claim": "All entry equalities B_{ij} = A_{ij} are equivalent to the exact operator identity B = A.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "Approximate embedding controls ||A - alpha B|| <= epsilon; small entry errors can sum to a large operator norm (up to d*delta).", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "13ff4d7410b8c1c651c0e8a652cb201ded626374ddf50a7f712d2a82784b3abf", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:matrix-analysis" ], "target_nodes": [ "concept:block-encoding", "concept:verification" ], "task_hash": "f385cf9ed4eb9d76b0b53de72cf3aa5588094a2f71f4c0f82743bff23b59ca56", "task_kind": "concept", "topic": "planned_scientific_computing" }, { "assumptions": [ "U unitary on the stated finite tensor product", "alpha>0 is a real scale; epsilon>=0; column-vector convention" ], "context_hash": "d274d63debbddcd2f11d0498a381b5a9373eddf7ddf06ab6bd6332c8a4a10f4f", "conventions": [ "Auxiliary register pinned to the all-zero value selects the clean block", "Exact case is the epsilon=0 instance of the approximate operator-norm bound" ], "core_id": "qcb-core-be-001", "difficulty": { "level": "intermediate", "rationale": "Requires reading the candidate record, the approximation-bound comment and the block-projection helper, and stating dimensions and the entrywise contract exactly." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-projected-block-contract", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-9aa2db7cc60e10000594", "key": "qcb-core-be-001-corrected-v2", "primary_sources": [], "question": "Let U be a unitary on n system qubits and a auxiliary qubits. For a strictly positive real alpha, define B=(<0^a| tensor I)U(|0^a> tensor I). State what it means for U to encode a system operator A exactly or with induced-operator-norm error at most epsilon>=0. Give the matrix dimensions and derive the scalar entry condition equivalent to exact encoding. Explain why division by alpha needs its stated premise.", "reference_answer": "U has dimension2^(n+a) square. B and A have dimension2^n square. Exact encoding means A=alpha B; approximate encoding means ||A-alpha B||<=epsilon in the induced Euclidean operator norm. Exact equality is equivalent to each clean entry U((0,i),(0,j))=A(i,j)/alpha because alpha>0, hence nonzero. The norm error is zero exactly for equal operators. Positivity is a scale convention, while nonzero is the logical premise needed for division; at alpha=0 the undivided equation reduces to A=0 and does not determine B, so the divided equivalent statement is not defined. The scale is not assumed optimal, and no efficient implementation follows from the existence of U.", "rubric": [ { "criterion": "Correct total and system block dimensions", "id": "dimensions", "weight": 1 }, { "criterion": "Exact equality and induced norm error with explicit scale convention", "id": "contract", "weight": 1 }, { "criterion": "Nonzero alpha required; correct alpha=0 boundary", "id": "division", "weight": 1 } ], "semantic_step_count": 2, "semantic_steps": [ { "claim": "The candidate's unitary has dimension fixed by system+auxiliary qubits, and the extracted block is the auxiliary-zero submatrix.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.OperatorBlockEncodingCandidate", "declaration:QuantumBlockEncoding.signalSystemBlockProjection" ] }, { "claim": "The exact contract A = alpha*block is the epsilon=0 case of the stored operator-norm approximation bound.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.ApproximateOperatorBlockEncodingCandidate" ] } ], "source_item_digest": "d3fcc5644825a903d365c5356b4e51c6754683966d400872169b733c2b7ed313", "source_refs": [ { "anchor": "structure OperatorBlockEncodingCandidate (lines 103-116); structure ApproximateOperatorBlockEncodingCandidate with approximationBound comment '||A - alpha * ((<0^a| tensor I) U (|0^a> tensor I))|| <= epsilon' (lines 136-153)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/BlockEncoding.lean", "sha256": "fe06f75a29822e28ccdbfedfb0b0e87f90872620fae7aed359085fad8fae64b4" }, { "anchor": "def signalSystemBlockProjection with doc 'The block ( tensor I)' (lines 741-772)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/CircuitSemantics.lean", "sha256": "761ac3a98ea5a82ad5030e280e3810821081d10cec3d0a0a1d3c2add3fcd4666" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingSpec", "declaration:QuantumBlockEncoding.signalSystemBlockProjection" ], "target_nodes": [ "declaration:QuantumBlockEncoding.OperatorBlockEncodingCandidate", "declaration:QuantumBlockEncoding.ApproximateOperatorBlockEncodingCandidate" ], "task_hash": "c13f2c50528cc25b7cffe4b00107cb0bf53d4a2f70e4fb149bb3a6b2fd0f3094", "task_kind": "concept", "topic": "block_encoding" }, { "assumptions": [ "Unitary circuit on S tensor W tensor D", "A clean workspace promise concerns all input data vectors on the signal-success subspace" ], "context_hash": "98a6455ff5b8016b9d5bd1ad1f24dd0309442e4ada3858108789ac9b12a39d02", "conventions": [ "auxiliaryQubits = signalQubits + pureAncillas", "Clean block uses the all-zero auxiliary bra and ket simultaneously" ], "core_id": "qcb-core-be-002", "difficulty": { "level": "intermediate", "rationale": "Semantics of clean vs flag registers and why uncleaned ancillas violate the projection contract; grounded in the layout and cost definitions." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "be-auxiliary-accounting", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-be0f58115df2d75d689f", "key": "qcb-core-be-002-corrected-v2", "primary_sources": [], "question": "A circuit acts on a signal register S, a workspace W and a data register D, starting S,W at zero. The signal value0 is called success. Explain the additional mathematical promise needed to call W clean on every successful branch. Does knowing only the smaller submatrix obtained by projecting BOTH S and W to zero imply that promise? Give a two-dimensional workspace counterexample, and explain which auxiliary qubits still occupy physical space.", "reference_answer": "Workspace cleanliness on signal success means that for every data input v, projecting the final state onto S=0 leaves a vector of the form |0>_S|0>_W tensor T v, with no component at S=0,W nonzero. This is an extra restriction on success-sector amplitudes. The doubly projected submatrix alone fixes only the W=0 component and does not constrain W nonzero. Counterexample: take signal and data one-dimensional and let W be a qubit with unitary R=[[cos t,-sin t],[sin t,cos t]],0=cos t|0>+sin t|1>, so workspace is not clean even though signal success is certain. Both signal and workspace qubits enlarge the physical unitary, regardless of whether a resource ledger counts them in different roles. Projection is not itself uncomputation; a separately proved clean-workspace promise cannot be inferred just from the scalar clean submatrix.", "rubric": [ { "criterion": "State success-sector workspace cleanup for every data input", "id": "promise", "weight": 1 }, { "criterion": "Valid rotation has correct projected block but dirty success sector", "id": "counterexample", "weight": 1 }, { "criterion": "Separate projection, cleanup, and allocated physical auxiliary space", "id": "cost", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The smaller clean submatrix constrains only the all-zero workspace component, not other successful workspace values.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RegisterLayout", "declaration:QuantumBlockEncoding.RegisterLayout.auxiliaryQubits" ] }, { "claim": "A workspace rotation has nonzero excited output while its all-zero subblock remains exactly cos t.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RegisterLayout", "declaration:QuantumBlockEncoding.RegisterLayout.auxiliaryQubits" ] }, { "claim": "Therefore a clean-workspace success promise is additional to subblock equality.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RegisterLayout", "declaration:QuantumBlockEncoding.RegisterLayout.auxiliaryQubits" ] } ], "source_item_digest": "0225404a2bcf4af0f22e5bf5145a57227181783389be19800bce939c379853e1", "source_refs": [ { "anchor": "structure RegisterLayout {systemQubits, signalQubits, pureAncillas}; def auxiliaryQubits 'combines the signal qubits selecting the block with pure ancillas that must be returned to a clean state' (lines 14-30); structure BlockEncodingCost (lines 50-55)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/BlockEncoding.lean", "sha256": "fe06f75a29822e28ccdbfedfb0b0e87f90872620fae7aed359085fad8fae64b4" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.OperatorBlockEncodingCandidate", "declaration:QuantumBlockEncoding.BlockEncodingCost" ], "target_nodes": [ "declaration:QuantumBlockEncoding.RegisterLayout", "declaration:QuantumBlockEncoding.RegisterLayout.auxiliaryQubits" ], "task_hash": "8ed915539093e751129d98e5c79b4d61c337609db19017e22ca1be7895375a32", "task_kind": "concept", "topic": "block_encoding" }, { "assumptions": [ "U,V unitary with unit normalization for simplicity", "Each clean block uses its own auxiliary bra/ket0; column-vector chronological order" ], "context_hash": "28c78077aed7f7b27e845004adf22ae3a2f45f363b7ace589d68cb19073c5a36", "conventions": [ "Product applies right factor first (B), then left (A), to the system", "Clean branch requires all-zero ancilla before each factor" ], "core_id": "qcb-core-be-010", "difficulty": { "level": "advanced", "rationale": "Separates the proved clean-block algebra from the physical ancilla requirement and refutes a plausible same-ancilla shortcut by describing the garbage branch; ties to the additive product resource cost." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "be-product-fresh-ancilla", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-997c309d1dba1dfd0304", "key": "qcb-core-be-010-corrected-v2", "primary_sources": [], "question": "Suppose unitaries U and V have clean blocks A and B on the same data space. Explain why applying V then U with DISJOINT auxiliary registers initialized at zero yields clean block AB. Show concretely why simply reusing a single auxiliary register can fail. Is resetting that register or applying its inverse automatically a free coherent repair?", "reference_answer": "With independent auxiliary indices, the final all-zero projection fixes each register only at its own factor; summing over the common data index gives sum_k A(i,k)B(k,j), hence AB. Reusing one register instead introduces paths that leave its zero sector under V and return under U. A minimal example has one-dimensional data and a qubit auxiliary, U=V=X. Each individual clean block is <0|X|0>=0, so AB=0, but <0|X X|0>=1. Resetting an entangled auxiliary is not unitary and may erase coherence; it does not turn the same-register product into the coherent operator AB for free. Applying V inverse cancels V on the entire state and can also erase the desired action, not merely its garbage. A different coherent cleanup construction must be proved under its own promises and charged; disjoint fresh registers give the straightforward product construction. The matrix product of already-extracted blocks alone is not a proof of a particular same-register circuit.", "rubric": [ { "criterion": "Correct independent-auxiliary finite sum gives AB", "id": "product", "weight": 1 }, { "criterion": "U=V=X zero individual blocks but composite clean block1", "id": "witness", "weight": 1 }, { "criterion": "Reset nonunitary and inverse can erase desired action; no free cleanup claim", "id": "cleanup", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Independent clean auxiliary indices leave a data-index product sum AB.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.productCleanBlockCertificate", "declaration:QuantumBlockEncoding.BlockEncodingClassics.matrix_mul_congr_pointwise" ] }, { "claim": "The X/X same-register witness creates a garbage-return path missed by AB.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.productCleanBlockCertificate", "declaration:QuantumBlockEncoding.BlockEncodingClassics.matrix_mul_congr_pointwise" ] }, { "claim": "Reset/inverse is not an automatically valid coherent product repair.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.productCleanBlockCertificate", "declaration:QuantumBlockEncoding.BlockEncodingClassics.matrix_mul_congr_pointwise" ] } ], "source_item_digest": "ee013faa49782085ff133a942fb0e639d3357fa9a4eed212f42f0bb7096d2323", "source_refs": [ { "anchor": "theorem matrix_mul_congr_pointwise (lines 792-797); def productCleanBlockCertificate 'cleanBlock := Matrix.mul left.cleanBlock right.cleanBlock' (lines 800-808); def productResourceCost adding auxiliaryQubits and gateCount, depth adds (lines 831-839)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/BlockEncodingClassics.lean", "sha256": "94ac6632924df0a49b1ebd99bb5d9ed69b2ea463ec527033f38bd73deeb1346a" }, { "anchor": "comment that any normalized orthogonal failure branch is additional unitarity evidence (lines 271-294)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/ConcreteSemantics.lean", "sha256": "454c0c5e215ca197c5883bedf74882f392d259c7afeadc63912230437698ebe5" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.productResourceCost", "declaration:QuantumBlockEncoding.BlockEncodingClassics.LCUCertificate", "declaration:QuantumBlockEncoding.ConcreteSemantics.cleanBasisActionAmplitude" ], "target_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.productCleanBlockCertificate", "declaration:QuantumBlockEncoding.BlockEncodingClassics.matrix_mul_congr_pointwise" ], "task_hash": "19fd86c68eec8d7cf232cf58ab3624e5087ce215e6d768aa916a70bd43bed992", "task_kind": "counterexample", "topic": "block_encoding" }, { "assumptions": [ "a and b are arbitrary real sibling amplitudes; the claim concerns one binary split, not the whole recursive loader", "the real Y-rotation convention is the standard orthogonal rotation used by the library backend" ], "context_hash": "63a9e13685297c3239f4453a1232a6d7ffb8753fabdcad60fb05e1c868071927", "conventions": [ "wires are little-endian; the split acts on one new highest wire", "the pair norm is the Euclidean length sqrt(a^2 + b^2) and the all-zero pair is assigned angle zero" ], "core_id": "qcb-core-sp-001", "difficulty": { "level": "intermediate", "rationale": "Requires reading the signed polar-angle construction and reasoning about the range of arccos and the sine sign, not merely recalling an angle formula." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "signed-amplitude-angles", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-92caf678d8f22ccf1e54", "key": "qcb-core-sp-001", "primary_sources": [], "question": "In the low-wire-first recursive loader for a finite real amplitude table, each binary split of a sibling pair of real amplitudes (a, b) is realised by a single uniformly-controlled real Y-rotation. The rotation angle is twice a signed polar angle: its magnitude is the inverse cosine of a divided by the pair norm sqrt(a^2 + b^2), and its sign is taken from the sign of b (with the all-zero pair assigned angle zero). Explain the role of the sign branch, and why using the bare inverse-cosine magnitude alone would fail for a general real amplitude table.", "reference_answer": "The sibling pair (a, b) lies on a circle of radius r = sqrt(a^2 + b^2). A real Y-rotation by angle phi sends the all-zero column to (r*cos(phi/2), r*sin(phi/2)). Choosing phi/2 = arccos(a/r) reproduces the first child exactly, since r*cos(arccos(a/r)) = a (valid because a/r lies in [-1, 1] when r > 0). The second entry is then r*sin(arccos(a/r)) = sqrt(r^2 - a^2) = sqrt(b^2) = |b|, which is nonnegative because arccos lands in [0, pi] where sine is nonnegative. So the bare magnitude reproduces b only when b >= 0; for a table with a negative amplitude below a split it would emit |b| and silently drop the sign. The sign branch repairs this: when b < 0 the half-angle is negated, flipping the lower entry to r*sin(-arccos(a/r)) = -|b| = b. The all-zero pair (r = 0) carries no direction and is assigned angle zero. Thus the signed polar angle recovers the full signed real pair with exact real angles, avoiding a numerically unverified atan2 primitive.", "rubric": [ { "criterion": "States that a real Y-rotation sends the zero column to (r*cos(phi/2), r*sin(phi/2)) with r the pair norm, and that the inverse-cosine magnitude recovers the first child a exactly.", "id": "r1", "weight": 1 }, { "criterion": "Explains that the bare inverse-cosine magnitude yields only |b| (nonnegative) for the second child, because arccos ranges over [0, pi] with nonnegative sine.", "id": "r2", "weight": 1 }, { "criterion": "Identifies the sign branch (negating the half-angle when b < 0) as the mechanism that flips the lower entry to the correct negative value, so general signed tables are handled.", "id": "r3", "weight": 1 }, { "criterion": "Notes the zero-pair angle-zero convention and that the angles are exact real expressions (no atan2), without claiming a gate-count or whole-circuit result.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A real Y-rotation by phi sends the zero column to (r*cos(phi/2), r*sin(phi/2)); with phi/2 = arccos(a/r) the first entry times r is exactly a.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.splitAngle", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.pairNorm" ] }, { "claim": "The second entry equals r*sin(arccos(a/r)) = |b|, nonnegative because arccos lands in [0, pi]; a bare magnitude angle therefore mis-signs negative children.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.splitAngle_firstColumn" ] }, { "claim": "Negating the half-angle when b < 0 flips the lower entry to -|b| = b, so the signed pair is recovered exactly for all real a, b, with the zero pair carrying angle zero.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.splitAngle", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.splitAngle_firstColumn" ] } ], "source_item_digest": "0131ad1ec8cff86035bb3b2ac0369a3fda6efe1bb41e47ed0f2a41355733e4ce", "source_refs": [ { "anchor": "lines 150-201: noncomputable def splitAngle (a b : R) : ExactAngle := if pairNorm a b = 0 then .rational 0 else .real (2 * if b < 0 then -Real.arccos (a / pairNorm a b) else Real.arccos (a / pairNorm a b)); and theorem splitAngle_firstColumn (a b : R) (v : Fin 2) : standardRyMatrix (splitAngle a b).eval v 0 * (pairNorm a b : C) = if v = 0 then (a : C) else (b : C)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RealAmplitudePreparation.lean", "sha256": "a6ded7e65a7e42d5724529befb48eefc9505814136206dd96248fb59b3c45efa" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.splitAngle", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.splitAngle_firstColumn", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.pairNorm" ], "task_hash": "3b01bce98dc5e4b867928b6da39f57c27771e1d4a21126344927db3f3228201e", "task_kind": "concept", "topic": "state_preparation" }, { "assumptions": [ "the train has scalar left and right boundaries (both dimension one) and real cores", "the sole numeric hypothesis is that the sum of squared contracted amplitudes is one" ], "context_hash": "6fa3a8186cbb6a2091d87fff705a6e8203148bf9c8201cec2b194a2d2addf22d", "conventions": [ "right-canonical means every individual core has orthonormal rows", "amplitudes are read off by full contraction of the chain for each bit word" ], "core_id": "qcb-core-sp-003", "difficulty": { "level": "advanced", "rationale": "Requires recognizing that normalization and amplitude recovery are derived output promises, not input hypotheses, which is the subtle point of the existence theorem." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "canonicalization-input-promise", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-b5cd6cf46bbdddaf1bde", "key": "qcb-core-sp-003", "primary_sources": [], "question": "A finite real tensor train with scalar (dimension-one) left and right boundaries defines its amplitudes by full contraction over all cores. Suppose the only hypothesis is that the sum of squared contracted amplitudes equals one. After right-canonicalization, what is guaranteed about (a) the new initial boundary vector and (b) each individual target amplitude, and what is deliberately NOT assumed?", "reference_answer": "Right-canonicalization produces an exactly equivalent right-canonical train D (each core has orthonormal rows) together with a new initial boundary vector u such that: (a) u is normalized, the sum over a of u_a squared equals one; and (b) every original amplitude is recovered exactly by contracting u against D, namely amplitude(x) = sum over a of u_a times contract(D, x)_a. What is deliberately not assumed is any equality to a prescribed target state: the normalization of u and the amplitude-recovery identity are consequences derived solely from the single scalar hypothesis that the squared amplitudes sum to one, together with the exact factorization produced by canonicalization. The terminal bond stays dimension one. In other words, normalization and recovery are provable output promises of the construction, not extra side conditions imposed on the input.", "rubric": [ { "criterion": "States that canonicalization yields a right-canonical train (orthonormal rows per core) plus a new initial boundary vector u.", "id": "r1", "weight": 1 }, { "criterion": "States that u is normalized (its squared entries sum to one).", "id": "r2", "weight": 1 }, { "criterion": "States that each original amplitude equals the contraction of u against the canonical train, and that the terminal bond stays dimension one.", "id": "r3", "weight": 1 }, { "criterion": "Emphasizes that no equality to a target state is assumed; normalization and recovery are derived from the single hypothesis sum of squared amplitudes = 1.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Canonicalization returns a right-canonical train and a residual initial vector while preserving every amplitude by an exact factorization.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_normalized_state" ] }, { "claim": "Under the hypothesis that the squared amplitudes sum to one, the residual initial vector is normalized.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_rightCanonical_normalized" ] }, { "claim": "Each original amplitude is recovered as the contraction of the normalized initial vector against the canonical train, with no target-state equality assumed.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_normalized_state" ] } ], "source_item_digest": "c4a49ccae5572ff4bf896d73c30f37c0c0c3b4caa98488b2efd4171679e36f72", "source_refs": [ { "anchor": "lines 183-195: theorem exists_normalized_state (C : Chain n 1 1) (hC : (sum x : Word n, (contract C x 0 0) ^ 2) = 1) : exists l' u D, RightCanonical D and RankReduced C D and mass u = 1 and for all x, contract C x 0 0 = sum a, u a * contract D x a 0; and lines 172-178 exists_rightCanonical_normalized", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/TensorTrainCanonical.lean", "sha256": "b8d1eb0651708ac2496be306e2d4b6f3025f7ea4d35a8c6907de02b89d1d981f" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.chainMass" ], "target_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_normalized_state", "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_rightCanonical_normalized" ], "task_hash": "2b3aa64ebcc133ee41f3c18db196966e2bf35e3808943ab5365b9fd8987ec7a5", "task_kind": "concept", "topic": "state_preparation" }, { "assumptions": [ "the input train is normalized (chain mass one) with a scalar terminal bond", "canonicalization uses orthonormal-row (right-canonical) cores" ], "context_hash": "065c5e00bf3b56abd0b734203921cd2fbc4d11d6d248aa0b2e4521929e6d9f7e", "conventions": [ "mass is the sum of squared entries of a real boundary vector", "vecMul v R denotes the row-vector action of the residual matrix on the boundary" ], "core_id": "qcb-core-sp-004", "difficulty": { "level": "intermediate", "rationale": "Requires linking the isometric action of canonical cores to the localization of mass in the residual boundary, beyond a single definitional fact." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "canonicalization-residual-norm", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-a9711a5ddf9a6dd05d29", "key": "qcb-core-sp-004", "primary_sources": [], "question": "A normalized finite real tensor train with a scalar terminal bond is right-canonicalized from the terminal end inward, absorbing each residual matrix into the preceding core. After this process, where does the total squared mass of the state reside, and what are the dimensions of the two boundary bonds?", "reference_answer": "Canonicalization factors the chain as a residual initial matrix R times a right-canonical train D. Every core of D has orthonormal rows, so D acts as a composition of row isometries that preserve the row-vector squared mass. Consequently the total squared mass of the original state equals the mass of the boundary vector after it is pushed through R: mass(vecMul v R) equals the chain mass. For a normalized input (chain mass one) the residual initial boundary is therefore itself normalized. The terminal bond stays exactly dimension one throughout, and the residual appears as a new initial boundary whose dimension is the reduced left rank. So after canonicalization all of the normalization is concentrated in the single residual initial boundary while the terminal bond is untouched at dimension one.", "rubric": [ { "criterion": "States the factorization chain = residual initial matrix R times a right-canonical train D.", "id": "r1", "weight": 1 }, { "criterion": "Explains that orthonormal-row cores act as row isometries preserving squared mass, so the mass equals mass(vecMul v R).", "id": "r2", "weight": 1 }, { "criterion": "Concludes that for a normalized input the residual initial boundary is normalized (all mass concentrated there).", "id": "r3", "weight": 1 }, { "criterion": "States the terminal bond stays dimension one and the residual initial boundary has the reduced left-rank dimension.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Canonicalization factors the chain as a residual initial matrix times a right-canonical train preserving every amplitude.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.residual_mass" ] }, { "claim": "Orthonormal-row cores compose to a mass-preserving map, so the chain mass equals the mass of the boundary pushed through the residual.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.chainMass_eq", "declaration:QuantumBlockEncoding.TensorTrainCanonical.residual_mass" ] }, { "claim": "For a normalized input the residual initial boundary is normalized, while the terminal bond stays dimension one.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_rightCanonical_normalized" ] } ], "source_item_digest": "1b31a5d5788641b5b7068c87fc829de70756a84e5d8acb6f2e049f0d12701aed", "source_refs": [ { "anchor": "lines 159-178: theorem residual_mass (C D R hD h v) : chainMass C v = mass (Matrix.vecMul v R); and theorem exists_rightCanonical_normalized (C : Chain n l 1) (v) (hv : chainMass C v = 1) : exists l' R D, RightCanonical D and RankReduced C D and (for all x, contract C x = R * contract D x) and mass (Matrix.vecMul v R) = 1", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/TensorTrainCanonical.lean", "sha256": "b8d1eb0651708ac2496be306e2d4b6f3025f7ea4d35a8c6907de02b89d1d981f" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.chainMass_eq" ], "target_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.residual_mass", "declaration:QuantumBlockEncoding.TensorTrainCanonical.exists_rightCanonical_normalized" ], "task_hash": "58e3a5fa8fa6db58861c4bccfed6a8de022fff43e0dc352201eaeb2980730d18", "task_kind": "concept", "topic": "state_preparation" }, { "assumptions": [ "f is an arbitrary real table on n+1 wires; no positivity or normalization is assumed", "the marginal is defined by the sibling-pair Euclidean length as in the source" ], "context_hash": "0c619f075d9fdac25aaed1917523540584742daeb085baae6080fcc19bc3bb0b", "conventions": [ "normSq g is the sum of squared entries; wires are little-endian with the last bit split off by snoc", "pairNorm is the nonnegative Euclidean length of a pair" ], "core_id": "qcb-core-sp-005", "difficulty": { "level": "intermediate", "rationale": "A short but genuine two-ingredient proof: the Pythagorean per-pair identity plus a last-bit reindexing of the full sum." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "marginal-norm-pythagorean", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-336534b6616f1544cef1", "key": "qcb-core-sp-005", "primary_sources": [], "question": "For a finite real amplitude table f on n+1 wires, define its low-wire marginal by taking, at each prefix word b, the Euclidean length of the sibling pair (f at b-with-last-bit-0, f at b-with-last-bit-1). Prove that the total squared Euclidean norm of the marginal equals the total squared Euclidean norm of f. Explain why this makes the recursive marginal pass norm-faithful.", "reference_answer": "Write normSq g = sum over basis words of g^2, and for a prefix word b let m(b) = pairNorm(f(snoc b 0), f(snoc b 1)) with pairNorm(a,b) = sqrt(a^2+b^2). First, pairNorm squares cleanly: pairNorm(a,b)^2 = a^2 + b^2, because pairNorm is the square root of the nonnegative quantity a^2+b^2 and squaring a square root of a nonnegative real returns the radicand. Hence m(b)^2 = f(snoc b 0)^2 + f(snoc b 1)^2 for every prefix b. Summing over all prefix words b: normSq(marginal f) = sum_b m(b)^2 = sum_b [ f(snoc b 0)^2 + f(snoc b 1)^2 ]. Now use the bijection between words on n+1 wires and (prefix word b, last bit v) given by w = snoc b v, which partitions every full word uniquely by its last bit. Reindexing the full sum normSq f = sum over w of f(w)^2 along this bijection and splitting the last bit into the two cases v=0 and v=1 gives exactly sum_b [ f(snoc b 0)^2 + f(snoc b 1)^2 ]. The two expressions coincide, so normSq(marginal f) = normSq f. Norm-faithfulness follows: the recursive loader first prepares the marginal on the lower wires and then splits each marginal entry with one rotation; because each marginal step reproduces the exact squared norm, the induction hypothesis about the normalized lower state transfers without any loss or rescaling, which is precisely the invariant the first-column proof consumes.", "rubric": [ { "criterion": "Establishes pairNorm(a,b)^2 = a^2 + b^2 (square of a square root of a nonnegative radicand), hence m(b)^2 = f(snoc b 0)^2 + f(snoc b 1)^2.", "id": "r1", "weight": 1 }, { "criterion": "Uses the (prefix word, last bit) bijection w = snoc b v to reindex the full squared-norm sum.", "id": "r2", "weight": 1 }, { "criterion": "Splits the last-bit sum into v=0 and v=1 to match the marginal sum, concluding normSq(marginal f) = normSq f.", "id": "r3", "weight": 1 }, { "criterion": "Explains that exact squared-norm preservation per level is what carries the normalization invariant through the recursion.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "pairNorm(a,b)^2 = a^2 + b^2, so each marginal entry squared equals the sum of its two children squared.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.pairNorm_sq", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.marginal" ] }, { "claim": "The map (prefix word, last bit) to full word via snoc is a bijection partitioning full words by their last bit.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normSq" ] }, { "claim": "Reindexing the full squared-norm sum along this bijection and splitting the last bit into 0 and 1 reproduces the marginal sum, giving normSq(marginal f) = normSq f.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normSq_marginal" ] } ], "source_item_digest": "4748bf7c81283496b115e778c1f365bd503d4384829e85ceb0959f348a5a97d4", "source_refs": [ { "anchor": "lines 142-222: theorem pairNorm_sq (a b : R) : pairNorm a b ^ 2 = a ^ 2 + b ^ 2; noncomputable def marginal f := fun b => pairNorm (f (Fin.snoc b 0)) (f (Fin.snoc b 1)); noncomputable def normSq f := sum b, f b ^ 2; theorem normSq_marginal (f : PrimitiveBasis (n+1) -> R) : normSq (marginal f) = normSq f", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RealAmplitudePreparation.lean", "sha256": "a6ded7e65a7e42d5724529befb48eefc9505814136206dd96248fb59b3c45efa" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normSq_marginal", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.pairNorm_sq", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.marginal", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normSq" ], "task_hash": "8ac3123a4b2b6a5c48387dfd57f08a7691bbda2fd7217289ca8382f197be0e31", "task_kind": "proof", "topic": "state_preparation" }, { "assumptions": [ "the only hypothesis is clean-column equality of each stage with the corresponding tensor core; no global state action is assumed", "the boundary is an explicit bond vector supplied to the run" ], "context_hash": "43ed171220c70901b1d674b992fc8b135d7063608619a21bf050ae0ba0714b37", "conventions": [ "the last emitted core multiplies on the left (same order as circuit evaluation)", "transfer at length 0 is the identity matrix" ], "core_id": "qcb-core-sp-007", "difficulty": { "level": "advanced", "rationale": "A genuine induction composing the per-step clean-column collapse with the chronological transfer recursion, and isolating which stage columns are free." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "sequential-transfer-composition", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-38379234beb95f4a335c", "key": "qcb-core-sp-007", "primary_sources": [], "question": "In the sequential bond construction, a complete state after n emitted bits is produced by applying stage matrices in order, starting from an explicit bond boundary vector. Assume for each stage index t and each clean input column (fresh bit input = 0, bond input a) the stage entry equals the corresponding tensor core entry K t. Prove by induction on n that the full run amplitude at bit word x and final bond label b equals the matrix-vector product of the chronological transfer contraction of the cores with the boundary, (transfer K n x) applied to the boundary, read at b. State precisely which columns of the stages are therefore irrelevant.", "reference_answer": "Induct on n. Base n = 0: run at length 0 is just the boundary read at the bond label, and transfer K 0 x is the identity matrix, so (transfer K 0 x) applied to boundary at b equals boundary b; both sides agree. Inductive step from n to n+1: by definition the length-(n+1) run is one step applied to the length-n run. Expanding the step via the clean-column collapse (only fresh-bit-input-0 columns act), the amplitude at (x, b) is sum over a of (stage entry at output bit x(last), bond b, clean column a) times (length-n run at (init x, a)). The clean-column hypothesis rewrites each stage entry to K n (x(last), b) a, i.e. the (b,a) entry of the emitted-bit slice coreSlice (K n) (x(last)). The inductive hypothesis rewrites the length-n run at (init x, a) to (transfer K n (init x)) applied to boundary at a. So the amplitude becomes sum over a of coreSlice(K n)(x(last)) at (b,a) times [(transfer K n (init x)) applied to boundary] at a, which is exactly the matrix-vector product coreSlice(K n)(x(last)) times (transfer K n (init x) applied to boundary), read at b. By the recursive definition of transfer, coreSlice(K n)(x(last)) * transfer K n (init x) = transfer K (n+1) x, and composing a matrix with a matrix-vector product equals the product matrix applied to the vector. Hence the amplitude equals (transfer K (n+1) x) applied to boundary at b, closing the induction. The irrelevant columns are exactly the fresh-bit-input-1 columns of every stage: they never appear in the collapsed sum, so arbitrary unitary completions on those columns leave the emitted state unchanged.", "rubric": [ { "criterion": "Handles the base case using transfer at length 0 = identity and run at length 0 = boundary.", "id": "r1", "weight": 1 }, { "criterion": "Expands the inductive step via the clean-column collapse of one step into a sum over the bond label a.", "id": "r2", "weight": 1 }, { "criterion": "Applies the clean-column hypothesis and the inductive hypothesis, then recognizes the sum as coreSlice(K n)(x(last)) times the length-n transfer-applied-to-boundary, i.e. transfer K (n+1) x applied to boundary.", "id": "r3", "weight": 1 }, { "criterion": "States that only clean (fresh-bit-input-0) columns matter, so arbitrary completions on fresh-bit-input-1 columns are irrelevant.", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Base case: run at length 0 equals the boundary, and transfer at length 0 is the identity, so both sides agree.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.transfer" ] }, { "claim": "The length-(n+1) run is one step on the length-n run; the clean-column collapse writes its amplitude as a sum over the bond label of clean-column entries times prior amplitudes.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_apply", "declaration:QuantumBlockEncoding.SequentialBondPreparation.step" ] }, { "claim": "Clean-column equality plus the inductive hypothesis identify the sum with coreSlice(K n)(x(last)) times the length-n transfer applied to the boundary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_eq_transfer" ] }, { "claim": "The recursive definition of transfer folds this into transfer K (n+1) x applied to the boundary, and fresh-bit-input-1 columns never enter, so completions there are irrelevant.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.transfer", "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_eq_transfer" ] } ], "source_item_digest": "f47a739f7ea08d5acd553764322f8c24c2c6b1e7c40abe98923f3bceb2f8963a", "source_refs": [ { "anchor": "lines 86-110: noncomputable def transfer K | 0,_ => 1 | n+1,x => coreSlice (K n) (x (Fin.last n)) * transfer K n (Fin.init x); theorem run_eq_transfer (U K) (cleanColumns : for all t bit b a, U t (bit,b) (0,a) = K t (bit,b) a) (boundary) (n x b) : run U boundary n (x,b) = (transfer K n x).mulVec boundary b, proved by induction with step_apply, cleanColumns, ih, mulVec_mulVec", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SequentialBondPreparation.lean", "sha256": "ad4542b108d44f196f2a02f78e59b58f43d45b1e890709cf4c22e1222805d5c5" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step" ], "target_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_eq_transfer", "declaration:QuantumBlockEncoding.SequentialBondPreparation.transfer", "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_apply" ], "task_hash": "e7e82f8a936619121ecf0bb750eddb33eaea84902a858c41e6db14c34a142534", "task_kind": "proof", "topic": "state_preparation" }, { "assumptions": [ "each stage is a full unitary on the fresh-bit-and-bond register", "the run starts from a single bond basis label" ], "context_hash": "5cb171fb889f0eb559c900255600df6b258beb632a5694df31c93f0aaa72a603", "conventions": [ "amplitude mass is sum of star(amplitude) times amplitude; a basis boundary has one unit coordinate", "the lifted stage is the Kronecker product of identity on passive wires with the stage" ], "core_id": "qcb-core-sp-008", "difficulty": { "level": "intermediate", "rationale": "Chains three preservation facts (padding, lifted-unitary action, induction) and interprets them as coherent unit-probability success rather than a measurement outcome." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "coherent-success-probability", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-938d6925b4acd7eb28a9", "key": "qcb-core-sp-008", "primary_sources": [], "question": "In the sequential bond construction, suppose every stage matrix is a full unitary on the fresh-bit-and-bond register and the run starts from a single bond basis label (one bond coordinate equal to one, the rest zero). Prove that the total probability (sum of squared amplitude moduli over all basis labels) of the emitted state is exactly one at every length n. Explain why this is a coherent success statement with no postselection, and what role the fresh-zero padding plays.", "reference_answer": "Define amplitudeMass v = sum over labels i of star(v_i) * v_i (a complex scalar equal to the total squared modulus). Two facts drive the proof. First, a single step preserves amplitude mass whenever its stage is unitary: the step is the composition of (a) appending a fresh zero bit, which preserves mass because it only adjoins zero components, and (b) applying the stage tensored with the identity on all passive emitted wires. A Kronecker product of unitaries is unitary, so the lifted stage is unitary, and a full unitary matrix-vector action preserves amplitude mass since star(Mv) dot (Mv) = star(v) (M^dagger M) v = star(v) dot v. A final basis regrouping is a bijection of labels and does not change the sum. Hence amplitudeMass(step U v) = amplitudeMass(v). Second, by induction on n the full run preserves mass: at length 0 the run is the boundary and the inductive step applies the one-step invariant, so amplitudeMass(run U boundary n) = amplitudeMass(boundary). Finally, for a basis boundary the mass is sum over labels of the indicator squared = 1. Therefore the emitted state has total probability one at every length. This is a coherent success statement because the entire register is retained: no amplitude is projected out and no renormalization is performed. The fresh-zero padding is essential and benign: it enlarges the register by a provably mass-neutral embedding, so unitarity of each stage is enough to carry total probability one through every emission.", "rubric": [ { "criterion": "Defines amplitude mass as the sum of star times value and shows a single step factors as fresh-zero padding then a lifted (Kronecker-with-identity) unitary action.", "id": "r1", "weight": 1 }, { "criterion": "Shows fresh-zero padding is mass-preserving and that a full unitary action preserves mass via star(Mv).(Mv) = star(v).v, hence one step preserves mass.", "id": "r2", "weight": 1 }, { "criterion": "Inducts on n to conclude the run preserves mass, then evaluates the basis-boundary mass to 1.", "id": "r3", "weight": 1 }, { "criterion": "Explains this is coherent success with no postselection/renormalization and that fresh-zero padding is a mass-neutral register enlargement.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "One step factors as fresh-zero padding followed by the lifted stage action and a label regrouping; the lifted stage is unitary because a Kronecker product of unitaries is unitary.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.liftStage_unitary", "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_step" ] }, { "claim": "Fresh-zero padding preserves mass and a full unitary action preserves mass, so a single step preserves amplitude mass.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_freshZero", "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_step" ] }, { "claim": "Induction on n propagates mass preservation to the whole run; the basis-boundary mass equals one, giving total probability one at every length with no postselection.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_run", "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_run_basisBoundary" ] } ], "source_item_digest": "d403cce404da75f2931579b020692a9cf08143dfd7b2a59eeec69fce4c59f771", "source_refs": [ { "anchor": "lines 288-340: noncomputable def amplitudeMass v := sum i, star (v i) * v i; theorem amplitudeMass_step (hU : U in unitaryGroup) : amplitudeMass (step U v) = amplitudeMass v; theorem amplitudeMass_run (unitary : for all t, U t in unitaryGroup) : amplitudeMass (run U boundary n) = amplitudeMass boundary; theorem amplitudeMass_run_basisBoundary : amplitudeMass (run U (basisBoundary initial) n) = 1", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SequentialBondPreparation.lean", "sha256": "ad4542b108d44f196f2a02f78e59b58f43d45b1e890709cf4c22e1222805d5c5" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_freshZero", "declaration:QuantumBlockEncoding.SequentialBondPreparation.liftStage_unitary" ], "target_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_run_basisBoundary", "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_step", "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_run" ], "task_hash": "4867425811d27d299b922c549a2e0237310cdd910db2feabc98b9d9491eb84e2", "task_kind": "proof", "topic": "state_preparation" }, { "assumptions": [ "A is a real core with orthonormal rows A * A^T = 1", "the complex embedding uses (bit, new bond) as output rows and old bond as input columns" ], "context_hash": "263caedce0b3dc6c261774a1284b653271e5d590a1d42e3ef51e83b46259685d", "conventions": [ "conjTranspose is the complex conjugate transpose; realness means entrywise conjugation is the identity", "the isometry condition is (complexCore A)^dagger (complexCore A) = 1 on the input-bond index" ], "core_id": "qcb-core-sp-009", "difficulty": { "level": "intermediate", "rationale": "Requires carefully distinguishing output rows from input columns and handling the real/imaginary split of the complex isometry identity." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "core-circuit-isometry", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-8b38b213fb13632febd2", "key": "qcb-core-sp-009", "primary_sources": [], "question": "A real tensor-train core carries orthonormal rows (its row-canonical condition). Show that embedding such a core into a complex matrix with the physical bit and new bond as output rows and the old bond as input columns yields a genuine circuit isometry: the conjugate-transpose times the matrix is the identity on the input-bond index. Explain why real orthonormal rows translate exactly to orthonormal complex input columns under this orientation.", "reference_answer": "Let A be a real core with A * A^T = 1 (orthonormal rows). The complex embedding places (bit, new bond) as the output row index and the old bond a as the input column, with entry the real number A_{a,(bit,newbond)} viewed as a complex number. The isometry claim is (complexCore A)^dagger * (complexCore A) = 1 on the old-bond index. Fix input columns a and c; the (a,c) entry of the product is sum over all outputs (bit, newbond) of conj(entry at output,(a)) times entry at output,(c). Because every entry is a real number embedded in the complex field, its conjugate is itself, so this equals the real sum over outputs of A_{a,out} * A_{c,out}, which is precisely the (a,c) entry of A * A^T. By hypothesis A * A^T = 1, so the sum is 1 when a = c and 0 otherwise. Splitting the complex equality into real and imaginary parts: the real part reproduces the (a,c) entry of A * A^T (equal to the Kronecker delta), and the imaginary part vanishes because every summand is a product of reals and hence has zero imaginary component. Therefore (complexCore A)^dagger (complexCore A) is the identity. The conceptual point: the row-canonical condition on the real core, summation over the physical bit together with the new bond, is literally the Gram matrix of the complex input-bond columns under this output-row/input-column orientation; realness makes conjugation trivial so the real orthonormality transfers verbatim to complex column orthonormality.", "rubric": [ { "criterion": "States the isometry target (complexCore A)^dagger (complexCore A) = 1 and expands the (a,c) entry as a sum over outputs (bit, new bond) of conj times entry.", "id": "r1", "weight": 1 }, { "criterion": "Uses realness of the entries (conjugate = itself) to reduce the sum to the real sum A_{a,out} A_{c,out}, i.e. the (a,c) entry of A * A^T.", "id": "r2", "weight": 1 }, { "criterion": "Invokes A * A^T = 1 to get the Kronecker delta, splitting the complex identity into a real part (= delta) and a vanishing imaginary part.", "id": "r3", "weight": 1 }, { "criterion": "Explains the orientation point: real orthonormal rows are exactly the Gram matrix of complex input-bond columns, so orthonormality transfers verbatim.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The (a,c) entry of the Gram product is a sum over outputs of conj(entry at a) times entry at c.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.complexCore_isometry" ] }, { "claim": "Realness makes conjugation trivial, reducing the sum to the real sum A_{a,out} A_{c,out} = (A * A^T)_{a,c}.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.complexCore_isometry" ] }, { "claim": "A * A^T = 1 gives the Kronecker delta; the real part matches it and the imaginary part is zero, so the complex Gram matrix is the identity.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.complexCore_isometry" ] } ], "source_item_digest": "3f133557d4caeab060a7631ee2641d0019434fd496c4f8fa413895aaf318e981", "source_refs": [ { "anchor": "lines 209-226: def complexCore A := fun out a => (A a out : C); theorem complexCore_isometry (A : Core l r) (hA : A * A.transpose = 1) : (complexCore A).conjTranspose * complexCore A = 1, proved via congrFun hA, Complex.ext on real and imaginary parts of the mul_apply / conjTranspose_apply / one_apply expansion", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/TensorTrainCanonical.lean", "sha256": "b8d1eb0651708ac2496be306e2d4b6f3025f7ea4d35a8c6907de02b89d1d981f" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.TensorTrainCanonical.complexCore_isometry" ], "task_hash": "a787511806f6461accade6977a07384e5a474f0d819cb595b1c6608f7aa38ac4", "task_kind": "proof", "topic": "state_preparation" }, { "assumptions": [ "ranks change inside a fixed padded register, with an explicit active-label predicate per stage", "stages agree with the tensor cores only on active input columns; inactive completions are arbitrary" ], "context_hash": "406a8d7283eb3f2bf194df8845a14e74448229bda6c28a59f59257dd8996e057", "conventions": [ "active a means the predicate holds for bond label a at that stage; inactive labels are the complement", "transfer is the chronological contraction of the cores" ], "core_id": "qcb-core-sp-012", "difficulty": { "level": "advanced", "rationale": "Requires the support-propagation induction and the insight that inactive completion columns are gauge freedom for the emitted state though not for unitarity." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "rank-change-support-gauge", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-7670f03c241e05e1459d", "key": "qcb-core-sp-012", "primary_sources": [], "question": "In the sequential bond construction with rank changes inside a fixed padded register, a reader claims: 'If the completion columns of the stages (the inputs for inactive, zero-padded bond labels) are chosen differently, the emitted state could change, so the construction is ambiguous.' Refute this using the support-propagation argument, and state precisely the minimal hypotheses under which the emitted state is uniquely the tensor contraction regardless of the inactive completion.", "reference_answer": "The claim is false. The emitted state depends only on amplitudes carried on ACTIVE bond labels, and inactive labels provably carry zero amplitude throughout the run, so columns acting on inactive inputs multiply zero and never contribute. Concretely, assume: (initial support) the boundary vanishes on inactive labels at stage 0; (core support) each tensor core is zero outside the next active label set; and (clean columns) each stage equals the tensor core only on active input columns. Then one proves, by induction, that the run has no leakage: at any intermediate stage the amplitude on an inactive label is zero. The induction splits the step sum over bond labels a into active and inactive: on active a the clean-column hypothesis supplies the correct core entry, while on inactive a the inductive support invariant makes the prior amplitude zero, killing the term regardless of what the stage's inactive column contains. This yields the rank-changing identity run = (transfer K n x) applied to boundary with no constraint on inactive completion columns. Because the right-hand side involves only the cores K and the boundary, two stage families that agree on active columns but differ arbitrarily on inactive completion columns produce the SAME emitted state. Minimal hypotheses: initial support, core support, and active-only clean-column equality. Under these, the emitted state is uniquely the tensor contraction, so the inactive completions are genuine gauge freedom, not a source of ambiguity. (The completions still matter for full unitarity of each stage, but not for the emitted amplitudes.)", "rubric": [ { "criterion": "States that only active-label amplitudes drive the emitted state and that inactive labels carry zero amplitude throughout.", "id": "r1", "weight": 1 }, { "criterion": "Gives the three hypotheses (initial support, core support, active-only clean columns) and the no-leakage induction splitting the step sum into active/inactive.", "id": "r2", "weight": 1 }, { "criterion": "Shows inactive terms vanish because prior amplitude is zero there, independent of the stage's inactive column entries.", "id": "r3", "weight": 1 }, { "criterion": "Concludes run = transfer applied to boundary with no inactive-column constraint, so differing completions give the same state (gauge freedom), refuting the ambiguity claim.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Under initial support, core support, and active-only clean columns, the run has no leakage: inactive-label amplitudes are zero at every stage.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_supported", "declaration:QuantumBlockEncoding.SequentialBondPreparation.transfer_boundary_supported" ] }, { "claim": "The step sum splits into active terms (fixed by clean columns) and inactive terms (killed by the zero prior amplitude), so inactive stage columns never contribute.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_apply", "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_eq_transfer_of_supported" ] }, { "claim": "Hence run equals the tensor contraction applied to the boundary with no inactive-column constraint, so differing completions give the identical emitted state.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_eq_transfer_of_supported" ] } ], "source_item_digest": "071ab0623ad314e9dec69f07d951817d7eba436b0cdbfe21a38c52f4ec7b8e30", "source_refs": [ { "anchor": "lines 125-174: theorem transfer_boundary_supported (initialSupport, coreSupport) (inactive : not active n b) : (transfer K n x).mulVec boundary b = 0; theorem run_eq_transfer_of_supported (initialSupport, coreSupport, cleanColumns : for all t bit b a, active t a -> U t (bit,b) (0,a) = K t (bit,b) a) : run U boundary n (x,b) = (transfer K n x).mulVec boundary b; theorem run_supported : run U boundary n (x,b) = 0 when inactive", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SequentialBondPreparation.lean", "sha256": "ad4542b108d44f196f2a02f78e59b58f43d45b1e890709cf4c22e1222805d5c5" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_apply" ], "target_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_eq_transfer_of_supported", "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_supported", "declaration:QuantumBlockEncoding.SequentialBondPreparation.transfer_boundary_supported" ], "task_hash": "f4c67b740c47b4c5325831da68cdd8b960696985035e9874f675429b499af9e2", "task_kind": "counterexample", "topic": "state_preparation" }, { "assumptions": [ "the stage is a genuine matrix-vector action on the fresh-bit-and-bond register, followed only by basis regrouping", "no claim is made that the stage's clean columns implement any particular tensor core" ], "context_hash": "48cde0f747fd7076bbf60f2c0d4b5eb191939c210a9048da18c2126bc937caca", "conventions": [ "the fresh output bit is appended at the highest wire; already-emitted bits are passive", "amplitude mass is the sum of star(amplitude) times amplitude over all basis labels" ], "core_id": "qcb-core-sp-002", "difficulty": { "level": "intermediate", "rationale": "Combines the clean-column collapse with the norm-preservation fact to reject a tempting postselection reading; needs two independent source facts." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "sequential-bond-input-promise", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-fa62bbbddde61635a3b8", "key": "qcb-core-sp-002-textbook-v1-index-repair", "primary_sources": [], "question": "A sequential bond-dimension construction prepares a state one output bit at a time: each stage acts on a freshly introduced bit set to zero together with the current bond register, appending the new bit at the highest wire while previously emitted bits stay passive. (i) Which input columns of a stage's local matrix can actually influence an emitted amplitude, and (ii) does introducing the fresh zero bit amount to a postselection or measurement?", "reference_answer": "(i) Let x denote the full emitted bit string after this stage, let c be its newly appended highest-wire bit, let p be the previous prefix obtained by removing that last bit, let b be the output bond label, and let a range over the prior input bond labels. The emitted amplitude is step(U,v)(x,b) = sum_a U((c,b),(0,a)) * v(p,a). The fresh input bit is fixed to zero, so the fresh-bit-input-1 columns multiply zero and cannot affect this clean-input action. The output bond b is free; only the prior bond a is summed. These unused columns are completion data not determined by the clean-input action, but they are not arbitrary if an additional admissibility condition such as unitarity is imposed. (ii) For the padding operation alone, freshZero(v)(p,(c,a)) is v(p,a) when c=0 and zero when c=1. Summing conjugate(amplitude) times amplitude over p,c,a therefore gives exactly the prior sum over p,a. This introduces a fresh zero register deterministically, with no measurement, discarded outcome weight, postselection or renormalization. Norm preservation of this padding does not by itself assert norm preservation of an arbitrary full stage matrix U; the latter requires an appropriate condition such as unitarity.", "rubric": [ { "criterion": "Identifies that only fresh-bit-input-0 columns affect the clean-input action; fresh-bit-input-1 columns are irrelevant to that action, not unconstrained under additional matrix admissibility requirements.", "id": "r1", "weight": 1 }, { "criterion": "States the collapsed amplitude with free output bond b and summed input bond a: sum_a U((c,b),(0,a))*v(p,a), with p the old prefix and c the newly appended highest-wire bit.", "id": "r2", "weight": 1 }, { "criterion": "Shows that fresh-zero padding preserves total squared-modulus mass by the fresh-bit 0/1 sum; does not claim arbitrary U is norm preserving.", "id": "r3", "weight": 1 }, { "criterion": "Concludes the padding is deterministic register enlargement, not postselection or measurement, with no discarded weight or renormalization.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The fresh bit starts in zero, so after padding only the fresh-bit-input-zero columns of the stage multiply nonzero prior amplitudes.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.freshZero", "declaration:QuantumBlockEncoding.SequentialBondPreparation.step" ] }, { "claim": "The emitted amplitude collapses to a sum over the prior bond label of clean-column entries times prior amplitudes; other columns never contribute.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_apply" ] }, { "claim": "Adding the zero bit leaves the total squared-modulus mass unchanged, so it is not a postselection or measurement.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_freshZero" ] } ], "source_item_digest": "913146f35e58a18dbdca2df520ed09bfef9b31fc3bddfaf331881d33a6cb06bf", "source_refs": [ { "anchor": "lines 39-68: def freshZero v := fun x => if x.2.1 = 0 then v (x.1, x.2.2) else 0; noncomputable def step U v := fun x => ((liftStage n U).mulVec (freshZero v)) (Fin.init x.1, (x.1 (Fin.last n), x.2)); theorem step_apply : step U v (x, b) = sum a, U (x (Fin.last n), b) (0, a) * v (Fin.init x, a); and lines 302-306 theorem amplitudeMass_freshZero : amplitudeMass (freshZero v) = amplitudeMass v", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SequentialBondPreparation.lean", "sha256": "ad4542b108d44f196f2a02f78e59b58f43d45b1e890709cf4c22e1222805d5c5" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step", "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_apply", "declaration:QuantumBlockEncoding.SequentialBondPreparation.freshZero", "declaration:QuantumBlockEncoding.SequentialBondPreparation.amplitudeMass_freshZero" ], "task_hash": "1f436afe4aab304d6508230d1475e704c28149f127e27c1729114a23c736220e", "task_kind": "concept", "textbook_successor": { "predecessor_item_digest": "7f1b2a1988e00771f6ae01653a655d04982f664f22fab24482075424eb8cbed5", "predecessor_key": "qcb-core-sp-002", "repair_record": "shared/repairs/sp-002-index-repair.json", "repair_record_digest": "427b454740badec4b182fabaef2468f74c2a27c2779f396d2541a520136453aa", "review_digest": "86ea63ca682f4098c9fd71863a65a6f6f3f957c28e20d3d5fa9ded1f06195b61", "review_record": "reports/textbook-v1/repairs/sp-002-index-repair-review.json" }, "topic": "state_preparation" }, { "assumptions": [ "the total squared norm S is strictly positive", "entries are real; normalization is by division by sqrt(S)" ], "context_hash": "bb5c27a5a112ee8427a6f9735af891e4ccd6be13b068d31fd2de938cb6dbf63a", "conventions": [ "normSq is the sum of squared entries; sqrt is the real nonnegative square root" ], "core_id": "qcb-core-sp-006", "difficulty": { "level": "basic", "rationale": "A direct algebraic normalization proof whose only subtlety is isolating where nonvanishing of the norm is required." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "normalization-identity", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-db9c364791c82872f5dc", "key": "qcb-core-sp-006-textbook-v1-normalization-boundary-repair", "primary_sources": [], "question": "Let f be a finite real amplitude table whose total squared Euclidean norm S = sum_b f(b)^2 is strictly positive. Prove that dividing every entry by sqrt(S) yields an exactly normalized state, i.e. sum_b (f(b)/sqrt(S))^2 = 1, and identify exactly where strict positivity of S is used.", "reference_answer": "Let S = sum_b f(b)^2 > 0. The common denominator is independent of b, so sum_b (f(b)/sqrt(S))^2 = (sum_b f(b)^2)/(sqrt(S))^2 = S/S = 1. The square-root identity (sqrt(S))^2 = S uses nonnegativity of S, which already follows from its being a finite sum of real squares. Strict positivity is needed for nonvanishing: sqrt(S) is nonzero and cancellation S/S = 1 is justified. It does not require the individual f(b) to be nonnegative. At S=0 every entry is zero, so there is no normalized state proportional to that table. Ordinary division by zero is undefined; in Lean's total real division, 0/0 is defined as zero, and the corresponding sum is zero rather than one. This norm identity alone is not a first-column correctness theorem for a particular preparation circuit. The library's prepareCircuit_firstColumn additionally requires every f(b) to be nonnegative. For example, the zero-qubit table with its sole entry -1 has positive S and normalizes to -1, while the empty preparation circuit has first column +1. Thus one must not transfer the circuit theorem to arbitrary signed real tables under the assumptions of this question.", "rubric": [ { "criterion": "Factors the constant squared denominator out of the finite sum to obtain S/(sqrt(S))^2.", "id": "r1", "weight": 1 }, { "criterion": "Uses (sqrt(S))^2 = S from S >= 0, independently of signs of individual entries.", "id": "r2", "weight": 1 }, { "criterion": "Uses S > 0 to ensure the nonzero denominator and S/S = 1; identifies nonvanishing as the load-bearing condition.", "id": "r3", "weight": 1 }, { "criterion": "Keeps the claim at normalization of the real table, without requiring or asserting arbitrary-signed circuit loading. If discussing S=0, distinguishes absence of a normalized proportional state from Lean's total division. No circuit discussion is required for full credit.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Factoring the common squared denominator out of the finite sum gives S/(sqrt(S))^2.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normSq" ] }, { "claim": "The square-root identity for nonnegative S reduces this expression to S/S.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normalized_sum_sq" ] }, { "claim": "Strict positivity makes S nonzero, hence S/S = 1, for arbitrary real entries. This algebraic conclusion does not remove the separate nonnegative-entry premise of prepareCircuit_firstColumn.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normalized_sum_sq", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.prepareCircuit_firstColumn" ] } ], "source_item_digest": "80d8cf9a6ab1179417909a57f88236985358dda61923d365c37edbbe5f4df69d", "source_refs": [ { "anchor": "lines 310-317: theorem normalized_sum_sq (f : PrimitiveBasis n -> R) (positive : 0 < normSq f) : (sum b, (f b / Real.sqrt (normSq f)) ^ 2) = 1, proved by simp_rw [div_pow]; rw [Finset.sum_div]; Real.sq_sqrt (normSq_nonneg f); div_self (ne_of_gt positive)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/RealAmplitudePreparation.lean", "sha256": "a6ded7e65a7e42d5724529befb48eefc9505814136206dd96248fb59b3c45efa" } ], "split": "test_public", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.prepareCircuit_firstColumn" ], "target_nodes": [ "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normalized_sum_sq", "declaration:QuantumBlockEncoding.RealAmplitudePreparation.normSq" ], "task_hash": "e3baa20abda38cc9c392803ecfb44a26e6f63424821ab3efe7575763444b3b77", "task_kind": "proof", "textbook_successor": { "predecessor_item_digest": "44c37085b4bd6ee2044b283002c7cb3d2067205bc96c3b84b5cb04a60a4002a6", "predecessor_key": "qcb-core-sp-006", "repair_record": "shared/repairs/sp-006-normalization-boundary-repair.json", "repair_record_digest": "fafbf96e8efc8d116beffec203eb79caf60dc5d551b63df44e3baca73cb3b5a3", "review_digest": "ee93b57c7dea37159597a7c3f2d5ac86f81f09019452d70a62da62acad058f6f", "review_record": "reports/textbook-v1/repairs/sp-006-normalization-boundary-repair-review.json" }, "topic": "state_preparation" }, { "assumptions": [ "the physical bond register is padded strictly larger than the active bond, with active dimension greater than two", "the full terminal stage is required to be unitary" ], "context_hash": "a29aa7e72aec0656c4f2200d8ebb526b02b725bd014bffb94f1bd2ba90ffb2ba", "conventions": [ "terminal core sends inputs to a single clean output bond label carrying a 2-row emitted-bit block", "active columns are the occupied input bond labels; inactive columns are free completion data" ], "core_id": "qcb-core-sp-011", "difficulty": { "level": "advanced", "rationale": "A genuine impossibility argument from a dimension-count against unitarity, paired with recognizing the exact restricted hypothesis that repairs it." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "terminal-cleanup-unitarity-obstruction", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-851f7fc3e0bce3ae8d2e", "key": "qcb-core-sp-011-textbook-v1-terminal-feasibility-repair", "primary_sources": [], "question": "In the sequential bond construction with a padded physical bond register larger than the active bond, a tempting simplification is to demand that the terminal cleanup stage send EVERY padded input column (not just the active ones) into the single clean output label. Show by a dimension/orthogonality argument that for an active bond dimension greater than two this demand is incompatible with the stage being unitary, and state the correct restricted condition the library uses instead.", "reference_answer": "Write D for the padded input bond dimension and k for the number of active input bond basis labels. The fresh input bit is zero, so the constrained columns of the stage U are indexed by (0,a). Requiring terminal cleanup confines each such column to the output subspace H spanned by (0,clean) and (1,clean), which has dimension two. If U is unitary, these columns must be orthonormal. Therefore no more than two columns can satisfy this action. An all-padded-column requirement is impossible whenever D > 2. In the question k > 2 and D > k, so this obstruction certainly applies. The library's step_terminalCore_of_supported instead assumes (1) v(p,a)=0 for every inactive label a and every prefix p, and (2) U(bit,b)(0,a)=terminalCore(clean,T)(bit,b)(a) only for active a. The inactive summands then vanish in step_apply, yielding step(U,v)(x,b)=0 when b differs from clean and sum_a T(lastBit(x),a)*v(prefix(x),a) when b=clean. This is a conditional state-action identity; it does not assume or prove that U is unitary or construct such a U. Crucially, under the stated k > 2 assumption, even the active-only condition is incompatible with unitarity: it still forces k > 2 orthonormal columns into H. Thus active-only restriction is not a feasible unitary remedy for this question's setting. A physically feasible terminal step of this form must also have k <= 2 and orthonormal prescribed active columns (equivalently the active columns of T have Gram identity); then those finitely many columns can be completed to a unitary on the full finite-dimensional stage space. This last linear-algebra existence observation is not a circuit synthesis or resource theorem. Without changing the stated k > 2 setting there is no such unitary terminal cleanup, and a correct answer must say so rather than claiming the support condition solves the dimension obstruction.", "rubric": [ { "criterion": "Uses distinct symbols D and k for padded and active input dimensions and identifies the two-dimensional output span (bit,clean). The columns in question are the fresh-input-bit-zero columns (0,a).", "id": "r1", "weight": 1 }, { "criterion": "Uses unitarity to force column orthonormality and proves all-padded cleanup impossible for D > 2.", "id": "r2", "weight": 1 }, { "criterion": "States exactly the active-only column identity plus zero inactive-support condition, and derives clean output factorization as a conditional state-action result, not a unitary-existence result.", "id": "r3", "weight": 1 }, { "criterion": "Explicitly recognizes that k > 2 also makes active-only unitary cleanup impossible. Does not claim a feasible remedy under the unchanged assumptions; any discussion of feasible alternatives must disclose k <= 2 and the Gram/isometry condition, without a circuit-cost claim.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Each terminal-action column lies in the two-dimensional output span indexed by emitted bit and the clean bond label.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_terminalCore" ] }, { "claim": "Unitary columns are orthonormal, so requiring this action on more than two padded or active input columns is impossible.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_terminalCore" ] }, { "claim": "Active-only terminal equality plus zero inactive support yields a conditional clean state-action factorization. It does not prove unitarity or its feasibility, and cannot yield a unitary in the question's active-dimension-greater-than-two case.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_terminalCore_of_supported", "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_terminal_clean" ] } ], "source_item_digest": "3dc8ebd018ba2a658d973683c272426bb6cf72ff07f489e20ed26fdb8a249818", "source_refs": [ { "anchor": "lines 201-222: comment 'Requiring this terminal action on all padded input columns would be incompatible with unitarity for bond dimension greater than two.'; theorem step_terminalCore_of_supported (support : for all p a, not active a -> v (p,a) = 0) (cleanColumns : for all bit b a, active a -> U (bit,b) (0,a) = terminalCore clean T (bit,b) a) : step U v (x,b) = if b = clean then sum a, T (x (Fin.last n)) a * v (Fin.init x, a) else 0", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SequentialBondPreparation.lean", "sha256": "ad4542b108d44f196f2a02f78e59b58f43d45b1e890709cf4c22e1222805d5c5" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.run_terminal_clean" ], "target_nodes": [ "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_terminalCore_of_supported", "declaration:QuantumBlockEncoding.SequentialBondPreparation.step_terminalCore" ], "task_hash": "f32459794bd9854cfb9a138ed4c381873857d91d9a312991fd7576089a0042e9", "task_kind": "counterexample", "textbook_successor": { "predecessor_item_digest": "8e29ce90772af81328f7d4b8e817f6183b46c52d96801e394ac4287c4be24a6f", "predecessor_key": "qcb-core-sp-011", "repair_record": "shared/repairs/sp-011-terminal-feasibility-repair.json", "repair_record_digest": "dbc81770f93ebf540ba98fa159e7b3552d4722fa04143ffd78163c75aa64457d", "review_digest": "26456a1e859051ac81eaa8a067abf7e48f04d6c643527c237aa7023d01423710", "review_record": "reports/textbook-v1/repairs/sp-011-terminal-feasibility-repair-review.json" }, "topic": "state_preparation" }, { "assumptions": [ "the environment assigns a rational to every symbol name" ], "context_hash": "695cd5f749e71a9b3b0647cd1e5c65a826aee44d754d5e5da3f7d73dcdf45bdd", "conventions": [ "coefficients are symbolic expressions over the rationals", "evaluation is relative to a fixed environment" ], "core_id": "qcb-core-la-003", "difficulty": { "level": "easy", "rationale": "Recall of a recursive evaluation and recognition of the homomorphism property." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "symbolic algebra evaluation homomorphism", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-b4c7f9d4d3176563d728", "key": "qcb-core-la-003-textbook-v1-syntax-semantics-repair", "primary_sources": [], "question": "A stencil coefficient is written in a small symbolic language whose expressions are built from rational constants, named symbols, and the operations addition, multiplication, and negation. Given an environment that assigns a rational number to each symbol name, the coefficient is turned into a concrete rational number by a single evaluation map. State how this map treats each of the five kinds of expression, and state precisely which algebraic structure the evaluation preserves (that is, which equalities relate the value of a compound expression to the values of its parts).", "reference_answer": "Fix an environment env:String->Rat and write ev for evalWith env. The five recursive clauses are ev(rat(q))=q, ev(symbol(s))=env(s), ev(add(a,b))=ev(a)+ev(b), ev(mul(a,b))=ev(a)*ev(b), and ev(neg(a))=-ev(a). In particular the constant expressions rat(0) and rat(1) evaluate to rational 0 and 1. Every rational q is attained by rat(q), so this fixed-environment interpretation is surjective.\n\nThese equations say exactly that evaluation preserves the specified expression constructors as rational operations: it is an interpretation, or homomorphism of the corresponding operation-signature term algebra. They do not make the raw inductive Coeff syntax a ring or a rational algebra under those constructors. For example, add(rat(0),rat(0)) and rat(0) are different syntax trees even though both evaluate to zero. Ring laws such as x+0=x hold for the evaluated rational values, not as literal equality of these trees. To speak of a ring homomorphism one would first need an appropriate semantic quotient or another separately defined ring of expressions satisfying the ring laws; the raw source definition supplies neither that quotient nor such a ring instance. A symbolic rewrite is justified when its evaluations agree under the environments in scope, not merely because an arbitrary syntactic identity has been asserted. The five displayed equations are the exact preserved structure requested here.", "rubric": [ { "criterion": "States constant and symbol evaluation exactly: ev(rat(q))=q and ev(symbol(s))=env(s), with env fixed.", "id": "r1", "weight": 1 }, { "criterion": "States all three recursive preservation identities for add, mul and neg, including the correct rational operations on evaluated subexpressions.", "id": "r2", "weight": 1 }, { "criterion": "Identifies precisely this constructor-preserving interpretation and 0/1 preservation. Distinguishes raw syntax from rational ring equality instead of claiming an unqualified ring/algebra homomorphism from constructor syntax. If surjectivity is claimed, justifies it using rat(q); no quotient construction is required.", "id": "r3", "weight": 1 } ], "semantic_step_count": 2, "semantic_steps": [ { "claim": "Rational constants evaluate to themselves, symbols to the fixed environment's values; rat(0) and rat(1) give rational zero and one.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Coeff.evalWith_rat", "declaration:QuantumBlockEncoding.Coeff.evalWith_symbol" ] }, { "claim": "Evaluation commutes with add, mul and neg as rational operations. This interprets raw constructor syntax; it does not assert that the syntax carrier itself satisfies ring equations.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Coeff.evalWith_add", "declaration:QuantumBlockEncoding.Coeff.evalWith_mul", "declaration:QuantumBlockEncoding.Coeff.evalWith_neg" ] } ], "source_item_digest": "9cf3d03948eddb4134368e2d31a81f2605928c2acee82aaae66e8f7bd5302a77", "source_refs": [ { "anchor": "declarations Coeff.evalWith (lines 175-181) and the evalWith_rat/evalWith_symbol/evalWith_add/evalWith_mul/evalWith_neg simp lemmas (lines 183-196)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Core.lean", "sha256": "c7e76d91c7fcfd844177f99c0399a3465a7d8360d3a522acb5d4c4e2f8277894" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Coeff.evalWith_rat", "declaration:QuantumBlockEncoding.Coeff.evalWith_symbol" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Coeff.evalWith", "declaration:QuantumBlockEncoding.Coeff.evalWith_add", "declaration:QuantumBlockEncoding.Coeff.evalWith_mul", "declaration:QuantumBlockEncoding.Coeff.evalWith_neg" ], "task_hash": "fc08a7ea66752f3f3f39e8804e459f26bad37b706cc8ea1ba6d3ff31c294d9ec", "task_kind": "concept", "textbook_successor": { "predecessor_item_digest": "33122c5bd9eba24d73f1abe0de97a9f05e8a48e6e2bdcb636e8e2ac68388a5e9", "predecessor_key": "qcb-core-la-003", "repair_record": "shared/repairs/la-003-syntax-semantics-repair.json", "repair_record_digest": "1bf531ffaa22350d6c4a2d855728a260eca2b6d1c2ba88e7e94cc12a426c1c11", "review_digest": "19a9943a100bea461bdfdc6eb862c9ef0652e2d045e1ecd4e8892c9e9f096ba0", "review_record": "reports/textbook-v1/repairs/la-003-syntax-semantics-repair-review.json" }, "topic": "linear_algebra" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-sp-105", "difficulty": { "level": "intermediate", "rationale": "Requires constructing the minimal nonzero element of the vanishing ideal and reasoning about coset dimension, not merely citing uniqueness." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "hermite-interpolant-uniqueness", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-b923b7a43f74b99e67a1", "key": "qcb-core-sp-105-textbook-v1-affine-dimension-repair", "primary_sources": [], "question": "A colleague claims that the two-point jet data 'all derivatives through order k agree at -1 and at 0' pins down a real polynomial uniquely, with no restriction on its degree. Show this is false by exhibiting an explicit nonzero polynomial of degree 2k+2 all of whose derivatives through order k vanish at both -1 and 0. Conclude that the degree bound 2k+1 in the uniqueness statement cannot be dropped, and state the exact dimension of the space of polynomials sharing a fixed set of such endpoint jets.", "reference_answer": "Let k be a nonnegative integer and set E(x)=(x+1)^(k+1)*x^(k+1). This is monic of degree 2k+2, hence nonzero. Its zero of multiplicity k+1 at each endpoint makes every derivative of order 0 through k vanish at -1 and at 0. Thus if p is any solution to given endpoint data, p+E is a distinct solution with the same data: uniqueness fails without a degree restriction.\n\nFor the exact solution space, first choose the degree-at-most-2k+1 Hermite interpolant p0 of the fixed data. Such a p0 exists for arbitrary real endpoint jets: use the Taylor coefficients v_j/j! at the two distinct nodes and the Chinese remainder theorem for the coprime moduli (x+1)^(k+1) and x^(k+1). A polynomial p has those same jets exactly when p-p0 is divisible by both moduli, equivalently by their product E. Therefore the complete solution set is the affine coset p0+E*R[x]. Its translation space E*R[x] has the countably infinite real basis {E*x^m : m=0,1,2,...}: it spans because every polynomial multiplier has finitely many monomials, and it is linearly independent because E is nonzero and R[x] is an integral domain. The unrestricted real affine dimension is consequently countably infinite, not a finite number and not the cardinality of the set of real coefficients. In general the solution set is affine rather than a vector subspace; it is a vector subspace precisely for all-zero prescribed jets.\n\nFor comparison, if the degree is capped at 2k+1+d with an integer d>=0, then the possible multipliers of E have degree =0; this dimension reasoning is an NL extension, not a new compiled Lean theorem.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.sourceInterpolant_unique" ] } ], "source_item_digest": "7a9a04f7d42afb7e8df78d995ac2f8c887d0e54dd1ce5d16e5bf558014d278c4", "source_refs": [ { "anchor": "lines 119-166: leftModulus k = (X-C(-1))^(k+1), rightModulus k = (X-C 0)^(k+1); interpolant degree <= 2k+1; product (leftModulus*rightModulus).natDegree = 2k+2 used in uniqueness degree count. lines 394-417: sourceInterpolant_unique requires hp : natDegree <= 2k+1", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/HermitePolynomial.lean", "sha256": "f8a3e1d5f29aed1228b6ede3577a9af1e76301559172ee67a266fcb0d56fee1e" } ], "split": "train", "supporting_nodes": [], "target_nodes": [ "declaration:QuantumBlockEncoding.HermitePolynomial.endpointModuli_coprime", "declaration:QuantumBlockEncoding.HermitePolynomial.sourceInterpolant_unique" ], "task_hash": "2bf0cf630eddd3f71451b9cab5ce567a2be95e25f29dadd02762bba72b4047c7", "task_kind": "counterexample", "textbook_successor": { "predecessor_item_digest": "fd8583539b598362720ad73856121d268dbc99230408318a2d7a311621b422dc", "predecessor_key": "qcb-core-sp-105", "repair_record": "shared/repairs/sp-105-affine-dimension-repair.json", "repair_record_digest": "3e6bc43d7c1c1f10a8a83c16b1fa833dcdb3f8a55662ec5273d302f520b1063f", "review_digest": "4819b114fed21a7058e879f3f5bda48c3efd08126aefc3edb1b45c633d772a8e", "review_record": "reports/textbook-v1/repairs/sp-105-affine-dimension-repair-review.json" }, "topic": "structured_preparation" }, { "assumptions": [ "Materialization charges supplier work plus a dimension-fixed storage surcharge", "Suppliers may perform data-dependent amounts of counted work" ], "context_hash": "e86938ab977c4e9e5d7aa7f1c11b0c2b5ee9537444288e54364c163dd92779b5", "conventions": [ "The input callback cost is an explicit boundary not certified by the constructor; counts are exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-214", "difficulty": { "level": "intermediate", "rationale": "A counterexample isolating the supplier-dependent term from the dimension-fixed storage surcharge." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-214", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-c2bb67ca8a9595e17a99", "key": "qcb-core-la-214-textbook-v1-uniform-supplier-bound-repair", "primary_sources": [], "question": "A stored N-by-M matrix is built by materializing an entry supplier: for each of the N*M entries the supplier is run to produce a value, and the construction charges, on top of the supplier's own counted work, two reads and two writes per entry plus two reads and two writes per row to store and project. The proved cost identity is that the materialized matrix costs the double sum of the supplier's per-entry costs plus (N*M + N) times the two-read-two-write surcharge. A colleague claims this identity proves that building any N-by-M stored matrix always costs a fixed polynomial in N and M alone. Give a precise counterexample showing the claim is false, and state the exact condition under which a clean polynomial-in-dimensions bound does hold.", "reference_answer": "The claimed implication is false. The issue is a uniform bound over the allowed suppliers and their inputs, not whether two suppliers can have different costs. Fix N=M=1. For each natural number K, let the sole entry supplier perform exactly K charged field additions, adding zero to a stored accumulator each time, and then return zero. This is a concrete terminating counted computation for every K, and all these suppliers return the same one-by-one zero matrix. Its field cost is K. The storage surcharge contributes no field operations, so the materialized result also has field cost K. Any proposed polynomial p(N,M) that is to bound every allowed supplier has the fixed value p(1,1). Choosing a natural K greater than that value contradicts the bound. An unbounded data parameter controlling the same loop gives the same obstruction for one supplier family at fixed dimensions. No growth in the dimensions is needed for this counterexample.\n\nFor each counter op the source identity is C_total(op)=S(op)+(N*M+N)*s(op), where S(op) is the double sum of entry-supplier costs and s(op)=2*tick(read)(op)+2*tick(write)(op). Since costs are nonnegative, a uniform polynomial-in-dimensions bound on C_total over a specified supplier/input class exists exactly when such a bound on S exists: S<=C_total, and the remaining surcharge is already polynomial. A convenient sufficient interface is a bound cost(f(i,j),op)<=B_op(N,M) for every entry and every admitted input, where B_op itself is bounded by a fixed polynomial in N,M uniformly over that class. Then C_total(op)<=N*M*B_op(N,M)+(N*M+N)*s(op). The zero-entry case has S=0 and only the stated surcharge. A bound written using an unrestricted additional parameter B is merely conditional; it is not a dimensions-only polynomial theorem until B is itself controlled in terms of the dimensions. Per-entry uniformity is a useful interface, not a claim that every particular supplier has constant cost.\n\nThe old example costing j operations per entry gives the polynomial N*M*(M-1)/2 and therefore does not refute the colleague's assertion of some polynomial upper bound. Nor does observing different costs at equal dimensions refute an upper bound. The arbitrary-K construction supplies the missing quantifier argument. All counts here belong to the declared exact-real operation model; they do not establish a finite-bit runtime bound.", "rubric": [ { "criterion": "Gives a genuinely unbounded supplier/input family at fixed positive dimensions, or another rigorous counterexample defeating every claimed uniform dimensions-only polynomial. Explains why a merely varying polynomial cost such as j per entry is insufficient.", "id": "c1", "weight": 1 }, { "criterion": "Uses the exact additive identity and separates nonnegative supplier work S from the dimension-fixed read/write surcharge. The concrete fixed one-by-one K-addition example has materialized field cost K.", "id": "c2", "weight": 1 }, { "criterion": "States the necessary-and-sufficient aggregate condition that S be uniformly polynomial over the specified supplier/input class, and the sufficient per-entry polynomial interface yielding N*M*B plus surcharge. Does not call an uncontrolled parameter B a dimensions-only polynomial bound or claim finite-bit certification.", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "The exact materialization identity retains the nonnegative double sum S of supplier per-entry costs, plus the specified dimension-fixed storage surcharge.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.materialize_cost" ] }, { "claim": "At N=M=1, a supplier returning zero after K charged additions gives field cost K for arbitrary K. This defeats a bound uniform over all suppliers at fixed dimensions; a j-cost polynomial example alone would not.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.materialize" ] }, { "claim": "A uniform polynomial bound on total supplier work S is necessary and sufficient after adding the polynomial surcharge. A uniform polynomial per-entry bound B(N,M) is a sufficient interface giving N*M*B plus surcharge.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.materialize_cost" ] } ], "source_item_digest": "c6459088608c600882dee195ea922797caf27d36a904e26ea3e2a0e0028f224a", "source_refs": [ { "anchor": "materialize75-78; materialize_cost369-374", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredGivens.lean", "sha256": "87a1f69d5b138f23c96a6f77a438aa81edef27427cb2581c4defe019079d2ee4" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.materialize", "declaration:QuantumBlockEncoding.StoredTensorTrainNorm.update_cost_le" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.materialize_cost" ], "task_hash": "dc04662395dcadd4a5561e77f41d870ef2027346a4afd75ea9c373cc53a5405a", "task_kind": "counterexample", "textbook_successor": { "predecessor_item_digest": "c676dd23f7a68be8627059eb6e12cfefe9910f62c9a635b3495675187dfae459", "predecessor_key": "qcb-core-la-214", "repair_record": "shared/repairs/la-214-uniform-supplier-bound-repair.json", "repair_record_digest": "ce7d672987cfbd1a176bae4bd3bfeb8ae9da448b0ed80c73d7a5303761d98bb0", "review_digest": "89a325bec0a4abff3223b7fdba84237fcb1f45f00d42b03512ed2bfde8002b83", "review_record": "reports/textbook-v1/repairs/la-214-uniform-supplier-bound-repair-review.json" }, "topic": "stored_cost_refinement" }, { "assumptions": [ "The angle routine returns zero early when the computed norm is zero", "The angle budget bounds every counter for all inputs" ], "context_hash": "c87688cbe20945ad72e303bc885941617c53bbb116888d8978a3cab56c545adc", "conventions": [ "The budget is a branch-independent upper bound, not an exact per-input count; exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-215", "difficulty": { "level": "intermediate", "rationale": "A branch-based counterexample clarifying the inequality-versus-equality status of a cost budget." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-215", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-b47eef925a7f499a6579", "key": "qcb-core-la-215-textbook-v1-branch-cost-repair", "primary_sources": [], "question": "The elimination-angle routine in the exact-real model computes a norm from two inputs x and y: it squares both, adds them, takes a square root, and tests whether that root is zero. If the root is zero it returns the angle zero immediately; otherwise it divides, takes an inverse cosine, tests the sign of y, conditionally negates, doubles, and negates again. A branch-independent angle budget of seven field operations, one square root, one inverse-cosine and two comparisons is proved to be an upper bound on the angle cost for all inputs. A colleague concludes that every call to the angle routine performs exactly one inverse-cosine call. Refute this with a specific input, and explain the precise logical relationship between the proved budget and the actual per-input cost.", "reference_answer": "Take x=y=0. The routine executes two field multiplications and one field addition, one square-root call and one zero comparison. The root is zero, so it returns the literal angle zero immediately. It executes no division, inverse cosine or sign comparison. Thus the inverse-cosine count is zero, disproving the claim that every input makes exactly one such call.\n\nTo distinguish an upper bound from an exact cost, order the relevant counters as (field,sqrt,angle,compare). The zero-root branch has exact tuple (3,1,0,1). On a nonzero-root branch, division, inverse cosine, the sign comparison, doubling and final negation are performed. If y is nonnegative the intermediate sign is passed through without a charged subtraction, so the tuple is (6,1,1,2). If y is negative the extra subtraction 0-a is performed and the tuple is (7,1,1,2). In angle itself all trig/read/write/emit counters are zero. Concrete inputs (1,0) and (0,-1) realize the two respective nonzero cases. A charged subtraction still counts when its numerical result happens to be zero; this is a counted execution trace, not an algebraically optimized expression.\n\nThe branch-independent budget is (7,1,1,2), componentwise, with the other counters zero. Every actual tuple is bounded by it. The full budget is attained on the negative-y nonzero branch, not on every nonzero input: the nonnegative-y nonzero branch uses only six field operations. One inverse cosine is exact for either nonzero branch, whereas the zero branch uses none. Summing componentwise inequalities gives a valid budget for many calls; it does not turn those inequalities into per-input equalities. Source counters that separately prove an exact emit count do not make this angle budget exact. These are exact-real operation counts, not a finite-bit or numerical-stability result.", "rubric": [ { "criterion": "Exhibits x=y=0, traces the early return and obtains zero inverse-cosine calls, with no division or sign test.", "id": "c1", "weight": 1 }, { "criterion": "Explains componentwise upper-bound semantics and distinguishes full-budget equality from inverse-cosine equality. Does not claim the full budget is attained on every nonzero input; the negative-y branch uses seven field operations and the nonnegative-y branch six.", "id": "c2", "weight": 1 }, { "criterion": "Justifies summing the valid cost inequalities without inferring exact per-call counts. Keeps charged exact-real operation counts distinct from simplification, finite-bit runtime and unrelated exact emit theorems.", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "At x=y=0 the early return has (field,sqrt,angle,compare)=(3,1,0,1), so the universal one-inverse-cosine assertion is false.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.angle" ] }, { "claim": "Nonzero-root inputs have (6,1,1,2) for nonnegative y and (7,1,1,2) for negative y; both are bounded by the declared (7,1,1,2) budget, which need not equal the actual tuple.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.angle_cost_le", "declaration:QuantumBlockEncoding.StoredGivens.angleBudget" ] }, { "claim": "Componentwise bounds can be summed across actual calls, without upgrading the budget to an exact per-call arithmetic count or a finite-bit theorem.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.angle_cost_le" ] } ], "source_item_digest": "8853829f9e72f7f223d2931d8d7c112de8c4eaacfc1853f545a2401b7d432dbb", "source_refs": [ { "anchor": "angle88-100; angleBudget253-255; angle_cost_le257-261", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredGivens.lean", "sha256": "87a1f69d5b138f23c96a6f77a438aa81edef27427cb2581c4defe019079d2ee4" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.angle", "declaration:QuantumBlockEncoding.StoredGivens.angleBudget" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredGivens.angle_cost_le" ], "task_hash": "62bd4c7a118278d509084f10e3606781e8eb2ab6fd43b9cda5f14003b5a14ed3", "task_kind": "counterexample", "textbook_successor": { "predecessor_item_digest": "5a8b2818c7e3028baaecd34a3473244b7bc65f2379c41f21d8fbabf8c03de06a", "predecessor_key": "qcb-core-la-215", "repair_record": "shared/repairs/la-215-branch-cost-repair.json", "repair_record_digest": "414a8f8ffaaf04545b35a1af1e917621f9e85b4522fc218304f1ed4bc0da949d", "review_digest": "04938fb0237333e396dd2bee3c87c52d17ae2c31f4242a70d7faffec39200d75", "review_record": "reports/textbook-v1/repairs/la-215-branch-cost-repair-review.json" }, "topic": "stored_cost_refinement" }, { "assumptions": [ "The work wire is a reusable scratch wire borrowed by the modular adder", "The clean-action contract is asserted only under the hypothesis work=0" ], "context_hash": "1ff91f56dd5b20017514537fde144ea483df92b34dae5f3f9f92029a1ac49dfe", "conventions": [ "Wire order is address[0..2], row[0..2], work", "The clean branch is the subspace where the work wire enters at 0" ], "core_id": "qcb-core-be-102", "difficulty": { "level": "intermediate", "rationale": "Requires reading the conjunctive clean-action theorem, isolating three independent guarantees, and reasoning about why a dirty-ancilla input falls outside the certified subspace." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "be-clean-branch-work-hypothesis", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-d39c6f84448343f35db5", "key": "qcb-core-be-102-textbook-v1-dirty-work-boundary-repair", "primary_sources": [], "question": "A fixed seven-wire reversible access circuit stores an address word on wires 0-2, a row word on wires 3-5, and one reusable work wire 6. On any input whose work wire is already 0, the circuit outputs the address word replaced by (loader(address)+row) mod 8, leaves the row word unchanged, and returns the work wire to 0. State precisely why the hypothesis 'work wire starts at 0' is required rather than cosmetic, what three separate guarantees the clean-branch contract bundles, and what can fail if the circuit is run on an input whose work wire starts at 1.", "reference_answer": "The clean-branch contract specifies three different properties of the same output: the address is overwritten by (loader(address)+row) mod 8, the row is unchanged, and the work bit ends at zero. The hypothesis work=0 is needed to guarantee that conjunction for every allowed input; it does not mean that every individual property must fail whenever work=1. In particular, the existing named clean-action theorem supplies only a sufficient clean-input contract. What actually happens outside it must be read from the program, not inferred by negating the hypothesis.\n\nFor this fixed circuit there is a stronger direct source-level fact: the row bits and the entering work bit are preserved on every basis input, clean or dirty. The loader flips only address bits 0 and 1. The first and third adder gates both toggle work bit 6 by the same product of row bit 3 and loaded address bit 0; between them only address bit 2 is changed, so these two work toggles cancel. No later gate targets work. No gate targets row wires 3,4,5 at all. Thus a dirty work bit returns as 1, not as 0: restoration to the stipulated clean state fails, but restoration to its entering value holds. Reversible borrowed-bit preservation and the known-zero clean-work promise are different interfaces.\n\nLet v be loaded address bit 1 and d the entering work bit. Relative to the clean-input computation on the same address and row, the dirty-input trace differs by an extra toggle of address bit 2 precisely when d=1 and v=1. The second adder gate uses the scratch bit before the third gate restores it. All subsequent address updates have controls independent of address bit 2, so this extra toggle persists to the output. Consequently the arithmetic sum can be wrong, but need not be wrong on every dirty input.\n\nTwo exact basis examples distinguish the claims. Start with address3, row0, work1. The XOR3 loader produces address0; the dirty-work high-bit control has v=0, and the output is address0,row0,work1. The required sum is correct while the return-to-zero promise fails. Start instead with address1,row0,work1. The loader produces address2, whose bit1 is1; the output address is6 rather than the required2, with row0 and work1 unchanged. Thus dirty inputs can lose arithmetic correctness and always fail this known-zero restoration promise, while row preservation survives. The whole seven-wire map remains a reversible permutation on all basis inputs. With a known clean quantum input, linear extension of the clean-action map returns a factored |0> work register; preservation of basis labels alone must not be rephrased as a no-entanglement guarantee for arbitrary unknown dirty quantum inputs. These claims concern this finite witness, not an arbitrary-width adder or a new resource bound.", "rubric": [ { "criterion": "Identifies the carry-scratch use of work and the exact clean-input conjunction. Does not infer universal dirty-input failure merely from the clean theorem's missing hypothesis.", "id": "c1", "weight": 1 }, { "criterion": "Separates correct address sum, unchanged row and known-zero work restoration. Derives unconditional row/input-work preservation for this actual circuit from its targets and paired work toggles.", "id": "c2", "weight": 1 }, { "criterion": "Shows what can fail and what still holds for work=1, using a dirty input with correct sum and a dirty input with incorrect sum. Keeps borrowed-value preservation, return-to-zero, reversibility and arbitrary-input entanglement distinct.", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "The clean-action theorem certifies the three-property output conjunction when the entering carry-scratch bit is zero; its converse is not a dirty-input failure theorem.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3_cleanAction" ] }, { "claim": "The certified clean-branch address equals (loader(address)+row) mod8.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3_cleanAction" ] }, { "claim": "The row is unchanged on the clean branch; direct gate-target inspection gives the stronger all-input row preservation for this witness.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3_preserves_row", "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3BasisEquiv" ] }, { "claim": "The paired scratch toggles restore the entering work bit on all basis inputs, so a clean input returns zero and a dirty input returns one.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3_workspaceClean", "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3BasisEquiv" ] }, { "claim": "Dirty address3,row0,work1 retains a correct sum while dirty address1,row0,work1 produces6 instead of2; neither returns a known-zero work bit, and the full map is still reversible.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3BasisEquiv" ] } ], "source_item_digest": "ae8623f2384d5350708cbe0dba53b4b300a965e6f3eb80d4bfde18e331df7839", "source_refs": [ { "anchor": "theorem primitiveAccess3_cleanAction (state : PrimitiveBasis 7) (workClean : state 6 = 0) : let output := primitiveAccess3BasisEquiv state; primitiveWord3 output 0 1 2 = ⟨((primitiveOffset3 (primitiveWord3 state 0 1 2)).val + (primitiveWord3 state 3 4 5).val) % 8, Nat.mod_lt _ (by decide)⟩ ∧ primitiveWord3 output 3 4 5 = primitiveWord3 state 3 4 5 ∧ output 6 = 0 (lines 48-57); theorem primitiveAccess3_preserves_row ... (lines 60-64); theorem primitiveAccess3_workspaceClean ... (lines 67-70)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/BandedSparseAccessPrimitive.lean", "sha256": "e4f35bfdd690445de12d5a7e6692b9ab86d1a096b9ed3f4d1a41e07b99181ca7" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3_preserves_row", "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3_workspaceClean", "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3BasisEquiv" ], "target_nodes": [ "declaration:QuantumBlockEncoding.BandedSparseAccess.primitiveAccess3_cleanAction" ], "task_hash": "9440e3412536170fcdd7f127e2d24acbd690b127c41bc1a7d36fa4199df5534e", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "ccce000279db59f25b24bf92dc5768b74e732a097f2cb4da9eb98c4851df6165", "predecessor_key": "qcb-core-be-102", "repair_record": "shared/repairs/be-102-dirty-work-boundary-repair.json", "repair_record_digest": "e2392b76510772802787bd9cfa7f7d814a04db9173e58655f4fccf2c7a789238", "review_digest": "1be83d6da782444784c317644691597b54806047ece8ae56702a35ff5544143b", "review_record": "reports/textbook-v1/repairs/be-102-dirty-work-boundary-repair-review.json" }, "topic": "oracle_composition" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-107", "difficulty": { "level": "intermediate", "rationale": "Combines a commutation check, Schur's corollary, and a trace normalization; conceptually richer than a coordinate check and tests why the result is rho-independent." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-schur-lemma", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-e404512f1cbf16b0e616", "key": "qcb-core-plan-107-textbook-v1-pauli-adjoint-repair", "primary_sources": [ { "anchors": "Example 3.19 and Eq. (3.27): (1/4)(rho + X rho X + Y rho Y + Z rho Z) = (1/2)1 following from irreducibility of the Pauli representation and Schur's Lemma (Proposition 3.17), cf. Corollary 3.18", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "20-21", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let rho be any single-qubit density matrix (a 2x2 positive semidefinite operator with trace 1). Using the fact that the Pauli operators {1, X, Y, Z} generate an irreducible representation of the single-qubit Pauli group acting on C^2, prove that (1/4)(rho + X rho X + Y rho Y + Z rho Z) = (1/2) 1. Base your argument on Schur's Lemma / the commutant-is-scalars corollary and on the trace, not on a brute-force coordinate computation.", "reference_answer": "Let S={I,X,Y,Z} and T=(1/4) sum over P in S of P*rho*P^dagger. Each of these four matrices is Hermitian and unitary, so this is the operator in the question. For Q in {X,Y,Z}, the conjugation expansion is Q*T*Q^dagger=(1/4) sum_P (Q*P)*rho*(Q*P)^dagger. The rightmost factor is P^dagger*Q^dagger=P*Q, not Q*P. Products of distinct Hermitian Paulis need not themselves be Hermitian, so that adjoint cannot be dropped.\n\nLeft multiplication by Q permutes the four Pauli classes modulo scalar phases. More explicitly, for each P there are a unit-modulus scalar omega_P and a uniquely associated R_P in S such that Q*P=omega_P*R_P, and P->R_P is a permutation. In each conjugation term the scalar cancels with its complex conjugate: (omega_P R_P)*rho*(omega_P R_P)^dagger=R_P*rho*R_P^dagger. Relabeling this finite sum therefore proves Q*T*Q^dagger=T. Since Q is unitary, multiplying on the right by Q gives Q*T=T*Q. The same follows for all products of the generators and scalar central phases, so T commutes with the irreducible Pauli representation.\n\nNow use Schur's commutant-is-scalars corollary on the finite-dimensional complex space C^2: T=lambda*I. The hypotheses needed for its scalar conclusion are present, not silently assumed for a general field or an infinite-dimensional representation. Cyclicity of trace and P^dagger*P=I give tr(P*rho*P^dagger)=tr(rho)=1 for every P. Thus tr(T)=1. On the other hand tr(lambda*I)=2*lambda, so lambda=1/2 and T=I/2. This follows from conjugation invariance, irreducibility and trace normalization, without computing the entries of rho. Positivity supplies the density-state interpretation but is not needed for the linear-operator identity T(A)=tr(A)*I/2. That extension is a mathematical consequence of the same proof, not a newly compiled local Lean theorem or a circuit-resource claim.", "rubric": [ { "criterion": "Expands Q*T*Q^dagger with right factor (QP)^dagger=PQ, then uses a bijection of Pauli classes modulo unit phases and adjoint phase cancellation. Does not treat every product QP as Hermitian.", "id": "r1", "weight": 1 }, { "criterion": "Derives commutation with the whole representation and applies the scalar commutant corollary on finite-dimensional C^2 with the given irreducibility.", "id": "r2", "weight": 1 }, { "criterion": "Uses trace cyclicity and P^dagger*P=I to get tr(T)=1=2*lambda and lambda=1/2.", "id": "r3", "weight": 1 }, { "criterion": "Concludes T=I/2 for every input density matrix by the Schur/averaging argument, not brute-force coordinates or a new formal/circuit claim.", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Each P in {I,X,Y,Z} is Hermitian-unitary, so P*rho*P^dagger=P*rho*P; a scalar phase cancels only against its adjoint phase.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Q*T*Q^dagger=(1/4)sum_P (QP)*rho*(QP)^dagger; QP permutes the Pauli classes modulo unit phases, so the correctly conjugated sum is T.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Unitary Q then gives Q*T=T*Q, and hence commutation with the entire irreducible Pauli representation.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The scalar Schur corollary applies on finite-dimensional C^2 and gives T=lambda*I.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Cyclicity gives tr(T)=1=2*lambda, so T=I/2 independently of the density matrix entries.", "supporting_nodes": [ "concept:matrix-analysis" ] } ], "source_item_digest": "27dbc029f5e7856c2ef078f782564d1e1b35c8c9a283a9394c8617aba39196c0", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "1b5756f54feb3e5f7b9d3a594c210c78bbe153db151fca6e6ad9c54f58261d11", "task_kind": "proof", "textbook_successor": { "predecessor_item_digest": "78bd0325d3ec88da16fbd93359b46049553ad3e088445cbb3f077baccc1e530a", "predecessor_key": "qcb-core-plan-107-pauli-twirl-maximally-mixed", "repair_record": "shared/repairs/plan-107-pauli-adjoint-repair.json", "repair_record_digest": "c6c7b731c68107c1b125b7d98580ee09d3c9db16660e823deb4d5275bc67c8e7", "review_digest": "c12fbdecd508c640ab02ca021efa49d5ac82294b14e717e481cdb3c40e0a5725", "review_record": "reports/textbook-v1/repairs/plan-107-pauli-adjoint-repair-migration-revalidation-review.json" }, "topic": "planned_representation_qit" }, { "assumptions": [ "0 < |c| < 1 (states overlap but are distinct)", "The ambient space is span{|psi>,|phi>}, two-dimensional", "Unambiguity means a conclusive label never fires on the wrong state" ], "context_hash": "2d2639b45a9de46f1950eff03d50c910b22c5d0a3b223c8c861d58a94b73e494", "conventions": [ "|psi_perp>, |phi_perp> are the normalized complements defined in the problem", "A POVM must have PSD elements summing to I on the relevant space" ], "core_id": "qcb-core-plan-204", "difficulty": { "level": "advanced", "rationale": "A genuine impossibility argument: it must turn unambiguity+positivity into a rank/weight constraint and expose a trace contradiction, then show the inconclusive outcome is the exact repair." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-state-discrimination", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-7120d70a933022bc0a07", "key": "qcb-core-plan-204-textbook-v1-complex-conjugation-repair", "primary_sources": [ { "anchors": "Example 9.14; Eq. (9.53) definitions of psi_perp, phi_perp; Eq. (9.54) POVM P_psi,P_phi,P_?", "edition": "29 April 2026", "inspection": "full bounded primary text read", "pages": "152-153", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text or external relicensing", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Two normalized but possibly non-orthogonal pure states |psi> and |phi> are given with overlap c := , |c| < 1, and we are promised the unknown state is one of them. Consider the three-element POVM with P_psi = (1/(1+|c|)) |phi_perp> := (|phi> - c|psi>)/sqrt(1-|c|^2) and |phi_perp> := (|psi> - c*|phi>)/sqrt(1-|c|^2). Demonstrate by a concrete counterexample-style argument that NO two-element POVM {E_psi, E_phi} (with only the two conclusive labels and no inconclusive outcome) can be unambiguous, i.e. cannot simultaneously satisfy = 0 and = 0 while being a valid POVM, whenever 0 < |c| < 1. Then verify that the given three-element POVM is unambiguous and explain why introducing P_? is exactly the structural escape that restores validity.", "reference_answer": "Use the bra-ket convention of the question: c=, conjugate-linear in the bra and linear in the ket. Put r=|c| and d=1-r^2>0. Let u=phi_perp=(psi-conj(c)*phi)/sqrt(d) and v=psi_perp=(phi-c*psi)/sqrt(d). Expanding the inner products gives ||u||=||v||=1, =0 and =0. In particular =(1-conj(c)*c)/sqrt(d)=sqrt(d), and =sqrt(d). The coefficient here is conj(c), not c; replacing it by c would incorrectly equate 1-c^2 with 1-|c|^2 for a complex overlap. No rephasing assumption is imposed.\n\nSuppose a two-element POVM with just the two conclusive labels existed. Positivity and =0 imply E_psi*phi=0, since that expectation is ||sqrt(E_psi)*phi||^2. Self-adjointness then places its range in phi's orthogonal complement. In the specified two-dimensional span this forces E_psi=a*|u>=0. Likewise E_phi=b*|v>2, whereas the identity on the two-dimensional space has trace2. This is a contradiction for every 0==0, contradicting the same assumption.\n\nFor the proposed three effects, let F be the linear map whose two columns are u and v. Their overlap is =-c, so F^dagger*F has diagonal entries1 and off-diagonal entries -c,-conj(c). Its characteristic polynomial is (1-lambda)^2-r^2; the eigenvalues are 1+r and 1-r. Since r<1, F is invertible on the two-dimensional span and F*F^dagger has the same eigenvalues. Therefore S=P_psi+P_phi=(F*F^dagger)/(1+r) has eigenvalues1 and (1-r)/(1+r). It is positive and at most the identity, so P_?=I-S is positive, with eigenvalues0 and 2r/(1+r). Each conclusive effect is positive and completeness holds by definition. The orthogonality relations above give zero probability for either wrong conclusive label.\n\nOn input psi, the correct conclusive probability is d/(1+r)=1-r, the wrong probability is zero and the inconclusive probability is r; the same holds with psi and phi exchanged. Introducing P_? supplies exactly the positive slack missing from a conclusive-only complete measurement. This verifies the stated construction, not uniqueness or optimality among every possible prior-weighted discrimination scheme. At r=0 the obstruction disappears and two orthogonal projectors suffice; r=1 is excluded because the complement formulas divide by zero. The question's actual regime is 0=(1-conj(c)c)/sqrt(1-|c|^2)=sqrt(1-|c|^2), with the actual complex conjugate, and derives a=b=1/(1-|c|^2) from completeness without restricting c to real.", "id": "r2", "weight": 1 }, { "criterion": "Obtains trace2/(1-|c|^2)>2, or an equivalent self-adjointness contradiction forcing c=0, to rule out every conclusive-only two-element POVM for overlapping states.", "id": "r3", "weight": 1 }, { "criterion": "Verifies the given three effects are PSD and complete, with zero wrong-label probabilities. Derives the rank-one-sum spectrum1±|c| and explains the positive inconclusive slack; does not assert a unique measurement or arbitrary-prior optimum.", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "PSD zero expectation implies E_psi annihilates phi; in the two-dimensional span its support is the normalized phi_perp line, and similarly for the other effect.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "With phi_perp=(psi-conj(c)phi)/sqrt(1-|c|^2), the correct overlap is sqrt(1-|c|^2); completeness forces both weights to1/(1-|c|^2).", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The forced two-effect sum has trace2/(1-|c|^2)>2 for0<|c|<1, contradicting completeness on the two-dimensional space.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "Thus overlapping distinct states cannot have a valid unambiguous measurement with only the two conclusive labels.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "The given smaller weights make the conclusive sum a positive contraction with spectrum1 and(1-|c|)/(1+|c|); its positive complement restores completeness while preserving unambiguity.", "supporting_nodes": [ "concept:block-encoding" ] } ], "source_item_digest": "978453b6eb90931fa2ea361e9bf82185e9a84e1604d41632767f0242a529507e", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:matrix-analysis" ], "target_nodes": [ "concept:operator-theory", "concept:probability", "concept:block-encoding" ], "task_hash": "ed3bccf607a3ce7d5b70680494a80a9dcfd78a0cfd395453ac95e6844313e76a", "task_kind": "counterexample", "textbook_successor": { "predecessor_item_digest": "d925946e833792c3cf95b2b993cd3b820be35abfd5c259fdb68384b3f29d3558", "predecessor_key": "qcb-core-plan-204-unambiguous-discrimination-inconclusive-necessity", "repair_record": "shared/repairs/plan-204-complex-conjugation-repair.json", "repair_record_digest": "bf3c01e5a75f04338bc73b1803fe7dbd7447f815747aee41380175ead74b16fb", "review_digest": "79dfb46418a93cef899a657194add6a16e21e6257ffd1ce97b40ae92fa68cde9", "review_record": "reports/textbook-v1/repairs/plan-204-complex-conjugation-repair-migration-revalidation-review.json" }, "topic": "planned_scientific_algorithms" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-004", "difficulty": { "level": "intermediate", "rationale": "A clean isometry computation plus the structural conclusion about non-uniqueness; needs careful index bookkeeping." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-state-foundations", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-f8cc6c059ae77d06f6b7", "key": "qcb-core-plan-004-textbook-v1-ensemble-zero-and-genuine-variation-repair", "primary_sources": [ { "anchors": "Definition 2.3 pure-state ensembles and Eq. (2.7); page 4 remark that every mixed state has infinitely many pure-state ensembles; spectral decomposition Eq. (2.8); Exercise 2.1 isometry construction Eqs. (2.37)-(2.38) with V^dagger V = I_r, q_j = , |phi_j> = q_j^{-1/2}|phi'_j>", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "3-12", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "A mixed state can be written as a probabilistic mixture of pure states |psi_i> with weights p_i; such a list is called a realizing ensemble. Prove that a mixed state has more than one realizing ensemble by the following explicit construction. Set unnormalized vectors |psi'_i> = sqrt(p_i) |psi_i>, so that the state equals sum_i |psi'_i>= r rows and r columns whose columns are orthonormal (so V with its conjugate transpose on the left gives the r-by-r identity). Define |phi'_j> = sum_l V_{jl} |psi'_l> for j = 1, ..., k. Show that the k outer products |phi'_j> and w_j=sum_l V_jl u_l. Expanding the outer products gives sum_j |w_j>=0 and sum_j q_j=1.\n\nIf q_j>0 define |phi_j>=w_j/sqrt(q_j). If q_j=0, positivity of the Hilbert norm implies w_j=0, so discard that term; alternatively retain its zero weight and assign it any unit vector, without dividing by zero. Thus one obtains a valid ensemble with at most k positive-weight entries, not necessarily exactly k. For example appending a zero row to V appends a zero output. Also different isometries need not give different physical ensembles: multiplying rows of V by phases changes w_j only by phases and leaves both q_j and |phi_j>r alone does not guarantee new rays. For a concrete genuine length increase from the rank-s positive spectral ensemble, replace the two coefficient columns above by the three rows (1/sqrt(2),0), (1/2,1/sqrt(2)), (-1/2,1/sqrt(2)), leaving the other s-2 columns in untouched identity rows. The two displayed columns have norm one and inner product zero, so the resulting (s+1)-by-s matrix is an isometry. Its three outputs are u/sqrt(2), u/2+v/sqrt(2), and -u/2+v/sqrt(2), with strictly positive weights lambda_1/2, lambda_1/4+lambda_2/2, and lambda_1/4+lambda_2/2. Their rays are pairwise distinct because u,v are independent, and are distinct from the untouched eigenvector rays, so this really has s+1 positive distinct-ray components instead of s. By contrast one may increase a list length just by splitting a positive weight among repeated copies of its ray, or pad V with zero rows; neither operation alone proves a new set of pure states. The identity for every V and the explicit genuine witnesses establish exactly the claimed nonuniqueness, without an injectivity claim for the isometry parameterization.", "rubric": [ { "criterion": "Expands the output outer-product sum with correct complex conjugation and identifies its coefficient as (V^dagger V)_{m,l}.", "id": "r1", "weight": 1 }, { "criterion": "Uses V^dagger V=I_r to recover rho exactly, and obtains q_j>=0 and sum_j q_j=1 by norms and trace.", "id": "r2", "weight": 1 }, { "criterion": "Normalizes only q_j>0 outputs; discards zero outputs or explicitly gives zero-weight slots arbitrary unit vectors. Does not claim an arbitrary k-row isometry necessarily gives k positive components.", "id": "r3", "weight": 1 }, { "criterion": "Proves genuine physical nonuniqueness for every mixed state by an explicit varying positive-component construction, not merely infinitely many isometries. Distinguishes phase/permutation/duplicate/zero-row redundancy and makes any longer-ensemble claim existential with a valid positive-output witness rather than automatic from k>r.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Expanding sum_j |w_j>= 0 for the separable direction, valid for any local dimension", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "13-16", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "A two-qubit state invariant under all local transformations of the form U tensor U can be written with a single real parameter x in [-1,1] as rho(x) = ((2 - x)/6) I + ((2x - 1)/6) F, where I is the identity and F is the swap operator (F|a>|b> = |b>|a>). It is given that x equals the trace overlap tr(F rho). Determine the exact threshold in x at which rho(x) becomes entangled, prove the entangled direction using the partial-transpose test (you may use that the partial transpose of F is twice the projector onto the maximally entangled state, and that the identity resolves into the four Bell projectors), and explain in one or two sentences why the separable direction holds without appealing to any low-dimension sufficiency shortcut.", "reference_answer": "The exact threshold is x=0: rho(x) is entangled for -1<=x<0 and separable for 0<=x<=1. Write a=(2-x)/6 and b=(2x-1)/6. Since swap F has eigenvalue +1 on the three-dimensional symmetric subspace and -1 on the antisymmetric line, rho itself has eigenvalues a+b=(1+x)/6 with multiplicity three and a-b=(1-x)/2 with multiplicity one; these are nonnegative on the stipulated interval and sum to one.\n\nFor the entangled direction, F^{T_B}=2P_+ and I^{T_B}=I give rho(x)^{T_B}=a I+2b P_+. In the Bell basis its eigenvalue on P_+ is a+2b=(2-x)/6+(2x-1)/3=x/2, while each of the three orthogonal Bell states has eigenvalue a=(2-x)/6. Thus x<0 produces a negative partial-transpose eigenvalue, so rho(x) is entangled: partial transpose of every product positive operator is positive, and convex mixtures preserve positivity. For x>=0 this calculation shows only PPT, not separability; no PPT-sufficiency theorem is used.\n\nThe swap identity tr[F(omega tensor sigma)]=tr(omega sigma)>=0 extends to mixtures and proves separable implies nonnegative swap overlap. It does not prove the converse for arbitrary bipartite states; Werner symmetry must be used to establish the required converse.\n\nFor each x in [0,1], choose |a>=|0> and |b>=sqrt(x)|0>+sqrt(1-x)|1>, and twirl the product projector tau=|a>|^2=x because F commutes with U tensor U, and hence equals the specified rho(x), which is the unique two-qubit Werner state with that overlap. Each integrand is a product state and finite convex approximations converge to the average; the finite-dimensional separable set is closed (the convex hull of the compact set of pure product projectors is compact), so this proves separability including x=0 and x=1 without the low-dimensional PPT shortcut, and the product-twirl argument, unlike these particular qubit coefficients and eigenvalue multiplicities, extends to local dimension at least two.", "rubric": [ { "criterion": "States the exact restricted-family threshold: entangled for -1<=x<0 and separable for 0<=x<=1, including the x=0 boundary.", "id": "r1", "weight": 1 }, { "criterion": "Computes rho^{T_B}=((2-x)/6)I+((2x-1)/3)P_+, with Bell eigenvalues x/2 once and (2-x)/6 three times; does not confuse this spectrum with rho's original spectrum.", "id": "r2", "weight": 1 }, { "criterion": "Uses the negative eigenvalue for x<0 and necessary positivity of partial transpose for separable states to prove entanglement. Does not infer separability merely from nonnegative partial-transpose eigenvalues.", "id": "r3", "weight": 1 }, { "criterion": "For x>=0 supplies a product state with squared local overlap x and twirls it; establishes preserved swap overlap and Werner invariance/uniqueness, and uses convexity plus closedness of finite-dimensional separable states to conclude the twirl equals rho(x) and is separable. Swap-overlap positivity extended to mixtures alone proves only the opposite implication and is not a valid converse proof; no PPT-sufficiency shortcut is accepted.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Partial transpose gives rho(x)^{T_B}=((2-x)/6)I+((2x-1)/3)P_+, using F^{T_B}=2P_+.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The Bell-basis partial-transpose eigenvalues are x/2 and three copies of (2-x)/6; negative x therefore proves entanglement by the necessary partial-transpose test.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "For every x in [0,1], the product of |0> and sqrt(x)|0>+sqrt(1-x)|1> has swap overlap x; its U tensor U twirl preserves that overlap, is the unique Werner rho(x), and stays in the closed convex separable set. This proves the separable converse, whereas overlap positivity alone proves only separable implies x>=0.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "07e6df25bb07479718e7bef8a988827de01d56b685ab5fb591320d335c56ad67", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "2bcb273c18a5006328b5033d1e7118c33bfe18ffbfd45cf7601c5d8768356928", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "e04084fe7f2ef7232f2d953cffe7f86e11665744ae6669b27b5f0ff5fa232312", "predecessor_key": "qcb-core-plan-007-werner-entanglement-threshold", "repair_record": "shared/repairs/plan-007-werner-converse-repair.json", "repair_record_digest": "f6638980d6a9548f92155a420a8362adb0e3b5f9a6e3e6f7840e36dd3c167999", "review_digest": "efb45db65c7902d860705bec4851463e25f1c957716c59c835b3b56becf8eeb7", "review_record": "reports/textbook-v1/repairs/plan-007-werner-converse-repair-migration-revalidation-review.json" }, "topic": "planned_quantum_information" }, { "assumptions": [ "The alphabet of tensor labels has exactly d letters." ], "context_hash": "e93976312fe2a996c24b2f375f18ed4172e2c9f39d0f5390401a9927db0f1746", "conventions": [ "A filled tableau places tensor labels into boxes according to the standard labels of T." ], "core_id": "qcb-core-plan-307", "difficulty": { "level": "intermediate", "rationale": "A boundary/vanishing phenomenon driven by pigeonhole and the sign structure of column antisymmetrization." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-young-row-bound", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-186d73f4b20342023567", "key": "qcb-core-plan-307-textbook-v1-young-source-order-repair", "primary_sources": [ { "anchors": "Section 5.4.2 remark that tableaux with more than d rows are annihilated by the Young symmetrizer, p.48", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "48", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "In the Schur-Weyl decomposition of the n-fold tensor power of C^d, only Young diagrams with at most d rows contribute. Using the realization of the unitary-group irreducible pieces by Young symmetrizers acting on tensor basis vectors, explain precisely why a Young diagram with more than d rows contributes nothing, and identify the exact algebraic mechanism responsible.", "reference_answer": "Fix a standard tableau T of shape lambda and write Q_pi for the source's permutation action on tensor factors, so Q_pi*Q_sigma=Q_(pi*sigma). Let R_T and C_T be the subgroups permuting positions within each row and column of T. Define the row sum r_T=sum_{rho in R_T} Q_rho and the signed column sum c_T=sum_{kappa in C_T} sgn(kappa)*Q_kappa. The source convention is e_T=r_T*c_T. On column vectors this means e_T(v)=r_T(c_T(v)): COLUMN antisymmetrization acts first, then ROW symmetrization. No commutation of these two sums is assumed.\n\nA tensor basis word v=e_{i_1} tensor ... tensor e_{i_n} fills the box carrying the number j in T with the label i_j. Suppose lambda has p>d nonempty rows. Its first column has p boxes, whose labels come from an alphabet of size d, so some two first-column positions have the same label. Let tau be their transposition. It lies in C_T, has sign-1, and fixes this basis word: Q_tau*v=v. Reindexing the entire signed column sum by kappa -> kappa*tau gives c_T*Q_tau=-c_T. Consequently c_T*v=c_T*Q_tau*v=-c_T*v. Over C,2 is invertible, so c_T*v=0.\n\nThe repeated pair may depend on the chosen basis word, but the preceding argument applies to EVERY word in the tensor basis. By linearity c_T is therefore the ZERO OPERATOR on the entire tensor power. Only after establishing this whole-space statement do we multiply by the row sum: e_T=r_T*c_T=r_T*0=0. Its image e_T((C^d) tensor n) is the zero space, so the Young-symmetrizer realization contributes no unitary-group irreducible piece of that shape. This is the precise algebraic mechanism: forced repetition in a column, a fixing transposition with negative sign, pair cancellation, and linear extension from a spanning basis. If d=0 and n>0 the tensor space is already zero; when n=0 the empty diagram has no row count exceeding a natural d, so there is no excluded case to analyze.\n\nThe earlier phrase 'row symmetrization followed by column antisymmetrization' was ambiguous: written factors r_T*c_T are not the temporal row-then-column action c_T*r_T. These products do not commute in general. For example, with T=[1,2;3] and v=e_1 tensor e_2 tensor e_1, let r=I+Q_(12) and c=I-Q_(13). Then r*c*v=0, whereas c*r*v=e_2 tensor e_1 tensor e_1-e_1 tensor e_1 tensor e_2 is nonzero. Thus one must preserve the source order, not interchange factors. In the excluded p>d regime, however, c_T has already been proved zero on the WHOLE space, so both r_T*c_T and c_T*r_T happen to vanish. That is why the old wording did not falsify the row-bound conclusion. Nothing here asserts that e_T acts nontrivially on every basis word when p<=d; absence of the forced long-column obstruction is not a pointwise survival theorem.", "rubric": [ { "criterion": "Describes the filled tableau of a tensor basis word by placing the label of factor j in T's box numbered j.", "id": "r1", "weight": 1 }, { "criterion": "Defines row symmetrization r_T and signed column antisymmetrization c_T and states the pinned source convention e_T=r_T*c_T. Explicitly gives column-first, then row action on column vectors; does not assume r_T and c_T commute. Treats the old row-then-column wording as ambiguous rather than refuting the valid row-bound conclusion.", "id": "r2", "weight": 1 }, { "criterion": "For p>d rows, applies pigeonhole to the first column to obtain equal labels and a column transposition tau fixing the input word with sign-1.", "id": "r3", "weight": 1 }, { "criterion": "Uses c_T*Q_tau=-c_T to show c_T kills each basis word, then by linearity proves c_T=0 on the ENTIRE tensor space before concluding e_T=r_T*c_T=0 and the Young-image piece is absent. Does not claim every allowed p<=d filling survives.", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Each tensor basis word fills T by recording its factor labels in the boxes carrying the corresponding position numbers.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "The source Young symmetrizer is e_T=r_T*c_T: signed COLUMN antisymmetrization acts first on column vectors, then ROW symmetrization; the factors are not generally interchangeable.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "If the first column has p>d boxes, pigeonhole supplies equal labels and a column transposition tau that fixes the input word and has sign-1.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "The identity c_T*Q_tau=-c_T kills every basis word, so c_T is zero on the entire space before the row product is taken. Thus e_T=0 and its image vanishes; this proves exclusion, not nonzero action on every allowed filling.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "9ad464165e637fe901e26fa853df3655cd2381a13c4fd50ba68aeb231ddf7d9c", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:tensor-networks", "concept:operator-theory" ], "task_hash": "3f6d51a33d5513489a678f201da0c30cbf19796a7cda47849ccb2e760b46aee4", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "5fd70ebf8a2f530a3753626e1424b902ca8e06be8d61426c16be0ec8ef234ec3", "predecessor_key": "qcb-core-plan-307-young-row-bound", "repair_record": "shared/repairs/plan-307-young-order-repair.json", "repair_record_digest": "4f130e179b86191d1b5651ee6eb1e94b6817708cd33791efe8a236d459b7e89f", "review_digest": "dc029b2911e10d0c293c83de43d48eacfcb7a9cda5da7931f24a21b145a7b35d", "review_record": "reports/textbook-v1/repairs/plan-307-young-order-repair-migration-revalidation-review.json" }, "topic": "planned_tensor_symmetry" }, { "assumptions": [ "The density operator is fixed with spectrum a probability vector; n grows large with d fixed." ], "context_hash": "29e7cf09ecfa27bf967c3bf861e7056afbe4a914f2f8b0de3fb6b352e001510e", "conventions": [ "The normalized diagram divides each row length by n; relative entropy of p against q is the sum of p times log of p over q." ], "core_id": "qcb-core-plan-319", "difficulty": { "level": "advanced", "rationale": "A concentration-of-measure boundary statement requiring the relative-entropy rate and the dominance of exponential over polynomial factors." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-spectrum-concentration", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-4e63c887bf087d93449e", "key": "qcb-core-plan-319-textbook-v1-prefactor-and-neighborhood-repair", "primary_sources": [ { "anchors": "Concentration of measure Eq. (5.35)-(5.36) with the relative-entropy bound, p.50-51", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "50-51", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Let a density operator on C^d have spectrum given by a probability vector in non-increasing order, and consider the n-fold tensor power of this density operator decomposed by Schur-Weyl into blocks indexed by Young diagrams. Explain precisely why the total weight carried by the block of a diagram concentrates, for large n, on those diagrams whose normalized row-length vector is close to the spectrum, and identify the exact rate governing the decay away from the spectrum.", "reference_answer": "Let n>=1, let lambda=(lambda_1,...,lambda_d) be a partition of n padded by zeros, and put p=lambda/n. Write s_1>=...>=s_d>=0 for the fixed spectrum and P_lambda for the ORTHOGONAL Schur-Weyl isotypical projection. Its probability is w_lambda=tr(P_lambda*rho^tensor(n)); these weights are nonnegative and sum to1. The Schur-Weyl form rho^tensor(n)=directsum_lambda I_(V_lambda) tensor omega_lambda(rho) gives w_lambda=f_lambda*tr(omega_lambda(rho)), with f_lambda=dim(V_lambda). A general unnormalized Young symmetrizer is not substituted for this orthogonal measurement projection.\n\nHere is a precise finite-n statement of the exponential rate, with natural logarithms. Put r=d(d-1)/2 and D(p||s)=sum_i p_i log(p_i/s_i), taking zero-p terms as0 and D=infinity if p has positive mass outside the support of s. Then\n(n+d)^(-r-d)*exp(-n*D(p||s)) <= w_lambda <= (n+d)^r*exp(-n*D(p||s)).\nBoth polynomial factors matter: the coefficient-one upper inequality is false. For rho=I_2/2,n=2,lambda=(2,0), the symmetric subspace has rank3, so w_lambda=3/4, whereas exp(-nD((1,0)||(1/2,1/2)))=1/4.\n\nFor completeness the polynomial-qualified bound can be reconstructed without dropping its prefactors. First suppose s is strictly positive. The Schur character tr(omega_lambda(diag(s))) is a sum of m_lambda=dim(U_lambda^d) monomials s^nu indexed by semistandard tableaux. The row-constant tableau has weight lambda. Every tableau weight has sorted content majorized by lambda: any k labels occupy at most min(k,h) cells of each height-h column, whose sum is lambda_1+...+lambda_k. Rearranging nu against the decreasing log(s_i), then summing by parts with these partial-sum inequalities, gives s^nu<=s^lambda. Consequently s^lambda <= tr(omega_lambda(rho)) <= m_lambda*s^lambda. This uses only the necessary content restriction, not a false fixed-word survival converse.\n\nLet C_lambda=n!/(product_i lambda_i!). Since each standard tableau is determined by which numbers occupy each increasing row, f_lambda<=C_lambda. The padded hook formula also gives\nf_lambda=n!*product_(i= C_lambda/(n+d)^r:\nevery numerator factor is at least1, and the r extra factors after the lambda_i! terms are each at most n+d. The Weyl dimension formula gives m_lambda=product_(i0 the coordinate likelihood-ratio factor lambda_i^(mu_i-lambda_i)*lambda_i!/mu_i! is at most1: for an increase it is a product lambda_i/(lambda_i+t), and for a decrease a product (lambda_i-t)/lambda_i. If lambda_i=0 but mu_i>0, that type has probability0. Thus lambda is a mode; at most(n+1)^d types have total probability1, so its probability is at least(n+1)^(-d). This proves the lower bound, including zero entries of p with the usual0^0=1 product convention. Combining the character, hook and multinomial bounds with exp(nH(p))*s^lambda=exp(-nD(p||s)) proves the displayed lower bound, since n+1<=n+d, and proves the upper bound as well.\n\nIf s has rank q=1; n=0 is irrelevant to normalized row lengths, which are only defined here for n>=1.\n\nTo pass from one block to concentration of the TOTAL probability, fix epsilon>0. On the probability simplex restricted to the positive support of s, D(p||s) is continuous and vanishes only at p=s. On the compact closed set ||p-s||_1>=epsilon it therefore has a strictly positive minimum delta_epsilon, unless the set is empty and the event is impossible. Diagrams outside the support already have weight0. There are at most(n+1)^d diagrams, so a union bound gives\nPr(||lambda/n-s||_1>=epsilon) <= (n+1)^d*(n+d)^r*exp(-n*delta_epsilon) ->0.\nThe count-of-diagrams factor cannot be omitted from this finite-n argument. Polynomial growth cannot defeat a fixed positive exponential rate when d is fixed.\n\nThe exact exponential scale is D: the two-sided inequalities give -log(w_lambda)/n = D(lambda/n||s)+O_d(log(n+d)/n) for nonzero supported weights. Along supported sequences with lambda/n tending to p, the rate tends to D(p||s); unsupported diagrams have zero weight and infinite rate. This is neighborhood concentration, not eventual equality lambda/n=s, and it does not say one near-spectrum block keeps a probability bounded below by a positive constant. The public question concerns mathematical probabilities, not an efficiently implemented Schur transform, a finite-bit resource certificate or a new local Lean proof.", "rubric": [ { "criterion": "Defines orthogonal Schur-Weyl block probabilities, nonnegativity and their total normalization; does not use a general unnormalized Young symmetrizer as a measurement projector.", "id": "r1", "weight": 1 }, { "criterion": "Derives w_lambda=f_lambda tr(omega_lambda(rho)) and the polynomial-qualified exponential scale from appropriate character/dimension bounds, with natural-log entropy and zero-spectrum conventions explicit.", "id": "r2", "weight": 1 }, { "criterion": "Retains polynomial prefactors in every finite-n claim; identifies D(lambda/n||s) as the exponential rate, not a universally valid coefficient-one upper bound. Does not assume full rank or a false content-only survival iff.", "id": "r3", "weight": 1 }, { "criterion": "Proves concentration in every fixed neighborhood with the count-of-diagrams union factor and fixed d. Does not claim an exact matching diagram or a constant lower probability for each near-spectrum block, and distinguishes mathematical concentration from an implemented measurement/cost certificate.", "id": "r4", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "The orthogonal Schur-Weyl projections give nonnegative per-diagram weights summing to one.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Each weight is f_lambda times the Schur-character trace; necessary tableau-content majorization, hook/Weyl and explicit multinomial bounds control it without a pointwise survival converse.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "The weight lies between (n+d)^(-d(d+1)/2) exp(-nD) and (n+d)^(d(d-1)/2) exp(-nD), not below the bare coefficient-one exponential; unsupported diagrams have zero weight.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "For every fixed neighborhood of the spectrum, sum over at most(n+1)^d diagrams to bound the outside mass by a polynomial times exp(-n delta_epsilon), tending to zero. Individual blocks or exact equality are not required.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "23ce008524da81fbc6003a3f740baf01af7753a835ba3a0e9657729b698f43bb", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:probability", "concept:tensor-networks" ], "task_hash": "27a93b1afe1f5a23c073aee8407c87db7425a69c334c936715124073fed9237b", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "f1c75541f386d843fd18d4f5da821e8f4752a5f188cfeef498618fbb468b386f", "predecessor_key": "qcb-core-plan-319-spectrum-concentration", "repair_record": "shared/repairs/plan-319-concentration-prefactor-repair.json", "repair_record_digest": "2fcc7f4bda646993c9b2da0b02c61c075fe06acbea9d840480a5643486142182", "review_digest": "7b209c5dc290d3a6304d89d1276f96ba7ee0df18d30db2756cc2a5afab40be40", "review_record": "reports/textbook-v1/repairs/plan-319-concentration-prefactor-repair-migration-revalidation-review.json" }, "topic": "planned_tensor_symmetry" }, { "assumptions": [ "A is non-normal; eigenvalues and singular values differ" ], "context_hash": "c9ce7e241dd76eac29b725045b6a093cc431e44f3cf62aa8dbb32422c8ba9f18", "conventions": [ "Degree-two singular value transformation uses right singular vectors on both sides (even parity)", "T_2(x) = 2 x^2 - 1" ], "core_id": "qcb-core-plan-505", "difficulty": { "level": "intermediate", "rationale": "A concrete falsification requiring the explicit singular value decomposition of a non-normal matrix and recognition that the construction is singular-value based." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-chebyshev-singular-value-transformation", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-bb6be08fe749ae275bf3", "key": "qcb-core-plan-505-textbook-v1-nilpotent-svt-normality-boundary-repair", "primary_sources": [ { "anchors": "Example 10.8: the 2x2 nilpotent A = [[0,1],[0,0]], its SVD W = I, Sigma = diag(1,0), V = X, and T^{triangleright}_2(A) = V T_2(Sigma) adjoint(V) = diag(-1,1), with the explicit (1,1)-block-encoding check Eqs. (10.79)-(10.80)", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "173-174", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Consider the two-by-two operator A that sends the first basis vector to zero and the second basis vector to the first (the strictly upper-triangular nilpotent). A tempting inference says: since the only eigenvalue of A is zero, any Chebyshev transformation of A should equal the Chebyshev value at zero times a projector pattern, so the degree-two singular value transformation of A should be the degree-two Chebyshev value at zero, namely minus one, times the identity. Show by explicit computation that this inference is FALSE: compute the singular value decomposition of A, compute the degree-two singular value transformation, and exhibit the discrepancy. State the correct principle that resolves the paradox.", "reference_answer": "Use the ordered standard basis e_1,e_2. The operator is A=[[0,1],[0,0]], so A^2=0 and both eigenvalues are zero. A singular value decomposition is A=W Sigma V† with W=I_2, Sigma=diag(1,0), and V=X=[[0,1],[1,0]]. Indeed Sigma X=A. Equivalently A†A=diag(0,1), so the right singular vector for singular value1 is e_2 and that for singular value0 is e_1.\n\nThe stipulated even singular-value transformation uses the right singular vectors on both sides. Since T_2(x)=2x^2-1, T_2(Sigma)=diag(1,-1). Therefore T_2^SV(A)=V T_2(Sigma)V†=X diag(1,-1)X=diag(-1,1). Its second diagonal entry is+1, so it is not -I_2. The same result follows directly from the even-transform identity T_2^SV(A)=2A†A-I_2.\n\nFor comparison, the ordinary polynomial evaluated on A is T_2(A)=2A^2-I_2=-I_2. Thus the alleged output -I_2 is the ordinary polynomial in this particular example, not the output of the specified singular-value transform. One must not generally determine a function of a defective matrix merely from its eigenvalue list: this A is not diagonalizable, and for example T_1(A)=A is nonzero even though T_1(0)=0. Here the ordinary degree-two result happens to reduce to -I because A^2=0. The resolution is that an even singular-value transformation is V f(Sigma)V†, whereas ordinary polynomial evaluation forms f(A); these are different operations for general A.\n\nA sufficient agreement statement is: for a Hermitian H and a function of definite parity, the parity-defined singular-value transformation agrees with the spectral function f(H). To see the sign boundary correctly, write H=Q D Q†, Sigma=|D| and choose a diagonal unitary J with J_jj=sign(D_jj) at nonzero eigenvalues and J_jj=+1 at zero. Then H=(QJ)Sigma Q† is an SVD with unitary left factor. For even f, f(|D|)=f(D), so the right transformation agrees. For odd f, f(0)=0 and J f(|D|)=f(D), so the balanced transformation agrees too. No zero-eigenvalue division or nonunitary sign(0)=0 factor is needed.\n\nNormality alone is not such a sufficient condition. For the normal contraction B=(i/2)I_2, the singular values are1/2 and T_2^SV(B)=(-1/2)I_2, while the ordinary polynomial T_2(B)=2B^2-I_2=(-3/2)I_2. These differ in magnitude, not just a sign or global phase. General normal matrices may have complex eigenphases, whereas Hermitian eigenvalues have only real signs; definite parity handles those real signs, not arbitrary complex phases. Conversely normality is not necessary for every individual coincidence, since the odd degree-one transform equals A for any square A. The relevant principle for the question is therefore the precise right-even singular-value formula, which produces diag(-1,1), not a blanket rule equating eigenvalue and singular-value transformations for normal matrices.\n\nThese are exact matrix identities and the projected-block meaning of the stated transformation. No deterministic normalized-state preparation, gate compilation, query bound, or approximation-error guarantee is inferred from the matrix calculation.", "rubric": [ { "criterion": "Exhibits and verifies A=I_2 diag(1,0)X†, with singular values1 and0 and X as the right-singular-vector matrix.", "id": "r1", "weight": 1 }, { "criterion": "Computes T_2(Sigma)=diag(1,-1) from T_2(1)=1 and T_2(0)=-1, using the stipulated even singular-value convention.", "id": "r2", "weight": 1 }, { "criterion": "Pulls back by the right singular vectors to obtain diag(-1,1), explicitly differing from -I_2; distinguishes this from the ordinary polynomial T_2(A)=2A^2-I_2=-I_2.", "id": "r3", "weight": 1 }, { "criterion": "Resolves the paradox by distinguishing singular-value/right-vector transformation from ordinary polynomial evaluation, without inferring general defective-matrix functions from eigenvalues alone or asserting that normality alone ensures agreement.", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "The nilpotent A=[[0,1],[0,0]] has SVD W=I, Sigma=diag(1,0), V=X, since Sigma X=A.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The scalar polynomial T_2(x)=2x^2-1 gives T_2(Sigma)=diag(1,-1).", "supporting_nodes": [ "concept:approximation" ] }, { "claim": "The stipulated right-even transform is X diag(1,-1)X†=diag(-1,1)=2A†A-I.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The ordinary polynomial equals T_2(A)=2A^2-I=-I because A^2=0; this is not a general rule determining defective-matrix functions from their eigenvalue list.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The actual singular-value output has a+1 second diagonal entry, so it differs from the proposed -I output.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Definite-parity SVT agrees with spectral evaluation for Hermitian matrices, with unit-modulus sign completion at zero; normality alone does not ensure agreement, since B=(i/2)I has T_2^SV(B)=-I/2 but T_2(B)=-3I/2.", "supporting_nodes": [ "concept:operator-theory", "concept:approximation" ] } ], "source_item_digest": "a973d61b0a43ba5ce483b75042576d8cffdfac75f5b85302df219bb85102afdf", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:approximation" ], "target_nodes": [ "concept:matrix-analysis", "concept:operator-theory" ], "task_hash": "dfd56eb6f1ad9121c41f48923c31b14e382d5731a433023e5f32dedb52f6ec2b", "task_kind": "counterexample", "textbook_successor": { "predecessor_item_digest": "fe853c4c836e167a6501e0cc7a8ea48915d4b78c8ff212e0d9c55ebe42f2c5f2", "predecessor_key": "qcb-core-plan-505-nilpotent-sv-vs-eigenvalue-counterexample", "repair_record": "shared/repairs/plan-505-nilpotent-transform-repair.json", "repair_record_digest": "79e18e22fe2156e12718911abc8e93c27c352ffbdd373021155f061835f6c7d6", "review_digest": "8f9716364925692f0142af5ebfec9a40d0657508478ef96207a69f9916585682", "review_record": "reports/textbook-v1/repairs/plan-505-nilpotent-transform-repair-review.json" }, "topic": "planned_polynomial_signals" }, { "assumptions": [], "context_hash": "4ed847ac7b15b344ee50eb0d580d88a829bc7026d458496d9b827323fe82616b", "conventions": [ "C = diag(c_j) are singular values of the top-left block; S = diag(s_j) nonnegative", "Chebyshev first/second kind evaluated on [-1,1] via cosine/sine of a real angle" ], "core_id": "qcb-core-plan-506", "difficulty": { "level": "intermediate", "rationale": "Combines the embedding bound on sub-blocks of a unitary with the Pythagorean cosine-sine identity and connects it to the admissible Chebyshev argument range." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-cosine-sine-decomposition", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-9435bc623d4d9c3b918d", "key": "qcb-core-plan-506-textbook-v1-unitary-compression-gram-repair", "primary_sources": [ { "anchors": "Theorem 10.9 cosine-sine decomposition: U = diag(W1,W2) [[C,S,0],[S,-C,0],[0,0,I]] diag(adjoint(V1),adjoint(V2)) with C, S real nonnegative diagonal and C^2 + S^2 = I_p, including the proof remark that singular values C of the embedded block lie in [0,1] (Eqs. 10.81-10.89)", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "174-176", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "A unitary U is written in two-by-two block form with top-left block B occupying the first p coordinates (p at most q, total p+q). Prove that the singular values of the top-left block B all lie in the closed interval from zero to one, and that there is a cosine-sine structure in which the diagonal cosine matrix C (the singular values of B) and a nonnegative diagonal sine matrix S satisfy C squared plus S squared equals the identity. Explain concretely why this forces every value fed into the Chebyshev polynomials during qubitization to lie in the interval from minus one to one.", "reference_answer": "Work over the complex field and partition U=[[B,F],[G,H]], where B is p×p, G is q×p and U is unitary on dimension p+q, with p≤q. The first p columns of U are an isometry, so the entire Gram identity is B†B+G†G=I_p. In particular, for every x∈C^p, ||Bx||^2=||x||^2-||Gx||^2≤||x||^2. Hence ||B||≤1, not merely a bound on each original column separately. Individual column lengths alone would not imply this spectral bound: [[1,1],[0,0]] has both columns of norm1 but largest singular value sqrt(2), and cannot be such a unitary compression.\n\nTake an SVD B=W_1 C V_1†, where W_1,V_1 are p×p unitary and C=diag(c_1,...,c_p) is nonnegative. Conjugating the Gram identity by V_1 gives (GV_1)†(GV_1)=I_p-C^2. The diagonal entries are squared norms, so1-c_j^2≥0; together with c_j≥0 this proves0≤c_j≤1. The columns of GV_1 are mutually orthogonal, with lengths s_j=sqrt(1-c_j^2). Thus S=diag(s_1,...,s_p) is real nonnegative and C^2+S^2=I_p.\n\nHere is an explicit endpoint-safe construction of the CS structure, including the zero-sine cases. For every j with s_j>0, let w_j=GV_1 e_j/s_j. These vectors form an orthonormal family in C^q. For indices with s_j=0, the corresponding GV_1 e_j is zero, so do not divide by s_j. Instead choose additional orthonormal w_j from the orthogonal complement of the already chosen nonzero-sine vectors. Since q≥p, there is room to choose all first p vectors and then complete them to a q-dimensional orthonormal basis. Let W_2 be the unitary with these vectors as columns. Then W_2†GV_1=[S;0_(q-p)×p]. This construction also covers c_j=1 (s_j=0), c_j=0 (s_j=1), and repeated singular values; no representative is asserted unique.\n\nThe transformed unitary U_hat=diag(W_1†,W_2†) U diag(V_1,I_q) therefore has its first p columns equal to L=[C;S;0], where the row blocks have sizes p,p,q-p. Define the (p+q)×q matrix K=[[S,0],[-C,0],[0,I_(q-p)]]. Because C and S are commuting real diagonal matrices with C^2+S^2=I, one has L†L=I_p, K†K=I_q and L†K=0. Thus [L,K] is a complete unitary basis. The remaining q columns Y of U_hat are another orthonormal basis of L's orthogonal complement, so there is a q×q unitary V_2 with Y=K V_2†. Consequently\nU=diag(W_1,W_2) [[C,S,0],[S,-C,0],[0,0,I_(q-p)]] diag(V_1†,V_2†).\nThis is the claimed cosine-sine form with nonnegative C,S. When q=p the final row/column blocks are absent; empty p=0 blocks, if allowed, reduce to an arbitrary q-dimensional unitary and the same basis-completion statement.\n\nFor qubitization applied to this encoded block B, the scalar singular-value arguments are precisely the c_j∈[0,1], hence lie in[-1,1]. In the Hermitian/eigenvalue variant, the real eigenvalues may be negative, but |lambda_j|≤||B||≤1, so the signed arguments also lie in[-1,1]. For a general nonnormal block, its possibly complex eigenvalues are not substituted into the singular-value construction. If B represents a separately scaled target A/alpha, this range conclusion concerns the actual normalized block's arguments, sigma_j(A)/alpha, not automatically unscaled sigma_j(A).\n\nFor each real argument t∈[-1,1], choose theta=arccos(t)∈[0,pi]. Then T_m(t)=cos(m theta), and in the interior U_l(t)=sin((l+1)theta)/sin(theta), with polynomial endpoint values U_l(1)=l+1 and U_l(-1)=(-1)^l(l+1). Thus no nonreal angle or division by a zero endpoint sine is needed to evaluate the polynomials on the encoded spectrum. Outside[-1,1] the polynomials are still defined, but this real-angle parametrization is unavailable and hyperbolic growth can occur. Staying inside the interval is a domain guarantee, not a uniform bound of1 on second-kind polynomials or a numerical-error/finite-bit stability theorem. The SVD and basis-completion argument proves mathematical existence of the factorization; it does not certify an efficient circuit to implement its basis changes or any query/gate count.", "rubric": [ { "criterion": "Uses the full identity B†B+G†G=I_p (or its all-vector norm inequality), not individual column lengths alone, to justify that B is a contraction and its singular values are at most1.", "id": "r1", "weight": 1 }, { "criterion": "After B=W_1 C V_1†, derives (GV_1)†GV_1=I_p-C^2 and C^2+S^2=I_p with S nonnegative; justifies the CS basis structure by orthonormal completion, explicitly avoiding division by zero when s_j=0.", "id": "r2", "weight": 1 }, { "criterion": "Identifies the actual encoded block's singular-value arguments c_j∈[0,1], and, for a Hermitian eigenvalue variant, its signed real eigenvalues in[-1,1]; does not substitute arbitrary nonnormal eigenvalues or unnormalized target values.", "id": "r3", "weight": 1 }, { "criterion": "Explains why these arguments admit the real-angle Chebyshev formulas on[-1,1], with polynomial endpoint values rather than undefined sine quotients, without claiming a uniform second-kind bound, compiled basis-change cost or numerical precision guarantee.", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "The first p columns of the partitioned unitary satisfy the full Gram identity B†B+G†G=I_p, which bounds ||Bx||≤||x|| for every vector x.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "For B=W_1 C V_1†, the rotated Gram identity (GV_1)†GV_1=I_p-C^2 implies 0≤c_j≤1 and mutually orthogonal lower-column vectors with squared lengths1-c_j^2.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Set s_j=sqrt(1-c_j^2). Normalize GV_1 e_j only when s_j>0, choose orthonormal complement vectors at s_j=0, and complete W_2 using q≥p, obtaining W_2†GV_1=[S;0].", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "The column isometry L=[C;S;0] and complement K=[[S,0],[-C,0],[0,I]] form a unitary; expressing the remaining transformed columns in K supplies V_2 and the full CS form, with C^2+S^2=I_p.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Qubitization's actual singular-value arguments c_j lie in[0,1]; in a Hermitian eigenvalue variant the signed eigenvalues lie in[-1,1], with any target scaling applied to the encoded block first.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Real-angle Chebyshev formulas therefore apply on[-1,1], using polynomial limits at zero endpoint sine; this is a domain statement, not a uniform second-kind bound or a circuit/finite-bit resource guarantee.", "supporting_nodes": [ "concept:approximation" ] } ], "source_item_digest": "3952e37c50db165ef6c493b022e8cdfb651aa4b11c09b5c8a045bc4f26233def", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:matrix-analysis", "concept:operator-theory", "concept:approximation" ], "task_hash": "e5a0c0327d3daca7da1f672142f490e5cda1c299c367fd0149374515e910cf0f", "task_kind": "proof", "textbook_successor": { "predecessor_item_digest": "f77a7a84ab1bb6ed5de47988c4b7da7369747254ab22e347e0bd4192d0cc9913", "predecessor_key": "qcb-core-plan-506-cosine-sine-constraint-and-range", "repair_record": "shared/repairs/plan-506-unitary-compression-gram-repair.json", "repair_record_digest": "0b41bdab4b8ec2575cbb444bce29200125e83fad05ccfed50e8a427842bf9a93", "review_digest": "d03e4749f18992790076ccc692dad76db29176d1215659a9a0db78b99d9ef246", "review_record": "reports/textbook-v1/repairs/plan-506-unitary-compression-gram-repair-review.json" }, "topic": "planned_polynomial_signals" }, { "assumptions": [ "A is Hermitian (indeed can be taken diagonal) with entries in (0,1)", "A block encoding only constrains the ancilla-0^m sub-block" ], "context_hash": "81a2c8277bff3081964cf538ac2f30d5d509b6b295a0c6381e444ca838322b8c", "conventions": [ "BE_{alpha,m}(A,0) fixes (<0^m| tensor I)U_A(|0^m> tensor I) = A/alpha", "HBE additionally demands U_A = U_A^dagger" ], "core_id": "qcb-core-plan-214", "difficulty": { "level": "intermediate", "rationale": "A boundary/diagnosis task separating what the block-encoding definition constrains from Hermiticity, with a concrete non-Hermitian example and the qubitization payoff." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-hermitian-block-encoding", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-8ead1fa41cebb5caa4d9", "key": "qcb-core-plan-214-textbook-v1-qubitization-source-and-sector-clarification", "primary_sources": [ { "anchors": "Section 9.8; Definition 9.23 (HBE); Example 9.17; qubitization Eq. (9.30)-(9.40)", "edition": "29 April 2026", "inspection": "full bounded primary text read", "pages": "159, 163-168", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text or external relicensing", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "A Hermitian matrix A does NOT automatically have a Hermitian block encoding: even for a Hermitian diagonal A and even when the approximation error is exactly 0, a generic (alpha, m, 0)-block-encoding U_A need not satisfy U_A = U_A^dagger -- only the upper-left n-qubit block of U_A is guaranteed Hermitian. (i) Explain precisely what the block-encoding definition does and does not constrain about U_A, so that Hermiticity of the encoded block does not propagate to the whole unitary. (ii) Give a concrete small example illustrating that the amplitude-oracle block encoding of a Hermitian diagonal matrix (Example 9.17) is generally not a Hermitian matrix. (iii) State what additional structural requirement defines a Hermitian block encoding (Definition 9.23) and why it is a strictly stronger condition useful for qubitization.", "reference_answer": "(i) Write P=|0^m><0^m| tensor I. For positive normalizer alpha, an exact block encoding constrains only the clean compression: (<0^m| tensor I) U_A (|0^m> tensor I)=A/alpha. The other blocks are not independently arbitrary: they must jointly complete a unitary. But that unitarity constraint does not require the whole completion to be self-adjoint. Thus a Hermitian encoded block need not imply U_A=U_A^dagger.\n\n(ii) A concrete completion consistent with Example9.17 is the following. Take A=(1/2)I_2 and one ancilla qubit, set b=sqrt(3)/2, and choose R=[[1/2,-b],[b,1/2]], U_A=R tensor I_2, with the ancilla first. Then R^T R=I_2, so U_A is unitary, and its clean block is A. On input |0>|i> it produces ((1/2)|0>+b|1>)|i>, exactly the source amplitude-oracle action for this diagonal A. Yet R^dagger=R^T has the opposite off-diagonal signs and differs from R, so U_A is not Hermitian. More generally use a real a in(0,1) and b=sqrt(1-a^2). The specified |0> input action alone does not uniquely determine the completion on |1>: [[a,b],[b,-a]] is a different, Hermitian-unitary completion with the same first column. The point is that a non-Hermitian completion is allowed, not that every amplitude oracle must be non-Hermitian.\n\n(iii) Definition9.23 in Section9.8, printed159, calls a block encoding Hermitian when it also satisfies the GLOBAL condition U_A=U_A^dagger. This is strictly stronger than merely encoding Hermitian A, as the preceding example proves. The nontrivial qubitization closure is in Section10.2.1, printed167-168, Eqs10.30-10.40; the old references9.30/9.34 (and9.30-9.40 in historical primary metadata) are incorrect equation locators, not alternative versions of this proof.\n\nHere is the precise exact-encoding motivation. If U_A is an exact Hermitian block encoding with scale alpha, put C=A/alpha. Compression of a unitary implies ||C||<=1. For an eigenvector v of C with eigenvalue lambda, let a_vec=|0^m>v. Then U_A a_vec=lambda a_vec+e_perp, with P e_perp=0 and ||e_perp||^2=1-lambda^2, as in Eq10.30-10.33. If |lambda|<1, define b_vec=e_perp/sqrt(1-lambda^2). Hermitian unitarity gives U_A^2=I; applying U_A a second time gives U_A b_vec=sqrt(1-lambda^2)a_vec-lambda b_vec (Eq10.34). Consequently span{a_vec,b_vec} is invariant and U_A has matrix [[lambda,sqrt(1-lambda^2)],[sqrt(1-lambda^2),-lambda]] there (Eq10.35), a reflection. Z_P=2P-I has matrix diag(1,-1) on this basis, so the iterate Q=U_A Z_P has matrix [[lambda,-sqrt(1-lambda^2)],[sqrt(1-lambda^2),lambda]], a rotation (Eq10.38-10.39). Its kth power has clean diagonal entry T_k(lambda), giving the exact clean block T_k(C), with source Eq10.40-10.41. When alpha=1 this is T_k(A).\n\nIf lambda=+1 or-1, e_perp=0 and the clean vector already spans a ONE-dimensional invariant subspace. No normalized leakage vector is needed or divided by zero; the same clean-block identity follows from lambda^k=T_k(lambda). In particular one must not claim every eigenvector necessarily yields a nontrivial2D sector. For an approximate block encoding, self-adjointness still defines HBE, but the preceding exact formula applies to its actual compression; it is not automatically an exact T_k(A/alpha) identity or an error bound for the requested A. These distinctions explain the structural usefulness of Hermiticity without claiming it is necessary for every more general qubitization construction. This is source/NL algebra, not a freshly executed circuit, finite-bit bound or local Lean certificate.", "rubric": [ { "criterion": "States the exact clean compression A/alpha and distinguishes coupled unitary-completion constraints from global self-adjointness; does not infer whole-unitary Hermiticity from the encoded block.", "id": "r1", "weight": 1 }, { "criterion": "Gives a unitary non-Hermitian rotation completion matching the amplitude-oracle |0> action and a Hermitian diagonal clean block, without claiming the first-column specification uniquely forces that completion.", "id": "r2", "weight": 1 }, { "criterion": "Correctly states Section9.8/Definition9.23's extra global requirement U_A=U_A^dagger, and identifies Section10.2.1/Eq10.30-10.40 as the actual qubitization proof locators.", "id": "r3", "weight": 1 }, { "criterion": "Explains exact normalized closure through U_A^2=I and two reflections for |lambda|<1; handles lambda=±1 as1D, gives T_k(A/alpha) at general scale, and does not extend an exact identity to arbitrary approximate HBE without an error argument.", "id": "r4", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Only the clean compression A/alpha is prescribed by an exact block encoding; the remaining coupled unitary-completion conditions do not force U_A=U_A^dagger.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "An admissible rotation completion for A=(1/2)I_2 has the required amplitude-oracle first column but is not Hermitian; other completions of that first column may be Hermitian.", "supporting_nodes": [ "concept:matrix-analysis" ] }, { "claim": "Section9.8/Definition9.23 adds global U_A=U_A^dagger, strictly stronger than Hermiticity of the encoded block.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "For exact HBE and eigenvalues of C=A/alpha strictly inside[-1,1], U_A^2=I closes the2D clean/leakage sector as in Eq10.30-10.35; at ±1 the clean sector is1D.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Combining U_A with Z_P yields the rotation and exact Chebyshev clean block T_k(C) in Eq10.38-10.41; no unproved approximate or finite-bit guarantee is inferred.", "supporting_nodes": [ "concept:block-encoding" ] } ], "source_item_digest": "2e436d4e4e3bae65ed15e372cb7a9ab95a132a11f3c1b8be76f1852988f10f1c", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:block-encoding", "concept:operator-theory", "concept:matrix-analysis" ], "task_hash": "a597d2efbc856facf32a069e5884b7c2c33b24c2008bea6414df4ea514d98d61", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "ee7b6ab22e7d5bafb9e15e983138f7ede1ac7947e99241fe3924ad09a543ebfc", "predecessor_key": "qcb-core-plan-214-hermitian-matrix-nonhermitian-block-encoding-boundary", "repair_record": "shared/repairs/plan-214-qubitization-source-clarification-repair.json", "repair_record_digest": "4554f42e4b47e86a076bf97e782baa34f6ae271884d2cc834ef7a1190551ab8d", "review_digest": "35d8a02e41a5e7d193e85191db3b4934473641f9c783d466f1b0dc0b5d6dd4a1", "review_record": "reports/textbook-v1/repairs/plan-214-qubitization-source-clarification-repair-review.json" }, "topic": "planned_scientific_algorithms" }, { "assumptions": [ "V = C^d is finite-dimensional; the span of tensor powers is the known commutant of the permutation algebra." ], "context_hash": "a0135c1652621d12a26819e25780b55a569e6c474a959ce03cf742016a396fb1", "conventions": [ "The n-fold tensor power of X denotes X applied to each of the n factors." ], "core_id": "qcb-core-plan-304", "difficulty": { "level": "advanced", "rationale": "The crux is the Lagrange interpolation trick to include non-invertible X, plus correct use of the double commutant theorem." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-schur-weyl-mutual-commutant", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-03c7d22ee23c4c76aa3c", "key": "qcb-core-plan-304-textbook-v1-gl-action-clarification", "primary_sources": [ { "anchors": "Proposition 4.10 with the subalgebras Eq. (4.15)-(4.16) and the Lagrange-interpolation span argument, p.41-42", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "41-42", "redistribution": "not redistributed", "reuse": "Original questions/derivations only; source text not redistributed/relicensed", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "On the n-fold tensor power of V = C^d let the symmetric group S_n act by permuting tensor factors, spanning an algebra A, and let the general linear group GL(V) act diagonally by sending U to the n-fold tensor power of U, spanning an algebra B. Assume it is already known that the commutant A' equals the span of all operators of the form the n-fold tensor power of a single endomorphism X of V. Prove that B equals A', and conclude using the double commutant theorem that the commutant of B equals A.", "reference_answer": "Use the question's exact algebra B=span_C{g^tensor(n):g in GL(V)}; 'diagonal action' here means the SAME g acts on every tensor factor, not that g is a diagonal matrix or necessarily a unitary. Write Q_pi for the factor-permutation action. For every endomorphism X, Q_pi X^tensor(n)=X^tensor(n) Q_pi: on a pure tensor both sides apply X to each factor and put the factors in the same permuted positions. Pure tensors span, so this is an operator equality for all X. In particular it holds for every invertible g, not merely unitary g. Hence every generator of B commutes with A, giving B subset A'.\n\nFor the reverse inclusion use the GIVEN description A'=span_C{X^tensor(n):X in End(V)}. Fix any such X, including a singular one. If d>0, det(X+tI) is a monic scalar polynomial of degree d, so only finitely many complex t make X+tI singular. Choose n+1 distinct complex numbers t_0,...,t_n outside that finite set. Every F(t_j)=(X+t_j I)^tensor(n) then belongs to B by its definition using GL(V). The operator-valued polynomial F(t) has degree AT MOST n: expand the n factors, collecting the finitely many terms with each power of t. Lagrange interpolation applies coefficientwise in this finite-dimensional operator space, and gives F(0)=sum_{j=0}^n ell_j(0)F(t_j), where ell_j(t)=product_{k!=j}(t-t_k)/(t_j-t_k). The denominators are nonzero because the sample points are distinct. B is a complex linear subspace, so F(0)=X^tensor(n) lies in B. Thus the span of all such powers, A', is contained in B. Together the inclusions prove B=A'.\n\nFinally A is the finite-dimensional unital star algebra spanned by the permutation representation of the finite group S_n: products stay in the group image, the identity is present, and Q_pi^dagger=Q_(pi^-1). Thus the finite-dimensional double-commutant theorem applies to THIS algebra and gives A''=A; it is not asserted for arbitrary non-self-adjoint subalgebras. Since B=A', we obtain B'=A''=A, as requested.\n\nFor n=0 the tensor power is C, both algebras are the scalar algebra and F is constant; one interpolation sample suffices. If V=0 and n>0 the ambient tensor power and endomorphism algebra are zero and the conclusion is trivial; this avoids asserting an exact nonzero polynomial degree in a zero operator space. No d>=n premise is needed. The source Prop4.10/Eq4.15-4.16 proof is a GL(V) argument. The historical word 'unitary' in its first inclusion is insufficient wording for the defined B, but the pure-tensor computation above establishes the full GL inclusion directly. A separate unitary-only spanning theorem can be true, but it is NOT proved merely because X+tI is invertible: those interpolation samples need not be unitary. No such bridge is assumed or needed here. This is an algebraic mutual-commutant proof, not an efficient Schur-transform algorithm, fresh Lean result or source-publication approval.", "rubric": [ { "criterion": "Uses B=span{g^tensor(n):g in GL(V)} and proves factor-permutation commutation for every g (indeed every endomorphism), not only unitary generators.", "id": "r1", "weight": 1 }, { "criterion": "Uses the explicitly given A' characterization to reduce the reverse inclusion to every endomorphism power, including singular X.", "id": "r2", "weight": 1 }, { "criterion": "Uses det(X+tI) to choose n+1 distinct invertible shifts over C and places their powers in the GL-defined algebra B, without treating these shifts as necessarily unitary.", "id": "r3", "weight": 1 }, { "criterion": "Treats F(t) as an operator-valued polynomial of degree at most n and applies valid Lagrange interpolation at distinct points to obtain X^tensor(n) in B.", "id": "r4", "weight": 1 }, { "criterion": "Combines both inclusions and applies the finite-dimensional group-image/unital-star bicommutant theorem to A to get B'=A; preserves zero/empty boundaries and adds no unitary-only or d>=n premise.", "id": "r5", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Every common endomorphism action X^tensor(n) commutes with factor permutations, so in particular the GL-generated B lies in A'.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "The given A'=span{X^tensor(n)} reduces the reverse inclusion to arbitrary endomorphism powers, including singular X.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Outside finitely many complex shift values X+tI is invertible, so choose n+1 distinct such points with their tensor powers in the GL algebra B.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "The operator-valued polynomial F(t)=(X+tI)^tensor(n) has degree at most n; Lagrange interpolation at those distinct points puts F(0)=X^tensor(n) in B.", "supporting_nodes": [ "concept:tensor-networks" ] }, { "claim": "Hence B=A'; A is the finite-group image unital star algebra, so A''=A and B'=A, with empty/zero tensor cases understood separately.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "63d0ffabef4e6aad5e2c9a64947cf393005f177424421361257150e9326c07b2", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:operator-theory", "concept:tensor-networks" ], "task_hash": "2f266327d4f87feaa608a303e58e9ba5deb271b9a95f1a20822fa72070318d7e", "task_kind": "proof", "textbook_successor": { "predecessor_item_digest": "8c71082c4fb736eefc753e15be5f10fee4e9608aa150fb873a1b2c6cec31a677", "predecessor_key": "qcb-core-plan-304-schur-weyl-mutual-commutant", "repair_record": "shared/repairs/plan-304-gl-action-clarification-repair.json", "repair_record_digest": "10c7a95815b55413a3f292826b15174f06ce558e4d3a47c26304f56f57c5adf2", "review_digest": "88afa4390d9eff4c0d240f1965c6da67d9082c26e0790774f0d34a783132c5be", "review_record": "reports/textbook-v1/repairs/plan-304-gl-action-clarification-repair-review.json" }, "topic": "planned_tensor_symmetry" }, { "assumptions": [ "A may be non-normal and non-diagonalizable; only its singular value decomposition is used" ], "context_hash": "2f22e20643b90d3bb302cb1d451be11b6f0e9ddeadfa8cb05615b865aca9235b", "conventions": [ "sigma_i are the singular values; v_i right singular vectors, w_i left singular vectors", "Chebyshev first-kind polynomials T_m" ], "core_id": "qcb-core-plan-504", "difficulty": { "level": "advanced", "rationale": "Needs the generalized-matrix-function viewpoint and careful bookkeeping of which singular-vector subspace the trajectory ends in as a function of order parity." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-chebyshev-singular-value-transformation", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-34b6305593a028541e78", "key": "qcb-core-plan-504-textbook-v1-projected-svt-directed-maps-repair", "primary_sources": [ { "anchors": "Section 10.3 Chebyshev singular value transformation: Eqs. (10.61)-(10.72) even output T^{triangleright}_{2k}(A) = sum_i v_i T_{2k}(sigma_i) adjoint(v_i) (Eq. 10.73) and odd output T^{diamond}_{2k+1}(A) = sum_i w_i T_{2k+1}(sigma_i) adjoint(v_i) (Eq. 10.75), Proposition 10.7", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "172-174", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "Let A be a general (not necessarily normal) n-qubit operator with singular value decomposition sending each right singular vector v_i to sigma_i times the left singular vector w_i. A block-encoding unitary U realizes A in its flag block, and the qubitization construction attaches to each index i a plane spanned by the flagged right vector and a flag-orthogonal companion, mapped by U to a plane spanned by the flagged LEFT vector and its companion. Prove that the EVEN qubitization output assembles into the operator sum over i of v_i times T_{2k}(sigma_i) times the adjoint of v_i, whereas the ODD output assembles into sum over i of w_i times T_{2k+1}(sigma_i) times the adjoint of v_i. Explain why the parity of the Chebyshev order decides whether the left factor is a right singular vector or a left singular vector.", "reference_answer": "Interpret the requested output operators as the flag-to-flag blocks of the qubitization unitaries, as in the given block-encoding model, not as their entire output states. Let J map a system vector into the clean flag state, let P=JJ† and R=2P-I, and use the question's exact clean block J†UJ=A. Hence A is a contraction: ||Ax||≤||UJx||=||x||, so each singular value sigma_i lies in[0,1]. No normality, diagonalizability or Hermiticity of A is assumed. If one instead had only a scaled or approximate encoding of a separate target, this exact premise would have to be replaced by its actual normalized block and an error analysis; the present identities do not automatically transform the unscaled target.\n\nChoose orthonormal right/left singular bases satisfying Av_i=sigma_i w_i and A†w_i=sigma_i v_i. Put p_i=Jv_i and p'_i=Jw_i. For sigma_i<1 let s_i=sqrt(1-sigma_i^2), q_i=(U†p'_i-sigma_i p_i)/s_i, and q'_i=(Up_i-sigma_i p'_i)/s_i. Exact compression and unitarity imply that q_i,q'_i are unit vectors orthogonal to the entire clean subspace. The directed input/output planes are H_i=span{p_i,q_i} and H'_i=span{p'_i,q'_i}. Directly,\nUp_i=sigma_i p'_i+s_i q'_i, Uq_i=s_i p'_i-sigma_i q'_i,\nU†p'_i=sigma_i p_i+s_i q_i, U†q'_i=s_i p_i-sigma_i q_i.\nThus U:H_i→H'_i and U†:H'_i→H_i have matrix S_i=[[sigma_i,s_i],[s_i,-sigma_i]] between the respective ordered bases. These are directed maps: U† is the guaranteed reverse, not a second application of U. The geometric planes may coincide or differ; neither possibility changes the argument.\n\nIn both bases R has matrix Z=diag(1,-1). Set E=U†RUR and M_i=S_i Z=[[sigma_i,-s_i],[s_i,sigma_i]]. Then E on H_i has matrix M_i^2, while UR maps H_i to H'_i with matrix M_i. Matrix powers of this scalar rotation yield the full state actions\nE^k p_i=T_{2k}(sigma_i)p_i+s_i U_{2k-1}(sigma_i)q_i for k≥1,\nUR E^k p_i=T_{2k+1}(sigma_i)p'_i+s_i U_{2k}(sigma_i)q'_i for k≥0.\nHere U_l is the second-kind polynomial, not the bare encoding unitary U, and E^0 p_i=p_i is treated directly. These formulas retain the generally nonzero flag-orthogonal components. In the odd expression the R in the extra UR matters on the leakage; it must not simply be omitted as an unqualified extra-U step.\n\nApplying the clean projection kills q_i and q'_i. For every right basis vector v_i we therefore have\nJ†E^kJ v_i=T_{2k}(sigma_i)v_i,\nJ†UR E^kJ v_i=T_{2k+1}(sigma_i)w_i.\nSince the right vectors form an orthonormal basis, summing these actions gives exactly\nJ†E^kJ=sum_i |v_i>T_{2k}(sigma_i)T_{2k+1}(sigma_i)v+(sqrt(3)/2)|1>v. Its clean block coefficient is -1/2, but the unit-norm output is not the unnormalized clean vector alone.\n\nFor a unit input psi, if F denotes one of the displayed clean operators, projection succeeds with probability ||F psi||^2; conditional on a nonzero success the normalized system vector is F psi/||F psi||. A zero branch cannot be normalized. The clean block identities alone are not deterministic state-preparation claims or gate/query/finite-bit resource bounds. The source's general-SVD derivation in Eqs10.61–10.75 establishes these formulas directly; its printed Proposition10.7 has a narrower Hermitian label, which is not used as an extra premise or as a substitute for this derivation.", "rubric": [ { "criterion": "Uses E=U†RUR and shows the even clean block is sum_i |v_i>T_{2k}(sigma_i)T_{2k+1}(sigma_i)T_{2k}(sigma_i)T_{2k+1}(sigma_i)k after columnSweep_zeroed. For an old column c=k>c. Consequently all entries of old column c in that window are zero by the prefix hypothesis. Each two-row rotation combines two zero entries of that old column into two zeros, while rows outside the window are unchanged. Equivalently columnSweep_preserves_zero_column applies with lo=k and count=N-1-k. Thus old entries retain their previous values everywhere and in particular retain their required below-diagonal zeros; the old diagonal row c is outside the swept window and need not be zero. Combining old columns with fresh column k gives UpperPrefix at k+1.\n\nIf k>=N, no pivot row exists and the actual procedure skips the sweep. Every old ck in Fin N, so its required clause is vacuous. Hence the same one-index advance holds without a hidden k=k>c; zero-column preservation and unchanged outside rows retain the old invariant without zeroing A[c,c].", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.columnSweep_preserves_zero_column" ] }, { "claim": "Fresh zeroing and old-column preservation give UpperPrefix(k+1); when k>=N the fresh clause is vacuous and the skipped sweep also advances it.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.RectangularGivens.sweep_upperPrefix" ] }, { "claim": "Induction from the vacuous index0 gives UpperPrefix(M), hence all entries with c=1, N=2^(q+1)=2S, and L=N(N-1)/2. AdjacentGivens.decomposeSO always EMITS L planes, including harmless zero-angle identity rotations. It is a list-length equality, not a theorem that a particular SO matrix optimally needs L nonidentity gates. Each full-control selected plane on the existing q+1 wires has q controls and emits S RY plus2(S-1) CX records, total3S-2. The exact generated counts are RY=L S, CX=L*2(S-1), G=L(3S-2), with zero opaque-oracle calls. These follow from the actual count equalities, not just a cubic upper bound.\n\nSubstituting S=N/2 gives G/N=((N-1)/2)*(3N/2-2). Along N=2,4,8,... this ratio is unbounded, so no fixed constant C bounds THIS backend's emitted count by C N for all widths. More quantitatively, for N>=4, L>=N^2/4 and3N/2-2>=N, hence G>=N^3/4; also G<=3N^3/4=6S^3. Thus this generated family has Theta(N^3) records. The lower comparison was derived from the exact equality for this backend; the source upper-bound theorem G<=6S^3 by itself would not prove nonlinearity or any lower bound.\n\nAt q=0,N=2,L=1 the counts are one RY and zero CX, G=1. At q=1,N=4,L=6 they are12 RY and12 CX, G=24, while one selected plane costs only four records. This datum refutes treating one-plane cost four as the whole generated list cost, but does NOT refute every asymptotic linear bound:24=6N at this size. Only the growing-width ratio above supplies that family argument.\n\nNor does any of this prove a lower bound on optimal SO synthesis or preparation of one chosen state/column. The identity matrix, for example, is SO and has an empty exact circuit, although this unsimplified backend still emits its fixed list and identity-angle records. A different optimizer or architecture is not constrained by the backend's count equality. Full-control coverage must actually enumerate all non-target wires, and plane realization/decomposition are extra semantic obligations beyond the per-plane resource line. In this current source those obligations are supplied separately by decomposeSO_action, selectedStep_matrix/Gray transport and compileSO_eval; they are not absent globally and are not proved merely by reading gate counts.\n\nThe count concerns the actual finite exact-angle RY/CX list on q+1 wires and no opaque oracle. It is not arithmetic preprocessing time, a finite-bit angle-synthesis/error bound, hardware routing cost, or an optimal gate lower bound. The correct lesson is to aggregate the generated family under its actual model and keep that claim distinct from optimal necessity.", "rubric": [ { "criterion": "Separates per-plane and whole generated-backend counts: L=N(N-1)/2 emitted planes, RY=L2^q, CX=2L(2^q-1), G=L(3*2^q-2), zero oracle, with identity rotations retained.", "id": "c1", "weight": 1 }, { "criterion": "Uses the exact family equality to prove G/N unbounded (or a valid Theta(N^3) comparison), not a cubic upper bound alone or one finite-size datum. Qualifies all lower comparisons as THIS unsimplified backend, not optimal SO synthesis.", "id": "c2", "weight": 1 }, { "criterion": "Checks q=0 and q=1; assigns the N4=24 datum only its legitimate per-stage/whole-list role. Notes full-control/decomposition semantics, their actual separate current proofs, and the identity empty-circuit counterexample to necessity.", "id": "c3", "weight": 1 } ], "semantic_step_count": 4, "semantic_steps": [ { "claim": "Each full-control selected rotation on q+1 wires emits exactly2^q RY and2(2^q-1) CX gates with no opaque oracle.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.compileSelectedRy_ryCount" ] }, { "claim": "The fixed decomposeSO list emits N(N-1)/2 planes, including identity angles; this is not a minimal necessary plane count for each target.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.AdjacentGivens.decomposeSO_length" ] }, { "claim": "Exact aggregation G=N(N-1)/2*(3N/2-2) has unbounded G/N along powers of two, unlike a linear backend family; an upper bound alone would not establish that.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.compileSO_cubic_bound" ] }, { "claim": "The q1 check gives24 total versus4 per plane, not a universal asymptotic refutation. Distinct semantics/decomposition proofs and optimal/finite-bit/resource limits remain explicit.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.compileSelectedRySteps_noOracle" ] } ], "source_item_digest": "a9967b91c8189fd5fee0735e021c093ddca3092baf82b778254b0426617d066c", "source_refs": [ { "anchor": "compileSelectedRy_ryCount143-148; compileSelectedRy_cxCount150-156; compileSelectedRySteps_noOracle225-227", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/SelectedRyPlane.lean", "sha256": "0098a9ef2bfc1af9eeaa44357b0818628084be35f88854232496e14d715c2f40" }, { "anchor": "compileSO_cubic_bound196-199", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/GrayGivensCompiler.lean", "sha256": "21bb92938d2d0d81341700123e66195fdebb12e72db7faf066a897479fe942ce" }, { "anchor": "decomposeSO_length611-613", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/AdjacentGivens.lean", "sha256": "a43c3b4f543738cb317be01fa664fd466bfa28e21c33d10c7a90e74037a58d9c" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.GrayGivensCompiler.compileSO_cubic_bound", "declaration:QuantumBlockEncoding.AdjacentGivens.decomposeSO_length", "declaration:QuantumBlockEncoding.compileSelectedRySteps_noOracle" ], "target_nodes": [ "declaration:QuantumBlockEncoding.compileSelectedRy_ryCount", "declaration:QuantumBlockEncoding.compileSelectedRy_cxCount" ], "task_hash": "e088b52d1a863e5bc95a16a3d9f2578804b608921fb0f3c5efc0d8bfbcd69d59", "task_kind": "counterexample", "textbook_successor": { "predecessor_item_digest": "2e11be7304fdf2cc58f23e09e0f22e1e42c339a4b78bf1f957a432a7a8aa9155", "predecessor_key": "qcb-core-la-119", "repair_record": "shared/repairs/la-119-generated-backend-count-repair.json", "repair_record_digest": "07269a311d3585f1f7888bf1c54ac0d8e1b9b8dd16fee466dde740b0819db8e4", "review_digest": "21277909d5def61bf93c7b0fec8375162c385327bb080dcae480242776f8d50e", "review_record": "reports/textbook-v1/repairs/la-119-generated-backend-count-repair-review.json" }, "topic": "matrix_constructions" }, { "assumptions": [ "Cores have bond D; the chain is closed to scalar boundaries of dimension one", "Maximum bond is the maximum bond dimension over the chain" ], "context_hash": "100ec3be3b93e62182397a2d49d4773702fe9a6f48ed36749805e43e820f6611", "conventions": [ "The bond bound is structural, separate from construction cost; exact-real, not finite-bit", "Original derivation from inspected local source; no fresh Lean verification" ], "core_id": "qcb-core-la-220", "difficulty": { "level": "intermediate", "rationale": "A boundary interpretation of a bond bound separating structure from cost." }, "evidence_relation": "nl_bridge_from_inspected_source", "export_role": "open_reader_not_stable_training_release", "family": "native-stored-cost-220", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-086683e3c3c993b0d27a", "key": "qcb-core-la-220-textbook-v1-actual-closed-chain-bond-repair", "primary_sources": [], "question": "A stored matrix-product chain is assembled from n+1 stored cores, each of bond size D, closed against left and right boundary vectors of length D to form a scalar-boundary chain. It is proved that the maximum bond of the resulting chain is at most the maximum of D and one. Interpret this bond bound as a statement about the represented object, explain why the maximum with one appears, and determine the boundary values of D for which the bond equals one rather than D. Also state what this does and does not say about the cost of building the chain.", "reference_answer": "The theorem maxBond(result)<=max(D,1) is a structural upper bound for the returned dependent chain, not an equality by itself and not the minimal rank needed to represent its amplitudes. maxBond includes both outer bond dimensions. Closing the left and right boundaries makes those dimensions one, so maxBond is always at least one even when an internal typed bond has dimension zero.\n\nTo answer the question about the ACTUAL bond, inspect length as well as D. If n=0 there is exactly ONE stored core. tailChain first contracts the right length-D boundary into that core, giving shape D->1; closeLeft then contracts the left vector into the SAME core, giving shape1->1 followed by nil(1). Thus actual maxBond=1 for EVERY natural D. In particular n=0,D=2 gives actual maximum1 even though the displayed bound is2; D alone cannot characterize when the returned maximum equals one.\n\nIf n>=1 there are at least two cores. The last is contracted to terminal bond1, the first to initial bond1, but one or more internal bonds remain dimensionD in the returned type. The producer does not canonicalize/compress them in response to zero or rank-deficient entries. Its typed maximum is therefore exactly max(1,D). Hence for multi-core outputs actual maxBond=1 when D=0 orD=1, and equalsD when D>=1 (at D=1 those descriptions coincide). For D>=2 it is greater than one. At D=0 there ARE zero-dimensional internal bonds in a multi-core chain, not 'no interior bonds'; all corresponding scalar amplitudes are zero because contraction includes an empty bond sum. The positive maximum is maintained by the outer unit bonds. In the single-core D=0 case its boundary contractions are empty sums as well, giving zero amplitudes but still a typed1->1 core. No normalization was assumed anywhere.\n\nThus the exact structural cases are: n=0 gives1 for allD; n>=1 gives max(D,1). These follow by unfolding the actual terminal/initial/tailChain/closeLeft producer and the maxBond recursion, and agree with its denotation/refinement theorem. They do not say that the represented tensor has minimal TT rankD: a chain with all-zero entries may retain typedD bonds while representing zero.\n\nFinally bond size by itself says nothing about construction traffic or runtime. There is a SEPARATE same-run theorem giving the exact-real stored-word total upper budget20D^2+30D+5n^2+7n+31, including boundary arithmetic/materialization, copied table references and nodes; ofTable_certified combines its value refinement with that bound. It is not derived from maxBond alone, not an exact attained cost identity, and not full arbitrary-precision finite-bit runtime. A static source reading neither executes that producer nor rechecks its kernel proofs.", "rubric": [ { "criterion": "Separates the structural upper bound from actual typed maxBond/minimal tensor rank, explains outer unit boundaries and retained internal dimensions.", "id": "c1", "weight": 1 }, { "criterion": "Derives actual maxBond=1 for n=0 and every D; for n>=1 derives max(D,1), hence1 at D0/1 andD for D>=1. Correctly retains zero-dimensional internal bonds at D0 and notes zero amplitudes/no normalization premise.", "id": "c2", "weight": 1 }, { "criterion": "Keeps actual bond structure separate from the source's separately proved same-run exact-real cost upper bound and finite-bit runtime; does not infer cost or minimal rank from maxBond.", "id": "c3", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Closing both boundaries on one core gives shape1->1; with multiple cores, only outer bonds collapse while typed internalD bonds remain.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_refines" ] }, { "claim": "The actual maximum is1 for n0 regardless ofD, and max(D,1) for n>=1; evaluating the upper bound alone cannot distinguish these length cases.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_maxBond" ] }, { "claim": "Actual bond dimensions, minimal representation rank and the separately bounded same-run exact-real construction work are distinct quantities.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_certified" ] } ], "source_item_digest": "69e87347564a6112eb1c794b7338c946298bd98b196261c6aeb81cfb2c1c1099", "source_refs": [ { "anchor": "ofTable_maxBond249-254; ofTable_certified257-265", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/StoredMatrixProductChain.lean", "sha256": "7a697d064a5fddd909bba048d6adc36cb645421bd565e2357588fef1ff0449f2" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_certified", "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_refines" ], "target_nodes": [ "declaration:QuantumBlockEncoding.StoredMatrixProductChain.ofTable_maxBond" ], "task_hash": "b075aea442229464760567200fff4f8c092e393938961c4a8bbf46334a3d99c8", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "89824ba88947fc98d212a79d3b548e480a272b0577fdce101305eb146bf3125c", "predecessor_key": "qcb-core-la-220", "repair_record": "shared/repairs/la-220-closed-chain-bond-repair.json", "repair_record_digest": "3f12d99d6c59e323aade2735a0015f43d0be7b4971c8f905cc4aaabdd28f24eb", "review_digest": "8f5c6ea9bac9fc2ce90b72ddb40f5d81385f75669b20a832f3dfe17db05aa596", "review_record": "reports/textbook-v1/repairs/la-220-closed-chain-bond-repair-review.json" }, "topic": "stored_cost_refinement" }, { "assumptions": [ "Real rational entries; normalizer fixed to 1", "Clean block is the single top-left entry x" ], "context_hash": "b0b4544945478b169f436b197018c92c0de67b1afe388e0a7b3d848968f40a18", "conventions": [ "Dilation layout [[x,y],[y,-x]] with clean entry at (0,0)", "Orthogonality tested entrywise by row Gram overlaps" ], "core_id": "qcb-core-be-008", "difficulty": { "level": "intermediate", "rationale": "Explicit 2x2 orthogonality computation yielding the sharp |x|<=1 boundary and the contraction/normalizer lower-bound interpretation; grounded in the scalarDilation lemmas." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-scalar-dilation-contraction", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-e7fb00b4c9fdf63a4ea3", "key": "qcb-core-be-008-textbook-v1-contraction-carrier-answer-repair", "primary_sources": [], "question": "Consider the 2x2 scalar dilation whose top-left clean entry is x, whose two off-diagonal entries are both y, and whose bottom-right entry is -x. Treat the single top-left entry as the one-dimensional clean block encoding the scalar x with normalizer 1. Determine the exact condition on (x,y) under which this dilation is orthogonal (a valid real one-qubit-signal block encoding), compute the two row self-overlaps and the row cross-overlap, and explain the boundary: what is the largest |x| for which a real y exists making the dilation orthogonal, and what does exceeding it mean for block-encoding a scalar of magnitude greater than 1 with normalizer 1?", "reference_answer": "Use the inherited rational-entry backend for the initial matrix: x,y∈Q, regarded as real numbers. For D=[[x,y],[y,-x]], the two row self-overlaps are x²+y² and y²+(-x)²=x²+y². The cross-overlaps are xy+y(-x)=0 and yx+(-x)y=0. Consequently DD^T=(x²+y²)I2, and D is orthogonal exactly when x²+y²=1; the same equation gives D^T D=I2. These identities are polynomial algebra valid overQ andR. Automatic zero cross-overlap alone is not unit normalization.\n\nThe question separately asks whether a REAL y exists. For a fixed rational x viewed inR (indeed for any real x under this scalar extension), a real y with y²=1-x² exists iff|x|≤1. Necessity is y²≥0; sufficiency is y=±sqrt(1-x²). This is real-existence mathematics, not a claim that those witnesses have rational entries. Remaining within the inherited rational matrix backend, a rational y exists exactly when1-x² is a square of a rational number. For rational x=1/2 the required y=±sqrt(3)/2 is irrational: if sqrt3=a/b in lowest positive integer terms, a²=3b² forces3 to divide a and then b, a contradiction. Thus|x|≤1 alone is not a rational-completion theorem. Conversely x=3/5,y=4/5 is rational; x0 admits y±1. At x=1 or-1, y0 works over both carriers, so the largest possible |x| is1 in either carrier, although the interior rational-existence criterion is stricter.\n\nThe normalizer-one obstruction is independent of auxiliary dimension. Let J be any clean isometric embedding, J†J=I, and U any unitary (real orthogonal in the real case). Then ||J†UJ z||≤||UJz||=||z||. Hence every clean compression is a contraction. In particular a scalar c with|c|>1 cannot appear as a normalizer-one clean block in ANY signal-register dimension. More signal qubits do not remedy this. This also follows entrywise because an entry of a unit vector has magnitude≤1.\n\nWith conventional positive normalizer alpha and exact target A=alpha B, it is B=J†UJ=A/alpha that must have norm≤1; for a scalar c this requires alpha≥|c|. Raising alpha is the necessary normalization remedy, not enlarging the signal space at fixed alpha1. Over real matrices, a2x2 dilation then realizes c/alpha; over rational entries its rational completion condition still applies and no arbitrary real square-root witness is silently imported. The Gram lemmas prove conditional algebraic identities, not a general rational-existence, finite-bit synthesis, success-probability or efficient circuit theorem.", "rubric": [ { "criterion": "Computes both row norms x²+y² and both cross terms0, and derives full orthogonality iff x²+y²=1.", "id": "r1", "weight": 1 }, { "criterion": "Distinguishes real-y existence |x|≤1 from rational-y existence1-x² a rational square, including the x1/2 obstruction and endpoint y0.", "id": "r2", "weight": 1 }, { "criterion": "Proves clean compression is a contraction for every auxiliary dimension; rejects added signal qubits as a remedy at alpha1 and states the positive-normalizer convention A=alpha B correctly.", "id": "r3", "weight": 1 } ], "semantic_step_count": 2, "semantic_steps": [ { "claim": "For rational x,y embedded inR, both row norms are x²+y² and the cross overlap0, so D is orthogonal iff x²+y²=1.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_row0_normSq", "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_row1_normSq", "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_rows01_orthogonal" ] }, { "claim": "The question's real-completion existence has boundary|x|≤1, but rational witnesses need a rational square; universal clean-compression contraction rules out|c|>1 at normalizer1 in every auxiliary dimension.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_row0_unit_norm_of" ] } ], "source_item_digest": "ab4a63627ae3bf951da2437aeaee8a566a6610f4487fbdbb7ba3c8344a2f4974", "source_refs": [ { "anchor": "def scalarDilation [[x,y],[y,-x]] (lines 532-537); scalarDilationRowDot and theorems scalarDilation_row0_normSq / row1_normSq = x*x+y*y (lines 556-571); scalarDilation_rows01_orthogonal = 0 (lines 583-588); scalarDilation_row0_unit_norm_of (lines 573-581)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/BlockEncodingClassics.lean", "sha256": "94ac6632924df0a49b1ebd99bb5d9ed69b2ea463ec527033f38bd73deeb1346a" } ], "split": "dev", "supporting_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_row1_normSq", "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_rows01_orthogonal", "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_row0_unit_norm_of" ], "target_nodes": [ "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation", "declaration:QuantumBlockEncoding.BlockEncodingClassics.scalarDilation_row0_normSq" ], "task_hash": "49191506c656a12fd921fc1030d2e4070a8d03d5929df8054c13818d31e70a40", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "f7551d5420a1e61e7be02a212873d5cc4384b08c62628605282e64b91c4a7181", "predecessor_key": "qcb-core-be-008", "repair_record": "shared/repairs/be-008-contraction-carrier-answer-repair.json", "repair_record_digest": "aa3d8dd0d2ca5db0dfb2a2f84a045d398df7860fd834759d424d61ae753e6d17", "review_digest": "4df56bcbfedbf0f37ed4c8fada83351f35f71cef958488dd11e17fb30c3b7aaf", "review_record": "reports/textbook-v1/repairs/be-008-contraction-carrier-answer-repair-review.json" }, "topic": "block_encoding" }, { "assumptions": [ "The operator O is unitary before relabelling", "Rows and columns are reindexed by one and the same bijection" ], "context_hash": "657d6de459797febd0ce33f485a623fe5a2eb1656f8e8c1a8d8b1f521dfb807c", "conventions": [ "Reindexing is an algebra equivalence, hence multiplicative", "The identity matrix reindexes to the identity" ], "core_id": "qcb-core-be-108", "difficulty": { "level": "intermediate", "rationale": "Requires knowing that conjugate transpose and matrix multiplication both commute with a single-equivalence reindexing, and seeing why consistent relabelling of the contracted index is indispensable." }, "evidence_relation": "local_source_supported", "export_role": "open_reader_not_stable_training_release", "family": "be-reindex-preserves-unitary", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-51ec8b28464c9e4dd782", "key": "qcb-core-be-108-textbook-answer-reindex-isomorphism-boundary", "primary_sources": [], "question": "Suppose an operator on a finite register is unitary, and we relabel the register's basis by a bijection (reindexing both rows and columns through the same equivalence of index sets). Prove that the reindexed operator is again unitary. Make explicit how the conjugate transpose commutes with reindexing and why using the SAME equivalence on rows and columns is essential for the argument to close.", "reference_answer": "Let I,K be finite index sets, e:I->K a bijection, and O a complex unitary matrix on I. Write U_e for its basis-relabeling matrix, U_e[e(i),i]=1 and all other entries zero; it is a unitary map from C^I to C^K. Define T_e(A)=U_e A U_e†, so T_e(A)[e(i),e(j)]=A[i,j], equivalently T_e(A)[a,b]=A[e^{-1}(a),e^{-1}(b)]. This is the simultaneous pullback along e^{-1} on the two matrix axes.\n\nConjugate transpose commutes with this SAME-axis relabeling. Entrywise, T_e(A)†[a,b]=conjugate(A[e^{-1}(b),e^{-1}(a)])=T_e(A†)[a,b]; equivalently (U_e A U_e†)†=U_e A† U_e†. Multiplication also commutes: T_e(A)T_e(B)=U_e A(U_e†U_e)B U_e†=U_e AB U_e†=T_e(AB). In entries the contracted label z in K is replaced bijectively by e^{-1}(z) in I. Finally T_e(I_I)=I_K. Therefore\nT_e(O)† T_e(O)=T_e(O†)T_e(O)=T_e(O†O)=T_e(I_I)=I_K.\nThe analogous reverse product is also I_K (or follows for square finite-dimensional isometries); hence the reindexed operator is unitary. This is exactly the same-map algebra-equivalence proof implemented by reindex_unitary.\n\nThe question's word 'essential' must be read with its explicit qualification 'for the argument': using the same bijection is what makes this single transformation T_e a unital star-algebra isomorphism between the two square-matrix algebras, commuting with adjoint. It is NOT a necessary condition for the resulting matrix to remain square or unitary. To see the boundary, consider arbitrary bijections e,f:I->K and the two-axis map T_{e,f}(A)=U_e A U_f†. It still gives a square K×K matrix. Its adjoint is T_{f,e}(A†), with the axes exchanged, rather than in general T_{e,f}(A†). Also\nT_{e,f}(A)T_{e,f}(B)=U_e A(U_f†U_e)B U_f†,\nwhich is generally not T_{e,f}(AB), and T_{e,f}(I)=U_e U_f† need not be I. Thus one cannot repeat the SAME fixed-map algebra-homomorphism proof when e differs from f.\n\nUnitarity nevertheless survives by a different cancellation:\nR=U_e O U_f† implies R†R=U_f O†(U_e†U_e)O U_f†=U_f I_I U_f†=I_K,\nand RR†=I_K similarly. More generally the correctly matched transformation rule T_{e,f}(A)T_{f,g}(B)=T_{e,g}(AB) explains how the swapped axes in the adjoint give this cancellation. Consistent contracted indices are needed for composition; that does not require the output and input basis bijections to be equal in a general two-sided relabeling.\n\nFor an explicit two-dimensional counterexample to any stronger necessity claim, take O=I_2, e=identity and f=swap. Then U_e=I_2, U_f=X=[[0,1],[1,0]], and R=X is square and unitary despite e!=f. Here T_{e,f}(I)=X!=I and T_{e,f}(I)T_{e,f}(I)=I!=X=T_{e,f}(I*I), displaying failure of the one-map unital/product rules without failure of unitarity. Adjoint commutation with that same map can fail too: for Z=diag(1,-1), T_{e,f}(Z)=ZX has adjoint -ZX, whereas T_{e,f}(Z†)=ZX. In zero- or one-dimensional spaces the same-axis proof still holds; there may simply be no distinct bijections available to instantiate this two-dimensional boundary example.\n\nThese comparisons do not replace the original standing hypothesis: the requested proof remains the same-bijection theorem. They correct only an overstatement about why that proof works. The result is exact finite-dimensional complex algebra; it makes no primitive-circuit or finite-bit cost claim and does not by itself preserve a separately chosen clean projected block unless its register/clean basis identifications are transported consistently.", "rubric": [ { "criterion": "Defines simultaneous reindexing by T_e(A)[e(i),e(j)]=A[i,j] and states O†O=I and the goal T_e(O)†T_e(O)=I.", "id": "c1", "weight": 1 }, { "criterion": "Proves same-axis adjoint commutation, multiplicativity and identity preservation, then uses these to prove unitarity (with finite square dimensions or both unitary identities understood).", "id": "c2", "weight": 1 }, { "criterion": "Explains SAME is essential to this fixed-map algebra-equivalence proof, not to squareness or unitarity in general. Gives R=U_e O U_f† for independent axes and a valid cancellation/counterexample, explicitly rejecting the old 'not square-compatible' explanation without changing the standing same-axis theorem.", "id": "c3", "weight": 1 } ], "semantic_step_count": 5, "semantic_steps": [ { "claim": "Define R=T_e(O) by R[e(i),e(j)]=O[i,j], i.e. pullback along e^{-1} on both axes, with the same finite source/target bases.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.reindex_unitary" ] }, { "claim": "For simultaneous reindexing, swapping matrix axes under adjoint leaves the same bijection on each axis, so T_e(O)†=T_e(O†).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.reindex_unitary" ] }, { "claim": "The same-axis transformation preserves products by consistently relabeling the contracted index and is unital: T_e(A)T_e(B)=T_e(AB).", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.reindex_unitary" ] }, { "claim": "Apply the original O†O=I identity, same-map adjoint/product rules and T_e(I)=I to get R†R=I and hence finite-dimensional unitarity.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.reindex_unitary" ] }, { "claim": "The same-bijection restriction belongs to this fixed-map algebra-equivalence proof, not a universal necessity for unitarity: independent-axis relabeling remains square/unitary as U_e O U_f†, while its same-map product/identity/adjoint rules can fail.", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.reindex_unitary" ] } ], "source_item_digest": "5eee24b8ce50209e2e4427318cd14e08055a01afadd92c2fbdd3e5aa2bb61e32", "source_refs": [ { "anchor": "theorem reindex_unitary {ι κ : Type*} [Fintype ι] [DecidableEq ι] [Fintype κ] [DecidableEq κ] (equiv : ι ≃ κ) (operator : _root_.Matrix ι ι ℂ) (unitary : operator ∈ _root_.Matrix.unitaryGroup ι ℂ) : _root_.Matrix.reindexAlgEquiv ℂ ℂ equiv operator ∈ _root_.Matrix.unitaryGroup κ ℂ (lines 105-119)", "license": "MIT (repository LICENSE)", "path": "QuantumBlockEncoding/Robin/ComplexLCU.lean", "sha256": "aa51d33a6ca788b17879e1d88030b525f4f035c97d74588636ccf25f8c04b843" } ], "split": "train", "supporting_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.equivPermutationMatrix_conjugates_reindex" ], "target_nodes": [ "declaration:QuantumBlockEncoding.Robin.ComplexLCU.reindex_unitary" ], "task_hash": "b54297a99a53b332bfed76d44a3b4847e614852d7cc23d0a60c81be24b6a191a", "task_kind": "proof", "textbook_successor": { "predecessor_item_digest": "a6acd83107961bd9360a0c2c55995f194064e1416340d73260285f5dbcb73c02", "predecessor_key": "qcb-core-be-108", "repair_record": "shared/repairs/be-108-reindex-isomorphism-boundary-repair.json", "repair_record_digest": "4c19bd5b346da36b690a62c8c64cb1293de7db96437a66becae0054e511a1e69", "review_digest": "2ff580871e5f7433c15b32425921046217aa207b97600fc0622e36544df4076e", "review_record": "reports/textbook-v1/repairs/be-108-reindex-isomorphism-boundary-repair-review.json" }, "topic": "oracle_composition" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-414", "difficulty": { "level": "intermediate", "rationale": "A Haar-twirl covariance proof via the invariance substitution; it is the symmetrization step enabling the cloning optimality argument, distinct from the specific commutant/projector computations done by the other author." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "haar-twirl-covariance", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-6e73875d984b502be8de", "key": "qcb-core-plan-414-textbook-v1-invariant-output-block-repair", "primary_sources": [ { "anchors": "Eq. (8.10) in the proof of Proposition 8.2 (PDF pages 66-67) and Exercise 8.4 (PDF page 69): the twirled channel is U^{tensor n}-covariant by left-invariance of the Haar measure.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "66-67, 69", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "Given a quantum channel T : L(H^{tensor N}) -> L(H^{tensor M}), define its twirled version by averaging over the Haar measure on U_d: Tbar(X) := integral_{U_d} dU (U^dagger)^{tensor M} T( U^{tensor N} X (U^dagger)^{tensor N} ) U^{tensor M}. Prove that Tbar is covariant, i.e. Tbar( V^{tensor N} X (V^dagger)^{tensor N} ) = V^{tensor M} Tbar(X) (V^dagger)^{tensor M} for all V in U_d, using the left-invariance of the Haar measure. Then explain in one sentence why covariance is the property that lets Schur-Weyl duality constrain the output on the symmetric subspace.", "reference_answer": "Put U_r=U^{tensor r}, and take normalized left Haar measure mu on the compact group U(d). First justify the translation actually needed. For fixed V, the right-translated measure nu(A)=mu(A V^{-1}) is a probability measure and satisfies nu(gA)=mu(g A V^{-1})=mu(A V^{-1})=nu(A). Uniqueness of normalized left Haar therefore gives nu=mu. Thus normalized compact left Haar is also right invariant. The substitution W=UV below is RIGHT translation, derived from left invariance plus normalization/uniqueness; it is not directly a left translation.\n\nFor any operator X, expand\nTbar(V_N X V_N^dagger)\n= integral dU (U^dagger)_M T((UV)_N X ((UV)^dagger)_N) U_M.\nSet W=UV, so U=W V^dagger and U^dagger=V W^dagger. Right invariance just established leaves dU=dW. Consequently this equals\nintegral dW V_M (W^dagger)_M T(W_N X (W^dagger)_N) W_M (V^dagger)_M\n= V_M Tbar(X) (V^dagger)_M.\nThis proves the specified covariance, for every X, not merely density operators or symmetric inputs. Every integrand map is a unitary conjugation before and after T; positivity of every matrix amplification and trace preservation survive finite-dimensional averaging, so Tbar is again a quantum channel.\n\nThe requested one-sentence Schur-Weyl explanation is: covariance converts invariance of an INPUT into invariance of its OUTPUT, so the multiplicity-one irreducible symmetric output block is scalar by Schur's lemma, without asserting that an arbitrary input has a scalar output.\n\nMore explicitly, write Pi_r for the projector onto Sym^r(C^d) and q_d(r)=dim Sym^r(C^d). Whenever q_d(N)>0, tau_N=Pi_N/q_d(N) is an invariant density operator. Covariance implies Y=Tbar(tau_N) commutes with every V_M. The compression Pi_M Y Pi_M also commutes with the irreducible U(d) action on Sym^M. Thus, if q_d(M)>0,\nPi_M Y Pi_M = (lambda/q_d(M)) Pi_M,\nlambda=tr(Pi_M Y), 0<=lambda<=1.\nThis is the statement used in the source's cloning argument. It describes only that compressed block, not all of Y or a channel determined by one parameter. Nor does unitary commutation force scalars on every isotypic component: in the Schur-Weyl order V_lambda tensor U_lambda^d it permits End(V_lambda) tensor I on the multiplicity factor. Additional permutation commutation is a different condition.\n\nFor a direct boundary witness, d=2,N=M=1,T=identity gives Tbar=identity and Tbar(|0><0|)=|0><0|, not a scalar on Sym^1(C^2)=C^2. Covariance is an intertwining identity for a SUPEROPERATOR, not permission to apply the ordinary irreducible-endomorphism version of Schur directly to that superoperator.\n\nN,M may be zero, with tensor power zero the one-dimensional tensor unit and U_0=1; the same proof applies. For d=1 the nonzero spaces and symmetric blocks are one-dimensional. For d=0 the covariance algebra still makes sense whenever such a channel exists, but positive tensor powers of C^0 are zero spaces with no trace-one state: tau_r and division by q_0(r)=0 must not be used there. For r=0 the tensor unit and its projector have dimension one. These cases do not enlarge the scalar conclusion. No optimal-cloning fidelity or efficient implementation is proved by this covariance identity alone.", "rubric": [ { "criterion": "Expands the rotated input, groups U_N V_N=(UV)_N, and keeps input/output tensor powers and adjoints in the displayed order.", "id": "r1", "weight": 1 }, { "criterion": "Correctly identifies W=UV as right translation and derives its measure invariance from uniqueness/normalization of compact left Haar; uses U=W V^dagger and U^dagger=V W^dagger without the original inverse-substitution typo.", "id": "r2", "weight": 1 }, { "criterion": "Pulls the fixed V_M and (V^dagger)_M outside the integral, recognizes Tbar(X), and proves covariance for every X. Does not add symmetric-input or scalar-output premises to that identity.", "id": "r3", "weight": 1 }, { "criterion": "Gives a qualified Schur-Weyl explanation: invariant input yields invariant output, whose irreducible symmetric compression is scalar when the normalized objects exist. Does not infer scalarity for arbitrary input, for multiplicity factors, or for the entire channel. The invariant tau_N calculation, identity-channel witness and zero-normalization discussion in the reference elaborate this boundary; they are not additional premises for the covariance equation.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "Expanding a rotated input groups its inside factors as (UV)_N X ((UV)^dagger)_N.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "Uniqueness of normalized compact left Haar implies right invariance, validating W=UV with U=W V^dagger.", "supporting_nodes": [ "concept:harmonic-analysis" ] }, { "claim": "The outside V factors give covariance; at the invariant input tau_N its symmetric output block is scalar, whereas arbitrary inputs and other multiplicities need not be.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "752a79aa877ed168dd52cc60b3828093ab27d127c04242a326ecbe936e29dc7c", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:harmonic-analysis" ], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "85ae0bd9981f9ea61a80c3a9154955f283262b283d434e027cd8d5879158bf25", "task_kind": "proof", "textbook_successor": { "predecessor_item_digest": "7655ddef61c1ba6c3690219bfc2fe0d1480ae1f152b262040ab453308f3159d0", "predecessor_key": "qcb-core-plan-414-twirled-channel-covariance-haar", "repair_record": "shared/repairs/plan-414-invariant-output-block-repair.json", "repair_record_digest": "4ebee5ff0c5ce67a2f1cf5ec3b53ceacafc6aaec41a99ce501937dcc8ed4d184", "review_digest": "e0260d51e579beb45173a44551f82f643fa82f8e60f0684ac836956d13f8b35b", "review_record": "reports/textbook-v1/repairs/plan-414-invariant-output-block-repair-review.json" }, "topic": "planned_quantum_information_apps" }, { "assumptions": [ "A is Hermitian so eigenvalue transformation f(A) is well defined" ], "context_hash": "0a2157475d0d9f2e81342f911f2e17d3a31d18e0c01bc94c6f2c1346a381c042", "conventions": [ "Subnormalization factor is the constant by which the encoded operator is divided inside the flag block", "Even/odd parts defined by x to minus x symmetry" ], "core_id": "qcb-core-plan-509", "difficulty": { "level": "intermediate", "rationale": "Combines the elementary parity decomposition with a quantitative subnormalization accounting for a two-term balanced linear combination." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qsc-lcu-chebyshev-synthesis", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-131531a12a6a62078c20", "key": "qcb-core-plan-509-textbook-v1-balanced-selected-port-boundary", "primary_sources": [ { "anchors": "Eq. (10.105) f_even(x) = (f(x)+f(-x))/2, f_odd(x) = (f(x)-f(-x))/2, and the statement that combining the two block encodings with one more ancilla to realize f(A) = f_even(A) + f_odd(A) multiplies the subnormalization factor by two", "edition": "29 April 2026 public 447-page edition", "inspection": "full bounded primary text read", "pages": "179", "redistribution": "not redistributed", "reuse": "Original questions/derivations; no copied source text; external source rights not relicensed", "sha256": "56825000d53025dea4c2998dd6a061592b8fb2b9a446c3194c5f408183327599", "url": "https://math.berkeley.edu/~linlin/qasc/live_notes_0429.pdf" } ], "question": "The pure qubitization-plus-linear-combination construction directly produces block encodings of polynomials of DEFINITE parity (all-even or all-odd Chebyshev content). Suppose instead you are given a general real polynomial f with no parity symmetry and a Hermitian operator A, and you want a block encoding of f(A). Prove that f splits uniquely into an even part and an odd part, that each part is separately implementable by the parity-respecting construction, and that combining them with one additional ancilla qubit to form f(A) multiplies the subnormalization factor by two. Explain why the factor two is unavoidable in this naive combination.", "reference_answer": "The polynomial split is f_e(x)=(f(x)+f(-x))/2 and f_o(x)=(f(x)-f(-x))/2. Direct substitution shows f_e is even, f_o is odd, and f_e+f_o=f. If f=e+o with e even and o odd, then f(-x)=e(x)-o(x), so adding/subtracting recovers these same formulas; uniqueness is as polynomials, not a claim that their evaluations on one spectrum uniquely determine them.\n\nFor the finite-dimensional Hermitian operator in the quantum circuit setting, write A=V diag(lambda_i)V†. An even polynomial's right singular-value transform equals its eigenvalue function because f_e(|lambda|)=f_e(lambda); an odd polynomial's balanced transform supplies the sign of lambda, giving f_o(lambda). Choose any unitary sign completion at zero eigenvalues; f_o(0)=0 makes that choice immaterial. Hence the parity-respecting construction implements the desired two functions, not an arbitrary mixed-parity singular-value function of a non-Hermitian matrix.\n\nInput access and scale are real obligations. If A is not already given as an exact normalized block, choose a known rho>0 with ||A||<=rho and let B=A/rho. Obtain an exact block encoding of B and use the definite-parity polynomials g_e(x)=f_e(rho*x) and g_o(x)=f_o(rho*x). They retain their parities and g_e(B)=f_e(A), g_o(B)=f_o(A). A finite Hermitian matrix has such a rho. Concretely, for S=sqrt(I-B^2), the block matrix U_B=[[B,S],[S,-B]] is unitary: spectral functional calculus gives BS=SB and B^2+S^2=I, so its square is I and its clean block is B. Expanding each g in its finite definite-parity Chebyshev basis, implementing the corresponding qubitized blocks, and using phase-aware coefficient LCU gives each parity branch at a positive coefficient-norm factor (with a separate zero-block construction when necessary). This is existence/construction mathematics, not an efficient oracle-synthesis guarantee from the word Hermitian alone. Chebyshev coefficients, coefficient preparation, SELECT and input-oracle work must actually be supplied and charged in any implementation; approximation would require an additional error contract. Nothing here assumes ||A||<=1 without this normalization step.\n\nPut E=f_e(A) and O=f_o(A). Implement the two parity branches at compatible clean signal registers s and a common positive scale c, meaning J†U_e J=E/c and J†U_o J=O/c with J|psi>=|0^r>_s|psi>. The scale need not be minimal. Workspace padding preserves a clean block. If existing nonzero-branch factors are c_e,c_o, take c>=max(c_e,c_o) and explicitly attenuate their accepted blocks by c_e/c or c_o/c, for example by a clean-qubit rotation whose00 matrix element is that ratio. Its clean register is included in s before combining; equalization/padding is not free and is not part of the assertion that the COMBINER adds just one qubit. A zero branch can be represented by a unitary with zero accepted block, for instance X on a signal qubit, at any chosen positive c; never divide by its zero coefficient norm.\n\nIntroduce exactly one new selector b and set SELECT=|0><0|_b tensor U_e+|1><1|_b tensor U_o, with V=(H_b tensor I)SELECT(H_b tensor I). For every normalized input |psi>, the complete projected-port transcript is\nV|0>_b|0^r>_s|psi>\n = |0>_b|0^r>_s (E+O)|psi>/(2c)\n + |1>_b|0^r>_s (E-O)|psi>/(2c) + |G_psi>,\nwhere (I_b tensor <0^r|_s tensor I)|G_psi>=0. Thus projection onto the FIXED b=0,s=0 port gives f(A)/(2c), a block encoding with declared factor2c. It is a projected block, not an assertion that the full unitary output has no garbage. Its success probability is ||f(A)psi||^2/(4c^2); if f(A)psi=0 it is0 and there is no normalized conditional output. Otherwise conditioning on this port gives f(A)psi/||f(A)psi||, with the exact coherent phase retained.\n\nThe coefficient1/2 is forced by this particular balanced H-SELECT-H and this chosen projector: each branch contributes the product of two Hadamard entries1/sqrt(2). It cannot be changed while leaving those operations and that accepted port fixed. This is the precise sense of unavoidable in the naive combination. It is not a universal lower bound on block factors or a state-independent claim about probability loss. The alternative control port contains (E-O)/(2c), not generically the sum, but it need not be useless or distinct on every allowed spectrum. For f(x)=1+x and A=0, a genuinely parity-mixed polynomial has E=I,O=0, and BOTH signal-clean control ports carry I/(2c). For E=0 they differ only by a minus sign. Conversely E=O gives a zero difference port. Polynomial parity alone does not preclude any of these operator degeneracies.\n\nOther schemes can use different accepted projectors, exploit known zero/equal branches, or use coefficient-weighted LCU: factors c_e,c_o give sum factor c_e+c_o using amplitudes sqrt(c_e/(c_e+c_o)) and sqrt(c_o/(c_e+c_o)), rather than2*max(c_e,c_o). That comparison does not remove the literal1/2 from the fixed common-c balanced selected block. A claimed minimal normalization or a complete gate, input-preparation, postselection/amplification or readout complexity theorem would require separate analysis.", "rubric": [ { "criterion": "Derives and proves uniqueness of the polynomial even/odd split, distinguishing polynomial uniqueness from degeneracies after evaluating on spec(A).", "id": "r1", "weight": 1 }, { "criterion": "Explains why definite-parity SVT gives the corresponding eigenvalue function for Hermitian A, including null eigenvalues; makes any A/rho input normalization, polynomial rescaling and oracle/preparation/equalization work explicit rather than silently assuming a unit-normalized available input.", "id": "r2", "weight": 1 }, { "criterion": "Uses compatible clean signal registers and common c>0 to calculate H-SELECT-H: chosen b0,s0 block f(A)/(2c), other clean control block (f_e(A)-f_o(A))/(2c), flag-orthogonal garbage and the actual input-dependent success probability/zero-output boundary.", "id": "r3", "weight": 1 }, { "criterion": "Explains the forced1/2 only for this fixed balanced common-c selected-port scheme. Refutes exclusive-success-port/global-optimality overclaims with an allowed degenerate-spectrum example and separates weighted LCU, workspace equalization, implementation and amplification costs.", "id": "r4", "weight": 1 } ], "semantic_step_count": 6, "semantic_steps": [ { "claim": "The parity averages are even/odd polynomials, sum to f, and are uniquely recovered by adding/subtracting f(x) and f(-x).", "supporting_nodes": [ "concept:approximation" ] }, { "claim": "Definite-parity transforms equal the desired functions of Hermitian A, with input normalization/rescaling and compatible common-c branch encodings made explicit.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "A new balanced selector controls the two completed parity encodings on the same signal workspace.", "supporting_nodes": [ "concept:operator-theory" ] }, { "claim": "After recombination, the clean b0 port is (E+O)/(2c), the clean b1 port is (E-O)/(2c), and remaining terms are orthogonal to the clean signal subspace.", "supporting_nodes": [ "concept:block-encoding" ] }, { "claim": "The fixed chosen block is f(A)/(2c) with input-dependent success probability; zero target output cannot be normalized conditionally.", "supporting_nodes": [ "concept:approximation" ] }, { "claim": "The1/2 follows from fixed Hadamard matrix elements, not exclusive-success-port or global optimality; allowed degenerate spectra and alternative weighted schemes delimit the claim.", "supporting_nodes": [ "concept:operator-theory" ] } ], "source_item_digest": "89f82db5a14621049574d7c581ff4b01fe7e90ffbc0280bb0d9290a0476cad14", "source_refs": [ { "anchor": "lin-wiebe-2026-qasc", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [], "target_nodes": [ "concept:approximation", "concept:block-encoding", "concept:operator-theory" ], "task_hash": "40fb3cf5fcb9403c296156eefe135d9515b24d2b54959999727b40024712f0ee", "task_kind": "proof", "textbook_successor": { "predecessor_item_digest": "2ee2566f02d779dbf9ff9fc536ecc37a6a7942b6547a893063cd89d3d3bc6052", "predecessor_key": "qcb-core-plan-509-even-odd-split-subnormalization-doubling", "repair_record": "shared/repairs/plan-509-balanced-selected-port-boundary-repair.json", "repair_record_digest": "fdada9695f430dfb21067da2a448e81ad393672d4d39e74cea469443c46c44b5", "review_digest": "dac3cff280624f300bbcd4cbfaa8e0498f3dd77766a3edcc8f6bcbe9ae50db22", "review_record": "reports/textbook-v1/repairs/plan-509-balanced-selected-port-boundary-repair-review.json" }, "topic": "planned_polynomial_signals" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-404", "difficulty": { "level": "intermediate", "rationale": "A boundary/asymptotic comparison of the two de Finetti constants, tracing the d-vs-d^2 gap to the purification dimension and extracting the vanishing regime in n and k." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "definetti-bound-scaling", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-61e1389c896a31eeb7ad", "key": "qcb-core-plan-404-textbook-v1-excess-copy-count-repair", "primary_sources": [ { "anchors": "Propositions 7.4 and 7.6 on PDF pages 62-65: pure bound sqrt(dk/(n-k)) (Eq. 7.15) vs mixed bound sqrt(d^2 k/(n-k)) (Eq. 7.57) via the symmetric purification Lemma 7.5.", "edition": "10 April 2026 edition (79 pages)", "inspection": "full bounded primary text read", "pages": "62-65", "redistribution": "not redistributed", "reuse": "Original mathematical questions/explanations only, no source-text redistribution/relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "The pure-symmetric de Finetti theorem bounds the trace distance between the k-body marginal of an n-party symmetric pure state and a convex mixture of i.i.d. pure states by sqrt(dk/(n - k)). The permutation-invariant mixed-state version bounds the analogous distance by sqrt(d^2 k/(n - k)), with mixed i.i.d. states and a dimension factor d^2 instead of d. (a) Explain quantitatively why the mixed-state bound carries d^2 where the pure-state bound carries d, in terms of the symmetric purification used to derive it. (b) For the bound to be non-trivial (strictly below 1) determine the required relationship between n, k, and the dimension in each case, and (c) describe the regime of k (as a function of n) in which the bound still tends to zero as n -> infinity, contrasting the two cases.", "reference_answer": "Use the source's trace distance D(rho,sigma)=||rho-sigma||_1/2, whose universal state-space cap is 1. Let d>=1, 0<=k0, and define the displayed numerical upper bounds\nB_p=sqrt(d k/m), B_m=sqrt(d^2 k/m).\nThey compare different state classes and approximating ensembles: pure vectors in Sym^n(C^d) versus conjugation-invariant density operators on (C^d)^{tensor n}. Invariance of a density operator does not imply original-space symmetric support. When a displayed bound exceeds 1 one can still use D<=min(1,B); the question(b) asks when the DISPLAYED numerical bound strictly improves that universal cap.\n\n(a) For a permutation-invariant rho choose matching PRODUCT bases on A^n and R^n, each R_i=C^d, and |Omega>=sum_w |w>_A|w>_R, unnormalized. The vector |Psi>=(sqrt(rho) tensor I)|Omega> has norm squared tr(rho)=1 and marginal rho. Since rho commutes with each factor permutation Q_pi, so does sqrt(rho). The real permutation matrices satisfy (Q_pi tensor Q_pi)|Omega>=|Omega>; hence the same simultaneous permutations fix |Psi>. Regroup A_1...A_n R_1...R_n into (A_1 R_1)...(A_n R_n). Thus |Psi> belongs to Sym^n(C^d tensor C^d), with LOCAL PAIR dimension d^2. Applying the pure theorem to this chosen symmetric purification replaces d by d^2, giving B_m. Tracing out R_1...R_k contracts D and sends each pure tensor power phi_AR^{tensor k} to (tr_R phi_AR)^{tensor k}, generally a mixed tensor power. Its pushforward probability measure supplies the mixed i.i.d. ensemble. The ancilla is a sufficient uniform construction, not the minimum required for every rho; an arbitrary purification need not be symmetric. No universal ancilla-resource lower bound is being asserted.\n\nFor the quantitative source route, put q_d(t)=binom(t+d-1,t). The Haar-projector identity gives conditional unnormalized k-system vectors v_phi=sqrt(q_d(m))(I tensor , with p(phi)=||v_phi||^2, integral p=1, and psi_k=integral p(phi)|e_phi>^{tensor k} equals q_d(m)/q_d(n). The exact ratio is product_{j=1}^{d-1}(m+j)/(n+j), with empty product 1. The elementary product/Bernoulli estimate bounds it below by 1-(d-1)k/(n+1), hence by 1-dk/m. Convexity of trace distance, the pure-state identity D=sqrt(1-F^2) with UNSQUARED fidelity F=||, and Jensen give D<=sqrt(1-q_d(m)/q_d(n))<=B_p. The mixed proof uses this same route with d^2 and partial trace; it is not a statement that the original mixed rho itself lies in Sym^n(C^d).\n\n(b) Squaring nonnegative bounds gives B_p<1 iff m>d k iff n>(1+d)k, and B_m<1 iff m>d^2 k iff n>(1+d^2)k. Equality gives the cap 1, not strict improvement. For k>0 the least integer total counts meeting these strict inequalities are (1+d)k+1 and (1+d^2)k+1, respectively.\n\nTo make the copy comparison precise, fix a retained k>0 and target normalized trace-distance tolerance 0=a k/epsilon^2. Therefore the least integer total n CERTIFIED BY THIS RAW BOUND is\nn_a^{<=}=k+ceil(a k/epsilon^2).\nFor a strict bound B_a1. Integer rounding need not preserve even the exact excess ratio. For d=2,k=1,epsilon=1/2 the least totals are 9 and 17, not 9 and 18; for epsilon^2=4/5 the least excesses are 3 and 5, not 3 and 6. At the SAME n,k>0 the two displayed errors instead obey B_m=sqrt(d) B_p. The phrase 'd-times larger total n at any fixed accuracy' is thus not an exact consequence. The real threshold ratio approaches d in the high-accuracy epsilon->0 regime at fixed d,k, but that is a qualified asymptotic statement.\n\n(c) At any FIXED finite d>=1, as n->infinity with 0<=k(n)0. Since k/(n-k)=[k/n]/[1-k/n], this is equivalent to k=o(n). A fixed retained k satisfies it. If k/n->c with 0sqrt(d c/(1-c)) and B_m->sqrt(d^2 c/(1-c)); neither numerical bound vanishes. If d=d(n) varies, the exact conditions are instead d(n)k/(n-k)->0 and d(n)^2 k/(n-k)->0, respectively (equivalently d(n)k=o(n) and d(n)^2 k=o(n) for d(n)>=1). Thus k=o(n) alone is insufficient uniformly in dimension: k=1,d(n)=floor(sqrt(n)) makes the pure bound tend to zero but the mixed bound tend to 1.\n\nBoundary and interpretation checks: k=0,n>0 gives both numerical bounds and the actual empty-marginal distance zero; within the displayed theorem domain its least allowed n is 1. At n=k, including n=k=0, the fractions are undefined and the source bound is not asserted. The zero-copy tensor unit is C and its unique normalized state is nevertheless exact; a singular bound does not itself prove impossibility of approximation. At d=1 every actual state and marginal is the unique scalar state, so the approximation is exact even where these loose numerical bounds do not vanish. At d=0 and n>0 there is no trace-one input state, not a zero-cost de Finetti problem. Epsilon=0 cannot be certified by a positive raw B at finite m when k>0, though special states and d=1 can be exactly i.i.d.; tolerances >=1 in normalized distance already follow from the universal cap. For an UNNORMALIZED trace-norm target ||rho-sigma||_1<=epsilon_1, use epsilon=epsilon_1/2, so m>=4a k/epsilon_1^2, and the universal cap is 2. These are existence/approximation guarantees from source algebra, not efficient ensemble computation, preparation circuits, gate counts or finite-bit complexity.", "rubric": [ { "criterion": "Explains the chosen permutation-invariant state's symmetric purification on n paired A_iR_i systems, with local dimension d^2 and a normalized marginal rho. Does not confuse mixed-state conjugation invariance with original symmetric support or call every purification symmetric/minimal.", "id": "r1", "weight": 1 }, { "criterion": "Applies the pure theorem with local dimension d^2 and contracts normalized trace distance under partial trace to obtain a probability mixture of mixed tensor powers. Keeps D=||rho-sigma||_1/2 and distinguishes a sufficient ancillary construction from a resource lower bound.", "id": "r2", "weight": 1 }, { "criterion": "Derives the strict numerical nontriviality conditions n-k>dk and n-k>d^2k (equality is not strict), interpreting them as sufficient bound thresholds, not necessary physical copy costs. Any fixed-accuracy copy comparison must distinguish total n from excess n-k and integer ceiling/floor rounding; the reference's exact formulas elaborate this distinction.", "id": "r3", "weight": 1 }, { "criterion": "Identifies k=o(n) as precisely the fixed-finite-d numerical-bound vanishing regime, with nonzero limits at k/n->c in (0,1); if dimension varies gives the actual d(n)k/(n-k) and d(n)^2k/(n-k) criteria rather than silently holding it fixed. Does not demand or assert exactly d-times larger total n: only the unrounded excess thresholds scale by d. Handles zero-retained-copy, one-dimensional and excluded zero-denominator/state-carrier cases without mistaking loose-bound failure for an actual distance lower bound or gate-complexity theorem.", "id": "r4", "weight": 1 } ], "semantic_step_count": 3, "semantic_steps": [ { "claim": "A specially chosen symmetric purification on A_iR_i pairs replaces the pure theorem's local dimension d by d^2; tracing references contracts the normalized trace distance and yields mixed tensor powers.", "supporting_nodes": [ "concept:tensor-networks", "concept:operator-theory" ] }, { "claim": "Strict improvement of the cap requires n-k>dk or n-k>d^2k. At accuracy epsilon the least raw-bound-certified total is k+ceil(a k/epsilon^2), a=d or d^2; only unrounded excess counts have exact factor d.", "supporting_nodes": [ "concept:probability" ] }, { "claim": "At fixed finite d the displayed bounds vanish iff k=o(n), while proportional retained fractions leave positive constants; varying-d bounds instead require d k/(n-k)->0 or d^2 k/(n-k)->0, with zero-copy/one-dimensional and source-domain boundaries explicit.", "supporting_nodes": [ "concept:probability" ] } ], "source_item_digest": "dd6da348bd8d27c7f6c8dbcad4757906f9c2d5c65952dced139686874a3c7ee1", "source_refs": [ { "anchor": "leditzky-2025-repth-qit", "license": "MIT metadata only", "path": "website/research/sources.json", "sha256": "9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae" } ], "split": "test_public", "supporting_nodes": [ "concept:tensor-networks" ], "target_nodes": [ "concept:operator-theory", "concept:probability" ], "task_hash": "12f546785cec8c579be25b30f1a3c09076585ff8b0e890807df23c4dd6d4be02", "task_kind": "boundary", "textbook_successor": { "predecessor_item_digest": "4130143bb729437029128decdc18b250157a590ff27650b756eb4be353884f70", "predecessor_key": "qcb-core-plan-404-definetti-bound-scaling-dk-vs-d2k", "repair_record": "shared/repairs/plan-404-excess-copy-count-answer-repair.json", "repair_record_digest": "6d4ec330241c853e12a67ae74a436c0748647e839c4fe9f7a6bcc81a78a95e31", "review_digest": "95b802b7dc8b4dd34283b8915a1173cd3312b6e9eeb2a385f9cc4bdfb6b7a746", "review_record": "reports/textbook-v1/repairs/plan-404-excess-copy-count-answer-repair-review.json" }, "topic": "planned_quantum_information_apps" }, { "assumptions": [], "context_hash": "8823ce7b9870c6b79352e468fd789ff96ceb9d125fd3a32a9438fbac2ec41501", "conventions": [], "core_id": "qcb-core-plan-115", "difficulty": { "level": "intermediate", "rationale": "A clean counterexample built from the symmetric/antisymmetric decomposition with explicit dimension counts; conceptually crisp and tests understanding that tensoring breaks irreducibility." }, "evidence_relation": "primary_source_nl_only", "export_role": "open_reader_not_stable_training_release", "family": "qit-tensor-representations", "formal_context": null, "graph_version": "qcb-snapshot-ffa644f0da3031a1ce754b9a", "id": "qcb-textbook-426a70f9d0202393cd63", "key": "qcb-core-plan-115-textbook-v1-universal-tensor-square-character-repair", "primary_sources": [ { "anchors": "Example 3.25 (symmetric and antisymmetric square): V (x) V = Sym^2(V) (+) Lambda^2(V), dim Sym^2(V) = d(d+1)/2, dim Lambda^2(V) = d(d-1)/2 (Eq. 3.77), with the stated conclusion that V (x) V is reducible whenever deg V >= 2", "edition": "10 April 2026 edition (79 pages); course directory FT25", "inspection": "full bounded primary text read", "pages": "26-28", "redistribution": "not redistributed", "reuse": "Original mathematical Q/A only; source text not redistributed; no external license relicensing", "sha256": "90784b02b568b8c7bb6a1ec20588d801541d047779333bd2b2b8e63749dc0fe3", "url": "https://www.felixleditzky.info/teaching/FT25/math595-repth-qit.pdf" } ], "question": "A student asserts: the tensor product of two irreducible representations of a group G is again irreducible. Show this is false by a counterexample. Specifically, take any representation (phi, V) of G with dim V >= 2 (you may take V irreducible) and consider the tensor-square representation on V (x) V. Exhibit a proper nonzero G-invariant subspace of V (x) V, and compute the dimensions to show V (x) V cannot be irreducible when dim V >= 2.", "reference_answer": "The assertion that an internal tensor product of two irreducible representations is always irreducible is false. First give the general tensor-square obstruction, without adding finite-dimensional or characteristic-not-two hypotheses to the question. Let F be the common field of the vector space, use the ordinary ALGEBRAIC tensor product V tensor_F V, and set T_g=phi(g) tensor phi(g). It is a representation because T_g T_h=T_{gh} and T_e=I.\n\nLet S(v tensor w)=w tensor v, extended F-linearly. Pure tensors span the tensor product, so checking them proves S^2=I and S T_g=T_g S for every g. Define A=im(I-S). This is a linear subspace and is G-invariant, since T_g(I-S)x=(I-S)T_g x. Group inverses also give equality T_g A=A if that convention for invariance is used. This route never divides by2 or presumes distinct eigenvalues +1 and -1.\n\nChoose a basis {e_i}_{i in I} of V, with two distinct elements e_1,e_2. The ordered tensors e_i tensor e_j form a basis of the algebraic tensor product. In a chosen ordering of I,\nA has basis a_ij=e_i tensor e_j-e_j tensor e_i, i=2, with no need to assume the original V is irreducible.\n\nFor finite d, dim(V tensor V)=d^2 and the basis count gives dim A=d(d-1)/2. This is an INTEGER count, not division by2 in the field. The fixed subspace B=ker(S-I) has a basis of the d diagonal tensors e_i tensor e_i and the d(d-1)/2 sums e_i tensor e_j+e_j tensor e_i for i=2, both counts are positive and strictly less than d^2. These supply the question's requested finite-dimensional comparison.\n\nIf char(F)!=2, then A=ker(S+I), B=ker(S-I), the projections are (I-S)/2 and (I+S)/2, and V tensor V=B direct_sum A. This is the source's symmetric/antisymmetric proof in finite complex dimension. In characteristic2, however, +1=-1, so the two eigenspaces just written coincide with B, not with A in general. In that case A is a proper subspace of B (the diagonal tensors witness it), and the displayed half-projections/direct-sum proof is unavailable. For example, at d=2, dim A=1 and dim B=3, although the +1 and -1 eigenspaces are the SAME B. The invariant image A still proves the requested reducibility. No source notation for an antisymmetric eigenspace is silently redefined to avoid this distinction.\n\nFor infinite algebraic dimension kappa=|I|, the ordered basis has cardinality |I times I|=kappa, and the basis of A has cardinality |{(i,j):i