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Exercise PLAN-007

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primary source nl only. Additive answer successor with distinct content-bound review; historical rejected answer retained only for audit. No new Lean verification is claimed for this NL view.

Assumptions and conventions

      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.

      Hint: mathematical route

      The two-qubit Werner state is entangled exactly for negative swap overlap. Its partial transpose has eigenvalue x/2, while the separable converse follows by twirling an explicit product state with overlap x and identifying the unique Werner state, not by reversing swap-overlap positivity.

      1. Partial transpose gives rho(x)^{T_B}=((2-x)/6)I+((2x-1)/3)P_+, using F^{T_B}=2P_+.
      2. 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.
      3. 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.
      Reference solution

      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.

      For 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.

      The 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.

      For each x in [0,1], choose |a>=|0> and |b>=sqrt(x)|0>+sqrt(1-x)|1>, and twirl the product projector tau=|a><a| tensor |b><b| by normalized Haar averaging of (U tensor U)tau(U tensor U)^dagger: the average is Werner by Haar invariance, preserves swap overlap |<a|b>|^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.

      Assessment rubric
      1. States the exact restricted-family threshold: entangled for -1<=x<0 and separable for 0<=x<=1, including the x=0 boundary. (1 weight)
      2. 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. (1 weight)
      3. 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. (1 weight)
      4. 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. (1 weight)

      Graph-grounded prerequisites

      These are support/source anchors, not an automatically extracted theorem-dependency proof.

      • Operator and semigroup theory (frozen curriculum anchor; see graph registry)
      • Matrix analysis (frozen curriculum anchor; see graph registry)

      Download the shared core registry

      Lean view and evidence boundary

      No aligned Lean view has been admitted for this native core in this textbook checkpoint. This is not a claim that the underlying mathematics lacks a formal proof. Contributions must preserve this exact target contract and split.

      Read the Lean track and existing diagnostic warmups

      Sources and provenance

      Contract: d2196fd968d517f6f9d4f870dab30d7bd70e0344eddb4b3ad19e6ff0464a6874

      Item digest: 07e6df25bb07479718e7bef8a988827de01d56b685ab5fb591320d335c56ad67

      Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a

      Authored support steps: 3. This is not measured graph distance or model difficulty.

      Answer revision lineage

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