QuantumComputinglib learn · inspect · formalize
Checked on this commit 4,524 public declarations commit 4e2c5b2c0be5 Build record

State Preparation / State Preparation

State Preparation

Exercise SP-008

intermediate · proof · train

nl bridge from inspected source. Current content-bound mathematical second pass accepted. No new Lean verification is claimed for this NL view.

Assumptions and conventions

  • each stage is a full unitary on the fresh-bit-and-bond register
  • the run starts from a single bond basis label
  • 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

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.

Hint: mathematical route

Unitary sequential stages from a basis boundary keep total probability one at every length, a coherent success with mass-neutral fresh-zero padding and no postselection.

  1. 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.
  2. Fresh-zero padding preserves mass and a full unitary action preserves mass, so a single step preserves amplitude mass.
  3. 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.
Reference solution

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.

Assessment rubric
  1. 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. (1 weight)
  2. 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. (1 weight)
  3. Inducts on n to conclude the run preserves mass, then evaluates the basis-boundary mass to 1. (1 weight)
  4. Explains this is coherent success with no postselection/renormalization and that fresh-zero padding is a mass-neutral register enlargement. (1 weight)

Graph-grounded prerequisites

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

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
  • QuantumBlockEncoding/SequentialBondPreparation.lean

    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

    Source SHA-256: ad4542b108d44f196f2a02f78e59b58f43d45b1e890709cf4c22e1222805d5c5 · MIT (repository LICENSE)

Contract: b292fd14d37907ff3e7aba598779c0fd83b21c819fa4be460397ff0fc38cfa02

Item digest: d403cce404da75f2931579b020692a9cf08143dfd7b2a59eeec69fce4c59f771

Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a

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