Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Acceptance-gap lower bound on final-state distance

Statement

If a common final two-outcome measurement accepts one unit state with probability at least 2/3 and another with probability at most 1/3, then the distance between their state vectors is at least 1/6.

Facts & Assumptions

Given: Unit vectors ψ,ϕ in the same finite-dimensional register and one common final acceptance measurement.

[F1]

A common two-outcome POVM has an acceptance effect E with 0⪯E⪯I and acceptance weight tr⁡(PuE) on a unit state u (Finite POVMs and their outcome law, Finite positive effect family).

[F5]

A unit-vector outer product is a density operator: for a unit u, Pu:w↦⟨w,u⟩u is its density operator. The root pairing is linear in the first argument.

[F6]
[F3]

Worst-case bounded-error correctness: bounded-error decision correctness gives acceptance probability at least 2/3 on a yes input and at most 1/3 on a no input.

Proof

1.1givenF1F4F5F6algebra

By [F1] and [F4], choose an orthonormal eigenbasis (ej) for E. Positivity of E and I−E puts each eigenvalue λj in [0,1]. For w=∑jwjej, ∥Ew∥2=∑jλj2∣wj∣2≤∑j∣wj∣2=∥w∥2. For a unit u=∑jujej, [F5]–[F6] give tr⁡(PuE)=∑jλj∣uj∣2=⟨Eu,u⟩. Thus the same acceptance expectation and contraction estimate hold for every two-outcome POVM.

2.1givenF2F3step 1.1algebra∎

Put δ=ψ−ϕ. The acceptance-probability difference equals ⟨Eδ,ψ⟩+⟨Eϕ,δ⟩. By [F2], unit normalization, and step 1.1, its absolute value is at most ∥Eδ∥∥ψ∥+∥Eϕ∥∥δ∥≤2∥δ∥. The probability gap in [F3] is at least 1/3, so ∥ψ−ϕ∥≥1/6.

Depends on

Used by

Dependency tree · two levels

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Sources