Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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.

Exact phase recovery on the QPE grid

Statement

Let m≥1, Q=2m, and θ∈[0,1). If Qθ=t∈Z, then t∈{0,…,Q−1}, phase estimation returns y=t with probability 1, and every other output has probability 0.

Facts & Assumptions

Given: The eigenphase input, m≥1, Q=2m, and Qθ=t∈Z.

[F1]

The outcome probability is Pθ(y)=Q−2∣∑j<Qexp⁡(2πij(θ−y/Q))∣2 (Phase-estimation finite outcome law).

[F2]

For Q≥1 and integers k,ℓ, ∑j<Qexp⁡(2πi(k−ℓ)j/Q) is Q when k≡ℓ(modQ) and 0 otherwise (Orthogonality of the characters x↦e2πikx/N on Z/NZ).

Proof

1.1givenF1algebra

Since 0≤θ<1, the integer t=Qθ satisfies 0≤t<Q. At y=t, every summand in [F1] is 1, so Pθ(t)=Q−2Q2=1.

2.1givenF1F2step 1.1algebra∎

If y∈{0,…,Q−1} and y≠t, then t−y is not divisible by Q, since ∣t−y∣<Q and t−y≠0. Substituting θ=t/Q in [F1], the sum is the nontrivial cyclic-character sum in [F2], hence zero; therefore Pθ(y)=0.

Depends on

Used by

Dependency tree · two levels

40 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources