ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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.

Nearest-grid and tail estimates in phase estimation

Example

Take a valid eigenphase-estimation input with $m = 3$ , $Q = 8$ , and $θ = 3 / 16$ . Compute the nearest-label probabilities and the exact circular tail probability at threshold $η = 1 / 4$ .

Facts & Assumptions

Given: A valid QPE eigenpair with $m = 3$ , $Q = 8$ , and $θ = 3 / 16$ .

[F1]

The finite outcome law is $P_{θ} (y) = Q^{- 2} ∣ \sum_{j < Q} \exp (2 π i j (θ - y / Q)) ∣^{2}$ (Phase-estimation finite outcome law).

[F2]

The geometric sum is $(1 - z^{Q}) / (1 - z)$ when $z \neq 1$ (Finite geometric sum for phase-estimation amplitudes).

[F3]

The circular distance is $d_{T} (x) = \min (∣ x ∣, 1 - ∣ x ∣)$ for $x \in [- 1, 1]$ (Circular phase distance modulo one).

[F4]

The success-and-tail theorem gives every nearest label probability at least $4 / π^{2}$ , and for $K = ⌊ Q η ⌋ \geq 2$ bounds the circular tail by $1 / (2 (K - 1))$ (Phase-estimation nearest-grid and tail bounds).

Verification

1.1givenF1F2F4algebra

Since $Q θ = 3 / 2$ , the grid labels $1$ and $2$ are tied nearest labels, each at circular distance $1 / 16$ . For $y = 1, 2$ , $Q (θ - y / 8) = \pm 1 / 2$ , so the geometric quotient in [F1]–[F2] has numerator modulus $2$ and denominator modulus $2 \sin (π / 16)$ . Thus $P_{θ} (1) = P_{θ} (2) = 1 / (64 \sin^{2} (π / 16))$ , which is at least $4 / π^{2}$ by the nearest-grid part of [F4].

1.2F1F2F3algebra

The circular distances for $y = 0, \dots, 7$ are $(3 / 16, 1 / 16, 1 / 16, 3 / 16, 5 / 16, 7 / 16, 7 / 16, 5 / 16)$ , by [F3]. Hence the tail event $d_{T} (θ - y / 8) \geq 1 / 4$ consists exactly of $y = 4, 5, 6, 7$ . For these labels the exact outcome probabilities are $P_{θ} (4) = P_{θ} (7) = 1 / (64 \sin^{2} (5 π / 16))$ and $P_{θ} (5) = P_{θ} (6) = 1 / (64 \sin^{2} (7 π / 16))$ , again by the finite geometric quotient.

2.1F4step 1.2algebra∎

Therefore the exact total tail is $\frac{1}{32} (\sin^{- 2} (5 π / 16) + \sin^{- 2} (7 π / 16))$ . Here $K = ⌊ 8 (1 / 4) ⌋ = 2$ , so [F4] gives the proven upper bound $1 / 2$ for this total. Numerically the exact expression is approximately $0.07769$ , well below $1 / 2$ .

Depends on

Used by

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, §4.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, version 6, §4.6 (standard reference, not scraped)
Lin Lin, Lecture Notes on Quantum Algorithms for Scientific Computation, §§16.1–16.3 (standard reference, not scraped)