ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

Spectral distribution and phase resolution for a two-level matrix

Example

Take $H = diag (- 1 / 2, 1 / 3)$ on $C^{2}$ and $ψ = \sqrt{3 / 5} ∣ 0 ⟩ + \sqrt{2 / 5} ∣ 1 ⟩$ . The state-dependent spectral distribution has masses $3 / 5$ at $- 1 / 2$ and $2 / 5$ at $1 / 3$ . With $U = e^{i H}$ and $Q = 128$ , the measured phase law is the mixture of the two exact QPE kernels with those weights, at the canonical phase representatives $1 - 1 / (4 π)$ and $1 / (6 π)$ . For additive eigenvalue accuracy $ϵ = 1 / 4$ , the total probability of an estimate within $1 / 4$ of its eigenvalue is at least $3 / 4$ ; the two success intervals are disjoint. With only controlled- $U$ access, the phase register uses $127$ controlled- $U$ calls per run.

Facts & Assumptions

Given: The diagonal matrix, normalized input, exact QPE access to $U = e^{i H}$ , and phase-register size $m = 7$ ( $Q = 2^{7} = 128$ ).

[F1]

State-dependent spectral distribution and QPE output law and conditional state for a spectral input: each eigenspace has weight $∥ P_{λ} ψ ∥^{2}$ and the sampler law is the corresponding mixture of phase kernels.

[F2]

Phase-estimation nearest-grid and tail bounds: if $K = ⌊ Q η ⌋ \geq 2$ , QPE has circular phase-error tail at most $1 / (2 (K - 1))$ .

[F3]

Pi is the first positive zero of sine, Sine is positive and cosine is strictly decreasing on (0,2), with cos 2 at most -1/3, and The exponential definitions of complex sine, cosine, hyperbolic sine, and hyperbolic cosine equal their entire power series: $\sin π = 0$ , sine is positive on $(0, π)$ , $π > 2$ because sine has no zero on $(0, 2]$ , and sine has its stated Taylor series.

[F4]

The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ : for an alternating series with nonincreasing nonnegative term magnitudes tending to zero, the series converges and the remainder after a partial sum is bounded by the first omitted term.

[F5]

Eigenvalue-estimation access and aliasing convention: with only controlled- $U$ access, implementing powers $U^{2^{j}}$ by repeated calls costs $\sum_{j = 0}^{m - 1} 2^{j} = 2^{m} - 1$ controlled- $U$ calls.

Verification

1.1F1givenalgebra

The two diagonal entries are distinct, so their spectral projections are $∣ 0 ⟩ ⟨ 0 ∣$ and $∣ 1 ⟩ ⟨ 1 ∣$ . Their action on $ψ$ gives squared norms $3 / 5$ and $2 / 5$ , respectively; the squares sum to one. The encoded phase classes have canonical representatives $1 - 1 / (4 π)$ for eigenvalue $- 1 / 2$ and $1 / (6 π)$ for eigenvalue $1 / 3$ . By [F1], the squared norms are exactly the masses and the measured phase law is $\frac{3}{5} P_{1 - 1 / (4 π)} + \frac{2}{5} P_{1 / (6 π)}$ .

1.2F3F4givenalgebra

We certify $π < 4$ . The sine Taylor partial sum through $- 4^{7} / 7!$ is $S_{3} = 4 - 4^{3} / 3! + 4^{5} / 5! - 4^{7} / 7! = - 6976 / 5040 = - 502272 / 362880$ . Starting with $4^{9} / 9!$ , the absolute Taylor terms decrease because the next-to-current ratio is $16 / ((n + 1) (n + 2)) < 1$ for odd $n \geq 7$ . These magnitudes tend to zero, since the sine series at $4$ converges by [F3]. Applying [F4] to this tail bounds the alternating remainder by $4^{9} / 9! = 262144 / 362880$ . Hence $\sin 4 \leq - 240128 / 362880 < 0$ . If $π > 4$ , [F3] would give $\sin 4 > 0$ ; if $π = 4$ , [F3] would give $\sin 4 = 0$ . Both contradict the negative bound, so $π < 4$ .

2.1F1F2F3step 1.2givenalgebra

Put $η = ϵ / (2 π) = 1 / (8 π)$ . For each eigenvalue $λ$ , let $θ_{0} = λ / (2 π)$ and let $\overset{ˉ}{θ} \in [0, 1)$ be its canonical encoded phase. Then $Q η = 16 / π > 4$ by $π < 4$ , so $K = ⌊ Q η ⌋ \geq 4$ ; [F2] bounds the circular tail at $\overset{ˉ}{θ}$ by at most $1 / 6 < 1 / 4$ . On this success event, the circular-distance definition gives an integer $k$ with $∣ y / Q + k - \overset{ˉ}{θ} ∣ < η$ . Writing $\overset{ˉ}{θ} = θ_{0} + n$ for an integer $n$ , the lift $y / Q + k - n = θ_{0} + e$ has $∣ e ∣ < η$ . Since $∣ θ_{0} ∣ \leq 1 / (2 π)$ , we have $∣ θ_{0} + e ∣ < 5 / (8 π) < 1 / 2$ by $π > 2$ , so it is the principal representative and the decoded eigenvalue error is at most $ϵ$ . Thus each eigenstate has success at least $3 / 4$ . By [F1], the total success probability is at least $(3 / 5) (3 / 4) + (2 / 5) (3 / 4) = 3 / 4$ . The eigenvalue gap is $1 / 3 - (- 1 / 2) = 5 / 6 > 1 / 2 = 2 ϵ$ , so the two intervals are disjoint.

3.1F5givenalgebra∎

The seven phase bits require the controlled powers with exponents $1, 2, 4, 8, 16, 32, 64$ . Repeating controlled- $U$ for each power uses $1 + 2 + 4 + 8 + 16 + 32 + 64 = 127$ calls.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

69 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

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)