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.

Fourier sampling for a cyclic subgroup

Example

Let $G = Z / 12 Z$ and $H = 3 Z / 12 Z = {0, 3, 6, 9}$ . Its annihilator has the three character labels $0, 4, 8$ , so a Fourier sample is uniform on those labels.

Facts & Assumptions

Given: The additive cyclic group $Z / 12 Z$ , its subgroup $H$ , and the negative-character QFT convention.

[F1]

The characters are $χ_{k} (x) = \exp (2 π i k x / 12)$ for $k \in Z / 12 Z$ , and these are all twelve characters. (A cyclic group of order n has exactly n additive characters)

[F2]

$H^{⊥}$ consists of the characters trivial on every element of $H$ . (Character annihilator of a subgroup)

[F3]

The normalized transform of a coset state has equal-modulus amplitudes exactly on $H^{⊥}$ . (Normalized Fourier transform on a finite abelian group, Fourier samples are uniform in the annihilator)

[F4]

Character values on finite groups have modulus one. (Values of additive characters of finite abelian groups have modulus one)

Verification

Given: The coset state $∣ a + H ⟩ = \frac{1}{2} \sum_{j = 0}^{3} ∣ a + 3 j ⟩$ for any $a \in Z / 12 Z$ .

1.1givenF1F2algebra

A character $χ_{k}$ is trivial on $H$ exactly when $χ_{k} (3) = \exp (2 π i k / 4) = 1$ , which holds exactly when $4 ∣ k$ . Among residues modulo $12$ this gives $k \in {0, 4, 8}$ , so $∣ H^{⊥} ∣ = 3$ .

1.2F1F3F4algebra

The amplitude of label $k$ after the normalized QFT is $\frac{1}{\sqrt{48}} \sum_{j = 0}^{3} \overline{χ_{k} (a + 3 j)} = \frac{\overline{χ_{k} (a)}}{\sqrt{48}} \sum_{j = 0}^{3} \exp (- 2 π i k j / 4)$ . If $4 ∣ k$ , the sum is $4$ and the amplitude has modulus $4 / \sqrt{48} = 1 / \sqrt{3}$ by [F4]. If $4 ∤ k$ , the ratio $q = \exp (- 2 π i k / 4)$ is a nontrivial fourth root of unity, so $(1 - q) \sum_{j = 0}^{3} q^{j} = 1 - q^{4} = 0$ and the sum vanishes.

2.1F2F3step 1.2∎

Squaring the amplitudes in step 1.2 gives probability $1 / 3$ for each of $k = 0, 4, 8$ and zero for the other nine labels. These probabilities sum to one and agree with the uniform annihilator law.

Depends on

Used by

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms (17 April 2025) (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)