ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Product block error in a two-factor instance

Example

Let $A = diag (1 / 2, 0)$ and $B = diag (1, 0)$ , with $α_{A} = 1$ , $α_{B} = 2$ , $ε_{A} = 1 / 10$ , and $ε_{B} = 1 / 5$ . Use the compressed blocks $C_{A} = diag (3 / 5, 0)$ and $C_{B} = diag (7 / 10, 0)$ . The example evaluates both the normalized product-error formula and its unnormalized version.

Facts & Assumptions

Given: The matrices and compressed blocks in the Example section, and the normalized product theorem Product block-encoding error bound.

[F1]

Product block-encoding error bound: the normalized two-factor product error is bounded by $ε_{A} ∥ B ∥ / α_{B} + ε_{B} ∥ A ∥ / α_{A} + ε_{A} ε_{B}$ .

[F2]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: the operator norm of a diagonal matrix on the standard inner product space is its maximum action on a unit vector.

[F3]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ .

[F4]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: a unitary is an invertible linear isometry.

Verification

1.1givenF2F3F4constructalgebra

The errors are $E_{A} = C_{A} - A / α_{A} = diag (1 / 10, 0)$ and $E_{B} = C_{B} - B / α_{B} = diag (1 / 5, 0)$ . For a diagonal $D = diag (d_{1}, d_{2})$ , $∥ D x ∥^{2} = ∣ d_{1} ∣^{2} ∣ x_{1} ∣^{2} + ∣ d_{2} ∣^{2} ∣ x_{2} ∣^{2}$ , so the maximum on unit vectors is $\max (∣ d_{1} ∣, ∣ d_{2} ∣)$ , attained at the corresponding basis vector. Hence $∥ E_{A} ∥ = 1 / 10$ , $∥ E_{B} ∥ = 1 / 5$ , $∥ A ∥_{o p} = 1 / 2$ , and $∥ B ∥_{o p} = 1$ . For each of $C_{A}, C_{B}$ , put $S = \sqrt{I - C^{2}}$ and $D (C) = (\begin{array}{cc} C & S \\ S & - C \end{array})$ . The diagonal matrices $C, S$ commute and satisfy $C^{2} + S^{2} = I$ , so block multiplication gives $D (C)^{*} = D (C)$ and $D (C)^{2} = I$ . Then [F3] gives $∥ D (C) x ∥^{2} = ⟨ x, D (C)^{*} D (C) x ⟩ = ∥ x ∥^{2}$ , and [F4] makes $D (C)$ unitary; its top-left block is $C$ . Thus both chosen compressed blocks have exact unitary dilations.

2.1F1F2step 1.1algebra

The theorem's normalized bound is $(1 / 10) (1 / 2) + (1 / 5) (1 / 2) + (1 / 10) (1 / 5) = 17 / 100$ . The actual product block is $C_{A} C_{B} = diag (21 / 50, 0)$ , whereas $A B / (α_{A} α_{B}) = diag (1 / 4, 0)$ ; their difference has norm $21 / 50 - 1 / 4 = 17 / 100$ , so this instance attains that bound.

3.1F1step 2.1algebra∎

Multiplying the normalized error by $α_{A} α_{B} = 2$ gives unnormalized error $17 / 50$ . The looser universal normalized estimate is $ε_{A} + ε_{B} + ε_{A} ε_{B} = 8 / 25$ ; after scaling by $2$ , its unnormalized counterpart is $16 / 25$ . Neither bound is tight here.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 29 (17 April 2025) (standard reference, not scraped)
András Gilyén, Yuan Su, Guang Hao Low, and Nathan Wiebe, Quantum singular value transformation and beyond, Lemma 53 (standard reference, not scraped)