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.

A three-stage precision budget

Example

Take three qubit-to-qubit stage compilers, so $H_{0} = H_{1} = H_{2} = H_{3} = Q_{2}$ , a global normalized diamond-error target $ε = 3 / 100$ , and the reference register $R = Q_{2}$ . Let $ρ$ be any density operator on $H_{0} \otimes R$ , including an entangled state. Write $e_{0}, e_{1}$ for the computational basis of $Q_{2}$ and put $u_{a b} = e_{a} \otimes e_{b}$ . On the final register $H_{3} \otimes R$ , use the four-outcome POVM $M = {E_{a b} : a, b \in {0, 1}}$ with $E_{a b} x = ⟨ x, u_{a b} ⟩ u_{a b}$ . Assume each compiler returns a channel whose local normalized diamond error is at most the precision budget it receives; the allocation procedure passes $1 / 100$ to each compiler.

Facts & Assumptions

Given: Three matching pairs of ideal and implemented finite-dimensional channels, global target $ε = 3 / 100$ , and the stated local compiler guarantee.

[F1]

Allocate local error budgets through a pipeline: for $L \geq 1$ , the allocated local precision is $ε_{j} = ε / L$ .

[F2]

Hybrid bound for a composed channel circuit: the normalized diamond error of a finite channel composition is at most the sum of its local normalized errors, and every final POVM outcome law on a finite reference input differs by at most that sum in total variation.

[F3]

Density operator: $ρ$ is a nonnegative trace-one operator on the nonzero finite-dimensional input-reference register.

[F4]

Finite registers and their ordered computational bases, The computational basis is orthonormal, Ordered tensor registers, their inner product, and wire permutations, and The product basis is orthonormal and tensor dimensions multiply: $(u_{a b})_{a, b \in {0, 1}}$ is an orthonormal basis of $H_{3} \otimes R$ .

[F5]

Non-negative and positive operators: an endomorphism is nonnegative when it is self-adjoint and its quadratic form is nonnegative; Finite POVMs and their outcome law defines a finite POVM as nonnegative effects summing to the identity.

Verification

1.1givenF1algebra

With $L = 3$ , the allocator sets $ε_{j} = (3 / 100) / 3 = 1 / 100$ for each $j \in {1, 2, 3}$ .

1.2F4F5algebra

For each $a, b$ , the tensor pairing in [F4] gives $⟨ E_{a b} x, y ⟩ = ⟨ x, u_{a b} ⟩ ⟨ u_{a b}, y ⟩ = ⟨ x, E_{a b} y ⟩$ , so $E_{a b}$ is self-adjoint. Since $∥ u_{a b} ∥ = 1$ , $E_{a b}^{2} = E_{a b}$ and $⟨ E_{a b} x, x ⟩ = ∣ ⟨ x, u_{a b} ⟩ ∣^{2} \geq 0$ . The basis expansion from [F4] gives $\sum_{a, b} E_{a b} = I$ ; hence the family is a finite POVM by [F5].

2.1givenstep 1.1F2algebra

Each local error is therefore at most $1 / 100$ . Applying [F2], the composed channel error is at most $1 / 100 + 1 / 100 + 1 / 100 = 3 / 100 = ε$ .

3.1givenF2F3step 2.1step 1.2∎

The input $ρ$ and the four-outcome POVM meet the reference and measurement hypotheses of [F2] by [F3] and step 1.2. Their outcome laws for the ideal and implemented composites therefore differ in total variation by at most $3 / 100$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms (standard reference, not scraped)
John Watrous, The Theory of Quantum Information (standard reference, not scraped)