ExampleConstruction: AI-adaptedVerification: 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.

Exact simulation of two commuting terms

Example

If two finite-dimensional self-adjoint matrices $H_{1}, H_{2}$ commute, then every first-order product step is exact: $e^{- i h H_{1}} e^{- i h H_{2}} = e^{- i h (H_{1} + H_{2})} (h \in R) .$ Consequently $S_{1} (t / r)^{r} = U_{H_{1} + H_{2}} (t)$ for every $t \in R$ and $r \geq 1$ .

Facts & Assumptions

Given: Self-adjoint endomorphisms $H_{1}, H_{2}$ of a finite-dimensional complex inner-product space, with $H_{1} H_{2} = H_{2} H_{1}$ ; real $t$ and integer $r \geq 1$ .

[F1]

First-order Lie product formula defines $S_{1} (h) = e^{- i h H_{1}} e^{- i h H_{2}}$ in the displayed operator order.

[F2]

Unitary and differential identities for finite matrix dynamics gives $U_{K}^{'} (s) = - i K U_{K} (s)$ , uniqueness for that initial-value problem, and $U_{K} (s + u) = U_{K} (s) U_{K} (u)$ .

Verification

1.1givenF2algebra

Put $A = - i H_{1}$ and $B = - i H_{2}$ , so $A B = B A$ . Define $Y (s) = e^{s A} e^{s B}$ . By [F2] and the product rule, $Y^{'} (s) = A e^{s A} e^{s B} + e^{s A} B e^{s B} = (A + B) Y (s)$ , where commutation lets $B$ pass through $e^{s A}$ . Also $Y (0) = I$ .

2.1step 1.1F1F2

The unique solution of $Z^{'} = (A + B) Z$ , $Z (0) = I$ , is $U_{H_{1} + H_{2}} (s)$ by [F2]. Hence $Y (h) = U_{H_{1} + H_{2}} (h)$ for every real $h$ , which proves the one-step identity in [F1].

3.1step 2.1F1F2algebra∎

Taking $h = t / r$ and composing the exact identity $r$ times gives $S_{1} (t / r)^{r} = U_{H_{1} + H_{2}} (t / r)^{r} = U_{H_{1} + H_{2}} (t)$ by the group law in [F2] and the product definition [F1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes (standard reference, not scraped)