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.

A two-site spin-chain local solve reaches the exact ground state

Example

Take two spin- $1 / 2$ sites with computational basis $∣ 00 ⟩, ∣ 01 ⟩, ∣ 10 ⟩, ∣ 11 ⟩$ , Pauli matrix $Z = diag (1, - 1)$ , and finite Hamiltonian $H = - Z \otimes I - I \otimes Z - Z \otimes Z .$ Supply the initial product state $∣ 11 ⟩$ by the rank-one open-boundary TT cores $G_{k} (1, i_{k}, 1) = 1_{{i_{k} = 2}}$ for $k = 1, 2$ . Starting from this presentation, one completed width-two exact sweep of Finite alternating tensor optimization with effective eigenproblems returns $∣ 00 ⟩$ and lowers the energy from $1$ to $- 3$ .

Facts & Assumptions

Given: The two-site state, Hamiltonian, and width-two exact sweep just specified.

[F1]

Finite matrix-product states and Hermitian Hamiltonians: a normalized finite state is a tensor array, and a finite Hamiltonian input is Hermitian.

[F2]

Finite alternating tensor optimization with effective eigenproblems: each window is updated by the exact lowest eigenvector of its effective Hermitian matrix; with two sites the width-two window is the entire coefficient space.

Verification

1.1givenF1algebra

In the displayed basis, $H$ is the diagonal matrix $diag (- 3, 1, 1, 1)$ : the three summands contribute respectively $(- 1, - 1, 1, 1)$ , $(- 1, 1, - 1, 1)$ , and $(- 1, 1, 1, - 1)$ . Thus $∣ 00 ⟩$ is the unique unit ground vector with energy $- 3$ , while $∣ 11 ⟩$ has energy $1$ .

2.1givenF2step 1.1algebra∎

For the only two-site window, the fixed environments are identities, so the embedding $E$ and effective Hamiltonian are $I_{4}$ and $H$ . The exact solve in [F2] therefore selects $∣ 00 ⟩$ . The backward pass sees the same $H$ and leaves this unique ground vector unchanged, and its exact SVD split has rank one and no discarded tail. Hence the completed sweep returns $∣ 00 ⟩$ with energy $- 3 < 1$ .

Depends on

Used by

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