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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.

QMA Verification and Local Hamiltonian — Examples

1 · Prerequisites

2 · Summary

For one Hadamard gate on input ∣0⟩, the example writes the two-clock history state, evaluates the propagation quadratic form, and checks that the scaled output penalty equals the rejection probability 1/2.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

One-gate history Hamiltonian energy check

Example

Take one work qubit in ∣0⟩, one gate U1=H, and declare outcome ∣1⟩ accepting. Then the history is ∣η⟩=12(∣0⟩∣c0⟩+∣+⟩∣c1⟩),∣+⟩=(∣0⟩+∣1⟩)/2. Verify that propagation energy is zero and that the scaled output penalty has energy 1/2, equal to the verifier's rejection probability.

Facts & Assumptions

Given: The T=1 history Hamiltonian and the stated one-qubit circuit.

[F1]

Feynman–Kitaev history Hamiltonian: at T=1, propagation is the half-weight edge penalty and output penalty coefficient is T+1=2.

[F2]

Propagation kernel and path-Laplacian gap: the edge penalty vanishes exactly when the second computation component is U1 applied to the first.

[F3]

Input and output penalties on history states: the scaled output penalty equals the terminal rejection probability.

Proof

1.1F1F2givenalgebra

The propagation quadratic form on an arbitrary vector ∣u⟩∣c0⟩+∣v⟩∣c1⟩ is 12∥v−Hu∥2. Here v=H∣0⟩=∣+⟩, so the norm is zero and propagation energy vanishes.

2.1F1F3step 1.1algebra∎

The terminal state is ∣+⟩, whose probability of rejecting the accept-on-∣1⟩ predicate is ∣⟨0∣+⟩∣2=1/2. Directly, the clock-c1 history component has squared norm 1/2, its reject projection has probability 1/2 within that component, and the coefficient 2 cancels the clock factor 1/2; the total output energy is 1/2. This matches [F3].

Sources