LemmaStatement: Literature-sourcedProof: 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.

Hadamard pauli action

Statement

For a Hadamard gate $H_{j}$ on factor $j$ , conjugation swaps the binary coordinates $x_{j}, z_{j}$ and changes the phase exactly by $2 x_{j} z_{j}$ modulo $4$ . In particular $H X H = Z$ , $H Z H = X$ , and $H Y H = - Y$ .

Facts & Assumptions

Given: n is a positive integer, j is a factor index, the phase-aware Pauli convention is fixed, and the register factors are ordered.

[F1]

The Pauli matrices: the explicit X,Y,Z matrices are defined in the ordered computational basis.

[F2]

Pauli matrix algebra and trace identities: X,Y,Z are self-adjoint involutions, Y=iXZ, ZX=-XZ, and YZ=iX.

[F3]

Exact one-qubit gates and rotations: H=(X+Z)/sqrt(2); its verification gives H*=H and H^2=I.

[F4]

Tensor products of finitely many linear maps and Tensor maps preserve composition, identities, and adjoints: tensor maps act factorwise on elementary tensors, preserve composition, and have factorwise adjoints.

[F5]

Ordered tensor registers, their inner product, and wire permutations: the register factors and product basis have the declared order.

Proof

1.1F1F2F3algebra

Direct multiplication of the displayed $2 \times 2$ matrices gives $H X H = Z$ and $H Z H = X$ . Since $Y = i X Z$ , conjugation preserves product order and gives $H Y H = i (Z X) = - i X Z = - Y$ .

2.1F1F3F4F5step 1.1algebra∎

On factor $j$ , $H X^{x_{j}} Z^{z_{j}} H = Z^{x_{j}} X^{z_{j}} = (- 1)^{x_{j} z_{j}} X^{z_{j}} Z^{x_{j}}$ ; all other tensor factors are unchanged by [F4] and [F5]. As $(- 1)^{x_{j} z_{j}} = i^{2 x_{j} z_{j}}$ , this is $P (k + 2 x_{j} z_{j}, x^{'}, z^{'})$ with $x_{j}^{'} = z_{j}$ , $z_{j}^{'} = x_{j}$ , and all other coordinates unchanged.

Depends on

Used by

Dependency tree · two levels

37 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

Ronald de Wolf, Quantum Computing: Lecture Notes (standard reference, not scraped)
John Watrous, Understanding Quantum Information and Computation, Lesson 14 (standard reference, not scraped)
Scott Aaronson and Daniel Gottesman, Improved Simulation of Stabilizer Circuits (standard reference, not scraped)