Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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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.

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Determinants of one-qubit Clifford+T words are eighth roots

Statement

Every exact one-qubit Clifford+T matrix word, including a free scalar Clifford phase ωjI, has determinant in {ωk:k∈Z}.

Facts & Assumptions

Given: ω=eiπ/4 and a finite word from the one-qubit Clifford+T matrix alphabet.

[F1]

Clifford+T gate library and exact matrices: H has its displayed Hadamard matrix, T=diag⁡(1,ω), T†=diag⁡(1,ω−1), S=T2, Z=T4, X,Y,Z are the displayed Pauli matrices, and ωI is an allowed scalar Clifford phase.

Proof

1.1F1algebra

Direct 2×2 determinants give det⁡H=−1=ω4, det⁡T=ω, det⁡T†=ω−1, det⁡S=ω2, det⁡S†=ω−2, and det⁡X=det⁡Y=det⁡Z=−1=ω4. Also det⁡(ωjI)=ω2j. Thus every one-qubit generator has determinant an integer power of ω.

2.1F1F2step 1.1algebra∎

By [F2], the determinant of a finite product is the product of the generator determinants from step 1.1. The product is again an integer power of ω, including for the empty word, whose determinant is 1.

Depends on

Used by

Dependency tree · two levels

11 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