Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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Pauli commutator is symplectic pairing

Statement

For a=(x,z) and b=(x′,z′) in Vn and any phases k,k′,

P(k,x,z)P(k′,x′,z′)=(−1)ω(a,b)P(k′,x′,z′)P(k,x,z).

Thus the Pauli operators commute exactly when their labels are symplectically orthogonal.

Facts & Assumptions

Given: the exact phase-aware product formula and the form ω((x,z),(x′,z′))=x⋅z′+z⋅x′.

[F1]

Pauli product and adjoint: the ordered product phase is k+k′+2(z⋅x′) modulo 4.

[F2]

Binary symplectic form: ω(a,b)=x⋅z′+z⋅x′ over F2.

Proof

1.1givenF1F2algebra

The forward product has phase k+k′+2(z⋅x′) by [F1]; the reversed product has phase k+k′+2(z′⋅x). Their binary labels agree. The difference of the exponents modulo 4 is 2((z⋅x′)−(z′⋅x)), and subtraction equals addition in F2, so this difference is 2ω(a,b) modulo 4.

2.1givenstep 1.1algebra∎

Multiplication by the phase ratio i2ω(a,b)=(−1)ω(a,b) gives the displayed identity. If ω=0 it is commutation; if ω=1, equality with the negative is impossible because a Pauli is a nonzero invertible matrix over C.

Depends on

Used by

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Sources