Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Modular frequency requires a physical time identification

Example

Assume the Axiom of Choice. Let 0<p<1 and ρ=diag⁡(p,1−p) on C2. Its modular automorphism group acts on the off-diagonal matrix unit by σt(E12)=eitlog⁡(p/(1−p))E12. The modular frequency log⁡(p/(1−p)) is dimensionless and does not specify a physical energy or clock. If one separately supplies a Hamiltonian H=diag⁡(E1,E2) with energies in joules, inverse energy β>0 in joules−1, and action ℏ>0 in joule-seconds with ρ=e−βH/Z, then p/(1−p)=e−β(E1−E2) and the physical angular frequency (E1−E2)/ℏ has units seconds−1. For p=12 the modular frequency is zero; for p≠12 it is nonzero but still has no physical time unit before this identification.

Facts & Assumptions

Given: The Axiom of Choice and a faithful two-level density matrix ρ=diag⁡(p,1−p) with 0<p<1.

[A1]

Assume the Axiom of Choice (The Axiom of Choice).

[F1]

For a finite-dimensional faithful density matrix, σt(A)=ρitAρ−it; if ρ=e−βH/Z, physical time is identified by αs=σ−s/(βℏ) (Faithful matrix modular flow and Gibbs time).

Verification

Given: [F1], E12=∣1⟩⟨2∣, and the displayed ρ.

1.1A1F1givenalgebra

Since ρit=diag⁡(pit,(1−p)it), conjugating E12 multiplies it by pit(1−p)−it=eitlog⁡(p/(1−p)). Thus the modular parameter has coefficient log⁡(p/(1−p)), which is dimensionless.

2.1A1F1step 1.1algebra∎

If ρ=e−βH/Z with H=diag⁡(E1,E2), then p/(1−p)=e−β(E1−E2) because the common partition function cancels. Hence log⁡(p/(1−p))=−β(E1−E2), and substituting t=−s/(βℏ) yields eis(E1−E2)/ℏ. The energy gap alone gives no time frequency until ℏ is specified, and the state alone gives no energy gap until H and β are specified.

Depends on

Dependency tree · 2 levels

Sources