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Modular automorphisms and physical-time KMS conversion — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Modular frequency requires a physical time identification

Example

Assume the Axiom of Choice. Let $0 < p < 1$ and $ρ = diag (p, 1 - p)$ on $C^{2}$ . Its modular automorphism group acts on the off-diagonal matrix unit by $σ_{t} (E_{12}) = e^{i t \log (p / (1 - p))} E_{12} .$ The modular frequency $\log (p / (1 - p))$ is dimensionless and does not specify a physical energy or clock. If one separately supplies a Hamiltonian $H = diag (E_{1}, E_{2})$ with energies in joules, inverse energy $β > 0$ in joules $^{- 1}$ , and action $ℏ > 0$ in joule-seconds with $ρ = e^{- β H} / Z$ , then $p / (1 - p) = e^{- β (E_{1} - E_{2})}$ and the physical angular frequency $(E_{1} - E_{2}) / ℏ$ has units seconds $^{- 1}$ . For $p = \frac{1}{2}$ the modular frequency is zero; for $p \neq \frac{1}{2}$ it is nonzero but still has no physical time unit before this identification.

Example references

Faithful matrix modular flow and Gibbs time

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, $σ_{t} (A) = ρ^{i t} A ρ^{- i t}$ ; if $ρ = e^{- β H} / Z$ , physical time is identified by $α_{s} = σ_{- s / (β ℏ)}$ (Faithful matrix modular flow and Gibbs time).

Verification

Given: [F1], $E_{12} = ∣ 1 ⟩ ⟨ 2 ∣$ , and the displayed $ρ$ .

1.1A1F1givenalgebra

Since $ρ^{i t} = diag (p^{i t}, (1 - p)^{i t})$ , conjugating $E_{12}$ multiplies it by $p^{i t} (1 - p)^{- i t} = e^{i t \log (p / (1 - p))}$ . Thus the modular parameter has coefficient $\log (p / (1 - p))$ , which is dimensionless.

2.1A1F1step 1.1algebra∎

If $ρ = e^{- β H} / Z$ with $H = diag (E_{1}, E_{2})$ , then $p / (1 - p) = e^{- β (E_{1} - E_{2})}$ because the common partition function cancels. Hence $\log (p / (1 - p)) = - β (E_{1} - E_{2})$ , and substituting $t = - s / (β ℏ)$ yields $e^{i s (E_{1} - E_{2}) / ℏ}$ . 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.