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Trace-Gibbs bounded correlation measure and balance

Statement

Assume the Axiom of Choice. Let H≠{0} be separable, K=K∗ densely defined and bounded below with energy unit J, and let e−sK be trace class for every s in an open interval I⊂(0,∞) measured in J−1. Fix β∈I, Zβ=Tr⁡(e−βK), σβ=Zβ−1e−βK, and U(t)=e−itK/ℏ for ℏ>0 in J s and t in s. For bounded A,B, let CAB and SAB be the correlation function and finite spectral measure defined in Absolute convergence of trace-Gibbs bounded correlation series. Write SBA(−⋅) for the pushforward under ω↦−ω. Then for every bounded Borel test function f with compact support, ∫f(ω) SBA(−dω)=∫f(ω)e−βℏω SAB(dω). The weighted measure on the right extends to a finite complex measure: the reindexed coefficient family is absolutely summable and has the same total-variation bound as SBA. For Hermitian A=B, SAA is a positive finite measure. This is a discrete-spectrum equilibrium correlation identity; it asserts no temporal decay, continuum limit, or ordinary zero-frequency transport coefficient.

Facts & Assumptions

Given: The Axiom of Choice, the self-adjoint trace-Gibbs hypotheses, the bounded observables, and the correlation measure from the Statement.

[A1]

The Axiom of Choice: full Choice is the exact assumption of the unbounded spectral theorem and trace-Gibbs supplier.

[F1]

Self-adjoint trace-Gibbs normalization and form-energy optimization: the stated open-interval trace-class hypothesis yields 0<Zβ<∞, a faithful normal Gibbs density, and a complete discrete energy basis.

[F2]

Absolute convergence of trace-Gibbs bounded correlation series: the bounded correlation has an absolutely summable atomic spectral measure with coefficients riAijBji at ωji=(Ej−Ei)/ℏ.

Proof

technique · reindex the absolutely convergent spectral measure

Given: [A1], [F1]–[F2] and bounded Borel f with compact support.

1.1givenA1F1F2algebra

Under the unbounded spectral convention [A1], apply [F2]: ∫f(ω)SBA(−dω)=2π∑i,jriBijAjif(−ωji). The coefficient family is absolutely summable after multiplication by ∥f∥∞, so interchanging indices i and j is permitted.

2.1givenF1F2step 1.1algebra

Reindexing the sum in step 1.1 gives 2π∑i,jrjAijBjif(ωji). Since rj/ri=e−β(Ej−Ei)=e−βℏωji, this is exactly ∫f(ω)e−βℏωSAB(dω). The absolute sum of the reindexed coefficients equals that defining SBA and is finite by [F2]; compact support also makes the test-function weight bounded. Thus the weighted measure is finite and equals the reflected measure.

3.1F1F2algebra∎

If A=B=A∗, then Bji=Aij‾, so each atom coefficient ri∣Aij∣2 of SAA is nonnegative. The measure is finite by [F2], proving positivity.

Depends on

Dependency tree · 2 levels

Sources