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Self-adjoint trace-Gibbs normalization and form-energy optimization

Statement

Assume the Axiom of Choice. Let $H \neq {0}$ be a separable complex Hilbert space and $A = A^{*}$ a densely defined Hamiltonian bounded below by $a \in R$ . Assume $e^{- β A}$ is trace class for every $β$ in an open interval $I \subset (0, \infty)$ . For each $β \in I$ , put $Z_{β} = Tr (e^{- β A})$ and $ρ_{β} = e^{- β A} / Z_{β}$ . Then $Z_{β} \in (0, \infty)$ and $ρ_{β}$ is a faithful normal density. For the entropy and optimization clause, fix $k_{B} > 0$ (in SI units J/K), put $T = (k_{B} β)^{- 1}$ and $F_{β} = - β^{- 1} \log Z_{β}$ , and let $ρ$ be any normal density with finite spectral form energy $E_{A} (ρ) = \int_{R} λ d μ_{ρ} (λ), \int_{R} ∣ λ ∣ d μ_{ρ} (λ) < \infty .$ Then $ρ$ has finite entropy and satisfies $E_{A} (ρ) - T S (ρ) \geq F_{β},$ with equality exactly at $ρ = ρ_{β}$ . The prepared Gibbs state also has $S (ρ_{β}) = k_{B} (\log Z_{β} + β E_{A} (ρ_{β})) .$ The trace-class and open-interval hypotheses are essential; a lower bound on $A$ alone does not imply finite $Z_{β}$ .

Facts & Assumptions

Given: The Axiom of Choice, the exact self-adjoint spectral PVM of $A$ , trace-class $e^{- β A}$ on the open interval $I$ , finite form energy of the competitor, and $k_{B} > 0$ for the entropy and optimization clause.

[A1]

The Axiom of Choice: the full Axiom of Choice is assumed for the unbounded self-adjoint spectral theorem and the trace-Gibbs supplier.

[F1]

Spectral theorem for unbounded self-adjoint operators (PVM form) and Bounded borel pvm integral: $A$ has its regular spectral PVM; bounded Borel functions of $A$ are defined by spectral integration, and the spectral moment law uses its actual form domain.

[F2]

Trace-class Gibbs normalization for a semibounded self-adjoint operator, fixed- $β$ clause: under AC, a nonzero Hilbert space, self-adjoint $A$ bounded below and trace-class $e^{- β A}$ , its injective compact exponential yields a complete energy eigenbasis, $0 < Z_{β} < \infty$ and a faithful normal density. Its interval moment clause is not used without the envelope proved below.

[F3]

Summable canonical differentiation: a nonempty finite/countable canonical model with energies $E_{j}$ , weights $g_{j} = 1$ and compact- $β$ summable $(1 + E_{j}^{2}) e^{- β E_{j}}$ envelopes has twice differentiable $Z$ , finite moments and $S = k_{B} (\log Z + β U)$ .

[F4]

Trace-class cross-log entropy inequality: for finite cross-log moment, $D (ρ ∥ σ) \geq 0$ , and its equality case is $ρ = σ$ .

Proof

1.1givenA1F1F2

By [F2], $Z_{β}$ is finite and positive and $ρ_{β}$ is a faithful trace-class density with a complete eigenbasis $u_{j}$ of $A$ , $A u_{j} = E_{j} u_{j}$ , and eigenvalues $q_{j} = e^{- β E_{j}} / Z_{β} > 0$ . The spectral logarithm is $- \log ρ_{β} = β A + (\log Z_{β}) I$ on the corresponding spectral form domain.

1.2givenF1F2algebra

Let $ρ = \sum_{i} p_{i} ∣ v_{i} ⟩ ⟨ v_{i} ∣$ be a normal density with the stated finite absolute form energy. Its cross-log moment against $ρ_{β}$ is $C (ρ, ρ_{β}) = \sum_{i} p_{i} \sum_{j} ∣ ⟨ v_{i}, u_{j} ⟩ ∣^{2} (- \log q_{j}) = β E_{A} (ρ) + \log Z_{β},$ because $- \log q_{j} = β E_{j} + \log Z_{β}$ , and the absolute first-moment hypothesis permits the signed sum. This value is finite.

2.1step 1.1F2F3algebra

Let $K = [b, B] \subset I$ be compact and choose $β_{0} \in I$ with $β_{0} < b$ . For $E \geq 0$ , the function $(1 + E^{2}) e^{- (b - β_{0}) E}$ is bounded; for $a \leq E \leq 0$ , $(1 + ∣ E ∣^{2}) e^{- (β - β_{0}) E}$ is uniformly bounded over $β \in K$ by compactness. Thus for a finite $C_{K}$ , $(1 + ∣ E_{j} ∣^{2}) e^{- β E_{j}} \leq C_{K} e^{- β_{0} E_{j}}$ for every $j$ and $β \in K$ . The right side is summable because $Z_{β_{0}} < \infty$ by [F2]. This is the exact uniform moment envelope required by [F3]; applying that supplier yields finite energy, finite entropy, and $S (ρ_{β}) = k_{B} (\log Z_{β} + β U_{β})$ , with $U_{β} = E_{A} (ρ_{β})$ .

3.1step 1.2F2F4algebra∎

Since $ρ_{β}$ is faithful by [F2], $supp ρ \subseteq supp ρ_{β}$ . Apply [F4] to step 1.2: $S (ρ) < \infty$ and $β E_{A} (ρ) + \log Z_{β} - \frac{S (ρ)}{k_{B}} = D (ρ ∥ ρ_{β}) \geq 0,$ with equality exactly when $ρ = ρ_{β}$ . Rearranging and using $T = (k_{B} β)^{- 1}$ , $F_{β} = - β^{- 1} \log Z_{β}$ gives the claimed inequality and equality case.

Self-adjoint trace-Gibbs normalization and form-energy optimization

Statement

Facts & Assumptions

Proof

Depends on

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