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Nonzero pure-point equilibrium covariance is not integrable

Statement

Assume the Axiom of Choice and the trace-class Gibbs hypotheses of Trace-Gibbs bounded correlation measure and balance for a self-adjoint Hamiltonian $K$ in J with a complete discrete energy basis and faithful Gibbs density $σ_{β}$ at $β > 0$ in J $^{- 1}$ . Let $ℏ > 0$ be in J s, $t$ in s, and let $A = A^{*}$ be a nonzero bounded observable with $Tr (σ_{β} A) = 0$ . Define $C_{A A} (t) = Tr (σ_{β} U (t)^{*} A U (t) A), G_{A A} (t) = Re C_{A A} (t), U (t) = e^{- i t K / ℏ} .$ Then neither $C_{A A}$ nor $G_{A A}$ belongs to $L^{1} ([0, \infty))$ . In particular, the ordinary absolutely convergent isolated-system correlation integral used in a dc Green–Kubo formula does not exist under these hypotheses. This claim concerns discrete-spectrum trace-Gibbs dynamics only; it asserts no failure of temporal decay or transport limits in an infinite-volume or open-system model.

Facts & Assumptions

Given: The Axiom of Choice, a faithful trace-class Gibbs state with a complete energy eigenbasis, a nonzero bounded self-adjoint centered $A$ , and $ℏ > 0$ .

[A1]

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

[F1]

Trace-Gibbs bounded correlation measure and balance: for the Gibbs energy basis, $C_{A A} (t)$ has an absolutely summable atomic expansion with nonnegative weights $r_{i} ∣ A_{i j} ∣^{2}$ , whose total is $C_{A A} (0) = \sum_{i} r_{i} ∥ A u_{i} ∥^{2} \leq ∥ A ∥^{2}$ ; $σ_{β}$ is faithful.

[F2]

Dominated convergence: on any measure space, pointwise convergence under an integrable majorant permits passage of limits through integrals.

Proof

technique · grouped pure-point frequencies have a nonzero mean-square limit, which contradicts absolute integrability

Given: [A1], [F1], [F2], and the centered nonzero observable in the Statement.

1.1givenA1F1algebra

Group the nonnegative coefficients in [F1] by equal frequency, writing $C_{A A} (t) = \sum_{ω \in Ω} w_{ω} e^{- i ω t}$ with $w_{ω} \geq 0$ and $\sum_{ω} w_{ω} = C_{A A} (0) = Tr (σ_{β} A^{2})$ . This sum is strictly positive: in the complete energy basis, $Tr (σ_{β} A^{2}) = \sum_{i} r_{i} ∥ A u_{i} ∥^{2}$ , every $r_{i} > 0$ , and $A \neq 0$ prevents all $A u_{i}$ from vanishing. It is bounded by $∥ A ∥^{2}$ because each $∥ A u_{i} ∥^{2} \leq ∥ A ∥^{2}$ and $\sum_{i} r_{i} = 1$ .

2.1givenF1F2step 1.1algebra

Absolute summability gives $\sum_{ω, ν} w_{ω} w_{ν} < \infty$ and allows termwise integration of $∣ C_{A A} (t) ∣^{2}$ . For each pair, $T^{- 1} \int_{0}^{T} e^{- i (ω - ν) t} d t$ tends to $1$ if $ω = ν$ and $0$ otherwise, and its modulus is at most $1$ . Apply [F2] to the countable pair set with summable majorant $w_{ω} w_{ν}$ ; then $\lim_{T \to \infty} \frac{1}{T} \int_{0}^{T} ∣ C_{A A} (t) ∣^{2} d t = \sum_{ω} w_{ω}^{2} > 0.$ If $C_{A A} \in L^{1} ([0, \infty))$ , the left side is at most $∥ A ∥^{2} T^{- 1} \int_{0}^{T} ∣ C_{A A} (t) ∣ d t \to 0$ , because step 1.1 gives $∣ C_{A A} (t) ∣ \leq \sum_{ω} w_{ω} \leq ∥ A ∥^{2}$ . This contradiction proves $C_{A A} \notin L^{1}$ .

3.1F1F2step 1.1step 2.1algebra∎

The real part has grouped nonnegative weights $g_{ω} = (w_{ω} + w_{- ω}) / 2$ , with absent frequencies assigned weight zero. They satisfy $\sum_{ω} g_{ω} = \sum_{ω} w_{ω} > 0$ and $G_{A A} (t) = \sum_{ω} g_{ω} e^{- i ω t}$ . Repeating step 2.1 with $g_{ω}$ gives $\lim_{T \to \infty} T^{- 1} \int_{0}^{T} G_{A A} (t)^{2} d t = \sum_{ω} g_{ω}^{2} > 0$ . If $G_{A A} \in L^{1}$ , then this average is at most $∥ A ∥^{2} T^{- 1} \int_{0}^{T} ∣ G_{A A} (t) ∣ d t \to 0$ , again a contradiction.

Nonzero pure-point equilibrium covariance is not integrable

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources