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Infinite spin thermal trace diverges despite a unique KMS state

Statement refuted

For an infinite spin lattice, a unique onsite $β$ -KMS state implies that the global onsite Gibbs operator in the finite-excitation vacuum representation has finite trace. The statement is false even for one-dimensional two-level spins.

Facts & Assumptions

Given: ${A C}_{ω}$ , $Γ = Z$ , $q = 2$ , $ϵ > 0$ in J, $β > 0$ in J $^{- 1}$ , and $ℏ > 0$ in J s.

[A1]

Countable Choice and the finite-excitation Hilbert basis are as in The Axiom of Countable Choice ( ${A C}_{ω}$ ) and Finite-excitation vacuum spin representation and normality.

[F1]

Local spin matrix algebras and their factor inclusions and Finite-excitation vacuum spin representation and normality: $Γ = Z$ has finite two-state local algebras, and $H_{v a c} = ℓ^{2} (S_{f i n})$ has the specified orthonormal basis $(e_{s})$ .

[F2]

Adjoint of a densely defined operator: $y \in D (T^{*})$ exactly when $x \mapsto ⟨ T x, y ⟩$ is bounded on $D (T)$ ; then $T^{*} y$ is characterized by $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ for all $x \in D (T)$ .

[F3]

Trace class operator and Trace of a trace class operator: a trace-class operator has an absolutely summable diagonal family in every supplied Hilbert basis, and its trace is that finite absolutely convergent sum.

[F4]

Unique onsite product Gibbs KMS state: the finite onsite Hamiltonian $h = ϵ ∣ 1 ⟩ ⟨ 1 ∣$ has a unique quasi-local $β$ -KMS product state with one-site density $r = e^{- β h} / Tr (e^{- β h})$ .

[F5]

Thermal product state has no normal vacuum extension: this full-rank product KMS state has no normal extension to $B (H_{v a c})$ .

Counterexample

For $s \in S_{f i n}$ , let $N (s) : = ∣ {x \in Z : s_{x} = 1} ∣ < \infty$ . Define the diagonal operator $H_{v a c} e_{s} = ϵ N (s) e_{s}, D (H_{v a c}) : = {ψ \in ℓ^{2} (S_{f i n}) : \sum_{s \in S_{f i n}} ϵ^{2} N (s)^{2} ∣ ψ_{s} ∣^{2} < \infty} .$ For $β > 0$ , define the bounded positive diagonal Gibbs weight $G_{β} e_{s} : = e^{- β ϵ N (s)} e_{s}$ . Its extended vacuum-basis partition sum is defined by $Z_{v a c} (β) : = \sup_{\begin{array}{c} J \subseteq S_{f i n} \\ J f i n i t e \end{array}} \sum_{s \in J} ⟨ G_{β} e_{s}, e_{s} ⟩ = \sup_{\begin{array}{c} J \subseteq S_{f i n} \\ J f i n i t e \end{array}} \sum_{s \in J} e^{- β ϵ N (s)} \in [0, \infty] .$

1.1A1F1construct

The finite linear span of $(e_{s})$ is contained in $D (H_{v a c})$ and is dense, so $H_{v a c}$ is densely defined. Its diagonal eigenvalues $ϵ N (s)$ are real, hence $⟨ H_{v a c} ψ, φ ⟩ = ⟨ ψ, H_{v a c} φ ⟩$ for $ψ, φ \in D (H_{v a c})$ by Cauchy–Schwarz. Thus it is symmetric.

1.2F1F4algebra

Define $U_{t} e_{s} = e^{i t ϵ N (s) / ℏ} e_{s}$ ; this is unitary because every diagonal coefficient has modulus one. A local matrix coefficient between $e_{s}$ and $e_{t}$ can be nonzero only if $s$ and $t$ agree outside its finite support $Λ$ . In that case $N (t) - N (s) = \sum_{x \in Λ} (1_{{t_{x} = 1}} - 1_{{s_{x} = 1}})$ , so conjugation by $U_{t}$ multiplies the coefficient by the same phase as conjugation by the finite onsite Hamiltonian $H_{Λ} = ϵ \sum_{x \in Λ} ∣ 1 ⟩ ⟨ 1 ∣_{x}$ . Thus $U_{t} π (A) U_{t}^{*} = π (α_{t} (A))$ for every local $A$ ; this operator identity extends to $A$ by norm continuity.

1.3F1constructalgebra

For each $x \in Z$ , the one-excitation configuration $s^{(x)}$ with $s_{x}^{(x)} = 1$ and all other labels $0$ contributes $e^{- β ϵ}$ to $Z_{v a c} (β)$ . These configurations are distinct, so for any $m$ distinct sites the finite subsum is $m e^{- β ϵ}$ . Since $ϵ, β > 0$ , this grows without bound as $m \to \infty$ , and $Z_{v a c} (β) = + \infty$ .

2.1F2step 1.1algebra

If $φ \in D (H_{v a c}^{*})$ , then by [F2] there is $η \in ℓ^{2} (S_{f i n})$ such that $⟨ H_{v a c} ψ, φ ⟩ = ⟨ ψ, η ⟩$ for every $ψ \in D (H_{v a c})$ . Taking $ψ = e_{s}$ gives $η_{s} = ϵ N (s) φ_{s}$ for every $s$ , so $(ϵ N (s) φ_{s})_{s} \in ℓ^{2} (S_{f i n})$ and $φ \in D (H_{v a c})$ . Conversely, if $φ \in D (H_{v a c})$ , the diagonal sums satisfy $⟨ H_{v a c} ψ, φ ⟩ = ⟨ ψ, H_{v a c} φ ⟩$ for all $ψ \in D (H_{v a c})$ , hence $H_{v a c}^{*} φ = H_{v a c} φ$ . Therefore $D (H_{v a c}^{*}) = D (H_{v a c})$ and the two operators agree: $H_{v a c}$ is self-adjoint.

3.1F3F4F5step 1.3∎

If $G_{β}$ were trace class, [F3] applied to the supplied basis $(e_{s})$ would make the nonnegative diagonal sum defining $Z_{v a c} (β)$ finite, contrary to step 1.3. Thus no normalized trace-class Gibbs density is obtained from $G_{β}$ . Yet [F4] gives a unique quasi-local KMS state, and [F5] confirms that it has no normal vacuum-representation extension.

Infinite spin thermal trace diverges despite a unique KMS state

Statement refuted

Facts & Assumptions

Counterexample

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