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Identical bosonic modes do not have a finite infinite-mode Gibbs trace

Physical scope

A countable family of independent identical bosonic modes at fixed positive excitation energy; the divergent positive trace rules out this particular normalized Gibbs density and says nothing about finite mode sets or renormalized models.

Example

Assume AC and Countable Choice as required by Finite bosonic and fermionic mode partition and occupation formulas and the trace-class suppliers. Take modes indexed by $j \in N$ , each with excitation energy $ε > 0$ joules, and let $q = e^{- β ε} \in (0, 1)$ . Let $F$ be the set of finite-support occupation sequences $n = (n_{j})_{j \geq 1}$ with $n_{j} \in N_{0}$ . Enumerate $F$ by increasing $R (n) = \sum_{j : n_{j} > 0} (j + n_{j})$ and lexicographic order of the finite list of nonzero pairs $(j, n_{j})$ within each level. Each level is finite because $j, n_{j} \leq R (n)$ on its support, and every configuration occurs; skipping empty levels gives a bijection $N \to F$ . Define $ℓ^{2} (F) : = {c : F \to C : \sum_{n \in F} ∣ c (n) ∣^{2} < \infty}, ⟨ c, d ⟩ : = \sum_{n \in F} c (n) \overline{d (n)} .$ The vector $∣ n ⟩$ is the indicator of ${n}$ ; these canonical occupation vectors form an orthonormal basis. Relabeling by the displayed enumeration is an isometry with $ℓ^{2} (N, C)$ , whose completeness and dense finite-support subspace are supplied by Completeness and linear domain of the diagonal sequence model; the coordinate pairing makes the displayed indicator vectors orthonormal, and their finite linear span is exactly that dense subspace, so they form an orthonormal basis; the definition of square-summable families and their inner product is Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ . Define the bounded positive diagonal Gibbs weight $K_{β} ∣ n ⟩ = q^{\sum_{j} n_{j}} ∣ n ⟩$ . On the subspace with $n_{j} = 0$ for every $j > M$ , the corresponding finite-mode partition sum is the trace of that restriction.

Facts & Assumptions

Given: The independent identical-mode Fock model, positive $β ε$ , and the canonical trace convention.

[F1]

Finite bosonic and fermionic mode partition and occupation formulas: for each finite collection of independent bosonic modes the Gibbs trace is the product of the one-mode traces $(1 - q)^{- 1}$ .

[F2]

Canonical sampling assumption for finite and normalized statistical models: the A-page countable canonical law is adopted only for a specified countable model with a positive finite partition sum.

[F3]

The Axiom of Countable Choice ( ${A C}_{ω}$ ), Trace class operator, and Trace of a positive operator is the sum of its eigenvalues: under Countable Choice, a positive trace-class operator has finite trace equal to the sum of its nonnegative eigenvalues with multiplicity.

[F4]

Countable canonical model: the nonnegative countable partition sum is the supremum of its finite subsums, so a finite normalizer bounds the sum over every subset of the listed states.

[F5]

SI quantities and unit conventions: $ε$ is in joules and $β$ in $J^{- 1}$ , so $β ε$ is dimensionless.

[F6]

Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ defines $ℓ^{2} (F)$ and its inner product. The explicit enumeration identifies it isometrically with $ℓ^{2} (N, C)$ , which is complete with a dense finite-support subspace by Completeness and linear domain of the diagonal sequence model. The indicator vectors are orthonormal by the coordinate pairing, and their finite linear span is that dense subspace, so they form an orthonormal basis.

[A1]

The basis indices are all finite-support occupation sequences with physical energy $E_{n} = ε \sum_{j} n_{j}$ joules, so their Gibbs eigenvalues are $e^{- β E_{n}} = q^{\sum_{j} n_{j}}$ ; every finite-mode subset is a subset of this full basis.

[A2]

AC and Countable Choice are assumed through the finite-mode Gibbs result and the trace-class definitions/theorems cited above.

Verification

Given: $0 < q < 1$ and the finite-mode subspaces described above.

1.1F1A1F5algebra

The trace contribution from states supported on the first $M$ modes is the finite product $Z_{M} = (1 - q)^{- M}$ by [F1]. Since $c = (1 - q)^{- 1} > 1$ , $Z_{M} = c^{M}$ is unbounded: writing $c = 1 + r$ with $r > 0$ , induction gives $(1 + r)^{M} \geq 1 + M r$ . The unit convention [F5] makes $β ε$ dimensionless.

2.1A1F1F2F3F4F6step 1.1algebra∎

If $K_{β}$ were trace class, [F3] would make its positive eigenvalue sum finite and equal to its trace. For each $M$ , the configurations with $n_{j} = 0$ for $j > M$ form a subset of the enumerated states [A1]; [F4] therefore makes their finite-mode partition sum $Z_{M}$ no larger than the full normalizer. [F1] identifies that subset sum as $Z_{M}$ , and step 1.1 makes $Z_{M}$ unbounded, a contradiction. Hence $K_{β}$ is not trace class and the countable partition sum has no finite value. The preparation condition [F2] fails, so this model has no normalized countable canonical law of the stated form.

Identical bosonic modes do not have a finite infinite-mode Gibbs trace

Example

Facts & Assumptions

Verification

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