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Infinite identical-mode divergence

Example

Assume the Axiom of Countable Choice and the Axiom of Choice. Let $H$ be a separable infinite-dimensional complex Hilbert space with complete orthonormal family $(e_{j})_{j \geq 1}$ , let $ε_{0} > 0$ , $β > 0$ , and consider the degenerate mode list $ε_{j} : = ε_{0} (j \geq 1)$ (infinitely many modes of one and the same positive frequency). Adopt the notation $F (H)$ , $M$ , $v_{I}$ , $λ_{I}$ , $u_{I}$ , $D_{β}$ of Countable-mode product criterion. This example verifies concretely that:

(a) the one-particle weight sum diverges for every $β > 0$ , $\sum_{j \geq 1} e^{- β ε_{j}} = \sum_{j \geq 1} e^{- β ε_{0}} = + \infty,$ so the product criterion of Countable-mode product criterion (d) is violated;

(b) the grand-canonical weights do not converge: for every $N \geq 1$ $\sum_{\begin{array}{c} I \in M \\ I \subseteq {1, \dots, N} \end{array}} e^{- β λ_{I}} = \prod_{j = 1}^{N} \frac{1}{1 - e^{- β ε_{0}}} = (1 - e^{- β ε_{0}})^{- N},$ and these partial products increase to $+ \infty$ ; quantitatively $(1 - e^{- β ε_{0}})^{- N} \geq e^{N e^{- β ε_{0}}}$ for every $N$ ;

(c) consequently $\sum_{I \in M} e^{- β λ_{I}} = + \infty,$ the diagonal operator $D_{β}$ is not trace class, and there is no finite grand-canonical partition sum $tr D_{β}$ for this mode list;

(d) the divergence is caused by the infinite multiplicity of the frequency, not by its value: for the one-mode space $H = C$ with frequency $ω = ε_{0} > 0$ , Single-mode thermal trace gives the finite value $Z (β) = 1 / (1 - e^{- β ω})$ of the trace of the weight operator, whereas the countable product of these one-mode factors diverges here.

No thermodynamic interpretation of $β$ , of $D_{β}$ or of the divergence is asserted; the example is a statement about the convergence of the displayed series for a degenerate mode list.

Verification

Given: the separable infinite-dimensional $H$ with complete orthonormal family $(e_{j})_{j \geq 1}$ , the constant mode list $ε_{j} = ε_{0} > 0$ , $β > 0$ , and the objects $M, v_{I}, λ_{I}, D_{β}$ of Countable-mode product criterion.

[F1] For every $N \geq 1$ , $\sum_{I \in M, I \subseteq {1, \dots, N}} e^{- β λ_{I}} = \prod_{j = 1}^{N} (1 - e^{- β ε_{j}})^{- 1}$ , and $\sum_{I \in M} e^{- β λ_{I}} < + \infty$ if and only if $\sum_{j \geq 1} e^{- β ε_{j}} < + \infty$ , in which case the sum equals the infinite product; $0 < e^{- β ε_{j}} < 1$ (Countable-mode product criterion (c), (d)).

[F2] $D_{β}$ is trace class exactly when $\sum_{I \in M} e^{- β λ_{I}} < + \infty$ , and then $tr D_{β}$ equals that sum (Countable-mode product criterion (b), (e)).

[F3] If a series of reals converges then its terms tend to $0$ ; contrapositively, a series whose terms do not tend to $0$ diverges (If a series converges then its terms tend to $0$ , Series and absolute convergence in a normed space).

[F4] For $0 < x < 1$ , put $y = (1 - x)^{- 1} > 1$ . The integral inequalities give $x = 1 - 1 / y \leq \log y$ , and the reciprocal law of the natural logarithm gives $\log (1 - x) = - \log y \leq - x$ ; the supplier also gives the product law $\log (x y) = \log x + \log y$ ; $\log$ is increasing and is inverse to the real exponential (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[F5] For the one-mode space $F (C)$ with frequency $ω > 0$ and $β > 0$ , $Z (β) = \sum_{n \geq 0} e^{- β ω n} = 1 / (1 - e^{- β ω})$ and the weight operator $G_{β} e_{n} = e^{- β ω n} e_{n}$ is trace class with $tr G_{β} = Z (β) < \infty$ (Single-mode thermal trace (b), (c)).

1.1F3algebra

[Divergence of the one-particle weights] By the definition of the mode list, $e^{- β ε_{j}} = e^{- β ε_{0}} > 0$ for every $j$ , a constant strictly positive sequence that does not converge to $0$ ; by [F3] the series $\sum_{j} e^{- β ε_{j}}$ diverges, and since all its terms are positive its value is $+ \infty$ . This is (a).

1.2F1F4algebra

[Divergent partial products and the finite- $N$ identity] By [F1] applied to the constant list, for every $N \geq 1$ , $\sum_{I \subseteq {1, \dots, N}} e^{- β λ_{I}} = \prod_{j = 1}^{N} (1 - e^{- β ε_{0}})^{- 1} = (1 - e^{- β ε_{0}})^{- N}$ . Since $0 < 1 - e^{- β ε_{0}} < 1$ , these finite powers strictly increase. Put $x : = e^{- β ε_{0}} \in (0, 1)$ . By [F4] and the logarithm product and reciprocal laws, $\log ((1 - x)^{- N}) = - N \log (1 - x) \geq N x$ , so monotonicity of $\log$ gives $(1 - x)^{- N} \geq e^{N x}$ . For any $M > 0$ , choose $r : = \log M / x$ ; if $N > r$ , then $N x > \log M$ , hence $e^{N x} > M$ because the exponential is the increasing inverse of $\log$ . Thus the partial products tend to $+ \infty$ , proving the quantitative bound in (b).

2.1F1F2step 1.1step 1.2

[Divergence of the grand-canonical weights, no trace] By step 1.1 the series $\sum_{j} e^{- β ε_{j}}$ is not finite, so the equivalence in [F1] gives $\sum_{I \in M} e^{- β λ_{I}} = + \infty$ ; by step 1.2 its finite partial sums are finite and increasing, consistently. By [F2], $D_{β}$ is not trace class and $tr D_{β}$ does not exist as a finite number. This is (c).

2.2F5step 1.2algebra

[The single-mode contrast] For $H = C$ with $ω = ε_{0}$ , [F5] gives $Z (β) = 1 / (1 - e^{- β ω}) < \infty$ and a trace-class weight operator with trace $Z (β)$ ; the same closed-form expression appears in step 1.2 with each of the infinitely many factors equal to it, and step 1.2 proves that these constant-factor products diverge by the quantitative bound $(1 - e^{- β ω})^{- N} \geq e^{N e^{- β ω}} \to + \infty$ . Thus the infinite multiplicity, not the value of the positive frequency, produces the divergence. This is (d).

3.1step 1.1step 1.2step 2.1step 2.2∎

[Conclusion and scope] Steps 1.1-1.2 prove (a) and (b), step 2.1 proves (c), and step 2.2 proves (d). The scope paragraph of the example records that no thermodynamic interpretation of $β$ , $D_{β}$ or the divergence is asserted. This completes the verification.

Infinite identical-mode divergence

Example

Verification

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