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Finite ground-state degeneracy entropy limit

Statement

Let $I$ be a finite nonempty set, take the canonical model Countable canonical model on $J = (0, \infty)$ , and let its energies be $E_{i}$ and positive weights be $g_{i}$ . Write $E_{0} : = \min_{i \in I} E_{i}$ , let $G : = {i \in I : E_{i} = E_{0}}$ have cardinality $m \geq 1$ , and set $W_{0} : = \sum_{i \in G} g_{i}$ . As $β \to + \infty$ , each ground-state probability tends to $g_{i} / W_{0}$ and each excited-state probability tends to zero. The Shannon entropy $H (p (β)) : = - \sum_{i \in I} p_{i} (β) \log p_{i} (β)$ tends to $- \sum_{i \in G} (g_{i} / W_{0}) \log (g_{i} / W_{0})$ . In particular, if all $g_{i} = 1$ , the ground-state probabilities tend to $1 / m$ and the entropy tends to $\log m$ . This is a finite-model statement only.

Facts & Assumptions

Given: As stated in the item.

[A1]

$I$ is finite and nonempty, the parameter interval is $J = (0, \infty)$ , $g_{i} > 0$ for every $i$ , $E_{0} = \min_{i} E_{i}$ , $G = {i : E_{i} = E_{0}}$ is nonempty, and every gap $E_{i} - E_{0}$ for $i \notin G$ is positive.

[F1]

Countable canonical model: finite canonical weights are $p_{i} = g_{i} e^{- β E_{i}} / Z$ with natural logarithms.

[F2]

The real exponential function and the number $e$ by a power series, The natural logarithm as the inverse of the exponential function, and Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: exponential is positive; logarithms convert products to sums and are continuous on positive reals.

[F3]

The exponential dominates every fixed nonnegative integer power at $+ \infty$ : $β e^{- β Δ} \to 0$ for every fixed $Δ > 0$ .

Proof

Given: A finite energy list with ground energy $E_{0}$ , degeneracy $g \geq 1$ , and positive gaps for all other states.

1.1A1F1F2

Set $D (β) : = W_{0} + \sum_{i \notin G} g_{i} e^{- β (E_{i} - E_{0})}$ , so factoring the ground energy gives $Z = e^{- β E_{0}} D (β)$ . Each gap is positive, and there are finitely many, so $D (β) \to W_{0} > 0$ . Thus $p_{i} (β) = g_{i} / D (β) \to g_{i} / W_{0}$ for $i \in G$ , and $p_{i} (β) = g_{i} e^{- β (E_{i} - E_{0})} / D (β) \to 0$ for $i \notin G$ .

2.1A1F2F3step 1.1∎

For $i \in G$ , step 1.1 and continuity of $x \mapsto - x \log x$ on $(0, \infty)$ give $- p_{i} \log p_{i} \to - (g_{i} / W_{0}) \log (g_{i} / W_{0})$ . For an excited state with gap $Δ_{i} > 0$ , step 1.1 gives $p_{i} = O (e^{- β Δ_{i}})$ and $\log p_{i} = \log g_{i} - β Δ_{i} - \log D (β)$ ; since $D (β) \to W_{0} > 0$ , $\log D (β)$ is bounded, so $∣ p_{i} \log p_{i} ∣ = O ((1 + β) e^{- β Δ_{i}}) \to 0$ by [F3]. There are finitely many states, hence the entropy tends to the stated ground-state entropy. If all $g_{i} = 1$ , then $W_{0} = m$ and the limit is $- m (1 / m) \log (1 / m) = \log m$ .

Finite ground-state degeneracy entropy limit

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources