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Single-mode Bose and Fermi occupation fluctuations

Statement

For $q \in (0, 1)$ , the bosonic geometric law on $n \in N_{0}$ with weights $(1 - q) q^{n}$ has mean $q / (1 - q)$ and variance $q / (1 - q)^{2} = E [n] (1 + E [n])$ ; its relative standard deviation, defined as $\sqrt{Var (n)} / E [n]$ , tends to $1$ as $q ↑ 1$ . The fermionic two-state law on $n \in {0, 1}$ with weights $1 / (1 + q)$ and $q / (1 + q)$ has mean $q / (1 + q)$ and variance $q / (1 + q)^{2} = E [n] (1 - E [n])$ . Here Bose and Fermi name these explicitly specified counting laws; no claim about a physical preparation is made.

Facts & Assumptions

Given: The geometric and two-state laws with $q \in (0, 1)$ .

[A1]

The bosonic states are $N_{0}$ with weights proportional to $q^{n}$ ; the fermionic states are ${0, 1}$ with weights proportional to $q^{n}$ .

[F1]

Countable canonical model and Summable canonical differentiation: countable canonical weights normalize when the partition sum is finite, and under the envelope the mean and variance follow from derivatives of $Z$ .

[F2]

Finite probability spaces, outcome weights, events, and event probabilities, Expectation of a real random variable on a finite probability space, and Variance, standard deviation, and covariance on a finite probability space: finite laws normalize and define means and variances by finite sums.

[F3]

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: every $q \in (0, 1)$ has the form $e^{- γ}$ with $γ = - \log q > 0$ .

[F4]

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : derivatives of the parameterized exponential and logarithm compositions obey the chain rule.

[F5]

The exponential dominates every fixed nonnegative integer power at $+ \infty$ : $n^{k} e^{- b n} \to 0$ for each fixed integer $k$ and $b > 0$ .

Proof

Given: The geometric and two-state laws with $q \in (0, 1)$ .

1.1A1F1F3F5algebra

Let $S_{N} = \sum_{n = 0}^{N} q^{n}$ . Multiplying this finite sum by $1 - q$ gives $(1 - q) S_{N} = 1 - q^{N + 1}$ . Since $q = e^{- γ}$ for $γ > 0$ , [F5] gives $q^{N + 1} \to 0$ ; hence $Z = (1 - q)^{- 1}$ and the weights $(1 - q) q^{n}$ sum to one. On a compact $γ$ -interval $[b, B] \subset (0, \infty)$ , set $a_{n} = (1 + n^{2}) e^{- b n}$ . Then $a_{n + 1} / a_{n} = e^{- b} (1 + (n + 1)^{2}) / (1 + n^{2}) \to e^{- b} < 1$ . Choose $c$ with $e^{- b} < c < 1$ ; for all sufficiently large $n$ , $a_{n + 1} \leq c a_{n}$ , so the tail is bounded by a convergent geometric series. Thus $\sum_{n} (1 + n^{2}) e^{- γ n} \leq \sum_{n} a_{n} < \infty$ , which is the local envelope required by [F1].

1.2A1F2F3

The fermionic partition sum is $1 + q$ ; its two probabilities are $1 / (1 + q)$ and $q / (1 + q)$ . Their finite weighted sums give mean $q / (1 + q)$ and variance $q / (1 + q)^{2}$ , equal to the mean times one minus the mean.

2.1F1F3F4step 1.1

With $Z (γ) = (1 - e^{- γ})^{- 1}$ , [F1] gives $E n = - \partial_{γ} \log Z = q / (1 - q)$ and $Var (n) = \partial_{γ}^{2} \log Z = q / (1 - q)^{2}$ . Substitution $q = e^{- γ}$ and [F4] justify the parameter derivatives. Therefore $\sqrt{Var (n)} / E n = 1 / \sqrt{q} \to 1$ as $q ↑ 1$ .

3.1A1step 2.1step 1.2∎

These are calculations for the two specified probability laws only; no independence, physical preparation, or measured fluctuation claim is used.

Single-mode Bose and Fermi occupation fluctuations

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources