Second moment of occupation number for bosons

[Rayan]

Homework Statement

Show that for bosons

$$ \overline{(\Delta \eta)^2} \quad=\quad \overline{ \eta_{BE}} (1 + \overline{ \eta_{BE}} ) $$

where

$$ (\Delta \eta) \quad=\quad \eta - \overline{ \eta } $$

Relevant Equations

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I tried to show this equality by explicitly determining what

$$ \overline{(\Delta \eta)^2} $$

is, but I got a totally different answer for some reason, here is my attempt to solve it, what did I miss?

 

[vanhees71]

Note that
$$\Delta n^2 = \overline{n^2}-\overline{n}^2=\langle n^2 \rangle - \langle n \rangle^2.$$
Then indeed you have in the grand-canonical ensemble
$$Z(\beta,\alpha)=\prod_{\vec{p}} \sum_{n=0}^{\infty} \exp[n (-\beta \omega_{\vec{p}}+\alpha)] = \prod_{\vec{p}} \frac{1}{1-\exp(-\beta \omega_{\vec{p}}+\alpha)}.$$
It's easier to work with
$$\Omega=\ln Z=-\sum_{\vec{p}} \ln [ 1-\exp(-\beta \omega_{\vec{p}}+\alpha)].$$
Now
$$\langle n \rangle=\frac{1}{Z} \partial_{\alpha} Z =\partial_{\alpha} \ln Z=\sum_{\vec{p}} f_{\text{B}}(\omega_{\vec{p}})=\sum{\vec{p}} \frac{1}{\exp(\beta \epsilon-\alpha)}.$$
Further you have
$$\partial_\alpha^2 \Omega =\partial_{\alpha} \left ( \frac{1}{Z} \partial_{\alpha Z} \right) =\frac{1}{Z} \partial_{\alpha}^2 Z -\left (\frac{1}{Z} \partial_{\alpha} Z \right)=\langle n^2 \rangle -\langle n \rangle^2=\Delta n^2.$$
Then you get
$$\Delta n^2 = \partial_{\alpha}^2 \ln Z=\sum_{\vec{p}} \partial_{\alpha} f_{\text{B}}(\omega_{\vec{p}}).$$
It's easy to check, that this gives what you try to prove.