Variance of the occupation Number of a quantum gas

Question

For a quantum gas (according to the grand canonical ensemble) the average occupation number is $$\langle n_{i}\rangle = \frac{1}{e^{\beta(\epsilon_i - \mu) } \pm 1}.$$ Since this is an average, that means we can find its standard deviation and also variance. For the variance we know: $\Delta x= \langle x{^2}\rangle - \langle x\rangle^{2}$.

How can we calculate the first term? In the grand canonical ensemble, at no point have we calculated the square of the occupation number. But logically speaking the variance should exist, no?

Answer 1

More of a hint, but it should guide you in the right direction.

The expectation of any observable $O$ in the grand-canonical potential is given by

$$ \bar{O} = \frac{1}{Z}\sum O\exp\left(-\beta\sum_i\epsilon_i n_i -\mu \sum_i n_i\right)$$ where the big sum in front extends over all particle number configurations.

The partition function $Z$ is retrieved by setting $O=1$. The grand-canonical potential is defined as $\Phi \equiv -T\ln Z$.

A bit of calculus shows that for example (do it!)

$$ \partial_{\epsilon_i} \Phi = \bar{n_i} $$

I leave it up to you to figure out how to obtain the variance of $n_i$.