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?

  • $\begingroup$ Can you modify the derivation of $n_i$ in such a way that it yields $n_i^2$ maybe? $\endgroup$
    – Nephente
    May 9, 2021 at 17:20
  • $\begingroup$ I don't know how you can do that. The only way i can think of is double derivation of the grand canonical potential according to the chemical potential. But I don't know. $\endgroup$
    – imbAF
    May 9, 2021 at 17:25
  • $\begingroup$ That is certainly a good idea! Although if you take the derivative wrt $\mu$, it will yield the total particle number, won't it? Maybe take derivates of the grand canonical potential wrt $\epsilon_i$... I suppose you have an expression for the potential. $\endgroup$
    – Nephente
    May 9, 2021 at 17:30
  • $\begingroup$ On what ground do i take a derivation wrt $\epsilon_i$. To me this looks like it falls out of the sky. $\endgroup$
    – imbAF
    May 9, 2021 at 17:34

1 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 $n_i$.

  • $\begingroup$ I kind of get it now. I was able to find the $\bar n_i$ but i used the partial derivative of the grand canonical potential with regards to the chemical potential. How is it that i got a similar result ? $\endgroup$
    – imbAF
    May 9, 2021 at 17:49
  • $\begingroup$ @imbAF If you do that, you will get the sum over all occupation numbers. You can read of one from the other. $\endgroup$
    – Nephente
    May 9, 2021 at 17:50
  • $\begingroup$ I wasn't able to get anything that had meaning, probably because I have no idea of what is happening. But regardless thx for the help, much appreciated ! $\endgroup$
    – imbAF
    May 9, 2021 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.