Variance in particle number

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Express the mean partice number in therms of the grandcanonical partition function.
[itex]\langle N \rangle[/itex] is the Fermi-Dirac distribution, which is derived on that wikipedia page. So, you can perform the derivative yourself and verify the second equality.

The first equality can be derived as follows. First,

[itex]\displaystyle \langle ( \Delta N )^2 \rangle = \langle (N - \langle N \rangle )^2 \rangle = \langle N^2 - 2 \langle N \rangle N + \langle N \rangle^2 \rangle = \langle N^2 \rangle - \langle N \rangle^2[/itex].

Next, at constant volume and temperature, the grand-canonical partition function is given as the sum over all states, [itex]s[/itex] of the Gibbs factors, [itex]e^{- (e_s - \mu n_s ) / k_B T}[/itex] (check out the wikipedia page on partition function if this is unfamiliar):

[itex]\displaystyle Z = \sum_s e^{- (e_s - \mu n_s )/ k_B T}[/itex].

Here [itex]e_s[/itex] and [itex]n_s[/itex] are the state energy and occupation number. The Gibbs factor of a state measures the relative probability that that state is occupied. Hence, by definition,

[itex]\displaystyle \langle N \rangle = \frac{\sum_s n_s e^{-(e_s - \mu n_s)/k_B T}}{\sum_s e^{- (e_s - \mu n_s ) / k_B T}}[/itex].

Given these formulas for [itex]Z[/itex] and [itex]\langle N \rangle[/itex], you should be able to show that

[itex]\displaystyle \langle N \rangle = k_B T \frac{1}{Z} \frac{dZ}{d \mu}[/itex].

I've taken it for granted that the derivatives are taken at constant volume and temperature.

Analogously, show that

[itex] \displaystyle \langle N^2 \rangle = (k_B T)^2 \frac{1}{Z} \frac{d^2 Z}{d \mu^2} [/itex].

Okay. At this point, I think you have all of the formulas you need to derive the first equality that you quoted from wikipedia. Just a little bit of ingenuity left. Good luck.
Thank you for the answer!

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