# Variance in particle number

#### cryptist

I saw an equation on wikipedia: (http://en.wikipedia.org/wiki/Fermi-Dirac_statistics) Does anybody know how this is derived?

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#### DrDu

Science Advisor
Express the mean partice number in therms of the grandcanonical partition function.

#### krome

$\langle N \rangle$ 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,

$\displaystyle \langle ( \Delta N )^2 \rangle = \langle (N - \langle N \rangle )^2 \rangle = \langle N^2 - 2 \langle N \rangle N + \langle N \rangle^2 \rangle = \langle N^2 \rangle - \langle N \rangle^2$.

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

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

Here $e_s$ and $n_s$ 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,

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

Given these formulas for $Z$ and $\langle N \rangle$, you should be able to show that

$\displaystyle \langle N \rangle = k_B T \frac{1}{Z} \frac{dZ}{d \mu}$.

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

Analogously, show that

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

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.

#### cryptist

Thank you for the answer!

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