How to get mean occupation numbers by Grand partition function?

[hokhani]

How to calculate $<n_i ^2>$ for an ideal gas by the grand partition function ($<n_i>$ is the occupation number)? In other words, I like to know how do we get to the formula $<n_i>=-1/\beta (\frac{\partial q}{\partial\epsilon})$ and $<n_i ^2>=1/Z_G [-(1/\beta \frac{\partial }{\partial\epsilon})^2 Z_G]$?

$Z_G$ is grand partition function , q=$ln Z_G$ and $\epsilon$ is the energy of the corresponding level.

 

[vanhees71]

Define
$$Z(\beta,\alpha)=\mathrm{Tr} \exp(-\beta \hat{H}+\alpha \hat{N}O.$$
Here, I assume we have a non-relativistic system with conserved particle number $\hat{N}$.
Then you get
$$\frac{\partial}{\partial \alpha} Z=Z \langle N \rangle, \quad \frac{\partial^2}{\partial \alpha^2} Z=Z \langle N^2 \rangle.$$
Now you have
$$\frac{\partial}{\partial \alpha} \ln Z=\frac{1}{Z} \frac{\partial Z}{\partial \alpha}=\langle N \rangle$$
and then
$$\frac{\partial^2}{\partial \alpha^2} \ln Z=\frac{1}{Z^2} \left (\frac{\partial Z}{\partial\alpha} \right )^2+\frac{1}{Z} \frac{\partial^2 Z}{\partial \alpha^2}=\langle N^2 \rangle -\langle N \rangle^2=\sigma_N^2.$$
So the 2nd derivative of the GK potential wrt. to $\alpha$ is the standard deviation $\sigma_N^2$ of the particle number, not the expectation value of the particle number squared!

More conventional is to write $\alpha = \beta \mu$, where $\mu$ is the chemical potential, but then it's a bit inconvenient for evaluating expectation values of the particle number and its powers (or equivalently cumulants of the particle number).

 

[hokhani]

Define
$$Z(\beta,\alpha)=\mathrm{Tr} \exp(-\beta \hat{H}+\alpha \hat{N}O.$$
Here, I assume we have a non-relativistic system with conserved particle number $\hat{N}$.
Then you get
$$\frac{\partial}{\partial \alpha} Z=Z \langle N \rangle, \quad \frac{\partial^2}{\partial \alpha^2} Z=Z \langle N^2 \rangle.$$
Now you have
$$\frac{\partial}{\partial \alpha} \ln Z=\frac{1}{Z} \frac{\partial Z}{\partial \alpha}=\langle N \rangle$$
and then
$$\frac{\partial^2}{\partial \alpha^2} \ln Z=\frac{1}{Z^2} \left (\frac{\partial Z}{\partial\alpha} \right )^2+\frac{1}{Z} \frac{\partial^2 Z}{\partial \alpha^2}=\langle N^2 \rangle -\langle N \rangle^2=\sigma_N^2.$$
So the 2nd derivative of the GK potential wrt. to $\alpha$ is the standard deviation $\sigma_N^2$ of the particle number, not the expectation value of the particle number squared!

More conventional is to write $\alpha = \beta \mu$, where $\mu$ is the chemical potential, but then it's a bit inconvenient for evaluating expectation values of the particle number and its powers (or equivalently cumulants of the particle number).

Thank you very much for your good response but what I meant by $n_i$ is the number of particles in the ith single particle state with energy $\epsilon_i$ and not the total number of particles, N.
I study the book "statistical mechanics by Pathria". Reading this book is somewhat difficult. Also I think this book (although is a very good book) hasn't well set apart the borders between quantum and classical treatment. Could you please tell me, if there is any, another good reference in that level instead?