If $Z=Z(z,V,T)$ is the Grand canonical Partition function, $\beta =\frac{1}{k_BT}$,$z=e^{\beta \mu }$ is the fugacity and $\epsilon_{\vec p}$ is the energy of a single particle in pth momentum state, then how can we derive the relation for the average occupation number:$$<n_{\vec p}>= -\frac{1}{\beta}\frac{\partial lnZ(z,V,T) }{\partial \epsilon_{\vec p}}$$ Here, z and T are constant
[Physics] Average Occupation Number in Bose Einstein Statistics using Grand Canonical Ensemble
statistical mechanics
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The reason for using microstates is that it is the only way to come up with quantum statistics, but the grand-canonical potential is defined also for a classical system. And you are right, one should take into acount the state with zero particles, let me show why on a simple example.
Let us take for instance a system in which $N$ indistiguishable independent classical particles can be in two states, one of zero energy called $a$ and one of energy $\epsilon$ called $b$. The canonical partition function for one particle is $$ z=\mathrm{e}^{-\beta E_a}+\mathrm{e}^{-\beta E_b}=1+\mathrm{e}^{-\beta\epsilon}.$$ The canonical partition function is $$Z_{\mathrm{c}}(N)=z^N$$ and the grand-canonical partition function is $$\mathcal{Z}=\sum_{N=0}^\infty\frac{1}{N!}Z_{\mathrm{c}}(N)\mathrm{e}^{\beta\mu N}=\exp\left(z\mathrm{e}^{\beta\mu}\right).$$ The factor $\frac1{N!}$ comes from the indistinguishability of the particles in the grand-canonical ensemble. Consider now the two possible microstates $a$ and $b$. For the zero-energy state, with chemical potential $\mu$, the microstates's grand-canonical functions are $$Z_a=\sum_{N=0}^\infty\frac1{N!}\mathrm{e}^{\beta\mu N}=\exp\left(\mathrm{e}^{\beta\mu}\right)$$ and $$Z_b=\sum_{N=0}^\infty\frac{1}{N!}\mathrm{e}^{-\beta N\epsilon+\beta\mu N} =\exp\left(\mathrm{e}^{\beta\mu-\beta\epsilon}\right).$$ Now computing $\mathcal{Z}=Z_aZ_b$, we find $$\mathcal{Z}=Z_aZ_b=\exp\left(\mathrm{e}^{\beta\mu}+\mathrm{e}^{\beta\mu-\beta\epsilon}\right)=\exp\left(z\mathrm{e}^{\beta\mu}\right),$$ the same result as before. If we did not take the 0 particle state into account in our calculations of $\mathcal{Z}$, $Z_a$ and $Z_b$, we could not obtain the same result from these two approaches because on the one hand we would get $\mathcal{Z}-1$ and on the other hand $$(Z_a-1)(Z_b-1)=\mathcal{Z}-Z_a-Z_b+1\neq\mathcal{Z}-1.$$
As a conclusion, it seems that this is a typo in your textbook. The role of the state with no particles is very important, it should always be there in a grand-canonical approach, because it accounts not only for the state with no particles, but also, if you consider the microstates, for the possibility that a microstate ($a$ or $b$ in the above example) has no particle in it. This is particularly important if the number of microstates is infinite because the number of particles is always finite and therefore all states cannot be occupied.
The difference lies in the way we count the number of states of the system in quantum and classical cases.
The formulas you wrote are actually for the grand canonical partition functions for a single energy state, not for the whole system including all the energy states. The total grand canonical partition function is $$\mathcal{Z} = \sum_{all\ states}{e^{-\beta(E-N\mu)}} = \sum_{N=0}^\infty\sum_{\{E\}}{e^{-\beta(E-N\mu)}}$$
Now, if the particles are bosons, then the energy eigenstates are countable as $\{\epsilon_i\}$ and $\mathcal{Z}$ would be $$\mathcal{Z} = \sum_{\{n_i\}}e^{-\beta\sum_{i}{n_i(\epsilon_i-\mu)}} = \sum_{\{n_i\}}\prod_ie^{-\beta n_i(\epsilon_i-\mu)}=\prod_i \mathcal{Z_i}^{B-E}$$ where $n_i$ is the number of particles in $i$-th energy state, thus $\sum_i{n_i}=N$, and $$\mathcal{Z_i}^{B-E} = \sum_{n=0}^\infty e^{-n\beta(\epsilon_i-\mu)}$$ Here, $\mathcal{Z_i}^{B-E}$ is the grand canonical partition function for one energy eigenstate with energy $\epsilon_i$ in Bose-Einstein statistics.
On the other hand, in classical regime the energy of a particle can take any energy. In this case, one point in the $6N$-dimensional phase space denotes one state of system. Therefore, the energy states are not countable as there is an infinite number of points in the phase space within any phase space volume. To count the states we take the "semi-classical" approach by taking the phase space volume of one state of the system to be $(2\pi\hbar)^{3N}$. We can then integrate over the whole phase space and divide the integral by this unit volume to get the number of states. However, as the particles are assumed to be indistinguishable, any permutation of the system configuration (the set of $\{\vec{x}_n,\ \vec{p}_n\}$) would actually be the same state of the system. Therefore, when we integrate over the whole phase space volume we overcount the total number of states by $N!$. That's why we need to divide the integral by Gibbs factor $N!$. For a system of non-interacting particles, the N-particle canonical partition function then can be written as $Z_N = \frac{Z_1^N}{N!}$ where $Z_1$ is the canonical partition function for one particle.
Now, the grand canonical partition function for a classical system would be $$\begin{align} \mathcal{Z}&=\sum_{N=0}^\infty{\int_0^\infty dE\ \Omega(E,N)e^{-\beta(E-\mu N)}} =\sum_{N=0}^\infty e^{\beta\mu N} Z_N = \sum_{N=0}^\infty e^{\beta\mu N} \frac{Z_1^N}{N!}\\ \end{align}$$
If we want to derive the partition function $\mathcal{Z_i}$ for a single energy state similar to the Bose-Einstein statistics, we can assume the energy of a single particle to be discrete and countable as $\{\epsilon_i\}$. This can be achieved by dividing the one particle phase space into s of unit volume and assigning one representative energy to every unit volume section. Then, the grand canonical partition function is, $$ \begin{align} \mathcal{Z} &= \sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} (\sum_i e^{-\beta\epsilon_i})^N\\ &=\sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} \sum_{\{n_i\},\sum n_i = N} \frac{N!}{\prod_i n_i!} e^{-\beta\sum_i n_i\epsilon_i}\\ &=\sum_{\{n_i\}}\prod_i \frac{1}{n_i!} e^{-\beta n_i(\epsilon_i-\mu)}\\ &=\prod_i \sum_{n=0}^\infty \frac{1}{n!} e^{-\beta n(\epsilon_i-\mu)} = \prod_i \mathcal{Z_i}^{M-B} \end{align}$$
This $\mathcal{Z_i}^{M-B}$ is the single energy state grand canonical partition function in Maxwell-Boltzmann statistics.
Maxwell-Boltzmann statistics is the classical limit for Bose-Einstein statistics. The condition for the system to be classical is the single state occupation number $\bar{n}$ to satisfy $\bar{n} \ll 1$, in other words, the total number of single particle states $M$ should satisfy $N \ll M$. As, the single particle partition function $Z_1$ is actually a weighted sum over all the states, $N\ll Z_1$ will satisfy $N \ll M$ for the system to be classical.
Best Answer
The grand partition function is given by $$Z(z,V,T)=\sum_{N=0}^{\infty}[z^N\sum_{\{n_{\vec{p}}\}}e^{-\beta\sum_{\vec{p}}n_{\vec{p}}\epsilon_{\vec{p}}}]$$ where $N$ is the total number of particles and $\{n_{\vec{p}}\}$ means sum over the distributed occupation numbers that subject to the constraint: $$\sum_{\vec{p}}n_{\vec{p}}=N$$ Then as usual, the average occupation number for a state labeled by $\vec{p}$ would be given by: $$<n_{\vec{p}}>=\frac{\sum_{N=0}^{\infty}[z^N\sum_{\{n_{\vec{p}}\}}n_{\vec{p}}e^{-\beta\sum_{\vec{p}}n_{\vec{p}}\epsilon_{\vec{p}}}]}{\sum_{N=0}^{\infty}[z^N\sum_{\{n_{\vec{p}}\}}e^{-\beta\sum_{\vec{p}}n_{\vec{p}}\epsilon_{\vec{p}}}]}$$ which is exactly $$-\frac{1}{\beta}\frac{1}{Z}\frac{\partial Z}{\partial \epsilon_{\vec{p}}}=-\frac{1}{\beta}\frac{\partial \mathrm{ln}Z}{\partial \epsilon_{\vec{p}}}$$