The partition function of the grand canonical ensemble can be generally stated as
$$ \mathcal{Z} = \sum_{r} e^{-\beta(E_{r} – N_r\mu)}\tag{1}$$
where $E_{r}$ is the energy of the micro-state $r$ of $N_r$ particles, $\beta$ the usual inverse temperature, and where $\mu$ is the chemical potential.
This is not the only way to state $\mathcal{Z}$.
Quoting Gould and Tobochnik,
[…] it is possible to distinguish the subset of all particles in a given single particle microstate from the particles in all other single particle microstates. For this reason we divide the system of interest into subsystems each of which is the set of all particles that are in a given single particle microstate. Because the number of particles in a given microstate varies,
we need to use the grand canonical ensemble and assume that each subsystem is coupled to a heat bath and a particle reservoir independently of the other single particle microstates.
From this definition it arises that one can also compute
$$\mathcal{Z} = \prod_{k} Z_{k}\tag{2}$$
with $Z_k$ the partition function for a subsystem (one particle microstate).
And lastly, one can make the dependencies on the number of particle explicit by using the variation (Gould and Tobochnik sec. 6.11 )
$$ \mathcal{Z} = \sum_{N=1}^\infty e^{\beta\mu N} Z_c(N)\tag{3}$$
where $Z_c(N)$ is the canonical partition function of the system of $N$ particles.
Now my question is , shouldn't equation (3) be summed from $0$ to $\infty$ rather than 1 to $\infty$? This would yield
$$ \mathcal{Z} = \sum_{N=0}^\infty e^{\beta\mu N} Z_c(N)=1+\sum_{N=1}^\infty e^{\beta\mu N} Z_c(N)\tag{*}$$
Stated more clearly: should we exclude the 0 particle state from the summation? The rational for excluding it is that
- the textbook says so,
- it will end up being a very important term in most situation, since everything else is a decreasing exponential.
- The interpretation of $Z_c(0)=1$ is rather ambiguous.
The reason for including it are:
- it makes sense physically (all particle in the reservoir), and
- it is essential for the coherence between eq. (2) and (3). i.e. in the fermion case, $Z_k = 1 +e^{-\beta(e_k-\mu)}$ and one has $\mathcal{Z}=1+….$
Best Answer
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.