# Grand canonical partition function: factorization

The grand canonical ensemble partition function is defined to be $$\mathcal{Z} := \sum_{\forall |n\rangle} e^{\beta \mu N_{|n\rangle} - \beta E_{|n\rangle}}$$ being $|n\rangle$ a notation for each microstate (not necessarily quantum, just a microstate), $N_{|n\rangle}$ the number of particles of that microstate and $E_{|n\rangle}$ the energy corresponding to this microstate. This is the formal definition.

Now I have from my lecture notes that when we are dealing with non-interacting and indistinguishable particles (e.g. an ideal gas) the partition function can be written as $$\mathcal{Z} = \prod_{\forall \epsilon_i} \sum_{\forall \text{ allowed } n}(z e^{-\beta \epsilon _i})^n$$ now being $\epsilon _i$ each "monopartiuclar state" and "$\forall \text{ allowed } n$" is $\{0,1\}$ for fermions and $\{0,1,2,\dots\}$ for bosons. The "monoparticular states" are the energy levels that each particle can be in.

The question: How do we go from the definition to the second formula?

# My approach (wrong, or at least incomplete)

If particles do not interact between one another then the energy of the microstate $|n\rangle$ can be written as a summation $$E_{|n\rangle} = \epsilon_1 + \epsilon_2 + \dots + \epsilon_{|n\rangle} = \sum_{i=1}^{N_{|n\rangle}} \epsilon_i$$ where $\epsilon_i$ is the energy of each particle. Thus the partition function writes $$\mathcal{Z} = \sum_{\forall |n\rangle} e^{\beta \mu N_{|n\rangle}} \prod_{i=1}^{N_{|n\rangle}} e^{-\beta \epsilon _i}$$ where I have already expanded the exponential of a summation as a product of exponentials.

Now, if particles are identical then the allowed values for $\epsilon _i$ are for all the particles the same, say $\epsilon _i \in \{\varepsilon_0, \varepsilon_1, \dots\}$. Using this notation, $\epsilon_8 = \varepsilon_3$ reads as "particle number 8 is in the third energy level". This allows to arrange the microstates of the system as follows (sorry for changing the language in the pic):

Now I don't know how to go on... Any help is appreciated.

You are misunderstanding what is meant by a product over monopartiuclar states. This is not a product over the states of the $N$ particles in the system, it is a product over all possible single particles states. The sum is then over the occupation number of those states, i.e. the number of particles actually in that state (which may be $0$). The advantage of this occupation number approach to over keeping track of individual particles is that it automatically takes account of particle indistinguishably.
The total number of particles in a state $\gamma$ is clearly simply the sum of the number of particles in each single particle state $$N_\gamma = \sum_i n_i$$ The total energy is the energy of each single particle state, times the number of particles in that state $$E_\gamma = \sum_i \epsilon_i n_i$$
• Thanks for your answer. There is one last part I am not understanding. It is when yo switch from one summation over $\gamma$ to many summations over the $n_i$. What values do this $n_i$ acquire? In $N_\gamma = \sum_i n_i$ it should be $N_\gamma = \sum_i n_{i,\gamma}$ because in each different microstate $\gamma$ you have a different $n_i$, thus $n_{i,\gamma}$. – user171780 Feb 8 '18 at 21:35
• yes, the $n_i$ are determined by which microstate, $\gamma$, the system is in (although writing out $n_{i,\gamma}$ gets cumbersome very quickly). Equally the list of occupation numbers $(n_0,n_1\dots)$ for a given microstate completely determine the state of the system, so we can view microstates and sets of occupation numbers as essentially interchangeable i.e. $\gamma = (n_0,n_1\dots)$. This means we can write $\sum_\gamma = \sum_{ (n_0,n_1\dots)} = \sum_{n_0}\sum_{n_1}\dots$. The possible values for $n_i$ can take in general are set by particle statistics, as you said in the question. – By Symmetry Feb 9 '18 at 13:51