Grand Canonical Partition Function for Simple System

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The discussion focuses on calculating the grand canonical partition function (GCPF) for a system with m lattice sites, where each site can either be occupied or unoccupied. The participants clarify that a "state" refers to a specific configuration of occupation numbers for the sites. The energy and particle number for a given state are defined, but the complexity of summing over all states raises concerns about the approach. It is suggested that due to the system's symmetry, one can simplify the calculation by focusing on the number of occupied sites rather than enumerating every configuration individually. The key takeaway is that the GCPF can be derived more efficiently by considering the density of states rather than all possible states.
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Homework Statement



I would like to calculate the grand canonical partition function (GCPF) for a system in which there are are m lattice sites. A configuration may be specified by the numbers (n_1, n_2, ... , n_m), where n_k = 1 if a particle occupies site k and n_k = 0 if no particle occupies site k. Occupied sites have an associated energy \epsilon (constant) and unoccupied sites have zero associated energy.

Homework Equations



The general form of the GCPF in my book (Chandler) is given like this:

\Xi = \sum_{\nu} e^{ -\beta E_{\nu} + \beta \mu N_{\nu} }

where \nu indicates a summation over all states. (I am confused as to what, exactly, is meant by a "state" in the context of this problem.)

The Attempt at a Solution



For a given state j the number of particles is given by N_j = \sum_{j=1}^{m} n_i (summing over all sites). For the same state j the energy is given by E_j = \epsilon \sum_{i=1}^{m} n_i.

I'm unsure of the correct direction from here. Inserting the expressions for N_j and E_j into \Xi creates a mess of summations. Is that the only way? Is it simplify-able?

I feel that I should be able to calculate the GCPF for just one site and then extend the result to m sites since the sites are independent of one another. Is this possible? If so, how?

Thank you all.
 
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A state is a set of occupation numbers (n_1,n_2,...,n_m) so when you sum over all possible states, you sum over configurations (0,0,...,0), (1,0,0,...,0), ... , (1,1,...,1). Luckily the system is quite symmetric, so you do not need to sum the states one by one. Instead, you can write the sum over the number of occupied sites like so:

\sum_{\nu} \rightarrow \sum_{n=0}^{m} \omega(n)

where \omega(n) is the density of states, ie. the number of microstates (n1,n2,...,nm) corresponding to the macrostate.
 

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