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I'm trying to work on this problem, which is supposed to be a very simplified model of an interacting fluid.

Suppose there is a recipient of volume $V$ divided into cells of volume $v_0$, each of which can be occupied by up to 2 particles. If, and only if, a cell is occupied by 2 particles, the energy associated with it is $\epsilon > 0$. Otherwise it is zero. The particles must be treated as indistinguishable, but are not subjected to a fermionic exclusion principle.

I tried to define a variable $t_i$, which is 1 if there are 2 particles in cell i and zero otherwise. This way, the Hamiltonian should be

$$\begin{equation} \epsilon\sum_{i=1}^{V/v_0}t_i \end{equation}$$

What I get for the partition function $Z$ is

$$\begin{equation} Z = \sum_{t_i\in(0,1)}^{}\exp\left(-\beta\epsilon\sum_{i = 1}^{V/v_0}t_i\right)\\ Z = \left(\sum_{t_i\in(0,1)}^{}\exp\left(-\beta\epsilon t_i\right)\right)^{V/v_0} = (1 + \exp(-\beta\epsilon))^{V/v_0} \end{equation}$$

and for the grand canonical partition function

$$\begin{equation} \Xi = \sum_{N = 0}^{V/v_0}z^N (1 + \exp(-\beta\epsilon))^{V/v_0} \end{equation}$$

which does not yield the correct answer, which I know is $\Xi = (1 + z(1+z\exp(-\beta\epsilon)))^{V/v_0}$.

I have no clue where it went wrong. I think my approach is somewhat correct, although I do find it weird I'm not taking into account the number $N$ of particles when calculating the partition function.

Any help is appreciated!

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You should not sum over $N$ separately.

The contribution from a cell that you have labelled $t_i=0$ is $1+z$ (since it contains $0$ or $1$ particle), the contribution from a cell that you have labelled $t_i=1$ is $z^2e^{-\beta\epsilon}$ (since it contains exactly $2$ particles). Thus the partition function for a single cell is $1+z+z^2e^{-\beta\epsilon} = 1+z(1+ze^{-\beta\epsilon})$ and you get what you want by taking the product over the cells.

[For this problem, it is thus not really helpful to label the cells with the variables $t_i$ as you do. Just work out directly the partition function associated to a cell, by summing over all $3$ possibilities: $0$, $1$ or $2$ particles.]


As per your question below, let me explain why you can take the product over cells of the single-cell partition function.

Probably, the easiest way to see this is to make the computation explicitly in the case of two cells. In this case, there can be 0, 1, 2, 3 or 4 particles in the system. Here are the corresponding contributions for each of these values:

$N=0$: $1$

$N=1$: $2z$ (since the particle can be in any of the two cells)

$N=2$: $2z^2e^{-\beta\epsilon} + z^2$ (the first term accounts for the two configurations in which both particles are in the same cell, while the second term accounts for the configuration in which both cells contain 1 particle)

$N=3$: $2z^3e^{-\beta\epsilon}$

$N=4$: $z^4e^{-2\beta\epsilon}$

This gives an overall partition function $1 + 2z + (1+2e^{-\beta\epsilon})z^2 + 2e^{-\beta\epsilon}z^3 + e^{-2\beta\epsilon}z^4$.

You can then check that this coincides with the product of the two single-cell partition functions: $$ (1+z+z^2e^{-\beta\epsilon})(1+z+z^2e^{-\beta\epsilon}) = 1 + 2z + (1+2e^{-\beta\epsilon})z^2 + 2e^{-\beta\epsilon}z^3 + e^{-2\beta\epsilon}z^4. $$


More generally, denote by $N_i \in \{0,1,2\}$ the number of particles in cell $i$. Then the full partition function is \begin{align*} \Xi &= \sum_{N_1,\dots,N_K} z^{N_1+\cdots+N_K} \prod_{i=1}^K (1_{N_i=0} + 1_{N_i=1} + 1_{N_i=2}e^{-\beta\epsilon})\\ &= \prod_{i=1}^K \sum_{N_i} z^{N_i}(1_{N_i=0} + 1_{N_i=1} + 1_{N_i=2}e^{-\beta\epsilon})\\ &= \prod_{i=1}^K (1 + z + z^2e^{-\beta\epsilon})\\ &= (1 + z + z^2e^{-\beta\epsilon})^K \end{align*} where I have denoted by $K$ the number of cells.

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