# Cluster Expansion

In the cluster expansion (section 5.2 in M. Kardar "Statistical Physics of Particles") we write the grand canonical partition function. During the expansion, we do the following switch between a sum and a product: $$\sum\limits_{\{n_l\}} \prod_l \frac{1}{n_l!}\left(\frac{e^{l\beta\mu} b_l}{\lambda^{3l}l!}\right)^{n_l} = \prod_l \sum\limits_{n_l=0}^\infty \frac{1}{n_l!}\left(\frac{e^{l\beta\mu} b_l}{\lambda^{3l}l!}\right)^{n_l}$$ ($l$ is the size of the cluster, and $n_l$ is the number of clusters)

I'm trying to understand why this is fine. I figured that $\sum\limits_{\{n_l\}} \prod_l$ means that we're going over all possible sets of cluster numbers (e.g 5 clusters of size 1, 3 of size 2, etc..) but in this case, what is the meaning of the product? How is the connection between $n_l$ and $l$ apparent?

$\prod_l \sum\limits_{n_l=0}^\infty$ is much clearer - for every cluster size, we look at all numbers of it possible.

• it's restricted to the linked clusters, such that $\sum_l ln_l = N$ , to evaluate 5.24
– user46925
Commented Jan 21, 2016 at 13:25
• this is right after the restriction has been lifted. We got rid of the summation on $N$ by applying the restriction. Commented Jan 21, 2016 at 13:59
• yes, it is in 5.25 . Very interesting trick anyway. I'll read again the ch 4 this night ( on a normal screen ) and come back.
– user46925
Commented Jan 21, 2016 at 14:46
• This is just linearity of the sum. Maybe it will become obvious if you look at the following particular case: $\sum_{i,j} a_i b_j = \sum_i \sum_j a_i b_j = \Bigl(\sum_i a_i\Bigr)\Bigl(\sum_j b_j\Bigr)$, which follows by pulling out $a_i$ from the inner sum and then pulling out $\sum_j b_j$ from the outer sum. Now, in your formula $\{n_\ell\}$ denotes the number of clusters of each possible length (so $\sum_{\{n_\ell\}} = \sum_{n_1\geq 0}\sum_{n_2\geq 0}\cdots = \prod_{\ell\geq 1} \sum_{n_\ell\geq 0}$). Commented Jan 21, 2016 at 17:25
• Yes you sum over all possible values of $n_\ell$ for each possible values of $\ell$ (so $\{n_\ell\}$ specifies the values of $n_1, n_2, n_3, \ldots$). Then, once these are fixed, you take the product of all the functions $f_\ell(n_\ell)$ (for these specific values of $n_1,n_2,n_3,\ldots$). Commented Jan 22, 2016 at 10:56

First, let me explain what the notation means. The sum over $\{n_\ell\}$ is a sum over all possible values of $n_\ell$, for each possible values of $\ell$ (in other words, $\{n_\ell\}$ specifies the values of $n_1,n_2,n_3,\ldots)$. Then, once these values are fixed, you take the product of all the functions $f_\ell(n_\ell) = \frac{1}{n_\ell!}\bigl( \frac{e^{\ell\beta\mu}b_\ell}{\lambda^{3\ell}\ell!} \bigr)^{n_\ell}$ (for these specific values of $n_1,n_2,n_3,\ldots$).
The identity is then essentially linearity of the sum. Let me explain it assuming (for ease of notation) that the only allowed cluster sizes are $1$ and $2$. Then: \begin{eqnarray} \sum_{\{n_1,n_2\}} f_1(n_1)f_2(n_2) &=& \sum_{n_1\geq 0} \sum_{n_2\geq 0} f_1(n_1) f_2(n_2)\\ &=& \Bigl(\sum_{n_1\geq 0} f_1(n_1)\Bigr) \Bigl(\sum_{n_2\geq 0} f_2(n_2)\Bigr)\\ &=& \prod_{\ell=1}^2 \sum_{n_\ell\geq 0} f_\ell(n_\ell), \end{eqnarray} where the second identity follows by first pulling out $f_1(n_1)$ outside the sum over $n_2$ and then pulling out $\sum_{n_2} f_2(n_2)$ outside the sum over $n_1$.