Cluster Expansion

Question

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.

Answer 1

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 $n_1$.