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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.

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  • $\begingroup$ it's restricted to the linked clusters, such that $\sum_l ln_l = N$ , to evaluate 5.24 $\endgroup$
    – user46925
    Jan 21, 2016 at 13:25
  • $\begingroup$ this is right after the restriction has been lifted. We got rid of the summation on $N$ by applying the restriction. $\endgroup$
    – golanor
    Jan 21, 2016 at 13:59
  • $\begingroup$ 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. $\endgroup$
    – user46925
    Jan 21, 2016 at 14:46
  • $\begingroup$ 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 $\sum_{\{n_\ell\}} = \sum_{n_1\geq 0}\sum_{n_2\geq 0}\cdots = \prod_{\ell\geq 1} \sum_{n_\ell\geq 0}$). $\endgroup$ Jan 21, 2016 at 17:25
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    $\begingroup$ 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 $n_1,n_2,n_3,\ldots$). $\endgroup$ Jan 22, 2016 at 10:56

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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 $n_1$.

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