6

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.

Qmechanic
  • 201,751
golanor
  • 143
  • it's restricted to the linked clusters, such that $\sum_l ln_l = N$ , to evaluate 5.24 –  Jan 21 '16 at 13:25
  • this is right after the restriction has been lifted. We got rid of the summation on $N$ by applying the restriction. – golanor Jan 21 '16 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. –  Jan 21 '16 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 $\sum_{{n_\ell}} = \sum_{n_1\geq 0}\sum_{n_2\geq 0}\cdots = \prod_{\ell\geq 1} \sum_{n_\ell\geq 0}$). – Yvan Velenik Jan 21 '16 at 17:25
  • I see, that does make sense. I'm still not sure I understand the first notation, do you know what it means? – golanor Jan 21 '16 at 19:04
  • Sure. Imagine that the only allowed cluster sizes are $\ell=1$ and $\ell=2$ (so that I can write things down explicitely). Let us use the notation $f_\ell(n_\ell) = \frac{1}{n_\ell!}\bigl( \frac{e^{\ell\beta\mu}b_\ell}{\lambda^{3\ell}\ell!} \bigr)^{n_\ell}$. Then your identity becomes $\sum_{{n_1,n_2} \in{0,1,2,\ldots}^2} \prod_{\ell=1}^2 f_\ell(n_\ell) = \sum_{n_1\geq 0} \sum_{n_2\geq 0} f_1(n_1) f_2(n_2)= \sum_{n_1\geq 0} f_1(n_1) \sum_{n_2\geq 0} f_2(n_2)= \prod_{\ell=1}^2 \sum_{n_\ell\geq 0} f_\ell(n_\ell)$ – Yvan Velenik Jan 22 '16 at 07:20
  • Sorry I wasn't clear - I meant that I don't understand the physical meaning of it. Are we summing over all possible configurations of $n_{\ell}$ and then we iterate over all possible $\ell$'s? – golanor Jan 22 '16 at 08:05
  • 2
    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 $n_1,n_2,n_3,\ldots$). – Yvan Velenik Jan 22 '16 at 10:56
  • Related: https://physics.stackexchange.com/q/107049/2451 and links therein. – Qmechanic Mar 17 '23 at 20:32
  • Beautiful lectures on cluster expansion: https://personal.math.ubc.ca/~db5d/Seminars/les_houches_84.pdf (A SHORT COURSE ON CLUSTER EXPANSIONS, David C. BRYDGES). Related: https://physics.stackexchange.com/q/119854/226902, https://physics.stackexchange.com/q/690142/226902, https://physics.stackexchange.com/q/780321/226902. – Quillo Sep 15 '23 at 16:26

1 Answers1

3

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 $n_1$.

Yvan Velenik
  • 10,371