# Summing graphs in the partition function (statistical physics)

I am looking at Tong's lecture notes on statistical physics, and I wanted to understand a step in his cluster expansion better.

The goal here is to calculate the partition function in the canonical ensemble as a sum over some graphs. The result is to find (after equation (2.35), on page 58)

\begin{align} Z(N,V,T)=\frac{1}{\lambda^{3N}}\sum_{\{m_l\}}\prod_l\frac{U_l^{m_l}}{(l!)^{m_l}m_l!},\tag{p.58} \end{align} where we have the condition \begin{align} \sum_{l=1}^Nm_ll=N.\tag{2.33} \end{align}

I'm happy with everything at this point, but the next step is where I don't exactly follow.

The argument is that the constraint means the sum is difficult to compute, and so a way to avoid this is to look at the grand canonical ensemble \begin{align} \mathcal{Z}(\mu,V,T)=\sum_Nz^NZ(N,V,T),\tag{p.58} \end{align} with $$z=e^{\beta\mu}$$. Substituting in the above definition of $$Z(N,V,T)$$, we find \begin{align} \mathcal{Z}(\mu,V,T)&=\sum_Nz^N\frac{1}{\lambda^{3N}}\sum_{\{m_l\}}\prod_l\frac{U_l^{m_l}}{(l!)^{m_l}m_l!}\\ &=\sum_N\sum_{\{m_l\}}\prod_l\left(\frac{z}{\lambda^{3}}\right)^N\frac{1}{m_l!}\left(\frac{U_l}{l!}\right)^{m_l}\\ &=\cdots\\ &=\sum_{m_l=0}^\infty\prod_{l=1}^\infty\left(\frac{z}{\lambda^{3}}\right)^{m_ll}\frac{1}{m_l!}\left(\frac{U_l}{l!}\right)^{m_l}.\tag{p.58} \end{align}

What I would like to understand is how to go from before the $$\cdots$$ to after. As far as I can tell, because the summation is over all $$N$$, which in turn dictates the values of $$l,m_l$$, there is some way to claim this is equivalent to the sum over all $$m_l$$, and the product over all $$l$$. I would like to make this a bit more rigorous, but I'm a bit stuck, so would appreciate an explanation.

• There's still a typo in your last line: you cannot sum over the value of $m_l$ before you have fixed the value of $l$... So either you permute the sum and the product, or you sum over $\{m_l\}$. Mar 17 at 20:22
• Related: physics.stackexchange.com/q/107049/2451 and links therein. Mar 17 at 20:31

To understand this type of computation, you should write down the details explicitly. To lighten the notation, I set $$a=z/\lambda^3$$ and $$b_l=U_l/l!$$. Then, writing down explicitly the constraint as an indicator function (so that the sum over $$\{m_l\}$$ is now unrestricted), \begin{align} \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} a^N \prod_{l}\frac{1}{m_l!} b_l^{m_l} &= \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \Bigr)\\ &= \sum_{\{m_l\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l}\Bigr) \sum_N 1_{\{\sum_l lm_l = N\}} \\ &= \sum_{\{m_l\}} \prod_{l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} \frac{1}{m_l!} (a^lb_l)^{m_l} \\ &= \prod_l \exp( a^lb_l ). \end{align} For the first identity, I replaced $$N$$ by $$\sum_l lm_l$$ (I can then write $$a^N$$ as a product over $$l$$). The third identity follows from the fact that exactly one term in the sum over $$N$$ is equal to $$\sum_l lm_l$$. The fourth identity is a consequence of the fact that the summand is completely factorized over $$l$$, so that $$m_l$$ can be summed separately for each $$l$$. The last identity is just the Taylor series of the exponential.
• Thanks for you answer. I'm still not quite clear how the 4th equality works - I don't understand why the sum is now over all $m_l$? Mar 17 at 20:34
• Let me explain on a particular case: $\sum_{m_1,m_2} f(m_1)f(m_2) = \bigl(\sum_{m_1}f(m_1)\bigr)\bigl(\sum_{m_2}f(m_2)\bigr) = \prod_{l=1}^2 \sum_{m_l}f(m_l)$. Does this help? Mar 17 at 20:37
• In that case, is this understanding correct: On the LHS of 4th equality (i.e. line 3) the sum over $\{m_l\}$ is no longer subject to the constraint? Mar 17 at 20:39
• Since I wrote explicitly the constraint as an indicator function, the sum over $\{m_l\}$ is never restricted (of course the summand vanishes when the constraint is not satisfied, because of the indicator). Mar 17 at 20:40