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.
Best Answer
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.