I have this expression: $$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\prod_{i=1}^{N}\sum_{S_{i}\in\{-1,1\}}e^{\beta HS_{i}} \qquad (1)$$ Where $\sum_{\{\vec{S}\}}$ means a sum over all possible vectors $\vec{S}=(S_1,...,S_N)$ with the restriction that $S_i$ can only take the values $\{-1,+1\}$, i.e. the sum is over $2^N$ different vectors: $\{\vec{S}\}$.

My question is: How can I be sure that (1) is right? Is there a criteria to interchange sums and products or it's always valid?

  • $\begingroup$ have you found any general rule of interchange sums and products? would be great help, thanks! $\endgroup$
    – sunxd
    Commented Jul 29, 2019 at 15:59

2 Answers 2


The problem was in how to write down the sum $\sum_{\{\vec{S}\}}$. Since this sum is over all the possible vectors $\vec{S}=(S_1,...,S_N)$, (where $S_i\in\{-1,1\}$), we can rewrite this sum like

$$\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}$$ i.e. $$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}\prod_{i=1}^{N}e^{\beta HS_{i}} \qquad(1)$$

Clearly this sum has $2^N$ elements of the type $\prod_{i=1}^{N}e^{\beta HS_{i}}$. Now, since

$$\prod_{i=1}^{N}e^{\beta HS_{i}}=e^{\beta HS_{1}}\cdots e^{\beta HS_{N}},$$ then (1) turns

$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}e^{\beta HS_{1}}\cdots e^{\beta HS_{N}} \qquad(2)$$ Since each $S_i$ is independent of the others, we can "factorize" the $\Sigma$'s: $$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\left(\sum_{S_{1}\in\{-1,1\}}e^{\beta HS_{1}}\right)\cdots\left(\sum_{S_{N}\in\{-1,1\}}e^{\beta HS_{N}}\right)$$ And finally: $$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\prod_{j=1}^{N}\sum_{S_{j}\in\{-1,1\}}e^{\beta HS_{j}} \qquad Q.E.D.$$

Note. This result can be generalized for vectors $\vec{S}=(S_1,...,S_N)$ with components $S_i\in\{1,...,k\}$ for some integer $k$.


Try to go from the right formula to the left and use distributivity. If you're not sure, try with N=2 to convince yourself.

  • $\begingroup$ I tried to do this but I failed. I found an answer though. $\endgroup$
    – Ana S. H.
    Commented Mar 11, 2014 at 18:49

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