Prove: $\sum_{\sigma_1,\cdots,\sigma_N} e^{a\sigma_1}\cdots e^{a\sigma_N}=2\prod_{j=1}^{N}\sum_{\sigma_i}e^{a\sigma}$ where $\sigma_i \in \{-1,1\}$

Question

In this Wikipedia article on the one dimensional Ising model they use the following factorization

$$ \displaystyle \sum_{\sigma_1,\sigma_2,\cdots,\sigma_N} e^{a\sigma_1} e^{a\sigma_2}\cdots e^{a\sigma_N} = 2\displaystyle\prod_{j=1}^{N}\sum_{\sigma_i}e^{a\sigma_i} = 2 \left[e^{-a}+e^{a}\right]^N \,. $$ The first sum is over all possible sequences ($\sigma_1,\sigma_2,\cdots, \sigma_N)$ where $\sigma_i \in \{-1,1\}$.

Does somebody have a reference for this factorization and/or able to give a proof?

Answer 1

The factor $2$ is not plausible. Taking $N=1$ we have \begin{align*} \sum_{\sigma_1}e^{a\sigma_1}=e^{a}+e^{-a} \end{align*}

In general we obtain \begin{align*} \color{blue}{\sum_{(\sigma_1,\sigma_2,\ldots,\sigma_N)\in\{-1,1\}^N}}&\color{blue}{e^{a\sigma_1}e^{a\sigma_2}\cdots e^{a\sigma_N}}\\ &=\sum_{\left(\sigma_1,\sigma_2,\ldots,\sigma_N\right)\in\{-1,1\}^N}\prod_{j=1}^Ne^{a\sigma_j}\tag{1}\\ &=\sum_{k=0}^N\sum_{{S\subseteq[N]}\atop{|S|=k}}e^{ak}e^{-a(N-k)}\tag{2}\\ &=\sum_{k=0}^N\binom{N}{k}e^{ak}e^{-a(N-k)}\tag{3}\\ &\,\,\color{blue}{=\left(e^a+e^{-a}\right)^N}\tag{4} \end{align*}

Comment:

• In (1) we use the product symbol $\prod_{j=1}^N$ to write the product of $N$ factors.

• In (2) we reorder the sum by selecting all subsets $S\subseteq [N]=\{1,2,\ldots,N\}$ for which the sign is positive and $[N]\setminus S$ for which the sign is negative.

• In (3) we use the fact that there are $\binom{N}{k}$ subsets $S\subseteq [N]$ with size $k$.

• in (4) we apply the binomial theorem.

Answer 2

It's simpler than you're probably thinking. Each sum $\sum_{\sigma_i} e^{a \sigma_i}$, over the set $\sigma_i \in \{-1,1\}$ is just a sum of two terms $e^{-a} + e^a$. This is the same value for every $i$, so it just gets multiplied $N$ times.

$$ \prod_{i=1}^N \sum_{\sigma_i} e^{a \sigma_i} = (e^{-a} + e^a)^N $$