Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers 2

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.

share|improve this answer
    
I tried to do this but I failed. I found an answer though. –  Anuar Mar 11 at 18:49
add comment
up vote 1 down vote accepted

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 $k$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.