11
$\begingroup$

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?

$\endgroup$
12
$\begingroup$

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 $k$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I tried to do this but I failed. I found an answer though. $\endgroup$ – Ana S. H. Mar 11 '14 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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