0
$\begingroup$

I was taking a look to a book of statistical mechanics, many equations show something as follows: $$ Q(K,N) = \Sigma_{s_{1},s_{2},...,s_{N}=\pm 1}[ e^{K(...+s_1s_2+s_2s_3+s_3s_4...)} ]$$ then they partition the sum as follows $$ Q(K,N) = \Sigma_{s_{1},s_{2},...,s_{N}}e^{K(s_1s_2+s_2s_3)}e^{K(s_3s_4+s_4s_5)} ... $$ After Summing over even numbered S's $$Q(K,N) = \Sigma_{s_{odd}} (e^{K(s_1+s_3)}+e^{-K(s_1+s_2)})(e^{K(s_3+s_5)}+e^{-K(s_3+s_5)})$$ How can I interpret this Summation with many lower indices.

$\endgroup$

1 Answer 1

1
$\begingroup$

$Q(K,N) = \Sigma_{s_{1},s_{2},...,s_{N}=\pm 1}[ e^{K(...+s_1s_2+s_2s_3+s_3s_4...)} ] $

To me, it looks like the summation is over the $2^N$ sets $\{s_{1},s_{2},...,s_{N}\} $ where each $s_i$ is either $1$ or $-1$.

For example, for $N=2$, these are

$\{1, 1\}, \{1, -1\}, \{-1, 1\}, \{-1, -1\} $.

$\endgroup$
1
  • $\begingroup$ So would we sum in the following way: $$\Sigma_{S_1} F(S_i)=F(1) + F(-1)$$ and $$\Sigma_{S_1,S_2,...} F(S_i)=(F(1) + F(-1))+(F(1) + F(-1))+...$$ $\endgroup$
    – RMS
    Mar 29, 2020 at 21:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .