Htykuut

This might seem stupid, but I'm really stuck. I don't understand how to calculate the following explicitly:

$$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}$$

(it's the Ising model for 3 lattice sites).

I don't understand how I can calculate this in a brute force way, since the sums for $s_1$ and $s_3$ only apply to one part of the equation to be summed over.

(I know you can simplify it and end up with a much nicer expression in terms of cosh)

Help! Thanks.

Proceeding from right to left:

$$begin{align*}sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =sum_{s_1=pm1} sum_{s_2=pm1}e^{-{s_1s_2}}left[ e^{{s_2}}+e^{-{s_2}} right]\
&=sum_{s_1=pm1}left( e^{{s_1}}left[ e^{{-1}}+e^{1} right]+e^{-{s_1}}left[ e^{1}+e^{-{1}} right]right)\
&=sum_{s_1=pm1}left( [e^{{s_1}}+e^{-{s_1}}]left[ e^{{-1}}+e^{1} right]right)\
&=[e^{{-1}}+e^{1}]left[ e^{{-1}}+e^{1} right]+[e^{{1}}+e^{-1}]left[ e^{{-1}}+e^{1} right]\
&=2[e^{{-1}}+e^{1}]^2\
&=4(cosh(2)+1)
end{align*}$$

A slightly alternate approach than in the other answer (dividing up the sum differently):

Starting with the inside sum of
$$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$
we can factor out the term that doesn't depend on $s_3$ to get
$$sum_{s_1=pm1} sum_{s_2=pm1} sum_{s_3=pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}=
sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}.
$$
Now, let's look at the exponent of the internal sum, since $s_2$ is either $+1$ or $-1$, then $-s_2$ is either $+1$ or $-1$. Multiplying by $s_3$ which is either $+1$ or $-1$ results in one of each of $+1$ and $-1$. Therefore, the sum simplifies to
$$
sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}sum_{s_3=pm1} e^{-{s_2s_3}}=
sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}.
$$
By applying the same argument as above, we see that the exponent of this $e$ is one of each of $+1$ and $-1$, so we get
$$
(e+e^{-1})sum_{s_1=pm1} sum_{s_2=pm1} e^{-{s_1s_2}}=(e+e^{-1})sum_{s_1=pm1} (e+e^{-1})=(e+e^{-1})^2sum_{s_1=pm1}1.
$$
Since there are only two values for $s_1$, we get that this simplifies to $2(e+e^{-1})^2$, which can then be simplified in terms of $cosh$.