# A Summation formula from statistical mechanics

#### hilbert2

Gold Member
Summary
About proving a summation rule that appears in theory of spin chains.
I ran into this kind of expression for a sum that appears in the theory of 1-dimensional Ising spin chains

$\displaystyle\sum\limits_{m=0}^{N-1}\frac{2(N-1)!}{(N-m-1)!m!}e^{-J(2m-N+1)/kT} = \frac{2e^{2J/kT-J(1-N)/kT}\left(e^{-2J/kT}(1+e^{2J/kT})\right)^N}{1+e^{2J/kT}}$

where the $k$ is the Boltzmann constant, $J$ is strength of magnetic interaction and $T$ is the absolute temperature.

It's quite simple to check that this holds for $N=1$, $N=2$ and $N=3$. However, it seems to be a bit more difficult to actually prove by induction than some other sums like

$\displaystyle\sum\limits_{m=1}^{N}m^2 = \frac{N(1+N)(1+2N)}{6}$,

where only the last term of the sum changes when $N\mapsto N+1$.

If anyone wants to try to prove this as an exercise, be my guest and post it here... It's probably just about applying the binomial theorem.

#### fresh_42

Mentor
2018 Award
First we write $\alpha =\dfrac{J}{kT}$ and cancel $2e^{(N-1)\alpha}$ on both sides. Then the equation reads
$$\sum_{m=0}^{N-1} \binom{N-1}{m}e^{2m\alpha} = \left( 1+e^{2\alpha} \right)^{N-1}$$

Last edited:
• hilbert2

#### hilbert2

Gold Member
Thanks,

so it's just the known formula

$\displaystyle (1+x)^n = \sum\limits_{k=0}^n \binom{n}{k}x^k$

with substitutions $n=N-1$, $k=m$ and $x=e^{2\alpha}$.

#### fresh_42

Mentor
2018 Award
Thanks,

so it's just the known formula

$\displaystyle (1+x)^n = \sum\limits_{k=0}^n \binom{n}{k}x^k$

with substitutions $n=N-1$, $k=m$ and $x=e^{2\alpha}$.
Yes, although the right hand side is written a bit more complicated than necessary:
$$2x^{-\frac{n}{2}}(1+x)^n =2e^{-(N-1)\alpha} \cdot (1+e^{2\alpha})^{N-1} = \dfrac{2e^{2\alpha + (N-1)\alpha }\left( e^{-2\alpha} \left( 1+e^{2\alpha} \right)\right)^N}{1+e^{2\alpha}}$$
I think I have found a sign error. On the left hand side we have
\begin{align*}
\sum_{m=0}^{N-1} \dfrac{ 2(N-1)! }{ (N-1-m)! m! } e^{ -2m \alpha + (N-1) \alpha } &= 2 e^{ (N-1) \alpha } \sum_{m=0}^{N-1} \binom{N-1}{m} \left( e^{ -2 \alpha } \right)^m \\ & = 2(e^{\alpha})^{N-1} \left( 1+e^{-2\alpha} \right)^{N-1}\\
&= 2 \left( e^\alpha + e^{-\alpha}\right)^{N-1} \\
&= 2 e^{-(N-1)\alpha}e^{+(N-1)\alpha} \left( e^\alpha + e^{-\alpha}\right)^{N-1}\\
&= 2 e^{-(N-1)\alpha}\left( e^{2\alpha} + 1 \right)^{N-1}
\end{align*}
Ok, no sign error. One just has to be very cautious.

• hilbert2

#### hilbert2

Gold Member
Actually, there was a sign error somewhere when I calculated these sums with Mathematica to get an equation for the heat capacity of an infinite spin chain... It shouldn't be left in the equation above, but it's not impossible either.

"Summation formula from statistical mechanics"

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