A Summation formula from statistical mechanics

hilbert2

Science Advisor
Insights Author
Gold Member
1,230
372
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
Insights Author
2018 Award
10,724
7,339
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

Science Advisor
Insights Author
Gold Member
1,230
372
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
Insights Author
2018 Award
10,724
7,339
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

Science Advisor
Insights Author
Gold Member
1,230
372
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.
 

Want to reply to this thread?

"Summation formula from statistical mechanics" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top