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Good day to you all,

I'm currently reading Goldendfeld's lectures on phase transitions, and I'm a bit perplexed by a formula appearing in section 3.1.3 of the book. He starts with the partition function for a chain of $N$ spins, which is given by

$Z(N) = \sum_{s_1}\dots\sum_{s_N} e^{K(s_1s_2 + s_2s_3 + \dots +s_{N-1}s_N)}\tag{1}$

where $K \equiv \beta J $ is the . The idea is then to go to a chain of $N+1$ spins, so as to get a recursion between both:

$Z(N+1) = \sum_{s_1}\dots\sum_{s_{N+1}} e^{K(s_1s_2 + s_2s_3 + \dots +s_Ns_{N+1})}\tag{2}$

Performing the sum over $s_{N+1}$ yields

$Z(N+1) = \sum_{s_1}\dots\sum_{s_{N}} e^{K(s_1s_2 + s_2s_3 + \dots +s_{N-1}s_{N})}(e^{Ks_N}+e^{-Ks_N})\tag{3}\label{3}$

From there Goldenfeld proceeds to state that therefore

$Z(N+1) = Z(N)2\cosh(K) \tag{4}$

So at long last my question: since we are summing over the last term in $\eqref{3}$, should we not get $2e^K + 2e^{-K}$ i.e. $4\cosh(K)$?

The resolution is probably rather trivial, but I get confused by sums sometimes^^

Best regards

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  • $\begingroup$ the term you write in (3) already has the sum in it. The last sum you write (the one with index $S_{N+1}$ should not be there), as the term in parenthesis is explicitly it. $\endgroup$
    – user245141
    Jan 8, 2020 at 11:27
  • $\begingroup$ oops that was a typo, I've edited that out. But in the last term, one is still summing over both individual terms ($e^{Ks_N}$ and $e^{-Ks_N}$), should we not therefore have two contributions from each term, i.e. four terms total? $\endgroup$
    – Trirac
    Jan 8, 2020 at 11:34
  • $\begingroup$ These two terms are already part of the original sum, which includes $s_N$. If you were to sum over them individually, you would do over-counting. For each value of $s_N$ you get $\exp(K)+\exp(-K)$, so the entire of $Z_N$ (which includes the sum over both values of $s_N$) is multiplied by this factor. $\endgroup$
    – user245141
    Jan 8, 2020 at 11:41
  • $\begingroup$ you can maybe try it with a small chain with $N=2$ and then expand to $N=3$ to convince yourself of that. $\endgroup$
    – user245141
    Jan 8, 2020 at 11:42
  • $\begingroup$ Ahhh I think I see it now, if I perform the sum directly in the parenthesis, I'm actually attributing two extra terms to the partition function, since for the spin up in the parenthesis, there corresponds two different terms in the original partition function... I don't know if that makes any sense, but I think I see it now. Thanks a lot for the tip, writing it out for a few spins made it clear! $\endgroup$
    – Trirac
    Jan 8, 2020 at 11:57

1 Answer 1

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A short derivation is: $$\eqalign{ Z(N+1)&=\sum_{s_1\ldots s_N}e^{K(s_1s_2+s_2s_3+⋯+s_{N−1}s_N)} \Big(e^{Ks_N}+e^{−Ks_N}\Big)\cr &=\sum_{s_1\ldots s_N} e^{K(s_1s_2+s_2s_3+⋯+s_{N−1}s_N)}\times 2\cosh Ks_N }$$ since $\cosh(-x)=\cosh x$, one can write $\cosh Ks_N=\cosh K$ and therefore $$Z(N+1)=Z(N)\times 2\cosh K$$

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  • $\begingroup$ Yeah I got that, I don't know why I thought you would have to sum over those $s_N$ which would yield $4\cosh(K)$, but as yu-v pointed out, that would be overcounting... $\endgroup$
    – Trirac
    Jan 8, 2020 at 20:03

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