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I am having a bit of trouble with basic combinatorics pertaining to the Ising model and mean field theory. Specifically, I get that the Hamiltonian can be written

$H=-\frac{J}{2}\Sigma_{i,j} s_is_{i+j} - \mu \Sigma_i s_i B$

Where the notation in the first sum denotes sum over all spins, i, and it nearest neigbours, indexed by j. The factor of 1/2 comes from double counting.

However then in my lectures notes, and other texts, when the mean spin is replaced by s, and the 2D nearest neighbours are explicitly put in, this somehow becomes

$H=-\Sigma_i s_i (\mu B + 2DJs)$

But this seems to be off by a factors of 2!

If s is the mean spin, then the sum over the nearest neighbours is

$\Sigma_{j} s_{i+j} = 2Ds$

i.e. number of nearest neighbours multiplied by the mean spin. The factor of 2 then cancels with the factor of $\frac{1}{2}$ in the first form of the Hamiltonian, so I get that the new Hamiltonian should be

$H=-\Sigma_i s_i (\mu B + DJs)$ ?

Where has the factor of 2 gone?

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  • $\begingroup$ The current answer is right, but also be careful, lots of sources really do change $J$ by factors of $2$! There isn't a universal convention for this. $\endgroup$ – knzhou Dec 24 '18 at 11:58
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This can be a bit confusing. But actually the Wikipedia page explains the counting fairly clearly. I prefer their notation (summing over nearest-neighbour pairs, i.e. edges of the lattice) to the double-counting notation. So the interaction term is written $$ -J\sum_{\langle i,j\rangle} s_i s_j \tag{1} $$ summing over all distinct nearest-neighbour pairs $\langle i,j\rangle$, so no factor $1/2$. The approximation is obtained by writing $$ s_i = s + (s_i-s) $$ and similarly for $s_j$. Then every pair term becomes $$ s_is_j = s^2 + s(s_i-s) + s(s_j-s) + (s_i-s)(s_j-s) . $$ The approximation consists of dropping the last term. Also, all the terms in $s^2$ are constant, and can be dropped. (Actually, more precisely, I should say that they can be taken out of the problem, and put back in again if we wish, once the value of $s$ has been determined).

So we are left with $$ -J\sum_{\langle i,j\rangle} s\,s_i + s\,s_j = -2J\sum_{\langle i,j\rangle} s\,s_i . $$ Notice that there are two terms here, not one, for every edge. Compare with eqn (1). This is where the factor of $2$ comes in. Physically we are counting the fluctuation of $i$ in the mean field provided by $j$, and the fluctuation of $j$ in the mean field provided by $i$. The $i$ and $j$ terms are equivalent, of course, hence the second form of the equation. If the dimensionality is $D$, there are $2D$ edges per site $i$, but we take care to avoid double counting, so $$ \sum_{\langle i,j\rangle} \rightarrow \frac{1}{2} \, 2D \sum_i . $$ So the result may also be written $$ -2DJs\, \sum_{i} s_i $$ as you have found in notes and text books.

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  • $\begingroup$ Dear LonelyProf, thank you for your answer. I delayed in accepting it as I wasn't satisfied with the fact that I couldn't see the logical inconsistency in my derivation that is off by a factor of 2, although I thin forgot about this concern and accepted your answer. I follow the steps in your answer but was wondering if you could see where mine is wrong? I feel it has to do with not treating the situation symmetrically in that I take the neighbours as having their average value and only allow the spin s_i to fluctuate, whereas the neighbours should fluctuate also. But I am not convined by this $\endgroup$ – Meep Dec 24 '18 at 12:07
  • $\begingroup$ Yes, I think that this is indeed the false step in your derivation. Your "picture" is of a single spin in the mean field of its neighbours. And indeed, the target of the derivation is an expression that looks exactly like that, and is described in exactly those words. But by starting with the unsymmetrical formula (sum over $i$, and then sum over the neighbours of $i$) one is led astray. Just asserting that the neighbours may be given their average values is not the same as making a specific mathematical approximation. .... $\endgroup$ – user197851 Dec 24 '18 at 12:49
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    $\begingroup$ .... So, I'm afraid that my equation for $s_is_j$ is about as clear as I can be! You could say that you are not treating the situation "symmetrically", but I think that it is better to say that preserving the symmetry just enables you to see (better) the correct mathematical formula for the approximation. Having made the approximation, then the symmetric form of the double sum may be converted (rigorously) into the unsymmetrical form where we just sum over $i$ (and the sum over nearest neighbours becomes trivial because they are all equivalent). Hope this helps, sorry if it is still unclear. $\endgroup$ – user197851 Dec 24 '18 at 12:53

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