# Mean field approximation in the Ising model

I am studying statistical physics for an exam scheduled next week and there's something I really do not get about mean field approximation in the Ising model.

The situation

In the lesson, we defined the Ising model as $$N$$ spins $$S_i \in \{-1,1\}$$ ; we consider the case where $$B=0$$ (no external magnetic field), and thus the Hamiltonian of the system is : $$H = -J \sum\limits_{i,j \: neighbors} S_i S_j$$

The teacher then said "this is an $$N$$-body problem which we cannot solve ; let's make an approximation to have $$N$$ independent $$1$$-body problems instead. In $$H$$, the term in $$S_i$$ is : $$-S_i \left( J \sum\limits_{j \: neighbors \: of \: i} S_j \right)$$ We replace the $$S_j$$ by the magnetization $$m= \frac{1}{N} \sum\limits_i \langle S_i \rangle$$ : $$-S_i \left( J \sum\limits_{j \: neighbors \: of \: i} m \right)$$ Assuming the number of neighbors of any two spins to be the same, that we denote $$z$$, we thus get: $$-S_i \times Jzm$$ Defining $$h = Jzm$$, we get: $$-S_i \times h$$ Which is the Hamiltonian of a spin in a field $$h$$. Therefore, we can apply statistical physics (canonical ensemble) and it follows that: $$\langle S_i \rangle = \text{tanh}(\beta h)$$ This result being independent of $$i$$, we can sum it over all possible $$i$$ and divide it by $$N$$, which yields: $$m = \text{tanh}(\beta J z m)$$ "

My problem

I don't understand why we say that the Hamiltonian for $$S_i$$ is $$-S_i \left( J \sum\limits_{j \: neighbors \: of \: i} S_j \right)$$ and not $$-\frac{1}{2}S_i \left( J \sum\limits_{j \: neighbors \: of \: i} S_j \right)$$. What leads me to think of this are the following arguments:

1. By using the professor's formula, we give all the energy of the interaction between two neighbors spins $$i$$ and $$j$$ to spin $$i$$, leaving no energy for spin $$j$$. But then, we proceed on summing over all possible spins, so we will sum counting each energy twice: once for $$i$$ and once for $$j$$.
2. The basic idea is to transform our $$N$$-body problem into $$N$$ $$1$$-body independent problems. To do such thing, I could start by saying that $$H$$ the Hamiltonian of the $$N$$-body problem must be the sum of the $$N$$ $$1$$-body problems, because these $$1$$-body problems are independent. So, something like : $$H = \sum\limits_{i=1}^N H_i$$ Back to our situation, I notice that: $$H = -J \sum\limits_{i,j \: neighbors} S_i S_j = \sum\limits_i - \frac{1}{2} S_i \left( J \sum\limits_{j \: neighbors \: of \: i} m \right)$$ So my factor $$1/2$$ arises naturally from not counting twice the same pair of spins (i.e. summing the $$(i,j)$$ term and after that summing it again by naming it $$(j,i)$$).

But I must be wrong somewhere, because doing what I propose would lead to the equation: $$m = \text{tanh}(\beta \frac{J}{2} z m)$$ Instead of the correct equation (for what it's worth, I know that it is correct "for sure" because Wikipedia says so, had it been wrong someone would have fixed such a big Wikipedia page already): $$m = \text{tanh}(\beta J z m)$$

Thanks a lot in advance for your help, that'd be greatly appreciated - and happy new year !

The starting point, $$H = -J \sum\limits_{i,j \: neighbors} S_i S_j \tag{1}$$ is sometimes used in the textbooks, but, without an explicit indication of what is meant by summing on $$i,j \: neighbors$$ is misleading.

If the pair interaction between two neighbor spins is $$-JS_i S_j$$, the formula $$(1)$$ should be intended as $$H = -J \sum\limits_{i that can always be rewritten as $$H = -\frac{J}{2} \sum\limits_{i,j \: restricted \: to \: nearest \:neighbors} S_i S_j. \tag{3}$$ These forms do not hamper to get the correct equation for $$m$$ in the absence od an external field. The most convincing way to see it, in my opinion goes as follows:

1. Introduce the average magnetization and introduce a fluctuation $$\delta S_i$$ such that $$S_i=m+\delta S_i$$.
2. Rewrite the product $$S_iS_j$$ neglecting the $$\delta S_i \delta S_j$$ term: $$m^2 + m (\delta S_i + \delta S_j)$$.
3. add and subtract an $$m^2$$ term to recover the original $$S_k$$ degrees of freedom: $$m(S_i + S_j) -m^2$$.
4. Use the previous approximate expression in place of $$S_i S_j$$ in formula $$(3)$$.

You'll end up with the following mean-field Hamiltonian: $$H = -\frac{J}{2} \sum\limits_{i,j \: restricted \: to \: nearest \:neighbors} [ m(S_i + S_j) -m^2 ]. \tag{4}$$ The second term in the square bracket does not depend on the summation indices. Therefore, when summed, it provides a spin-independent term in the Hamiltonian equal to $$\frac{J}{2}m^2Nz$$ because $$Nz$$ is the number of the ordered nearest neighbors pairs. The factor $$\frac12$$ correctly counts the independent ones.

In the remaining sum, the role of $$i$$ and $$j$$ is symmetric. Therefore, by renaming the indices in one of the two summations, we get twice one of them: $$-\frac{J}{2} \sum\limits_{i,j \: restricted \: to \: nearest \:neighbors} m(S_i + S_j) = -J m \sum\limits_{i,j \: restricted \: to \: nearest \:neighbors} S_i = -J m z \sum\limits_{i } S_i .$$ In the last passage, I have used the independence on $$j$$ to get a factor equal to the number of first neighbors $$z$$. The final form of the equation $$(4)$$ is $$H= \frac{J}{2}m^2Nz -J m z \sum\limits_{i } S_i.$$ From this expression for the mean-field Hamiltonian, the usual equation for $$m$$ can be recovered.

Notice that the spin-independent term is necessary both to get the equation for $$m$$ from the minimum condition on the free-energy and show that below the critical point, the solution $$m=0$$ is not stable below the critical point.

• Thanks a lot for that fast answer ! I understood how the derivation works. My mistake was to not take into account the fact that expanding the $S_j$ around $m$ implied doing so with $S_i$ as well, therefore giving the missing $2$ factor. That's clear now. Thanks ! Jan 1 at 19:23