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 !


1 Answer 1


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 $$H = -J \sum\limits_{i<j \: restricted \: to \: nearest \:neighbors} S_i S_j, \tag{2}$$ 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.

  • $\begingroup$ 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 ! $\endgroup$
    – Kal8578
    Jan 1, 2022 at 19:23

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