# Bragg-Williams approximation for the Potts model

We now apply the same approximation that we have just seen to a slightly more complex situation: the Potts model. This is defined exactly as the Ising model, but with an essential difference: the degrees of freedom of the system, which we now call $\sigma _{i}$ , instead of just two values can assume $q$ integer values: $\sigma _{i}\in \lbrace 1,\dots ,q\rbrace$ . We can therefore write the Hamiltonian of such a system as:

${\mathcal {H}}=-J\sum _{\left\langle i,j\right\rangle }\delta _{\sigma _{i},\sigma _{j}}-H\sum _{i}\delta _{\sigma _{i},1}$ (where we have supposed that the external magnetic field tends to favour the situation where the degrees of freedom assume the value 1; of course we could have done otherwise). As can be expected the Potts model with $q=2$ is equivalent to an Ising model, as can be seen from the following equivalence:
$\delta _{\sigma ,\sigma '}={\frac {1}{2}}(1+\sigma \sigma ')$ where $\sigma$ and $\sigma '$ are the degrees of freedom of the Ising model ($\sigma ,\sigma '=\pm 1$ ). However, a Potts model with $q=3$ is not equivalent to an Ising model where $S_{i}=-1,0,+1$ , as one could have expected. In fact it is not possible in this case to write a $\delta$ of the three-state variable $\sigma$ with "simple" terms involving $\sigma$ -s (namely, only quadratic terms); in particular it turns out that:
$\delta _{\sigma ,\sigma '}=1+{\frac {1}{2}}\sigma \sigma '+{\frac {3}{2}}\sigma ^{2}{\sigma '}^{2}-(\sigma ^{2}+{\sigma '}^{2})$ We therefore want to apply the Bragg-Williams approximation to a $q$ -state Potts model. First of all, we call $x=\left\langle \delta _{\sigma ,1}\right\rangle$ our order parameter, and write the probability distribution of a single degree of freedom as:

$\rho _{\sigma }=a\delta _{\sigma ,1}+b(1-\delta _{\sigma ,1})$ Therefore, from:
$\operatorname {Tr} \rho _{\sigma }=a+(q-1)b=1\quad \qquad \operatorname {Tr} (\rho _{\sigma }\delta _{\sigma ,1})=\sum _{\sigma =1}^{q}\left[\delta _{\sigma ,1}\cdot a\delta _{\sigma ,1}^{}+\delta _{\sigma ,1}\cdot b(1-\delta _{\sigma ,1})\right]=x$ we get:
$a=x\quad \qquad b={\frac {1-x}{q-1}}$ From now on one can proceed like we have previously seen.

1. As we have stated in the footnote on page , we write $\rho$ in general as the sum of the probability that the degree of freedom assume a particular value and of that of all the remaining values; in this case we have chosen $\sigma =1$ as this particular value, but of course we could have done otherwise.