# 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 ${\displaystyle \sigma _{i}}$, instead of just two values can assume ${\displaystyle q}$ integer values: ${\displaystyle \sigma _{i}\in \lbrace 1,\dots ,q\rbrace }$. We can therefore write the Hamiltonian of such a system as:

${\displaystyle {\mathcal {H}}=-J\sum _{\left\langle i,j\right\rangle }\delta _{\sigma _{i},\sigma _{j}}-H\sum _{i}\delta _{\sigma _{i},1}}$
(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 ${\displaystyle q=2}$ is equivalent to an Ising model, as can be seen from the following equivalence:
${\displaystyle \delta _{\sigma ,\sigma '}={\frac {1}{2}}(1+\sigma \sigma ')}$
where ${\displaystyle \sigma }$ and ${\displaystyle \sigma '}$ are the degrees of freedom of the Ising model (${\displaystyle \sigma ,\sigma '=\pm 1}$). However, a Potts model with ${\displaystyle q=3}$ is not equivalent to an Ising model where ${\displaystyle S_{i}=-1,0,+1}$, as one could have expected. In fact it is not possible in this case to write a ${\displaystyle \delta }$ of the three-state variable ${\displaystyle \sigma }$ with "simple" terms involving ${\displaystyle \sigma }$-s (namely, only quadratic terms); in particular it turns out that:
${\displaystyle \delta _{\sigma ,\sigma '}=1+{\frac {1}{2}}\sigma \sigma '+{\frac {3}{2}}\sigma ^{2}{\sigma '}^{2}-(\sigma ^{2}+{\sigma '}^{2})}$

We therefore want to apply the Bragg-Williams approximation to a ${\displaystyle q}$-state Potts model. First of all, we call ${\displaystyle x=\left\langle \delta _{\sigma ,1}\right\rangle }$ our order parameter, and write the probability distribution of a single degree of freedom as[1]:

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

1. As we have stated in the footnote on page , we write ${\displaystyle \rho }$ 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 ${\displaystyle \sigma =1}$ as this particular value, but of course we could have done otherwise.