Suppose that each atom of a crystal lattice can be in one of several internal
states (which may be denoted by the symbol \(\sigma\) ) and the interaction
energy between an atom in state \(\sigma^{\prime}\) and its nearest neighbor in
state \(\sigma^{\prime \prime}\) is denoted by \(u\left(\sigma^{\prime},
\sigma^{\prime \prime}\right)\left\\{=u\left(\sigma^{\prime \prime},
\sigma^{\prime}\right)\right\\}\). Let \(f(\sigma)\) be the probability of an
atom being in a particular state \(\sigma\), independently of the states in
which its nearest neighbors are. The interaction energy and the entropy of the
lattice may then be written as
$$
E=\frac{1}{2} q N \sum_{\sigma^{\prime}, \sigma^{\prime \prime}}
u\left(\sigma^{\prime}, \sigma^{\prime \prime}\right)
f\left(\sigma^{\prime}\right) f\left(\sigma^{\prime \prime}\right)
$$
and
$$
S / N k=-\sum_{\sigma} f(\sigma) \ln f(\sigma),
$$
respectively. Minimizing the free energy \((E-T S)\), show that the equilibrium
value of the function \(f(\sigma)\) is determined by the equation
$$
f(\sigma)=C \exp \left\\{-(q / k T) \Sigma_{\sigma^{\prime}} u\left(\sigma,
\sigma^{\prime}\right) f\left(\sigma^{\prime}\right)\right\\},
$$
where \(C\) is the constant of normalization. Further show that, for the special
case \(u\left(\sigma^{\prime}, \sigma^{\prime \prime}\right)=-J \sigma^{\prime}
\sigma^{\prime \prime}\), where the \(\sigma\) can be either \(+1\) or \(-1\), this
equation reduces to the Weiss equation (12.5.11), with
\(f(\sigma)=\frac{1}{2}\left(1+\bar{L}_{0} \sigma\right)\).