Consider an Ising model system where the total energy is $E = −J \sum_{<ij>} S_iS_j $, $S_i = \pm 1$ and $< ij >$ implies sum over nearest neighbours. For $J < 0$ the ground state of this system at $T = 0$ is antiferromagnetic. (All adjacent spins misaligned so net magnetisation zero and thus antiferromagnetic).

Let $m_{1,2}$ be the magnetisations in the two sublattices of an antiferromagnetic Ising model. Define the order parameter $\psi ≡ m_1−m_2$. The Landau free energy for this system should have the form, $$F = at\psi^2 + b\psi^4 + ...,$$ where $t \equiv (T − T_c)/T_c $ and $a, b > 0$ are constants. Why is this the case?

I know that $F$ is some polynomial expansion in order parameter $\psi$ and as far as I understand from the Landau theory, it is constructed based solely on the symmetries of the system, particularly that obeyed by the order parameter. $m_i \rightarrow -m_i$ can be viewed as a rotation of the system by $\pi$ so F should be invariant under $m_i \rightarrow -m_i$ which is to say $\psi \rightarrow -\psi$ is a symmetry. So we construct $$F = C + A(T)\psi^2 + B(T)\psi^4 + ....,$$ where $C$ is a constant, can be set to $0$. I am just a bit confused as to how they obtained the expressions for $A(T)$ and $B(T)$? Imposing that at equilibrium, $\partial F/\partial \psi = 0$ then this means $$2\psi(A(T) + 2B(T) \psi^2) = 0$$ ie $\psi = 0$ or $\psi^2 = -A(T)/2B(T)$. The case $\psi = 0$ corresponds to the case when $m_1 = m_2$ so that this could be realised in the ground state. I just would like the argument as to why we infer the dependence of $A$ on $T$ and that $B$ is a constant. Thanks!

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    $\begingroup$ Notice that Landau theory is a phenomenological theory, so the dependence of $A$ and $B$ on $T$ is postulated to describe the phase transition. You can also derive these coefficients from a microscopic model. $\endgroup$ – Meng Cheng Apr 19 '15 at 15:36
  • $\begingroup$ @MengCheng: Thanks, but could you please elaborate? At the critical temperature (or temperature at which we reach criticality) the first term vanishes which implies the equilibrium situation is one in which we have $\psi = 0$. Similarly for the case $T > T_c$. (We must choose $\psi=0$ otherwise we get an imaginary solution for $\psi$ which is unphysical. For $T<T_c$ we get two minima. This is based on the form of the free energy function given. Is this correct understanding? Thanks. $\endgroup$ – CAF Apr 19 '15 at 17:29
  • $\begingroup$ @MengCheng: seems reasonable I guess, although the above analysis is the same as if $J > 0$. $\endgroup$ – CAF Apr 20 '15 at 11:35

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