# Construction of free energy using Landau theory

1. Apr 22, 2015

### CAF123

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 beinvariant 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.

Or by using the given expression, 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<Tc we get two minima. Is this correct understanding? Thanks :)

2. Apr 25, 2015

### BruceW

That's right. At equilibrium state, the magnitisation will be such that we have the derivative of the Free energy (with respect to magnetisation) be equal to zero. Below the critical temperature, we want to have antiferromagnetic states possible. But above the critical temperature, we want to have only the ground state of random spins. Therefore, we choose A's dependence on temperature so that this is true.

3. Apr 29, 2015

### CAF123

Hi BruceW, :)
What do you mean by above $T_c$ we only have the ground state of random spins?

Another question I had is if we apply an external magnetic field, $h$ to the system , how will this affect the Landau free energy expression? I thought that if this was the case then we would no longer have the symmetry $\psi = -\psi$ since there is now a preferred direction in the system so the lowest order term that should be added is $\psi h$ but apparantly this is not correct. Any thoughts here? thanks!

4. Apr 29, 2015

### BruceW

Hi :) yes, I was just confirming what you said "we must choose psi=0 or we get an imaginary solution". In other words, above $T_c$ we must choose psi=0 or we get an imaginary solution. and psi=0 is the state where spins are just random. I guess by 'ground state' I just mean a state which is minimum of free energy. But maybe this is not correct terminology.

Even if there is a preferred direction to the system, we still have the symmetry transform $\psi$ to $- \psi$ Remember how $\psi$ is defined, as the difference between magnetisation of the two sublattices. In other words, if we paint a checkerboard pattern on to our lattice, it doesn't matter which squares we call black and which squares we call white. And this is still true even if the system does not have a directional symmetry. The two sublattices are not related to the directional symmetry of the lattice, but to the translational symmetry.

5. Apr 30, 2015

### CAF123

Ok thanks

If we have a system where the order parameter is just $m$ say, the magnetisation of the whole system, then the correct addition to the free energy upon imposing a magnetic field is the $\psi h$. Why is this? Sorry maybe I didn't quite understand what you said.

6. Apr 30, 2015

### BruceW

If the order parameter was magnetisation of the whole system, then we would be trying to describe a ferromagnet, not an antiferromagnet. For a ferromagnet, yes the symmetry of $m$ to $-m$ is broken by the external magnetic field. But for an antiferromagnet, the symmetry of $\psi$ to $- \psi$ is not broken by the external magnetic field. To see why this is true, draw an 8by8 grid of squares, and color them black and white like a chessboard. The order parameter for the antiferromagnet is the spins on the white squares minus the spins on the black squares. So for $T<T_c$, the state which minimises free energy is the state where all spins on white squares are 'up' and all spins on black squares are 'down'. So overall, the total magnetisation is zero. And for this reason, the state is not affected by an external magnetic field. The symmetry related to our definitions of black squares and white squares is not affected by an external magnetic field.

7. Apr 30, 2015

### CAF123

Hi BruceW, sorry I am still not understanding a few things.
Why?

This makes sense since in this configuration, the Ising energy is minimised so constitutes the ground state which is antiferromagnetic because the magnetisation vanishes. But did I not show in the above that for $T < T_c$ we have two solutions? Are these solutions one where all spins on white are up and the all spins on black down and vice versa for the other solution? I am still not sure why the application of the field doesn't change the symmetry, is it because whatever the field does to the spins on the white squares it does the opposite to the ones on the black, so overall the system is unchanged? Thanks!

8. Apr 30, 2015

### DrDu

$at$ is the general first order Taylor expansion of A(T). For B(T), only the lowest order, i.e. a constant, is taken into account as B has to stay positive to guarantee the stability of the system ( at least as long as there are no higher order terms in the field).

9. Apr 30, 2015

### BruceW

that's ok! For the ferromagnet, when $T<T_c$ we have a nonzero net magnetisation (up or down). Therefore, if we add an external magnetic field (up or down), there would be some energy preference of one state over the other. i.e. the symmetry between the states is lost.

yes. those are the two solutions. yeah, well the system has changed, because of the external magnetic field. But we still have the symmetry operation of turning black squares into white squares and vice versa. Yeah, you're right. the external field would affect the system in the same way if the (white squares were 'up' and black were 'down') or (black square were 'up' and white squares were 'down'). Therefore the symmetry remains.