# Landau theory for the Ising model

To make things more clear, let us now consider the Ising model without any external field and see how we can determine the form of ${\displaystyle {\mathcal {L}}}$ from the general assumptions we have introduced.

## Construction of L

First of all, since the equilibrium configurations of the system must be minima of ${\displaystyle {\mathcal {L}}}$:

${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial \eta }}=a_{1}+2a_{2}\eta +3a_{3}\eta ^{2}+4a_{4}\eta ^{3}=0}$
where we have chosen to stop the expansion at the fourth order[1]. Now, since this equation must hold for all ${\displaystyle T}$ and for ${\displaystyle T>{\overline {T}}}$ we have ${\displaystyle \eta =0}$, we see that ${\displaystyle a_{1}=0}$.

Considering now the constraint on the symmetries of the system, in Absence of phase transitions for finite systems we have seen that the Ising model is invariant under parity, i.e. its Hamiltonian is simultaneously even in ${\displaystyle H}$ and ${\displaystyle \lbrace S_{i}\rbrace }$: ${\displaystyle {\mathcal {H}}(H,\lbrace S_{i}\rbrace )={\mathcal {H}}(-H,\lbrace -S_{i}\rbrace )}$. Thus, in absence of external fields the Hamiltonian of the Ising model is even; this means that also ${\displaystyle {\mathcal {L}}}$ must be invariant under parity, namely an even function of ${\displaystyle \eta }$: ${\displaystyle {\mathcal {L}}(-\eta )={\mathcal {L}}(\eta )}$. Therefore all the odd terms of the expansion are null:

${\displaystyle a_{2k+1}=0\quad \quad \forall k\in \mathbb {N} }$
Finally, since we have assumed that ${\displaystyle {\mathcal {L}}}$ is an analytic function of ${\displaystyle \eta }$ then its expansion cannot contain terms like ${\displaystyle |\eta |}$.

In conclusion, the minimal form of the Landau free energy for the Ising model is:

${\displaystyle {\mathcal {L}}=a_{0}+a_{2}\eta ^{2}+a_{4}\eta ^{4}}$
where ${\displaystyle a_{0}}$, ${\displaystyle a_{2}}$ and ${\displaystyle a_{4}}$ can in general be functions of ${\displaystyle J_{ij}}$ (the coupling constants) and ${\displaystyle T}$. However, ${\displaystyle {\mathcal {L}}}$ can be further simplified and we can also explicitly show its dependence on the temperature. In fact, first of all we can note that ${\displaystyle a_{0}}$ is the value of ${\displaystyle {\mathcal {L}}}$ in the paramagnetic state (when ${\displaystyle T>{\overline {T}}}$, ${\displaystyle \eta ={\overline {\eta }}=0}$ and ${\displaystyle {\mathcal {L}}({\overline {\eta }})=a_{0}}$), and so for simplicity we can set ${\displaystyle a_{0}=0}$ (it's just a constant shift in the energy). Then, we must also require ${\displaystyle a_{4}>0}$ because otherwise ${\displaystyle {\mathcal {L}}}$ can be minimized by ${\displaystyle \eta \to \pm \infty }$, which makes no sense physically. Finally, expanding ${\displaystyle a_{2}}$ and ${\displaystyle a_{4}}$ in ${\displaystyle T}$ near ${\displaystyle {\overline {T}}}$:
${\displaystyle a_{2}=a_{2}^{0}+t{\frac {a}{2}}+\cdots \quad \qquad a_{4}={\frac {b}{4}}+\cdots }$
where we have renamed the constants this way for later convenience, and ${\displaystyle t=(T-{\overline {T}})/{\overline {T}}}$ is the reduced temperature; in the expansion of ${\displaystyle a_{4}}$ we have neglected any explicit dependence on ${\displaystyle T-{\overline {T}}}$ because as we will see it will not dominate the behaviour of the thermodynamics near ${\displaystyle {\overline {T}}}$[2]. Taking ${\displaystyle a_{2}^{0}=0}$ we finally have that the form of the Landau free energy for the Ising model is:
${\displaystyle {\mathcal {L}}={\frac {a}{2}}t\eta ^{2}+{\frac {b}{4}}\eta ^{4}}$
If an external field ${\displaystyle H}$ is present, then:
${\displaystyle {\mathcal {L}}={\frac {a}{2}}t\eta ^{2}+{\frac {b}{4}}\eta ^{4}-H\eta }$

## Critical properties of the Landau theory for the Ising model

Let us now see what does the Landau theory for the Ising model predict. First of all, in the absence of external fields we have that the equilibrium states are determined by:

${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial \eta }}=a\eta t+b\eta ^{3}=\eta (at+b\eta ^{2})=0\quad \Rightarrow \quad \eta ={\overline {\eta }}={\begin{cases}0&t>0{\text{ (i.e. }}T>{\overline {T}}{\text{)}}\\\pm {\sqrt {-at/b}}&t<0{\text{ (i.e. }}T<{\overline {T}}{\text{)}}\end{cases}}}$
We therefore see that a phase transition really occurs: when the temperature is greater than the critical one, the only global minimum of ${\displaystyle {\mathcal {L}}}$ is ${\displaystyle {\overline {\eta }}=0}$, while for ${\displaystyle T<{\overline {T}}}$ two possible new equilibrium states appear (which of course are the two possible values that the spontaneous magnetization can assume in the magnetic phase).

Let us therefore see what critical exponents does the Landau theory for the Ising model predict.

### Exponent beta

This is immediately determined from what we have just seen:

${\displaystyle \eta =\pm \left(-{\frac {a}{b}}t\right)^{1/2}\sim (-t)^{1/2}\quad {\text{when}}\;t<0\quad \Rightarrow \quad \beta ={\frac {1}{2}}}$

### Exponent alpha

The specific heat at constant external field of the system is ${\displaystyle C_{H}=-T\partial ^{2}{\mathcal {L}}/\partial T^{2}}$; as we have seen, ${\displaystyle {\mathcal {L}}=0}$ for ${\displaystyle t>0}$ (since we have set ${\displaystyle a_{2}^{0}=0}$) while ${\displaystyle {\mathcal {L}}={\mathcal {L}}({\overline {\eta }})=-a^{2}t^{2}/4b}$ for ${\displaystyle t<0}$. Therefore:

${\displaystyle C_{H}={\begin{cases}0&t>0\\{\frac {a^{2}}{2b{\overline {T}}^{2}}}T&t<0\end{cases}}}$
We thus see that ${\displaystyle C_{H}}$ has a jump discontinuity at ${\displaystyle T={\overline {T}}}$, so:
${\displaystyle \alpha =0}$

### Exponent delta

Considering now also an external field, the state equation of the system will be given by the differentiation of ${\displaystyle {\mathcal {L}}}$:

${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial \eta }}=at\eta +b\eta ^{3}-H=0\quad \Rightarrow \quad at\eta +b\eta ^{3}=H}$
At the critical point ${\displaystyle t=0}$ and so we have ${\displaystyle H\propto \eta ^{3}}$. Therefore:
${\displaystyle \delta =3}$

### Exponent gamma

If we now differentiate the state equation with respect to ${\displaystyle H}$ we get:

${\displaystyle at{\frac {\partial \eta }{\partial H}}+3b\eta ^{2}{\frac {\partial \eta }{\partial H}}=1}$
Since ${\displaystyle \partial \eta /\partial H=\chi _{T}}$, then:
${\displaystyle \chi _{T}={\frac {1}{at+3b\eta ^{2}}}}$
If we now set ${\displaystyle H=0}$ then for ${\displaystyle t>0}$ we will have ${\displaystyle \eta ={\overline {\eta }}=0}$ and thus ${\displaystyle \chi _{T}=(at)^{-1}}$; on the other hand, if ${\displaystyle t<0}$ then ${\displaystyle \eta ={\overline {\eta }}=\pm (-at/b)^{1/2}}$ and therefore ${\displaystyle \chi _{T}=(-2at)^{-1}}$. In both cases ${\displaystyle \chi _{T}\sim t^{-1}}$, and thus:
${\displaystyle \gamma =1}$

Therefore, the Landau theory for the Ising model gives the critical exponents:

${\displaystyle \alpha =0\quad \qquad \beta ={\frac {1}{2}}\quad \qquad \gamma =1\quad \qquad \delta =3}$
which, as we expected, are identical to those we have found within Weiss mean field theory.

1. This choice has been made because we suppose ${\displaystyle \eta }$ to be small and also because in both the Weiss and Van der Waals theories we have seen that the relevant physics involves third and fourth powers of the order parameter. Of course, the choice of where to stop the expansion depends on the system considered (in particular, it depends also on the dimensionality of the system).
2. In other words, the occurrence of the phase transition at ${\displaystyle T={\overline {T}}}$ depends only on how ${\displaystyle a_{2}}$ depends on ${\displaystyle t}$.