screenshot of the relevant part

Above is a screenshot of Kadanoff's review article "more is the same". The free energy in Landau's theory is very well known, but the highlighted sentence seems to be quite confusing. First of all, why would the order parameter go to zero when criticality is approached?


At the critical point, we have $\Psi(\vec r) = 0$ because that's the basic way in which this whole Landau theory works.

First of all, it's important to realize that while making such statements, we consider the case of zero external field, so the term $Bh(\vec r) \Psi(\vec r)$ drops.

Without this term, considering the constant configurations (in space) $\Psi(\vec r)=\Psi$ for a while, the terms with $\nabla\Psi$ drop as well and the integrand of $F$ is $$ A+C \Psi^2 + D \Psi^4$$ Now, the key observation of the Landau theory is that this function of $\Psi$ – which is bounded from below because we assume $D\gt 0$ everywhere – has differently distributed minima for $C\gt 0$ and $C\lt 0$. For a positive $C$, the only minimum is at $\Psi=0$. The value of the minimum (of the integrand) is simply $A$.

For a negative $C$, the term $C\Psi^2$ makes it energetically preferred to have a nonzero $\Psi$. So the minima appear at two points $\pm \Psi_0$, which are easy to calculated to be $\pm \sqrt{-2D/C}$.

The behavior is therefore qualitatively different for positive and negative $C$. There are one or two minima for positive or negative $C$, respectively, and the minima preserve or violate the $\Psi\to-\Psi$ symmetry, respectively. The boundary case in between these two qualitative situations, $C=0$, is the critical point. The minimum of the integrand still occurs for $\Psi=0$.

  • $\begingroup$ I see... I should have realised that the Landau theory is for second order phase transitions, also called continuous phase transitions. The order parameter is 0 on one side and continuously approach 0 on the other side. But how to explain that $\Psi$ goes to 0 more rapidly than ${\Psi}^2$? $\endgroup$ – M. Zeng Aug 9 '15 at 11:09
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    $\begingroup$ Hi, the terms he means are $C\Psi^2$ and $D\Psi^4$. $\Psi^4\ll \Psi^2$ for $\Psi\to 0$, so this quartic term goes to zero more quickly than $\Psi^2$. The term $C\Psi^2$ goes to zero more quickly than $\Psi^2$ as well because the limit near criticality has $C\to 0$ as well, as I mentioned above, so there is an extra suppression by the small factor $C\to 0$ on top of $\Psi^2\to 0$. $\endgroup$ – Luboš Motl Aug 9 '15 at 19:48

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