# Simplified Ginzburg-Landau equation

So assume $a(T)= a_{0}(\frac{T}{T_{c}}-1)$. Then, I start with a simplified version of the free energy density that is used for the Ginzburg-Landau equations (Assume $\vec{A} = 0$ and spatial variation of $\psi$ is negligible): $$f = f_{n} + a(T)\left|\psi(\vec{r})\right|^2 + b(T)\left|\psi(\vec{r})\right|^4$$ Then I want to find minimum of free energy density with respect to the order parameter, which I believe is $\psi$.

Then my first instinct is:

$$\frac{\partial f}{\partial \psi} = a(T)\psi^{*} + b(T)\left|\psi(\vec{r})\right|^2 \psi^{*}$$

Then set $\frac{\partial f}{\partial \psi} = 0$, and obtain:

$$\left|\psi\right|^2 = -\frac{a(T)}{b(T)}$$

Now I want to prove that this is a minimum, however I am not sure how to do this since taking second derivative of $f$ gives:

$$\frac{\partial^2 f }{\partial \psi^2} = b(T)\left( \psi^{*} \right)^2$$

And the RHS is complex, any ideas?

Order parameter $\psi$ and its hermitian conjugate $\psi^*$ are not independent variables, so your method to find the minimum is not correct.
The free energy is actually a function of the magnitude of the order parameter $|\psi|$. So it doesn't matter which phase your parameter get. Now first assume the order parameter is uniform in real space, say $\psi(r) = \psi$, then write the parameter in the form of magnitude and phase $\psi = \sqrt{\rho}e^{i\phi}$. The free energy (density) then will be: $$f(T,\rho)=a(T) \rho + b(T) \rho^2$$ Treat the free energy $f$ as the function of $\rho$, we get $$\frac{\partial f}{\partial \rho} = a(T) + 2b(T)\rho=0\quad \Rightarrow \quad |\psi|_{min}^2 = -\frac{a(T)}{2b(T)}>0\quad (a_0>0~and~T<T_c)$$ Then we can show that this solution is a minimum: $$\frac{\partial^2 f}{\partial \rho^2} = 2b(T)>0$$ Hope my answer is helpful for you :)