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?


1 Answer 1


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 :)

  • $\begingroup$ They can be treated like independent parameters, actually it's what people do in QFT when working on complex scalar fields. $\endgroup$
    – Hossein
    Feb 26, 2017 at 6:51
  • $\begingroup$ I think the two cases are different: people treat them as independent parameters to derive the Euler-Lagrange equation in field theory, not to find the minimum of a function. $\endgroup$
    – fan9x13
    Feb 26, 2017 at 6:58
  • $\begingroup$ They are similar. The Euler-Lagrange equation for a functional of a field -not the field's derivatives- is equivalent to putting the functional's derivative with respect to the field equal to zero. $\endgroup$
    – Hossein
    Feb 26, 2017 at 13:35

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