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When deriving the Ginzburg-Landau equations, we minimize the following free energy over the complex function $\psi$:

$$F = \int dV \left \{\alpha |\psi|^2 + \frac{\beta}{2}|\psi|^4 + \frac{1}{2m^*} \mid (\frac{\hbar}{i}\nabla - \frac{e^*}{c}A)\psi \mid^2 + \frac{h^2}{8\pi}\right \} .$$

We take a variation over $\psi$. I know there are two ways to do that: (1) write $\psi$ as $\mid \psi \mid e^{i\phi}$, then vary $\mid \psi (r)\mid$ and $\phi(r)$ separately; (2) vary $\psi^*$ as if $\psi$ remained unchanged. The former gives two differential equations, the latter gives one complex equation.

My question is: why is the second way valid? And how to show the two ways are equivalent?

Qmechanic
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Kirill Moskovtsev
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    The main question How can we treat $\psi$ and $\psi^{\ast}$ as independent variables? is essentially a duplicate of http://physics.stackexchange.com/q/89002/2451 – Qmechanic Feb 09 '15 at 21:14

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Both ways are equally valid and yield identical answers. If you wanted to convince yourself, get the single complex $\psi$ equation, write $\psi = |\psi|e^{i\phi}$, and you'll find that you've reproduced the $|\psi|, \phi$ equations of motion.

Varying w.r.t. $\psi$ or the real components are the same because of the product rule applied to $\psi$:

$\frac{\delta}{\delta \psi} = \frac{\delta |\psi|}{\delta \psi} \frac{\delta }{\delta |\psi|} + \frac{\delta \phi}{\delta \psi} \frac{\delta }{\delta \phi}$

So one variation is simply a linear combination of the other two (and similarly for the complex conjugate variation)

Surgical Commander
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