How is the current equation calculated from Ginzburg-Landau (GL) free energy?

Question

In the Ginzburg-Landau theory, we can get the current expression from GL free energy:

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

The corresponding current is (see Tinkham introduction to superconductivity eqn(4.14) or this pdf for example):

$$J=\frac{c}{4\pi}\mathrm{curl}h=\frac{e^*\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*)-\frac{e^{*2}}{mc}\psi\psi^* A$$

I want to know exactly how this equation is derived, I think it is from $J=c\frac{\delta F}{\delta A}$, but the third term in $F$ seems already give the result of the above equation. How about the fourth term's $F$ variation w.r.t A?

and why does this equation $J=\frac{c}{4\pi}\mathrm{curl}h$ holds?

This post imported from StackExchange Physics at 2015-03-15 09:48 (UTC), posted by SE-user buzhidao

Comments

Please define notation used within the post - what is $h$, and how does it depend on $A$ and how is your current $J$ defined in the first place?

This post imported from StackExchange Physics at 2015-03-15 09:48 (UTC), posted by SE-user ACuriousMind

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The current can be defined as $J=\frac{e^*}{2mi}(\psi^*p\psi-\psi p\psi^*)$, where $p=-i\nabla-2e^*A$ is the canonical momentum.

This post imported from StackExchange Physics at 2015-03-15 09:48 (UTC), posted by SE-user Meng Cheng