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The Ginzburg-Landau free energy is defined as $$F = \int dV \left \{\alpha |\psi|^2 + \frac{\beta}{2}|\psi|^4 + \frac{1}{2m^*} \left| \left(\frac{\hbar}{i}\nabla - \frac{e^*}{c}\vec{A}\right)\psi \right|^2 + \frac{h^2}{8\pi}\right \},$$ where $\vec{h}=\nabla\times \vec{A}$. The current equation is derived by taking the variation of the vector field $\vec{A}$. My concern is mainly about the variation of the last term in the free energy: $$\delta \int dV\,\left(\nabla\times \vec{A}\right)^2 = 2\oint_S \left(\delta \vec{A} \times \vec{h}\right)\cdot d\vec{S} + 2\int dV \,\delta\vec{A}\cdot \left(\nabla\times \vec{h}\right).$$ However, in arriving at the final current equation, usually the surface integral is ignored. Why is that so?

In the original paper of Ginzburg (above equation 9), he said:

... by varying this expression with respect to $\psi^*$ and $A$, we obtain the equations defining $\psi$ and $A$ (it must be assumed that ${\rm div} A = 0$)....

Is he implying that the surface term is ignored because $\nabla\cdot \vec{A}=0$? If so, why?

I can also think of other two possibilities:

  1. We manually require $\delta\vec{A}$ to vanish at surface boundary. If so, why is a similar requirement not applied to $\delta \psi^*$?
  2. We can also ignore the surface integral by an appropriate choice of gauge, which is complicated but exist.

What is your opinion?

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  • $\begingroup$ I am not able to answer your question but I find very nice your nickname and the associated image "un'offerta che non puoi rifiutare". Very funny! $\endgroup$ Dec 7 '20 at 10:45
  • $\begingroup$ @ValterMoretti :) $\endgroup$ Dec 7 '20 at 10:54
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    $\begingroup$ When computing variations of the action (e.g. deriving the Euler-Lagrange equations) it is typically assumed that boundary terms vanish. Is there a particular reason why it should be different here? $\endgroup$
    – fqq
    Dec 10 '20 at 2:13
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Short answer: the key is that $\delta \vec{A}$ is arbitrary, so you can set it to whatever you need.

In the equation $$\delta \int dV\,\left(\nabla\times \vec{A}\right)^2 = 2\oint_S \left(\delta \vec{A} \times \vec{h}\right)\cdot d\vec{S} + 2\int dV \,\delta\vec{A}\cdot \left(\nabla\times \vec{h}\right),$$ you get rid of the surface term in the same breath as you get rid of $\int dV$ to pass from an integral expression to an expression valid point by point:

$\delta \vec{A}$ is arbitrary, so you set it equal to zero everywhere apart from an infinitesimal volume $dV$ around an arbitrary point $x$.

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