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?

  • 1
    $\begingroup$ 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? $\endgroup$
    – ACuriousMind
    Mar 14, 2015 at 14:41
  • $\begingroup$ 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. $\endgroup$
    – Meng Cheng
    Mar 14, 2015 at 22:29

1 Answer 1


The full Maxwell equation $j=\nabla\times H+\partial_{t}D$ ($j$ current, $H$ magnetic field and $D$ electric induction) is recover from the action

$$S=\int dx\left[L\left(A_{\mu}\right)\right]=\int dx\left[-\dfrac{F_{\mu\nu}F^{\mu\nu}}{4}-j_{\mu}A^{\mu}\right]$$

in a standard way, provided we define $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ (Abelian case) with $\mu=\left\{ 0,1,2,3\right\}$, $E_{i}=F_{0i}$ is the electric field, $B_{\left|i\right|}=F_{\left|i+1\right|\left|i+2\right|}$ the magnetic field with the modulus notation $\left|i+3\right|=\left|i\right|=i$ and with latin index $i=\left\{ 1,2,3\right\}$. It is the Euler-Lagrange equation for the gauge-degree of freedom:

$$\dfrac{\delta S}{\delta A_{\nu}}=0\Rightarrow\partial_{\mu}\dfrac{\delta L}{\delta\partial^{\mu}A_{\nu}}-\dfrac{\delta L}{\delta A_{\nu}}=0$$

and you should obtain $j_{\mu}=\partial^{\nu}F_{\nu\mu}$ in covariant notations. It is the same equation as the Maxwell-Ampère one, provided you define properly the magnetic and electric field-to-induction constitutive relation: usually, it is $B\propto H$ and $E\propto D$ in any system of dimension. This is a nice exercice to start with. More details can be found in any text-book on field-theory (see e.g. Itzykson and Zuber, Quantum-field-theory, chapter 1).

Here, you have a simpler system, since you do not have time-dependency in your effective free-energy, which plays the role of an effective Lagrangian. Defining as usual $B=\nabla\times A$ for the definition of the magnetic induction in term of the potential, you obtain the desired equation. More precisely, the equations of motion are given by

$$\dfrac{\delta F}{\delta\Psi}=0\;;\;\dfrac{\delta F}{\delta\Psi^{\ast}}=0\;;\;\dfrac{\delta F}{\delta\boldsymbol{A}}=\boldsymbol{0}$$

since $\left\{ \Psi,\Psi^{\ast},\boldsymbol{A}\right\}$ is the full set of degrees of freedom in your system. Usually, the two first equations have the same meaning (the second one is really the complex-conjugate of the first one): they tell you how the $\Psi$ degree-of-freedom evolves in a superconductor close to the critical temperature. The third equation gives you the equation of motion of the gauge-field, the one you are looking for.

You obtain something like

$$\sim e\hbar\Im\left\{ \Psi^{\ast}\left(\nabla-\mathbf{i}eA/\hbar\right)\Psi\right\} -\nabla\times\nabla\times A=0$$

(I hope I discarded the irrelevant prefactors correctly ; perhaps it's in a strange unit for the magnetic-field, in SI the second term $\nabla\times B$ has a $\mu^{-1}$ prefactor (called the inverse of the magnetic permeability), it seems you're using Gaußian units) and you recognise the Maxwell equation $j=\nabla\times H$ provided you define $H\propto B$, and $j$ as you did in your question.

You can verify that the current has the correct form given by the alternative Noether theorem, which is also a nice exercice. Nöther's theorem gives the expression for the current, the Euler-Lagrange equation gives the Maxwell's equations.

Addendum: Suppose there is a magnetic action $$S_{B}\propto\int dx\left[B^{2}\right]=\int dx\left[\left(\nabla\times A\right)^{2}\right]$$ then we calculate easily $$\delta S_{B}\propto2\int dx\left[\left(\nabla\times A\right)\cdot\left(\nabla\times\delta A\right)\right]=2\left[\delta A\times\left(\nabla\times A\right)\right]+2\int dx\left[\left(\nabla\times\nabla\times A\right)\cdot\delta A\right]$$ using an integration by part. The boundary term $\left[\delta A\times\left(\nabla\times A\right)\right]$ disappears by generic argument. Do it component-wise if you have difficulties. It's no more complicated than the demonstration of the Euler-Lagrange equation.


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