1
$\begingroup$

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?

$\endgroup$
2
  • 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 '15 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 '15 at 22:29
2
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.