We define the free energy functional in the Ginzburg-Landau theory as

$$ \mathcal{F} = F_n + \alpha |\psi|^2 + \frac{\beta}{2}|\psi|^4 + \frac{1}{2m^*}\big|(-i \hbar \nabla + e^*\mathbf{A})\psi \big |^2 + \frac{|\mathbf{B}|^2}{2 \mu_0}, $$


  • $\psi$ is a macroscopic wavefunction,
  • $\alpha$ and $\beta$ are phenomenological parameters,
  • $m^*$ is the effective mass of the charge carrier,
  • $e^*$ is the effective charge of the charge carrier,
  • $\mathbf{A}$ is a vector potential.

Moreover, we assume a static magnetic field, which leads to

$$ \nabla \times \mathbf{B} = \mu_0 \mathbf{j}. $$

In the book that I'm reading it's said, that the minimization of the free energy with respect to the vector potential

$$ \frac{d \mathcal{F}}{dA} = 0 $$

leads to the following equation for the dissipation-less electric current density

$$ \mathbf{j} = \frac{e^*}{2m^*} [\psi^*(-i \hbar \nabla - e^* \mathbf{A})\psi + c.c].$$

Could you explain please, how is it derived?


1 Answer 1


You use the usual Euler-Lagrange equations using calculus of variation on the free energy functional: $$ F = \int \mathcal F d^3x $$ which gives: $$ \frac{\delta F}{\delta A} := \frac{\partial\mathcal F}{\partial A_i} -\partial_j\frac{\partial\mathcal F}{\partial (\partial_jA_i)}\\ \frac{\delta F}{\delta A} = 0 $$

You have three contributions. Using $B = \nabla\times A$, the "harmonic" contribution is: $$ \frac{\delta}{\delta A}\int \frac{1}{2\mu_0}B^2 d^3x = \frac{1}{\mu_0}\nabla\times B $$ the first terms give a trivial contribution: $$ \frac{\delta}{\delta A}\int \left(\mathcal F_n+\alpha|\psi| 2+\frac{\beta}{2}|\psi|^4\right)d^3x = 0 $$ and the final interacting term gives your current: $$ \frac{\delta}{\delta A}\int \frac{1}{2m^*}\left|(-i\hbar\nabla +e^*A)\psi\right|^2d^3x = -\frac{e^*}{2m^*}(\psi^*(-i\hbar\nabla +e^*A)\psi+h.c.) $$ so you do get: $$ \nabla \times B = \mu_0 j\\ j = \frac{e^*}{2m^*}(\psi^*(-i\hbar\nabla +e^*A)\psi+h.c.) $$

Hope this helps.


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