# Calculating the derivative of the free energy over the vector potential in the Ginzburg-Landau theory

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

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

where:

• $$\psi$$ 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?

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: $$\nabla \times B = \mu_0 j\\ j = \frac{e^*}{2m^*}(\psi^*(-i\hbar\nabla +e^*A)\psi+h.c.)$$