Why is the supercurrent the functional derivative of the Ginzberg Landau free energy with respect to the vector potential?

In the book "Superconductivity, Superfluids and Condensates", they write down the macroscopic free energy to be

$$F_s(T) = F_n(T) + \int\left(\frac{1}{2m^*}\left|\left(\frac{\hbar}{i}\nabla + 2e \boldsymbol{A}\right)\psi\right|^2 + a|\psi|^2+\frac{b}{2}|\psi|^4 \right) d^3r + \frac{1}{2\mu_0} \int B(\boldsymbol{r})^2d^3r$$

where $$F_n$$ is the free energy of the normal state. Then they claim that the supercurrent can be written by

$$\boldsymbol{j}_s = \frac{\partial F_s}{\partial \boldsymbol{A}(\boldsymbol{r})}$$

My question is, how did you get this equation rigorously? It 'seems' to me somehow that one can vaguely say that the current is always couple to the vector potential in the Hamiltonian. But it looks very hand wavy to me if one wants to explain it like that. Is there any better way to derive this expression?

• Quick hint: Vector potential acts like a source field in partition function and free energy is proportional to logarithm of partition function. Jun 24, 2020 at 19:17

The transparent way is to just take the variational derivative of free energy with respect to $$A$$ and put it equal to zero (NOT equal to current) and see that one will get the Ampere's law for $$B$$ with a current given by the minus of the functional derivative of the first term in the free energy with respect to $$A$$.