Phase gradient, vector potential and current in a superconductor For a superconductor we may write a wave function $\psi$ as a function of a density $\rho(\vec{r})$ of particles as
$$\psi(\vec{r})=\rho^{1/2}(\vec{r})\,e^{i\theta} \, .$$
This leads to the probability density current
$$\vec{J}=\frac{\hbar}{m} \left(\nabla\theta-\frac{q}{\hbar}\vec{A} \right)\rho \, ,$$
which, as I understood from reading Vol.3, Chap. 24 of the Feynman Lectures, can be seen as the wave function of all cooper pairs and the actual charge current in the material.
I've been struggling to understand how this same expression for the current leads to various different results:


*

*If we want to deduce flux quantization, the current equation should be integrated in a closed curve around the magnetic flux in which there is no current, equating phase change over the closed curve and the magnectic flux itself by the integral of $\vec{A}$.

*In order to deduce the second London equation we must assume $\nabla\theta=0$. How can we justify this, considering that phase gradient is fundamental for explaining flux quantizaion? We needed $\nabla\theta$ before and now it is $=0$. What is different in this situation?

*If we assume London equation is correct in the case of flux quantization, there should be current all around every vortex (for the whole bulk), where there is non-zero vector potential. Why is this reasoning wrong?

*In a Josephson junction, current is related to phase difference between the two superconductors, again we need $\nabla\theta$, however London equations tell us current should be only related to vector potential $\vec{A}$. How can we make sense of this?

 A: The main point here is that the London equation in the form $\vec J(\vec r)=-\frac{\hbar}{m}\rho(\vec r)\vec A(\vec r)$ is not gauge invariant, hence it is valid only in a specific gauge, where you can set an homogeneous electric potential $\nabla\phi(\vec r)=0$. This is possible when you consider a simply connected superconductor domain in a magnetic field, but it is inappropriate when you describe multi-connected domains, like superconducting rings. In the latter case, you have indeed to use the more general expression
$\vec J(\vec r)=\frac{\hbar}{m}(\nabla\theta(\vec r)-\vec A(\vec r))\rho(\vec r)$, which is gauge invariant. Similarly, I guess that you cannot describe Josephson effect in this gauge, because you cannot take a uniform field $\phi(\vec r)$, due to the phase gradient across the junction. If you want a more detailed and unified presentation, you can look at this link https://www.wmi.badw.de/teaching/Lecturenotes/SLTTP%20I/SL_Chapter%203_2014.pdf. If you want more details about the specific questions, we can discuss about it. However I think that the origin of your confusion is mostly related to this gauge fixing.
