# Supercurrent phase and gauge change: why a specific choice for the quantum phase?

I am following the following lecture notes: http://web.mit.edu/6.763/www/FT03/Lectures/Lecture9.pdf

In the last slide, we see how a gauge change for the EM field impact the phase of the wavefunction. I remind:

$$\psi(x,t)=\sqrt{n(x,t)}e^{i \theta(x,t)}$$ $$\mathbf{J}=qn(x,t) \left( \frac{\hbar}{m} \mathbf{\nabla}(\theta(x,t))-\frac{q}{m} A(x,t) \right)$$

If we change the E.M potential like the following, the physical description is the same:

$$A'= A+\mathbf{\nabla} \chi$$ $$\phi'= \phi - \frac{\partial \chi}{\partial t}$$

Thus in the Gauge "prime", the physical quantities are the same: $$n(x,t)=n'(x,t)$$ and $$\mathbf{J}=\mathbf{J'}$$.

From those equalities, we find:

$$qn(x,t)\left( \frac{\hbar}{m} \mathbf{\nabla}(\theta'(x,t))-\frac{q}{m} A'(x,t) \right)=qn(x,t)\left( \frac{\hbar}{m} \mathbf{\nabla}(\theta(x,t))-\frac{q}{m} A(x,t) \right)$$

It implies:

$$\mathbf{\nabla}(\theta'-\theta-\frac{q}{\hbar} \chi)=0$$

A particular solution for this is: $$\theta'=\theta+\frac{q}{\hbar} \chi$$, but we could expect other. Why is it this particular solution that is only considered in the slides ?

Consider first a time-independent gauge transformation. Then the vanishing of the gradient implies only that $$\theta'-\theta+\frac{q}{\hbar}\chi= {\rm constant}.$$ But we also know that $$\theta-\theta'=0$$ if $$\chi=0$$. Therefore the constant is zero. For a time dependent gauge transformation, one needs to include the gauge covariance of the Josephson acceleration equation to make sure that the "constant" is also time independent