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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 ?

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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

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  • $\begingroup$ Hello. Thank you for your answer. I totally agree with the first part (time independant). It mostly answers my question. However I am not sure to understand your explanation for the "time dependance". Could you elaborate more ? Why doesn't the first answer works for time dependance ? I'm not sure to follow. Thanks ! $\endgroup$
    – StarBucK
    Jul 9 '20 at 20:20

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