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In electromagnetism there is one vector potential, $\vec{A}$. So the correct way to write the Hamiltonian is $$H = \sum_n \frac{(p_n-q_n \vec{A})^2}{2m_n} + V(\vec{r}_1, \vec{r}_2,...\vec{r}_N)$$ This Hamiltonian is invariant under gauge transformations of the form (apologies if I get a sign wrong) \begin{eqnarray} \Psi_n &\...

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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 ...

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