It is simple to show that under the gauge transformation $$\begin{cases}\vec A\to\vec A+\nabla\chi\\ \phi\to\phi-\frac{\partial \chi}{\partial t}\\ \psi\to \psi \exp\left(\frac{iq\chi}{\hbar}\right)\end{cases}$$ The Schrodinger equation $$\left[-\frac{\hbar^2}{2m}\left(\nabla-\frac{iq\vec A}{\hbar}\right)^2+q\phi \right]\psi=i\hbar\frac{\partial}{\partial t}\psi$$ gives back the same equation.
How does it follow that the probability current is gauge invariant?
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