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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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  • $\begingroup$ Calculate the probability current, the result will depend on the vector potential, but the combination is gauge-invariant. $\endgroup$
    – Hydro Guy
    May 7, 2013 at 22:27
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    $\begingroup$ Wikipedia cites this question and answer. $\endgroup$
    – HDE 226868
    Mar 21, 2015 at 14:40

1 Answer 1

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Note that the probability current in the presence of a EM field is given by

$$ \mathbf{j}=\frac{1}{2m}\left(\psi^{*}\mathbf{p}\psi-\psi\mathbf{p}\psi^{*}-2q\mathbf{A}\psi^{*}\psi\right) $$

As you note a local phase shift

$$ \psi^{\prime}=e^{iq\chi(\mathbf{r},t)/\hbar}\psi $$

leads to a gauge transformation of the vector potential

$$ \mathbf{A}^{\prime}=\mathbf{A}+\nabla\chi $$

Substituting these into the expression for the probability current gives

$$\begin{multline} \mathbf{j}^{\prime}=\frac{1}{2m}\left(e^{-iq\chi(\mathbf{r},t)/\hbar}\psi^{*}\mathbf{p}e^{iq\chi(\mathbf{r},t)/\hbar}\psi-e^{iq\chi(\mathbf{r},t)/\hbar}\psi\mathbf{p}e^{-iq\chi(\mathbf{r},t)/\hbar}\psi^{*}\right.\\\left.-2q\mathbf{A}e^{-iq\chi(\mathbf{r},t)/\hbar}\psi^{*}e^{iq\chi(\mathbf{r},t)/\hbar}\psi-2q\nabla\chi(\mathbf{r},t) e^{-iq\chi(\mathbf{r},t)/\hbar}\psi^{*}e^{iq\chi(\mathbf{r},t)/\hbar}\psi\right) \end{multline}$$

Operating with $\mathbf{p}\rightarrow-i\hbar\nabla$ one obtains

$$\begin{multline} \mathbf{j}^{\prime}=\frac{1}{2m}\left(\psi^{*}\left(-i\hbar\right)\nabla\psi+\psi^{*}\psi\frac{iq}{\hbar}(-i\hbar)\nabla\chi(\mathbf{r},t)-\psi\left(-i\hbar\right)\nabla\psi^{*}\right.\\\left.-\psi^{*}\psi\left(-\frac{iq}{\hbar}\right)(-i\hbar)\nabla\chi(\mathbf{r},t)-2q\mathbf{A}\psi^{*}\psi-2q\nabla\chi(\mathbf{r},t)\psi^{*}\psi\right) \end{multline}$$

Sorting it out one obtains

$$ \mathbf{j}^{\prime}=\frac{1}{2m}\left(\psi^{*}\mathbf{p}\psi-\psi\mathbf{p}\psi^{*}-2q\mathbf{A}\psi^{*}\psi\right)=\mathbf{j} $$ thus the probability current is gauge invariant.

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  • $\begingroup$ Great prove ! +1 $\endgroup$ Mar 6, 2014 at 3:39

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