6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Calculate the probability current, the result will depend on the vector potential, but the combination is gauge-invariant. $\endgroup$ – Hydro Guy May 7 '13 at 22:27
  • $\begingroup$ Wikipedia cites this question and answer. $\endgroup$ – HDE 226868 Mar 21 '15 at 14:40
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Great prove ! +1 $\endgroup$ – xslittlegrass Mar 6 '14 at 3:39

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.