Without electromagnetic coupling, the QM charged particle wave function is not invariant under a local gauge transformation — one with a phase that depends on space (space-time):

\begin{equation} \psi \mapsto e^{-i\alpha(x)}\phi \end{equation}

In order to turn on the electromagnetic coupling, we make the replacement

$$\hat{\vec{p}} \mapsto \hat{\vec{p}} - \frac{ie}{c}\vec{A}$$

This means that, in the Schrodinger equation is gauge invariant because the term arising from $\vec{A}$ is cancelled through the action of $\hat{\vec{p}} = i\hbar\partial$ on the phase factor $e^{-i\alpha(x)}$ in the wave function.

The math trick is clear. But physics cannot be understood from it, namely, why the requirement of local gauge invariance requires 1) the charge conservation 2) and the emergence of some kind of carrier of interaction.

  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/266992, physics.stackexchange.com/q/13870 $\endgroup$ Commented Jan 18, 2021 at 17:00
  • $\begingroup$ There is no physical meaning at all. You can do without electromagnetic gauge invariance. Read my peer reviewed paper at arxiv.org/abs/physics/0106078. $\endgroup$
    – my2cts
    Commented Jan 18, 2021 at 17:47
  • $\begingroup$ A good physical explanation I also found here feynmanlectures.caltech.edu/III_07.html Here the case is considered when the gauge occurs not at every point in space, but in two regions of space. It seems to me that such a simplified approach is precisely the physical minimum from which to move on. $\endgroup$
    – Sergio
    Commented Jan 18, 2021 at 18:38
  • $\begingroup$ Related: physics.stackexchange.com/q/48305/2451 $\endgroup$
    – Qmechanic
    Commented Jan 18, 2021 at 19:50
  • $\begingroup$ @Qmechanic Unfortunately, the Noether's theorem is also math trick. As I wrote in comment, good physical explanation I saw in Faynman Lectures on Physics. All mathematical tricks does not explain what is phase of wave function, remain it as abstract value. $\endgroup$
    – Sergio
    Commented Jan 18, 2021 at 19:53

1 Answer 1


Perhaps the most well known example of gauge invariance might shine some light on this: the potential. If you have the gravitational potential $U(r)=-\frac{GM}{r}$ then $\tilde U(r)=-\frac{GM}{r}+C$ is just as good a potential. They provide the same equations of motion so they are both equally valid. So which constant do we pick for $C$? The one that's most convenient. For the gravitational potential it turns out $C=0$ is the most convenient choice. For Maxwell's equations there are a couple gauges each with their own benefits as can be seen by looking at the wikipedia page on gauge fixing.

So a Mathematical trick would be the best description from your options, but it's not a trick. It's excessive mathematical freedom that we have to get rid off before we can do calculations. It's kind of similar to an integration constant.

They are not just an annoyance though. In theoretical physics gauge invariance occurs often and is sometimes even necessary to get a good theory. For example gauge bosons, which describe forces, arise from the gauge invariance of some interaction. The photon is the gauge boson of the electromagnetic interaction.


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