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Why within classical field-theory the electromagnetic four-potential (usually $A_{\mu}$) not an observable?

In classical mechanics we don't have problems with energy measurements and in quantum mechanics we talk all the time about the Hamiltonian which "is the observable of energy".

So why is $A_{\mu}$ not also considered as an observable? If we don't have a smart way to measure it how can we be 100% certain that tomorrow some dude won't figure out how to do it?

This confuses me greatly, especially with respect to the Aharonov–Bohm effect which is within quantum mechanics "a way to measure $A$".

The non-observable nature of $A$ seems to be the reason why we "gauge" it in order to do all kinds of stuff. That is why I'm interested in it.

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The four-potential is not an observable because it is not invariant under a change of gauge. And no predictions of any physical theory are dependent on the choice of gauge, so the four-potential is not observable.

What is gauge invariant and observable is the integral of the four potential around a loop, and that is what is observed in the AB effect. However, it should be noted that the AB effect can be explained entirely in terms of local action by fields. As such, there is no reason to invoke the four potential to explain what's going on. The potential may sometimes be useful for doing calculations, but that is a different issue from whether it is an observable.

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  • $\begingroup$ Maybe this should be a new question, but are differences in the four potential observable in the sense that differences in potential energy are observable? $\endgroup$ – John Rennie Jun 17 '15 at 10:51
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    $\begingroup$ The difference in potential between two points will be dependent on the gauge. The amount of energy required to move a charge between two points will not be gauge dependent. There may be choices of gauge that look natural in a particular situation. For example, in electrostatics it may be useful for calculation to use the Coulomb gauge: physics.stackexchange.com/a/73344/28512. $\endgroup$ – alanf Jun 17 '15 at 11:35
  • $\begingroup$ I thought the reasoning is that is is not an observable and thus we can gauge it. If you reason that you can gauge it and thus it is not an observable how do you justify your "gauging"? $\endgroup$ – Thomas Elliot Jun 17 '15 at 16:43
  • $\begingroup$ No. Any given theory implies that some things are observable and others are not. In this particular case, the quantity in question can't be observable because there is a family of vectors satisfying the same constraints for any given solution, so there is no physical way to distinguish between them and you can't say you're observing one of that family rather than another. There is no way to decide that X is observable without an explanation of why it is not observable. $\endgroup$ – alanf Jun 18 '15 at 9:52

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