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In gauge theories such as QED or QCD, are the rest energies, and therefore the invariant masses, of various bound states, such as those of positronium or charmonium, gauge-invariant observable quantities?

It seems obvious to me that they must be, but I need confirmation from reliable experts in quantum field theory to refute arguments in other threads that energy is in general gauge-dependent and that only energy differences are observable.

If these energies are not gauge-invariant observables, then I would like to understand why not.

For context, the arguments have been in the following threads:

The reasons I have for believing that rest energies of bound states are gauge-independent observables are:

  • It is common to talk about the invariant mass of such states, implying well-defined-ness, gauge-invariance, and observability of the rest-energy.

  • There are well-known gauge-invariant energy-momentum-stress tensors for gauge field theories.

  • Energy curves spacetime and therefore is obviously well-defined, gauge-invariant, and observable.

Once we settle how this works in full quantum field theories, I will post a follow-on question about the gauge-invariance of energies of bound states in non-relativistic quantum mechanics, where I think there are more subtleties since the electromagnetic field isn't quantized.

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  • $\begingroup$ As was explained to you in several of those other threads, masses are gauge invariant $\textit{energy differences}$. The mass of the electron is (half) the gap between the vacuum (no particles) and the first excited state (an $e^-e^+$ pair). $\endgroup$ – user213887 Dec 7 '18 at 5:51
  • $\begingroup$ @JulianIngham When I apply a known force to an electron and observe how it moves, I can measure its invariant mass and its rest energy without having to consider any energy differences involving the quantum vacuum. So your statement in one of the threads that “Energy is not an observable quantity” makes no sense to me. Particle physicists observe the energy of particles every day. Fundamental relativistic relationships like $m^2 = E^2 - p^2$ deal with energies, not energy differences. I’m not going to continue arguing wth you; we’ve done enough of that. I want to hear from new voices. $\endgroup$ – G. Smith Dec 7 '18 at 6:56
  • $\begingroup$ And, yes, I took quantum field theory as part of my Ph.D. If “energy is not an observable” got taught, I must have been sick that day. $\endgroup$ – G. Smith Dec 7 '18 at 7:01
  • $\begingroup$ Then I don't understand why you're asking if mass is gauge invariant since we agree. "When I apply a known force to an electron and observe how it moves, I can measure its invariant mass and its rest energy without having to consider any energy differences involving the quantum vacuum." Like, yeah - that's another energy difference, i.e. the energy difference between when an electron moves and when it doesn't. $m^2=E^2-p^2$, again, is a dispersion relation - it's not the energy of the many body state that defines the for e.g. vacuum. $\endgroup$ – user213887 Dec 7 '18 at 15:46
  • $\begingroup$ You agreed it is $p_\mu-eA_\mu$ that's gauge invariant, but then $p_\mu$ (and in particular $p_0$) cannot be gauge invariant unless you think $A_\mu$ is gauge invariant when it absolutely obviously is not. You must have been sick for a few classes - not just QFT but a few QM and probably classical mech classes too. $\endgroup$ – user213887 Dec 7 '18 at 15:48

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