I have seen explanations that canonical momentum for charged particles $p = mv + qA/c$ is not a measurable quantity/observable because it is not gauge invariant. However, there are many quantities that also depend on an "arbitrary choice", for example even the Hamiltonian (which corresponds to the energy observable) involves an arbitrary choice of where the zero point of the potential energy is.

What's the difference between the two quantities? One might further argue that "ok, it is the change in energy that is observable, not absolute energy" -> In this case, can I not look at change in canoncial momentum? The arbitrary choice in the gauge transformation $\nabla f$ will similarly "cancel out" when I look at change in canonical momentum.

  • 2
    $\begingroup$ I would certainly be among those saying that you can't observe absolute energy, only changes in energy. Those changes are invariant under shifting the Hamiltonian (or, perhaps more relevant, the Lagrangian) by a constant. In contrast, the change in the canonical momentum would retain a gauge-dependent term, right? That seems like a pretty big difference to me. $\endgroup$
    – David
    Jul 11, 2021 at 5:14
  • $\begingroup$ I see, the $\nabla f$ term will not be identical in the initial and final momentum because $\nabla f$ depends on $r$ and $t$, correct? $\endgroup$
    – suncup224
    Jul 11, 2021 at 9:52
  • $\begingroup$ The Hamiltonian is not necessarily an "observable" in the sense of being meaningfully measurable either, so the question is ill-formed - there is no difference, the Hamiltonian is not always the total energy, see e.g. physics.stackexchange.com/q/194772/50583 $\endgroup$
    – ACuriousMind
    Jul 11, 2021 at 10:54
  • $\begingroup$ Physical observables in classical electromagnetism should be in a way "covariant", like $E$,$B$ fields, after all this is a relativistic theory. But EM potentials satisfying Maxwell's eqn may not be covariant themselves, it depends on choice of gauge (see Griffiths for more info ). It's just that this particular combination of EM potentials giving rise to $E$ and $B$ is covariant, which is also gauge invariant. So all observables should respect gauge invariance. The canonical momentum $p$ is not gauge invariant, so is the associated Hamiltonian ... hence neither of them are observables $\endgroup$
    – KP99
    Jul 11, 2021 at 15:59
  • $\begingroup$ You have to be careful with the language here. The momentum of a single quantum (what you call "charged particle") is perfectly well defined. Only the momentum of the ensemble of these quanta isn't because you are not in a momentum conserving potential. This is no different from classical mechanics, though. The momentum of a beam of charged particles in a magnetic field is not well defined. The momentum (density) of the beam in any one infinitesimal volume elements, however, may be. $\endgroup$ May 28 at 19:15

1 Answer 1


The "change in canonical momentum" is not gauge-invariant.

The $A$ in $p(x,\dot{x},t) = m\dot{x} + qA(x,t)$ is a function, not a constant, just like the $\nabla f$ in the gauge transformation is a function $f(x,t)$. So if you have two canonical momenta, their difference $$ p(x_1,\dot{x}_1, t_1) - p(x_2, \dot{x}_2,t_2) = m(\dot{x}_1 - \dot{x}_2) + q(A(x_1,t_1) - A(x_2, t_2)) + q (\nabla f(x_1,t_1) - \nabla f(x_2,t_2))$$ is in general not the same under choices of different $f$ - the first two summands don't change, but the latter is essentially arbitrary.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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