In Quantum Mechanics, there seem to exist two types of momenta, namely kinematic momentum and canonical momentum, both of which are $$ p(t) = \mathscr{U}^\dagger i\hbar \frac{\partial}{\partial x}\mathscr{U} \\ \Pi(t) = m\frac{d{x}}{dt} = \frac{m}{i\hbar} \left[x(t),H(t)\right] $$ in the Heisenberg picture. From my understanding, the kinematic momentum is the "physical momentum" and the canonical momentum is the generator of infinitesimal translations (as well as being the conjugate to position with Lagrangians, but that doesn't seem applicable in this context).

Both of these are observables, due to being self-adjoint operators, so they would seem to be mensurable in a laboratory. However, while $\langle\Pi\rangle$ is gauge invariant, $\langle p\rangle$ is not, which is not something that I would expect from a physically observable quantity.

The question is, therefore, what is the unambiguous physical interpretation of the expectation value of $p$? Can it truly be measured in a lab? If not, is there any other insight that can be obtained from $\langle p\rangle$?

  • 1
    $\begingroup$ How can it be a physically observable quantity if it's not gauge invariant? $\endgroup$
    – fqq
    Jul 17 '19 at 20:55
  • $\begingroup$ @fqq Energy of a system is also not invariant under the transformation $V\to V+c$, but why is the energy a physcially observable quantity then? $\endgroup$
    – suncup224
    Jul 11 '21 at 4:43
  • $\begingroup$ @suncup224 Ah right, so not every physical quantity has to be gauge invariant, but then how do physical measuring devices choose a gauge? $\endgroup$
    – John K
    Jul 11 '21 at 19:26
  • $\begingroup$ Only differences of energy can be measured. $\endgroup$
    – fqq
    Jul 12 '21 at 7:41

Your Answer

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

Browse other questions tagged or ask your own question.