I think the answer to this question is "no", but the reason is that the question itself is based on an assumption: that kinetic energy is always an observable. I don't think that's true; I think kinetic energy is only an observable for the case of a free particle, for which the energy spectrum is continuous, not discrete, meaning kinetic energy is not quantized. The energy spectrum is discrete for the case of a particle in a bound system, but for such a particle, kinetic energy by itself is not an observable: only total energy is (i.e., including both kinetic and potential energy). So while total energy is quantized (has a discrete spectrum) for this case, you can't say that kinetic energy is, because you can't observe kinetic energy for this case.Is the kinetic energy of a particle quantized?
A charged particle in a magnetic field has quantized kinetic energy. It also has x and y velocity operators which do not commute with other nor their corresponding position operators. Ballentine treats this special case (where the particle is not free but has no potential energy) very readably.Is the kinetic energy of a particle quantized? Does it increase and decrease in discreet amounts?
How about a particle in a square well? The boundary conditions lead to quantized total energies, but the potential is zero.The energy spectrum is discrete for the case of a particle in a bound system, but for such a particle, kinetic energy by itself is not an observable: only total energy is (i.e., including both kinetic and potential energy).
Maybe there's some reason why the kinetic energy operator is not defined in a rigorous sense in that system (as is the case with the momentum operator). But you can get around that problem by considering a "particle on a ring" instead. Then there should be quantized kinetic energy eigenstates.How about a particle in a square well? The boundary conditions lead to quantized total energies, but the potential is zero.
No, the KE operator is well-defined and self-adjoint on the following subset ofMaybe there's some reason why the kinetic energy operator is not defined in a rigorous sense in that system (as is the case with the momentum operator). [...]
Hm, ok, yes, in this case the particle is bound but the total energy is the same as the kinetic energy.How about a particle in a square well? The boundary conditions lead to quantized total energies, but the potential is zero.
Not just the "energy" operator--the kinetic energy operator, since that's specifically what the OP was asking about.Finding the energy operator.