Recent content by Muninn

Ok! That is identical to Ballentine's treatment, only that he works with the velocity operator $$\vec{v} = \frac{d\vec{x}}{dt}$$ instead of introducing the kinematic momentum. Thanks a lot!

Thank you for your excellent and very clear answers, PeroK and vanhees71. This has been an enlightening discussion. Sorry for replying this late. Indeed, but it's still a concept I would like to be more familiar with. Do you know of any textbooks which discuss it? Thanks.

This makes a lot of sense. Just as you're saying, in the classical case we would expect to generate centrifugal terms in the Hamiltonian as we switch to the rotating coordinate system. Then we should expect this to happen also in quantum mechanics. Since the form of the Hamiltonian changes...

Hi! I recently came across a quantum mechanics problem involving a change of basis to a rotating basis. As part of the solution, I wanted to transform the Hamiltonian operator into the rotating basis. Since the new basis is rotating, the basis change operator is time-dependent. This led to a...