I always wondered, during my QM courses, if we don't explore enough of the freedom that the Lagrangian and Hamiltonian Classical Dynamics give us.
Classically, we can always make canonical transformations and/or describe a system with a better suited set of variables.
Quantically, though, it's customary to work with operators $(\hat{x},\hat{y},\hat{z})$ and $(\hat{p_x},\hat{p_y},\hat{p_z})$ and only start working with variables $(r,\theta,\phi)$ after taking the projection of the Hamiltonian in the $|\vec{r}\rangle$ basis.

Given the quantization postulates, one could think that its possible to take a classical Hamiltonian such as $$ H=\frac{1}{2m}\left({p_r}^2+\frac{{p_\theta}^2}{r^2}+\frac{{p_\phi}^2}{r^2\sin^2(\theta)}\right)+V(r,\theta,\phi) $$ use the Poisson brackets to determine $$ [\hat{r},\hat{P_r}]=i\hbar\hat1\\ [\hat{\theta},\hat{P_\theta}]=i\hbar\hat1\\ [\hat{\phi},\hat{P_\phi}]=i\hbar\hat1 $$ and then use the "translation" operators to get $$ e^{-i\frac{\hat{P_r}}{\hbar}r'}|r\rangle=|r+r'\rangle\\ ... $$ This should make possible to determine the representation of the generalized momenta in the $|r,\theta,\phi\rangle$ basis. After some concern about the limits of $|\theta\rangle$ and $|\phi\rangle$ this would look like a good formulation to directly work with some problems.

I haven't developed this fully, but is this a possible framework to work with QM? If so, does anyone know how to exactly deal with this? Am I wrong in any assumptions?

  • $\begingroup$ When dealing with states in QM, you are not bound to a basis. In fact, many problems (e.g. central force) are solved in spherical coordinates. I don't see the point of your question. $\endgroup$ – noah May 17 '17 at 19:22
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    $\begingroup$ Cf. e.g. this answer and the mentioned exercise in Shankar - changing to non-Cartesian coordinates leads to "naive" canonical quantization (just replacing Poisson brackets with commutators) failing, see also physics.stackexchange.com/q/216226/50583, $\endgroup$ – ACuriousMind May 17 '17 at 19:25
  • $\begingroup$ @noah , yes! that's the point. If we are not bound to a basis why shouldn't we quantize the problem in any basis we'd like? $\endgroup$ – João Vitor Ignácio Costa May 17 '17 at 19:30
  • $\begingroup$ @ACuriousMind , thanks for the answer so much, that's been bugging me for ages. $\endgroup$ – João Vitor Ignácio Costa May 17 '17 at 19:31

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