Direct Hamiltonian Operator computation in polar coordinate

Question

Suppose a classical Hamiltonian of the form $$\mathcal{H}=\frac{1}{2m}(p^2_x+p^2_y)+a(x^2+y^2)^{1/2}$$ We know that this change to following quantum operator

$$\hat{H}\rightarrow -\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial x^2}\right)+a(x^2+y^2)^{1/2}$$ in coordinate basis. But due to rotational symmetry I want to work in polar coordinate for which I need to do the transformation from one coordinate to another.

But why I can not do the following directly from starting from Hamiltonian to be in polar coordinate. So that $$\mathcal{H}=\frac{p^2_\rho}{2m}+\frac{p^2_\phi}{2m\rho^2}+a\rho$$ and then use the directly use operators from here: $$\hat{P}_\rho\rightarrow -i\hbar \frac{\partial}{\partial \rho} $$ $$\hat{P}_\phi\rightarrow -i\hbar \frac{\partial}{\partial \phi} $$

as theny follow the canonical commutation rules but on doing these two both leads to different Hamiltonian operators. What's wrong with this reasoning?

Answer 1

I'll elide your question, betting you'd prefer to bypass its subtleties: I have linked one question on curvilinear coordinate quantization, and you'd find lots of answers on the subject. I think it's more important to recall the drill, employed, e.g., in spherical coordinates for the Hydrogen atom.

That is, eschew interpretation, and convert directly your fine hamiltonian $$\hat{H} = -\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)+a(x^2+y^2)^{1/2}$$to polar coordinates, $$\hat{H}= -\frac{\hbar^2}{2m}\left( \frac{1}{r} \partial_r ~r ~\partial_r +\frac{1}{r^2} \partial_\phi ^2\right)+a r,$$ for which you may find factorized eigenfunctions, etc.

Recall the symmetry operator commuting with your hamiltonian is $L= \partial_\phi$.

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NB. (Geeky) The polar variables' measure $r~dr ~d\phi$ is not flat, so your proposed radial momentum operator is not hermitian. You may consider the hermitian operator $\Pi= -i\hbar(\partial_r +1/2r)$, instead, inadvisably, in which case $ \Pi^2= -\hbar^2((1/r) \partial_r ~r ~\partial_r-1/4r^2)$.