Derivation of the radial momentum operator

Question

I have been studying Quantum Mechanics and when my book was going through the Hydrogen wave equation, it was talking about this equation: $$ \frac{p_r^2}{2\mu} +\frac{L^2}{2\mu r^2}+V(r)=E$$ I completely understand how they got to this formula, but then they said they could use the above equation to write Schrodinger's equation for spherical coordinates using de Brogie's relations and appropriate operators. My book then said finding said operators is a "lengthy though not particularly difficult exercise." It then said that the appropriate operator for p_r² is the next equation: $$(p_r^2)_{op}=-\hbar^2\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})$$ I am fairly new to operators so I tried to derive this operator. My result was: $$(p_r^2)_{op}=-\hbar^2 r\frac{\partial}{\partial r}(r\frac{\partial}{\partial r})$$ I realize this result is wrong, and from reading other forums, I'm pretty sure I got the above result from finding the radial component of the momentum operator and then just operating on itself or in essence "squaring" it. This leaves me completely lost on how to get the correct equation from my book. If somebody could help me understand how I can derive the proper operator that would be great.

Answer 1

The correct radial momentum operator is in fact

$$p_{r} = -i\hbar\left(\frac{\partial}{\partial r} + \frac{1}{r}\right)$$

This is hermitian. And as you rightly point out, squaring it does indeed yield the radial part of the Laplacian (times $-\hbar^2$):

$$p_r^2 \psi = -\hbar^2\left(\frac{\partial}{\partial r} + \frac{1}{r}\right)\left(\frac{\partial\psi}{\partial r} + \frac{\psi}{r}\right) = -\hbar^2\left(\frac{\partial^2\psi}{\partial r^2} + \frac{2}{r}\frac{\partial\psi}{\partial r}\right) $$

You might find this related post useful. The derivation of the above radial momentum operator can be found in this paper, though I should add that this particular result is well known (see for instance, the QM textbook by R. Shankar, e.g. the excerpt at https://books.google.com/books?id=sDvrBwAAQBAJ&lpg=PP1&pg=PA216#v=onepage&q=radial%20momentum&f=false).

Edit: The point here is that the ''usual'' operator $-i\hbar\frac{\partial}{\partial r}$ is not hermitian.