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 pr2 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.


1 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.

  • $\begingroup$ If that is indeed the correct operator, then where do you think my book got the one in my question? In the book's context it makes perfect sense, I just don't know how to derive it. $\endgroup$
    – Phantom101
    Commented Aug 18, 2016 at 19:37
  • $\begingroup$ I'm confused. What you cite is not what we usually call the radial momentum operator. Which one is "actually" the physical radial momentum? $\endgroup$
    – knzhou
    Commented Aug 18, 2016 at 19:39
  • $\begingroup$ @knzhou, please have a look at the paper I referred to in my post (the URL is arxiv.org/abs/0706.0924). What do you ``usually'' call the radial momentum operator? More crucially, is that object Hermitian? [I have just corrected a reference in my post to Griffiths's textbook. It should actually be Shankar's textbook.] $\endgroup$ Commented Aug 18, 2016 at 19:45
  • $\begingroup$ @user120233, remember the Laplacian has a radial part $\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)$ (in 3D spherical coordinates). The expression I wrote for $p_r^2$ matches this. Which book are you referring to? $\endgroup$ Commented Aug 18, 2016 at 19:56
  • $\begingroup$ Wow! I learned something new today. I'm pretty sure Griffiths must have glossed over this point entirely, since I don't remember seeing this in there. $\endgroup$
    – knzhou
    Commented Aug 18, 2016 at 20:10

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