I Commutator of p and x/r: Elegant Derivation in Position Basis

[VKint]

This question came up in this thread: <https://www.physicsforums.com/threa...ydrogen-atom-hamiltonian.933842/#post-5898454>

In the course of answering the OP's question, I came across the commutator
$$ \left[ p_k, \frac{x_k}{r} \right] $$
where $r = (x_1 + x_2 + x_3)^{1/2}$ and $p_k$ is the momentum operator conjugate to $x_k$. It's easy to show that the commutator is
$$ -i \hbar \left( \frac{1}{r} - \frac{x_k^2}{r^3} \right) $$
by working in the position basis. My question is: Is there a more elegant way (i.e., independent of basis) of deriving this commutator?

 

[thephystudent]

It seems to me that some squares are missing in the definition of r.

apart from that, a useful relation can be used

if $[A,B]=C$ and $[A,C]=[B,C]=0$ then
$$[A,f(B)]=f'(B).C$$

This can be proven from filling in the series expansion of f.

 

[VKint]

Yeah, it seems I typed out the question a bit too quickly :P

Thanks!