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I Commutator of p and x/r

  1. Dec 8, 2017 #1
    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?
     
  2. jcsd
  3. Dec 9, 2017 #2
    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.
     
    Last edited by a moderator: Dec 9, 2017
  4. Dec 9, 2017 #3
    Yeah, it seems I typed out the question a bit too quickly :P

    Thanks!
     
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