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Commutator of L^2 with x_i

  1. Feb 6, 2010 #1
    Task: The task is to compute the commutator of L^2 with all components of the r-vector. It seems to be an unusual task for I was unable to find it in any book.

    Known stuff: I know that [tex][L_i,x_j]=i \hbar \epsilon_{ijk} x_k[/tex] ([tex]\epsilon_{ijk}[/tex] being the Levi-Civita symbol). Now I would go about as follows (summation implied):

    My attempt: [tex][L^2,x_j]=[L_iL_i,x_j]=L_i[L_i,x_j]+[L_i,x_j]L_i=i \hbar \epsilon_{ijk} (L_i x_k + x_k L_i)[/tex]

    Question: Is this correct? Is there a physical meaning to this result other than that they do not commute (with the usual implications)?

    Regards, Matti from Germany
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2
    Yes, it's correct.

    When two operators commute their eigenvectors coincide. In quantum mechanics it means that the corresponding physical magnitudes can be measured simultaneously.
  4. Feb 6, 2010 #3
    Alright, thanks for the reply.

    So, does that mean that the only real information contained in commutators as such (and not their application, as in constructing angular momentum eigenstates) is in it being zero or not? It's clear to me that zero commutators imply instantaneous measurability but I thought maybe there was something to be seen from the "value" that the commutator takes...
  5. Feb 6, 2010 #4
    Not instantaneous measurements, but simultaneous diagonalization--you will be able to find eigenstates of both A and B if [A,B]=0.
    Having a non-zero commutators means the observables cannot be simultaneously known (cf. Heisenberg's uncertainty principle).
  6. Feb 6, 2010 #5
    Sorry, simultaneous measurements is what I was trying to say, of course. Thanks.
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