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Commutator [x, p^2]

  1. Sep 10, 2008 #1
    i've been trying to evaluate this commutator the 'easy' way--that is, without using the definition of the momentum operator. the farthest i got was trying to use this rule..

    [A, BC] = [A, B]C + B[A, C]

    [x, p^2] = [x, p]p + p[x, p]

    so i guess i get 2ihp. but that doesn't make sense, b/c there's an operator in that result. so i don't get what else i'm supposed to do. can anyone help me out?
  2. jcsd
  3. Sep 10, 2008 #2

    James R

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    There's no reason you shouldn't have an operator in the result.

    What you have done is fine.
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