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Show operator can be an eigenfunction of another operator given commutation relation

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that two operators P and Q satisfy the commutation relation: [P,Q]=P. Suppose that psi is an eigenfunction of the operator P with eigenvalue p. Show that Qpsi is also an eigenfunction of P, and find its eigenvalue.


    2. Relevant equations



    3. The attempt at a solution
    First off, I know that if psi is an eigenfunction of P it means that P(psi)=p*psi. If Qpsi is also an eigenfunction of P it means that P(Qpsi)=q*Qpsi. p and q would be the eigenvalues. I also know that I have to use the commutation relation to manipulate these two equations. What I don't understand is how [P,Q] can equal Q. I thought [P,Q]=PQ-QP=0 if the two operators commute.
     
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  3. Oct 16, 2008 #2

    Hootenanny

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    Re: Show operator can be an eigenfunction of another operator given commutation relat

    The question doesn't say that P & Q commute does it?
     
  4. Oct 16, 2008 #3
    Re: Show operator can be an eigenfunction of another operator given commutation relat

    Correction: The two operators P and Q satisfy the commutation relation [P,Q]=Q.
    It doesn't say that they commute but that they satisy the relation. How else can they satisfy the relation if they don't commute?
     
  5. Oct 16, 2008 #4

    Hootenanny

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    Re: Show operator can be an eigenfunction of another operator given commutation relat

    What condition must two operators satisfy to be said to commute?

    (HINT: You said it yourself in your first post)

    Edit: Perhaps I'm being a little too cryptic here. My point was merely that to commute P and Q must satisfy [P,Q] = 0, since they don't they do not commute. However, does because they do not commute doesn't mean they cannot satisfy a general commutation relation.

    Does that make sense?
     
    Last edited: Oct 16, 2008
  6. Oct 16, 2008 #5
    Re: Show operator can be an eigenfunction of another operator given commutation relat

    So P and Q satisfy the given relation...this means that PQ-QP=Q? Is this the correct expression I should be using to evaluate the eigenvalues?
     
  7. Oct 16, 2008 #6

    Avodyne

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    Re: Show operator can be an eigenfunction of another operator given commutation relat

    Yep. (Except that, in your first post, you say [P,Q]=P, not Q; you switched to Q in a later post ...)
     
  8. Nov 16, 2010 #7
    Re: Show operator can be an eigenfunction of another operator given commutation relat

    Hey I'm working on the same problem. Are you saying that Q=0? I don't understand why P and Q 'must' commute to 0.
     
  9. Nov 16, 2010 #8

    vela

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    Re: Show operator can be an eigenfunction of another operator given commutation relat

    They don't. If P and Q commute, that means [P,Q]=0. You're given that [P,Q]=P (or [P,Q]=Q), so P and Q obviously don't commute.
     
  10. Nov 16, 2010 #9
    Re: Show operator can be an eigenfunction of another operator given commutation relat

    Ok so here's my thinking:

    Let's say Y is Psi--

    [P,Q] = PQ - QP = Q
    = PQY - QPY = QY plug in (PY=pY)
    = PQY - QpY = QY
    PQY = QY + QpY

    is the eigenvalue of QY then QY + QpY? i'm pretty sure the answer to that question is no, but I don't know where to go from here.
     
  11. Nov 16, 2010 #10

    vela

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    Re: Show operator can be an eigenfunction of another operator given commutation relat

    The eigenvalue p is just a number, so it commutes with Q in the last term. Then you can factor QY out on the RHS of the equation.
     
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