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Quantum and Commutation - Help me start

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data

    http://www.bravus.com/question.jpg [Broken]

    2. Relevant equations

    See below

    3. The attempt at a solution

    Below are my scribbles toward a solution. The point is that the two expressions are different *unless* either the operators A and B or the operators B and C commute.

    Not really looking for someone to solve it completely, but this is one of those questions I'm just looking at blankly to figure out how to start... yeah, I suspect I'm not smart... ;-)

    http://www.bravus.com/scribbles.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 28, 2012 #2
    Hi! In my opinion it is easier to proceed in the following way: consider the case in which A and B commute (the other case is analogous); in this case there exists a common set of eigenstates; this induce in the formula you want to prove a lot fo deltas (eigenstates corresponding to different eigenvalues are orthogonal); this equation can now be easily verified.
    Best,
    Francesco
     
  4. Apr 28, 2012 #3
    Thanks, Francesco. That kinda makes sense to me. But I'm a little unclear.

    The fact that A and B commute means there *exist* simultaneous eigenkets, |ab> of both A and B.

    I don't think that means, though, that |a> and |b> are *necessarily* all simultaneous eigenkets.

    And they would need to be, for the solution you propose to work, yes?
     
  5. Apr 28, 2012 #4
    You are welcome; in any case, the correct statement is: let A and B be commuting operators with nondegenerate eigenvalues; then, if [itex]|a\rangle[/itex] is an eigenstate of A, [itex]|a\rangle[/itex] is also an eigenstate of B; moreover, if [itex]|a\rangle[/itex] and [itex]|a'\rangle[/itex] are two eigenstates of A corresponding to different eigenvalues, then they are eigenstates of B corresponding to different eigenvalues.
    I hope I have been clear.
    Francesco
     
  6. Apr 28, 2012 #5
    Thanks, yes, excellent!
     
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