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Degeneracy and commuting observables

  1. Dec 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Theorem 5 of a text I've been reading that I downloaded from online (for interested parties, the link (a pdf) is http://bohr.physics.berkeley.edu/classes/221/0708/notes/hilbert.pdf) says that
    "If two observables A and B commute, [A, B] = 0, then any nondegenerate eigenket of A is also an eigenket of B. (It may be a degenerate eigenket of B.)"​

    What I don't understand is how the eigenket could be degenerate in B. I am assuming, I hope correctly, that by "degenerate eigenket" the author means "an eigenket in a degenerate eigenspace". My line of thought is as follows:

    Since [A,B]=0, A and B have the same eigenbasis and therefore the same eigenspaces. But then because the degeneracy of an eigenspace is defined to be its dimension, any degenerate eigenspace of A must also be a degenerate eigenspace of B, and vice-versa.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 30, 2009 #2
    Well, say A is nondegenerate, then B's matrix elements are all diagonal in A's representation

    You can see it quite fast, by sandwiching the commutator between two basekets.

    Well considering A as nondegenerate might not help, because it switches A and B in the problem...
     
    Last edited: Dec 30, 2009
  4. Dec 30, 2009 #3
    You can see what quite fast?
     
  5. Dec 30, 2009 #4

    Physics Monkey

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    Hi foxjwill,

    I'm not totally sure I understand the question and the setup. As you say, the wording is a bit poor, but let me give it a shot.

    I believe what you're looking for can be illustrated with the simple example of a particle moving on a line. Let A = p (momentum) and B = p^2 so that [A,B] = 0. Eigenvalues of A are all non-degenerate, but eigenvalues of B are all two fold degenerate except for p = 0.

    Thus any eigenstate of A belonging to a non-degenerate eigenvalue (in this case, all of them) is an eigenstate of B. However, an eigenstate of B like |p>+|-p> (which is degenerate with |p> - |-p>, for example) is NOT an eigenstate of A, hence the non-degeneracy condition is important.

    Hope this helps.
     
  6. Dec 30, 2009 #5
    That definitely helps. What I think is holding my back is that I had taken "simultaneous eigenbases" to mean "the same eigenbasis". Which brings me to the next question: What does "simultaneous eigenbases" mean?
     
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