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Mistake in proof regarding degeneracy property

  1. Apr 11, 2013 #1
    Greetings,

    I was trying to prove a theorem regarding degeneracy, and I succeeded. However, I also proved the converse of the if-then part of the theorem (underlined below), which I know is wrong. I can't spot my mistake though.

    The theorem and my proof are written below - could someone please point out my mistake?

    Theorem:

    Let [A, B]=0 and let A|[itex]\alpha[/itex]>=a|[itex]\alpha[/itex]> (capital letters are operators, lowercase letters are scalars). If the eigenkets of A are not degenerate, then |[itex]\alpha[/itex]> is also an eigenket of B.

    Proof:

    Using the given relations,

    AB|[itex]\alpha[/itex]>=BA|[itex]\alpha[/itex]>=Ba|[itex]\alpha[/itex]>=aB|[itex]\alpha[/itex]>,

    so that |[itex]\alpha[/itex]> and B|[itex]\alpha[/itex]> are both eigenkets of A with eigenvalue a.

    CASE 1 - the theorem: Assume B|[itex]\alpha[/itex]>[itex]\neq[/itex]b|[itex]\alpha[/itex]>. Then B|[itex]\alpha[/itex]> is not writeable as a constant b times |[itex]\alpha[/itex]>, so that |[itex]\alpha[/itex]> and B|[itex]\alpha[/itex]> are degenerate eigenkets of A. Therefore...

    ...If |[itex]\alpha[/itex]> is not an eigenket of B, then the eigenkets of A are degenerate,

    or equivalently,

    ...If the eigenkets of A are not degenerate, then |[itex]\alpha[/itex]> is also an eigenket of B.

    CASE 2 - the converse: Assume B|[itex]\alpha[/itex]>[itex]=[/itex]b|[itex]\alpha[/itex]>. Then the eigenkets of A, |[itex]\alpha[/itex]> and B|[itex]\alpha[/itex]>, are the same except for multiplication by a constant, and so do not qualify as degenerate eigenkets of A. Therefore...

    ...If |[itex]\alpha[/itex]> is an eigenket of B, then the eigenkets of A are not degenerate,

    or equivalently,

    ...If the eigenkets of A are degenerate, then |[itex]\alpha[/itex]> is not an eigenket of B.

    The conclusion I reached in CASE 2 is just wrong. I can think of several examples where it is contradicted. For instance, letting A=J2 and B=Jz, eigenkets of A can be written as |j,mi>. These are degenerate eigenkets of A (one fore each -j\leqmi\leqj, and are simultaneous eigenkets of B.

    What did I do wrong in CASE 2?

    By the way, I am just looking for a description of my mistake, not a proof of something contradicting my conclusion (ie., I know the correct answer, I don't know what's wrong about my wrong answer). It's one of those obnoxious situations where I know proofs and examples that contradict my answer, but I'm not sure what the probably-obvious mistake I made in finding my answer is.

    Actually, at the end of writing this post, it occured to me that my mistake is probably that in CASE 2, I construct eigenkets of A using B and generalize my result to all eigenkets of A, which might not be constructible using B in this manner. I've already got the whole post written, though, so I may as well put it up. Could someone confirm, is that my mistake?

    Thanks for any help you can give.

    -HJ Farnsworth
     
  2. jcsd
  3. Apr 11, 2013 #2
    Yeah, looks to me like the flaw is in case 2. There may be eigenkets of A which cannot be written as [itex]B|\alpha\rangle[/itex], so you can't prove that there is no degeneracy.
     
    Last edited: Apr 11, 2013
  4. Apr 12, 2013 #3
    Great, thanks for the response.
     
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