Mistake in proof regarding degeneracy property

[HJ Farnsworth]

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=J_z, eigenkets of A can be written as |j,m_i>. These are degenerate eigenkets of A (one fore each -j\leqm_i\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 occurred 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

 

[Chopin]

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.

 

[HJ Farnsworth]

Great, thanks for the response.