Degeneracy and commuting observables

[foxjwill]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[White]

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...

 

[foxjwill]

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

You can see what quite fast?

 

[Physics Monkey]

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.

 

[foxjwill]

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.

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?