Quantum and Commutation - Help me start

1. Apr 28, 2012

Bravus

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. Apr 28, 2012

francesco85

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

3. Apr 28, 2012

Bravus

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?

4. Apr 28, 2012

francesco85

You are welcome; in any case, the correct statement is: let A and B be commuting operators with nondegenerate eigenvalues; then, if $|a\rangle$ is an eigenstate of A, $|a\rangle$ is also an eigenstate of B; moreover, if $|a\rangle$ and $|a'\rangle$ 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

5. Apr 28, 2012

Bravus

Thanks, yes, excellent!