Quantum and Commutation - Help me start

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Homework Help Overview

The discussion revolves around a problem in quantum mechanics related to the commutation of operators A, B, and C. The original poster expresses difficulty in starting the problem and seeks guidance on how to approach it, particularly regarding the implications of operator commutation.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of operator commutation, particularly the existence of simultaneous eigenstates. There is a discussion on the conditions under which the original poster's proposed solution may hold true, including the need for nondegenerate eigenvalues.

Discussion Status

Some participants have provided clarifications regarding the relationship between commuting operators and their eigenstates. The original poster has acknowledged understanding some points but still expresses uncertainty about specific details. The conversation is ongoing, with multiple interpretations being explored.

Contextual Notes

The original poster indicates a desire for guidance rather than a complete solution, suggesting a focus on understanding the foundational concepts rather than resolving the problem outright.

Bravus
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Homework Statement



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Homework Equations



See below

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
 
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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.

Francesco
 
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?
 
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
 
Thanks, yes, excellent!
 

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