Quantum and Commutation - Help me start

In summary, the question is asking how to prove a formula involving operators A, B, and C. The solution involves considering the case where A and B commute and using eigenstates to easily verify the formula. It is also noted that in this case, there must exist simultaneous eigenstates of A and B for the solution to work. The correct statement is that if A and B are commuting operators with nondegenerate eigenvalues, then any eigenstate of A is also an eigenstate of B, and any two eigenstates of A with different eigenvalues are also eigenstates of B with different eigenvalues.
  • #1
Bravus
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0

Homework Statement



http://www.bravus.com/question.jpg

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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  • #2
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
 
  • #3
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
You are welcome; in any case, the correct statement is: let A and B be commuting operators with nondegenerate eigenvalues; then, if [itex]|a\rangle[/itex] is an eigenstate of A, [itex]|a\rangle[/itex] is also an eigenstate of B; moreover, if [itex]|a\rangle[/itex] and [itex]|a'\rangle[/itex] 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
Thanks, yes, excellent!
 

Related to Quantum and Commutation - Help me start

1. What is quantum mechanics?

Quantum mechanics is a branch of physics that studies the behavior of matter and energy at a very small scale, such as atoms and subatomic particles. It explains how particles interact with each other and how they behave in different situations.

2. What is the difference between classical mechanics and quantum mechanics?

Classical mechanics is based on Newton's laws of motion and describes the behavior of macroscopic objects, while quantum mechanics describes the behavior of microscopic objects. Classical mechanics follows deterministic principles, whereas quantum mechanics is probabilistic in nature.

3. What is the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle states that it is impossible to know both the position and momentum of a particle with absolute certainty. The more precisely we know one of these quantities, the less precisely we can know the other.

4. What is commutation in quantum mechanics?

In quantum mechanics, commutation refers to the ordering of operators in a mathematical expression. The order in which operators are applied can affect the outcome of a measurement, and commutation rules help determine the order in which operators should be applied.

5. How is quantum mechanics applied in real-world technologies?

Quantum mechanics has many practical applications, such as in the development of transistors, lasers, and computer memory. It is also used in cryptography and quantum computing, which could potentially revolutionize information processing and communication in the future.

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