Need help understanding Commuting Operators

  • I
  • Thread starter expos4ever
  • Start date
  • Tags
    Operators
In summary, the conversation discusses a question about a claim made in a teaching video on commuting operators in quantum mechanics. The question pertains to the understanding of why B-alpha, an eigenvector of A, can only be a scaled alpha when both have the same eigenvalue lambda. The conversation also delves into the concept of non-degenerate eigenvalues and their corresponding eigenspaces. The speaker clarifies that in QM, states are rays rather than vectors and therefore the direction matters more than the length. There is a discussion on whether the speaker in the video could have been more clear in their explanation and an apology for not being able to use symbols in the conversation. Overall, the main takeaway is the proof for the non-degenerate
  • #1
expos4ever
21
5
TL;DR Summary
I do not understand a claim in a teaching video. Subject = commuting operators.
Here again with another question about the Quantum Sense video series. Thanks for all the useful feedback to my last question. My question concerns a very short chunk of about 20 seconds beginning at 4:25 of this link:



At around 4:34, he says "B-alpha has to be the same eigenvector as alpha since they both have eigenvalue lambda". Fine, no problem - I think I understand this completely. But then he immediately says "B-alpha can only be a scaled alpha". I must have watched this bit 15 times and I simply do not get it. If B-alpha is the same eigenvector as alpha, how is it not equal to alpha? For some reason, I suspect I have a weird mental block here as I will bet the answer will be obvious to the rest of you. The less likely possibility is that the speaker made an error (an unlikely possibility, I concede with humility). Thanks.
 
Physics news on Phys.org
  • #2
If ##v## is an eigenvector of an operator corresponding to eigenvalue ##a##, then so is ##bv## for any non-zero scalar ##b##.

The proof I'll leave as an exercise.
 
  • Like
Likes malawi_glenn, topsquark and DrClaude
  • #3
expos4ever said:
TL;DR Summary: I do not understand a claim in a teaching video. Subject = commuting operators.

Here again with another question about the Quantum Sense video series. Thanks for all the useful feedback to my last question. My question concerns a very short chunk of about 20 seconds beginning at 4:25 of this link:



At around 4:34, he says "B-alpha has to be the same eigenvector as alpha since they both have eigenvalue lambda". Fine, no problem - I think I understand this completely. But then he immediately says "B-alpha can only be a scaled alpha". I must have watched this bit 15 times and I simply do not get it. If B-alpha is the same eigenvector as alpha, how is it not equal to alpha? For some reason, I suspect I have a weird mental block here as I will bet the answer will be obvious to the rest of you. The less likely possibility is that the speaker made an error (an unlikely possibility, I concede with humility). Thanks.

To geometrically expand a bit on PeroK's answer, states in QM are rays, not vectors: it is the direction that matters, not so much the length.

-Dan
 
  • Like
Likes malawi_glenn
  • #4
expos4ever said:
At around 4:34, he says "B-alpha has to be the same eigenvector as alpha since they both have eigenvalue lambda". Fine, no problem - I think I understand this completely. But then he immediately says "B-alpha can only be a scaled alpha".
If ##|\alpha\rangle## is an eigenvector of ##A## with eigenvalue ##\lambda## i.e. ##A|\alpha\rangle=\lambda|\alpha\rangle##, then so is ##B|\alpha\rangle##, because the operators comute, ##AB|\alpha\rangle=BA|\alpha\rangle=B\lambda|\alpha\rangle=\lambda B|\alpha\rangle##. Because ##\lambda## is nondegenerate, which means the space of eigenvectors with that eigenvalue is one dimensional, the two vectors ##B|\alpha\rangle## and ##|\alpha\rangle##, which belong to it, have to be multiples of each other i.e. ## B|\alpha\rangle= \mu|\alpha\rangle##.
 
  • Like
Likes topsquark and malawi_glenn
  • #5
martinbn said:
If ##|\alpha\rangle## is an eigenvector of ##A## with eigenvalue ##\lambda## i.e. ##A|\alpha\rangle=\lambda|\alpha\rangle##, then so is ##B|\alpha\rangle##, because the operators comute, ##AB|\alpha\rangle=BA|\alpha\rangle=B\lambda|\alpha\rangle=\lambda B|\alpha\rangle##. Because ##\lambda## is nondegenerate, which means the space of eigenvectors with that eigenvalue is one dimensional, the two vectors ##B|\alpha\rangle## and ##|\alpha\rangle##, which belong to it, have to be multiples of each other i.e. ## B|\alpha\rangle= \mu|\alpha\rangle##.
Great. Got it, very clear explanation. However, do you not agree that the speaker misleads us just a little when he says "B-alpha has to be the same eigenvector as alpha since they both have eigenvalue lambda". Shouldn't he have said something like this after introducing the nondegenerate case: "The action of A on its eigenvector alpha is the same as the action of A on this other eigenvector of A, that is B-alpha: both these eigenvectors get "stretched by lambda when acted on by A. Therefore, both these vectors, that is alpha and B-alpha, have to be multiples of each other". Do you see the difference? Is my version, although wordier, not fundamentally more correct?

By the way, apologies for not yet figuring out how to get the symbols to work. Can you point me in the right direction on this, please?
 
  • #6
expos4ever said:
Great. Got it, very clear explanation. However, do you not agree that the speaker misleads us just a little when he says "B-alpha has to be the same eigenvector as alpha since they both have eigenvalue lambda". Shouldn't he have said something like this after introducing the nondegenerate case: "The action of A on its eigenvector alpha is the same as the action of A on this other eigenvector of A, that is B-alpha: both these eigenvectors get "stretched by lambda when acted on by A. Therefore, both these vectors, that is alpha and B-alpha, have to be multiples of each other". Do you see the difference? Is my version, although wordier, not fundamentally more correct?

By the way, apologies for not yet figuring out how to get the symbols to work. Can you point me in the right direction on this, please?
Perhaps the important point is that if ##A## commutes with ##B## and ##A## has a non-degenerate eigenvalue ##\lambda##, then:

1) ##\lambda## is also a non-degenerate eigenvector of ##B##

2) ##A## and ##B## share the same (one-dimensional) eigenspace corresponding to ##\lambda##.

In QM all states/vectors have unit norm, which restricts the vectors we are interested in. Whether it's allowable to talk about the eigenvector in this context is a moot point. I wouldn't get hung up about it.
 
  • Like
Likes expos4ever and topsquark
  • #7
The proof for the non-degenerate case is much less trivial. I'd focus on that, as there is some good mathematics to learn there.
 
  • Like
Likes topsquark

Similar threads

  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
6
Views
1K
Replies
16
Views
546
  • Calculus and Beyond Homework Help
Replies
2
Views
944
Replies
16
Views
2K
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top