Need help understanding Commuting Operators

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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
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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.
 
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  • #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.
 
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  • #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
 
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  • #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##.
 
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  • #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.
 
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  • #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.
 
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1. What is a commuting operator?

A commuting operator is a mathematical concept used in linear algebra and quantum mechanics. It refers to two operators that can be applied in any order and still produce the same result. In other words, the order in which the operators are applied does not affect the outcome.

2. How do commuting operators relate to quantum mechanics?

In quantum mechanics, commuting operators are used to describe physical observables, such as position and momentum, that can be measured simultaneously. This is because the operators representing these observables commute with each other, meaning they have a common set of eigenvectors.

3. What is the significance of commuting operators in linear algebra?

Commuting operators are important in linear algebra because they can be simultaneously diagonalized. This means that they share a set of eigenvectors, which makes it easier to solve problems involving these operators.

4. Can non-commuting operators be transformed into commuting operators?

No, non-commuting operators cannot be transformed into commuting operators. This is because the commutator, which measures the non-commutativity of two operators, is a fundamental property that cannot be changed.

5. How are commuting operators used in quantum computing?

In quantum computing, commuting operators are used in quantum algorithms to perform operations on quantum states. This allows for efficient computation and can help solve problems that are difficult for classical computers.

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