Need help understanding Commuting Operators

[expos4ever]

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.

 

[PeroK]

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.

 

[topsquark]

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

 

[martinbn]

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. $B|\alpha\rangle= \mu|\alpha\rangle$.

 

[expos4ever]

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?

 

[PeroK]

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.

 

[PeroK]

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.