Useful notation for angular momentum eigenvalues

[@stefanhill2193]

I'm currently struggling with a lot of things in my life, I had no motivation to study this topics in order to approve my QM courses, but for some reason I find your videos beautiful and motivating. I'm binge watching them hoping to recover the interest in such an important and mesmerizing subject. Thanks for these videos, at least you can be sure that you're making a person happy while living through a hard time.

[@ProfessorMdoesScience]

This is great to hear, and hoping you leave the hard times behind soon!

[@j.abrahamhernandez3629]

already binge watched pretty much all of your vids and they're absolutely brilliant! keep up the amazing work quick question, which software do you use for capturing your notes? thanks!

[@ProfessorMdoesScience]

Glad you like them! :) We write our notes on Explain Everything.

[@guoxinxin693]

Thanks, great video as always! Could you please explain how the eigenvectors transformed from lambda, mu to j,m at 8:35 ?

[@ProfessorMdoesScience]

Good question! The eigenvectors are always the same, the only thing that really changes is how we "label" them. We typically label an eigenvector by the corresponding eigenvalue, and when we started with angular momentum we called the eigenvalues lambda (for J^2) and mu (for Jz), so we labelled the corresponding eigenvectors |lambda,mu>. When we started with angular momentum, we didn't know anything about lambda and mu, so they could in general take any value. However, in the video on angular momentum eigenvalues, we demonstrate that lambda and mu can only take very specific values, in particular lambda=j(j+1)*hbar^2 (where j=0,1,2,...) and mu=m*hbar (where m=-j,-j+1,...,j). This means we can update the eigenvectors from |lambda,mu> to |j(j+1)*hbar^2,m*hbar>, we are simply changing their "label". The final step is simply a convention: we can construct the full eigenvalue of J^2 from only knowing j (all we need to do is then to calculate j(j+1)*hbar^2), and similarly we can construct the full eigenvalue of Jz from only knowing m (all we need to do is to calculate m*hbar). So, for simplicity, we simply "label" the eigenvectors with |j,m>. The key is that this label is sufficient to identify the correct eigenvector for any given eigenvalue. I hope this helps (and sorry for the long message)!

[@guoxinxin693]

​@@ProfessorMdoesScience Thanks for your kind and very clear explanation. I really appreciate your time. I majored in civil engineering and started to learn quantumn mechanics this year. I am often afraid that I probablly asked a silly question and waste your time. Just like some guy commented in your another video: Physics are beautiful because of brilliant people like you.

[@ProfessorMdoesScience]

We're glad to help, and there are never silly questions!

[@jangky97]

I must say thank you for these amazing explanations! I am taking Quantum Mechanics I this semester, and you literally saved me from failing:):)

[@ProfessorMdoesScience]

Glad you find our videos useful! May I ask what university you attend?

[@jangky97]

@@ProfessorMdoesScience Georgia Institute of Technology! It’s in Atlanta, Georgia of US:)

[@oraange]

What an excellent video ! I can really understand faster the concepts in QM books. I look forward !

[@ProfessorMdoesScience]

Glad you like it! :)

[@afaq7564]

Please make a video on Angular momentum in spherical coordinates.

[@ProfessorMdoesScience]

Stay tuned and check again in 2 or 3 weeks! ;-)

[@ProfessorMdoesScience]

We now have a series of videos on "orbital" angular momentum using spherical coordinates: kzbin.info/www/bejne/e6qqe2aAep52nac kzbin.info/www/bejne/a5mYnpWCateWpZI kzbin.info/www/bejne/fZyViYGjfq2JrdE Hope you like them!

[@arezaajouneghani3082]

Your are very clear and a great teacher,!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! THANKS.

[@ProfessorMdoesScience]

Glad you like this!

[@zoubirbecer6529]

Good luck guys, you are doing good jobs

[@ProfessorMdoesScience]

Thanks for your support!

[@vrendus522]

DanielJ Blatecky: Interesting. staying with it. Thanks prof :) USA

[@goopyt267]

i request u to please make such lectures of statistical mechanics also....i rarely found and good lecture on internet on this .... plzzz

[@ProfessorMdoesScience]

Thanks for the suggestion, we do hope to move to other topics after we finish with quantum mechanics!

[@gnnrclvrt]

@@ProfessorMdoesScience I suggest quantum field theory!

[@mathewvincent692]

Which all books do you recommend, Ma'am?

[@ProfessorMdoesScience]

Books we like include: Shankar, Sakurai, Cohen-Tannoudji. Hope this helps!

[@ajilbabu13]

Is there any operator represention for the spin momentum in position basis?

[@ProfessorMdoesScience]

Very good question! The answer is no: spin is not a "spatial" degree of freedom and there is no position representation. We plan a series on spin 1/2 in a few months, where these ideas should become clear.

[@ajilbabu13]

@@ProfessorMdoesScience thanks prof. Waiting for those lects

[@josfus]

Good afternoon. What is please the sequence of classes because i'm in 2024. Thanks.

[@ajilbabu13]

Everyone is saying the spin momentum havv no classical analogue, I dont understand why? So can you please give an example to validate the statement prof.

[@ProfessorMdoesScience]

All we mean is that "spin" is not a concept that exists in classical mechanics. It was introduced with the development of quantum mechanics, as physicists realised that they needed it to explain some quantum phenomena -- for example the Stern-Gerlach experiment. I hope that makes sense!

[@quantum4everyone]

One way to understand the issue is that an electron has spin. But, if we think of the angular momentum as arising from the spinning of the electron about some axis, there are two issues: (i) the electron has not been measured to have any spatial size and (ii) if we assumed it was a spinning ball of charge with its classical radius, then we would find it has to move much faster than the speed of light at its equator to produce the observed spin value. Neither of these two situations makes any sense, so there is no classical way to understand the intrinsic angular momentum of an elementary particle.

[@edoardocasotto6731]

i need the spin, please i need it....pleaseeeeeeeeee

[@ProfessorMdoesScience]

We have been very busy latetly, but when we can re-start videos spin is one of our top priorities, thanks for your patience!

[@amaljeevk3950]

❤

[@snjy1608]

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