Orbital angular momentum eigenvalues

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Professor M does Science

Professor M does Science

Күн бұрын

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📚 From our study of general angular momentum in quantum mechanics, we know that angular momentum eigenvalues are quantized. They are labelled by the quantum numbers j and m, which can take integer or half-integer values. The eigenvalues of orbital angular momentum must also follow these rules, because orbital angular momentum is just a particular instance of angular momentum. However, in this video we will learn that not all allowed values of j and m can actually exist for orbital angular momentum: only integer values are possible.
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⏮️ BACKGROUND
General angular momentum: • Angular momentum in qu...
Ladder operators: • Ladder operators in an...
General angular momentum eigenvalues: • Angular momentum eigen...
Orbital angular momentum: • Orbital angular moment...
⏭️ WHAT NEXT?
Orbital angular momentum eigenfunctions: • Orbital angular moment...
~
Director and writer: BM
Producer and designer: MC

Пікірлер: 56
@dutonic
@dutonic Жыл бұрын
A few weeks ago I became frustrated with all the resources I was reading about angular momentum in QM. I was looking for a fundamental understanding of why quantum systems were modeled as they were. Not simply accepting the theories. Angular momentum was irking me because I had this idea that the ladder operators were using themselves to prove quantization. That if we approach angular momentum with the ladder operators it just works. This wasn't good enough for me, so I kept looking deeper. Why is angular momentum actually quantized? I kept digging and digging and became almost hopeless and exasperated as I began thinking I needed to master group theory and lie algebras to claw my way to the bottom of this well of endless questions. Then I had the silliest realization. It somehow took weeks to dawn on me: angular momentum is quantized because energy is quantized. Angular momentum is literally just a form of kinetic energy. And if that parameter is not quantized, then energy is no longer quantized. A non-infinite divisibility of total energy demands a model that quantizes angular momentum. After weeks of digging through math papers I could hardly read and watching hours of lectures, to realize how simple the answer was all along was an experience that made me reflect on my approach to learning. This idea that I must delve all the way to the bottom of scientific knowledge and slowly build my way back up from quantum physics to chemistry to biology; it almost seems like I'm pulling a Descartes, trying to throw everything away and build a world from "I think therefore I am". Physics can become overwhelming when I have the notion that I must always know "why". But accepting that I won't know why is equally outrageous. So in the end I'm not really sure. This comment is more of a thought log as I continue to learn and grow. But I get the feeling that all physicists confront this notion eventually.
@ProfessorMdoesScience
@ProfessorMdoesScience Жыл бұрын
Thanks for sharing your thoughts!
@aeishahameera6579
@aeishahameera6579 3 жыл бұрын
The explanation is so elegant. This series has helped me to better understand quantum mechanics. I am really grateful for this. Have you posted the video on hydrogen atom and spin angular momentum? I'm looking forward to it!
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
Glad to hear you've found the videos helpful! We are plannig to start with the videos on the hydrogen atom in the next few weeks, so stay tuned!
@quantum4everyone
@quantum4everyone 2 жыл бұрын
I think it is somewhat confusing to use an ket when computing the wavefunction here. The position bra associated with the angular momentum states should only depend on the angles theta and phi. A more correct way would be to have a tensor product of a bra of . I don’t think many textbooks get this right, but you can consult J. Math. Phys. 62, 072102 (2021) for a proper description of how to develop the separation of variables, based on the translation operator in cartesian versus spherical coordinates.
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
Thanks for the suggestion! We do "abuse" notation (in several places), as also done by many textbooks as you point out. Will try to be clearer on this point in the future.
@MrAnuragprasad
@MrAnuragprasad 2 жыл бұрын
At 1:10 I think there is a small error. There should be a (h/2π) factor multiplied to the RHS of the equation for [Ji,Jj].
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
You are absolutely correct, thanks for pointing out the typo!
@workerpowernow
@workerpowernow 2 жыл бұрын
thank you. This series is helping me a lot
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
Glad to hear it! :)
@miguelmicallef1415
@miguelmicallef1415 3 жыл бұрын
Very well explained. Thank you
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
Thanks for watching! :)
@ajilbabu13
@ajilbabu13 3 жыл бұрын
Thanks for this❤ lect.
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
Glad you liked it! :)
@viplovedhoke6597
@viplovedhoke6597 2 жыл бұрын
Really nice content...
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
Thanks a lot 😊
@richardthomas3577
@richardthomas3577 Жыл бұрын
Thank you! Gee your videos are so clear, but they make one think. Don't you find it strange that a particle can have an intrinsic quality (quantum number), spin, which actually can "add" to a quantity it picks up from its behavior in space (orbital angular momentum), with the combination determining how it reacts to, e.g., a B field? S and L are two quantities with entirely different origins (apparently, as far as we know), but they are somehow the same in how they interact with the world. (I can just picture they guys doing the Stern Gerlach experiment looking at the results and thinking "OMG" over and over again at different layers as all the implications sank in.) I guess in a sense mass is maybe similar -- sort of an intrinsic energy that adds to energy picked up from extrinsic circumstances in determining how a particle behaves. But I gather there are other quantum numbers, such as electric charge, where there is no way a fundamental particle can pick up extra charge from extrinsic circumstances. Somehow angular momentum seems special . . . Anyway, obviously I need to learn a lot more. Thank you again; you guys are great!
@ProfessorMdoesScience
@ProfessorMdoesScience Жыл бұрын
Thanks for your thoughts! Coincidentaly we are now working on the next series of videos which will cover spin :)
@quantum4everyone
@quantum4everyone 2 жыл бұрын
The condition used to determine integral m is actually incorrect. Many textbooks get this wrong. The wavefunction need not be periodic. It is the probability density and the probability current that must be continuous, not the wavefunction. This is discussed well in both Green’s textbook and Ballentine’s. It is true, of course that m must be an integer, but periodicity of the wavefunction is not the reason why. It is more fundamental than that. And periodicity of the wavefunction is actually not required. This problem is closely related to the particle in a box, when you make the box a circle. There, you can pick any boundary condition desired at the endpoints, and all is well. But for angular momentum, one does require m to be integral for other reasons. For similar reasons, there are no pure angle operators in quantum mechanics, but cosine or sine of these angles are well defined, because these can be determined in terms of the Cartesian operators. Just thought you should know.
@angelmendez-rivera351
@angelmendez-rivera351 7 ай бұрын
What you are doing is misrepresent the context and content of the sources you allege to be citing (despite the fact that ad hoc name-dropping is not a citation of any kind at all). Periodicity follows from the differentiability condition. Also, the wavefunction trivially must be differentiable, since it satisfies Schrödinger's equation, which is a differential equation. Claiming many texbooks get this incorrect, only to allude to a single textbook that allegedy does not get it incorrect, is nothing more than revisionism/denialism.
@quantum4everyone
@quantum4everyone 7 ай бұрын
@@angelmendez-rivera351 I am sorry to hear you feel this way. But the wavefunction is not physical, only the probability and probability current are, and this is the origin of the issue. I suggest you look into the texts I suggested before you accuse me of ad hoc name dropping. The best sources are Herbert Green's book (Matrix Mechanics), Ballentine's book (Quantum Mechanics: A modern development), and Arno Bohm's book (Quantum Mechanics: Foundations and Applications). You can also look at the careful discussion of the particle on a circle by Konishi and Paffuti (Quantum Mechanics), which discusses the boundary condition carefully. If you look at them you can learn something If you prefer other actions, I cannot help you. But you accomplish nothing by saying I am misrepresenting these sources. For example, as Ballentine says on page 171, after using the L=rxp operator form to show one must have integer z-component of spin eigenvalues "Thus we have shown, directly from the properties of the position and momentum operators, that the orbital angular momentum must be integer multiples of hbar. This approach avoids any problematic discussion of boundary conditions on the state vector Psi." I think what I said represents what is discussed in these books completely faithfully. An issue not discussed in most textbooks either is that one should never think of the angle variables theta or phi as being well defined operators. Only trig functions of them are well defined operators. But perhaps that is something else you will say I am getting wrong...
@psiphizone2020
@psiphizone2020 Жыл бұрын
Where is the spin angular momentum video? Is it available?
@ProfessorMdoesScience
@ProfessorMdoesScience Жыл бұрын
Not yet unfortunately. We've been very busy but hoping to get back to publishing more regularly soon!
@passionisinspiration1912
@passionisinspiration1912 7 ай бұрын
Hi, as you have mentioned in the video to checkout the video on spin, I can't find it on ur channel? Can you please provide the link?
@ProfessorMdoesScience
@ProfessorMdoesScience 7 ай бұрын
It's not ready yet, but coming soon. We are working on a series on spin at the moment. Stay tuned!
@snjy1619
@snjy1619 8 ай бұрын
@rodrigoappendino
@rodrigoappendino 3 жыл бұрын
6:31 I'm pretty confused. In another video, the one where you talk about representations, I think, the wave function was the coefficient that multiplied the eigenstate, while here, the wave function is the eigenstate itself.
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
Good question! The wave function is a "representation" of the state. Let the state of the system be |psi>. To define a representation, we first need a basis, which in the case of the wave function is the position basis, {|r>}. We then write the state in this basis: |psi>=\int dr psi(r) |r> (equation 1) where the expansion coefficients psi(r)= are what we call the wave function. This means that, by specifying the wave function psi(r), you can immediately build the full state |psi> using equation 1 above. What we are doing in this video is to use the position representation, which means that we are trying to determine the expansion coefficients of the state in the position basis, and what you are referring to is the equation that allows us to do so. Physicists often abuse language and call the wave function the "state" of the system, and this is probably what I did here, but to be more precise you are correct and we should say "the position representation of the state of the system". For more details about this, you can follow our series on the position representation, which is often also called "wave mechanics": kzfaq.info/sun/PL8W2boV7eVfnHHCwSB7Y0jtvyWkN49UaZ I hope this helps!
@ILsupereroe67
@ILsupereroe67 2 жыл бұрын
Where is the video on spin?
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
Unfortunately we haven't published it yet, but it is on our to-do list!
@dhananjaykapse9221
@dhananjaykapse9221 2 жыл бұрын
I couldn't find video on spin angular momentum on channel u were referring to in this video...??
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
We haven't got to that point yet, we hope to publish videos on spin over the next few months. We'll keep you informed!
@ashishchhimpa2533
@ashishchhimpa2533 2 жыл бұрын
@@ProfessorMdoesScience Ma'am please try to make the videos on spin ang. momentum available as soon as possible. your teaching is great!
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
@@ashishchhimpa2533 Glad you like it! We are currently preparing a series on the hydrogen atom, and we'll move on to spin shortly after that!
@harshmishra8189
@harshmishra8189 2 жыл бұрын
Please make a video on Hydrogen atom. Thanks!
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
Thanks for the suggestion! We are planning to publish a series of videos on hydrogen, and coincidentally our latest video builds up towards it: kzfaq.info/get/bejne/obWlo6x2zb_IiWw.html
@subhajitsadhukhan8521
@subhajitsadhukhan8521 3 жыл бұрын
Hello professor. As I'm totally new in this subject I don't know this as much. I just read the position and momentum representations of state vectors and followed your videos. What I understand is the inner product of r with state vector psi ......that means gives us the so called wave function. But here you took the Inner product of r with eigenket |l, m> that is and called it wave function. But |psi> and |l, m> are not same right? |l, m> is just eigenkets of angular momentum operator just like |p> are eigenkets of momentum operator. So I can't understand that..... I know this is a weird question but as I said I'm new in this subject so don't know as much. If you help me a bit that would be great for me.....
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
The quantity |(something)> denotes a quantum state (also called a ket). "(something)" is simply a label that allows us to identify the specific quantum state we are working with, so all of |psi>, |l,m>, and |p> are quantum states. |psi> is typically used to denote a general quantum state, |l,m> in this context denotes a very specific quantum state: an eigenstate of the angular momentum operators L^2 and L_z. And |p> also denotes a very specific quantum state: the eigenstate of the linear momentum operator. Calculating always gives the wave function associated with the state |(something)>. So all of , , and are wave functions, but they are wave functions of differnet states. I hope this helps!
@subhajitsadhukhan8521
@subhajitsadhukhan8521 3 жыл бұрын
Ok professor... But to represent an abstract state vector we choose a set of basis..... if we take the basis as the eigenstates of position operator then |psi> can be expanded in terms of position eigen basis and that is what we call position representation right? Now after expressing |psi> in position eigenbasis if we take this gives us wave function psi(r). So as you said and both are wave function of corresponding states what are the position representation of |p> and |l, m>....? Can we expand the kets |p> and |l, m> in the position eigenbasis |r>?
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
You are correct that when we want to work with a quantum state we typically write it down in a particular basis (also called representation). The position representation indeed corresponds to expanding a state in the basis of position eigenstates |r>. And you can indeed expand the kets |p> and |l,m> in the position basis. For example, when we calculate the representation of the momentum eigenstate |p> in the position basis, we get a wave function that is a plane wave. We go over this derivation in the following video: kzfaq.info/get/bejne/aNKiZsRymdzFcqc.html I hope this helps!
@subhajitsadhukhan8521
@subhajitsadhukhan8521 3 жыл бұрын
Thanks a lot professor......
@ajilbabu13
@ajilbabu13 3 жыл бұрын
Hai prof, why do these observables quantise their values? Who is prohibiting them to take any value they like? I understood the theory(with minimum uncertainty) but couldnt figure it out this. SE may be the reason, but do we have any other reasons Regards
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
The best way to think about this is to simply follow the maths, which we explain in detail for a general angular momentum in this video: kzfaq.info/get/bejne/qp2oadmlm97PcX0.html
@ajilbabu13
@ajilbabu13 3 жыл бұрын
@@ProfessorMdoesScience thanks prof
@shayanmoosavi9139
@shayanmoosavi9139 3 жыл бұрын
That's just how nature works. We're only discovering these laws one by one. We make observations (experiments) and try to model them with mathematics and explain the phenomenon (theory). One example of these observations is the Stern-Gerlach experiment that showed spin angular momentum is quantized and can only have certain values. Math follows afterwards and explain exactly why (in our opinion of course). Nature just is, it's we humans who give it a reason.
@12388696
@12388696 2 жыл бұрын
Cool
@ProfessorMdoesScience
@ProfessorMdoesScience 2 жыл бұрын
Thanks for watching!
@ajilbabu13
@ajilbabu13 3 жыл бұрын
Why do we take the third component of J, and J² as compatible observables?
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
Let me start by saying why we don't take two (or three) components of J as compartible observables: because they don't commute. Given this, we can only use one of the components. The choice of J_3 is simply a convention, we could equally well have chosen J_1 or J_2 and the resulting theory would be equivalent. You can find more details in our video introducing angular momentum in quantum mechanics: kzfaq.info/get/bejne/eNVlpNKSsdmmf3k.html
@ajilbabu13
@ajilbabu13 3 жыл бұрын
@@ProfessorMdoesScience thanks prof. Can you please suggest me a ref.book. I have gone through Zettili and Griffith
@ajilbabu13
@ajilbabu13 3 жыл бұрын
@@ProfessorMdoesScience thanks prof. Can you please suggest me a ref.book. I have gone through Zettili and Griffith
@ProfessorMdoesScience
@ProfessorMdoesScience 3 жыл бұрын
The main books we are using for reference are Sakurai, Cohen-Tannoudji, and Shankar, and I think they are all a very good starting point!
@ajilbabu13
@ajilbabu13 3 жыл бұрын
@@ProfessorMdoesScience thank you so much
@sensorer
@sensorer 6 ай бұрын
I've heard before that "spin 1/2 means that you have to turn a particle two times to get back where you started". This derivation gives a glimpse as to why this explanation might have come about. If you let phi be [0,4*pi], m can be a half-integer
@ProfessorMdoesScience
@ProfessorMdoesScience 6 ай бұрын
We are currently working on a full series on spin-1/2 particles, hopefully will come out soon :)
@Puentezz
@Puentezz 6 ай бұрын
Came back here just to check for this specific comment. I'm wondering exactly the same thing.
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