Your videos bring tears of joy to my eyes. With college tuition skyrocketing these days, your videos are like manna from heaven. Not only are they free but they are substantially of a higher didactic quality than the classes I took in college. Incredible and amazing.
I'd bet this is because you already have a background in the subject. These videos make parts your existing knowledge cohere with themselves better since you are given an overall picture concisely.
This is my third time working through your angular momentum playlist. I did it once for fun. A second time in undergrad quantum. And now a third time for grad quantum. This time I'm challenging myself to stop and work through most of your derivations as an exercise before I watch you do it. Truly a phenomenal resource. I love this channel, and I love you guys. You've really made a hugely positive impact on my learning experience in all stages of development.
I've jumped into the video since I saw Mire's face. I didn't know about this channel, good to know. I am really close to finish my PhD and it went from pure chemistry to more-in-detail molecular magnetism. I am not capable of understand the whole video, so I checked your previous videos and there is material useful for me, for the defense. Thank you, you have gained a follower, I will check your previous videos in the during the following weeks. Maybe then, I might be capable to follow this one also!
These videos are so good. Thanks Professor M. I have been learning quantum mechanics this semester and these videos with textbook questions (I am referring to Griffiths QM) are great learning experience.
I am eagerly waiting for your videos on addition of angular momentum, Clebsch-Gordan coefficients, Wigner-Eckart theorem. I love your videos. They are my first and last resort for clearing of any confusion. Thank you very much.
We are slowly working towards that. We first plan to cover spin angular momentum, and then move on to addition of angular momentum. And glad you find the videos helpful!
This is such a clear exposition and wonderfully explained of the angular momentum in quantum mechanics. I want to go through this more carefully so that I understand all the quantum mechanical calculations. This definitely helps me understand spin better. You show in the video how orbital angular momentum is defined in quantum mechanics and then show calculations of all the commutation relations. Then spin angular momentum is defined as a triple of three Hermitian operators that satisfy these same commutation relations. This is very nicely explained.
Glad to see that you like our videos. We are actually working on a full series focusing on spin-1/2, so we'll dive into this topic further over the coming months!
Thank you very much. I do enjoy your videos and you have done so much work to make quantum mechanics accessible and very enjoyable. I am not sure if you delve into the topic of addition of angular momenta, but that is one thing I would like to understand. I see you explain much about spherical harmonics, 3d harmonic oscillator, the hydrogen atom and so many other topics.
I didn't really understand angular-momentum until a I wrote some python to split rank "N", for 0 to 16, into irreducible representations. It took about 3 years as a hobby, but it started with N, say "3", partitioned it into 3, 2+1, 1+1+1,....then made those into young tableaux, then normal young diagrams , of which there are 1, 2, and 1, from those 4 diagrams, you get representations of the permutation group on three indices (ijk), and then make some kind of product (I forgot what it was called) that permutes ijk into stuff, e.g. for 3=1+1+1 is +ijk, +jki, +kij, -kji, -jik, -ikj so add those up on the tensor and get L=0 M=0, which is the famed Levi-Civitta thing. It's scalar. for 3=3, all the signs are + and you get the total symmetric traceless pure rank-3 tensor, with, wait for it, 7 degrees of freedom because its L=3, M=-3,-2,-1,0,1,2,3 (breaking out the M's requires Clebsch-Gordanry, and is hard), The middle ones have mixed symmetry, but they ALL look just like Spherical Harmonics in cartesian coordinates made out of not x, y, z coordinates, but x, y, z unit vectors in outer-product mode. So what I learned is that angular momentum is conserved b/c space is isotropic, and scalars, vectors, tensor, or |l, m>, |j, m> are things that can be rotated but are also eigenvectors of Z rotations, and then to go from a cartesian dyads to spherical tensors is the same as CG going from |j,m>|j',m'> --> |J, M> + |J', M'> + ...., and because rotations don't commute, angular momentum operators don't commute, and the math is easy to understand but hard to do. *Oh btw: said tensor are classical systems (e.g. the rank-4 generalized Hooke's tensor), but in the spherical basis they behave EXACTLY like quantum systems with L-= 4, 3, 2, 1, 0. (though I suspect the odd L are all zero for that example).*
I could never imagine this quality of content on youtube. You're just doing so amazing ❤❤. I love your explanations and your videos.❤ Please continue 🙏🙏. PS: sorry for my poor english
Pure and clear presentation, great work. Both presenters are really excellent . They kept the same behaviour and the two face looks similar, I am right.?
We do hope to cover statistical mechanics, and indeed our videos on the density operator are a first step in that direction: ruclips.net/p/PL8W2boV7eVflL73N8668N0EQUnID1XaEU
@@ProfessorMdoesScience thanks for the response, and waiting so very eagerly to see your videos on statistical mechanics.....never missed a single video of your channel....
Thank you soo much for these wonderful content. Watching your videos makes me feel that, I am on a journey and the way you explain is just incredible. Wish I would have found these videos during my post-graduation. Still, they will make a huge impact on learning quantum mechanics. Keep going. 🙃
Ma'am i have three questions Q1. magnitude of L: for n=1, l=0 ...L=0h(bar)----------it says that in the ground state, angular momentum is zero...How L is zero in the ground state? Q2. why we take always the direction of L Along the z-Axis? Q3. for n=2, l=0,1, mz=0,+-1 for that Lz= 0h(bar), Lz=1h(bar), Lz=-1h(bar)...How three directions of angular momentum is possible at a time?
Here are some thoughts/comments: Q1. I don't quite follow your question. Are you asking about the eigenvalues of angular momentum? If so, we go over them in this video: ruclips.net/video/t7x6vt6xkAI/видео.html Q2. L has three components, L=(Lx,Ly,Lz). We don't take the angular momentum operator to be along the z direction. What we do is to realize that the various angular momentum components do not commute, so they cannot form a set of compatible observables (which is what we need to specify quantum states). We therefore can only pick one component, and we usually pick Lz (together with L^2). But the choice of Lz is simply a convention, you could equally well choose Lx or Ly, and still develop the quantum theory of angular momentum. Q3. I am not sure what your "n" is. Is it a quantum number not associated with angular momentum? For example, in the hydrogen atom n labels energy eigenstates and is typically used together with angular momentum to specify quantum states, but in this video we are only discussing angular momentum and therefore do not consider additional quantum numbers. Having said this, what you are quoting for Lz are the possible eigenvalues (when the eigenvalue of L^2 is labelled by l=1). A quantum state in any basis (angular momentum or not) can be in a superposition of basis states. Even though the state is a superposition, when you actually measure the relevant property, you can only get one possible value. We go over measurements in this video: ruclips.net/video/u1R3kRWh1ek/видео.html I hope this helps!
Awesome and great presentation video. Now,, back to basics...what makes two observables commute? I didn't quite understand your reasoning as you spoke so fast. :) So, [ri , pj] do not commute because ...?
Good question! The commutator of position and momentum is typically taken as an axiom of quantum mechanics, so no justification there. With this, you can then derive results like the reason why the momentum operator in the position representation calculates a derivative, or why the position and momentum wave functions are related by a Fourier transform, as we explain in this playlist: ruclips.net/p/PL8W2boV7eVfnHHCwSB7Y0jtvyWkN49UaZ Alternatively, you could use other relations as axioms, and then derive the canonical commutation relation from those; but you would always have to "postulate" something to get started. For example, Sakurai starts with the translation operator instead, and then derives the commutation relation. (I should also add that [ri,pj] in your example actually commute as they are along different axes.) This is an interesting question without an easy answer, but hope this helps!
@@ProfessorMdoesScience Thank you, Professor M for your answer. My understanding is that [ri , pj] = ih_bar, hence not commuting. But you just said that they commute, meaning [ri , pj] = 0, which isn't the case. Am I confused? And this is really fascinating ... you remind me of my excellent Intro to Quantum Mechanics professor, Johannes (Jos) Zwanikken, who sadly left our University last semester (Dec 2020) in the US after 10 years and is back home in the Netherlands (Holland?) at the Technische Universiteit Delft. He was homesick -:) but I swear you are fraternal twins in every way. Message
In your notation, I understand "i" and "j" as indices labelling the different directions (e.g. r1=x, r2=y, r3=z). With this notation, then [ri,pj]=ihbar*delta_ij, which means that [ri,pj]=0 if i is different from j, and [ri,pj]=ihbar only if i=j. And thanks for your lovely compliment!
Position and momentum don't commute if they are both along the same direction, so x and px don't commute. However, if they are along different directions, they do commute, so x and py will commute. I hope this helps!
Question: The second postulate of QM says that all physical observables are described by their corresponding operators acting on quantum states. I assume this is why we can change from classical r and p to quantum operators. But I'm not logically convinced we can simply say that their cross products (for angular momentum) will also produce the quantum operator for angular momentum. Is there a way to prove that cross products (or other operations) of quantum operators will always give a new viable hermitian operator corresponding to a physical observable? Or is this just part of a postulate? Classically, L = r x p, but what tells us that this relationship holds for the quantum operator counterparts?
This is a rather subtle question, so I'll just provide a few thoughts. Promoting the classical r and p to operators is actually another assumption in the basic theory of quantum mechanics in that it tells us how to actually build the corresponding operators. However, this is not typically treated as a postulate as not all observables can be obtained by such a "promotion" from classical mechanics (e.g. spin operators). The more fundamental definition of angular momentum in quantum mechanics is actually through the commutation relations (which are obeyed by both orbital L and spin S angular momentum). It then turns out that using the L=rxp definition leads to the correct commutation relations. I hope this helps!
Hey once again 😅 I have a question ,The reason why spin and orbital angular momentum are not conserved in atom is that spin breaks the rotational symmetry of Hamiltonian? But then why is total angular momentum conserved?
In an isolated system, the total angular momentum must be conserved, and if we have contributions from both spin and orbital angular momenta, then the total angular momentum is their sum. More practically, if the Hamiltonian contains spin-dependent terms, then spin angular momentum by itself will likely not be conserved. An example is the spin-orbit interaction, which means that orbital and spin angular momentum are not individually conserved. I hope this helps!
What is the complementary property of ang mom or spin? We have position-momentum. We have energy-time which are two complementary pairs. We also have ang monentum-angle complementary pair. What is the angle property of ang mom?
It went very well listening to your lecture, but you suddenly cancelled two terms and I didn't understand the rest of your lecture. If you can help me understanding more of it?
You explained it very clearly for sure. I could understand well even if I paused the video so often. I've been digging into quantum mechanics for around 1 year. recently, I think I started to catch a sense. I'm happy :D thanks!
We don't use a single specific book, although we do like the approaches of Sakurai, Shankar, and Cohen-Tannoudji, which we get inspiration from. I hope this helps!
Glad you find them helpful! We don't have slides available at present, but we are working on creating more content to complement the videos, hopefully coming out soon!
Wonderful video thank you for taking the time to better educate these complicated topics, I just have a question regarding 6.16 where the two pieces vanish. I am having trouble understanding how we can cancel them. I saw a comment bellow similar to mine but I am still confused on the explanation. I should say I do understand if the mixed commutations are along different axis then they are zero, I am particularly confused on how you get it into this form where you can clearly see they are along different axis. To me all four pieces are along different axis. Thank you!
If we take the second term, it is: [r2p3,r1p3] We can then use a general property of commutators, that [AB,C]=A[B,C]+[A,C]B (proved in this video: ruclips.net/video/57xgSIV9PY0/видео.html) to re-write it as (A=r2, B=p3, C=r1p3): [r2p3,r1p3] = r2[p3,r1p3] + [r2,r1p3]p3 For the two commutators we now have, we can again use a similar general property of commutators, [A,BC]=B[A,C]+[A,B]C, to get: [p3,r1p3] = r1[p3,p3] + [p3,r1]p3 [r2,r1p3] = r1[r2,p3] + [r2,r1]p3 Putting everything together, we end up with: [r2p3,r1p3] = r1[p3,p3] + [p3,r1]p3 + r1[r2,p3] + [r2,r1]p3 Now we can see that all commutators vanish because they are either between two momentum (first) or position (fourth) operators, or because they mix position and momentum but along different axes (second and third). I do realize that if you haven't seen this before, it is somewhat lengthy to derive. However, once you've worked through a few of those, you'll immediately see which commutators vanish without having to explicitly check. I hope this helps!
@@ProfessorMdoesScience Extremely helpful, thank you very much! I have shared your guy's channel with my entire department. You guys are outstanding educators. Please don't stop ahahaha!
I skipped a few steps in the derivation. The reason the commutators vanish is because they don't involve any position or momentum operators along the same direction. If we use the first as an example: [r2p3,p1] = r2[p3,p1] + [r2,p1]p3 This step is a general property of commutators which, if you are not familiar with, you can find more details in this video: ruclips.net/video/57xgSIV9PY0/видео.html In the expression above, we then have two commutators. The first is [p3,p1]=0, and it vanishes because the momentum components commute amongst themselves. The second is [r2,p1]=0, and it vanishes because the position and momentum are along different directions (if the latter was something like [r2,p2], then it wouldn't vanish). This means that [r2p3,p1]=0, and similarly for the other one. Overall, any commutator of products of position and momentum operators that are not along the same direction will vanish based on a similar argument. I hope this helps!
We do cover the Coulomb potential in the series of videos on the hydrogen atom. You can find it here: ruclips.net/p/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa Note that we are still not done with this series, and we'll add more videos over the next few weeks. I hope this helps!
@@ProfessorMdoesScience honestly I don't know what is actually solution for Coulomb problem. We have an assignment to solve the solution for Coulomb problem. And I don't think it's a numerical concern. That's why I asked to you. Could you help me out to find this.
@@Cans9594 I'm afraid that without more information, I don't know either what your professor means by "the Coulomb problem". I referred you to the hydrogen atom, as the potential in the hydrogen atom is given by the Coulomb potential. Without more details, I'm afraid this is as far as we can get...
Use use this form to write the nine different possible commutators between L1, L2, and L3 in a compact form. This is accomplished by using the Levi-Civita symbol, that is defined in that slide, and the sum over k that runs from 1 to 3. To give you an example, let's imagine we consider the expression in the slide with i=1 and j=2. We would then write, expanding the sum: [L_1,L_2] = ihbar(e_121*L1 + e_122*L2 + e_123*L3) As e_121 and e_122 are both zero, then the sum only has one term, and the expression reduces to our desired form: [L_1,L_2]= ihbar*L3. To give you another example, consider i=1 and j=1. We then get: [L_1,L_1] = ihbar(e_111*L1 + e_112*L2 + e_113*L3) Now all of e_111, e_112 and e_113 vanish, so we end up with the result that L_1 commutes with itself: [L_1,L_1]=0. Overall, from that compact expression you can get all possible commutators by using the appropriate values of i and j. I hope this helps!
@@ProfessorMdoesScience now i get it. Thanks so much! I had another question, regarding the Einstein notation. I thought in Einstein notation we only summed over a superscript/subscript pair, but here it's all subscript. When is this alternative convention used?
@@hershyfishman2929 This is a somewhat subtle point, and I think the wiki entry provides a good overview: en.wikipedia.org/wiki/Einstein_notation#Superscripts_and_subscripts_versus_only_subscripts
Thanks So So So much for such an awesome explanation . Would you mind providing a pdf of the class so that we can follow through during the class, if its not a problem . That would really be of a very great help to me
Glad you like it! Unfortunately, we don't have pdf copies at the moment, but we are working on trying to share more material associated with the videos, so stay tuned! :)
Thanks for the suggestion! Formulas are needed to understand where the different concepts come from, but we are hoping to complement them with more graphs/pictures. For example, we'll (hopefully soon) release a video visualising the spherical harmonics of orbital angular momentum!
Your videos bring tears of joy to my eyes. With college tuition skyrocketing these days, your videos are like manna from heaven. Not only are they free but they are substantially of a higher didactic quality than the classes I took in college. Incredible and amazing.
Thanks for your comment! This positive reception really motivates us to keep going :)
I agreed
Thanks a lot
I understood more in 17 minutes of video than in a month of school teaching. I'm from Italy and I subscribed to your channel! Bravissima!
Thanks for your kind words, and glad you like it! :)
I'd bet this is because you already have a background in the subject. These videos make parts your existing knowledge cohere with themselves better since you are given an overall picture concisely.
There are more hidden to be known in this physics is complex
This is my third time working through your angular momentum playlist. I did it once for fun. A second time in undergrad quantum. And now a third time for grad quantum. This time I'm challenging myself to stop and work through most of your derivations as an exercise before I watch you do it. Truly a phenomenal resource. I love this channel, and I love you guys. You've really made a hugely positive impact on my learning experience in all stages of development.
Wow, thanks for your trust in us, comments like this really motivate us to keep going :)
I've jumped into the video since I saw Mire's face. I didn't know about this channel, good to know. I am really close to finish my PhD and it went from pure chemistry to more-in-detail molecular magnetism. I am not capable of understand the whole video, so I checked your previous videos and there is material useful for me, for the defense. Thank you, you have gained a follower, I will check your previous videos in the during the following weeks. Maybe then, I might be capable to follow this one also!
Small world, and hope you find the other videos useful!
Molecular magnetism? sounds hard...like Racha algebra all day, every day!
Thank you for making these videos. Very few resources are available for studying these topics online , your videos really helped :)
Really happy that you find them helpful! :)
These videos are so good. Thanks Professor M. I have been learning quantum mechanics this semester and these videos with textbook questions (I am referring to Griffiths QM) are great learning experience.
Glad you find them helpful! :)
I am eagerly waiting for your videos on addition of angular momentum, Clebsch-Gordan coefficients, Wigner-Eckart theorem. I love your videos. They are my first and last resort for clearing of any confusion. Thank you very much.
We are slowly working towards that. We first plan to cover spin angular momentum, and then move on to addition of angular momentum. And glad you find the videos helpful!
Wish I knew about this earliar!! This has made life for us University students soo easiar!! Thank you so much for these wonderful playlists
Glad you like them!
Just awesome series on QM! Thank you, this helps a lot to make sense of the general structure of this theory!
Really glad you like it, comments like yours motivate us to keep going! :)
@@ProfessorMdoesScience your vedios really helps a lott!!! Best wishes
@@sivaprasadas3033 Thanks for watching! :)
This is such a clear exposition and wonderfully explained of the angular momentum in quantum mechanics. I want to go through this more carefully so that I understand all the quantum mechanical calculations. This definitely helps me understand spin better. You show in the video how orbital angular momentum is defined in quantum mechanics and then show calculations of all the commutation relations. Then spin angular momentum is defined as a triple of three Hermitian operators that satisfy these same commutation relations. This is very nicely explained.
Glad to see that you like our videos. We are actually working on a full series focusing on spin-1/2, so we'll dive into this topic further over the coming months!
Thank you very much. I do enjoy your videos and you have done so much work to make quantum mechanics accessible and very enjoyable. I am not sure if you delve into the topic of addition of angular momenta, but that is one thing I would like to understand. I see you explain much about spherical harmonics, 3d harmonic oscillator, the hydrogen atom and so many other topics.
@@noelwass4738 We do hope to cover addition of angular momentum in the future, after we cover spin-1/2 systems :)
Your style of teaching is very good. Sad to see so few likes. I hope your channel becomes more popular.
Thanks for your support! We are slowly growing, and it always help if you tell your friends! ;)
I didn't really understand angular-momentum until a I wrote some python to split rank "N", for 0 to 16, into irreducible representations. It took about 3 years as a hobby, but it started with N, say "3", partitioned it into 3, 2+1, 1+1+1,....then made those into young tableaux, then normal young diagrams , of which there are 1, 2, and 1, from those 4 diagrams, you get representations of the permutation group on three indices (ijk), and then make some kind of product (I forgot what it was called) that permutes ijk into stuff, e.g. for 3=1+1+1 is +ijk, +jki, +kij, -kji, -jik, -ikj so add those up on the tensor and get L=0 M=0, which is the famed Levi-Civitta thing. It's scalar. for 3=3, all the signs are + and you get the total symmetric traceless pure rank-3 tensor, with, wait for it, 7 degrees of freedom because its L=3, M=-3,-2,-1,0,1,2,3 (breaking out the M's requires Clebsch-Gordanry, and is hard), The middle ones have mixed symmetry, but they ALL look just like Spherical Harmonics in cartesian coordinates made out of not x, y, z coordinates, but x, y, z unit vectors in outer-product mode.
So what I learned is that angular momentum is conserved b/c space is isotropic, and scalars, vectors, tensor, or |l, m>, |j, m> are things that can be rotated but are also eigenvectors of Z rotations, and then to go from a cartesian dyads to spherical tensors is the same as CG going from |j,m>|j',m'> --> |J, M> + |J', M'> + ...., and because rotations don't commute, angular momentum operators don't commute, and the math is easy to understand but hard to do.
*Oh btw: said tensor are classical systems (e.g. the rank-4 generalized Hooke's tensor), but in the spherical basis they behave EXACTLY like quantum systems with L-= 4, 3, 2, 1, 0. (though I suspect the odd L are all zero for that example).*
I could never imagine this quality of content on youtube. You're just doing so amazing ❤❤. I love your explanations and your videos.❤ Please continue 🙏🙏. PS: sorry for my poor english
Glad you like the videos!
These are great videos! Thank you! Liked and subscribed!
I am looking forward to your future videos on other areas of Physics. Keep up the good work!
Glad you like the videos and joined us, and don't forget to tell your friends! ;)
Great explanation. Keep posting more videos
Glad you like it! :)
Thank you for the video,i was losing hope thinking about the upcoming exams,you saved my life❤
Glad you found us, and good luck with the exams!
@@ProfessorMdoesScience Thank you
You just gave me a huge boost on passing my final exams... thanks a million
Glad to be helpful! :)
Excellent teaching, it's really wonderful.
Glad you like it!
Pure and clear presentation, great work. Both presenters are really excellent . They kept the same behaviour and the two face looks similar, I am right.?
Thanks! As to the two faces looking similar... interesting guess :)
It will be absolutely awesome to hear from you on statistical mechanics too.
We do hope to cover statistical mechanics, and indeed our videos on the density operator are a first step in that direction: ruclips.net/p/PL8W2boV7eVflL73N8668N0EQUnID1XaEU
@@ProfessorMdoesScience thanks for the response, and waiting so very eagerly to see your videos on statistical mechanics.....never missed a single video of your channel....
@@paulbk2322 Thanks for your continued support! :)
U guys just saved me from the madness of second language written QM textbook. 감사합니다 ㅠㅠ
Glad to hear! :)
Thank you soo much for these wonderful content. Watching your videos makes me feel that, I am on a journey and the way you explain is just incredible. Wish I would have found these videos during my post-graduation. Still, they will make a huge impact on learning quantum mechanics. Keep going. 🙃
This is great to hear, thanks for the comment! :)
absolutely beautiful explanation
Glad you like it!
Ma'am i have three questions
Q1. magnitude of L: for n=1, l=0 ...L=0h(bar)----------it says that in the ground state, angular momentum is zero...How L is zero in the ground state?
Q2. why we take always the direction of L Along the z-Axis?
Q3. for n=2, l=0,1, mz=0,+-1
for that Lz= 0h(bar), Lz=1h(bar), Lz=-1h(bar)...How three directions of angular momentum is possible at a time?
Here are some thoughts/comments:
Q1. I don't quite follow your question. Are you asking about the eigenvalues of angular momentum? If so, we go over them in this video: ruclips.net/video/t7x6vt6xkAI/видео.html
Q2. L has three components, L=(Lx,Ly,Lz). We don't take the angular momentum operator to be along the z direction. What we do is to realize that the various angular momentum components do not commute, so they cannot form a set of compatible observables (which is what we need to specify quantum states). We therefore can only pick one component, and we usually pick Lz (together with L^2). But the choice of Lz is simply a convention, you could equally well choose Lx or Ly, and still develop the quantum theory of angular momentum.
Q3. I am not sure what your "n" is. Is it a quantum number not associated with angular momentum? For example, in the hydrogen atom n labels energy eigenstates and is typically used together with angular momentum to specify quantum states, but in this video we are only discussing angular momentum and therefore do not consider additional quantum numbers. Having said this, what you are quoting for Lz are the possible eigenvalues (when the eigenvalue of L^2 is labelled by l=1). A quantum state in any basis (angular momentum or not) can be in a superposition of basis states. Even though the state is a superposition, when you actually measure the relevant property, you can only get one possible value. We go over measurements in this video: ruclips.net/video/u1R3kRWh1ek/видео.html
I hope this helps!
Thanks Madam. Your videos so informative, doubt clearly, and helpful.
Glad you like them! :)
Awesome and great presentation video. Now,, back to basics...what makes two observables commute? I didn't quite understand your reasoning as you spoke so fast. :) So, [ri , pj] do not commute because ...?
Good question! The commutator of position and momentum is typically taken as an axiom of quantum mechanics, so no justification there. With this, you can then derive results like the reason why the momentum operator in the position representation calculates a derivative, or why the position and momentum wave functions are related by a Fourier transform, as we explain in this playlist:
ruclips.net/p/PL8W2boV7eVfnHHCwSB7Y0jtvyWkN49UaZ
Alternatively, you could use other relations as axioms, and then derive the canonical commutation relation from those; but you would always have to "postulate" something to get started. For example, Sakurai starts with the translation operator instead, and then derives the commutation relation.
(I should also add that [ri,pj] in your example actually commute as they are along different axes.)
This is an interesting question without an easy answer, but hope this helps!
@@ProfessorMdoesScience Thank you, Professor M for your answer. My understanding is that [ri , pj] = ih_bar, hence not commuting. But you just said that they commute, meaning [ri , pj] = 0, which isn't the case. Am I confused?
And this is really fascinating ... you remind me of my excellent Intro to Quantum Mechanics professor, Johannes (Jos) Zwanikken, who sadly left our University last semester (Dec 2020) in the US after 10 years and is back home in the Netherlands (Holland?) at the Technische Universiteit Delft. He was homesick -:) but I swear you are fraternal twins in every way.
Message
In your notation, I understand "i" and "j" as indices labelling the different directions (e.g. r1=x, r2=y, r3=z). With this notation, then [ri,pj]=ihbar*delta_ij, which means that [ri,pj]=0 if i is different from j, and [ri,pj]=ihbar only if i=j.
And thanks for your lovely compliment!
Very clear and comprehensive
Glad you like this!
Thank you, ma'am, these videos helped me a lott!!!
Glad to hear! :)
Thank you for your time and effort.
Thanks for watching!
why in the angular momentum the position and momentum commute? thanks for the video !!
Position and momentum don't commute if they are both along the same direction, so x and px don't commute. However, if they are along different directions, they do commute, so x and py will commute. I hope this helps!
Question: The second postulate of QM says that all physical observables are described by their corresponding operators acting on quantum states. I assume this is why we can change from classical r and p to quantum operators. But I'm not logically convinced we can simply say that their cross products (for angular momentum) will also produce the quantum operator for angular momentum. Is there a way to prove that cross products (or other operations) of quantum operators will always give a new viable hermitian operator corresponding to a physical observable? Or is this just part of a postulate? Classically, L = r x p, but what tells us that this relationship holds for the quantum operator counterparts?
This is a rather subtle question, so I'll just provide a few thoughts. Promoting the classical r and p to operators is actually another assumption in the basic theory of quantum mechanics in that it tells us how to actually build the corresponding operators. However, this is not typically treated as a postulate as not all observables can be obtained by such a "promotion" from classical mechanics (e.g. spin operators). The more fundamental definition of angular momentum in quantum mechanics is actually through the commutation relations (which are obeyed by both orbital L and spin S angular momentum). It then turns out that using the L=rxp definition leads to the correct commutation relations. I hope this helps!
Hey once again 😅
I have a question ,The reason why spin and orbital angular momentum are not conserved in atom is that spin breaks the rotational symmetry of Hamiltonian? But then why is total angular momentum conserved?
In an isolated system, the total angular momentum must be conserved, and if we have contributions from both spin and orbital angular momenta, then the total angular momentum is their sum. More practically, if the Hamiltonian contains spin-dependent terms, then spin angular momentum by itself will likely not be conserved. An example is the spin-orbit interaction, which means that orbital and spin angular momentum are not individually conserved. I hope this helps!
Very well explained
Glad you like it!
Can you make video about momentum angular momentum and spin for electromagnetic field
Thanks for the suggestion!
Great Lecture video! Thank you !
Glad you like it! :)
Your doing phd in physics ?
@@sahiljangra4479 not quite! we both finished our PhDs in Physics a long time ago, we are now university profs :)
Great
I am in masters final year I start my phd in material science after complete my masters
Thanks to this EXCELLENT video!
Glad you like it! :)
What is the complementary property of ang mom or spin? We have position-momentum. We have energy-time which are two complementary pairs. We also have ang monentum-angle complementary pair. What is the angle property of ang mom?
It would be something like "angular position" or orientation. Wikipedia has a good intro to this: en.wikipedia.org/wiki/Conjugate_variables
thank for uploading this video
Glad you like it!
Excellent!
Thanks for watching!
Professor M looks different. Jokes aside, well done video!
Thanks :)
It went very well listening to your lecture, but you suddenly cancelled two terms and I didn't understand the rest of your lecture. If you can help me understanding more of it?
Glad to help! Can you please give the time in the video of the step that is unclear?
You explained it very clearly for sure. I could understand well even if I paused the video so often. I've been digging into quantum mechanics for around 1 year. recently, I think I started to catch a sense. I'm happy :D thanks!
Glad this was helpful, and thanks for your comment!
Thanks a lot!!!
Glad you like the video!
Great video! Thank you!
Glad you like it!
WHAT MATHS IS REQUIRED FOR QUANTUM MECHANICS AND QUANTUM COMPUTING
Hi! Great video
Which book are you specifically following for the lec?
We don't use a single specific book, although we do like the approaches of Sakurai, Shankar, and Cohen-Tannoudji, which we get inspiration from. I hope this helps!
@@ProfessorMdoesScience Thankyou For the info! 😁
I hope you get rich from these videos, they are great
Hahahaha, fingers crossed ;)
Thanks ❤
Thanks for watching!
So much help!
Glad you find them helpful! We don't have slides available at present, but we are working on creating more content to complement the videos, hopefully coming out soon!
top line
Thanks for watching!
Wonderful video thank you for taking the time to better educate these complicated topics,
I just have a question regarding 6.16 where the two pieces vanish. I am having trouble understanding how we can cancel them. I saw a comment bellow similar to mine but I am still confused on the explanation. I should say I do understand if the mixed commutations are along different axis then they are zero, I am particularly confused on how you get it into this form where you can clearly see they are along different axis. To me all four pieces are along different axis.
Thank you!
If we take the second term, it is:
[r2p3,r1p3]
We can then use a general property of commutators, that [AB,C]=A[B,C]+[A,C]B (proved in this video: ruclips.net/video/57xgSIV9PY0/видео.html) to re-write it as (A=r2, B=p3, C=r1p3):
[r2p3,r1p3] = r2[p3,r1p3] + [r2,r1p3]p3
For the two commutators we now have, we can again use a similar general property of commutators, [A,BC]=B[A,C]+[A,B]C, to get:
[p3,r1p3] = r1[p3,p3] + [p3,r1]p3
[r2,r1p3] = r1[r2,p3] + [r2,r1]p3
Putting everything together, we end up with:
[r2p3,r1p3] = r1[p3,p3] + [p3,r1]p3 + r1[r2,p3] + [r2,r1]p3
Now we can see that all commutators vanish because they are either between two momentum (first) or position (fourth) operators, or because they mix position and momentum but along different axes (second and third).
I do realize that if you haven't seen this before, it is somewhat lengthy to derive. However, once you've worked through a few of those, you'll immediately see which commutators vanish without having to explicitly check. I hope this helps!
@@ProfessorMdoesScience Extremely helpful, thank you very much!
I have shared your guy's channel with my entire department. You guys are outstanding educators. Please don't stop ahahaha!
@@alessandrotripoli9268 Thanks for your support! What university are you at?
@@ProfessorMdoesScience StonyBrook University in New York!
this is amazingly helpful, i still feel dumb, but maybe a little less so now
Glad you like it!
Thanks!
Thanks for watching!
can not understand why these terms in 6.59 vanish. Do I miss some knowledge or insight ? Am I stupid or dull ?
I skipped a few steps in the derivation. The reason the commutators vanish is because they don't involve any position or momentum operators along the same direction. If we use the first as an example:
[r2p3,p1] = r2[p3,p1] + [r2,p1]p3
This step is a general property of commutators which, if you are not familiar with, you can find more details in this video: ruclips.net/video/57xgSIV9PY0/видео.html
In the expression above, we then have two commutators. The first is [p3,p1]=0, and it vanishes because the momentum components commute amongst themselves. The second is [r2,p1]=0, and it vanishes because the position and momentum are along different directions (if the latter was something like [r2,p2], then it wouldn't vanish). This means that [r2p3,p1]=0, and similarly for the other one. Overall, any commutator of products of position and momentum operators that are not along the same direction will vanish based on a similar argument. I hope this helps!
@@ProfessorMdoesScience Thank you very much ! Now I get it.
Ngl i watch a lot of science videos but this channel i'm having trouble understanding, guess i'm here because the narrator is very pretty (and smart)
What is going in at 7:50? sunds like a bucket of water just hit your mouth. Very good video nonetheless :) thank you
Oh well, I guess the audio recording has the occasional blip... ;)
I'm jumping in joy
I hope this is a good thing! :)
Hey
Do u have any lecture video on Solution for Columb prblm?
We do cover the Coulomb potential in the series of videos on the hydrogen atom. You can find it here:
ruclips.net/p/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa
Note that we are still not done with this series, and we'll add more videos over the next few weeks. I hope this helps!
@@ProfessorMdoesScience thanks, I already cover this series, but particularly, I'm searching solution for Coulomb problem.
@@Cans9594 Could you please clarify what you mean by the "Coulomb problem"?
@@ProfessorMdoesScience honestly I don't know what is actually solution for Coulomb problem. We have an assignment to solve the solution for Coulomb problem. And I don't think it's a numerical concern. That's why I asked to you. Could you help me out to find this.
@@Cans9594 I'm afraid that without more information, I don't know either what your professor means by "the Coulomb problem". I referred you to the hydrogen atom, as the potential in the hydrogen atom is given by the Coulomb potential. Without more details, I'm afraid this is as far as we can get...
never clicked so fast
I hope you liked it :)
hello! How to find representation of angular momentum in position representation professor ?
We look into this in our videos on orbital angular momentum, starting with this one:
ruclips.net/video/EyGJ3JE9CgE/видео.html
I hope this helps!
Amazing
Thanks for watching!
I don't understand what is being summed over k at 8:28. I only see a single L_k for any [L_i, L_j] in the equation.
Use use this form to write the nine different possible commutators between L1, L2, and L3 in a compact form. This is accomplished by using the Levi-Civita symbol, that is defined in that slide, and the sum over k that runs from 1 to 3. To give you an example, let's imagine we consider the expression in the slide with i=1 and j=2. We would then write, expanding the sum:
[L_1,L_2] = ihbar(e_121*L1 + e_122*L2 + e_123*L3)
As e_121 and e_122 are both zero, then the sum only has one term, and the expression reduces to our desired form:
[L_1,L_2]= ihbar*L3.
To give you another example, consider i=1 and j=1. We then get:
[L_1,L_1] = ihbar(e_111*L1 + e_112*L2 + e_113*L3)
Now all of e_111, e_112 and e_113 vanish, so we end up with the result that L_1 commutes with itself:
[L_1,L_1]=0.
Overall, from that compact expression you can get all possible commutators by using the appropriate values of i and j. I hope this helps!
@@ProfessorMdoesScience now i get it. Thanks so much! I had another question, regarding the Einstein notation. I thought in Einstein notation we only summed over a superscript/subscript pair, but here it's all subscript. When is this alternative convention used?
@@hershyfishman2929 This is a somewhat subtle point, and I think the wiki entry provides a good overview: en.wikipedia.org/wiki/Einstein_notation#Superscripts_and_subscripts_versus_only_subscripts
Thank you!
Thanks So So So much for such an awesome explanation . Would you mind providing a pdf of the class so that we can follow through during the class, if its not a problem . That would really be of a very great help to me
Glad you like it! Unfortunately, we don't have pdf copies at the moment, but we are working on trying to share more material associated with the videos, so stay tuned! :)
Good
Thanks!
This is very amazing explaintion
I m from Pakistan🇵🇰🇵🇰🇵🇰
Glad you like it! :)
@@ProfessorMdoesScience you are the best one.......
Τhis nice
Thanks!
❤
A new face to me. Nice to see you.
Thanks! Had kept behind the scenes but it was about time to join the action :)
@@ProfessorMdoesScience Great teachers and a great channel. Welcome to the action!
3:36
Less formulas and more graphs/pictures please.
Thanks for the suggestion! Formulas are needed to understand where the different concepts come from, but we are hoping to complement them with more graphs/pictures. For example, we'll (hopefully soon) release a video visualising the spherical harmonics of orbital angular momentum!