I'm refreshing on Sakurai and Napolitano as I dive into more advanced physics like QFT but it has been a few years since I've been out of college so I'm pretty rusty with it. Seeing the derivations worked through, simply, without too much convolution, is absolutely critical, but there haven't really been any resources on KZbin in short video form that is this rigorous and also this easy to follow. Thank you very much for making these videos. I'm sure I will use them as reference as often if not more than any text I own
@ProfessorMdoesScience3 жыл бұрын
This is great to hear, so glad you find our videos helpful! :)
@mohamedsaadaldin8239 Жыл бұрын
you guys are doing us a huge favor, especially for people like us who are from chemistry background and would like to gain solid knowledge in quantum mechanics. thanks a lot and please continue.
@ProfessorMdoesScience Жыл бұрын
Thanks for your kind words! :)
@smashedpotatoe443 жыл бұрын
thank you soooo much for making these videos... you are doing a great job...prayers and best wishes for your channel to grow big🖤
@ProfessorMdoesScience3 жыл бұрын
Thanks for your support! :)
@JohnVKaravitis3 жыл бұрын
This and the prior video need a Playlist.
@ProfessorMdoesScience3 жыл бұрын
Thanks for pointing this out! Playlist created :) kzbin.info/aero/PL8W2boV7eVfkqnDmcAJTKwCQTsFQk1Air
@TheWingEmpire3 жыл бұрын
Awesome, I needed this one a lot, thank you very much
@ProfessorMdoesScience3 жыл бұрын
Glad this was helpful! :)
@canyadigit62743 жыл бұрын
I know that this is kind of unrelated to this video but you guys answer questions very well. Currently in QM I am learning about identical particles. Where you have psi(x_1,x_2). Say we have psi(x1,x2)=sin(x1)sin(x2). How would we plot this on a graph? Do we have 2 axis for x1 and x2? Are they on the same axis? Thanks!
@ProfessorMdoesScience3 жыл бұрын
I think the most convenient way to plot these would be with two axes, one per particle. Plotting multi-particle states is in general tricky because in a real 3D system you already need 6 axes for 2 particles (more for more particles), which is impossible to visualize. Something that is typically done for two-particle systems (e.g. excitons) is to integrate the degrees of freedom of one of the particles and only show the other.
@canyadigit62743 жыл бұрын
@@ProfessorMdoesScience thanks! For some reason my textbook never talked about actually graphing these kind of functions
@rodrigoappendino3 жыл бұрын
I understand that comutators show us if both operator commute or not, but what are the implications of two operator, observables or not, something? The comutator of X and P is important to prove the Heisenberg uncertainty principle, right? But what about other operators? For example, you say in the beginning that H and L^2 comute, but why is it important for us to know this information? I still don't get it completely.
@ProfessorMdoesScience3 жыл бұрын
Good question! The fact that two operators commute has huge implications, because it means that we can find eigenstates that are simultaneous eigenstates of the two operators. This helps dramatically when solving quantum problems, for example in the case of central potentials it means that the angular part of the problem is already solved from our knowledge of orbital angular moment. We describe the importance of commuting observables (also called "compatible observables") in this video: kzbin.info/www/bejne/f5mtp4tqfZyroaM I hope this helps!
@b6kf367 Жыл бұрын
around 14:06 mark there was a mention of l-th derivative of the delta function as laplacian. Can you point some resources for correlating these? btw thanks for the awesome lectures
@ibrahimtemiza8947 Жыл бұрын
did u find the resources ??
@b6kf367 Жыл бұрын
@@ibrahimtemiza8947 no I took it as it is I guess, will have to check my notes. But you can proceed further on to general r case
@oraange2 жыл бұрын
14:16 I don't understand, why is it related to the delta function? Could you please explain me in another way? Also why for l > 0, the wavefunction doesn't exist (when r->0). I can see it on the graph (end of the video with Potential Plot) but I haven't gotten it by the mathematical description. Thank you!
@ProfessorMdoesScience2 жыл бұрын
This is a rather subtle point. The kinetic energy term of the Hamiltonian involves the Laplacian operator (Del^2). For the tentative solution proportional to r^{-l-1}, this term requires the calculation of the Laplacian of (r^{-l-1}*Y_lm), and it can be proven (we did not in the video) that this term is then proportional to the lth derivative of the delta function. The proof is similar to that used in electrostatics that the Laplacian of (1/r) is equal to -4*pi*delta(r) if you are familiar with that one. The wave function does exist for r-->0, it just vanishes at r=0 for l>0. I hope this helps!
@nirbhayjithara8973 жыл бұрын
plz make video on (1)Hydrogen atom and (2) effect of Magnetic fields on central potentials ...
@ProfessorMdoesScience3 жыл бұрын
Hydrogen atom coming over the next few months, magnetic field will come a little later... Will let you know when these happen!
@sheetukhan3 жыл бұрын
Appreciate your efforts. When will you upload hydrogen atom's radial equation?
@ProfessorMdoesScience3 жыл бұрын
The next videos will first cover the 3D harmonic oscillator from a radial equation point of view, and right after that we'll cover hydrogen, so hopefully over the next few months!
@armalify3 жыл бұрын
Thank you. Central potential, is it the same if we call it a spherical potential in 3 D?
@ProfessorMdoesScience3 жыл бұрын
It is a potential that does not change under a rotation about the origin, so I guess that you could indeed call it a spherical potential in 3D. Hope this helps!
@debankurbasak35623 жыл бұрын
Why Ekl term vanishes for both small and large r
@ProfessorMdoesScience3 жыл бұрын
In the video we discuss the limit of small r, where the term proportional to 1/r^2 becomes very large, whereas the term E_kl stays constant. This means that, for sufficiently small r, the term E_kl becomes negligible compared to the 1/r^2 term, and we can therefore ignore the constant E_kl term in that limit. Why do you think that the E_kl term vanishes for large r?
@debankurbasak35623 жыл бұрын
@@ProfessorMdoesScience i think that will depend on the form of potential function like if its of the form (1/r) ^n it will tend to 0 while energy term remains constant. Like what happens for hydrogen atom case. But for isotropic harmonic oscillator case where V~r^2 the energy term will be negligible as compared to V
@ProfessorMdoesScience3 жыл бұрын
We are actually preparing videos on both the isotropic harmonic oscillator and the hydrogen atom, so we'll get to discuss this in much more detail there :)
@debankurbasak35623 жыл бұрын
@@ProfessorMdoesScience wow thats great!!
@ManojKumar-cj7oj3 жыл бұрын
Waiting for video on hidrozen atom
@ProfessorMdoesScience3 жыл бұрын
It's in the pipeline, hopefully in a few month's time!
@Lilliana12 жыл бұрын
So I had a doubt regarding application of this. If we were to put - V0 into radial equation we get just bessel function by the reasoning that solutions are boud and oscillatory if E
@ProfessorMdoesScience2 жыл бұрын
Could you please clarify what you mean by -V0? Is this a constant potential with a finite range? Spherical?
@Lilliana12 жыл бұрын
@@ProfessorMdoesScience yes exactly that Finite, cosntant spherical potential.
@ProfessorMdoesScience2 жыл бұрын
@@Lilliana1 I haven't solve this problem directly, so you should definitely double-check this. But in the case of E
@oraange2 жыл бұрын
Hi! One question, why did you choose though "r^2 * V(r)" as an approximation? Could it be another different form? or is it just written in textbook ?
@ProfessorMdoesScience2 жыл бұрын
The form of V(r) could in general be anything, so we cannot make a concrete analysis of a limit for r-->0 if we don't restrict its form somewhat. We are ultimately interested in studying central potentials that feature in interesting physical situations, and these include the Coulomb potential where V(r) is proportional to 1/r, or the harmonic oscillator, where V(r) is proportional to r^2. Making the choice r^2V(r) means that we can include all of these examples (essentially any potential that approaches 0 at or slower than 1/r). I hope this helps!
@debankurbasak35623 жыл бұрын
Can you make a video explaining hydrogen atom ?
@ProfessorMdoesScience3 жыл бұрын
It is on our list, and we'll hopefully cover hydrogen in the next few months :)
@ibrahimtemiza8947 Жыл бұрын
thanks so much prodessor but i have aquestion is R when r equal to a real 0 is define and also for l=0
@angelthomas20682 жыл бұрын
mam can u please upload the notes too ... if possible?
@ProfessorMdoesScience2 жыл бұрын
We don't have the notes available for upload, but we are working on it right now (together with more material such as problems and solutions). We'll hopefully provide an update soon!
@RohitSaini-xn2vd2 жыл бұрын
"Ekl is atleast 2l+1 fold degenerate. " Why this called accidental degeneracy?,?
@ProfessorMdoesScience2 жыл бұрын
The name accidental degeneracy is typically used for degeneracies that do not arise due to symmetry. The 2l+1-fold degeneracy here arises because orbital angular momentum is a conserved quantity for central potentials, so I would not call this an accidental degeneracy. I hope this helps!
@RohitSaini-xn2vd2 жыл бұрын
@@ProfessorMdoesScience can u plzz 🙏 elaborate more on accidental and normal degeneracy with example
@ProfessorMdoesScience2 жыл бұрын
@@RohitSaini-xn2vd This is too long for a comment, but we do hope to cover these ideas in several future videos, such as those on the hydrogen atom and also on condensed matter physics.