Very clear, didactical explanation of Operators in Quantum Mechanics Professor M with beautiful edited math and very interesting stuff. I just came up with your channel 2 days ago and I will definitively follow You.
I have test of quantum mechanics this week, and we are already learning about spinors and approximation methods. I use your videos a lot, because the subjects I was supposed to know already are not entirely clear in my mind. So thanks a lot.
I am unsure whether we're actually using the fact that A commutes with [A,B] in the proof that [A, F(B)] = [A,B] F'(B). It seems that we're only using that B commutes with [A,B] at 15:14, so the relation should still hold even if A does not commute with [A,B]. Obviously both commutative relations are needed for the example of x and p, but I figure this somewhat more general result should be noted.
As always, your explanations are just brilliant. This is quite an involved topic, but it's redeemed by having a friendly face explain it in a clear way. Thank you for posting this, and for your hard work.
Hello, for the proof that [A,B^n]=n[A,B]B^(n-1) if [A,[A,B]]=0 and [B,[A,B]]=0 ( 15:39 ), I don’t understand why we need [A,[A,B]]=0 ??? It seems we do not use this condition, so does it works if we only suppose [B,[A,B]]=0 ? Thanks for yours videos, those are great!
We don't use any specific books for creating the videos, but instead use a variety of books and try to create our own approach based on those. Books we like include those by Shankar, Sakurai, Cohen-Tannoudji, and Merzbacher. I hope this helps!
do we have some constraints on these functions? if they are not bijective then we can't invert them so if our measure is related to an non invertible function of an operator, how can we knoow which of the base state of the operator we are in?
Unfortunately we don't have those available at the moment; but we are working on sharing more content (including PDFs and problems+solutions). Hopefully soon!
Thanks a lot for the video. It seems to me the addition of operators is not defined? Or maybe I am missing something? Is (A1 + A2)(f(x)) defined as A1(f(x)) + A2(f(x))? The definition of addition is needed for the expansion? Oh, perhaps it is just a given? I can sort of understand it if I map it to matrices and vectors. Oh, I see from the later equations that it is the case.......stupid question then....Thanks for the video.
You can in principle use the recipe in the video. However, this operator features in several problems associated with central potentials, in which it is more natural to work in the position representation, and then to turn from Cartesian to spherical coordinates. A few examples include: 1. 3D quantum harmonic oscillator: th-cam.com/video/jOQThICjLlw/w-d-xo.html 2. Hydrogen atom playlist: th-cam.com/play/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa.html I hope this helps!
The starting point is the power n of an operator, which means the application of that operator n times. From this, it is then natural to extend the definition to more general functions that can be represented as power series. I hope this helps!
Very clear, didactical explanation of Operators in Quantum Mechanics Professor M with beautiful edited math and very interesting stuff. I just came up with your channel 2 days ago and I will definitively follow You.
Glad you like it and many thanks!
Oh wow, the timing, that's what I needed today
Nice! And thanks for watching!
I have test of quantum mechanics this week, and we are already learning about spinors and approximation methods. I use your videos a lot, because the subjects I was supposed to know already are not entirely clear in my mind. So thanks a lot.
We are preparing a series on approximation methods, but it will only be ready in a few months time... good luck with your test!
Amazing explanation and great English pronunciation. Many thanks
Thank you! 😃
I am unsure whether we're actually using the fact that A commutes with [A,B] in the proof that [A, F(B)] = [A,B] F'(B).
It seems that we're only using that B commutes with [A,B] at 15:14, so the relation should still hold even if A does not commute with [A,B].
Obviously both commutative relations are needed for the example of x and p, but I figure this somewhat more general result should be noted.
I think you are correct, we don't need that A commutes with [A,B]; thanks for pointing this out!
Can we say that If A and B commute then [A^n , B^n] and [ B^n , A^n] are always commute ? Thanks !
Yes, because A^n and B^n are simply AAAA...A and BBBBB...B, and as you can exchange any A with any B, then you can also exchange the whole sequence.
Your videos are always great , I can say that [ ALWAYS , GREAT ] commute ;-)
Thanks!! :)
@@ProfessorMdoesScience please keep posting more videos
@@tanishqkaur2525 We are working on new videos, and we'll try to get them posted as soon as possible!
As always, your explanations are just brilliant. This is quite an involved topic, but it's redeemed by having a friendly face explain it in a clear way. Thank you for posting this, and for your hard work.
Thanks for your kind words of support! :)
thnx i am following Zettili and your videos are helping me a lot
Glad to hear this!
Hello, for the proof that [A,B^n]=n[A,B]B^(n-1) if [A,[A,B]]=0 and [B,[A,B]]=0 ( 15:39 ), I don’t understand why we need [A,[A,B]]=0 ??? It seems we do not use this condition, so does it works if we only suppose [B,[A,B]]=0 ? Thanks for yours videos, those are great!
Thanks!
Thanks for watching!
Thank you sir. I was searching for such a video. Please sir tell me from which book you are explaining this topic.
We don't use any specific books for creating the videos, but instead use a variety of books and try to create our own approach based on those. Books we like include those by Shankar, Sakurai, Cohen-Tannoudji, and Merzbacher. I hope this helps!
Wonderful sir! Really you are the fantastic teacher. Where are you from?...
Thank you!!!
Thanks for watching!
do we have some constraints on these functions?
if they are not bijective then we can't invert them
so if our measure is related to an non invertible function of an operator, how can we knoow which of the base state of the operator we are in?
CAN YOU PLEASE PROVIDE THE PDF FORM OF THE NOTE IN DESCRIPTION?
Unfortunately we don't have those available at the moment; but we are working on sharing more content (including PDFs and problems+solutions). Hopefully soon!
Thanks a lot for the video. It seems to me the addition of operators is not defined? Or maybe I am missing something? Is (A1 + A2)(f(x)) defined as A1(f(x)) + A2(f(x))? The definition of addition is needed for the expansion? Oh, perhaps it is just a given? I can sort of understand it if I map it to matrices and vectors.
Oh, I see from the later equations that it is the case.......stupid question then....Thanks for the video.
Yes, in quantum mechanics we work with linear operators where addition is defined as you correctly identify :)
Thank u so much sir
Awsm videos as well as explaination
Glad you like the videos! :)
nice
Thanks for watching!
What is the operator sqrt{ x^2 + y^2 + z^2 } ?
You can in principle use the recipe in the video. However, this operator features in several problems associated with central potentials, in which it is more natural to work in the position representation, and then to turn from Cartesian to spherical coordinates. A few examples include:
1. 3D quantum harmonic oscillator: th-cam.com/video/jOQThICjLlw/w-d-xo.html
2. Hydrogen atom playlist: th-cam.com/play/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa.html
I hope this helps!
amazing sir
why we write function of an operator in a power series
The starting point is the power n of an operator, which means the application of that operator n times. From this, it is then natural to extend the definition to more general functions that can be represented as power series. I hope this helps!
@@ProfessorMdoesScience ok thanks
Thank you so much
Thanks for watching!
The last property is just the chain rule in disguise since the commutator is just a Lie derivative :p
Thanks for the insight! :)