I have never seen the subject of symmetry introduced so naturally and so simply in an introduction to QM. It makes it compelling and simple. Just brilliant.
That's cool and all, but did you notice that the Schroedinger equation does not even obey local conservation laws? Exactly. Most students don't and I have not seen many QM 101 lecturers who are pointing out the necessary reasons for that.
He mentioned Noether's theorem and invariance. It would be interesting to have a whole lecture on that among other philosophical issues in the history of physics. Perhaps that is more of a Relativity class but it would still be fun.
If x and px don't commute, it means one cannot find complete set of common eigen state, but one may find some (not all, to form basis) common eigen state. For example, Lx and Ly don't commute, but we have a state designated by l=0 (s-state) which is eigen state of both Lx and Ly (but other states are not simultaneous eigen states of these two operators.
So with regard to "simultaneously eigenfunctions of both operators" AND picking the separable solutions to the differential equations... In this lecture we separate the x, y, and z factors and the solutions are simultanious eigenfunctions of all three variables. In the next lecture the same thing is done less explicitly with the angular momentum in the z direction and the magnitude of the total angular momentum. There is definitely a connection here. The connection being that the solutions that separate in such a way, ARE THE SOLUTIONS THAT ARE SIMULTANIOUSLY EIGENFUNCTIONS OF BOTH OPERATORS. It can easily be seen that this is the case. And THAT is why we only take those solutions. They are complete with regard to one another. Now I want to say that commuting with the hamiltonian makes these separable solutions also eigenfunctions of the hamiltonian(meaning that they form a complete basis, since the eigenfunctions of the hamtonian create a complete basis), but it doesn't if the hamiltonian has degenerate states. It only means that a linear combination of them must be an eigenfunction of the hamiltonian (in the case that the hamiltonian is degenerate). However, this still makes them a complete basis. If the hamiltonian isn't degenerate, the former is true, and you have found the energy states explicitly.
Mistake at 17:30. Having two non-commuting operators A and B means that there is not a complete set of states that has the certain values of A and B. Yes, in the case where the commutator is a constant, no simultaneous eigenstates exist, but in the general case, it certainly does NOT mean that we cannot get certain values of the two operators simultaneously. This only means that there is at least one state that we cannot have a certain value of the two operators. This is a widespread misconception.
When [A, B] equals a non-zero constant in the complex field, there exists no simultaneous eigenfunction of both A and B. A non-zero constant can't annihilate any wavefunction. However, if [A, B] equals another nontrivial operator, then it's possible to have simultaneous eigenfunctions of A and B.
siriuslh you are right, I have edited my comment to make my objection clearer. Operators whose commutators have a zero-dimensional null space cannot have a simultaneous eigenvector. But in general not having simultaneous eigenstates does not directly follow from "not commuting". I think one has to be careful about this.
what is the cameraman even doing all the time? zooms in the face when not required. when the prof runs to the other board after like 10 seconds he finally follows. when the prof is explaining contents in the lower board he zooms to the upper board. Man it screws this great lecture
Chanakya Tiwari it’s a difficult topic, as is most of this course. It is just being presented here with a lot more clarity than many other places/lecturers. It’s still difficult. As for possible reasons as to why you didn’t understand the lecture - more context is needed - this is lecture 15 in this course. Did you understand all the previous ones but found this one above your head, or did you just land in this lecture directly... it’s a difficult enough topic as it is... While many understand this one, rest assured, many also don’t. Just like you. Hope this improves for you in the future :)
@@grolich This is not a difficult topic, at all. Nonrelativistic quantum mechanics is some of the most straightforward math you will find in any of physics. There are next to no pitfalls here.
His lectures are highly enthusiastic and informative but the mathematician in me worries when he plays fast and loose with unproved assertions, uniqueness and existance. Is there somewhere in the MIT series that looks a bit more rigorously at the mathematical underpinnings?
8.05 Quantum Physics II ruclips.net/video/QI13S04w8dM/видео.html (course materials: ocw.mit.edu/8-05F13 ) might be more interesting for you. Topics covered in 8.05 include the general formalism of quantum mechanics, harmonic oscillator, quantum mechanics in three-dimensions, angular momentum, spin, and addition of angular momentum. Best wishes on your studies!
Your assertion that there are "mathematical underpinnings" to non-relativistic quantum mechanics is misguided. You can easily check that the Schroedinger equation does not even obey local conservation laws. Neither does the projection operator formalism. Nonrelativistic QM is an ad-hoc approximation that works for a very small number of toy systems and nothing else. If anything, the major weakness of most introductory physics courses is to over-emphasize the mathematics of the formalism while making essentially no contact to actual physics.
The camera is automated. It locks onto the doo-hickey he has in one of his pockets that allows the camera to track him instantaneously. If something is screwed up, it's a fault of the programming or the camera settings, etc.
Why is he calling these states by numbers? Is it alphabetical, like Alabama=(000) and Alaska=(001)... etc? And the degeneracy are how many families with multiple generations raised in the state?this isn't what i thought quantum mechanics would be.
Professor Adams has great presentation skills, however his method of trying to express every single detail while speaking is very misleading, at least for me.
I have never seen the subject of symmetry introduced so naturally and so simply in an introduction to QM. It makes it compelling and simple. Just brilliant.
That's cool and all, but did you notice that the Schroedinger equation does not even obey local conservation laws? Exactly. Most students don't and I have not seen many QM 101 lecturers who are pointing out the necessary reasons for that.
He mentioned Noether's theorem and invariance. It would be interesting to have a whole lecture on that among other philosophical issues in the history of physics. Perhaps that is more of a Relativity class but it would still be fun.
Incredible lecture. Very helpful in refreshing my understanding in preparation for the PGRE
If x and px don't commute, it means one cannot find complete set of common eigen state, but one may find some (not all, to form basis) common eigen state. For example, Lx and Ly don't commute, but we have a state designated by l=0 (s-state) which is eigen state of both Lx and Ly (but other states are not simultaneous eigen states of these two operators.
This is just awesome , thanks MIT.
So with regard to "simultaneously eigenfunctions of both operators" AND picking the separable solutions to the differential equations... In this lecture we separate the x, y, and z factors and the solutions are simultanious eigenfunctions of all three variables. In the next lecture the same thing is done less explicitly with the angular momentum in the z direction and the magnitude of the total angular momentum. There is definitely a connection here. The connection being that the solutions that separate in such a way, ARE THE SOLUTIONS THAT ARE SIMULTANIOUSLY EIGENFUNCTIONS OF BOTH OPERATORS. It can easily be seen that this is the case. And THAT is why we only take those solutions. They are complete with regard to one another. Now I want to say that commuting with the hamiltonian makes these separable solutions also eigenfunctions of the hamiltonian(meaning that they form a complete basis, since the eigenfunctions of the hamtonian create a complete basis), but it doesn't if the hamiltonian has degenerate states. It only means that a linear combination of them must be an eigenfunction of the hamiltonian (in the case that the hamiltonian is degenerate). However, this still makes them a complete basis. If the hamiltonian isn't degenerate, the former is true, and you have found the energy states explicitly.
Someone, please teach the cameraman how to handle a camera...i'm getting dizzy every 10s due to excessive zoom in and out, and movement of the camera
15:40 it's stone's theorem if somebody wants to know!
Mistake at 17:30. Having two non-commuting operators A and B means that there is not a complete set of states that has the certain values of A and B. Yes, in the case where the commutator is a constant, no simultaneous eigenstates exist, but in the general case, it certainly does NOT mean that we cannot get certain values of the two operators simultaneously. This only means that there is at least one state that we cannot have a certain value of the two operators. This is a widespread misconception.
When [A, B] equals a non-zero constant in the complex field, there exists no simultaneous eigenfunction of both A and B. A non-zero constant can't annihilate any wavefunction. However, if [A, B] equals another nontrivial operator, then it's possible to have simultaneous eigenfunctions of A and B.
siriuslh you are right, I have edited my comment to make my objection clearer.
Operators whose commutators have a zero-dimensional null space cannot have a simultaneous eigenvector. But in general not having simultaneous eigenstates does not directly follow from "not commuting". I think one has to be careful about this.
@@siriuslh why in the quantum harmonic oscillator a+ =(x/xo)-i(p/po) and a =(x/xo)+i(p/po) but in the angular momentum L+ =Lx + iLy and L- =Lx - iLy
Complete Set Of Commuting Observables @1:15:00
what is the cameraman even doing all the time? zooms in the face when not required. when the prof runs to the other board after like 10 seconds he finally follows. when the prof is explaining contents in the lower board he zooms to the upper board. Man it screws this great lecture
nice lecture. interesting and helpful at times
It seems all genius here . Am I the only fool who didn't understood the lecture ?
Chanakya Tiwari it’s a difficult topic, as is most of this course. It is just being presented here with a lot more clarity than many other places/lecturers.
It’s still difficult.
As for possible reasons as to why you didn’t understand the lecture - more context is needed - this is lecture 15 in this course. Did you understand all the previous ones but found this one above your head, or did you just land in this lecture directly... it’s a difficult enough topic as it is...
While many understand this one, rest assured, many also don’t. Just like you. Hope this improves for you in the future :)
@@grolich This is not a difficult topic, at all. Nonrelativistic quantum mechanics is some of the most straightforward math you will find in any of physics. There are next to no pitfalls here.
At the start, he mentions previous years exams on OCW. Is this correct? Where could we find them?? :)
Thomas Moore Yes, this is correct. See ocw.mit.edu/8-04S13 for the complete course materials.
Should i read zettili?
When should you use the lowering operator?
This is helpful ❤️🤍
Are the notes for this lecture available anywhere? I couldn't find them on the OCW site :(
Sorry, we only have lecture notes for the first eight lectures.
@@mitocw why?
At 1:03:25 it should be del^2phi
What is the fourrier theorem he mention in min 9:34 ?
Camera guy buggin
can somebody tell me , what was the question that is asked at 46:40.
The zoom job is bonkers
he is so, so, so great. I just wish he would talk a LITTLE bit slower :(
Thank you.
His lectures are highly enthusiastic and informative but the mathematician in me worries when he plays fast and loose with unproved assertions, uniqueness and existance. Is there somewhere in the MIT series that looks a bit more rigorously at the mathematical underpinnings?
8.05 Quantum Physics II ruclips.net/video/QI13S04w8dM/видео.html (course materials: ocw.mit.edu/8-05F13 ) might be more interesting for you. Topics covered in 8.05 include the general formalism of quantum mechanics, harmonic oscillator, quantum mechanics in three-dimensions, angular momentum, spin, and addition of angular momentum. Best wishes on your studies!
Your assertion that there are "mathematical underpinnings" to non-relativistic quantum mechanics is misguided. You can easily check that the Schroedinger equation does not even obey local conservation laws. Neither does the projection operator formalism. Nonrelativistic QM is an ad-hoc approximation that works for a very small number of toy systems and nothing else. If anything, the major weakness of most introductory physics courses is to over-emphasize the mathematics of the formalism while making essentially no contact to actual physics.
Thanks 🤍❤️
In mit, even the camera has adhd
Is the camera shooter a new guy? He screwed up the course!
The camera is automated. It locks onto the doo-hickey he has in one of his pockets that allows the camera to track him instantaneously. If something is screwed up, it's a fault of the programming or the camera settings, etc.
Why is he calling these states by numbers? Is it alphabetical, like Alabama=(000) and Alaska=(001)... etc? And the degeneracy are how many families with multiple generations raised in the state?this isn't what i thought quantum mechanics would be.
+ardweaden it was a joke...>_>
Professor Adams has great presentation skills, however his method of trying to express every single detail while speaking is very misleading, at least for me.
'Misleading' is not right word. Distracting perhaps, but not misleading. at all.
💐
THERMODYNAMIC SYSTEMS
so just for fun i am going to comment this comment
Wait a minute, is he wearing socks? 11:18
He always does.