Advanced Quantum Mechanics Lecture 2

(September 30, 2013) Leonard Susskind presents an example of rotational symmetry and derives the angular momentum operator as the generator of this symmetry. He then discusses symmetry groups and Lie algebras, and shows how these concepts require that magnetic quantum numbers - i.e. spin - must have whole- or half-integer values.

Originally presented by the Stanford Continuing Studies Program.

Stanford University:

Continuing Studies Program:

Stanford University Channel on YouTube:

[joabrosenberg]

Energy degeneracy 5:00; Particle on a circle, Rotational invariance, Angular momentum 7:00; Angular momentum is an integer 21:00; Magnetic field breaks the degeneracy 24:30; Mirror/reflection symmetry 26:00; Reflection and Rotation do not commute 35:00; The commutator of two symmetries is a symmetry 41:30; Commutator Algebra (Lie Algebra), Symmetry group 46:30; Commuting symmetries are Abelian 50:00; Angular momentum 53:15; Creation and Annihilation operators 1:19:30; Lz can be either integer or half integer 1:38:00; All eigen states of Lz has the same energy (degeneracy) 1:42:30;

[qbtc]

I consider this Advanced QM series and the first 10 QM lectures together to be one course. Same with the General Relativity and Cosmology lectures. These lectures are pure gold. He covers some topics in depth that textbooks only gloss over like the derivation of the Pauli spin matrices. Awesome.

[MARILYNANDERSON]

Thanks to the University for taking the opportunity to share real coursework with those of us who love the intrigue of  Dr. Susskind's mental atheletics. A subject made fascinating and applicable. Believe me, Leonard could not work any harder to get this understanding in my brain-  nice that he cares about our "learning something" even if it is "more difficult than learning nothing" . Obvious: he likes his job..

[hasanshirazi]

"p" is used for linear momentum because it comes from Latin "Petere" meaning "impetus" which was used in place of "momentum" in earlier times. "i" & "m" were already taken so "p" was the next best choice..
Angular momentum in terms of vector cross product is L = r x p. Since the two vectors "r" and "p" are perpendicular to each other in x-y plane hence their cross product is represented by "L" i.e. two lines at 90 deg. to each other.

[TheMathestar]

Take a shot each time he talks about dropping h bar

[TheVivek]

52:09

After he presented a lecture at a conference, one colleague raised his hand and said: "I don't understand the equation on the top-right-hand corner of the blackboard". After a long silence, the moderator asked Dirac if he wanted to answer the question, to which Dirac replied: "That was not a question, it was a comment. [Wikipedia]

[simoncrase]

I took a course an Quantum Field Theory 50 years ago. It's a pity we didn't have access to Leonard Susskind's lectures on YouTube back then, as I would have understood the earlier course. Better late than never.

[MikeRoePhonicsMusic]

An upper case L makes perfect sense for angular momentum because the L is made of 2 lines intersecting at a 90 degree angle.

[KipIngram]

9:43 - He has it right. The idea is that you've rotated slightly counterclockwise (i.e., increased theta). So (new theta) - (epsilon) = (old theta). I.e., if (new theta) is larger than (old theta), then you have to subtract something from (new theta) to get the function to pump out the previous value.

[zphuo]

@52:00 G in formula U=1-ieG means generator which include all group information..

[sijojosephdr]

Modern Dirac says "That is not a question" !!!

[fryztv]

tfw you feel special because Mr.Susskind says "It's in our book" and that you remember "It" in the book

[jy_decipherer_]

I don't know if anyone's raised this question yet, but I still don't understand why the z-component of angular momentum is quantized from the ladder operator argument (I mean I do agree with the derivation where we use the fact that the wave function has to be the same when θ = 2π, but not so much with the ladder operator argument). The confusion I'm having is even if we have ladder operators that raise or lower Lz by an increment of ℏ (or an increment of 1 since Prof. S drops ℏ), how does that guarantee that there are no other states in with Lz between 2 integers? It's not like there's a rule that all states that exist have to be reached by the ladder operator right?

[IronCharioteer]

How the heck do you get to "Advanced" QM without having covered the harmonic oscillator?

[petergreen]

❤beautiful lesson thank you very much professor and class

[englishinenglish]

1:24:00 LOL . Me trying to find where I’ve written creation and annihilation operators : “Bloody hell where could I write it down “ .
Me 1 minute later : Huh, I thought I would die 😂😂😂 from laugh

[filiprunjic]

Algebra (from Arabic "al-jabr", literally meaning "reunion of broken parts")

[ChristopherAquilo]

I feel that gravity may be accommodating. I wonder if I changes under observation or that we may properly be able to interpret I. There are many types of energy.

[ZHIYIC]

How many people in this classes, I like the atmosphere there, envy them!

[ashoknaganur]

Understood symmetry of angular momentum and generacy of energy