Angular momentum operators and their algebra

[@thechaoswolf2941]

just wanna say i have my QM final tomorrow and this saved me, tysm

[@RJRyan]

It's so beautiful it hurts.

[@knowbeautiful2615]

RJ Skerry-Ryan I want to talk to you about an important issue related to quantum mechanics, I hope you will respond to my message 🙏🏻🙏🏻

[@knowbeautiful2615]

RJ Skerry-Ryan Please

[@Kalumbatsch]

@@knowbeautiful2615 I am the press secretary, please tell me all about your important issue.

[@erikgonzales3178]

I dropped out in 11th grade because I had no interest in school but as I get older I gain more interest in things like this. It sucks that my interest in knowledge didn't gain traction until a decade after I dropped out and now I feel like I lost out

[@willplays7954]

The universe is your real school

[@MerinaShow]

Community college will always accept you and there are often times grants for students of all kinds of backgrounds. I joined CC "late" and took my time learning many concepts and enjoying the CC life, then transferred to university just last year. I am now studying physics at Berkeley. I spent years working and never thought I would join school, but I was also really bored and found myself learning quantum mechanics from KZbin at 4am. You can do it!

[@chumuheha]

This guy is a phenomenal lecturer but I would recommend listening to this in 1.25 or 1.5 speed.

[@chaitanaymittal9075]

I watched it in 2x dude

[@Boooommerang]

Yes! He is a phenomenal lecturer and Professor!

[@aniksarkar8658]

What happen when value of z component of angular momentum become zero as m quantum number become zero. Other components seems to commute at this state then does uncertainty principle fail?

[@adityakapdi939]

The actual value of angular momentum doesn't matter when talking of uncertainty, only the magnitude of the operator Lz=hbar. So no matter what the value of 'm' is, the uncertainty principle still holds true

[@ianbrown6639]

I believe you still have uncertainty in the angular momentum in the x and y directions

[@davidsoncheng6905]

You are right that when the value of the z component is zero, it is possible to simultaneously measure both Lx, and Ly with certainty. In fact, since you already know Lz = 0 and you can measure both Lx and Ly exactly, you can know all 3 at the same time! But I wouldn't call this the uncertainty principle failing, because the uncertainty principle only tells you the relationship between the product of the variances of two observables and the commutation, and here, this commutation happens to be 0.

[@Boooommerang]

Thanks, Professor!

[@jacobvandijk6525]

ONCE AGAIN, mr Zwiebach is very interested in the mathematical side of QM. But what about the physical side? According to QM, angular momentum L = (l . (l + 1))^2, with the quantum number l = n - 1. So when n = 1 = the ground state, we have l = 0. But l = 0 results in no angular momentum L !!! This means that in its ground state, the H-atom has an electron that isn't revolving around the nucleus. Ask mr Zwiebach how he explains this. Because in the entire course, he won't tell you about this physical problem.

[@jamshedsaeed9445]

Sir How can you solve or find out from first stepof [lx,ly] that ypz commute with zpx etc? .

[@davidsoncheng6905]

In the first step of [lx,ly], he took out px because it is independent of z,pz, and y. To see this, remember how if you take the derivative of a function with respect to x, you can treat all z,y-related terms in the function as constants. Here, he took out everything unrelated to x, which is [ypz, z], and treated it as a constant.

[@Boooommerang]

Aos 6': comutadores: [LxLy], e assim por diante. Fazer

[@thendomatshisevhe4365]

Wow thanks so much🤲🏾

[@not_amanullah]

This is helpful ❤️🤍

[@not_amanullah]

Thanks ❤️🤍

[@wondererasl]

Why does Pz and y commute ?

[@scalesconfrey5739]

Pz = -ihbar d/dz, and y = y. d/dz yf = y d/dz f, so the order doesn't matter ( [y, Pz] = 0 ).

[@beetlesstrengthandpower1890]

@@scalesconfrey5739 Bro thank you sm

[@sukdipkar8876]

What does it mean two operators commutes?

[@joycelynlongdon6933]

[A,B]=A*B-B*A

[@ysaackfranco2825]

Thank you

[@rohanmathew5728]

The commutation relation of two operators strikes right at the nature of their physical observability. That is, if two operators commute, it implies that they have a common Eigen vector to act upon and hence produce an Eigen value. In contrast, if they don't commute, these operators find no Eigen vectors that are common to both and hence cannot act simultaneously. To make things a bit more simple, we can, on a lighter note, say that when two operators commute, the measurements corresponding to each of them (made on the system) can be taken simultaneously and if they do not commute, they cannot be measured simultaneously. Hope it was helpful.

[@rubahasan8107]

@@rohanmathew5728 at the end you must say about two operators that don’t commute that you can’t measure them simultaneously ” with high precision “. because you actually can measure them simultaneously but not with high accuracy.

[@souravbhowmik781]

great

[@قتيبةالصالح-ز5د]

Please prove that L2L-+=o how this With big welcome from iraq

[@Musfiqur_anik]

He did really go from basic math operations to real analyses 💀

[@schoob69]

when did he go to real analysis lol, he just started talking about algebras

[@emad-phy]

hello , prove [Lz, r2]=0

[@cesareduardogarza3446]

[Lz,x^2]+[Lz,y^2]+[Lz,z^2] [Lz,x]x+x[Lz,x]+[Lz,y]y+y[Lz,y]+[Lz,z]z+z[Lz,z] (ihxy)+(ihxy)+(-ihxy)+(-ihxy)+0+0 0

[@erick.gudino]

great