Anyone that can walk me through the "check your understanding questions" at the end? For the last 2, I've come to L^2f=0 for the first one and 6h^2f on the second (h being h-bar), am I correct? How does one answer the other 2?
Answer to the last one I think will be (15/4)h(bar)^2 f. No way its going to be 6h(bar)^2. regarding question 1 and 2, I got the answer 5 for Q1 and 7 for Q2. Am i right ?
manish singh - The answer to Q1 should not be 5. m is 2 and so l can range from -2 to 2. Since l is a half integer, you have to include more than just -2, -1, 0, 1, and 2. You also have to include half integers like 1/2. Therefore, the answer must be 9.
Well, the ladder operators bump you up and down in units of h bar right? Given that and the fact that the spectrum has to be symmetric about the origin, you would have to be either half or whole integer. That's where the distinction between half and whole integer spin particles come from in nature, since the spin of a particle is given by the maximum value of the z-component for instance, depending on your choice of basis right? And regarding the other questions. Since 15/4 is not an integer nor halfinteger, I believe the answer is 0. Regarding the last two, in the first case the possible values must be integers since it includes 0, and in the other case halfintegers by a simular argument. Do I have it right? Please correct me if not
Hey, I don't know if the comments reach you but your lecturing style is fantastic! Thanks for the help with my revision - I've always found that seeing the derivation of something helps me instantly understand and commit to memory.
check your understanding: . . . 1. L2|f> = 2hbar^2|f>, L = 1, Lz could be -1, 0 , 1 ( all * hbar ) . A total of 3 states. 2. L2|f> = 15/4 hbar^2|f>, L = 3/2, Lz could be -3/2, -1/2, 1/2, 3/2 ( all * hbar ). A total of 4 states. 3. Lz|f> = 0, L2|f> could be any (hbar)^2 * l * ( l+1 ), l = 0, 1/2, 1, 3/2, .... 4. Lz|f> = 3/2 * hbar |f> could be any (hbar)^2 * l * ( l+1 ), l = 3/2, 2, 5/2, 3, ... Shouldn't there be a |f> at the end for the 4th question?
Question 1&2 Answers: For the first one, l would need to equal 1 for l*(l+1) to equal 2. And there are always 2*l+1 states, so 3 states. For the second question, l=3/2 (again 3/2*(3/2+1)=15/4), and there are 2*l+1 states, so 3/2*2+1=4 states.
I do understand that L± acting on an eigenfunction can find new eigenfunctions. However I do not understand why there can't be more eigenfunctions. Why can't there be an other operators that finds finds other eigenfunctions? What did I miss?
Great video. But I had a doubt. At 29:26 you say that L+ft=0. But you also mentioned that the highest value that Lz can take is L. So, why exactly is L+ft=0 and not L(total angular momentum)? Sorry if I am being naive. I am new to Quantum Mechanics.
+Krishna Jha I would like to see Dr. Carlson answer this too, but I believe the answer is that (L_+)(f_t)=0 because an eigenstate that is zero everywhere is non-normalizable, which means it is not physically possible.
41:58 There is a small detail I would want to highlight: after concluding that l_(l_-1)=l(l+1), you argue that the ONLY solution must be l_=-l As I was watching the video, I was expecting l_(l_-1)=(l_-1)l_=l(l+1), thus having l_=l+1 Of course, as it is a second degree equation on l_ (or l) there are two solution; the other one being l_=-l So, unless you add the condition l_
Very good video. I was wondering about the step at around 16:10. You mention adding Lz into the middle of the two Lx's and then go on to say "this is essentially an identity". How/where can I find a way to carry out this identity?
Basically he just added 0 to the equation, by adding and subtracting the same thing allowing him to later use the terms added to define the commutator. Hope that helped.
please correct me if I am wrong but I believe that at 9:07 it should be zzPyPy, not zzPyPx. This then means that the two terms wont cancel later at 9:35
+Kalyan jyoti Kalita No, they're not. The hamiltonian is time dependent, as defined by the schrodinger equation. The time independent form of the schrodinger equation has eigen-functions that are stationary states. These solutions were examined previously and is what he's referring to. In general if you know the stationary state, you know (can calculate ) the state at time = t (any time later on) by tacking on the time dependent term psi(t)=exp(iEt/hbar) as long as the potential is time invariant. The time dependent term is only interesting if we're looking at superpositions/linear combinations of wave functions, in which case the phase cancelations give interesting results (as seen at the end of the 19th video/lesson). If the potential varies over time then time-dependent purturbation theory is used, and from my understanding as of so far, thats only used when the time dependent portion is small in comparison to the time independent portion. I'm also new to this, so im probably not 100% correct.
No. This terms comes from Linear Algebra. In comparison, E.state is similar to a vector and E.value to a scalar. In QM it is mostly used because of Schrodinger equation.
After watching this video, I kind of started understanding the quantum mechanics ..Thank you so much Mr Carlson.
hh kidn of XD
Anyone that can walk me through the "check your understanding questions" at the end?
For the last 2, I've come to L^2f=0 for the first one and 6h^2f on the second (h being h-bar), am I correct? How does one answer the other 2?
Answer to the last one I think will be (15/4)h(bar)^2 f. No way its going to be 6h(bar)^2.
regarding question 1 and 2, I got the answer 5 for Q1 and 7 for Q2. Am i right ?
@@manishsingh-vk8if I think the answer for Q2 must be 6, because N has to be an even number and l goes in steps of hbar, so it cannot be zero.
manish singh - The answer to Q1 should not be 5. m is 2 and so l can range from -2 to 2. Since l is a half integer, you have to include more than just -2, -1, 0, 1, and 2. You also have to include half integers like 1/2. Therefore, the answer must be 9.
Well, the ladder operators bump you up and down in units of h bar right? Given that and the fact that the spectrum has to be symmetric about the origin, you would have to be either half or whole integer. That's where the distinction between half and whole integer spin particles come from in nature, since the spin of a particle is given by the maximum value of the z-component for instance, depending on your choice of basis right? And regarding the other questions. Since 15/4 is not an integer nor halfinteger, I believe the answer is 0. Regarding the last two, in the first case the possible values must be integers since it includes 0, and in the other case halfintegers by a simular argument. Do I have it right? Please correct me if not
your videos are great. This adds so much value to youtube, having such good education available to the whole world for free.
You explained this infinitely better than my professor. Thanks, my dude.
Hey, I don't know if the comments reach you but your lecturing style is fantastic!
Thanks for the help with my revision - I've always found that seeing the derivation of something helps me instantly understand and commit to memory.
30:49 ladder puns "rung by rung"... This man is a real legend
It’s almost as if they call them ladder operators for a reason, weird.
check your understanding:
.
.
.
1. L2|f> = 2hbar^2|f>, L = 1, Lz could be -1, 0 , 1 ( all * hbar ) . A total of 3 states.
2. L2|f> = 15/4 hbar^2|f>, L = 3/2, Lz could be -3/2, -1/2, 1/2, 3/2 ( all * hbar ). A total of 4 states.
3. Lz|f> = 0, L2|f> could be any (hbar)^2 * l * ( l+1 ), l = 0, 1/2, 1, 3/2, ....
4. Lz|f> = 3/2 * hbar |f> could be any (hbar)^2 * l * ( l+1 ), l = 3/2, 2, 5/2, 3, ...
Shouldn't there be a |f> at the end for the 4th question?
Question 1&2 Answers: For the first one, l would need to equal 1 for l*(l+1) to equal 2. And there are always 2*l+1 states, so 3 states. For the second question, l=3/2 (again 3/2*(3/2+1)=15/4), and there are 2*l+1 states, so 3/2*2+1=4 states.
Shouldn't it be 5 and 7? as each step is seperated by only 1/2
@@michaelchang2012No. The answers will be 3 and 4 respectively.
@@soniabiswas3703Yup, just realized that for m it’s always separated by 1 not 1/2
@@michaelchang2012 I dont Understand the first two questions but the thrid should be 0 and the fourth schould be 15/4 if i understood it correctly?
I'm a british physics student and this massively helped me at university. Your playlist is thorough and explainative. Many many thanks
I do understand that L± acting on an eigenfunction can find new eigenfunctions. However I do not understand why there can't be more eigenfunctions. Why can't there be an other operators that finds finds other eigenfunctions? What did I miss?
thank you thank you thank you thank you thank you!!!!!!!!!!!!!!!!!!
simple and effective explanation indeed.
Very useful video. Thank you for your clear explanation.
really thank you finally I understand the whole angular momentum 🌹🌹🌹
Great video.
But I had a doubt. At 29:26 you say that L+ft=0. But you also mentioned that the highest value that Lz can take is L. So, why exactly is L+ft=0 and not L(total angular momentum)?
Sorry if I am being naive. I am new to Quantum Mechanics.
+Krishna Jha I would like to see Dr. Carlson answer this too, but I believe the answer is that (L_+)(f_t)=0 because an eigenstate that is zero everywhere is non-normalizable, which means it is not physically possible.
+Krishna Jha Lz cannot equal L because they do not commute.
What? @19:51 they definitely commute.
@@Winium L^2 and Lz commute.
I be watching this on my pz
Thanks for the video. Is there any video on the calculation ΔLx= sqrt( - ^2 ) ?!?
What is the result of [ Lz,L^2+_] ? PLEASE I NEED TO KNOW
thank you for the explain but you should adding some questios for every section
>>>>>>realy very good lecture
Thanks Dr Brant carlson
Fantastic video, this is helping me with my Mathematical Physics classes. Thanks!
Can you cancel out terms that have double Z and Pz operators since ZZ and PzPz commute?
classrooms are very outdated at this point. I'd rather listen to the best lecturers anytime I want to.
Thank u sir...it really helps me to understand q.m. better.
At 17.32 shouldn't it be lx(lz,lx)
Your video is life saving, thank you !
Thank you so much for these videos! We follow Griffiths four our college and this is indeed very very helpful for my exams!! Thank you so much! 💜
Thanks!
I could not be more thankful...
From Iraq, thank you so much🌱
Very clear and helpful explanation, Thank you!
41:58 There is a small detail I would want to highlight: after concluding that l_(l_-1)=l(l+1), you argue that the ONLY solution must be l_=-l
As I was watching the video, I was expecting l_(l_-1)=(l_-1)l_=l(l+1), thus having l_=l+1
Of course, as it is a second degree equation on l_ (or l) there are two solution; the other one being l_=-l
So, unless you add the condition l_
He should have said this explicitly, but the condition l_
nice thanks was looking for it
Very good video. I was wondering about the step at around 16:10. You mention adding Lz into the middle of the two Lx's and then go on to say "this is essentially an identity". How/where can I find a way to carry out this identity?
Basically he just added 0 to the equation, by adding and subtracting the same thing allowing him to later use the terms added to define the commutator. Hope that helped.
So. Thank you very much nice
Very well explanations ...
please correct me if I am wrong but I believe that at 9:07 it should be zzPyPy, not zzPyPx. This then means that the two terms wont cancel later at 9:35
Ricardo Morin his x looks like y. Thats all.☺
very useful series. I have a doubt regrading the eigenstate of Hamiltonian operator you mentioned. Is it always a stationary state?
+Kalyan jyoti Kalita No, they're not. The hamiltonian is time dependent, as defined by the schrodinger equation. The time independent form of the schrodinger equation has eigen-functions that are stationary states. These solutions were examined previously and is what he's referring to. In general if you know the stationary state, you know (can calculate ) the state at time = t (any time later on) by tacking on the time dependent term psi(t)=exp(iEt/hbar) as long as the potential is time invariant. The time dependent term is only interesting if we're looking at superpositions/linear combinations of wave functions, in which case the phase cancelations give interesting results (as seen at the end of the 19th video/lesson). If the potential varies over time then time-dependent purturbation theory is used, and from my understanding as of so far, thats only used when the time dependent portion is small in comparison to the time independent portion. I'm also new to this, so im probably not 100% correct.
so nice .. thank you so much
thank you so much
you're so smart.
حلو جدا
Really like this, thanks
l = 0 , + or - 1/2 , + or - 1 ..............
is eigenstate and eigenfunction synonymous?
No. This terms comes from Linear Algebra. In comparison, E.state is similar to a vector and E.value to a scalar. In QM it is mostly used because of Schrodinger equation.
how do we confirm the answers for check your understanding?
Nice math but no insights. No physics. No explanation about why the operator behave in that way.
Shut up bitch u ever try to talk shit on Brant Carlson and I’ll find you.
that's what I neeeeeeed!