at 19:05, i think, the anti-parallel alignment(l-1/2) is more “favourable”, has a lower energy. That is why l+1/2(maximum J) appears at higher binding energy in photoemission spectrum (XPS).
Thank you very much for these excellent videos. At about 18.08 you explain how the positive LS term depresses the energy level. I,m confused why the positive LS term reacts this way with the negative values of the Potential Well . Again, many thanks.
When LS is positive there is some energy "trapped inside" the coupling, and when it is negative it has "released the energy". In other words, when LS is positive it takes more energy to rip off the electron.
Thank you for the lecture. It helps me a lot! But I still have some questions, hope that someone could answer me, please. 1. Why does the j term have only one value when L=0? Don't we take account to the other j value when s=-1/2? 2. I don't quite understand how does the LS term work on the binding energy of the levels. What does the the value in LS term mean? And why the negative term means less contribution to the potential well? I have watched another video (shell model video) mentioned in this video, but still can't find these answers. Please answer my questions if you could. Thank you very much for your kindness!
Actually, S has only one value (1/2); it's ms(components of s) which takes positive or negative values and make +1/2,-1/2. So j would still have only one value if l is zero.
+Stefan Epler - Snow however OP didnt said anything if n=3 or not. So i assume, he knew that much. If n=3, l = 2,1,0 and each j is equal l+s. so, there is 3/2.
for the operators on these vectors wouldnt it be more proper (or more easily destinguished from other values) for those tems to have a "hat" or ( ^ ) above them?
verumINscientia A lot of book don't even bother about putting those '^' above the operator, as people assume once you able to reach to this video, you already well known about operator.
when s=1/2 the term of -s(s+1)=-3/4 and when s=-1/2 the term of -s(s+1)=+1/4,then you calculate LS term in l=0 will give you both 0. then when calculate l=1 LS term should be l*(h bar sqr/2) and -l*(h bar sqr/2). That should be right answer. I think professor make a small mistake here.
s is always 1/2, there is a point where he says that s=+1/2 or -1/2 in explaining why j can be l+s or l-s, but this is because j can take values from modulus(l-s) up to l+s in integer steps. As l=1 and s=1/2, j can be either 1/2 or 3/2. Hope that helps :)
quantum number s for the electron is always a half. But the z component projection m_s may be plus or minus a half. the s^2 operator only acts on the quantum number s and has eigenvalue s(s+1), the s^2 operator does not act on m_s which is why we do not consider the case for m_s equal to minus a half.
Are you sure +1/2 decreases the energy level and -1/2 increases it? Because everywhere else I look it is the other way round. Or maybe I get mixed up with notations? Does j-j coupling and LS coupling lead to different energy directions?
I feel the same way..when it's positive it should reduce the energy level since it's negative therefore less energy is needed to release the electrons not more
I need someone who know how to use it CCFULL programe for the japanese scientists [ K. Hagino , N. Rowley and A. T. Kruppa ] called : A FORTRAN77 program for coupled-channels calculations with all order couplings for heavy-ion fusion reactions .
Why call the video spin orbit coupling then go through really basic physics at the beginning that anyone who's watching this video will already know? and you don't even explain the physics of spin orbit coupling. what a waste of my time
I love how you sound a little bit like a kids tv presenter - much more engaging than regular tutorials! Even for undergrads! :D
This is great stuff. Would have been very useful during my degree.
Thanks man real easy to follow and clear, helped a lot to work it out in my head
I really should thank you sir, you helped me to pass my exam.
Congratulations. Well done!
Sir you are awesome, please keep making videos... ur knowledge is much needed, thanks alot :)
Thanks for this nicely presented lecturer.
great explanation, thanks
Brilliant explination!
What a beauty!
at 19:05, i think, the anti-parallel alignment(l-1/2) is more “favourable”, has a lower energy. That is why l+1/2(maximum J) appears at higher binding energy in photoemission spectrum (XPS).
thank you for this great video!
Best ever explanation
hey , finaly i understand what spin-orbit-coupling is . thanks
really wonderfull thank you so much
Thank you very much for these excellent videos. At about 18.08 you explain how the positive LS term depresses the energy level. I,m confused why the positive LS term reacts this way with the negative values of the Potential Well . Again, many thanks.
When LS is positive there is some energy "trapped inside" the coupling, and when it is negative it has "released the energy".
In other words, when LS is positive it takes more energy to rip off the electron.
thanks for the video
Thank you for the lecture. It helps me a lot!
But I still have some questions, hope that someone could answer me, please.
1. Why does the j term have only one value when L=0? Don't we take account to the other j value when s=-1/2?
2. I don't quite understand how does the LS term work on the binding energy of the levels. What does the the value in LS term mean? And why the negative term means less contribution to the potential well?
I have watched another video (shell model video) mentioned in this video, but still can't find these answers. Please answer my questions if you could. Thank you very much for your kindness!
Actually, S has only one value (1/2); it's ms(components of s) which takes positive or negative values and make +1/2,-1/2.
So j would still have only one value if l is zero.
It’s harder to locate the math on one side as a crystal.
I always watch your videos
Thank you very much
your always the best
im a little confused, why have you said that s(s+1) is always 3/4? when s could be +/- 1/2?
Your voice resembles that of Richard Dawkins. (in my ears at least)
Good lecture.
Ikr I couldn't stop thinking about that
Thankyou
Awesome!!!
thanks!
Please keep it up
Brilliant video. I wanted to ask, for l=2, what are the possible values of j?
Dipanshu Gupta 5/2
+aknelkaiser Could it b 3/2 as well?
+Stefan Epler - Snow Yes.
+Stefan Epler - Snow however OP didnt said anything if n=3 or not. So i assume, he knew that much. If n=3, l = 2,1,0 and each j is equal l+s. so, there is 3/2.
for the operators on these vectors wouldnt it be more proper (or more easily destinguished from other values) for those tems to have a "hat" or ( ^ ) above them?
verumINscientia A lot of book don't even bother about putting those '^' above the operator, as people assume once you able to reach to this video, you already well known about operator.
when s=1/2 the term of -s(s+1)=-3/4 and when s=-1/2 the term of -s(s+1)=+1/4,then you calculate LS term in l=0 will give you both 0. then when calculate l=1 LS term should be l*(h bar sqr/2) and -l*(h bar sqr/2). That should be right answer. I think professor make a small mistake here.
Wait, if s = +- 1/2, why is s(s+1) = 3/4? s(s+1) = 3/4 is only the case for when s = 1/2, what about s = -1/2??
s is always 1/2, there is a point where he says that s=+1/2 or -1/2 in explaining why j can be l+s or l-s, but this is because j can take values from modulus(l-s) up to l+s in integer steps. As l=1 and s=1/2, j can be either 1/2 or 3/2. Hope that helps :)
quantum number s for the electron is always a half. But the z component projection m_s may be plus or minus a half. the s^2 operator only acts on the quantum number s and has eigenvalue s(s+1), the s^2 operator does not act on m_s which is why we do not consider the case for m_s equal to minus a half.
What happens if you raise an electron and fill it’s previous energy level with a new electron
What about the jj coupling? why it occurs? and in wich way is it different from the LS one?
Are you sure +1/2 decreases the energy level and -1/2 increases it? Because everywhere else I look it is the other way round. Or maybe I get mixed up with notations? Does j-j coupling and LS coupling lead to different energy directions?
I feel the same way..when it's positive it should reduce the energy level since it's negative therefore less energy is needed to release the electrons not more
19.47 spin is -1/2 so there should be 1/4
Sometimes you sound like Kane Williamson!
I need someone who know how to use it CCFULL programe for the japanese scientists [ K. Hagino , N. Rowley and A. T. Kruppa ] called :
A FORTRAN77 program for coupled-channels calculations with all order couplings for heavy-ion fusion reactions .
Not a neat lecture. I think there is a mix up.
Why call the video spin orbit coupling then go through really basic physics at the beginning that anyone who's watching this video will already know?
and you don't even explain the physics of spin orbit coupling. what a waste of my time