Spin orbit coupling/orbital degeneracy

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  • #1
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Main Question or Discussion Point

disregarding the hyperfine structure and the lamb shift for the moment, spin orbit coupling says that 2S1/2 and 2P1/2 have the same energy. do 2S0 and 2P0 have the same energy?

if 2P splits into 3/2 and 1/2 does the 3D orbital split into 5/2, 3/2, and 1/2 and the 4F orbital into 7/2, 5/2, 3/2, and 1/2? google gets plenty of hits but they only talk about 2P.


this is just my own personal interest but you can move it to homework if you are so inclined.
 

Answers and Replies

  • #2
clem
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The subscript1/2 is the total J=L+S (QM addition). That limits J to L+1/2 and L-1/2.
There is no 0 subscrlpt for one electron.
A D orbital (L=2) can have only 3/2 or 5/2.
An F orbital (L=3) can have only 5/2 or 7/2.
The G...
 
  • #3
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thank you for responding.

the web says that those numbers only hold if the number of electrons is odd. I was wondering what happens if the number of electrons is even. that is why I asked about 2S0.
 
  • #4
malawi_glenn
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"the web says" is a pretty non specific statement, why not state where you read it? Maybe you have misunderstood something?

Which atoms are you considering?

nLj referes to single electron-orbitals, a single electron orbital can not have j = half integer. 2So does not exists as single orbital, are you suggesting 2So to be notation for what??
 
  • #5
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because I read it on many sites. they all say the same thing. that the formula only works of odd number of electrons. I'll try to find the exact source.
 
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  • #6
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I know that the zeeman effect is different for even and odd and all the sites talk about 2P1/2 and 2S1/2. maybe I assumed it was different. if the number of electrons was even then why would the S orbital have spin 1/2?
 
  • #7
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from the web:
The nlj quantum numbers are, then, again appropriate for a single electron outside closed subshells.

so what do I use for closed shells or nonclosed shells with more than one electron? especially for calculating fine structure.
 
  • #8
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https://www.physicsforums.com/showthread.php?t=114363

They are still valid. But that's not the point. The point is whether the *perturbation* hamiltonian commutes with these operators. One finds that the spin orbit hamiltonian commutes with L^2, S^2, J^2 and J_z, but not with L_z and S_z. Therefore, m_s and m_l are not good quantum numbers but must be replaced by m_j and j. So the states of definite energy when the spin-orbit interaction are taken into account are the states labelled by the quantum numbers l,s,j, m_j (instead of the usual l,m_l,s,m_s that one uses to label the unperturbed hydrogenic wavefunctions).
unperturbed hydrogenic wavefunctions?
 
  • #9
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http://www.pha.jhu.edu/~rt19/hydro/node9.html [Broken]

http://www.pha.jhu.edu/~rt19/hydro/img194.gif [Broken]
 
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  • #10
malawi_glenn
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yes what about them, and what about the picture? what do you want?
 
  • #11
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I thought others might find it helpful.
 
  • #12
malawi_glenn
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but you explicit ask "unperturbed hydrogenic wavefunctions?"
 
  • #13
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that was 3 days ago. I figured you just werent going to answer.

I guess that 'unperturbed hydrogenic wavefunction' just means the simplified hydrogen model without taking spin-orbit coupling into account? or does it mean something else completely?
 
  • #14
malawi_glenn
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yes, that is correct.
 
  • #15
nrqed
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nLj referes to single electron-orbitals, a single electron orbital can not have j = half integer.
?? L is an integer and S= 1/2 so j must be a half integer for a single electron orbital
 
  • #16
malawi_glenn
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?? L is an integer and S= 1/2 so j must be a half integer for a single electron orbital
LOL correct, i did too many "not" in one sentence ;-) Sorry
 

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