Is There Spin-Orbital Coupling in S-Orbitals with L=0?

[Mortendk]

Hi everyone.
Im reading quantum chemistry due to an exam, but there are a few things, that I am not sure of:

1. I'm not sure, but as far as i understand, there is coupling between spin and orbital angular momentum, but if L = 0, in an S-orbital: is there then any coupling ?


2. Does energisplit happen without interference from an foreign magnetic field?

Hope you can help me.

 

[bartek2009]

1. If L=0 then J=1/2 only as S=1/2 for an electron and so there is no splitting.

2. In the above case there would be no splitting and magnetic field would be required to
cause it (by Zeeman effect) but if L>0 then J can have more than one value and each will
have different energy due to LS-coupling so splitting will occur without magnetic field present.

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[Mortendk]

Is it the nucleus' magnetic field that gives rise to the different energy states? - which makes the different LS-couplings have different energies, and futhermore makes splitting occur

 

[Mortendk]

Another question concerning spin-orbit coupling.

I have read that phosphorescense might be caused by spin-orbit coupling. How can that be explained?

Thanks a lot

 

[bartek2009]

In classical consideration, an electron orbits the nucleus and in electron's frame of reference it looks like the nucleus is orbiting the electron. A moving nucleus could essentially be regarded as a current loop and you can then calculate the magnetic field generated by this motion. This is how the LS-coupling arises, basically the magnetic field depends on momentum of the nucleus (L), higher momentum means more revolutions around the electron and greater current. The magnetic moment of the electron depends on its spin (S) and the usual interaction of magnetic field with a magnetic moment proportional to m.B (greek mu is usually used for m) becomes proportional to L.S where L and S are replaced with quantum operators.

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