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Good Quantum numbers in $L$-$S$ coupling

Physics Asked by 1MegaMan1 on December 6, 2020

I’m having a hard time understanding a few things about L-S and j-j coupling in 2 (or more) electron atoms. What I picked up from our lectures is the following:

The Hamiltonian can be “broken up” into 3 parts, a Central field part, a residual electrostatic part and a Spin-Orbit part:

$$hat{H} = hat{H}_{CF}+Deltahat{H}_{RE} + Deltahat{H}_{SO}$$

where we can treat $Deltahat{H}_{RE}$ and $Deltahat{H}_{SO}$ as small perturbations. Which of these terms has to be tackled first is decided by their relative size, so that $Deltahat{H}_{RE} > Deltahat{H}_{SO}$ leads to what we call the LS-coupling scheme, where we treat $Deltahat{H}_{RE}$ first, and $Deltahat{H}_{RE} < Deltahat{H}_{SO}$ gives us the jj-coupling scheme, where $Deltahat{H}_{SO}$ is treated first.

In the LS-coupling scheme we find that in order to diagonalise $Deltahat{H}_{RE}$ we have to pick a basis of eigenstates labeled by the quantum numbers $|L,M_L,S,M_Srangle$ as these diagonalise the perturbation $Deltahat{H}_{RE}$. This then leads to energy shift, which depend on $L$ and $M$, $Delta E_{RE} = Delta E_{RE}left(L,Sright)$, and the introduction of term symbols.

Now we treat the smaller perturbation $Deltahat{H}_{SO}$. According to our lectures the following is true:
$$Deltahat{H}_{SO} = sum_i beta_i underline{hat{l}}_i cdot underline{hat{s}}_i$$
and also that:
$$ langle underline{hat{l}}_i cdot underline{hat{s}}_i rangle propto langle underline{hat{L}}cdot underline{hat{S}}rangle = frac{1}{2} langle hat{J}^2 – hat{L}^2 – hat{S}^2 rangle $$

This seems to imply to me that, actually, the basis states of $|L,S,J,M_J rangle$ would diagonalise both the Residual Electrostatic perturbation and also the Spin-Orbit interaction at the same time.

This must obviously be wrong as if it was possible to diagonalise both perturbations at the same time using the same set of eigenfunctions, there would be no need to distinguish between the two cases of L-S coupling and j-j coupling.

I’d greatly appreciate it, if someone could explain to me where I went to wrong, or was able to point me to some resources explaining this in more detail.

One Answer

Spin-orbit coupling is not coupling the total $L$ and $S ,$ but rather it is coupling the (lowercase) $l$ and $s ;$ i.e. it couples the spin angular momentum and the orbital angular momentum of each electron individually, not the total $L$ and $S .$

Answered by The Notorious on December 6, 2020

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