I was looking at grotrian diagrams for helium and I see that there is no splitting of energy levels due to spin-orbit coupling. In my book it is said that spin-orbit coupling in helium is small and can be neglected but no further explanation is given. At the same time we do spin-spin coupling between electrons in He which obviously is not negligible.

If there is no spin-orbit coupling then there is no fine structure in helium spectral lines like in alkali metals, right? So why is that we can neglect spin-orbit coupling in He?


Who says it is neglected? Having a look at the NIST Atomic Spectra Database (enter He I to see the levels, or go here), the first few levels read

Configuration    Term    J   Level (eV)

1s^2             (^1)S   0    0.0000000             

1s2s             (^3)S   1    [19.81961467579]    

1s2s             (^1)S   0    [20.6157749605]       

1s2p             (^3)P°  2    [20.96408703092]       
                         1    [20.96409650646]       
                         0    [20.96421899233]       

As you can see, there is a definite nonzero spin-orbit splitting between the states of different $J$ at the very first state that can support it. This looks pretty small, as it's on the seventh significant figure, but that's roughly the same $J$ splitting that you get in the hydrogen $2p$ state, so it feels about par for the course. If you care about the spectrum to enough precision, everything contributes.

That said, there is of course the question of at what level the spin-spin coupling on the electron contributes, and here I'm less sure. However, since $Z$ is small here (and therefore the electron speeds are low, so the currents and orbital dipole magnetic moments are also low) it wouldn't be surprising to see that it was significantly higher than the spin-orbit coupling.

However, it's important to keep in mind that the spin-sector dynamics, particularly through their symmetry, can have effects on the energy that are unrelated to an explicit spin-spin coupling, because a symmetric spin state must correspond to an antisymmetric spatial wavefunction that has the electrons farther on average, so the electron-electron electrostatic repulsion goes down. This effect is very important (and likely to drive the lion's share of the difference between the $^3S$ and $^1S$ levels above) but it should not be confused with an actual spin-spin coupling.


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