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I already know about spin-orbit coupling in QM. Spin-orbit coulping is that spin of electron and it`s orbital(angular momentum) are considered for getting Hamiltonian which can cause energy splitting. Based on this concept, I have troubled with understanding strong spin-orbit coupling in TMDCs.

Is there anyone who can let me know why spin-orbit coupling in TMDCs is strong?

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    $\begingroup$ What is TMDC? The only situation with strong spin-orbit splitting that I am aware of is nuclear physics. $\endgroup$
    – Lewis Miller
    Commented Feb 11, 2017 at 14:55
  • $\begingroup$ @LewisMiller TMDC is a transition metal dichalcogenides which is important in valleytronics and spintronics. $\endgroup$
    – skyhj105
    Commented Feb 12, 2017 at 1:46
  • $\begingroup$ Thanks. I have no idea why spin-orbit coupling would be strong in that situation. $\endgroup$
    – Lewis Miller
    Commented Feb 12, 2017 at 3:46

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Spin-orbit splittings increase with atomic number because it is a relativistic effect. So it is higher for Te and Se than for sulphides and oxides.

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    $\begingroup$ for hydrogenic atoms it scales as $Z^4$ (since Bohr radius decreases as $1/Z$ and SO depends on $\langle 1/r^4 \rangle$ $\endgroup$
    – wcc
    Commented Feb 22, 2019 at 2:41
  • $\begingroup$ +1 this is the right answer. For the record, spin orbit coupling in these materials isn't even that strong, it's just much stronger than graphene or silicon. $\endgroup$
    – KF Gauss
    Commented Feb 22, 2019 at 5:37
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    $\begingroup$ @IamAStudent Yes. But the relevant orbitals here are the valence orbitals, and those have approximately the same size for oxygen, sulfur, selenium and tellurium. $\endgroup$
    – user137289
    Commented Feb 22, 2019 at 8:56
  • $\begingroup$ really? I thought they would differ because of different principal quantum numbers. Anyway, focusing on length scale may be a little misleading since SO really comes from $B = v \times E$ and for hydrogenic atom it's written as $L/(mr^2) \times e/r^2$ but I guess the r-scaling could be different in a more complex situation $\endgroup$
    – wcc
    Commented Feb 22, 2019 at 10:16
  • $\begingroup$ and also, my comments about scalingare actually wrong...the r-scaling is $1/r^3$, not $1/r^4$ since $v \sim L/mr$. So you get your $Z^3$ from there and the last factor of $Z$ from the strength of the electric field $Ze/r^2$ $\endgroup$
    – wcc
    Commented Feb 22, 2019 at 10:35

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