# Why spin-orbit coupling in TMDCs is strong?

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?

• What is TMDC? The only situation with strong spin-orbit splitting that I am aware of is nuclear physics. – Lewis Miller Feb 11 '17 at 14:55
• @LewisMiller TMDC is a transition metal dichalcogenides which is important in valleytronics and spintronics. – skyhj105 Feb 12 '17 at 1:46
• Thanks. I have no idea why spin-orbit coupling would be strong in that situation. – Lewis Miller Feb 12 '17 at 3:46

• for hydrogenic atoms it scales as $Z^4$ (since Bohr radius decreases as $1/Z$ and SO depends on $\langle 1/r^4 \rangle$ – wcc Feb 22 '19 at 2:41
• 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 – wcc Feb 22 '19 at 10:16
• 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 $Ze/r^2$ – wcc Feb 22 '19 at 10:35