Separation of valence bands in transition metal dichalcogenides (TMDs)

Question

The image below refers to a phenomenon that occurs in TMDCs (transition metal dichalcogenides) monolayers that allowed the development of valleytronics. Why are there separate bands of different colors in this image (some with red on top and blue on the bottom and others with blue on top and red on the bottom)?

Figure is taken from: Nature chemistry, 5(4):263–275, 2013

Answer 1

I should start by saying that I am no expert in MoS$_2$, so this answer is my guess from looking at the reference you provide, and would be happy if someone corrects me.

The general things to keep in mind when looking at such band structures are:

1. If the system has time reversal symmetry, then if there is an electron with quantum numbers $(\mathbf{k},\uparrow)$, then there is another electron with quantum numbers $(-\mathbf{k},\downarrow)$ with the same energy.
2. If the system has inversion symmetry, then if there is an electron with quantum numbers $(\mathbf{k},\uparrow)$, then there is another electron with quantum numbers $(-\mathbf{k},\uparrow)$ with the same energy.
3. If the system has both time reversal and inversion symmetry, then both conditions above apply, which together mean that each electron state is doubly degenerate.

Therefore, to have splitting of energies between electrons of opposite spins, we need to break one of the two symmetries above. In MoS$_2$, the symmetry that is broken is inversion symmetry. However, inversion symmetry breaking per se is not enough to split the bands energetically, you also need a spin-dependent term in the Hamiltonian. This is provided by the spin-orbit interaction.

Based on these general comments, this is my guess for what happens in the diagram you sent:

Green valleys. The green valleys correspond to the conduction band. Although the system breaks inversion symmetry, the spin-orbit coupling in these bands is very weak, so to a good approximation there is no energy splitting of electrons of opposite spin, so the green valleys are doubly-degenerate.

Red and blue valleys. These valleys correspond to the valence band. Spin-orbit coupling is strong for these bands, so together with inversion symmetry breaking they lead to a splitting of bands with opposite spin (the blue bands are spin "down" and the red bands spin "up"). However, time reversal symmetry is still present, so property 1 above still needs to be obeyed. To see what the implications of this are, let the two energies in valley K be $E_1$ and $E_2$, where $E_2>E_1$. This means that the up electron (red) has energy $E_2$, and the down electron (blue) has energy $E_1$ at K. Time reversal symmetry then says that there is another electron at $-$K with energy $E_2$ but with opposite spin to the electron with energy $E_2$ in K (so down electron, blue). Similarly, the $E_1$ down electron at K has a time reversal partner at $-$K with the same energy but up spin. Hence, at $-$K the colors switch.

This discussion assumes that the $z$-component of spin is a good quantum number (so I can say "up" and "down"). This is not strictly true when spin-orbit coupling is present, because now we need to consider the total angular moment (sum of spin and orbital components). However, in many systems spin is very close to a good quantum number, so we allow for this language.

Answer 2

ProfM's argument is absolutely right. Here I support a more detailed explanation based on first-principles calculations.

The spin-resolved band structure of monolayer MoS$_2$ with the consideration of spin-orbit coupling is shown below:

• You can first find the two split valence bands around $K$ and $-K$ valleys. In particular, spin-$z$ is a good quantum number at these valleys.

• The conduction band edges around $K$ and $-K$ are almost degenerated.

• The first Brillouin zone of monolayer MoS$_2$ is hexagonal, there are three $K$ and $-K$.

With the above information, I believe you can understand the image you posted.