How to derive the spin-orbit term in Kane Mele model?

Question

The Kane mele model is a famous quantum spin hall model on honeycomb lattice (C.L. Kane and E.J. Mele, Phys. Rev. Lett. 95, 226801 (2005)). The Hamiltonian is

$$H = - t\sum\limits_{\left\langle {i,j} \right\rangle \alpha } {c_{i\alpha }^\dagger {c_{j\alpha }}} + i{\lambda _{SO}}\sum\limits_{\left\langle {\left\langle {i,j} \right\rangle } \right\rangle \alpha \beta } {{\nu _{ij}}c_{i\alpha }^\dagger \sigma _{\alpha \beta }^z{c_{j\beta }}} $$

The first term is just the nearest hopping term for graphene, while the second term represents the spin-orbit interaction, which seems quite peculiar to me.

1. Must it be in the z direction? Why?
2. Why does the spin-orbit term only have the second nearest hopping term, but no nearest hopping term? How does it relate to the symmetry of honeycomb lattice?
3. Is there an intuitive way to understand this term?

Can somebody tell me the derivation for the second term?

Answer 1

The spin-orbit interaction can be written:

$$ H=\mathscr{E} \hat{L} \cdot \hat{S}, $$

where $\hat{L}=\hat{r} \times \hat{p}$.

Consider an electron hopping to its next nearest neighbour on the honeycomb lattice. During its hop, the atom closest to it (not counting the atom it's hopping from or toward) has a coulombic potential which affects the hopping electron. You can use the right-hand rule to calculate the sign of this term (since $\hat{r}$ points towards the atom, $\hat{p}$ is the direction the electron is travelling, and $\hat{S}$ corresponds to spin-up or spin-down).

If you consider the electron hopping back along the same path, the momentum is now $-\hat{p}$, so the overall sign of H will pick up a minus sign. This explains the antisymmetry of $\nu_{ij}$.

For nearest neighbour jumps, $\mathscr{E}=0$, so no spin-orbit interaction occurs.