As I have seen it in the Fu-Kane-Mele model context of a bipartite lattice, it is represented as a coupling between second-nearest neighbors $\langle\langle i,j\rangle\rangle$, like so:

$i \lambda \sum\limits_{\langle\langle i,j\rangle\rangle}(c_i^\dagger~\mathbf{e}_{ij}\cdot\mathbf{s}~c_j-H.c.)$,

where $\mathbf{e}_{ij}=\frac{\mathbf{e}_{i}\times\mathbf{e}_{i}}{|\mathbf{e}_{i}\mathbf{e}_{j}|}$, and the $\mathbf{e}_{i}$ are vectors connecting second nearest neighbors. $\mathbf{s}=(s_x,s_y,s_z)$ are Pauli Matrices acting on spin.

I just wanted to know why it is not nearest neighbor? Is it connected to the physical origin of Dresselhaus SOC?


1 Answer 1


Because the nearest neighbor bond is inversion symmetric on the lattice (i.e. it must map to itself under inversion), the Dresselhaus SOC is forbidden by the inversion symmetry on any inversion symmetric bond.

Under inversion: $\mathbf{e}_{ij}\to - \mathbf{e}_{ij}$ because the displacement operator is a vector, and $\mathbf{s}\to \mathbf{s}$ because the spin (angular momentum) operator is a pseudo-vector, so the Dresselhaus SOC term $\mathbf{e}_{ij}\cdot \mathbf{s}$ is odd (changing sign) under inversion, and can not be defined on inversion symmetric bonds.

  • $\begingroup$ Would you mind elaborating on why the inversion symmetry excludes neareast-neighbor spin-orbit coupling? $\endgroup$
    – Frank
    Commented Nov 14, 2023 at 21:40

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