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Rashba spin-orbit coupling Hamiltonian in free space can be written as: $H_{\text{so}}=\int d^3r \Psi^{\dagger}(\mathbf{r}) \gamma (p_{x}\sigma _{y}-p_{y}\sigma _{x})\Psi(\mathbf{r})$.

I expand $\Psi(\mathbf{r})=\sum_{i} c_{i}w(\mathbf{r}-\mathbf{R}_{i})$ in Wannier basis. But how can I get the final answer $H^{'}_{\text{so}}=i\lambda c_{i}^{\dagger }\mathbf{e}_{z}\cdot (\boldsymbol{\sigma} \times \mathbf{d})c_{j}^{\phantom{\dagger}}+h.c.$, where $\mathbf{d}$ is the displacement vector from site j to i. Can someone help me fill the gap?

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  • $\begingroup$ The standard second-quantization procedure combined with symmetry analysis would yield the final Rashba SOC Hamiltonian. By the way, what's the underlying lattice considered by you? $\endgroup$ – Kai Li Feb 13 '15 at 7:14
  • $\begingroup$ @Kai Li I am considering a square lattice. $\endgroup$ – Timothy Feb 13 '15 at 16:20
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Lets start from the beginning. I will drop $r$ (to prove my lazyness).

By definition

$p_x \Psi = -i \sum_i c_i \frac{x}{a}(w(R_i+\hat{x})-w(R_i)) $

$p_y \Psi = -i \sum_i c_i \frac{y}{a}(w(R_i+\hat{y})-w(R_i)) $

Ignore the $\hbar$. I keep $x$ and $y$ to make the expression general. Now,

$\Psi(\sigma_y p_x - \sigma_x p_y) \Psi\\ = \sum_{ij} c_i^\dagger w^*(R_i) (\sigma_y p_x - \sigma_x p_y) c_j w(R_j) \\ = -\frac{i}{a} \sum_{ij} c_i^\dagger w^*(R_i) c_j[\sigma_x (w(R_i+\hat{y})-w(R_i))y - \sigma_y (w(R_i+\hat{x})-w(R_i))x]$

Now we have an integration over space. We are going to use the the fact that Wannier functions are localised. So

$\int dr^3 w^*(R)w(R+l) = \delta(l)$ where $\delta$ is Dirac delta function.

and after the integration we are left with

$\int d^3r \Psi(\sigma_y p_x - \sigma_x p_y) \Psi = \frac{i}{a} \sum_{ij} c_i^\dagger c_j[\sigma_x y - \sigma_y x]$

which is

$i c_i^\dagger \hat{e_z}\cdot (\vec{\sigma} \times \vec{d_{ij}}) c_j$

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