# Derivation of Rashba spin-orbit coupling in tight-binding model

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?

• 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? – Kai Li Feb 13 '15 at 7:14
• @Kai Li I am considering a square lattice. – Timothy Feb 13 '15 at 16:20

From definition we have

$$\Psi(\vec{r}) = \sum_i c_i w(\vec{R}_i-\vec{r})$$

where $$w(\vec{R_i}-\vec{r})$$ is a Wannier function centred at $$\vec{R}_i$$ and follows the ortho-normality condition

$$\int d^3r \left[ w^*(\vec{R_i}-\vec{r}) w(\vec{R_j}-\vec{r}) \right] = \delta_{ij}$$

From first principle, one can define a derivative as,

$$\vec{\nabla}w(\vec{R}_i) = \frac{\vec{d}}{d} [w(\vec{R}_i + \vec{d}) - w(\vec{R}_i)]/d$$

In principle this definition is valid only for $$d\rightarrow 0$$. In practice we use it when $$d$$ is the distance between two nearest neighbours. ($$\vec{d} = d_x \hat{e}_x + d_y \hat{e}_y + d_z \hat{e}_z$$)

Now we can write the momentum operators as

$$\hat{p}_x \Psi = -i \partial_x \sum_i c_i w(\vec{R}_i) = -i \sum_i c_i \hat{e}_x.\vec{\nabla}w(\vec{R}_i) = -i \sum_i c_i \frac{d_x}{d^2}(w(\vec{R}_i+\vec{d})-w(\vec{R}_i))$$

Here I am not writing the variable $$\vec{r}$$ and $$\hbar$$. Using this to evaluate the inner product

$$\int d^3r \left[ \Psi^\dagger(\vec{r}) \hat{p}_x \Psi(\vec{r})\right] = -i \sum_{i,j} c_i^\dagger c_j \int d^3r \frac{d_x}{d^2}\left[ w^*(\vec{R}_i)(w(\vec{R}_j+\vec{d})-w(\vec{R}_j) \right]$$

Due to the orthonormality of Wannier functions this gives a finite contribution only when $$\vec{R}_j= \vec{R}_i + \vec{d}$$. Therefore the sum over all $$i,j$$ effectively reduced to summation over all $$i$$ and its first nearest neighbours. To denote that I use $$\langle i,j\rangle$$ as the summation index.

Therefore we end up with

$$\int d^3r \left[ \Psi^\dagger(\vec{r}) \hat{p}_x \Psi(\vec{r})\right] = -i \sum_{\langle i,j\rangle} c_i^\dagger c_j \frac{d_x}{d^2}$$

$$\int d^3r \left[ \Psi^\dagger(\vec{r}) \hat{p}_y \Psi(\vec{r})\right] = -i \sum_{\langle i,j\rangle} c_i^\dagger c_j \frac{d_y}{d^2}$$

Combining them

$$i \gamma \int d^3 r \left[\Psi(\vec{r})(\sigma_y p_x - \sigma_x p_y) \Psi (\vec{r})\right]\\ = \frac{\gamma}{d^2} \sum_{\langle i,j\rangle} c_i^\dagger (\sigma_y d_x - \sigma_x d_y) c_j \\ =\lambda \sum_{\langle i,j\rangle} c_i^\dagger \hat{e}_z.(\vec{d}\times \sigma) c_j$$

where $$\hat{e}_z = (0,0,1)$$ is the unit vector along $$z$$ axis and

$$\hat{e}_z.(\vec{d}\times \vec{\sigma}) = \begin{vmatrix} 0 & 0 & 1 \\ d_x & d_y & d_z \\ \sigma_x & \sigma_y & \sigma_z \end{vmatrix} = (d_x \sigma_y - d_y \sigma_x)$$

• $\frac{df}{dx}=\lim_{\delta \to 0}\frac{1}{\delta}(f(x+\delta)-f(x))$ – Sumit Jan 21 at 7:27
• Sorry, could someone explain or link to me how the last equation comes from the second to last? – Andrew Hardy Jun 2 at 20:27

I was stucked by this problem a few weeks ago and think it's an easy question, but it turns out that the derivation of Rashba Effect in real space is quite complicated. The exact Hamiltonian you put forward is just the nearest neighbor hopping approximation and there does exist other high order terms, such as the next neighbor hopping term. The derivation is highly dependent on which orbitals you have in the system(s, p, d...) and a specific example of how to obtain real space Rashba Hamiltonian in graphene could be found in Chap 4 of the following article:

Tight-binding theory of spin-orbit coupling in graphynes