这是一份warwick华威大学MA260-10的成功案例

In a two-dimensional electron gas, due to the relativistic effects, an electric field $\mathbf{E}$ is seen by moving electrons as a magnetic field that couples to the spin in the $x y$ plane. This fact is described by the Bychkov-Rashba Hamiltonian
$$
\mathrm{H}{R}=\alpha{R}(|E|) \boldsymbol{\sigma} \cdot\left(\boldsymbol{k} | \times \boldsymbol{e}{z}\right), $$ where $\boldsymbol{k}{|}$is the momentum of an electron, $\boldsymbol{\sigma}$ the vector of Pauli matrices, and $\boldsymbol{e}{z}=(0,0,1)$. Thus, the spin degeneracy is lifted and the energy dispersion has the form $$ E{\mathrm{RSO}}=\frac{\hbar^{2}}{2 m^{2}}\left(k_{|} \pm \Delta k\right)-E_{\mathrm{SO}}
$$

MA260-10 COURSE NOTES ：

$$
\left[\frac{d \sigma(\mathrm{z})}{d z}\right]^{2}-\frac{2 \beta_{0}}{d} \sigma^{2}(\mathrm{z})=\frac{2 \beta_{0}}{d} \sigma^{2}(D)
$$
with the boundary conditions
$$
\left.\frac{d \sigma(\mathrm{z})}{d z}\right|{z=0}=2 \beta{0} \sigma_{0},\left.\quad \frac{d \sigma(\mathrm{z})}{d z}\right|{z=D}=0 $$ Here we have defined $\beta{0}=e^{2} /\left(4 \varepsilon_{0} k_{\perp} A_{0}\right)$. One can rewrite Eq. (6.31) in the form
$$
\int_{r_{D}}^{1} \frac{d u}{\left(u^{2}-r_{D}^{2}\right)^{\frac{1}{2}}}=\sqrt{\frac{2 \beta_{0}}{d} D}
$$
where $r_{D}=\sigma(D) / \sigma(0)$. Therefore, the potential difference $\Delta V(D)$ between the two ends of a sample of thickness $D$ is
$$
\Delta V(D)=2 \mathrm{~A}{0} \sigma{0} \sqrt{2 \beta_{0} d} \frac{1-r_{D}}{\left(1-r_{n}^{2}\right)^{1 / 2}}
$$