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Currently, I am reading the original paper about spin field effect transistor proposed by Supriyo Datta and Biswajit Das. In the last part of this paper, to obtain a larger overall current modulation the authors introduced the following Hamiltonian:

\begin{equation} H= \begin{bmatrix} -\dfrac{\hbar^2}{2m^*} \left( \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial z^2} \right) + V(z) -i\eta \dfrac{\partial}{\partial x} & i\eta \dfrac{\partial}{\partial z} \\ -i\eta \dfrac{\partial}{\partial z} & -\dfrac{\hbar^2}{2m^*} \left( \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial z^2} \right) + V(z) + i\eta \dfrac{\partial}{\partial x} \end{bmatrix} \end{equation} in which $\eta$ is the spin-orbit coupling constant and $V(z)$ is a confining potential that confines the electrons in a waveguide.

To find the eigenstates of $H$ one can use the unperturbed ($\eta=0$) eigenstates as a basis set. These eigenstates can be labeled with three indices: the subband index $m$, the wave vector $k$ in the $x$ direction, and the spin. For unperturbed Hamiltonian $H_0$:

\begin{align} & H_0|m,k\rangle = E_{m,k}|m,k\rangle \\ & E_{m,k}=\epsilon_m+\dfrac{\hbar^2 k^2}{2m^*} \end{align} the two spins $[1,0]^T$ and $[0,1]^T$ are degenerate. The subband energy $ \epsilon_m $ is obtained by solving the eigenequation

\begin{equation} \left( -\dfrac{\hbar^2}{2m^*}\dfrac{d^2}{dz^2} + V(z) \right)\phi_m(z)=\epsilon_m \phi_m(z) \end{equation}

The Rashba term $H_R$ ($\eta$ related terms in $H$ matrix) leads to matrix elements coupling the eigenstates of $H_0$ as follows: ($ + \leftrightarrow [1,0]^T ;- \leftrightarrow [0,1]^T $):

\begin{align} \langle m',k',+|H_R|m,k,+ \rangle & = + \eta k \delta_{m',m} \delta_{k',k} \\ \langle m',k',-|H_R|m,k,- \rangle & = - \eta k \delta_{m',m} \delta_{k',k} \\ \langle m',k',+|H_R|m,k,- \rangle & = \dfrac{\eta}{\hbar} \langle m'|p_z|m\rangle \delta_{k',k} \\ \end{align}

My question is how to derive these matrix elements? Or what's the mathematical form of the state $|m,k,+\rangle$ ?

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You can't know what's exactly the mathematical form of $|m,k,+\rangle$ without knowing what $V(z)$ is. Nevertheless, you can still obtain the matrix elements without knowing the details of it. You should have realized that $|m\rangle = \phi_m(z) $ and need to know that $ |m,k,+\rangle = |m\rangle \otimes |k\rangle \otimes |+\rangle$

Also, guess what? $|k\rangle$ are just plane-waves because no potential in the x-direction: $|k\rangle = A e^{ikx}$.

That said now you can determine the first element: \begin{align} \langle m',k',+|H_R|m,k,+ \rangle & = \left[ \langle m',k'|, 0 \right] \begin{bmatrix} -i\eta \dfrac{\partial}{\partial x} & i\eta \dfrac{\partial}{\partial z} \\ -i\eta \dfrac{\partial}{\partial z} & i\eta \dfrac{\partial}{\partial x} \end{bmatrix} \left[ |m,k\rangle, 0 \right]^T \\ & = \langle m',k'| \left(-i\eta \dfrac{\partial}{\partial x} \right) |m,k\rangle \\ & = -i\eta\ \langle m' | m\rangle \ \langle k'|\dfrac{\partial}{\partial x}|k\rangle\\ & = -i\eta\ \delta_{m',m}\ (ik) \ \langle k'|k\rangle\\ & = \eta\ k\ \delta_{m',m} \ \delta_{k,k'}\\ \end{align}

Please note that $|m\rangle$ are not eigenstates of $\dfrac{\partial}{\partial x}$.

I trust you can now derive the results for other elements! Good luck!

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