一维器件: kp哈密顿量离散化过程(价带和导带)

参考 https://www.cnblogs.com/ghzhan/articles/17368238.html
待填入

价带 Valence

\[H(k_x,k_y,k_z) = \begin{pmatrix} Mk_y^2+Mk_z^2 & 0 & 0\\ 0 &Lk_y^2+Mk_z^2 & Nk_yk_z\\ 0 &Nk_yk_z&Mk_y^2 + Lk_z^2 \end{pmatrix}+k_x^2 \begin{pmatrix} L & 0 & 0\\ 0 & M & 0\\ 0 & 0 & M \end{pmatrix}+k_x \begin{pmatrix} 0 & Nk_y& Nk_z\\ Nk_y& 0 & 0\\ Nk_z& 0 & 0 \end{pmatrix}\\ = H_c + k_x^2* H_{k^2} + k_x* H_{k} \]

沿着 \(x\) 方向离散化得到

\[H(k_y,k_z) = \begin{pmatrix} H_{00} & H_{01} & \\ H_{10} & H_{00} & H_{01}\\ &H_{10} & H_{00} & H_{01}\\ & & \ddots &\ddots &\ddots \end{pmatrix}\\ H_{00} = H_c + 2/a^2*H_{k^2}\\ H_{01} = -1/a^2*H_{k^2} - i/(2a)*H_k\\ H_{10} = -1/a^2*H_{k^2} + i/(2a)*H_k \]

导带 Conduction

\[H = \begin{pmatrix} H_1 & H_3\\ H_3^{\dagger} & H_2 \end{pmatrix} \]

沿着 y, z 方向的能带

\[H_1 = \frac{1}{m_t}(k_x^2+k_y^2) + \frac{1}{m_l}k_z^2-\frac{2}{m_l}k_0k_z+U(z)\\ H_2 = \frac{1}{m_t}(k_x^2+k_y^2) + \frac{1}{m_l}k_z^2+\frac{2}{m_l}k_0k_z+U(z)\\ H_3 = -\frac{2k_xk_y}{M}.\]

where \(M^{-1}=m_t^{-1}-1\)

\[H(k_x,k_y,k_z) = \begin{pmatrix} m_{t}^{-1}k_y^2+m_{l}^{-1}k_z^2-2m_{l}^{-1}k_0k_z + E_c & 0 \\ 0 &m_{t}^{-1}k_y^2+m_{l}^{-1}k_z^2+2m_{l}^{-1}k_0k_z + E_c \end{pmatrix}+k_x^2 \begin{pmatrix} m_{t}^{-1} & 0\\ 0 & m_{t}^{-1} \end{pmatrix}+k_x \begin{pmatrix} 0 & -2M^{-1}k_y\\ -2M^{-1}k_y& 0 \end{pmatrix}\\ = H_c + k_x^2* H_{k^2} + k_x* H_{k} \]

沿着x方向的能带.

\[H_1 = \frac{1}{m_t}(k_y^2+k_z^2) + \frac{1}{m_l}k_x^2-\frac{2}{m_l}k_0k_x+U(x)\\ H_2 = \frac{1}{m_t}(k_y^2+k_z^2) + \frac{1}{m_l}k_x^2+\frac{2}{m_l}k_0k_x+U(x)\\ H_3 = -\frac{2k_yk_z}{M}.\]

where \(M^{-1}=m_t^{-1}-1\)

\[H(k_x,k_y,k_z) = \begin{pmatrix} m_{t}^{-1}k_y^2+m_{t}^{-1}k_z^2 + E_c & -2M^{-1}k_yk_z \\ -2M^{-1}k_yk_z &m_{t}^{-1}k_y^2+m_{t}^{-1}k_z^2 + E_c \end{pmatrix}+k_x^2 \begin{pmatrix} m_{l}^{-1} & 0\\ 0 & m_{l}^{-1} \end{pmatrix}+k_x \begin{pmatrix} -2m_{l}^{-1}k_0 & 0\\ 0& 2m_{l}^{-1}k_0 \end{pmatrix}\\ = H_c + k_x^2* H_{k^2} + k_x* H_{k} \]