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積層2重量子井戸のポテンシャル

Last updated at Posted at 2023-11-11

$$\left ( \begin{array}{cccccccccccccc}
\Delta E_c & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & \ddots& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & \Delta E_c & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & \ddots & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & \Delta E_c & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \Delta E_c & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \ddots & 0
\end{array} \right )
$$

$$\frac{d}{dz}\left(\frac{1}{m(z)}\right)\frac{d}{dz} \Psi(z) = \frac{d}{dz}\left(\frac{1}{m_m}\right)
\frac{\Psi_{m+0.5} - \Psi_{m-0.5}}{\Delta z}$$

$$= \left(\frac{1}{m_{m+0.5}}\right)
\frac{\Psi_{m+1} - \Psi_{m}}{\left( \Delta z \right )^2}- \left(\frac{1}{m_{m-0.5}}\right)
\frac{\Psi_{m} - \Psi_{m-1}}{\left (\Delta z \right)^2}$$

$$=\frac{\hbar^2}{\left (\Delta z \right)^2} \left\{
-\frac{\Psi_{m+1}}{m_{m+1}} + \left(\frac{1}{m_{m+1}}+\frac{1}{m_{m-1}}\right) \Psi_m -\frac{\Psi_{m-1}}{m_{m-1}}
\right \} $$

$X=\frac{\hbar^2}{em_e\left (\Delta z \right)^2}$として、

$$H=X\times \left (
\begin{array}{ccccc}
\frac{\Delta E_c}{2X}+\frac{1}{m_W}+\frac{1}{m_B} & -\frac{1}{m_B} &0&0&0&0&0&0&0&0&\cdots&-\frac{1}{m_W}\\
-\frac{1}{m_B} & \frac{\Delta E_c}{X}+\frac{2}{m_B} & -\frac{1}{m_B} & 0 &0 &0&0&0&0&0&\cdots&0\\
0 & \ddots & \ddots & \ddots & 0&0 &0&0&0&0&\cdots&0\\
0& 0& -\frac{1}{m_B} & \frac{\Delta E_c}{X}+\frac{2}{m_B} & -\frac{1}{m_B} &0 &0&0&0&0&\cdots&0\\
0&0&0&-\frac{1}{m_B}&\frac{\Delta E_c}{2X}+\frac{1}{m_W}+\frac{1}{m_B} & -\frac{1}{m_W}&0&0&0&0&\cdots&0\\
0&0&0&0&-\frac{1}{m_W} & \frac{2}{m_W} & -\frac{1}{m_W} & 0 &0 &0&\cdots&0\\
0&0&0&0&0& \ddots & \ddots & \ddots & 0&0&\cdots&0 \\
0&0&0&0&0&0& -\frac{1}{m_W} & \frac{2}{m_W} & -\frac{1}{m_W} &0 &\cdots &0 \\
0&0&0&0&0&0&0& -\frac{1}{m_W} & \frac{\Delta E_c}{2X}+\frac{1}{m_W}+\frac{1}{m_B} & -\frac{1}{m_B}&\cdots&0\\
0&0&0&0&0&0&0& \ddots & \ddots & \ddots & \cdots&0 \\
-\frac{1}{m_W} &0&0&0&0&0&0&0&0 & \cdots &-\frac{1}{m_W} & \frac{2}{m_W}
\end{array} \right )
$$

$$\frac{d}{dz}\left ( \epsilon (z) \frac{d\phi(z)}{dz}\right ) = \frac{\rho (z)}{\epsilon_0}$$
$$\epsilon (z) \frac{d\phi(z)}{dz} = \frac{1}{\epsilon_0}\int_0^z \rho(z)\; dz $$
$$\frac{d\phi(z)}{dz} = \frac{1}{\epsilon_0}\int_0^z\left [ \frac{1}{\epsilon(z_1) } \int_0^{z_1} \rho(z_0) \; d z_0 \right ] $$

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