写在前面: 5年前的笔记，再次做个备份.
假设器件长度为 \(L\), 均匀分成 \(N+1\) 份, 网格spacing 为 \(a = L/(N+1)\).
\[H\varphi = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\varphi = E\varphi
\]
因为 \(\varphi(0) = \varphi(N+1) = 0\), 所以
\[\varphi(0) + \varphi(2) - 2\varphi(1) = E\varphi(1)\\
\varphi(1) + \varphi(3) - 2\varphi(2) = E\varphi(2)\\
\cdots\\
\varphi(N-1) + \varphi(N+1) - 2\varphi(N) = E\varphi(N)\\
\]
写成矩阵形式
\[\begin{pmatrix}
-2 & 1 & & & & &\\
1 & -2 & 1 & & & &\\
& 1 & -2 & 1 & & &\\
& & \cdots & \cdots & \cdots & &\\
& & & 1 & -2 & 1 & \\
& & & & 1 & -2 & 1\\
& & & & & 1 & -2\\
\end{pmatrix}\begin{pmatrix}
\varphi(1)\\
\varphi(2)\\
\varphi(3)\\
\cdots \\
\varphi(N-2)\\
\varphi(N-1)\\
\varphi(N)\\
\end{pmatrix}=E\begin{pmatrix}
\varphi(1)\\
\varphi(2)\\
\varphi(3)\\
\cdots \\
\varphi(N-2)\\
\varphi(N-1)\\
\varphi(N)\\
\end{pmatrix}
\]
求解该哈密顿量, 得到本征值为
\[E=2t+2t\cos\frac{m\pi}{N+1}.\\
=2t+2t\cos k.
\]
这里 \(m=1,2,3,...,N\), 且 \(k\) 也可以写成:
\[k = \frac{x\pi}{L} = \frac{ma\pi}{(N+1)a} = \frac{m\pi}{(N+1)}.
\]