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:
\[k = \frac{x\pi}{L} = \frac{ma\pi}{(N+1)a} = \frac{m\pi}{(N+1)}. \] 标签：2t,frac,varphi,cdots,实例,pmatrix,一维,pi,量子 From： https://www.cnblogs.com/ghzhan/p/17338104.html