---
tags: Numerical method
---
# Given method
Definition
$k$ is index
$\Phi_{k} =\Phi(x_k)$
$\Delta =x_{k+1}-x_k$

Discrete Taylor
$\Phi_{k\pm1} =\Phi_{k} \pm \Delta\Phi_k' +\frac{\Delta^2}{2!}\Phi_k'' \pm \frac{\Delta^3}{3!}\Phi_k'''+O(\Delta^4)$

The odd term will cancel out
$\Phi_{k+1}+\Phi_{k-1} =2\Phi_{k} +\Delta^2\Phi_k'' +O(\Delta^4)$

Use Schrödinger equation replace $\Phi_{k}''$
$(\frac{-\hbar^2}{2\mu}\nabla^2+V)\Phi=E\Phi$

$\nabla^2\Phi=\Phi''$

$\Phi_k''=\frac{2\mu}{\hbar^2}(V_k-E)\Phi_k$

$\Phi_{k+1}+\Phi_{k-1} =2\Phi_{k} +\frac{2\mu\Delta^2}{\hbar^2}(V_k-E)\Phi_k +O(\Delta^4)$

Assume $O(\Delta^4)=0$
$\Phi_{k+1}+\Phi_{k-1} -(2+\frac{2\mu\Delta^2}{\hbar^2})V_k\Phi_k=-\frac{2\mu\Delta^2}{\hbar^2}E\Phi_k$
Change to matrix
$H_k=-(2+\frac{2\mu\Delta^2}{\hbar^2})V_k$
$\lambda=-(\frac{2\mu\Delta^2}{\hbar^2})E$
$$ \left(
\begin{array}{ccc}
H_1&1&0&\cdots&0\\
1&H_2&1&\cdots&0\\
0&1&H_3&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&H_n
\end{array}
\right)\left(
\begin{array}{ccc}
\Phi_1\\
\Phi_2\\
\Phi_3\\
\vdots\\
\Phi_n\\
\end{array}
\right)=\lambda \left(
\begin{array}{ccc}
1&0&0&\cdots&0\\
0&1&0&\cdots&0\\
0&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&1
\end{array}
\right)\left(
\begin{array}{ccc}
\Phi_1\\
\Phi_2\\
\Phi_3\\
\vdots\\
\Phi_n\\
\end{array}
\right)$$
Because the matris is tridiagnoal, there are efficient algorithms for diagonalizing and get eigenvalue.