柱坐标系中的薛定谔方程

贡献者：待更新

预备知识　球坐标系中的定态薛定谔方程

\begin{matrix} (1) & u (r) = \sqrt{r} R (r), \end{matrix}

\begin{matrix} (2) & H = K_{r} + \frac{L_{z}^{2}}{2 m r^{2}}, \end{matrix}

\begin{matrix} (3) & K_{r} R = - \frac{1}{2 m} \frac{1}{r} \frac{d}{d r} (r \frac{d R}{d r}) = - \frac{1}{2 m} \frac{1}{\sqrt{r}} (\frac{d^{2} u}{d r^{2}} + \frac{u}{4 r^{2}}), \end{matrix}

\begin{matrix} (4) & \frac{L_{z}^{2}}{2 m r^{2}} ψ = \frac{1}{2 m} \frac{m_{z}^{2}}{r^{2}} ψ . \end{matrix}

所以径向方程为

\begin{matrix} (5) & - \frac{1}{2 m} \frac{d^{2} u}{d r^{2}} + [V (r) + \frac{1}{2 m} (\frac{m_{z}^{2}}{r^{2}} - \frac{1}{4 r^{2}})] u = E u . \end{matrix}