柱坐标系中的薛定谔方程

\begin{equation} u(r) = \sqrt r R(r)~, \end{equation}

\begin{equation} H = K_r + \frac{L_z^2}{2m r^2}~, \end{equation}

\begin{equation} K_r R = -\frac{1}{2m} \frac1r \frac{\mathrm{d}}{\mathrm{d}{r}} \left(r \frac{\mathrm{d}{R}}{\mathrm{d}{r}} \right) = - \frac{1}{2m} \frac{1}{\sqrt r} \left( \frac{\mathrm{d}^{2}{u}}{\mathrm{d}{r}^{2}} + \frac{u}{4 r^2} \right) ~, \end{equation}

\begin{equation} \frac{L_z^2}{2m r^2}\psi = \frac{1}{2m} \frac{m_z^2}{r^2}\psi ~. \end{equation}

\begin{equation} - \frac{1}{2m} \frac{\mathrm{d}^{2}{u}}{\mathrm{d}{r}^{2}} + \left[V(r) + \frac{1}{2m} \left(\frac{m_z^2}{r^2} - \frac{1}{4 r^2} \right) \right] u = Eu~. \end{equation}