需要求解的定态薛定谔方程$\hat H\psi=E\psi$：
$$ \left[-\frac{\hbar^2}{2\mu}\nabla^2+V(r)\right]\psi(r)=(E_t-E_c)\psi(r)=E \psi(r) $$
哈密顿量，忽略质心部分运动，只考虑中心力场部分：
$$ \begin{aligned} \hat{H} & = -\frac{\hbar^2}{2m_1}\nabla_1^2-\frac{\hbar^2}{2m_2}\nabla_2^2+V(\vec{r}_1-\vec{r}_2)\\ & = -\frac{\hbar^2}{2M}\nabla_R^2-\frac{\hbar^2}{2\mu }\nabla^2+V(\vec{r})\\ & =-\frac{\hbar^2}{2\mu }\nabla^2+V(\vec{r}) \end{aligned} $$
标量的拉普拉斯
$$ \nabla^2f=\nabla\cdot(\nabla f)=\frac{1}{\sqrt{g}}\partial_{j}(\sqrt{g}(\nabla f)^{j})=\frac{1}{\sqrt{g}}\partial_{j}(\sqrt{g}g^{ji}\nabla i f)\\=\frac{1}{\sqrt{g}}\partial{j}(\sqrt{g}g^{-1}_{jj}\nabla _j f) $$
$$ \begin{aligned} {\nabla}^{2}{\psi}& =\frac{1}{h_{1}h_{2}h_{3}}\bigg[\frac{\partial}{\partial u_{1}}\bigg(\frac{h_{2}h_{3}}{h_{1}}\frac{\partial\psi}{\partial u_{1}}\bigg)+\frac{\partial}{\partial u_{2}}\bigg(\frac{h_{3}h_{1}}{h_{2}}\frac{\partial\psi}{\partial u_{2}}\bigg)+\frac{\partial}{\partial u_{3}}\bigg(\frac{h_{1}h_{2}}{h_{3}}\frac{\partial\psi}{\partial u_{3}}\bigg)\bigg]. \end{aligned} $$
具有球对称性中心力场可以使用球坐标展开哈密顿量算符
$$ \begin{aligned} \nabla^2&=\frac 1{r^2\sin\theta}\bigg[\partial _r\bigg(\frac {r^2\sin\theta}{1}\partial _r\bigg) +\partial _\theta\bigg(\frac {r\sin\theta}{r}\partial _\theta\bigg) +\partial \varphi\bigg(\frac{r}{r\sin\theta}\partial\varphi\bigg)\bigg] \\ &=\frac 1{r^2\sin\theta}\bigg[\partial _r\bigg({r^2\sin\theta}\partial _r\bigg) +\partial _\theta\bigg( {\sin\theta}\partial _\theta\bigg) +\frac{1}{\sin\theta}\partial _\varphi^2\bigg] \end{aligned} $$
带入球坐标展开到定态薛定谔方程，分离角向变量和径向变量（球对称$V(\vec r)\to V(r),\partial _{\theta , \varphi}V=0$）
$$ \hat H\psi=E\psi\to\bigg[-\frac{\hbar^2}{2\mu }\nabla^2+V(\vec{r})\bigg]R(r)Y(\theta,\varphi)=ER(r)Y(\theta,\varphi) $$
Laplace算符中径向偏导部分与$Y(\theta,\varphi)$无关，可以提出偏导；角向部分与$R(r)$无关，同理提出偏导：
$$ -\frac{\hbar^2}{2\mu }\frac 1{r^2\sin\theta}\bigg[\frac {\partial _r\bigg({r^2\sin\theta}\partial _rR\bigg)}{R} +\frac{\partial _\theta\bigg( {\sin\theta}\partial _\theta Y\bigg)}{Y} +\frac{1}{Y\sin\theta}\partial _\varphi^2Y\bigg]+V(r)-E=0 $$
分离变量，引入分离常数$-\frac {\lambda \hbar^2}{2\mu r^2}$
$$ \begin{aligned} &-\frac{\hbar^2}{2\mu }\frac 1{r^2\sin\theta}\bigg[\frac {\partial _r\bigg({r^2\sin\theta}\partial _rR\bigg)}{R}\bigg]+V(r)-E\\ =&\frac{\hbar^2}{2\mu }\frac 1{r^2\sin\theta}\bigg[\frac{\partial _\theta\bigg( {\sin\theta}\partial _\theta Y\bigg)}{Y} +\frac{1}{Y\sin\theta}\partial _\varphi^2Y\bigg]\\ =&-\frac {\lambda \hbar^2}{2\mu r^2} \end{aligned} $$
集中注意力得到，径向部分仅为Laplace径向分量，可以使用径向动量算符的平方代替；角向部分实际上是角动量平方算符：
$$ \frac {1}{2\mu}(-\hbar^2\nabla^2_r)R+\frac {\lambda \hbar^2}{2\mu r^2}R+VR=ER $$
$$ \bigg[\frac {\hat p^2_r}{2\mu}+\frac {\lambda \hbar^2}{2\mu r^2}+V\bigg]R(r)=ER(r) $$