Step 1: 代入与归一化
将 $\psi = RY$ 代入原方程,并两边同时除以 $R(r)Y(\theta, \phi)$,利用偏导数的性质,方程变形为:
$$ \frac{1}{R r^2}\frac{d}{dr}\left(r^2 \frac{dR}{dr}\right) + \frac{1}{Y r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial Y}{\partial \theta}\right) + \frac{1}{Y r^2\sin^2\theta}\frac{\partial^2 Y}{\partial \phi^2} + \frac{2\mu}{\hbar^2}(E - V(r)) = 0 $$
Step 2: 变量解耦
两边乘以 $r^2$ 并移项,使等式左边只含 $r$,右边只含角度:
$$ \underbrace{\frac{1}{R}\frac{d}{dr}\left(r^2 \frac{dR}{dr}\right) + \frac{2\mu r^2}{\hbar^2}(E - V(r))}_{\text{仅依赖 } r} = \underbrace{-\frac{1}{Y}\left[ \dots \right]}_{\text{仅依赖 } \theta, \phi} = l(l+1) $$
由于变量独立,两边必须等于同一个常数 $l(l+1)$。