Polar Coordinates

It is useful to introduce polar coordinates

$$x=r\sin \theta \cos \varphi,\quad y=r\sin \theta \sin \varphi,\quad z=r\cos \theta$$ (282)

and to re-write the Laplacian in polar coordinates,

$$\Delta \Psi= \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \f... ...ight) +\frac{1}{\sin^2 \theta}\frac{\partial^2\Psi}{\partial \varphi^2}\right].$$ (283)

The wave function $\Psi=\Psi(r,\theta,\varphi)$ now depends on polar coordinates. We multiply the Schrödinger equation with $r^2$,

$$\frac{\partial}{\partial r}\left(r^2 \frac{\partial \Psi}{\partia... ...ht) +\frac{1}{\sin^2 \theta}\frac{\partial^2\Psi}{\partial \varphi^2}\right]=0,$$

which can be written with two operators $\hat{h}$ and $\hat{\Omega}$ as

$$\hat{h}\Psi + \hat{\Omega}\Psi =$$ 0

$$\hat{h}\Psi := \frac{\partial}{\partial r}\left(r^2 \frac{\partial \Psi}{\partial r}\right)+\frac{2m}{\hbar^2}r^2[E-V(r)]\Psi$$

$$\hat{\Omega}\Psi := \left[\frac{1}{\sin \theta}\frac{\partial}{\partial \theta} \left... ...ight) +\frac{1}{\sin^2 \theta}\frac{\partial^2\Psi}{\partial \varphi^2}\right].$$

The separation of $r$-dependences and angle dependences suggests a separation Ansatz, that is a wave function of the form

$$\Psi(r,\theta,\varphi) = R(r)Y(\theta,\varphi).$$ (284)

Then, $\hat{h}\Psi + \hat{\Omega}\Psi =0$ means

$$\hat{h}R(r)Y(\theta,\varphi) + \hat{\Omega}R(r)Y(\theta,\varphi) = Y(\theta,\varphi)\hat{h}R(r) + R(r)\hat{\Omega}Y(\theta,\varphi)=0$$

$$\leadsto \frac{1}{R(r)}\hat{h}R(r) = - \frac{1}{Y(\theta,\varphi)}\hat{\Omega}Y(\theta,\varphi)=:-c.$$ (285)

Here, we have used the fact that $\hat{h}$ performs a differentiation with respect to $r$ so that $Y(\theta,\varphi)$ can be pulled in front of it. In the same way, $\hat{\Omega}$ performs a differentiation with respect to $\theta$ and $\varphi$ only so that $R(r)$ can be pulled in front of it. We thus have succeeded to completely seperate the radial part $R(r)$ from the angular part $Y(\theta,\varphi)$. The left side in (4.62) depends only on $r$, the right side only on $\theta,\varphi$ whence both side must be a constant that we have denoted for convenience as $-c$ here.

We first investigate the angular part as it can be solved exactly. The radial part can not be solved exactly for an arbitrary potential $V(r)$.