4.2.3.1 So­lu­tion an­guc-a

Ques­tion:

The gen­eral wave func­tion of a state with az­imuthal quan­tum num­ber $l$ and mag­netic quan­tum num­ber $m$ is $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R(r)Y_l^m(\theta ,\phi)$, where $R(r)$ is some fur­ther ar­bi­trary func­tion of $r$. Show that the con­di­tion for this wave func­tion to be nor­mal­ized, so that the to­tal prob­a­bil­ity of find­ing the par­ti­cle in­te­grated over all pos­si­ble po­si­tions is one, is that

$$\int_{r=0}^\infty R(r)^* R(r) r^2 { \rm d}r = 1.$$

An­swer:

You need to have $\langle\Psi\vert\Psi\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\int\Psi^*\Psi{ \rm d}^3{\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 for the wave func­tion to be nor­mal­ized. Now the vol­ume el­e­ment ${\rm d}^3{\skew0\vec r}$ is in spher­i­cal co­or­di­nates given by $r^2\sin\theta{ \rm d}{r}{\rm d}\theta{\rm d}\phi$, so you must have

$$\int_{r=0}^\infty\int_{\theta =0}^{\pi}\int_{\phi =0}^{2\pi}... ...ta ,\phi) r^2\sin\theta{ \rm d}r{\rm d}\theta{\rm d}\phi = 1.$$

Tak­ing this apart into two sep­a­rate in­te­grals:

$$\int_{r=0}^\infty R(r)^* R(r) r^2 { \rm d}r \int_{\theta =0... ... Y_l^m(\theta ,\phi) \sin\theta{ \rm d}\theta{\rm d}\phi = 1.$$

The sec­ond in­te­gral is one on ac­count of the nor­mal­iza­tion of the spher­i­cal har­mon­ics, so you must have

$$\int_{r=0}^\infty R(r)^* R(r) r^2 { \rm d}r = 1.$$