4.3.2.1 So­lu­tion hydb-a

Ques­tion:

Use the ta­bles for the ra­dial wave func­tions and the spher­i­cal har­mon­ics to write down the wave func­tion

$$\psi_{nlm} = R_{nl}(r) Y_l^m(\theta ,\phi)$$

for the case of the ground state $\psi_{100}$.

Check that the state is nor­mal­ized. Note: $\int_0^{\infty}e^{-2u}u^2{ \rm d}{u}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac 14$.

An­swer:

The ta­bles show that $R_{10}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2e^{-r/a_0}$$\raisebox{.5pt}{$/$}$​$\sqrt{a_0^3}$ and that $Y_0^0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1/$\sqrt{4\pi}$, so

$$\psi_{100}=\frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$$

The to­tal prob­a­bil­ity of find­ing the par­ti­cle in­te­grated over all pos­si­ble po­si­tions is, us­ing the tech­niques of vol­ume in­te­gra­tion in spher­i­cal co­or­di­nates:

$$\int\vert\psi_{100}\vert^2 { \rm d}^3 {\skew0\vec r}= \int_... ...} e^{-2r/a_0} r^2 \sin\theta{ \rm d}r{\rm d}\theta{\rm d}\phi$$

or re­ar­rang­ing

$$\frac{1}{\pi} \int_{r/a_0=0}^\infty e^{-2r/a_0} \frac{r^2}{a... ...\sin\theta{ \rm d}\theta\int_{\phi =0}^{2\pi} 1 { \rm d}\phi$$

giv­ing

$$\frac{1}{\pi} \times\frac 14 \times 2 \times 2\pi$$

which is one as re­quired.