3.2.2.1 Solution

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

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

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

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

Answer: The tables show that $R_{10}=2e^{-r/a_0}/\sqrt{a_0^3}$ and that $Y_0^0=1/\sqrt{4\pi}$ , so

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

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

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

or rearranging

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

giving

$\begin{displaymath} \frac{1}{\pi} \times\frac 14 \times 2 \times 2\pi \end{displaymath}$

which is one as required.