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{\bf Question}

Let

$$\Psi_{1s}=\frac{1}{\sqrt{\pi a_0^3}}e^{\frac{-r}{a_0}},\ \
\Psi_{2s}=\frac{1}{4\sqrt{2\pi a_0^3}}\left (2-\frac{r}{a_0}
\right ) e^{\frac{-r}{2a_0}},$$

and

$$\Psi_{2p_z}=\frac{1}{4\sqrt{2\pi
a_0^5}}re^{\frac{-r}{a_0}}\cos(\theta).$$

Show that each of $\Psi_{1s}$, $\Psi_{2s}$ and $\Psi_{2p_z}$ is
{\sl normalised}, i.e. that

$$\int \!\!\! \int \!\!\! \int \Psi^2\,dV = 1$$

where the integration is taken over all of space. Show also that

$$\int \!\!\! \int \!\!\! \int \Psi_{1s} \Psi_{2s} \,dV = \int
\!\!\! \int \!\!\! \int \Psi_{1s} \Psi_{2p_s} \,dV =0.$$



{\bf Answer}

In the spherical polar coordinates a volume element is  $$dv = r^2
\sin \theta dr \, d\theta \, d\phi$$  Integrating over all of
space means that $\left. \begin{array}{ccccc} 0 & \leq & \phi &
\leq & 2\pi \\ 0 & \leq & \theta & \leq & \pi \\ 0 & \leq & r
\end{array} \right\}$  We will require the following integrals:

$\displaystyle \int_0^\pi sin \theta \, d\theta = 2$,

$\displaystyle \int_0^{2\pi} \, d\phi = 2\pi$

$\displaystyle \int_0^\pi \sin \theta \cos \theta \, d\theta =
\int_0^\pi \frac{1}{2}\sin2\theta \, d\theta = \left[
-\frac{1}{4}\cos 2\theta \right]_0^\pi = -\frac{1}{4}+\frac{1}{4}
= 0$

Note that $\displaystyle \frac{d}{d\theta}\left( -\frac{1}{3}cos^3
\theta\right) = \cos^2 \theta \sin \theta$ so that $\displaystyle
\int_0^\pi \cos^2 \theta \sin \theta \, d\theta =
\left[-\frac{1}{3}\cos^3 \theta \right]_0^\pi = \frac{1}{3} +
\frac{1}{3} = \frac{2}{3}$

Also note we have $\displaystyle \int_0^\infty r^ne^{-\lambda r}\,
dr = \frac{n!}{\lambda^{n+1}}$

To see this let $\displaystyle I_n = \int_0^\infty r^ne^{-\lambda
r} \, dr$ (n a positive integer)

As a special case we have \begin{eqnarray*} I_0 & = &
\int_0^\infty e^{-\lambda r} dr  \\ & = &
\left[-\frac{1}{\lambda}e^{-\lambda r}\right]_0^\infty \\ & = &
\left( 0 + \frac{1}{\lambda} \right) \\ & = & \frac{1}{\lambda}
\end{eqnarray*}

In general \begin{eqnarray*} I_n & = & \int_0^\infty
r^ne^{-\lambda r}dr \\ & = & \left[
\frac{r^n}{-\lambda}e^{-\lambda r} \right]_0^\infty +
\int_0^\infty \frac{nr^{n-1}}{\lambda}e^{-\lambda r} dr \\  & = &
0 + \frac{n}{\lambda} \int_0^\infty r^{n -1}e^{-\lambda r}dr \\ &
= & \frac{n}{\lambda}I_{n-1} \end{eqnarray*}

So $\displaystyle I_n = \frac{n}{\lambda}I_{n-1}$.  We find that:

\begin{eqnarray*} I_n & = & \left( \frac{n}{\lambda} \right)I_{n-1} \\
& = & \left( \frac{n}{\lambda} \right)\left( \frac{n-1}{\lambda}
\right)I_{n-2} \\ & = & \left( \frac{n}{\lambda} \right)\left(
\frac{n-1}{\lambda} \right)\left( \frac{n-2}{\lambda}
\right)I_{n-3}\\ &  & ..... \\& = & \left( \frac{n}{\lambda}
\right)\left( \frac{n-1}{\lambda} \right)\left(
\frac{n-2}{\lambda}\right).... \left( \frac{1}{\lambda}
\right)I_{0}\\ & = & \frac{n!}{\lambda^n}I_0 \\ & = & \left(
\frac{n!}{\lambda^n} \right) \left( \frac{1}{\lambda} \right) \\ &
= & \frac{n!}{\lambda^{n+1}} \end{eqnarray*} \begin{flushright}
[QED] \end{flushright}
\begin{eqnarray*}
\int \!\!\! \int \!\!\! \int \Psi_{1s}^2 dv & = & \frac{1}{\pi
a_0^3} \int_0^\infty r^2e^{-\frac{2r}{a_0}} dr \int_0^\pi sin
\theta d\theta \int_0^{2\pi} d\phi \\ & = & \frac{4}{a_0^3}
\int_0^\infty r^2e^{-\frac{2r}{a_0}} dr \\ & = & \left(
\frac{4}{a_0^3} \right) \left( \frac{2!}{\left( \frac{2}{a_0}
\right)^3} \right) \\  & = & \frac{8a_0^3}{8a_0^3} \\ & = & 1
\end{eqnarray*}
\begin{eqnarray*}
\int \!\!\! \int \!\!\! \int \Psi_{2s}^2 dv & = & \frac{1}{32\pi
a_0^3} \int_0^\infty r^2\left(2 -
\frac{r}{a_0}\right)^2e^{-\frac{r}{a_0}} dr \int_0^\pi sin \theta
d\theta \int_0^{2\pi} d\phi \\ & = & \frac{1}{8a_0^3}
\int_0^\infty \left(4r^2 - \frac{4r^3}{a_0} + \frac{r^4}{a_0^2}
\right)e^{-\frac{r}{a_0}} dr  \\ & = & \frac{1}{8a_0^3} \left\{
4\left( \frac{2!}{\left( \frac{1}{a_0} \right)^3} \right) -
\frac{4}{a_0}\left( \frac{3!}{\left( \frac{1}{a_0} \right)^4}
\right) + \left(\frac{1}{a_0^2}\right)\left( \frac{4!}{\left(
\frac{2}{a_0} \right)^5} \right) \right\} \\ & = &
\frac{1}{8a_0^3} \left\{ 8a_0^3 - 24a_0^3 + 24a_0^3 \right\}\\  &
= & \frac{8a_0^3}{8a_0^3}
\\ & = & 1
\end{eqnarray*}
\begin{eqnarray*}
\int \!\!\! \int \!\!\! \int \Psi_{2p_z}^2 dv & = & \frac{1}{23\pi
a_0^5} \int_0^\infty r^4e^{-\frac{r}{a_0}} dr \int_0^\pi sin
\theta \cos^2 \theta d\theta \int_0^{2\pi} d\phi \\ & = &
\frac{\frac{4\pi}{3}}{32\pi a_0^5} \int_0^\infty
r^4e^{-\frac{r}{a_0}} dr \\ & = & \frac{1}{24a_0^5}(4!a_0^5)
\\ & = & 1
\end{eqnarray*}
\begin{eqnarray*}
\int \!\!\! \int \!\!\! \int \Psi_{1s} \Psi_{2s} dv & = &
\frac{1}{4\pi a_0^3 \sqrt2} \int_0^\infty r^2\left(2 -
\frac{r}{a_0}\right)e^{-\frac{r}{a_0}}e^{-\frac{r}{2a_0}} dr
\int_0^\pi sin \theta d\theta \int_0^{2\pi} d\phi \\ & = &
\frac{1}{a_0^3 \sqrt2} \int_0^\infty
\left(2r^2e^{-\frac{3r}{2a_0}} -
\frac{r^3}{a_0}e^{-\frac{3r}{2a_0}}\right) dr
\\ & = & \frac{1}{a_0^3 \sqrt2} \left\{ 2\left( \frac{2!}{\left(
\frac{3}{2a_0} \right)^3} \right) - \frac{1}{a_0}\left(
\frac{3!}{\left( \frac{3}{2a_0} \right)^4} \right) \right\} \\ & =
& \frac{1}{a_0^3 \sqrt2} \left\{ \frac{2^5a_0^3}{3^3} -
\left(\frac{1}{a_0} \right)\left( \frac{2^5a_0^4}{3^3}\right)
\right\}\\ & = & 0
\end{eqnarray*}
\begin{eqnarray*}
\int \!\!\! \int \!\!\! \int \Psi_{1s} \Psi_{2P_z} dv & = &
\frac{1}{4\pi a_0^4 \sqrt2} \int_0^\infty r^3e^{-\frac{3r}{2a_0}}
dr \int_0^\pi sin \theta \cos \theta d\theta \int_0^{2\pi} d\phi
\\ & = & 0
\end{eqnarray*}
Since $\displaystyle \int_0^\infty \sin \theta \cos \theta d\theta
= 0$ by above.


\end{document}
