= Solution
The attractive <Coulomb potential energy> is $V(r)=-e^2/(4\pi\epsilon_0r)$, so
$$
\frac1r\frac{dV}{dr}=\frac{e^2}{4\pi\epsilon_0r^3}.
$$
Use precisely the radial normalization stated in the PDF: the radial probability measure is $4\pi r^2|\psi|^2dr$. The supplied function has
$$
|\psi|^2=\frac{r^2}{96\pi a_0^5}e^{-r/a_0},
\qquad4\pi r^2|\psi|^2dr=\frac{r^4e^{-r/a_0}}{24a_0^5}dr.
$$
Its integral is one, since $\int_0^\infty r^4e^{-r/a_0}dr=24a_0^5$. In particular,
$$
\langle r^{-3}\rangle=\frac1{24a_0^5}\int_0^\infty re^{-r/a_0}dr
=\frac1{24a_0^3}.
$$
This calculation uses the supplied convention, rather than multiplying a separately normalized standard radial function by an extra $4\pi$.
The <addition of angular momentum> bracket is $j(j+1)-l(l+1)-s(s+1)=3/4-2-3/4=-2$. Substitution in the <spin-orbit energy shift in a central potential> therefore gives
$$
\boxed{\langle H_1\rangle
=-\frac{\hbar^2}{2m^2c^2}\frac{e^2}{4\pi\epsilon_0}\frac1{24a_0^3}
=-\frac{\hbar^2e^2}{192\pi\epsilon_0m^2c^2a_0^3}
\simeq-3.02\times10^{-5}\,\mathrm{eV}.}
$$
Here $a_0=5.29\times10^{-11}\,\mathrm m$ is the supplied <Bohr radius>. The negative sign means the $j=1/2$ member of the $l=1$ pair is shifted downward by this <spin-orbit interaction>; it is not the total hydrogen binding energy.
Back to article page
