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The Hydrogen Atom

Finally, consider the hydrogen atom as a proton fixed at the origin, orbited by an electron of reduced mass $\mu$. The potential due to electrostatic attraction is
\begin{displaymath}
V(r) = - \frac{e^2}{4 \pi \epsilon_0 r}
\end{displaymath} (126)

in SI units. The kinetic energy term in the Hamiltonian is
\begin{displaymath}
\hat{T} = - \frac{\hbar^2}{2 \mu} \nabla^2
\end{displaymath} (127)

so we write out the Schrödinger equation in spherical polar coordinates as
\begin{displaymath}
- \frac{\hbar^2}{2 \mu} \left[ \frac{1}{r^2} \frac{\partial}...
...\epsilon_0 r} \psi(r, \theta, \phi) =
E \psi(r, \theta, \phi)
\end{displaymath} (128)

It happens that we can factor $\psi(r, \theta, \phi)$ into $R(r)_{nl}Y_l^m(\theta, \phi)$, where $Y_l^m(\theta, \phi)$ are again the spherical harmonics. The radial part $R(r)$ then can be shown to obey the equation
\begin{displaymath}
- \frac{\hbar^2}{2 \mu r^2} \frac{d}{dr} \left( r^2 \frac{dR...
...[ \frac{\hbar^2 l(l+1)}{2 \mu r^2} + V(r) - E \right] R(r) = 0
\end{displaymath} (129)

which is called the radial equation for the hydrogen atom. Its (messy) solutions are
\begin{displaymath}
R_{nl}(r) = - \left[ \frac{(n - l - 1)!}{2n[(n+l)!]^3} \righ...
...^l e^{-r/na_0} L_{n+l}^{2l+1}
\left( \frac{2r}{n a_0} \right)
\end{displaymath} (130)

where $0 \leq l \leq n - 1$, and $a_0$ is the Bohr radius, $\epsilon_0 h^2 /
\pi \mu e^2$. The functions $L_{n+l}^{2l+1}(2r/na_0)$ are the associated Laguerre functions. The hydrogen atom eigenvalues are
\begin{displaymath}
E_n = - \frac{e^2}{8 \pi \epsilon_0 a_0 n^2} \hspace{1.0cm} n=1,2,\ldots
\end{displaymath} (131)

There are relatively few other interesting problems that can be solved analytically. For molecular systems, one must resort to approximate solutions.


next up previous contents
Next: Approximate Methods Up: Some Analytically Soluble Problems Previous: The Rigid Rotor   Contents
David Sherrill 2006-08-15