D.21 Solution of the hydrogen molecular ion

The key to the variational approximation to the hydrogen molecular ion is to be able to accurately evaluate the expectation energy

\begin{displaymath}
\big\langle E\big\rangle
= \langle a \psi_{\rm {l}} + b...
...r}}\vert
H\vert a \psi_{\rm {l}} + b \psi_{\rm {r}} \rangle
\end{displaymath}

This can be multiplied out and simplified by noting that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are eigenfunctions of the partial Hamiltonians. For example,

\begin{displaymath}
H\psi_{\rm {l}} = E_1 \psi_{\rm {l}}
- \frac{e^2}{4\pi\epsilon_0}\frac{1}{r_{\rm {r}}} \psi_{\rm {l}}
\end{displaymath}

where $E_1$ is the -13.6 eV hydrogen atom ground state energy. The expression can be further simplified by noting that by symmetry

\begin{displaymath}
\langle \psi_{\rm {r}}\vert r_{\rm {l}}^{-1}\psi_{\rm {r}}...
...gle \psi_{\rm {r}}\vert r_{\rm {r}}^{-1}\psi_{\rm {l}}\rangle
\end{displaymath}

and that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are real, so that the left and right sides of the various inner products can be reversed. Also, $a$ and $b$ are related by the normalization requirement

\begin{displaymath}
a^2 + b^2 + 2 ab \langle\psi_{\rm {l}}\vert\psi_{\rm {r}}\rangle = 1
\end{displaymath}

Cleaning up the expectation energy in this way, the result is

\begin{eqnarray*}
\big\langle E\big\rangle &= & E_1 - \\
&& \frac{e^2}{4 \p...
...\rm {r}}^{-1} \psi_{\rm {l}} \right\rangle
\right\}
\right]
\end{eqnarray*}

which includes the proton to proton repulsion energy (the 1$\raisebox{.5pt}{$/$}$$d$). The energy $E_1$ is the $\vphantom0\raisebox{1.5pt}{$-$}$13.6 eV amount of energy when the protons are far apart.

Numerical integration is not needed; the inner product integrals in this expression can be done analytically. To do so, take the origin of a spherical coordinate system $(r,\theta,\phi)$ at the left proton, and the axis towards the right one, so that

\begin{displaymath}
r_{\rm {l}} = \vert{\skew0\vec r}-{\skew0\vec r}_{\rm {lp}...
...w0\vec r}_{\rm {rp}}\vert = \sqrt{d^2+r^2-2 d r\cos(\theta)}.
\end{displaymath}

In those terms,

\begin{displaymath}
\psi_{\rm {l}} = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}
\...
...sqrt{\pi a_0^3}}
e^{-\sqrt{d^2+r^2-2 d r\cos(\theta)}/a_0}.
\end{displaymath}

Then integrate angles first using ${\,\rm d}^3{\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r^2\sin(\theta){\rm d}{r}{\,\rm d}\theta{\,\rm d}\phi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-r^2{\rm d}{r}{\,\rm d}\cos(\theta){\,\rm d}\phi$. Do not forget that $\sqrt{x^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert x\vert$, not $x$, e.g. $\sqrt{(-3)^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, not $\vphantom0\raisebox{1.5pt}{$-$}$3. More details are in [25, pp. 305-307].

The overlap integral turns out to be

\begin{displaymath}
\left\langle \psi_{\rm {l}} \vert\psi_{\rm {r}} \right\ran...
...+ \frac{d}{a_0} + \frac13 \left(\frac{d}{a_0}\right)^2\right]
\end{displaymath}

and provides a measure of how much the regions of the two wave functions overlap. The direct integral is

\begin{displaymath}
\left\langle \psi_{\rm {l}}\vert r_{\rm {r}}^{-1} \psi_{\r...
...ac{1}{d} - \left[\frac{1}{a_0}+\frac{1}{d}\right] e^{-2d/a_0}
\end{displaymath}

and gives the classical potential of an electron density of strength $\vert\psi_{\rm {l}}\vert^2$ in the field of the right proton, except for the factor $\vphantom0\raisebox{1.5pt}{$-$}$$e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0$. The “exchange integral” is

\begin{displaymath}
\left\langle \psi_{\rm {l}} \vert r_{\rm {l}}^{-1} \psi_{\...
...
=
\left[\frac{1}{a_0}+\frac{d}{a_0^2}\right] e^{-d/a_0}.
\end{displaymath}

and is somewhat of a twilight term, since $\psi_{\rm {l}}$ suggests that the electron is around the left proton, but $\psi_{\rm {r}}$ suggests it is around the right one.