D.23 Solution of the hydrogen molecule

To find the approximate solution for the hydrogen molecule, the key is to be able to find the expec­tation energy of the approximate wave functions $a\psi_{\rm {l}}\psi_{\rm {r}}+b\psi_{\rm {r}}\psi_{\rm {l}}$.

First, for given $a$$\raisebox{.5pt}{$/$}$$b$, the individual values of $a$ and $b$ can be computed from the normal­ization requirement

$$a^2+b^2 + 2ab \langle \psi_{\rm {l}} \vert \psi_{\rm {r}} \rangle^2 = 1$$ (D.11)

where the value of the overlap integral $\langle\psi_{\rm {l}}\vert\psi_{\rm {r}}\rangle$ was given in derivation {D.21}.

The inner product

$$\langle a\psi_{\rm {l}}\psi_{\rm {r}}+b\psi_{\rm {r}}\psi_... ...\rm {l}}\psi_{\rm {r}}+b\psi_{\rm {r}}\psi_{\rm {l}}\rangle_6$$

is a six-di­mensional integral, but when multi­plied out, a lot of it can be factored into products of three-di­mensional integrals whose values were given in derivation {D.21}. Cleaning up the inner product, and using the normal­ization condition, you can get:

$$\big\langle E\big\rangle = 2 E_1 - {\displaystyle\frac{e^2... ...b\langle\psi_{\rm {l}}\vert\psi_{\rm {r}}\rangle^2 A_2\Biggr]$$

using the abbrevi­ations

$$A_1 = 2\langle\psi_{\rm {l}}\vert r_{\rm {r}}^{-1}\psi_{\r... ...rm {r}}\vert r_{12}^{-1}\psi_{\rm {l}}\psi_{\rm {r}}\rangle$$

$$A_2 = {\displaystyle\frac{2\langle\psi_{\rm {l}}\vert r... ...rm {r}}\vert r_{12}^{-1}\psi_{\rm {l}}\psi_{\rm {r}}\rangle$$

Values for several of the inner products in these expressions are given in derivation {D.21}. Unfortunately, these involving the distance $r_{12}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert{\skew0\vec r}_1-{\skew0\vec r}_2\vert$ between the electrons cannot be done analyti­cally. And one of the two cannot even be reduced to a three-di­mensional integral, and needs to be done in six dimensions. (It can be reduced to five dimensions, but that introduces a nasty singularity and sticking to six dimensions seems a better idea.) So, it gets really elaborate, because you have to ensure numerical accuracy for singular, high-di­mensional integrals. Still, it can be done with some perseverance.

In any case, the basic idea is still to print out expec­tation energies, easy to obtain or not, and to examine the print-out to see at what values of $a$$\raisebox{.5pt}{$/$}$$b$ and $d$ the energy is minimal. That will be the ground state.

The results are listed in the main text, but here are some more data that may be of interest. At the 1.62 $a_0$ nuclear spacing of the ground state, the anti­symmetric state $a$$\raisebox{.5pt}{$/$}$$b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1 has a positive energy of 7 eV above separate atoms and is therefore unstable.

The nucleus to electron attraction energies are 82 eV for the symmetric state, and 83.2 eV for the anti­symmetric state, so the anti­symmetric state has the lower potential energy, like in the hydrogen molecular ion case, and unlike what you read in some books. The symmetric state has the lower energy because of lower kinetic energy, not potential energy.

Due to electron cloud merging, for the symmetric state the electron to electron repulsion energy is 3 eV lower than you would get if the electrons were point charges located at the nuclei. For the anti­symmetric state, it is 5.8 eV lower.

As a consequence, the anti­symmetric state also has less potential energy with respect to these repulsions. Adding it all together, the symmetric state has quite a lot less kinetic energy than the anti­symmetric one.