The \bgroup\color{black}$\mathrm{H}_2$\egroup Molecule

The \bgroup\color{black}$\mathrm{H}_2$\egroup molecule consists of four particles bound together: \bgroup\color{black}$e_1$\egroup, \bgroup\color{black}$e_2$\egroup, proton \bgroup\color{black}$_A$\egroup, and proton \bgroup\color{black}$_B$\egroup. The Hamiltonian can be written in terms of the \bgroup\color{black}$\mathrm{H}_2^+$\egroup Hamiltonian, the repulsion between electrons, plus a correction term for double counting the repulsion between protons.

\begin{displaymath}\bgroup\color{black}H=H_1 + H_2 + {e^2\over{r_{12}}} - {e^2\over{R_{AB}}} \egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black}H_1 = {p^2_1\over{2m}} - {e^2\over {r_{A1}}} - {e^2\over {r_{B1}}} + {e^2\over {R_{AB}}} \egroup\end{displaymath}

We wish to compute variational upper bound on \bgroup\color{black}$R_{AB}$\egroup and the energy.

We will again use symmetric electron wavefunctions,

= {1\over {2[1+S(R_{AB})]}...
...t[ \psi_A(\vec{r_2}) + \psi_B(\vec{r_2})\right] \chi_{s}\egroup\end{displaymath}

where the spin singlet is required because the spatial wfn is symmetric under interchange.

The space symmetric state will be the ground state as before.

\begin{displaymath}\bgroup\color{black}\left< \psi\vert H\vert\psi\right>=2E_{H_...\left\vert {e^2\over{r_{12}}} \right\vert \psi\right> \egroup\end{displaymath}

From this point, we can do the calculation to obtain

Distance Energy
Calculated 0.85 Å -2.68 eV
Actual 0.74 Å -4.75 eV.

wIth a multiterm wavefunction, we could get good agreement.

Jim Branson 2013-04-22