The H $_2$ Molecule

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

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

$$\bgroup\color{black}H_1 = {P^2_1\over{2m}} - {e^2\over {r_{A1}}} - {e^2\over {R_{AB}}} \cdots\egroup$$

We wish to compute variational upper bound on $R_{AB}$ and the energy.

We will again use symmetric (and antisymmetric) electron wavefunctions,

$$\bgroup\color{black}\psi(r_1,r_2) = {1\over {2[1+S(R_{AB})]}... ...\psi_A(\vec{r_2})\pm \psi_B(\vec{r_2})\right] \chi_{s,t}\egroup$$

where the spin singlet corresponds to the plus sign.

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

$$\bgroup\color{black}\left< \psi_+\vert H\vert\psi_+\right>=2E... ...\left\vert {e^2\over{r_{12}}} \right\vert \psi_+\right> \egroup$$

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.