# Molecular Hamiltonian of Helium hydride ion

Currently, I am going through Quantum Simulation of Helium Hydride Cation in a Solid-State Spin Register. I am not a chemist but rather a computer scientist hence having trouble following the paper.

I am using this easy to access note on Hartree-Fock approximation.

So, I am not sure how the authors created the molecular Hamiltonian in Eq. 2 of the supplementary material.

$$H = \begin{pmatrix} -2.85404 & 0.130671 & 0.0 \\ 0.130671 & -0.760916 & -0.323568\\ 0.0 & -0.323568 & -1.91238 \end{pmatrix}$$

From the note, I understand that the Born-Oppenheimer approximation (Eq. 3) is the starting point.

$$H = -\sum^{N}_{i = 1} \frac{1}{2} \nabla^2_i - \sum^{N}_{i = 1} \sum^{M}_{A = 1} \frac{Z_A}{r_i A} + \sum^{N}_{i = 1} \sum^{N}_{j > 1} \frac{1}{r_{ij}}$$

What should be my following steps to reach at the molecular Hamiltonian.

I am extremely sorry if I am asking for a really long answer.

UPDATE 1: The molecular Hamiltonian is:

\begin{align} \hat{H} &= \underbrace{-\frac{\hbar^2}{2 M_{He}} \nabla^2_{He} - \frac{\hbar^2}{2 M_H}}_{\text{Nuclear kinetic energy}} \underbrace{- \frac{\hbar^2}{2 m_e} \nabla_1^2 - \frac{\hbar^2}{2 m_e} \nabla^2_2}_{e^-\text{'s kinetic energy}} \nonumber\\ & \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{He}}|} -\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 1\text{\textsuperscript{st} } e^- \text{ and nuclei}} \nonumber\\ & \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{He}}|} -\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 2\text{\textsuperscript{nd} } e^- \text{ and nuclei}} \nonumber\\ & \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{r_2}|} }_{e^-\text{ - } e^- \text{ repulsion}} \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{R_{He}} - \overrightarrow{R_H}|} }_{\text{Nuclear - Nuclear repulsion}} \end{align}

After Born–Oppenheimer approximation:

\begin{align} \hat{H} &= \underbrace{- \frac{\hbar^2}{2 m_e} \nabla_1^2 - \frac{\hbar^2}{2 m_e} \nabla^2_2}_{e^-\text{'s kinetic energy}} \nonumber\\ & \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{He}}|} -\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 1\text{\textsuperscript{st} } e^- \text{ and nuclei}} \nonumber\\ & \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{He}}|} -\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 2\text{\textsuperscript{nd} } e^- \text{ and nuclei}} \nonumber\\ & \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{r_2}|} }_{e^-\text{ - } e^- \text{ repulsion}} \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{R_{He}} - \overrightarrow{R_H}|} }_{\text{Now a constant since nuclei positions are fixed}} \end{align}

Your provided note is about the HF Hamiltonian in the one-electron basis set (atomic orbitals). Since a minimal STO-3G basis set is used here, the corresponding matrix representation of this Hamiltonian is a $2\times 2$ matrix.
The $3\times 3$ Hamiltonian from the SI is the FCI Hamiltonian in the many-electron basis set. These basis functions are Configuration State Functions, arising from linear combinations of (excited) Slater determinants (electron configurations), which in turn are built from the HF molecular orbitals.