What Problem Are We Solving?

It is always important to remember the context of a theory. Hartree-Fock theory was developed to solve the electronic Schrödinger equation that results from the time-independent Schrödinger equation after invoking the Born-Oppenheimer approximation. In atomic units, and with ${\mathbf r}$ denoting electronic and ${\mathbf R}$ denoting nuclear degrees of freedom, the electronic Schrödinger equation is

$$\left[ - \frac{1}{2} \sum_{i} \nabla^2_i - \sum_{A,i} \fra... ...\right] \Psi({\mathbf{r; R}}) = E_{el} \Psi({\mathbf{r; R}}),$$ (1)

or, in our previous more compact notation,

$$\left[ {\hat T}_e(\mathbf{r}) + {\hat V}_{eN}(\mathbf{r; R}) + {\hat V}_{NN}(\mathbf{R}) + {\hat V}_{ee}(\mathbf{r}) \right] \Psi({\mathbf{r; R}}) = E_{el} \Psi({\mathbf{r; R}}).$$ (2)

Recall from the Born-Oppenheimer approximation that $E_{el}$ (plus or minus ${\hat V}_{NN}(\mathbf{R})$, which we include here) will give us the potential energy experienced by the nuclei. In other words, $E_{el}({\bf R})$ gives the potential energy surface (from which we can get, for example, the equilibrium geometry and vibrational frequencies). That's one good reason why we want to solve the electronic Schrödinger equation. The other is that the electronic wavefunction $\Psi({\mathbf{r; R}})$ contains lots of useful information about molecular properties such as dipole (and multipole) moments, polarizability, etc.