Energy Expression

Now that we have a form for the wavefunction and a simplified notation for the Hamiltonian, we have a good starting point to tackle the problem. Still, how do we obtain the molecular orbitals?

We state that the Hartree-Fock wavefunction will have the form of a Slater determinant, and that the energy will be given by the usual quantum mechanical expression (assuming the wavefunction is normalized):

$$E_{el} = \langle \Psi \vert {\hat H}_{el} \vert \Psi \rangle .$$ (14)

For symmetric energy expressions, we can employ the variational theorem, which states that the energy is always an upper bound to the true energy. Hence, we can obtain better approximate wavefunctions $\Psi$ by varying their parameters until we minimize the energy within the given functional space. Hence, the correct molecular orbitals are those which minimize the electronic energy $E_{el}$! The molecular orbitals can be obtained numerically using integration over a grid, or (much more commonly) as a linear combination of a set of given basis functions (so-called ``atomic orbital'' basis functions, usually atom-centered Gaussian type functions).

Now, using some tricks we don't have time to get into, we can re-write the Hartree-Fock energy $E_{el}$ in terms of integrals of the one- and two-electron operators:

$$E_{HF} = \sum_i \langle i \vert h \vert i \rangle + \frac{1}{2} \sum_{ij} [ii \vert jj] - [ij \vert ji],$$ (15)

where the one electron integral is

$$\langle i \vert h \vert j \rangle = \int d{\bf x}_1 \chi_i^*({\bf x}_1) h({\bf r}_1) \chi_j({\bf x}_1)$$ (16)

and a two-electron integral (Chemists' notation) is

$$[ij\vert kl]= \int d{\bf x}_1 d{\bf x}_2 \chi_i^*({\bf x}_1) ... ...x}_1) \frac{1}{r_{12}} \chi_k^*({\bf x}_2) \chi_l({\bf x}_2).$$ (17)

There exist efficient computer algorithms for computing such one- and two-electron integrals.