Computational Chemistry 4.14 - Hartree-Fock Approximation

Short lecture on the Hartree-Fock approximation for the Hamiltonian operator of molecular systems.
Even after applying the Born-Oppenheimer approximation the molecular Schrodinger equation is still an N-body problem for all electrons due to the non-separable two-electron terms. One solution is to use the Hartree-Fock approximation to average the repulsion of electrons over the average position of all other electrons, rather than an explicit pairwise interaction. This allows the Hamiltonian to be expressed as a sum of one-electron operators and is finally separable. This would result in a straightforward eigenvalue equation, however the form of the mean-field operator depends on all of the other N-1 occupied spin orbitals, thus it is a non-linear pseudo-eigenvalue equation.
Notes Slide: i.imgur.com/Ky5DTD9.png
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[@user-yt3fx8zb6g]

Your content is so touching

[@caterinacevallos9822]

AMAZING EXPLANATION! Please keep doing videos. I'm truly interested in learning DFT as well.

[@TMPChem]

Though at a slow pace, progress continues. DFT is approximately 4 chapters down the line, as the next three will be CI, MP2, and basis sets.

[@hellmrf]

Maybe I'm too late (3 years), but your videos are really great. Thank you.

[@pianoten]

Only 3 comments? This video is amazing!

[@ashwinvenkat1239]

why does the orbital changes while solving 1 electron equation and also other orbitals?

[@TMPChem]

Hi Ashwin. The key issue is the interactions of the electrons with each other. WIthout that, you can diagonalize the matrix, and the 1-electron orbitals are all eigenfunctions / eigenvectors of the 1-electron operators. When you include the interactions of the electrons with each other, things are no longer cleanly separable, and it requires a much more sophisticated method to find the orbitals. One way is via Hartree-Fock, where we assume they're only interacting with the average of the other electrons, rather than all electrons exactly. This lets us solve these equations iteratively, as each time we try to solve for it we can get closer and closer. A similar kind of example is Newton's method for numerically approximating the roots of a function. Hartree-Fock is like that, but for fancy matrices.

[@apburner1]

First.