CompChem.03.01 Foundations of Molecular Orbital Theory: LCAO Wave Functions

University of Minnesota Chem 4021/8021 Computational Chemistry, as taught by Professor Christopher J. Cramer (pdf slide decks at pollux.chem.umn...)

[@cafeinst]

I can see the method of taking the determinant and then solving for the roots as failing in practice due to numerical stability problems.

[@mihirsahoo4538]

Excellent lecture

[@sull5307]

For the last question, I would assume that H and S would be zero if the orbitals are infinitely separated by each other (the atom centers are very far away).

[@ChemProfCramer]

That is certainly one good example of a situation where both H_ij and S_ij would be expected to be zero.

[@evgeniinekhoroshev8204]

Thank you so much for these lectures, there are really helpful. I have a question for which I cannot find an answer for. I understand that the variational principle allows us to approximate the ground state if we have a parametric form for a wavefunction. There is only one true solution with a minimal energy and the lower the energy we get, the better. However, LCAO finds n solutions for a one-electron molecule and only one of them (usually) has the lowest energy. How can we imply that all other eigenvectors are also good approximations of the exited states of a one-electron molecule? According to the variational principle, only the one with the lowest energy would be and the rest of the solutions could be meaningless?

[@ChemProfCramer]

A great question! My memory (I haven't looked at this in a looooong time) is that there is a generalization of the variational principle that applies to CI wave functions (and HF is a special case of CI for one electron) that says that all CI roots are bounded from below by exact roots. Mind you, since the field that defines the energy of the higher roots is largely defined by the optimization of orbitals that are occupied in the LOWEST root (for HF, that is), improvement in the estimate of an excited-state energy may converge much less quickly than improvement in the ground-state energy as one takes steps to better the quality of the latter wave function.

[@evgeniinekhoroshev8204]

@@ChemProfCramer thank you)

[@zulqarnainchaughtai]

In case of nitrogen molecule due to sigma pi crossover, pi MO's have low energy as compared to sigma MO. Can we say that pi bonds are stronger than sigma bond in case of nitrogen molecule?