Why is orthogonality crucial in basis sets despite atomic orbital overlap?

[raman]

1: Why are the elements of a basis set taken to be orthogonal? But in real sense atomic orbitals do overlap.

 

[JPRitchie]

The AO's found as solutions to the Schroedinger equation for the hydrogen atom are, in fact, all orthogonal to one another. That is, the overlap integrals <s|px>, <px|py> etc. are zero.
In molecular systems, atomic orbitals on different atoms are, in general, not orthogonal to one another. Also, it's possible to use more than one set of s, p, d, etc. basis functions on a single center that aren't orthogonal. In such cases, one forms linear combinations of atomic orbitals (LCAO's) to make MO's or AO's for atoms that are orthogonal to one another. What happens here is that, for systems with more than 2 electrons, one starts with a set of so-called basis functions that are not necessarily orthogonal, and then uses them to obtain the LCAO's that are solutions to the Hartree-Fock equations. These LCAO's are orthogonal.

 

[quetzalcoatl9]

Is is for the reasons that JPRitchie mentions that solving the electronic structure for heavy metals is difficult: for elements in the fifth row and lower the inner core electrons have an average momentum that it well within the relativistic limit. By orthogonality, this affects ALL of the electrons of the atom, including the valence (the ones we are typically most interested in as chemists). In short, the assumption of hydrogenic orbitals breaks down.

One way around this is to parameterize an Effective Core Potential that lumps these relativistic effects into a single, fit potential that is "felt" by the remaining electrons. The other way is to perform a fully relativistic simulation (DK, etc.) using a Dirac-like Hamiltonian. Both methods work very well. I am not aware of any other techniques for handling this practical difficulty.

 

[JPRitchie]

The number of basis functions involved in heavy atom electronic structure calculations are small compared to those needed for proteins. The smallest known protein has about 45 residues, and a couple of hundred atoms. This results in thousands of basis functions.
Now if you add a transition metal or two, then you've really got a lot. Not only do you have to compute a lot of integrals, but you have to form and diagonalize the Fock matrix or something like it.
Post-HF treatments are needed, in any case, for reliable results, and that's really out there for these systems.
-Jim Ritchie

 

[quetzalcoatl9]

The number of basis functions involved in heavy atom electronic structure calculations are small compared to those needed for proteins. The smallest known protein has about 45 residues, and a couple of hundred atoms. This results in thousands of basis functions.
Now if you add a transition metal or two, then you've really got a lot. Not only do you have to compute a lot of integrals, but you have to form and diagonalize the Fock matrix or something like it.
Post-HF treatments are needed, in any case, for reliable results, and that's really out there for these systems.
-Jim Ritchie

i realize that. i wasn't referring to the computation time required in solving the electronic structure of heavy metals, but rather the complications with regards to basis set and not getting a garbage answer.

i have had nightmares of indium 2+ chasing me for the last year...most people take it for granted that there is extensive experimental data on their systems.

most folks have gone semi-empirical or QM/MM DFT for protein systems.

 

[JPRitchie]

Yes, the QM/MM method is used a lot for biochemical systems, because the number of basis functions otherwise becomes large.

I thought the problems with metals atoms (transition or heavy) wasn't in the basis set. It's not a big problem to go as high as "g" or "h" in the angular momentum and thousands of basis functions can be handled more or less routinely.
Rather, the biggest problems lie in that 1. simple LS spin-orbit coupling is no longer valid and 2. relativistic effects become important.
These problems mean that, for even qualitative calculations, some sort of multi-reference wavefunction is needed and that you've got to include relativistic effects in the Hamiltonian. These effects lead to the otherwise strongly forbidden E1 transition in CaI, and the E2 transition in SrII, YbII, and HgII, for example.
Although not very familiar with work in this area, I found Phys. Rev. A 1999, 59, 230 by Dzuba, Flambaum and Webb provacative. They cite some spectroscopic studies of InII as a standard for analyzing atomic spectra from quasars. Although they didn't look at InII, their computed results for the other atomic spectra looked quite reasonable.
-Jim Ritchie