When creating molecular orbitals via the linear combination of atomic orbitals method I am told that Slater orbitals provide a more accurate representation whilst Gaussian orbitals are computationally more straightforward to integrate (two-electron integrals for instance), since the product of Gaussians is another Gaussian.

However, the Slater-type orbital can itself be approximated by a linear combination of a few Gaussians. This means when we represent the integrals we do so via many more basis functions than if we were to use Slater-type orbitals, however my supervisor assures me that despite there being a higher number of functions when using Gaussian-type orbitals (compared to Slater) the Gaussian-type orbitals' far simpler integration makes it substantially more computationally efficient.

I was wondering if there was anyone who could provide me with a source or some data that quantifies just how much more computationally efficient doing Hartree-Fock calculations with Gaussian-type orbitals is than using Slater-type orbitals.

According to Wikipedia:

The speedup by 4—5 orders of magnitude compared to Slater orbitals more than outweighs the extra cost entailed by the larger number of basis functions generally required in a Gaussian calculation.

However, the Wikipedia article does not give a reference for this and I was just wondering if this is actually an accurate statement and any evidence for this would be really helpful. I could not find any sources that verified this and was wondering if anyone knew of such a source.


1 Answer 1


Your supervisor is correct that in almost all practical cases in quantum chemistry, the cost savings of using Gaussian-type orbitals far outweigh the disadvantages of needing more orbitals.

First of all, there is a bit of a misconception that Slater-type orbitals are far more accurate. The motivation for using Slater-type orbitals is due to their resemblance to the known shape of the wavefunction for the hydrogen atom which is essentially the only system for which we can access the solution to the non-relativistic Schroedinger equation (without other bells and whistles) analytically. Slater-type orbitals are not much better than Gaussian-type orbitals for calculating the frozen-core electron correlation energy of benzene, for example, where the theoretical motivation for using them based on the analytic solution to the H-atom Schroedinger equation is not nearly as relevant.

Perhaps the best data that justifies the use of Gaussian-type orbitals over Slater-type orbitals, is the fact that almost everyone in quantum chemistry and even in atomic physics, uses them. The ADF software uses Slater-type orbitals and is aimed at fairly large systems with DFT, but since you mentioned Hartree-Fock I'll speak from the perspective of smaller systems (containing a few atoms) where we aim for .

In the realm of high-accuracy wavefunction-based ab initio calculations, Michal Lesiuk and co-workers have in recent years put some serious effort into "re-examining" the option of using Slater-type orbitals. At first sight, their paper "Reexamination of the calculation of two-center, two-electron integrals over Slater-type orbitals. III. Case study of the beryllium dimer" seemed to show evidence that Slater-type orbitals gave a better result for the well depth of the $\ce{Be2}$ potential than any previous calculation: their calculated energy with Slater-type orbitals was 929.0±1.9 cm$^{−1}$ and the empirical result that they compared to was 929.7±2.0 cm$^{−1}$. However, that empirical result was based on an older paper that had a major error (the author's didn't know that $\ce{Be2}$ has one more vibrational level before the molecule dissociated). A more careful analysis by two different groups, who both used different models to fit the experimental data, arrived at empirical values for this energy to be 934.8±0.3 cm$^{−1}$ and 935.0±0.3 cm$^{−1}$, and guess what: That's the answer that Gaussian-type orbitals predicted several years earlier when Jacek Koput used an aug-cc-pCV7Z(i) basis set on the same system and got an energy of 935 cm$^{−1}$.

I have worked with the authors of the Slater-type orbital paper, and can say that they are an extremely competent group when it comes to high-precision computational spectroscopy and their work has a tendency to be truly state-of-the-art, so when their series of three consecutive papers on reexamining the use of Slater-type orbitals ends in a paper that gives a lower-accuracy result than a calculation from several years earlier using GTOs, I'm confident in saying that GTOs are the way to go.

There's much more that can be said on this topic, and I don't want this answer to get too much longer, but I'll conclude with a few more points in this paragraph. Not all quantum chemistry software has implemented STOs, so it might be less convenient to work with them even if they have some advantages. Furthermore, the integrals (whether GTO or STO) are often not the biggest issue for a quantum chemistry calculation, since the FCI problem is even harder, so we can pay the price to switch from GTOs to STOs since that price is smaller than the price of FCI, but we will gain very little compared to the other errors we'll have in the calculation (due to approximating FCI, due to still being far from the basis-set limit, due to ignoring Born-Oppenheimer breakdown, etc.), and this is a major reason to avoid the inconvenience of switching to STOs.

  • $\begingroup$ Well the STO paper only had 5Z basis sets while Koput's work used 7Z. The angular expansion of the correlation energy converges slowly anyhow. $\endgroup$ Commented Jan 24 at 21:48
  • $\begingroup$ In 2011 Koput single-handedly used 7Z with GTOs and in 2015 it took a group of 5 people to use 5Z with STOs for the same system, in a 3-paper series that attempted to bring back STOs. I wonder why they didn't use 7Z. Embarrassing that Koput's GTO results were just as good? $\endgroup$ Commented Jan 25 at 10:37

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