On basis set exchange, the only correlation consistent (cc) basis sets for potassium, are relativistic (labelled by X2C). Very recently (I consider 2017 quite recent, because most Dunning basis sets date back to the late 1980s) the ccRepo has provided a basis set for potassium that "was constructed by re-contracting the DK (relativistic Douglas-Kroll) set for a non-relativistic Hamiltonian".

I am curious why there was never any basis set for potassium that was originally contracted for a non-relativistic Hamiltonian?

I understand that potassium is not a light element, however non-relativistic cc basis sets have existed all along for the elements all the way up to krypton, which is much heavier than potassium. Non-relativistic cc basis sets have even been made recently by the group of Angela Wilson, up to xenon, which is almost triple as heavy (so in some sense, triple as relativistic) as potassium.


2 Answers 2


The answer in part is that not all the elements up through $\ce{Xe}$ have (or had until fairly recently) a nonrelativistic correlation consistent basis set. Specifically, the alkali and alkaline earth elements (columns 1 and 2 of the periodic table) have been much slower to get nonrelativistic basis sets. The cc-pVnZ basis sets for $\ce{Li}, \text{ }\ce{Be}, \text{ }\ce{Na}, \text{ and } \ce{Mg}$ were only published in 2011.[1] As you mentioned, $\ce{K}$ was very recent, but $\ce{Ca}$ only got cc-pVnZ basis sets in 2002.[2] Neither $\ce{Rb}$ nor $\ce{Sr}$ have a nonrelativistic cc-basis set. This discrepancy was mentioned in a recent article[3] that defined relativistic cc-basis for the alkali and alkaline earth metals, it was not further explained.

As to why these took so long, it may just have been a lack of interest in these atoms. Both the p-block and transition metal elements are widely studied in chemistry, but the alkali and alkaline earth metals have generally not been as studied (at least, not in contexts requiring insight from electronic structure theory). The previously discussed $\ce{Ca}$ basis was made specifically so they could better characterize the potential energy surface of a compound that had seen limited experimental or theoretical interest. So it doesn't seem to be due to any inherent difficulty in making basis sets for these atoms. Rather they seemed to just hold off until there enough potential applications to warrant it.


[1] Prascher, B.P., Woon, D.E., Peterson, K.A. et al. Gaussian basis sets for use in correlated molecular calculations. VII. Valence, core-valence, and scalar relativistic basis sets for Li, Be, Na, and Mg. Theor Chem Acc 128, 69–82 (2011). DOI: 10.1007/s00214-010-0764-0

[2] Koput, J. & Peterson K.A. J. Phys. Chem. A 2002, 106, 41, 9595-9599 DOI:10.1021/jp026283u

[3] Hill, J.G. & Peterson K.A. Gaussian basis sets for use in correlated molecular calculations. XI. Pseudopotential-based and all-electron relativistic basis sets for alkali metal (K–Fr) and alkaline earth (Ca–Ra) elements J. Chem. Phys. 147, 244106 (2017); DOI: 10.1063/1.5010587


Like many others, the correlation consistent sets start from an MCHF (i.e. small CASSCF) ground state calculation.

For Li and Be the MCHF ground state has S orbitals only, but there's also a low lying P state where you promote one s electron to the p shell. The P configuration is so low that you typically include a p shell for Li and Be basis sets.

Now, when you go down in the periodic table to Ca, you're next to the transition metals, and you might expect a state with electrons in the D orbitals to appear.

If I remember correctly what Kirk told me half a decade ago when I was visiting him at WSU, the issue was that while in Ca the state with occupations on the 3d shell is easily optimized, but the one in K is a Rydberg-like state, which you don't want to use in a general-use molecular basis set.

The problem with the D orbitals was the reason why he hadn't produced a basis set for K back then, but I guess they have figured a way around the problem :)

  • $\begingroup$ +1. Very useful insight indeed. I thought cc basis sets were made usually at CISD level on top of HF (not MCHF), but in any case, an answer referring to Kirk's wisdom, is a useful answer indeed! Now I just hope Kirk can come and give THE answer to this question :) $\endgroup$ Aug 7, 2020 at 21:25
  • $\begingroup$ @NikeDattani yeah I'm not 100% sure; it's just that single-reference HF is very bad for atoms since you tend to get symmetry breaking whenever you have open shells. (Actually, it can happen even in closed-shell cases like F- and Ne!!) I can't say I've ever fully understood the procedures that are used to generate basis sets, they're such specialized expert knowledge that it easily feels like black magic to an outsider :P $\endgroup$ Aug 8, 2020 at 15:40
  • $\begingroup$ Yes I would be skeptical of a single-reference treatment for the ground state of carbon, but it turns out to work quite fine, maybe because there's such a small number of total e-. I made an aug-cc-pCV8Z basis set for carbon in this paper: arxiv.org/abs/2006.13453 and we did ROHF followed by CISD. David Feller & Jacek Koput were the co-authors that brought the basis-set experience and wisdom to the table though, and both of them gained from doing papers with Kirk Peterson in the past, who was a post-doc with Dunning, so I am a few levels below where the true experts of black magic are. $\endgroup$ Aug 8, 2020 at 15:51
  • $\begingroup$ @NikeDattani "These were then treated as free parameters to minimize the difference between the frozen core and all-electron CISD energies of the carbon atom with all other exponent functions fixed" sounds like the exact opposite of what you want to do, since the CISD energies are negative ;) $\endgroup$ Aug 8, 2020 at 17:01
  • $\begingroup$ The CV correction (to go from cc-pVXZ $\rightarrow$ cc-pCVXZ) was done in a way consistent with how CV corrections were made for all other Dunning basis sets, and Feller and Koput lead that part of the project. Maybe we meant to say "maximize" instead of "minimize", will have to look. Thanks for spotting this! Yes: if we minimize the difference, we may just end up with 0. I think we meant "maximize". $\endgroup$ Aug 8, 2020 at 17:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .