Dear all,

I am recently learning excited state calculations (TDDFT or EOM-CC). When I compared the results using basis sets cc-pvtz or aug-cc-pvtz, I have following observations (excitation energies were always calculated up to 9.92eV, which is the limit I can reach experimentally using vacuum uv):

- with aug-cc-pvtz, more excitations (for transitions <9.92eV) were found.

From what I read the reason seems to be that a lot of Rydberg transition can only be found with diffuse functions. So a lot of Rydberg transitions are missing when using cc-pvtz. Is this explanation correct?

- with aug-cc-pvtz, the total sum of oscillator strengths (for all transitions <9.92eV) increase significantly as well.

There is sum rule that \sigma_j f_{ij}=1 and \sigma_{ij} f_{ij}=N (total number of electrons). Ideally if all transitions were found, sum of f should be the same. So I think if the number of transitions is big enough, the sum of f should tend to be the same. For many molecules I calculated, the number of transitions is several hundred. So why the sum of f with aug-cc-pvtz is always much bigger than the one with cc-pvtz? I am interested in this question because I want to correlate the total oscillator strength to experimental total response from vacuum uv, but I not sure which calculated results I sould trust more.

Thank you very much!

M.M

Hmm, think this question may be for @crawdad.

The answer to question (1) is “yes”, you’re accessing Rydberg states when you include diffuse functions in the basis set.

The second question is more complicated, but I believe the answer relates to the fact that both of the basis sets are far from complete. The Kuhn sum rule will hold only if the set of transitions is truly complete – including continuum states, doubly (and higher) excited states, etc. Your one-electron and N-electron basis sets are not sufficient to describe the complete space, so the sum rule is hard to evaluate perfectly.

And I would say that aug-cc-pVTZ results will be preferable to cc-pVTZ, but I don’t know that either is sufficient for comparison of the integrated oscillator strength with experiment. It depends on the nature of the transitions you’re measuring.

I hope this helps!

-TDC

Dear Dr. Crawford,

Thank you very much! It helps a lot!

Can you tell a lit bit more about “comparison of the integrated oscillator strength with experiment depends on the nature of the transitions”? What types of the transitions are not suitable to compare with experiment and why?

Thank you again!

Best regards,

Michael

One cannot always, for example, directly compare the vertical transitions computed by default in quantum chemistry codes with spectroscopic measurements, as the latter correspond to vibronic transitions. Also, doubly excited states (shake-up processes) would be very poorly described by EOM-CCSD and non-existent in TDDFT.

-TDC