4
$\begingroup$

Referring to previous questions asked here, I would like to pose a more general question about the calculation of ΔE(S1-T1) for organic molecules with the software Gaussian16.

In particular, I have seen that the most spread methodology, showed to be better than simple TD-DFT, is TDA-DFT. Consequently, I have used an input file like this:

# opt M062X/6-311g* TDA=(singlet,root=1,nstates=6) scrf=(solvent=dichloromethane) Geom=AllCheck int=ultrafine Nosymm 

molecule1

0 1

to calculate the state S1, while I used exactly the same (but with the TDA=triplet keyword) to calculate the T1 state. From these, I extract the difference in energy.

I understand that this methodology is particularly important when the Franck-Condon approximation can't be used anymore, thus allowing for the optimization of each state.

Having said that, is this correct or one should use a different methodology? Keeping in mind that the molecules I want to study are way to big to apply any coupled cluster methodology.

$\endgroup$

1 Answer 1

3
$\begingroup$

Note, TDA is an approximation of full linear response TD-DFT. While it's possible to compute excited state energies by just running SCF DFT, it generally requires some specialized approach like Delta SCF to converge to higher energy states.

Having said all that, your TDA approach will work.

There are potentially faster (and potentially less accurate) ways to run it depending on your system if the cost of TDA geopts starts to become prohibitive. As mentioned in this prior answer, if you know your ground state is a singlet, you can just run a regular SCF calculation. If you knew the geometry didn't vary much between the various electronic states, you can just optimize the ground state and run a single TDA=(50-50,states=12) calculation without the geopt to get all the singlet/triplet excited energies from that initial geometry.

$\endgroup$
2
  • $\begingroup$ Thank you. Just to better understand your answer: you say 'it is possible to compute excited state energies by running SCF DFT but it generally requires a specialized approch'. This mean that it could be more accurate/appropriate than TDA? $\endgroup$
    – Laura
    Dec 13, 2023 at 8:41
  • 1
    $\begingroup$ TDA will likely be easier and of similar accuracy to Delta SCF. There are some special cases cases where Delta SCF would be more accurate, but to run Delta SCF, you would need to have a pretty good sense of what the excited state looked like (e.g. excite from HOMO to LUMO) whereas a TDA or TDDFT will just find the N lowest energy excited states. $\endgroup$
    – Tyberius
    Dec 13, 2023 at 13:46

You must log in to answer this question.

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