15
$\begingroup$

For the first time I'm doing TD-DFT calculations (wB97XD functional) in Gaussian 09 for an open-shell system and the results look like the hell of a mess for me. The molecule is rather big, so I started with just 10 excited states to see how it would work.

First off, the ground electronic state is $^2A_2$, orbital 453 is HOMO and 454 is SOMO. Now to the excited states. In the output I found, for example, the following piece that looks strange to my eyes

Excited State   3:  2.139-B2     0.9431 eV 1314.65 nm  f=0.0816 <S**2>=0.894    
453B -> 454B       0.98895  
453B <- 454B       0.17190
  1. The first line is fine: it says that the third excited state is $^2B_2$ (taking spin contamination into account).
  2. The second line is also fine: it tells that the determinant corresponding to excitation from HOMO to SOMO has the largest coefficients in the CI expansion.
  3. But what is the meaning of the third line?
$\endgroup$
11
+100
$\begingroup$

Time-dependent DFT can be used to predict excitation energies through a linear-response formulation.

In this Gaussian result, beyond the first line, you are looking at the largest coefficients in the configuration-interaction (CI) style expansion.

(It's not strictly CI, but the implementation of time-dependent HF or RPA is essentially the same for TDDFT or Tamm-Dancoff Approximation TDDFT.)

Anyway, since you're performing a calculation on an open-shell system, the formulation for the state will include all one-electron excitations AND all one-electron de-excitations. That state has a de-excitation from 454B into 453B (i.e., the SOMO will fill the hole left by the excitation). Now you may think "but that doesn't accomplish anything." Remember that Gaussian only prints the dominant coefficients.

For this reason, Rich Martin created the Natural Transition Orbitals method as a nice way to express the excitation, rather than a set of coefficients and the ground-state orbitals.

Oftentimes, there is no dominant configuration in the list of excitation amplitudes, thereby making a straightforward interpretation of the excited state difficult. This is particularly unsatisfactory when attempting to determine the qualitative nature of an excited state. An additional complication in the DFT case is that in principle all orbitals in DFT but the HOMO are devoid of physical significance. However, chemical intuition is built on the orbital construct, and a simple orbital interpretation of "what got excited to where" is important.

This is available in Gaussian from the Population keyword and in other codes.

Incidentally, there's a nice review of TDDFT in: "Progress in Time-Dependent Density-Functional Theory" Annual Review of Physical Chemistry Vol. 63: 287-323.

$\endgroup$
  • $\begingroup$ I spent almost the week reading about RPA here and there, and as far as I understand your are right: "de-excitations" (as well as "excitations") are involved in the wave function expansion in RPA. And I simply did not know that. $\endgroup$ – Wildcat Oct 31 '14 at 19:03
  • $\begingroup$ @Wildcat I don't think I've ever seen a case where there's an excitation and a deexcitation from the same states, but I think NTOs would help to understand what's going on. The "minor" contributors are probably important. $\endgroup$ – Geoff Hutchison Oct 31 '14 at 19:05
  • $\begingroup$ NTO seems to be quite interesting, but so far I decided to do calculations within Tamm-Dancoff approximation in which "de-excitations" are neglected. $\endgroup$ – Wildcat Oct 31 '14 at 19:11
  • $\begingroup$ I'd do some benchmarks if I were you. Things may have changed, but when I benchmarked TDA vs. TDDFT, full TDDFT won. Granted, that was on closed-shell systems, but if possible, I'd test out some experimental systems. $\endgroup$ – Geoff Hutchison Oct 31 '14 at 19:29
  • $\begingroup$ Sure. I'm doing some benchmarks right now. Luckily we have experimental UV-vis spectra for the molecules in question, so I can benchmark TDA approximation vs. full TD-DFT, as well as different functionals/basis sets/ECPs combinations. $\endgroup$ – Wildcat Oct 31 '14 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.