When a theoretical UV-Vis spectrum is computed by TD-DFT method in Gaussian the electronic transitions are detailed in an output file giving the excited states, transitions energies, assignments and coefficients. Sometimes for an excited state we have two or three transitions and each transition has a coefficient. What are the meaning of these coefficients? How to calculate the contribution (in percentage %) of the electronic transitions based on these coefficients?
The "transitions" as you call them are not really transitions, but the electronic configurations that span the excited state wavefunction. The fact that they look like transitions is because they show between which orbitals of the ground state (Hartree Fock) configuration (occupied and unoccupied) the electrons are transfered to get the given excited configuration.
The excited state wavefunction is never a single excited configuration (just as the true ground state is not a single HF configuration). It can be represented, though, as the linear combination of the single-determinant configurations.
TDDFT lists the most important configurations, meaning those with the largest coefficients in the linear combination.
Hope it helps!
Marcin
Percentage is not a coorect word, these are coefficients in linear combination. Though, something similar to percentage you can get taking squares of these coefs (or maybe gaussian yet gives squares, check it). Summ of squares will yield unity
Mr. Cojocaru,
You can use GaussSum. It is free of charge and can be downloaded by:
http://gausssum.sourceforge.net/
Dear Bojidarka B. Ivanova,
Many thanks for useful link!
Regards,
--CC
The "transitions" as you call them are not really transitions, but the electronic configurations that span the excited state wavefunction. The fact that they look like transitions is because they show between which orbitals of the ground state (Hartree Fock) configuration (occupied and unoccupied) the electrons are transfered to get the given excited configuration.
The excited state wavefunction is never a single excited configuration (just as the true ground state is not a single HF configuration). It can be represented, though, as the linear combination of the single-determinant configurations.
TDDFT lists the most important configurations, meaning those with the largest coefficients in the linear combination.
Hope it helps!
Marcin
Thank you Marcin Andrzejak for the useful information!
Regards,
--CC
By using Gaussian, you can compute the percentage of an excitation as follows:
ci*ci*2, where ci is the coefficient a specific configuration. See the explanation given by Andrzejak.
Hope it helps.
Aggelos
Many thanks Aggelos!
The formula provided matches with the results from GaussSum program.
Regards,
--CC
Dear Aggelos Avramopoulos,
Does exist a reference providing this formula? (for appropriate citation)
Regards,
--CC
Mr. Gromov,
The answer to your question you may find in [Ref.1], accounting for that MCSCF, SCF and full CI belong to the variational methods, where the computing the expectation E value the expansion coeficients are obtained variationally by linear variation theory.
[ref.1]T. Helgaker, P. Jorgensen, J. Olsen, Molecular Electronic-Structure Theory, ISBN: 978-0-471-96755-2, Wiley, 2000, page 1 - 938. (Configuration-interaction and multi-configurational self-consistent theory)
Dear Corneliu,
I found this in a exercise pdf file, provided for gaussian course
If you are interested for a citation, better send a message to GAUSSIAN developers.
Aggelos
Aggelos, never used gaussian, programms i used print spin up and spin down separately (or its because of uhf reference, never used rhf one so far), consequently no need in additional coefficients, so i was surprised =)
Bojidarka, thanks, it was a purely technical question, just differencies in program printout
after completion of uv-vis calculation in gaussian program go to Swizard program and open the uv-vis log file here..................u will get the contributions.
Guys, in my case, some of these coefficients coming out to be negative. What do those negatives coefficients mean? An example is below
Excited State 3: Singlet-A 3.6577 eV 338.97 nm f=1.1324 =0.000
74 -> 78 -0.11341
76 -> 77 0.68923
This isn't a problem, just as coordinates of a vector can be negative. The vector itself does not have negative length because of it.
The contributions of a configuration are suqares of the coefficients (if the coefficients are real - non imaginary :), so their negativity is not an issue, really :)
Correct me if I am wrong. Do I square each coefficient then multiply by 2? Or do I just use the two biggest coefficients.
You should square each coefficient and then multiply with 2.
The multiplication with 2 is for RHF case, not for UHF.
Thank you. Below is an example.
Coefficient
53 --> 55 -0.21431
53 --> 60 -0.12102
54 --> 55 0.63420
If i just want to find out the major MO contribution for 54 --> 55 transition, do i just calculate 2 x 0.63420^2?
Thank you for your help
Corneliu Cojocaru
In the example given by Faizah Islam, If you just want to calculate the contribution for 54 and 55 separately (%) 54 --> 55 (%), how to proceed?
Hi,
You can also use GaussSum software. It will directly give the corresponding excited state energies, HOMO-LUMOs, and percentage contributions. You can download freely here.
https://gausssum.software.informer.com/3.0/
Best regards,
- Badges
- Science topic