Hello!
I have a few questions about a TDDFT calculation that I ran ( # td b3lyp/6-31g(d,p) scrf=(iefpcm,solvent=chloroform) guess=mix) and when I calculation the % probability of some of the excitation states I am getting >100%. What I remember from statistics is that we cannot actually have >100% probability so I am trying to figure out why I have that occurring in my data.
I calculated %probability by 2*(coefficient^2). I have included one of data's oscillator strength information below.
"Excitation energies and oscillator strengths:
Excited State 1: 2.047-A 0.5492 eV 2257.37 nm f=0.1453 =0.797
339B -> 341B 0.20758 (8.62%)
340B -> 341B 0.97366 (189.6%)
This state for optimization and/or second-order correction.
Total Energy, E(TD-HF/TD-DFT) = -6185.76906590
Copying the excited state density for this state as the 1-particle RhoCI density.
Excited State 2: 2.048-A 0.6312 eV 1964.25 nm f=0.0730 =0.798
339B -> 341B 0.97645 (190.69%)
340B -> 341B -0.20706 (8.57%)
Excited State 3: 2.037-A 0.7499 eV 1653.42 nm f=0.0000 =0.787
331B -> 341B 0.98349 (193.45%)
SavETr: write IOETrn= 770 NScale= 10 NData= 16 NLR=1 NState= 3 LETran= 64.
Hyperfine terms turned off by default for NAtoms > 100."
The other three questions I have are:
- what the ####-A means (in bold above) as I have some calculations with various numbers-A and others that have singlet-A. (ZnbisPEB file)
- I obtained a negative wavelength what should I do? I have already read a question on here about something similar, but the only suggestion was to remove the +, which I do not have in my initial gjf file. Should I solve for more states or completely eliminate (d,p)? (Trimer file)
- In another calculation I obtained a negative oscillator strength which I know from some web searches is not theoretically possible and indicates that there is a lower energy state (is that correct?) - how would I fix that? I have included it below, the same basis set as above is used. (1ZnTrimer file)
"Excitation energies and oscillator strengths:
Excited State 1: 4.010-A -0.2239 eV -5538.35 nm f=-0.0004 =3.771"
Any clarification would be super helpful. I have also included the out files for the three compounds I am talking about.
Thank you so much!
Hello!
The probability for transition is most likely proportional only to the coefficient^2; that way you won't have numbers above 100%.
Concerning the '####-A' - I don't really know what the numbers mean. The letter is connected to the symmetry of the state - 'A' is an irreducible representation showing states that are not degenerate.
Please, don't remove the polarization functions (d,p) from your basis set as this will worsen the quality of the results. Usually, the computational protocol (the combination of method and basis set) should be the same for the geometry optimization and the other properties that are to be calculated.
As you point out, the negative values for the wavelength and the oscillator strength are probably indicators of a lower lying state. I visualized the molecules and in each of the three there is a C≡C group missing a hydrogen - is it intentional? That changes the multiplicity and the produced radical is certainly unstable which can lead to not very sensible results.
Kind regards,
Hristo Rasheev
Hristo Georgiev Rasheev Thank you for the clarification! I was told that to find the % probability the equation to use was supposed to be 2*(coefficent^2) and did not know there was another option. Was I told the wrong equation or is there specific requirements for when to use coefficent^2 vs the other one? The trimer compounds are fused together with the porphyrin system and so there should be no hydrogens on the alkyne linker group but my thiophene moiety for the trimer did seem to be messed up so that could be the problem. Thank you so much for the help!
2*(coefficient^2) is only correct for closed-shell case, this is because coefficients of orbital transitions of alpha and beta spins are exactly the same, therefore Gaussian only prints one of them, and thus the factor of 2 should be introduced. For open-shell cases like your example, alpha and beta transition information are printed separatedly, so it should be directly calculated as coefficient^2.
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