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I would like to calculate the energy difference between the first excited singlet and the first excited triplet of a given molecule in solvent, on Gaussian. In this publication, they state that using TDA-DFT can be a good level of theory for this type of calculation. For this reason, I have built two inputs. To calculate the singlet:

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

and to calculate the triplet

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

After running these calculations, I obtain two outputs that contain, for the singlet:

    Excited State   1:   Singlet-?Sym    3.0913 eV  401.07 nm  f=0.7556  <S**2>=0.000
     110 ->111         0.67231
 This state for optimization and/or second-order correction.
 Total Energy, E(CIS/TDA) =  -1315.38267036
 Copying the excited state density for this state as the 1-particle RhoCI density.
 
 Excited state symmetry could not be determined.
 Excited State   2:  Singlet-?Sym    3.6009 eV  344.32 nm  f=0.0405  <S**2>=0.000
     106 ->111         0.14329
     107 ->111        -0.15242
     108 ->111         0.32533
     110 ->112         0.51705
     110 ->113         0.18032
 
 Excited state symmetry could not be determined.
 Excited State   3:  Singlet-?Sym    3.9376 eV  314.87 nm  f=0.1058  <S**2>=0.000
     106 ->111        -0.22741
     106 ->112         0.10392
     108 ->111         0.11810
     108 ->115         0.11499
     109 ->111         0.50685
     110 ->112         0.12586
     110 ->113        -0.28556

and for the triplet:

Excited State   1:   Triplet-?Sym    2.1716 eV  570.93 nm  f=0.0000  <S**2>=2.000
     109 ->111        -0.10733
     110 ->111         0.66438
 This state for optimization and/or second-order correction.
 Total Energy, E(CIS/TDA) =  -1315.41646875
 Copying the excited state density for this state as the 1-particle RhoCI density.
 
 Excited state symmetry could not be determined.
 Excited State   2:  Triplet-?Sym    2.9918 eV  414.41 nm  f=0.0000  <S**2>=2.000
     109 ->111         0.34352
     109 ->113        -0.11747
     110 ->111         0.13157
     110 ->112         0.55107
 
 Excited state symmetry could not be determined.
 Excited State   3:  Triplet-?Sym    3.3346 eV  371.81 nm  f=0.0000  <S**2>=2.000
     107 ->111         0.14468
     108 ->111        -0.25678
     109 ->111        -0.34944
     109 ->112        -0.21840
     110 ->112         0.22031
     110 ->113         0.37233

I am a little confused by the output itself. Is the energy that one should consider for the difference in energy the one reported after 'Total energy' (for example, -1315.38267036 for the singlet) or the one reported after the 'Excited state 1' (3.0913eV for the singlet)?

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    $\begingroup$ Just a heads-up, Gaussian includes a TDA=(50-50) (and also TD=(50-50)) option to calculate the singlets and triples at the same time, which is easier and should save time. $\endgroup$
    – leeman
    Mar 2 at 16:39

1 Answer 1

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The excitation energies are 3.09 and 2.17 eV, respectively, both relative to the (same) ground state, and thus the energy difference is 3.09 - 2.17 = 0.92 eV. The Total Energy is the ground state energy + the excitation energy, now in Hartree units. Subtracting these, -1315.38267 - (-1315.41647) and multiplying with 27.21 to get eV, again gives 0.92 eV.

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    $\begingroup$ Thank you Frank. I don't know if I should add another question or I can ask here since it's strictly related. Would it be correct to take into consideration the solvent for these calculations as reported in the input? $\endgroup$
    – Laura
    Dec 8, 2023 at 15:40
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    $\begingroup$ The solvent model will take the 'fast' component of the dielectric medium into consideration, and thus account for the difference in the polarity between the ground and the two excited states, and it should be better than no solvent. But DCM has a low dielectric constant, so likely it makes only a small difference relative to the gas-phase results $\endgroup$ Dec 9, 2023 at 9:13
  • $\begingroup$ Thank you very much :) $\endgroup$
    – Laura
    Dec 10, 2023 at 10:17

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