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I am studying a system with TD-DFT method implemented in Gaussian. I am not sure whether this two inputs are adequate by the mean of output - optimization of T1?

#p b3lyp/6-31G* TD=(triplets,root=1) opt

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0 1
<geometry>

#p b3lyp/6-31G* TD=(triplets,root=1) opt

title

0 3
<geometry>
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At least when the ground state of the molecule is a singlet state, the first one is correct, and the second one is wrong. Note that when $S_0$ is the ground state of the molecule, a triplet ground state calculation, such as from the following input file

#p b3lyp/6-31G* opt

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0 3
<geometry>

actually calculates the $T_1$ state, because in this case the lowest triplet state is $T_1$, not $T_0$ (which does not exist). Therefore, your second input file calculates the first triplet state that is above the triplet ground state, i.e. it calculates $T_2$.

An immediate follow-up question is: should you use your first input file or the input file that I just mentioned, given that both calculate the $T_1$ state? Actually, when the exact functional and an infinite basis set are used, the two calculations are equivalent. So the best choice is the one that provides the best cost/performance ratio for approximate functionals. Specifically:

  1. Use your first input file when your $S_0$ state is single-referential and your $T_1$ state is multireferential. Only by this way can you have a single-referential reference state, and minimize errors due to static correlation.
  2. Use my input file when either (1) your $T_1$ state is single-referential and your $S_0$ state is multireferential; (2) your $S_0$ state has a triplet instability or a near triplet instability; (3) your $T_1$ state has significant charge-transfer character, but for some reason you have to use B3LYP which is known to perform poorly for charge transfer excitations; or (4) your $T_1$ state has significant double excitation character when viewed from a $S_0$ reference. Either one of these four situations make the linear response TDDFT calculation of $T_1$ unreliable, or at least much less reliable than it would otherwise be.
  3. When the two methods are equally good in terms of accuracy, use my input file, since it is computationally faster.
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  • $\begingroup$ Thanks a lot for the extensive explanation! Indeed, I didn't consider the fundamental aspect that there's no "T0" and hence can be sufficiently computed by mean of just 0 3. $\endgroup$
    – farmaceut
    Commented Jul 9, 2022 at 19:36
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    $\begingroup$ I think what wzkchem5 means is that the T1 is the lowest energy triplet state by definition (S1 is the lowest singlet excited state, T1 is the lowest triplet excited state and the ground state is a singlet, so T1 is the lowest triplet state). $\endgroup$
    – Greg
    Commented Jul 10, 2022 at 15:00
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    $\begingroup$ Greg is correct. As a side note, for molecules whose ground state is a triplet state, the ground state is T0, the first excited triplet state is T1, and the lowest singlet state is S1; in this case, S0 does not exist, and a "0 1" calculation without the TD keyword yields the S1 state, while a "0 3" calculation without the TD keyword yields the T0 state. $\endgroup$
    – wzkchem5
    Commented Jul 10, 2022 at 16:39

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