How to get vertical excitation energy S0 to S1 using SF-TDDFT?

Questions
#1

Hi,
I am using the SF-TD-DFT method and have already optimized my geometry at S0 (triplet state). After optimization, I want excitation to a singlet excited state. In the input, I am giving the following keywords
$rem
BASIS = GEN
PURECART = 1
CIS_N_ROOTS = 8
EXCHANGE = omegaB97X-D
GUI = 2
JOB_TYPE = SP
SCF_CONVERGENCE = 7
MAX_SCF_CYCLES = 100
SPIN_FLIP = 1
MAX_CIS_CYCLES = 100
SOLVENT_METHOD = PCM
MEM_STATIC = 6000
MEM_TOTAL = 30000
$end

$solvent
DIELECTRIC 32.613000
OPTICALDIELECTRIC 1.765709
$end

As for output, I am getting these energy values. Here the strength value is not coming from which I get to know about singlets and triplets. So, I want to know which is my first singlet excited state. Or are there any other keywords for this calculation?

       SF-DFT Excitation Energies

(The first “excited” state might be the ground state)

Excited state 1: excitation energy (eV) = -3.6675
Total energy for state 1: -748.90152786 au
<S**2> : 0.0532
S( 2) → S( 1) amplitude = 0.9897 alpha

Excited state 2: excitation energy (eV) = 0.1466
Total energy for state 2: -748.76136030 au
<S**2> : 1.0234
S( 2) → S( 2) amplitude = 0.2136 alpha
S( 2) → V( 1) amplitude = 0.2426 alpha
S( 2) → V( 2) amplitude = 0.6574 alpha
S( 2) → V( 3) amplitude = -0.3417 alpha
S( 2) → V( 4) amplitude = -0.5664 alpha

Excited state 3: excitation energy (eV) = 0.5351
Total energy for state 3: -748.74708582 au
<S**2> : 2.0242
S( 1) → S( 1) amplitude = 0.5959 alpha
S( 2) → S( 2) amplitude = 0.2296 alpha
S( 2) → V( 1) amplitude = 0.1966 alpha
S( 2) → V( 5) amplitude = -0.7010 alpha

Excited state 4: excitation energy (eV) = 0.7735
Total energy for state 4: -748.73832475 au
<S**2> : 0.1273
S( 1) → S( 1) amplitude = 0.7600 alpha
S( 2) → V( 1) amplitude = -0.2109 alpha
S( 2) → V( 5) amplitude = 0.5264 alpha

Excited state 5: excitation energy (eV) = 1.1563
Total energy for state 5: -748.72425650 au
<S**2> : 1.0295
S( 2) → S( 2) amplitude = 0.7982 alpha
S( 2) → V( 1) amplitude = -0.5803 alpha

Excited state 6: excitation energy (eV) = 1.4911
Total energy for state 6: -748.71195347 au
<S**2> : 1.0336
S( 2) → S( 2) amplitude = 0.4835 alpha
S( 2) → V( 1) amplitude = 0.7058 alpha
S( 2) → V( 2) amplitude = -0.2904 alpha
S( 2) → V( 5) amplitude = 0.3977 alpha

Excited state 7: excitation energy (eV) = 1.8386
Total energy for state 7: -748.69918338 au
<S**2> : 1.0367
D( 64) → S( 1) amplitude = 0.1816
S( 2) → V( 2) amplitude = 0.5298 alpha
S( 2) → V( 3) amplitude = 0.7843 alpha

Excited state 8: excitation energy (eV) = 1.9797
Total energy for state 8: -748.69399616 au
<S**2> : 1.0473
D( 64) → S( 1) amplitude = 0.9135
S( 2) → V( 3) amplitude = -0.2964 alpha
S( 2) → V( 4) amplitude = 0.1732 alpha

#2

Is the reference state a triplet? If so then S0 is probably the first excitation energy.

#3

Yes, the reference state is the triplet state. Using spin flip DFT i optimized at the triplet state, that geometry i am saying S0. Excitation of my molecule is at singlet excited state. So, how i will get to know my singlet excited state for optimization?

#4

If you want to optimize S0, using SF-TDDFT with a triplet reference state, then that is an excited-state optimization, presumably on the first excited-state surface. Need to use CIS_STATE_DERIV=1. Please be aware that excited-state geometry optimizations can be troublesome, due to state-switching.

#5

Dear John,

Thanks for me updating me. Using CIS_STATE_DERIV = 1, I already optimized my geometry at triplet state.

My query is, I did a TDDFT calculation there I found 7th excited state was the first singlet state, and I want to optimize at first singlet excited state using SF-TDDFT to get the S0 to S1 excitation.

I optimized my geometry using SF-DFT at triplet state (S0) (CIS_STATE_DERIV = 1, SPIN_FLIP = 1). So, now what will I do for the first singlet excited state optimization?

#6

I guess I don’t understand. What is the multiplicity of the reference state? You did not provide a $molecule section so I can’t tell. I assumed it was a triplet, but then your statement that you used CIS_STATE_DERIV = 1 to optimize the triplet does not make sense. Regarding the 7th excited state being S0, if you are referring to the output above then I don’t understand. What makes you conclude that this particular state is S0?