Keywords for finding the strength value using SF-TDDFT method

Questions
#1

Hi,
I want to know what keywords are used to know the strength value of excited states in the SF-TDDFT method.
For eg: when I do the single-point calculation using the TD-DFT method. In the output file, I am getting the strength value of each excited state.
Excited state 1: excitation energy (eV) = 2.2934
Total energy for state 1: -890.67525215 au
Multiplicity: Triplet
Trans. Mom.: 0.0000 X 0.0000 Y 0.0000 Z
Strength : 0.0000000000
D( 65) → V( 1) amplitude = 0.2583
D( 67) → V( 1) amplitude = 0.8981
D( 67) → V( 2) amplitude = -0.2258

Excited state 2: excitation energy (eV) = 3.4836
Total energy for state 2: -890.63151547 au
Multiplicity: Singlet
Trans. Mom.: 0.1612 X -2.8646 Y -0.4745 Z
Strength : 0.7217704923
D( 67) → V( 1) amplitude = 0.9670

But when I am doing SF-TDDFT calculations. I am not getting strength values in the output.
input:
$rem
BASIS = cc-pVDZ
EXCHANGE = omegaB97XD
OMEGA = 104
JOB_TYPE = sp
MAX_CIS_CYCLES = 100
SCF_CONVERGENCE = 8
MAX_SCF_CYCLES = 100
THRESH = 14
UNRESTRICTED = true
SPIN_FLIP = 1
CIS_N_ROOTS = 4
SYMMETRY_IGNORE = true
SYMMETRY = false
CIS_STATE_DERIVATIVE = 1
SOLVENT_METHOD = PCM
$end

$solvent
DIELECTRIC 32.613000
OPTICALDIELECTRIC 1.765709
$end

output:Excited state 1: excitation energy (eV) = -2.1850
Total energy for state 1: -899.25802480 au
<S**2> : 0.0293
S( 2) → S( 1) amplitude = 0.9945 alpha

Excited state 2: excitation energy (eV) = 0.4820
Total energy for state 2: -899.16001222 au
<S**2> : 2.0048
S( 1) → S( 1) amplitude = 0.6614 alpha
S( 2) → S( 2) amplitude = 0.7411 alpha

Excited state 3: excitation energy (eV) = 1.1392
Total energy for state 3: -899.13585999 au
<S**2> : 0.0714
S( 1) → S( 1) amplitude = 0.7192 alpha
S( 2) → S( 2) amplitude = -0.6520 alpha

Excited state 4: excitation energy (eV) = 1.8313
Total energy for state 4: -899.11042750 au
<S**2> : 1.0375
D( 67) → S( 1) amplitude = -0.2834
S( 2) → V( 1) amplitude = 0.9448 alpha
Can anyone tell me which keywords help me to get strength values in the SF-TDDFT method?

#2

try STS_MOM = True in $rem

#3

Hi @jherbert
I want to ask one thing: I am using the SF-TDDFT method and used STS_MOM = True in $rem to check the oscillator strength value.
I got results like this:

 States      X                  Y            Z            Strength(a.u.)

1    2  -0.010946   0.303695  -0.093218    0.006447822
1    3   0.161319  -3.963231   1.181059       1.365597
1    4   0.679974  -0.299475   0.024505     0.05474773
2    3  -0.113581   0.079795  -0.002898   0.0003067573
2    4   0.572152   0.008468  -0.057749     0.01165924
3    4  -0.236962  -0.202548   0.087508    0.002027044

The vertical excitation occurs from states 1 to 3 with a 1.365 strength value.
Then, I want to check the vertical de-excitation. I optimized the 3rd excited state by adding STS_MOM = True in $rem and checking the emission strength value. From the output, I got this:

States X Y Z Strength(a.u.)

1    2   0.015297  -0.021027   0.012800   4.529821E-05
1    3  -0.785044   2.894757  -2.796066       1.163463
1    4   0.297492   0.159075  -0.052959     0.01097873
2    3   0.133987   0.030784  -0.001106   0.0002885858
2    4  -0.312708   1.410063  -1.339372      0.1560525
3    4   0.180262  -0.946430   0.869080     0.04200707

Here, states are written like 1 to 3 instead of 3 to 1. Can I consider emission from states 3 to 1 with a strength value of 1.163463?

#4

can you please provide a complete input file? I do not understand what it means to say that you “optimized the 3rd excited state by adding STS_MOM = True”.

#5

I want to say that I optimized my structure at excited state 3, which is S1, with the listed input:
$molecule
0 3
coordinates
$end

$rem
BASIS = cc-pVDZ
METHOD = wB97XD
OMEGA = 104
JOB_TYPE = optimization
MAX_CIS_CYCLES = 700
SCF_CONVERGENCE = 8
MAX_SCF_CYCLES = 700
THRESH = 14
UNRESTRICTED = true
SPIN_FLIP = 1
CIS_N_ROOTS = 4
SYMMETRY_IGNORE = true
SYMMETRY = false
Geom_opt_max_cycles = 100
CIS_STATE_DERIVATIVE = 3
$end

and then do the Single point energy calculation, taking optimized coordinates:
$rem
BASIS = cc-pVDZ
METHOD = wB97XD
OMEGA = 104
JOB_TYPE = sp
MAX_CIS_CYCLES = 700
SCF_CONVERGENCE = 8
MAX_SCF_CYCLES = 700
THRESH = 14
UNRESTRICTED = true
SPIN_FLIP = 1
CIS_N_ROOTS = 4
STS_MOM = true
SYMMETRY_IGNORE = true
SYMMETRY = false
Geom_opt_max_cycles = 100
CIS_STATE_DERIVATIVE = 3
$end

#6

So those state-to-state (STS) transition moments should be between pairs of states at the geometry that you provided. There’s no distinction between excitation and de-excitation because what is output are matrix elements of the dipole moment operator between states |n> and |m>, and those are symmetric for real basis functions.

#7

Thanks @jherbert
In the TD-DFT method, each excited state has a strength value. From there, we can say that S0 to S1 vertical excitation (absorption), wavelength (nm) has f (strength) = some value. Similarly, after excited state optimization, S1 to S0 vertical de-excitation (emission), wavelength (nm) has f (strength) = some value.

Can we also say this thing here in the SF-TDDFT method?

#8

In principle yes, but be careful that the first state in spin-flip is the reference state with a different multiplicity.

You should probably take a look at some of the literature that’s cited in the Q-Chem manual.