Conceptual question regarding the applicability of SF-TDDFT method

Hi everyone,

I have a conceptual question regarding the applicability of SF-TDDFT method.

I am studying a benzene + O(¹D) system, where benzene is in its closed-shell singlet ground state, and the oxygen atom is in its excited singlet ¹D state. At large separation, the overall system is therefore singlet–singlet.

My question is:
Can this system be meaningfully treated using the SF-TDDFT framework?

As I understand it, SF-TDDFT requires a high-spin reference state (e.g., a triplet), from which lower-spin states are accessed by spin-flip excitations. However, in the benzene + O(¹D) case, both fragments are singlets.

Is it inappropriate to use SF-TDDFT for this system, at least in the non-interacting or weakly interacting regime? If so, would conventional TDDFT or a multireference approach be more appropriate for describing excited-state dynamics?

I would appreciate any clarification or guidance on how one should properly set up excited-state calculations for this type of system in Q-Chem.

Thank you very much.

I tried to put a job as well:
input:

$molecule
0 3
  H      1.954110     1.501681    -0.000017
  C      1.099407     0.844699    -0.000021
  C      1.281178    -0.530526    -0.000028
  H      2.277161    -0.943033    -0.000032
  C      0.181436    -1.373577    -0.000037
  H      0.322351    -2.442191    -0.000046
  C     -1.099388    -0.844754    -0.000030
  H     -1.954087    -1.501741    -0.000034
  C     -1.281160     0.530470    -0.000016
  H     -2.277141     0.942980    -0.000009
  C     -0.181409     1.373524    -0.000016
  H     -0.322324     2.442138    -0.000008
  O      0.000022    -0.000010    -3.000025
$end

$rem
   BASIS  =  6-31G*
   JOB_TYPE  =  sp
   METHOD  =  BHHLYP
   SYMMETRY_IGNORE = TRUE
   MAX_CIS_CYCLES = 500
   MAX_SCF_CYCLES = 500
   THRESH = 14
   SPIN_FLIP = true
   CIS_N_ROOTS = 6
   CIS_STATE_DERIVATIVE = 1
   sts_mom = true
   UNRESTRICTED = true
   SCF_CONVERGENCE  =  8
$end

and output is:

        SF-DFT Excitation Energies              

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

Excited state 1: excitation energy (eV) = 1.6066
Total energy for state 1: -307.08509720 au
<S**2> : 2.0054
D( 16) → S( 1) amplitude = 0.6923
D( 17) → S( 2) amplitude = -0.6924

Excited state 2: excitation energy (eV) = 2.6689
Total energy for state 2: -307.04605844 au
<S**2> : 0.0190
D( 16) → S( 1) amplitude = -0.2698
D( 16) → S( 2) amplitude = 0.6312
D( 17) → S( 1) amplitude = -0.6326
D( 17) → S( 2) amplitude = -0.2700

Excited state 3: excitation energy (eV) = 2.6689
Total energy for state 3: -307.04605836 au
<S**2> : 0.0190
D( 16) → S( 1) amplitude = -0.6320
D( 16) → S( 2) amplitude = -0.2697
D( 17) → S( 1) amplitude = 0.2701
D( 17) → S( 2) amplitude = -0.6317

Excited state 4: excitation energy (eV) = 3.7159
Total energy for state 4: -307.00758397 au
<S**2> : 0.2174
D( 16) → S( 2) amplitude = -0.5899
D( 17) → S( 1) amplitude = -0.5868
S( 1) → S( 2) amplitude = 0.3663 alpha
S( 2) → S( 1) amplitude = 0.3726 alpha

Excited state 5: excitation energy (eV) = 4.1273
Total energy for state 5: -306.99246574 au
<S**2> : 1.0257
S( 1) → S( 1) amplitude = -0.6000 alpha
S( 1) → S( 2) amplitude = -0.1699 alpha
S( 2) → S( 1) amplitude = -0.2185 alpha
S( 2) → S( 2) amplitude = 0.7367 alpha

Excited state 6: excitation energy (eV) = 4.1631
Total energy for state 6: -306.99114905 au
<S**2> : 1.0124
S( 1) → S( 1) amplitude = 0.2073 alpha
S( 1) → S( 2) amplitude = -0.6416 alpha
S( 2) → S( 1) amplitude = 0.6714 alpha
S( 2) → S( 2) amplitude = 0.2201 alpha