I am doing TDDFT calculation for the C30N44H12Li2P molecule. I have calculated the excited-state calculation for only one state. But it is showing negative transition energy and oscillator strength. What does it mean? I have used
# td=(nstates=1) cam-b3lyp/6-31g(d) scf=qc maxdisk=2GB geom=connectivity polar
for my calculation.
The output is as follows:
Excited states from singles matrix:
***********************************************************************
ExtSpn: IState= 1 X2= 1.075 Y2= 0.075 X2-Y2-1= 2.89D-15
S2Grnd= 0.782 Ra= 0.490 Rb= 0.501 T= 0.992 S2Ext= 1.983 S2= 2.766
Ground to excited state transition densities written to RWF 633
Ground to excited state transition electric dipole moments (Au):
state X Y Z Dip. S. Osc.
1 0.0103 0.0000 -0.0007 0.0001 0.0000
Ground to excited state transition velocity dipole moments (Au):
state X Y Z Dip. S. Osc.
1 -0.0005 0.0000 -0.0002 0.0000 0.0000
Ground to excited state transition magnetic dipole moments (Au):
state X Y Z
1 -0.0058 0.0000 0.0239
Ground to excited state transition velocity quadrupole moments (Au):
state XX YY ZZ XY XZ YZ
1 0.0000 0.0000 0.0000 0.0129 0.0000 -0.0010
* + *
Rotatory Strengths (R) in cgs (10**-40 erg-esu-cm/Gauss)
state XX YY ZZ R(velocity)
1 0.0744 -0.0098 -0.0225 0.0140
1/2[* + (*)*]
Rotatory Strengths (R) in cgs (10**-40 erg-esu-cm/Gauss)
state XX YY ZZ R(length)
1 0.0418 0.0000 0.0126 0.0181
1/2[* + (*)*] (Au)
state X Y Z Dip. S. Osc.(frdel)
1 0.0000 0.0000 0.0000 0.0000 0.0000
Excitation energies and oscillator strengths:
Excited State 1: 3.473-A -1.0859 eV -1141.74 nm f=-0.0000 =2.766
261A -> 262A 0.60543
261A -> 264A -0.19278
261A -> 266A -0.13951
261A -> 268A -0.14660
261A -> 270A -0.14086
261A -> 271A -0.17846
261A -> 274A -0.11546
260B -> 261B -0.57253
260B -> 262B 0.30767
260B -> 265B 0.16695
260B -> 268B 0.24517
260B -> 274B -0.11678
261A
Negative excitation energy means the wavefunction of reference state (ground state in current context) is unstable, therefore electron can be excited from this artificial state to lower state. Using scf=QC is often dangerous, because this keyword significantly increases the probability of SCF convergence to unstable state.
Current result is meaningless, you should remove scf=QC.
By the way, even if you are only interested in the first excited state, you should not use "nstates=1", because if totally only one excited state is calculated, this state may be not quite accurate (you will naturally understand the reason if you are familiar with Davidson diagonalization algorithm).
Note that using "TD" in combination with "polar" is highly deprecated, Gaussian may be confused.
It is also possible to have physically meaningful negative transition energies in TDDFT in the event that you converge to a physically meaningful excited state solution during your reference state SCF calculation. This can happen when applying symmetry or other constraints. In these cases, the transition energies correspond to negative (deexcitation) energies from the excited reference state to the ground or lower energy states.
That said, I think Tian Lu's answer likely covers the specifics of your situation as described.
Thank you so much Tian Lu and Magnus Hanson-Heine, for your kind suggestions and valuable time. It will be a great help to me.
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