# Excited state PES

Hi,
I am doing the constrained optimization at the excited state PES using the SF-TDDFT method. I am doing PES at 3rd excited state.
Here are my output energy values:

Excited state 1: excitation energy (eV) = 0.3297
Total energy for state 1: -1156.42642021 au
<S**2> : 0.1252
S( 1) → S( 1) amplitude = -0.4674 alpha
S( 1) → S( 2) amplitude = 0.2223 alpha
S( 2) → S( 1) amplitude = 0.7602 alpha
S( 2) → S( 2) amplitude = 0.3448 alpha

Excited state 2: excitation energy (eV) = 0.4440
Total energy for state 2: -1156.42221997 au
<S**2> : 0.1957
S( 1) → S( 1) amplitude = 0.7508 alpha
S( 2) → S( 1) amplitude = 0.5919 alpha
S( 2) → S( 2) amplitude = -0.2450 alpha

Excited state 3: excitation energy (eV) = 0.4455
Total energy for state 3: -1156.42216540 au
<S**2> : 1.8423
S( 1) → S( 1) amplitude = 0.4313 alpha
S( 1) → S( 2) amplitude = 0.1669 alpha
S( 2) → S( 1) amplitude = -0.1749 alpha
S( 2) → S( 2) amplitude = 0.8531 alpha

Excited state 4: excitation energy (eV) = 0.7877
Total energy for state 4: -1156.40958838 au
<S**2> : 0.0572
S( 1) → S( 2) amplitude = 0.9473 alpha
S( 2) → S( 2) amplitude = -0.2452 alpha

From value, I am thinking that E.S-1 and E.S-2 are singlet states. Can we say that E.S-1 is S0 and E.S-2 is S1 states?
According to my understanding If the E.S-1 excitation energy value is negative then we will call it the S0 state but near the CI points I observed that E.S-1 does not have a negative excitation energy value.
Can anybody please clarify this?

If the excitation energy for a state is negative, all it means is that the state is lower lying than the high-spin reference state used in the spin-flip calculation. In your case, ES1 and ES2 are indeed S0 and S1 at the present geometry. ES3 is a triplet.