How to interpret orbital transition in TDDFT?

Question

Consider the following TDDFT run with GAMESS:

!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=N311 NGAUSS=6 $END
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END
 $CONTRL ICHARG=0 MULT=1 $END
 $TDDFT NSTATE=9 $END
 $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END
 $SYSTEM MWORDS=1000 PARALL=.TRUE. $END
 $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END

 $DATA 
Title
C1
O     8.0     0.00000    -1.27900     0.00300
C     6.0    -0.00000    -0.05800     0.00100
C     6.0     1.29700     0.69100    -0.00000
C     6.0    -1.29800     0.69000    -0.00000
H     1.0     1.35900     1.32900    -0.90600
H     1.0     1.35900     1.33200     0.90300
H     1.0     2.15700    -0.01300     0.00100
H     1.0    -2.15700    -0.01400     0.00100
H     1.0    -1.35900     1.32900    -0.90600
H     1.0    -1.35900     1.33200     0.90300
 $END

From this we obtain the following part of the output. (The complete calculation can be found here.) How can we interpret the orbital transitions, in terms of for example $\mathrm{n}\to\sigma^*$ or $\pi\to\pi^*$?

Answer 1

To visualise the orbitals of your calculation, use a program of your choice. For Gamess, there are a few options availabile. I use ChemCraft and Molden, and they work quite well. Here is a compilation of some with the former mentioned:

You can further use the summary to identify the most interesting, strongest transitions:

                          SUMMARY OF TDDFT RESULTS

   STATE             ENERGY     EXCITATION  TRANSITION DIPOLE, A.U.  OSCILLATOR
                    HARTREE          EV         X       Y       Z     STRENGTH
   0  A         -193.0290234748    0.000
   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000
   2  A         -192.7831335626    6.691     0.4051 -0.0004  0.0000    0.027
   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002
   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042
   5  A         -192.7210047911    8.382    -0.0121  0.0000 -0.0006    0.000
   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001
   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000
   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022
   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290

For the calculated 9 states that is the last one, with an oscillator strength of 0.290. Now skip back one section to where you find:

      -------------------
      SINGLET EXCITATIONS
      -------------------

Look for excited state number 9:

 STATE #   9  ENERGY =    9.326387 EV
 OSCILLATOR STRENGTH =    0.290423
 LAMBDA DIAGNOSTIC   =    0.561 (RYDBERG/CHARGE TRANSFER CHARACTER)
 SYMMETRY OF STATE   =    A   
                 EXCITATION  DE-EXCITATION
     OCC     VIR  AMPLITUDE      AMPLITUDE
      I       A     X(I->A)        Y(A->I)
     ---     ---   --------       --------
      9      17    0.137515       0.018404
     15      17    0.694034      -0.078753
     13      18   -0.085768      -0.006093
     14      19   -0.059549      -0.004510
     16      19   -0.289828       0.018031
     15      20    0.051487       0.010566
     11      21    0.053380       0.009416
     13      21   -0.060120      -0.009392
     16      22   -0.607830       0.005676
     11      24    0.052546       0.012226
     13      25   -0.039117      -0.008987
     16      26    0.112895       0.016712
     16      30   -0.041659      -0.014274
     16      33    0.052880       0.019973
     16      36   -0.031783      -0.011699
     13      40   -0.030352      -0.013004

There you find the orbital transitions. You can also have a look at the amplitudes to identify the most dominant one. In this case it is probably 15 to 17 and therefore corresponds to $\pi\to\pi^*$.

The more complicated the molecules get, the more confusing will this process be. Also the higher the level of theory, the more transitions you will need to consider. In this case Natural Transition Orbitals will certainly become very helpful, see Richard L. Martin, J. Chem. Phys., 2003, 118, 4775-4777.

Answer 2

Form this output you dont.

Find this in the output:

             EXCITATION  DE-EXCITATION
 OCC     VIR  AMPLITUDE      AMPLITUDE
  I       A     X(I->A)        Y(A->I)
 ---     ---   --------       --------
 10      17   -0.039945       0.001926
 16      17    0.996619      -0.032612
 16      20   -0.070681       0.003443

it shows you the involved orbitals in the transitions. This a dominant transition from the 16ths orbital to the 17ths (HOMO to LUMO). You need to visualize the MOs and decide on the nature (sigma,pi,n) of the orbitals. Since they are MOs, it can be difficult with large molecules. You can look up natural transition orbitals that aim to simplify the visualization.

How you can look at orbitals with gamess? I have no clue.