Thanks a lot for your kind replies.
I have done the calculation for benzene and i get for instance for the 11th excitonic state :
Code: Select all
# Electron-Hole pairs that contributes to Excitonic State 11 more than 5%
#
#             K-point [iku]                      Weight
#   0.000000   0.000000   0.000000    0.127363
#   0.500      -.306E-16  -.306E-16      0.286    
#   0.00      0.500      -.306E-16          0.289    
#   0.500      0.500      -.612E-16        1.00    
#   0.000000   0.000000   0.500000    0.109865
#   0.500      -.306E-16  0.500            0.204    
#   0.000000   0.500000   0.500000    0.207410
#   0.500000   0.500000   0.500000    0.652437
#  
#  Band_V     Band_C     K  ibz     Symm.      Weight     Energy
#  
  15.00000   21.00000    4.00000    1.00000    0.24443    6.63483
  15.00000   21.00000    8.00000    1.00000    0.16787    6.79957
  15.00000   25.00000    3.00000    1.00000    0.06640    7.15014
  15.00000   25.00000    2.00000    1.00000    0.06576    7.15347
  14.00000   21.00000    4.00000    1.00000    0.06174    6.63822
  15.00000   23.00000    7.00000    1.00000    0.05596    7.24592
  15.00000   23.00000    6.00000    1.00000    0.05504    7.24994Code: Select all
 15.00000   21.00000    4.00000    1.00000    0.24443    6.63483How to deduce it now from symmetries of KS-states ???
To your information, benzene has 15 occupied states : with following irreducible representations from state 1 to 30 :
Code: Select all
        1A1g      1E1u      1E1u      2A1g      1E2g      1E2g      1B2u      1A2u      1B1u      2E1u      2E1u    
        2E2g      2E2g      1E1g      1E1g      1E2u      1E2u      2B1u      3A1g      1B2g      3E2g      3E2g
        3E1u      3E1u      4E2g      4E2g      4E1u      4E1u      3B1u      1A2g
Vito