Thanks a lot for your kind replies.
I have done the calculation for benzene and i get for instance for the 11th excitonic state :
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# 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 7.24994
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15.00000 21.00000 4.00000 1.00000 0.24443 6.63483
How 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 :
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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