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example_Mulliken_ethane_UFF_pbe_cc-pVDZ.txt
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35 lines (22 loc) · 1.63 KB
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Output of Mulliken.out :
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""
Atom number 1:
State eigenvalue occ.number total l=0 l=1 l=2
k point number: 1: ( 0.00000000 0.00000000 0.00000000 ); weight: 1.00000000
1 -269.36573 2.0000000 0.40690 0.40697 0.00001 -0.00008
2 -269.36354 2.0000000 0.59324 0.59293 0.00031 0.00000
. .. .. .. .. .. ..
. .. .. .. .. .. ..
Atom number 2:
State eigenvalue occ.number total l=0 l=1 l=2
k point number: 1: ( 0.00000000 0.00000000 0.00000000 ); weight: 1.00000000
1 -269.36573 2.0000000 0.59310 0.59321 -0.00005 -0.00006
2 -269.36354 2.0000000 0.40705 0.40669 0.00037 -0.00001
. .. .. .. .. .. ..
. .. .. .. .. .. ..
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""
The energy are assigned based on the maximum value in "total" . For example:
For Atom number 1, the highest total is 0.59324 and it corresponds to eigenvalue -269.36354, so
atom number 1 has energy -269.36354.
Similarly, for Atom number 2, the highest total is 0.59310 and it corresponds to eigenvalue
-269.36573, so atom number 2 has energy -269.36573.