DFT with exact exchange
Peter REINHARDT
reinh at lct.jussieu.fr
Fri Mar 12 07:29:15 GMT 2004
Dear Molpro users,
trying to calculate a total energy with the B3LYP functional made
me wondering about the sum of the individual terms: (Ne dimer, reasonable basis,
molpro2002.7)
!RKS STATE 1.1 ENERGY -257.85537304
Nuclear energy 20.35297112
One-electron energy -405.46184934
Two-electron energy 147.34548467
SCF exchange energy -24.10956394 Factor=0.2000
Density functional -20.09197948 B88=-24.20368458 S=-22.00316260 LYP= -0.76813157 VWN= -1.48887896
Virial quotient -1.00418838
!RKS STATE 1.1 DIPOLE MOMENTS: 0.00000000 0.00000000 0.00000000
If I sum up Nuclear energy, One-electron energy, Two-electron energy and Density functional
20.35297112-405.46184934+147.34548467-20.09197948 I arrive at -257.85537303,
which is the total RKS energy, as well printed at the very end of the output. However,
0.72*B88+0.08*S+0.81*LYP88+0.19*VWN as is the density functional combination for B3LYP
WITHOUT exact exchange gives
-0.72*24.20368458-0.08*22.00316260-0.81*0.76813157-0.19*1.48887896 = -20.0919794797
Do I have to conclude that the total energy is the printed RKS energy + 0.2 * exact exchange?
Which gives -262.677285818 a.u. ?
Yours,
Peter Reinhardt
--
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