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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