Example: Grimme Molecular Mechanics dispersion-corrected functionals (DFT-D3-BJ)

Download MM_Dispersion.run

In this example a structure with 2 benzene molecules and a hydrogen molecule is optimized with the Grimme dispersion corrected PBE, in this case the (so far) latest variant the DFT-D family, which is DFT-D3-BJ. Needed is the subkey DISPERSION in the key XC, and the arguments Grime3 and BJDAMP. If one starts with atomic fragments the part of the bond energy that is due to the Grimme dispersion corrected functional is both inter-molecular as well as intra-molecular.

$ADFBIN/adf << eor
Title Geometry optimization with Grimme3-BJ dispersion correction for GGA
basis
  type TZP
  core small
End
XC
  GGA PBE
  DISPERSION Grimme3 BJDAMP
End
geometry
  converge grad=0.001
  iterations 50
end
Atoms   cartesians
C  0.000000000000     3.050000000000     1.391500000000
H  0.000000000000     3.050000000000     2.471500000000
C  1.205074349366     3.050000000000     0.695750000000
H  2.140381785453     3.050000000000     1.235750000000
C  1.205074349366     3.050000000000    -0.695750000000
H  2.140381785453     3.050000000000    -1.235750000000
C -0.000000000000     3.050000000000    -1.391500000000
H -0.000000000000     3.050000000000    -2.471500000000
C -1.205074349366     3.050000000000    -0.695750000000
H -2.140381785453     3.050000000000    -1.235750000000
C -1.205074349366     3.050000000000     0.695750000000
H -2.140381785453     3.050000000000     1.235750000000
C -1.205074349366    -3.050000000000    -0.695750000000
H -2.140381785453    -3.050000000000    -1.235750000000
C -0.000000000000    -3.050000000000    -1.391500000000
H -0.000000000000    -3.050000000000    -2.471500000000
C  1.205074349366    -3.050000000000    -0.695750000000
H  2.140381785453    -3.050000000000    -1.235750000000
C  1.205074349366    -3.050000000000     0.695750000000
H  2.140381785453    -3.050000000000     1.235750000000
C -0.000000000000    -3.050000000000     1.391500000000
H -0.000000000000    -3.050000000000     2.471500000000
C -1.205074349366    -3.050000000000     0.695750000000
H -2.140381785453    -3.050000000000     1.235750000000
H  0.0                0.35               0.0
H  0.0               -0.35               0.0
End
End Input

The next calculations in this example demonstrate dispersion corrections when using non-atomic fragments. First three molecules (2 benzene molecules and a hydrogen molecule) are calculated . Needed again is the subkey DISPERSION in the key XC. The one for H2 is given below:

$ADFBIN/adf << eor
Title Grimme dispersion-corrected GGA
basis
  type TZP
  core small
End
XC
  GGA PBE
  DISPERSION Grimme3 BJDAMP
End
SCF
  Iterations  60
  Converge  1.0E-06  1.0E-6
End
Atoms
H         0.000000    0.000000   -0.377906
H         0.000000    0.000000    0.377906
End
End Input
eor
mv TAPE21 h2.t21

Note that even for such a molecule there is a contribution from the so called Dispersion energy in the bonding energy (although it will be very small in this case). Next a structure is calculated in which the three calculated molecules in it. If one starts with molecular fragments the part of the bond energy that is due to the Grimme dispersion corrected functional is only inter-molecular.

$ADFBIN/adf << eor
Title Grimme dispersion-corrected GGA
Fragments
  b1 benzene1.t21
  b2 benzene2.t21
  h2 h2.t21
End
XC
  GGA PBE
  DISPERSION Grimme3 BJDAMP
End
Atoms
C         0.000000    1.398973   -3.054539  f=b1
H         0.000000    2.490908   -3.049828  f=b1
C         1.211546    0.699486   -3.054539  f=b1
H         2.157190    1.245454   -3.049828  f=b1
C         1.211546   -0.699486   -3.054539  f=b1
H         2.157190   -1.245454   -3.049828  f=b1
C         0.000000   -1.398973   -3.054539  f=b1
H         0.000000   -2.490908   -3.049828  f=b1
C        -1.211546   -0.699486   -3.054539  f=b1
H        -2.157190   -1.245454   -3.049828  f=b1
C        -1.211546    0.699486   -3.054539  f=b1
H        -2.157190    1.245454   -3.049828  f=b1
C        -1.211546   -0.699486    3.054539  f=b2
H        -2.157190   -1.245454    3.049828  f=b2
C         0.000000   -1.398973    3.054539  f=b2
H         0.000000   -2.490908    3.049828  f=b2
C         1.211546   -0.699486    3.054539  f=b2
H         2.157190   -1.245454    3.049828  f=b2
C         1.211546    0.699486    3.054539  f=b2
H         2.157190    1.245454    3.049828  f=b2
C         0.000000    1.398973    3.054539  f=b2
H         0.000000    2.490908    3.049828  f=b2
C        -1.211546    0.699486    3.054539  f=b2
H        -2.157190    1.245454    3.049828  f=b2
H         0.000000    0.000000   -0.377906  f=h2
H         0.000000    0.000000    0.377906  f=h2
End
End Input
eor