Sample directory: adf/MM_Dispersion/
Summary:
First example shows a geometry optimization of a van der Waals complex of two benzene molecules, connected to each other with a hydrogen molecule. With the MMDISPERSION keyword an extra empirical force (of similar form as in molecular mechanics) is added to the interaction between the three fragments, where one benzen molecule is fragment 1 (FD=1), the other benzene molecule is fragment 2 (FD=2), and the hydrogen molecule is fragment 3 (FD=3).
The atomic parameters are read from the file $ADFRESOURCES/MMDispersion/disp-param. The PBE functional and the TZP basis set are used, which is necessary if one wants to use the TZ parameters for the damping function, which are optimized for this combination of functional and basis set.
$ADFBIN/adf << eor basis type TZP core small End XC GGA PBE End geometry converge grad=0.001 iterations 5 end Integration 4.5 SCF Iterations 60 Converge 1.0E-06 1.0E-6 End mmdispersion damping sigm damp_param tz combi s-k file_name $ADFRESOURCES/MMDispersion/disp-param nodefault end noprint sfo Atoms cartesians C.ctr 0.000000000000 3.050000000000 1.391500000000 FD=1 H.h 0.000000000000 3.050000000000 2.471500000000 FD=1 C.ctr 1.205074349366 3.050000000000 0.695750000000 FD=1 H.h 2.140381785453 3.050000000000 1.235750000000 FD=1 C.ctr 1.205074349366 3.050000000000 -0.695750000000 FD=1 H.h 2.140381785453 3.050000000000 -1.235750000000 FD=1 C.ctr -0.000000000000 3.050000000000 -1.391500000000 FD=1 H.h -0.000000000000 3.050000000000 -2.471500000000 FD=1 C.ctr -1.205074349366 3.050000000000 -0.695750000000 FD=1 H.h -2.140381785453 3.050000000000 -1.235750000000 FD=1 C.ctr -1.205074349366 3.050000000000 0.695750000000 FD=1 H.h -2.140381785453 3.050000000000 1.235750000000 FD=1 C.ctr -1.205074349366 -3.050000000000 -0.695750000000 FD=2 H.h -2.140381785453 -3.050000000000 -1.235750000000 FD=2 C.ctr -0.000000000000 -3.050000000000 -1.391500000000 FD=2 H.h -0.000000000000 -3.050000000000 -2.471500000000 FD=2 C.ctr 1.205074349366 -3.050000000000 -0.695750000000 FD=2 H.h 2.140381785453 -3.050000000000 -1.235750000000 FD=2 C.ctr 1.205074349366 -3.050000000000 0.695750000000 FD=2 H.h 2.140381785453 -3.050000000000 1.235750000000 FD=2 C.ctr -0.000000000000 -3.050000000000 1.391500000000 FD=2 H.h -0.000000000000 -3.050000000000 2.471500000000 FD=2 C.ctr -1.205074349366 -3.050000000000 0.695750000000 FD=2 H.h -2.140381785453 -3.050000000000 1.235750000000 FD=2 H.h 0.0 0.35 0.0 FD=3 H.h 0.0 -0.35 0.0 FD=3 End End Input
The part of the bond energy that is due to the Grimme dispersion corrected functional is only inter-molecular (atom-atom contributions for which the fragment numbers FD are different).
In the second example a structure with 2 benzene molecules and a hydrogen molecule is optimized with the Grimme dispersion corrected PBE. Needed is the subkey DISPERSION in the key XC. 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. In this case the subargument FD= in the ATOMS block key word is not used, which was only used in the old MM dispersion calculation.
$ADFBIN/adf << eor Title Geometry optimization with Grimme dispersion correction for GGA basis type TZP core small End XC GGA PBE DISPERSION End geometry converge grad=0.001 Branch OLD iterations 50 end Integration 4.5 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
In the last example first three molecules (2 benzene molecules and a hydrogen molecule) are calculated with the Grimme dispersion corrected PBE. 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 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 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




