13.11.4 Examples

XPol on its own is not a useful method because it neglects all intermolecular interactions except for polarization, so the two examples below demonstrate the use of XPol in conjunction with a Lennard-Jones and a Buckingham potential, respectively.

Example 13.24  An XPol single point calculation on the water dimer using a Lennard-Jones potential. Löwdin atomic embedding charges (which is the default) are used.

$molecule 0 1 -- water 1 0 1 O -1.364553 .041159 .045709 H -1.822645 .429753 -.713256 H -1.841519 -.786474 .202107 -- water 2 0 1 O 1.540999 .024567 .107209 H .566343 .040845 .096235 H 1.761811 -.542709 -.641786$end

$rem METHOD HF BASIS 3-21G XPOL TRUE$end

$xpol_mm 1 O -1.364553 .041159 .045709 1 2 3 2 H -1.822645 .429753 -.713256 2 1 3 H -1.841519 -.786474 .202107 2 1 4 O 1.540999 .024567 .107209 1 5 6 5 H .566343 .040845 .096235 2 4 6 H 1.761811 -.542709 -.641786 2 4$end

$xpol_params 1 0.16 3.16 2 0.00 0.00$end


Example 13.25  An XPol single point calculation on the water dimer using a Buckingham potential.

$molecule 0 1 -- water 1 0 1 O -1.364553 .041159 .045709 H -1.822645 .429753 -.713256 H -1.841519 -.786474 .202107 -- water 2 0 1 O 1.540999 .024567 .107209 H .566343 .040845 .096235 H 1.761811 -.542709 -.641786$end

$rem METHOD HF BASIS 3-21G XPOL TRUE$end

$xpol_mm 1 O -1.364553 .041159 .045709 1 2 3 2 H -1.822645 .429753 -.713256 2 1 3 H -1.841519 -.786474 .202107 2 1 4 O 1.540999 .024567 .107209 1 5 6 5 H .566343 .040845 .096235 2 4 6 H 1.761811 -.542709 -.641786 2 4$end

$xpol_params BUCKINGHAM 500000.0 12.5 2.25 1 0.16 3.16 2 0.00 0.00$end