[gmx-developers] how to retrieve the electrostatic potential from pme?

[Berk Hess]

Hi,

 From what you write I seems to be much simpler.

If you have charges for the qm atoms you can simply do PME as usual,
you would only need to subtract the interactions between the qm atoms,
which would be done in the same way as normal exclusions.
Or am I missing something?

Cheers,

Berk

On 05/30/2012 10:11 AM, Gerrit Groenhof wrote:
>  Dear All,
>
> To include also the effect of coulomb interactions between MM and QM 
> atoms, I need the electrostatic potential V(r). For the moment I need 
> it only at the positions of the QM nuclei.
>
> But, I do not understand what is going on in pme.c.
>
> What I learned from browsing pme.c is that after computing the 
> potential on the grid in k space, it is transformed back on the real 
> space grid and then forces are interpolated  by gather_f_bsplines.
>
> However, I do not understand if there are already forces on the grid 
> points that are interpolated, or that the potential is interpolated at 
> the site of the atom, after which forces on that atom are computed, 
> and the interpolated potential is not further used. The comment after 
> the routine suggests the latter is done, but that's a comment of 13 
> years ago.  Furhtermore, in the spline interpolation, the output looks 
> like a force already: it is different in x,y and z, and is called f?
>
> Or should I be using the gather_energy_bslpine instead? Is it possible 
> then to vary the position within the gridcell? Or shoud I stick to the 
> position of atom n?
>
> Then, if we manage to get v(r), I thoughT of testing it the following 
> way:
>
> 1) We compute the total (MM) electrostatic energy, with the QM atoms 
> exluded among themselved. In that case, I think that the QM atoms see 
> their periodic counterparts (charges are assiggend to the atoms at 
> every step of the SCF), but not the QM atoms in the central box.
>
> 2) we add a probe charge of +1 at one of the QM nuclei (and exclude 
> its interaction from the QM atoms) and recompute the total MM 
> electrostatic energy.
>
> 3) we remove all atoms (MM + QM) from the box, and only keep the probe 
> charge, and comute the total electrostatic energy of the probe charge 
> in the periodic system
>
> 4) The potential at the position of the probe charge is then: E(2) 
> -E(1)-E(3)?
>
> Are there any objections for this?
>
>
> Best wishes,
>
> Gerrit
>