[CP2K:4542] Ewald and charged simulation box

[Teodoro Laino]

Any Ewald scheme (for quantum calculation this is also known as Poisson Solver) for periodical charged systems, includes automatically the compensating background charge. 
The energy of a periodical charged system, without background charge, is undefined (the series 1/r is not converging for systems with a net charge different from zero - actually, in general,  this series is only conditionally convergent but not absolutely convergent).

I don't really understand your claim (and didn't check the paper you mentioned) - but there is nothing special in cp2k compared to other codes doing Ewald (cp2k, cpmd, quantum espresso, etc..).. So the answer is:

 there is no section to activate to have the compensating background charge to have an overall neutral simulation box : simply run a standard setup for a 3D periodic system (without decoupling) and you will get automatically this term in the energy.



On Aug 8, 2013, at 10:29 AM, Gianni Carifaccia <gian... at gmail.com> wrote:

> Dear All,
> 
> I need to simulate a charged QM system (I need to see what happens after protonation of my molecule) and I cannot make the whole box neutral by adding a counterion. Looking into literature, I've seen that it is possible to solve the charged box issue in cp2k through Ewald, by smearing a compensating background charge to have an overall neutral simulation box (one for all, see Sprik et al JCTC 2011, 7, 1951. http://pubs.acs.org/doi/pdf/10.1021/ct100715).
> 
> My question is: which exact input section should I look for ? I looked around the Poisson solvers, but I don't feel confident I found the right place. Other Ewald is in the MM section, but I don't think I've to run a QM/MM if a I have a pure QM box.  In case it is in the Poisson section, which keywords ? Further, I've not seen how to input the amount of the background charge. Is this "understood" automatically by cp2k ?
> 
> Of course, any help on how to handle this issue, and how to check things are done properly, is wellcome.
> 
> Thank you in advance,
> Giancar
> 
> 
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