6
$\begingroup$

I'm doing a polarization calculation for the first time. To specify one of the input parameters (the "reference point") correctly, I have to figure out if my system is "charged." If it is not charged, I don't have to worry about the reference point.

Could anyone please help ensure I understand what "charged" means in this context?

On one hand, I understand that a system made up of water molecules (for example) is not charged. But what if some of the water molecules dissociate or we add salt that ends up as Cl- and a corresponding +1 ion? I.e., does "charged system" refer to the entire system or does it mean that we need to account for what's going on locally inside the system?

Improve this question
$\endgroup$
3
  • $\begingroup$ Could you add some information about what type of system you are trying to model, the software, type of calculation etc.? In molecular dynamics for example, the charged system would mean the charge of the main molecule (e.g. protein) is non-zero, even if the total system (i.e. the simulation system with solvents) may have zero charge. $\endgroup$ – S R Maiti Apr 6 at 17:02
  • 1
    $\begingroup$ +1 but I agree with NTS. You are asking what "charged means in this context" but what precisely is the context? Who said that you have to figure out if the system is "charged" in order to specify one of the input parameters correctly? Which software, and where did it say you have to do that? $\endgroup$ – Nike Dattani Apr 6 at 17:08
  • 2
    $\begingroup$ @ShoubhikRMaiti, thank you. I would like to do MD in CP2K on systems that contain water with and without metal and dissolved salt. The documentation only shows how to switch on calculation of dipole moments. There's a subsection that mentions "reference point." A forum post I found elsewhere says that all choices of reference point should lead to the same polarization (for a Berry phase calculation) for a non-charged system. I have no idea if my system is considered to be charged or not. It is charge-neutral overall but like I said, I don't know if it's the overall charge that matters. $\endgroup$ – NTS Apr 6 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.