9
$\begingroup$

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 if some of the water molecules dissociate or we add salt that ends up as Cl- and a corresponding +1 ion, I'm not sure. 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?

$\endgroup$
5
  • 1
    $\begingroup$ 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? $\endgroup$ 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
  • $\begingroup$ @SRMaiti was the response by NTS helpful? $\endgroup$ Oct 30 at 20:24
  • $\begingroup$ @NikeDattani Yeah it was helpful, but I have only experience with classical MD, not much on material science, so I left it to others to answer the question. $\endgroup$
    – S R Maiti
    Nov 1 at 10:42
3
$\begingroup$

I have not used CP2K, but based on the documentation, this should refer to the charge of the whole system. So if you don't have a reason to believe your system is charged, you should treat it as neutral.

The reason for the reference keyword when computing the dipole/polarization (or any higher multipole of a molecule/solid) is that for charged systems, multipoles are dependent on your choice of coordinate origin. The dipole is still well-defined/calculable and you can compute derived properties from it like the polarizability, but you will need to report the reference point that you used for someone to be able to reproduce your specific value of the dipole. For consistency, the center of charge is often used as the origin in these cases.

As a simple example of the origin dependence for charged systems, consider the dipole of $\ce{H2+}$, which should have a bond length of roughly $\pu{2bohr}$ and the charge split evenly among the two hydrogens. If we put the origin at one of the atoms: $$|\mu_\text{atom_origin}|=(\pu{0.5 e} \times \pu{0 bohr})+(\pu{0.5 e} \times \pu{2 bohr})=\pu{1 e*bohr}$$

However, if we put the origin at the center of mass (COM) (which is also the center of charge): $$|\mu_\text{COM_origin}|=(\pu{0.5 e} \times \pu{-1 bohr})+(\pu{0.5 e} \times \pu{1 bohr})=\pu{0 e*bohr}$$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.