units of electric field in VASP; conversion to volts/length

[sophie_weber]

Hello,

I had another question related to units of electric field in VASP. A user asked the same question a couple years ago:
https://www.vasp.at/forum/viewtopic.php ... V%2F%C3%85.

However, I wasn't able to parse the moderator's answer as it seemed contradictory to me; "Because VASP is using (mostly) atomic units then q is unity. In that case, the potential V and energy Energy are in eV.
The numeric value you want is EFIELD=0.25V/Å which corresponds to EFIELD=1.60217663e-19*0.25 eV/Å."

The statement that q is unity seems to directly contradict the statement the EFIELD should be given with the charge of the electron in coulombs. (I don't understand the EFIELD=0.25V/angstrom bit, since EFIELD is explicitly in energy/length units.

So to ask again really explicitly; Does an EFIELD setting of EFIELD=1 eV/angstrom in VASP correspond to a "real" or "lab" electric field of 1 V/angstrom, OR of (1.6*10^(-19) V/angstrom?)
And to go the other way, if in my VASP potential energy, I have a potential energy drop of 1 eV, does that correspond to applying an electric field of 1 V/angstrom, or 1.6*10^-19 V/angstrom?

This is rather important to understand in order to determine whether the electric field strength as calculated for the system of interest is remotely relevant experimentally.

Thank you in advance!

[sophie_weber]

Sorry, it just occurred to me that since eV should be divide by charge to convert to volt, the alternate possibility of the "true" electric field if EFIELD=1 eV/angstrom should actually probably be 1.602*10^+19 eV/angstrom (via dividing by 1.602*10^-19 coulombs), as opposed to the 1.602*10^-19 option that the moderator wrote.

If you could definitively clarify this it would be greatly appreciated (I now have one possible value, moduo 10^+19 or 10^-19 )

[martin.schlipf]

Here is the way I see it: Electric fields have SI units of V/m which can easily be transferred to V/Å. Since VASP measures energies in eV, you should see the slope of the potential agreeing with the EFIELD that you set in your INCAR file. For example, you could try to use your recent post and measure the slope in the vacuum region and compare it against the EFIELD value.

The elemental charge enters in the equation when you want to convert the energies in that potential to SI units, e.g, the energy required to move an electron out of the crystal or detach a molecule from a surface.

To summarize, in VASP the potential gradient and the field are equivalent because we measure energies in units of eV. Depending on which quantity you want to compare to experiment you need to include (for the energy) the factor e (1.602e-19) when converting to SI units or not (for the field).

I hope this clarifies the issue.

[sophie_weber]

Hello, thanks a lot for the followup. I just want to be sure I'm absolutely clear because it's still a bit confusing to me. I'll try to be very specific about my problem. In relation to my earlier post which you flagged, I have a heterostructure with a large potential energy change at the interface, which leads to an electric dipole if you average across the plane. That potential energy drop, as plotted in the LOCPOT, is say 4 eV. And that drop over a distances of about 4 angstrom (this is the region near the interface which appears as a crossover region in the LOCPOT, with the electrostatic potential assuming the profile of the two bulk crystal constituents beyond this crossover region.

So 4eV/4 angstrom= 10 eV/nm. I want to compare the strength of this "effective" electric field due to the dipole to a "reasonable" electric field that you would apply in a lab (which tend to be on the order of 1 V/nm for bulk crystals). My question is, would my "effective" electric field be 10 V/nm, 1.6*10^-19 V/nm, or 1.6*10^19 V/nm? The first is comparable or slightly larger than applied electric fields, the second, is completely negligible from an experimental perspective, and the third is presumably well beyond the Zener breakdown limit.

If you could explicitly say which of these three options it would be (from your post I understand it to be 10 V/nm if I care just about the electric field) I'd be really grateful for the clarification. Apologies for my slow understanding. Thanks already for your help!

[martin.schlipf]

Yes, the first solution 10 V/nm is the correct interpretation. It is still a large field compared to experimental fields but only a 2 or 3 orders of magnitude and not 20

[sophie_weber]

Perfect, thank you so much! I really appreciate it