[CP2K-user] [CP2K:14983] What are the default units of hartree potential and electric field in cp2k?

[Fangyong Yan]

Hi, Kai,

I think you mean 5GV/cm = 500 V/Angstrom.

I think you may need to do integration to obtain the electric field,
because it is a cube file, so what is the unit for the numbers
of your cube file?

Fangyong

On Mon, Mar 22, 2021 at 2:25 PM Fangyong Yan <fyya... at gmail.com> wrote:

> Hi, Kai,
>
> 50000 MV/cm = 50 V / (10E8) Angstrom = 5E-7 V/Angstrom, which is a very
> very small electric field.
>
> Best,
>
> Fangyong
>
> On Mon, Mar 22, 2021 at 1:07 PM Kaixuan Chen <kaix... at gmail.com> wrote:
>
>>
>> Dear all,
>> I have generated the cube files of hartree potential (CP2K_INPUT
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT.html> / FORCE_EVAL
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL.html> /
>> DFT
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT.html>
>> / PRINT
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT.html>
>> / V_HARTREE_CUBE
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT/V_HARTREE_CUBE.html>)
>> and electric field (CP2K_INPUT
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT.html> / FORCE_EVAL
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL.html> /
>> DFT
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT.html>
>> / PRINT
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT.html>
>> / EFIELD_CUBE
>> <https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT/EFIELD_CUBE.html>)
>> from cp2k. I don't see an explicit description on the units that are used
>> in these cube files. If I take the a.u. as the default unit (hartree/e for
>> potential, and hartree/e/bohr for electric field), the value seems pretty
>> large. For example, I study the single water molecule system. The largest
>> electric field at some density grid is 10~15 hartree/e/bohr, that is,
>> ~50000 MV/cm. Please correct me if I am wrong, but the value seems
>> unreasonable to me.
>> Any suggestion will be welcome, thanks in advance.
>> Best,
>> Kai
>>
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