[CP2K-user] [CP2K:16940] What are the default units of hartree potential and electric field in cp2k?

[Fangyong Yan]

Hi, Kai,

I think the electric field gradient unit is hartree/e/bohr.

10 hartree/e/bohr is a large electric field, may I ask what is the location
for these large electric field in the cube file?

Thanks!

Fangyong

On Wed, Apr 13, 2022 at 10:42 AM gary Washington <lindagary0 at gmail.com>
wrote:

> I find that we can test for the units of the Hartree potential by using a
> test molecule. I selected Li2 and applied an external electric field
> (Volts/Angstroms). First obtain the Hartree Potential cube file for a zero
> external potential
>
> # electric field along the z direction at efieldmag eV/Angstrom or
> efieldmag V/Angstrom
>     &EXTERNAL_POTENTIAL
>       FUNCTION (A/B)*Z
>       VALUES [eV] ${efieldmag} [angstrom] 1.0
>       PARAMETERS A B
>     &END EXTERNAL_POTENTIAL
>
>
> Cutoff = 400, RelCutoff=80
> so we set efieldmag = 0.0
>
> Now using a box/cell **without periodic boundary conditions** we can
> optimize the geometry ( an energy calculation should also work ) now we use
> cubecruncher to obtain the potential along the z axis. We expect the
> initial slope, which is equal to the initial electric field magnitude in
> the z-direction. Now molecules will have an electric field around them due
> to their charge distribution of nuclei and electrons so even when
> efieldmag=0.0 the electric field around the molecule will be nonzero and
> will diminish as you move away from the molecule. If the box is large
> enough the influence of the molecule on the initial electric field will
> nearly vanish far from the molecule. I used {Lx, Ly, Lz} of {12,12,15} ,
> {12,12,20}, {12,12,25} and the initial slopes are -0.019047, -0.00027563,
> -0.0000032. The initial slope is obtained on the left in the limit as z -->
> 0. We see the influence of the molecule can be greatly reduced in a
> sufficiently large box and this is expected to be true for any molecule in
> a sufficiently large box/cell.
>
> Now we set efieldmag=0.005 Volts/Angstrom, using a box with dimensions of
> {12,12,15}, {12,12,20}, {12,12, 25} and the initial slopes are -0.01932522,
> -0.00008413067, -0.0001806136
> Next we set Cutoff=600, relCutoff=80 and box size {12,12, 25} and the
> initial slope is -0.0001811421
> Next we set Cutoff=800, relCutoff=80  and box size {12,12, 25} and the
> initial slope is -0.0001815272
> Next we set Cutoff=800, relCutoff=80  and box size {12,12, 35} and the
> initial slope is -0.0001837458
>
> * assumption 1:                                   Hartree potential has
> the units of hartree/e = 27.211386 Volts / a.u.
> *assumption 2:                                    Electric Field Magnitude
> (Volts/Angstroms) = 27.211386 * (Hartree Potential at a point)
>
> Check: Given an electric field magnitude of 0.005 V/Ang we expect that the
> observed Hartree potential sufficiently far from the molecule will yield a
> slope along the z axis, Ez (electric field in the z direction) of the same
> magnitude.                We expect that the result should approach (0.005
> V/Ang)/27.211386 = 0.000183747 a.u. ( the expected slope of the Hartree
> potential due only to the external potential)
>
> We see that the initial slope is equal to efieldmag/27.211386 (Volts/a.u.)
> which shows that assumption 1 holds.
>
>
>
>
>
> On Monday, March 22, 2021 at 2:25:52 PM UTC-4 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 <kaixu... 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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