[CP2K-user] [CP2K:16941] What are the default units of hartree potential and electric field in cp2k?

Fangyong Yan fyyan2019 at gmail.com
Tue May 10 16:40:22 UTC 2022

Previous message (by thread): [CP2K-user] [CP2K:16940] What are the default units of hartree potential and electric field in cp2k?
Next message (by thread): [CP2K-user] [CP2K:16942] What are the default units of hartree potential and electric field in cp2k?
Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]

Hi, Kai,

Regards the strength of eclectic field, an external applied periodic
electric field of 0.30 V/Angstrom is able to dissociate water. (see paper,
G. Cassone et al., pccp, 2019,  DOI: 10.1039/C9CP03101D (Paper) Phys. Chem.
Chem. Phys., 2019, 21, 21205-21212)

Fangyong





On Tue, May 10, 2022 at 12:31 PM Fangyong Yan <fyyan2019 at gmail.com> wrote:

> 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
>>>>
>>>> --
>>>> You received this message because you are subscribed to the Google
>>>> Groups "cp2k" group.
>>>> To unsubscribe from this group and stop receiving emails from it, send
>>>> an email to cp2k+uns... at googlegroups.com.
>>>> To view this discussion on the web visit
>>>> https://groups.google.com/d/msgid/cp2k/dc6eef99-9c9e-4c32-a57e-0dbd4065fff3n%40googlegroups.com
>>>> <https://groups.google.com/d/msgid/cp2k/dc6eef99-9c9e-4c32-a57e-0dbd4065fff3n%40googlegroups.com?utm_medium=email&utm_source=footer>
>>>> .
>>>>
>>> --
>> You received this message because you are subscribed to the Google Groups
>> "cp2k" group.
>> To unsubscribe from this group and stop receiving emails from it, send an
>> email to cp2k+unsubscribe at googlegroups.com.
>> To view this discussion on the web visit
>> https://groups.google.com/d/msgid/cp2k/d493d5ae-5afd-44e3-a632-b05b65f6de79n%40googlegroups.com
>> <https://groups.google.com/d/msgid/cp2k/d493d5ae-5afd-44e3-a632-b05b65f6de79n%40googlegroups.com?utm_medium=email&utm_source=footer>
>> .
>>
>

-- 
You received this message because you are subscribed to the Google Groups "cp2k" group.
To unsubscribe from this group and stop receiving emails from it, send an email to cp2k+unsubscribe at googlegroups.com.
To view this discussion on the web visit https://groups.google.com/d/msgid/cp2k/CAJjmc%2BKd%3DQ-dtnFRQ%3DA2wnuD5dSjvQAg%2BViFqny2szyV5A%2B9pA%40mail.gmail.com.
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <https://lists.cp2k.org/archives/cp2k-user/attachments/20220510/6be42e64/attachment-0001.htm>

Previous message (by thread): [CP2K-user] [CP2K:16940] What are the default units of hartree potential and electric field in cp2k?
Next message (by thread): [CP2K-user] [CP2K:16942] What are the default units of hartree potential and electric field in cp2k?
Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]

More information about the CP2K-user mailing list