[CP2K-user] [CP2K:14991] What are the default units of hartree potential and electric field in cp2k?

[Fangyong Yan]

Hi, Kai,

We havenot gotten an answer from cp2k developers, so it is still unclear.

Dear cp2k developers, what is the unit for the electric field in the
EFIELD_CUBE
<https://manual.cp2k.org/cp2k-8_1-branch/CP2K_INPUT/FORCE_EVAL/DFT/PRINT/EFIELD_CUBE.html>
file?
Thanks!

Fangyong

On Tue, Mar 23, 2021 at 5:02 PM Kaixuan Chen <kaix... at gmail.com> wrote:

> Hi Fangyong,
> Thanks a lot for the answer. That is really helpfule.
> Kai
>
> 在2021年3月22日星期一 UTC-7 下午3:24:45<fy... at gmail.com> 写道：
>
>> Hi, Kai,
>>
>> The electric field near the point charge, which is the nuclei, can be
>> very large.
>>
>> I think cp2k does use Hartree/Bohr, and for the large values, such as 10
>> Hartree/Bohr = 500 V/Angstrom, this may be due to the fact that the
>> eclectic field point is very close to the nuclei.
>>
>> For the unit, you can ask cp2k developer to confirm, but I think it is
>> Hartree/Bohr, and I also think 500 V/Angstrom is correct, when it is
>> close to the nuclei.
>>
>> Fangyong
>>
>> On Mon, Mar 22, 2021 at 4:34 PM Kaixuan Chen <ka... at gmail.com> wrote:
>>
>>> Dear Fangyong,
>>> Thank you very much for the response. The units for the numbers of the
>>> cube files (both potential and electric field) is exactly what I want to
>>> know. They are not stated, neither in the cp2k manual nor in the cube
>>> files. So I try to treat the numbers in a.u. units. For example, the
>>> electric field cube file shows that at some positions, the number is 10
>>> hartree/e/bohr if the a.u. unit is the correct one. I think it is really
>>> huge for electric field, isn't it?
>>> By the way, when I mean 50000 MV/cm, I think it is equal to 50000 (10E6
>>> V)/(10E8 Angstrom) = 500 V/Angstrom.
>>> Because 1 hartree/e/bohr = 5142.20652 MV/cm, correct?
>>> Any suggestion on what is the default unit?
>>> Best,
>>> Kai
>>>
>>> 在2021年3月22日星期一 UTC-7 下午12:03:03<fy... at gmail.com> 写道：
>>>
>>>> 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 <fy... 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 <ka... 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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>>>>>>
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