What are the default units of hartree potential and electric field in cp2k?

[Kaixuan Chen]

Dear all,

I have generated the cube files of hartree potential (CP2K_INPUT / FORCE_EVAL / DFT / PRINT / V_HARTREE_CUBE) and electric field (CP2K_INPUT / FORCE_EVAL / DFT / PRINT / EFIELD_CUBE) 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

[Fangyong Yan]

Hi, Kai,

50000 MV/cm = 50 V / (10E8) Angstrom = 5E-7 V/Angstrom, which is a very very small electric field. 

Best,

Fangyong

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[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

[Kaixuan Chen]

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

[Fangyong Yan]

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

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[Kaixuan Chen]

Hi Fangyong,

Thanks a lot for the answer. That is really helpfule.

Kai

[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 file? Thanks!

Fangyong

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[gary Washington]

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.

[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

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[Fangyong Yan]

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

[Fangyong Yan]

Hi, Kai,

I think eclectic field with 10 Hartree/e/Bohr is 514 V/Angstrom. But I guess this is electric field can be balanced by the nearby eclectic field with opposite directions, so the system can be stable. 

For water, when you apply an external periodic electric field where the direction is fixed, so under such electric field, water will become more ordered like ice structure because electric field forces to be ordered, and since the electric field is along the same direction, this can result in a break of bond when the force is large enough because there is a persistent force along the same direction. So far this is what I understand. 

Thanks!

Fangyong