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Electric polarization of BiFeO3 (procedure in obtaining net polarization)
Date: 2024/10/29 17:41
Name: Amran Yatmeidhy   <yatmeidhy.amran@nims.go.jp>

Dear all,

I tried to calculate the electric polarization of BiFeO3 along [111] direction.
Here are my results using polB of OpenMX:

1) Electric polarization

***************************************************************
              Electric polarization (muC/cm^2) : Berry phase         
***************************************************************

              Background        Core            Electron          Total

Px          -0.00000000      -0.00000000      -0.00000093      -0.00000093
Py          -0.00000000      -0.00000000      -0.00000172      -0.00000172
Pz          -0.00000000    7492.03744942      12.95572100    7504.99317042

I used this input for polB calculation:

11 11 11 #k-grid
1 1 1 #target direction


And this is the quantum polarization based on my cell structure:

2) Quantum polarization (muC/cm^2)

eRix/V 35.41736768 20.44822676 59.40232721
eRiy/V -35.41736768 20.44822676 59.40232721
eRiz/V 0       -40.89645353 59.40232721


I want to know how to properly obtain the net electric polarization by considering the electric polarization from polB calculation and the quantum polarization.

A previous thread (https://www.openmx-square.org/forum/patio.cgi?mode=view&no=2690) discussed the relation between electric and quantum polarizations for obtaining net electric polarization. However, I still do not understand how to get the arbitrary parts of N1, N2, and N3.

Px(net)=Px(Total)+N1*eR1x/V+N2*eR2x/V+N3*eR3x/V
Py(net)=Py(Total)+N1*eR1y/V+N2*eR2y/V+N3*eR3y/V
Pz(net)=Pz(Total)+N1*eR1z/V+N2*eR2z/V+N3*eR3z/V

I hope someone can answer my question.
Thank you very much.

*)
For additional reference, I used this input file of BiFeO3 structure to obtain the *.scfout file:

3) input.dat for the SCF calculation

#
# File Name     
#

System.CurrrentDirectory        ./    # default=./
System.Name                    bifeo3
level.of.stdout                  1    # default=1 (1-3)
level.of.fileout                  0    # default=1 (0-2)
DATA.PATH /home/amranmy/openmx3.9/DFT_DATA19

#
# Definition of Atomic Species
#

Species.Number      3
<Definition.of.Atomic.Species
Fe  Fe10.0H-s2p2d2  Fe_PBE19H
Bi  Bi10.0-s2p2d2  Bi_PBE19
O  O7.0-s2p2      O_PBE19
Definition.of.Atomic.Species>

#
# Atoms
#

Atoms.Number        10
Atoms.SpeciesAndCoordinates.Unit  Frac # Ang|AU
<Atoms.SpeciesAndCoordinates
1  Bi 0.0036899999999989  0.0036899999999989  0.0036899999999989  7.5  7.5 off
2  Bi 0.5036899999999989  0.5036899999999989  0.5036899999999989  7.5  7.5 off
3  Fe 0.2232600000000033  0.2232600000000033  0.2232600000000033 10.5  5.5 off
4  Fe 0.7232600000000033  0.7232600000000033  0.7232600000000033  5.5 10.5 off
5  O  0.8891899999999993  0.4349199999999982  0.0330399999999997  3.0  3.0 off
6  O  0.4349199999999982  0.0330399999999997  0.8891899999999993  3.0  3.0 off
7  O  0.0330399999999997  0.8891899999999993  0.4349199999999982  3.0  3.0 off
8  O  0.9349199999999982  0.3891899999999993  0.5330399999999997  3.0  3.0 off
9  O  0.5330399999999997  0.9349199999999982  0.3891899999999993  3.0  3.0 off
10 O  0.3891899999999993  0.5330399999999997  0.9349199999999982  3.0  3.0 off
Atoms.SpeciesAndCoordinates>
Atoms.UnitVectors.Unit            Ang # Ang|AU
<Atoms.UnitVectors
    2.7903465736272248    1.6110073454160283    4.6799943377191253
    -2.7903465736272248    1.6110073454160283    4.6799943377191253
    0.0000000000000000  -3.2220146908320566    4.6799943377191253
Atoms.UnitVectors>

#
# SCF or Electronic System
#

scf.XcType                GGA-PBE    # LDA|LSDA-CA|LSDA-PW|GGA-PBE
scf.SpinPolarization        on        # On|Off|NC
scf.ElectronicTemperature  300.0      # default=300 (K)
scf.energycutoff          400.0      # default=150 (Ry)
scf.maxIter                300        # default=40
scf.EigenvalueSolver        band    # DC|GDC|Cluster|Band
scf.Kgrid                  11 11 11    # means n1 x n2 x n3
scf.ProExpn.VNA            off        # default=on
scf.Mixing.Type          rmm-diish    # Simple|Rmm-Diis|Gr-Pulay|Kerker|Rmm-Diisk
scf.Init.Mixing.Weight    0.001        # default=0.30
scf.Min.Mixing.Weight      0.0001      # default=0.001
scf.Max.Mixing.Weight      0.300      # default=0.40
scf.Mixing.History          40          # default=5
scf.Mixing.StartPulay      30          # default=6
scf.Mixing.EveryPulay      1          # default=5
scf.criterion            1.0e-8      # default=1.0e-6 (Hartree)
#scf.lapack.dste            dstevx      # dstevx|dstedc|dstegr,default=dstevx

HS.fileout              on    # on|off, default=off

scf.Hubbard.U              on  # on|off, default=off
scf.DFTU.Type                1  # 1:Simplified(Dudarev)|2:General, default=1

<Hubbard.U.values  # eV
  Fe 1s 0.0 2s 0.0 1p 0.0 2p 0.0 1d 4.0 2d 0.0
  Bi 1s 0.0 2s 0.0 1p 0.0 2p 0.0 1d 0.0 2d 0.0
  O  1s 0.0 2s 0.0 1p 0.0 2p 0.0
Hubbard.U.values>

#
# MD or Geometry Optimization
#

MD.Type                    Nomd      # Nomd|Opt|NVE|NVT_VS|NVT_NH
#MD.Opt.DIIS.History          3        # default=7
#MD.Opt.StartDIIS            20        # default=5
#MD.Opt.EveryDIIS          1000        # default=10
MD.maxIter                  1          # default=1
MD.Opt.criterion        1.0e-4        # default=1.0e-4 (Hartree/bohr)
ƒƒ“ƒe
Page: [1]

Re: Electric polarization of BiFeO3 (procedure in obtaining net polarization) ( No.1 )
Date: 2024/10/30 14:29
Name: Naoya Yamaguchi

Dear Amran-san,

It is not clear what the circumstances are since the original case was not left in the previous thread, but my explanation of the earlier discussion seems to lack generality, but basically there is a 2ƒÎ arbitrariness in the inner product of the reciprocal lattice vector and the electronic contribution to the electric polarization: G_iEP_e=2ƒÎe/V, that is, (R_j~R_k)/|R_j~R_k|EP_e=e/|R_j~R_k|=e|R_i||cosƒÆ|/V, where ƒÆ is the angle between R_i and R_j~R_k.

Also, the electric polarization in solids can be well defined only by its difference.

Regards,
Naoya Yamaguchi
ƒƒ“ƒe
Re: Electric polarization of BiFeO3 (procedure in obtaining net polarization) ( No.2 )
Date: 2024/10/31 15:54
Name: Amran Yatmeidhy

Dear Yamaguchi-san,

Thank you for your response. Here is the calculation results comparing ferroelectric and paraelectric phases of BiFeO3:

1) Ferroelectric (I use slightly different atomic position than I previously posted)

***************************************************************
              Electric polarization (muC/cm^2) : Berry phase         
***************************************************************

              Background        Core            Electron          Total

Px          -0.00000000      -0.00000000      -0.00000096      -0.00000096
Py          -0.00000000        0.00000000      -0.00000018      -0.00000018
Pz          -0.00000000    7473.32992459      23.97013715    7497.30006174

cell structure:
Atoms.SpeciesAndCoordinates.Unit  Frac # Ang|AU
<Atoms.SpeciesAndCoordinates
1  Bi 0.0000000000000000  0.0000000000000000  0.0000000000000000  7.5  7.5 off
2  Bi 0.5000000000000000  0.5000000000000000  0.5000000000000000  7.5  7.5 off
3  Fe 0.2233892803913956  0.2233892803913956  0.2233892803913956 10.5  5.5 off
4  Fe 0.7233892803913956  0.7233892803913956  0.7233892803913956  5.5 10.5 off
5  O  0.8820430721883523  0.4365085820904617  0.0387306473898750  3.0  3.0 off
6  O  0.4365085820904617  0.0387306473898750  0.8820430721883523  3.0  3.0 off
7  O  0.0387306473898750  0.8820430721883523  0.4365085820904617  3.0  3.0 off
8  O  0.9365085820904617  0.3820430721883524  0.5387306473898750  3.0  3.0 off
9  O  0.5387306473898750  0.9365085820904617  0.3820430721883524  3.0  3.0 off
10 O  0.3820430721883524  0.5387306473898750  0.9365085820904617  3.0  3.0 off
Atoms.SpeciesAndCoordinates>
Atoms.UnitVectors.Unit            Ang # Ang|AU
<Atoms.UnitVectors
    2.7903465736272248    1.6110073454160283    4.6799943377191253
    -2.7903465736272248    1.6110073454160283    4.6799943377191253
    0.0000000000000000  -3.2220146908320566    4.6799943377191253
Atoms.UnitVectors>

2) Paraelectric

***************************************************************
              Electric polarization (muC/cm^2) : Berry phase         
***************************************************************

              Background        Core            Electron          Total

Px          -0.00000000        0.00000000      -0.00000007      -0.00000007
Py          -0.00000000        0.00000000        0.00000000        0.00000000
Pz          -0.00000000    6861.03272492      -0.00000000    6861.03272492

cell structure:
Atoms.SpeciesAndCoordinates.Unit  Frac # Ang|AU
<Atoms.SpeciesAndCoordinates
1  Bi 0.0000000000000000  0.0000000000000000  0.0000000000000000 7.5  7.5 off
2  Bi 0.5000000000000000  0.5000000000000000  0.5000000000000000 7.5  7.5 off
3  Fe 0.2500000000000000  0.2500000000000000  0.2500000000000000 10.5  5.5 off
4  Fe 0.7500000000000000  0.7500000000000000  0.7500000000000000 5.5 10.5 off
5  O  0.9185700000000000  0.5000000000000000  0.0814300000000000 3.0  3.0 off
6  O  0.5000000000000000  0.0814300000000000  0.9185700000000000 3.0  3.0 off
7  O  0.0814300000000000  0.9185700000000000  0.5000000000000000 3.0  3.0 off
8  O  0.0000000000000000  0.4185700000000000  0.5814300000000000 3.0  3.0 off
9  O  0.5814300000000000  0.0000000000000000  0.4185700000000000 3.0  3.0 off
10 O  0.4185700000000000  0.5814300000000000  0.0000000000000000 3.0  3.0 off
Atoms.SpeciesAndCoordinates>
Atoms.UnitVectors.Unit            Ang # Ang|AU
<Atoms.UnitVectors
    2.7903465736272248    1.6110073454160283    4.6799943377191253
    -2.7903465736272248    1.6110073454160283    4.6799943377191253
    0.0000000000000000  -3.2220146908320566    4.6799943377191253
Atoms.UnitVectors>

If I just simply take the difference of the total polarization of the two phases, I got 636.27 muC/cm^2.
And considering the quantum polarization, the net polarization will be 106.04 muC/cm^2. Is this correct?

PS: the results I got from VASP calculation is about 103.12 muC/cm^2.

Sincerely,
Amran
ƒƒ“ƒe
Re: Electric polarization of BiFeO3 (procedure in obtaining net polarization) ( No.3 )
Date: 2024/10/31 22:31
Name: Naoya Yamaguchi

Dear Amran-san,

As the calculation process is not written, it is hard to say, but if the calculated value matches the reference value well, I think it is probably correct.

Regards,
Naoya Yamaguchi
ƒƒ“ƒe
Re: Electric polarization of BiFeO3 (procedure in obtaining net polarization) ( No.4 )
Date: 2024/11/01 13:09
Name: Amran Yatmeidhy

Dear Yamaguchi-san,

> And considering the quantum polarization, the net polarization will be 106.04 muC/cm^2.

I just divided 636.27 muC/cm^2 by 6 (~lattice = 5.68 &#197;).

Could you explain how to handle quantum polarization in openmx?
(in step by step manner)

(In vasp, quantum polarization = integer * lattice)


Sincerely,
Amran
ƒƒ“ƒe
Re: Electric polarization of BiFeO3 (procedure in obtaining net polarization) ( No.5 )
Date: 2024/11/01 21:56
Name: Naoya Yamaguchi

Dear Amran-san,

I was a little confused, but it seems that my explanation in the previous thread was based on eR_i/V and was correct.
I was stuck with the notation in k-space, but indeed, it seems that eR_i/V is simply fine, since P_e is a quantum of eR_i/V from G_i P_e=2ƒÎe/V.
Since polB up to 3.8 calculated the quantum of electric polarization as follows, after I commented it out, it was released without any modification as it was, so 3.9 does not output anything. However, it is something that should be corrected.


```
Parb[i] = AU2Mucm*2.0*PI/Gabs[i]/Cell_Volume
```

>I just divided 636.27 muC/cm^2 by 6 (~lattice = 5.68 &#197;).

At the very least, if 636.27 is the difference between the two phase values, I don't see the point of your process. It would be better to study "the modern theory of polarization". Instead, An integer multiple of eR_i/V should be added.


Regards,
Naoya Yamaguchi
ƒƒ“ƒe

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