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HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc
Date: 2020/06/13 03:09
Name: Maedeh   <rasekh.maede@gmail.com>

Dear OpenMX developers,
Hi,

I am using the HS.out file. and this is my input:(antimonene)

---------------------------------------------------------------

# Definition of Atomic Species#

Species.Number 1
<Definition.of.Atomic.Species
Sb Sb7.0-s1p1d1 Sb_PBE13
Definition.of.Atomic.Species>

Atoms.Number 2
Atoms.SpeciesAndCoordinates.Unit FRAC
<Atoms.SpeciesAndCoordinates
1 Sb 0.3333330 0.6666660 0.5343910 7.5 7.5
2 Sb 0.6666660 0.3333330 0.6105270 7.5 7.5
Atoms.SpeciesAndCoordinates>

Atoms.UnitVectors.Unit Ang
<Atoms.UnitVectors
4.3853100 0.0000000 0.0000000
-2.1926550 3.7977899 0.0000000
0.0000000 0.0000000 20.0000000
Atoms.UnitVectors>

#
# SCF or Electronic System
#

scf.XcType GGA-PBE # LDA|LSDA-CA|LSDA-PW
scf.SpinPolarization NC # On|Off|NC
scf.SpinOrbit.Coupling on # On|Off, default=off
scf.ElectronicTemperature 300.0 # default=300 (K)
scf.energycutoff 220 # default=150 (Ry)
scf.maxIter 100 # default=40
scf.EigenvalueSolver band # Recursion|Cluster|Band
scf.Kgrid 11 11 1 # means 4x4x4
scf.Mixing.Type rmm-diish # Simple|Rmm-Diis|Gr-Pulay
scf.Init.Mixing.Weight 0.030 # default=0.30
scf.Min.Mixing.Weight 0.001 # default=0.001
scf.Max.Mixing.Weight 0.400 # default=0.40
scf.Kerker.factor 1.0 # default=1.0
scf.Mixing.History 14 # default=5
scf.Mixing.StartPulay 6 # default=6
scf.Mixing.EveryPulay 1 # default=5
scf.criterion 1.0e-6 # default=1.0e-6 (Hartree)


HS.Fileout on


scf.dftD on # on|off, default=off
version.dftD 2 # 2|3, default=2
DFTD3.damp bj # zero|bj, default=bj
DFTD.Unit AU # Ang|AU
DFTD.rcut_dftD 100.0 # default=100 (DFTD.Unit)
DFTD.cncut_dftD 40 # default=40 (DFTD.Unit)
DFTD.IntDirection 0 0 1 # default=1 1 1 (1:on 0:off)

---------------------------------------------------------------

I have included SOC in the Calculation. and I expect nonzero imaginary parts of spin mixing output (ie. Kohn-Sham Hamiltonian spin=3). but it is all zero. it should be zero or I am doing something incorrectly in my calculation with OpenMX?


Best regards,
Maedeh

メンテ
Page: [1]

Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.1 )
Date: 2020/06/13 15:41
Name: Naoya Yamaguchi

Dear Maedeh,

I have reproduced it, and I think that it is normal.

The following are a part to set the Hamiltonian written in "Set_Hamiltonian.c":

if (SpinP_switch==3){
for (Mc_AN=1; Mc_AN<=Matomnum; Mc_AN++){
Gc_AN = M2G[Mc_AN];
Cwan = WhatSpecies[Gc_AN];
for (h_AN=0; h_AN<=FNAN[Gc_AN]; h_AN++){
Gh_AN = natn[Gc_AN][h_AN];
Hwan = WhatSpecies[Gh_AN];
for (i=0; i<Spe_Total_NO[Cwan]; i++){
for (j=0; j<Spe_Total_NO[Hwan]; j++){

if (ProExpn_VNA==0){
H[0][Mc_AN][h_AN][i][j] = F_Kin_flag*H0[0][Mc_AN][h_AN][i][j]
+ F_NL_flag*HNL[0][Mc_AN][h_AN][i][j];
H[1][Mc_AN][h_AN][i][j] = F_Kin_flag*H0[0][Mc_AN][h_AN][i][j]
+ F_NL_flag*HNL[1][Mc_AN][h_AN][i][j];
H[2][Mc_AN][h_AN][i][j] = F_NL_flag*HNL[2][Mc_AN][h_AN][i][j];
H[3][Mc_AN][h_AN][i][j] = 0.0;
}
else{
H[0][Mc_AN][h_AN][i][j] = F_Kin_flag*H0[0][Mc_AN][h_AN][i][j]
+ F_VNA_flag*HVNA[Mc_AN][h_AN][i][j]
+ F_NL_flag*HNL[0][Mc_AN][h_AN][i][j];
H[1][Mc_AN][h_AN][i][j] = F_Kin_flag*H0[0][Mc_AN][h_AN][i][j]
+ F_VNA_flag*HVNA[Mc_AN][h_AN][i][j]
+ F_NL_flag*HNL[1][Mc_AN][h_AN][i][j];
H[2][Mc_AN][h_AN][i][j] = F_NL_flag*HNL[2][Mc_AN][h_AN][i][j];
H[3][Mc_AN][h_AN][i][j] = 0.0;
}

/* Effective Hubbard Hamiltonain --- added by MJ */

if( (Hub_U_switch==1 || 1<=Constraint_NCS_switch) && F_U_flag==1 && 2<=SCF_iter ){
H[0][Mc_AN][h_AN][i][j] += H_Hub[0][Mc_AN][h_AN][i][j];
H[1][Mc_AN][h_AN][i][j] += H_Hub[1][Mc_AN][h_AN][i][j];
H[2][Mc_AN][h_AN][i][j] += H_Hub[2][Mc_AN][h_AN][i][j];
}

/* core hole Hamiltonain */

if (core_hole_state_flag==1){
H[0][Mc_AN][h_AN][i][j] += HCH[0][Mc_AN][h_AN][i][j];
H[1][Mc_AN][h_AN][i][j] += HCH[1][Mc_AN][h_AN][i][j];
H[2][Mc_AN][h_AN][i][j] += HCH[2][Mc_AN][h_AN][i][j];
}

}
}
}
}
}

And, the following are comments written in "Band_DFT_NonCol.c":

in case of SO_switch==0 && Hub_U_switch==0 && Constraint_NCS_switch==0
&& Zeeman_NCS_switch==0 && Zeeman_NCO_switch==0

H[i ][j ].r = RH[0];
H[i ][j ].i = 0.0;
H[i+NUM][j+NUM].r = RH[1];
H[i+NUM][j+NUM].i = 0.0;
H[i ][j+NUM].r = RH[2];
H[i ][j+NUM].i = RH[3];

in case of SO_switch==1 or Hub_U_switch==1 or 1<=Constraint_NCS_switch
or Zeeman_NCS_switch==1 or Zeeman_NCO_switch==1

H[i ][j ].r = RH[0];
H[i ][j ].i = IH[0];
H[i+NUM][j+NUM].r = RH[1];
H[i+NUM][j+NUM].i = IH[1];
H[i ][j+NUM].r = RH[2];
H[i ][j+NUM].i = RH[3] + IH[2];

Because RH[3] (H[3] in OpenMX) always looks to include only zero, changes can appear only in the imaginary part of the Hamiltonian (i.e. "iHks" in a scfout file).

Regards,
Naoya Yamaguchi
メンテ
Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.2 )
Date: 2020/06/14 15:38
Name: Maedeh  <rasekh.maede@gmail.com>

Dear Naoya,
Thank you so much. your explanation gave me the point.

I also found some explanations in the Analysis_example file of OpenMX ver.3.9.
I was using version 3.8.
Now everything makes sense.

kind regards,
Maedeh
メンテ
Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.3 )
Date: 2020/06/16 16:28
Name: Maedeh  <rasekh.maede@gmail.com>

Dear Naoya,
Hi,

I was trying to build the Hamiltonians of the same example that I posted before using the recipe we have in "analysis_example" file:

For SpinP_switch=3, the matrix elements are given by

up-up: Hks[0] + I*iHks[0]
up-down: Hks[2] + I*(Hks[3]+iHks[2])
down-up: Hks[2] - I*(Hks[3]+iHks[2])
down-down: Hks[1] + I*iHks[1]

Following this recipe will not give a Hermitian H00 (onsite Hamiltonian) for sure, one must transpose-conjugate, not only conjugate. Following this recipe for hopping Hamiltonians like H10 (hopping to the right) and H_10 (hopping to the left), for a simple example on paper, I can get a Hermitian Bloch Hamiltonian at the end, ie: Hk=H10*exp(ik*a)+H_10*exp(-(iK*a)).

However, looking into the "Hamiltonian_Band_NC" file, I see that you guys first build the H11, H12, H22 blocks of the Bloch Hamiltonian H(k) and then calculate H21 block by the conjugate transpose of H12. This seems to me as forcing the Hamiltonian to be Hermitian while missing hopping elements like down-up hopping for H10 and H_10. Or is it the same in the end?.

On the other hand, in case of real space H10, for instance, if we build the spin-mixing left-down (down-up) block of Hamiltonian by transpose conjugate of the right-up block (up-down), in the left-down block of Hamiltonian (again for a simple example like the cell I have tried to show bellow) we will have hopping between atom 1 in center cell and atom 2 of right neighbor which does not make sense to me. So, in summary, I would like to know how to build down-up blocks for the real space hopping Hamiltonians so I do not have to force hermiticity on the final Bloch Hamiltonian H(k).

-------- -------- --------
| .1 .2 | .1 .2 | .1 .2 |
-------- -------- --------

Thanks for your help in advance.
kind regards,
Maedeh
メンテ
Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.4 )
Date: 2020/06/16 17:47
Name: Naoya Yamaguchi

Dear Maedeh,

>up-up: Hks[0] + I*iHks[0]
>up-down: Hks[2] + I*(Hks[3]+iHks[2])
>down-up: Hks[2] - I*(Hks[3]+iHks[2])
>down-down: Hks[1] + I*iHks[1]

>Following this recipe will not give a Hermitian H00 (onsite Hamiltonian) for sure, one must transpose-conjugate, not only conjugate.

According to the explanation in "analysis_example.c", although a Hamiltonian seems to satisfy only complex conjugate but this explanation is done in terms of spin indices, each element must satisfy the transpose conjugate condition in the Hamiltonian matrix "H". As you may know, the Hamiltonian "H" is prepared by following Eq. (28) in the following technical note:
http://www.openmx-square.org/tech_notes/tech1-1_2.pdf
And, it is clear that H in Eq. (28) satisfies the transpose conjugate condition (Hks or iHks corresponds "h" in Eq. (28).).


>However, looking into the "Hamiltonian_Band_NC" file, I see that you guys first build the H11, H12, H22 blocks of the Bloch Hamiltonian H(k) and then calculate H21 block by the conjugate transpose of H12. This seems to me as forcing the Hamiltonian to be Hermitian while missing hopping elements like down-up hopping for H10 and H_10. Or is it the same in the end?.

Do you mean "H11, H12, H22" is H_{up-up}, H_{up-down} and H_{down-down}? I think that the treatment that H_{down-up} is prepared by taking the transpose conjugate of H_{up-down} has no influence on getting the correct solution while considering that the Hamiltonian is Hermitian.


>On the other hand, in case of real space H10, for instance, if we build the spin-mixing left-down (down-up) block of Hamiltonian by transpose conjugate of the right-up block (up-down), in the left-down block of Hamiltonian (again for a simple example like the cell I have tried to show bellow) we will have hopping between atom 1 in center cell and atom 2 of right neighbor which does not make sense to me. So, in summary, I would like to know how to build down-up blocks for the real space hopping Hamiltonians so I do not have to force hermiticity on the final Bloch Hamiltonian H(k).

I think that the operation of the conjugate transpose is done only for H, not h in Eq. (28), and the Hamiltonian you mean such as H10 corresponds h. You can get down-up (up-down) elements from Hks[2] and iHks[2] by checking correspondence.

Regards,
Naoya Yamaguchi
メンテ
Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.5 )
Date: 2020/06/16 20:57
Name: T. Ozaki

Hi,

How to construct the Hamiltonian for the non-collinear calculation can be found
in Hamiltonian_Band_NC.c.

In the routine, one can confirm the cases for
1. /* non-spin-orbit coupling and non-LDA+U */
2. /* spin-orbit coupling or LDA+U */

and at the nearly end of code, you see the following part:


/******************************************************
the full complex matrix of H
******************************************************/

for (i=1; i<=NUM; i++){
for (j=1; j<=NUM; j++){
H[j+NUM][i].r = H[i][j+NUM].r;
H[j+NUM][i].i = -H[i][j+NUM].i;
}
}

which sets the down-up block elements and guarantees the Hermicity.

Regards,

TO
メンテ
Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.6 )
Date: 2020/06/17 15:30
Name: Maedeh  <rasekh.maede@gmail.com>

Dear Naoya and prof. Ozaki,

Thanks for your quick response.

let me try to make my saying more clear:

let's forget H(k) for a moment.
I want to talk about sub-matrices:

each h matrix should have this form:

h=
(up-up up-down)
(down-up down-down)

we have the information about these three spin-blocks: up-up, up-down, down-down
and "analysis_example" file says that I should build down-up block so:
down-up: Hks[2] - I*(Hks[3]+iHks[2])

following this recipe is not giving me a hermitian h00 Hamiltonian (ie. onsite Hamiltonian).

let's go further, following this recipe to build h10 (hopping matrix to the right in a one-dimensional system) and h_10 (hopping matrix to the left) is not giving me a Hermitian Hk=h10*exp(ik*a)+h_10*exp(-(iK*a)).

so my question is: how should I build the spin down-up block for hopping matrices before creating the Bloch Hamiltonian? This must be possible and I have the impression that the result at the end is not the same.

Best regards,
Maedeh
メンテ
Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.7 )
Date: 2020/06/17 19:44
Name: Naoya Yamaguchi

Dear Maedeh,

In the non-collinear case,
As the spin index doesn't appear in the full Hamiltonian matrix "H", down-up elements are obtained just by taking the transpose conjugate.
H(k)ji=H(k)ij*=(Σexp(iRn・k)h(Rn)ij)*=Σexp(-iRn・k)h(Rn)ij*
And, H(k)ji=Σexp(-iRn・k)h(-Rn)ji
Then, h(-Rn)ji=h(Rn)ij*
As such, you may get down-up elements of "h" by the above relation.

Regards,
Naoya Yamaguchi
メンテ
Re: HS.Fileout-zero imaginary part of spin mixing (spin=3) while including soc ( No.8 )
Date: 2020/06/18 04:09
Name: Maedeh  <rasekh.maede@gmail.com>

Dear Naoya;

Yes, you are right. Thank you.

Best regards,
Maedeh
メンテ

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