Qeustion about magntoctystalline anisotropy of iron

 

Nan AO - 2021-01-11

Hello.

I am dealing with the magntoctystalline anisotropy with elk.
For the first method, I am using subtracting total energy of models including magnetic moment in x and z directions, you can find the input file as :
tasks
0

xctype
20

stype
1

spinpol
.true.

spinorb
.true.

fsmtype
-2

taufsm
0.01

bfieldc
0.01 0.0 0.0

mommtfix
1 1 4 0 0
1 2 4 0 0
1 3 4 0 0
1 4 4 0 0
1 5 4 0 0
1 6 4 0 0
1 7 4 0 0
1 8 4 0 0
1 9 4 0 0
1 10 4 0 0
1 11 4 0 0
1 12 4 0 0
1 13 4 0 0
1 14 4 0 0
1 15 4 0 0
1 16 4 0 0

epspot
1.e-8

epsengy
1.e-8

rgkmax
9

ngridk
14 14 14

nempty
30

scale
1.0

avec
10.7091119654 0.00000000000 0.00000000000
0.00000000000 10.7091119654 0.00000000000
0.00000000000 0.00000000000 10.7091119654

sppath
'/imec/other/abinitio/pseudos/Elk/species/'

atoms
1 : nspecies
'Fe.in' : spfname
16 : natoms; atpos, bfcmt below
1.00000000 0.00000000 0.00000000
0.75000000 0.25000000 0.25000000
0.50000000 0.00000000 0.00000000
0.25000000 0.25000000 0.25000000
1.00000000 0.50000000 0.00000000
0.75000000 0.75000000 0.25000000
0.50000000 0.50000000 0.00000000
0.25000000 0.75000000 0.25000000
1.00000000 0.00000000 0.50000000
0.75000000 0.25000000 0.75000000
0.50000000 0.00000000 0.50000000
0.25000000 0.25000000 0.75000000
1.00000000 0.50000000 0.50000000
0.75000000 0.75000000 0.75000000
0.50000000 0.50000000 0.50000000
0.25000000 0.75000000 0.75000000

The second method is magnetic force theorem, which I find in this forum. I used tasks 29 and npmae 3. The input file is as following:
tasks
29

npmae
3

xctype
20

stype
1

spinpol
.true.

spinorb
.true.

bfieldc
0.0 0.0 0.01

epspot
1.e-8

epsengy
1.e-8

rgkmax
9

ngridk
14 14 14

nempty
30

scale
1.0

avec
10.7091119654 0.00000000000 0.00000000000
0.00000000000 10.7091119654 0.00000000000
0.00000000000 0.00000000000 10.7091119654

sppath
'/imec/other/abinitio/pseudos/Elk/species/'

atoms
1 : nspecies
'Fe.in' : spfname
16 : natoms; atpos, bfcmt below
1.00000000 0.00000000 0.00000000
0.75000000 0.25000000 0.25000000
0.50000000 0.00000000 0.00000000
0.25000000 0.25000000 0.25000000
1.00000000 0.50000000 0.00000000
0.75000000 0.75000000 0.25000000
0.50000000 0.50000000 0.00000000
0.25000000 0.75000000 0.25000000
1.00000000 0.00000000 0.50000000
0.75000000 0.25000000 0.75000000
0.50000000 0.00000000 0.50000000
0.25000000 0.25000000 0.75000000
1.00000000 0.50000000 0.50000000
0.75000000 0.75000000 0.75000000
0.50000000 0.50000000 0.50000000
0.25000000 0.75000000 0.75000000

The different of results of these 2 methoids are bit large and the MCA value is larger than the reference value and MCA results from other dft codes, such as openmx and jdftx.
Could anyone help me to check if there is any problem in my input files?
And anyone one know the mechanism of tasks 29 for MAE calculation?

Thank you and best wishes!
Nan Ao

 

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  J. K. Dewhurst - 2021-01-13

  Hi Nan Ao,

  Calculating the MAE for iron is particularly challenging with LAPW because the energy changes are so small. This is because it has relatively small spin-orbit coupling and cubic symmetry.

  I think the way to do it is to scale the spin-orbit coupling and spin exchange-correlation field using (for example)

  socscf
    4.0
  
  sxcscf
    1.02
  

  ...and calculate the MAE as a function of the scaling factor. This can then be extrapolated to 1.0. The spin exchange-correlation has to be scaled because large spin-orbit coupling reduces the moment. The value can be chosen to reproduce the correct moment of iron.

  Also, why are you using such a large unit cell? bcc Fe needs one atom per primitive cell. You can automatically reduce a conventional cell to a primitive cell with:

  primcell
   .true.
  

  Regards,
  Kay.

   

• 
  

Nan AO - 2021-01-15

Hello Kay

Thank you for your suggestion! I wil try to add socscf in my file.

Best wishes!
Nan

 

https://sourceforge.net/p/elk/discussion/897820/thread/318278b75d/