The problem of using second-order perturbation method to calculate magnetic anisotropy performance |
- Date: 2024/03/20 18:00
- Name: jack
- Dear all:
When I was using the second-order perturbation method to calculate MAE according to manual 3.9, I found that the spin angle of the result was offset from the direction I initially set, and the angles I set were all 0. However, when I checked the output file, I found that my theta angle remained unchanged, but phi changed to around 120 °. What should I do? Is the result of this calculation correct?Here are my. dat files and the results of my last scf：
System.CurrrentDirectory ./ # default=./
System.Name Fe_Bulk_jx
level.of.stdout 1 # default=1 (1-3)
level.of.fileout 0 # default=1 (1-3)
HS.fileout on # on|off, default=off
Species.Number 1
<Definition.of.Atomic.Species
Fe Fe6.0H-s2p2d2 Fe_PBE13H
Definition.of.Atomic.Species>
Atoms.Number 2
Atoms.SpeciesAndCoordinates.Unit Ang
<Atoms.SpeciesAndCoordinates
1 Fe 0.000000 0.000000 0.000000 9.00 7.00
2 Fe 1.433000 1.433000 1.433000 9.00 7.00
Atoms.SpeciesAndCoordinates>
Atoms.UnitVectors.Unit Ang
<Atoms.UnitVectors
2.866 0.000 0.000
0.000 2.866 0.000
0.000 0.000 2.866
Atoms.UnitVectors>
scf.XcType GGA-PBE # LDA|LSDA-CA|LSDA-PW
scf.SpinPolarization nc # On|Off
scf.spinorbit.coupling on # on|off, default=off
scf.ElectronicTemperature 300.0 # default=300 (K)
scf.energycutoff 600.0 # default=150 (Ry)
scf.maxIter 100 # default=40
scf.EigenvalueSolver band # Recursion|Cluster|Band
scf.Kgrid 27 27 27
scf.Mixing.Type rmm-diisk # Simple|Rmm-Diis|Gr-Pulay
scf.Init.Mixing.Weight 0.300 # default=0.30
scf.Min.Mixing.Weight 0.001 # default=0.001
scf.Max.Mixing.Weight 0.3000 # default=0.40
scf.Mixing.History 40 # default=5
scf.Mixing.StartPulay 20 # default=6
scf.criterion 1.0e-6 # default=1.0e-6 (Hartree)
MD.Type nomd
MD.maxIter 1
MD.Opt_criterion 0.000300
scf.restart.filename Fe_Bulk_jx
scf.restart c2n
scf.Restart.Spin.Angle.Theta 0.0
scf.Restart.Spin.Angle.Phi 0.0
******************* MD= 1 SCF=30 *******************
<Poisson> Poisson's equation using FFT...
<Set_Hamiltonian> Hamiltonian matrix for VNA+dVH+Vxc...
<Band> Solving the eigenvalue problem...
KGrids1: -0.48148 -0.44444 -0.40741 -0.37037 -0.33333 -0.29630 -0.25926 -0.22222 -0.18518 -0.14815 -0.11111 -0.07407 -0.03704 0.00000 0.03704 0.07408 0.11111 0.14815 0.18519 0.22222 0.25926 0.29630 0.33333 0.37037 0.40741 0.44445 0.48148
KGrids2: -0.48148 -0.44445 -0.40741 -0.37037 -0.33333 -0.29630 -0.25926 -0.22222 -0.18519 -0.14815 -0.11111 -0.07408 -0.03704 -0.00000 0.03704 0.07407 0.11111 0.14815 0.18518 0.22222 0.25926 0.29630 0.33333 0.37037 0.40741 0.44444 0.48148
KGrids3: -0.48148 -0.44444 -0.40741 -0.37037 -0.33333 -0.29629 -0.25926 -0.22222 -0.18518 -0.14815 -0.11111 -0.07407 -0.03704 0.00000 0.03704 0.07408 0.11111 0.14815 0.18519 0.22222 0.25926 0.29630 0.33334 0.37037 0.40741 0.44445 0.48148
<Band_DFT> Eigen, time=62.540168
<Band_DFT> DM, time=154.828903
1 Fe MulP 9.13 6.87 sum 16.00 diff 2.26 ( 0.00 114.51) Ml 0.06 ( 0.00 117.36) Ml+s 2.32 ( 0.00 114.58)
2 Fe MulP 9.13 6.87 sum 16.00 diff 2.26 ( 0.00 121.89) Ml 0.06 ( 0.00 120.72) Ml+s 2.32 ( 0.00 121.86)
Sum of MulP: up = 18.26407 down = 13.73593
total= 32.00000 ideal(neutral)= 32.00000
<DFT> Total Spin Moment (muB) 4.528141495 Angles 0.000003912 118.196476851
<DFT> Total Orbital Moment (muB) 0.110204429 Angles 0.000004353 119.018677950
<DFT> Total Moment (muB) 4.638345924 Angles 0.000003912 118.218154687
<DFT> Mixing_weight= 0.030000000000
<DFT> Uele = -43.806320372067 dUele = 0.000000056186
<DFT> NormRD = 0.000011154003 Criterion = 0.000001000000
<MD= 1> Force calculation
Force calculation #1
Force calculation #2
Force calculation #3
Force calculation #4
Force calculation #5
<MD= 1> Total Energy
Force calculation #6
Force calculation #7
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Re: The problem of using second-order perturbation method to calculate magnetic anisotropy performance