Magnetic ground state geometry

Elk Users
2013-02-05
2013-06-11
  • Giovanni Baez

    Giovanni Baez - 2013-02-05

    Hi all,
    I am trying to learn to calculate the magnetic ground state geometry of a system. For example I am testing a well known non-colinear system, IrMn. I have randomized the spins in the Mn and I wish to see if the system returns to the stable anti ferromagnetic state. Here is the input that I am using:
    tasks
      0
      2

    ! Groundstate calculation from scratch
    ! Geometric optimization

    spinpol
    .true.

    xctype
    3

    !lsda

    mixtype
    2

    ! fairly large number of empty states required for magnetic cases
    nempty
    10

    swidth
    0.001

    scale
    10.1153

    avec
    0.866025 -0.5 0.0
    0.0 1.0 0.0
    0.0 0.0 1.2247795

    sppath
    './'

    atoms
    2
    'Ir.in'
    3
    0.666666 0.666666 0.0 0.0 0.0 0.0
    0.0 0.0 .66664 0.0 0.0 0.0
    0.33333 0.33333 0.33332 0.0 0.0 0.0
    'Mn.in'
    9
    0.16667 0.16667 0.0  0.0 1.0  0.0
    0.16667 0.66667 0.0  0.5 0.0 -0.5
    0.66667 0.16667 0.0  -1.0 0.0 0.0
    0.0 0.5 0.66664  0.0 0.0 -1.0
    0.5 0.0 0.66664  0.555564444 -0.666666 0.0
    0.5 0.5 0.66664  1.0 0.0 0.0
    0.33333 0.83333 0.33332   -1.0 0.0 0.0
    0.83333 0.33332 0.33332   0.5 -0.866025 0.0
    0.83333 0.83333 0.33332   0.5 0.0 0.866025

    ! the large magnetic local field on Ni is halved at every iteration
    ! this speeds up the convergence of magnetic calculations

    reducebf
      0.5

    ngridk
    4 4 4

    vkloff
    0 0 0

     
  • Anton

    Anton - 2013-02-05

    It will not "move" away from the initial magnetic configuration unless you turn on spin-orbit interaction.

     
  • Lars Nordström

    Lars Nordström - 2013-02-08

    Dear Giovanni,

    If you are only interested in the magnetic ordering you do not need tasks=2 which optimise the atomic structure.
    Rather I would run with a larger field, say 0.9, and higher empty say 50, to ensure a magnetic solution. A larger swidth (0.01) will help to get a stable convergence and a smaller ngridk=2 2 2 will make the calculations faster.
    However, expect a relatively slow convergence to the AF structure.
    Since Ir has a large spin orbit coupling it might be needed to put on spin-orbit as suggested by Toxa.

    Good luck,
       Lars

     
  • Giovanni Baez

    Giovanni Baez - 2013-02-11

    Thank you Lars. I will try as sugested and let you know how it goes.
    Giovanni

     
  • Anton

    Anton - 2013-02-11

    Dear Lars!
    Please correct me if I'm wrong here: you must(!) turn on the spin-orbit coupling to "rotate" initial magnetisation  to some other preferable direction not just because the Ir has a large spin-orbit coupling, but because spin and orbital degrees must be coupled.  Then (and only then) the total energy depends on the spin orientation.

     
  • Lars Nordström

    Lars Nordström - 2013-02-11

    Dear Anton,

    in these calculations  it is  the angle in between different moments that come into play. The total energy will be minimized when these angles reach convergence (at least hopefully, there might be many minima). But yes, there is still a symmetry with respect to a a global rotation of all spins, so the direction of *one* spin has no meaning, but only the angles in between the different spins have a meaning.

    When spin orbit coupling is switched on, the individual directions get  a meaning, but also if the spin orbit coupling is large it will influence the exchange coupling of the moments which might lead to another ground state magnetic order.

    Best,
        Lars

     

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