## Different moment and energy for QSPIRAL (0,0,0) and (0,0,1)

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tpaudel1
Newbie Posts: 1
Joined: Wed May 19, 2010 5:08 pm

### Different moment and energy for QSPIRAL (0,0,0) and (0,0,1)

In the spin spiral calculation and I found some confusing results even for very simple system

According to the examples in https://cms.mpi.univie.ac.at/wiki/index ... in_spirals, QSPIRAL = (0, 0, 0) means no spin rotation between unit cells while QSPIRAL = (0, 0, 0.5) means spin flip between adjacent cells. Then, QSPIRAL = (0, 0, 1) should be the same as QSPIRAL = (0, 0, 0) because 360-degree rotation is equal to no rotation.

However, QSPIRAL = (0, 0, 0) and (0, 0, 1) give me different results when I check with the example given in the website. I use the simple cubic Fe supercell (1×1×2, 2 Iron atoms). Initial mag-mom of the two Fe atoms are both in y direction. However, the calculated magnetic moments and total energies printed in OUTCAR for these two case are quite different:

QSPIRAL = 0 0 0 calculation gives ferromagnetic result (as we expected). Both atoms have moment 2.4 muB in y direction, the same as the initial magnetic moments.

QSPIRAL = 0 0 1 calculation gives anti-ferromagnetic result. Two iron atoms have spins 2.4 muB and -2.4 muB in y direction, which is not what we expected.

Additionally, two calculations have 1.164 meV energy difference which should not account for calculation error.

So, my question is, why QSPIRAL = (0, 0, 0) and QSPIRAL = (0, 0, 1) lead to different outputs. Is there any special way we can constrain the spin direction in spin spiral calculation to avoid this?

I listed POSCAR and INCAR below. Do you have any suggestions?

POSCAR-----------------
fe1
1.00000000000000
2.3669077402514498 0.0000000000000000 -0.0000000000000000
-0.0000000000000000 2.3669077402514498 -0.0000000000000000
-0.0000000000000000 0.0000000000000000 4.7338154805028996
Fe
2
Direct
0.0000000000000000 0.0000000000000000 0.0000000000000000
0.0000000000000000 0.0000000000000000 0.5000000000000000
------------------------

INCAR-----------------
NPAR=6
ENCUT = 450 eV
ENINI = 320 eV #for spiral

LMAXMIX = 4
PREC = Accurate

ICHARG=1
ISYM = -1
LASPH = .TRUE.
LSPIRAL = .TRUE.
QSPIRAL = 0 0 0 # or 0 0 1
SAXIS=0 0 1
LSORBIT = .TRUE.
LNONCOLLINEAR = .TRUE.

MAGMOM = 0 3 0 0 3 0

LORBIT = 11
LREAL = .FALSE.
LCHARG=.TRUE.
LWAVE=.FALSE.

NELM=200
NELMIN=2

EDIFF=1E-6

ISMEAR = -5
SIGMA = 0.2
------------------------