[Pw_forum] lda+u low spin question
sasha
ackollias at gmail.com
Thu Jun 14 18:48:27 CEST 2007
Greetings,
For a low spin (non-magnetic) LDA+U calculation of MgFeO, the following
Hubbard matrix occupations were obtained:
z2-1/3 zx zy x2-y2 xy
0.384 0.000 0.000 0.000 0.002
0.000 0.993 0.000 0.001 0.000
0.000 0.000 0.993 -0.001 0.000
0.000 0.001 -0.001 0.380 0.000
0.002 0.000 0.000 0.000 0.993
For the low-spin solution the eg orbitals should have near zero
occupancies however above it's clear that while the t2g have the
correction occupancy (near 1) the occupancy of the eg seems to be too
large. The basic question I have is am I computing the low-spin
solution properly? Any pointers are greatly appreciated. The input file
used to obtain the above occupations is given below.
regards,
Sasha
-------------------------------------------------
FeO
FeO Wustite whithin LDA+U using standard initial ns matrices
&control
calculation = 'scf'
restart_mode='from_scratch',
verbosity='high',
tstress=.true.,
prefix='mgfeo-a-7.28-low-initial-444-0.6',
pseudo_dir = '/home/akollias/espresso-3.2/pseudo/',
outdir='/home/scratch/akollias'
/
&system
ibrav= 0, celldm(1)=7.28, nat= 16, ntyp= 6,
ecutwfc = 60.0, ecutrho = 480.0, nbnd=53
occupations='smearing', smearing='gauss', degauss=0.01,
nspin=1,
lda_plus_u=.true. Hubbard_U(1)=5.0, Hubbard_U(2)=5.0,
Hubbard_U(3)=5.0, Hubbard_U(4)=5.0,
report=1
/
&electrons
mixing_mode = 'plain'
mixing_beta = 0.30
mixing_ndim = 8
mixing_fixed_ns = 0
conv_thr = 1.0d-6
electron_maxstep = 300
/
CELL_PARAMETERS
0.50 -0.50 0.00
1.00 0.50 -0.50
1.00 1.00 2.00
ATOMIC_SPECIES
Fe1 1. Fe.pz-nd-rrkjus.UPF
Fe2 1. Fe.pz-nd-rrkjus.UPF
Fe3 1. Fe.pz-nd-rrkjus.UPF
Fe4 1. Fe.pz-nd-rrkjus.UPF
Mg1 1. Mg.pz-n-vbc.UPF
O1 1. O.LDA.US.RRKJ3.UPF
ATOMIC_POSITIONS {alat}
Fe1 0.5 0.5 1.0
Fe2 1.0 0.0 0.0
Fe3 1.5 1.0 1.5
Fe4 0.0 0.0 0.0
Mg1 1.5 0.5 0.0
Mg1 1.0 0.5 1.5
Mg1 1.0 0.5 0.5
Mg1 1.5 0.5 1.0
O1 1.0 0.5 0.0
O1 1.5 0.5 1.5
O1 0.5 0.0 0.0
O1 2.0 1.0 1.5
O1 1.5 0.5 0.5
O1 1.0 0.5 1.0
O1 1.5 1.0 1.0
O1 1.0 0.0 0.5
K_POINTS {automatic}
4 4 4 0 0 0
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