[Pw_forum] Symmetry operation identification for supercell
Krishnendu Mukherjee
krishnendu.mukherjee789 at gmail.com
Fri Feb 16 14:54:42 CET 2018
Dear Expert,
I have created a Zr supercell with 16 atoms (the positions of the atoms
are given in the input file below). Zr has the spacegroup P 63/m m c (No.
194).
However, in output I notice,
Found symmetry operation: I + ( -0.5000 0.5000 0.0000)
This is a supercell, fractional translations are disabled
Found symmetry operation: I + ( -0.5000 0.5000 0.0000)
This is a supercell, fractional translations are disabled
Now, although the space group has no fractional translational along a and
b, I think the fractional translations are identified as it is a supercell.
But why there is no fractional translation identified along c? There is a
fractional transformation along c in this spacegroup.
I will be grateful for your kind explanation. I am attaching the input
below and some part of the output.
-------------------------------------------
cat > thermo_control << EOF
&INPUT_THERMO
what='mur_lc_elastic_constants',
frozen_ions=.FALSE.
/
EOF
cat > zr.elastic.in << EOF
&control
calculation = 'scf'
restart_mode='from_scratch',
prefix='zr',
tstress = .true.,
tprnfor = .true.,
pseudo_dir = '$PSEUDO_DIR/',
outdir='$TMP_DIR/'
/
&system
ibrav= 4,
celldm(1) =12.241645,
celldm(3) = 1.59185,
nat= 16,
ntyp= 1,
ecutwfc=50.0,
ecutrho = 430,
occupations='smearing',
smearing='marzari-vanderbilt',
degauss=0.02
starting_magnetization(1) = 0.7,
use_all_frac = .true.
/
&electrons
conv_thr = 1.0d-10
/
ATOMIC_SPECIES
Zr 91.22 Zr.pz-spn-kjpaw_psl.1.0.0.UPF
ATOMIC_POSITIONS (angstrom)
Zr 0.000000 1.870038 1.289000
Zr 3.239000 3.740075 9.023001
Zr 1.619500 4.675094 1.289000
Zr 1.619500 0.935019 9.023001
Zr -1.619500 4.675094 1.289000
Zr 4.858500 0.935019 9.023001
Zr 3.239000 3.740075 3.867000
Zr 1.619500 0.935019 3.867000
Zr 4.858500 0.935019 3.867000
Zr 0.000000 1.870038 6.445000
Zr 1.619500 4.675094 6.445000
Zr -1.619500 4.675094 6.445000
Zr 3.239000 1.870038 1.289000
Zr 0.000000 3.740075 9.023001
Zr 0.000000 3.740075 3.867000
Zr 3.239000 1.870038 6.445000
K_POINTS AUTOMATIC
5 5 3 0 0 0
EOF
---------------------------------------------------------------------
Info: using nr1, nr2, nr3 values from input
Found symmetry operation: I + ( -0.5000 0.5000 0.0000)
This is a supercell, fractional translations are disabled
Found symmetry operation: I + ( -0.5000 0.5000 0.0000)
This is a supercell, fractional translations are disabled
Computing the elastic constants at the minimum volume
FFT mesh: ( 81, 81, 135 )
Bravais lattice:
ibrav= 4: hexagonal
Cell parameters:
alat= 12.241645 a.u., c/a= 1.591850
Starting primitive lattice vectors:
crystal axes: (cart. coord. in units of alat)
a(1) = ( 1.000000 0.000000 0.000000 )
a(2) = ( -0.500000 0.866025 0.000000 )
a(3) = ( 0.000000 0.000000 1.591850 )
Starting reciprocal lattice vectors:
reciprocal axes: (cart. coord. in units 2 pi/alat)
b(1) = ( 1.000000 0.577350 -0.000000 )
b(2) = ( 0.000000 1.154701 0.000000 )
b(3) = ( 0.000000 -0.000000 0.628200 )
Starting atomic positions in Cartesian axes:
site n. atom positions (alat units)
1 Zr tau( 1) = ( 0.0000000 0.2886752 0.1989812
)
2 Zr tau( 2) = ( 0.5000000 0.5773503 1.3928684
)
3 Zr tau( 3) = ( 0.2500000 0.7216879 0.1989812
)
4 Zr tau( 4) = ( 0.2500000 0.1443376 1.3928684
)
5 Zr tau( 5) = ( -0.2500000 0.7216879 0.1989812
)
6 Zr tau( 6) = ( 0.7500001 0.1443376 1.3928684
)
7 Zr tau( 7) = ( 0.5000000 0.5773503 0.5969435
)
8 Zr tau( 8) = ( 0.2500000 0.1443376 0.5969435
)
9 Zr tau( 9) = ( 0.7500001 0.1443376 0.5969435
)
10 Zr tau( 10) = ( 0.0000000 0.2886752 0.9949059
)
11 Zr tau( 11) = ( 0.2500000 0.7216879 0.9949059
)
12 Zr tau( 12) = ( -0.2500000 0.7216879 0.9949059
)
13 Zr tau( 13) = ( 0.5000000 0.2886752 0.1989812
)
14 Zr tau( 14) = ( 0.0000000 0.5773503 1.3928684
)
15 Zr tau( 15) = ( 0.0000000 0.5773503 0.5969435
)
16 Zr tau( 16) = ( 0.5000000 0.2886752 0.9949059
)
Starting atomic positions in crystallographic axes:
site n. atom positions (cryst. coord.)
1 Zr tau( 1) = ( 0.1666667 0.3333334 0.1250000 )
2 Zr tau( 2) = ( 0.8333334 0.6666667 0.8749998 )
3 Zr tau( 3) = ( 0.6666667 0.8333334 0.1250000 )
4 Zr tau( 4) = ( 0.3333334 0.1666667 0.8749998 )
5 Zr tau( 5) = ( 0.1666667 0.8333334 0.1250000 )
6 Zr tau( 6) = ( 0.8333334 0.1666667 0.8749998 )
7 Zr tau( 7) = ( 0.8333334 0.6666667 0.3749999 )
8 Zr tau( 8) = ( 0.3333334 0.1666667 0.3749999 )
9 Zr tau( 9) = ( 0.8333334 0.1666667 0.3749999 )
10 Zr tau( 10) = ( 0.1666667 0.3333334 0.6249998 )
11 Zr tau( 11) = ( 0.6666667 0.8333334 0.6249998 )
12 Zr tau( 12) = ( 0.1666667 0.8333334 0.6249998 )
13 Zr tau( 13) = ( 0.6666668 0.3333334 0.1250000 )
14 Zr tau( 14) = ( 0.3333334 0.6666667 0.8749998 )
15 Zr tau( 15) = ( 0.3333334 0.6666667 0.3749999 )
16 Zr tau( 16) = ( 0.6666668 0.3333334 0.6249998 )
The energy minimization will require 9 scf calculations
The point group 118 D_3d (-3m) is compatible with the Bravais lattice.
The rotation matrices with the order used inside thermo_pw are:
12 Sym. Ops., with inversion, found
s frac. trans.
isym = 1 identity
cryst. s( 1) = ( 1 0 0 )
( 0 1 0 )
( 0 0 1 )
cart. s( 1) = ( 1.000 0.000 0.000 )
( 0.000 1.000 0.000 )
( 0.000 0.000 1.000 )
isym = 2 180 deg rotation - cart. axis [1,0,0]
cryst. s( 2) = ( 1 0 0 )
( -1 -1 0 )
( 0 0 -1 )
cart. s( 2) = ( 1.000 0.000 0.000 )
( 0.000 -1.000 0.000 )
( 0.000 0.000 -1.000 )
isym = 3 120 deg rotation - cryst. axis [0,0,1]
cryst. s( 3) = ( 0 1 0 )
( -1 -1 0 )
( 0 0 1 )
cart. s( 3) = ( -0.500 -0.866 0.000 )
( 0.866 -0.500 0.000 )
( 0.000 0.000 1.000 )
isym = 4 120 deg rotation - cryst. axis [0,0,-1]
cryst. s( 4) = ( -1 -1 0 )
( 1 0 0 )
( 0 0 1 )
cart. s( 4) = ( -0.500 0.866 0.000 )
( -0.866 -0.500 0.000 )
( 0.000 0.000 1.000 )
isym = 5 180 deg rotation - cryst. axis [0,1,0]
cryst. s( 5) = ( -1 -1 0 )
( 0 1 0 )
( 0 0 -1 )
cart. s( 5) = ( -0.500 -0.866 0.000 )
( -0.866 0.500 0.000 )
( 0.000 0.000 -1.000 )
isym = 6 180 deg rotation - cryst. axis [1,1,0]
cryst. s( 6) = ( 0 1 0 )
( 1 0 0 )
( 0 0 -1 )
cart. s( 6) = ( -0.500 0.866 0.000 )
( 0.866 0.500 0.000 )
( 0.000 0.000 -1.000 )
isym = 7 inversion
cryst. s( 7) = ( -1 0 0 )
( 0 -1 0 )
( 0 0 -1 )
cart. s( 7) = ( -1.000 0.000 0.000 )
( 0.000 -1.000 0.000 )
( 0.000 0.000 -1.000 )
isym = 8 inv. 180 deg rotation - cart. axis [1,0,0]
cryst. s( 8) = ( -1 0 0 )
( 1 1 0 )
( 0 0 1 )
cart. s( 8) = ( -1.000 0.000 0.000 )
( 0.000 1.000 0.000 )
( 0.000 0.000 1.000 )
isym = 9 inv. 120 deg rotation - cryst. axis [0,0,1]
cryst. s( 9) = ( 0 -1 0 )
( 1 1 0 )
( 0 0 -1 )
cart. s( 9) = ( 0.500 0.866 0.000 )
( -0.866 0.500 0.000 )
( 0.000 0.000 -1.000 )
isym = 10 inv. 120 deg rotation - cryst. axis [0,0,-1]
cryst. s(10) = ( 1 1 0 )
( -1 0 0 )
( 0 0 -1 )
cart. s(10) = ( 0.500 -0.866 0.000 )
( 0.866 0.500 0.000 )
( 0.000 0.000 -1.000 )
isym = 11 inv. 180 deg rotation - cryst. axis [0,1,0]
cryst. s(11) = ( 1 1 0 )
( 0 -1 0 )
( 0 0 1 )
cart. s(11) = ( 0.500 0.866 0.000 )
( 0.866 -0.500 0.000 )
( 0.000 0.000 1.000 )
isym = 12 inv. 180 deg rotation - cryst. axis [1,1,0]
cryst. s(12) = ( 0 -1 0 )
( -1 0 0 )
( 0 0 1 )
cart. s(12) = ( 0.500 -0.866 0.000 )
( -0.866 -0.500 0.000 )
( 0.000 0.000 1.000 )
point group D_3d (-3m)
there are 6 classes
the character table:
E 2C3 3C2' i 2S6 3s_d
A_1g 1.00 1.00 1.00 1.00 1.00 1.00
A_2g 1.00 1.00 -1.00 1.00 1.00 -1.00
E_g 2.00 -1.00 0.00 2.00 -1.00 0.00
A_1u 1.00 1.00 1.00 -1.00 -1.00 -1.00
A_2u 1.00 1.00 -1.00 -1.00 -1.00 1.00
E_u 2.00 -1.00 0.00 -2.00 1.00 0.00
the symmetry operations in each class and the name of the first
element:
E 1
identity
2C3 3 4
120 deg rotation - cryst. axis [0,0,1]
3C2' 2 5 6
180 deg rotation - cart. axis [1,0,0]
i 7
inversion
2S6 9 10
inv. 120 deg rotation - cryst. axis [0,0,1]
3s_d 8 11 12
inv. 180 deg rotation - cart. axis [1,0,0]
Space group identification, 12 symmetries:
Bravais lattice 4 hexagonal
Point group number 25 / 118 D_3d (-3m)
Nonsymmorphic operations not found: All fractional translations vanish
Symmetries of the point group in standard order
1 E 1
2 3z 27
3 3-z 28
4 2x 4
5 2110 32
6 2010 31
7 i 33
8 i3z 59
9 i3-z 60
10 i2x 36
11 i2110 64
12 i2010 63
Space group nymber 164
Space group P-3m1 (group number 164).
The origin coincides with the ITA tables.
The Laue class is D_3d (-3m)
In this class the elastic tensor is
( c11 c12 c13 c14 . . )
( c12 c11 c13 -c14 . . )
( c13 c13 c33 . . . )
( c14 -c14 . c44 . . )
( . . . . c44 c14 )
( . . . . c14 X )
X=(c11-c12)/2
It requires three strains: e1, e3, and e4
for a total of 12 scf calculations
----------------------------------------------------------------------
Ions are relaxed in each calculation
----------------------------------------------------------------------
--------------------------------------------------------
Thanks,
Best regards,
Krishnendu
--
Dr. Krishnendu Mukherjee,
Principal Scientist,
CSIR-NML,
Jamshedpur.
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