[Pw_forum] Proper Symmetry Representation
Dr. Robert Molt Jr.
r.molt.chemical.physics at gmail.com
Mon Jul 10 19:02:52 CEST 2017
Good evening,
This is my first solid-state electronic structure calculation; please
forgive if my questions are obvious. I am not confident in the proper
symmetry representation of my system. I attach my input file and
associated questions. Don't worry on my value of ecutwfc; I am trying to
do cheap calculations at the moment for pedagogy, and do not care how
quality a value this is right now. Same goes for my k space points.
&CONTROL
calculation = "vd-relax"
outdir='/p/work1/workspace/molt/Espresso_Test',
pseudo_dir = '/p/home/molt/Espresso_PP',
restart_mode = "from_scratch"
/
&SYSTEM
celldm(1) = 1.3454e+01
ecutwfc = 4.00000e+01
ibrav = 2
nat = 74
nspin = 1
ntyp = 2
/
&ELECTRONS
diagonalization = "david"
electron_maxstep = 100
mixing_beta = 7.00000e-01
mixing_mode = "plain"
startingpot = "atomic"
startingwfc = "atomic+random"
/
&IONS
/
&CELL
cell_dynamics=bfgs
/
K_POINTS {automatic}
2 2 2 0 0 0
ATOMIC_SPECIES
Si 28.085 Si.pbe-n-kjpaw_psl.0.1.UPF.TXT
O 15.999 O.pbe-n-kjpaw_psl.0.1.UPF.TXT
ATOMIC_POSITIONS {angstrom}
Si 0.000000 0.000000 0.000000
Si 0.000000 0.000000 7.120000
Si 0.000000 7.120000 0.000000
Si 0.000000 7.120000 7.120000
Si 7.120000 0.000000 0.000000
Si 7.120000 0.000000 7.120000
Si 7.120000 7.120000 0.000000
Si 7.120000 7.120000 7.120000
Si 0.000000 3.560000 3.560000
Si 7.120000 3.560000 3.560000
Si 3.560000 3.560000 0.000000
Si 3.560000 3.560000 7.120000
Si 3.560000 0.000000 3.560000
Si 3.560000 7.120000 3.560000
Si 5.340000 1.780000 5.340000
Si 1.780000 1.780000 1.780000
Si 1.780000 5.340000 5.340000
Si 5.340000 5.340000 1.780000
O 0.890000 0.890000 0.890000
O 6.230000 2.670000 4.450000
O 2.670000 4.450000 6.230000
O 4.450000 6.230000 2.670000
O 0.890000 4.450000 4.450000
O 6.230000 6.230000 0.890000
O 2.670000 0.890000 2.670000
O 4.450000 2.670000 6.230000
O 4.450000 0.890000 4.450000
O 2.670000 2.670000 0.890000
O 6.230000 4.450000 2.670000
O 0.890000 6.230000 6.230000
O 4.450000 4.450000 0.890000
O 2.670000 6.230000 4.450000
O 6.230000 0.890000 6.230000
O 0.890000 2.670000 2.670000
O -0.890000 -0.890000 0.890000
O -0.890000 0.890000 -0.890000
O 0.890000 -0.890000 -0.890000
O -0.890000 -0.890000 8.010000
O -0.890000 0.890000 6.230000
O 0.890000 -0.890000 6.230000
O 0.890000 0.890000 8.010000
O -0.890000 6.230000 0.890000
O -0.890000 8.010000 -0.890000
O 0.890000 6.230000 -0.890000
O 0.890000 8.010000 0.890000
O -0.890000 6.230000 8.010000
O -0.890000 8.010000 6.230000
O 0.890000 8.010000 8.010000
O 6.230000 -0.890000 0.890000
O 6.230000 0.890000 -0.890000
O 8.010000 -0.890000 -0.890000
O 8.010000 0.890000 0.890000
O 6.230000 -0.890000 8.010000
O 8.010000 -0.890000 6.230000
O 8.010000 0.890000 8.010000
O 6.230000 8.010000 -0.890000
O 8.010000 6.230000 -0.890000
O 8.010000 8.010000 0.890000
O 6.230000 6.230000 8.010000
O 6.230000 8.010000 6.230000
O 8.010000 6.230000 6.230000
O 8.010000 8.010000 8.010000
O -0.890000 2.670000 4.450000
O -0.890000 4.450000 2.670000
O 8.010000 2.670000 2.670000
O 8.010000 4.450000 4.450000
O 2.670000 4.450000 -0.890000
O 4.450000 2.670000 -0.890000
O 2.670000 2.670000 8.010000
O 4.450000 4.450000 8.010000
O 2.670000 -0.890000 4.450000
O 4.450000 -0.890000 2.670000
O 2.670000 8.010000 2.670000
O 4.450000 8.010000 4.450000
1.) As far as I can tell, SiO2 is a cubic fcc lattice. The Espresso
website (http://www.quantum-espresso.org/faq/input-data/#3.1) states
that Espresso uses the primitive, rather than conventional, unit cell,
unless one specifies otherwise. As I have elected ibrav=2, I infer I am
now in the latter category. The directions also state that an all-atom,
rather than symmetry-unique atom, representation is chosen. Would I thus
specify the Cartesians of 18 Silicon atoms, for an fcc lattice? The code
is such that it can recognize the symmetry of the atoms put in, parse
out which ones are the unique atoms, and use that symmetry to not
re-calculate the same integrals over-and-over, etc.? If it's helpful to
visualize, I pulled this structure from a cif file from
http://www.crystallography.net/cod/1010954.html
2.) I have specified the lattice constant as celldm(1)=13.454. I assume
that the default, Bohr radius, is not affected by selecting the
"Angstrom" flag later on.
3.) nat, according the manual, specifies the number of atoms in the unit
cell. Does this mean the total number of atoms, i.e., 18+56 for SiO2 if
I have selected ibrav=2, or does the code expect the primitive cell atom
number since that's the Espresso default?
Thank you for your time,
Dr. Robert Molt
Indiana-University Perdue University
ENSCO
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