Dear Siesters,
I am working with b-cristobalite, a silica polymorph. I am puzzled by two
different equilibrium lattice constant i obtain using a 2-atom primitive
cell (7.14A) and a 24-atom one complete unit cell (7.57A). Maybe I doing
something fundamentally wrong, appreciate if anyone can point out the error.
Here's my fdf for 2atom version:
SystemName b-cristobalite
SystemLabel bcris_7.14A
NumberOfAtoms 2
NumberOfSpecies 2
%block ChemicalSpeciesLabel
1 14 Si.gga
2 8 O.gga
%endblock ChemicalSpeciesLabel
LatticeConstant 7.14 Ang
%block LatticeParameters
1.0 1.0 1.0 90. 90. 90.
%endblock LatticeParameters
BandLinesScale pi/a
%block BandLines
1 1.5 1.5 0.0 K # Begin at K
38 0.0 0.0 0.0 \Gamma # 38 points from K to Gamma
36 0.0 2.0 0.0 X # 36 points from Gamma to X
18 1.0 2.0 0.0 W # 18 points from X to W
26 1.0 1.0 1.0 L # 26 points from W to L
31 0.0 0.0 0.0 \Gamma # 31 points from L to Gamma
%endblock BandLines
AtomicCoordinatesFormat Fractional
%block AtomicCoordinatesAndAtomicSpecies
0.000 0.000 0.000 1 Si
0.125 0.125 0.125 2 O
%endblock AtomicCoordinatesAndAtomicSpecies
%block kgrid_Monkhorst_Pack
4 0 0 0.5
0 4 0 0.5
0 0 4 0.5
%endblock kgrid_Monkhorst_Pack
XC.functional GGA
XC.authors PBE
PAO.BasisSize DZP
PAO.EnergyShift 300 meV
MeshCutoff 100. Ry
and for the 24 atom version, I only change the add the extra coordinates:
AtomicCoordinatesFormat Fractional
%block AtomicCoordinatesAndAtomicSpecies
0.125 0.125 0.125 2 O1 1
0.125 0.625 0.625 2 O2 2
0.625 0.625 0.125 2 O3 3
0.625 0.125 0.625 2 O4 4
0.125 0.875 0.875 2 O5 5
0.125 0.375 0.375 2 O6 6
0.625 0.375 0.875 2 O7 7
0.625 0.875 0.375 2 O8 8
0.875 0.125 0.875 2 O9 9
0.875 0.625 0.375 2 O10 10
0.375 0.625 0.875 2 O11 11
0.375 0.125 0.375 2 O12 12
0.875 0.875 0.125 2 O13 13
0.875 0.375 0.625 2 O14 14
0.375 0.375 0.125 2 O15 15
0.375 0.875 0.625 2 O16 16
0 0 0 1 Si1 17
0 0.5 0.5 1 Si2 18
0.5 0.5 0 1 Si3 19
0.5 0 0.5 1 Si4 20
0.25 0.25 0.25 1 Si5 21
0.25 0.75 0.75 1 Si6 22
0.75 0.75 0.25 1 Si7 23
0.75 0.25 0.75 1 Si8 24
%endblock AtomicCoordinatesAndAtomicSpecies
Thank you for your time,
Prat