ABINIT Discussion Forums

Hello,

I have a problem understanding how symmetry is implemented in the code.
I calculated a structure with a hexagonal symmetry group. The group has 12 symmetry elements, and the corresponding irreducible Brillouin zone (BZ) in ABINIT was 1/12 of the whole BZ(Gamma-M-K-Gamma on the figure).
I added an atom to the unit cell of the structure, lowering the symmetry to trigonal (but the translational vectors and reciprocal unit cell stayed unaltered). This group has 6 symmetry elements, and I would expect this irreducible BZ as twice as large. Basically, I'm saying that point K and K' on the figure should not be physically equivalent anymore.
But ABINIT gives exactly the same irreducible BZs for both structures. The same amount of points for equal grids with equal positions (the same Gamma-M-K-Gamma). The symmetry is lowered, and ABINIT recognises it, so I don't understand why would ABINIT create the same irreducible BZ then.
What I consider is only the space group symmetries (translations+crystal point group). I also calculated the electronic band structure in points K and K', and it was equal.

Apparently, there is some additional symmetry present in the system which brings the irreducible BZ to small 1/12 of the whole hexagon (which ABINIT recognises)
What symmetries are taken into account in the ABINIT? what should I check?
If I calculate everything on the whole BZ the output gives me different type of reciprocal unit cell (rhombic instead of hexagonal), and I cannot compare them easily.

Thank you!
Best regards, Mikhail

Dear Mikhail,
To be sure I fully understand your problem, could you provide the Abinit output file symmetry information for your two cases as well as the definition of cell parameters and positions (or give directly the output files if it is fine with you)?
Best wishes,
Eric

Dear Eric,

Thank you for your response!

1.This is the first (high symmetry) structure:
1.1 From the log file
symlatt : the Bravais lattice is hP (primitive hexagonal)
xred is defined in input file
ingeo : takes atomic coordinates from input array xred
symfind : exit, nsym= 12
symrel matrices, symafm and tnons are :
1 1 0 0 0 1 0 0 0 1 1 0.0000 0.0000 0.0000
2 0 1 0 1 0 0 0 0 1 1 -0.0000 0.0000 0.0000
3 1 1 0 -1 0 0 0 0 1 1 0.0000 0.0000 0.0000
4 -1 0 0 1 1 0 0 0 1 1 -0.0000 0.0000 0.0000
5 0 1 0 -1 -1 0 0 0 1 1 0.0000 0.0000 0.0000
6 -1 -1 0 0 1 0 0 0 1 1 0.0000 0.0000 0.0000
7 -1 0 0 0 -1 0 0 0 1 1 0.0000 0.0000 0.0000
8 0 -1 0 -1 0 0 0 0 1 1 0.0000 0.0000 0.0000
9 -1 -1 0 1 0 0 0 0 1 1 -0.0000 0.0000 0.0000
10 1 0 0 -1 -1 0 0 0 1 1 0.0000 0.0000 0.0000
11 0 -1 0 1 1 0 0 0 1 1 -0.0000 0.0000 0.0000
12 1 1 0 0 -1 0 0 0 1 1 0.0000 0.0000 0.0000

symlatt : the Bravais lattice is hP (primitive hexagonal)

symlatt : the Bravais lattice is hP (primitive hexagonal)
symspgr : spgroup= 183 P6 m m (=C6v^1)

1.2.From out file

DATASET 1 : space group P6 m m (#183); Bravais hP (primitive hexag.)
================================================================================
Values of the parameters that define the memory need for DATASET 1.
intxc = 0 ionmov = 0 iscf = 7 lmnmax = 4
lnmax = 4 mgfft = 216 mpssoang = 2 mqgrid = 3001
natom = 3 nloc_mem = 1 nspden = 1 nspinor = 1
nsppol = 1 nsym = 12 n1xccc = 2501 ntypat = 2
occopt = 3 xclevel = 2
- mband = 16 mffmem = 1 mkmem = 154
mpw = 20042 nfft = 437400 nkpt = 154
For the susceptibility and dielectric matrices, or tddft :
mgfft = 120 nbnd_in_blk= 6 nfft = 69120 npw = 109
================================================================================
1.3. Input positions
rprim
5.0000000000E-01 -8.6602540378E-01 0.0000000000E+00
5.0000000000E-01 8.6602540378E-01 0.0000000000E+00
0.0000000000E+00 0.0000000000E+00 1.0000000000E+00
xred
8.4847175841E-26 5.3480052276E-25 -2.1516874730E-04
6.6666666667E-01 3.3333333333E-01 1.1160758437E-01
3.3333333333E-01 6.6666666667E-01 1.1160758437E-01

2.Structure with less symmetry :
2.1 From the log file
symlatt : the Bravais lattice is hP (primitive hexagonal)
xred is defined in input file
ingeo : takes atomic coordinates from input array xred
symfind : exit, nsym= 6
symrel matrices, symafm and tnons are :
1 1 0 0 0 1 0 0 0 1 1 0.0000 0.0000 0.0000
2 -1 0 0 1 1 0 0 0 1 1 -0.0000 0.0000 0.0000
3 0 1 0 -1 -1 0 0 0 1 1 -0.0000 -0.0000 0.0000
4 0 -1 0 -1 0 0 0 0 1 1 -0.0000 -0.0000 0.0000
5 -1 -1 0 1 0 0 0 0 1 1 -0.0000 -0.0000 0.0000
6 1 1 0 0 -1 0 0 0 1 1 0.0000 -0.0000 0.0000

symlatt : the Bravais lattice is hP (primitive hexagonal)

symlatt : the Bravais lattice is hP (primitive hexagonal)
symspgr : spgroup= 156 P3 m 1 (=C3v^1)

2.2.From out file

DATASET 1 : space group P3 m 1 (#156); Bravais hP (primitive hexag.)
================================================================================
Values of the parameters that define the memory need for DATASET 1.
intxc = 0 ionmov = 0 iscf = 7 lmnmax = 4
lnmax = 4 mgfft = 256 mpssoang = 2 mqgrid = 3001
natom = 4 nloc_mem = 1 nspden = 1 nspinor = 1
nsppol = 1 nsym = 6 n1xccc = 2501 ntypat = 3
occopt = 3 xclevel = 2
- mband = 16 mffmem = 1 mkmem = 3
mpw = 24318 nfft = 518400 nkpt = 154
For the susceptibility and dielectric matrices, or tddft :
mgfft = 144 nbnd_in_blk= 6 nfft = 82944 npw = 137
================================================================================

2.3. Input positions

rprim 5.0000000000E-01 -8.6602540379E-01 0.0000000000E+00
5.0000000000E-01 8.6602540379E-01 0.0000000000E+00
0.0000000000E+00 0.0000000000E+00 1.0000000000E+00

xred -1.0652527319E-21 -1.8225602508E-21 -1.2332333074E-03
6.6666666667E-01 3.3333333333E-01 9.5793821882E-02
3.3333333333E-01 6.6666666667E-01 9.8107033068E-02
3.3333333333E-01 6.6666666667E-01 -5.5899469392E-02
3. Discussion
The first structure is described by space group #183 which is a hexagonal system, whereas the second structure is #156 which is trigonal. (Source: https://en.wikipedia.org/wiki/List_of_space_groups)
The corresponding symmetry elements are in 1.1 and 2.1. There are 12 for space group #183 and 6 for #156; everything is logical.
In both cases I took k-grid 40x40x1. And from 1.2 and 2.2 you can see that irreducible BZ contain 154 (nkpt=154) points. And these are the same points, the same 1/12 of the hexagon (triangle GAMMA-K-M-GAMMA)
But what I can't understand is how it can be that two space groups so different in symmetry can have the similar irreducible BZ?

If you need anything else, please let me know!

Best regards, Mikhail

Dear Mikhail,

Note that Abinit can take advantage of spatial symmetries as well as time-reversal symmetry.
Actually this is the default behaviour for k-point sampling (kptopt 1)
while kptopt 4 allows one to remove time-reversal symmetry when generating the IBZ.
See also https://docs.abinit.org/variables/basic/#kptopt

K and K' are connected by TR: K = -K' + G
so the eigenvalues at K and K' must be equal if TR is used in the calculation
even if there's no point group operation connecting the two points.