Chksymbreak=1 error though structure is symmetric
Posted: Fri Sep 13, 2019 7:21 pm
Dear all,
I am running a simple self-consistent ground-state calculation with FCC cubic lattice (original space group 227 but primitive cell is fcc). My structure is symmetric as it should be for 227 space group. But when I run the abinit it shows error,
--- !ERROR
src_file: symkpt.F90
src_line: 231
mpi_rank: 36
message: |
Chksymbreak=1 . It has been observed that the k point grid is not symmetric :
for the symmetry number 3
with symrec= 0 0 1 -1 -1 -1 1 0 0
the symmetric of the k point number 64 with components -1.250000E-01 -1.250000E-01 -1.250000E-01
does not belong to the k point grid.
Read the description of the input variable chksymbreak,
You might switch it to zero, or change your k point grid to one that is symmetric.
The ngkpt1 4 4 4 is set in the input which is correct as I have a cubic cell.
Then why I am getting this error?
Can anyone please help me with this?
Thanks in advance,
SantuB
I am running a simple self-consistent ground-state calculation with FCC cubic lattice (original space group 227 but primitive cell is fcc). My structure is symmetric as it should be for 227 space group. But when I run the abinit it shows error,
--- !ERROR
src_file: symkpt.F90
src_line: 231
mpi_rank: 36
message: |
Chksymbreak=1 . It has been observed that the k point grid is not symmetric :
for the symmetry number 3
with symrec= 0 0 1 -1 -1 -1 1 0 0
the symmetric of the k point number 64 with components -1.250000E-01 -1.250000E-01 -1.250000E-01
does not belong to the k point grid.
Read the description of the input variable chksymbreak,
You might switch it to zero, or change your k point grid to one that is symmetric.
The ngkpt1 4 4 4 is set in the input which is correct as I have a cubic cell.
Then why I am getting this error?
Can anyone please help me with this?
Thanks in advance,
SantuB