Chksymbreak problem although cubic structure is symmetric
Posted: Tue Sep 17, 2019 11:10 pm
Dear all,
I have been using abinit to do ground state calculation for a nonspinpolarized situation. My structure is perfectly cubic symmetric. My kpoint grid in the input file is chosen as 6 6 6 which should be symmetric for a cubic symmetric system. But after running it shows me error.
--- !ERROR
src_file: symkpt.F90
src_line: 231
mpi_rank: 38
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 512 with components -6.250000E-02 -6.250000E-02 -6.250000E-02
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.
...
Please tell if there is any reason for this or I should always switch off the cheksymbreak.
Thank you,
Santu
I have been using abinit to do ground state calculation for a nonspinpolarized situation. My structure is perfectly cubic symmetric. My kpoint grid in the input file is chosen as 6 6 6 which should be symmetric for a cubic symmetric system. But after running it shows me error.
--- !ERROR
src_file: symkpt.F90
src_line: 231
mpi_rank: 38
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 512 with components -6.250000E-02 -6.250000E-02 -6.250000E-02
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.
...
Please tell if there is any reason for this or I should always switch off the cheksymbreak.
Thank you,
Santu