[Pw_forum] Run QE with space_group and "ATOMIC_POSITIONS crystal_sg"

[Tsung-Lung Li]

Dear QE Friends:

An excerpt of my QE input is appended by the end of this message.

I ran QE with "space_group = 15" and "ATOMIC_POSITIONS crystal_sg"
to do 'relax' calculation.

It seems to me QE does the following.

1. Initially, QE uses space_group and the 7 inequivalent atoms in
    ATOMIC_POSITIONS to produce 24 atoms in the crystal unit cell.
2. In each subsequent geometric step of the 'relax' calculation,
    the 24 atoms are moved independently according to the forces
    acting on each atom.

Therefore, at the end of the 'relax' calculation, the space_group=15
symmetry is not always preserved.

My question is that

Can QE only move the 7 inequivalent atoms and use space_group symmetry
to construct the other 17 atoms for next geometric step?

This way, the space_group symmetry will be preserved at the end of
the simulation.

Thank you in advance.

Best Regards,
Tsung-Lung Li

------------------------------------------------------------------
&control
   calculation = 'relax'
   restart_mode = 'from_scratch'
/
&system
   space_group = 15
   celldm(1) = 11.86181088389489 ! bohr
   celldm(2) = 2.41851202803887
   celldm(3) = 0.903616377250279
   celldm(5) = -0.408489773780522
   ntyp = 4
   nat = 7
/

ATOMIC_SPECIES
    Ca   40.0780000000  Ca_pbe_v1.uspp.F.UPF
     S   32.0650000000  S_pbe_v1.2.uspp.F.UPF
     O   15.9994000000  O.pbe-n-kjpaw_psl.0.1.UPF
     H    1.0079000000  H.pbe-rrkjus_psl.0.1.UPF

ATOMIC_POSITIONS crystal_sg
   Ca  0.00000   0.17050   0.25000
   S   0.00000   0.32727   0.75000
   O   0.08319   0.27218   0.59103
   O   0.19997   0.38195   0.91298
   O  -0.20823   0.06826  -0.07831
   H  -0.25800   0.08700  -0.23400
   H  -0.24400   0.02000  -0.07700