In the case of ISIF=3, different free energy
Posted: Fri Dec 02, 2011 7:11 am
I used ISIF=3 to relax stress and unit-cell. After the finish of the calculation, I calculated again with the final unit-cell and atomic positions in ISIF=0. Then, the final free energies are different each other. I don't know reason why. Which free energy is right ? I guess the difference comes from symmetry. But in both calculation, I used ISYM = 0. Could any master let me know the origin of the difference?
These are INPUTs in my calculation.
INPUT in first calculation
Startparameter for this Run:
INIWAV = 1
ISTART = 0
LREAL = .TRUE.
ENCUT = 286.567
Electronic Relaxation 1
EDIFF = 1E-4
ICHARG = 2
NELM = 200
Ionic Relaxation
NSW = 550
IBRION = 2
ISIF = 3
POTIM = 0.30
ISYM = 0 0-nonsym 1-usesym 2-fastsym
DOS related values:
ISMEAR = 0;
SIGMA = 0.07
.....................................
INPUT in second calculation
Startparameter for this Run:
INIWAV = 1
ISTART = 0
LREAL = .TRUE.
ENCUT = 286.567
Electronic Relaxation 1
EDIFF = 1E-4
ICHARG = 2
NELM = 200
Ionic Relaxation
NSW = 550
IBRION = 2
ISIF = 0
POTIM = 0.30
DOS related values:
ISMEAR = 0;
SIGMA = 0.07
~
These are INPUTs in my calculation.
INPUT in first calculation
Startparameter for this Run:
INIWAV = 1
ISTART = 0
LREAL = .TRUE.
ENCUT = 286.567
Electronic Relaxation 1
EDIFF = 1E-4
ICHARG = 2
NELM = 200
Ionic Relaxation
NSW = 550
IBRION = 2
ISIF = 3
POTIM = 0.30
ISYM = 0 0-nonsym 1-usesym 2-fastsym
DOS related values:
ISMEAR = 0;
SIGMA = 0.07
.....................................
INPUT in second calculation
Startparameter for this Run:
INIWAV = 1
ISTART = 0
LREAL = .TRUE.
ENCUT = 286.567
Electronic Relaxation 1
EDIFF = 1E-4
ICHARG = 2
NELM = 200
Ionic Relaxation
NSW = 550
IBRION = 2
ISIF = 0
POTIM = 0.30
DOS related values:
ISMEAR = 0;
SIGMA = 0.07
~