My Community

Hi everyone,
I tried to find out energy of single Ir, Os, Pt, W, Re, and Ta atoms.
I use PBE. The results looks strange.
For example for Ir atom:
INCAR:
SYSTEM = Ir atom in a box
ISMEAR = 0 ! Gaussian smearing
SIGMA=0.1
ISPIN =2
MAGMOM=2
DIPOL = 0.0 0.0 0.0
IDIPOL = 4

POSCAR:
sc Ir
22
1.0 0.0 0.0
0.0 0.9 0.0
0.0 0.0 1.1
1
direct
0.0 0.0 0.0
I tried different size of box.

The POTCAR is:
TITEL = PAW_PBE Ir 06Sep2000

KPOINTS:
Gamma-point only
1 ! one k-point
rec ! in units of the reciprocal lattice vector
0 0 0 1 ! 3 coordinates and weight

The results are:
BOX-Size=12A: 1 F= -.16468923E+01 E0= -.16242881E+01 d E =-.452083E-01 mag= 3.0001
BOX-Size=15A: 1 F= -.15934184E+01 E0= -.15901338E+01 d E =-.656917E-02 mag= 3.0105
BOX-Size=18A: 1 F= -.16540231E+01 E0= -.16316841E+01 d E =-.446780E-01 mag= 3.0009
BOX-Size=20A: 1 F= -.13303948E+01 E0= -.12582512E+01 d E =-.144287E+00 mag= 3.0003
BOX-size=22A: 1 F= -.13093349E+01 E0= -.12708378E+01 d E =-.769941E-01 mag= 2.9508

Then I change SIGMA to 0.05

Results are:
BOX-size=12A: 1 F= -.16288060E+01 E0= -.16221862E+01 d E =-.132397E-01 mag= 2.9989
BOX-size=15A: 1 F= -.16331467E+01 E0= -.16264235E+01 d E =-.134464E-01 mag= 3.0002
BOX-size=18A: 1 F= -.16361334E+01 E0= -.16296340E+01 d E =-.129988E-01 mag= 3.0000
BOX-size=20A: 1 F= -.15298670E+01 E0= -.15167329E+01 d E =-.262683E-01 mag= 2.9999
BOX-size=22A: 1 F= -.13777921E+01 E0= -.13708755E+01 d E =-.138331E-01 mag= 2.7327

Would you please give me a hand, thanks?
Min

1) the box must not be chosen too large, usually 15Ã… is more than sufficient for a free atom.
2) please decrease SIGMA even further to obtain a sharp line spectrum of the atomic levels. (SIGMA=0.0001)
3) please also check in OUTCAR whether the occupancies of the levels are correct (for the free atom limit). Mind that the POTCARs are generated for the bulk electronic ground state configuration, which may differ form the free atom configuration for transition metals.
The online manual gives good hints how to calculate the atomic ground state (in the Examples section)

I changed INCAR to :
SYSTEM = Ag atom in a box
ISMEAR = 0 ! Gaussian smearing
SIGMA=0.0001
ISPIN =2
MAGMOM=3
DIPOL = 0.0 0.0 0.0
IDIPOL = 4
NELM=500

Results are:
OSZICAR_10: 1 F= -.15463493E+01 E0= -.15463493E+01 d E =0.000000E+00 mag= 3.0000
OSZICAR_12: 1 F= -.14600465E+01 E0= -.14600465E+01 d E =-.104916E-09 mag= 3.0000
OSZICAR_15: 1 F= -.14577524E+01 E0= -.14577516E+01 d E =-.164303E-05 mag= 3.0000
OSZICAR_18: 1 F= -.14648536E+01 E0= -.14648536E+01 d E =-.155658E-39 mag= 3.0000
OSZICAR_20: 1 F= -.14601335E+01 E0= -.14601295E+01 d E =-.807779E-05 mag= 3.0000

Occupancies in OUTCAR for box-size =10A are:
k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -7.8672 1.00000
2 -7.4236 1.00000
3 -7.3564 1.00000
4 -6.8408 1.00000
5 -6.6418 1.00000
6 -6.0699 1.00000
7 -1.2013 0.00000
8 -1.0473 0.00000
9 -0.8277 0.00000
10 -0.1424 0.00000

spin component 2

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -5.8677 1.00000
2 -5.5849 1.00000
3 -5.1982 1.00000
4 -5.0626 0.00000
5 -5.0277 0.00000
6 -4.9621 0.00000
7 -0.6544 0.00000
8 -0.4965 0.00000
9 -0.4892 0.00000
10 -0.0814 0.00000

Occupancies in OUTCAR for box-size =12A are:
spin component 1

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -7.7440 1.00000
2 -7.7430 1.00000
3 -7.4474 1.00000
4 -6.9959 1.00000
5 -6.7843 1.00000
6 -6.3635 1.00000
7 -1.2032 0.00000
8 -1.1218 0.00000
9 -1.0998 0.00000
10 -0.0639 0.00000

spin component 2

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -6.0122 1.00000
2 -5.7597 1.00000
3 -5.2781 1.00000
4 -5.2774 0.00000
5 -5.1557 0.00000
6 -5.1557 0.00000
7 -0.7348 0.00000
8 -0.7098 0.00000
9 -0.6692 0.00000
10 -0.0850 0.00000

Occupancies in OUTCAR for box-size =15A are:

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -7.8022 1.00000
2 -7.8021 1.00000
3 -7.5003 1.00000
4 -7.0595 1.00000
5 -6.8432 1.00000
6 -6.4285 1.00000
7 -1.2707 0.00000
8 -1.1987 0.00000
9 -1.1971 0.00000
10 -0.0619 0.00000

spin component 2

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -6.0755 1.00000
2 -5.8185 1.00000
3 -5.3391 0.99609
4 -5.3388 0.00391
5 -5.2150 0.00000
6 -5.2149 0.00000
7 -0.7978 0.00000
8 -0.7872 0.00000
9 -0.7841 0.00000
10 -0.0948 0.00000
Occupancies in OUTCAR for box-size =18A are:
k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -7.8308 1.00000
2 -7.8301 1.00000
3 -7.5274 1.00000
4 -7.0902 1.00000
5 -6.8719 1.00000
6 -6.4579 1.00000
7 -1.3102 0.00000
8 -1.2283 0.00000
9 -1.2282 0.00000
10 -0.0631 0.00000

spin component 2

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -6.1055 1.00000
2 -5.8472 1.00000
3 -5.3697 1.00000
4 -5.3673 0.00000
5 -5.2439 0.00000
6 -5.2438 0.00000
7 -0.8455 0.00000
8 -0.8168 0.00000
9 -0.8112 0.00000
10 -0.0970 0.00000


Questions:
1. why orbital energies correspond to different box-sizes are quite different?
2. why energies correspond to size 10A and 12 A are quite different?
3. Are these results reliable and which one should I pick?
Thanks a lot.
Min

as the energy zero is not fixed due to the periodicity, the absolute numbers of the KS-eigenvalues are essentially meaningless, because the energies are not taken with respect to a zero-level.
increasing the accuracy (PREC) and the basis set size (ENCUT) may give better convergence of the results with respect to delta (V). Again, please note that the boxes need not be chosen so large. As a hint, check the table in the online manual (chapter examples : Atoms)
Pick the results of the calculation with the set of parameters (basis functions,...) which matches the calculation you want to compare with best.