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Hi, everyone. I'm trying to calculate the energy of oxygen atom in a box, and wondering how I can choose the accurate one when it is conifned to have degeneracy or not.
Handson Session I by VASP forum showed that the POSCAR with symmetric unit cell would be more unstable and have the degenerated spin state(three level with occupancy of 0.333), whereas asymmetric unit cell would be more stable and posess the undegenerated spin state as followed.(And also coupled level is more unstable than that of unpaired level.)

[O in a box - un-degenerated / asymmetric unit cell]

spin component 1

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -25.2092 1.00000
2 -10.8834 1.00000
3 -10.8832 1.00000
4 -8.7840 1.00000
5 -0.3111 0.00000
6 0.3966 0.00000
7 0.5964 0.00000
8 0.6990 0.00000

spin component 2

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -21.5438 1.00000
2 -7.5296 1.00000
3 -6.4232 0.00000
4 -6.4230 0.00000
5 -0.1702 0.00000
6 0.5343 0.00000
7 0.6857 0.00000
8 0.8000 0.00000


[O in a box - degenerated / symmetric unit cell]

spin component 1

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -25.2080 1.00000
2 -10.2010 1.00000
3 -10.2010 1.00000
4 -10.2010 1.00000
5 -0.3033 0.00000
6 0.4148 0.00000
7 0.7143 0.00000
8 0.7143 0.00000

spin component 2

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -21.9597 1.00000
2 -7.1731 0.33333
3 -7.1731 0.33333
4 -7.1731 0.33333
5 -0.2192 0.00000
6 0.5245 0.00000
7 0.7505 0.00000
8 0.7505 0.00000

Which one would be more relevant if I want to use the oxygen energy for calculating binding energy of O/Ni system?





<span class='smallblacktext'>[ Edited ]</span>

I think that if you use a large enough box and you put your atom in that box, setting ISPIN = 2 and ISYM = 0 (off) usually results the lowest possible energy. however, i usually use these settings for adsorption on the the surfaces.

Thank you, Kamibz. But I have tried that setting already, ISPIN=2 and ISYM=0. The results are same as above. In shorts, which one is more relevant, energy with a high symmetric unit cell(cubic), or a lower symmetric unit cell(orthorhomibc)? And even a lower symmetric unit cell is more stable, but is the paired 2p level more unstable? Please show your opinion. Thanks.

I dont have enough knowledge about your answer. However, from what I ve learnt, for me it seems that different unit cells shouldn't change the energy of a gaseous system, if KBANDS are chosen highly enough.

As far as I remember Ni has a FCC unit cell, so you may use a cubic unit cell for Ads of O on Ni. however, I m not sure about this.

Kambiz

Hi,

as always in quantum mechanics a state with lower energy wins.
You have to use a non-cubic cell in the case of O atom (and O2 as well). Otherwise you are trapped by the symmetry of your system (like in your example #2). Example #1 is fine from the occupation point of view. The binding energy of O2 should be about -6 eV for e.g. PBE and the standard O PAW.

Cheers,

alex

Hi,

However, this is my opinion:

If you use a large-enough box and ISYM = 0 (off), the shape of the box does not affect the energy.

Because DFT is not suitable for isolated gaseous molecules, I always try to use high enough values of NBANDS and ENCUT.

Besides, when you find the lattice parameter for your FCC surface you may just add some space (vacuum) to one of the directions (z direction).

Let s say that you find that your lattice parameter is 8(A). it s reasonable to choose a cell with size of (8 * 8 * 15). But it s not reasonable to choose (10 * 8 * 15).

Cheers
Kambiz

I _strongly_ disagree with:
"Because DFT is not suitable for isolated gaseous molecules ..."

alex