KPOINT effect in surface energy calculations

[ccccc]

Hi all,
I'm confused when calculating the surface energy of a surface.
I created a surface of 6-layer thickness from a supercell and relaxed it.

For 6-layer surface relaxation i used 6x6x1 kpoints, because i optimised lattice parameter with 6x6x6.

For surface energy calculation, as far as i know, i need to find the total energy of the whole supercell without vacuum.

For this calculation should i use 6x6x6 or 6x6x1 ?

I tried both of them, although it is an ideal continuous supercell with no vaccum, the results are different.
Here is my second question. Should i relax it or just do a single point calculation?

Many thanks in advance.

C.

[alex]

Hi ccccc,

calculating the surface energy is a complex thing. I'd suggest some reading:
V. Fiorentini and M. Methfessel, J. Phys.: Condens. Matter, 1996 (8) 6525.

The mesh for the supercell bulk needs adjustment to the lenght of c in your case. E.g. c is 3 times larger than the bulk, you'd need to reduce the 6 kpoints by a factor of 3 ending up with 2 kpoints. You can check both total energies, they should be exactly the same (well, if you multiply one by 3 in that example). Since this might work only in special cases, where an integer multiplier is possible, you'll end up reading the paper given above. :-)


Cheers,

alex

[ccccc]

Dear Alex

When consider those lattice paramaters as:
a=0.7, b=1, c=4.25

Kpoints must be 6x4x1 according to that you said.

But "1" here at the end of the kpoint set refers to a surface as mentioned in the manual. But my system is not a surface. It is a continuous bulk structure of a specially oriented surface.

[alex]

If supercell c is a multiple of bulk c it's fine. It's called Brillouin zone folding and takes at least part of the job of multiple k-points. Check a solid state physics book on that.

Hth,
alex

[ccccc]

I apologise for taking your time but it is not clear for me again. Let me to tell you what i did.

Iused a 12-atom cubic cell oriented in 001 direction just to optimize lattice parameter.

Then, i created a 110 oriented slab structure of 10 layers apart from this cubic cell whose c parameter is more bigger than the primary cell. To create a 110 surface, i deleted 5 layers. To calculate the surface energy, i relaxed both of 10 and 5 layer structures and got the total energies.

For 5-layer geometry, i used 6x4x1 as relevant with the a and b lattice parameter lenghts. For 10 layer structure (without vacuum) i used 6x6x6. If i use 6x4x1 for this; doesnt it serve as a surface although it is not? It doesnt make sense to me.

Thanks

[ccccc]

I mean that, for the surfaces we use kz=1 because for getting a two dimesional surface brillouin zone we only need 1 kpoint in teh c direction. But the supercell (which is used to form the surface) is a infinitive slab structure. So i am confused. Will 1 kpoint in c direction be really sufficient?

Thanks again

[alex]

Hi,

the questions, which are open, are as follows:
- how many atoms are in one 'layer' of the 12-atom cubic cell in the [001] direction?
- how many atoms are in one layer of the later slab?

You are welcome!

alex

[ccccc]

Hi Alex,

Do you mean that if i calculate the energy per layer in 12 atom cell, i can reach to the 110 continious structure's energy by multiplying it with the number of layers.

But;

This is a flourite AB2 structure. For 12 atom cell in the first layer there are 2 A atoms and in the second layer 4 B atoms.

In the 110 slab, in one layer there is 1 A and 2 B.

In this situation should i calculate the energy per 3 atoms (A+2B) in 12-atom cell than evaulate it by multiplying it with the layer number of 110 cell?

PS: Even if it is, this not a sufficient solution for me. Because i need that infinitive geometry to compare the displacement of the atoms in the new surface structure according to the original. As far as i know they had to be in the ideal bulk positions unless i creat a vacuum but when i did the calculation for 110 (1x1) infintive slabs for both of 6x6x1 and 6x6x6, there has been some little changes in the positions.

For 110 (2x2):

with 3x3x1, no change, the atoms stayed in the bulk positions
with 3x3x6, some little changes have been

May be i have to do a single point calculation instead of relaxing them. But there must not be changes even i relax these ideal bulk structures. Because thet are bulk!

So not just for this calculation i want to understand the logic of this.

Thank you.

[alex]

Hi there,

I think we have to start from scratch. I sense some confusion here and there.

a) surface slab: k-points perpendicular to the surface: just one, because no periodicty of matter here.
b) filled 'bulk' slab: (c is perpendicular to the surface)
k-points(c-direction of the bulk) / int(c_surface/c_bulk) would be appropriate, but this is rarely achieved, because in general the result of this little exercise is rarely a number. :-(
c) read V. Fiorentini and M. Methfessel, J. Phys.: Condens. Matter, 1996 (8) 6525. It's a great paper (and short! :-) ). It shows you how to calculate reliable surface energies and where the traps are. :O

And the logic behind k-points and doubling a cell parameter is simple: half k-points only. And the total energy is exactly(!) the same (don't forget to tighten your criteria if you are on the quest). Do it for once and convince yourself. It lights the famous bulb.

Cheers,

alex

[ccccc]

Thank you Alex for all help.
I will read this paper