Kpt boundaries for cubic supercell

Running Abinit Other tools and postprocessing in ABINIT
1

Dear Users,

I am trying to calculate the EBS of a cubic perovskite supercell (2x2x2 cubic STO). The geometry information can be found on the MP here. I need the supercell to simulate a doped system.

I have followed the abinit documentation regarding the supercell of Hydrogen here and the unfolding of the Brillouin zone here, although I find it not complete.

I start from the standard k-path for a cubic structure in the primitive cell.

kptopt2        -5   
kptbounds2
            0.0   0.0   0.0  #  Gamma
            0.0   0.5   0.0  #  X
            0.5   0.5   0.0  #  M
            0.0   0.0   0.0  #  Gamma
            0.5   0.5   0.5  #  R
            0.0   0.5   0.0  #  X

I have tried several paths, such as the following, trying to include all the “half” points and making sure to substitute G->X with -X/2->G->X/2 and I get a bunch of spikes.

kptopt2    -9 
kptbounds2
           -0.25 -0.25  0.0  # -M/2
            0.0  -0.25  0.0  # -X/2
            0.0   0.0   0.0  #  Gamma
            0.0   0.25  0.0  #  X/2
            0.25  0.25  0.0  #  M/2
            0.0  -0.25  0.0  # -X/2
           -0.25 -0.25 -0.25 # -R/2
            0.0   0.0   0.0  #  Gamma
            0.25  0.25  0.25 #  R/2
            0.0   0.25  0.0  #  X/2

Could it be that this is still incomplete and I am missing points such as 0.25 0.5 0.0 and 0.5 0.25 0.0 for M and 0.5 0.25 0.25 for R (etc…?).

Is there any automatic tool to produce the kpt path?
unnamed (1)

Thank you

Marco

2

Hi,

You can use this website to help you find high-symmetry points and thus the path along which you want to go.

Did you compute the band structure of the unit cell? The band structure of the super cell should be the same in the reduced Brillouin zone, if you simply expand the cell. This is a good check to perform before relaxing the super cell. In this case, there are a lot of bands, so you do not recover the unit cell band structure.

Hope this help,
Best,
Olivier

3

I calculated the unit cell, and I agree with you that the supercell should give the same result. The unit cell gives the expected result:

image
image1370×1176 159 KB

I set more nbands in the supercell (about 8 times more), hence a large number of states.

Thank you
Marco

4

Hi,

It is normal to have more states indeed, but they should be folded. I think you see spikes because you take half the position of the points. You don’t have to do this, since when you double the unit cell, the reciprocal zone calculated will already be twice as small. Also, can you use the same number of points along the path? On the first figure you have 10 points, and 6 on the second.
Olivier