12

The y-axis corresponds to the direct eigenvalues from your DFT calculation. You need to set the valence band maximum to '0' manually. The DFT output of your calculation will give the highest occupied state energy (or the Fermi energy, if you use occupations smearing). So whatever that energy is, subtract that from the energies in your band plot. A note on ...


11

It is not necessary to care about the total energy versus the number of k points in irreducible reciprocal space, which is closely related to the symmetry of the investigated structure. Of course, you can always read it from IBZKPT with a self-consistent run or any other run using VASP. Also, the printed number in IBZKPT is related to the ISYM tag. I assume ...


11

If the crystal unit cell is in a format readable by ASE, then you can use code that looks approximately like so: from ase.io import read atoms = read("myfilename.xyz") bandpath = atoms.cell.bandpath() This bandpath object will have the relevant attributes to play with (kpoint coordinates, special point labels, special point coordinates, etc). ...


11

Is this calculated indirect bandgap at room temp. or at 0K? QE is based on the density functional theory (DFT). DFT is a ground-state (0K) theory and hence the calculated bandstructure is 0K. If it is at room temperature the indirect bandgap should be around 1.1 eV as from the literature. So is there any energy correction factor that has to be performed to ...


10

I would like to add a few clarifications to Jack's answer: Standard DFT calculations with fixed ionic positions are actually not even 0 K. A better way to describe them is that they are static lattice calculations. The difference between static lattice and 0 K is the contribution from quantum zero-point fluctuations. This contribution is generally small, ...


8

If you are really interested in learning how to generate the path, I strongly advice you to avoid using any automatic tool like suggested in previous answers. In the Wiki page entitle Brillouin zone you can find the first Brillouin zone for each one of the Bravais lattice. This Wiki is based on the following paper: Setyawan, Wahyu; Curtarolo, Stefano (2010)....


6

The reason why even number k-point grids are prefered in the case of some symmetries like FCC lattices is mainly due to the concept of the Irreducible Brillouin Zone (IBZ). Here is an example for FCC: Here the number of points in the k-point grid in the IBZ is the same for even as well as odd k-points (M value). Here the points in the IBZ determine the ...


4

Pymatgen can definitely do what you are asking for. You can find several algorithms under the pymatgen.symmetry.kpath module here. I include a minimal example below using the new method from Munro et al. implemented in Pymatgen: from pymatgen.core import Structure from pymatgen.symmetry.kpath import KPathLatimerMunro struc = Structure.from_file('myfilename....


3

Here are some possible changes to your input that may resolve the problem: Your EDIFF is very small; use the default and then increase precision from there. Use a smaller SIGMA, start with 0.15 or 0.1. Decrease the number of k-points, start with 5x5x5, then increase to reach the desired precision. Good luck!


3

The electronic band structure is a concept for periodic system. Heterostructures are not periodic system (in the direction of heterostructure grow). What is done is an approximation where the bulk's band of each bulk system is combined using the band off-sets. Some reference about how to calculate band off-sets are: Theoretical calculations of ...


3

I think I figured this out. For most practical purposes, I think it is fine to just choose $-\pi\leq k_x\leq \pi$ and $0\leq k_y\leq \pi$ (half-BZ, with exact ranges depending on the model). I think I over-complicated the authors' work. It is unlikely that the vortices/singularities will occur near the boundary of the chosen region. So, I can just take the ...


3

I've never done this before but here is my guess of how you can do it. If the numbers you posted are the unit cell axes vectors with a vector for each column then their direct matrices are. $$\tag{1}\mathbf{A} = \begin{bmatrix}3.31 & 0.00 & 0.00 \\ 0.00 & 10.47 & 0.00 \\ 0.00 & 0.00 & 4.37\end{bmatrix}...


1

I think Camps gives a mostly correct answer for this. I will say though, it may be possible in some cases to generate a band structure in the direction of the heterostructure by making a periodic bulk heterojunction. By layering two materials with minimal (or large if intended) strain in a bulk system with bulk crystal structure, you may get some ...


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