11
$\begingroup$

How to generate or create the correct KPOINTS ( K-PATH) file for a 4x4x1 supercell bandstructure calculation ? Is there a software compatible with vasp can do that ?

$\endgroup$
2
  • $\begingroup$ +1 but can you clarify the question? Do you have a k-point path for a primitive cell and want to convert it to the corresponding path for a 4x4x1 supercell? $\endgroup$
    – ProfM
    Jan 19, 2021 at 19:52
  • $\begingroup$ Yes, that's it brother. $\endgroup$
    – Chi Kou
    Jan 20, 2021 at 16:14

2 Answers 2

7
$\begingroup$

Is there a software compatible with VASP that can do that?

You can use VASPKIT tool, which can read your POSCAR to generate a k-path for the band structure plot. [vaspkit$\rightarrow$3$\rightarrow$302].

How to generate or create the correct KPOINTS ( K-PATH) file for a 4x4x1 supercell bandstructure calculation?

Except for the VASPKIT tool, you can use this online k-path generator (Of course, you should tell it what your structure is):

$\endgroup$
6
  • $\begingroup$ What confused me is that when I use VASPKIT with 4x4x1 supercell it gives me a PRIMCELL.vasp that is the unit cell, and asks me to replace my orginal 4x4x1 supercell with it. But when I use I doped 4x4x1 supercell structure it keeps it 4x4x1 supercell. $\endgroup$
    – Chi Kou
    Jan 21, 2021 at 6:59
  • $\begingroup$ @ChiKou First of all, your structure is two-dimensional? Secondly, do you want to generate a k-path for band structure calculations? $\endgroup$
    – Jack
    Jan 21, 2021 at 7:18
  • $\begingroup$ Yes, it is 2D and I want to generate a k-path for band structure calculations for the supercell not the unit cell. $\endgroup$
    – Chi Kou
    Jan 21, 2021 at 7:35
  • 1
    $\begingroup$ @ChiKou Because you are considering a 4x4x1 supercell, the high symmetry point will keep unchanged, which means you can use the k-path of the primitive cell to plot the band structure. For example, the 4x4x1 supercell of 2H-phase MoS2 is still $\Gamma-M-K-\Gamma$ $\endgroup$
    – Jack
    Jan 21, 2021 at 8:12
  • $\begingroup$ So you mean I keep my 4x4x1 supercell structure, and use the same k-pash of the unit cell right ? What about if I use a doped structure would it be the same ? $\endgroup$
    – Chi Kou
    Jan 21, 2021 at 8:26
9
$\begingroup$

Imagine that we have a supercell that is related to a primitive cell by the following supercell matrix:

$$ \begin{pmatrix}\tag{1} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \\ \end{pmatrix}. $$

Consider a $\mathbf{k}$-point with coordinates $(k_{p_1},k_{p_2},k_{p_3})$ in terms of primitive cell reciprocal lattice vectors. Then the same $\mathbf{k}$-point in terms of the supercell reciprocal lattice vectors has the following coordinates:

$$ \begin{pmatrix}\tag{2} k_{s_1} \\ k_{s_2} \\ k_{s_3} \end{pmatrix}= \begin{pmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \\ \end{pmatrix} \begin{pmatrix} k_{p_1} \\ k_{p_2} \\ k_{p_3} \end{pmatrix}. $$

Therefore, if you want to have the same path as you had in the primitive cell, you need to apply the correct supercell matrix to the $\mathbf{k}$-points in the original path. In your case, the supercell matrix is:

$$ \begin{pmatrix}\tag{3} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \\ \end{pmatrix} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}, $$ so all you need to do is to multiply the first two coordinates of every $\mathbf{k}$-point by $4$, and leave the third one unchanged. From a more conceptual point of view, when you build a supercell in real space, then the Brillouin zone in reciprocal space is correspondingly smaller. This means that in terms of the reciprocal lattice vectors of the supercell Brillouin zone, you need to travel further to get to the same point.

The resulting band structure will now cover the same path as that covered by the original primitive cell band structure. You will get the original bands you had and, due to band folding in the supercell, you will get a large number of additional bands.

$\endgroup$
9
  • 2
    $\begingroup$ +1 as always, and welcome back after so long! I've just added equation numbers since, even if there's no reference to specific equations in your answer, if someone else is citing this answer for something, they can now refer to "Eq. 2 of [...]" instead of "the second equation in [...]". $\endgroup$ Jan 20, 2021 at 18:57
  • 2
    $\begingroup$ Thanks for this! It's been a busy few weeks... $\endgroup$
    – ProfM
    Jan 20, 2021 at 21:45
  • 1
    $\begingroup$ @ChiKou when I use VASP I typically only give the high symmetry points of the path as then VASP adds the extra points in-between. So the transformation can almost be done manually or with a small script. I imagine that there is some software that automates it. $\endgroup$
    – ProfM
    Jan 21, 2021 at 11:34
  • 1
    $\begingroup$ @ParmeetSinghEP066 imagine you have the point $(1/2,0,0)$, with coordinates given with respect to the primitive BZ. As you correctly say, the BZ corresponding to the supercell will be smaller. Therefore, to get to the same point, the coordinates with respect to the supercell BZ will have to be larger, as they are referenced agains a smaller "unit". If the real space supercell is $4$ times longer in that particular direction, then the reciprocal space supercell BZ is $4$ times shorter in that direction, so the same point now has coordinates $(2,0,0)$. $\endgroup$
    – ProfM
    Jun 14 at 7:51
  • 1
    $\begingroup$ @ParmeetSinghEP066 what will happen when you calculate the band structure for a supercell is that you get band folding. For example, if you double the cell size in one direction, then you'll get twice as many bands at each $\mathbf{k}$-point. If you stayed within the BZ of the supercell, you would cover a smaller region of reciprocal space compared to the BZ of the primitive cell, but you would still have the same band information with the caveat that it would be folded in. An option when that happens is to use some "band unfolding" method. I hope this helps! $\endgroup$
    – ProfM
    Jun 15 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.