Locating iku on BZ

[Daniele Varsano]

Dear Haseeb,
if time reversal holds, it is mentioned in the report file. After the symmetries list you will have something like:
[SYMs] Time-reversal derived K-space symmetries:

Having all the symmetries listed (including the time reversal related) you can rotate the k points in the negative quadrant to the one you considered in your band structure.
Best,
Daniele

[haseebphysics1]

Dear Daniele, thanks for the response!
Yes, time-reversal does hold in the system and total SYMs are also mentioned below.

Code: Select all

[05.02] Symmetries
==================

DL (S)ymmetries [cc]
[S1] 1.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 1.000000

[SYMs] Time-reversal derived K-space symmetries: 2 2
[SYMs] Spatial inversion 2 is NOT a symmetry
[SYMs] Group table built correctly

[05.03] RL shells
=================
...
...
...

[05.04] K-grid lattice
======================

Compatible Grid is 3D
B1 [rlu]= 0.000000 -0.500000 0.000000
B2 = 0.000000 0.000000 0.166667
B3 =-0.166667 0.000000 0.000000
Grid dimensions : 2 6 6
K lattice UC volume [au]:0.9788E-3

The other thing which I was worried about, in QE I did 6 2 6, k-mesh! So, why here it is swapping kz with kx (possibly)?

My task: Basically, I have found the k-points of my interest from Yambo in rlu (for the optical transitions from ypp), the only thing I want is to relate them on QE band structure. Since BSE was calculated on the regular MP k-grid while QE bands were using a high symmetry path! So, I'm finding it hard to find all the k-points + these negative values in k-points from Yambo!!

Can you suggest any solution to match the k-points? I want to avoid yambopy (also it is producing some errors)!

Thanks,

[Daniele Varsano]

Dear Haseeb,

1.

why here it is swapping kz with kx (possibly)?

Sounds strange, anyway you have the entire electronic structure (eigenvalues) reported in QE and Yambo for each k point, so you can easily recognise them.

2. Can you suggest any solution to match the k-points?

Converting all the k points (regular grid and selected path) in cartesian unit or rlu, rotate by symmetry the k points belonging to a different quadrant, next you can assign the k points coming from the regular grid belonging to the band structure path, of course there will be k points that do not belong to the path.

Best,
Daniele

[haseebphysics1]

Dear Daniele, thanks for the useful method!

rotate by symmetry the k points belonging to a different quadrant

just as an example of this:
if I have,
*X* K [6] : 0.000000 -0.500000 0.166667 (rlu)

from Yambo, then in my case, would it be like

0.166667 0.000000 0.500000 (rlu), (negative value in k was taken as +ve due to TR Symm.)

in the QE bands plot? (this might not be in the path as you suggested, but to look for it have I done the correct conversion?)

Best wishes and thanks,

[Daniele Varsano]

Dear Haseeb,
you have just the identity, and the TR.
In your example the k point transforms in:
0.000000 0.500000 -0.166667

If you do not have this point in the path you cannot use it.
...unless the symmetry you considered has been not recognised in QE, and you can verify by looking if the eigenvalues at a the two points are the same.

Best,
Daniele