Yambo Community Forum

Hi,

I am working on calculating the dielectric function ε(q,ω) for bulk gold and encountered some confusion regarding the maximum wavevector values obtained from QE and Yambo.

System setup:

Supercell size: 3×3×3 (27 atoms)

No symmetry applied, so IBZ = BZ (same number of q-points)

K-point grid in QE:

K_POINTS automatic
19 19 19 0 0 0


Total q-points: 6859

Observation:

From QE (in Cartesian coordinates, cc units), the maximum |q| is 0.4272 1/a.u.

After running Yambo and converting q-points using ypp (iku → cc), I notice that some q-points (around indices 40–50 q of 6859 q) differ totally. In Yambo, the maximum |q| is 0.2941 1/a.u.

I am indicating the the q-points in which ε(q,ω) is printed.

Question:
Why does the maximum wavevector appear smaller in Yambo (~0.2941) compared to QE (~0.4272)? I was expecting the q-point ranges to remain consistent.

I would be grateful if anyone could clarify this behavior. Please let me know if you need further details about my setup.

Thank you in advance for your help!

Best regards,
M J Hasan
PhD student, Mechanical Engineering
University of Maine

Dear M J Hasan,

can you post in attachment the relevant files? QE output and yambo report?
Best,
Daniele

Hi Professor,

I have uploaded the necessary documents for your kind perusal.

The maximum magnitude comes from 2081-th q point in case of yambo. Thank you

Dear M J Hasan,

the ordering between Yambo and QE can be different as Yambo reorder the K/q points, so you should not compare the same indexes.
Looking at your files at least the first points seems to be ok.

Best,
Daniele

Hi Professor,

Thank you for your comment. I understand that the index can be different. However, when I calculate the maximum magnitude of the q-point, Yambo gives me 0.294055 (a.u), while QE gives me 0.427201 (a.u). This time, I have attached the full set of q-vectors with their magnitudes and also included a comparison in the Excel sheet. So that it is easy for you to look at it. Could you please check it?

Thank you.

Best regards,
M J Hasan
PhD student, Mechanical Engineering
University of Maine

Dear Hasan,

there should be some issue in the conversion (maybe in the ypp step?), as I can see:

Au.nscf QE number of points 6859
r_setup Yambo number of points 6859

Bz dimension:

QE in alat unit:
b(1) = ( -0.707107 -0.707107 0.707107 )
b(2) = ( 0.707107 0.707107 0.707107 )
b(3) = ( -0.707107 0.707107 -0.707107 )

in cc units: 0.707107*2Pi/alat=0.2719239


Yambo
Recip. lattice vectors: B[ 1 ] B[ 2 ] B[ 3 ]
B[ 1 ] -0.271924 -0.271924 0.271924 [cc]
B[ 2 ]: 0.271924 0.271924 0.271924 [cc]
B[ 3 ]: -0.271924 0.271924 -0.271924 [cc]

Being the sampling uniform, the two have to be the same.

Best,

Daniele

Hi Professor,

I have noticed that if I add or subtract a G vector to the q point in Yambo, I can obtain the corresponding q point in QE for those cases where they don’t directly match.

1. Can we say that q and q + G are physically the same vector?


2. Is it possible to generate additional q vectors using Yambo? For example, I currently have 6859 points along with the SAVE and r-setup files. By not running QE again, can I obtain more q points (e.g., 8000 points) through Yambo? I do not need the dielectric function for those q points; I only need the values of the q points themselves.

Thank you.

Best
M J Hasan
PhD Student, Mechanical Engineering
University of Maine

Dear M J Hasan,


1. Of course, q and q+G are not the same vectors. Instead, if you have a q outside the BZ it can be expressed as q=q_bz+G.

2. Unfortunately it is not possible as q points as defined as q=k-k' so you will need to run a nscf calculation with a larger grid.

Best,

Daniele