Yambo Community Forum

Dear Daniele:
Thank you very much for your continuous help. Over the years using yambo, I have gained a lot of knowledge from your and other developers' replies, which makes me very grateful！
1. I want to ensure that:Even if I adopt the truncated coulomb potential and only that redefine the response function,is it true? And the imaginary part of the dielectric function o * exc * eps still satisfies this equation: Eq.(23) in the Yambo's articlehttps://www.sciencedirect.com/science/a ... 5509000472，is it true?

2. I read the "Dielectric screening in two-dimensional insulators: Implications for excitonic and impurity states in graphane":https://journals.aps.org/prb/abstract/1 ... .84.085406,I could see that:
\epsilon(q)=1+2\pi*\alpha_2D|q|,which is the Eq.6 in that article，so the difference between the imaginary part of the dielectric function and the imaginary part of the polarization function is only 2 \pi, but I cannot match the data when I substitute it

3. I know that the data of the imaginary part of the dielectric function is directly proportional to the data of the imaginary part of the polarizability function.
So I want to know what the exact expressions for the dielectric function and response function are, as well as the exact expressions for the polarizability and response function, which makes me very confused.

4.What do I need to multiply to obtain the imaginary part of the polarizability function from the imaginary part of the dielectric function？
Thanks for your help again!
Best wishes!
Quxiao

Dear Quxiao,

1) No, you have the truncated Coulomb potential instead of the 1/q^2 term.
2) In that equation there is a formal definition for eps in 2D, which is not the eps calculated by yambo, you can calculated it by using the alpha.
3/4) Eps and alpha are related by a constant factor that depends on the cutoff potential V_cut(q~0)/q^2. You do not have to look at eps when using the cutoff coulomb potential. The meaningful quantity is alpha and you can build the observables you are interested in using alpha.

Best,
Daniele

Dear Daniele：
Thank you very much for your help!
Combined with your previous help, I have resolved all my doubts.
Best wishes！
Quxiao

Dear Daniele：
I want to calculate dielectric function from the two-dimensional polarizabilby alpha.
According to the ref. https://doi.org/10.1039/D0TC01322F (J. Mater. Chem. C, 2020, 8, 8856), it shows effective dielectric tensor eps=1+4x(pi)x(alpha)/d (on page 2).
My question is :
1. Is this equation is reseanable for all 2D materials?
2. What is the unit of alpha calculated by yambo (data in the file xxx.alpha_q1_diago_bse) and d?
Best wishes,

Dear Yimin,
this is an effective dielectric function and it should be well defined for all 2D materials: see also: PHYSICAL REVIEW B 84, 085406 (2011)

The Yambo units are atomic units, if not explicitly stated differently in the output (d is in bohr as well as alpha_2d):

Best,
Daniele

Dear Daniele：
Thanks for your help.
Best,