Real part of the dielectric function

[Zafer Kandemir]

Dear Developers,
Why is the real part of the dielectric function 1.0? How can I get the real part of the dielectric function?

Best regards.

I attached the output files of o.eps_q1_diago_bse for yambo 4.5 and 4.4 versions here

[Daniele Varsano]

Dear Zafer,
this is because you are using a coulomb cutoff potential, so I presume you are dealing with a reduced dimentsional system.
In this case, the macroscopic epsilon is not well defined.
You can have a look at the FAQ, pointing to this thread

Best,
Daniele

[Zafer Kandemir]

Dear Daniele,

Thank you for replying. The problem is solved with CutGEO="none"

Code: Select all

#
#    E/ev[1]      EPS-Im[2]    EPS-Re[3]    EPSo-Im[4]   EPSo-Re[5]
#
    0.000000     0.000000     2.485797     0.000000     1.968009
   0.6006E-2    0.1662E-3     2.486       0.4159E-4     1.968   
   0.1201E-1    0.3324E-3     2.486       0.8318E-4     1.968   
   0.1802E-1    0.4987E-3     2.486       0.1248E-3     1.968   
   0.2402E-1    0.6650E-3     2.486       0.1664E-3     1.968   
   0.3003E-1    0.8314E-3     2.486       0.2080E-3     1.968   
   0.3604E-1    0.9979E-3     2.486       0.2496E-3     1.968   
   0.4204E-1    0.1164E-2     2.486       0.2912E-3     1.968 

Best regards.
Zafer

[Daniele Varsano]

Dear Zafer,
yes, note now you have the epsilon of a 3D system, that depends on the volume of the supercell (or the vacuum region, or interlayer distance if you are dealing with a 2D system) and goes to 1 in the limit on infinite supercell volume . You can anyway extract the alpha_2D, see e.g. Eq.16 of Cudazzo et al. PHYSICAL REVIEW B 84, 085406 (2011).

Best,
Daniele

[Zafer Kandemir]

Dear Daniele,

Thank you for replying. I saw this paper. I obtained the real part of the dielectric function by using the alpha data. The eps1(real part) values are exactly the same. However, the y-axes of the eps2(imaginary part) were at different scales. Why did that happen?
What are the units of eps1(real part), eps2(imaginary part), and alpha?

Best regards.
Zafer

[Daniele Varsano]

Dear Zafer,

However, the y-axes of the eps2(imaginary part) were at different scales. Why did that happen?

This is explained in the FAQ. Eps it is not well defined in 2D, and it its imaginary part goes to zero for infinite volume supercell, and it is what happens when using the coulomb cutoff potential.

What are the units of eps1(real part), eps2(imaginary part), and alpha?

eps has not units, and can be found in textbooks.

Daniele