Yambo Community Forum

Hi all,

Can I obtain the infinite frequency permittivity \epsilon_{\infty} from Yambo? It is worth noticing that \epsilon_{\infty} is not the permittivity at infinite frequency, which should be 1. Static function comes from two contributions from electrons and ions. The summation of these two should be the lattice static dielectric constant. The only contribution of electrons should be the infinite dielectric constant.

Best,
Weiqing

Dear Weiqing,
the epsilon yambo calculates provides the electronic contribution only as ionic part it is not taken into account.
Best,
Daniele

Dear Daniele,

Thanks for reply!

So, the first value of yambo output Re[\epsilon(\omega=0)] is \epsilon_{\infty} ? From experiment, \epsilon_{\infty} is estimated as ~ 7. But in Yambo, Re[\epsilon(\omega=0, q=0)] is ~ 300. Can yambo calculate static dielectric function of metal correctly?
Notice that : The problem is not about Drude term, \epsilon_{\infty} should be estimated without Drude term. and I use 80 Ry and 49*49*21 k-points grid in DFT, and I don't think the problem is about DFT part.

b.t.w. can I plot non-interacting response function χ(q,ω= 0) and coulomb interaction v(q) separately?

Best,
Weiqing

Dear Weiqing,

you can calculate the dielectric constant for different q using the QpntsRXd variable.
Yambo will produce different files, one for each q-point.
Then in columns 3 and 4, you can find the imaginary and real part of the non-interacting dielectric constant
that is related to the x(q,w=0)

best
Claudio

Dear Weiking,
let me add that I've the impression that Re[\epsilon(\omega=0)] for metals, even if omitting the Drude term will strongly depend on the q point sampling.
Best,
Daniele

Dear Daniele，

If I don't add DrudeWd, only interband transition would be taken into account even if the system is a metal? Is that right?

Best,
Weiqing

Yes this is right

Claudio

Dear Weiqing Zhou,
I would say no, if you do not include the Drude contribution you do not have the intraband contribution in the q->0 limit, but for finite q vectors intraband contributions of the metallic bands are included.

Daniele

Dear Daniele,

if you do not include the Drude contribution you do not have the intraband contribution in the q->0 limit, but for finite q vectors intraband contributions of the metallic bands are included.

I check the asymptotics of \epsilon_{infty} where \epsilon_{infty} should be \frac{1}{q^{2}} at q→0 in 3D. From the asymptotics of finite q (see fig in attachment), I can extrapolate the \epsilon_{infty} at q=0 is 230.

I doubt that intraband contributions of the metallic bands is the reason why \epsilon_{infty} calculated is much higher than that of experiment.
Can I eliminate all intraband contributions and obtain a "clean" interband dielectric function?

Weiqing

Dear Weiqing,

Can I eliminate all intraband contributions and obtain a "clean" interband dielectric function?

No this is not possible, what you can do is to reduce the electronic temperature to somehow reduce the metallicity but I doubt it will work.
Just a question, but are you looking at the IP reposnse funcion or the irreducible X:
X=X^0+X^0+vX
as in this case the macrsocopic dielectric function would be :
1/eps_00^-1 which is different from eps_00

In other words, have you tried to include local field effects?

Best,
Daniele