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### Re: the infinite frequency permittivity

Posted: Wed Jun 05, 2019 3:37 am
Dear Daniele,
1/eps_00^-1 which is different from eps_00
I understand non-diagonal in \epsilon_{G,G'} would also contribute into eps_00^-1 when the local field effect is included, and I added the local field effect before. No obvious difference in eps_{infty} is found. (see attachment)
Also the effect of electron temperature has been checked. the result is almost same.

b.t.w, have you read any work to determine \eps_{infty}} of metal accurately based on RPA+DFT ?

Best,
Weiqing

### Re: the infinite frequency permittivity

Posted: Wed Jun 05, 2019 8:38 am
Dear Weiking,
Also the effect of electron temperature has been checked. the result is almost same.
Sure, it make sense.
b.t.w, have you read any work to determine \eps_{infty}} of metal accurately based on RPA+DFT ?
No, but maybe the PhD thesis of Andrea Marini on optical properties of metals can give you some insight:
http://www.yambo-code.org/people/andrea ... thesis.pdf

Best,
Daniele

### Re: the infinite frequency permittivity

Posted: Wed Jun 05, 2019 1:48 pm
Dear Daniele,

I know the dielectric function eps^-1=1+vX. Can I split eps(q,\omega) into v and X, and obtain v(q,\omega) and X(q,\omega) as output. If I can't, can I know where is corresponding source code ? I can modify by myself.

Best,
Weiqing

### Re: the infinite frequency permittivity

Posted: Wed Jun 05, 2019 2:27 pm
Dear Weiqing

have a look to the file

src/pol_function/OPTICS_driver.F

at the line 222

Epsilon_ii(:,1)=1.0_SP-X_par(1)%blc(i_G_shift,i_G_shift,:)*4.0_SP*pi/Q_plus_G_sq_modulus

and line 240

Epsilon_ii(:,2)=1._SP/(X_par(1)%blc(i_G_shift,i_G_shift,:)+1._SP)

one is the independent particle and another the RPA one

best
Claudio

### Re: the infinite frequency permittivity

Posted: Fri Jun 07, 2019 4:05 am
Dear Claudio,

Many thanks for your guidance, and I have already obtained X and V as output.

If I don't misunderstand the code, I take X_par(1)%blc(i_G_shift,i_G_shift,:) and 4.0_SP*pi/Q_plus_G_sq_modulus as IP response function X(q,\omega) and Coulomb interaction V(q,\omega), respectively. Theoretically speaking, eps^-1 = 1 + xV. However, I found Yambo use eps = 1 - xV to calculate the independent particle permittivity.
Q1: I cannot see why these two formulas are equivalent with independent particle approximation.
As for RPA permittivity, Yambo use eps = 1/( xV + 1 )

if (Macroscopic_CPU) Epsilon_ii(:,2)=1._SP/(X_par(1)%blc(i_G_shift,i_G_shift,:)*
bare_qpg(iq,i_G_shift)**2/Q_plus_G_sq_modulus+1._SP)

Q2: V(q) should be 4\pi/(|q+G|^2). what is bare_qpg(iq,i_G_shift)**2 ?

Best,
Weiqing

### Re: the infinite frequency permittivity

Posted: Fri Jun 07, 2019 9:27 am
Dear Weiqing,

1) in RPA we have eps_M macroscopic = 1/eps_00^-1
eps^-1= 1+vX for each component G,G' where X is calculated via a Dyson equation X=Xo(1-vX0)^-1. matrix expression.
In IP it is a scalar quantity so you have eps^-1=1+vX0/1-vX0=1/(1-vX0) so eps=1-vX0

2) In the latest version of the code, the lines you mention should be commented and you should have:
if (Macroscopic_CPU) Epsilon_ii(:,2)=1._SP/(X_par(1)%blc(i_G_shift,i_G_shift,:)+1._SP)

Here in X_par it is alrady contained the symmetrized Coulomb potential: you can find it in X_redux.F
BUFFER%blc(ig1,ig2,iw_par)=BUFFER%blc(ig1,ig2,iw_par)*4._SP*pi/bare_qpg(iq,ig1)/bare_qpg(iq,ig2)

which is for each GG' component: X*4*pi/|q+G||q+G'|

when dealing with the macroscopic part it reduces to:
X*4*pi/q^2

In any case, the lines you were mentioned were related to the fact that the vector bare_qpg is modified when dealing with Coulomb truncation cutoff technique and the X was multiplied and divided for that quantities to recover the correct divergence of the Coulomb potential. If you are dealing with bulk you do not need to worry about that as in this case you are multiplying and dividing for the same quantity.

Best,
Daniele