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[Fabien Bruneval <fabien.bruneval@polytechnique.fr>]

If the formula comes from my PhD thesis, it is likely to be correct. ;-)

What do you label k_1 and k_2 ? The Fourier transform of 2 positions 
quantity has only one k index (by translational symmetry). Do you want 
to calculate
A(k + G_1, k + G_2, omega) ?
where k is in the BZ and G_1 and G_2 are reciprocal lattice vectors?
In that case, it is going to be tough.

In the GW subroutine, one calculates immediately G_ii from Sigma_ii. We 
never store or evaluate the Green's function directly on the PW basis 
set. Try to imagine the number of elements this would represent...

On the contrary, the KS Green's function projected on KS wavefunctions 
is simply

G^{KS}_ii(omega) = 1 / ( omega - e^{KS}_i )

Much simpler, no?

I strongly suggest you to try to express the quantity you need in the 
basis set of KS wavefunctions...


Fabien

PS: if you want to take a look inside the source code, look at
src/11drive/sigma.F90
src/08seqpar/csigme.F90







mperez@mpi-halle.mpg.de wrote:
> Dear Fabien,
>
>  thanks for your reply to my question on spectral functions in the GW 
> implementation.
>
> According to the formula you wrote for the spectral function A_ii=Im(G_ii),
> you have to build first G(r1,r2,omega), from the FT of G(k1,k2,omega).
>
> that is the quantity i need in order to get A(k1,k2,omega)  !!!!
>
>
> Unless.., it is possible that you computed A_ii from S_ii (self-energy)
>                   ___
>            S_ii = \
>                   /__ii
>
> as
>                                         1
>                 A_ii = Im { __________________________________ }
>                              (omega -  epsilon_i) - S_ii(omega)           
>
>
> where in the code can i find then G(k1,k2,omega)?, (interacting and 
> Kohn-Sham, if available)
>
> thanks
>
> Manolo
>
>
> PD the formula is from your PhD thesis!