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TDDFT: a question on the kernel


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  • From: "Davide Sangalli" <davide.sangalli@gmail.com>
  • To: forum@abinit.org
  • Subject: TDDFT: a question on the kernel
  • Date: Mon, 26 Mar 2007 17:20:44 +0200
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It's the fist time I'm using the forum and I'm not sure that this
is exactly a question to be posted here.
If not sorry...anyway:

according to what is written in Abinit (rhohxc subroutine)
the K_xc kernel should be computed in this way:
________________________________________________________________________
  kxc(nfft,nkxc)=exchange and correlation kernel
(returned only if nkxc/=0 and abs(option)=2 )
allowed if LDAs ixc=0...9 :
if nspden==1 and option==2 : return kxc(:,1)= d2Exc/drho2
that is 1/2 ( d2Exc/drho_up drho_up + d2Exc/drho_up drho_dn )
if nspden==1 and option=-2 : also return kxc(:,2)= d2Exc/drho_up drho_dn
if nspden==2, return kxc(:,1)=d2Exc/drho_up drho_up
kxc(:,2)=d2Exc/drho_up drho_dn
kxc(:,3)=d2Exc/drho_dn drho_dn
________________________________________________________________________
Now I was trying to understand how
"K^{xc}_{up,up}=\frac{d^2 E_{xc}}{d\rho_{up} d\rho_{up}}"
is different form
"K^{xc}_{up,dn}=
\frac{d^2 E_{xc}{d\rho_{up} d\rho_{dn}}"
when a non polaryzed system is studied (like N_2 for exemple)
within LDA. (So nspden==1)

Now according to me as in LDA the energy is just a functional of the density
"E_{xc}=E_{xc}[\rho]=E_{xc}[\rho_{up}+\rho_{dn}]"
if I try to evaluate the second order derivatives written using the rule
"\frac{d f(y(x))}{d x}=\frac{d f(y)}{d y}\frac{d y(x)}{d x}"
and the fact that
"\frac{d \rho}{d \rho_{up}}=
\frac{d \rho}{d \rho_{dn}}=1"
I got
"K^{xc}_{up,up}=K^{xc}_{up,dn}=\frac{d^2 E_{xc}{d\rho^2}

So where is it that I'm doing a mistake?

Davide




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