Dear colleagues,
I am puzzled because calculating the dielectric tensor with local fields and using different options of the GW routines, I obtain different values of the HEAD of the dielectric tensor. I had expected to obtain differences only in the inverse dielectric tensor, but not in the HEAD.
Using the LOPTICS approach I get these values
OUTCAR2.960: frequency dependent REAL DIELECTRIC FUNCTION (independent particle, no local field effects)
OUTCAR2.960- E(ev) X Y Z XY YZ ZX
OUTCAR2.960- --------------------------------------------------------------------------------------
OUTCAR2.960- 0.000000 4.436184 4.784793 4.469953 -0.000092 0.000068 -0.001236
Using
ALGO=CHI; LSPECTRAL = .FALSE.
LRPA=.TRUE.
ENCUTGW = 186.5
NOMEGA = 1
NKRED=2
the I get
OUTCAR4.960: HEAD OF MICROSCOPIC DIELECTRIC TENSOR (INDEPENDENT PARTICLE)
OUTCAR4.960- -------------------------------------
OUTCAR4.960- w= 0.000 0.000
OUTCAR4.960- 4.4355 0.0000 -0.0001 0.0000 -0.0012 0.0000
OUTCAR4.960- -0.0001 0.0000 4.7841 0.0000 0.0001 0.0000
OUTCAR4.960- -0.0012 0.0000 0.0001 0.0000 4.4693 0.0000
the above result is consistent with the calculation using LOPTICS.
However, when I set LRPA=.FALSE. , then I get different values
OUTCAR5.960: HEAD OF MICROSCOPIC DIELECTRIC TENSOR (INDEPENDENT PARTICLE)
OUTCAR5.960- -------------------------------------
OUTCAR5.960- w= 0.000 0.000
OUTCAR5.960- 4.8265 0.0000 -0.0001 0.0000 -0.0015 0.0000
OUTCAR5.960- -0.0001 0.0000 5.2302 0.0000 0.0001 0.0000
OUTCAR5.960- -0.0015 0.0000 0.0001 0.0000 4.8590 0.0000
The values of the above HEAD are different from the previous values.
What is the reason of this difference ?
I have noticed in the last calculation OUTCAR5.960, the following message
For the Nanoquanta kernel the diagonal of the BSE matrix causes
a conduction band shift by 0.491
this shift is now applied to the conduction band states
Is this shift responseible of the difference? Can I unset this shift ?
Thanks
Eduardo Menendez
Universidad de Chile
www.gnm.cl/emenendez
how to obtain the HEAD of the dielectric tensor
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- eariel
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how to obtain the HEAD of the dielectric tensor
Last edited by eariel on Fri Dec 20, 2013 8:42 pm, edited 1 time in total.
- eariel
- Newbie
- Posts: 16
- Joined: Sat Sep 13, 2008 6:45 pm
- License Nr.: 490
how to obtain the HEAD of the dielectric tensor
Hi,
Just to essay a self-answer, I guess it can be found in the slide presentation
http://www.vasp.at/mmars/day2.pdf
Following the above source (pages 3 and 4),
the HEAD of the dielectric tensor is computed from
epsilon(0,0)(q->0,omega=0), where
epsilon= 1 - (4 pi e^3 / q^2) P,
When LRPA=.TRUE., then P=X0 ,
where X0 is the irreducible polarizability computed from the Adler and Wiser formulae.
On the other hand, when LRPA=.FALSE., P is obtained from the Dyson equation
P = X0 + X0 fxc P, or equivalently
P=(1-X0 fxc)^-1 X0
hence, epsilon(0,0)(q->0) must give different values for both settings of LRPA.
I still have the doubt about the shift applied to the conduction band states.
Best regards
Eduardo
Just to essay a self-answer, I guess it can be found in the slide presentation
http://www.vasp.at/mmars/day2.pdf
Following the above source (pages 3 and 4),
the HEAD of the dielectric tensor is computed from
epsilon(0,0)(q->0,omega=0), where
epsilon= 1 - (4 pi e^3 / q^2) P,
When LRPA=.TRUE., then P=X0 ,
where X0 is the irreducible polarizability computed from the Adler and Wiser formulae.
On the other hand, when LRPA=.FALSE., P is obtained from the Dyson equation
P = X0 + X0 fxc P, or equivalently
P=(1-X0 fxc)^-1 X0
hence, epsilon(0,0)(q->0) must give different values for both settings of LRPA.
I still have the doubt about the shift applied to the conduction band states.
Best regards
Eduardo
Last edited by eariel on Tue Jan 07, 2014 1:10 pm, edited 1 time in total.