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Posted: **Wed Jun 20, 2012 11:10 pm**

Dear all,
I trying to use vasp to calculate f-dependent epsilon for a hexagonal system using both triogonal unit cell and hexagonal unit cell.

The INCAR options are:

Code: Select all

ENCUT=400
PREC=medium

ISMEAR=-5
SIGMA=0.01

LOPTICS=.TRUE.
CSHIFT=0.0001
NEDOS=9001

#NPAR=1
NSIM=1

with NBANDS = 2*NELECT

The primitive vectors for them are:

Code: Select all

Triagonal
4.37800000000000
?0.468582?0.270536?0.840963
?-0.468582??0.270536?0.840963
?0.000000?-0.541072?0.840963

Hexagonal

?4.10500000000000
?1.0000000000000000?0.0000000000000000?0.0000000000000000
?-0.5000000000000000?0.8660000000000000?0.0000000000000000
?0.0000000000000000?0.0000000000000000?2.6920000000000002

I have too questions:

1. Why did I get different result for triogonal and hexagonal unit cell? I would expect the same results (at least for z component).

2. Why does decreasing CSHIFT increase epsilon? In my understanding CSHIFT just change the smoothness of the curve.

Any suggestion is grateful.
Best,
Dat Do

[ Edited ]

Posted: **Thu Jun 21, 2012 6:55 pm**

Just to keep you updated:
1. I've found that the peaks in spectrum is similar for both trigonal and hexagonal unitcell but the magnitudes.
2. changing CSHIFT changes the imaginary part dramatically (magnitude) but the real part is not affected much.

Cheers
[ Edited Thu Jun 21 2012, 06:56PM ]

Posted: **Mon Jun 25, 2012 10:57 pm**

Hi everyone,
I've check the code and found out that when doing the Kramers-Kronig transformation, the integral

\epsilon1=1+2/pi P\int_0^{\infty}\epsilon(w')w'/(w'^2-w^2+i\nu)dw'

blows up at w'=w, and the value depends on \nu, i.e the less \nu the greater integral.

And in the code the imaginary part of epsilon is assigned as AIMAG(\epsilon1).

Can anyone explain that for me?
Thanks
[ Edited Mon Jun 25 2012, 11:03PM ]

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LOPTICS in hexagonal system & CSHIFT problem

LOPTICS in hexagonal system & CSHIFT problem

LOPTICS in hexagonal system & CSHIFT problem

LOPTICS in hexagonal system & CSHIFT problem