The ABINIT Users Mailing List ( CLOSED )

["Chun Li " <chun.li@mf.mpg.de>]

Dear Marek Veithen,

Thanks a lot for you kind help. I encountered one more question about
the polarization calculation using Berry phase method in abinit.

I wonder if the positive or negative of the obtained polarization
indicates the real polarization direction of the model. The reason why I
ask this is: according to the routine of berry phase, if the absolute
value of the obtained total phase is larger than 1, even a little
bigger, the code will give a positive value (suppose the value of the
total berry is negative), like below:

Summary of the results
 Electronic Berry phase    -8.483634287E-01
            Ionic phase    -1.531124789E-01
            Total phase    -1.001475908E+00
    Remapping in [-1,1]     9.985240924E-01

           Polarization     3.101809093E-02 (a.u. of charge)/bohr^2
           Polarization     1.774692648E+00 C/m^2

But it is very possible that the absolute value of the obtained total
phase is a little smaller than 1 (for example, change the dense of the
k-point). Then one will get the completely different result (I mean the
different directions), like below:

Summary of the results
 Electronic Berry phase    -8.350116714E-01
            Ionic phase    -1.531124789E-01
            Total phase    -9.881241503E-01
    Remapping in [-1,1]    -9.881241503E-01

           Polarization    -3.069502777E-02 (a.u. of charge)/bohr^2
           Polarization    -1.756208667E+00 C/m^2

I think the polarization of the same cell in the same direction should
be a certain value, at least its direction. So do you have some ideas
about this problem? Thank you.

Best regards.

Chun Li

-----Original Message-----
From: Marek Veithen [mailto:Marek.Veithen@ulg.ac.be] 
Sent: Tuesday, November 09, 2004 2:58 PM
To: forum@abinit.org
Subject: Re: [abinit-forum] about berry phase

Dear Chun Li,

>1) What is the indeed difference between berryopt=1 and -1? Which is
>better?
>  
>
berryopt = 1 only works for kptopt = 2 or 3 while berryopt = -1 also
works
for kptopt = 1. The option berryopt = -1 is more efficient because it is
sufficient to sample the irreductible part of the Brillouin zone.

>2) Why I obtained much different polarizations for different berryopt?
>For example, the polarizations of zincblende structure are
>1.642499562E-01 C/m^2 and 8.674776685E-01 C/m^2 for berryopt=1 and -1
>respectively.
>  
>
In your input file, you use bdberry  1 24. That means, you compute
the Berry phase for 24 bands while the number of occupied bands
of ZnO is equal to 9. I modified your input file and I added a few
comments
(look for "MV"). You will see that berryopt = 1 and -1 provide exactly
the
same result.

Best regards,
Marek

-- 
Marek VEITHEN
Universite de Liege
Institut de Physique, Bat. B5
Allee du 6 aout, 17
B- 4000 Sart Tilman
BELGIUM
 

Phone : ++(32) (0)4-366.36.12
Fax   : ++(32) (0)4-366.29.90
E-mail: Marek.Veithen@ulg.ac.be