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- From: Marc Torrent <marc.torrent@cea.fr>
- To: forum@abinit.org
- Subject: Re: [abinit-forum] PAW-Problem, Al
- Date: Wed, 25 Apr 2007 14:09:32 +0200
Dear Pio,
1) Concerning the convergence behaviour with respect to ecut:
It seems that the results you get are the same as those published (for pwpaw code) at http://www.wfu.edu/%7Enatalie/papers/pwpaw/periodictable/atoms/Al/Al_dat,
provided that you use the following translation formula:
ecut_abinit=gcut_pwpaw**2/2
So, this atomic data file, created by PWPAW group seems to have a slow convergence vs ecut; it could be improved...
2) Concerning the energy difference:
a) did you try with a norm-conserving pseudopotential ? is the behaviour the same in that case...
b) did you have a look at the decomposition of energy (at the end of each datasry in output file) ? you could perhaps identify which term of energy is responsible for the non-zero value...
Marc Torrent
CEA-Bruyeres-la-Chatel
France
Pio Baettig a écrit :
Dear Abinit-users,
I have a small problem using PAW implemented in Abinit.
I downloaded the Al-PAWs (LDA and GGA) from
http://www.wfu.edu/%7Enatalie/papers/pwpaw/periodictable/periodictable.html
and started doing convergence tests with respect to number of K-points and
the cutoff-energy. I am using Abinit 5.3.4
The first thing I noticed, is that the convergence with respect to the
energy-cutoff is quite slow, for Al around 35 Ha are needed to converge the
energy to one mHa. (8*8*8 k-points seem also to be sufficient.)
I tried the same calculations for a cubic unit cell
(a=4.05Angst, pos. 0.0 0.0 0.0;0.0 0.5 0.5;0.5 0.0 0.5;0.5 0.5;0.0 , nband=40)
and for a rhombohedral unit cell
(a=4.05/SQRT(2), alpha=60 deg, pos:0.0 0.0 0.0, nband=10)
and then took the difference of
E(cubic)-(4*E(rhomb)) which should converge to zero quite fast. (even though
the number of k-points in the irreducible Brillouin-zone is not the same
for the same mesh in the full BZ, the convergence should still be good),
but the energy-difference converges to about 0.28 Ha, which I interpret,
that I cannot compare energies of different structures.
I did not yet repeat the same calculations with another element in order to
check whether the problem is with the Al-PAWs or with the method. I hope not
to have forgotten something important in the inputs.
Please find attached the inputs used for the cubic and the rhombohedral case.
I also attach a table with the energies. Lines are for the different k-point-
meshes, columns for the different cutoff energies.
First comes LDA, then GGA,
for each I put a table with the energies of the cubic unit cell, then the
rhombohedral unit cell, then the difference (c-4r)
If someone has some insight into the behaviour, I would be very grateful
Best regards,
Pio Baettig
- how can i define the symmetry ?, taoj0003, 04/25/2007
- Re: [abinit-forum] how can i define the symmetry ?, Masayoshi Mikami, 04/25/2007
- PAW-Problem, Al, Pio Baettig, 04/25/2007
- Re: [abinit-forum] PAW-Problem, Al, Marc Torrent, 04/25/2007
- Re: [abinit-forum] PAW-Problem, Al, Pio Baettig, 04/26/2007
- Re: [abinit-forum] PAW-Problem, Al, Natalie Holzwarth, 04/26/2007
- Re: [abinit-forum] PAW-Problem, Al, Marc Torrent, 04/26/2007
- Re: [abinit-forum] PAW-Problem, Al, Pio Baettig, 04/26/2007
- Re: [abinit-forum] PAW-Problem, Al, Josef Zwanziger, 04/25/2007
- Re: [abinit-forum] PAW-Problem, Al, Marc Torrent, 04/25/2007
- PAW-Problem, Al, Pio Baettig, 04/25/2007
- Re: [abinit-forum] how can i define the symmetry ?, Masayoshi Mikami, 04/25/2007
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