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Re: [abinit-forum] spacegroup-question, geometry optimization


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  • From: Dan Ricinschi <ricinski@semi.ee.es.osaka-u.ac.jp>
  • To: forum@abinit.org
  • Subject: Re: [abinit-forum] spacegroup-question, geometry optimization
  • Date: Mon, 02 Jul 2007 21:44:17 +0900

Hi!

I cannot confidently point out the cause of this behavior, but in my experience this happens when the optimization goes on for quite a large number of Broyden steps, without being able to find the absolute minimum of the total energy with the symmetry strictly enforced. It could also mean that the imposed symmetry is not that of the ground state, (although in many cases it could be possible to optimize the crystal under other desired symmetry). If you don't suspect a serious physical error, you could restart the optimization with the R3c structure obtained after adjusting those small deviations by hand and see what happens.

On the other hand, by looking at your atomic coordinates I think that it is likely that you have found a local minimum of the total energy surface, as the Bi atoms might be "fixed" to their centro-symmetrical positions. If you have not already done so, I suggest you to restart the optimization from scratch by displacing the Bi atoms slightly from their symmetrical positions, in the starting configuration. That is, use 0.01 0.01 0.01 and 0.51 0.51 0.51 instead of 0 0 0 and 0.5 0.5 0.5. In this way the algorithm might find the optimized structure easier, without breaking the symmetry for lengths and angles.

I hope this helps.

Dan Ricinschi

At 06:34 PM 7/2/2007, you wrote:
Dear abinit users,

I have found a small inconsistency when doing geometry optimizations of a rhombohedral (R3c) system.
The angles and vectors stay the same (as given by the space group) for about ten geometry iterations, and then one starts to diverge slightly.
Within the limits of the space group, they should be the same.
I guess, this is a problem of the floating-point precision?
Below the worst case, I encountered so far, the first angle deviates by 0.5 mdegrees from the two others.


Angles (23,13,12)= 5.96739156E+01 5.96739156E+01 5.96739156E+01 degrees
angles (23,13,12)= 59.67391564 59.67391564 59.67391564 degrees
angles (23,13,12)= 59.65429950 59.65429950 59.65429950 degrees
angles (23,13,12)= 59.60987717 59.60987717 59.60987717 degrees
angles (23,13,12)= 59.59774016 59.59774016 59.59774016 degrees
angles (23,13,12)= 59.58954710 59.58954710 59.58954710 degrees
angles (23,13,12)= 59.58994126 59.58994126 59.58994126 degrees
angles (23,13,12)= 59.59640191 59.59640191 59.59640191 degrees
angles (23,13,12)= 59.60556657 59.60556657 59.60556657 degrees
angles (23,13,12)= 59.61517982 59.61517982 59.61517982 degrees
angles (23,13,12)= 59.62547486 59.62547486 59.62547486 degrees
angles (23,13,12)= 59.64340038 59.64340037 59.64340037 degrees ! start of divergence
angles (23,13,12)= 59.66163347 59.66163344 59.66163344 degrees
angles (23,13,12)= 59.67145659 59.67145646 59.67145646 degrees
angles (23,13,12)= 59.67329734 59.67329688 59.67329688 degrees
angles (23,13,12)= 59.67356472 59.67356266 59.67356266 degrees
angles (23,13,12)= 59.67416017 59.67414948 59.67414948 degrees
angles (23,13,12)= 59.67575367 59.67570578 59.67570578 degrees
angles (23,13,12)= 59.67768192 59.67750735 59.67750735 degrees
angles (23,13,12)= 59.67851421 59.67798716 59.67798716 degrees
^ ^ ^

acell 1.0086679784E+01 1.0086679784E+01 1.0086679784E+01 Bohr
acell= 1.0086679784E+01 1.0086679784E+01 1.0086679784E+01
acell= 1.0095845448E+01 1.0095845448E+01 1.0095845448E+01
acell= 1.0107077039E+01 1.0107077039E+01 1.0107077039E+01
acell= 1.0105341284E+01 1.0105341284E+01 1.0105341284E+01
acell= 1.0100122543E+01 1.0100122543E+01 1.0100122543E+01
acell= 1.0097132505E+01 1.0097132505E+01 1.0097132505E+01
acell= 1.0093418047E+01 1.0093418047E+01 1.0093418047E+01
acell= 1.0092339678E+01 1.0092339678E+01 1.0092339678E+01
acell= 1.0092419651E+01 1.0092419651E+01 1.0092419651E+01
acell= 1.0092238158E+01 1.0092238159E+01 1.0092238159E+01 ! start of divergence
acell= 1.0090904084E+01 1.0090904084E+01 1.0090904084E+01
acell= 1.0088374760E+01 1.0088374763E+01 1.0088374763E+01
acell= 1.0086037737E+01 1.0086037745E+01 1.0086037745E+01
acell= 1.0084928271E+01 1.0084928299E+01 1.0084928299E+01
acell= 1.0084184900E+01 1.0084185025E+01 1.0084185025E+01
acell= 1.0083734213E+01 1.0083734862E+01 1.0083734862E+01
acell= 1.0084085342E+01 1.0084088246E+01 1.0084088246E+01
acell= 1.0084919226E+01 1.0084929815E+01 1.0084929815E+01
acell= 1.0085332361E+01 1.0085364332E+01 1.0085364332E+01
acell 1.0085332361E+01 1.0085364332E+01 1.0085364332E+01 Bohr
^ ^ ^



The optimized atomic positions remain in the initially determined space group, so if three equal angles and acell are given for a subsequent scf, R3c is the space group obtained.
(The same behaviour is also observed for optcell 3)

BAO.in
------
kptopt 1
ngkpt 6 6 6
ecut 45.0
pawecutdg 60

optcell 2
ionmov 2
dilatmx 1.05
ecutsm 0.5
ntime 25

getwfk 1

acell 10.086679784 10.086679784 10.086679784 bohr
angdeg 59.67391564 59.67391564 59.67391564
ntypat 3
znucl 83 13 8
natom 10
typat 1 1 2 2 3 3 3 3 3 3
xred
0.00000000000000 0.00000000000000 0.00000000000000
0.50000000000320 0.50000000000320 0.50000000000320
0.27141788893320 0.27141788893320 0.27141788893320
0.77141788893320 0.77141788893320 0.77141788893320
0.54462841053320 0.98371566192320 0.07314091094120
0.07314091094120 0.54462841053320 0.98371566192320
0.98371566192320 0.07314091094120 0.54462841053320
0.48371566192320 0.04462841053420 0.57314091094320
0.04462841053420 0.57314091094320 0.48371566192320
0.57314091094320 0.48371566192320 0.04462841053420

nstep 100
toldfe 1.0d-10
diemac 12.0
nsppol 1
-----

Which set of angles and vectors should be taken for subsequent scf-runs?

Thank you very much for any hints.

Best regards,

Pio Baettig




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