Dear Stefan Cottenier Sir,
Thank you very much for your reply.
Yes you are right about your guess and Sir your answer really pulled us out
from lots of worries. Thank you very much Sir.
with best regards,
Shamik Chakrabarti
On Tue, Oct 26, 2010 at 5:21 PM, S
Dear Peter Blaha Sir and wien2k users,
We have implemented your suggestion and was
able to remove the space group tolerence problem. But still I have a
question regarding the structure. In the energy and force optimized
structure there are some atomic networks and ato
My guess might be wrong, but it looks like you call two atoms
'bonded/unbonded' if there is/isn't a connecting rod between them in a
3D display? If so, then you shouldn't worry. This is just a way of
drawing (determined by the interatomic distance and the Z-value of the
atoms involved). It has
Dear Peter Blaha Sir,
Thank you very much for your reply. I have
understand what you have said and will try to implement the same. Yes it
should solve the problem. Thank you sir, thank you very much!!
with best regards,
Shamik Chakrabarti
On Sat, Oct 23, 2010 at 1
Dear Peter Blaha Sir,
I am using wien2k_08. No it is not an
hexagonal lattice. Actually the space group is unknown. We have taken a
structure A2BCO4 which has space group Pnma and then *replace atom C with
atom D *considering *primitive space group*. We have used the
Dear wien2k users,
we have done structure optimization (both volume
and force) of a A2BCO4 type material. After optimization the maximum force
on an atom is *2.036 mRy/a.u.* After optimization if we view the structure
with xcrysden we can see that some atoms remain unbo
Then it simply means, that after min_lapw your atoms arrived at
positions, which made them nearly equivalent, so that sgroup does not
know within its accuracy limits if two atoms are equivalent or not.
Eventually "correct" positions manually, eg. setting atoms at
0.50002 to 0.5 and so on.
Are you using WIEN2k_10 ??? It should not happen anymore.
Hexagonal lattice ??? min_lapw could have produced atomic positions,
which are not completely equivalent anymore (like 0.333 and
0.66) because of rounding errors.
Check your atomic positions manually.
Am 23.10.2010 06:22, schr
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