dear Navaneeth,
as noted previously IFC should be symmetric w.r.t exchange of the
two atoms. this include the vector R that joins them.
in your case IFC Phi(na=1,nb=1)_yx(R=a1) ==
Phi(nb=1,na=1)_xy(R=-a1) because when you exchange the two atoms you
need to invert the connecting vector as well.
indeed line 280 is equal to line 802 and line 282 is equal to 800
as they should.
stefano
On 11/11/2014 08:23 AM, Navaneetha Krishnan wrote:
Hi Stefano,
Thanks for your reply. I am particularly looking to address an anomaly
in my simulation. I've attached my scf input, ph input and q2r input
files along with the generated force constant file.
The force constants are not symmetric (in fact anti-symmetric, for
certain direction pairs) with respect to the displacement directions.
For eg., if you look at line number 280 and line 800, the fcs are
supposed to be equal as they are Phi_{XY} and Phi_{YX} for the same
pair of atoms. I have tried this with several combinations of
k-sampling and Q-sampling and I get the same anti-symetry (except when
no. of Q's is very small, like Q=2). I also used a large ecutwfc so
that there is no convergence issue. This problem was not there with Al
lattice. The force constants were symmetric in this sense. Is there
bug in the code or am I making a mistake in one of the input files?
These are not negligible IFCs. Please have a look.
Thanks in advance!
On Mon, Nov 10, 2014 at 2:44 AM, stefano de gironcoli
<[email protected] <mailto:[email protected]>> wrote:
the IFC should be symmetric w.r.t. exchange of the two atoms and
displacement directions as a consequence of being second order
derivatives. other symmetries may apply in specific cases.
difficult to comment more without more info on what you are
looking for.
stefano
On 11/10/2014 09:16 AM, Navaneetha Krishnan wrote:
Hello QE users,
I was investigating the IFCs generated by q2r.x after scf and ph
calculations. It looks like the force constant matrix for certain
pairs of atoms is not symmetric (with respect to the cartesian
coordinates). For silicon, the IFC matrices have to be symmetric
with respect to the cartesian coordinates - right? For aluminum,
the output is symmetric, as expected. Any thoughts?
Thanks a lot!
--Navaneeth
Caltech.
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--
Navaneetha Krishnan Ravichandran,
Graduate Student,
Mechanical Engineering,
Caltech.
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