Thank you sir
On Mon, 7 Nov, 2022, 7:04 pm Kazume NISHIDATE,
wrote:
> Dear Jayraj Anadani
>
> > AIMD procedure, i am only taking 54 atomic supercell according to
> > CPMD procedure. And also , for taking 'one MD step' of 1 atoms
>
> Than, take the time averaged RDF of the cp.x calculation,
Dear Jayraj Anadani
> AIMD procedure, i am only taking 54 atomic supercell according to
> CPMD procedure. And also , for taking 'one MD step' of 1 atoms
Than, take the time averaged RDF of the cp.x calculation, 1000 to
2000 MD steps may be enough. You can get a smooth RDF graph thanks
to
Hello sir,
Actually, i am evaluating the RDF in LAMMPS for 1 atoms in a box. and
for cp.x code in QE which is AIMD procedure, i am only taking 54 atomic
supercell according to CPMD procedure. And also , for taking 'one MD step'
of 1 atoms will take too much time to complete the simulation
Dear Jayraj Anadani
> in both cases my g(r) RDF does not look smooth compared to the g(r)
> of LAMMPS MD simulation.
Did you use the same super-cell for these calculations?
Look at the atomic positions of the LAMMPS you evaluated the RDF and
convert these into the QE input format. And calculate
Dear Dhilshada V N (please sign always your posts with your scientific
affiliation, too)
Ferromagnetic and antiferromagnetic coupling is related to (at least)
two spin centers, e.g., two different metal ions. I suppose you want
to simulate high-spin and low-spin states of a single Fe
Hello QE Community,
I am trying to simulate the metals at high temperatures (~2500K) using the
cpmd (cp.x) method. I followed this two ways, For *electron cg dynamics*
electron kinetic energy completely attaches to BO surfaces (i.e. ekinc=0)
which i run upto 1.2 ps and for *electron damp
Hello,
first thing, I would plot the phonon dispersion sat each volume to
verify that they are actually nice and positive. If you have internal
degrees of freedom, and you did not relax them, it is likely that you
got negative phonons, which definitely breaks QHA (more or less
Hello Xin,
Thank you for your reply. Sorry for the late reply from my side. Below is the
k-points obtained from kpoints.x for Simple Cubic Bravais Lattice, for a 4 4 4
0 0 0 mesh, considering the 48 symmetries. Let’s call it Table 4.
TABLE 4
10
1 0.000 0.000 0.000 1.00
2
