Finding lattice constants using EOS and the stress tensor
See also
HCP
Let’s try to find the EMT
potential.
First, we make a good initial guess for
import numpy as np
from ase.build import bulk
from ase.calculators.emt import EMT
from ase.io import Trajectory, read
a0 = 3.52 / np.sqrt(2)
c0 = np.sqrt(8 / 3.0) * a0
and create a trajectory for the results:
traj = Trajectory('Ni.traj', 'w')
Finally, we do the 9 calculations (three values for
eps = 0.01
for a in a0 * np.linspace(1 - eps, 1 + eps, 3):
for c in c0 * np.linspace(1 - eps, 1 + eps, 3):
ni = bulk('Ni', 'hcp', a=a, c=c)
ni.calc = EMT()
ni.get_potential_energy()
traj.write(ni)
Analysis
Now, we need to extract the data from the trajectory. Try this:
>>> from ase.build import bulk
>>> ni = bulk('Ni', 'hcp', a=2.5, c=4.0)
>>> ni.cell
array([[ 2.5 , 0. , 0. ],
[-1.25 , 2.165, 0. ],
[ 0. , 0. , 4. ]])
So, we can get ni.cell[0, 0]
and ni.cell[2,
2]
:
configs = read('Ni.traj@:')
energies = [config.get_potential_energy() for config in configs]
a = np.array([config.cell[0, 0] for config in configs])
c = np.array([config.cell[2, 2] for config in configs])
We fit the energy to this expression:
The best fit is found like this:
functions = np.array([a**0, a, c, a**2, a * c, c**2])
p = np.linalg.lstsq(functions.T, energies, rcond=-1)[0]
and we can find the minimum like this:
p0 = p[0]
p1 = p[1:3]
p2 = np.array([(2 * p[3], p[4]),
(p[4], 2 * p[5])])
a0, c0 = np.linalg.solve(p2.T, -p1)
with open('lattice_constant.csv', 'w') as fd:
fd.write(f'{a0:.3f}, {c0:.3f}\n')
Results:
a |
c |
---|---|
2.466 |
4.023 |
Using the stress tensor
One can also use the stress tensor to optimize the unit cell. For this we cannot use the EMT calculator.:
from ase.optimize import BFGS
from ase.constraints import StrainFilter
from gpaw import GPAW, PW
ni = bulk('Ni', 'hcp', a=a0,c=c0)
calc = GPAW(mode=PW(200),xc='LDA',txt='Ni.out')
ni.calc = calc
sf = StrainFilter(ni)
opt = BFGS(sf)
opt.run(0.005)
If you want the optimization path in a trajectory, add these lines
before calling the run()
method:
traj = Trajectory('path.traj', 'w', ni)
opt.attach(traj)