## Finding the volume of a unit cell at a fixed pressure

Posted June 12, 2013 at 04:17 PM | categories: uncategorized | tags:

A typical unit cell optimization in DFT is performed by minimizing the total energy with respect to variations in the unit cell parameters and atomic positions. In this approach, a pressure of 0 GPa is implied, as well as a temperature of 0K. For non-zero pressures, the volume that minimizes the total energy is not the same as the volume at P=0.

Let \(x\) be the unit cell parameters that can be varied. For P ≠ 0, and T = 0, we have the following

\(G(x; p) = E(x) + p V(x)\)

and we need to minimize this function to find the groundstate volume. We will do this for fcc Cu at 5 GPa of pressure. We will assume there is only one degree of freedom in the unit cell, the lattice constant. First we get the \(E(x)\) function, and then add the analytical correction.

from jasp import * from ase import Atom, Atoms from ase.utils.eos import EquationOfState LC = [3.5, 3.55, 3.6, 3.65, 3.7, 3.75] volumes, energies = [], [] ready = True P = 5.0 / 160.2176487 # pressure in eV/ang**3 for a in LC: atoms = Atoms([Atom('Cu',(0, 0, 0))], cell=0.5 * a*np.array([[1.0, 1.0, 0.0], [0.0, 1.0, 1.0], [1.0, 0.0, 1.0]])) with jasp('../bulk/Cu-{0}'.format(a), xc='PBE', encut=350, kpts=(8,8,8), atoms=atoms) as calc: try: e = atoms.get_potential_energy() energies.append(e) volumes.append(atoms.get_volume()) except (VaspSubmitted, VaspQueued): ready = False if not ready: import sys; sys.exit() import numpy as np energies = np.array(energies) volumes = np.array(volumes) eos = EquationOfState(volumes, energies) v0, e0, B = eos.fit() print 'V0 at 0 GPa = {0:1.2f} ang^3'.format(v0) eos5 = EquationOfState(volumes, energies + P * volumes) v0_5, e0, B = eos5.fit() print 'V0 at 5 GPa = {0:1.2f} ang^3'.format(v0_5)

V0 at 0 GPa = 12.02 ang^3 V0 at 5 GPa = 11.62 ang^3

You can see here that apply pressure decreases the equilibrium volume, and increases the total energy.

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