# Modeling materials using density functional theory

Copyright \copyright 2012--\the\year\ John Kitchin\par Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.3 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled "GNU Free Documentation License".

## 1 Introduction to this book

This book serves two purposes: 1) to provide worked examples of using DFT to model materials properties, and 2) to provide references to more advanced treatments of these topics in the literature. It is not a definitive reference on density functional theory. Along the way to learning how to perform the calculations, you will learn how to analyze the data, make plots, and how to interpret the results. This book is very much "recipe" oriented, with the intention of giving you enough information and knowledge to start your research. In that sense, many of the computations are not publication quality with respect to convergence of calculation parameters.

You will read a lot of python code in this book. I believe that computational work should always be scripted. Scripting provides a written record of everything you have done, making it more probable you (or others) could reproduce your results or report the method of its execution exactly at a later time.

This book makes heavy use of many computational tools including:

The DFT code used primarily in this book is VASP.

Similar code would be used for other calculators, e.g. GPAW, Jacapo, etc… you would just have to import the python modules for those codes, and replace the code that defines the calculator.

Review all the hyperlinks in this chapter.

## 2 Introduction to DFT

A comprehensive overview of DFT is beyond the scope of this book, as excellent reviews on these subjects are readily found in the literature, and are suggested reading in the following paragraph. Instead, this chapter is intended to provide a useful starting point for a non-expert to begin learning about and using DFT in the manner used in this book. Much of the information presented here is standard knowledge among experts, but a consequence of this is that it is rarely discussed in current papers in the literature. A secondary goal of this chapter is to provide new users with a path through the extensive literature available and to point out potential difficulties and pitfalls in these calculations.

A modern and practical introduction to density functional theory can be found in Sholl and Steckel sholl-2009-densit-funct-theor. A fairly standard textbook on DFT is the one written by Parr and Yang parr-yang. The Chemist's Guide to DFT koch2001 is more readable and contains more practical information for running calculations, but both of these books focus on molecular systems. The standard texts in solid state physics are by Kittel kittel and Ashcroft and Mermin ashcroft-mermin. Both have their fine points, the former being more mathematically rigorous and the latter more readable. However, neither of these books is particularly easy to relate to chemistry. For this, one should consult the exceptionally clear writings of Roald Hoffman hoffmann1987,RevModPhys.60.601, and follow these with the work of N\o rskov and coworkers hammer2000:adv-cat,greeley2002:elect.

In this chapter, only the elements of DFT that are relevant to this work will be discussed. An excellent review on other implementations of DFT can be found in Reference freeman1995:densit, and details on the various algorithms used in DFT codes can be found in Refs. payne1992:iterat,Kresse199615.

One of the most useful sources of information has been the dissertations of other students, perhaps because the difficulties they faced in learning the material are still fresh in their minds. Thomas Bligaard, a coauthor of Dacapo, wrote a particularly relevant thesis on exchange/correlation functionals bligaard2000:exchan-correl-funct and a dissertation illustrating the use of DFT to design new alloys with desirable thermal and mechanical properties bligaard2003:under-mater-proper-basis-densit. The Ph.D. thesis of Ari Seitsonen contains several useful appendices on k-point setups, and convergence tests of calculations, in addition to a thorough description of DFT and analysis of calculation output seitsonen2000:phd. Finally, another excellent overview of DFT and its applications to bimetallic alloy phase diagrams and surface reactivity is presented in the PhD thesis of Robin Hirschl hirschl2002:binar-trans-metal-alloy-their-surfac.

### 2.1 Background

In 1926, Erwin Schrödinger published the first accounts of his now famous wave equation pauling1963. He later shared the Nobel prize with Paul A. M. Dirac in 1933 for this discovery. Schrödinger's wave function seemed extremely promising, as it contains all of the information available about a system. Unfortunately, most practical systems of interest consist of many interacting electrons, and the effort required to find solutions to Schrödinger's equation increases exponentially with the number of electrons, limiting this approach to systems with a small number of relevant electrons, $$N \lesssim O(10)$$ RevModPhys.71.1253. Even if this rough estimate is off by an order of magnitude, a system with 100 electrons is still very small, for example, two Ru atoms if all the electrons are counted, or perhaps ten Pt atoms if only the valence electrons are counted. Thus, the wave function method, which has been extremely successful in studying the properties of small molecules, is unsuitable for studies of large, extended solids. Interestingly, this difficulty was recognized by Dirac as early as 1929, when he wrote "The underlying physical laws necessary for the mathematical theory of a large part of physics and the whole of chemistry are thus completely known, and the difficulty is only that the application of these laws leads to equations much too complicated to be soluble." dirac1929:quant-mechan-many-elect-system.

In 1964, Hohenberg and Kohn showed that the ground state total energy of a system of interacting electrons is a unique functional of the electron density PhysRev.136.B864. By definition, a function returns a number when given a number. For example, in $$f(x)=x^2$$, $$f(x)$$ is the function, and it equals four when $$x=2$$. A functional returns a number when given a function. Thus, in $$g(f(x))=\int_0^\pi f(x) dx$$, $$g(f(x))$$ is the functional, and it is equal to two when $$f(x)=\sin(x)$$. Hohenberg and Kohn further identified a variational principle that appeared to reduce the problem of finding the ground state energy of an electron gas in an external potential (i.e., in the presence of ion cores) to that of the minimization of a functional of the three-dimensional density function. Unfortunately, the definition of the functional involved a set of 3N-dimensional trial wave functions.

In 1965, Kohn and Sham made a significant breakthrough when they showed that the problem of many interacting electrons in an external potential can be mapped exactly to a set of noninteracting electrons in an effective external potential PhysRev.140.A1133. This led to a set of self-consistent, single particle equations known as the Kohn-Sham (KS) equations:

with

where $$v(\mathbf{r})$$ is the external potential and $$v_{xc}(\mathbf{r})$$ is the exchange-correlation potential, which depends on the entire density function. Thus, the density needs to be known in order to define the effective potential so that Eq. \eqref{eq:KS} can be solved. $$\varphi_j(\mathbf{r})$$ corresponds to the $$j^{th}$$ KS orbital of energy $$\epsilon_j$$.

The ground state density is given by:

To solve Eq. \eqref{eq:KS} then, an initial guess is used for $$\varphi_j(r)$$ which is used to generate Eq. \eqref{eq:density}, which is subsequently used in Eq. \eqref{eq:veff}. This equation is then solved for $$\varphi_j(\mathbf{r})$$ iteratively until the $$\varphi_j(\mathbf{r})$$ that result from the solution are the same as the $$\varphi_j(\mathbf{r})$$ that are used to define the equations, that is, the solutions are self-consistent. Finally, the ground state energy is given by:

where $$E_{xc}[n(\mathbf{r})]$$ is the exchange-correlation energy functional. Walter Kohn shared the Nobel prize in Chemistry in 1998 for this work RevModPhys.71.1253. The other half of the prize went to John Pople for his efforts in wave function based quantum mechanical methods RevModPhys.71.1267. Provided the exchange-correlation energy functional is known, Eq. (eq:dftEnergy) is exact. However, the exact form of the exchange-correlation energy functional is not known, thus approximations for this functional must be used.

### 2.2 Exchange correlation functionals

The two main types of exchange/correlation functionals used in DFT are the local density approximation (LDA) and the generalized gradient approximation (GGA). In the LDA, the exchange-correlation functional is defined for an electron in a uniform electron gas of density $$n$$ PhysRev.140.A1133. It is exact for a uniform electron gas, and is anticipated to be a reasonable approximation for slowly varying densities. In molecules and solids, however, the density tends to vary substantially in space. Despite this, the LDA has been very successfully used in many systems. It tends to predict overbonding in both molecular and solid systems fuchs1998:pseud, and it tends to make semiconductor systems too metallic (the band gap problem) perdew1982:elect-kohn-sham.

The generalized gradient approximation includes corrections for gradients in the electron density, and is often implemented as a corrective function of the LDA. The form of this corrective function, or "exchange enhancement" function determines which functional it is, e.g. PBE, RPBE, revPBE, etc. hammer1999:improv-pbe. In this book the PBE GGA functional is used the most. N\o{}rskov and coworkers have found that the RPBE functional gives superior chemisorption energies for atomic and molecular bonding to surfaces, but that it gives worse bulk properties, such as lattice constants compared to experimental data hammer1999:improv-pbe.

Finally, there are increasingly new types of functionals in the literature. The so-called hybrid functionals, such as B3LYP, are more popular with gaussian basis sets (e.g. in Gaussian), but they are presently inefficient with planewave basis sets. None of these other types of functionals were used in this work. For more details see Chapter 6 in Ref. koch2001 and Thomas Bligaard's thesis on exchange and correlation functionals bligaard2000:exchan-correl-funct.

### 2.3 Basis sets

Briefly, VASP utilizes planewaves as the basis set to expand the Kohn-Sham orbitals. In a periodic solid, one can use Bloch's theorem to show that the wave function for an electron can be expressed as the product of a planewave and a function with the periodicity of the lattice ashcroft-mermin:

$$\psi_{n\mathbf{k}}(\mathbf{r})=\exp({i\mathbf{k}\cdot\mathbf{r}})u_{n\mathbf{k}}(\mathbf{r})$$

where $$\mathbf{r}$$ is a position vector, and $$\mathbf{k}$$ is a so-called wave vector that will only have certain allowed values defined by the size of the unit cell. Bloch's theorem sets the stage for using planewaves as a basis set, because it suggests a planewave character of the wave function. If the periodic function $$u_{n\mathbf{k}}(\mathbf{r})$$ is also expanded in terms of planewaves determined by wave vectors of the reciprocal lattice vectors, $$\mathbf{G}$$, then the wave function can be expressed completely in terms of a sum of planewaves payne1992:iterat:

$$\psi_i(\mathbf{r})=\sum_\mathbf{G} c_{i,\mathbf{k+G}} \exp(i\mathbf{(k+G)\cdot r}).$$

where $$c_{i,\mathbf{k+G}}$$ are now coefficients that can be varied to determine the lowest energy solution. This also converts Eq. \eqref{eq:KS} from an integral equation to a set of algebraic equations that can readily be solved using matrix algebra.

In aperiodic systems, such as systems with even one defect, or randomly ordered alloys, there is no periodic unit cell. Instead one must represent the portion of the system of interest in a supercell, which is then subjected to the periodic boundary conditions so that a planewave basis set can be used. It then becomes necessary to ensure the supercell is large enough to avoid interactions between the defects in neighboring supercells. The case of the randomly ordered alloy is virtually hopeless as the energy of different configurations will fluctuate statistically about an average value. These systems were not considered in this work, and for more detailed discussions the reader is referred to Ref. makov1995:period-bound-condit. Once a supercell is chosen, however, Bloch's theorem can be applied to the new artificially periodic system.

To get a perfect expansion, one needs an infinite number of planewaves. Luckily, the coefficients of the planewaves must go to zero for high energy planewaves, otherwise the energy of the wave function would go to infinity. This provides justification for truncating the planewave basis set above a cutoff energy. Careful testing of the effect of the cutoff energy on the total energy can be done to determine a suitable cutoff energy. The cutoff energy required to obtain a particular convergence precision is also element dependent, shown in Table tab:pwcut. It can also vary with the "softness" of the pseudopotential. Thus, careful testing should be done to ensure the desired level of convergence of properties in different systems. Table tab:pwcut refers to convergence of total energies. These energies are rarely considered directly, it is usually differences in energy that are important. These tend to converge with the planewave cutoff energy much more quickly than total energies, due to cancellations of convergence errors. In this work, 350 eV was found to be suitable for the H adsorption calculations, but a cutoff energy of 450 eV was required for O adsorption calculations.

Table 1: Planewave cutoff energies (in eV) required for different convergence precisions for two different elements with different pseudopotential setups.
Precision Low High
Mo 168 293
O 300 520
O_sv 1066 1847

Bloch's theorem eliminates the need to calculate an infinite number of wave functions, because there are only a finite number of electrons in the unit (super) cell. However, there are still an infinite number of discrete k points that must be considered, and the energy of the unit cell is calculated as an integral over these points. It turns out that wave functions at k points that are close together are similar, thus an interpolation scheme can be used with a finite number of k points. This also converts the integral used to determine the energy into a sum over the k points, which are suitably weighted to account for the finite number of them. There will be errors in the total energy associated with the finite number of k, but these can be reduced and tested for convergence by using higher k-point densities. An excellent discussion of this for aperiodic systems can be found in Ref. makov1995:period-bound-condit.

The most common schemes for generating k points are the Chadi-Cohen scheme PhysRevB.8.5747, and the Monkhorst-Pack scheme PhysRevB.13.5188. The use of these k point setups amounts to an expansion of the periodic function in reciprocal space, which allows a straight-forward interpolation of the function between the points that is more accurate than with other k point generation schemes PhysRevB.13.5188.

### 2.4 Pseudopotentials

The core electrons of an atom are computationally expensive with planewave basis sets because they are highly localized. This means that a very large number of planewaves are required to expand their wave functions. Furthermore, the contributions of the core electrons to bonding compared to those of the valence electrons is usually negligible. In fact, the primary role of the core electron wave functions is to ensure proper orthogonality between the valence electrons and core states. Consequently, it is desirable to replace the atomic potential due to the core electrons with a pseudopotential that has the same effect on the valence electrons PhysRevB.43.1993. There are essentially two kinds of pseudopotentials, norm-conserving soft pseudopotentials PhysRevB.43.1993 and Vanderbilt ultrasoft pseudopotentials PhysRevB.41.7892. In either case, the pseudopotential function is generated from an all-electron calculation of an atom in some reference state. In norm-conserving pseudopotentials, the charge enclosed in the pseudopotential region is the same as that enclosed by the same space in an all-electron calculation. In ultrasoft pseudopotentials, this requirement is relaxed and charge augmentation functions are used to make up the difference. As its name implies, this allows a "softer" pseudopotential to be generated, which means fewer planewaves are required to expand it.

The pseudopotentials are not unique, and calculated properties depend on them. However, there are standard methods for ensuring the quality and transferability (to different chemical environments) of the pseudopotentials PhysRevB.56.15629.

TODO PAW description
nil

VASP provides a database of PAW potentials PhysRevB.50.17953,PhysRevB.59.1758.

### 2.5 Fermi Temperature and band occupation numbers

At absolute zero, the occupancies of the bands of a system are well-defined step functions; all bands up to the Fermi level are occupied, and all bands above the Fermi level are unoccupied. There is a particular difficulty in the calculation of the electronic structures of metals compared to semiconductors and molecules. In molecules and semiconductors, there is a clear energy gap between the occupied states and unoccupied states. Thus, the occupancies are insensitive to changes in the energy that occur during the self-consistency cycles. In metals, however, the density of states is continuous at the Fermi level, and there are typically a substantial number of states that are close in energy to the Fermi level. Consequently, small changes in the energy can dramatically change the occupation numbers, resulting in instabilities that make it difficult to converge to the occupation step function. A related problem is that the Brillouin zone integral (which in practice is performed as a sum over a finite number of k points) that defines the band energy converges very slowly with the number of k points due to the discontinuity in occupancies in a continuous distribution of states for metals gillan1989:calcul,Kresse199615. The difficulty arises because the temperature in most DFT calculations is at absolute zero. At higher temperatures, the DOS is smeared across the Fermi level, resulting in a continuous occupation function over the distribution of states. A finite-temperature version of DFT was developed PhysRev.137.A1441, which is the foundation on which one solution to this problem is based. In this solution, the step function is replaced by a smoothly varying function such as the Fermi-Dirac function at a small, but non-zero temperature Kresse199615. The total energy is then extrapolated back to absolute zero.

### 2.6 Spin polarization and magnetism

There are two final points that need to be discussed about these calculations, spin polarization and dipole corrections. Spin polarization is important for systems that contain net spin. For example, iron, cobalt and nickel are magnetic because they have more electrons with spin "up" than spin "down" (or vice versa). Spin polarization must also be considered in atoms and molecules with unpaired electrons, such as hydrogen and oxygen atoms, oxygen molecules and radicals. For example, there are two spin configurations for an oxygen molecule, the singlet state with no unpaired electrons, and the triplet state with two unpaired electrons. The oxygen triplet state is lower in energy than the oxygen singlet state, and thus it corresponds to the ground state for an oxygen atom. A classically known problem involving spin polarization is the dissociation of a hydrogen molecule. In this case, the molecule starts with no net spin, but it dissociates into two atoms, each of which has an unpaired electron. See section 5.3.5 in Reference koch2001 for more details on this.

In VASP, spin polarization is not considered by default; it must be turned on, and an initial guess for the magnetic moment of each atom in the unit cell must be provided (typically about one Bohr-magneton per unpaired electron). For Fe, Co, and Ni, the experimental values are 2.22, 1.72, and 0.61 Bohr-magnetons, respectively kittel and are usually good initial guesses. See Reference PhysRevB.56.15629 for a very thorough discussion of the determination of the magnetic properties of these metals with DFT. For a hydrogen atom, an initial guess of 1.0 Bohr-magnetons (corresponding to one unpaired electron) is usually good. An oxygen atom has two unpaired electrons, thus an initial guess of 2.0 Bohr-magnetons should be used. The spin-polarized solution is sensitive to the initial guess, and typically converges to the closest solution. Thus, a magnetic initial guess usually must be provided to get a magnetic solution. Finally, unless an adsorbate is on a magnetic metal surface, spin polarization typically does not need to be considered, although the gas-phase reference state calculation may need to be done with spin-polarization.

The downside of including spin polarization is that it essentially doubles the calculation time.

The original papers on DFT are PhysRev.136.B864,PhysRev.140.A1133.

Kohn's Nobel Lecture RevModPhys.71.1253 and Pople's Nobel Lecture RevModPhys.71.1267 are good reads.

This paper by Hoffman RevModPhys.60.601 is a nice review of solid state physics from a chemist's point of view.

All calculations in this book were performed using VASP Kresse199615,PhysRevB.54.11169,PhysRevB.49.14251,PhysRevB.47.558 with the projector augmented wave (PAW) potentials provided in VASP.

## 3 Molecules

In this chapter we consider how to construct models of molecules, how to manipulate them, and how to calculate many properties of molecules. For a nice comparison of VASP and Gaussian see paier:234102.

### 3.1 Defining and visualizing molecules

We start by learning how to define a molecule and visualize it. We will begin with defining molecules from scratch, then reading molecules from data files, and finally using some built-in databases in ase.

#### 3.1.1 From scratch

When there is no data file for the molecule you want, or no database to get it from, you have to define your atoms geometry by hand. Here is how that is done for a CO molecule (Figure fig:co-origin). We must define the type and position of each atom, and the unit cell the atoms are in.

from ase import Atoms, Atom
from ase.io import write

# define an Atoms object
atoms = Atoms([Atom('C', [0., 0., 0.]),
Atom('O', [1.1, 0., 0.])],
cell=(10, 10, 10))

print('V = {0:1.0f} Angstrom^3'.format(atoms.get_volume()))

write('images/simple-cubic-cell.png', atoms, show_unit_cell=2)

V = 1000 Angstrom^3


There are two inconvenient features of the simple cubic cell:

1. Since the CO molecule is at the corner, its electron density is spread over the 8 corners of the box, which is not convenient for visualization later (see Visualizing electron density).
2. Due to the geometry of the cube, you need fairly large cubes to make sure the electron density of the molecule does not overlap with that of its images. Electron-electron interactions are repulsive, and the overlap makes the energy increase significantly. Here, the CO molecule has 6 images due to periodic boundary conditions that are 10 Å away. The volume of the unit cell is 1000 Å$$^3$$.

The first problem is easily solved by centering the atoms in the unit cell. The second problem can be solved by using a face-centered cubic lattice, which is the lattice with the closest packing. We show the results of the centering in Figure fig:co-fcc, where we have guessed values for $$b$$ until the CO molecules are on average 10 Å apart. Note the final volume is only about 715 Å$$^3$$, which is smaller than the cube. This will result in less computational time to compute properties.

from ase import Atoms, Atom
from ase.io import write

b = 7.1
atoms = Atoms([Atom('C', [0., 0., 0.]),
Atom('O', [1.1, 0., 0.])],
cell=[[b, b, 0.],
[b, 0., b],
[0., b, b]])

print('V = {0:1.0f} Ang^3'.format(atoms.get_volume()))

atoms.center()  # translate atoms to center of unit cell
write('images/fcc-cell.png', atoms, show_unit_cell=2)

V = 716 Ang^3


At this point you might ask, "How do you know the distance to the neighboring image?" The ag viewer lets you compute this graphically, but we can use code to determine this too. All we have to do is figure out the length of each lattice vector, because these are what separate the atoms in the images. We use the numpy module to compute the distance of a vector as the square root of the sum of squared elements.

from ase import Atoms, Atom
import numpy as np

b = 7.1
atoms = Atoms([Atom('C', [0., 0., 0.]),
Atom('O', [1.1, 0., 0.])],
cell=[[b, b, 0.],
[b, 0., b],
[0., b, b]])

# get unit cell vectors and their lengths
(a1, a2, a3) = atoms.get_cell()
print('|a1| = {0:1.2f} Ang'.format(np.sum(a1**2)**0.5))
print('|a2| = {0:1.2f} Ang'.format(np.linalg.norm(a2)))
print('|a3| = {0:1.2f} Ang'.format(np.sum(a3**2)**0.5))

|a1| = 10.04 Ang
|a2| = 10.04 Ang
|a3| = 10.04 Ang


#### 3.1.2 Reading other data formats into a calculation

ase.io.read supports many different file formats:

    Known formats:

=========================  ===========
format                     short name
=========================  ===========
GPAW restart-file          gpw
Dacapo netCDF output file  dacapo
Old ASE netCDF trajectory  nc
Virtual Nano Lab file      vnl
ASE pickle trajectory      traj
ASE bundle trajectory      bundle
GPAW text output           gpaw-text
CUBE file                  cube
XCrySDen Structure File    xsf
Dacapo text output         dacapo-text
XYZ-file                   xyz
VASP POSCAR/CONTCAR file   vasp
VASP OUTCAR file           vasp_out
SIESTA STRUCT file         struct_out
ABINIT input file          abinit
V_Sim ascii file           v_sim
Protein Data Bank          pdb
CIF-file                   cif
FHI-aims geometry file     aims
FHI-aims output file       aims_out
VTK XML Image Data         vti
VTK XML Structured Grid    vts
VTK XML Unstructured Grid  vtu
TURBOMOLE coord file       tmol
exciting input             exi
AtomEye configuration      cfg
WIEN2k structure file      struct
DftbPlus input file        dftb
CASTEP geom file           cell
CASTEP output file         castep
CASTEP trajectory file     geom
ETSF format                etsf.nc
DFTBPlus GEN format        gen
CMR db/cmr-file            db
CMR db/cmr-file            cmr
LAMMPS dump file           lammps
Gromacs coordinates        gro
=========================  ===========


You can read XYZ file format to create ase.Atoms objects. Here is what an XYZ file format might look like:

#+include: molecules/isobutane.xyz


The first line is the number of atoms in the file. The second line is often a comment. What follows is one line per atom with the symbol and Cartesian coordinates in Å. Note that the XYZ format does not have unit cell information in it, so you will have to figure out a way to provide it. In this example, we center the atoms in a box with vacuum on all sides (Figure fig:isobutane).

from ase.io import read, write

atoms.center(vacuum=5)
write('images/isobutane-xyz.png', atoms, show_unit_cell=2)


#### 3.1.3 Predefined molecules

ase defines a number of molecular geometries in the ase.data.molecules database. For example, the database includes the molecules in the G2/97 database curtiss:1063. This database contains a broad set of atoms and molecules for which good experimental data exists, making them useful for benchmarking studies. See this site for the original files.

The coordinates for the atoms in the database are MP2(full)/6-31G(d) optimized geometries. Here is a list of all the species available in ase.data.g2. You may be interested in reading about some of the other databases in ase.data too.

from ase.data import g2
keys = g2.data.keys()
# print in 3 columns
for i in range(len(keys) / 3):
print('{0:25s}{1:25s}{2:25s}'.format(*tuple(keys[i * 3: i * 3 + 3])))

isobutene                CH3CH2OH                 CH3COOH
COF2                     CH3NO2                   CF3CN
CH3OH                    CCH                      CH3CH2NH2
PH3                      Si2H6                    O3
O2                       BCl3                     CH2_s1A1d
Be                       H2CCl2                   C3H9C
C3H9N                    CH3CH2OCH3               BF3
CH3                      CH4                      S2
C2H6CHOH                 SiH2_s1A1d               H3CNH2
CH3O                     H                        BeH
P                        C3H4_C3v                 C2F4
OH                       methylenecyclopropane    F2O
SiCl4                    HCF3                     HCCl3
C3H7                     CH3CH2O                  AlF3
CH2NHCH2                 SiH2_s3B1d               H2CF2
SiF4                     H2CCO                    PH2
OCS                      HF                       NO2
SH2                      C3H4_C2v                 H2O2
CH3CH2Cl                 isobutane                CH3COF
HCOOH                    CH3ONO                   C5H8
2-butyne                 SH                       NF3
HOCl                     CS2                      P2
C                        CH3S                     O
C4H4S                    S                        C3H7Cl
H2CCHCl                  C2H6                     CH3CHO
C2H4                     HCN                      C2H2
C2Cl4                    bicyclobutane            H2
C6H6                     N2H4                     C4H4NH
H2CCHCN                  H2CCHF                   cyclobutane
HCl                      CH3OCH3                  Li2
Na                       CH3SiH3                  NaCl
CH3CH2SH                 OCHCHO                   SiH4
C2H5                     SiH3                     NH
ClO                      AlCl3                    CCl4
NO                       C2H3                     ClF
HCO                      CH3CONH2                 CH2SCH2
CH3COCH3                 C3H4_D2d                 CH
CO                       CN                       F
CH3COCl                  N                        CH3Cl
Si                       C3H8                     CS
N2                       Cl2                      NCCN
F2                       CO2                      Cl
CH2OCH2                  H2O                      CH3CO
ClF3                     Li                       PF3
B                        CH3SH                    CF4
C3H6_Cs                  C2H6NH                   N2O
LiF                      H2COH                    cyclobutene
LiH                      SiO                      Si2
C2H6SO                   C5H5N                    trans-butane
Na2                      C4H4O                    SO2
NH3                      NH2                      CH2_s3B1d
ClNO                     C3H6_D3h                 Al
CH3SCH3                  H2CO                     CH3CN


Some other databases include the ase.data.s22 for weakly interacting dimers and complexes, and ase.data.extra_molecules which has a few extras like biphenyl and C60.

Here is an example of getting the geometry of an acetonitrile molecule and writing an image to a file. Note that the default unit cell is a 1 Å × Å × Å cubic cell. That is too small to use if your calculator uses periodic boundary conditions. We center the atoms in the unit cell and add vacuum on each side. We will add 6 Å of vacuum on each side. In the write command we use the option show_unit_cell =2 to draw the unit cell boundaries. See Figure fig:ch3cn.

from ase.structure import molecule
from ase.io import write

atoms = molecule('CH3CN')

atoms.center(vacuum=6)
print('unit cell')
print('---------')
print(atoms.get_cell())

write('images/ch3cn.png', atoms, show_unit_cell=2)

unit cell
---------
[[ 13.775328   0.         0.      ]
[  0.        13.537479   0.      ]
[  0.         0.        15.014576]]


It is possible to rotate the atoms with ase.io.write if you wanted to see pictures from another angle. In the next example we rotate 45 degrees about the $x$-axis, then 45 degrees about the $y$-axis. Note that this only affects the image, not the actual coordinates. See Figure fig:ch3cn-rot.

from ase.structure import molecule
from ase.io import write

atoms = molecule('CH3CN')

atoms.center(vacuum=6)
print('unit cell')
print('---------')
print(atoms.get_cell())

write('images/ch3cn-rotated.png', atoms,
show_unit_cell=2, rotation='45x,45y,0z')

unit cell
---------
[[ 13.775328   0.         0.      ]
[  0.        13.537479   0.      ]
[  0.         0.        15.014576]]


If you actually want to rotate the coordinates, there is a nice way to do that too, with the ase.Atoms.rotate method. Actually there are some subtleties in rotation. One rotates the molecule an angle (in radians) around a vector, but you have to choose whether the center of mass should be fixed or not. You also must decide whether or not the unit cell should be rotated. In the next example you can see the coordinates have changed due to the rotations. Note that the write function uses the rotation angle in degrees, while the rotate function uses radians.

from ase.structure import molecule
from ase.io import write
from numpy import pi

atoms = molecule('CH3CN')
atoms.center(vacuum=6)
p1 = atoms.get_positions()

atoms.rotate('x', pi/4, center='COM', rotate_cell=False)
atoms.rotate('y', pi/4, center='COM', rotate_cell=False)

write('images/ch3cn-rotated-2.png', atoms, show_unit_cell=2)
print('difference in positions after rotating')
print('atom    difference vector')
print('--------------------------------------')
p2 = atoms.get_positions()

diff = p2 - p1
for i, d in enumerate(diff):
print('{0} {1}'.format(i, d))

difference in positions after rotating
atom    difference vector
--------------------------------------
0 [-0.65009456  0.91937255  0.65009456]
1 [ 0.08030744 -0.11357187 -0.08030744]
2 [ 0.66947344 -0.94677841 -0.66947344]
3 [-0.32532156  0.88463727  1.35030756]
4 [-1.35405183  1.33495444 -0.04610517]
5 [-0.8340703   1.33495444  1.2092413 ]


Note in this last case the unit cell is oriented differently than the previous example, since we chose not to rotate the unit cell.

#### 3.1.4 Combining Atoms objects

It is frequently useful to combine two Atoms objects, e.g. for computing reaction barriers, or other types of interactions. In ase, we simply add two Atoms objects together. Here is an example of getting an ammonia and oxygen molecule in the same unit cell. See Figure fig:combined-atoms. We set the Atoms about three Å apart using the ase.Atoms.translate function.

from ase.structure import molecule
from ase.io import write

atoms1 = molecule('NH3')

atoms2 = molecule('O2')
atoms2.translate([3, 0, 0])

bothatoms = atoms1 + atoms2
bothatoms.center(5)

write('images/bothatoms.png', bothatoms, show_unit_cell=2, rotation='90x')


### 3.2 Simple properties

Simple properties do not require a DFT calculation. They are typically only functions of the atom types and geometries.

#### 3.2.1 Getting cartesian positions

If you want the $$(x,y,z)$$ coordinates of the atoms, use the ase.Atoms.get_positions. If you are interested in the fractional coordinates, use ase.Atoms.get_scaled_positions.

from ase.structure import molecule

atoms = molecule('C6H6')  # benzene

# access properties on each atom
print(' #  sym   p_x     p_y     p_z')
print('------------------------------')
for i, atom in enumerate(atoms):
print('{0:3d}{1:^4s}{2:-8.2f}{3:-8.2f}{4:-8.2f}'.format(i,
atom.symbol,
atom.x,
atom.y,
atom.z))

# get all properties in arrays
sym = atoms.get_chemical_symbols()
pos = atoms.get_positions()
num = atoms.get_atomic_numbers()

atom_indices = range(len(atoms))

print()
print('  # sym    at#    p_x     p_y     p_z')
print('-------------------------------------')
for i, s, n, p in zip(atom_indices, sym, num, pos):
px, py, pz = p
print('{0:3d}{1:>3s}{2:8d}{3:-8.2f}{4:-8.2f}{5:-8.2f}'.format(i, s, n,
px, py, pz))

 #  sym   p_x     p_y     p_z
------------------------------
0 C      0.00    1.40    0.00
1 C      1.21    0.70    0.00
2 C      1.21   -0.70    0.00
3 C      0.00   -1.40    0.00
4 C     -1.21   -0.70    0.00
5 C     -1.21    0.70    0.00
6 H      0.00    2.48    0.00
7 H      2.15    1.24    0.00
8 H      2.15   -1.24    0.00
9 H      0.00   -2.48    0.00
10 H     -2.15   -1.24    0.00
11 H     -2.15    1.24    0.00
()
# sym    at#    p_x     p_y     p_z
-------------------------------------
0  C       6    0.00    1.40    0.00
1  C       6    1.21    0.70    0.00
2  C       6    1.21   -0.70    0.00
3  C       6    0.00   -1.40    0.00
4  C       6   -1.21   -0.70    0.00
5  C       6   -1.21    0.70    0.00
6  H       1    0.00    2.48    0.00
7  H       1    2.15    1.24    0.00
8  H       1    2.15   -1.24    0.00
9  H       1    0.00   -2.48    0.00
10  H       1   -2.15   -1.24    0.00
11  H       1   -2.15    1.24    0.00


#### 3.2.2 Molecular weight and molecular formula

molecular weight We can quickly compute the molecular weight of a molecule with this recipe. We use ase.Atoms.get_masses to get an array of the atomic masses of each atom in the Atoms object, and then just sum them up.

from ase.structure import molecule

atoms = molecule('C6H6')
masses = atoms.get_masses()

molecular_weight = masses.sum()
molecular_formula = atoms.get_chemical_formula(mode='reduce')

# note use of two lines to keep length of line reasonable
s = 'The molecular weight of {0} is {1:1.2f} gm/mol'
print(s.format(molecular_formula, molecular_weight))

The molecular weight of C6H6 is 78.11 gm/mol


Note that the argument reduce=True for ase.Atoms.get_chemical_formula collects all the symbols to provide a molecular formula.

#### 3.2.3 Center of mass

center of mass The center of mass (COM) is defined as:

COM = $$\frac{\sum m_i \cdot r_i}{\sum m_i}$$

The center of mass is essentially the average position of the atoms, weighted by the mass of each atom. Here is an example of getting the center of mass from an Atoms object using ase.Atoms.get_center_of_mass.

from ase.structure import molecule
import numpy as np

# ammonia
atoms = molecule('NH3')

# cartesian coordinates
print('COM1 = {0}'.format(atoms.get_center_of_mass()))

# compute the center of mass by hand
pos = atoms.positions
masses = atoms.get_masses()

COM = np.array([0., 0., 0.])
for m, p in zip(masses, pos):
COM += m*p
COM /= masses.sum()

print('COM2 = {0}'.format(COM))

# one-line linear algebra definition of COM
print('COM3 = {0}'.format(np.dot(masses, pos) / np.sum(masses)))

COM1 = [  0.00000000e+00   5.91843349e-08   4.75457009e-02]
COM2 = [  0.00000000e+00   5.91843349e-08   4.75457009e-02]
COM3 = [  0.00000000e+00   5.91843349e-08   4.75457009e-02]


You can see see that these centers of mass, which are calculated by different methods, are the same.

#### 3.2.4 Moments of inertia

moment of inertia The moment of inertia is a measure of resistance to changes in rotation. It is defined by $$I = \sum_{i=1}^N m_i r_i^2$$ where $$r_i$$ is the distance to an axis of rotation. There are typically three moments of inertia, although some may be zero depending on symmetry, and others may be degenerate. There is a convenient function to get the moments of inertia: ase.Atoms.get_moments_of_inertia. Here are several examples of molecules with different types of symmetry.:

from ase.structure import molecule

print('linear rotors: I = [0 Ia Ia]')
atoms = molecule('CO2')
print('  CO2 moments of inertia: {}'.format(atoms.get_moments_of_inertia()))
print('')

print('symmetric rotors (Ia = Ib) < Ic')
atoms = molecule('NH3')
print('  NH3 moments of inertia: {}'.format(atoms.get_moments_of_inertia()))

atoms = molecule('C6H6')
print('  C6H6 moments of inertia: {}'.format(atoms.get_moments_of_inertia()))
print('')

print('symmetric rotors Ia < (Ib = Ic)')
atoms = molecule('CH3Cl')
print('CH3Cl moments of inertia: {}'.format(atoms.get_moments_of_inertia()))
print('')

print('spherical rotors Ia = Ib = Ic')
atoms = molecule('CH4')
print('  CH4 moments of inertia: {}'.format(atoms.get_moments_of_inertia()))
print('')

print('unsymmetric rotors Ia != Ib != Ic')
atoms = molecule('C3H7Cl')
print('  C3H7Cl moments of inertia: {}'.format(atoms.get_moments_of_inertia()))

linear rotors: I = [0 Ia Ia]
CO2 moments of inertia: [  0.          44.45384271  44.45384271]

symmetric rotors (Ia = Ib) < Ic
NH3 moments of inertia: [ 1.71012426  1.71012548  2.67031768]
C6H6 moments of inertia: [  88.77914641   88.77916799  177.5583144 ]

symmetric rotors Ia < (Ib = Ic)
CH3Cl moments of inertia: [  3.20372189  37.97009644  37.97009837]

spherical rotors Ia = Ib = Ic
CH4 moments of inertia: [ 3.19145621  3.19145621  3.19145621]

unsymmetric rotors Ia != Ib != Ic
C3H7Cl moments of inertia: [  19.41351508  213.18961963  223.16255537]


If you want to know the principle axes of rotation, we simply pass vectors=True to the function, and it returns the moments of inertia and the principle axes.

from ase.structure import molecule

atoms = molecule('CH3Cl')
moments, axes = atoms.get_moments_of_inertia(vectors=True)
print('Moments = {0}'.format(moments))
print('axes = {0}'.format(axes))

Moments = [  3.20372189  37.97009644  37.97009837]
axes = [[ 0.  0.  1.]
[ 0.  1.  0.]
[ 1.  0.  0.]]


This shows the first moment is about the z-axis, the second moment is about the y-axis, and the third moment is about the x-axis.

#### 3.2.5 Computing bond lengths and angles

A typical question we might ask is, "What is the structure of a molecule?" In other words, what are the bond lengths, angles between bonds, and similar properties. The Atoms object contains an ase.Atoms.get_distance method to make this easy. To calculate the distance between two atoms, you have to specify their indices, remembering that the index starts at 0.

from ase.structure import molecule

# ammonia
atoms = molecule('NH3')

print('atom symbol')
print('===========')
for i, atom in enumerate(atoms):
print('{0:2d} {1:3s}' .format(i, atom.symbol))

# N-H bond length
s = 'The N-H distance is {0:1.3f} angstroms'
print(s.format(atoms.get_distance(0, 1)))

atom symbol
===========
0 N
1 H
2 H
3 H
The N-H distance is 1.017 angstroms


Bond angles are a little trickier. If we had vectors describing the directions between two atoms, we could use some simple trigonometry to compute the angle between the vectors: $$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\theta)$$. So we can calculate the angle as $$\theta = \arccos\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$$, we just have to define our two vectors $$\vec{a}$$ and $$\vec{b}$$. We compute these vectors as the difference in positions of two atoms. For example, here we compute the angle H-N-H in an ammonia molecule. This is the angle between N-H$$_1$$ and N-H$$_2$$. In the next example, we utilize functions in numpy to perform the calculations, specifically the numpy.arccos function, the numpy.dot function, and numpy.linalg.norm functions.

from ase.structure import molecule

# ammonia
atoms = molecule('NH3')

print('atom symbol')
print('===========')
for i, atom in enumerate(atoms):
print('{0:2d} {1:3s}'.format(i, atom.symbol))

a = atoms.positions[0] - atoms.positions[1]
b = atoms.positions[0] - atoms.positions[2]

from numpy import arccos, dot, pi
from numpy.linalg import norm

print('theta = {0:1.1f} degrees'.format(theta_rad * 180./pi))

atom symbol
===========
0 N
1 H
2 H
3 H
theta = 106.3 degrees


Alternatively you could use ase.Atoms.get_angle. Note we want the angle between atoms with indices [1, 0, 2] to get the H-N-H angle.

from ase.structure import molecule
from numpy import pi
# ammonia
atoms = molecule('NH3')

print('theta = {0} degrees'.format(atoms.get_angle([1, 0, 2]) * 180. / pi))

theta = 106.334624232 degrees

##### 3.2.5.1 Dihedral angles

There is support in ase for computing dihedral angles. Let us illustrate that for ethane. We will compute the dihedral angle between atoms 5, 1, 0, and 4. That is a H-C-C-H dihedral angle, and one can visually see (although not here) that these atoms have a dihedral angle of 60° (Figure fig:ethane-dihedral).

# calculate an ethane dihedral angle
from ase.structure import molecule
import numpy as np

atoms = molecule('C2H6')

print('atom symbol')
print('===========')
for i, atom in enumerate(atoms):
print('{0:2d} {1:3s}'.format(i, atom.symbol))

da = atoms.get_dihedral([5, 1, 0, 4]) * 180. / np.pi
print('dihedral angle = {0:1.2f} degrees'.format(da))

atom symbol
===========
0 C
1 C
2 H
3 H
4 H
5 H
6 H
7 H
dihedral angle = 60.00 degrees


In this section we covered properties that require simple calculations, but not DFT calculations, to compute.

### 3.3 Simple properties that require single computations

There are many properties that only require a single DFT calculation to obtain the energy, forces, density of states, electron density and electrostatic potential. This section describes some of these calculations and their analysis.

#### 3.3.1 Energy and forces

Two of the most important quantities we are interested in are the total energy and the forces on the atoms. To get these quantities, we have to define a calculator and attach it to an ase.Atoms object so that ase knows how to get the data. After defining the calculator a DFT calculation must be run.

Here is an example of getting the energy and forces from a CO molecule. The forces in this case are very high, indicating that this geometry is not close to the ground state geometry. Note that the forces are only along the $x$-axis, which is along the molecular axis. We will see how to minimize this force in Manual determination and Automatic geometry optimization with VASP.

from ase import Atoms, Atom
from vasp import Vasp

co = Atoms([Atom('C', [0, 0, 0]),
Atom('O', [1.2, 0, 0])],
cell=(6., 6., 6.))

calc = Vasp('molecules/simple-co',  # output dir
xc='pbe',  # the exchange-correlation functional
nbands=6,    # number of bands
encut=350,    # planewave cutoff
ismear=1,    # Methfessel-Paxton smearing
sigma=0.01,  # very small smearing factor for a molecule
atoms=co)

print('energy = {0} eV'.format(co.get_potential_energy()))
print(co.get_forces())

energy = -14.69111507 eV
[[ 5.09138064  0.          0.        ]
[-5.09138064  0.          0.        ]]


We can see what files were created and used in this calculation by printing the vasp attribute of the calculator.

from vasp import Vasp
print(Vasp('molecules/simple-co').vasp)

INCAR
-----
INCAR created by Atomic Simulation Environment
ENCUT = 350
LCHARG = .FALSE.
NBANDS = 6
ISMEAR = 1
LWAVE = .FALSE.
SIGMA = 0.01

POSCAR
------
C  O
1.0000000000000000
6.0000000000000000    0.0000000000000000    0.0000000000000000
0.0000000000000000    6.0000000000000000    0.0000000000000000
0.0000000000000000    0.0000000000000000    6.0000000000000000
1   1
Cartesian
0.0000000000000000  0.0000000000000000  0.0000000000000000
1.2000000000000000  0.0000000000000000  0.0000000000000000

KPOINTS
-------
KPOINTS created by Atomic Simulation Environment
0
Monkhorst-Pack
1 1 1
0.0 0.0 0.0

POTCAR
------
cat $VASP_PP_PATH/potpaw_PBE/C/POTCAR$VASP_PP_PATH/potpaw_PBE/O/POTCAR > POTCAR

##### 3.3.1.1 Running a job in parallel
from ase import Atoms, Atom
from vasp import Vasp
from vasp.vasprc import VASPRC

VASPRC['queue.ppn'] = 4

co = Atoms([Atom('C', [0, 0, 0]),
Atom('O', [1.2, 0, 0])],
cell=(6., 6., 6.))

calc = Vasp('molecules/simple-co-n4',  # output dir
xc='PBE',  # the exchange-correlation functional
nbands=6,    # number of bands
encut=350,    # planewave cutoff
ismear=1,    # Methfessel-Paxton smearing
sigma=0.01,  # very small smearing factor for a molecule
atoms=co)

print('energy = {0} eV'.format(co.get_potential_energy()))
print(co.get_forces())

energy = -14.69072754 eV
[[ 5.09089107  0.          0.        ]
[-5.09089107  0.          0.        ]]

##### 3.3.1.2 Convergence with unit cell size

There are a number of parameters that affect the energy and forces including the calculation parameters and the unit cell. We will first consider the effect of the unit cell on the total energy and forces. The reason that the unit cell affects the total energy is that it can change the distribution of electrons in the molecule.

from vasp import Vasp
from ase import Atoms, Atom
import numpy as np
np.set_printoptions(precision=3, suppress=True)

atoms = Atoms([Atom('C', [0, 0, 0]),
Atom('O', [1.2, 0, 0])])

L = [4, 5, 6, 8, 10]

energies = []
for a in L:
atoms.set_cell([a, a, a], scale_atoms=False)
atoms.center()
calc = Vasp('molecules/co-L-{0}'.format(a),
encut=350,
xc='PBE',
atoms=atoms)

energies.append(atoms.get_potential_energy())

print(energies)
calc.stop_if(None in energies)

import matplotlib.pyplot as plt
plt.plot(L, energies, 'bo-')
plt.xlabel('Unit cell length ($\AA$)')
plt.ylabel('Total energy (eV)')
plt.savefig('images/co-e-v.png')

[-15.35943747, -14.85641864, -14.68750595, -14.63202061, -14.65342838]


Here there are evidently attractive interactions between the CO molecules which lower the total energy for small box sizes. We have to decide what an appropriate volume for our calculation is, and the choice depends on the goal. We may wish to know the total energy of a molecule that is not interacting with any other molecules, e.g. in the ideal gas limit. In that case we need a large unit cell so the electron density from the molecule does not go outside the unit cell where it would overlap with neighboring images.

It pays to check for convergence. The cost of running the calculation goes up steeply with increasing cell size. Doubling a lattice vector here leads to a 20-fold increase in computational time! Note that doubling a lattice vector length increases the volume by a factor of 8 for a cube. The cost goes up because the number of planewaves that fit in the cube grows as the cube gets larger.

from vasp import Vasp

L = [4, 5, 6, 8, 10]

for a in L:
calc = Vasp('molecules/co-L-{0}'.format(a))
print('{0} {1} seconds'.format(a, calc.get_elapsed_time()))

4 10.748 seconds
5 11.855 seconds
6 15.613 seconds
8 28.346 seconds
10 45.259 seconds


Let us consider what the pressure in the unit cell is. In the ideal gas limit we have $$PV = nRT$$, which gives a pressure of zero at absolute zero. At non-zero temperatures, we have $$P=n/V RT$$. Let us consider some examples. In atomic units we use $$k_B$$ instead of $$R$$.

from ase.units import kB, Pascal
import numpy as np
import matplotlib.pyplot as plt

atm = 101325 * Pascal

L = np.linspace(4, 10)
V = L**3

n = 1  # one atom/molecule per unit cell

for T in [298, 600, 1000]:
P = n / V * kB * T / atm  # convert to atmospheres

plt.plot(V, P, label='{0}K'.format(T))

plt.xlabel('Unit cell volume ($\AA^3$)')
plt.ylabel('Pressure (atm)')
plt.legend(loc='best')
plt.savefig('images/ideal-gas-pressure.png')

##### 3.3.1.3 Convergence of ENCUT

The total energy and forces also depend on the computational parameters, notably ENCUT.

from ase import Atoms, Atom
from vasp import Vasp
import numpy as np
np.set_printoptions(precision=3, suppress=True)

atoms = Atoms([Atom('C', [0, 0, 0]),
Atom('O', [1.2, 0, 0])],
cell=(6, 6, 6))
atoms.center()

ENCUTS = [250, 300, 350, 400, 450, 500]

calcs = [Vasp('molecules/co-en-{0}'.format(en),
encut=en,
xc='PBE',
atoms=atoms)
for en in ENCUTS]

energies = [calc.potential_energy for calc in calcs]
print(energies)
calcs[0].stop_if(None in energies)

import matplotlib.pyplot as plt
plt.plot(ENCUTS, energies, 'bo-')
plt.xlabel('ENCUT (eV)')
plt.ylabel('Total energy (eV)')
plt.savefig('images/co-encut-v.png')

[-14.95250419, -14.71808896, -14.68750595, -14.66725733, -14.65604528, -14.65012078]


You can see in this figure that it takes a cutoff energy of about 400 eV to achieve a convergence level around 10 meV, and that even at 500 meV the energy is still changing slightly. Keep in mind that we are generally interested in differences in total energy, and the differences tend to converge faster than a single total energy. Also it is important to note that it is usually a single element that determines the rate of convergence. The reason we do not just use very high ENCUT all the time is it is expensive.

grep "Elapsed time (sec):" molecules/co-en-*/OUTCAR

molecules/co-en-250/OUTCAR:                         Elapsed time (sec):       11.634
molecules/co-en-300/OUTCAR:                         Elapsed time (sec):       14.740
molecules/co-en-350/OUTCAR:                         Elapsed time (sec):       13.577
molecules/co-en-400/OUTCAR:                         Elapsed time (sec):       16.310
molecules/co-en-450/OUTCAR:                         Elapsed time (sec):       17.704
molecules/co-en-500/OUTCAR:                         Elapsed time (sec):       11.658

##### 3.3.1.4 Cloning

You may want to clone a calculation, so you can change some parameter without losing the previous result. The clone function does this, and changes the calculator over to the new directory.

from ase import Atoms, Atom
from vasp import Vasp

calc = Vasp('molecules/simple-co')
print('energy = {0} eV'.format(calc.get_atoms().get_potential_energy()))

# This creates the directory and makes it current working directory
calc.clone('molecules/clone-1')
calc.set(encut=325)  # this will trigger a new calculation

print('energy = {0} eV'.format(calc.get_atoms().get_potential_energy()))

energy = -14.69111507 eV
energy = -14.77117554 eV


#### 3.3.2 Visualizing electron density

The electron density is a 3$$d$$ quantity: for every $$(x,y,z)$$ point, there is a charge density. That means we need 4 numbers for each point: $$(x,y,z)$$ and $$\rho(x,y,z)$$. Below we show an example (Figure fig:cd1) of plotting the charge density, and we consider some issues we have to consider when visualizing volumetric data in unit cells with periodic boundary conditions. We will use the results from a previous calculation.

from vasp import Vasp
from enthought.mayavi import mlab
from ase.data.colors import cpk_colors

calc = Vasp('molecules/simple-co')
calc.clone('molecules/co-chg')
calc.set(lcharg=True)
calc.stop_if(calc.potential_energy is None)

atoms = calc.get_atoms()
x, y, z, cd = calc.get_charge_density()

# make a white figure
mlab.figure(1, bgcolor=(1, 1, 1))

# plot the atoms as spheres
for atom in atoms:
mlab.points3d(atom.x,
atom.y,
atom.z,
#this determines the size of the atom
resolution=20,
# a tuple is required for the color
color=tuple(cpk_colors[atom.number]),
scale_mode='none')

# draw the unit cell - there are 8 corners, and 12 connections
a1, a2, a3 = atoms.get_cell()
origin = [0, 0, 0]
cell_matrix = [[origin,  a1],
[origin,  a2],
[origin,  a3],
[a1,      a1 + a2],
[a1,      a1 + a3],
[a2,      a2 + a1],
[a2,      a2 + a3],
[a3,      a1 + a3],
[a3,      a2 + a3],
[a1 + a2, a1 + a2 + a3],
[a2 + a3, a1 + a2 + a3],
[a1 + a3, a1 + a3 + a2]]

for p1, p2 in cell_matrix:
mlab.plot3d([p1[0], p2[0]], # x-positions
[p1[1], p2[1]], # y-positions
[p1[2], p2[2]], # z-positions

# Now plot the charge density

mlab.contour3d(x, y, z, cd)
mlab.view(azimuth=-90, elevation=90, distance='auto')

mlab.savefig('images/co-cd.png')


If we take care to center the CO molecule in the unit cell, we get a nicer looking result.

from vasp import Vasp
from enthought.mayavi import mlab
from ase.data.colors import cpk_colors
from ase import Atom, Atoms

atoms = Atoms([Atom('C', [2.422, 0.0, 0.0]),
Atom('O', [3.578, 0.0, 0.0])],
cell=(10,10,10))

atoms.center()

calc = Vasp('molecules/co-centered',
encut=350,
xc='PBE',
atoms=atoms)
calc.set(lcharg=True,)
calc.stop_if(calc.potential_energy is None)

atoms = calc.get_atoms()
x, y, z, cd = calc.get_charge_density()

mlab.figure(bgcolor=(1, 1, 1))

# plot the atoms as spheres
for atom in atoms:
mlab.points3d(atom.x,
atom.y,
atom.z,
resolution=20,
# a tuple is required for the color
color=tuple(cpk_colors[atom.number]),
scale_mode='none')

# draw the unit cell - there are 8 corners, and 12 connections
a1, a2, a3 = atoms.get_cell()
origin = [0, 0, 0]
cell_matrix = [[origin,  a1],
[origin,  a2],
[origin,  a3],
[a1,      a1 + a2],
[a1,      a1 + a3],
[a2,      a2 + a1],
[a2,      a2 + a3],
[a3,      a1 + a3],
[a3,      a2 + a3],
[a1 + a2, a1 + a2 + a3],
[a2 + a3, a1 + a2 + a3],
[a1 + a3, a1 + a3 + a2]]

for p1, p2 in cell_matrix:
mlab.plot3d([p1[0], p2[0]], # x-positions
[p1[1], p2[1]], # y-positions
[p1[2], p2[2]], # z-positions

# Now plot the charge density
mlab.contour3d(x, y, z, cd, transparent=True)

# this view was empirically found by iteration
mlab.view(azimuth=-90, elevation=90, distance='auto')

mlab.savefig('images/co-centered-cd.png')
mlab.show()


TODO: how to make this figure

#### 3.3.3 Visualizing electron density differences

Here, we visualize how charge moves in a benzene ring when you substitute an H atom with an electronegative Cl atom.

#!/usr/bin/env python
from ase import *
from ase.structure import molecule
from vasp import Vasp

### Setup calculators
benzene = molecule('C6H6')
benzene.set_cell([10, 10, 10])
benzene.center()

calc1 = Vasp('molecules/benzene',
xc='PBE',
nbands=18,
encut=350,
atoms=benzene)
calc1.set(lcharg=True)

chlorobenzene =  molecule('C6H6')
chlorobenzene.set_cell([10, 10, 10])
chlorobenzene.center()
chlorobenzene[11].symbol ='Cl'

calc2 = Vasp('molecules/chlorobenzene',
xc='PBE',
nbands=22,
encut=350,
atoms=chlorobenzene)
calc2.set(lcharg=True)
calc2.stop_if(None in (calc1.potential_energy, calc2.potential_energy))

x1, y1, z1, cd1 = calc1.get_charge_density()
x2, y2, z2, cd2 = calc2.get_charge_density()

cdiff = cd2 - cd1

print(cdiff.min(), cdiff.max())
##########################################

##### set up visualization of charge difference
from enthought.mayavi import mlab
mlab.contour3d(x1, y1, z1, cdiff,
contours=[-0.02, 0.02])

mlab.savefig('images/cdiff.png')

(-2.0821159999999987, 2.9688999999999979)

#!/usr/bin/env python
from ase import *
from ase.structure import molecule
from vasp import Vasp
import bisect
import numpy as np

def vinterp3d(x, y, z, u, xi, yi, zi):
"Interpolate the point (xi, yi, zi) from the values at u(x, y, z)"
p = np.array([xi, yi, zi])

#1D arrays of coordinates
xv = x[:, 0, 0]
yv = y[0, :, 0]
zv = z[0, 0, :]

# we subtract 1 because bisect tells us where to insert the
# element to maintain an ordered list, so we want the index to the
# left of that point
i = bisect.bisect_right(xv, xi) - 1
j = bisect.bisect_right(yv, yi) - 1
k = bisect.bisect_right(zv, zi) - 1

if i == len(x) - 1:
i = len(x) - 2
elif i < 0:
i = 0

if j == len(y) - 1:
j = len(y) - 2
elif j < 0:
j = 0

if k == len(z) - 1:
k = len(z) - 2
elif k < 0:
k = 0

# points at edge of cell. We only need P1, P2, P3, and P5
P1 = np.array([x[i, j, k],
y[i, j, k],
z[i, j, k]])
P2 = np.array([x[i + 1, j, k],
y[i + 1, j, k],
z[i + 1, j, k]])
P3 = np.array([x[i, j + 1, k],
y[i, j + 1, k],
z[i, j + 1, k]])
P5 = np.array([x[i, j, k + 1],
y[i, j, k + 1],
z[i, j, k + 1]])

# values of u at edge of cell
u1 = u[i, j, k]
u2 = u[i+1, j, k]
u3 = u[i, j+1, k]
u4 = u[i+1, j+1, k]
u5 = u[i, j, k+1]
u6 = u[i+1, j, k+1]
u7 = u[i, j+1, k+1]
u8 = u[i+1, j+1, k+1]

# cell basis vectors, not the unit cell, but the voxel cell containing the point
cbasis = np.array([P2 - P1,
P3 - P1,
P5 - P1])

# now get interpolated point in terms of the cell basis
s = np.dot(np.linalg.inv(cbasis.T), np.array([xi, yi, zi]) - P1)

# now s = (sa, sb, sc) which are fractional coordinates in the vector space
# next we do the interpolations
ui1 = u1 + s[0] * (u2 - u1)
ui2 = u3 + s[0] * (u4 - u3)

ui3 = u5 + s[0] * (u6 - u5)
ui4 = u7 + s[0] * (u8 - u7)

ui5 = ui1 + s[1] * (ui2 - ui1)
ui6 = ui3 + s[1] * (ui4 - ui3)

ui7 = ui5 + s[2] * (ui6 - ui5)

return ui7

### Setup calculators
calc = Vasp('molecules/benzene')
benzene = calc.get_atoms()
x1, y1, z1, cd1 = calc.get_charge_density()

calc = Vasp('molecules/chlorobenzene')
x2, y2, z2, cd2 = calc.get_charge_density()

cdiff = cd2 - cd1

#we need the x-y plane at z=5
import matplotlib.pyplot as plt
from scipy import mgrid

X, Y = mgrid[0: 10: 25j, 0: 10: 25j]

cdiff_plane = np.zeros(X.shape)
ni, nj = X.shape

for i in range(ni):
for j in range(nj):
cdiff_plane[i, j] = vinterp3d(x1, y1, z1, cdiff,
X[i, j], Y[i, j], 5.0)

plt.imshow(cdiff_plane.T,
vmin=-0.02,  # min charge diff to plot
vmax=0.02,  # max charge diff to plot
cmap=plt.cm.gist_heat,  # colormap
extent=(0., 10., 0., 10.))  # axes limits

# plot atom positions. It is a little tricky to see why we reverse the x and y coordinates. That is because imshow does that.
x = [a.x for a in benzene]
y = [a.y for a in benzene]
plt.plot(x, y, 'bo')

plt.savefig('images/cdiff-imshow.png')
plt.show()


#### 3.3.4 Dipole moments

dipole moment The dipole moment is a vector describing the separation of electrical (negative) and nuclear (positive) charge. The magnitude of this vector is the dipole moment, which has units of Coulomb-meter, or more commonly Debye. The symmetry of a molecule determines if a molecule has a dipole moment or not. Below we compute the dipole moment of CO. We must integrate the electron density to find the center of electrical charge, and compute a sum over the nuclei to find the center of positive charge.

from vasp import Vasp
from vasp.VaspChargeDensity import VaspChargeDensity
import numpy as np
from ase.units import Debye
import os

calc = Vasp('molecules/co-centered')
atoms = calc.get_atoms()
calc.stop_if(atoms.get_potential_energy() is None)

vcd = VaspChargeDensity('molecules/co-centered/CHG')

cd = np.array(vcd.chg[0])

n0, n1, n2 = cd.shape

s0 = 1.0 / n0
s1 = 1.0 / n1
s2 = 1.0 / n2

X, Y, Z = np.mgrid[0.0:1.0:s0,
0.0:1.0:s1,
0.0:1.0:s2]

C = np.column_stack([X.ravel(),
Y.ravel(),
Z.ravel()])

atoms = calc.get_atoms()
uc = atoms.get_cell()
real = np.dot(C, uc)

# now convert arrays back to unitcell shape
x = np.reshape(real[:, 0], (n0, n1, n2))
y = np.reshape(real[:, 1], (n0, n1, n2))
z = np.reshape(real[:, 2], (n0, n1, n2))

nelements = n0 * n1 * n2
voxel_volume = atoms.get_volume() / nelements
total_electron_charge = -cd.sum() * voxel_volume

electron_density_center = np.array([(cd * x).sum(),
(cd * y).sum(),
(cd * z).sum()])
electron_density_center *= voxel_volume
electron_density_center /= total_electron_charge

electron_dipole_moment = -electron_density_center * total_electron_charge

# now the ion charge center. We only need the Zval listed in the potcar
from vasp.POTCAR import get_ZVAL

LOP = calc.get_pseudopotentials()
ppp = os.environ['VASP_PP_PATH']

zval = {}
for sym, ppath, hash in LOP:
fullpath = os.path.join(ppp, ppath)
z = get_ZVAL(fullpath)
zval[sym] = z
ion_charge_center = np.array([0.0, 0.0, 0.0])
total_ion_charge = 0.0

for atom in atoms:
Z = zval[atom.symbol]
total_ion_charge += Z
pos = atom.position
ion_charge_center += Z * pos

ion_charge_center /= total_ion_charge
ion_dipole_moment = ion_charge_center * total_ion_charge

dipole_vector = (ion_dipole_moment + electron_dipole_moment)

dipole_moment = ((dipole_vector**2).sum())**0.5 / Debye
print('The dipole vector is {0}'.format(dipole_vector))
print('The dipole moment is {0:1.2f} Debye'.format(dipole_moment))

The dipole vector is [ 0.02048406  0.00026357  0.00026357]
The dipole moment is 0.10 Debye


Note that a function using the code above exists in vasp which makes it trivial to compute the dipole moment. Here is an example of its usage.

from vasp import Vasp
from ase.units import Debye

calc = Vasp('molecules/co-centered')

dipole_moment = calc.get_dipole_moment()
print('The dipole moment is {0:1.2f} Debye'.format(dipole_moment))

The dipole moment is 0.10 Debye


#### 3.3.5 The density of states (DOS)

The density of states (DOS) gives you the number of electronic states (i.e., the orbitals) that have a particular energy. We can get this information from the last calculation we just ran without having to run another DFT calculation.

from vasp import Vasp
from ase.dft.dos import DOS
import matplotlib.pyplot as plt

calc = Vasp('molecules/simple-co')   # we already ran this!
dos = DOS(calc)
plt.plot(dos.get_energies(), dos.get_dos())
plt.xlabel('Energy - $E_f$ (eV)')
plt.ylabel('DOS')

# make sure you save the figure outside the with statement, or provide
# the correct relative or absolute path to where you want it.
plt.savefig('images/co-dos.png')


#### 3.3.6 Atom-projected density of states on molecules

Let us consider which states in the density of states belong to which atoms in a molecule. This can only be a qualitative consideration because the orbitals on the atoms often hybridize to form molecular orbitals, e.g. in methane the $$s$$ and $$p$$ orbitals can form what we call $$sp^3$$ orbitals. We can compute atom-projected density of states in VASP, which is done by projecting the wave function onto localized atomic orbitals. Here is an example. We will consider the CO molecule. To get atom-projected density of states, we must set RWIGS for each atom. This parameter defines the radius of the sphere around the atom which cuts off the projection. The total density of states and projected density of states information comes from the DOSCAR file.

Note that unlike the DOS, here we must run another calculation because we did not specify the atom-projected keywords above. Our strategy is to get the atoms from the previous calculation, and use them in a new calculation. You could redo the calculation in the same directory, but you risk losing the results of the first step. That can make it difficult to reproduce a result. We advocate our approach of using multiple directories for the subsequent calculations, because it leaves a clear trail of how the work was done.

The RWIGS is not uniquely determined for an element. There are various natural choices, e.g. the ionic radius of an atom, or a value that minimizes overlap of neighboring spheres, but these values can change slightly in different environments.

You can also get spin-polarized atom-projected DOS, and magnetization projected DOS. See http://cms.mpi.univie.ac.at/vasp/vasp/DOSCAR_file.html#doscar for more details.

from vasp import Vasp
from ase.dft.dos import DOS
import matplotlib.pyplot as plt

# get the geometry from another calculation
calc = Vasp('molecules/simple-co')
atoms = calc.get_atoms()

encut=300,
xc='PBE',
rwigs={'C': 1.0, 'O': 1.0},     # these are the cutoff radii for projected states
atoms=atoms)

calc.stop_if(calc.potential_energy is None)

# now get results
dos = DOS(calc)
plt.plot(dos.get_energies(), dos.get_dos() + 10)

plt.plot(energies, c_s + 6, energies, o_s + 5)
plt.plot(energies, c_p + 4, energies, o_p + 3)
plt.plot(energies, c_d, energies, o_d + 2)
plt.xlabel('Energy - $E_f$ (eV)')
plt.ylabel('DOS')
plt.legend(['DOS',
'C$_s$', 'O$_s$',
'C$_p$', 'O$_p$',
'C$_d$', 'O$_d$'],
ncol=2, loc='best')


#### 3.3.7 Electrostatic potential

This is an example of the so-called σ hole in a halogen bond. The coordinates for the CF3Br molecule were found at http://cccbdb.nist.gov/exp2.asp?casno=75638.

from vasp import Vasp
from ase import Atom, Atoms
from ase.io import write

from enthought.mayavi import mlab
from ase.data.colors import cpk_colors

atoms = Atoms([Atom('C',  [ 0.0000,     0.0000,         -0.8088]),
Atom('Br', [ 0.0000,     0.0000,          1.1146]),
Atom('F',  [ 0.0000,     1.2455,         -1.2651]),
Atom('F',  [ 1.0787,    -0.6228,         -1.2651]),
Atom('F',  [-1.0787,    -0.6228,         -1.2651])],
cell=(10, 10, 10))
atoms.center()

calc = Vasp('molecules/CF3Br',
encut=350,
xc='PBE',
ibrion=1,
nsw=50,
lcharg=True,
lvtot=True,
lvhar=True,
atoms=atoms)
calc.set_nbands(f=2)
calc.stop_if(calc.potential_energy is None)

x, y, z, lp = calc.get_local_potential()
x, y, z, cd = calc.get_charge_density()

mlab.figure(1, bgcolor=(1, 1, 1)) # make a white figure

# plot the atoms as spheres
for atom in atoms:
mlab.points3d(atom.x,
atom.y,
atom.z,
resolution=20,
# a tuple is required for the color
color=tuple(cpk_colors[atom.number]),
scale_mode='none')
# plot the bonds. We want a line from C-Br, C-F, etc...
# We create a bond matrix showing which atoms are connected.
bond_matrix = [[0, 1],
[0, 2],
[0, 3],
[0, 4]]

for a1, a2 in bond_matrix:
mlab.plot3d(atoms.positions[[a1,a2], 0], # x-positions
atoms.positions[[a1,a2], 1], # y-positions
atoms.positions[[a1,a2], 2], # z-positions
[2, 2],
colormap='Reds')

mlab.contour3d(x, y, z, lp)
mlab.savefig('images/halogen-ep.png')
mlab.show()


See http://www.uni-due.de/~hp0058/?file=manual03.html&dir=vmdplugins for examples of using VMD for visualization.

Note: Thanks to @prtkm for helping improve this section (https://github.com/jkitchin/dft-book/issues/2). Bader analysis is a charge partitioning scheme where charge is divided by surfaces of zero flux that define atomic basins of charge. The most modern way of calculating the Bader charges is using the bader program from Graeme Henkelmen's group Henkelman2006354,doi.10.1021/ct100125x. Let us consider a water molecule, centered in a box. The strategy is first to run the calculation, then run the bader program on the results.

We have to specify laechg to be true so that the all-electron core charges will be written out to files. Here we setup and run the calculation to get the densities first.

from vasp import Vasp

from ase.structure import molecule
atoms = molecule('H2O')
atoms.center(vacuum=6)

xc='PBE',
encut=350,
lcharg=True,
laechg=True,
atoms=atoms)
print calc.potential_energy

-14.22250648


Now that the calculation is done, get the bader code and scripts from http://theory.cm.utexas.edu/henkelman/code/bader/.

We use this code to see the changes in charges on the atoms.

from vasp import Vasp

atoms = calc.get_atoms()
for atom in atoms:
print('|{0} | {1} |'.format(atom.symbol, atom.charge))

|O | -1.2326 |
|H | 0.6161 |
|H | 0.6165 |


You can see some charge has been "transferred" from H to O.

### 3.4 Geometry optimization

#### 3.4.1 Manual determination of a bond length

The equilibrium bond length of a CO molecule is approximately the bond length that minimizes the total energy. We can find that by computing the total energy as a function of bond length, and noting where the minimum is. Here is an example in VASP. There are a few features to point out here. We want to compute 5 bond lengths, and each calculation is independent of all the others. vasp is set up to automatically handle jobs for you by submitting them to the queue. It raises a variety of exceptions to let you know what has happened, and you must handle these to control the workflow. We will illustrate this by the following examples.

from vasp import Vasp
from ase import Atom, Atoms

bond_lengths = [1.05, 1.1, 1.15, 1.2, 1.25]
energies = []

for d in bond_lengths:  # possible bond lengths

co = Atoms([Atom('C', [0, 0, 0]),
Atom('O', [d, 0, 0])],
cell=(6, 6, 6))

calc = Vasp('molecules/co-{0}'.format(d),  # output dir
xc='PBE',
nbands=6,
encut=350,
ismear=1,
sigma=0.01,
atoms=co)

energies.append(co.get_potential_energy())
print('d = {0:1.2f} ang'.format(d))
print('energy = {0:1.3f} eV'.format(energies[-1] or 0))
print('forces = (eV/ang)\n {0}'.format(co.get_forces()))
print('')  # blank line

if None in energies:
calc.abort()
else:
import matplotlib.pyplot as plt
plt.plot(bond_lengths, energies, 'bo-')
plt.xlabel(r'Bond length ($\AA$)')
plt.ylabel('Total energy (eV)')
plt.savefig('images/co-bondlengths.png')

d = 1.05 ang
energy = -14.216 eV
forces = (eV/ang)
[[-14.93017486   0.           0.        ]
[ 14.93017486   0.           0.        ]]

d = 1.10 ang
energy = -14.722 eV
forces = (eV/ang)
[[-5.81988086  0.          0.        ]
[ 5.81988086  0.          0.        ]]

d = 1.15 ang
energy = -14.841 eV
forces = (eV/ang)
[[ 0.63231023  0.          0.        ]
[-0.63231023  0.          0.        ]]

d = 1.20 ang
energy = -14.691 eV
forces = (eV/ang)
[[ 5.09138064  0.          0.        ]
[-5.09138064  0.          0.        ]]

d = 1.25 ang
energy = -14.355 eV
forces = (eV/ang)
[[ 8.14027842  0.          0.        ]
[-8.14027842  0.          0.        ]]


Before continuing, it is worth looking at some other approaches to setup and run these calculations. Here we consider a functional approach that uses list comprehensions pretty extensively.

from vasp import Vasp
from ase import Atom, Atoms

bond_lengths = [1.05, 1.1, 1.15, 1.2, 1.25]

ATOMS = [Atoms([Atom('C', [0, 0, 0]),
Atom('O', [d, 0, 0])],
cell=(6, 6, 6))
for d in bond_lengths]

calcs = [Vasp('molecules/co-{0}'.format(d),  # output dir
xc='PBE',
nbands=6,
encut=350,
ismear=1,
sigma=0.01,
atoms=atoms)
for d, atoms in zip(bond_lengths, ATOMS)]

energies = [atoms.get_potential_energy() for atoms in ATOMS]

print(energies)

[-14.21584765, -14.72174343, -14.84115208, -14.69111507, -14.35508371]


We can retrieve data similarly.

from vasp import Vasp

bond_lengths = [1.05, 1.1, 1.15, 1.2, 1.25]
calcs = [Vasp('molecules/co-{0}'.format(d)) for d in bond_lengths]

energies = [calc.get_atoms().get_potential_energy() for calc in calcs]

print(energies)

[-14.21584765, -14.72174343, -14.84115208, -14.69111507, -14.35508371]

from vasp import Vasp
from ase.db import connect

bond_lengths = [1.05, 1.1, 1.15, 1.2, 1.25]
calcs = [Vasp('molecules/co-{0}'.format(d)) for d in bond_lengths]

con = connect('co-database.db', append=False)
for atoms in [calc.get_atoms() for calc in calcs]:
con.write(atoms)


Here we just show that there are entries in our database. If you run the code above many times, each time will add new entries.

ase-db co-database.db

id|age|user    |formula|calculator| energy|  fmax|pbc| volume|charge|  mass| smax|magmom
1|12s|jkitchin|CO     |vasp      |-14.216|14.930|TTT|216.000| 0.000|28.010|0.060| 0.000
2|10s|jkitchin|CO     |vasp      |-14.722| 5.820|TTT|216.000| 0.000|28.010|0.017| 0.000
3| 9s|jkitchin|CO     |vasp      |-14.841| 0.632|TTT|216.000| 0.000|28.010|0.017| 0.000
4| 9s|jkitchin|CO     |vasp      |-14.691| 5.091|TTT|216.000| 0.000|28.010|0.041| 0.000
5| 7s|jkitchin|CO     |vasp      |-14.355| 8.140|TTT|216.000| 0.000|28.010|0.060| 0.000
Rows: 5


This database is now readable in Python too. Here we read in all the results. Later we will learn how to select specific entries.

from ase.io import read

print([a[0].x - a[1].x for a in ATOMS])
print([atoms.get_potential_energy() for atoms in ATOMS])

[-1.0499999999999998, -1.09999998, -1.15000002, -1.2000000000000002, -1.2499999800000001]
[-14.21584765, -14.72174343, -14.84115208, -14.69111507, -14.35508371]


Now, back to the goal of finding the minimum. To find the minimum we could run more calculations, but a simpler and faster way is to fit a polynomial to the data, and find the analytical minimum. The results are shown in Figure fig:co-bondlengths.

from vasp import Vasp
import numpy as np
import matplotlib.pyplot as plt

bond_lengths = [1.05, 1.1, 1.15, 1.2, 1.25]
energies = []

for d in bond_lengths:  # possible bond lengths

calc = Vasp('molecules/co-{0}'.format(d))
atoms = calc.get_atoms()
energies.append(atoms.get_potential_energy())

# Now we fit an equation - cubic polynomial
pp = np.polyfit(bond_lengths, energies, 3)
dp = np.polyder(pp)  # first derivative - quadratic

# we expect two roots from the quadratic eqn. These are where the
# first derivative is equal to zero.
roots = np.roots(dp)

# The minimum is where the second derivative is positive.
dpp = np.polyder(dp)  # second derivative - line
secd = np.polyval(dpp, roots)

minV = roots[secd > 0]
minE = np.polyval(pp, minV)

print('The minimum energy is {0[0]} eV at V = {1[0]} Ang^3'.format(minE, minV))

# plot the fit
x = np.linspace(1.05, 1.25)
fit = np.polyval(pp, x)

plt.plot(bond_lengths, energies, 'bo ')
plt.plot(x, fit, 'r-')
plt.plot(minV, minE, 'm* ')
plt.legend(['DFT', 'fit', 'minimum'], numpoints=1)
plt.xlabel(r'Bond length ($\AA$)')
plt.ylabel('Total energy (eV)')
plt.savefig('images/co-bondlengths.png')

The minimum energy is -14.8458440947 eV at V = 1.14437582331 Ang^3


#### 3.4.2 Automatic geometry optimization with VASP

It is generally the case that the equilibrium geometry of a system is the one that minimizes the total energy and forces. Since each atom has three degrees of freedom, you can quickly get a high dimensional optimization problem. Luckily, VASP has built-in geometry optimization using the IBRION and NSW tags. Here we compute the bond length for a CO molecule, letting VASP do the geometry optimization for us.

Here are the most common choices for IBRION.

IBRION value algorithm
1 quasi-Newton (use if initial guess is good)

VASP applies a criteria for stopping a geometry optimization. When the change in energy between two steps is less than 0.001 eV (or 10*EDIFF), the relaxation is stopped. This criteria is controlled by the EDIFFG tag. If you prefer to stop based on forces, set EDIFFG=-0.05, i.e. to a negative number. The units of force is eV/Å. For most work, a force tolerance of 0.05 eV/Å is usually sufficient.

from ase import Atom, Atoms
from vasp import Vasp

co = Atoms([Atom('C',[0, 0, 0]),
Atom('O',[1.2, 0, 0])],
cell=(6, 6, 6))

calc = Vasp('molecules/co-cg',
xc='PBE',
nbands=6,
encut=350,
ismear=1,
sigma=0.01, # this is small for a molecule
nsw=5,      # do at least 5 steps to relax
atoms=co)

print('Forces')
print('=======')
print(co.get_forces())

pos = co.get_positions()
d = ((pos[0] - pos[1])**2).sum()**0.5
print('Bondlength = {0:1.2f} angstroms'.format(d))

Forces
=======
[[-0.8290116  0.         0.       ]
[ 0.8290116  0.         0.       ]]
Bondlength = 1.14 angstroms


#### 3.4.3 Relaxation of a water molecule

It is not more complicated to relax more atoms, it just may take longer because there are more electrons and degrees of freedom. Here we relax a water molecule which has three atoms.

from ase import Atoms, Atom
from vasp import Vasp

atoms = Atoms([Atom('H', [0.5960812, -0.7677068, 0.0000000]),
Atom('O', [0.0000000,  0.0000000, 0.0000000]),
Atom('H', [0.5960812,  0.7677068, 0.0000000])],
cell=(8, 8, 8))

atoms.center()

calc = Vasp('molecules/h2o-relax-centered',
xc='PBE',
encut=400,
ismear=0,  # Gaussian smearing
ibrion=2,
ediff=1e-8,
nsw=10,
atoms=atoms)

print("forces")
print('=======')
print(atoms.get_forces())

[[ 4.2981572   3.23149312  4.        ]
[ 3.70172616  4.          4.        ]
[ 4.2981572   4.76850688  4.        ]]
forces
=======
[[ -3.49600000e-05   5.06300000e-05   0.00000000e+00]
[  6.99200000e-05   0.00000000e+00   0.00000000e+00]
[ -3.49600000e-05  -5.06300000e-05   0.00000000e+00]]

from vasp import Vasp
calc = Vasp('molecules/h2o-relax-centered')

from ase.visualize import view
view(calc.traj)


### 3.5 Vibrational frequencies

#### 3.5.1 Manual calculation of vibrational frequency

The principle idea in calculating vibrational frequencies is that we consider a molecular system as masses connected by springs. If the springs are Hookean, e.g. the force is proportional to the displacement, then we can readily solve the equations of motion and find that the vibrational frequencies are related to the force constants and the masses of the atoms. For example, in a simple molecule like CO where there is only one spring, the frequency is:

$$\nu = \frac{1}{2\pi}\sqrt{k/\mu}$$ where $$\frac{1}{\mu} = \frac{1}{m_C} + \frac{1}{m_O}$$ and $$k$$ is the spring constant. We will compute the value of $$k$$ from DFT calculations as follows:

$$k = \frac{\partial^2E}{\partial x^2}$$ at the equilibrium bond length. We actually already have the data to do this from Manual determination. We only need to fit an equation to the energy vs. bond-length data, find the minimum energy bond-length, and then evaluate the second derivative of the fitted function at the minimum. We will use a cubic polynomial for demonstration here. Polynomials are numerically convenient because they are easy to fit, and it is trivial to get the roots and derivatives of the polynomials, as well as to evaluate them at other points using numpy.polyfit, numpy.polyder, and numpy.polyval.

from vasp import Vasp
import numpy as np
from ase.units import *

bond_lengths = [1.05, 1.1, 1.15, 1.2, 1.25]
energies = []

for d in bond_lengths:
calc = Vasp('molecules/co-{0}'.format(d))
atoms = calc.get_atoms()
energies.append(atoms.get_potential_energy())

# fit the data
pars = np.polyfit(bond_lengths, energies, 3)
xfit = np.linspace(1.05, 1.25)
efit = np.polyval(pars, xfit)

# first derivative
dpars = np.polyder(pars)
# find where the minimum is. chose the second one because it is the
# minimum we need.
droots = np.roots(dpars)

# second derivative
ddpars = np.polyder(dpars)

d_min = droots[np.polyval(ddpars, droots) > 0]

# curvature at minimum = force constant in SI units
k = np.polyval(ddpars, d_min) / (J / m**2)

# mu, reduced mass
from ase.data import atomic_masses
C_mass = atomic_masses[6] / kg
O_mass = atomic_masses[8] / kg

mu = 1.0 / (1.0 / C_mass + 1.0 / O_mass)

frequency = 1. / (2. * np.pi) * np.sqrt(k / mu)
print('The CO vibrational frequency is {0} Hz'.format(*frequency))
print('The CO vibrational frequency is {0[0]} cm^{{-1}}'.format(frequency / 3e10))

import matplotlib.pyplot as plt
plt.plot(bond_lengths, energies, 'bo ')
plt.plot(xfit, efit, 'b-')
plt.xlabel('Bond length ($\AA$)')
plt.ylabel('Total energy (eV)')
plt.savefig('images/co-freq.png')

The CO vibrational frequency is 6.43186126691e+13 Hz
The CO vibrational frequency is 2143.95375564 cm^{-1}


This result is in good agreement with experiment. The procedure described above is basically how many vibrational calculations are performed. With more atoms, you have to determine a force constant matrix and diagonalize it. For more details, see wilson1955. In practice, we usually allow a packaged code to automate this, which we cover in Automated vibrational calculations.

We now consider how much energy is in this vibration. This is commonly called zero-point energy (ZPE) and it is defined as $$E_{ZPE} = \frac{1}{2} h \nu$$ for a single mode, and $$h$$ is Planck's constant (4.135667516e-15 eV/s).

c = 3e10  # speed of light cm/s
h = 4.135667516e-15  # eV*s

nu = 2143.6076625*c  # 1/s

E_zpe = 0.5*h*nu

print('E_ZPE = {0:1.3f} eV'.format(E_zpe))

E_ZPE = 0.133 eV


This is a reasonable amount of energy! Zero-point energy increases with increasing vibrational frequency, and tends to be very important for small atoms.

A final note is that this analysis is in the "harmonic approximation". The frequency equation is the solution to a harmonic oscillator. If the spring is non-linear, then there are anharmonic effects that may become important, especially at higher temperatures.

#### 3.5.2TODO Automated vibrational calculations

VASP has built-in capability for performing vibrational calculations. We access the capability by using a new value for IBRION. The values of 5 and 6 calculated the Hessian matrix using finite differences. For IBRION=5, all atoms that are not constrained are displaced. For IBRION=6, only symmetry inequivalent displacements are considered, which makes the calculations slightly cheaper. You can specify the number of displacements with NFREE. The default number of displacements is 2. You can also specify the size of the displacement with POTIM (the default is 0.015 Å).

# <<water-vib>>
from ase import Atoms, Atom
from vasp import Vasp
import ase.units

atoms = Atoms([Atom('H', [0.5960812,  -0.7677068,   0.0000000]),
Atom('O', [0.0000000,   0.0000000,   0.0000000]),
Atom('H', [0.5960812,   0.7677068,   0.0000000])],
cell=(8, 8, 8))
atoms.center()

calc = Vasp('molecules/h2o_vib',
xc='PBE',
encut=400,
ismear=0,     # Gaussian smearing
ibrion=6,     # finite differences with symmetry
nfree=2,      # central differences (default)
potim=0.015,  # default as well
ediff=1e-8,   # for vibrations you need precise energies
nsw=1,        # Set to 1 for vibrational calculation
atoms=atoms)

print('Forces')
print('======')
print(atoms.get_forces())
print('')
calc.stop_if(calc.potential_energy is None)

# vibrational energies are in eV
energies, modes = calc.get_vibrational_modes()
print('energies\n========')
for i, e in enumerate(energies):
print('{0:02d}: {1} eV'.format(i, e))

Forces
======
[[ 0.01810349 -0.03253721 -0.00127275]
[-0.03620698  0.          0.0025455 ]
[ 0.01810349  0.03253721 -0.00127275]]

energies
========
00: 0.475855773 eV
01: 0.46176517 eV
02: 0.196182182 eV
03: 0.007041992 eV
04: 0.002445078 eV
05: (0.000292003+0j) eV
06: (0.012756432+0j) eV
07: (0.01305212+0j) eV
08: (0.015976377+0j) eV


Note we get 9 frequencies here. Water has 3 atoms, with three degrees of freedom each, leading to 9 possible combinations of collective motions. Three of those collective motions are translations, i.e. where all atoms move in the same direction (either $$x$$, $$y$$ or $$z$$) and there is no change in the total energy of the molecule. Another three of those motions are rotations, which also do not change the total energy of the molecule. That leaves 3N-6 = 3 degrees of vibrational freedom where some or all of the bonds are stretched, resulting in a change in the total energy. The modes of water vibration are (with our calculated values in parentheses):

1. a symmetric stretch at 3657 cm-1 (3723)
2. an asymmetric stretch at 3756 cm-1 (3836)
3. and a bending mode at 1595 cm-1 (1583)

The results are not too far off, and more accurate frequencies may be possible using tighter tolerance on POTIM, or by using IBRION=7 or 8.

Let us briefly discuss how to determine which vectors are vibrations and which are rotations or translations. One way is to visualize the modes. The vibrations are easy to spot. The rotations/translations are not always cleanly separable. This is an issue of accuracy and convergence. We usually do not worry about this because these modes are usually not important.

1. mode 0 is an asymmetric stretch
2. mode 1 is a symmetric stretch
3. mode 2 is a bending mode
4. mode 3 is a mixed translation/rotation
5. mode 4 is a rotation
6. mode 5 is a translation
7. mode 6 is a rotation
8. mode 7 is a partial translation
9. mode 8 is a rotation
# <<h2o-vib-vis>>
from vasp import Vasp
import numpy as np
calc = Vasp('molecules/h2o_vib')
energies, modes = calc.get_vibrational_modes(mode=0, massweighted=True,
show=True)


See http://www.gaussian.com/g_whitepap/vib.htm for a more quantitative discussion of these modes, identifying them, and a method to project the rotations and translations out of the Hessian matrix.

##### 3.5.2.1 Zero-point energy for multiple modes

For a molecule with lots of vibrational modes the zero-point energy is defined as the sum over all the vibrational modes:

$$E_{ZPE} = \sum_i \frac{1}{2} h \nu_i$$

Here is an example for water. Note we do not sum over the imaginary modes. We should also ignore the rotational and translational modes (some of those are imaginary, but some are just small).

from vasp import Vasp
import numpy as np
c = 3e10  # speed of light cm/s
h = 4.135667516e-15  # eV/s

# first, get the frequencies.
calc = Vasp('molecules/h2o_vib')
freq = calc.get_vibrational_frequencies()

ZPE = 0.0
for f in freq:
if not isinstance(f, float):
continue  # skip complex numbers
nu = f * c  # convert to frequency
ZPE += 0.5 * h * nu

print(np.sum([0.5 * h * f * c for f in freq if isinstance(f, float)]))

print('The ZPE of water is {0:1.3f} eV'.format(ZPE))

# one liner
ZPE = np.sum([0.5 * h * f * c for f in freq if isinstance(f, float)])
print('The ZPE of water is {0:1.3f} eV'.format(ZPE))


Note the zero-point energy of water is also fairly high (more than 0.5 eV). That is because of the high frequency O-H stretches.

### 3.6 Simulated infrared spectra

infrared intensity At http://homepage.univie.ac.at/david.karhanek/downloads.html#Entry02 there is a recipe for computing the Infrared vibrational spectroscopy intensities in VASP. We are going to do that for water here. First, we will relax a water molecule.

from ase import Atoms, Atom
from vasp import Vasp

atoms = Atoms([Atom('H', [0.5960812,  -0.7677068,   0.0000000]),
Atom('O', [0.0000000,   0.0000000,   0.0000000]),
Atom('H', [0.5960812,   0.7677068,   0.0000000])],
cell=(8, 8, 8))

calc = Vasp('molecules/h2o_relax',
xc='PBE',
encut=400,
ismear=0,  # Gaussian smearing
ibrion=2,
ediff=1e-8,
nsw=10,
atoms=atoms)
print('Forces')
print('===========================')
print(atoms.get_forces())

Forces
===========================
[[ -3.80700000e-05   5.32200000e-05   0.00000000e+00]
[  7.61400000e-05   0.00000000e+00   0.00000000e+00]
[ -3.80700000e-05  -5.32200000e-05   0.00000000e+00]]


Next, we instruct VASP to compute the vibrational modes using density functional perturbation theory with IBRION=7. Note, this is different than in 3.5 where finite differences were used.

from vasp import Vasp

calc = Vasp('molecules/h2o_relax')
atoms = calc.get_atoms()

# now define a new calculator
calc = Vasp('molecules/h2o_vib_dfpt',
xc='PBE',
encut=400,
ismear=0,  # Gaussian smearing
ibrion=7,  # switches on the DFPT vibrational analysis (with
# no symmetry constraints)
nfree=2,
potim=0.015,
lepsilon=True,  # enables to calculate and to print the BEC
# tensors
lreal=False,
nsw=1,
nwrite=3,  # affects OUTCAR verbosity: explicitly forces
# SQRT(mass)-divided eigenvectors to be printed
atoms=atoms)

print(calc.potential_energy)

-14.22662275


To analyze the results, this shell script was provided to extract the results.

#!/bin/bash
# A utility for calculating the vibrational intensities from VASP output (OUTCAR)
# (C) David Karhanek, 2011-03-25, ICIQ Tarragona, Spain (www.iciq.es)

# extract Born effective charges tensors
BORN_NROWS=grep NIONS OUTCAR | awk '{print $12*4+1}' if [ grep 'BORN' OUTCAR | wc -l = 0 ] ; then \ printf " .. FAILED! Born effective charges missing! Bye! \n\n" ; exit 1 ; fi grep "in e, cummulative" -A$BORN_NROWS OUTCAR > born.txt

# extract Eigenvectors and eigenvalues
if [ grep 'SQRT(mass)' OUTCAR | wc -l != 1 ] ; then \
printf " .. FAILED! Restart VASP with NWRITE=3! Bye! \n\n" ; exit 1 ; fi
EIG_NVIBS=grep -A 2000 'SQRT(mass)' OUTCAR | grep 'cm-1' | wc -l
EIG_NIONS=grep NIONS OUTCAR | awk '{print $12}' EIG_NROWS=echo "($EIG_NIONS+3)*$EIG_NVIBS+3" | bc grep -A$(($EIG_NROWS+2)) 'SQRT(mass)' OUTCAR | tail -n$(($EIG_NROWS+1)) | sed 's/f\/i/fi /g' > eigenvectors.txt printf " ..done\n" # set up a new directory, split files - prepare for parsing printf "..splitting files" mkdir intensities ; mv born.txt eigenvectors.txt intensities/ cd intensities/ let NBORN_NROWS=BORN_NROWS-1 let NEIG_NROWS=EIG_NROWS-3 let NBORN_STEP=4 let NEIG_STEP=EIG_NIONS+3 tail -n$NBORN_NROWS born.txt > temp.born.txt
tail -n $NEIG_NROWS eigenvectors.txt > temp.eige.txt mkdir inputs ; mv born.txt eigenvectors.txt inputs/ split -a 3 -d -l$NEIG_STEP temp.eige.txt temp.ei.
split -a 3 -d -l $NBORN_STEP temp.born.txt temp.bo. mkdir temps01 ; mv temp.born.txt temp.eige.txt temps01/ for nu in seq 1$EIG_NVIBS ; do
let nud=nu-1 ; ei=printf "%03u" $nu ; eid=printf "%03u"$nud ; mv temp.ei.$eid eigens.vib.$ei
done
for s in seq 1 $EIG_NIONS ; do let sd=s-1 ; bo=printf "%03u"$s ; bod=printf "%03u" $sd ; mv temp.bo.$bod borncs.$bo done printf " ..done\n" # parse deviation vectors (eig) printf "..parsing eigenvectors" let sad=$EIG_NIONS+1
for nu in seq 1 $EIG_NVIBS ; do nuu=printf "%03u"$nu
tail -n $sad eigens.vib.$nuu | head -n $EIG_NIONS | awk '{print$4,$5,$6}' > e.vib.$nuu.allions split -a 3 -d -l 1 e.vib.$nuu.allions temp.e.vib.$nuu.ion. for s in seq 1$EIG_NIONS ; do
let sd=s-1; bo=printf "%03u" $s; bod=printf "%03u"$sd; mv temp.e.vib.$nuu.ion.$bod e.vib.$nuu.ion.$bo
done
done
printf " ..done\n"

# parse born effective charge matrices (born)
printf "..parsing eff.charges"
for s in seq 1 $EIG_NIONS ; do ss=printf "%03u"$s
awk '{print $2,$3,$4}' borncs.$ss | tail -3 > bornch.$ss done mkdir temps02 ; mv eigens.* borncs.* temps02/ printf " ..done\n" # parse matrices, multiply them and collect squares (giving intensities) printf "..multiplying matrices, summing " for nu in seq 1$EIG_NVIBS ; do
nuu=printf "%03u" $nu int=0.0 for alpha in 1 2 3 ; do # summing over alpha coordinates sumpol=0.0 for s in seq 1$EIG_NIONS ; do   # summing over atoms
ss=printf "%03u" $s awk -v a="$alpha" '(NR==a){print}' bornch.$ss > z.ion.$ss.alpha.$alpha # summing over beta coordinates and multiplying Z(s,alpha)*e(s) done by the following awk script paste z.ion.$ss.alpha.$alpha e.vib.$nuu.ion.$ss | \ awk '{pol=$1*$4+$2*$5+$3*$6; print$0,"  ",pol}' > matr-vib-${nuu}-alpha-${alpha}-ion-${ss} done sumpol=cat matr-vib-${nuu}-alpha-${alpha}-ion-* | awk '{sum+=$7} END {print sum}'
int=echo "$int+($sumpol)^2" | sed 's/[eE]/*10^/g' |  bc -l
done
freq=awk '(NR==1){print $8}' temps02/eigens.vib.$nuu
echo "$nuu$freq $int">> exact.res.txt printf "." done printf " ..done\n" # format results, normalize intensities printf "..normalizing intensities" max=awk '(NR==1){max=$3} $3>=max {max=$3} END {print max}' exact.res.txt
awk -v max="$max" '{printf "%03u %6.1f %5.3f\n",$1,$2,$3/max}' exact.res.txt > results.txt
printf " ..done\n"

# clean up, display results
printf "..finalizing:\n"
mkdir temps03; mv bornch.* e.vib.*.allions temps03/
mkdir temps04; mv z.ion* e.vib.*.ion.* temps04/
mkdir temps05; mv matr-* temps05/
mkdir results; mv *res*txt results/
let NMATRIX=$EIG_NVIBS**2 printf "%5u atoms found\n%5u vibrations found\n%5u matrices evaluated" \$EIG_NIONS $EIG_NVIBS$NMATRIX > results/statistics.txt
# fast switch to clean up all temporary files
rm -r temps*
cat results/results.txt


Note that the results above include the rotational and translational modes (modes 4-9). The following shell script removes those, and recalculates the intensities. Note that it appears to just remove the last 6 modes and req compute the intensities. It is not obvious that will always be the right way to do it as the order of the eigenvectors is not guaranteed.

#!/bin/bash
# reformat intensities, just normal modes: 3N -> (3N-6)
printf "..reformatting and normalizing intensities"
cd intensities/results/
nlns=wc -l exact.res.txt | awk '{print $1}' ; let bodylns=nlns-6 head -n$bodylns exact.res.txt > temp.reform.res.txt
max=awk '(NR==1){max=$3}$3>=max {max=$3} END {print max}' temp.reform.res.txt awk -v max="$max" '{print $1,$2,$3/max}' temp.reform.res.txt > exact.reform.res.txt awk -v max="$max" '{printf "%03u %6.1f %5.3f\n",$1,$2,$3/max}' temp.reform.res.txt > reform.res.txt printf " ..done\n..normal modes:\n" rm temp.reform.res.txt cat reform.res.txt cd ../..  ..reformatting and normalizing intensities ..done ..normal modes:  The interpretation of these results is that the mode at 3713 cm-1 would be nearly invisible in the IR spectrum. Earlier we interpreted that as the symmetric stretch. In this mode, there is only a small change in the molecule dipole moment, so there is a small IR intensity. See also giannozzi:8537. For HREELS simulations see 0953-8984-22-26-265006. The shell script above has been translated to a convenient python function in vasp. from vasp import Vasp calc = Vasp('molecules/h2o_vib_dfpt') print('mode Relative intensity') for i, intensity in enumerate(calc.get_infrared_intensities()): print('{0:02d} {1:1.3f}'.format(i, intensity))  mode Relative intensity 00 0.227 01 0.006 02 0.312 03 1.000 04 0.002 05 0.000 06 0.006 07 0.000 08 0.350  ### 3.7 Thermochemical properties of molecules thermochemistry ase.thermochemistry can be used to estimate thermodynamic properties of gases in the ideal gas limit. The module needs as input the geometry, the total energy, the vibrational energies, and some information about the molecular symmetry. We first consider an N2 molecule. The symmetry numbers are determined by the molecular point group springerlink-10.1007/s00214-007-0328-0. Here is a table of the most common ones. Table 2: Symmetry numbers for common point groups point group $$\sigma$$ examples C1 1 Cs 1 C2 2 C2v 2 H2O C3v 3 NH3 $$C_{\infty v}$$ 1 CO D2h 4 D3h 6 D5h 10 $$D_{\infty h}$$ 2 CO2, H2 D3d 6 Td 12 CH4 Oh 24 from ase.structure import molecule from ase.thermochemistry import IdealGasThermo from vasp import Vasp atoms = molecule('N2') atoms.set_cell((10,10,10), scale_atoms=False) # first we relax a molecule calc = Vasp('molecules/n2-relax', xc='PBE', encut=300, ibrion=2, nsw=5, atoms=atoms) electronicenergy = atoms.get_potential_energy() # next, we get vibrational modes calc2 = Vasp('molecules/n2-vib', xc='PBE', encut=300, ibrion=6, nfree=2, potim=0.15, nsw=1, atoms=atoms) calc2.wait() vib_freq = calc2.get_vibrational_frequencies() # in cm^1 #convert wavenumbers to energy h = 4.1356675e-15 # eV*s c = 3.0e10 #cm/s vib_energies = [h*c*nu for nu in vib_freq] print('vibrational energies\n====================') for i,e in enumerate(vib_energies): print('{0:02d}: {1} eV'.format(i,e)) # # now we can get some properties. Note we only need one vibrational # energy since there is only one mode. This example does not work if # you give all the energies because one energy is zero. thermo = IdealGasThermo(vib_energies=vib_energies[0:0], potentialenergy=electronicenergy, atoms=atoms, geometry='linear', symmetrynumber=2, spin=0) # temperature in K, pressure in Pa, G in eV G = thermo.get_gibbs_energy(temperature=298.15, pressure=101325.)  vibrational energies ==================== 00: 0.281619180732 eV 01: 0.0302718194691 eV 02: 0.0302718194691 eV 03: 6.20350125e-10 eV 04: 4.962801e-10 eV 05: 0.0 eV Enthalpy components at T = 298.15 K: =============================== E_pot -16.484 eV E_ZPE 0.000 eV Cv_trans (0->T) 0.039 eV Cv_rot (0->T) 0.026 eV Cv_vib (0->T) 0.000 eV (C_v -> C_p) 0.026 eV ------------------------------- H -16.394 eV =============================== Entropy components at T = 298.15 K and P = 101325.0 Pa: ================================================= S T*S S_trans (1 atm) 0.0015579 eV/K 0.464 eV S_rot 0.0007868 eV/K 0.235 eV S_elec 0.0000000 eV/K 0.000 eV S_vib 0.0000000 eV/K 0.000 eV S (1 atm -> P) -0.0000000 eV/K -0.000 eV ------------------------------------------------- S 0.0023447 eV/K 0.699 eV ================================================= Free energy components at T = 298.15 K and P = 101325.0 Pa: ======================= H -16.394 eV -T*S -0.699 eV ----------------------- G -17.093 eV =======================  Let us compare this to what is in the Nist webbook via the Shomate equations. import numpy as np A = 28.98641 B = 1.853978 C = -9.647459 D = 16.63537 E = 0.000117 F = -8.671914 G = 226.4168 H = 0.0 T = 298.15 t = T/1000. S = A*np.log(t) + B*t + C*t**2/2 + D*t**3/3 - E/(2*t**2) + G print('-T*S = {0:1.3f} eV'.format(-T*S/1000/96.4853))  -T*S = -0.592 eV  This is reasonable agreement for the entropy. You will get different results if you use different exchange correlation functionals. ### 3.8 Molecular reaction energies #### 3.8.1 O2 dissociation The first reaction we consider is a simple dissociation of oxygen molecule into two oxygen atoms: O2 → 2O. The dissociation energy is pretty straightforward to define: it is the energy of the products minus the energy of the reactant. $$D = 2*E_O - E_{O_2}$$. It would appear that we simply calculate the energy of an oxygen atom, and the energy of an oxygen molecule and evaluate the formula. Let us do that. ##### 3.8.1.1 Simple estimate of O2 dissociation energy from vasp import Vasp from ase import Atom, Atoms atoms = Atoms([Atom('O', [5, 5, 5])], cell=(10, 10, 10)) calc = Vasp('molecules/O', xc='PBE', encut=400, ismear=0, atoms=atoms) E_O = atoms.get_potential_energy() # now relaxed O2 dimer atoms = Atoms([Atom('O', [5, 5, 5]), Atom('O', [6.22, 5, 5])], cell=(10, 10, 10)) calc = Vasp('molecules/O2', xc='PBE', encut=400, ismear=0, ibrion=2, nsw=10, atoms=atoms) E_O2 = atoms.get_potential_energy() if None not in (E_O, E_O2): print('O2 -> 2O D = {0:1.3f} eV'.format(2 * E_O - E_O2))  O2 -> 2O D = 8.619 eV  The answer we have obtained is way too high! Experimentally the dissociation energy is about 5.2 eV (need reference), which is very different than what we calculated! Let us consider some factors that contribute to this error. We implicitly neglected spin-polarization in the example above. That could be a problem, since the O2 molecule can be in one of two spin states, a singlet or a triplet, and these should have different energies. Furthermore, the oxygen atom can be a singlet or a triplet, and these would have different energies. To account for spin polarization, we have to tell VASP to use spin-polarization, and give initial guesses for the magnetic moments of the atoms. Let us try again with spin polarization. ##### 3.8.1.2 Estimating O2 dissociation energy with spin polarization in triplet ground states To tell VASP to use spin-polarization we use ISPIN=2, and we set initial guesses for magnetic moments on the atoms with the magmom keyword. In a triplet state there are two electrons with spins of the same sign. from vasp import Vasp from ase import Atom, Atoms atoms = Atoms([Atom('O', [5, 5, 5], magmom=2)], cell=(10, 10, 10)) calc = Vasp('molecules/O-sp-triplet', xc='PBE', encut=400, ismear=0, ispin=2, # turn spin-polarization on atoms=atoms) E_O = atoms.get_potential_energy() print('Magnetic moment on O = {0} Bohr' ' magnetons'.format(atoms.get_magnetic_moment())) # now relaxed O2 dimer atoms = Atoms([Atom('O', [5, 5, 5], magmom=1), Atom('O', [6.22, 5, 5], magmom=1)], cell=(10, 10, 10)) calc = Vasp('molecules/O2-sp-triplet', xc='PBE', encut=400, ismear=0, ispin=2, # turn spin-polarization on ibrion=2, # make sure we relax the geometry nsw=10, atoms=atoms) E_O2 = atoms.get_potential_energy() # verify magnetic moment print('Magnetic moment on O2 = {0} Bohr' ' magnetons'.format(atoms.get_magnetic_moment())) if None not in (E_O, E_O2): print('O2 -> 2O D = {0:1.3f} eV'.format(2 * E_O - E_O2))  Magnetic moment on O = 2.0000072 Bohr magnetons Magnetic moment on O2 = 2.0000084 Bohr magnetons O2 -> 2O D = 6.746 eV  This is much closer to accepted literature values for the DFT-GGA O$$_2$$ dissociation energy. It is still more than 1 eV above an experimental value, but most of that error is due to the GGA exchange correlation functional. Some additional parameters that might need to be checked for convergence are the SIGMA value (it is probably too high for a molecule), as well as the cutoff energy. Oxygen is a "hard" atom that requires a high cutoff energy to achieve high levels of convergence. ###### 3.8.1.2.1 Looking at the two spin densities In a spin-polarized calculation there are actually two electron densities: one for spin-up and one for spin-down. We will look at the differences in these two through the density of states. from vasp import Vasp from ase.dft.dos import * calc = Vasp('molecules/O2-sp-triplet') dos = DOS(calc, width=0.2) d_up = dos.get_dos(spin=0) d_down = dos.get_dos(spin=1) e = dos.get_energies() ind = e <= 0.0 # integrate up to 0eV print('number of up states = {0}'.format(np.trapz(d_up[ind], e[ind]))) print('number of down states = {0}'.format(np.trapz(d_down[ind], e[ind]))) import pylab as plt plt.plot(e, d_up, e, -d_down) plt.xlabel('energy [eV]') plt.ylabel('DOS') plt.legend(['up', 'down']) plt.savefig('images/O2-sp-dos.png')  number of up states = 6.11729553486 number of down states = 5.00000794208  You can see in Figure fig:o2-sp-dos that there are two different densities of states for the two spins. One has 7 electrons in it (the blue lines), and the other has 5 electrons in it (the green line). The difference of two electrons leads to the magnetic moment of 2 which we calculated earlier. Remember that only peaks in the DOS below the Fermi level are occupied. It is customary to set the Fermi level to 0 eV in DOS plots. The peaks roughly correspond to electrons. For example, the blue peak between -25 and -30 eV corresponds to one electron, in a 1s orbital, where as the blue peak between -5 and -10 eV corresponds to three electrons. ##### 3.8.1.3 Convergence study of the O2 dissociation energy from vasp import Vasp from ase import Atom, Atoms encuts = [250, 300, 350, 400, 450, 500, 550] D = [] for encut in encuts: atoms = Atoms([Atom('O', [5, 5, 5], magmom=2)], cell=(10, 10, 10)) calc = Vasp('molecules/O-sp-triplet-{0}'.format(encut), xc='PBE', encut=encut, ismear=0, ispin=2, atoms=atoms) E_O = atoms.get_potential_energy() # now relaxed O2 dimer atoms = Atoms([Atom('O', [5, 5, 5], magmom=1), Atom('O', [6.22, 5, 5], magmom=1)], cell=(10, 10, 10)) calc = Vasp('molecules/O2-sp-triplet-{0}'.format(encut), xc='PBE', encut=encut, ismear=0, ispin=2, # turn spin-polarization on ibrion=2, # this turns relaxation on nsw=10, atoms=atoms) E_O2 = atoms.get_potential_energy() if None not in (E_O, E_O2): d = 2*E_O - E_O2 D.append(d) print('O2 -> 2O encut = {0} D = {1:1.3f} eV'.format(encut, d)) if not D or None in D: calc.abort() import matplotlib.pyplot as plt plt.plot(encuts, D) plt.xlabel('ENCUT (eV)') plt.ylabel('O$_2$dissociation energy (eV)') plt.savefig('images/O2-dissociation-convergence.png')  O2 -> 2O encut = 250 D = 6.774 eV O2 -> 2O encut = 300 D = 6.804 eV O2 -> 2O encut = 350 D = 6.785 eV O2 -> 2O encut = 400 D = 6.746 eV O2 -> 2O encut = 450 D = 6.727 eV O2 -> 2O encut = 500 D = 6.725 eV O2 -> 2O encut = 550 D = 6.727 eV  Based on these results (Figure fig:o2-encut), you could argue the dissociation energy is converged to about 2 meV at a planewave cutoff of 450 eV, and within 50 meV at 350 eV cutoff. You have to decide what an appropriate level of convergence is. Note that increasing the planewave cutoff significantly increases the computational time, so you are balancing level of convergence with computational speed. It would appear that planewave cutoff is not the cause for the discrepancy between our calculations and literature values. encuts = [250, 300, 350, 400, 450, 500, 550] print('encut (eV) Total CPU time') print('--------------------------------------------------------') for encut in encuts: OUTCAR = 'molecules/O2-sp-triplet-{0}/OUTCAR'.format(encut) f = open(OUTCAR, 'r') for line in f: if 'Total CPU time used (sec)' in line: print('{0} eV: {1}'.format(encut, line)) f.close()  encut (eV) Total CPU time -------------------------------------------------------- 250 eV: Total CPU time used (sec): 1551.338 300 eV: Total CPU time used (sec): 2085.191 350 eV: Total CPU time used (sec): 2795.841 400 eV: Total CPU time used (sec): 2985.064 450 eV: Total CPU time used (sec): 5155.562 500 eV: Total CPU time used (sec): 4990.818 550 eV: Total CPU time used (sec): 5262.052  ##### 3.8.1.4 Illustration of the effect of SIGMA The methodology for extrapolation of the total energy to absolute zero is only valid for a continuous density of states at the Fermi level Kresse199615. Consequently, it should not be used for semiconductors, molecules or atoms. In VASP, this means a very small Fermi temperature (SIGMA) should be used. The O2 dissociation energy as a function of SIGMA is shown in Figure fig:sigma-o2-diss. A variation of nearly 0.2 eV is seen from the default Fermi temperature of $$k_bT=0.2$$ eV and the value of $$k_bT=0.0001$$ eV. However, virtually no change was observed for a hydrogen atom or molecule or for an oxygen molecule as a function of the Fermi temperature. It is recommended that the total energy be calculated at several values of the Fermi temperature to make sure the total energy is converged with respect to the Fermi temperature. We were not careful in selecting a good value for SIGMA in the calculations above. The default value of SIGMA is 0.2, which may be fine for metals, but it is not correct for molecules. SIGMA is the broadening factor used to smear the electronic density of states at the Fermi level. For a metal with a continuous density of states this is appropriate, but for molecules with discrete energy states it does not make sense. We are somewhat forced to use the machinery designed for metals on molecules. The solution is to use a very small SIGMA. Ideally you would use SIGMA=0, but that is not practical for convergence reasons, so we try to find what is small enough. Let us examine the effect of SIGMA on the dissociation energy here. from vasp import Vasp from ase import Atom, Atoms sigmas = [0.2, 0.1, 0.05, 0.02, 0.01, 0.001] D = [] for sigma in sigmas: atoms = Atoms([Atom('O',[5, 5, 5], magmom=2)], cell=(10, 10, 10)) calc = Vasp('molecules/O-sp-triplet-sigma-{0}'.format(sigma), xc='PBE', encut=400, ismear=0, sigma=sigma, ispin=2, atoms=atoms) E_O = atoms.get_potential_energy() # now relaxed O2 dimer atoms = Atoms([Atom('O',[5, 5, 5],magmom=1), Atom('O',[6.22, 5, 5],magmom=1)], cell=(10, 10, 10)) calc = Vasp('molecules/O2-sp-triplet-sigma-{0}'.format(sigma), xc='PBE', encut=400, ismear=0, sigma=sigma, ispin=2, # turn spin-polarization on ibrion=2, # make sure we relax the geometry nsw=10, atoms=atoms) E_O2 = atoms.get_potential_energy() if None not in (E_O, E_O2): d = 2 * E_O - E_O2 D.append(d) print('O2 -> 2O sigma = {0} D = {1:1.3f} eV'.format(sigma, d)) import matplotlib.pyplot as plt plt.plot(sigmas, D, 'bo-') plt.xlabel('SIGMA (eV)') plt.ylabel('O$_2$dissociation energy (eV)') plt.savefig('images/O2-dissociation-sigma-convergence.png')  O2 -> 2O sigma = 0.2 D = 6.669 eV O2 -> 2O sigma = 0.1 D = 6.746 eV O2 -> 2O sigma = 0.05 D = 6.784 eV O2 -> 2O sigma = 0.02 D = 6.807 eV O2 -> 2O sigma = 0.01 D = 6.815 eV O2 -> 2O sigma = 0.001 D = 6.822 eV  Clearly SIGMA has an effect, but it does not move the dissociation energy closer to the literature values! ##### 3.8.1.5 Estimating singlet oxygen dissociation energy Finally, let us consider the case where each species is in the singlet state. from vasp import Vasp from ase import Atom, Atoms atoms = Atoms([Atom('O', [5, 5, 5], magmom=0)], cell=(10, 10, 10)) calc = Vasp('molecules/O-sp-singlet', xc='PBE', encut=400, ismear=0, ispin=2, atoms=atoms) E_O = atoms.get_potential_energy() print('Magnetic moment on O = {0} Bohr' ' magnetons'.format(atoms.get_magnetic_moment())) # now relaxed O2 dimer atoms = Atoms([Atom('O', [5, 5, 5], magmom=1), Atom('O', [6.22, 5, 5], magmom=-1)], cell=(10, 10, 10)) calc = Vasp('molecules/O2-sp-singlet', xc='PBE', encut=400, ismear=0, ispin=2, # turn spin-polarization on ibrion=2, # make sure we relax the geometry nsw=10, atoms=atoms) E_O2 = atoms.get_potential_energy() # verify magnetic moment print('O2 molecule magnetic moment = ', atoms.get_magnetic_moment()) if None not in (E_O, E_O2): print('O2 -> 2O D = {0:1.3f} eV'.format(2 * E_O - E_O2))  Magnetic moment on O = 0.0001638 Bohr magnetons ('O2 molecule magnetic moment = ', 0.0) O2 -> 2O D = 8.619 eV  Let us directly compare their total energies: from vasp import Vasp calc = Vasp('molecules/O2-sp-singlet') print('singlet: {0} eV'.format(calc.potential_energy)) calc = Vasp('molecules/O2-sp-triplet') print('triplet: {0} eV'.format(calc.potential_energy))  singlet: -8.77378302 eV triplet: -9.84832389 eV  You can see here the triplet state has an energy that is 1 eV more stable than the singlet state. ##### 3.8.1.6 Estimating triplet oxygen dissociation energy with low symmetry It has been suggested that breaking spherical symmetry of the atom can result in lower energy of the atom. The symmetry is broken by putting the atom off-center in a box. We will examine the total energy of an oxygen atom in a few geometries. First, let us consider variations of a square box. from vasp import Vasp from ase import Atom, Atoms # square box origin atoms = Atoms([Atom('O', [0, 0, 0], magmom=2)], cell=(10, 10, 10)) pars = dict(xc='PBE', encut=400, ismear=0, sigma=0.01, ispin=2) calc = Vasp('molecules/O-square-box-origin', atoms=atoms, **pars) print('Square box (origin): E = {0} eV'.format(atoms.get_potential_energy())) # square box center atoms = Atoms([Atom('O', [5, 5, 5], magmom=2)], cell=(10, 10, 10)) calc = Vasp('molecules/O-square-box-center', atoms=atoms, **pars) print('Square box (center): E = {0} eV'.format(atoms.get_potential_energy())) # square box random atoms = Atoms([Atom('O', [2.13, 7.32, 1.11], magmom=2)], cell=(10, 10, 10)) calc = Vasp('molecules/O-square-box-random', atoms=atoms, **pars) print('Square box (random): E = {0} eV'.format(atoms.get_potential_energy()))  Square box (origin): E = -1.51654778 eV Square box (center): E = -1.51654804 eV Square box (random): E = -1.5152871 eV  There is no significant difference in these energies. The origin and center calculations are identical in energy. The meV variation in the random calculation is negligible. Now, let us consider some non-square boxes. # calculate O atom energy in orthorhombic boxes from vasp import Vasp from ase import Atom, Atoms # orthorhombic box origin atoms = Atoms([Atom('O', [0, 0, 0], magmom=2)], cell=(8, 9, 10)) calc = Vasp('molecules/O-orthorhombic-box-origin', xc='PBE', encut=400, ismear=0, sigma=0.01, ispin=2, atoms=atoms) print('Orthorhombic box (origin): E = {0} eV'.format(atoms.get_potential_energy())) # orthorhombic box center atoms = Atoms([Atom('O', [4, 4.5, 5], magmom=2)], cell=(8, 9, 10)) calc = Vasp('molecules/O-orthorhombic-box-center', xc='PBE', encut=400, ismear=0, sigma=0.01, ispin=2, atoms=atoms) print('Orthorhombic box (center): E = {0} eV'.format(atoms.get_potential_energy())) # orthorhombic box random atoms = Atoms([Atom('O', [2.13, 7.32, 1.11], magmom=2)], cell=(8, 9, 10)) calc = Vasp('molecules/O-orthorhombic-box-random', xc='PBE', encut=400, ismear=0, sigma=0.01, ispin=2, atoms=atoms) print('Orthorhombic box (random): E = {0} eV'.format(atoms.get_potential_energy()))  Orthorhombic box (origin): E = -1.89375092 eV Orthorhombic box (center): E = -1.89375153 eV Orthorhombic box (random): E = -1.87999536 eV  This is a surprisingly large difference in energy! Nearly 0.4 eV. This is precisely the amount of energy we were in disagreement with the literature values. Surprisingly, the "random" position is higher in energy, similar to the cubic boxes. Finally, we put this all together. We use a non-symmetric box for the O-atom. from vasp import Vasp from ase import Atom, Atoms atoms = Atoms([Atom('O', [5.1, 4.2, 6.1], magmom=2)], cell=(8, 9, 10)) calc = Vasp('molecules/O-sp-triplet-lowsym', xc='PBE', encut=400, ismear=0, sigma=0.01, ispin=2, atoms=atoms) E_O = atoms.get_potential_energy() print('Magnetic moment on O = {0} Bohr magnetons'.format(atoms.get_magnetic_moment())) # now relaxed O2 dimer atoms = Atoms([Atom('O', [5, 5, 5], magmom=1), Atom('O', [6.22, 5, 5], magmom=1)], cell=(10, 10, 10)) calc = Vasp('molecules/O2-sp-triplet', xc='PBE', encut=400, ismear=0, sigma=0.01, ispin=2, # turn spin-polarization on ibrion=2, # make sure we relax the geometry nsw=10, atoms=atoms) E_O2 = atoms.get_potential_energy() # verify magnetic moment print('Magnetic moment on O2 = {0} Bohr magnetons'.format(atoms.get_magnetic_moment())) if None not in (E_O, E_O2): print('E_O: ', E_O) print('O2 -> 2O D = {0:1.3f} eV'.format(2 * E_O - E_O2))  Magnetic moment on O = 2.0000073 Bohr magnetons Magnetic moment on O2 = 2.0000084 Bohr magnetons ('E_O: ', -1.89307116) O2 -> 2O D = 6.062 eV  This actually agrees within 30-50 meV of reported literature values, although still nearly an eV greater than the experimental dissociation energy. Note that with a different "random" position, we get the lower energy for the O atom. All the disagreement we had been seeing was apparently in the O atom energy. So, if you do not need the dissociation energy in your analysis, you will not see the error. Also note that this error is specific to there being a spherical atom in a symmetric cell. This is not a problem for most molecules, which are generally non-spherical. ##### 3.8.1.7 Verifying the magnetic moments on each atom It is one thing to see the total magnetic moment of a singlet state, and another to ask what are the magnetic moments on each atom. In VASP you must use LORBIT = 11 to get the magnetic moments of the atoms written out. from vasp import Vasp calc = Vasp('molecules/O2-sp-singlet') calc.clone('molecules/O2-sp-singlet-magmoms') calc.set(lorbit=11) atoms = calc.get_atoms() magmoms = atoms.get_magnetic_moments() print('singlet ground state') for i, atom in enumerate(atoms): print('atom {0}: magmom = {1}'.format(i, magmoms[i])) print(atoms.get_magnetic_moment()) calc = Vasp('molecules/O2-sp-triplet') calc.clone('molecules/O2-sp-triplet-magmoms') calc.set(lorbit=11) atoms = calc.get_atoms() magmoms = atoms.get_magnetic_moments() print() print('triplet ground state') for i, atom in enumerate(atoms): print('atom {0}: magmom = {1}'.format(i, magmoms[i])) print(atoms.get_magnetic_moment())  singlet ground state atom 0: magmom = 0.0 atom 1: magmom = 0.0 0.0 () triplet ground state atom 0: magmom = 0.815 atom 1: magmom = 0.815 2.0000083  Note the atomic magnetic moments do not add up to the total magnetic moment. The atomic magnetic moments are not really true observable properties. The moments are determined by a projection method that probably involves a spherical orbital, so the moments may be over or underestimated. ##### 3.8.1.8 Using a different potential It is possible we need a higher quality potential to get the 6.02 eV value quoted by many in the literature. Here we try the O_sv potential, which treats the 1s electrons as valence electrons. Note however, the ENMIN in the POTCAR is very high! grep ENMIN$VASP_PP_PATH/potpaw_PBE/O_sv/POTCAR

from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('O', [4, 4.5, 5], magmom=2)],
cell=(8, 9, 10))

calc = Vasp('molecules/O-sp-triplet-lowsym-s',
xc='PBE',
ismear=0,
ispin=2,
sigma=0.01,
setups=[['O', '_s']],
atoms=atoms)

E_O = atoms.get_potential_energy()
print(E_O)

-1.57217591


In the following calculation, we let VASP select an appropriate ENCUT value.

from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('O', [4, 4.5, 5], magmom=2)],
cell=(8, 9, 10))

calc = Vasp('molecules/O-sp-triplet-lowsym-s',
xc='PBE',
ismear=0,
ispin=2,
sigma=0.01,
setups=[['O', '_s']],
atoms=atoms)

E_O = atoms.get_potential_energy()

print('Magnetic moment on O = {0} Bohr'
' magnetons'.format(atoms.get_magnetic_moment()))

# now relaxed O2 dimer
atoms = Atoms([Atom('O', [5,    5, 5], magmom=1),
Atom('O', [6.22, 5, 5], magmom=1)],
cell=(10, 10, 10))

calc = Vasp('molecules/O2-sp-triplet-s',
xc='PBE',
ismear=0,
sigma=0.01,
ispin=2,   # turn spin-polarization on
ibrion=2,  # make sure we relax the geometry
nsw=10,
setups=[['O', '_s']],
atoms=atoms)

E_O2 = atoms.get_potential_energy()

# verify magnetic moment
print('Magnetic moment on O2 = {0} Bohr'
' magnetons'.format(atoms.get_magnetic_moment()))

if None not in (E_O, E_O2):
print('O2 -> 2O  D = {0:1.3f} eV'.format(2*E_O - E_O2))

Magnetic moment on O = 1.9999982 Bohr magnetons
Magnetic moment on O2 = 2.0000102 Bohr magnetons
O2 -> 2O  D = 6.120 eV


This result is close to other reported values. It is possibly not converged, since we let VASP choose the ENCUT value, and that value is the ENMIN value in the POTCAR. Nevertheless, the point is that a harder potential does not fix the problem of overbinding in the O2 molecule. That is a fundamental flaw in the GGA exchange-correlation functional.

#### 3.8.2 Water gas shift example

We consider calculating the reaction energy of the water-gas shift reaction in this example.

CO + H2O $$\leftrightharpoons$$ CO2 + H2

We define the reaction energy as the difference in energy between the products and reactants.

$$\Delta E = E_{CO_2} + E_{H_2} - E_{CO} - E_{H_2O}$$

For now, we compute this energy simply as the difference in DFT energies. In the next section we will add zero-point energies and compute the energy difference as a function of temperature. For now, we simply need to compute the total energy of each molecule in its equilibrium geometry.

from ase.structure import molecule
from vasp import Vasp

# first we define our molecules. These will automatically be at the coordinates from the G2 database.

CO = molecule('CO')
CO.set_cell([8, 8, 8], scale_atoms=False)

H2O = molecule('H2O')
H2O.set_cell([8, 8, 8], scale_atoms=False)

CO2 = molecule('CO2')
CO2.set_cell([8, 8, 8], scale_atoms=False)

H2 = molecule('H2')
H2.set_cell([8, 8, 8], scale_atoms=False)

# now the calculators to get the energies
c1 = Vasp('molecules/wgs/CO',
xc='PBE',
encut=350,
ismear=0,
ibrion=2,
nsw=10,
atoms=CO)

eCO = CO.get_potential_energy()

c2 = Vasp('molecules/wgs/CO2',
xc='PBE',
encut=350,
ismear=0,
ibrion=2,
nsw=10,
atoms=CO2)

eCO2 = CO2.get_potential_energy()

c3 = Vasp('molecules/wgs/H2',
xc='PBE',
encut=350,
ismear=0,
ibrion=2,
nsw=10,
atoms=H2)

eH2 = H2.get_potential_energy()

c4 = Vasp('molecules/wgs/H2O',
xc='PBE',
encut=350,
ismear=0,
ibrion=2,
nsw=10,
atoms=H2O)

eH2O = H2O.get_potential_energy()

if None in (eCO2, eH2, eCO, eH2O):
pass
else:
dE = eCO2 + eH2 - eCO - eH2O
print('Delta E = {0:1.3f} eV'.format(dE))
print('Delta E = {0:1.3f} kcal/mol'.format(dE * 23.06035))
print('Delta E = {0:1.3f} kJ/mol'.format(dE * 96.485))

Delta E = -0.723 eV
Delta E = -16.672 kcal/mol
Delta E = -69.758 kJ/mol


We estimated the enthalpy of this reaction at standard conditions to be -41 kJ/mol using data from the NIST webbook, which is a fair bit lower than we calculated here. In the next section we will examine whether additional corrections are needed, such as zero-point and temperature corrections.

It is a good idea to verify your calculations and structures are what you expected. Let us print them here. Inspection of these results shows the geometries were all relaxed, i.e., the forces on each atom are less than 0.05 eV/Å.

from vasp import Vasp

print('**** Calculation summaries')
print('***** CO')
calc = Vasp('molecules/wgs/H2O')
print('#+begin_example')
print(calc)
print('#+end_example')

##### 3.8.2.1 Calculation summaries
###### 3.8.2.1.1 CO
Vasp calculation in /home-research/jkitchin/dft-book-new-vasp/molecules/wgs/H2O

INCAR created by Atomic Simulation Environment
ENCUT = 350
LCHARG = .FALSE.
IBRION = 2
ISMEAR = 0
LWAVE = .TRUE.
SIGMA = 0.1
NSW = 10

O  H
1.0000000000000000
8.0000000000000000    0.0000000000000000    0.0000000000000000
0.0000000000000000    8.0000000000000000    0.0000000000000000
0.0000000000000000    0.0000000000000000    8.0000000000000000
1   2
Cartesian
0.0000000000000000  0.0000000000000000  0.1192620000000000
0.0000000000000000  0.7632390000000000 -0.4770470000000000
0.0000000000000000 -0.7632390000000000 -0.4770470000000000


#### 3.8.3 Temperature dependent water gas shift equilibrium constant

To correct the reaction energy for temperature effects, we must compute the vibrational frequencies of each species, and estimate the temperature dependent contributions to vibrational energy and entropy. We will break these calculations into several pieces. First we do each vibrational calculation. After those are done, we can get the data and construct the thermochemistry objects we need to estimate the reaction energy as a function of temperature (at constant pressure).

##### 3.8.3.1 CO vibrations
from vasp import Vasp

# get relaxed geometry
calc = Vasp('molecules/wgs/CO')
CO = calc.get_atoms()

# now do the vibrations
calc = Vasp('molecules/wgs/CO-vib',
xc='PBE',
encut=350,
ismear=0,
ibrion=6,
nfree=2,
potim=0.02,
nsw=1,
atoms=CO)
calc.wait()
vib_freq = calc.get_vibrational_frequencies()
for i, f in enumerate(vib_freq):
print('{0:02d}: {1} cm^(-1)'.format(i, f))

00: 2064.699153 cm^(-1)
01: 170.409559 cm^(-1)
02: 170.409559 cm^(-1)
03: (1.171397+0j) cm^(-1)
04: (6.354831+0j) cm^(-1)
05: (6.354831+0j) cm^(-1)


CO has only one vibrational mode (3N-5 = 6 - 5 = 1). The other 5 modes are 3 translations and 2 rotations.

##### 3.8.3.2 CO2 vibrations
from vasp import Vasp

# get relaxed geometry
calc = Vasp('molecules/wgs/CO2')
CO2 = calc.get_atoms()

# now do the vibrations
calc = Vasp('molecules/wgs/CO2-vib',
xc='PBE',
encut=350,
ismear=0,
ibrion=6,
nfree=2,
potim=0.02,
nsw=1,
atoms=CO2)
calc.wait()
vib_freq = calc.get_vibrational_frequencies()
for i, f in enumerate(vib_freq):
print('{0:02d}: {1} cm^(-1)'.format(i, f))

00: 2339.140984 cm^(-1)
01: 1309.517832 cm^(-1)
02: 639.625419 cm^(-1)
03: 639.625419 cm^(-1)
04: (0.442216+0j) cm^(-1)
05: (1.801034+0j) cm^(-1)
06: (1.801034+0j) cm^(-1)
07: (35.286745+0j) cm^(-1)
08: (35.286745+0j) cm^(-1)


CO$$_2$$ is a linear molecule with 3N-5 = 4 vibrational modes. They are the first four frequencies in the output above.

##### 3.8.3.3 H2 vibrations
from vasp import Vasp

# get relaxed geometry
H2 = Vasp('molecules/wgs/H2').get_atoms()

# now do the vibrations
calc = Vasp('molecules/wgs/H2-vib',
xc='PBE',
encut=350,
ismear=0,
ibrion=6,
nfree=2,
potim=0.02,
nsw=1,
atoms=H2)
calc.wait()
vib_freq = calc.get_vibrational_frequencies()
for i, f in enumerate(vib_freq):
print('{0:02d}: {1} cm^(-1)'.format(i, f))

00: 4484.933386 cm^(-1)
01: 0.0 cm^(-1)
02: 0.0 cm^(-1)
03: (1.5e-05+0j) cm^(-1)
04: (586.624928+0j) cm^(-1)
05: (586.624928+0j) cm^(-1)


There is only one frequency of importance (the one at 4281 cm$$^{-1}$$) for the linear H2 molecule.

##### 3.8.3.4 H2O vibrations
from vasp import Vasp

# get relaxed geometry
H2O = Vasp('molecules/wgs/H2O').get_atoms()

# now do the vibrations
calc = Vasp('molecules/wgs/H2O-vib',
xc='PBE',
encut=350,
ismear=0,
ibrion=6,
nfree=2,
potim=0.02,
nsw=1,
atoms=H2O)
calc.wait()
vib_freq = calc.get_vibrational_frequencies()
for i, f in enumerate(vib_freq):
print('{0:02d}: {1} cm^(-1)'.format(i, f))

00: 3846.373652 cm^(-1)
01: 3734.935388 cm^(-1)
02: 1573.422217 cm^(-1)
03: 16.562103 cm^(-1)
04: 8.00982 cm^(-1)
05: (0.375952+0j) cm^(-1)
06: (225.466583+0j) cm^(-1)
07: (271.664033+0j) cm^(-1)
08: (286.859818+0j) cm^(-1)


Water has 3N-6 = 3 vibrational modes.

##### 3.8.3.5 Thermochemistry

Now we are ready. We have the electronic energies and vibrational frequencies of each species in the reaction. ase.thermochemistry.IdealGasThermo

from ase.thermochemistry import IdealGasThermo
from vasp import Vasp
import numpy as np
import matplotlib.pyplot as plt

# first we get the electronic energies
c1 = Vasp('molecules/wgs/CO')
E_CO = c1.potential_energy
CO = c1.get_atoms()

c2 = Vasp('molecules/wgs/CO2')
E_CO2 = c2.potential_energy
CO2 = c2.get_atoms()

c3 = Vasp('molecules/wgs/H2')
E_H2 = c3.potential_energy
H2 = c3.get_atoms()

c4 = Vasp('molecules/wgs/H2O')
E_H2O = c4.potential_energy
H2O = c4.get_atoms()

# now we get the vibrational energies
h = 4.1356675e-15  # eV * s
c = 3.0e10  # cm / s

calc = Vasp('molecules/wgs/CO-vib')
vib_freq = calc.get_vibrational_frequencies()
CO_vib_energies = [h * c * nu for nu in vib_freq]

calc = Vasp('molecules/wgs/CO2-vib')
vib_freq = calc.get_vibrational_frequencies()
CO2_vib_energies = [h * c * nu for nu in vib_freq]

calc = Vasp('molecules/wgs/H2-vib')
vib_freq = calc.get_vibrational_frequencies()
H2_vib_energies = [h * c * nu for nu in vib_freq]

calc = Vasp('molecules/wgs/H2O-vib')
vib_freq = calc.get_vibrational_frequencies()
H2O_vib_energies = [h * c * nu for nu in vib_freq]

# now we make a thermo object for each molecule
CO_t = IdealGasThermo(vib_energies=CO_vib_energies[0:0],
potentialenergy=E_CO, atoms=CO,
geometry='linear', symmetrynumber=1,
spin=0)

CO2_t = IdealGasThermo(vib_energies=CO2_vib_energies[0:4],
potentialenergy=E_CO2, atoms=CO2,
geometry='linear', symmetrynumber=2,
spin=0)

H2_t = IdealGasThermo(vib_energies=H2_vib_energies[0:0],
potentialenergy=E_H2, atoms=H2,
geometry='linear', symmetrynumber=2,
spin=0)

H2O_t = IdealGasThermo(vib_energies=H2O_vib_energies[0:3],
potentialenergy=E_H2O, atoms=H2O,
geometry='nonlinear', symmetrynumber=2,
spin=0)

# now we can compute G_rxn for a range of temperatures from 298 to 1000 K
Trange = np.linspace(298, 1000, 20)  # K
P = 101325. # Pa
Grxn = np.array([(CO2_t.get_gibbs_energy(temperature=T, pressure=P)
+ H2_t.get_gibbs_energy(temperature=T, pressure=P)
- H2O_t.get_gibbs_energy(temperature=T, pressure=P)
- CO_t.get_gibbs_energy(temperature=T, pressure=P)) * 96.485
for T in Trange])

Hrxn = np.array([(CO2_t.get_enthalpy(temperature=T)
+ H2_t.get_enthalpy(temperature=T)
- H2O_t.get_enthalpy(temperature=T)
- CO_t.get_enthalpy(temperature=T)) * 96.485
for T in Trange])

plt.plot(Trange, Grxn, 'bo-', label='$\Delta G_{rxn}$')
plt.plot(Trange, Hrxn, 'ro:', label='$\Delta H_{rxn}$')
plt.xlabel('Temperature (K)')
plt.ylabel(r'$\Delta G_{rxn}$ (kJ/mol)')
plt.legend(loc='best')
plt.savefig('images/wgs-dG-T.png')

plt.figure()
R = 8.314e-3  # gas constant in kJ/mol/K

Keq = np.exp(-Grxn/R/Trange)
plt.plot(Trange, Keq)
plt.ylim([0, 100])
plt.xlabel('Temperature (K)')
plt.ylabel('$K_{eq}$')
plt.savefig('images/wgs-Keq.png')


You can see a few things here. One is that at near 298K, the Gibbs free energy is about -75 kJ/mol. This is too negative compared to the experimental standard free energy, which we estimated to be about -29 kJ/mol from the NIST webbook. There could be several reasons for this disagreement, but the most likely one is errors in the exchange-correlation functional. The error in energy has a significant effect on the calculated equilibrium constant, significantly overestimating it.

### 3.9 Molecular reaction barriers

We will consider a simple example of the barrier for NH3 inversion. We have to create an NH3 molecule in the initial and inverted state (these have exactly the same energy), and then interpolate a band of images. Then, we use the NEB method sheppard:134106 to compute the barrier to inversion. The NEB class of methods are pretty standard, but other algorithms for finding barriers (saddle-points) exist that may be relevant olsen:9776.

#### 3.9.1 Get initial and final states

# compute initial and final states
from ase import Atoms
from ase.structure import molecule
import numpy as np
from vasp import Vasp
from ase.constraints import FixAtoms

atoms = molecule('NH3')
constraint = FixAtoms(mask=[atom.symbol == 'N' for atom in atoms])
atoms.set_constraint(constraint)

Npos = atoms.positions[0]

# move N to origin
atoms.translate(-Npos)
atoms.set_cell((10, 10, 10), scale_atoms=False)

atoms2 = atoms.copy()
pos2 = atoms2.positions

for i,atom in enumerate(atoms2):
if atom.symbol == 'H':
# reflect through z
pos2[i] *= np.array([1, 1, -1])
atoms2.positions = pos2

#now move N to center of box
atoms.translate([5, 5, 5])
atoms2.translate([5, 5, 5])

calcs = [Vasp('molecules/nh3-initial',
xc='PBE',
encut=350,
ibrion=1,
nsw=10,
atoms=atoms),
Vasp('molecules/nh3-final',
xc='PBE',
encut=350,
ibrion=1,
nsw=10,
atoms=atoms2)]

print [c.potential_energy for c in calcs]


#### 3.9.2 Run band calculation

Now we do the band calculation. nudged elastic band

# Run NH3 NEB calculations
from vasp import Vasp
from ase.neb import NEB

atoms = Vasp('molecules/nh3-initial').get_atoms()
atoms2 = Vasp('molecules/nh3-final').get_atoms()

# 5 images including endpoints
images = [atoms]   # initial state
images += [atoms.copy() for i in range(3)]
images += [atoms2]  # final state

neb = NEB(images)
neb.interpolate()

calc = Vasp('molecules/nh3-neb',
xc='PBE',
ibrion=1, encut=350,
nsw=90,
spring=-5.0,
atoms=images)

#calc.write_db(atoms, 'molecules/nh3-neb/00/DB.db')
#calc.write_db(atoms2, 'molecules/nh3-neb/04/DB.db')
images, energies = calc.get_neb()
calc.stop_if(None in energies)

print images
print energies
p = calc.plot_neb(show=False)
import matplotlib.pyplot as plt
plt.savefig('images/nh3-neb.png')

[Atoms(symbols='NH3', positions=..., magmoms=..., cell=[10.0, 10.0, 10.0], pbc=[True, True, True], constraint=FixAtoms(indices=[0]), calculator=Vasp(...)), Atoms(symbols='NH3', positions=..., magmoms=..., cell=[10.0, 10.0, 10.0], pbc=[True, True, True], constraint=FixAtoms(indices=[0])), Atoms(symbols='NH3', positions=..., magmoms=..., cell=[10.0, 10.0, 10.0], pbc=[True, True, True], constraint=FixAtoms(indices=[0])), Atoms(symbols='NH3', positions=..., magmoms=..., cell=[10.0, 10.0, 10.0], pbc=[True, True, True], constraint=FixAtoms(indices=[0])), Atoms(symbols='NH3', positions=..., magmoms=..., cell=[10.0, 10.0, 10.0], pbc=[True, True, True], constraint=FixAtoms(indices=[0]), calculator=Vasp(...))]
[  0.00000000e+00   1.26688520e-01   2.25038820e-01   1.26688620e-01
9.99999727e-09]
Optimization terminated successfully.
Current function value: -0.225039
Iterations: 15
Function evaluations: 30


The calculator view function shows you the band.

from vasp import Vasp

calc = Vasp('molecules/nh3-neb')
calc.view()


#### 3.9.3 Make a movie of the animation

animation It is helpful sometimes to animate the Nudged elastic band path. Here is a script to do that. I have not figured out how to embed the movie in this document

# make neb movie
from ase.io import write
from ase.visualize import view
from vasp import Vasp

calc = Fasp('molecules/nh3-neb') as calc:
images, energies = calc.get_neb()

# this rotates the atoms 90 degrees about the y-axis
[atoms.rotate('y', np.pi/2.) for atoms in images]

for i,atoms in enumerate(images):
write('images/00{0}-nh3.png'.format(i), atoms, show_unit_cell=2)

# animated gif
os.system('convert -delay 50 -loop 0 images/00*-nh3.png images/nh3-neb.gif')

# Shockwave flash
os.system('png2swf -o images/nh3-neb.swf images/00*-nh3.png ')


## 4 Bulk systems

See http://arxiv.org/pdf/1204.2733.pdf for a very informative comparison of DFT codes for computing different bulk properties.

### 4.1 Defining and visualizing bulk systems

#### 4.1.1 Built-in functions in ase

As with molecules, ase provides several helper functions to create bulk structures. We highlight a few of them here. Particularly common ones are:

For others, see https://wiki.fysik.dtu.dk/ase/ase/lattice.html

We start with a simple example, fcc Ag. By default, ase knows Ag is an fcc metal, and knows the experimental lattice constant. We have to specify the directions (vectors along each axis) to get something other than the default output. Here, the default fcc cell contains four atoms.

from ase.io import write
from ase.lattice.cubic import FaceCenteredCubic

atoms = FaceCenteredCubic('Ag')

write('images/Ag-fcc.png', atoms, show_unit_cell=2)

print(atoms)

Lattice(symbols='Ag4', positions=..., cell=[4.09, 4.09, 4.09], pbc=[True, True, True])


A ase.lattice.bravais.Lattice object is returned! This is practically the same as as an ase.atoms.Atoms object.

Here we specify the primitive unit cell, which only has one atom in it.

from ase.io import write
from ase.lattice.cubic import FaceCenteredCubic

atoms = FaceCenteredCubic('Ag', directions=[[0, 1, 1],
[1, 0, 1],
[1, 1, 0]])

write('images/Ag-fcc-primitive.png', atoms, show_unit_cell=2)

print atoms

Lattice(symbols='Ag', positions=..., cell=[[2.892066735052979, 0.0, 0.0], [1.4460333675264898, 2.5046032619957996, 0.0], [1.4460333675264896, 0.8348677539985997, 2.3613626009855695]], pbc=[True, True, True])

Lattice(symbols='Ag', positions=..., cell=[[2.892066735052979, 0.0, 0.0], [1.4460333675264898, 2.5046032619957996, 0.0], [1.4460333675264896, 0.8348677539985997, 2.3613626009855695]], pbc=[True, True, True])


We can use these modules to build alloy unit cells. The basic strategy is to create the base unit cell in one element and then selectively change some atoms to different chemical symbols. Here we examine an Ag3Pd alloy structure.

from ase.io import write
from ase.lattice.cubic import FaceCenteredCubic

atoms = FaceCenteredCubic(directions=[[1, 0, 0],
[0, 1, 0],
[0, 0, 1]],
size=(1, 1, 1),
symbol='Ag',
latticeconstant=4.0)

write('images/Ag-bulk.png', atoms, show_unit_cell=2)

# to make an alloy, we can replace one atom with another kind
atoms[0].symbol = 'Pd'
write('images/AgPd-bulk.png', atoms, show_unit_cell=2)


To create a graphite structure we use the following code. Note that we have to specify the lattice constants (taken from http://www.phy.ohiou.edu/~asmith/NewATOMS/HOPG.pdf) because ase has C in the diamond structure by default. We show two views, because the top view does not show the spacing between the layers.

from ase.lattice.hexagonal import Graphite
from ase.io import write

atoms = Graphite('C', latticeconstant={'a': 2.4612,
'c': 6.7079})
write('images/graphite.png',
atoms.repeat((2, 2, 1)),
rotation='115x', show_unit_cell=2)

write('images/graphite-top.png',
atoms.repeat((2, 2, 1)),
show_unit_cell=2)


To get a compound, we use the following code. We have to specify the basis atoms to the function generating the compound, and the lattice constant. For NaCl we use the lattice constant at (http://en.wikipedia.org/wiki/Sodium_chloride).

from ase.lattice.compounds import NaCl
from ase.io import write

atoms = NaCl(['Na', 'Cl'], latticeconstant=5.65)
write('images/NaCl.png', atoms, show_unit_cell=2, rotation='45x,45y,45z')

##### 4.1.1.1ase.spacegroup

A final alternative to setting up bulk structures is ase.spacegroup. This is a concise way to setup structures if you know the following properties of the crystal structure:

1. Chemical symbols
2. Coordinates of the non-equivalent sites in the unit cell
3. the spacegroup
4. the cell parameters (a, b, c, alpha, beta, gamma)
from ase.lattice.spacegroup import crystal
# FCC aluminum
a = 4.05
al = crystal('Al', [(0, 0, 0)],
spacegroup=225,
cellpar=[a, a, a, 90, 90, 90])
print(al)

Atoms(symbols='Al4', positions=..., cell=[[4.05, 0.0, 0.0], [2.4799097682733903e-16, 4.05, 0.0], [2.4799097682733903e-16, 2.4799097682733903e-16, 4.05]], pbc=[True, True, True])


Here is rutile TiO2.

from ase.lattice.spacegroup import crystal

a = 4.6
c = 2.95
rutile = crystal(['Ti', 'O'], basis=[(0, 0, 0), (0.3, 0.3, 0.0)],
spacegroup=136, cellpar=[a, a, c, 90, 90, 90])
print rutile


Atoms(symbols'Ti2O4', positions=…, cell=[[4.6, 0.0, 0.0], [2.816687638038912e-16, 4.6, 0.0], [1.806354028742346e-16, 1.806354028742346e-16, 2.95]], pbc=[True, True, True]) =Atoms(symbols'Ti2O4', positions=…, cell=[[4.6, 0.0, 0.0], [2.816687638038912e-16, 4.6, 0.0], [1.806354028742346e-16, 1.806354028742346e-16, 2.95]], pbc=[True, True, True]) =sho

#### 4.1.2 Using http://materialsproject.org/

The Materials Project offers web access to a pretty large number of materials (over 21,000 at the time of this writing), including structure and other computed properties. You must sign up for an account at the website, and then you can access the information. You can search for materials with lots of different criteria including formula, unit cell formula, by elements, by structure, etc… The website allows you to download the VASP files used to create the calculations. They also develop the pymatgen project (which requires python 2.7+).

For example, I downloaded this cif file for a RuO$$_2$$ structure (Material ID 825).

#\#CIF1.1
##########################################################################
#               Crystallographic Information Format file
#               Produced by PyCifRW module
#
#  This is a CIF file.  CIF has been adopted by the International
#  Union of Crystallography as the standard for data archiving and
#  transmission.
#
#  For information on this file format, follow the CIF links at
#  http://www.iucr.org
##########################################################################

data_RuO2
_symmetry_space_group_name_H-M          'P 1'
_cell_length_a                          3.13970109
_cell_length_b                          4.5436378
_cell_length_c                          4.5436378
_cell_angle_alpha                       90.0
_cell_angle_beta                        90.0
_cell_angle_gamma                       90.0
_chemical_name_systematic               'Generated by pymatgen'
_symmetry_Int_Tables_number             1
_chemical_formula_structural            RuO2
_chemical_formula_sum                   'Ru2 O4'
_cell_volume                            64.8180127062
_cell_formula_units_Z                   2
loop_
_symmetry_equiv_pos_site_id
_symmetry_equiv_pos_as_xyz
1  'x, y, z'

loop_
_atom_site_type_symbol
_atom_site_label
_atom_site_symmetry_multiplicity
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_attached_hydrogens
_atom_site_B_iso_or_equiv
_atom_site_occupancy
O  O1  1  0.000000  0.694330  0.694330  0  .  1
O  O2  1  0.500000  0.805670  0.194330  0  .  1
O  O3  1  0.000000  0.305670  0.305670  0  .  1
O  O4  1  0.500000  0.194330  0.805670  0  .  1
Ru  Ru5  1  0.500000  0.500000  0.500000  0  .  1
Ru  Ru6  1  0.000000  0.000000  0.000000  0  .  1


We can read this file in with ase.io.read. That function automatically recognizes the file type by the extension.

from ase.io import read, write

write('images/Ru2O4.png', atoms, show_unit_cell=2)


### 4.2 Computational parameters that are important for bulk structures

#### 4.2.1 k-point convergence

In the section on molecules, we learned that the total energy is a function of the planewave cutoff energy (ENCUT) used. In bulk systems that is true also. There is also another calculation parameter you must consider, the k-point grid. The k-point grid is a computational tool used to approximate integrals of some property, e.g. the electron density, over the entire unit cell. The integration is performed in reciprocal space (i.e. in the Brillouin zone) for convenience and efficiency, and the k-point grid is where the property is sampled for the integration. The higher the number of sampled points, the more accurately the integrals are approximated.

We will typically use a Monkhorst-Pack PhysRevB.13.5188 $k$-point grid, which is essentially a uniformly spaced grid in the Brillouin zone. Another less commonly used scheme is the Chadi-Cohen k-point grid PhysRevB.8.5747. The Monkhorst-Pack grids are specified as $$n1 \times n2 \times n3$$ grids, and the total number of k-points is $$n1 \cdot n2 \cdot n3$$. The computational cost is linear in the total number of k-points, so a calculation on a $$4 \times 4 \times 4$$ grid will be roughly 8 times more expensive than on a $$2 \times 2 \times 2$$ grid. Hence, one seeks again to balance convergence with computational tractability. Below we consider the k-point convergence of fcc Ag.

from ase.lattice.cubic import FaceCenteredCubic
from vasp import Vasp
import numpy as np

atoms = FaceCenteredCubic('Ag')

KPTS = [2, 3, 4, 5, 6, 8, 10]

TE = []

for k in KPTS:
calc = Vasp('bulk/Ag-kpts-{0}'.format(k),
xc='PBE',
kpts=[k, k, k],  # specifies the Monkhorst-Pack grid
encut=300,
atoms=atoms)
TE.append(atoms.get_potential_energy())

if None in TE:
calc.abort()

import matplotlib.pyplot as plt

# consider the change in energy from lowest energy state
TE = np.array(TE)
TE -= TE.min()

plt.plot(KPTS, TE)
plt.xlabel('number of k-points in each dimension')
plt.ylabel('Total Energy (eV)')
plt.savefig('images/Ag-kpt-convergence.png')


Based on this figure, we need at least a $$6 \times 6 \times 6$$ k-point grid to achieve a convergence level of at least 50 meV. Note: the k-point convergence is not always monotonic like it is in this example, and sometimes very dense grids (e.g. up to $$20 \times 20 \times 20$$) are needed for highly converged properties such as the density of states in smaller unit cells. Oscillations in the total energy are typical, and it can be difficult to get high levels of convergence. The best practices are to use the same k-point sampling grid in energy differences where possible, and dense (high numbers of k-points) otherwise. It is important to check for convergence in these cases.

As unit cells get larger, the number of k-points required becomes smaller. For example, if a $$1 \times 1 \times 1$$ fcc unit cell shows converged energies in a $$12 \times 12 \times 12$$ k-point grid, then a $$2 \times 2 \times 2$$ fcc unit cell would show the same level of convergence with a $$6 \times 6 \times 6$$ k-point grid. In other words, doubling the unit cell vectors results in a halving of the number of k-points.

Sometimes you may see k-points described as k-points per reciprocal atom. For example, a $$12 \times 12 \times 12$$ k-point grid for a primitive fcc unit cell would be 1728 k-points per reciprocal atom. A $$2 \times 2 \times 2$$ fcc unit cell has eight atoms in it, or 0.125 reciprocal atoms, so a $$6 \times 6 \times 6$$ k-point grid has 216 k-points in it, or 216/0.125 = 1728 k-points per reciprocal atom, the same as we discussed before.

In the k-point convergence example above, we used a $$6 \times 6 \times 6$$ k-point grid on a unit cell with four atoms in it, leading to 864 k-points per reciprocal atom. If we had instead used the primitive unit cell, we would need either a $$9 \times 9 \times 9$$ or $$10 \times 10 \times 10$$ k-point grid to get a similar level of accuracy. In this case, there is no exact matching of k-point grids due to the difference in shape of the cells.

#### 4.2.2TODO Effect of SIGMA

In the self-consistent cycle of a DFT calculation, the total energy is minimized with respect to occupation of the Kohn-Sham orbitals. At absolute zero, a band is either occupied or empty. This discrete occupation results in discontinuous changes in energy with changes in occupation, which makes it difficult to converge. One solution is to artificially broaden the band occupancies, as if they were occupied at a higher temperature where partial occupation is possible. This results in a continuous dependence of energy on the partial occupancy, and dramatically increases the rate of convergence. SIGMA and ISMEAR affect how the partial occupancies of the bands are determined.

Some rules to keep in mind:

1. The smearing methods were designed for metals. For molecules, semiconductors and insulators you should use a very small SIGMA (e.g. 0.01).
2. Standard values for metallic systems is SIGMA=0.1, but the best SIGMA may be material specific.

The consequence of this finite temperature is that additional bands must be included in the calculation to allow for the partially occupied states above the Fermi level; the number of extra bands depends on the temperature used. An example of the maximum occupancies of the bands for an Cu bulk as a function of SIGMA is shown in Figure fig:sigma-occ. Obviously, as SIGMA approaches 0, the occupancy approaches a step function. It is preferable that the occupancy of several of the highest bands be zero (or at least of order $$1\times 10^{-8}$$) to ensure enough variational freedom was available in the calculation. Consequently, it is suggested that fifteen to twenty extra bands be used for a SIGMA of 0.20. In any case, it should be determined that enough bands were used by examination of the occupancies. It is undesirable to have too many extra bands, as this will add computational time.

Below we show the effect of SIGMA on the band occupancies.

from vasp import Vasp
from ase import Atom, Atoms
import matplotlib.pyplot as plt
import numpy as np

a = 3.61
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]])).repeat((2, 2, 2))

SIGMA = [0.001, 0.05, 0.1, 0.2, 0.5]

for sigma in SIGMA:

calc = Vasp('bulk/Cu-sigma-{0}'.format(sigma),
xc='PBE',
encut=350,
kpts=[4, 4, 4],
ismear=-1,
sigma=sigma,
nbands=9 * 8,
atoms=atoms)

if calc.potential_energy is not None:
nbands = calc.parameters.nbands
nkpts = len(calc.get_ibz_k_points())

occ = np.zeros((nkpts, nbands))
for i in range(nkpts):
occ[i, :] = calc.get_occupation_numbers(kpt=i)

max_occ = np.max(occ, axis=0) #axis 0 is columns

plt.plot(range(nbands), max_occ, label='$\sigma = {0}$'.format(sigma))

plt.xlabel('band number')
plt.ylabel('maximum occupancy (electrons)')
plt.ylim([-0.1, 2.1])
plt.legend(loc='best')
plt.savefig('images/occ-sigma.png')


#### 4.2.3 The number of bands

In the last figure, it is evident that due to the smearing of the electronic states you need to have extra bands to accommodate the electrons above the Fermi level, and the higher the SIGMA value is, the more bands you need. You need enough bands so that the highest energy bands are unoccupied, and VASP will give you a warning that looks like this:

 -----------------------------------------------------------------------------
|                                                                             |
|  ADVICE TO THIS USER RUNNING 'VASP/VAMP'   (HEAR YOUR MASTER'S VOICE ...):  |
|                                                                             |
|      Your highest band is occupied at some k-points! Unless you are         |
|      performing a calculation for an insulator or semiconductor, without    |
|      unoccupied bands, you have included TOO FEW BANDS!! Please increase    |
|      the parameter NBANDS in file 'INCAR' to ensure that the highest band   |
|      is unoccupied at all k-points. It is always recommended to             |
|      include a few unoccupied bands to accelerate the convergence of        |
|      molecular dynamics runs (even for insulators or semiconductors).       |
|      Because the presence of unoccupied bands improves wavefunction         |
|      prediction, and helps to suppress 'band-crossings.'                    |
|      Following all k-points will be listed (with the Fermi weights of       |
|      the highest band given in paranthesis) ... :                           |
|                                                                             |
|                          6       (-0.01472)                                 |
|                          8       (-0.01413)                                 |
|                         13       (-0.01733)                                 |
|                         14       (-0.01838)                                 |
|                                                                             |
|      The total occupancy of band no.    49 is  -0.00932 electrons ...       |
|                                                                             |
-----------------------------------------------------------------------------


We tell VASP the number of bands to use with the NBANDS keyword. VASP will set the NBANDS automatically if you do not provide a value, but this is in general bad practice (even though it is often done in this book!). There are a few general guidelines for setting NBANDS. First we recognize that a band can only have two electrons in it (one spin up, and one spin down) in an calculation without spin-polarization, or one electron per band for a spin-polarized calculation (note that spin-polarization doubles the number of bands). There absolutely must be enough bands to accommodate all the electrons, so the minimum number of bands is int(ceil(nelectrons/2)).

Here is an example of what this equation does.

import numpy as np

print int(np.ceil(50 / 2.))
print int(np.ceil(51 / 2.))

25
26


However, due to the smearing, the minimum number of bands is almost never enough, and we always add more bands. The default behavior in VASP is:

 non-spin polarized NELECT/2 + NIONS/2 spin-polarized 0.6*NELECT + NMAGIONS

These do not always work, especially for small molecular systems where NIONS/2 may be only 1, or transition metals where it may be necessary to add up to 2*NIONS extra bands.

To figure out how many bands you need, it is necessary to know how many electrons are in your calculation. The Vasp.get_valence_electrons provides this for you. Alternatively, you can look in the Appendix for a table listing the number of valence electrons for each POTCAR file. Armed with this information you can set NBANDS the way you want.

from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('Cu',  [0.000, 0.000, 0.000])],
cell= [[1.818, 0.000, 1.818],
[1.818, 1.818, 0.000],
[0.000, 1.818, 1.818]])

calc = Vasp('bulk/alloy/cu',
xc='PBE',
encut=350,
kpts=[13, 13, 13],
nbands=9,
ibrion=2,
isif=4,
nsw=10,
atoms=atoms)

print(calc.get_valence_electrons())
print(calc.potential_energy)

11.0
-3.73436945


For this calculation we need at least 6 bands (11/2=5.5 which is rounded up to 6) and we need to include some extra bands. The default rule would only add half a band, which is not enough. We add three additional bands. This system is so small it does not substantially increase the computational cost.

If you are too trifling to do that much work, you can use the Vasp.set_nbands to automatically set the number of bands. This function takes an argument N to set the number of bands to N, or an argument f to set the NBANDS according to the formula $$nbands = int(nelectrons/2 + len(atoms)*f)$$. The default value of f is 1.5. If you want the default VASP behavior, set f=0.5. For transition metals, it may be required that f=2. This function does not consider whether the calculation is spin-polarized or not. Here is an example of using Vasp.set_nbands.

from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('Cu', [0.000, 0.000, 0.000])],
cell=[[1.818, 0.000, 1.818],
[1.818, 1.818, 0.000],
[0.000, 1.818, 1.818]])

calc = Vasp('bulk/alloy/cu',
xc='PBE',
encut=350,
kpts=[13, 13, 13],
ibrion=2,
isif=4,
nsw=10,
atoms=atoms)
calc.set_nbands(f=7)
calc.write_input()  # you have to write out the input for it to take effect
print calc

*************** VASP CALCULATION SUMMARY ***************
Vasp calculation directory:
---------------------------
[[/home-research/jkitchin/dft-book/bulk/alloy/cu]]

Unit cell:
----------
x       y       z             |v|
v0   1.818   0.000   1.818       2.571 Ang
v1   1.818   1.818   0.000       2.571 Ang
v2   0.000   1.818   1.818       2.571 Ang
alpha, beta, gamma (deg):  60.0  60.0  60.0
Total volume:                   12.017 Ang^3
Stress:    xx     yy     zz     yz     xz     xy
nan    nan    nan    nan    nan    nan GPa

ID  tag     sym    x         y         z        rmsF (eV/A)
0   0       Cu     0.000     0.000     0.000        nan
Potential energy: nan eV

INPUT Parameters:
-----------------
pp        : PBE
isif      : 4
xc        : pbe
kpts      : [13, 13, 13]
encut     : 350
lcharg    : False
ibrion    : 2
nbands    : 13
ismear    : 1
lwave     : True
sigma     : 0.1
nsw       : 10

Pseudopotentials used:
----------------------
Cu: potpaw_PBE/Cu/POTCAR (git-hash: 13fa889d46be8b12a676c1063c5e4faede17e89b)


Note the defaults that were set.

from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('Cu', [0.000, 0.000, 0.000])],
cell=[[1.818, 0.000, 1.818],
[1.818, 1.818, 0.000],
[0.000, 1.818, 1.818]])

calc = Vasp('bulk/alloy/cu-setnbands',
xc='PBE',
encut=350,
kpts=[13, 13, 13],
ibrion=2,
isif=4,
nsw=10,
atoms=atoms)
calc.set_nbands(f=3)
calc.write_input()
print calc

*************** VASP CALCULATION SUMMARY ***************
Vasp calculation directory:
---------------------------
[[/home-research/jkitchin/dft-book/bulk/alloy/cu-setnbands]]

Unit cell:
----------
x       y       z             |v|
v0   1.818   0.000   1.818       2.571 Ang
v1   1.818   1.818   0.000       2.571 Ang
v2   0.000   1.818   1.818       2.571 Ang
alpha, beta, gamma (deg):  60.0  60.0  60.0
Total volume:                   12.017 Ang^3
Stress:    xx     yy     zz     yz     xz     xy
nan    nan    nan    nan    nan    nan GPa

ID  tag     sym    x         y         z        rmsF (eV/A)
0   0       Cu     0.000     0.000     0.000        nan
Potential energy: nan eV

INPUT Parameters:
-----------------
pp        : PBE
isif      : 4
xc        : pbe
kpts      : [13, 13, 13]
encut     : 350
lcharg    : False
ibrion    : 2
nbands    : 9
ismear    : 1
lwave     : True
sigma     : 0.1
nsw       : 10

Pseudopotentials used:
----------------------
Cu: potpaw_PBE/Cu/POTCAR (git-hash: 13fa889d46be8b12a676c1063c5e4faede17e89b)


You are, of course, free to use any formula you want to set the number of bands. Some formulas I have used in the past include:

1. NBANDS = 0.65*NELECT + 10
2. NBANDS = 0.5*NELECT + 15
3. etc…

### 4.3 Determining bulk structures

What we typically mean by determining bulk structures includes the following:

• What is the most stable crystal structure for a material?
• What is the lattice constant of fcc Cu?
• What are the lattice parameters and internal atom parameters for TiO2?

All of these questions can often be addressed by finding the volume, shape and atomic positions that minimize the total energy of a bulk system. This is true at 0K. At higher temperatures, one must consider minimizing the free energy, rather than the internal energy.

#### 4.3.1 fcc/bcc crystal structures

The fcc and bcc structures are simple. They only have one degree of freedom: the lattice constant. In this section we show how to calculate the equilibrium volume of each structure, and determine which one is more stable. We start with the fcc crystal structure of Cu. We will manually define the crystal structure based on the definitions in Kittel kittel (Chapter 1).

from vasp import Vasp
from ase import Atom, Atoms
import numpy as np

# fcc
LC = [3.5, 3.55, 3.6, 3.65, 3.7, 3.75]
fcc_energies = []
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]]))

calc = Vasp('bulk/Cu-{0}'.format(a),
xc='PBE',
encut=350,
kpts=[8, 8, 8],
atoms=atoms)

e = atoms.get_potential_energy()
fcc_energies.append(e)

calc.stop_if(None in fcc_energies)

import matplotlib.pyplot as plt
plt.plot(LC, fcc_energies)
plt.xlabel('Lattice constant ($\AA$)')
plt.ylabel('Total energy (eV)')
plt.savefig('images/Cu-fcc.png')

print '#+tblname: cu-fcc-energies'
print r'| lattice constant ($\AA$) | Total Energy (eV) |'
for lc, e in zip(LC, fcc_energies):
print '| {0} | {1} |'.format(lc, e)

 lattice constant ($$\AA$$) Total Energy (eV) 3.5 -3.66182568 3.55 -3.70817569 3.6 -3.73109713 3.65 -3.73432446 3.7 -3.72094163 3.75 -3.69342783

Use the data in the table above to plot the total energy as a function of the lattice constant. Fit a cubic polynomial to the data, and find the volume that minimizes the total energy.

If you want to know the lattice constant that gives the lowest energy, you would fit an equation of state to the data. Here is an example using ase.utils.eos. See also the appendix equations of state.

from vasp import Vasp
from ase.utils.eos import EquationOfState
LC = [3.5, 3.55, 3.6, 3.65, 3.7, 3.75]
energies = []
volumes = []
for a in LC:
calc = Vasp('bulk/Cu-{0}'.format(a))
atoms = calc.get_atoms()
volumes.append(atoms.get_volume())
energies.append(atoms.get_potential_energy())

calc.stop_if(None in energies)

eos = EquationOfState(volumes, energies)
v0, e0, B = eos.fit()

print '''
v0 = {0} A^3
E0 = {1} eV
B  = {2} eV/A^3'''.format(v0, e0, B)

eos.plot('images/Cu-fcc-eos.png')

v0 = 11.9941760954 A^3
E0 = -3.73528237713 eV
B  = 0.862553823078 eV/A^3


3.63585568663 3.63585568663 3.63585568663

Before we jump into the bcc calculations, let us consider what range of lattice constants we should choose. The fcc lattice is close-packed, and the volume of the primitive cell is $$V = 1/4 a^3$$ or about 11.8 Å3/atom. The volume of the equilibrium bcc primitive cell will probably be similar to that. The question is: what bcc lattice constant gives that volume? The simplest way to answer this is to compute the answer. We will make a bcc crystal at the fcc lattice constant, and then compute the scaling factor needed to make it the right volume.

from ase import Atom, Atoms
import numpy as np
a = 3.61 # lattice constant

atoms = Atoms([Atom('Cu', [0,0,0])],
cell=0.5 * a*np.array([[ 1.0,  1.0, -1.0],
[-1.0,  1.0,  1.0],
[ 1.0, -1.0,  1.0]]))

print 'BCC lattice constant = {0:1.3f} Ang'.format(a * (11.8 / atoms.get_volume())**(1./3.))

BCC lattice constant = 2.868 Ang


Now we run the equation of state calculations.

from vasp import Vasp
from ase import Atom, Atoms
import numpy as np

LC = [2.75, 2.8, 2.85, 2.9, 2.95, 3.0]

for a in LC:
atoms = Atoms([Atom('Cu', [0, 0, 0])],
cell=0.5 * a * np.array([[ 1.0,  1.0, -1.0],
[-1.0,  1.0,  1.0],
[ 1.0, -1.0,  1.0]]))

calc = Vasp('bulk/Cu-bcc-{0}'.format(a),
xc='PBE',
encut=350,
kpts=[8, 8, 8],
atoms=atoms)
print(calc.potential_energy)

-3.59937543
-3.67930795
-3.71927399
-3.72637899
-3.70697046
-3.66645678


Finally, we will compare the two crystal structures.

from vasp import Vasp

# bcc energies and volumes
bcc_LC = [2.75, 2.8, 2.85, 2.9, 2.95, 3.0]
bcc_volumes = []
bcc_energies = []
for a in bcc_LC:
calc = Vasp('bulk/Cu-bcc-{0}'.format(a))
atoms = calc.get_atoms()
bcc_volumes.append(atoms.get_volume())
bcc_energies.append(atoms.get_potential_energy())

# fcc energies and volumes
fcc_LC = [3.5, 3.55, 3.6, 3.65, 3.7, 3.75]
fcc_volumes = []
fcc_energies =[]
for a in fcc_LC:
calc = Vasp('bulk/Cu-{0}'.format(a))
atoms = calc.get_atoms()
fcc_volumes.append(atoms.get_volume())
fcc_energies.append(atoms.get_potential_energy())

import matplotlib.pyplot as plt
plt.plot(fcc_volumes, fcc_energies, label='fcc')
plt.plot(bcc_volumes, bcc_energies, label='bcc')

plt.xlabel('Atomic volume ($\AA^3$/atom)')
plt.ylabel('Total energy (eV)')
plt.legend()
plt.savefig('images/Cu-bcc-fcc.png')

# print table of data
print '#+tblname: bcc-data'
print '#+caption: Total energy vs. lattice constant for BCC Cu.'
print '| Lattice constant (\AA$^3$) | Total energy (eV) |'
print '|-'
for lc, e in zip(bcc_LC, bcc_energies):
print '| {0} | {1} |'.format(lc, e)

Table 3: Total energy vs. lattice constant for BCC Cu.
Lattice constant (Å$$^3$$) Total energy (eV)
2.75 -3.59937543
2.8 -3.67930795
2.85 -3.71927399
2.9 -3.72637899
2.95 -3.70697046
3.0 -3.66645678

Use the data for FCC and BCC Cu to plot the total energy as a function of the lattice constant.

Note we plot the energy vs. atomic volume. That is because the lattice constants of the two crystal structures are very different. It also shows that the atomic volumes in the two structures are similar.

What can we say here? The fcc structure has a lower energy than the bcc structure, so we can conclude the fcc structure is more favorable. In fact, the fcc structure is the experimentally found structure for Cu. Some caution is in order; if you run these calculations at a $$4 \times 4 \times 4$$ $k$-point grid, the bcc structure is more stable because the results are not converged!

Compute the energy vs. volume for fcc and bcc Cu for different $k$-point grids. Determine when each result has converged, and which structure is more stable.

What can we say about the relative stability of fcc to hcp? Nothing, until we calculate the hcp equation of state.

#### 4.3.2 Optimizing the hcp lattice constant

The hcp lattice is more complicated than the fcc/bcc lattices because there are two lattice parameters: $$a$$ and $$c$$ or equivalently: $$a$$ and $$c/a$$. We will start by making a grid of values and find the set of parameters that minimizes the energy. See Figure fig:ru-e-ca.

from ase.lattice.hexagonal import HexagonalClosedPacked
from vasp import Vasp
import matplotlib.pyplot as plt

atoms = HexagonalClosedPacked(symbol='Ru',
latticeconstant={'a': 2.7,
'c/a': 1.584})

a_list = [2.5, 2.6, 2.7, 2.8, 2.9]
covera_list = [1.4, 1.5, 1.6, 1.7, 1.8]

for a in a_list:
energies = []
for covera in covera_list:

atoms = HexagonalClosedPacked(symbol='Ru',
latticeconstant={'a': a,
'c/a': covera})

wd = 'bulk/Ru/{0:1.2f}-{1:1.2f}'.format(a, covera)

calc = Vasp(wd,
xc='PBE',
# the c-axis is longer than the a-axis, so we use
# fewer kpoints.
kpts=[6, 6, 4],
encut=350,
atoms=atoms)

energies.append(atoms.get_potential_energy())
if not None in energies:
plt.plot(covera_list, energies, label=r'a={0} $\AA$'.format(a))

plt.xlabel('$c/a$')
plt.ylabel('Energy (eV)')
plt.legend()
plt.savefig('images/Ru-covera-scan.png')


It looks like there is a minimum in the a=2.7 Å curve, at a $$c/a$$ ratio of about 1.6. We can look at the same data in a contour plot which shows more clearly there is minimum in all directions near that point (Figure fig:ru-contourf).

from vasp import Vasp
import matplotlib.pyplot as plt
import numpy as np

x = [2.5, 2.6, 2.7, 2.8, 2.9]
y = [1.4, 1.5, 1.6, 1.7, 1.8]

X,Y = np.meshgrid(x, y)
Z = np.zeros(X.shape)

for i,a in enumerate(x):
for j,covera in enumerate(y):
wd = 'bulk/Ru/{0:1.2f}-{1:1.2f}'.format(a, covera)
calc = Vasp(wd)
Z[i][j] = calc.potential_energy

calc.stop_if(None in Z)

cf = plt.contourf(X, Y, Z, 20,
cmap=plt.cm.jet)

cbar = plt.colorbar(cf)
cbar.ax.set_ylabel('Energy (eV)')

plt.xlabel('$a$ ($\AA$)')
plt.ylabel('$c/a$')

plt.legend()
plt.savefig('images/ru-contourf.png')
plt.show()


#### 4.3.3 Complex structures with internal degrees of freedom

A unit cell has six degrees of freedom: the lengths of each unit cell vector, and the angle between each vector. There may additionally be internal degrees of freedom for the atoms. It is impractical to try the approach used for the hcp Ru on anything complicated. Instead, we rely again on algorithms to optimize the unit cell shape, volume and internal degrees of freedom. It is usually not efficient to make a wild guess of the geometry and then turn VASP loose on to optimize it. Instead, the following algorithm works pretty well.

1. Find the volume (at constant shape, with relaxed ions) that minimizes the total energy (ISIF=2). The goal here is to just get an idea of where the right volume is.
2. Using the results from step 1 as a starting point, perform a set of calculations at constant volume around the minimum from step 1, but the shape and internal atom positions are allowed to change (ISIF=4).
3. Finally, do a final calculation near the minimum energy allowing the volume to also change. (ISIF=3).

This multistep process is pretty reasonable to get a converged structure pretty quickly. It is not foolproof, however, and if you have materials such as graphite it may not work well. The problem with graphite is that it is a layered compound that is held together by weak van der waal type forces which are not modeled well by typical GGA functionals. Thus the change in energy due to a volume change is larger in the plane of the graphite sheet than in the direction normal to the sheet. With a typical GGA, the sheets may just move apart until they do not interact any more.

We will illustrate the process on a well-behaved system (rutile TiO2) which has two lattice parameters and one internal degree of freedom. There are a few subtle points to mention in doing these calculations. The VASP manual recommends that you set PREC to 'high', and that ENCUT be set to 1.3*max(ENMAX) of the pseudopotentials. This is necessary to avoid problems caused by small basis sets when the volume changes, and Pulay stress. It is important to ensure that the energies are reasonably converged with respect to k-point grids. Hence, it can be a significant amount of work to do this right! Let us start with determining the ENCUT value that is appropriate for TiO2.

grep ENMAX $VASP_PP_PATH/potpaw_PBE/Ti/POTCAR grep ENMAX$VASP_PP_PATH/potpaw_PBE/O/POTCAR

ENMAX  =  178.330; ENMIN  =  133.747 eV
ENMAX  =  400.000; ENMIN  =  300.000 eV


According to the manual, we should use ENCUT = 1.3*400 = 520 eV for good results.

Now we consider the k-point convergence. The lattice vectors of the rutile TiO2 structure are not all the same length, which means it is not essential that we use the same number of k-points in each direction. For simplicity, however, we do that here.

# step 1 frozen atoms and shape at different volumes
from ase import Atom, Atoms
import numpy as np
from vasp import Vasp
import matplotlib.pyplot as plt

'''
create a TiO2 structure from the lattice vectors at
http://cst-www.nrl.navy.mil/lattice/struk/c4.html
This site does not exist anymore.
'''
a = 4.59  # experimental degrees of freedom.
c = 2.96
u = 0.3  # internal degree of freedom!

#primitive vectors
a1 = a * np.array([1.0, 0.0, 0.0])
a2 = a * np.array([0.0, 1.0, 0.0])
a3 = c * np.array([0.0, 0.0, 1.0])

atoms = Atoms([Atom('Ti', [0., 0., 0.]),
Atom('Ti', 0.5 * a1 + 0.5 * a2 + 0.5 * a3),
Atom('O', u * a1 + u * a2),
Atom('O', -u * a1 - u * a2),
Atom('O', (0.5 + u) * a1 + (0.5 - u) * a2 + 0.5 * a3),
Atom('O', (0.5 - u) * a1 + (0.5 + u) * a2 + 0.5 * a3)],
cell=[a1, a2, a3])

KPOINTS = [2, 3, 4, 5, 6, 7, 8]
energies = []

for k in KPOINTS:
calc = Vasp('bulk/tio2/kpts-{0}'.format(k),
encut=520,
kpts=[k, k, k],
xc='PBE',
sigma=0.05,
atoms=atoms)

energies.append(atoms.get_potential_energy())

calc.stop_if(None in energies)

plt.plot(KPOINTS, energies)
plt.xlabel('number of k-points in each vector')
plt.ylabel('Total energy (eV)')
plt.savefig('images/tio2-kpt-convergence.png')


A k-point grid of $$5 \times 5 \times 5$$ appears suitable for reasonably converged results. Now we proceed with step 1: Compute the total energy of the unit cell allowing internal degrees of freedom to relax, but keeping a constant cell shape.

# step 1 frozen atoms and shape at different volumes
from ase import Atom, Atoms
import numpy as np
from vasp import Vasp
import matplotlib.pyplot as plt

'''
create a TiO2 structure from the lattice vectors at
http://cst-www.nrl.navy.mil/lattice/struk/c4.html
'''
a = 4.59  # experimental degrees of freedom.
c = 2.96
u = 0.3  # internal degree of freedom!

#primitive vectors
a1 = a * np.array([1.0, 0.0, 0.0])
a2 = a * np.array([0.0, 1.0, 0.0])
a3 = c * np.array([0.0, 0.0, 1.0])

atoms = Atoms([Atom('Ti', [0., 0., 0.]),
Atom('Ti', 0.5 * a1 + 0.5 * a2 + 0.5 * a3),
Atom('O', u * a1 + u * a2),
Atom('O', -u * a1 - u * a2),
Atom('O', (0.5 + u) * a1 + (0.5 - u) * a2 + 0.5 * a3),
Atom('O', (0.5 - u) * a1 + (0.5 + u) * a2 + 0.5 * a3)],
cell=[a1, a2, a3])

v0 = atoms.get_volume()
cell0 = atoms.get_cell()

factors = [0.9, 0.95, 1.0, 1.05, 1.1] #to change volume by

energies, volumes = [], []

for f in factors:
v1 = f * v0
cell_factor = (v1 / v0)**(1. / 3.)

atoms.set_cell(cell0 * cell_factor, scale_atoms=True)

calc = Vasp('bulk/tio2/step1-{0:1.2f}'.format(f),
encut=520,
kpts=[5, 5, 5],
isif=2, # relax internal degrees of freedom
ibrion=1,
nsw=50,
xc='PBE',
sigma=0.05,
atoms=atoms)

energies.append(atoms.get_potential_energy())
volumes.append(atoms.get_volume())

calc.stop_if(None in energies)

plt.plot(volumes, energies)
plt.xlabel('Vol. ($\AA^3)$')
plt.ylabel('Total energy (eV)')
plt.savefig('images/tio2-step1.png')

print '#+tblname: tio2-vol-ene'
print '#+caption: Total energy of TiO_{2} vs. volume.'
print '| Volume ($\AA^3$) | Energy (eV) |'
print '|-'
for v, e in zip(volumes, energies):
print '| {0} | {1} |'.format(v, e)

Table 4: Total energy of TiO2 vs. volume.
Volume ($$\AA^3$$) Energy (eV)
56.1254185488 -51.81932158
59.2434971663 -52.46118036
62.361576 -52.76017908
65.4796549456 -52.80043775
68.5977335623 -52.64628895

Now, we know the minimum energy volume is near 64 Å^3. You could at this point fit an equation of state to find that minimum. However, we now want to use these initial starting points for a second round of optimization where we allow the unit cell shape to change, at constant volume: ISIF=4.

from vasp import Vasp

calc = Vasp('bulk/tio2/step1-0.90')
calc.clone('bulk/tio2/step2-0.90')
#calc.set(isif=4)
print calc.set(isif=4)
print calc.calculation_required()

clone:  Atoms(symbols='Ti2O4', positions=..., magmoms=..., cell=[4.41041021, 4.41041021, 2.88537073], pbc=[True, True, True], calculator=SinglePointCalculator(...))
{}
False

from vasp import Vasp

factors = [0.9, 0.95, 1.0, 1.05, 1.1]  # to change volume by

energies1, volumes1 = [], []  # from step 1
energies, volumes = [], []  # for step 2
for f in factors:
calc = Vasp('bulk/tio2/step1-{0:1.2f}'.format(f))
atoms = calc.get_atoms()
energies1.append(atoms.get_potential_energy())
volumes1.append(atoms.get_volume())

calc.clone('bulk/tio2/step2-{0:1.2f}'.format(f))
calc.set(isif=4)
# You have to get the atoms again.
atoms = calc.get_atoms()

energies.append(atoms.get_potential_energy())
volumes.append(atoms.get_volume())

print(energies, volumes)
calc.stop_if(None in energies)

import matplotlib.pyplot as plt
plt.plot(volumes1, energies1, volumes, energies)
plt.xlabel('Vol. ($\AA^3)$')
plt.ylabel('Total energy (eV)')
plt.legend(['step 1', 'step 2'], loc='best')
plt.savefig('images/tio2-step2.png')

([-51.82715553, -52.46235848, -52.76127768, -52.80903199, -52.67597935], [56.125418401558292, 59.243497055444799, 62.36157600817976, 65.47965475812434, 68.597733708605929])


The take away message here is that the total energy slightly decreases when we allow the unit cell shape to change, especially for the larger unit cell deformation. This has little effect on the minimum volume, but would have an effect on the bulk modulus, which is related to the curvature of the equation of state. At this point, you could fit an equation of state to the step 2 data, and estimate the volume at the minimum volume, and recalculate the total energy at that volume.

An alternative is a final calculation with ISIF=3, which optimizes the unit cell volume, shape and internal coordinates. It looks like the calculation at bulk/tio2/step2-1.05 is close to the minimum, so we will use that as a starting point for the final calculation.

from vasp import Vasp

calc = Vasp('bulk/tio2/step2-1.05')
calc.clone('bulk/tio2/step3')

calc = Vasp('bulk/tio2/step3',
isif=3)
calc.wait()
print calc

from pyspglib import spglib
print '\nThe spacegroup is {0}'.format(spglib.get_spacegroup(calc.atoms))

*************** VASP CALCULATION SUMMARY ***************
Vasp calculation directory:
---------------------------
[[/home-research/jkitchin/dft-book/bulk/tio2/step3]]

Unit cell:
----------
x       y       z             |v|
v0   4.661   0.000   0.000       4.661 Ang
v1   0.000   4.661   0.000       4.661 Ang
v2   0.000   0.000   2.970       2.970 Ang
alpha, beta, gamma (deg):  90.0  90.0  90.0
Total volume:                   64.535 Ang^3
Stress:    xx     yy     zz     yz     xz     xy
-0.002 -0.002 -0.000 -0.000 -0.000 -0.000 GPa

ID  tag     sym    x         y         z        rmsF (eV/A)
0   0       Ti     0.000     0.000     0.000       0.00
1   0       Ti     2.331     2.331     1.485       0.00
2   0       O      1.420     1.420     0.000       0.00
3   0       O      3.241     3.241     0.000       0.00
4   0       O      3.751     0.910     1.485       0.00
5   0       O      0.910     3.751     1.485       0.00
Potential energy: -52.8176 eV

INPUT Parameters:
-----------------
pp        : PBE
isif      : 3
xc        : pbe
kpts      : [5, 5, 5]
encut     : 520
lcharg    : False
ibrion    : 1
ismear    : 1
lwave     : False
sigma     : 0.05
nsw       : 50

Pseudopotentials used:
----------------------
Ti: potpaw_PBE/Ti/POTCAR (git-hash: 39cac2d7c620efc80c69344da61b5c43bc16e9b8)
O: potpaw_PBE/O/POTCAR (git-hash: 592f34096943a6f30db8749d13efca516d75ec55)

The spacegroup is P4_2/mnm (136)


This is the final result. You can see that the forces on all the atoms are less than 0.01 eV/Å, and the stress is also very small. The final volume is close to where we expect it to be based on steps 1 and 2. The space group is still correct. The lattice vectors are close in length to the experimentally known values, and the angles between the vectors has not changed much. Looks good!

As a final note, the VASP manual recommends you do not use the final energy directly from the calculation, but rather run a final calculation with ISMEAR set to -5. Here we examine the effect.

from vasp import Vasp

calc = Vasp('bulk/tio2/step3')
atoms = calc.get_atoms()
print 'default ismear: ', atoms.get_potential_energy()

calc.clone('bulk/tio2/step4')
calc.set(ismear=-5,
nsw=0)
atoms = calc.get_atoms()
print 'ismear=-5:      ', atoms.get_potential_energy()

default ismear:  -52.81760338
ismear=-5:       -52.8004532


The difference here is on the order of a meV. That does not seem significant here. I suspect the recommended practice stems from early days when much smaller ENCUT values were used and changes in the number of basis functions were more significant.

#### 4.3.4 Effect of XC on bulk properties

The exchange correlation functional can significantly affect computed bulk properties. Here, we examine the effect on the bulk lattice constant of Pd (exp. 3.881). An excellent review of this can be found in mattsson-084714. We examine several functionals. The xc keyword in Vasp is used to select the POTCARs. Let us consider the LDA functional first.

from vasp import Vasp
from ase import Atom, Atoms
from ase.utils.eos import EquationOfState
import numpy as np

LC = [3.75, 3.80, 3.85, 3.90, 3.95, 4.0, 4.05, 4.1]

volumes, energies = [], []
for a in LC:
atoms = Atoms([Atom('Pd', (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]]))
calc = Vasp('bulk/Pd-LDA-{0}'.format(a),
encut=350,
kpts=[12, 12, 12],
xc='LDA',
atoms=atoms)

e = atoms.get_potential_energy()
energies.append(e)
volumes.append(atoms.get_volume())

calc.stop_if(None in energies)

eos = EquationOfState(volumes, energies)
v0, e0, B = eos.fit()
print('LDA lattice constant is {0:1.3f} Ang^3'.format((4*v0)**(1./3.)))

LDA lattice constant is 3.841 Ang^3


For a GGA calculation, it is possible to specify which functional you want via the GGA tag. This tag was designed to use the LDA POTCAR files, but with a GGA functional. We will consider four different functionals here.

from vasp import Vasp
from ase import Atom, Atoms
from ase.utils.eos import EquationOfState
import numpy as np

LC = [3.75, 3.80, 3.85, 3.90, 3.95, 4.0, 4.05, 4.1]

GGA = {'AM': 'AM05',
'PE': 'PBE',
'PS': 'PBEsol',
'RP': 'RPBE'}

for key in GGA:
volumes, energies = [], []
for a in LC:
atoms = Atoms([Atom('Pd', (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]]))
calc = Vasp('bulk/Pd-GGA-{1}-{0}'.format(a, key),
encut=350,
kpts=[12, 12, 12],
xc='LDA',
gga=key,
atoms=atoms)

e = atoms.get_potential_energy()
energies.append(e)
volumes.append(atoms.get_volume())

if None not in energies:
eos = EquationOfState(volumes, energies)
v0, e0, B = eos.fit()
print '{1:6s} lattice constant is {0:1.3f} Ang^3'.format((4*v0)**(1./3.),
GGA[key])
else:
print energies, LC

PBEsol lattice constant is 3.841 Ang^3
AM05   lattice constant is 3.841 Ang^3
RPBE   lattice constant is 3.841 Ang^3
PBE    lattice constant is 3.939 Ang^3


These results compare very favorably to those in mattsson-084714. It is typical that LDA functionals underestimate the lattice constants, and that GGAs tend to overestimate the lattice constants. PBEsol and AM05 were designed specifically for solids, and for Pd, these functionals do an exceptional job of reproducing the lattice constants. RPBE is particularly bad at the lattice constant, but it has been reported to be a superior functional for reactivity hammer1999:improv-pbe.

### 4.4TODO Using built-in ase optimization with vasp

ASE has some nice optimization tools built into it. We can use them in vasp too. This example is adapted from this test: https://trac.fysik.dtu.dk/projects/ase/browser/trunk/ase/test/vasp/vasp_Al_volrelax.py

First the VASP way.

from vasp import Vasp
from ase.lattice import bulk

Al = bulk('Al', 'fcc', a=4.5, cubic=True)
calc = Vasp('bulk/Al-lda-vasp',
xc='LDA', isif=7, nsw=5,
ibrion=1, ediffg=-1e-3,
lwave=False, lcharg=False,
atoms=Al)
print(calc.potential_energy)
print(calc)

-10.07430725

Vasp calculation in /home-research/jkitchin/dft-book/bulk/Al-lda-vasp

INCAR created by Atomic Simulation Environment
ISIF = 7
LCHARG = .FALSE.
IBRION = 1
EDIFFG = -0.001
ISMEAR = 1
LWAVE = .TRUE.
SIGMA = 0.1
NSW = 5

Al
1.0000000000000000
4.5000000000000000    0.0000000000000000    0.0000000000000000
0.0000000000000000    4.5000000000000000    0.0000000000000000
0.0000000000000000    0.0000000000000000    4.5000000000000000
4
Cartesian
0.0000000000000000  0.0000000000000000  0.0000000000000000
0.0000000000000000  2.2500000000000000  2.2500000000000000
2.2500000000000000  0.0000000000000000  2.2500000000000000
2.2500000000000000  2.2500000000000000  0.0000000000000000

#+BEGIN_SRC python
from vasp import Vasp
calc = Vasp('bulk/Al-lda-vasp')
calc.view()
print [atoms.get_volume() for atoms in calc.traj]
print [atoms.get_potential_energy() for atoms in calc.traj]

[91.124999999999986, 78.034123525818302, 72.328582812881763, 73.422437849114189, 73.368474506164134]
[-9.58448747, -10.02992063, -10.07180132, -10.07429962, -10.07430725]


Now, the ASE way. TODO

from vasp import Vasp
from ase.lattice import bulk
from ase.optimize import BFGS as QuasiNewton

Al = bulk('Al', 'fcc', a=4.5, cubic=True)

calc = Vasp('bulk/Al-lda-ase',
xc='LDA',
atoms=Al)

from ase.constraints import StrainFilter
sf = StrainFilter(Al)
qn = QuasiNewton(sf, logfile='relaxation.log')
qn.run(fmax=0.1, steps=5)
print('Stress:\n', calc.stress)
print('Al post ASE volume relaxation\n', calc.get_atoms().get_cell())
print(calc)


Now for a detailed comparison:

from vasp import Vasp

atoms = Vasp('bulk/Al-lda-vasp').get_atoms()

atoms2 = Vasp('bulk/Al-lda-ase').get_atoms()

import numpy as np

cellA = atoms.get_cell()
cellB = atoms2.get_cell()

print((np.abs(cellA - cellB) < 0.01).all())

False


This could be handy if you want to use any of the optimizers in ase.optimize in conjunction with ase.constraints, which are more advanced than what is in VASP.

### 4.5 Cohesive energy

cohesive energy The cohesive energy is defined as the energy to separate neutral atoms in their ground electronic state from the solid at 0K at 1 atm. We will compute this for rhodium. Rh is normally an fcc metal, so we will use that structure and let VASP find the equilibrium volume for us.

from vasp import Vasp
from ase.lattice.cubic import FaceCenteredCubic
from ase import Atoms, Atom

# bulk system
atoms = FaceCenteredCubic(directions=[[0, 1, 1],
[1, 0, 1],
[1, 1, 0]],
size=(1, 1, 1),
symbol='Rh')

calc = Vasp('bulk/bulk-rh',
xc='PBE',
encut=350,
kpts=[4, 4, 4],
isif=3,
ibrion=2,
nsw=10,
atoms=atoms)
bulk_energy = atoms.get_potential_energy()

# atomic system
atoms = Atoms([Atom('Rh',[5, 5, 5])],
cell=(7, 8, 9))

calc = Vasp('bulk/atomic-rh',
xc='PBE',
encut=350,
kpts=[1, 1, 1],
atoms=atoms)
atomic_energy = atoms.get_potential_energy()

calc.stop_if(None in (bulk_energy, atomic_energy))

cohesive_energy = atomic_energy - bulk_energy
print 'The cohesive energy is {0:1.3f} eV'.format(cohesive_energy)

The cohesive energy is 6.184 eV


According to Kittel kittel, the cohesive energy of Rh is 5.75 eV. There are a few reasons we may have discrepancy here:

1. The k-point grid used in the bulk state is not very dense. However, you can see below that the total energy is pretty converged by a $$6 \times 6 \times 6$$ $k$-point grid.
2. We did not check for convergence with the planewave cutoff.
3. We neglected spin on the atomic state. Rh in the atomic state has this electronic structure: [Kr] 4d8 5s1 and is a doublet.

First we consider the k-point convergence. convergence!KPOINTS

from vasp import Vasp

calc = Vasp('bulk/atomic-rh')
atomic_energy = calc.potential_energy

calc = Vasp('bulk/bulk-rh')
atoms = calc.get_atoms()

kpts = [3, 4, 6, 9, 12, 15, 18]

calcs = [Vasp('bulk/bulk-rh-kpts-{0}'.format(k),
xc='PBE',
encut=350,
kpts=[k, k, k],
atoms=atoms)
for k in kpts]

energies = [calc.potential_energy for calc in calcs]

calcs[0].stop_if(None in energies)

for k, e in zip(kpts, energies):
print('({0:2d}, {0:2d}, {0:2d}):'
' cohesive energy = {1} eV'.format(k,
e - atomic_energy))

( 3,  3,  3): cohesive energy = -4.76129426 eV
( 4,  4,  4): cohesive energy = -6.17915613 eV
( 6,  6,  6): cohesive energy = -6.20654198 eV
( 9,  9,  9): cohesive energy = -6.20118094 eV
(12, 12, 12): cohesive energy = -6.20897225 eV
(15, 15, 15): cohesive energy = -6.2091123 eV
(18, 18, 18): cohesive energy = -6.21007962 eV


Using only 1 k-point for the bulk energy is a terrible approximation! It takes at least a 6 × 6 × 6 grid to get the total energy converged to less than 10 meV. Note we do not need to check the k-point convergence of the atomic state because it is surrounded by vacuum on all sides, and so there should not be any dispersion in the bands.

We will examine the magnetic state next.

from vasp import Vasp
from ase.lattice.cubic import FaceCenteredCubic
from ase import Atoms, Atom
# bulk system
atoms = FaceCenteredCubic(directions=[[0, 1, 1],
[1, 0, 1],
[1, 1, 0]],
size=(1, 1, 1),
symbol='Rh')

calc = Vasp('bulk/bulk-rh',
xc='PBE',
encut=350,
kpts=[4, 4, 4],
isif=3,
ibrion=2,
nsw=10,
atoms=atoms)
bulk_energy = atoms.get_potential_energy()

# atomic system
atoms = Atoms([Atom('Rh',[5, 5, 5], magmom=1)],
cell=(7, 8, 9))

calc = Vasp('bulk/atomic-rh-sp',
xc='PBE',
encut=350,
kpts=[1, 1, 1],
ispin=2,
atoms=atoms)
atomic_energy = atoms.get_potential_energy()

calc.stop_if(None in [atomic_energy, bulk_energy])

cohesive_energy = atomic_energy - bulk_energy
print 'The cohesive energy is {0:1.3f} eV'.format(cohesive_energy)

The cohesive energy is 6.127 eV


Again, the value in Kittel kittel is 5.75 eV which is very close to this value. Finally, it is also possible there is a lower energy non-spherical atom energy; we did not check that at all (see Estimating triplet oxygen dissociation energy with low symmetry).

### 4.6 Elastic properties

See this reference PhysRevB.65.104104.

We seek the elastic constant tensor that relates stress (σ) and strain (ε) via $$\sigma = c \epsilon$$. The stress and strain are six component vectors, so $$c$$ will be a 6 × 6 symmetric matrix.

#### 4.6.1 Fe elastic properties

As with molecular vibrations, we need a groundstate geometry. Let us get one for BCC Fe.

from vasp import Vasp
from ase.lattice.cubic import BodyCenteredCubic

atoms = BodyCenteredCubic(symbol='Fe')
for atom in atoms:
atom.magmom = 3.0

from vasp.vasprc import VASPRC
VASPRC['mode'] = None

import logging
log = logging.getLogger('Vasp')
#log.setLevel(logging.DEBUG)

calc = Vasp('bulk/Fe-bulk',
xc='PBE',
kpts=[6, 6, 6],
encut=350,
ispin=2,
isif=3,
nsw=30,
ibrion=1,
atoms=atoms)
print(atoms.get_potential_energy())
print(atoms.get_stress())

-15.53472773
[ 0.00031141  0.00031141  0.00031141 -0.         -0.         -0.        ]


Ok, now with a relaxed geometry at hand, we proceed with the elastic constants. This is accomplished with IBRION=6 and ISIF > 3 in VASP. See this reference (from the VASP page) Y. Le Page and P. Saxe, Phys. Rev. B 65, 104104 (2002) doi:10.1103/PhysRevB.65.104104 for more details.

from vasp import Vasp

calc = Vasp('bulk/Fe-bulk')
calc.clone('bulk/Fe-elastic')

calc.set(ibrion=6,    #
isif=3,      # gets elastic constants
potim=0.05,  # displacements
nsw=1,
nfree=2)

print(calc.potential_energy)

-15.52764065


Now, the results are written out to the OUTCAR file. Actually, three sets of moduli are written out 1) the elastic tensor for rigid ions, 2) the contribution from allowing the atoms to relax, and 3) the total elastic modulus, all in kBar.

  SYMMETRIZED ELASTIC MODULI (kBar)
Direction    XX          YY          ZZ          XY          YZ          ZX
--------------------------------------------------------------------------------
XX        2803.5081   1622.6085   1622.6085      0.0000      0.0000      0.0000
YY        1622.6085   2803.5081   1622.6085      0.0000      0.0000      0.0000
ZZ        1622.6085   1622.6085   2803.5081      0.0000      0.0000      0.0000
XY           0.0000      0.0000      0.0000    866.8792      0.0000      0.0000
YZ           0.0000      0.0000      0.0000      0.0000    866.8792      0.0000
ZX           0.0000      0.0000      0.0000      0.0000      0.0000    866.8792
--------------------------------------------------------------------------------


and

 ELASTIC MODULI CONTR FROM IONIC RELAXATION (kBar)
Direction    XX          YY          ZZ          XY          YZ          ZX
--------------------------------------------------------------------------------
XX           0.0000      0.0000      0.0000      0.0000      0.0000      0.0000
YY           0.0000      0.0000      0.0000      0.0000      0.0000      0.0000
ZZ           0.0000      0.0000      0.0000      0.0000      0.0000      0.0000
XY           0.0000      0.0000      0.0000      0.0000      0.0000      0.0000
YZ           0.0000      0.0000      0.0000      0.0000      0.0000      0.0000
ZX           0.0000      0.0000      0.0000      0.0000      0.0000      0.0000
--------------------------------------------------------------------------------

TOTAL ELASTIC MODULI (kBar)
Direction    XX          YY          ZZ          XY          YZ          ZX
--------------------------------------------------------------------------------
XX        2803.5081   1622.6085   1622.6085      0.0000      0.0000      0.0000
YY        1622.6085   2803.5081   1622.6085      0.0000      0.0000      0.0000
ZZ        1622.6085   1622.6085   2803.5081      0.0000      0.0000      0.0000
XY           0.0000      0.0000      0.0000    866.8792      0.0000      0.0000
YZ           0.0000      0.0000      0.0000      0.0000    866.8792      0.0000
ZX           0.0000      0.0000      0.0000      0.0000      0.0000    866.8792
--------------------------------------------------------------------------------


Let us write a small code here to extract the Total elastic moduli from the OUTCAR file. First we get the line where the total elastic moduli start, then take the six lines that start three lines after that. Finally we parse out the matrix elements and cast them as floats.

import numpy as np
EM = []

with open('bulk/Fe-elastic/OUTCAR') as f:
for i, line in enumerate(lines):
if line.startswith(' TOTAL ELASTIC MODULI (kBar)'):
j = i + 3
data = lines[j:j+6]
break

for line in data:
EM += [[float(x) for x in line.split()[1:]]]

print np.array(EM)

[[ 1125.1405  3546.8135  3546.8135     0.         0.         0.    ]
[ 3546.8135  1125.1405  3546.8135     0.         0.         0.    ]
[ 3546.8135  3546.8135  1125.1405     0.         0.         0.    ]
[    0.         0.         0.      1740.2372     0.         0.    ]
[    0.         0.         0.         0.      1740.2372     0.    ]
[    0.         0.         0.         0.         0.      1740.2372]]


Fe is in a BCC crystal structure, which is high in symmetry. Consequently, many of the elements in the matrix are equal to zero.

See http://www.nist.gov/data/PDFfiles/jpcrd34.pdf for a lot of detail about Fe-Ni alloys and general theory about elastic constants. In the next section, we show how the code above is integrated into Vasp.

#### 4.6.2 Al elastic properties

First, the relaxed geometry.

from vasp import Vasp
from ase.lattice.cubic import FaceCenteredCubic

atoms = FaceCenteredCubic(symbol='Al')

calc = Vasp('bulk/Al-bulk',
xc='PBE',
kpts=[12, 12, 12],
encut=350,
prec='High',
isif=3,
nsw=30,
ibrion=1,
atoms=atoms)
print(calc.potential_energy)

-14.97511793


Ok, now with a relaxed geometry at hand, we proceed with the elastic constants. This is accomplished with IBRION=6 and ISIF ≥ 3 in VASP.

from vasp import Vasp

calc = Vasp('bulk/Al-bulk')
calc.clone('bulk/Al-elastic')

calc.set(ibrion=6,    #
isif=3,      # gets elastic constants
potim=0.015,  # displacements
nsw=1,
nfree=2)

calc.wait(abort=True)

EM = calc.get_elastic_moduli()

print(EM)

c11 = EM[0, 0]
c12 = EM[0, 1]
B = (c11 + 2 * c12) / 3.0
print(B)

[[ 110.17099   59.54652   59.54652    0.         0.         0.     ]
[  59.54652  110.17099   59.54652    0.         0.         0.     ]
[  59.54652   59.54652  110.17099    0.         0.         0.     ]
[   0.         0.         0.        11.52331    0.         0.     ]
[   0.         0.         0.         0.        11.52331    0.     ]
[   0.         0.         0.         0.         0.        11.52331]]
76.4213433333


This example shows the basic mechanics of getting the elastic constants. The C44 constant above is too low, and probably we need to check these constants for convergence with respect to kpoints, planewave cutoff, and maybe the value of POTIM.

#### 4.6.3 Manual calculation of elastic constants

It is possible to manually calculate single elastic constants; you just need to know what strain corresponds to the elastic constant.

For the C11 elastic constant in a cubic system, we simply strain the cell along the x-axis, and then evaluate the second derivative at the minimum to calculate C11 like this.

$$C_{11} = \frac{1}{V_C}\frac{\partial^2 E^{tot}}{\partial \gamma^2}$$

from vasp import Vasp
from ase.lattice.cubic import FaceCenteredCubic
import numpy as np
import matplotlib.pyplot as plt

DELTAS = np.linspace(-0.05, 0.05, 5)
calcs = []
volumes = []

for delta in DELTAS:
atoms = FaceCenteredCubic(symbol='Al')

cell = atoms.cell

T = np.array([[1 + delta, 0, 0],
[0,1, 0],
[0, 0, 1]])
newcell = np.dot(cell, T)
atoms.set_cell(newcell, scale_atoms=True)
volumes += [atoms.get_volume()]

calcs += [Vasp('bulk/Al-c11-{}'.format(delta),
xc='pbe',
kpts=[12, 12, 12],
encut=350,
atoms=atoms)]

Vasp.run()
energies =  [calc.potential_energy for calc in calcs]

# fit a parabola
eos = np.polyfit(DELTAS, energies, 2)

# first derivative
d_eos = np.polyder(eos)

print(np.roots(d_eos))

xfit = np.linspace(min(DELTAS), max(DELTAS))
yfit = np.polyval(eos, xfit)

plt.plot(DELTAS, energies, 'bo', xfit, yfit, 'b-')
plt.xlabel('$\delta$')
plt.ylabel('Energy (eV)')
plt.savefig('images/Al-c11.png')

[ 0.00727102]


### 4.7 Bulk thermodynamics

We can predict temperature dependent thermodynamic properties of bulk materials without too much effort. As with the thermochemical properties of ideal gases, we must use some simple models that we parameterize by DFT. Here we follow the example in Reference Shang20101040 for computing the thermal coefficient of expansion, heat capacity, enthalpy and entropy for Ni as a function of temperature.

We start by computing the equation of state for fcc Ni.

from vasp import Vasp
from ase import Atom, Atoms
import numpy as np
# fcc
LC = [3.5, 3.55, 3.6, 3.65, 3.7, 3.75]
volumes, energies = [], []
for a in LC:
atoms = Atoms([Atom('Ni', (0, 0, 0), magmom=2.5)],
cell=0.5 * a * np.array([[1.0, 1.0, 0.0],
[0.0, 1.0, 1.0],
[1.0, 0.0, 1.0]]))

calc = Vasp('bulk/Ni-{0}'.format(a),
xc='PBE',
encut=350,
kpts=[12, 12, 12],
ispin=2,
atoms=atoms)
energies.append(calc.potential_energy)
volumes.append(atoms.get_volume())

calc.stop_if(None in energies)

import matplotlib.pyplot as plt
plt.plot(LC, energies)
plt.xlabel('Lattice constant ($\AA$)')
plt.ylabel('Total energy (eV)')
plt.savefig('images/Ni-fcc.png')


### 4.8 Effect of pressure on phase stability

So far we have only considered relative stability at a pressure of 0 Pa. We now consider the relative stability of two phases under pressure. We will consider TiO$$_2$$ in the rutile and anatase phases.

The pressure is defined by: $$P = -\left(\frac{\partial E}{\partial V}\right)_T$$. So if we have an equation of state $$E(V)$$ we can calculate the pressure at any volume, or alternatively, given a pressure, compute the volume. Pressure can affect the energy of two phases differently, so that one may become stable under pressure. The condition where a phase transition occurs is when the pressure in the two phases is the same, which occurs at a common tangent.

To show this, we need $$E_{rutile}(V)$$ and $$E_{anatase}(V)$$.

# run the rutile calculations
from vasp import Vasp
from ase import Atom, Atoms
import numpy as np

B = 'Ti'; X = 'O'; a = 4.59; c = 2.958; u = 0.305;
'''
create a rutile structure from the lattice vectors at
http://cst-www.nrl.navy.mil/lattice/struk/c4.html

spacegroup: 136 P4_2/mnm
'''
a1 = a * np.array([1.0, 0.0, 0.0])
a2 = a * np.array([0.0, 1.0, 0.0])
a3 = c * np.array([0.0, 0.0, 1.0])

atoms = Atoms([Atom(B, [0., 0., 0.]),
Atom(B, 0.5*a1 + 0.5*a2 + 0.5*a3),
Atom(X,  u*a1 + u*a2),
Atom(X, -u*a1 - u*a2),
Atom(X, (0.5+u)*a1 + (0.5-u)*a2 + 0.5*a3),
Atom(X, (0.5-u)*a1 + (0.5+u)*a2 + 0.5*a3)],
cell=[a1, a2, a3])

nTiO2 = len(atoms) / 3.
v0 = atoms.get_volume()
cell0 = atoms.get_cell()

volumes = [28., 30., 32., 34., 36.]  #vol of one TiO2

for v in volumes:
atoms.set_cell(cell0 * ((nTiO2 * v / v0)**(1. / 3.)),
scale_atoms=True)

calc = Vasp('bulk/TiO2/rutile/rutile-{0}'.format(v),
encut=350,
kpts=[6, 6, 6],
xc='PBE',
ismear=0,
sigma=0.001,
isif=2,
ibrion=2,
nsw=20,
atoms=atoms)
calc.update()

# run the anatase calculations
import numpy as np
from vasp import Vasp
from ase import Atom, Atoms
# http://cst-www.nrl.navy.mil/lattice/struk/c5.html

B = 'Ti'; X = 'O'; a = 3.7842; c = 2*4.7573; z = 0.0831;

a1 = a * np.array([1.0, 0.0, 0.0])
a2 = a * np.array([0.0, 1.0, 0.0])
a3 = np.array([0.5 * a, 0.5 * a, 0.5 * c])

atoms = Atoms([Atom(B, -0.125 * a1 + 0.625 * a2 + 0.25 * a3),
Atom(B,  0.125 * a1 + 0.375 * a2 + 0.75 * a3),
Atom(X, -z*a1 + (0.25-z)*a2 + 2.*z*a3),
Atom(X, -(0.25+z)*a1 + (0.5-z)*a2 + (0.5+2*z)*a3),
Atom(X, z*a1 - (0.25 - z)*a2 + (1-2*z)*a3),
Atom(X, (0.25 + z)*a1 + (0.5 + z)*a2 + (0.5-2*z)*a3)],
cell=[a1,a2,a3])

nTiO2 = len(atoms) / 3.
v0 = atoms.get_volume()
cell0 = atoms.get_cell()

volumes = [30., 33., 35., 37., 39.]  #vol of one TiO2

for v in volumes:
atoms.set_cell(cell0 * ((nTiO2*v/v0)**(1./3.)),
scale_atoms=True)

calc = Vasp('bulk/TiO2/anatase/anatase-{0}'.format(v),
encut=350,
kpts=[6, 6, 6],
xc='PBE',
ismear=0,
sigma=0.001,
isif=2,
ibrion=2,
nsw=20,
atoms=atoms)
calc.update()


Now we will fit cubic polynomials to the data.

# fit cubic polynomials to E(V) for rutile and anatase
from vasp import Vasp
import matplotlib.pyplot as plt
import numpy as np
np.set_printoptions(precision=2)

# anatase equation of state
volumes = [30., 33., 35., 37., 39.]  # vol of one TiO2 formula unit
a_volumes, a_energies = [], []
for v in volumes:
calc = Vasp('bulk/TiO2/anatase/anatase-{0}'.format(v))
atoms = calc.get_atoms()
nTiO2 = len(atoms) / 3.0
a_volumes.append(atoms.get_volume() / nTiO2)
a_energies.append(atoms.get_potential_energy() / nTiO2)

# rutile equation of state
volumes = [28., 30., 32., 34., 36.]  # vol of one TiO2
r_volumes, r_energies = [], []
for v in volumes:
calc = Vasp('bulk/TiO2/rutile/rutile-{0}'.format(v))
atoms = calc.get_atoms()
nTiO2 = len(atoms) / 3.0
r_volumes.append(atoms.get_volume() / nTiO2)
r_energies.append(atoms.get_potential_energy() / nTiO2)

# cubic polynomial fit to equation of state E(V) = pars*[V^3 V^2 V^1 V^0]
apars = np.polyfit(a_volumes, a_energies, 3)
rpars = np.polyfit(r_volumes, r_energies, 3)

print 'E_anatase(V) = {0:1.2f}*V^3 + {1:1.2f}*V^2 + {2:1.2f}*V + {3:1.2f}'.format(*apars)
print 'E_rutile(V) =  {0:1.2f}*V^3 + {1:1.2f}*V^2 + {2:1.2f}*V + {3:1.2f}'.format(*rpars)
print 'anatase epars: {0!r}'.format(apars)
print 'rutile epars: {0!r}'.format(rpars)
# get pressure parameters P(V) = -dE/dV
dapars = -np.polyder(apars)
drpars = -np.polyder(rpars)

print 'anatase ppars: {0!r}'.format(dapars)
print 'rutile ppars: {0!r}'.format(drpars)

print
print 'P_anatase(V) = {0:1.2f}*V^2 + {1:1.2f}*V + {2:1.2f}'.format(*dapars)
print 'P_rutile(V) =  {0:1.2f}*V^2 + {1:1.2f}*V + {2:1.2f}'.format(*drpars)

vfit = np.linspace(28, 40)

# plot the equations of state
plt.plot(a_volumes, a_energies,'bo ', label='Anatase')
plt.plot(vfit, np.polyval(apars, vfit), 'b-')

plt.plot(r_volumes, r_energies,'gs ', label='Rutile')
plt.plot(vfit, np.polyval(rpars, vfit), 'g-')

plt.xlabel('Volume ($\AA^3$/f.u.)')
plt.ylabel('Total energy (eV/f.u.)')
plt.legend()
plt.xlim([25, 40])
plt.ylim([-27, -26])
plt.savefig('imag


To find the conditions where a phase transition occurs, we have to find the common tangent line between the rutile and anatase phases. In other words we have to solve these two equations:

$$(E_{anatase}(V1) - E_{rutile}(V2))/(V1-V2) = P_{anatase}(V1)$$

$$(E_{anatase}(V1) - E_{rutile}(V2))/(V1-V2) = P_{rutile}(V2)$$

This is a nonlinear algebra problem. We use the scipy.optimize.fsolve to solve this problem.

from ase.units import GPa
from numpy import array, linspace, polyval

# copied from polynomial fit above
anatase_epars = array([-1.06049246e-03,   1.30279404e-01,  -5.23520055e+00,
4.25202869e+01])
rutile_epars = array([-1.24680208e-03,   1.42966536e-01,  -5.33239733e+00,
3.85903670e+01])

# polynomial fits for pressures
anatase_ppars = array([3.18147737e-03,  -2.60558808e-01,   5.23520055e+00])
rutile_ppars = array([3.74040625e-03,  -2.85933071e-01,   5.33239733e+00])

def func(V):
V1 = V[0] # rutile volume
V2 = V[1] # anatase volume

E_rutile = polyval(rutile_epars, V1)
E_anatase = polyval(anatase_epars, V2)

P_rutile =  polyval(rutile_ppars, V1)
P_anatase = polyval(anatase_ppars, V2)

return [(E_anatase - E_rutile) / (V1 - V2) - P_anatase,
(E_anatase - E_rutile) / (V1 - V2) - P_rutile]

from scipy.optimize import fsolve
x0 = fsolve(func, [28, 34])
print 'The solutions are at V = {0}'.format(x0)
print 'Anatase pressure: {0} GPa'.format(polyval(anatase_ppars, x0[1]) / GPa)
print 'Rutile  pressure: {0} GPa'.format(polyval(rutile_ppars, x0[0]) / GPa)

# illustrate the common tangent
import matplotlib.pyplot as plt

vfit = linspace(28, 40)
plt.plot(vfit, polyval(anatase_epars,vfit), label='anatase')
plt.plot(vfit, polyval(rutile_epars,vfit), label='rutile')
plt.plot(x0, [polyval(rutile_epars,x0[0]),
polyval(anatase_epars,x0[1])], 'ko-', label='common tangent')
plt.legend()
plt.xlabel('Volume ($\AA^3$/f.u.)')
plt.ylabel('Total energy (eV/f.u.)')
plt.savefig('images/eos-common-tangent.png')

The solutions are at V = [ 31.67490656  34.60893508]
Anatase pressure: 4.5249494236 GPa
Rutile  pressure: 4.52494942374 GPa


At a pressure of 4.5 GPa, we expect that anatase will start converting into rutile. Along this common tangent, a mixture of the two phases will be more stable than either pure phase.

there is some controversy about the most stable phase. add discussion here.

### 4.9 Bulk reaction energies

#### 4.9.1 Alloy formation energies

In this section we will consider how to calculate the formation energy of an fcc Cu-Pd alloy and how to use that information to discuss relative stabilities. The kinds of questions we can easily answer are:

1. Is the formation of an alloy at a particular composition and structure energetically favorable?
2. Given two alloy structures at the same composition, which one is more stable?
3. Given a set of alloy structures at different compositions, which ones are stable with respect to phase separation?

Each of these questions is answered by calculating the formation energy of the alloy from the parent metals. Thus, we will need the total energies of fcc Cu and fcc Pd. To get started. We get those first. Rather than compute a full equation of state for these, we will rely on the built in unit cell optimization algorithm in VASP (ISIF=3).

##### 4.9.1.1 Basic alloy formation energy
# get bulk Cu and Pd energies. <<pure-metal-components>>
from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('Cu',  [0.000,      0.000,      0.000])],
cell=  [[ 1.818,  0.000,  1.818],
[ 1.818,  1.818,  0.000],
[ 0.000,  1.818,  1.818]])

cuc = Vasp('bulk/alloy/cu',
xc='PBE',
encut=350,
kpts=[13, 13, 13],
nbands=9,
ibrion=2,
isif=3,
nsw=10,
atoms=atoms)

cu = cuc.potential_energy

atoms = Atoms([Atom('Pd',  [0.000,      0.000,      0.000])],
cell=[[ 1.978,  0.000,  1.978],
[ 1.978,  1.978,  0.000],
[0.000,  1.978,  1.978]])

pd = Vasp('bulk/alloy/pd',
xc='PBE',
encut=350,
kpts=[13, 13, 13],
nbands=9,
ibrion=2,
isif=3,
nsw=10,
atoms=atoms).potential_energy

print 'Cu energy = {0} eV'.format(cu)
print 'Pd energy = {0} eV'.format(pd)

Cu energy = -3.73437194 eV
Pd energy = -5.22003433 eV


Note that the Pd energy is more negative than the Cu energy. This does not mean anything significant. We cannot say Pd is more stable than Cu; it is not like Cu could transmutate into Pd!

Next, we will consider a question like which of two structures with composition of CuPd is more stable. These coordinates for these structures came from research of the author. The approach is pretty general, you must identify the coordinates and unit cell of the candidate structure, and then run a calculation to find the optimized geometry and unit cell. This may take some work, as previously described in the multistep process for optimizing a bulk system. Here the geometry is pretty close to optimized, so we can use the VASP optimization routines. We consider two structures with composition CuPd.

from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('Cu',  [0.000,      0.000,      0.000]),
Atom('Pd',  [-1.652,     0.000,      2.039])],
cell=  [[0.000, -2.039,  2.039],
[0.000,  2.039,  2.039],
[-3.303,  0.000,  0.000]])

calc = Vasp('bulk/alloy/cupd-1',
xc='PBE',
encut=350,
kpts=[12, 12, 8],
nbands=17,
ibrion=2,
isif=3,
nsw=10,
atoms=atoms)
cupd1 = atoms.get_potential_energy()

atoms = Atoms([Atom('Cu',  [-0.049,     0.049,      0.049]),
Atom('Cu',  [-11.170,   11.170,     11.170]),
Atom('Pd',  [-7.415,     7.415,      7.415]),
Atom('Pd',  [-3.804 ,    3.804,      3.804])],
cell=[[-5.629,  3.701,  5.629 ],
[-3.701,  5.629,  5.629 ],
[-5.629,  5.629,  3.701 ]])

calc = Vasp('bulk/alloy/cupd-2',
xc='PBE',
encut=350,
kpts=[8, 8, 8],
nbands=34,
ibrion=2,
isif=3,
nsw=10,
atoms=atoms)
cupd2 = atoms.get_potential_energy()

print 'cupd-1 = {0} eV'.format(cupd1)
print 'cupd-2 = {0} eV'.format(cupd2)

cupd-1 = -9.17593835 eV
cupd-2 = -18.07779325 eV


Looking at these energies, you could be tempted to say cupd-2 is more stable than cupd-1 because its energy is much lower. This is wrong, however, because cupd-2 has twice as many atoms as cupd-1. We should compare the normalized total energies, that is the energy normalized per CuPd formula unit, or as an alternative the number of atoms in the unit cell. It does not matter which, as long as we normalize consistently. It is conventional in alloy calculation to normalize by the number of atoms in the unit cell.

from vasp import Vasp

calc = Vasp('bulk/alloy/cupd-1')
atoms = calc.get_atoms()
e1 = atoms.get_potential_energy()/len(atoms)

calc = Vasp('bulk/alloy/cupd-2')
atoms = calc.get_atoms()
e2 = atoms.get_potential_energy()/len(atoms)

print 'cupd-1: {0} eV/atom'.format(e1)
print 'cupd-2: {0} eV/atom'.format(e2)

cupd-1: -4.587969175 eV/atom
cupd-2: -4.5194483125 eV/atom


After normalizing by number of atoms, we can see that cupd-1 is a more stable structure. However, we are looking at total energies, and we might ask: is cupd-1 more stable than an unreacted mixture of the parent compounds, fcc Cu and Pd? In other words, is the following reaction exothermic:

Cu + Pd $$\rightarrow$$ CuPd for the two configurations we examined? Below, we show some pretty general code that computes these formation energies, and normalizes them by the number of atoms in the unit cell.

from vasp import Vasp

# bulk energy 1
calc = Vasp('bulk/alloy/cu')
atoms = calc.get_atoms()
cu = atoms.get_potential_energy()/len(atoms)

# bulk energy 2
calc = Vasp('bulk/alloy/pd')
atoms = calc.get_atoms()
pd = atoms.get_potential_energy()/len(atoms)

calc = Vasp('bulk/alloy/cupd-1')
atoms = calc.get_atoms()
e1 = atoms.get_potential_energy()
# subtract bulk energies off of each atom in cell
for atom in atoms:
if atom.symbol == 'Cu':
e1 -= cu
else:
e1 -= pd

e1 /= len(atoms)  # normalize by number of atoms in cell

calc = Vasp('bulk/alloy/cupd-2')
atoms = calc.get_atoms()
e2 = atoms.get_potential_energy()
for atom in atoms:
if atom.symbol == 'Cu':
e2 -= cu
else:
e2 -= pd
e2 /= len(atoms)

print 'Delta Hf cupd-1 = {0:1.2f} eV/atom'.format(e1)
print 'Delta Hf cupd-2 = {0:1.2f} eV/atom'.format(e2)

Delta Hf cupd-1 = -0.11 eV/atom
Delta Hf cupd-2 = -0.04 eV/atom


The answer is yes. Both structures are energetically more favorable than an equal composition mixture of the parent metals. The heat of formation for both structures is exothermic, but the cupd-1 structure is more stable than the cupd-2 structure. This is shown conceptually in Figure fig:alloy1.

#### 4.9.2 Metal oxide oxidation energies

We will consider here the reaction 2 Cu2O + O2 $$\rightleftharpoons$$ 4 CuO. The reaction energy is:

$$\Delta E = 4E_{CuO} - 2E_{Cu_2O} - E_{O_2}$$. We need to compute the energy of each species.

##### 4.9.2.1 Cu2O calculation
# run Cu2O calculation
from vasp import Vasp
from ase import Atom, Atoms

# http://phycomp.technion.ac.il/~ira/types.html#Cu2O
a = 4.27

atoms = Atoms([Atom('Cu', [0, 0, 0]),
Atom('Cu', [0.5, 0.5, 0.0]),
Atom('Cu', [0.5, 0.0, 0.5]),
Atom('Cu', [0.0, 0.5, 0.5]),
Atom('O', [0.25, 0.25, 0.25]),
Atom('O', [0.75, 0.75, 0.75])])

atoms.set_cell((a, a, a), scale_atoms=True)

calc = Vasp('bulk/Cu2O',
encut=400,
kpts=[8, 8, 8],
ibrion=2,
isif=3,
nsw=30,
xc='PBE',
atoms=atoms)

print atoms.get_potential_energy()
print atoms.get_stress()

-27.27469148
[-0.01018402 -0.01018402 -0.01018402 -0.         -0.         -0.        ]

##### 4.9.2.2 CuO calculation
# run CuO calculation
from vasp import Vasp
from ase import Atom, Atoms
import numpy as np

# CuO
# http://cst-www.nrl.navy.mil/lattice/struk/b26.html
# http://www.springermaterials.com/docs/info/10681727_51.html
a = 4.6837
b = 3.4226
c = 5.1288
beta = 99.54/180*np.pi
y = 0.5819

a1 = np.array([0.5*a, -0.5*b, 0.0])
a2 = np.array([0.5*a, 0.5*b, 0.0])
a3 = np.array([c*np.cos(beta), 0.0, c*np.sin(beta)])

atoms = Atoms([Atom('Cu', 0.5*a2),
Atom('Cu', 0.5*a1 + 0.5*a3),
Atom('O', -y*a1 + y*a2 + 0.25*a3),
Atom('O',  y*a1 - y*a2 - 0.25*a3)],
cell=(a1, a2, a3))

calc = Vasp('bulk/CuO',
encut=400,
kpts=[8, 8, 8],
ibrion=2,
isif=3,
nsw=30,
xc='PBE',
atoms=atoms)
print(atoms.get_potential_energy())

-19.61568557

##### 4.9.2.3TODO Reaction energy calculation
from vasp import Vasp

# don't forget to normalize your total energy to a formula unit. Cu2O
# has 3 atoms, so the number of formula units in an atoms is
# len(atoms)/3.
calc = Vasp('bulk/Cu2O')
atoms1 = calc.get_atoms()
cu2o_energy = atoms1.get_potential_energy()

calc = Vasp('bulk/CuO')
atoms2 = calc.get_atoms()
cuo_energy = atoms2.get_potential_energy()

# make sure to use the same cutoff energy for the O2 molecule!
calc = Vasp('molecules/O2-sp-triplet-400')
atoms3 = calc.get_atoms()
o2_energy = atoms3.get_potential_energy()

calc.stop_if(None in [cu2o_energy, cuo_energy, o2_energy])

cu2o_energy /= (len(atoms1) / 3)  # note integer math
cuo_energy /= (len(atoms2) / 2)
rxn_energy = 4.0 * cuo_energy - o2_energy - 2.0 * cu2o_energy
print 'Reaction energy = {0} eV'.format(rxn_energy)

Reaction energy = -2.11600154 eV


This is the reaction energy for 2 Cu2O $$\rightarrow$$ 4 CuO. In PhysRevB.73.195107, the experimental reaction is estimated to be about -3.14 eV.

There are a few reasons why our number does not agree with the experimental reaction energy. One reason is related to errors in the O2 dissociation energy, and another reason is related to localization of electrons in the Cu 3$$d$$ orbitals PhysRevB.73.195107. The first error of incorrect O2 dissociation error is a systematic error that can be corrected empirically PhysRevB.73.195107. Fixing the second error requires the application of DFT+U (see DFT+U).

The heat of reaction is reported to be 1000 J/g product at http://onlinelibrary.wiley.com/doi/10.1002/er.4440130107/pdf for the reaction 2CuO $$\rightleftharpoons$$ Cu2O + 1/2 O2.

from ase import Atoms
atoms = Atoms('Cu2O')
MW = atoms.get_masses().sum()

H = 1.  # kJ/g
print 'rxn energy = {0:1.1f} eV'.format(-2 * H * MW / 96.4)  # convert to eV

rxn energy = -3.0 eV


This is pretty close to the value in PhysRevB.73.195107 and might need a temperature correction to get agreement at 298K.

### 4.10 Bulk density of states

The density of states refers to the number of electronic states in a particular energy range.

The solution to Eq. \eqref{eq:KS} yields a set of Kohn-Sham (K-S) orbitals and an associated set of eigenvalues that correspond to the energies of these orbitals, neither of which have any known directly observable meaning RevModPhys.71.1253. The sum of the squared K-S orbitals, however, is equal to the electron density (Eq. \eqref{eq:density}), and the sum of the eigenvalues is a significant part of the total energy (Eq. \eqref{eq:dftEnergy}). Thus, it seems reasonable to suppose these quantities have other significant relationships to physical observables. Perdew et al. showed that the highest occupied eigenvalue is equal to the ionization energy of a system within an exact density functional theory perdew1982:elect-kohn-sham, but their interpretation has been vigorously debated in the literature kleinman1997:signif-kohn-sham,perdew1997,kleinman1997:reply-commen-kohn-sham, and is only true for the exact exchange/correlation functional, not the approximate ones used in practice koch2001. Stowasser and Hoffmann discussed an approach to using the K-S orbitals in more traditional molecular orbital interpretations, but the results were primarily qualitative stowasser1999:what-kohn-sham. More recently, a DFT analog of Koopmans' theorem has been developed that formally identifies the eigenvalues with vertical ionization potentials, which can be measured with photoelectron spectroscopy gritsenko2002.

Despite the arguments against ascribing physical meaning to the K-S orbitals and eigenvalues, it has become fairly standard, especially for solids, to use them to calculate the density of states (DOS) jones1989 [Sec. VI. B]. This has been found to yield reasonable results for the valence bands in metals, but poor results for tightly bound orbitals and band gaps perdew1982:elect-kohn-sham. A highly technical discussion of this issue can be found in Ref. perdew1981:self. The density of states can be calculated by a sum over the k-points seitsonen2000:phd:

$$\label{eq:dos} \rho(\epsilon)=\sum_\mathbf{\mathrm{k}} \omega_\mathbf{\mathrm{k}} \sum_i \beta(\epsilon -\epsilon_{i\mathbf{\mathrm{k}}})$$

where $$\omega_\mathbf{\mathrm{k}}$$ is the weight associated with the k-point, and $$\beta$$ is a broadening function, typically a gaussian function, to account for the finite number of k-points used in the calculations. The amount of broadening is arbitrary, and should tend to zero as the number of k-points approaches infinity.

from vasp import Vasp

npoints = 200
width = 0.5

def gaussian(energies, eik):
x = ((energies - eik) / width)
return np.exp(-x**2) / np.sqrt(np.pi) / width

calc = Vasp('bulk/pd-dos')

# kpt weights
wk = calc.get_k_point_weights()

# for each k-point there are a series of eigenvalues
# here we get all the eigenvalues for each k-point
e_kn = []
for i, k in enumerate(wk):
e_kn.append(calc.get_eigenvalues(kpt=i))

e_kn = np.array(e_kn) - calc.get_fermi_level()

# these are the energies we want to evaluate the dos at
energies = np.linspace(e_kn.min(), e_kn.max(), npoints)

# this is where we build up the dos
dos = np.zeros(npoints)

for j in range(npoints):
for k in range(len(wk)): # loop over all kpoints
for i in range(len(e_kn[k])): # loop over eigenvalues in each k
dos[j] += wk[k] * gaussian(energies[j], e_kn[k][i])

import matplotlib.pyplot as plt
plt.plot(energies, dos)
plt.savefig('images/manual-dos.png')
plt.show()


Here is a more convenient way to compute the DOS using ase.dft.

from vasp import Vasp
import matplotlib.pyplot as plt
from ase.dft import DOS

calc = Vasp('bulk/pd-dos')
dos = DOS(calc, width=0.2)
d = dos.get_dos()
e = dos.get_energies()

import pylab as plt
plt.plot(e, d)
plt.xlabel('energy (eV)')
plt.ylabel('DOS')
plt.savefig('images/pd-dos.png')


This DOS looks roughly like you would expect. The peak between -5 to 0 eV is the Pd d-band.

The VASP manual recommends a final run be made with ISMEAR=-5, which uses the tetrahedron method with Bl\"ochl corrections.

from vasp import Vasp
from ase.dft import DOS
calc = Vasp('bulk/pd-dos')
calc.clone('bulk/pd-dos-ismear-5')

bulk = calc.get_atoms()
calc.set(ismear=-5)

bulk.get_potential_energy()
dos = DOS(calc, width=0.2)
d = dos.get_dos()
e = dos.get_energies()

import pylab as plt
plt.plot(e, d)
plt.xlabel('energy [eV]')
plt.ylabel('DOS')
plt.savefig('images/pd-dos-ismear-5.png')


This not notably different to me.

We can test for convergence of the DOS. The k-points are most important.

from ase import Atoms, Atom
from vasp import Vasp
Vasp.vasprc(mode=None)
#Vasp.log.setLevel(10)

import matplotlib.pyplot as plt
import numpy as np
from ase.dft import DOS
import pylab as plt

a = 3.9  # approximate lattice constant
b = a / 2.
bulk = Atoms([Atom('Pd', (0.0, 0.0, 0.0))],
cell=[(0, b, b),
(b, 0, b),
(b, b, 0)])

kpts = [8, 10, 12, 14, 16, 18, 20]

calcs = [Vasp('bulk/pd-dos-k{0}-ismear-5'.format(k),
encut=300,
xc='PBE',
kpts=[k, k, k],
atoms=bulk) for k in kpts]

Vasp.wait(abort=True)

for calc in calcs:
# this runs the calculation
if calc.potential_energy is not None:
dos = DOS(calc, width=0.2)
d = dos.get_dos() + k / 4.0
e = dos.get_energies()

plt.plot(e, d, label='k={0}'.format(k))
else:
pass
plt.xlabel('energy (eV)')
plt.ylabel('DOS')
plt.legend()
plt.savefig('images/pd-dos-k-convergence-ismear-5.png')
plt.show()

got here
<ase.dft.dos.DOS instance at 0xec2d710>

from vasp import Vasp
from ase.dft import DOS
# This seems very slow...
calc = Vasp('bulk/pd-dos-k20-ismear-5')
print DOS(calc, width=0.2)

<ase.dft.dos.DOS instance at 0x168a1ea8>


### 4.11 Atom projected density of states

One major disadvantage of a planewave basis set is that it is difficult to relate the completely delocalized planewaves to localized phenomena such as bonding. Much insight into bonding has been gained by atomic/molecular orbital theory, which has carried over to the solid-state arena RevModPhys.60.601. Consequently, several schemes have been developed to project the one-electron Kohn-Sham wave functions onto atomic wave functions sanchez-portal1995:projec,segall1996:popul,segall1996:popul-mp. In VASP, the one electron wave functions can be projected onto spherical harmonic orbitals. The radial component of the atomic orbitals extends to infinity. In a solid, this means that the projection on one atom may overlap with the projection on a neighboring atom, resulting in double counting of electrons. Consequently, a cutoff radius was introduced, beyond which no contributions are included. It is not obvious what the best cutoff radius is. If the radius is too small, it might not capture all of the electrons associated with the atom. However, if it is too large, it may include electrons from neighboring atoms. One might want to use different cutoff radii for different atoms, which have different sizes. Furthermore, the ideal cutoff radius for an atom may change in different environments, thus it would require an iterative procedure to determine it. This difficulty arises because the orbital-band occupations are not observable, thus how the electrons are divided up between atoms is arbitrary and, as will be seen later, is sensitive to the cutoff radius (and in other DFT implementations, the basis set). However, Mulliken orbital populations have been used successfully for many years to examine the qualitative differences between similar systems, and that is precisely what these quantities are used for here. Thus, a discussion of the analysis and results is warranted.

The s and p states in a metal are typically delocalized in space and more like free-electrons, whereas the d-orbitals are fairly localized in space and have been treated successfully with tight-binding theories such as extended H\"u ckel theory RevModPhys.60.601, and linear muffin tin orbital theory ruban1997:surfac. Consequently, the remaining discussion will be focused on the properties of the projected d-states.

In this example, we consider how to get the atom-projected density of states (ADOS). We are interested in properties of the $d$-band on Pd, such as the $d$-band center and $d$-band width. You must set the RWIGS tag to get ADOS, and these are the Wigner-Seitz radii for each atom. By integrating the projected d-band up to the Fermi level, the d-band filling can be determined. It is not obvious what the electron count in the d-band should be for an atom in a metal. For a gas-phase, neutral metal atom in the ground state, however, the d-orbital electron count is well defined, so it will be used as an initial reference point for comparison kittel.

A powerful method for characterizing distributions is to examine various moments of the distribution (see Chapter 4 in Ref. cottrell1988 and Chapter 6 in Refs. ducastelle1991 and pettifor1992:elect-theor-alloy-desig). The $$n^{th}$$ order moment, $$\mu_n$$, of a distribution of states $$\rho(\epsilon)$$ with respect to a reference $$\epsilon_o$$ is defined by

$$\mu_n = \frac{\int_{-\infty}^\infty \epsilon^n \rho(\epsilon-\epsilon_o)d\epsilon} {\int_{-\infty}^\infty \rho(\epsilon-\epsilon_o)d\epsilon}$$

In this work, the reference energy is always the Fermi level. The zeroth moment is just the total number of states, in this case it will be normalized to unity. The first moment is the average energy of distribution, analogous to the center of mass for a mass density distribution. The second moment is the mean squared width of the distribution. The third moment is a measure of skewness and the fourth moment is related to kurtosis, but these moments are rarely used, and only the first and second moments are considered in this work.

It is important to note that these projected density of states are not physical observables. They are the wavefunctions projected onto atomic orbitals. For some situations this makes sense, e.g. the $$d$$ orbitals are fairly localized and reasonably approximated by atomic orbitals. The $$s$$ valence orbitals in a metal, in contrast, are almost totally delocalized. Depending on the cutoff radius (RWIGS) you choose, you can see very different ADOS.

from ase import Atoms, Atom
from vasp import Vasp

import matplotlib.pyplot as plt
import numpy as np

a = 3.9  # approximate lattice constant
b = a / 2.
bulk = Atoms([Atom('Pd', (0.0, 0.0, 0.0))],
cell=[(0, b, b),
(b, 0, b),
(b, b, 0)])

encut=300,
xc='PBE',
lreal=False,
kpts=[8, 8, 8],
atoms=bulk)

# this runs the calculation
calc.wait(abort=True)

# now get results

# we will select energies in the range of -10, 5
ind = (energies < 5) & (energies > -10)

energies = energies[ind]

Nstates = np.trapz(dos, energies)
occupied = energies <= 0.0
N_occupied_states = np.trapz(dos[occupied], energies[occupied])
# first moment
ed = np.trapz(energies * dos, energies) / Nstates

# second moment
wd2 = np.trapz(energies**2 * dos, energies) / Nstates

print 'Total # states = {0:1.2f}'.format(Nstates)
print 'number of occupied states = {0:1.2f}'.format(N_occupied_states)
print 'd-band center = {0:1.2f} eV'.format(ed)
print 'd-band width  = {0:1.2f} eV'.format(np.sqrt(wd2))

# plot the d-band
plt.plot(energies, dos, label='$d$-orbitals')

# plot the occupied states in shaded gray
plt.fill_between(x=energies[occupied],
y1=dos[occupied],
y2=np.zeros(dos[occupied].shape),
color='gray', alpha=0.25)

plt.xlabel('$E - E_f$ (eV)')
plt.ylabel('DOS (arbitrary units)')


Total # states = 9.29
number of occupied states = 7.95
d-band center = -1.98 eV
d-band width  = 2.71 eV


#### 4.11.1 Effect of RWIGS on ADOS

Here we examine the effect of changing RWIGS on the number of counted electrons, and properties of the d-band moments.

from ase import Atoms, Atom
from vasp import Vasp

import matplotlib.pyplot as plt
import numpy as np

a = 3.9  # approximate lattice constant
b = a / 2.
bulk = Atoms([Atom('Pd', (0.0, 0.0, 0.0))],
cell=[(0, b, b),
(b, 0, b),
(b, b, 0)])

RWIGS = [1.0, 1.1, 1.25, 1.5, 2.0, 2.5, 3.0, 4.0, 5.0 ]

ED, WD, N = [], [], []

for rwigs in RWIGS:
calc.set(rwigs={'Pd': rwigs})
if calc.potential_energy is None:
continue

# now get results

#we will select energies in the range of -10, 5
ind = (energies < 5) & (energies > -10)

energies = energies[ind]
dos = dos[ind]

Nstates = np.trapz(dos, energies)
occupied = energies <= 0.0
N_occupied_states = np.trapz(dos[occupied], energies[occupied])
ed = np.trapz(energies * dos, energies) / np.trapz(dos, energies)
wd2 = np.trapz(energies**2 * dos, energies) / np.trapz(dos, energies)

N.append(N_occupied_states)
ED.append(ed)
WD.append(wd2**0.5)

plt.plot(RWIGS, N, 'bo', label='N. occupied states')
plt.legend(loc='best')
plt.xlabel('RWIGS ($\AA$)')
plt.ylabel('# occupied states')

fig, ax1 = plt.subplots()
ax1.plot(RWIGS, ED, 'bo', label='d-band center (eV)')
ax1.set_xlabel('RWIGS ($\AA$)')
ax1.set_ylabel('d-band center (eV)', color='b')
for tl in ax1.get_yticklabels():
tl.set_color('b')

ax2 = ax1.twinx()
ax2.plot(RWIGS, WD, 'gs', label='d-band width (eV)')
ax2.set_ylabel('d-band width (eV)', color='g')
for tl in ax2.get_yticklabels():
tl.set_color('g')

plt.show()


You can see the number of occupied states increases approximately linearly here with RWIGS. This is due to overcounting of neighboring electrons.

q

The d-band center and width also change.

### 4.12 Band structures

band structure To compute a band structure we do two things. First, we compute the self-consistent band structure. Then we compute the band structure at the desired $k$-points. We will use Si as an example (adapted from http://bbs.sciencenet.cn/bbs/upload/20083418325986.pdf).

First, we get the self-consistent electron density in a calculation.

from vasp import Vasp
from ase import Atom, Atoms
from ase.visualize import view

a = 5.38936
atoms = Atoms([Atom('Si', [0, 0, 0]),
Atom('Si', [0.25, 0.25, 0.25])])

atoms.set_cell([[a / 2., a / 2., 0.0],
[0.0,  a / 2., a / 2.],
[a / 2., 0.0, a / 2.]], scale_atoms=True)

calc = Vasp('bulk/Si-selfconsistent',
xc='PBE',
prec='Medium',
lcharg=True,
lwave=True,
kpts=[4, 4, 4],
atoms=atoms)
calc.run()


Now, we run a new calculation along the k-point path desired. The standard VASP way of doing this is to modify the INCAR and KPOINTS file and rerun VASP. We will not do that. Doing that results in some lost information if you overwrite the old files. We will copy the old directory to a new directory, using code to ensure this only happens one time.

from vasp import Vasp

wd = 'bulk/Si-bandstructure'

calc = Vasp('bulk/Si-selfconsistent')
calc.clone(wd)

kpts = [[0.5, 0.5, 0.0],   # L
[0, 0, 0],         # Gamma
[0, 0, 0],
[0.5, 0.5, 0.5]]  # X

calc.set(kpts=kpts,
reciprocal=True,
kpts_nintersections=10,
icharg=11)

print calc.run()

-3.62224484


We will learn how to manually parse the EIGENVAL file here to generate the band structure. The structure of the EIGENVAL file looks like this:

head -n 20 bulk/Si-bandstructure/EIGENVAL

    2    2    1    1
0.1956688E+02  0.3810853E-09  0.3810853E-09  0.3810853E-09  0.5000000E-15
1.000000000000000E-004
CAR
unknown system
8     20      8

0.5000000E+00  0.5000000E+00  0.0000000E+00  0.5000000E-01
1       -1.826747
2       -1.826743
3        3.153321
4        3.153347
5        6.743989
6        6.744017
7       16.392596
8       16.393943

0.4444444E+00  0.4444444E+00  0.0000000E+00  0.5000000E-01
1       -2.669487
2       -0.918463


We can ignore the first five lines.

f = open('bulk/Si-bandstructure/EIGENVAL', 'r')

unknown, nkpoints, nbands = [int(x) for x in f.readline().split()]

band_energies = [[] for i in range(nbands)]

for i in range(nkpoints):
x, y, z, weight = [float(x) for x in f.readline().split()]

for j in range(nbands):
id, energy = int(fields[0]), float(fields[1])
band_energies[id - 1].append(energy)
f.close()

import matplotlib.pyplot as plt

for i in range(nbands):
plt.plot(range(nkpoints), band_energies[i])

ax = plt.gca()
ax.set_xticks([]) # no tick marks
plt.xlabel('k-vector')
plt.ylabel('Energy (eV)')
ax.set_xticks([0, 10, 19])
ax.set_xticklabels(['$L$', '$\Gamma$', '$X$'])
plt.savefig('images/Si-bandstructure.png')


Next we will examine the connection between band structures and density of states. In this example, we will compute the band structure of TiO2 using a function built into vasp to do the analysis described above.

from vasp import Vasp

calc = Vasp('bulk/tio2/step3')
print calc.get_fermi_level()
calc.abort()
n, bands, p = calc.get_bandstructure(kpts_path=[('$\Gamma$', [0.0, 0.0, 0.0]),
('X', [0.5, 0.5, 0.0]),
('X', [0.5, 0.5, 0.0]),
('M', [0.0, 0.5, 0.5]),
('M', [0.0, 0.5, 0.5]),
('$\Gamma$', [0.0, 0.0, 0.0])])

p.savefig('images/tio2-bandstructure-dos.png')


#### 4.12.1 create example showing band dispersion with change in lattice constant

In this section, we examine the effect of the lattice constant on the band structure. Since the lattice constant affects the overlap of neighboring atoms, we expect that smaller lattice constants will show more dispersion, i.e. broader bands. Larger lattice constants, in contrast, should show narrower bands. We examine this in silicon.

from vasp import Vasp
from ase import Atom, Atoms

calcs = []
for i, a in enumerate([4.7, 5.38936, 6.0]):

atoms = Atoms([Atom('Si', [0, 0, 0]),
Atom('Si', [0.25, 0.25, 0.25])])

atoms.set_cell([[a/2., a/2., 0.0],
[0.0,  a/2., a/2.],
[a/2., 0.0, a/2.]], scale_atoms=True)

calc = Vasp('bulk/Si-bs-{0}'.format(i),
xc='PBE',
lcharg=True,
lwave=True,
kpts=[4, 4, 4],
atoms=atoms)

print(calc.run())
calcs += [calc]

Vasp.wait(abort=True)

for i, calc in enumerate(calcs):
n, bands, p  = calc.get_bandstructure(kpts_path=[('L', [0.5,0.5,0.0]),
('$\Gamma$', [0, 0, 0]),
('$\Gamma$', [0, 0, 0]),
('X', [0.5, 0.5, 0.5])],
kpts_nintersections=10)

if p is not None:
png = 'images/Si-bs-{0}.png'.format(i)
p.savefig(png)

-7.55662509
-10.80024435
-10.13735105


You can see the band structure for a=6.0 is notably sharper than the band structure for a=4.0.

### 4.13 Magnetism

#### 4.13.1 Determining if a magnetic solution is energetically favorable

We can force a total magnetic moment onto a unit cell and compute the total energy as function of the total magnetic moment. If there is a minimum in the energy, then we know there is a lower energy magnetic solution than a non-magnetic solution. We use NUPDOWN to enforce the magnetic moment in the cell. Note that NUPDOWN can only be an integer. You cannot set it to be an arbitrary float.

from vasp import Vasp
from ase.lattice.cubic import BodyCenteredCubic

atoms = BodyCenteredCubic(directions=[[1, 0, 0],
[0, 1, 0],
[0, 0, 1]],
size=(1, 1, 1),
symbol='Fe')

calc = Vasp('bulk/Fe-bcc-fixedmagmom-{0:1.2f}'.format(0.0),
xc='PBE',
encut=300,
kpts=[4, 4, 4],
ispin=2,
nupdown=0,
atoms=atoms)

print(atoms.get_potential_energy())

-15.34226703

from vasp import Vasp
from ase.lattice.cubic import BodyCenteredCubic

atoms = BodyCenteredCubic(directions=[[1, 0, 0],
[0, 1, 0],
[0, 0, 1]],
size=(1, 1, 1),
symbol='Fe')

NUPDOWNS = [0.0, 2.0, 4.0, 5.0, 6.0, 8.0]
energies = []
for B in NUPDOWNS:
calc = Vasp('bulk/Fe-bcc-fixedmagmom-{0:1.2f}'.format(B),
xc='PBE',
encut=300,
kpts=[4, 4, 4],
ispin=2,
nupdown=B,
atoms=atoms)
energies.append(atoms.get_potential_energy())

if None in energies:
calc.abort()

import matplotlib.pyplot as plt
plt.plot(NUPDOWNS, energies)
plt.xlabel('Total Magnetic Moment')
plt.ylabel('Energy (eV)')
plt.savefig('images/Fe-fixedmagmom.png')


You can see here there is a minimum in energy at a total magnetic moment somewhere between 4 and 5. There are two Fe atoms in the unit cell, which means the magnetic moment on each atom must be about 2.5 Bohr-magnetons. This is a good guess for a real calculation. Note that VASP recommends you overestimate the magnetic moment guesses if you are looking for ferromagnetic solutions.

To run a spin-polarized calculation with initial guesses on each atom, we must set the magnetic moment on the atoms. Here we set it through the magmom attribute on the atom. In the example after this, we set it in the Atoms object.

from vasp import Vasp
from ase.lattice.cubic import BodyCenteredCubic

atoms = BodyCenteredCubic(directions=[[1, 0, 0],
[0, 1, 0],
[0, 0, 1]],
size=(1, 1, 1),
symbol='Fe')

# set magnetic moments on each atom
for atom in atoms:
atom.magmom = 2.5

calc = Vasp('bulk/Fe-bcc-sp-1',
xc='PBE',
encut=300,
kpts=[4, 4, 4],
ispin=2,
lorbit=11, # you need this for individual magnetic moments
atoms=atoms)

e = atoms.get_potential_energy()
B = atoms.get_magnetic_moment()
magmoms = atoms.get_magnetic_moments()

print 'Total magnetic moment is {0:1.2f} Bohr-magnetons'.format(B)
print 'Individual moments are {0} Bohr-magnetons'.format(magmoms)

Total magnetic moment is -0.01 Bohr-magnetons
Individual moments are [-0.013 -0.013] Bohr-magnetons


#### 4.13.2 Antiferromagnetic spin states

In an antiferromagnetic material, there are equal numbers of spin up and down electrons that align in a regular pattern, but pointing in opposite directions so that there is no net magnetism. It is possible to model this by setting the magnetic moments on each ase.Atom object. lreal

from vasp import Vasp
from ase import Atom, Atoms

atoms = Atoms([Atom('Fe', [0.00,  0.00,  0.00], magmom=5),
Atom('Fe', [4.3,   4.3,   4.3],  magmom=-5),
Atom('O', [2.15,  2.15,  2.15], magmom=0),
Atom('O', [6.45,  6.45,  6.45], magmom=0)],
cell=[[4.3,    2.15,    2.15],
[2.15,    4.3,     2.15],
[2.15,    2.15,    4.3]])

ca = Vasp('bulk/afm-feo',
encut=350,
prec='Normal',
ispin=2,
nupdown=0, # this forces a non-magnetic solution
lorbit=11, # to get individual moments
lreal=False,
atoms=atoms)
print 'Magnetic moments = ', atoms.get_magnetic_moments()
print 'Total magnetic moment = ', atoms.get_magnetic_moment()

Magnetic moments =  [-0.061 -0.061  0.063  0.063]
Total magnetic moment =  -5e-06


You can see that even though the total magnetic moment is 0, there is a spin on both Fe atoms, and they are pointing in opposite directions. Hence, the sum of spins is zero, and this arrangement is called anti-ferromagnetic.

phonopy

### 4.15TODO solid state NEB

Caspersen10052005 Carter paper sheppard:074103 recent Henkelman paper

## 5 Surfaces

### 5.1 Surface structures

As with molecules and bulk systems ase provides several convenience functions for making surfaces.

#### 5.1.1 Simple surfaces

ase provides many utility functions to setup surfaces. Here is a simple example of an fcc111 Al surface. There are built in functions for fcc111, bcc110, bcc111, hcp001 and diamond111.

from ase.lattice.surface import fcc111
from ase.io import write
from ase.visualize import view

slab = fcc111('Al', size=(2, 2, 3), vacuum=10.0)
from ase.constraints import FixAtoms
constraint = FixAtoms(mask=[atom.tag >= 2 for atom in slab])
slab.set_constraint(constraint)

view(slab)
write('images/Al-slab.png', slab, rotation='90x', show_unit_cell=2)


#### 5.1.2 Vicinal surfaces

The vast majority of surface calculations are performed on flat surfaces. This is partially because these surfaces tend to have the lowest surface energies, and thus are likely to be experimentally observed. The flat surfaces, also known as low Miller index surfaces, also have small unit cells, which tends to make them computationally affordable. There are, however, many reasons to model the properties of surfaces that are not flat. You may be interested in the reactivity of a step edge, for example, or you may use the lower cooridnation of steps as a proxy for nanoparticle reactivity. Many stepped surfaces are not that difficult to make now. The main idea in generating them is described here. ase provides a general function for making vicinal surfaces. Here is an example of a (211) surface.

from ase.lattice.surface import surface
from ase.io import write

# Au(211) with 9 layers
s1 = surface('Au', (2, 1, 1), 9)
s1.center(vacuum=10, axis=2)

write('images/Au-211.png',
s1.repeat((3, 3, 1)),
rotation='-30z,90x',  # change the orientation for viewing
show_unit_cell=2)


### 5.2TODO Surface calculation parameters

There is one important parameter that is different for surfaces than for bulk calculations, the k-point grid. Assuming you have followed the convention that the z-axis is normal to the surface, the k-point grids for slab calculations always have the form of $$M \times N \times 1$$. To illustrate why, consider this example:

from ase.lattice.surface import fcc111
from vasp import Vasp

slab = fcc111('Al', size=(1, 1, 4), vacuum=10.0)

calc = Vasp('surfaces/Al-bandstructure',
xc='PBE',
encut=300,
kpts=[6, 6, 6],
lcharg=True,  # you need the charge density
lwave=True,   # and wavecar for the restart
atoms=slab)

n, bands, p = calc.get_bandstructure(kpts_path=[(r'$\Gamma$', [0, 0, 0]),
('$K1$', [0.5, 0.0, 0.0]),
('$K1$', [0.5, 0.0, 0.0]),
('$K2$', [0.5, 0.5, 0.0]),
('$K2$', [0.5, 0.5, 0.0]),
(r'$\Gamma$', [0, 0, 0]),
(r'$\Gamma$', [0, 0, 0]),
('$K3$', [0.0, 0.0, 1.0])],
kpts_nintersections=10)

if p is None: calc.abort()

p.savefig('images/Al-slab-bandstructure.png')


### 5.3 Surface relaxation

When a surface is created, the bulk symmetry is broken and consequently there will be forces on the surface atoms. We will examine some consequences of this with a simple Al slab. First, we show there are forces on the slab atoms.

from vasp import Vasp
from ase.lattice.surface import fcc111

atoms = fcc111('Al', size=(1, 1, 4), vacuum=10.0)

calc = Vasp('surfaces/Al-slab-unrelaxed',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
atoms=atoms)

print(atoms.get_forces())

[[ 0.          0.         -0.01505445]
[ 0.          0.          0.18818605]
[ 0.          0.         -0.18818605]
[ 0.          0.          0.01505445]]


Some points to note. The forces on the atoms have symmetry to them. That is because the slab is centered. Had the slab had an odd number of atoms, it is likely the center atom would have no forces on it. Next we consider the spacing between each layer in the slab. We do this for comparison later.

from vasp import Vasp

calc = Vasp('surfaces/Al-slab-unrelaxed')
atoms = calc.get_atoms()
print 'Total energy: {0:1.3f} eV'.format(atoms.get_potential_energy())

for i in range(1, len(atoms)):
print '{0}  deltaz = {1:1.3f} angstroms'.format(i, atoms[i].z - atoms[i-1].z)

Total energy: -14.179 eV
1  deltaz = 2.338 angstroms
2  deltaz = 2.338 angstroms
3  deltaz = 2.338 angstroms


To reduce the forces, we can let VASP relax the geometry. We have to make some decisions about how to relax the slab. One choice would be to relax all the atoms in the slab. If we do that, then there will be no atoms with bulk like spacing unless we increase the slab thickness pretty dramatically. It is pretty common to freeze some atoms at the bulk coordinates, and let the others relax. We will freeze the bottom two layers (defined by tags 3 and 4) and let the first two layers relax. To do that we add constraints to the slab.

Note: the ase constraints are only partially used by Vasp. The ase.constraints.FixAtoms constraint gets written to the POSCAR file, and is then used internally in VASP. The only other constraint that VASP can use internally is ase.constraints.FixScaled. The other constraints are not written to the POSCAR and are not used by VASP.

from ase.lattice.surface import fcc111
atoms = fcc111('Al', size=(2, 2, 4), vacuum=10.0)
print([atom.z for atom in atoms])
print [atom.z <= 13 for atom in atoms]

[9.9999999999999982, 9.9999999999999982, 9.9999999999999982, 9.9999999999999982, 12.338268590217982, 12.338268590217982, 12.338268590217982, 12.338268590217982, 14.676537180435968, 14.676537180435968, 14.676537180435968, 14.676537180435968, 17.01480577065395, 17.01480577065395, 17.01480577065395, 17.01480577065395]
[True, True, True, True, True, True, True, True, False, False, False, False, False, False, False, False]

from vasp import Vasp
from ase.lattice.surface import fcc111
from ase.constraints import FixAtoms

atoms = fcc111('Al', size=(1, 1, 4), vacuum=10.0)

constraint = FixAtoms(mask=[atom.tag >= 3 for atom in atoms])
atoms.set_constraint(constraint)

calc = Vasp('surfaces/Al-slab-relaxed',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
ibrion=2,
isif=2,
nsw=10,
atoms=atoms)

print(calc.potential_energy)
print(calc)

-14.17963819

Vasp calculation directory:
---------------------------
[[/home-research/jkitchin/dft-book/surfaces/Al-slab-relaxed]]

Unit cell:
----------
x       y       z             |v|
v0   2.864   0.000   0.000       2.864 Ang
v1   1.432   2.480   0.000       2.864 Ang
v2   0.000   0.000  27.015      27.015 Ang
alpha, beta, gamma (deg):  90.0  90.0  60.0
Total volume:                  191.872 Ang^3
Stress:    xx     yy     zz     yz     xz     xy
0.006  0.006  0.002 -0.000 -0.000 -0.000 GPa

ID  tag     sym    x         y         z        rmsF (eV/A)
0   4       Al     0.000*    0.000*   10.000*      0.00
1   3       Al     1.432*    0.827*   12.338*      0.00
2   2       Al     2.864     1.653    14.677       0.19
3   1       Al     0.000     0.000    17.015       0.01
Potential energy: -14.1796 eV

INPUT Parameters:
-----------------
pp        : PBE
isif      : 2
xc        : pbe
kpts      : [6, 6, 1]
encut     : 350
lcharg    : False
ibrion    : 2
ismear    : 1
lwave     : True
sigma     : 0.1
nsw       : 10

Pseudopotentials used:
----------------------


You can see that atoms 2 and 3 (the ones we relaxed, because the have tags of 1 and 2, which are less than 3) now have very low forces on them and it appears that atoms 0 and 1 have no forces on them. That is because the FixAtoms constraint works by setting the forces on those atoms to zero. We can see in the next example that the z-positions of the relaxed atoms have indeed relaxed and changed, while the position of the frozen atoms did not change.

Note there are two versions of the forces. The true forces, and the forces when constraints are applied. ase.atoms.Atoms.get_forces

from vasp import Vasp

calc = Vasp('surfaces/Al-slab-relaxed')
atoms = calc.get_atoms()

print('Constraints = True: ', atoms.get_forces(apply_constraint=True))

print('Constraints = False: ', atoms.get_forces(apply_constraint=False))

('Constraints = True: ', array([[ 0.        ,  0.        ,  0.        ],
[ 0.        ,  0.        ,  0.        ],
[ 0.        ,  0.        , -0.00435222],
[ 0.        ,  0.        , -0.07264519]]))
('Constraints = False: ', array([[ 0.        ,  0.        ,  0.        ],
[ 0.        ,  0.        ,  0.        ],
[ 0.        ,  0.        , -0.00435222],
[ 0.        ,  0.        , -0.07264519]]))

Constraints = True:  [[ 0.     0.     0.   ]
[ 0.     0.     0.   ]
[ 0.     0.    -0.049]
[ 0.     0.    -0.019]]
Constraints = False:  [[ 0.     0.    -0.002]
[ 0.     0.     0.069]
[ 0.     0.    -0.049]
[ 0.     0.    -0.019]]

from vasp import Vasp
from ase.lattice.surface import fcc111

calc = Vasp('surfaces/Al-slab-relaxed')
atoms = calc.get_atoms()
print 'Total energy: {0:1.3f}'.format(atoms.get_potential_energy())

for i in range(1, len(atoms)):
print 'd_({0},{1}) = {2:1.3f} angstroms'.format(i, i-1,
atoms[i].z - atoms[i-1].z)

Total energy: -14.182
d_(1,0) = 2.338 angstroms
d_(2,1) = 2.309 angstroms
d_(3,2) = 2.370 angstroms


Depending on the layer there is either slight contraction or expansion. These quantities are small, and careful convergence studies should be performed. Note the total energy change from unrelaxed to relaxed is not that large in this case (e.g., it is about 5 meV). This is usually the case for metals, where the relaxation effects are relatively small. In oxides and semiconductors, the effects can be large, and when there are adsorbates, the effects can be large also.

### 5.4 Surface reconstruction

We previously considered how relaxation can lower the surface energy. For some surfaces, a more extreme effect can reduce the surface energy: reconstruction. In a simple surface relaxation, the basic structure of a surface is preserved. However, sometimes there is a different surface structure that may have a lower surface energy. Some famous reconstructions include: Si-$$\sqrt{7}\times\sqrt{7}$$, Pt(100) hex reconstruction PhysRevB.56.10518,PhysRevB.82.161418, and the Au(111) herringbone reconstruction.

We will consider the (110) missing row reconstruction PhysRevB.83.075415. For some metals, especially Pt and Au, it is energetically favorable to form the so-called missing row reconstruction where every other row in the surface is "missing". It is favorable because it lowers the surface energy. Let us consider how we might calculate and predict that. It is straightforward to compute the energy of a (110) slab, and of a (110) slab with one row missing. However, these slabs contain different numbers of atoms, so we cannot directly compare the total energies to determine which energy is lower.

We have to consider where the missing row atoms have gone, so we can account for their energy. We will consider that they have gone into the bulk, and so we to consider the energy associated with the following transformation:

slab110 $$\rightarrow$$ slabmissing row + bulk

Thus, if this change in energy: $$E_{bulk} + E_{slab_{missing row}} - E_{slab_{110}}$$ is negative, then the formation of the missing row is expected to be favorable.

#### 5.4.1 Au(110) missing row reconstruction

We first consider the Au(110) case, where the reconstruction is known to be favorable.

##### 5.4.1.1 Clean Au(110) slab
from ase.lattice.surface import fcc110
from ase.io import write
from ase.constraints import FixAtoms
from ase.visualize import view

atoms = fcc110('Au', size=(2, 1, 6), vacuum=10.0)
constraint = FixAtoms(mask=[atom.tag > 2 for atom in atoms])
atoms.set_constraint(constraint)
view(atoms)

from vasp import Vasp
from ase.lattice.surface import fcc110
from ase.io import write
from ase.constraints import FixAtoms

atoms = fcc110('Au', size=(2, 1, 6), vacuum=10.0)
constraint = FixAtoms(mask=[atom.tag > 2 for atom in atoms])
atoms.set_constraint(constraint)

write('images/Au-110.png', atoms.repeat((2, 2, 1)), rotation='-90x', show_unit_cell=2)

print Vasp('surfaces/Au-110',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
ibrion=2,
isif=2,
nsw=10,
atoms=atoms).potential_energy

-35.92440066

##### 5.4.1.2 Missing row in Au(110)
from vasp import Vasp
from ase.lattice.surface import fcc110
from ase.io import write
from ase.constraints import FixAtoms

atoms = fcc110('Au', size=(2, 1, 6), vacuum=10.0)
del atoms[11]  # delete surface row

constraint = FixAtoms(mask=[atom.tag > 2 for atom in atoms])
atoms.set_constraint(constraint)

write('images/Au-110-missing-row.png',
atoms.repeat((2, 2, 1)),
rotation='-90x',
show_unit_cell=2)

calc = Vasp('surfaces/Au-110-missing-row',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
ibrion=2,
isif=2,
nsw=10,
atoms=atoms)

calc.update()

##### 5.4.1.3 Bulk Au
from vasp import Vasp
from ase.visualize import view
from ase.lattice.cubic import FaceCenteredCubic

atoms = FaceCenteredCubic(directions=[[0, 1, 1],
[1, 0, 1],
[1, 1, 0]],
size=(1, 1, 1),
symbol='Au')

print Vasp('bulk/Au-fcc',
xc='PBE',
encut=350,
kpts=[12, 12, 12],
atoms=atoms).potential_energy

-3.19446244

##### 5.4.1.4 Analysis of energies
from vasp import Vasp

print 'dE = {0:1.3f} eV'.format(Vasp('surfaces/Au-110-missing-row').potential_energy
+ Vasp('bulk/Au-fcc').potential_energy
- Vasp('surfaces/Au-110').potential_energy)

natoms slab        = 12
natoms missing row = 11
natoms bulk        = 1
dE = -0.070 eV


The missing row formation energy is slightly negative. The magnitude of the formation energy is pretty small, but just slightly bigger than the typical convergence errors observed, so we should cautiously conclude that the reconstruction if favorable for Au(110). We made a lot of shortcuts in computing this quantity, including using the experimental lattice constant of Au, not checking for convergence in k-points or planewave cutoff, and not checking for convergence with respect to slab thickness or number of relaxed layers.

#### 5.4.2 Ag(110) missing row reconstruction

##### 5.4.2.1 Clean Ag(110) slab
from vasp import Vasp
from ase.lattice.surface import fcc110
from ase.io import write
from ase.constraints import FixAtoms

atoms = fcc110('Ag', size=(2, 1, 6), vacuum=10.0)
constraint = FixAtoms(mask=[atom.tag > 2 for atom in atoms])
atoms.set_constraint(constraint)

calc = Vasp('surfaces/Ag-110',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
ibrion=2,
isif=2,
nsw=10,
atoms=atoms)
calc.update()

##### 5.4.2.2 Missing row in Ag(110)
from vasp import Vasp
from ase.lattice.surface import fcc110
from ase.io import write
from ase.constraints import FixAtoms

atoms = fcc110('Ag', size=(2, 1, 6), vacuum=10.0)
del atoms[11]  # delete surface row

constraint = FixAtoms(mask=[atom.tag > 2 for atom in atoms])
atoms.set_constraint(constraint)

Vasp('surfaces/Ag-110-missing-row',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
ibrion=2,
isif=2,
nsw=10,
atoms=atoms).update()

##### 5.4.2.3 Bulk Ag
from vasp import Vasp
from ase.visualize import view
from ase.lattice.cubic import FaceCenteredCubic

atoms = FaceCenteredCubic(directions=[[0, 1, 1],
[1, 0, 1],
[1, 1, 0]],
size=(1, 1, 1),
symbol='Ag')

Vasp('bulk/Ag-fcc',
xc='PBE',
encut=350,
kpts=[12, 12, 12],
atoms=atoms).update()

##### 5.4.2.4 Analysis of energies
from vasp import Vasp

eslab = Vasp('surfaces/Ag-110').potential_energy

emissingrow = Vasp('surfaces/Ag-110-missing-row').potential_energy

ebulk = Vasp('bulk/Ag-fcc').potential_energy

print 'dE = {0:1.3f} eV'.format(emissingrow + ebulk - eslab)

dE = -0.010 eV


For Ag(110), the missing row formation energy is practically thermoneutral, i.e. not that favorable. This energy is so close to 0eV, that we cannot confidently say whether the reconstruction is favorable or not. Experimentally, the reconstruction is not seen on very clean Ag(110) although it is reported that some adsorbates may induce the reconstruction PhysRevLett.59.2307.

#### 5.4.3 Cu(110) missing row reconstruction

##### 5.4.3.1 Clean Cu(110) slab
from vasp import Vasp
from ase.lattice.surface import fcc110
from ase.constraints import FixAtoms

atoms = fcc110('Cu', size=(2, 1, 6), vacuum=10.0)
constraint = FixAtoms(mask=[atom.tag > 2 for atom in atoms])
atoms.set_constraint(constraint)

Vasp('surfaces/Cu-110',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
ibrion=2,
isif=2,
nsw=10,
atoms=atoms).update()

##### 5.4.3.2 Missing row in Cu(110)
from vasp import Vasp
from ase.lattice.surface import fcc110
from ase.constraints import FixAtoms

atoms = fcc110('Cu', size=(2, 1, 6), vacuum=10.0)
del atoms[11]  # delete surface row

constraint = FixAtoms(mask=[atom.tag > 2 for atom in atoms])
atoms.set_constraint(constraint)

Vasp('surfaces/Cu-110-missing-row',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
ibrion=2,
isif=2,
nsw=10,
atoms=atoms).update()

##### 5.4.3.3 Bulk Cu
from vasp import Vasp
from ase.visualize import view
from ase.lattice.cubic import FaceCenteredCubic

atoms = FaceCenteredCubic(directions=[[0, 1, 1],
[1, 0, 1],
[1, 1, 0]],
size=(1, 1, 1),
symbol='Cu')

Vasp('bulk/Cu-fcc',
xc='PBE',
encut=350,
kpts=[12, 12, 12],
atoms=atoms).update()

##### 5.4.3.4 Analysis
from vasp import Vasp

eslab = Vasp('surfaces/Cu-110').potential_energy

emissingrow = Vasp('surfaces/Cu-110-missing-row').potential_energy

ebulk = Vasp('bulk/Cu-fcc').potential_energy

print 'natoms slab        = {0}'.format(len(slab))
print 'natoms missing row = {0}'.format(len(missingrow))
print 'natoms bulk        = {0}'.format(len(bulk))

print 'dE = {0:1.3f} eV'.format(emissingrow + ebulk - eslab)


It is questionable whether we should consider this evidence of a missing row reconstruction because the number is small. That does not mean the reconstruction will not happen, but it could mean it is very easy to lift.

### 5.5 Surface energy

The easiest way to calculate surface energies is from this equation:

$$\sigma = \frac{1}{2}(E_{slab} - \frac{N_{slab}}{N_{bulk}} E_{bulk})$$

where $$E_{slab}$$ is the total energy of a symmetric slab (i.e. one with inversion symmetry, and where both sides of the slab have been relaxed), $$E_{bulk}$$ is the total energy of a bulk unit cell, $$N_{slab}$$ is the number of atoms in the slab, and $$N_{bulk}$$ is the number of atoms in the bulk unit cell. One should be sure that the bulk energy is fully converged with respect to $k$-points, and that the slab energy is also converged with respect to $k$-points. The energies should be compared at the same cutoff energies. The idea is then to increase the thickness of the slab until the surface energy $$\sigma$$ converges.

Unfortunately, this approach does not always work. The bulk system is treated subtly different than the slab system, particularly in the $z$-direction where the vacuum is (where typically only one $k$-point is used in slabs). Consequently, the $k$-point sampling is not equivalent in the two systems, and one can in general expect some errors due to this, with the best case being cancellation of the errors due to total $k$-point convergence. In the worst case, one can get a linear divergence in the surface energy with slab thickness PhysRevB.49.16798.

A variation of this method that usually results in better $k$-point error cancellation is to calculate the bulk unit cell energy using the slab unit cell with no vacuum space, with the same $k$-point mesh in the $$x$$ and $$y$$ directions, but with increased $k$-points in the $z$-direction. Thus, the bulk system and slab system have the same Brillouin zone in at least two dimensions. This maximizes the cancellation of $k$-point errors, but still does not guarantee convergence of the surface energy, as discussed in PhysRevB.49.16798,0953-8984-10-4-017.

For quick estimates of the surface energy, one of the methods described above is likely sufficient. The advantage of these methods is the small number of calculations required to obtain the estimate, one needs only a bulk calculation (which must be done anyhow to get the bulk lattice constant to create the slab), and a slab calculation that is sufficiently thick to get the estimate. Additional calculations are only required to test the convergence of the surface energy.

An alternative method for calculating surface energies that does not involve an explicit bulk calculation follows Ref. 0953-8984-10-4-017. The method follows from equation (ref{eq:se}) where for a N-atom slab, in the limit of N → ∞,

$$E_{slab} \approx 2\sigma + \frac{N_{slab}}{N_{bulk}} E_{bulk}$$

Then, we can estimate Ebulk by plotting the total energy of the slab as a function of the slab thickness.

$$\sigma = \lim_{N \rightarrow \infty} \frac{1}{2}(E_{slab}^N - N \Delta E_N)$$

where $$\Delta E_N = E_{slab}^N - E_{slab}^{N-1}$$.

We will examine this approach here. We will use unrelaxed slabs for computational efficiency.

from vasp import Vasp
from ase.lattice.surface import fcc111
import matplotlib.pyplot as plt

Nlayers = [3, 4, 5, 6, 7, 8, 9, 10, 11]
energies = []
sigmas = []

for n in Nlayers:

slab = fcc111('Cu', size=(1, 1, n), vacuum=10.0)
slab.center()

calc = Vasp('bulk/Cu-layers/{0}'.format(n),
xc='PBE',
encut=350,
kpts=[8, 8, 1],
atoms=slab)
calc.set_nbands(f=2)  # the default nbands in VASP is too low for Cu
energies.append(slab.get_potential_energy())

calc.stop_if(None in energies)

for i in range(len(Nlayers) - 1):
N = Nlayers[i]
DeltaE_N = energies[i + 1] - energies[i]
sigma = 0.5 * (-N * energies[i + 1] + (N + 1) * energies[i])
sigmas.append(sigma)
print 'nlayers = {1:2d} sigma = {0:1.3f} eV/atom'.format(sigma, N)

plt.plot(Nlayers[0:-1], sigmas, 'bo-')
plt.xlabel('Number of layers')
plt.ylabel('Surface energy (eV/atom)')
plt.savefig('images/Cu-unrelaxed-surface-energy.png')

nlayers =  3 sigma = 0.561 eV/atom
nlayers =  4 sigma = 0.398 eV/atom
nlayers =  5 sigma = 0.594 eV/atom
nlayers =  6 sigma = 0.308 eV/atom
nlayers =  7 sigma = 0.590 eV/atom
nlayers =  8 sigma = 0.332 eV/atom
nlayers =  9 sigma = 0.591 eV/atom
nlayers = 10 sigma = 0.392 eV/atom


One reason for the oscillations may be quantum size effects Fiolhais2003209. In PhysRevB.75.115131 the surface energy of Cu(111) is reported as 0.48 eV/atom, or 1.36 J/m$$^2$$. Here is an example showing a conversion between these two units. We use ase to compute the area of the unit cell from the norm of the cross-product of the vectors defining the surface unit cell.

from ase.lattice.surface import fcc111
from ase.units import J, m
import numpy as np

slab = fcc111('Cu', size=(1, 1, 3), vacuum=10.0)
cell = slab.get_cell()

area = np.linalg.norm(np.cross(cell[0], cell[1]))  # area per atom

sigma = 0.48  # eV/atom

print 'sigma = {0} J/m^2'.format(sigma / area / (J / m**2))

sigma = 1.36281400415 J/m^2


#### 5.5.1 Advanced topics in surface energy

The surface energies can be used to estimate the shapes of nanoparticles using a Wulff construction. See doi.10.1021/jp200950a for an example of computing Mo2C surface energies and particle shapes, and Inoglu2009188 for an example of the influence of adsorbates on surface energies and particle shapes of Cu.

For a classic paper on trends in surface energies see Vitos1998186.

### 5.6 Work function

work function To get the work function, we need to have the local potential. This is not written by default in VASP, and we have to tell it to do that with the LVTOT and LVHAR keywords.

from vasp import Vasp
import matplotlib.pyplot as plt
import numpy as np

calc = Vasp('surfaces/Al-slab-relaxed')
atoms = calc.get_atoms()

calc = Vasp('surfaces/Al-slab-locpot',
xc='PBE',
kpts=[6, 6, 1],
encut=350,
lvtot=True,  # write out local potential
lvhar=True,  # write out only electrostatic potential, not xc pot
atoms=atoms)
calc.wait()
ef = calc.get_fermi_level()
x, y, z, lp = calc.get_local_potential()

nx, ny, nz = lp.shape

axy = np.array([np.average(lp[:, :, z]) for z in range(nz)])
# setup the x-axis in realspace
uc = atoms.get_cell()
xaxis = np.linspace(0, uc[2][2], nz)

plt.plot(xaxis, axy)
plt.plot([min(xaxis), max(xaxis)], [ef, ef], 'k:')
plt.xlabel('Position along z-axis')
plt.ylabel('x-y averaged electrostatic potential')
plt.savefig('images/Al-wf.png')

ind = (xaxis > 0) & (xaxis < 5)
wf = np.average(axy[ind]) - ef
print ' The workfunction is {0:1.2f} eV'.format(wf)

The workfunction is 4.17 eV


The workfunction of Al is listed as 4.08 at http://hyperphysics.phy-astr.gsu.edu/hbase/tables/photoelec.html.

### 5.7 Dipole correction

A subtle problem can arise when an adsorbate is placed on one side of a slab with periodic boundary conditions, which is currently the common practice. The problem is that this gives the slab a dipole moment. The array of dipole moments created by the periodic boundary conditions generates an electric field that can distort the electron density of the slab and change the energy. The existence of this field in the vacuum also makes the zero-potential in the vacuum ill-defined, thus the work function is not well-defined. One solution to this problem is to use slabs with adsorbates on both sides, but then very thick (eight to ten layers) slabs must be used to ensure the adsorbates do not interact through the slab. An alternative solution, the dipole correction scheme, was developed by Neugebauer and Scheffler PhysRevB.46.16067 and later corrected by Bengtsson PhysRevB.59.12301. In this technique, an external field is imposed in the vacuum region that exactly cancels the artificial field caused by the slab dipole moment. The advantage of this approach is that thinner slabs with adsorbates on only one side can be used.

There are also literature reports that the correction is small morikawa2001:c2h2-si. Nevertheless, in the literature the use of this correction is fairly standard, and it is typical to at least consider the correction.

Here we will just illustrate the effect.

#### 5.7.1 Slab with no dipole correction

We simply run the calculation here, and compare the results later.

# compute local potential of slab with no dipole
from vasp import Vasp
import matplotlib.pyplot as plt
from ase.io import write

slab = fcc111('Al', size=(2, 2, 2), vacuum=10.0)

slab.center()
write('images/Na-Al-slab.png', slab, rotation='-90x', show_unit_cell=2)

print(Vasp('surfaces/Al-Na-nodip',
xc='PBE',
encut=340,
kpts=[2, 2, 1],
lcharg=True,
lvtot=True,  # write out local potential
lvhar=True,  # write out only electrostatic potential, not xc pot
atoms=slab).potential_energy)

-22.55264459


#### 5.7.2TODO Slab with a dipole correction

Note this takes a considerably longer time to run than without a dipole correction! In VASP there are several levels of dipole correction to apply. You can use the IDIPOL tag to turn it on, and specify which direction to apply it in (1=$$x$$, 2=$$y$$, 3=$$z$$, 4=$$(x,y,z)$$). This simply corrects the total energy and forces. It does not change the contents of LOCPOT. For that, you have to also set the LDIPOL and DIPOL tags. It is not efficient to set all three at the same time for some reason. The VASP manual recommends you first set IDIPOL to get a converged electronic structure, and then set LDIPOL to True, and set the center of electron density in DIPOL. That makes these calculations a multistep process, because we must run a calculation, analyze the charge density to get the center of charge, and then run a second calculation.

# compute local potential with dipole calculation on
from vasp import Vasp
import numpy as np

slab = fcc111('Al', size=(2, 2, 2), vacuum=10.0)

slab.center()

calc = Vasp('surfaces/Al-Na-dip',
xc='PBE',
encut=340,
kpts=[2, 2, 1],
lcharg=True,
idipol=3,   # only along z-axis
lvtot=True,  # write out local potential
lvhar=True,  # write out only electrostatic potential, not xc pot
atoms=slab)

calc.stop_if(calc.potential_energy is None)

x, y, z, cd = calc.get_charge_density()
n0, n1, n2 = cd.shape
nelements = n0 * n1 * n2
voxel_volume = slab.get_volume() / nelements
total_electron_charge = cd.sum() * voxel_volume

electron_density_center = np.array([(cd * x).sum(),
(cd * y).sum(),
(cd * z).sum()])
electron_density_center *= voxel_volume
electron_density_center /= total_electron_charge

print 'electron-density center = {0}'.format(electron_density_center)
uc = slab.get_cell()

# get scaled electron charge density center
sedc = np.dot(np.linalg.inv(uc.T), electron_density_center.T).T

# we only write 4 decimal places out to the INCAR file, so we round here.
sedc = np.round(sedc, 4)

calc.clone('surfaces/Al-Na-dip-step2')

# now run step 2 with dipole set at scaled electron charge density center
calc.set(ldipol=True, dipol=sedc)
print(calc.potential_energy)


#### 5.7.3 Comparing no dipole correction with a dipole correction

To see the difference in what the dipole correction does, we now plot the potentials from each calculation.

from vasp import Vasp
import matplotlib.pyplot as plt

calc = Vasp('surfaces/Al-Na-nodip')
atoms = calc.get_atoms()

x, y, z, lp = calc.get_local_potential()
nx, ny, nz = lp.shape

axy_1 = [np.average(lp[:, :, z]) for z in range(nz)]
# setup the x-axis in realspace
uc = atoms.get_cell()
xaxis_1 = np.linspace(0, uc[2][2], nz)

e1 = atoms.get_potential_energy()

calc = Vasp('surfaces/Al-Na-dip-step2')
atoms = calc.get_atoms()

x, y, z, lp = calc.get_local_potential()
nx, ny, nz = lp.shape

axy_2 = [np.average(lp[:, :, z]) for z in range(nz)]
# setup the x-axis in realspace
uc = atoms.get_cell()
xaxis_2 = np.linspace(0, uc[2][2], nz)

ef2 = calc.get_fermi_level()
e2 = atoms.get_potential_energy()

print 'The difference in energy is {0} eV.'.format(e2-e1)

plt.plot(xaxis_1, axy_1, label='no dipole correction')
plt.plot(xaxis_2, axy_2, label='dipole correction')
plt.plot([min(xaxis_2), max(xaxis_2)], [ef2, ef2], 'k:', label='Fermi level')
plt.xlabel('z ($\AA$)')
plt.ylabel('xy-averaged electrostatic potential')
plt.legend(loc='best')
plt.savefig('images/dip-vs-nodip-esp.png')


The key points to notice in this figure are:

1. The two deep dips are where the atoms are.
2. Without a dipole correction, the electrostatic potential never flattens out. there is near constant slope in the vacuum region, which means there is an electric field there.
3. With a dipole correction the potential is flat in the vacuum region, except for the step jump near 23 Å.
4. The difference between the Fermi level and the flat vacuum potential is the work function.
5. The difference in energy with and without the dipole correction here is small.

#### 5.8.1 Simple estimate of the adsorption energy

Calculating an adsorption energy amounts to computing the energy of the following kind of reaction:

slab + gas-phase molecule $$\rightarrow$$ slab_adsorbate + products

There are many variations of this idea. The slab may already have some adsorbates on it, the slab may reconstruct on adsorption, the gas-phase molecule may or may not dissociate, and the products may or may not stick to the surface. We have to decide where to put the adsorbates, i.e. what site to put them on, and some sites will be more stable than others. We will consider the dissociative adsorption of O$$_2$$ on three sites of a Pt(111) slab. We will assume the oxygen molecule has split in half, and that the atoms have moved far apart. We will model the oxygen coverage at 0.25 ML, which means we need to use a $$2 \times 2$$ surface unit cell. For computational speed, we will freeze the slab, but allow the adsorbate to relax.

$$\Delta H_{ads} (eV/O) = E_{slab+O} - E_{slab} - 0.5*E_{O_2}$$

##### 5.8.1.1 Calculations
###### 5.8.1.1.1 clean slab calculation
from vasp import Vasp
from ase.lattice.surface import fcc111
from ase.constraints import FixAtoms

atoms = fcc111('Pt', size=(2, 2, 3), vacuum=10.0)
constraint = FixAtoms(mask=[True for atom in atoms])
atoms.set_constraint(constraint)

from ase.io import write
write('images/Pt-fcc-ori.png', atoms, show_unit_cell=2)

print(Vasp('surfaces/Pt-slab',
xc='PBE',
kpts=[4, 4, 1],
encut=350,
atoms=atoms).potential_energy)

-68.23616204

###### 5.8.1.1.2 fcc site
from vasp import Vasp

from ase.constraints import FixAtoms

atoms = fcc111('Pt', size=(2, 2, 3), vacuum=10.0)

# note this function only works when atoms are created by the surface module.

constraint = FixAtoms(mask=[atom.symbol != 'O' for atom in atoms])
atoms.set_constraint(constraint)

from ase.io import write
write('images/Pt-fcc-site.png', atoms, show_unit_cell=2)

print(Vasp('surfaces/Pt-slab-O-fcc',
xc='PBE',
kpts=[4, 4, 1],
encut=350,
ibrion=2,
nsw=25,
atoms=atoms).potential_energy)

-74.23018764

###### 5.8.1.1.3 O atom on the bridge site
from vasp import Vasp
from ase.constraints import FixAtoms

atoms = fcc111('Pt', size=(2, 2, 3), vacuum=10.0)

# note this function only works when atoms are created by the surface module.

constraint = FixAtoms(mask=[atom.symbol != 'O' for atom in atoms])
atoms.set_constraint(constraint)

print(Vasp('surfaces/Pt-slab-O-bridge',
xc='PBE',
kpts=[4, 4, 1],
encut=350,
ibrion=2,
nsw=25,
atoms=atoms).potential_energy)

-74.23023073

###### 5.8.1.1.4 hcp site
from vasp import Vasp
from ase.constraints import FixAtoms

atoms = fcc111('Pt', size=(2, 2, 3), vacuum=10.0)

# note this function only works when atoms are created by the surface module.

constraint = FixAtoms(mask=[atom.symbol != 'O' for atom in atoms])
atoms.set_constraint(constraint)

from ase.io import write
write('images/Pt-hcp-o-site.png', atoms, show_unit_cell=2)

print(Vasp('surfaces/Pt-slab-O-hcp',
xc='PBE',
kpts=[4, 4, 1],
encut=350,
ibrion=2,
nsw=25,
atoms=atoms).potential_energy)

-73.76942127

##### 5.8.1.2 Analysis of adsorption energies
from vasp import Vasp
from ase.io import write

calc = Vasp('surfaces/Pt-slab')
atoms = calc.get_atoms()
e_slab = atoms.get_potential_energy()
write('images/pt-slab.png', atoms,show_unit_cell=2)

calc = Vasp('surfaces/Pt-slab-O-fcc')
atoms = calc.get_atoms()
e_slab_o_fcc = atoms.get_potential_energy()
write('images/pt-slab-fcc-o.png', atoms,show_unit_cell=2)

calc = Vasp('surfaces/Pt-slab-O-hcp')
atoms = calc.get_atoms()
e_slab_o_hcp = atoms.get_potential_energy()
write('images/pt-slab-hcp-o.png', atoms,show_unit_cell=2)

calc = Vasp('surfaces/Pt-slab-O-bridge')
atoms = calc.get_atoms()
e_slab_o_bridge = atoms.get_potential_energy()
write('images/pt-slab-bridge-o.png', atoms,show_unit_cell=2)

calc = Vasp('molecules/O2-sp-triplet-350')
atoms = calc.get_atoms()
e_O2 = atoms.get_potential_energy()

Hads_fcc = e_slab_o_fcc - e_slab - 0.5 * e_O2
Hads_hcp = e_slab_o_hcp - e_slab - 0.5 * e_O2
Hads_bridge = e_slab_o_bridge - e_slab - 0.5 * e_O2



You can see the hcp site is not as energetically favorable as the fcc site. Interestingly, the bridge site seems to be as favorable as the fcc site. This is not correct, and to see why, we have to look at the final geometries of each calculation. First the fcc (Figure fig:fcc and hcp (Figure fig:hcp sites, which look like we expect.

The bridge site (Figure fig:bridge, however, is clearly not at a bridge site!

Let us see what the original geometry and final geometry for the bridge site were. The POSCAR contains the initial geometry (as long as you haven't copied CONTCAR to POSCAR), and the CONTCAR contains the final geometry.

from ase.io import read, write

write('images/Pt-o-brige-ori.png', atoms, show_unit_cell=2)

write('images/Pt-o-brige-final.png', atoms, show_unit_cell=2)


You can see the problem. We should not call the adsorption energy from this calculation a bridge site adsorption energy because the O atom is actually in an fcc site! This kind of result can happen with relaxation, and you should always check that the result you get makes sense. Next, we consider how to get a bridge site adsorption energy by using constraints.

Some final notes:

1. We did not let the slabs relax in these examples, and allowing them to relax is likely to have a big effect on the adsorption energies. You have to decide how many layers to relax, and check for convergence with respect to the number of layers.
2. The slabs were pretty thin. It is typical these days to see slabs that are 4-5 or more layers thick.
3. We did not consider how well converged the calculations were with respect to $k$-points or ENCUT.
4. We did not consider the effect of the error in O$$_2$$ dissociation energy on the adsorption energies.
5. We did not consider coverage effects (see Coverage dependence).
##### 5.8.1.3 Adsorption on bridge site with constraints

To prevent the oxygen atom from sliding down into the fcc site, we have to constrain it so that it only moves in the $z$-direction. This is an artificial constraint; the bridge site is only metastable. But there are lots of reasons you might want to do this anyway. One is the bridge site is a transition state for diffusion between the fcc and hcp sites. Another is to understand the role of coordination in the adsorption energies. We use a ase.constraints.FixScaled constraint in ase to constrain the O atom so it can only move in the $z$-direction (actually so it can only move in the direction of the third unit cell vector, which only has a $z$-component).

from vasp import Vasp

from ase.constraints import FixAtoms, FixScaled
from ase.io import write

atoms = fcc111('Pt', size=(2, 2, 3), vacuum=10.0)

# note this function only works when atoms are created by the surface module.
constraint1 = FixAtoms(mask=[atom.symbol != 'O' for atom in atoms])
# fix in xy-direction, free in z. actually, freeze movement in surface
# unit cell, and free along 3rd lattice vector
constraint2 = FixScaled(atoms.get_cell(), 12, [True, True, False])

atoms.set_constraint([constraint1, constraint2])
write('images/Pt-O-bridge-constrained-initial.png', atoms, show_unit_cell=2)
print 'Initial O position: {0}'.format(atoms.positions[-1])

calc = Vasp('surfaces/Pt-slab-O-bridge-xy-constrained',
xc='PBE',
kpts=[4, 4, 1],
encut=350,
ibrion=2,
nsw=25,
atoms=atoms)
e_bridge = atoms.get_potential_energy()

write('images/Pt-O-bridge-constrained-final.png', atoms, show_unit_cell=2)
print 'Final O position  : {0}'.format(atoms.positions[-1])

calc = Vasp('surfaces/Pt-slab')
atoms = calc.get_atoms()
e_slab = atoms.get_potential_energy()

calc = Vasp('molecules/O2-sp-triplet-350')
atoms = calc.get_atoms()
e_O2 = atoms.get_potential_energy()

calc.stop_if(None in [e_bridge, e_slab, e_O2])

Hads_bridge = e_bridge - e_slab - 0.5*e_O2


Initial O position: [  1.38592929   0.          15.72642611]
Final O position  : [  1.38592929   0.          15.9685262 ]


You can see that only the $$z$$-position of the O atom changed. Also, the adsorption energy of O on the bridge site is much less favorable than on the fcc or hcp sites.

#### 5.8.2 Coverage dependence

The adsorbates on the surface can interact with each other which results in coverage dependent adsorption energies PhysRevB.82.045414. Coverage dependence is not difficult to model; we simply compute adsorption energies in different size unit cells, and/or with different adsorbate configurations. Here we consider dissociative oxygen adsorption at 1ML on Pt(111) in an fcc site, which is one oxygen atom in a $$1 \times 1$$ unit cell.

##### 5.8.2.1 clean slab calculation
from vasp import Vasp
from ase.io import write
from ase.lattice.surface import fcc111
from ase.constraints import FixAtoms

atoms = fcc111('Pt', size=(1, 1, 3), vacuum=10.0)
constraint = FixAtoms(mask=[True for atom in atoms])
atoms.set_constraint(constraint)

write('images/Pt-fcc-1ML.png', atoms, show_unit_cell=2)

print(Vasp('surfaces/Pt-slab-1x1',
xc='PBE',
kpts=[8, 8, 1],
encut=350,
atoms=atoms).potential_energy)

-17.05903301

##### 5.8.2.2 fcc site at 1 ML coverage
from vasp import Vasp

from ase.constraints import FixAtoms
from ase.io import write

atoms = fcc111('Pt', size=(1, 1, 3), vacuum=10.0)

# note this function only works when atoms are created by the surface module.

constraint = FixAtoms(mask=[atom.symbol != 'O' for atom in atoms])
atoms.set_constraint(constraint)

write('images/Pt-o-fcc-1ML.png', atoms, show_unit_cell=2)

print(Vasp('surfaces/Pt-slab-1x1-O-fcc',
xc='PBE',
kpts=[8, 8, 1],
encut=350,
ibrion=2,
nsw=25,
atoms=atoms).potential_energy)

-22.13585728

##### 5.8.2.3 Adsorption energy at 1ML
from vasp import Vasp

e_slab_o = Vasp('surfaces/Pt-slab-1x1-O-fcc').potential_energy

# clean slab
e_slab = Vasp('surfaces/Pt-slab-1x1').potential_energy

e_O2 = Vasp('molecules/O2-sp-triplet-350').potential_energy

hads = e_slab_o - e_slab - 0.5 * e_O2

Hads (1ML) = -0.099 eV


The adsorption energy is much less favorable at 1ML coverage than at 0.25 ML coverage! We will return what this means in Atomistic thermodynamics effect on adsorption.

#### 5.8.3 Effect of adsorption on the surface energy

There is a small point to make here about what adsorption does to surface energies. Let us define a general surface formation energy scheme like this:

Let us presume the surfaces are symmetric, and that each surface contributes half of the energy change. The overall change in energy:

$$\Delta E = E_{slab,ads} - E_{ads} - E_{bulk}$$

where the the energies are appropriately normalized for the stoichiometry. Let us rearrange the terms, and add and subtract a constant term $$E_{slab}$$.

$$\Delta E = E_{slab,ads} - E_{slab} - E_{ads} - E_{bulk} + E_{slab}$$

We defined $$\gamma_{clean} = \frac{1}{2A}(E_{slab} - E_{bulk})$$, and we defined $$H_{ads} = E_{slab,ads} - E_{slab} - E_{ads}$$ for adsorption on a single side of a slab. In this case, there are adsorbates on both sides of the slab, so $$E_{slab,ads} - E_{slab} - E_{ads} = 2 \Delta H_{ads}$$. If we normalize by $$2A$$, the area for both sides of the slab, we get

$$\frac{\Delta E}{2A} = \gamma = \gamma_{clean} + \frac{H_{ads}}{A}$$

You can see here that the adsorption energy serves to stabilize, or reduce the surface energy, provided that the adsorption energy is negative.

Some final notes about the equations above:

• We were not careful about stoichiometry. As written, it is assumed there are the same number of atoms (not including the adsorbates) in the slabs and bulk, and the same number of adsorbate atoms in the slab and $$E_{ads}$$. Appropriate normalization factors must be included if that is not true.
• It is not necessary to perform a symmetric slab calculation to determine the effect of adsorption on the surface energy! You can examine $$\gamma - \gamma_{clean}$$ with knowledge of only the adsorption energies!

Adsorbates also have vibrational modes. Unlike a free molecule, the translational and rotational modes of an adsorbate may actually have real frequencies. Sometimes they are called frustrated translations or rotations. For metal surfaces with adsorbates, it is common to only compute vibrational modes of the adsorbate on a frozen metal slab. The rationale is that the metal atoms are so much heavier than the adsorbate that there will be little coupling between the surface and adsorbates. You can limit the number of modes calculated with constraints (ase.constraints.FixAtoms or ase.constraints.FixScaled) if you use IBRION=5. The other IBRION settings (6, 7, 8) do not respect the selective dynamics constraints. Below we consider the vibrational modes of an oxygen atom in an fcc site on Pt(111).

from vasp import Vasp

calc = Vasp('surfaces/Pt-slab-O-fcc')
calc.clone('surfaces/Pt-slab-O-fcc-vib')

calc.set(ibrion=5,     # finite differences with selective dynamics
nfree=2,      # central differences (default)
potim=0.015,  # default as well
ediff=1e-8,
nsw=1)
atoms = calc.get_atoms()
f, v = calc.get_vibrational_modes(0)
print 'Elapsed time = {0} seconds'.format(calc.get_elapsed_time())
allfreq = calc.get_vibrational_modes()[0]

from ase.units import meV
c = 3e10  # cm/s
h = 4.135667516e-15  # eV*s

print 'vibrational energy = {0} eV'.format(f)
print 'vibrational energy = {0} meV'.format(f/meV)
print 'vibrational freq   = {0} 1/s'.format(f/h)
print 'vibrational freq   = {0} cm^{{-1}}'.format(f/(h*c))
print
print 'All energies = ', allfreq


There are three modes for the free oxygen atom. One of them is a mode normal to the surface (the one with highest frequency. The other two are called frustrated translations. Note that we did not include the surface Pt atoms in the calculation, and this will have an effect on the result because the O atom could be coupled to the surface modes. It is typical to neglect this coupling because of the large difference in mass between O and Pt. Next we look at the difference in results when we calculate all the modes.

from vasp import Vasp

calc = Vasp('surfaces/Pt-slab-O-fcc')
calc.clone('Pt-slab-O-fcc-vib-ibrion=6')
calc.set(ibrion=6,  # finite differences with symmetry
nfree=2,  # central differences (default)
potim=0.015,  # default as well
ediff=1e-8,
nsw=1)
calc.update()
print 'Elapsed time = {0} seconds'.format(calc.get_elapsed_time())

f, m = calc.get_vibrational_modes(0)
allfreq = calc.get_vibrational_modes()[0]

from ase.units import meV
c = 3e10  # cm/s
h = 4.135667516e-15  # eV*s

print 'For mode 0:'
print 'vibrational energy = {0} eV'.format(f)
print 'vibrational energy = {0} meV'.format(f / meV)
print 'vibrational freq   = {0} 1/s'.format(f / h)
print 'vibrational freq   = {0} cm^{{-1}}'.format(f / (h * c))
print
print 'All energies = ', allfreq

Elapsed time = 77121.015 seconds
For mode 0:
vibrational energy = 0.063537929 eV
vibrational energy = 63.537929 meV
vibrational freq   = 1.53634035507e+13 1/s
vibrational freq   = 512.113451691 cm^{-1}

All energies =  [0.06353792899999999, 0.045628623, 0.045628623, 0.023701702, 0.023701702, 0.023223747, 0.022978233, 0.022978233, 0.022190167, 0.021807461, 0.02040119, 0.02040119, 0.019677135000000002, 0.015452848, 0.015302098000000002, 0.015302098000000002, 0.0148412, 0.0148412, 0.014071851000000002, 0.012602063, 0.012602063, 0.012409611999999999, 0.012300973000000002, 0.011735683, 0.011735683, 0.011714521, 0.011482183, 0.011482183, 0.010824891, 0.010414177, 0.010414177, 0.009799697, 0.00932905, 0.00932905, 0.003859079, 0.003859079, (2.9894000000000002e-05+0j), (2.9894000000000002e-05+0j), (0.00012182999999999999+0j)]


Note that now there are 39 modes, which is 3*N where N=13 atoms in the unit cell. Many of the modes are low in frequency, which correspond to slab modes that are essentially phonons. The O frequencies are not that different from the previous calculation (497 vs 512 cm$$^{-1}$$. This is why it is common to keep the slab atoms frozen.

Calculating these results took 39*2 finite differences. It took about a day to get these results on a single CPU. It pays to use constraints to minimize the number of these calculations.

#### 5.9.1 Vibrations of the bridge site

Here we consider the vibrations of an O atom in a bridge site, which we saw earlier is a metastable saddle point.

from vasp import Vasp
from ase.constraints import FixAtoms

# clone calculation so we do not overwrite previous results
calc = Vasp('surfaces/Pt-slab-O-bridge-xy-constrained')
calc.clone('surfaces/Pt-slab-O-bridge-vib')

calc.set(ibrion=5,  # finite differences with selective dynamics
nfree=2,   # central differences (default)
potim=0.015,  # default as well
ediff=1e-8,
nsw=1)

atoms = calc.get_atoms()
del atoms.constraints
constraint = FixAtoms(mask=[atom.symbol != 'O' for atom in atoms])
atoms.set_constraint([constraint])

f, v = calc.get_vibrational_modes(2)
print(calc.get_vibrational_modes()[0])

from ase.units import meV
c = 3e10  # cm/s
h = 4.135667516e-15  # eV*s

print('vibrational energy = {0} eV'.format(f))
print('vibrational energy = {0} meV'.format(f/meV))
print('vibrational freq   = {0} 1/s'.format(f/h))
print('vibrational freq   = {0} cm^(-1)'.format(f/(h*c)))

[0.06691932, 0.047345270999999994, (0.020649715000000003+0j)]
vibrational energy = (0.020649715+0j) eV
vibrational energy = (20.649715+0j) meV
vibrational freq   = (4.99307909065e+12+0j) 1/s
vibrational freq   = (166.435969688+0j) cm^(-1)


Note that we have one imaginary mode. This corresponds to the motion of the O atom falling into one of the neighboring 3-fold sites. It also indicates this position is not a stable minimum, but rather a saddle point. This position is a transition state for hopping between the fcc and hcp sites.

### 5.10 Surface Diffusion barrier

See this review ANIE.ANIE200602223 of diffusion on transition metal surfaces.

#### 5.10.1 Standard nudged elastic band method

Here we illustrate a standard NEB method. You need an initial and final state to start with. We will use the results from previous calculations of oxygen atoms in an fcc and hcp site. then we will construct a band of images connecting these two sites. Finally, we let VASP optimize the band and analyze the results to get the barrier.

from vasp import Vasp
from ase.neb import NEB
import matplotlib.pyplot as plt

calc = Vasp('surfaces/Pt-slab-O-fcc')
initial_atoms = calc.get_atoms()

final_atoms = Vasp('surfaces/Pt-slab-O-hcp').get_atoms()

# here is our estimated transition state. we use vector geometry to
# define the bridge position, and add 1.451 Ang to z based on our
# previous bridge calculation. The bridge position is half way between
# atoms 9 and 10.
ts = initial_atoms.copy()
ts.positions[-1] = 0.5 * (ts.positions[9] + ts.positions[10]) + [0, 0, 1.451]

# construct the band
images = [initial_atoms]
images += [initial_atoms.copy()]
images += [ts.copy()]  # this is the TS

neb = NEB(images)
# Interpolate linearly the positions of these images:
neb.interpolate()

# now add the second half
images2 = [ts.copy()]
images2 += [ts.copy()]
images2 += [final_atoms]

neb2 = NEB(images2)
neb2.interpolate()

# collect final band. Note we do not repeat the TS in the second half
final_images = images + images2[1:]

calc = Vasp('surfaces/Pt-O-fcc-hcp-neb',
ibrion=1,
nsw=90,
spring=-5,
atoms=final_images)

images, energies = calc.get_neb()
p = calc.plot_neb(show=False)
plt.savefig('images/pt-o-fcc-hcp-neb.png')

Optimization terminated successfully.
Current function value: -26.953429
Iterations: 12
Function evaluations: 24


We should compare this barrier to what we could estimate from the simple adsorption energies in the fcc and bridge sites. The adsorption energy in the fcc site was -1.04 eV, and in the bridge site was -0.49 eV. The difference between these two is 0.55 eV, which is very close to the calculated barrier from the NEB calculation. In cases where you can determine what the transition state is, e.g. by symmetry, or other means, it is much faster to directly compute the energy of the initial and transition states for barrier determinations. This is not usually possible though.

#### 5.10.2 Climbing image NEB

One issue with the standard NEB method is there is no image that is exactly at the transition state. That means there is some uncertainty of the true energy of the transition state, and there is no way to verify the transition state by vibrational analysis. The climbing image NEB method henkelman:9901 solves that problem by making one image climb to the top. You set LCLIMB==True= in Vasp to turn on the climbing image method. Here we use the previous calculation as a starting point and turn on the climbing image method.

# perform a climbing image NEB calculation
from vasp import Vasp

calc = Vasp('surfaces/Pt-O-fcc-hcp-neb')
calc.clone('surfaces/Pt-O-fcc-hcp-cineb')
calc.set(ichain=0, lclimb=True)

images, energies = calc.get_neb()
calc.plot_neb(show=False)
import matplotlib.pyplot as plt
plt.savefig('images/pt-o-cineb.png')
plt.show()


#### 5.10.3 Using vibrations to confirm a transition state

A transition state should have exactly one imaginary degree of freedom which corresponds to the mode that takes reactants to products. See Vibrations of the bridge site for an example.

## 6 Atomistic thermodynamics

atomistic thermodynamics Let us consider how much the Gibbs free energy of an O2 molecule changes as a function of temperature, at 1 atm. We use the Shomate polynomials to approximate the temperature dependent entropy and enthalpy, and use the parameters from the NIST Webbook for O2.

from ase.units import *
K = 1.0

print J, mol, K

print 0.100 * kJ / mol / K

print 1 * eV / (kJ / mol)

6.24150912588e+18 6.022140857e+23 1.0
0.00103642695747
96.4853328825

import numpy as np
import matplotlib.pyplot as plt
from ase.units import *
K = 1.  # Kelvin not defined in ase.units!

# Shomate parameters
A = 31.32234; B = -20.23531; C = 57.86644
D = -36.50624; E = -0.007374; F = -8.903471
G = 246.7945; H = 0.0

def entropy(T):
'''entropy returned as eV/K
T in K
'''
t = T / 1000.
s = (A * np.log(t) + B * t + C * (t**2) / 2.
+ D * (t**3) / 3. - E / (2. * t**2) + G)
return s * J / mol / K

def enthalpy(T):
''' H - H(298.15) returned as eV/molecule'''
t = T / 1000.
h = (A * t + B * (t**2) / 2. + C * (t**3) / 3.
+ D * (t**4) / 4. - E / t + F - H)
return h * kJ / mol

T = np.linspace(10, 700)

G = enthalpy(T) - T * entropy(T)

plt.plot(T, G)
plt.xlabel('Temperature (K)')
plt.ylabel(r'$\Delta G^\circ$ (eV)')
plt.savefig('images/O2-mu.png')


This is clearly a big effect! Between 500-600K, the energy has dropped by nearly 1 eV.

Pressure also affects the free energy. In the ideal gas limit, the pressure changes the free energy by $$k T \ln P/P_0$$ where $$P_0$$ is the standard state pressure (1 atm or 1 bar depending on the convention chosen). Let us see how this affects the free energy at different temperatures.

import matplotlib.pyplot as plt
import numpy as np
from ase.units import *

atm = 101325 * Pascal #atm is not defined in units
K = 1 # Kelvin

# examine range over 10^-10 to 10^10 atm
P = np.logspace(-10, 10) * atm

plt.semilogx(P / atm, kB * (300 * K) * np.log(P / (1 * atm)), label='300K')
plt.semilogx(P / atm, kB * (600 * K) * np.log(P / (1 * atm)), label='600K')
plt.xlabel('Pressure (atm)')
plt.ylabel(r'$\Delta G$ (eV)')
plt.legend(loc='best')
plt.savefig('images/O2-g-p.png')


Similarly, you can see that simply changing the pressure has a large effect on the Gibbs free energy of an ideal gas through the term: $$kT\ln(P/P_0)$$, and that this effect is also temperature dependent. This leads us to the final formula we will use for the chemical potential of oxgyen:

$$\mu_{O_2} = E_{O_2}^{DFT} + E_{O_2}^{ZPE} + \Delta \mu (T) + kT \ln(P/P_0)$$

We can use $$\mu_{O_2}$$ in place of $$E_{O_2}$$ everywhere to include the effects of pressure and temperature on the gas phase energy. If T=0K, and P=1 bar, we are at standard state, and this equation reduces to the DFT energy (+ the ZPE).

### 6.1 Bulk phase stability of oxides

We will consider the effects of oxygen pressure and temperature on the formation energy of Ag2O and Cu2O. For now, we neglect the effect of pressure and temperature on the solid phases. Neglecting pressure is pretty reasonable, as the solids are not that compressible, and we do not expect the energy to change for small pressures. For neglecting the temperature, we assume that the temperature dependence of the oxide is similar to the temperature dependence of the metal, and that these dependencies practically cancel each other in the calculations. That is an assumption, and it may not be correct.

$$2Cu + 1/2 O_2 \rightarrow Cu_2O$$

In atomistic thermodynamics, we define the free energy of formation as:

$$G_f = G_{Cu_2O} -2G_{Cu} - 0.5 G_{O_2}$$

We will at this point assume that the solids are incompressible so that $$p\Delta V \approx 0$$, and that $$S_{Cu_2O} -2S_{Cu} \approx 0$$, which leads to $$G_{Cu_2O} -2G_{Cu} \approx E_{Cu_2O} -2E_{Cu}$$, which we directly compute from DFT. We express $$G_{O_2} = \mu_{O_2} = E_{O_2}^{DFT} + E_{O_2}^{ZPE} + \Delta \mu (T) + kT \ln(P/P_0)$$. In this example we neglect the zero-point energy of the oxygen molecule, and finally arrive at:

$$G_f \approx E_{Cu_2O} -2E_{Cu} - 0.5 (E_{O_2}^{DFT} + \Delta \mu (T) + kT \ln(P/P_0))$$

Which, after grouping terms is:

$$G_f \approx E_{Cu_2O} -2E_{Cu} - 0.5 (E_{O_2}^{DFT}) - 0.5*\Delta \mu_{O_2}(P,T)$$

with $$\Delta \mu_{O_2}(P,T) = \Delta \mu (T) + kT \ln(P/P_0)$$. We get $$\Delta \mu (T)$$ from the Janaf Tables, or the NIST Webbook.

• we are explicitly neglecting all entropies of the solid: configurational, vibrational and electronic
• we also neglect enthalpic contributions from temperature dependent electronic and vibrational states

You will recognize in this equation the standard formation energy we calculated in Metal oxide oxidation energies plus a correction for the non standard state pressure and temperature ($$\Delta \mu_{O_2}(P,T) = 0$$ at standard state).

$$G_f \approx H_f - 0.5*\Delta \mu_{O_2}(P, T)$$

The formation energy of Cu2O is -1.9521 eV/formula unit. The formation energy for Ag2O is -0.99 eV/formula unit. Let us consider what temperature the oxides decompose at a fixed oxygen pressure of 1$$\times 10^{-10}$$ atm. We need to find the temperature where:

$$H_f = 0.5*\Delta \mu_{O_2}(P, T)$$

which will make the formation energy be 0.

import numpy as np
import matplotlib.pyplot as plt
from ase.units import *
from scipy.optimize import fsolve

K = 1. #not defined in ase.units!
atm = 101325 * Pascal

# Shomate parameters valid from 100-700K
A = 31.32234; B = -20.23531; C = 57.86644
D = -36.50624; E = -0.007374; F = -8.903471
G = 246.7945; H = 0.0

def entropy(T):
'''entropy returned as eV/K
T in K
'''
t = T/1000.
s = (A * np.log(t) + B * t + C * (t**2) / 2.
+ D * (t**3) / 3. - E / (2. * t**2) + G)
return s * J / mol / K

def enthalpy(T):
''' H - H(298.15) returned as eV/molecule'''
t = T / 1000.
h = (A * t + B * (t**2) / 2. + C * (t**3) / 3.
+ D * (t**4) / 4. - E / t + F - H)
return h * kJ / mol

def DeltaMu(T, P):
'''
returns delta chemical potential of oxygen at T and P
T in K
P in atm
'''
return enthalpy(T) - T * entropy(T) + kB * T * np.log(P / atm)

P = 1e-10*atm

def func(T):
'Cu2O'
return -1.95 - 0.5*DeltaMu(T, P)

print 'Cu2O decomposition temperature is {0:1.0f} K'.format(fsolve(func,
900)[0])

def func(T):
'Ag2O'
return -0.99 - 0.5 * DeltaMu(T, P)

print 'Ag2O decomposition temperature is {0:1.0f} K'.format(fsolve(func,
470)[0])

T = np.linspace(100, 1000)
# Here we plot delta mu as a function of temperature at different pressures
# you have use \\times to escape the first \ in pyplot
plt.plot(T, DeltaMu(T, 1e10*atm), label=r'1$\times 10^{10}$ atm')
plt.plot(T, DeltaMu(T, 1e5*atm), label=r'1$\times 10^5$ atm')
plt.plot(T, DeltaMu(T, 1*atm), label='1 atm')
plt.plot(T, DeltaMu(T, 1e-5*atm), label=r'1$\times 10^{-5}$ atm')
plt.plot(T, DeltaMu(T, 1e-10*atm), label=r'1$\times 10^{-10}$ atm')

plt.xlabel('Temperature (K)')
plt.ylabel(r'$\Delta \mu_{O_2}(T,p)$ (eV)')
plt.legend(loc='best')
plt.savefig('images/O2-mu-diff-p.png')

Cu2O decomposition temperature is 917 K
Ag2O decomposition temperature is 478 K


Now, let us make a phase diagram that shows the boundary between silver oxide, and silver metal in P and T space.

import numpy as np
import matplotlib.pyplot as plt
from ase.units import *
from scipy.optimize import fsolve

K = 1. #not defined in ase.units!
atm = 101325*Pascal

# Shomate parameters valid from 100-700K
A = 31.32234; B = -20.23531; C = 57.86644
D = -36.50624; E = -0.007374; F = -8.903471
G = 246.7945; H = 0.0

def entropy(T):
'''entropy returned as eV/K
T in K
'''
t = T/1000.
s = (A*np.log(t) + B*t + C*(t**2)/2.
+ D*(t**3)/3. - E/(2.*t**2) + G)
return s*J/mol/K

def enthalpy(T):
''' H - H(298.15) returned as eV/molecule'''
t = T/1000.
h = (A*t + B*(t**2)/2. + C*(t**3)/3.
+ D*(t**4)/4. - E/t + F - H)
return h*kJ/mol

def DeltaMu(T, P):
'''
T in K
P in atm
'''
return enthalpy(T) - T * entropy(T) + kB * T * np.log(P / atm)

P = np.logspace(-11, 1, 10) * atm
T = []
for p in P:

def func(T):
return -0.99 - 0.5 * DeltaMu(T, p)
T.append(fsolve(func, 450)[0])

plt.semilogy(T, P / atm)
plt.xlabel('Temperature (K)')
plt.ylabel('Pressure (atm)')
plt.text(800, 1e-7, 'Ag')
plt.text(600, 1e-3, 'Ag$_2$O')
plt.savefig('images/Ag2O-decomposition.png')


This shows that at high temperature and low pO2 metallic silver is stable, but if the pO2 gets high enough, the oxide becomes thermodynamically favorable. Here is another way to look at it.

import numpy as np
import matplotlib.pyplot as plt
from ase.units import *

K = 1.  # not defined in ase.units!
atm = 101325*Pascal
Hf = -0.99

P = 1 * atm

Dmu = np.linspace(-4, 0)

Hf = -0.99 - 0.5*Dmu

plt.plot(Dmu, Hf, label='Ag$_2$O')
plt.plot(Dmu, np.zeros(Hf.shape), label='Ag')
plt.xlabel(r'$\Delta \mu_{O_2}$ (eV)')
plt.ylabel('$H_f$ (eV)')
plt.savefig('images/atomistic-thermo-hf-mu.png')


This graph shows graphically the $$\Delta \mu_{O_2}$$ required to make the metal more stable than the oxide. Anything less than about -2 eV will have the metal more stable. That can be achieved by any one of the following combinations (graphically estimated from Figure fig:mu-o2): About 500K at 1$$\times 10^{-10}$$ atm, 600K at 1$$\times 10^{-5}$$ atm, 900K at 1atm, etc…

We now consider the question: Given a pressure and temperature, what coverage would you expect on a surface? We saw earlier that adsorption energies depend on the site and coverage. We lso know the coverage depends on the pressure and temperature. Above some temperature, desorption occurs, and below some pressure adsorption will not be favorable. We seek to develop a quantitative method to determine those conditions.

We redefine the adsorption energy as:

$$\Delta G_{ads} \approx E_{slab, ads} - E_{slab} - \mu_{ads}$$

where again we neglect all contributions to the free energy of the slabs from vibrational energy and entropy, as well as configurational entropy if that is relevant. That leaves only the pressure and temperature dependence of the adsorbate, which we treat in the ideal gas limit.

We expand $$\mu_{ads}$$ as $$E_{ads}+\Delta \mu(T,p)$$, and thus:

$$\Delta G_{ads} \approx E_{slab, ads} - E_{slab} - E_{ads} -\Delta \mu(T,p)$$

or

$$\Delta G_{ads} \approx \Delta H_{ads} -\Delta \mu(T,p)$$

where $$\Delta H_{ads}$$ is the adsorption energy we defined earlier. Now we can examine the effect of $$\Delta \mu(T,p)$$ on the adsorption energies. We will use the adsorption energies for the oxygen on Pt(111) system we computed earlier:

Table 5: Adsorption site dependence of adsorption energies of oxygen on Pt(111).
system $$\Delta H (eV/O)$$
fcc (0.25 ML) -1.04
hcp (0.25 ML) -0.60
bridge (0.25 ML) -0.49
fcc(1ML) -0.10
import numpy as np
import matplotlib.pyplot as plt

fcc25 = -1.04
hcp25 = -0.60
bridge25 = -0.49
fcc1 = -0.10

Dmu = np.linspace(-4, 2)

plt.plot(Dmu, np.zeros(Dmu.shape), label='Pt(111)')
plt.plot(Dmu, 0.25 * (fcc25 - 0.5*Dmu), label='fcc - 0.25 ML')
plt.plot(Dmu, 0.25 * (hcp25 - 0.5*Dmu), label='hcp - 0.25 ML')
plt.plot(Dmu, 0.25 * (bridge25 - 0.5*Dmu), label='bridge - 0.25 ML')
plt.plot(Dmu, 1.0 * (fcc1 - 0.5*Dmu), label='fcc - 1.0 ML')

plt.xlabel(r'$\Delta \mu O_2$ (eV)')
plt.ylabel(r'$\Delta G_{ads}$ (eV/O)')
plt.legend(loc='best')


### 6.3 Atomistic therodynamics and multiple reactions

In Inoglu2009188 we considered multiple reactions in an atomistic thermodynamic framework. Let us consider these three reactions of dissociative adsorption of hydrogen and hydrogen disulfide, and consider how to compute the reaction energy for the third reaction.

1. $$H_2 + 2* \leftrightharpoons 2H*$$
2. $$H_2S + 2* \leftrightharpoons H* + SH*$$
3. $$SH* + * \leftrightharpoons S* + H*$$

The reaction energy of interest is $$E_{rxn} = \mu_{S*} + \mu{H*} - \mu{SH*}$$ The question is, what are these chemical potentials? We would like them in terms of pressures and temperature, preferrably of molecules that can be approximated as ideal gases. By equilibrium arguments we can say that $$\mu_{H*} = \frac{1}{2} \mu_{H_2}$$. It follows that at equilibrium:

$$\mu_{H*} + \mu_{SH*} = \mu_{H_2S}$$ and $$\mu_{H*} + \mu_{S*} = \mu_{SH*}$$.

From the first equation we have:

$$\mu_{SH*} = \mu_{H_2S} - \frac{1}{2}\mu_{H_2}$$

and from the second equation we have:

$$\mu_{S*} = \mu_{SH*} - \mu_{H*} = \mu_{H_2S} - \mu_{H_2}$$.

Thus, the chemical potentials of all these three adsorbed species depend on the chemical potentials of two gas-phase species. The chemical potentials of each of these gases can be defined as:

$$\mu_{gas}(T,p) = E_{gas}(0K) + \delta \mu + kT\ln\left (p/p^0\right )$$, as we have defined before, so that only simple DFT calculations are needed to estimate them.

## 7 Advanced electronic structure methods

### 7.1 DFT+U

It can be difficult to find the lowest energy solutions with DFT+U. Some strategies for improving this are discussed in PhysRevB.82.195128.

#### 7.1.1 Metal oxide oxidation energies with DFT+U

We will reconsider here the reaction (see Metal oxide oxidation energies) 2 Cu2O + O2 $$\rightleftharpoons$$ 4 CuO. We need to compute the energy of each species, now with DFT+U. In PhysRevB.73.195107 they use a U parameter of 4 eV for Cu which gave the best agreement with the experimental value. We will also try that.

##### 7.1.1.1 Cu2O calculation with U=4.0
from vasp import Vasp
from ase import Atom, Atoms
import logging

calc = Vasp('bulk/Cu2O')
calc.clone('bulk/Cu2O-U=4.0')

calc.set(ldau=True,   # turn DFT+U on
ldautype=2,  # select simplified rotationally invariant option
ldau_luj={'Cu':{'L':2,  'U':4.0, 'J':0.0},
'O':{'L':-1, 'U':0.0, 'J':0.0}},
ldauprint=1,
ibrion=-1,  #do not rerelax
nsw=0)
atoms = calc.get_atoms()

print(atoms.get_potential_energy())
#print calc

-22.32504781

grep -A 3 "LDA+U is selected, type is set to LDAUTYPE" bulk/Cu2O-U=4.0/OUTCAR

LDA+U is selected, type is set to LDAUTYPE =  2
angular momentum for each species LDAUL =     2   -1
U (eV)           for each species LDAUU =   4.0  0.0
J (eV)           for each species LDAUJ =   0.0  0.0

##### 7.1.1.2 CuO calculation with U=4.0
from vasp import Vasp
from ase import Atom, Atoms

calc = Vasp('bulk/CuO')
calc.clone('bulk/CuO-U=4.0')

calc.set(ldau=True,   # turn DFT+U on
ldautype=2,  # select simplified rotationally invariant option
ldau_luj={'Cu':{'L':2,  'U':4.0, 'J':0.0},
'O':{'L':-1, 'U':0.0, 'J':0.0}},
ldauprint=1,
ibrion=-1,  #do not rerelax
nsw=0)

atoms = calc.get_atoms()
print(atoms.get_potential_energy())

-16.91708676

##### 7.1.1.3TODO Reaction energy calculation with DFT+U
from vasp import Vasp

calc = Vasp('bulk/Cu2O-U=4.0')
atoms = calc.get_atoms()
cu2o_energy = atoms.get_potential_energy() / (len(atoms) / 3)

calc = Vasp('bulk/CuO-U=4.0')
atoms = calc.get_atoms()
cuo_energy = atoms.get_potential_energy() / (len(atoms) / 2)

# make sure to use the same cutoff energy for the O2 molecule!
calc = Vasp('molecules/O2-sp-triplet-400')
o2_energy = calc.results['energy']

calc.stop_if(None in [cu2o_energy, cuo_energy, o2_energy])

# don't forget to normalize your total energy to a formula unit. Cu2O
# has 3 atoms, so the number of formula units in an atoms is
# len(atoms)/3.

rxn_energy = 4.0 * cuo_energy - o2_energy - 2.0 * cu2o_energy

print('Reaction energy  = {0} eV'.format(rxn_energy))
print('Corrected energy = {0} eV'.format(rxn_energy - 1.36))

Reaction energy  = 7.36775847 eV
Corrected energy = 6.00775847 eV


This is still not in quantitative agreement with the result in PhysRevB.73.195107, which at U=4 eV is about -3.14 eV (estimated from a graph). We have not applied the O$$_2$$ correction here yet. In that paper, they apply a constant shift of -1.36 eV per O$$_2$$. After we apply that correction, we agree within 0.12 eV, which is pretty good considering we have not checked for convergence.

##### 7.1.1.4 How much does U affect the reaction energy?

It is reasonable to consider how sensitive our results are to the U parameter. We do that here.

from vasp import Vasp
for U in [2.0, 4.0, 6.0]:
## Cu2O ########################################
calc = Vasp('bulk/Cu2O')
calc.clone('bulk/Cu2O-U={0}'.format(U))

calc.set(ldau=True,   # turn DFT+U on
ldautype=2,  # select simplified rotationally invariant option
ldau_luj={'Cu':{'L':2,  'U':U, 'J':0.0},
'O':{'L':-1, 'U':0.0, 'J':0.0}},
ldauprint=1,
ibrion=-1,  # do not rerelax
nsw=0)
atoms1 = calc.get_atoms()
cu2o_energy = atoms1.get_potential_energy() / (len(atoms1) / 3)

## CuO ########################################
calc = Vasp('bulk/CuO')
calc.clone('bulk/CuO-U={0}'.format(U))

calc.set(ldau=True,   # turn DFT+U on
ldautype=2,  # select simplified rotationally invariant option
ldau_luj={'Cu':{'L':2,  'U':U, 'J':0.0},
'O':{'L':-1, 'U':0.0, 'J':0.0}},
ldauprint=1,
ibrion=-1,  # do not rerelax
nsw=0)
atoms2 = calc.get_atoms()
cuo_energy = atoms2.get_potential_energy() / (len(atoms2) / 2)

## O2 ########################################
# make sure to use the same cutoff energy for the O2 molecule!
calc = Vasp('molecules/O2-sp-triplet-400')
atoms = calc.get_atoms()
o2_energy = atoms.get_potential_energy()

if not None in [cu2o_energy, cuo_energy, o2_energy]:

rxn_energy = (4.0 * cuo_energy
- o2_energy
- 2.0 * cu2o_energy)

print 'U = {0}  reaction energy = {1}'.format(U, rxn_energy - 1.99)

U = 2.0  reaction energy = 3.32752349
U = 4.0  reaction energy = 5.37775847
U = 6.0  reaction energy = 5.71849513

U = 2.0  reaction energy = -3.876906
U = 4.0  reaction energy = -3.653819
U = 6.0  reaction energy = -3.397605


In PhysRevB.73.195107, the difference in reaction energy from U=2 eV to U=4 eV was about 0.5 eV (estimated from graph). Here we see a range of 0.48 eV from U=2 eV to U=4 eV. Note that for U=0 eV, we had a (corrected reaction energy of -3.96 eV). Overall, the effect of adding U decreases this reaction energy.

This example highlights the challenge of using an approach like DFT+U. On one hand, U has a clear effect of changing the reaction energy. On the other hand, so does the correction factor for the O$$_2$$ binding energy. In PhysRevB.73.195107 the authors tried to get the O$$_2$$ binding energy correction from oxide calculations where U is not important, so that it is decoupled from the non-cancelling errors that U fixes. See PhysRevB.84.045115 for additional discussion of how to mix GGA and GGA+U results.

In any case, you should be careful to use well converged results to avoid compensating for convergence errors with U.

### 7.2 Hybrid functionals

#### 7.2.1 FCC Ni DOS

This example is adapted from http://cms.mpi.univie.ac.at/wiki/index.php/FccNi_DOS HSE06

from vasp import Vasp
from ase.lattice.cubic import FaceCenteredCubic
from ase.dft import DOS

atoms = FaceCenteredCubic(directions=[[0, 1, 1],
[1, 0, 1],
[1, 1, 0]],
size=(1, 1, 1),
symbol='Ni')
atoms[0].magmom = 1

calc = Vasp('bulk/Ni-PBE',
ismear=-5,
kpts=[5, 5, 5],
xc='PBE',
ispin=2,
lorbit=11,
lwave=True, lcharg=True,  # store for reuse
atoms=atoms)

e = atoms.get_potential_energy()
print('PBE energy:   ',e)
calc.stop_if(e is None)

dos = DOS(calc, width=0.2)
e_pbe = dos.get_energies()
d_pbe = dos.get_dos()

calc.clone('bulk/Ni-PBE0')
calc.set(xc='pbe0')
atoms = calc.get_atoms()
pbe0_e = atoms.get_potential_energy()

if atoms.get_potential_energy() is not None:
dos = DOS(calc, width=0.2)
e_pbe0 = dos.get_energies()
d_pbe0 = dos.get_dos()

## HSE06
calc = Vasp('bulk/Ni-PBE')
calc.clone('bulk/Ni-HSE06')

calc.set(xc='hse06')
atoms = calc.get_atoms()
hse06_e = atoms.get_potential_energy()
if hse06_e is not None:
dos = DOS(calc, width=0.2)
e_hse06 = dos.get_energies()
d_hse06 = dos.get_dos()

calc.stop_if(None in [e, pbe0_e, hse06_e])

import pylab as plt
plt.plot(e_pbe, d_pbe, label='PBE')
plt.plot(e_pbe0, d_pbe0, label='PBE0')
plt.plot(e_hse06, d_hse06, label='HSE06')
plt.xlabel('energy [eV]')
plt.ylabel('DOS')
plt.legend()
plt.savefig('images/ni-dos-pbe-pbe0-hse06.png')


### 7.3 van der Waals forces

Older versions (5.2.11+) implement DFT+D2 JCC-JCC20495 with the LVDW tag.

The vdW-DF klimes-2011-van-waals is accessed with LUSE_VDW. See http://cms.mpi.univie.ac.at/vasp/vasp/vdW_DF_functional_Langreth_Lundqvist_et_al.html for notes on its usage.

In Vasp 5.3+, the IVDW tag turns van der Waal calculations on.

You should review the links below before using these

IVDW method
0 no correction
1 or 10 DFT-D2 method of Grimme (available as of VASP.5.2.11)
11 zero damping DFT-D3 method of Grimme (available as of VASP.5.3.4)
12 DFT-D3 method with Becke-Jonson damping (available as of VASP.5.3.4)
2 or 20 Tkatchenko-Scheffler method tkatchenko-2009-accur-molec (available as of VASP.5.3.3)

Van der Waal forces can play a considerable role in binding of aromatic molecules to metal surfaces (ref). Here we consider the effects of these forces on the adsorption energy of benzene on an Au(111) surface.First, we consider the regular PBE functional.

#### 7.3.1 PBE

##### 7.3.1.1 gas-phase benzene
from vasp import Vasp
from ase.structure import molecule

benzene = molecule('C6H6')
benzene.center(vacuum=5)

print(Vasp('molecules/benzene-pbe',
xc='PBE',
encut=350,
kpts=[1, 1, 1],
ibrion=1,
nsw=100,
atoms=benzene).potential_energy)

-76.03718564

##### 7.3.1.2 clean slab
# the clean gold slab
from vasp import Vasp
from ase.constraints import FixAtoms

atoms = fcc111('Au', size=(3,3,3), vacuum=10)

# now we constrain the slab
c = FixAtoms(mask=[atom.symbol=='Au' for atom in atoms])
atoms.set_constraint(c)

#from ase.visualize import view; view(atoms)

print(Vasp('surfaces/Au-pbe',
xc='PBE',
encut=350,
kpts=[4, 4, 1],
ibrion=1,
nsw=100,
atoms=atoms).potential_energy)

-81.22521492

##### 7.3.1.3 benzene on Au(111)
# Benzene on the slab
from vasp import Vasp
from ase.structure import molecule
from ase.constraints import FixAtoms

atoms = fcc111('Au', size=(3,3,3), vacuum=10)
benzene = molecule('C6H6')
benzene.translate(-benzene.get_center_of_mass())

# I want the benzene centered on the position in the middle of atoms
# 20, 22, 23 and 25
p = (atoms.positions[20] +
atoms.positions[22] +
atoms.positions[23] +
atoms.positions[25])/4.0 + [0.0, 0.0, 3.05]

benzene.translate(p)
atoms += benzene

# now we constrain the slab
c = FixAtoms(mask=[atom.symbol=='Au' for atom in atoms])
atoms.set_constraint(c)

#from ase.visualize import view; view(atoms)

print(Vasp('surfaces/Au-benzene-pbe',
xc='PBE',
encut=350,
kpts=[4, 4, 1],
ibrion=1,
nsw=100,
atoms=atoms).potential_energy)

/home-research/jkitchin/dft-book/surfaces/Au-benzene-pbe submitted: 1413525.gilgamesh.cheme.cmu.edu
None


resubmitted

/home-research/jkitchin/dft-book/surfaces/Au-benzene-pbe submitted: 1399668.gilgamesh.cheme.cmu.edu
None

from vasp import Vasp

e1, e2, e3 = [Vasp(wd).potential_energy
for wd in ['surfaces/Au-benzene-pbe',
'surfaces/Au-pbe',