Molecular Dynamics and Density Functional Theory

What do we need?

• An account in “pemfc” cluster:

Host name: pemfc.chem.sfu.ca

I will take care of that. This can be usually a common account for all of you but

please at the beginning of your work every group makes its own directory.

• pemfc cluster is a linux based cluster, therefore you should

know some linux commands. I will give you a list of important linux commands later.

• The following packages:

1. “putty.exe” a open shell program for SSH under the windows.

With that you can connect to pemfc cluster and run the programs/commands

of the cluster.2. “psftp.exe” a open shell program for ftp to transfer your files from your

computer to the pemfc cluster and vise versa.

you can download them from the following address (they are free):

http://www.chiark.greenend.org.uk/~sgtatham/putty/download.html

3. “rasmol” it is a graphic interface packege in which you can see the position of the atoms in your system

you can download it from the following address:

http://www.bernstein-plus-sons.com/software/rasmol/

(please download the latest version under the windows

The aim of this project

We want to calculate the first or possibility both of the following items:

• Hydrogen or CO adsorption energy.

• Molecular dynamics simulation of a hydrogen atom or a CO molecule

adsorbed on a Pd slab.

Hydrogen and CO adsorption energies

In order to calculate the CO adsorption energy on a Pd slab we have

to do:

1. Calculate the total energy of a Pd slab. The total energy

means all interaction energies between all the Pd-Pd atoms. We call it EslabNote that each atom has proton + electrons and the electrons can interact with each other (What is the sign of the total energy?) Fig.1 shows a

slab of Pd(111) with 4 layers of Pd.

Fig.1 shows a slab of Pd(111) with 4 layers of Pd.

When we say Pd(111) that means a Pd bulk

which is cut along the surface (111). We have

infinite number of Pd atoms along x and y

direction (in the picture it shows only part of that)

2. We have to calculate the total energy of a CO molecule in a gas phase (in the vacuum

and in the absence of everything). It is called ECO.

3. We have to calculate the total energy of a slab of Pd when CO molecule is

adsorbed on that (Eslab+CO). That means the interaction of all Pd atoms with each other

including CO molecule. Fig.2 shows a picture of the Pd slab when CO is adsorbed on top of it.

Fig.2 CO adsorption on Pd(111) slab. The

total energy contains all the interactions

including the interaction with CO.

Hence the CO adsorption energy Eads (or CO-Pd binding energy) can be obtained by:

Eads= Eslab+CO – Eslab - ECO

Now the question is how we can calculate the total energy of for example

a slab of Pd?

Ab-initio total energy calculation of a system of nucleons and electrons

Fig.3 System of nucleons (red) and

electrons (gray). The nucleons

consider as classical particles

while electrons consider as

quantum particles

When we say ab-initio calculation this means calculations in which we need no

experimental data for that as an input, i.e. atomic radius, binding energies, ionization

energies....

In order to calculate the total energy of a system of nucleons and electrons (see Fig.3

for example) we have to consider two facts:

1. In principal electrons and nucleons are very small therefore we have to consider them

as quantum particles. When we say quantum particles that means that at

time t=t0 we can not say exactly where they are and what energy they have.We have to consider them as a wave packet or wavefunction.

Since in our case nucleons are Pd and they are much bigger compare to electrons we

approximately consider the nucleons as classical particles.

Electrons -> quantum particles -> Wavefunction -> Charge density n(r) -> Schrödinger eq.

Nucleons -> classical particles -> Point charge -> Newton law

2. “Born-Oppenheimer approximation: is based on smallness of the electron mass as

compared to the nuclear mass. Within this approximation when we want to solve

the Schrödinger Eq. for electrons the nuclei are regarded as a fixed charges.

where Ψ

is the total wave function of the system and

HΨ

= E Ψ

H = T + Ve-e + Vext

is the Hamiltonian. Ve-e is electron-electron coulomb interaction and it is given by:

Fig. 4 The electronic charge

density n(r) of a system of electrons.

The charge density is larger in

the darker region. Vext is a source

of an external potential i.e. a proton.

∫ −=−

'rr

'dr)'r(n]n[V ee

xy

z

n(r’)r’

r

r-r’

Charge density of electrons n(r)

Vext

Vext

Vext

Vext

Schrödinger equation for system of electrons

Suppose that we would like to calculate the total energy of a system.

First we have to calculate the Schrödinger equation for electrons and then

determine the total energy of electrons Eel and then calculate the energy of the ions EionThe total energy will be given by: Etot = Eel + Eion

∑= −

=N

1i i

ext|Rr|

z]n[V

xy

z

zRi

r

r-Ri

T is the kinetic energy operator of the interacting system. The bad news is that

there no analytical form for T and Ψ

therefore there is no solution for

Schrödinger equation.

and Vext is the interaction with an external potential. In our case the external

potential is just electron-proton interaction:

Ri is the position of ions.

Based on the DFT and these two theories the many body problem of fully

interacting particles in an external potential Vext (or potential from the protons) is

simply replaced by a system of non-interacting particles in a effective potential Veff giving the same ground state charge density.

H = T + Ve-e+ Vext Hs= Ts + Veff

Interacting particles (electrons) Non interacting particles

Veff = Ve-e + Vext + Vxc

where Hs is the Hamiltonian of a non interacting system and Ts is now the

kinetic operator of a non-interacting system with N particles which can be

easily given by:

2

2N

1i

2

sdr

d

m2T ∑

=

−= h

The effective potential Veff is:

It has one extra term which contains all the energy contribution which were not

taking into account in the transition from the interacting system to the non-interacting system. Vxc is called exchange-correlation potential which introduce

quantum mechanical many-body electronic effects into the model and a portion

of the kinetic energy which is needed to correct Ts to obtain the true kinetic operator

of a real system T.

Density Functional Theory (DFT)

Theory1: External potential is a unique functional of the density n(r). That means

if we have a system of electrons, the ground state density n0(r)

corresponding to an external potential (like the potential from the protons) Vext can not be reproduced using any other potential V’ext.

This theory was introduced in 1964 by Hohenberg and Kohn*. It is based on two

mathematical theories:

* W. Kohn and L. J. Sham. Phys. Rev., 140: A1133, 1995

* P. Hohenberg, W. Kohn. Phys. Rev., 136: B864, 1964

Theory 2: The correct ground state density n0(r) minimizes the total energy

functional.

DFT

The exchange-correlation potential can be expressed as:

dn

]n[dE]n[V xcxc =

where Exc[n] is the exchange-corralation energy.

The above DFT is formally exact, but as such it is unless in practical applications

because all the difficulties related to the exchange correlation energy are still unsolved.

To proceed further it is necessary to find an approximation for the exchange correlation energy. The most common approach is the Local Density Approximation LDA which

describes quite well a large number of systems and has been successfully applied to

some ab-initio calculations. Within this approximation the exchange correlation energy

is given by:

dr)r(n)n(]n[E xcxc ∫ ε=

where ε xc(n) is the energy of a homogeneous electron gas and it is known. Thereforethe Exc[n] and hence Vxc[n] are all known.

This approximation has been shown capable of dealing on the same ground with atoms, molecules, clusters, surfaces and interfaces which have been successfully reproduced. Nevertheless, besides these successes there are also some drawbacks of the approximation. For instance, the cohesive energies of solids are systematically overestimates, while lattice constants are systematically underestimated Errors in the structural properties are usually small for crystals with covalent or metallic bonds, but it is well known that the hydrogen bond cannot be described accurately within LDA In the field of metals, the ground state structure of crystalline iron is predicted to be paramagnetic fcc, instead of a ferromagnetic bcc. Various approximations have been introduced in the course of the years to improve LDA. The generalized gradient approximation(GGA) is one of those approximations which is more or less commonly accepted to be an improvement over LDA. Another approach to obtain a more accurate results is to construct a functional which depends not only on charge density n(r) but also on the magnitude of the gradient of the charge density . This approach is called Generalized Gradient Approximation GGA.

|)n(| n∇

The choice of the functional form for the GGA is not unique and many differentfunctional have been proposed. Among the many types of functional, in this workwe used Perdew-Wang* 1991 functional (PW91).

John P. Perdew and et.al, phys., Rev. B, 46:6671, 1992

Implementation

What we have done so far was to express the effective potential Veff in terms of Ve-e[n],

Vext[n] and Vxc[n]:

Veff[n] = Ve-e[n] + Vext[n] + Vxc[n]

In order to calculate to total energy of the system of electrons we need to solve

the following Schrödinger equation.

HsΨ

= EelΨ →

(Ts + Ve-e[n] + Vext[n] + Vxc[n]) Ψ

= EelΨ

Since the particles in our system does not interact with each other the total wave

function can be written as the super position of the one particle wave function andtherefore the above equation can be given as a set of single-particle equations known

as one-particle Kohn-Sham equations.

(Ts + Ve-e[n] + Vext[n] + Vxc[n]) φ i = ε i φ i , i=1,….,Nwhere N in the total number of particles in the system, therefore we need to solve

N times of this equation. Note that in order to solve this equation we need to knowthe charge density of the system n(r). But on the other hand the charge density

of the system is given by:

∑=

ϕ=N

1i

2i ||)r(n

which depends on φ i. The effective potential depends on the density, the density depends on Kohn-Sham orbitals (φ i) and Kohn-Sham orbitals depends on the effective potential!!

This set of equations form a loop which has to be solved self consistently since the

Hamiltonian is a functional fo wavefunctions and wavefunctions

themselves are the solution of the Hamiltonian: One starts i.e. with the approximate density, constructs the effective potential and solve the Kohn-Sham equation to obtain

the orbitals which are then used to calculate the new density and new effective

potential and new orbitals etc. This procedure is continued until the density, orbitals

and hence total energy of electrons no longer change from previous interaction.

The result of solving the above self consistent equation are the total

energy of the electrons:

∑ε=i

ielE

and Kohn-Sham orbitals φ i and charge density n(r) (see flowchart 1)

Trial charge density n(r)

Set up Hamiltonian

H=Ts+Ve-e+Vext+Vxc

calculate the wavefunctions

and the total energy E from

one-particle Shrodinger Eq.

∑=

ϕ=N

1i

2newi ||)r(n

E,newiϕ

newiϕ

5i

newi 10||

−

Basis set-Plane wave

In this section we are going to briefly explain how we can solve the Kohn-Sham equation

and obtain the wavefunctions and energies. If we multiply both sides of the Kohn-Sham

Eq.? by φ *i(r) and take an integral from both sides we gets:

)r.ikexp(.C)r(1k

k,ii ∑∞

=

=ϕ

One way to solve the Kohn-Sham equation is to expand the wavefunction φ i in a plane wave basis set. Using plane wave as a basis set has some advantages and some disadvantage. The advantage of using the plane wave is that it which help us to simplify the above Kohn-Sham equation.

( ) dr)r()r(dr)r(]n[V]n[V]n[V)r(dr).r(dr

d).r(

m2i

*iiixcextee

*ii2

2*i

2

∫∫∫ ϕϕε=ϕ++ϕ+ϕϕ−

−h

where mathematically the sum is over a infinite number of plane wave but of course in practice we see that it already converges for a large number of plane wave. By substituting this equation into the above Shrodinger equation and using thefollowing mathematical principal:

( ) ( ) 'kkdrr'ikexp.rikexp δ=⋅⋅∫

We end up with a matrix with Nmax x Nmax elements where Nmax is the number ofthe plane waves.

Nmax depends very much to the size of your system. i.e. number of atoms in your system and also the elements you use. it can be as big as 1e+5 number of plane waves. In order to get φ i and ε i from the matrix we need to diagonal it which of course needs a powerful computer or even several computers working as parallel.

Reminding: When φ i and ε i are calculated in the next iteration a new charge density should be constructed and again a new matrix should be diagonalized. Usually 10-30 iterations in needed to get the total energy of the system self consistently.

Total energy of the system

The total energy of the system will be then the sum over the total energy of electrons

and protons:

Etot = Eel + Eion

where Eion is just the culomb interaction between ions

∑≠ −

=ji,j,i

2ji

2

ion|RR|

ZE

where Ri and Rj are the position of the ions.

After we calculated the total energy we can then calculate the force on each ion in thesystem

i

toti

dR

dEF =

Now if you want to perform a geometry optimization the ions will move along the force

Fi in the next ionic interactions until we approach to a situation where there is no

significant force on any ions. In this case you have found the relaxed position for ions. In this point the total energy of the last ionic interaction is what you need for

calculating the adsorption energy. Note the you have two types of loop (iteration):

inner loop or electronic iteraction and outer loop or ionic iteration. See flowchart 2.

What is the difference between Molecular dynamics and geometry optimization?

When you perform the geometry optimization in every ionic iteration the initial velocity

of ions are zero and therefore when you move the ions along the forces you always go

along the more minimum energy and finally you get to global minimum energy. It is like moving a spring toward to its equilibrium position very slowly, so in equilibrium point (or global

minimum energy) there is no force on ions and since you have to initial velocity you

stop there. When you perfome a molecular dynamics in every ionic iteration

you consider the initial velocity V0≠0 for the ions and therefore when you move the ion along

the force you will approach to the minimum energy (or equilibrum) but you will pass this pointin the next iterations due to having initial velocity (it is like a pendulum where the solid

oscillate around its minimum

Trial charge density n(r)

Set up Hamiltonian

H=Ts+Ve-e+Vext+Vxc

calculate the wavefunctions

and the total energy E from

one-particle Shrodinger Eq.

∑=

ϕ=N

1i

2newi ||)r(n

E,newiϕ

newiϕ

5i

newi 10||

−

Geometry optimization

Fion Fion

Next ionic

iteration

Next ionic

iteration

In each iteration the initial velocity is zero (T=0K)

Molecular dynamics

Fion FionNext ionic

iteration i=2

Next ionic

iteration i=3

In each iteration the initial velocity is not zero. It is calculated accordingto the temperature of the system

First configuration

Fion FionFion

i=4i=5i=6

Problem with Kohn-Sham equations:

The one-particle Kohn-Sham equations can be solved for an isolated

system with maximum few hundred number of particles. In a crystal or a slab however we are dealing with a order of 1023 number of atoms. Simulating

such an amount of particles is a impossible task. Therefore a simplification needs

to be done. If the system is fully periodic a appropriate choice would be to

use periodic boundary conditions. This can be done by using the

definition of a Bravais lattice.

Bravais lattices:

A Bravais lattice is a fundamental concept in the description of any crystalline solid,

which specifies the periodic array in which the repeated units of crystal are arranged. A Bravais lattice is an infinite array of discrete points with an arrangement and

orientation that appears exactly the same on all Bravais lattice points. It consist

of all points with position R of the form:

R = n1 a1 + n2 a2 + n3 a3

where a1, a2 and a3 are any three vectors not all in the same plane and n1, n2 and n3 are all integer values. Therefore if there is an atom or a specific electronic structure at position r the same atom or structure is repeated along threebasis vectors a1,2,3 (see Fig1 for two dimension). The calculation saving

comes since only the atoms and electrons inside the unit cell

of the calculation - called super-cell and shown in the Fig 1 with thick solid

lines - needs to be explicitly considered.

Fig 1. replication of super-cell c along x, y and z direction.

Thre are 6 atoms of type 1 and four atoms of type 2 exist in the super cell.

The super cell and the Bravais lattice vectors a1 and a2 in Fig. 2 are as following:

a1

a2

x

y

a

a1 = a i = (1 0 0) aa2 = a j = (0 1 0) a

There are two atoms in the super cell. The position

of these two atoms are (000) and (½ ½ 0). Note thatwe still consider a two dimension Bravais lattice

therefore the position of atoms along

z-direction are zero.

a1a2

x

y

Primitive unit cell: is a super cell in which contains only one atom. The Bravais lattice of a primitive super cell (or unit cell) of fig. 2 is given by:

Fig. 2

( ) ( )( )( )

+=+= j

2

2i

2

2a4545a oo sincosa1

( ) ( )( )( )

+−=+−= j

2

2i

2

2a4545a2

oo sincosa

Fig.3

Now with that overview we are going to three-dimension Bravais lattice. Lets assume

we have fcc structure which is cut along surface(100)..

First layer (a3=0)

a1

a2

a3

a1

a2

(0 0 0)

(½ ½ ½ )

If we consider the following Bravais

lattice vectors:

a1 = a i

a2 = a j (Equation. A)

a3 = a k

then the position of the atoms

in the first layer (where a3=0)will be:

Atom 1: (0 0 0)

Atom 2: ( ½ ½ ½ )

Fig.4

And according to Fig.4 the position of the atoms in second layer (where a3= ½ ) will be:

Atom1: ( ½ 0 ½ )

Atom2: (0 ½ ½ ) Second layer (a3= ½ a)

(0 ½ ½ )

(½ 0 ½ )

And since we have indicated a3 = a j therefore the 3rd layer (where a3 = a)

is the same as first layer (where a3 = 0). This is in agreement with Fig.4.

Exercise 1: Consider the same surface (100) and try to indicate the position of

the atoms and Bravais lattice vectors for a primitive unit cell. This result will bevery useful for your project.

Exercise 2: Do the same as exercise 1 for surface (110).

Exersice 3: Try to specify the Bravais lattice vectors for a primitive unit cell of

a (111) surface.

By considering the Barvais lattice vectors equation A we will have a big problem.

The problem is that they do not represent any surface. They just show a fcc bulk(see Fig.5)! However in order to show a surface we should be able to break the

symmetry along the z axes! How can we do that?

Top-view

a1

a3a2

Side view

1st layer (blue)

2nd layer (red)

+∞

-∞

a3

Fig. 5