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3a. Lattice Dynamics3a. Lattice Dynamics

� Introduction

� Lattice Vibrations of 1D crystals

� Monoatomic chain

� Diatomic chain

� Periodic boundary conditions

� Lattice Vibrations of 3D crystals

G. Bracco-Material Science SERP CHEM 1

Lattice Dynamics


In previous lectures we have assumed that the atoms were at rest at their

equilibrium position. This can not be entirely correct

→ Atoms vibrate about their equilibrium position even at absolute zero!

The energy they possess at T=0 is known as zero point energy.

The amplitude of the motion increases as the atoms gain more thermal energy

at higher temperatures.

In this chapter we discuss the nature of atomic motions, referred to as lattice

vibrations.

In crystal dynamics we will use the harmonic approximation → amplitude of

the lattice vibration is small.

At higher temperature some anharmonic effects occur.

Lattice Dynamics

• The atomic position is a lattice position only in average

• We exclude diffusion events: the atom vibrates always around the samelattice point

• Classical mechanics will be initially employed to describe the atomicmotion

• Motion of a system of atoms connected by springs


Lattice vibrations of 1D crystal

Chain of identical atoms

( )2 2

2( ) ( ) ...........

2r a

r a d VV r V a

dr =

− = + +

rR

V(R)

0 r0

Repulsive

Attractivemin

In equilibrium position first derivative is zero. This equation looks like as the potential energy associated of a spring with a spring constant :


Atoms interact with a potential V(r), for small vibrational amplitude it can

be expanded in Taylor’s series around the equilibrium position r=a.

Harmonic approximation: retain terms up to the quadratic one.

� � ��

��

Force= ��

Monoatomic Chain

• The simplest crystal is the one dimensional chain of identical atoms.

• Chain consists of a very large number of identical atoms with identical masses.

• Atoms are separated by a distance “a”, the lattice periodicity.

• Atoms move only in a direction parallel to the chain.

• As a further approximation: only nearest neighbours interact (short-range forces).

a a a a a a

Un-2 Un-1 Un Un+1 Un+2


If one expands the energy near the equilibrium point for the nth atom and use elastic approximation, Newton’s equation becomes

The force on the nth atom:• Force to the right F

r= �(�� − ��)

• Force to the leftFl= �(�� − ��)

• Total force = Fr- F

l

�� = �(�� − 2�� + ��)

a a

Un-1 Un Un+1

Eqn’s of motion of all atoms are of this form, only the value of

‘n’ varies


Monoatomic Chain

• All atoms oscillate with a same amplitude A and frequency ω. Then we can try

a solution

( ).

0expnn n

duu i A i kx t

dtω ω = = − −

( )0expn nu A i kx tω = −

( ) ( )2.. 2 2 0

2expn

n n

d uu i A i kx t

dtω ω = = −

..2

n nu uω= −

nax n =0nn unax +=Undisplaced

position

Displaced position


Monoatomic Chain

Cancel Common terms ��(��)


m�� = α(��-2��+��)

Monoatomic Chain

mω2��(��)

= α(��(��

� ��)-2��(��

�

��)+��

�(��)

0� = na, 0�� = (n + 1)a, 0�� = (n − 1)a

-mω2 =α(��-2+��)

With some algebra

ω(k)=α

sin(

��

�) → dispersion relation ωωωω(k)

Maximum value ωM=α

e e 2cosika ika ka−+ =

( ) 21 cos 2 sin2

xx

− =

• Dispersion relation: ω versus k relation


ω

k

π/a-π/a

(redundant)

Monoatomic Chain

-�

�� λ �

�

�

λ � ��

�λ �

�

�

• The waves with wave numbers k and k+2π/a describe the

same atomic displacement

• Therefore, we can restrict k to within the first BZ [-π/a, π/a]

• Function with periodicity 2π/a

�� 2�

�

v� � ��⁄ Phase velocity

Note that:

• In above equation n is cancelled out, this means that the eqn. of motion of allatoms leads to the same algebraic eqn. This shows that our trial function Un isindeed a solution of the eqn. of motion of n-th atom.

• We started from the eqn. of motion of N coupled harmonic oscillators (if oneatom starts vibrating it does not continue with constant amplitude, but transferenergy to the others)

• Our wavelike solutions are uncoupled oscillations called normal modes:each k has a definite ω given by above eqn. and oscillates independently of theother modes.

• the number of modes is expected to be the same as the number ofequations N.


Monoatomic Chain

� = 4�� sin �

2


Monoatomic Chain

For a finite chain of N atoms we have to

choose the boundary conditions:

Since the solution is a travelling wave, a

suitable condition is the

periodic boundary condition (PBC)

u0(t) = uN(t)

This conditions (generally in 3D) is called

Born–von Karman boundary condition

N=8

1 8

08

0

0

2, 1, 2

or 1,2 2

exp(

exp[ ( ,

)

)]

1n n

N

mk m N

N aN N

m

u A i kX t

u u ikNa

π

ω

∴ = =

= − +

== −= ⇒

L

L

The value of k is discrete

with spacing ∆k=2π/Na

Each k describes a normal

mode of the vibration

Pattern of vibration:

• k ～ 0, exp(ikXn) ～ 1.

Every atom move in unison→ Little restoring force.

• k ～ π/a,exp(ikXn) ～ (-1)n.

Adjacent atoms move in opposite directions→ Maximum restoring force.

( )( ) ,ni kX tn nu t Ae X naω−= =Displacement of the n-th atom

Velocities of wave (phase velocity, group velocity):

• k ～ 0, Linear dispersion ω = (ωM

a/2)k→ phase velocity = group velocity

This is the long wavelength condition for elastic sound wave

• k ～ π/a, group velocity ～ 0 → no transmission of energy

Bragg condition: in 1D, waves are reflected back (2θ=180°→θ=90°)

→ 2d sin(θ)=nλ → 2d=n(2π/k) at k= π/a→d=a, therefore at these points

Bragg condition is verified.

p

g

vkd

vdk

ω

ω

=

=


Monoatomic Chain


If k ∼0 →λ is very long and waves are not sensitive to

the lattice structure of the crystal → continuum

approximation of the material.

ka<<1 → sin(ka)≈ka →ω = (ωMax

a/2)k

Sound velocity Vs= (ω

Max a/2)=

�

�a

k

ωContinuum

Discrete

0

Monoatomic Chain

Or using macroscopic parameters Vs=α��

�⁄=

��

�, � � density Kb=bulk modulus

Longitudinal Waves

Transverse Waves

Remember that in 1D

Wave polarization

in 2D and 3D

Physical significance of wave numbers outside [-π/a, π/a]?

x

un

un

xa


Monoatomic Chain

But

�

�and +

�

�

�are not equivalent →

�

�- 2

�

�= -

�

�

�is equivalent →same pattern, ω, and velocity

ω

k

π/a-π/a

4λ=7a →λ=��

�→ ��

��/�=��

��

3λ=7a →λ=��

→ ��

��/=�

��

Equivalent vectors

�

�+ Gn =

�

�+ n

�

�n integer (-∞, …-2,-1,0,1,2,…,∞)

� there is only one possible propagation direction and one polarizationdirection → 1D crystal has only one sound velocity.

� In this calculation only nearest neighbor interaction although this is agood approximation for the inert-gas solids, its not a good assumption formany solids (depends on the range of interactions)

� Extending to a model beyond nearest neighbor interaction many of thefeatures in above calculation are preserved.

• Wave equation solution still satisfies.

• The detailed form of the dispersion relation is changed but ω is still periodic function of k with period 2π/a

• Group velocity vanishes at k=(±)π/a

• There are still N distinct normal modes

• Furthermore the motion at long wavelengths corresponds to sound waves


Monoatomic Chain

2�

�

number of modulations = number of further

neighbor interactions to be considered

Lattice with basis: diatomic chain

• Two different types of atoms of masses M1 and M2 are connected by

identical springs of spring constant α;

Vn-1Un Vn Un+1 Vn+1

α α α αM2 M2M1

M2M1a)

b)

(n-1) (n) (n) (n+1) (n+1) cells

a

• This is the simplest possible model of an ionic crystal.• Since a is the periodicity, the nearest neighbors separations is a/2


M2 M1 M2M1 M2

Vn-1Un Vn Un+1 Vn+1

Equation of motion for mass M1 (nth):mass x acceleration = restoring force

Equation of motion for mass M2 (nth):


Diatomic chain

We will consider only the first neighbour interaction although it is a poorapproximation in ionic crystals because electrostatic interaction is a long rangeinteraction.

M1� = � �� − � + � �� − � =� �� + �� − 2�

M2�� = � �� − �� + � � − �� =� �� + � − 2��

2

1 12

2

2 12

( 2 ),

( 2 ).

nn n n

nn n n

d uM v v u

dt

d vM u u v

dt

α

α

−

+

= + −

= + − a

vn-1

un

vn

un+1

Diatomic chain

22

21

2 2 cos( / 2)

2 cos( / 2) 2det 0.

M ka

ka M

α ω αα α ω

⇒

− − − −

=

2 22

1 2 1 2 1 2

1 1 1 1 4sin ( / 2).

ka

M M M M M Mω α α±

+ + −

⇒ = ±

212

221

2 2 cos( / 2)

2 cos( / 2) 2 0,

AM ka

Aka M

α ω αα α ω

− − − −

⇒ =The 2nd grade equation gives

two solutions with different

dispersion relations ω(k)

1( 1/2)

2

Assume ikna

n i t

ik n an

u A e

v A ee ω−

+

=

We need two non trivial independent

solutions that are travelling waves


Two branches of dispersion

curves: optical & acoustic

assume M2 > M1

Diatomic chain

Patterns of vibration:

similar

a

d

b

c

a

b

c

d



Diatomic chain

�2 =�(�

1+�

2)

�1�

2

1 ± 1 −�

1�

2)

2 �1+�

22 2�2

� 2�2

2(�1+�

2)

In long wavelength region (ka«1); sin(ka/2)≈ ka/2 in ω(k), using a Taylor expansion:

Optical modes: Atoms in the cell oscillate out of

phase (for k=0 the center of mass is at rest

M1un+M2vn=0)

��2 =

2�(�1+�

2)

�1�

2

��2 =

� 2�2

2(�1+�

2)

Acoustic modes: atoms in the cell oscillate with

a small phase difference.

The sound speed is

��= �

�

2(�1+�

2)

• The other limiting solutions of equation ω2 are for ka= π (k at BZ),

→ sin(ka/2)=1. In this case


Diatomic chain

��2 =

�(�1+�

2) ± �(�

2−�

1)

�1�

2

��2 =

2�

�1

��2 =

2�

�2

Optical modes: light atoms in the cell oscillate out

of phase with other light atoms, heavy atoms are at

rest

Acoustic modes: heavy atoms in the cell

oscillate out of phase with other heavy atoms,

light atoms are at rest

• The bandwidth ��

− ��

of both modes depends on α: in particular,decreasing the coupling α the curve for optical mode tends to become flat(dispersionless modes)

2a


a

If M1=M

2

• If M1<M

2

Change of Periodicity a →2a

Repeat the dispersion curve

with the new periodicity

• If M1<M

2

Opening of gaps

At the BZ boundary

Monotomic chain →Diatomic chain

k

in k

How many normal modes (k points) in each branch?

1( 1/2)

2

0

0

2, 1,2

or 1,2 2

Imposing PBC on

exp( ) 1

iknan i t

ik n an

N

N

u A e

v A e

u u

v v

mk m N

N aN N

m

ik

e

Na

ω

π

−+

∴ = =

= − +

=

= =⇒

L

L Same as before for the lattice with no basis

• The total number of k points is N, number of cells, the total Degrees

Of Freedom (DOF) is 2N equal to number of the atoms (this remains

true for complex crystals in higher dimensions)

Diatomic chain


Modes for the diatomic chain

Amplitude of vibration is strongly exaggerated!


Acoustic mode

Optical mode

• The acoustic branch has this name because it gives rise to long

wavelength vibrations which travel with the speed of sound and ω→0 fork→0. The 2 atoms generally vibrate with a small phase difference

• The optical branch is a higher energy vibration and they vibrate generally

in out of phase. If the two vibrating atoms are ions, a dipole is crated duringthe vibration that can interact with electromagnetic radiation giving opticalproperties to the crystal.


Diatomic chain

to understand the energy difference between acoustic & optical modes

Let’s examine the oscillation pattern for acoustic modes

k→0 (long wave length) shorter wavelength

The phase difference between neighbors is generally small

Instead for optical modes

The difference is always very close to the out of phase

The spring are less compress in acoustic mode → lower energy for acoustic

branches


Vibration of a 3D lattice

1x Longitudinal Waves

2x Transverse Waves

Wave polarization

In 3D the atomic displacement of the atom in a cell (l,m,n→R=la+mb+nc) has

3 components ��=(��, ��, ��) → R+u

The energy of interaction in harmonic approximation between atoms is given

by

V=�

∑ �� ′�� where ��

��

�� | !"

Equation of motion: M ��=�∑ �� ′��

assuming a sinusoidal solution , ��, �� #∙��%�, � polarization vector

the solutions are similar to 1D case but with 3 different polarizations

Sodium (BCC)

Monoatomic

Three dimensional vibration

Along a given direction of propagation, there are 1 longitudinal wave and 2

transverse waves, each may have different velocities,

• monoatomic systems have only acoustic branches


28

Energy scales:

Wavenumber (cm-1): A wavelength of energy that is also called a reciprocal

centimeter. Wavenumbers are obtained when frequency is expressed in Hertz

and the speed of light is expressed in cm/s (c= 29979245800 cm/s).

Electron Volt (eV): The electron volt is the energy that we would give an

electron (e=1.60217657 10-19 C) if it were accelerated by one volt potential

difference. 1 eV = 1.602 x 10-19 J.

Frequency f and angular frequency ω: the vibrational energy is �� con h

the Planck constant (h= 6.62606957 10-34 Js, � �&

�= 1.054571726 10-34 Js)

1 THz→��/� � ��/� �0.0041356 eV= 4.1356 meV

1 THZ→ (1 1012 Hz)/c =33.3564095 cm-1

Longitudinal Waves

Transverse Waves

Energy difference:

Longitudinal modes usually corresponds to a greater compression of the

springs with respect to transvers ones → higher energy

G. Bracco-Material Science SERP CHEM


Neon (FCC) monoatomic

All modes are acoustic (Monoatomic system)

Transverse mode are degenerate along [001]

and [111]. The maximum frequency is lower

than that of sodium since van der Waals

interaction in neon is weaker than metal

bond in sodium.

along [110] the longitudinal mode presents a non monotonic behavior different from a

sinusiodal one: this suggests that at least a next near neighbor interaction is necessary

This is related to the FCC structure: atom A of plane I

interacts with nearest neighbor C atoms in plane II and with

next nearest neighbor B in plane III.

3D crystal with atom basis

For a 3-dim crystal, if each unit cell has p atoms, the degrees of

freedom (DOF) are 3p for each cell (polarization L, T1

e T2)

3 acoustic branches, 3(p-1) optical branches

• If a crystal has N unit cells, then each branch has

N normal modes (number of k-points for each dispersion curve).

• As a result, the total number of normal modes of the whole

crystal is 3pN (= total DOF of this crystal).


P=2 atoms (Ga, As),

then there are

3 acoustic branches L, T1

e T2

3 optical branches L, T1

e T2

FCC lattice with 2-atom basis (diamond)

cm-1

K

∼9Thz

The covalent bond is stronger than previous bonds

therefore maximum frequency is higher.

Γ Is the center of the first BZ and X, K, W, and L

special points on BZ boundary.

3D crystal with atom basis



P=2 atoms (C, C),

then there are

3 acoustic branches L, T1

e T2

3 optical branches L, T1

e T2

Diamond FCC lattice with 2-atom basis (diamond)

∼39Thz

The covalent bond is even stronger than previous one (the strongest bond!)


NaCl (Sodium Chloride structure)

The structure corresponds to two

interpenetrating FCC lattices

two atoms in the unit cell

→ p=2 (6 vibrational branches)

The lattice vibration spectrum shows

acoustic (A) and optical (O) modes with

longitudinal (L) and transverse (T)

polarization.

The stronger interaction determines a

maximum frequency higher than the

case of neon and sodium


Even for the more complex

structure of α-alumina with 10

atoms in the primitive cell

(p=10, 30 vibrational branches),

there are 3 acoustic branches

and 27 optical branches

Conventional Hex.cell Primitive Trigonal cell

αααα-Al2O

3


PBC or Born–von Karman boundary condition

The concept of periodic boundary condition can extended to 2D and 3D

1 8

Moreover any finite system can be considered as

part of an infinite system

1D: un(t) = un+N(t)

3D: ul,m,n(t) = ul+N1,m,n(t)

ul,m,n(t) = ul,m+N2,n(t)

ul,m,n(t) = ul,m,n+N3(t)

81

A finite 2D area is mapped on a torus

X: u0,n(t) = uM,n (t) M,N integers

Y: um,0(t) = um,N (t)

In 3D it is not possible to make a

graphical representationM cells

N c

ell

s

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