Download Prof.P. Ravindran, Lattice Dynamics-1

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Lattice Dynamics-1
Prof.P. Ravindran,
Department of Physics, Central University of Tamil
Nadu, India
http://folk.uio.no/ravi/CMP2013
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
What is phonon?






Consider the regular lattice of atoms in a uniform solid
material.
There should be energy associated with the vibrations of
these atoms.
But they are tied together with bonds, so they can't vibrate
independently.
The vibrations take the form of collective modes which
propagate through the material.
Such propagating lattice vibrations can be considered to be
sound waves.
And their propagation speed is the speed of sound in that
material.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Phonon





The vibrational energies of molecules are quantized and
treated as quantum harmonic oscillators.
Quantum harmonic oscillators have equally spaced energy
levels with separation ΔE = h.
So the oscillators can accept or lose energy only in discrete
units of energy h.
The evidence on the behaviour of vibrational energy in
periodic solids is that the collective vibrational modes can
accept energy only in discrete amounts, and these quanta
of energy have been labelled "phonons".
Like the photons of electromagnetic energy, they obey
Bose-Einstein statistics.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Lattice Dynamics

Concern with the spectrum of characteristics vibrations of a
crystalline solid.

Leads to;
– understanding of the conditions for wave propagation in a
periodic lattice,
– the energy content,
– the specific heat of lattice waves,
– the particle aspects of quantized lattice vibrations
(phonons)
– consequences of an harmonic coupling between atoms.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Lattice Vibrations  Phonons
• Mechanical waves are waves which propagate through a material medium
at a wave speed which depends on the elastic and inertial properties of that
medium.
• There are two basic types of wave motion for mechanical waves:
longitudinal waves and transverse waves.
Longitudinal Waves
Transverse Waves
•
It corresponds to the atomic vibrations with a long λ.
•
Presence of atoms has no significance in this wavelength
limit, since λ>>a, so there will no scattering due to the
presence of atoms.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
5
Crystal Dynamics

In previous chapters we have assumed that the atoms were at rest at
their equilibrium position. This can not be entirely; Atoms vibrate about
their equilibrium position at absolute zero.

The energy they possess as a result of zero point motion is known as zero
point energy.

The amplitude of the motion increases as the atoms gain more thermal
energy at higher temperatures.

Here we discuss the nature of atomic motions, sometimes referred to as
lattice vibrations.

In crystal dynamics we will use the harmonic approximation , amplitude
of the lattice vibration is small. At higher amplitude some unharmonic
effects occur.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Crystal Dynamics

Our analysis will be restricted to lattice vibrations of small amplitude.
Since the solid is then close to a position of stable equilibrium, its
motion can be calculated by a generalization of the method used to
analyse a simple harmonic oscillator. The small amplitude limit is
known as harmonic limit.

In the linear region (the region of elastic deformation), the restoring
force on each atom is approximately proportional to its displacement
(Hooke’s Law).

There are some effects of nonlinearity or ‘anharmonicity’ for larger
atomic displacements.

Anharmonic effects are important for interactions between phonons
and photons.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Crystal Dynamics

Atomic motions are governed by the forces exerted on
atoms when they are displaced from their equilibrium
positions.

To calculate the forces it is necessary to determine the
wavefunctions and energies of the electrons within the
crystal. Fortunately many important properties of the
atomic motions can be deduced without doing these
calculations.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
PHONONS
• Quanta of lattice vibrations
• Energies of phonons are
quantized
E phonon 
h s

PHOTONS
• Quanta of electromagnetic
radiation
• Energies of photons are
quantized as well
E photon 
~a0=10-10m
hc

~10-6m
p phonon 
h

p photon 
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
h

Energy of harmonic oscillator
Obtained by in a classical way of considering the normal modes
that we have found are independent and harmonic.
1

 n   n  
2

•
Make a transition to Q.M.
•
Represents equally spaced energy
levels
Energy, E




Energy levels of atoms
vibrating at a single
frequency ω
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
It is possible to consider  n as constructed by adding n excitation
quanta each of energy  to the ground state.
1
 0  
2
A transition from a lower energy level to a higher energy level.
1
1


   n2     n1  
2
2


   n2  n1      
unity
absorption of phonon
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I





The converse transition results an emission of phonon
with an energy  .
Phonons are quanta of lattice vibrations with an
angular frequency of .
Phonons are not localized particles.
Its momentum is exact, but position can not be
determined because of the uncertainity principle.
However, a slightly localized wavepacket can be
considered by combining modes of slightly different 
and .
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

Assume waves with a spread of k of
; so this wavepacket
will be localized within 10 unit cells. 10a
This wavepacket will represent a fairly localized phonon moving
with group velocity d .
dk
Such phonons can be treated as localized particles within some limits.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Thermal energy and lattice vibrations
•Atoms vibrate about their equilibrium position.
•They produce vibrational waves.
•This motion is increased as the temperature is
raised.
In a solid, the energy associated with this vibration and perhaps also
with the rotation of atoms and molecules is called as thermal energy.
Note: In a gas, the translational motion of atoms and molecules
contribute to this energy.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Therefore, the concept of thermal energy is fundamental to an
understanding many of the basic properties of solids. We would
like to know:
•
What is the value of this thermal energy?
•
How much is available to scatter a conduction electron in a
metal; since this scattering gives rise to electrical resistance.
•
The energy can be used to activate a crystallographic or a
magnetic transition.
•
How the vibrational energy changes with temperature since
this gives a measure of the heat energy which is necessary to
raise the temperature of the material.
•
Recall that the specific heat or heat capacity is the thermal
energy which is required to raise the temperature of unit mass
or 1g mole by one Kelvin.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Hooke's Law

One of the properties of elasticity is that it takes about
twice as much force to stretch a spring twice as far. That
linear dependence of displacement upon stretching force is
called Hooke's law.
Fspring  k .x
Spring constant k
F
It takes twice
as much force
to stretch a
spring twice
as far.
2F 
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Hooke’s Law

The point at which the Elastic Region ends is called the inelastic
limit, or the proportional limit. In actuality, these two points are
not quite the same.

The inelastic limit is the point at which permanent deformation
occurs, that is, after the elastic limit, if the force is taken off the
sample, it will not return to its original size and shape, permanent
deformation has occurred.

The inelastic limit is the point at which the deformation is no
longer directly proportional to the applied force (Hooke's Law no
longer holds).
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
SOUND WAVES

Sound waves propagate through solids. This tells us that
wavelike lattice vibrations of wavelength long compared to the
interatomic spacing are possible. The detailed atomic
structure is unimportant for these waves and their
propagation is governed by the macroscopic elastic properties
of the crystal.

We discuss sound waves since they must correspond to the
low frequency, long wavelength limit of the more general
lattice vibrations considered later.

At a given frequency and in a given direction in a crystal it is
possible to transmit three sound waves, differing in their
direction of polarization and in general also in their velocity.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Elastic Waves

A solid is composed of discrete atoms, however when the
wavelength is very long, one may disregard the atomic nature
and treat the solid as a continous medium. Such vibrations are
referred to as elastic waves.
Elastic Wave Propagation (longitudinal) in a bar
• At the point x elastic displacement is
U(x) and strain ‘e’ is defined as the
change in length per unit length.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
A
x
x+dx
dU
e
dx
Elastic Waves

According to Hooke’s law stress S (force per unit area) is
proportional to the strain e.
S  C.e

A
C = Young modulus
x
x+dx
To examine the dynamics of the bar, we choose an arbitrary
segment of length dx as shown above. Using Newton’s second law,
we can write for the motion of this segment,
 2u
(  Adx) 2   S ( x  dx)  S ( x) A
t
Mass x Acceleration
Net Force resulting from stresses
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Elastic Waves

S  C.e
Equation of motion
 2u
(  Adx) 2   S ( x  dx)  S ( x) A
t
S
 S ( x  dx)  S ( x)  dx
x
 2u
 2u
(  Adx) 2  C 2 Adx
t
x
 2u
 2u
 2 C 2
t
x
i ( kx t )
u  Ae
du
dx
u
S  C.
x
S
 2u
 C. 2
x
x
e
Cancelling common terms of Adx;
Which is the wave eqn. with an offered
sol’n and velocity of sound waves ;
– k = wave number (2π/λ)
– ω = frequency of the wave
– A = wave amplitude
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
  vs k
vs 
C/
Dispersion Relation

The relation connecting the frequency and wave number is
known as the dispersion relation.
  vs k
ω
• At small λ
• At long λ
Continuum
Discrete
0
k
k → ∞ (scattering occurs)
k → 0 (no scattering)
• When k increases velocity decreases.
As k increases further, the scattering
becomes greater since the strength of
scattering increases as the wavelength
decreases, and the velocity decreases
even further.
* Slope of the curve gives
the velocity of the wave.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Speed of Sound Wave

The speed with which a longitudinal wave moves through a liquid
of density ρ is
VL   
C

C = Elastic bulk modulus
ρ = Mass density
• The velocity of sound is in general a function of the direction of
propagation in crystalline materials.
• Solids will sustain the propagation of transverse waves, which
travel more slowly than longitudinal waves.
• The larger the elastic modules and smaller the density, the more
rapidly can sound waves travel.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Sound Wave Speed
Speed of sound for some typical solids
Structure
Type
Nearest
Neighbour
Distance
(A°)
Density
ρ
(kg/m3)
Elastic bulk
modules
Y
(1010 N/m2)
Calculate
d Wave
Speed
(m/s)
Observed
speed of
sound
(m/s)
Sodium
B.C.C
3.71
970
0.52
2320
2250
Copper
F.C.C
2.55
8966
13.4
3880
3830
Aluminum
F.C.C
2.86
2700
7.35
5200
5110
Lead
F.C.C
3.49
11340
4.34
1960
1320
Silicon
Diamond
2.35
2330
10.1
6600
9150
Germanium
Diamond
2.44
5360
7.9
3830
5400
NaCl
Rocksalt
2.82
2170
2.5
3400
4730
Solid
•VL values are comparable with direct observations of speed of sound.
•Sound speeds are of the order of 5000 m/s in typical metallic, covalent
and ionic solids.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Sound Wave Speed

A lattice vibrational wave in a crystal is a repetitive and
systematic sequence of atomic displacements of
– longitudinal,
– transverse, or
– some combination of the two
• An equation of motion for any displacement can be produced by means
of considering the restoring forces on displaced atoms.
• They can be characterized by
– A propagation velocity, v
– Wavelength λ or wave vector
– A frequency  or angular frequency ω=2π
• As a result we can generate a dispersion relationship between
frequency and wavelength or between angular frequency and
wavevector.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

To illustrate the procedure for treating the interatomic potential in
the harmonic approximation, consider just two neighboring atoms.
Assume that they interact with a known potential V(r).
Expand V(r) in a Taylor’s series in displacements about the
equilibrium separation, keeping only up through quadratic
terms in the displacements:
V(R)
Repulsive
0
a
Attractive
r
R
This potential energy is the same as
that associated with a spring with
spring constant:
 d 2V 
K   2 
 dr r a
Force  K (r  a)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Normal Modes of Vibration
One dimensional model # 1: The Monatomic
Chain


Consider a Monatomic Chain of Identical Atoms
with
nearest-neighbor, “Hooke’s Law” type forces (F = - kx)
between the atoms.
This is equivalent to a force-spring model, with masses m &
spring constants K.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
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 of “a”.
Atoms move only in a direction parallel to the chain.
Only nearest neighbours interact (short-range forces).
a
a
Un-2
a
Un-1
a
Un
a
Un+1
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
a
Un+2
Monoatomic Chain

a
Start with the simplest case
of monoatomic linear chain
with only nearest neighbour
interaction
Un-1
a
Un
Un+1
If one expands the energy near the equilibrium point for the nth atom
and use elastic approximation, Then, the Newton’s 2nd Law
equation of motion becomes:
..
m u n  K (un1  2un  un1 )
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Monoatomic Chain
The force on the nth atom;
a
a
•The force to the right;
K (un1  un )
•The force to the left;
K (un  un1 )
Un-1
Un
Un+1
•The total force = Force to the right – Force to the left
..
m u n  K (2u n  u n1  u n1 )  0
Eqn’s of motion of all atoms are of this form, only the value of
‘n’ varies
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

Assume that all atoms oscillate with the same amplitude A &
the same frequency ω. Assume harmonic solutions for the
displacements un of the form:
un  A exp i  kx  t  u n  dun  i A exp i  kxn0  t 
0
n
.
d 2 un
2
u n  2   i   2 A exp i  kxn0  t  
dt
..
Undisplaced Position
xn0  na
dt
..
u n   2un
Displaced Position
xn  na  un
• Put all of this into the equation of motion:
..
• Now, carry out some simple mathematical manipulation:
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
th atom
Equation
of
motion
for
n
Monoatomic Chain
..
mu n  K (un1  2un  un1 )

2
 m A e
i kxn0 t


K


Ae
 m A e
2
 m A e
i  kna t 
i  kna t 
  2 A ei kx

 K Ae

 K Ae
t
  A eikx
0
n1 t
k ( n  1) a
kna
i  kna  ka t 
i  kna t 
0
n
k ( n  1) a
kna
2
i kxn10 t
e
ika
 2A e
 2A e
i  kna t 
i  kna t 
Cancel Common terms
m 2  K  eika  2  eika 
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
 Ae
 Ae
i  kna ka t 
i  kna t 
e
 ika




Monoatomic Chain
m 2  K  eika  2  eika 
eix  eix  2cos x
eika  eika  2cos ka
m 2  K  2cos ka  2 
 2 K (1  cos ka)
2
2  ka 
m  4 K sin  
 2 
2 
4K
 ka 
sin 2  
m
 2 
1  cos x   2 sin 2 
x

2
Maximum value of it is 1
4K
 ka 

sin  
m
 2 
max 
4K
m
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

Mathematical Manipulation finally gives:
• After more manipulation, this simplifies to
Solution to the Normal
Mode Eigenvalue Problem
for the monatomic chain.
• The maximum allowed frequency is:
• The physical significance of these results is that, for the monatomic
chain, the only allowed vibrational frequencies ω must be related to
the wavenumber k = (2π/λ) or the wavelength λ in this way.
• This result is often called the “Phonon Dispersion Relation” for
the chain, even though these are classical lattice vibrations &
there are no (quantum mechanical) phonons in the classical theory.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Monoatomic Chain

ω versus k relation;

max  2
K
m
Vs   / k
C
–л / a
B
0
0
A
л /a
2 л /a
k
•
Normal mode frequencies of a 1D chain
The points A, B and C correspond to the same frequency, therefore
they all have the same instantaneous atomic displacements.
The dispersion relation is periodic with a period of 2π/a.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Monoatomic Chain
Note that:
4K
ka

sin
m
2

In above equation n is cancelled out, this means that the eqn. of motion of all
atoms leads to the same algebraic eqn. This shows that our trial function Un is
indeed a solution of the eqn. of motion of nth atom.

We started from the eqn. of motion of N coupled harmonic oscillators. If one
atom starts vibrating it does not continue with constant amplitude, but transfer
energy to the others in a complicated way; the vibrations of individual atoms are
not simple harmonic because of this exchange energy among them.

Our wavelike solutions on the other hand are uncoupled oscillations called
normal modes; each k has a definite ω given by above eqn. and oscillates
independently of the other modes.

As we already said, these are called the Normal Modes of the system. They are a
collective property of the system as a whole & not a property of any of the individual
atoms. Each mode represented by ω(k) oscillates independently of the other
modes. Also, it can be shown that the number of modes is the same as the original
number of equations N. Proof of this follows.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Establish which wavenumbers are possible for our one dimensional chain. Not
all values are allowed because nth atom is the same as the (N+n)th as the chain
is joined on itself. This means that the wave eqn. of
un  A exp i  kxn0  t  
must satisfy the periodic boundary condition
un  uN n
which requires that there should be an integral number of wavelengths in the
length of our ring of atoms
Na  p
Thus, in a range of 2π/a of k, there are N allowed values of k.
Na  p   
Na 2
2

 Nk 
p
p
k
a
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Monoatomic Chain
What is the physical significance of wave numbers outside
the range of 2π/a?
Un
x
a
Un
x
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Monoatomic Chain
  2a;k  2  k  

a
This value of k corresponds to the
maximum frequency; alternate atoms
oscillate in antiphase and the waves at
this value of k are standing waves.
White line :
4  7a    7a  k  2  8 1.14
7a 7a
4
a
4
Green line :
3  7a    7a  k  2  6  0.85
7a 7a
3
a
3
un
x
un a
ω-k relation
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
• What is the physical significance of wave numbers k outside of the
First Brillouin Zone [-(π/a)  k  (π/a)]?
• At the Brillouin Zone edge:
• This k value corresponds to the maximum frequency. A detailed analysis
of the displacements shows that, in that mode, every atom is oscillating
π radians out of phase with it’s 2 nearest neighbors. That is, a wave at
this value of k is A STANDING WAVE.
Black
k = π/a
or
 = 2a
Green:
k = (0.85)π/a
or
 = 2.35 a
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
x
Monoatomic Chain
•The points A and C both have same
frequency and same atomic displacements
•They are waves moving to the left.
•The green line corresponds to the point B in
dispersion diagram.
•The point B has the same frequency and
displacement with that of the points A and C
with a difference.
w
C
K
m
Vs   / K
2
B
A
π/a 2π/a k
-π/a 0
ω-k relation (dispertion diagram)
•The points A and C are exactly
equivalent; adding any multiple of
2π/a to k does not change the
•The point B represents a wave moving to the frequency and its group velocity, so
right since its group velocity (dω/dk)>0.
point A has no physical significance.
•k=±π/a has special significance
un
x
un
a
ka
2
2
 m  4K sin
2
•θ=90o
Bragg reflection can be obtained at
k= ±nπ/a
n  n2  n 2n 
k
a
For the whole range of k (λ)
x
2a  2d sin90  d  a
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
1
At the beginning stated that, in
the long wavelength limit, the
velocity of sound waves has been
derived as
c  Ka


Vs  c 
m
 a
Using elastic properties, let’s see
whether the dispersion relation
leads to the same equation in the
long λ limit.
If λ is very long;
ka
2a 2
k
2
 m  4K
4
1 so
sin ka  ka
K

Vs  
a
k
m
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
K a2
m
Monoatomic Chain



Since there is only one possible propagation direction and
one polarization direction, the 1D crystal has only one
sound velocity.
In this calculation we only take nearest neighbor
interaction although this is a good approximation for the
inert-gas solids, its not a good assumption for many solids.
If we use a model in which each atom is attached by
springs of different spring constant to neighbors at
different distances many of the features 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 with a velocity given by (velocity formula)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
 (k ): Dispersion relation
Slope( d 
) = Group velocity
dk


1st Brillouin zone :
 k
a
a
Frequency, w



a
0

a
Wavevector, k
First Brillouin zone
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
2
a

Briefly look in more detail at the group velocity, vg.

The dispersion relation is:


4K
ka
sin
m
2
So, the group velocity is:
vg  (dω/dk) = a(K/m)½cos(½ka)
vg = 0 at the BZ edge [k =  (π/a)]
– This tells us that a wave with λ corresponding to a zone edge
wavenumber k =  (π/a) will not propagate.
That is, it must be a standing wave!
– At the BZ edge, the displacements have the form (for site n):
Un= Uoeinka = Uo ei(nπ/a) = Uo(-1)n
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Group Velocity, vg in the 1st BZ
The dispersion relation is:
4K

sin
m
vg  (dω/dk) = a(K/m)½cos(½ka)
At the 1st BZ Edge,
ka
vg = 0
2• This means that a wave
with λ corresponding to a
zone edge wavenumber
k =  (π/a)
Will Not
Propagate!
• That is, it must be a
Standing Wave!
vg = 0 at the BZ edge [k =  (π/a)]
1st BZ Edge
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
 Group velocity
transmission velocity of a wave packet
v g  d dK
or
v g  grad K  K 
K
1
2
| sin( ka ) |
m
2
vg 
K
1
a cos( ka )
m
2
- Beats
- Wave packet or Wave group
The velocity of energy propagation in the medium
This is zero at the edge of the zone
Standing wave
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The Monatomic Chain
k = (/a) = (2/);  = 2a
k  0;   
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Acoustic branch
n  0 exp(it )exp(inka),
vg 
K
1
a cos( ka )
m
2
-k  0 : Long wavelength or continuum limit
K
vg  a
 a0 : Velocity of sound in a crystal
m
a
vg  0
 n1  0 exp{i ( n  1)ka}

 exp( ika )
n
 0 exp( inka )
Frequency, w
Acoustic branch

K

- k
: Edge of first Brillouin zone
Speed of sound
Edge of First
Brillouin zone
0
Wavevector, k
Standing
wave
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

a
Chain of two types of atom

Two different types of atoms of masses M and m are connected
by identical springs of spring constant K;
(n-2)
(n-1)
(n)
K
M
(n+1)
K
m
(n+2)
K
K
m
M
M
a)
a
b)
Un-2
Un-1
Un
Un+1
Un+2
• This is the simplest possible model of an ionic crystal.
• Since a is the repeat distance, the nearest neighbors separations is a/2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom

We will consider only the first neighbour interaction although it
is a poor approximation in ionic crystals because there is a long
range interaction between the ions.

The model is complicated due to the presence of two different
types of atoms which move in opposite directions.
Our aim is to obtain ω-k relation for diatomic lattice
Two equations of motion must be written;
One for mass M, and
One for mass m.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
M
m
Un-2
Un-1
m
M
Un
Un+1
Equation of motion for mass M (nth):
mass x acceleration = restoring force
..
M u n  K (un1  un )  K (un  un1 )
 K (un1  2un  un1 )
Equation of motion for mass m (n-1)th:
..
mu n  K (un  un1 )  K (un1  un2 )
-1
..
mu n  K (un  2un1  un2 )
-1
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
M
Un+2
Chain of two types of atom
M
Un-1
Un-2
m
M
m
Un
Un+1
M
Un+2
Offer a solution for the mass M
un  A exp i  kxn0  t 
xn0  na / 2
For the mass m;
un -1  A exp i  kxn0  t 
α : complex number which determines the relative amplitude and phase of the
vibrational wave.
u n   2 A exp i  kxn0  t 
..
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
For nth atom (M):
..
M u n  K (un1  2un  un1 )
 2 MAe
 kna

i
t 
 2

 2 MAe
 kna

i
t 
 2

 k  n 1 a

 k  n 1 a

 kna


i
t 
i
t  
i
t 
2
2


 K   Ae 
 2 Ae  2    Ae 




 kna

 kna

 kna


i
t  i ka
i
t 
i
t   i ka 
 K   Ae  2  e 2  2 Ae  2    Ae  2 e 2 




Cancel common terms
ka
ka
i
i


2
2
2
 M  K   e  2   e 


ka 

 2 M  2 K 1   cos 
2 

eix  e  ix  2 cos x
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
For the (n-1)th atom (m)
..
mu n1  K (un  2un1  un2 )
 A 2 me
 2 mAe
 k  n 1 a

i
t 
2


 kna

i
t   i ka
 2

2
e
 k  n 1 a

 k  n  2a

 i kna t 
i
t 
i
t  
2
2


 K  Ae  2   2 Ae 
 Ae 





 kna

 kna

 i kna
t 
i
t   i ka
i
t   i 2 ka 
 K  Ae  2   2 Ae  2  e 2  Ae  2  e 2 




Cancel common terms
ka
i

2
 ika 
2
 me
 K 1  2 e  e 


ka
ka
i
i


2
2
2
 m  K  e  2  e 


eix  e  ix  2 cos x
i
ka
2
ka


 m  2 K  cos   
2


2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
ka 

 M  2 K 1   cos 
2 

ka 

2
2
 m  K    cos 
2 

2

for M
for m
Now we have a pair of algebraic equations for α and ω as a
function of k. α can be found as
2 K cos(ka / 2)
2K   2 M


2
2K   m
2 K cos(ka / 2)

A quadratic equation for ω2 can be obtained by crossmultiplication
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
2 K cos(ka / 2)
2K   2 M


2
2K   m
2 K cos(ka / 2)
ka
4 K cos ( )  4 K 2  2 K  2 ( M  m)   4 Mm
2
2
4 K 2 (1  cos 2 (
2
ka
))  2 K 2 (m  M )   4 Mm  0
2
2
m

M
sin
(ka / 2)
4
2
2
  2K (
)  4 K
0
mM
mM
The two roots are;

2
K (m  M )

mM
b  b 2  4ac
x1,2 
2a
m  M 2 4sin 2 (ka / 2) 1/ 2
K [(
) 
]
mM
mM
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom

ω versus k relation for diatomic chain;
w
A
–л/a
0
B
C
л/a
2л/a
k
•
If the crystal contains N unit cells we would expect to find 2N normal
modes of vibrations and this is the total number of atoms and hence the
total number of equations of motion for mass M and m.
•
Normal mode frequencies of a chain of two types of atoms.
• At A, the two atoms are oscillating in antiphase with their centre of
mass at rest;
• at B, the lighter mass m is oscillating and M is at rest;
• at C, M is oscillating and m is at rest.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom

As there are two values of ω for each value of k, the
dispersion relation is said to have two branches;

Optical Branch
A
Upper branch is due to the
+ve sign of the root.
B
C
Acoustical Branch
Lower branch is due to the
-ve sign of the root.
–л/a
0
л/a
2л/a
k
• The dispersion relation is periodic in k with a period
2 π /a = 2 π /(unit cell length).
• This result remains valid for a chain of containing an
arbitrary number of atoms per unit cell.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom


Let’s examine the limiting solutions at 0, A, B and C.
In long wavelength region (ka«1); sin(ka/2)≈ ka/2 in ω-k.
2
K
(
m

M
)
m

M
4sin
(ka / 2) 1/ 2
2
2
1,2 
K [(
) 
]
mM
mM
mM
 m  M 
K m  M 
4 k a 
 
 K 

 
mM
 mM  mM 4 
12



K m  M   
mM
2 2

1  1 
k
a  
2
 
  m  M 
mM
 
 
2
2
2
12
2
Use Taylor expansion:1  x 
12
 1  x 2 for small x
K m  M   
mM
2 2 
 
k a 
1  1 
2
mM

  2(m  M )
2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
Taking +ve root; sinka«1
max opt
(max value of optical branch)
2
2K  m  M 

mM
Taking -ve root; (min value of acoustical brach)
min acus.2
K  m  M   mMk 2 a 2 
Kk 2 a 2



2
mM
 2(m  M )  2(m  M )
By substituting these values of ω in α (relative amplitude)
equation and using cos(ka/2) ≈1 for ka«1 we find the
corresponding values of α as;
2K   2 M

2 K cos(ka / 2)
 1
OR
M
 
m
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
2

Substitute min

2
min ac
into relative amplitude α
ac
2K   2 M

2 K cos(ka / 2)
2 2
K(k a )

2(m  M)
 1
This solution represents long-wavelength sound waves in
the neighbourhood of point 0 in the graph; the two types
of atoms oscillate with same amplitude and phase, and the
velocity of sound is
w
1/ 2


w
K
vs   a 

k
2(
m

M
)


A
Optical
B
C
Acoustical
k
–π/a
0
π/a
2π/a
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom
Substitute maxop
2

into relative amplitude we obtain,
2K   2 M

2 K cos(ka / 2)
2K(m  M)

mM
2
max op
 
M
m
w
A
Optical
B
C
Acoustical
k
–π/a
0
π/a
This solution corresponds to point A
in dispersion graph. This value of α
shows that the two atoms oscillate in
antiphase with their center of mass at
rest.
2π/a
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Chain of two types of atom

The other limiting solutions of equation ω2 are for ka= π ,
i.e sin(ka/2)=1. In this case

2
max ac
K (m  M )

Mm
1/ 2
 M  m 
4 
K 

 
 Mm  Mm 
2
K (m  M ) K ( M  m)

Mm

2
max ac
2K
2K
2
(C)
OR min 

(B)
op
m
M
• At max.acoustical point C, M oscillates and m is at rest.
• At min.optical point B, m oscillates and M is at rest.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The Diatomic Chain Solution:
( = 2a)
Optic Modes
Near the BZ edge [q =  (π/a)]
(Optic Branch)
(Assuming m > M)
The Optic Mode becomes:
(ω+)2  2K/M
Gap
qa
Acoustic Modes
(Acoustic Branch)
The Acoustic Mode becomes:
(ω-)2  2K/m
So, at the BZ edge, the vibrations of
wavelength  = 2a for the 2 modes
behave as if there were 2 uncoupled
masses m & M , vibrating
independently with identical springs
of constant K.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Acoustic/Optical Branches

The acoustic branch has this name because it gives rise to
long wavelength vibrations - speed of sound.

The optical branch is a higher energy vibration (the
frequency is higher, and you need a certain amount of
energy to excite this mode). The term “optical” comes from
how these were discovered - notice that if atom 1 is +ve and
atom 2 is -ve, that the charges are moving in opposite
directions. You can excite these modes with
electromagnetic radiation (ie. The oscillating electric fields
generated by EM radiation)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Acoustic & Optic Branches
 Despite the fact that diatomic chain model is one-dimensional,
it’s results for the vibrational normal modes ω contain
considerable qualitative physics that carries over to the observed
vibrational frequencies for many real materials.
 So, much of the physics contained in the diatomic chain
results can teach us something about the physics contained in
the normal modes of many real materials.
 In particular, ALL MATERIALS with 2 atoms per unit cell are
observed to have two very different kinds of vibrational normal
modes. These are called The Acoustic Branch &
The
Optic Branch
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The Acoustic Branch

This branch received it’s name because it contains long
wavelength vibrations of the form ω = vsk, where vs is the
velocity of sound. Thus, at long wavelengths, it’s ω vs. k
relationship is identical to that for ordinary acoustic (sound)
waves in a medium like air.
The Optic Branch
 This branch is always at much higher vibrational frequencies than the
acoustic branch. So, in real materials, a probe at optical frequencies is
needed to excite these modes.
 Historically, the term “Optic” came from how these modes were
discovered. Consider an ionic crystal in which atom 1 has a positive
charge & atom 2 has a negative charge. As we’ve seen, in those modes,
these atoms are moving in opposite directions. (So, each unit cell contains
an oscillating dipole.) These modes can be excited with optical frequency
range electromagnetic radiation.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Transverse optical mode for
diatomic chain
Amplitude of vibration is strongly exaggerated!
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
A Longitudinal Optic Mode
The vibrational amplitude is highly exaggerated!
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Transverse acoustical mode for
diatomic chain
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Acoustic vs. Optic Phonons
Which has lower energy? Why?
Optic Mode
Acoustic Mode
Acoustic wave has Lower Energy
Less Compression of Springs
Optic mode invole bond stretching and
hence higher energy.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Phonon Dispersion and Scattering
 The 1-D Diatomic Chain
The chemical bonds between atoms are not rigid :
Act like spring
Vibration
Quanta of lattice vibration are called phonons.
Lattice vibrations are responsible for transport of energy in
many solids
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
 Transverse vs. Longitudinal polarization
Three modes of wave vectors for one atom per unit cell
Two transverse modes
One longitudinal mode
us-1
us-1
s-1
us+1
us
s
s+1
s+2
us+2
us+3
s+3
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
us
us+1
us+2
Dispersion Relation for Real Crystal
- Number of branches
If there are q atoms in the primitive cell, there are 3q
branches to the dispersion relation
3 acoustic branches : 1 longitudinal acoustic (LA)
2 transverse acoustic (TA)
3q - 3 optical branches : q - 1 longitudinal optical (LO)
2q - 2 transverse optical (TO)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Frequency gap
- Si[100] direction
- SiC
The group velocity of phonons in the optical branches is small
contribute little to the thermal conduction
At low temperatures : TA are dominant contributors to the heat
conduction
At high temperatures : LA are dominant contributors to the heat
conduction
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Diatomic Basis: Experimental Results
The optical modes generally have frequencies near  = 1013 1/s, which is in the infrared part
of the electromagnetic spectrum. Thus, when IR radiation is incident upon such a lattice it
should be strongly absorbed in this band of frequencies.
Transmission spectrum for IR radiation
incident upon a very thin NaCl film.
Note the sharp minimum in transmission
(maximum in absorption) at a wavelength of
about 61 10-4 cm. This corresponds to a
frequency  = 4.9 1012 1/s.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
77
 Phonon scattering
 Phonon-phonon scattering
Governs the thermal transport properties of dielectric
and semiconductor
Inelastic scattering : the phonon frequency before the
scattering event is different from that after the event
Normal (or N) – process : Inside the 1st brillouin zone
1  2  3
or
1  2  3 : energy conservation
k1  k2  k3
or
k1  k2  k3 : crystal momentum
conservation
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Umklapp(or U) process : Outside the 1st brillouin zone
k  k  G
or
k1  k2  k3  G
k1  G  k2  k3
kz
kz
k1

k1
a
k2
k3


a
a

k2
a
kx
k3
G
N process
kx
U process
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
k1  k2
N-process vs. U-process
N - process
U - process
Energy
Conservation
Conserved
Conserved
Momentum
Conservation
Conserved
Thermal
conductivity
Not dominant
Thermophysical
role
Distributing the
phonon energy
Net momentum not
conserved
Dominant
Act as a direct
resistance to heat flow
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Phonon scattering: Temperature dependence
 U  1/  :
Above room temperature
A, B :
 U  ( A  B 2 )T
 U
Positive constants
Scattering rate of the Uprocess
1
T  
T
Below room temperature
Cv  
p
K
 

T  e
 Cv  T , Cv  
1
 / k BT
1
 
1 2 
  T
At high temperature
specific heat does not change
significantly
Thermal conductivity of
silicon
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Phonon scattering – 4 phonon & defect
Four – phonon scattering
1
2
3
2
3
4
1
 four   2T 2
  four
4
1
3
2
4
(Temperature range : 300 K ~ 1000 K)
 U : Negligible
Phonon – defect scattering
Elastic scattering
Independent of temperature
Defendant on the phonon wavelength
 ph-d   4
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Phonon scattering- phonon-electron
Dominant at high temperature
Scattering by acoustic phonon is essentially elastic.
Eac
Ee  negligible
Scattering by optical phonon is inelastic : Polar scattering
Facilitates heat transfer between optical phonon
and electron (Joule heating)
Ef  Ei  ph : energy conservation
kf  G  ki  kph : momentum conservation
+ : phonon absorption
(i : initial state f : final
- : phonon emission
state)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Phonon scattering- Raman scattering
Phonon and photon inelastic scattering
called Raman scattering, X-ray scattering,
neutronscattering, and Brillouin scattering
s  i  ph
Stokes shift (phonon emission)
s  i  ph
anti-Stokes shift (phonon absorption)
i: incident photon
s: scattered photon
ph: phonon
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Energy of
phonons
1D crystals


k
Multiply by 
k
Crystal momentum
•Phonons are not conserved
•They can be created and destroyed during collisions .
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Heat capacity from Lattice vibrations
The energy given to lattice vibrations is the
dominant
contribution to the heat capacity in most solids. In non-magnetic
insulators, it is the only contribution.
Other contributions;
•
In metals from the conduction electrons.
•
In magnetic materials from magnetic ordering.
Atomic vibrations leads to band of normal mode frequencies from
zero up to some maximum value. Calculation of the lattice energy
and heat capacity of a solid therefore falls into two parts:
i) the evaluation of the contribution of a single mode, and
ii) the summation over the frequency distribution of the modes.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Energy and heat capacity of a harmonic
oscillator, Einstein Model
_
   Pn n
n
Avarage energy of a harmonic
oscillator and hence of a lattice
mode of angular frequency ω
at temperature T
The probability of the oscillator
being in this level as given by the
Boltzman factor exp( n / kBT )
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Energy of
oscillator


1
n   n   
2

1
 
1



 n    exp    n    / k BT 
_
2
2

n 0 




 
1

exp    n    / k BT 

2
n 0
 


_
   Pn n
n
(*)
1 
z   exp[(n  )
]
2 k BT
n 0

z  e
 / 2 k BT
 e 3
 / 2 k BT
 e 5
 / 2 k BT
z  e
 / 2 k BT
(1  e 
 / k BT
 e 2
 / k BT
z  e
 / 2 k BT
(1  e 
 / k BT 1
 .....
 .....
)
According to the Binomial expansion for x«1 where
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
x    / kBT
Eqn (*) can be written
1 z
2 
  k BT
 k BT
(ln z )
z T
T
_
 e   / 2 k BT 
2 
  k BT
ln 

T  1  e   / kBT 
_
2
    / 2 k BT
  / k BT

ln
e

ln
1

e




T
_
  

  
  / k BT
2
  k BT   
ln 1  e



T
2
k
T

T
B 
 

 k B   / k BT 


 2k  k 2T 2 e
 1
_

e
  k BT 2  B2 2  B   / k T    

B
2
1

e
1

e
 4 k BT






_
  k BT 2
_
1
  
2
e

x'
(ln x) 
x
x
 / k BT
 / k BT

 / k BT
1
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I


_
1
  
2
e
 / k BT
1
This is the mean energy of phonons.The first term in the above
equation is the zero-point energy. As we have mentioned before
even at 0ºK atoms vibrate in the crystal and have zero-point
energy. This is the minimum energy of the system.
The avarage number of phonons with a particular frequecy ω is
given by Bose-Einstein distribution as
_
(number of phonons) x (energy of phonon)=(second term in  )
1
n( ) 

e
kBT
1
The second term in the mean energy is the contribution of
phonons to the energy.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

k BT
Mean energy of a
harmonic oscillator
as a function of T
1

2
T
low temperature limit
  k BT
1

    
2
e kBT  1
_
Since exponential term gets
bigger
_
1
   Zero point energy
2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

Mean energy of a
harmonic oscillator as a
function of T
k BT
kBT
1

2
T
high temperature limit
• is independent of frequency of oscillation.
•This is the classical limit because the energy
steps are now small compared with the
energy of the harmonic oscillator.
2
x
e x  1  x   ..........
2!


k BT
e
 1
k BT
1

   

2
1
1
k BT
_
_
1
    kBT
2
•So that  = kBTis the thermal energy of the
classical 1D harmonic oscillator.
_
  kB T
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Heat Capacity C

Heat capacity C can be found by differentiating the average
energy of phonons of
_

d
Cv 

dT
Let
 
 kB
 kBT 
e

kBT


e
2

1

k
1

  
2
e kBT  1
2
kBT


Cv  k B
2
 kBT 

2

e

kBT
e

eT
 
Cv  k B   
 T  e T 1
2

P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

2
kBT

1
2
Plot of Cv
as a function of T

eT
 
Cv  k B   
 T  e T 1
2


where
2


k
Specific heat vanishes exponentially at
low T’s and tends to classical value at
high temperatures.
Cv
kB

2
Area=

kB
T
The features are common to all
quantum systems; the energy tends to the
zero-point-energy at low T’s and to the
classical value of Boltzmann constant at
high T’s.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Plot of Cv as a function of T
Specific heat at constant volume depends on temperature as shown in
figure below. At high temperatures the value of Cv is close to 3R, where
R is the universal gas constant. Since R is approximately 2 cal/K-mole, at
high temperatures Cv is app. 6 cal/K-mole.
Cv
3R
T, K
This range usually includes RT. From
the figure it is seen that Cv is equal to
3R at high temperatures regardless of
the substance. This fact is known as
Dulong-Petit law. This law states that
specific heat of a given number of atoms
of any solid is independent of
temperature and is the same for all
materials!
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Classical theory of
heat capacity of solids
The solid is one in which each atom is bound to its side
by a harmonic force. When the solid is heated, the atoms
vibrate around their
sites like a set of harmonic
oscillators. The average energy for a 1D oscillator is kT.
Therefore, the energy per atom, regarded as a 3D
oscillator, is 3kT, and consequently the energy per mole is
= 3NkBT  3RT
where N is Avagadro’s number, kB is Boltzmann constant
and R is the gas constant. The differentiation wrt
temperature gives;

d
Cv 
dT
Cv  3R  3  6.02  1023 (atoms / mole) 1.38 1023 ( J / K )
Cv  24.9
J
Cal
;1J  0.2388Cal  Cv = 6
( K  mole)
( K  mole)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Einstein heat capacity of solids

The theory explained by Einstein is the first quantum theory of solids. He
made the simplifying assumption that all 3N vibrational modes of a 3D
solid of N atoms had the same frequency, so that the whole solid had a
heat capacity 3N times
 
Cv  k B  
T 


2
e
e

T
T

1
2
In this model, the atoms are treated as independent oscillators, but the
energy of the oscillators are taken quantum mechanically. 
This refers to an isolated oscillator, but the atomic oscillators in a solid are
not isolated.They are continuslly exchanging their energy with their
surrounding atoms.

Even this crude model gave the correct limit at high temperatures, a heat
capacity of 3NkB  3R
Dulong-Petit law where R is universal gas constant.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
• At high temperatures, all crystalline solids have a specific heat
of 6 cal/K per mole; they require 6 calories per mole to raise
their temperature 1 K.
•This arrangement between observation and classical theory
break down if the temperature is not high.
•Observations show that at room temperatures and below the
specific heat of crystalline solids is not a universal constant.
Cv
6
cal
Kmol

kB
T
Cv  3R
In all of these materials (Pb,Al,
Si,and Diamond) specific heat
approaches constant value
asymptotically at high T’s. But
at low T’s, the specific heat
decreases towards zero which
is in a complete contradiction
with the above classical result.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The Discrepancy of Einstein model

Einstein model also gave correctly a specific heat tending
to zero at absolute zero, but the temperature dependence
near T=0 did not agree with experiment.

Taking into account the actual distribution of vibration
frequencies in a solid this discrepancy can be accounted
using one dimensional model of monoatomic lattice
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Having studied the structural arrangements of atoms in solids,
we now turn to properties of solids that arise from collective
vibrations of the atoms about their equilibrium positions.
A. Heat Capacity—Einstein Model (1907)
For a vibrating atom:
kz
E1  K  U
m
 12 mv x2  12 mv y2  12 mv z2  12 k x x 2  12 k y y 2  12 k z z 2
ky
kx
Classical statistical mechanics — equipartition theorem: in thermal
equilibrium each quadratic term in the E has an average energy 12 k BT , so:
E1  6(12 k BT )  3k BT
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Classical Heat Capacity
For a solid with N such atomic oscillators:
Total energy per mole:
Heat capacity at constant
volume per mole is:
E  NE1  3Nk BT
E 3Nk BT

 3N Ak BT  3RT
n
n
d E
   3R  25 molJ K
CV 
dT  n V
This law of Dulong and Petit (1819) is approximately obeyed
by most solids at high T ( > 300 K). But by the middle of the
19th century it was clear that CV  0 as T  0 for solids.
So…what was happening?
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Einstein Uses Planck’s Work
Planck (1900): vibrating oscillators (atoms) in a solid have quantized
energies En  n n  0, 1, 2, ...
[later QM showed En  n  12  
is actually correct]
Einstein (1907): model solid as collection of 3N independent 1-D
oscillators, all with same , and use Planck’s equation for energy levels
occupation of energy level n:
(probability of oscillator
being in level n)
f ( En ) 
e  En / kT

e
classical physics
(Boltzmann factor)
 En / kT
n 0

Average total
energy of solid:

E  U  3 N  f ( En ) En  3 N
n 0
 En / kT
E
e
 n
n 0

e
n 0
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
 En / kT
Some Nifty Summing

Using Planck’s equation:
U  3N
 n e
n 0

e
 n / kT
Now let x 
 n / kT

kT
n 0

U  3 N
ne
n 0

e
 nx
 nx
Which can
be rewritten: U  3N
n 0
Now we can use
the infinite sum:
So we obtain:
d   nx
 e
dx n 0

e
 3N
 nx
n 0

1
x


1 x
n 0
n
for x  1 To give:
 
d  x
 e
dx n 0
n
 e 

x n
n 0
 e 

x n
n 0
d  ex 
  x 
dx  e  1  3N
3N
U  3N


e x  1 e / kT  1
 ex 
 x 
 e 1 
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
1
ex

 x
x
1 e
e 1
At last…the Heat Capacity!
d U 
d  3N A 
CV 
  
  / kT

dT  n V dT  e
1 
Using our previous definition:
Differentiating:
CV 

   3R   e
e  1
 1
 3 N A e  / kT
Now it is traditional to define
an “Einstein temperature”:
So we obtain the prediction:
e
 / kT
E 
 
kT 2
 2
kT
 / kT
 / kT
2
2

k
CV (T ) 
3R
 e
E 2
e
T
E /T
 E /T

1
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
2
Limiting Behavior of CV(T)
High T limit:
E
T
 1
CV



1 
(T ) 
 3R
1  1
3R   e
(T ) 
 3R   e
e 
3R
E 2
E
T
T
2
E
T
Low T limit:
E
T
 1
E 2
CV
E /T
T
E 2
 E /T 2
T
These predictions are qualitatively correct: CV 
3R for large T and CV  0 as T  0:
CV
3R
T/E
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
 E / T
But Let’s Take a Closer Look:
High T behavior:
Reasonable
agreement with
experiment
Low T behavior:
CV  0 too quickly
as T  0 !
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
B. The Debye Model (1912)
Despite its success in reproducing the approach of CV  0 as T  0, the
Einstein model is clearly deficient at very low T. What might be wrong with
the assumptions it makes?
• 3N independent oscillators, all with frequency 
• Discrete allowed energies: En  n
n  0, 1, 2, ...
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Details of the Debye Model
Pieter Debye succeeded Einstein as professor of physics in Zürich, and soon
developed a more sophisticated (but still approximate) treatment of atomic
vibrations in solids.
Debye’s model of a solid:
• 3N normal modes (patterns) of oscillations
• Spectrum of frequencies from  = 0 to max
• Treat solid as continuous elastic medium (ignore details of atomic structure)
This changes the expression for CV
because each mode of oscillation
contributes a frequency-dependent
heat capacity and we now have to
integrate over all :
CV (T ) 
max
N ( ) C


E
( , T ) d
0
# of oscillators per
unit 
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Einstein function
for one oscillator
C. The Continuous Elastic Solid
We can describe a propagating vibration of amplitude u along a rod of
material with Young’s modulus E and density  with the wave equation:
 2u
E  2u

2
t
 x 2
for wave propagation along the x-direction
By comparison to the general form of the 1-D wave equation:
2
 2u
2  u
v
2
t
x 2
  2f  2
we find that
v

v
E

So the wave speed is independent of
wavelength for an elastic medium!
(k ) is called the

 kv
group velocity vg 
d
dk
dispersion relation of
the solid, and here it is
linear (no dispersion!)
k
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Dispersion Relations: Theory vs.
Experiment
In a 3-D atomic lattice we
expect to observe 3 different
branches of the dispersion
relation, since there are two
mutually perpendicular
transverse wave patterns in
addition to the longitudinal
pattern we have considered.
Along different directions in
the reciprocal lattice the
shape of the dispersion
relation is different. But
note the resemblance to the
simple 1-D result we found.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Density of States
According to Quantum Mechanics if a particle is constrained;
 the energy of particle can only have special discrete energy
values.
 it cannot increase infinitely from one value to another.
 it has to go up in steps.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I



These steps can be so small depending on the system that
the energy can be considered as continuous.
This is the case of classical mechanics.
But on atomic scale the energy can only jump by a discrete
amount from one value to another.
Definite energy levels
Steps get small
Energy is continuous
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I



In some cases, each particular energy level can be
associated with more than one different state (or
wavefunction )
This energy level is said to be degenerate.
 ( ) is the number of discrete
The density of states
states per unit energy interval, and so that the number
of states between  and   d will be  ( )d .
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
There are two sets of waves for solution;
 Running waves
 Standing waves
Running waves:

4
L

2
L
0
2
L
4
L
6
L
k
These allowed k wavenumbers corresponds to the running
waves; all positive and negative values of k are allowed. By
means of periodic boundary condition
an integer
Na 2
2
2
L  Na  p   

k 
pk 
p
p
k
Na
L
Length of
the 1D
chain
These allowed wavenumbers are uniformly distibuted in k at a
density of  R  k  between k and k+dk.
running waves
L
 R  k  dk  dk
2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
5
L
Standing waves:
0

L
2
L
k
3
L
6
L
7
L
4
L
0
3
L

L
2
L
In some cases it is more suitable to use standing waves,i.e.
chain with fixed ends. Therefore we will have an integral
number of half wavelengths in the chain;
n
2
2 n
n
L  ;k 
k 
k 
2

2L
L
These are the allowed wavenumbers for standing waves; only
positive values are allowed.
k
2
p
L
for
running waves
k

L
p
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
for
standing waves
These allowed k’s are uniformly distributed between k and
k+dk at a density of S (k )
L
DOS of standing wave
 S (k )dk  dk

 R  k  dk 
L
dk
2
DOS of running wave
•The density of standing wave states is twice that of the running
waves.
•However in the case of standing waves only positive values are
allowed
•Then the total number of states for both running and standing
waves will be the same in a range dk of the magnitude k
•The standing waves have the same dispersion relation as running
waves, and for a chain containing N atoms there are exactly N
distinct states with k values in the range 0 to  / a .
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
E. Counting Modes and Finding N()

A vibrational mode is a vibration of a given wave vector k (and thus ),
E   . How many
frequency  , and energy
are found in the
 modes


interval between (, E, k ) and (  d , E  dE , k  dk ) ?
# modes

dN  N ()d  N ( E)dE  N (k )d k
3
We will first find N(k) by examining allowed values of k. Then we will be
able to calculate N() and evaluate CV in the Debye model.
First step: simplify problem by using periodic boundary conditions for the
linear chain of atoms:
We assume atoms s
and s+N have the
same displacement—
the lattice has periodic
behavior, where N is
very large.
s+N-1
L = Na
s
s+1
x = sa
x = (s+N)a
s+2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
First: finding N(k)
Since atoms s and s+N have the same displacement, we can write:
us  us  N
uei ( ksa t )  uei ( k ( s  N ) at )
This sets a condition on
allowed k values:
kNa  2n 
So the separation between
allowed solutions (k values) is:
Thus, in 1-D:
k
1  eikNa
2n
Na
2
2
k 
n 
Na
Na
n  1, 2, 3, ...
independent of k, so
the density of modes
in k-space is uniform
# of modes
1
Na
L



interval of k  space k 2 2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Next: finding N()
Now for a 3-D lattice we can apply periodic boundary
conditions to a sample of N1 x N2 x N3 atoms:
N3c
# of modes
N a N 2b N 3c
V
 1
 3  N (k )
volume of k  space 2 2 2 8
Now we know from before
that we can write the
differential # of modes as:
We carry out the integration
in k-space by using a
“volume” element made up
of a constant  surface with
thickness dk:
N2b
N1a
 V 3
dN  N ( )d  N (k )d k  3 d k
8
3

d k  ( surface area) dk 
3
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
 dS dk

N() at last!
Rewriting the differential
number of modes in an interval:
We get the result:
N ( ) 
dN  N ( )d 
V
dS dk
3 
8
V
dk
V
1
dS

dS

 
8 3 
d 8 3 
k
A very similar result holds for N(E) using constant energy surfaces for the
density of electron states in a periodic lattice!
This equation gives the prescription for calculating the density of modes
N() if we know the dispersion relation (k).
We can now set up the Debye’s calculation of the heat capacity of a solid.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The density of states per unit frequency range
g():


The number of modes with frequencies  and +d will be
g()d.
g() can be written in terms of S(k) and R(k).
dR
dn
modes with frequency from  to +d corresponds
modes with wavenumber from k to k+dk
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
dn  S (k )dk  g ()d
;
dn  R (k )dk  g ()d
g ( )
Choose standing waves to obtain
g ()  S (k )
dk
d
Let’s remember dispertion relation for 1D monoatomic lattice
4K
2 ka
 
sin
m
2
2
2
K
ka
sin
m
2
1
d
2a K
ka

cos
dk
2 m
2
K
ka
a
cos
m
2
g ()  S (k )
K
ka
a
cos
m
2
1
a
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
m
K
1
 ka 
cos  
 2
1
g ()  S (k )
a
m
1
K cos  ka / 2 
sin 2 x  cos 2 x  1  cos x  1  sin 2 x
1
g ()  S (k )
a
g ()  S (k )
g ( ) 
2L
Na
N

m
K
1
a
1
 ka 
1  sin 2  
 2
4
4
4 K 4 K 2  ka 

sin  
m
m
 2



2 1/ 2
Multibly and divide
Let’s remember:
 S (k )dk 
2
2
max
 ka 
2  ka 
cos    1  sin  
 2
 2
2
1
2
a max
 2
L

dk
L  Na
4K
2
2  ka 
 
sin  
m
 2 
4K
2
max

m
True density of states
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
g ( )
g ( ) 
N

m
K
2N


2
max


2 1/ 2
True density of states by
means of above equation

max
K
2
m
K

m
True DOS(density of states) tends to infinity at max  2
K
,
m
since the group velocity d / dk goes to zero at this value of .
Constant density of states can be obtained by ignoring the dispersion
of sound at wavelengths comparable to atomic spacing.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The energy of lattice vibrations will then be found by
integrating the energy of single oscillator over the
distribution of vibration frequencies. Thus

1
    
2
e
0

 / kT

  g   d
1 
2N
Mean energy of a
harmonic oscillator


2
max


2 1/ 2 for 1D
One can obtain same expression of g ( ) by means of using
running waves.
It should be better to find 3D DOS in order to compare the
results with experiment.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
3D DOS
Let’s do it first for 2D
Then for 3D.
Consider a crystal in the shape of 2D box with crystal lengths of
L.



ky
y
+
L
0
-
-
+
+
L


L
L
kx
x
Standing wave pattern for a 2D
box
Configuration in k-space
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
•Let’s calculate the number of modes within a range of
wavevector k.
•Standing waves are choosen but running waves will lead
same expressions.
•Standing waves will be of the form
U  U 0 sin  k x x  sin  k y y 
• Assuming the boundary conditions of
•Vibration amplitude should vanish at edges of
x  0; y  0; x  L; y  L
Choosing
p
q
kx 
; ky 
L
L
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
positive integer
ky
y
+
+
L
0
+
L

L
L
kx
x
Standing wave pattern for
a 2D box

Configuration in k-space
•The allowed k values lie on a square lattice of side  / L in the
positive quadrant of k-space.
•These values will so be distributed uniformly with a density of
per unit area.  L /  2
• This result can be extended to 3D.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
L
Octant of the crystal:
kx,ky,kz(all have positive values)
The number of standing waves;
L
3
V 3
L 3
s  k  d k    d k  3 d k

 
1
 4 k 2 dk
L /
8
V 1
3
 s  k  d k  3  4 k 2 dk
 8
2
Vk
3

k
d
k  2 dk


ky
s
2
Vk 2
S  k   2
2
3
L
kz
dk
k
kx
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
2
Vk
•  k  
is a new density of states defined as the number
2
2
of states per unit magnitude of in 3D. This eqn can be
obtained by using running waves as well.
• (frequency) space can be related to k-space:
g   d     k  dk
g      k 
dk
d
Let’s find C at low and high temperature by means of using
the expression of g   .
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
High and Low Temperature Limits
Each of the 3N lattice
modes of a crystal
containing N atoms
  3NkBT
This result is true only if


w
d
C
dT
C  3NkB

T=

kB
At low T’s only lattice modes having low frequencies
can be excited from their ground states;
long 
Low frequency
sound waves
0

k
a
  vs k
vs 
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

k

k
1
dk 1
vs    

k
 vs
d  vs
 2 
V 2 
vs  1

g   
2 2 vs
and
Vk 2 dk
g    2
2 d
at low T’s
vs depends on the direction and there are two transverse, one longitudinal acoustic
branch:
V2 1
V2  1 2 
g   
 g   
 3
2
3
2  3
2 vs
2  vL vT 
Velocities of sound in
longitudinal and transverse
direction
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I

1
    
2
e
0


 / kT

  g   d
1 
Zero point energy=  z
1 2
 3  3  d
 / kT
 vL vT 



V  1 2  
3
   z  2  3  3      / kT
d 

2  vL vT    0  e
 1



3
k
T
 B  3
x

 

3
k BT



dx
0 e  / kT  1 d  0 e x  1
1
    
2
e
0

 V
 2
 1  2
2
V  1 2   k BT  
  z  3 2  3  3  4  3 3
2  vL  kvBTT
x 15
d 
dx
4
e 
/ kT
0
1
3
e
0
x
4


k BT
k BT
x
d 
k BT
dx
x
1
at low temperatures
1 2
d V   1 2  4 3
d 2
2


 3  k B 4T
Cv 
 V  kB  3  3
3 P.Ravindran,
3
PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
dT 30  vL vT 
dT 15
 vL vT
2
 4 15
  kBT


How good is the Debye approximation at low T?
 1
d
2
2  k T 
Cv 
 V  2 kB  3  3   B 
dT 15

 vL vT  
3
The lattice heat capacity of solids thus
varies as T 3 at low temperatures; this
is referred to as the Debye T 3
law.
Figure
illustrates
the
excellent
aggrement of this prediction with
experiment for a non-magnetic
insulator. The heat capacity vanishes
more slowly than the exponential
behaviour
of a single harmonic
oscillator because the vibration
spectrum extends down to zero
frequency.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The Debye interpolation scheme
The calculation of g ( ) is a very heavy calculation for 3D, so
it must be calculated numerically.
Debye obtained a good approximation to the resulting heat
capacity by neglecting the dispersion of the acoustic waves, i.e.
assuming
  s k
for arbitrary wavenumber. In a one dimensional crystal this is
equivalent to taking
as given by the broken line of
g ( )
density of states figure rather than full curve. Debye’s
approximation gives the correct answer in either the high and
low temperature limits, and the language associated with it is
still widely used today.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The Debye approximation has two
main steps:
1. Approximate the dispersion relation of any branch by a
linear extrapolation of the small k behaviour:
Einstein
approximation
to the
dispersion
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Debye
approximation
to the   vk
dispersion
Debye cut-off frequency D
2. Ensure the correct number of modes by imposing a cut-off
frequency D , above which there are no modes. The cut-off
freqency is chosen to make the total number of lattice modes
correct. Since there are 3N lattice vibration modes in a crystal
having N atoms, we choose D so that
D

g ( )d   3 N

0
V 1 2
g ( ) 
( 3  3)
2
2 vL vT
2
V 1 2 D 2
(  )  d  3N
2 2 vL3 vT3 0
V
1
2
3N
9N
( 3  3) 3 3 3
2
2 vL vT
D
D
V
1
2
(

) D3  3 N
2
3
3
6
vL
vT
g ( ) 
9N

3
D

2
g ( ) /  2
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
The lattice vibration energy of

1

E   (    / kBT )g ( )d
2
e
1
0
becomes
9N
E 3
D
D
D
3


1

9
N

3
2
0 ( 2   e  / kBT 1) d  D3  0 2 d  0 e  / kBT 1d 


D
and,
9
9N
E  N D  3
8
D
D
 3d
e
0
/ k BT
1
First term is the estimate of the zero point energy, and all T
dependence is in the second term. The heat capacity is obtained by
differentiating above eqn wrt temperature.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
C
The heat capacity is
9
9N
E  N D  3
8
D
D

0
dE
dT
 d
e  / k BT  1
3
dE 9 N
CD   3
dT D
D

0
 4 e  / k BT
d
2
2
kBT  e  / kBT  1
2
Let’s convert this complicated integral into an expression for
x
the specific heat changing variables to
x

k BT
d
kT

dx
and define the Debye temperature

kT
D
D
D 
kB
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
x
The Debye prediction for lattice specific heat
dE 9 N kBT  kBT 
CD 
 3


dT D


4
 T 
CD  9 Nk B 


 D
where
D 



2 
 kBT 
2
3  /T
D

0
D / T

0
x4e x
 e 1
x
x 4e x
e
x
 1
D
kB
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
2
2
dx
dx
How does CD limit at high and low temperatures?
High temperature T = D
X is always small
x2
x3
e  1 x 


2!
3!
x
x 4 (1  x)
2



x
2
2
2
x
x
1

x

1


 e 1
x 4e x
T
=
x 4 (1  x)
 T 
 D  CD  9 NkB 


 D
3  /T
D

x 2 dx  3Nk B
0
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
How does CD limit at high and low temperatures?
Low temperature
T < D
For low temperature the upper limit of the integral is infinite;
the integral is then a known integral of 4 4 /15 .
 T 
<
T  D  CD  9 Nk B 


 D
3  /T
D

0
x 4e x
e
x
 1
2
dx
We obtain the Debye T 3 law in the form
12 NkB 4  T 
CD 


5

 D
3
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Lattice heat capacity due to Debye interpolation
scheme
Figure shows the heat
capacity between the two limits of
high and low T as predicted by the
Debye interpolation formula.
 T 
CD  9 Nk B 

 D 
3 D / T

0
x 4e x
e
x
 1
2
C
3 Nk B T
1
dx
Because it is exact in both high and low T
limits the Debye formula gives quite a good
representation of the heat capacity of most solids, even
though the actual phonon-density of states curve may
differ appreciably from the Debye assumption.
1
T / D
Lattice heat capacity of a solid as
predicted by the Debye interpolation
scheme
 D Debye frequency and Debye temperature scale with the velocity of sound in the
solid. So solids with low densities and large elastic moduli have high  D . Values of  D for
various solids is given in table. Debye energy
D can be used to estimate
the maximum phonon energy in a solid.
Solid
Ar
Na
Cs
Fe
Cu
Pb
C
KCl
D ( K )
93
158
38
457
343
105
2230
235
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
F. The Debye Model Calculation
We know that we need to evaluate an upper limit for the heat capacity integral:
CV (T ) 
max
N ( ) C


E
( , T ) d
0
If the dispersion relation is known, the upper limit will be the maximum  value.
But Debye made several simple assumptions, consistent with a uniform, isotropic,
elastic solid:
• 3 independent polarizations (L, T1, T2) with equal propagation speeds vg
• continuous, elastic solid:  = vgk
• max given by the value that gives the correct number of modes per polarization
(N)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
N() in the Debye Model
d
vg 
dk
First we can evaluate
the density of modes:
N ( ) 
Since the solid is isotropic, all
directions in k-space are the same, so
the constant  surface is a sphere of
radius k, and the integral reduces to:
Giving:
V 2
N ( )  3 4k  2 3
8 vg
2 vg
V
2
V
1
V
dS

dS

3 
3

8
vg 8 vg
k
2
dS

4

k
 
for one polarization
Next we need to find the upper limit for the integral over the allowed range of
frequencies.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
max in the Debye Model
max
Since there are N atoms in the solid, there are N unique
modes of vibration for each polarization. This requires:
N ( )d  N


0
max
3
Vmax
2
 d  2 3  N
Giving:
2 3 
2 vg  0
6 vg
V
max
1/ 3
 6 N 

 vg 
 V 
2
 D
The Debye cutoff frequency
Now the pieces are in place to evaluate the heat capacity using the Debye
model! Remember that there are three polarizations, so you should add a factor
of 3 in the expression for CV. If you follow the instructions in the problem, you
should obtain:
T 
CV (T )  9 Nk B  
 D 
3  /T
D

0
4 z
z e dz
(e z  1) 2
And you should evaluate this
expression in the limits of low T
(T << D) and high T (T >> D).
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Debye Model:
Theory vs. Expt.
Better agreement
than Einstein
model at low T
Universal behavior
for all solids!
Debye temperature
is related to
“stiffness” of solid,
as expected
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Debye Model
at low T:
Theory vs.
Expt.
Quite impressive
agreement with
predicted CV  T3
dependence for Ar!
(noble gas solid)
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Anharmonic Effects




Any real crystal resists compression to a smaller volume than its equilibrium
value more strongly than expansion due to a larger volume.
This is due to the shape of the interatomic potential curve.
This is a departure from Hooke’s law, since harmonic application does not
produce this property.
This is an anharmonic effect due to the higher order terms in potential which
are ignored in harmonic approximation.
V (r )  V (a ) 



r  a
2
2
 d 2V 
 2   ....................
 dr r a
Thermal expansion is an example to the anharmonic effect.
In harmonic approximation phonons do not interact with each other, in the
absence of boundaries, lattice defects and impurities (which also scatter the
phonons), the thermal conductivity is infinite.
In anharmonic effect phonons collide with each other and these collisions
limit thermal conductivity which is due to the flow of phonons.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Phonon-phonon collisions
The coupling of normal modes by the unharmonic terms in the
interatomic forces can be pictured as collisions between the phonons
associated with the modes. A typical collision process of
phonon1
1 , k1
3 , k3
After collision another phonon is produced
phonon2
2 , k2
3  1  2
k3  k1  k2
3  1  2
conservation of energy
conservation of momentum
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
and
k3  k1  k2
Phonons are represented by wavenumbers with

If

a
k

a
k3 lies outside this range add a suitable multible of
2
a
to bring
it back within the range of    k   . Then, k3  k1  k2 becomes
a
This phonon is indistinguishable from a phonon
with wavevector
where
k1 , k2 , and k3

Longitudinal


are all in the above range.

3'

0
a
n  0  Normal process
k
k3
1
2
a
Phonon3 has
n2
k3 
 k1  k2
a
3
1
Transverse
a

a
k


0
3
2

k
a
a
n  0  Umklapp process
(due to anharmonic effects)
; Phonon3 has
k

and Phonon3=Phonon3’
a
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Thermal conduction by phonons
 A flow of heat takes place from a hotter region to a cooler





region when there is a temperature gradient in a solid.
The most important contribution to thermal conduction
comes from the flow of phonons in an electrically insulating
solid.
Transport property is an example of thermal conduction.
Transport property is the process in which the flow of some
quantity occurs.
Thermal conductivity is a transport coefficient and it describes
the flow.
The thermal conductivity of a phonon gas in a solid will be
calculated by means of the elementary kinetic theory of the
transport coefficients of gases.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Kinetic theory
In the elementary kinetic theory of gases, the steady state flux of a property
P
in the z direction is
1 _ dP
flux  l 
3 dz
Angular average
Constant average speed for
molecules
In the simplest case where P is the number density of particles the transport
1
coefficient obtained from above eqn. is the diffusion
D  coefficient
l
3
.
Mean free path
_
If P is the energy density E then the flux W is the heat flow per unit area
so that
1 _ dE 1 _ dE dT
W  l
 l
3 dz 3 dT dz
Now dE / dT is the specific heat C per unit volume, so that the thermal
conductivity;
K 
1 _
l C
3
Works well for a phonon gas
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Heat conduction in a phonon and real gas
The essential differences between the processes of heat
conduction in a phonon and real gas;
Phonon gas
Real gas
•Speed is approximately constant.
•No flow of particles
•Both the number density and energy
density is greater at the hot end.
•Average velocity and kinetic energy per
particle are greater at the hot end, but
the number density is greater at the cold
end, and the energy density is uniform
due to the uniform pressure.
•Heat flow is primarily due to phonon
flow with phonons being created at the
hot end and destroyed at the cold end
hot
cold
•Heat flow is solely by transfer of kinetic
energy from one particle to another in
collisions which is a minor effect in
phonon case.
hot
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
cold
Temperature dependence of thermal
conductivity K
Vanishes exponentially at low
T’s and tends to classical value
at high T’s k B
1 _
K  l C
3
Approximately equal to velocity
of sound and so temperature
independent.
?
•Temperature dependence of phonon mean free length is determined by
phonon-phonon collisions at low temperatures
•Since the heat flow is associated with a flow of phonons, the most
effective collisions for limiting the flow are those in which the phonon
group velocity is reversed. It is the Umklapp processes that have this
property, and these are important in limiting the thermal conductivity
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Conduction at high temperatures

At temperatures much greater then the Debye temperature  D the heat
capacity is given by temperature-independent classical result of
C  3NkB
 phonon density.

The rate of collisions of two phonons

If collisions involving larger number of phonons are important, however, then
the scattering rate will increase more rapidly than this with phonon density.

At high temperatures the average phonon density is constant and
the total lattice energy
T;
phonon number
T , so

Scattering rate  T
and

mean free length 
_
1
Then the thermal conductivity of K  l  C
3

T 1
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
T 1
.

Experimental results do tend towards this behaviour at
high temperatures as shown in figure (a).
10
10
0
0
1
T
10-1
T3
10-1
5 10 20 50 100
T (K )
(a)Thermal conductivity of a
quartz crystal
2
5 10 20 50 100
T (K )
(b)Thermal conductivity of artificial
sapphire rods of different diameters
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Conduction at intermediate temperatures
Referring figure a
At T<  D ; the conductivity rises more steeply with falling temperature,
although the heat capacity is falling in this region. Why?
This is due to the fact that Umklapp processes which will only occur if there
are phonons of sufficient energy to create a phonon with k   / a . So
3
Energy of phonon must be
 the Debye energy ( k D )
The energy of relevant phonons is thus not sharply defined but their
number is expected to vary roughly as
e D / bT when T D ,
where b is a number of order unity 2 or 3. Then
l  eD / bT
This exponential factor dominates any low power of T in thermal
conductivity,
such as a factor of T 3 from the heat capacity.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Conduction at low temperatures
l
for phonon-phonon collisions becomes very long at low T’s and eventually
exceeds the size of the solid, because
number of high energy phonons necessary for Umklapp processes decay
exponentially as
eD / bT
l
is then limited by collisions with the specimen surface, i.e.
l 
Specimen diameter
T dependence of K comes from
12 NkB  T 
CD 


5

 D
4
Cv
which obeys
T 3 law in this region
3
Temperature dependence of
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Cv dominates.
Size effect

When the mean free path becomes comparable to the dimensions of the sample,
transport coefficient depends on the shape and size of the crystal. This is known
as a size effect.

If the specimen is not a perfect crystal and contains imperfections such as
dislocations, grain boundaries and impurities, then these will also scatter
phonons. At the very lowest T’s the dominant phonon wavelength becomes so
long that these imperfections are not effective scatterers, so;
the thermal conductivity has a T 3 dependence at these temperatures.


3
The maximum conductivity between T
imperfections.
and
D / bT
e
region is controlled by
For an impure or polycrystalline specimen the maximum can be broad and low
[figure (a) on pg 59], whereas for a carefully prepared single crystal, as
illustrated in figure(b) on pg 59, the maximum is quite sharp and conductivity
reaches a very high value, of the order that of the metallic copper in which the
conductivity is predominantly due to conduction electrons.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I
Crystal Dynamics

Atomic motions are governed by the forces exerted on
atoms when they are displaced from their equilibrium
positions.

To calculate the forces it is necessary to determine the
wavefunctions and energies of the electrons within the
crystal. Fortunately many important properties of the
atomic motions can be deduced without doing these
calculations.
P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I