Phonons
The Quantum Mechanics of Lattice Vibrations
What is a Phonon?
• We’ve seen that the physics of lattice vibrations in a crystalline solid
reduces to a CLASSICAL normal mode problem.
The goal of the entire discussion has been to
find the normal mode vibrational frequencies of the solid.
• In the harmonic approximation, this is achieved by first writing the
solid’s vibrational energy as a system of coupled simple harmonic
oscillators & then finding the classical normal mode frequencies & ion
displacements for that system.
• Given the results of these classical normal mode calculations for
the vibrating solid, in order to treat some properties of the solid,
it is necessary to QUANTIZE these normal modes.
• These quantized normal modes of vibration are called
PHONONS
• PHONONS are massless quantum mechanical particles which
have no classical analogue.
– They behave like particles in momentum space or k space.
• Phonons are one example of many like this in many areas of physics. Such
quantum mechanical particles are often called
“Quasiparticles”
Examples of other Quasiparticles:
Photons: Quantized Normal Modes of electromagnetic waves.
Rotons: Quantized Normal Modes of molecular rotational excitations.
Magnons: Quantized Normal Modes of magnetic excitations in magnetic solids
Excitons: Quantized Normal Modes of electron-hole pairs
Polaritons: Quantized Normal Modes of electric polarization excitations in solids
+ Many Others!!!
Comparison of Phonons & Photons
PHONONS
PHOTONS
• Quantized normal modes of
• Quantized normal modes of
lattice vibrations. The energies electromagnetic waves. The
& momenta of phonons are
energies & momenta of photons
quantized
are quantized
E phonon 
p phonon
h s

h


Phonon wavelength:
λphonon ≈ a0 ≈ 10-10 m
E photon 
hc
p photon 
h


Photon wavelength (visible):
λphoton ≈ 10-6 m
Quantum Mechanical Simple Harmonic Oscillator
• Quantum mechanical results for a simple harmonic oscillator
with classical frequency ω: The energy is quantized
1

E
 nn   n  

• Energy levels are
equally spaced!
E




2
n = 0,1,2,3,..
Often, we consider En as being constructed by adding n excitation
quanta of energy  to the ground state.
1
E
 0  
2
Ground state energy of the
oscillator.
If the system makes a transition from a lower energy level to a higher
energy level, it is always true that the change in energy is an integer
multiple of 
Phonon absorption or
emission
ΔE = (n – n΄) 
n & n ΄ = integers
In complicated processes, such as phonons interacting with
electrons or photons, it is known that phonons are not conserved.
That is, they can be created and destroyed during such interactions.
Thermal Energy & Lattice Vibrations
As we’ve been discussing in detail, the atoms in a crystal vibrate
about their equilibrium positions.
This motion produces vibrational waves.
The amplitude of this vibrational motion increases
as the temperature increases.
In a solid, the energy associated with these vibrations is called
Thermal Energy
• A knowledge of the thermal energy is fundamental to obtaining an
understanding many of the basic properties of solids. A relevant
question is how do we calculate this thermal energy?
• Also, we would like to know how much thermal energy is available to
scatter a conduction electron in a metal or semiconductor. This is
important; this scattering contributes to electrical resistance in the material.
• Most important, though, this thermal energy plays a fundamental
role in determining the
Thermal Properties of a Solid
• A knowledge of how the thermal energy changes with temperature gives an
understanding of the heat energy which is necessary to raise the
temperature of the material.
• An important, measureable property of a solid is it’s
Specific Heat or Heat Capacity
Lattice Vibrational Contribution to the Heat Capacity
The thermal energy is the dominant contribution to the heat
capacity in most solids. In non-magnetic insulators, it is the only
contribution. Some other contributions:
Conduction Electrons in metals & semiconductors.
The magnetic ordering in magnetic materials.
Calculation of the vibrational contribution to the thermal energy & heat
capacity of a solid has 2 parts:
1. Evaluation of the contribution of a single vibrational mode.
2. Summation over the frequency distribution of the modes.
Vibrational Specific Heat of Solids
cp Data at T = 298 K
Thermal Energy & Heat Capacity
Einstein Model
_
   Pn n
Average energy of a harmonic
oscillator and hence of a lattice
mode of angular frequency at
temperature T
n
Energy of oscillator


1
n   n   
2
The probability of the oscillator being in
this level as given by the Boltzman factor
exp( n / kBT )

_
   Pn n
n
1
 
1



 n    exp    n    / k BT 
_
2
2

 

  n 0 
 
1

exp    n    / k BT 

2
n 0
 


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
x    / kBT
(*)
Eqn (*) can be written
1 z

 k BT 2
(ln z )
z T
T
_
 e   / 2 k BT 
2 
  k BT
ln 

T  1  e   / kBT 
_
  k BT 2
    / 2 k BT
  / k BT

ln
e

ln
1

e




T
_
  

  
  / k BT
  k BT 2   

ln
1

e




T
2
k
T

T
B

 

 k B   / k BT 

 2k  k 2T 2 e
 1
_
e
2
B
B

  k BT  2 2 

  / k BT
4
k
T
2
 B
1  e

1  e


_
  k BT 2
_
Finally, the
result is
1
  
2
e
 / k BT
 / k BT

 / k BT
1


x'
(ln x) 
x
x

_
1
  
2
e
 / k BT
1
This is the Mean Phonon Energy. The first term in the above
equation is the zero-point energy. As mentioned before even at 0ºK
atoms vibrate in the crystal and have zero-point energy. This is the
minimum energy of the system.
The average number of phonons is given by the 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 phonon contribution to the
thermal energy.
kBT

Mean energy of a
harmonic oscillator
as a function of T
1

2
T
Low Temperature Limit
  kBT
1

    
2
e kBT  1
_
_
1
2
  
Since exponential term
gets bigger
Zero Point Energy
kBT
Mean energy of a
harmonic oscillator
as a function of T

1

2
x2
e  1  x   ..........
2!
x
T

e
 1
High Temperature Limit
k BT
_
1

 << kBT
   

2
 is independent of frequency of oscillation.
1
1
k BT
This is the classical limit because the energy
_
1
steps are now small compared with the energy
    k BT
2
of the harmonic oscillator.

k BT
So that  is the thermal energy of the
classical 1D harmonic oscillator.
_
  kBT
Heat Capacity C
• Heat capacity C can be found by differentiating the
average phonon energy _ 1

    
2
e
1

kBT
d
Cv 

dT
Let
 
  kB
 k BT 
e


kBT
e
2

1

k

2
kBT



Cv  k B
2
k
T
 B 
2

e

kBT
e

eT
 
Cv  k B   
 T  e T 1
2


2
k BT

1
2

eT
 
Cv  k B   
 T  e T 1
2


2
where


k
Specific heat in this approximation
vanishes exponentially at low T and
tends to classical value at high
temperatures.
Cv
kB

2
Area =

kB
T
These features are common to all
quantum systems; the energy tends to
the zero-point-energy at low T and to
the classical value of Boltzmann
constant at high 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.
3R
Cv
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!
Cv vs T for Diamond
Points:
Experiment
Curve:
Einstein Model
Prediction
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
averaga 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  24.9
Cv  3R  3  6.02  1023 (atoms / mole)  1.38  1023 ( J / K )
J
;1J  0.2388Cal  Cv
( K  mole)
6
Cal
( K  mole)
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
2
a heat capacity 3N times
e
 

Cv  k B  
T 
T
e

T

1
2
• In this model, the atoms are treated as independent oscillators, but the energy of
the oscillators are taken quantum mechanically as

This refers to an isolated oscillator, but the atomic oscillators in a solid are not
isolated.They are continually exchanging their energy with their surrounding
atoms.
• Even this crude model gave the correct limit at high temperatures, a heat
capacity of the Dulong-Petit law where R is universal gas constant.
3NkB  3R
• 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
Cv  3R
In each of these materials (Pb,Al,
Si,and Diamond) specific heat
approaches constant value
asymptotically at high T. But at
low T’s, the specific heat
decreases towards zero which is
in a complete contradiction with
T the above classical result.
• 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
Thermal Energy & Heat Capacity
Debye Model
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.
• 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
• 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.
 ( )
• The density of states
is the number of discrete
states per unit energy interval, and so that the number
of states between  and   d will be  ( )d  .
There are two sets of waves for solution;
• Running waves
• Standing waves
Running waves:
0

4
L

2
L
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
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;
L
n
2
2 n
n
;k 
k 
k 
2

2L
L
These are the allowed wavenumbers for standing waves; only
positive values are allowed.
2
k
p
L
for
running waves
k

L
p
for
standing waves
These allowed k’s are uniformly distributed between k and k+dk
at a density of  S (k )
 S (k )dk 
 R  k  dk 
L

dk
L
dk
2
DOS of standing wave
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 .
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
dn  S (k )dk  g ( )d
dn   R (k )dk  g ( )d;
Choose standing waves to obtain
g ( )
g ( )   S (k )
dk
d
Let’s remember dispertion relation for 1D monoatomic lattice
4K
2 ka
 
sin
m
2
2
d
2a

dk
2
K
ka
cos
m
2
K
ka
2
sin
m
2
g ( )  S (k )
1
K
ka
a
cos
m
2
1
g ( )   S (k )
a
m
1
K cos  ka / 2 
sin x  cos x  1  cos x  1  sin x
2
2
2
1
g ( )   S (k )
a
m
K
1
 ka 
1  sin 2  
 2 
4
4
 ka 
2  ka 
cos    1  sin  
 2
 2
Multibly and divide
Let’s remember:
g ( )   S (k )
g ( ) 
1
a
2
4 K 4 K 2  ka 

sin  
m
m
 2
1
L 2
2
 a max
2
True density of states
 S (k )dk 
L

dk
L  Na
4K
 ka 
2 
sin 2  
m
 2 
4K
2
max 
m
g ( )
g ( ) 
N

m
K
2N


2
max

True density of states by
means of above equation

max
K
2
m
K

m

2 1/ 2
constant density of states
K
True DOS(density of states) tends to infinity at max  2 ,
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.
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

 for 1D
2 1/ 2
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.
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
•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
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
positive quadrant of k-space.
in
 /the
L
•These values will so be distributed uniformly with a density of
2
L
/

perunit area.

• This result can be extended to 3D.
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
 s  k  d 3k  2 dk
2
ky
Vk 2
S  k   2
2
3
L
kz
dk
k
kx
2
Vk
•  k  
is a new density of states defined as the number of
2
2
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 . 
High and Low Temperature Limits
  3NkBT
•
Each of the 3N lattice
modes of a crystal
containing N atoms
d
C
dT
C  3NkB

This result is true only ifT

kB
At low T’s only lattice modes having low frequencies can be
excited from their ground states;

long 
Low frequency
sound waves
  vs k
0

k
a
vs 

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
depends on the direction and there are two transverse, one
vs
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


1
    
2
e
0

 / kT
1
    
2
e
0

  g   d
1 

 / kT
Zero point energy =  z
2
 V  1 2 
  2  3  3  d
 1  2  vL vT 
x



V  1 2  
3
   z  2  3  3      / kT
d 


2  vL vT   0  e
 1



3
V  1 2   k BT   4
  z  2  3  3 
3
2  vL vT 
15
4

e 
0

e 
3
/ kT
0

/ kT
3
1
1

d  
d 
0
 k BT 
3
 k BT  3

 x kT


B
dx
x
e 1

k BT

k BT
x
d 
k BT
dx
4 
x3
0 e x  1dx
 4 15
 1 2   kBT 
d 2
2
Cv 
 V  kB  3  3  

dT 15
v
v

 L
T 
3
at low temperatures
How good is the Debye approximation at low T?
 1
d
2
2   k BT 
2
Cv 
 V  kB  3  3  

dT 15
v
v


 L
T 
The lattice heat capacity of solids
thus varies as T 3at low
temperatures; this is referred to
as the Debye T 3 law. Figure
illustrates the excellent
agreement 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.
3
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 g ( ) as given by the broken line of
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.
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
Debye
approximation
to the dispersion
  vk
Debye cut-off frequency  D
2. Ensure the correct number of modes by imposing a cutoff 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

V2 1 2
g ( ) 
( 3  3)
2
g ( )d  3N
2 vL vT
0
V
1
2
(

)D3  3 N
2
3
3
6
vL vT
g ( ) 
9N
 D3

V 1 2 D 2
(  )  d  3N
2 2 vL3 vT3 0
V
1
2
3N
9N
(

)

3

2 2 vL3 vT3
D3
D3
2
g ( ) /  2

The lattice vibration energy of
E  (
0
becomes
and,
9N
E 3
D
1

   / kBT )g ( )d
2
e
1
D
D
3


1

9
N

3
2
0 ( 2   e  / kBT  1) d  D3  0 2 d  0 e  / kBT  1d 


D
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.
C
The heat capacity is
9
9N
E  N D  3
8
D
D

0
 d
e  / k BT  1
3
dE
dT
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 the
specific heat changing variables to
x

k BT
d
kT

dx
and define the Debye temperature

D 
D
kB
kT
x
The Debye prediction for lattice specific heat

dE 9 N kBT  kBT  
CD 
 3

 
2 
dT D

  kBT 
4
 T 
CD  9 Nk B 

 D 
where
D 
2
3  /T
D
D
kB

0
D / T

0
x 4e x
 e 1
x
x 4e x
e
x
 1
2
2
dx
dx
How does C D limit at high and low temperatures?
High temperature
x is always small
T
D
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 Nk B 

 D 
3  /T
D

0
x 2 dx  3Nk B
How does C D 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
CD 
12 Nk B  T 


5

 D
4
3
Lattice heat capacity due to Debye interpolation scheme
 T 
CD  9 Nk B 


Figure shows the heat capacity  D 
3 D / T

0
x 4e x
e
x
between the two limits of high and
low T as predicted by the Debye
interpolation formula.
 1
2
dx
C
3 Nk B T
1
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.
Lattice heat capacity of a solid as
predicted by the Debye interpolation
scheme
1
T / 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  Dfor
various solids is given in table. Debye energy D can be used to estimate the
maximum phonon energy in a solid.
Solid
D (K )
Ar
Na
Cs
Fe
Cu
Pb
C
KCl
93
158
38
457
343
105
2230
235