Transcript Part I

Chapter 5: Phonons II
– Thermal Properties
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 so far has been to
find the normal mode vibrational frequencies of
the crystalline 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 the classical normal mode calculation for the
lattice vibrations, 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.
• 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”
• 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.
• 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
• 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!!!
• 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
• Quantized normal modes
of lattice vibrations. The
energies & momenta of
phonons are quantized
E phonon 
p phonon
PHOTONS
• Quantized normal modes
of electromagnetic waves.
The energies & momenta
of photons are quantized
h s

h


Phonon Wavelength:
λphonon ≈ a ≈ 10-10 m
E photon 
hc
p photon 
h


Photon Wavelength:
λphoton ≈ 10-6 m >> a
(visible)
Quantum Mechanical Simple Harmonic Oscillator
• Quantum mechanical results for a simple
harmonic oscillator with classical frequency ω:
The energy is quantized.
1

Enn   n  
2

E
The energy
levels are
equally spaced!




n = 0,1,2,3,..
Often, we consider En as being constructed by adding
n excitation quanta of energy  to the ground state.
1
E00 = 
2
Oscillator Ground State
(or “zero point”) Energy.
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
The number of phonons is NOT conserved.
That is, phonons can be created & destroyed during such interactions.
Thermal Energy & Lattice Vibrations
As was already discussed 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 the
Thermal Energy
• A knowledge of the thermal energy is fundamental
to obtaining an understanding many of the basic
properties (thermodynamic properties & others!) of solids.
Examples
Heat Capacity, Entropy, Helmholtz
Free Energy, Equation of State, etc....
• A relevant question is how is this thermal energy
calculated? We would like to know how much
thermal energy is available to scatter a conduction
electron in a metal or a semiconductor. This is
important because this scattering contributes to
electrical resistance & other transport properties.
• How is this thermal energy calculated? We would like to
know how much thermal energy is available to scatter a
conduction electron in a metal or a semiconductor. This is
important because this scattering contributes to electrical
resistance & other transport properties. Most important, the
thermal energy plays a fundamental role in determining the
Thermal (Thermodynamic)
Properties of a Solid
• Knowledge of how the thermal energy changes wih
temperature gives an understanding of heat energy
necessary to raise the temperature of the material. An
important, measureable property of a solid is its
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. Other contributions:
Conduction Electrons in metals & semiconductors.
Magnetic ordering in magnetic materials.
The 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
Historical Background
• In 1819, using room temperature data,
Dulong and Petit empirically found that
the molar heat capacity for solids is
approximately
CV  3R
Here, R is the gas constant.
This relationship is now known as the
Dulong-Petit “Law”
17
The Molar Heat Capacity
• Assume that the heat supplied to a solid is
transformed into the vibrational kinetic &
potential energies of the lattice.
• To explain the Dulong-Petit
“Law” theoretically, using
Classical, MaxwellBoltzmann Statistical
Mechanics, a knowledge of how the
heat is divided up among the degrees
of freedom of the solid is needed.
18
The Molar Heat Capacity
The Molar Energy of a Solid
• The Dulong-Petit “Law” can be explained using
Classical Maxwell-Boltzmann statistical physics.
Specifically,
The Equipartition Theorem can be used.
• This theorem states that, for a system in thermal equilibrium
with a heat reservoir at temperature T,
The thermal average energy per
degree of freedom is (½)kT
1 
If each atom has 6 degrees of E  N A 6  2 kT   3RT


freedom: 3 translational &
R = NA k
3 vibrational, then
19
The Molar Heat Capacity
Heat Capacity at Constant Volume
By definition, the heat capacity of a substance
at constant volume is
E 
CV 

T V
Classical physics therefore predicts:
CV  3R
A value independent of temperature
20
The Molar Heat Capacity
Experimentally, the Dulong-Petit Law, however, is
found to be valid only at high temperatures.
21
Einstein Model of a Vibrating Solid
In 1907, Einstein extended Planck’s ideas to matter:
he proposed that the energy values of atoms are
quantized and proposed the following simple model
of a vibrating solid: Each atom is independent
Each vibrates in 3-dimensions
Each vibrational normal mode
has energy:
En  n , n  0,1, 2,...
  hf
22
In effect, Einstein modeled one mole of a solid
as an assembly of 3NA distinguishable
oscillators. He used the Canonical Ensemble
to calculate the average energy of an oscillator
in this model.

E   En f B ( En )
n 0

 A En e
 En / kT
n 0
23
To compute the average, note that it can be
written as
1 dZ
E 
Z d
Z has the form:


n 0
n 0
Z   f B (n )   e
 n
Here,   [1/(kT)]
Z is called
The Partition Function
24
With   [1/(kT)] and En = n, , the partition
function Z for the Einstein Model is

Z  e
 n
n 0
1

 
1 e
This follows from the
geometric series result

1
S x 
, | x | 1
1 x
n 0
n
25
1 dZ
E 
Z d

Z   e  n  (1  e  ) 1
n 0
Differentiating Z with
respect to  gives:
dZ
 e

 2
d
(1  e )
Multiplying by –1/Z gives:
1 dZ

E 
 
Z d  e 1
This is the Einstein Model Result for the average
thermal energy of an oscillator. The total vibrational
energy of the solid is just 3NA times this result. 26
The Heat Capacity in the Einstein Model
is given by:
E 
CV 

T V
Do the derivative & define TE  /k.
TE is called The Einstein Temperature
Finally, in the Einstein Model, CV has
the form:
2
E
e
 TE 
CV 
 3R   TE / T
2
T
 1)
 T  (e
TE / T
27
The Einstein Model of a Vibrating Solid
Einstein, Annalen der Physik 22 (4), 180 (1907)
CV for Diamond
28
Thermal Energy & Heat Capacity:
The Einstein Model: Another Derivation
• The following assumes that you know enough
statistical physics to have seen the Cannonical
Ensemble & the Boltzmann Distribution!
• The Quantized Energy of a Single Oscillator has
the form:
• If the oscillator interacts with a heat reservoir at
absolute temperature T, the probability Pn of it
being in quantum level n is proportional to the
Boltzmann Factor:
Pn  exp( n / kBT )
Quantized Energy of a Single Oscillator:
• In the Cannonical Ensemble, a formal expression for the
average energy of the harmonic oscillator & therefore of
a lattice normal mode of angular frequency ω at
temperature T is given by: _
   Pn n
n
• The probability Pn of the oscillator being in quantum
level n has the form:
Pn  [exp (-β)/Z]
where the partition function Z is given by:


n 0
n 0
Z   f B (n )   e  n
Now, some straightforward math manipulation!
Average Energy:
_
   Pn n
n
Putting in the explicit form gives:
1
 
1


n


exp

n


/
k
T




B 
 
_
2
2

n 0 




 
1

exp

n


/
k
T


B 
 
2
n 0
 


x    / kBT
According to the Binomial expansion, for x << 1 where
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
)
 .....
 .....
The equation for ε can be rewritten:
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 BT
1  e
 2
1  e


_
  k BT 2
Finally, the
result is:
_
1
  
2
e

 / k BT
1
 / k BT
 / k BT

1

(1)
     / k BT
2
e
1
• This is the Mean Phonon Energy. The first term in
(1) is called the Zero-Point Energy. As mentioned
_
before, even at 0ºK the atoms vibrate in the crystal &
have a zero-point energy. This is the minimum energy
of the system.
• The thermal average number of phonons n(ω) at
Temperature T is given by The Bose-Einstein
Distribution, & the denominator of the second
term in (1) is often written:
1
n( ) 

e
kBT
1
_
1
  
2
e

 / k BT
1
(1)
1
n( ) 

e
kBT
1
(2)
• By using (2) in (1), (1) can be rewritten:
<> = ћω[n() + ½]
• In this form, the mean energy <> looks analogous to a
quantum mechanical energy level for a simple harmonic
oscillator. That is, it looks similar to:
• So the second term in the mean energy (1) is interpreted as
The number of phonons at temperature
T & frequency ω.
Temperature dependence of
the mean energy <> of a
quantum harmonic oscillator.

1

2
Taylor’s series
2
x
x
expansion of ex  e  1  x   ..........
_
2!
1

    /k T
for x << 1
_
1

2
e
1







k BT

2
e

1

T
1
1
k BT
k BT
kBT
B
High Temperature Limit:
ħω << kBT
At high T, <> is independent of ω. This high T
limit is equivalent to the classical limit, (the
energy steps are small compared to the total
energy). So, in this case, <> is the thermal
energy of the classical 1D harmonic
oscillator (given by the equipartition theorem).
_
1
    k BT
2
_
  kBT
Temperature dependence of
the mean energy <> of a
quantum harmonic oscillator.

1

2
kBT

_
1
  
2
e
 / k BT
1
T
Low Temperature Limit:
ħω > > kBT
“Zero Point
Energy”
At low T, the exponential in the denominator of
the 2nd term gets larger as T gets smaller. At small
enough T, neglect 1 in the denominator. Then,
the 2nd term is e-x, x = (ħω/(kBT). At very small
T, e-x  0. So, in this case, <> is
independent of T:
<> 
(½)ħω
_
1
  
2
Einstein Heat Capacity CV
• The heat capacity CV is found by differentiating
the average phonon energy:
_
1

     / k BT
2
e
1
d
Cv 

dT
 

  kB
 k BT 

e
kBT

e
2

1
2
kBT


Cv  k B
2
 kBT 
Let

k

k BT
e

e
kBT

1
2

eT
 
Cv  k B   
 T  e T 1
2


2


2
Einstein Heat Capacity CV

eT
 
Cv  k B   
 T  e T 1
2


where

2

k
Cv
kB
Area =

kB

2
T
The specific heat CV in this
approximation vanishes
exponentially at low T &
tends to the classical value
at high T. These features
are common to all
quantum systems; the
energy tends to the zeropoint-energy at low T & to
the classical value at high T.
• The specific heat at constant volume Cv depends
qualitatively on temperature T as shown in the
figure below. For high temperatures, Cv (per mole) is
close to 3R (R = universal gas constant. R  2 cal/K- mole).
So, at high temperatures Cv  6 cal/K-mole
The figure shows that
3R
Cv = 3R
Cv
T, K
at high temperatures for all
substances. This is the classical
Dulong-Petit law.
This states that specific heat of a
given number of atoms of any solid
is independent of temperature & is
the same for all materials!
Einstein Model for Lattice Vibrations in a Solid
Cv vs T for Diamond
Einstein, Annalen der Physik 22 (4), 180 (1907)
Points:
Experiment
Curve:
Einstein
Model
Prediction
Einstein Model of Heat Capacity of Solids
• The Einstein Model was the first quantum theory of lattice
vibrations in solids. He made the 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

eT
 
Cv  k B   
 T  e T 1
2


3NkB  3R
2
• In this model, the atoms are treated as independent oscillators,
but the energies of the oscillators are the quantum mechanical
energies. This assumes that the atoms are each isolated
oscillators, which is not at all realistic. In reality, they are a
huge number of coupled oscillators. Even this crude model
gives the correct limit at high temperatures, where it
reproduces the Dulong-Petit law of 3R per mole.
At high temperatures, all crystalline solids have a
vibrational 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
breaks 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
In each of these materials
(Pb,Al, Si,and Diamond)
specific heat approaches a
constant value asymptotically
at high T. But at low T, the
specific heat decreases
towards zero which is in a
T complete contradiction with
Cv  3R
the above classical result.
• The Einstein model also gives correctly a
specific heat tending to zero at absolute zero,
but the temperature dependence near T= 0 does
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 correspond to the running waves;
all positive and negative values of k are allowed. By means of
periodic boundary condition
Na 2
2
2
L  Na  p   

k 
pk 
p
p
k
Na
L
an integer
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
 &  + d will be g()d.
g() can be written in terms of S(k) and R(k).
dR modes with frequency from  to +d corresponds to
dn 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


2 1/ 2
True density of states by
means of above equation

max
K
2
m
K

m
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 sides 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  / L in the
positive quadrant of k-space.
These values will so be distributed uniformly with a density
of  L /  2 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
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  .
High and Low Temperature Limits
  3NkBT
Each of the 3N
lattice modes of a
crystal containing N
atoms
d
C
dT
T
=
C  3NkB


kB
This result is true only if 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
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


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
Debye Model of Heat Capacity of Solids
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 T3 at low temperatures; this
is referred to as the Debye T3 law.
The 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
approximatio
n to the
dispersion
  vk
Debye cut-off frequency  D
2. E nsure 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
x 4 (1  x)
 T 
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
D
T
For low temperature the upper limit of the integral is infinite; the
integral is then a known integral of
.
4 4 /15
 T 
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
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
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  D for 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