#### Transcript Slide 1

```Phonon spectroscopy =
Conditions for: in elastic scattering
Constraints:
Conservation laws of
Momentum
Energy
In all interactions involving phonons, energy must be
conserved and crystal momentum must be conserved to within
a reciprocal lattice vector.
Phonons have important effects on
Solar cells
Heat Capacity Revisited
(now with phonons)
Revisiting Heat Capacity
In Chapters 1&2,
g
f(E) =
1
𝐸−𝜇
exp(
)±1
𝑘 𝑇
Plus for fermions, minus for bosons
Planck distribution
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.
Classical Specific Heat (Spring Model)
For phonon gas, classically expect a temperature independent result.
Cv  3R  3  6.02 1023 (atoms / mole) 1.38 1023 ( J / K )
U= 3NkBT  3RT
Dulong-Petit 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 at high temperature!
At low T quantum phenomena are important and must be taken
into account.
Specific heat tends to classical value at high temperatures.
Thermal Energy & Heat Capacity
Quantum Model
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.
Treating Phonons like QM Oscillator
Einstein Model = Heat capacity C can be found by
differentiating the QM average phonon energy
   Pn n
The energy of a harmonic
oscillator and hence of a
lattice mode of angular
frequency  at temperature T
n
What happens to  as the temperature increases?
The probability of the
oscillator being in this level as
given by the Boltzman factor
exp(  n / k BT )
Energy of a single
oscillator
1

n   n   
2

Average energy
_

 Pn n
n
P
n
n
1
 
1



 n    exp    n    / kBT 
_
2
2

n 0 




 
1

exp    n    / kBT 

2
n 0
 


1 
z   exp[(n  )
]
2 kBT
n 0

Partition function
 What is dz/dT?
z  e
 / 2 k BT
 e3  / 2 kBT  e5
z  e
 / 2 k BT
(1  e
 / k BT
z  e
 / 2 k BT
(1  e
 / k BT 1
 e2
)
 / 2 k BT
 / k BT
 .....
 .....
 Will need soon
According to the Binomial expansion for x«1 where
xe

 h / k BT
1
x 

1 x
n 0
n
1
 
1


n


exp

n


/
k
T




B 
 
_
2
2

n 0 




 
1

exp

n


/
k
T


B 
 
2
n 0
 


Can be written as:
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
2
  k BT   
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

x'
(ln x) 
x
x
 / k BT
 / k BT


 / k BT
1

_
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.
The number of phonons is given by the Bose-Einstein distribution
as
Planck distribution =
1
f ( ) 

e
kBT
1
_
(number of phonons) x (energy of phonon) = (second term in  )
The second term in the mean energy is the phonon contribution to the
thermal energy.
k BT

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
High Temperature Limit
 << kBT
 is assumed independent of frequency of
oscillation by the Einstein method.
This is the classical limit because the energy
steps are now small compared with the energy
of the harmonic oscillator.
Notice that  is the thermal energy of the
classical 1D harmonic oscillator.

e
 1
k BT
_
1

   

2
1
1
k BT
_
1
    k BT
2

kBT
_
  kBT
Heat Capacity C (Einstein Model)
• Heat capacity C can be found by differentiating the
average phonon energy
1

    
2
e kBT  1
_
d
Cv 

dT
Let
 
 kB
 kBT 
e


kBT
e
2

1

k

2
kBT



Cv  kB
2
k
T
 B 
2
e
e

kBT

eT
 
Cv  k B   
 T  e T 1
2


2
kBT

1
2

eT
 
Cv  k B   
 T  e T 1
2


where
2


k
Specific heat in this approximation
vanishes exponentially at low T and
tends to classical value at high
temperatures.
Cv
kB

Area =
2

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 the Boltzmann
constant at high T.
The difference between classical and Einstein
models comes from zero point energy.
Einstein Model Fails at Low Temp
• 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. Why?
• Taking into account the actual distribution of vibration
frequencies in a solid this discrepancy can be accounted for
Einstein was aware that getting
the frequency of the actual
oscillations would be different.
He proposed this theory
because it was a simple
demonstration that QM
could solve the specific heat
problem.
CV of Diamond
Finding the 3d density of states D(E) for Photons
g (k ) 
k
2
Good analogy for low temp phonons. Why?
2
g (kg(k)
)dk  gg(E)
( E )dE
dE
g (k )  g ( E )
dk
g(k)
g(E)
Number of States N=gV (for Photons)
N
Factor of 3 from one
longitudinal and 2 transverse
When k is not small, DOS bigger
Total energy determination tricky
Gradient goes to zero at edges of Brillouin Zone
Like electronic density of
states, there will be a van
Hove singularity in the
phonon density of states.
Another View of the Low Temperature Limit
At low T’s only lattice modes having low frequencies can be
excited from their ground states, since the Bose-Einstein function
falls to zero for small values of 

Low frequency
sound waves
at low T
0

  vs k

vs 
k
k
a
Vk 2 dk
g    2
2 d

k 1
dk 1
vs    

k
 vs
d vs
 2 
V 2 
vs  1

g   
2 2 vs
v s depends on the direction
V2 1
V2  1 2 
g   
 g   
 3
2
3
2  3
2 vs
2  vL vT 
Debye approximation has two main steps
1. Approx. dispersion relation of any branch by a linear extrapolation
2. Ensure 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:

Einstein
Debye
D
approximation to
the dispersion
approximation
to the dispersion
  vk

g ( )d   3 N

0
V
1
2
(

)D3  3N
2
3
3
6
vL vT
g ( ) 
9N
D3
2
V 1 2 D 2
( 3  3 )   d  3N
2
2 vL vT 0
V 1
2
3N
9N
(

)

3

2 2 vL3 vT3
D3
D3
g () /  2
Accordingly, the Debye spectrum may be written as
9 N 2
 3
g( )    D
0

for
for
Einstein
Debye
  D
  D
Actual
Notes:
The Debye spectrum is only an idealization of the actual situation obtaining in a
solid; it may be compared with a typical spectrum.
While for low-frequency modes (the so called acoustical modes) the Debye
approximation is reasonably valid, there are serious discrepancies in the case
of high-frequency modes ( the optical modes).
For “averaged” quantities, such as the specific heat, the finer details of the
spectrum are not very important.
Though not as common, the longitudinal and the transverse modes of the solid
should have their own cut-off frequencies,
D,L and D,T say, rather than
D. Accordingly, we should have:
 D ,L
 D ,T
2
 d
 2 d
0 V 2 2cL3  N and
0 2 2cT3  2 N
having a common cut-off at
Debye Lattice Energy and Heat Capacity

1

E   (    / kBT )g ()d
2
e
1
0
The lattice vibration energy of
becomes
and,
9N
E 3
D
g ( ) 
9N
D3
2
D
D
3
3


1

9
N


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
/ k BT
0
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.
dE 9 N
CD   3
dT D
D

0
 4 e  / k BT
d
2
2
kBT  e  / kBT  1
2
3  /T
D
 T 
CD  9 NkB 


 D

0
4
x e
e
x
x
 1
d
kT

dx
k BT
kT

x define the Debye
temperature
x
2

dx
D 
D
kB
How does CD limit at high and low temperatures?
T >> D
High temperature
x2
x3
e  1 x 


2!
3!
3  /T
D
 T 
CD  9 NkB 

 D 
x
x (1  x)
x (1  x) ~ 2


x
2
2
2
x
x
 e 1 1  x 1
4 x
4
xe
T
>>
4
 T 
 D  CD  9 Nk B 


 D
x
3  /T
D

0


0
x 4e x
e
x
 1
2
dx
is small
k BT
x 2 dx  3Nk B
T << D
Low temperature
For low temperature the upper limit of the integral is ~infinite; the
integral is then a known integral of 4 4 /15 .
3  /T
D
T <<
 T 
D  CD  9 NkB 


 D
We obtain the Debye T
3

0
x 4e x
e
x
 1
2
dx
law in the form
12 Nk B 4  T 
CD 


5

 D
3
D 
D
kB
How good is the Debye approximation at low T?

 
1
       / kT   g   d
2
e
1 
0
 1
d
2
2  k T 
Cv 

V  2 kB  3  3   B 
dT 15

 vL vT  
3
The lattice heat capacity of solids varies
as T 3at low temperatures; Debye law.
Excellent agreement for
non-magnetic insulators!
Motivation for Debye: The exact calculation of DOS is difficult for 3D.
Debye obtained a good approximation to the heat capacity by neglecting
the dispersion of the acoustic waves, i.e. assuming   s k
for arbitrary wavenumber.
In a 1D crystal, Debye’s approximation gives the correct answer in both
the high and low temperature limits.
This approximation is still widely used today.
Lattice heat capacity due to Debye scheme
 T 
CD  9 Nk B 

 D 
3 D / T

0
x 4e x
e
x
 1
2
dx
C
3NkBT
Debye formula gives quite a good
representation of the heat capacity of most
solids, even though the actual phonondensity of states curve may differ
appreciably from the Debye assumption.
1
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 . Debye energy  D
can be used to estimate the maximum phonon energy in a solid.
Solid
Ar
D (K ) 93
Na
Cs
Fe
Cu
Pb
C
KCl
158
38
457
343
105
2230
235
Summary: Debye Specific Heat
1
c  3 Nk
For T>>D
V
Same as a classical non-interacting gas
For T<<D cT  
1
4  T 
3Nk
 
V
5 
4
3
Quantum effects important
The assumptions of Debye theory are:
• the crystal is harmonic
• elastic waves in the crystal are non-dispersive
• the crystal is isotropic (no directional dependence)
• there is a high-frequency cut-off determined by the number of
degrees of freedom
Comparing Phonon and Electron
contributions to heat capacity (for metals)
Phonon contribution
Electron contribution
 1
d
2
2   k BT 
2
Cv 

V  kB  3  3  

dT 15
v
v

 L
T 
3
Cv=
𝜋2𝑘𝐵 𝑇
2 𝑇𝐹
T > Debye temp
• If each unit cell contributes one or two electrons, then
the ratio is about kBT/EF (small for most metals)
• Thus, electrons do not contribute much to room
temperature heat capacity (significant at low temp)
Class Discussion
Experimentally, the specific heat of EuO at very
low temperatures is proportional to T3/2.
• Is EuO a metal or an insulator? Why?
• Any idea what type of elementary excitation this
behavior results from? (you may not know)
• What was the temperature dependence in metals?
• Remember what materials Debye was good for?
• Magnons with an energy dependence of k2
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  
The lattice heat capacity of solids thus
varies as T 3 at low temperatures; this is
3
referred to as the Debye T 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
Reminder: High and Low Temperature Limits
  3NkBT
•
Each of the 3N lattice
modes of a crystal
containing N atoms
d
C
dT
This result is true only if T >>
C  3NkB


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 
k
Calculating Specific Heat Capacity

 
1
       / kT   g   d
2
e
1 
0

1
    
2
e
0

 / 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
 kBT 
3
 kBT  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   k BT 
d
2
2
Cv 
 V  kB  3  3  

dT 15

 vL vT  
3
at low temperatures
Example
Calculate the temperature at which the photon
energy density in a solid will exceed the phonon
energy density in the same solid, if the solid has
a Debye temperature of 300K, a phonon velocity
of 400,000 cm/s and an index of refraction of
n=2
and assuming the solid does not melt.
Should we consider the high or low temperature regime?
Heat capacity of the free electron gas
• From the diagram of n(E,T) the change in the distribution of
electrons can be resembled into triangles of height ½ g(EF) and
a base of 2kBT so the area gives that ½ g(EF)kBT electrons
increased their energy by kBT.

The difference in thermal energy
from the value at T=0°K
E (T )  E (0) ~
1
g ( EF )(k BT ) 2
2
• Differentiating with respect to T gives the heat
capacity at constant volume,
E
Cv 
 g ( EF )k B 2T
T
g (E) 
E
E
2
V
N

g
(
E
)
dE

N  EF g ( EF )
0
0 2 2
3
3 N
3N
g ( EF ) 

2 EF 2k B TF
F
F
3N
Cv  g ( EF )kB T 
k B 2T
2kBTF
2
V
2 2
3/ 2 1/ 2
(2
m
)
E dE 
3
3/ 2 1/ 2
(2
m
)
E
3
V
3 2
3/ 2
(2
mE
)
F
3
T 
3
Cv  Nk B  
2
 TF 
Heat capacity of
Free electron gas
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