Examples where phonons are important
Energy loss is problem for solar cells
Indirect semiconductors
Describe differences.
Low energy transition can only be excited if phonons present
Real Phonon Spectra Might Look Slightly Different
Why?
We made several simplifications:
•Interactions beyond nearest neighbors are
not included
•Assumed harmonic potential
•Ignored electron-phonon coupling
Predict for Diamond
along one direction (try 001)
http://exciting-code.org/phonon-properties-in-diamond-structure-crystals
Longitudinal Optical
Transversal Optical
degenerate
Longitudinal Acoustic
(0,0,0)
1 1 1
( , , )
4 4 4
fcc lattice with basis
Transversal Acoustic
degenerate
P=2
2x3=6 branches
expected
Can you guess what this is?
What makes this look quite different?
How could we measure this?
All measurements before from neutrons.
Neutron Diffraction for Phonons
Neutrons were discovered in 1932 and their wave
properties was shown in 1936.
E = p2/2m p = h/λ
E=Energy λ=Wavelength
p=Momentum
mn=Mass of neutron = 1.67x10-27kg
λ ~1A°
Energy E~0.08 eV. This energy is of the same order
of magnitude as the thermal energy kT at room
temperature, 0.025 eV, and for this reason they are
good at probing phonons.
Diffraction Methods
Diffraction
X-ray
Neutron
Electron
The physical basis for the diffraction of electron and
neutron beams is the same as that for the diffraction of Xrays, the only difference is the mechanism of scattering.
Neutron Diffraction
• Neutron has no electrical charge. Interaction is with atomic
nuclei (strong force, short range)
• Neutron-matter interaction is weak (mean free path ~1 cm)
• Although uncharged, neutron has an intrinsic magnetic
moment, so it will interact strongly with atoms and ions in
the crystal which also have magnetic moments.
• Neutrons are more useful than X-rays for determining the
crystal structures of solids containing light (low atomic
number) elements.
• Neutron sources in the world are limited so neutron
diffraction is a very special tool.
Oak Ridge, Tennessee
Start with negatively charged hydrogen ions (1 proton and 2 electrons), produced by
an ion source. Ions injected into a linear accelerator.
The ions pass through a foil, which strips off the electrons, converting it to a proton.
The high-energy protons strike a target of liquid mercury, where spallation occurs.
The spalled neutrons are guided through beam lines.
Very expensive and involved experiments
(construction of Oak Ridge neutron source was ~ $1 billion)
Table top alternatives ?
Yes, infrared absorption and
inelastic light scattering (Raman and Brillouin)
However only q~0 accessible (atomic
scattering factors decrease as q2 for light).
See ch. 7 of http://www.slideshare.net/dasepbux/wavediffraction-53527785
Lonely scientist in the reactor hall
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.
Magnetic Applications of Neutron Diffraction
Establishing magnetic structure of materials
– Neutrons carry magnetic moments and their scattering
is affected by magnetic moments of atoms
Example:
Mn
Magnetism of MnO
Mn2+ ions
face-centered cube
O2- ions
in the centers of all edges
and in the body center
What crystal structure?
What happens to the unit
cell when you add the
magnetic structure?
O
Magnetic Super-Structures
(cubic symmetry)
c
c
a
Ferromagnetic (FM)=has a spontaneous
magnetization whose direction can be
changed with a magnetic field
Magnetic and charge have the same unit cell
a
Anti-Ferromagnetic=FM in one plane
but antiparallel in next plane, M=0
Magnetic and charge have the different
unit cells. Magnetic unit cell doubles
nuclear unit cell.
Compare the a can c lattice
parameters to the non-magnetic
version in the case of MnO.
Mn
cnonmagnetic
O
anon-magnetic
XRD of MnO
(from Kittel)
What do you expect to happen?
Chemical unit
cell (FCC)
(all odd or
even h,k,l)
Magnetic
unit cell
15
Like Telescope Time, Beamtime is Competitive
3 pages to describe why
your work is important,
what you expect to learn,
how you will do it and
why you can’t do it
anywhere else!
Ranking scheme:
1. Must do
2. High Priority
3. Medium Priority
4. Low priority
5. Don’t do
Due ~ 1 year before the
work is done!
Cutoff
If You Do Get Time
Magnetism experts
XAS expert
me
I = Installation/Maintanence
You might get three days to do all of your experiments for the year!
Heat Capacity Revisited
(now with phonons)
Revisiting Heat Capacity
• Dustin will discuss a summary of heat capacity
next time, so let me skip to the focus of today
• Drude (spring-like) -> constant heat capacity
• Sommerfield: linear T dependence
At low T quantum phenomena are
important and must be taken into account.
Specific heat tends to classical value at
high temperatures.
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.
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 / kBT )
Energy of a single
oscillator
1

n   n   
2

Average energy
_

 Pn n
n
P
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

Partition function
z  e
 / 2 k BT
 e 3
z  e
 / 2 k BT
z  e
 / 2 k BT
 What is dz/dT?
 / 2 k BT
 e 5
 / 2 k BT
(1  e 
 / k BT
 e 2
 / k BT
(1  e 
 / k BT 1
)
 .....
 .....
 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
  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

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
k BT
1
f ( ) 

e
1

2
T
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.
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
 k BT 

e

kBT
e
2

1

k

2
kBT



Cv  k B
2
 kBT 
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


where
2


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 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

2
Good analogy for low temp phonons. Why?
g (kg(k)
)dk  gg(E)
( E )dE
g(E) dE
g(k)
g (k )  g ( E )
dk
Number of States N=gV (for Photons)
How about for
Phonons?
V2 1
V2  1 2 
Ng   
 3
Ng   
2
3
2  3
2 vs
2  vL vT 
N
Factor of 3 from one
longitudinal and 2 transverse
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  3N
0
V
1
2
(

)D3  3 N
2
3
3
6
vL vT

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 ( ) 
9N
 D3
2
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
 V 2 2cL3  N and
 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
 T 
CD  9 Nk B 

 D 
3  /T
D

0
4
x e
e
x
x
k BT

d
kT

dx
kT
x define the Debye
temperature
x
 1

2
dx
D 
D
kB
How does C D limit at high and low temperatures?
T >> D
High temperature
x2
x3
e  1 x 


2!
3!
 T 
CD  9 Nk B 

 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

3  /T
D

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 .
T <<
 T 
 D  CD  9 Nk B 


 D
We obtain the Debye T
3  /T
D

0
3
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
    
2
e
0

 / kT

  g   d
1 
 1
d
2
2   k BT 
2
Cv 
 V  kB  3  3  

dT 15
v
v


 L
T 
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
3 Nk B T
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
1
4  T 
3Nk
 
V
5 
Quantum effects important
For T<<D cT  
4
3
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 BT 
2
Cv 
 V  kB  3  3  

dT 15
v
v


 L
T 
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
    
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
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 BTF
F
F
3N
Cv  g ( EF )k B T 
k B 2T
2k B TF
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
The energy of diffraction methods
Electron
X-Ray
Neutron
λ = 1A°
λ = 1A°
λ = 2A°
E ~ 104 eV
E ~ 0.08 eV
E ~ 150 eV
interact with electron
interact with nuclei
interact with electron
Penetrating
Highly Penetrating
Less Penetrating
~ Bulk Measurement Bulk Measurement
if sample is thin (<100s nm)
Surface Sensitive
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
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

k BT
_
  kBT
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
vs depends on the direction
V2 1
V2  1 2 
g   
 g   
 3
2
3
2  3
2 vs
2  vL vT 