Statistical and Low Temperature Physics (PHYS393)
3. Photons and phonons
Kai Hock
2011 - 2012
University of Liverpool
Contents
3.1 Phonons
3.2 Photons
3.3 Einstein’s Model
3.4 Useful Integrals
Photons and phonons
1
3.1 Phonons
Photons and phonons
2
Specific heat capacities of solids
Previously, we have looked at Einstein’s model of the 3-D
simple harmonic oscillator, in which each atom in a solid is in
its own potential well. The predicted heat capacity was
accurate at high temperature, but fails at low temperature.
In 1912, Peter Debye, a Dutch physicist working in Germany,
produced a theory which turns out to be very accurate both
hot and cold.
Photons and phonons
3
Atoms vibrating in a solid
In Debye’s model, we think of the atoms in a solid as
connected by springs. They vibrate in 3D in a complicated way.
The problem can be simplified by treating the vibration as a
superposition of waves of different frequencies.
If we suppose that the atoms are fixed at the edge of the solid,
the frequency would be quantised. A quantum of this vibration
energy is called a phonon.
Photons and phonons
4
Quantised thermal waves
We can count the number of states for the phonons, like we
did for atoms in an ideal gas, or electrons in a metal.
Unlike electrons, however, the exclusion principle does not
apply to phonons.
So each energy state can be occupied by any number of
phonons.
This means that we need yet another way to find the
macrostate. This was worked out by Bose and Einstein in the
1920s.
Photons and phonons
5
Bose-Einsten Statistics
The method is similar to what we have used for the ideal gas.
There, we have made the assumption that there are many
more energy levels than there are particles. This has meant
that it is unlikely for two gas atoms to occupy the same state.
We do not make this assumption now, since we would be
particularly interested in the low temperature behaviour.
Since the exclusion principle does not apply to phonons, more
than one phonons can occupy the same state.
Photons and phonons
6
Bose-Einsten Statistics
Suppose, in the energy bundle i, that there are gi levels and ni
phonons. The phonons are now completely free to arrange
themselves among the energy levels.
This means that we have ni + gi objects altogether in the
bundle. There are ni of one type, and gi of the other type. The
number of possible arrangements would be
Ωi =
(ni + gi)!
ni!gi!
The total number of arrangements for all bundles would be
obtained by multiplication of the number for every bundle:
Ω=
Y (ni + gi)!
ni!gi!
i
Photons and phonons
7
.
Bose-Einstein Distribution
Then we need to maximise ln Ω using the Lagrange multiplier
method.
Using the same constraints as before on the number of
particles N and the total energy U , the method would give the
following answer:
ni
1
.
=
gi
exp(−λ1 − λ2εi) − 1
where λ1 and λ2 are the Lagrange multipliers.
This is called the Bose-Einstein distribution function. It looks
almost the same as the Fermi-Dirac distribution function,
except that it has -1 in the deniminator instead of +1.
When we apply this to phonon, however, we need to make
some changes.
Photons and phonons
8
Bose-Einstein Distribution
Phonons are not real particles like atoms or electrons. The
number of phonons is related to the vibration energy. The
number would increase when the temperature increases. At 0
Kelvin, there is no vibration, so there would be no phonons.
Because of this, there would be no constraint on the number N .
We really only have one constraint - that of constant energy U .
The Lagrange function is ln Ω + λU . N does not appear.
Maximising this function, we get
ni
1
=
.
gi
exp(−λεi) − 1
With the help of entropy as before, we would find
1
λ=−
.
kB T
Photons and phonons
9
Phonon occupation number
Maximising Lagrange function, we have obtained
ni
1
.
=
gi
exp(εi/kB T ) − 1
Following the same notations that we have used for ideal gas
and electrons, we rewrite:
ni as n(ε)dε and
gi as g(ε)dε.
This gives
g(ε)dε
n(ε)dε =
exp(ε/kB T ) − 1
Next, we need a formula for the density of states g(ε).
Note that although we have previously used the same symbol
g(ε) for the particle in a 3-D box, the phonon density of states
would have a different formula.
Photons and phonons
10
density of states for phonons
The vibrations in a solid is described by a wave equation. The
solution for the displacement of the atoms ξ , can be given by
the same function
ξ = sin kxx
in each of the x, y and z directions. Assuming that the
displacements are zero at the edge of the solid, the
wavevectors are discretised in the same way:
nxπ
kx =
.
a
So the density of states can also be given by the same formula:
V k2
g(k) =
.
2π 2
However, two changes are needed from the way we have used it
for the ideal gas and the electrons.
Photons and phonons
11
density of states for phonons
The first change is in the change of variable to energy .
For atoms or electrons, the energy is
~2k2
ε=
.
2m
For vibrations, the energy is given by by the same formula as
for photons:
ε = ~ω,
where ω is the angular frequency 2πf and f is the frequency. ω
is in turn related to k by
ω
v= ,
k
where v is the velocity.
Photons and phonons
12
density of states for phonons
The variable of angular frequency ω is often used instead of
energy ε. For the change of variable, we then use
ω = vk.
We can now do the change of variable using
g(ω)dω = g(k)dk.
Substituting
V k2
g(k) =
,
2
2π
we get the density of states in ω:
V ω2
.
g(ω) =
2
3
2π v
Photons and phonons
13
About polarisation
The second change is the related to the polarisation. Recall
that for electrons, we have to multiply by 2 because of the 2
spin states.
For vibration, a state is a particular frequency, for a particular
set of wavevectors (kx, ky , kz ).
Each state correspond to a wave, and a wave in a solid can
have 2 transverse polarisations, and 1 longitudinal polarisation.
So we have to multiply by 3 to get the correct density of state:
V ω 2dω
g(ω)dω = 3 ×
2π 2v 3
Photons and phonons
14
The Debye Frequency
There is one more practical point to note. In a solid, there is a
limit to the highest phonon frequency.
This is related to the fact that the wavelength cannot possibly
be shorter than the distance between atoms.
Let this highest frequency be ωD .
In a solid with N identical atoms, there are 3N energy states, or
normal modes. This is when we count every possible frequency,
and every possible ”direction of vibration” for each frequency.
This can be proven mathematically, but we shall skip that. We
can find ωD by integrating the density of states:
Z ω
D
0
Photons and phonons
Z ω
2
D 3V ω dω
= 3N
g(ω)dω =
2
3
2π v
0
15
The Debye Frequency
This equation
Z ω
D
0
Z ω
2
D 3V ω dω
g(ω)dω =
= 3N
2
3
2π v
0
can then be solved to find the Debye frequency. Integrating
gives
3
V ωD
= 3N
2π 2v 3
Solving for ωD gives the Debye frequency:
ωD =
Photons and phonons
!1/3
2
3
6N π v
V
16
The phonon distribution
As we have shown earlier, the distribution of phonons is given
by
g(ε)dε
n(ε)dε =
exp(ε/kB T ) − 1
We shall express this in terms of frequency ω as well. The
phonon energy ε is given by
ε = ~ω.
This comes from assuming that the atoms vibrate in a
harmonic potential, so that the energy is quantised in steps of
~ω.
In terms of the frequency ω, the phonon distribution is then
given by
g(ω)dω
n(ω)dω =
.
exp(~ω/kB T ) − 1
Photons and phonons
17
Debye’s theory for Heat capacity of solid
The phonon distribution is given by
g(ω)dω
.
n(ω)dω =
exp(~ω/kB T ) − 1
We can multiply by ~ω to get the energy at this frequency
interval:
g(ω)dω
~ωn(ω)dω =
.
exp(~ω/kB T ) − 1
The total energy is then
Z ω
D
Z ω
D
~ωg(ω)dω
.
exp(~ω/kB T ) − 1
0
0
We have found earlier that the density of states is:
U =
~ωn(ω)dω =
V ω2
.
g(ω) = 3 ×
2
3
2π v
Substituting into the integral for the total energy above ...
Photons and phonons
18
Debye’s theory for Heat capacity of solid
We get:
ωD
3V ~
ω 3dω
U =
2π 2v 3 0 exp(~ω/kB T ) − 1
Z
At low temperature, it can be shown that this is proportional to
T 4. This would imply that the heat capacity C is proportional
to T 3. (This has been mentioned in the topic on the specific
heat of electrons in metal.)
At high temperatures, the heat capacity tends to 3N kB .
Photons and phonons
19
Is verified at high temperature
In 1912, Debye measured the high temperature behaviour of
copper.
He showed that the heat capacity did tend to 3N kB , as he had
predicted.
Photons and phonons
20
And verified at low temperature
In 1953, at Purdue University, Keesom and Pearlman measured
the low temperature behaviour of potassium chloride.
P. H. Keesom and N. Pearlman, Physical Review, vol. 91 (1953), pp. 1354-1355
As predicted by Debye, the heat capacity was indeed
proportional to T 3.
Photons and phonons
21
3.2 Photons
Photons and phonons
22
Radiation from a very hot object
When an object gets very hot, it can give out light.
The temperature of a volcano lava flow can be estimated by
observing its color. The result agrees well with the measured
temperatures of lava flows at about 1,000 to 1,200 deg C.
http://en.wikipedia.org/wiki/Black_body
Photons and phonons
23
An ideal black body
The amount of radiation emitted depends on the nature of the
object - its colour, whether it is smooth or rough, etc.
An ideal black body is one that absorbs all the radiation that
falls on it.
No real material can do this. Soot is about the best, absorbing
all but 3%.
In 1859 Gustav Kirchhoff had a good idea: ”a small hole in the
side of a large box is an excellent absorber, since any radiation
that goes through the hole bounces around inside, a lot getting
absorbed on each bounce, and has little chance of ever getting
out again.”
http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html
The energy of the absorbed radiation reaches equilibrium
among different frequencies in the cavity. This results in a
characteristic spectrum that would be emitted again through
the hole.
Photons and phonons
24
Black Body Spectrum
http://en.wikipedia.org/wiki/Black_body
As the temperature decreases, the peak of the black-body
radiation curve moves to lower intensities and longer
wavelengths. The black-body radiation graph is also compared
with the classical model of Rayleigh and Jeans.
Photons and phonons
25
Photons in a box
We can derive a formula for the black body radiation by
considering the energies of a free photon gas in a box
(following Kirchhoff’s idea).
Photons arise from oscillations in electromagnetic fields, which
is also governed by a wave equation.
The same ideas for phonons can be used, leading to the same
density of states:
V ω2
g(ω) =
,
2
3
2π c
where the speed v is now the speed of light c. This is the
formula before considering polarisations.
Photons and phonons
26
Photons in a box
Like phonons, the photon number is not fixed. We get few
photons when the box is cold, and more photons when it is hot.
Unlike phonons, there is no upper limit to the frequency (no
Debye frequency). The wavelength in a solid cannot be shorter
than the distance between atoms. The electromagnetic wave in
a box has no such limit.
Finally, phonons in a solid can have 3 polarisations: 2
transverse and 1 longitudinal (like sound). Photons can only
have 2: both transverse (e.g. light).
Photons and phonons
27
Counting photon states
Recall that for phonons, the density of states is give by:
V ω 2dω
g(ω)dω = 3 ×
2π 2v 3
where the factor of 3 comes from the 3 polarisations of a
phonon, 2 transverse and 1 longitudinal.
Since photons only have 2 polarisations, both transverse, the 3
should be replaced by a 2:
V ω 2dω
g(ω)dω = 2 ×
2π 2c3
where the sound speed v is replaced by light speed c.
Photons and phonons
28
Distribution of photon number
Like phonons, each energy state can be occupied by any
number of photons.
So it obeys Bose-Einstein statistics.
The number density is therefore given by the same formula:
n()d =
g()d
exp(ε/kB T ) − 1
So the number of photons in a given frequency interval is
V ω 2dω
dω
×
n(ω)dω = 2 ×
2π 2c3
exp(~ω/kB T ) − 1
Photons and phonons
29
Distribution of photon energy
The energy of a photon is ~ω. The energy in the frequency
interval dω is then
V ω 2dω
dω
~ωn(ω)dω = 2 × ~ω ×
×
2π 2c3
exp(~ω/kB T ) − 1
The energy density is given by:
V ~ω 3
1
u(ω) = ~ωn(ω) = 2 3
.
π c exp(~ω/kB T ) − 1
This is essentially Planck’s law for black body radiation.
Integrating gives the total energy:
4
π 2 V kB
4
U =
T
15~3c3
It can be shown that this leads to Stefan’s law of radiation:
η = σT 4
for radiation emitted by an object at a temperature T .
Photons and phonons
30
Distribution of photon energy
Berlin, 1899. In the Imperial Institute of Physics and
Technology, the scientists were just trying to help the German
lighting and heating industries.
Lummer and Pringsheim measured the radiation spectrum from
a blackbody very accurately.
Photons and phonons
31
Blackbody radiation
The results did not look at all like what the physicists thought
it should.
Then Max Planck came along and derived his formula for black
body radiation that agreed very well with these measurements.
And the rest is history.
Photons and phonons
32
What we have learnt so far.
1. The Bose-Einstein distribution function is given by
1
f (ε) =
.
exp[(ε − µ)/kB T ] − 1
2. For phonons and photons, the following form is used:
f (ε) =
1
.
exp(ε/kB T ) − 1
3. The density of states can take different forms:
phonons
V ω 2dω
g(ω)dω = 3 ×
2π 2v 3
photons
V ω 2dω
g(ω)dω = 2 ×
2π 2v 3
bosonic atoms
4mπV
1/2
g(ε) =
(2mε)
h3
Photons and phonons
33
What we have learnt so far.
electrons
4mπV
1/2
g(ε) = 2 ×
(2mε)
h3
classical ideal gas
4mπV
1/2
(2mε)
g(ε) =
h3
3. At low temperature, the specific heat capacity for the Debye
model tends to T 3. At high temperature, it tends to 3N kB .
4. The energy distribution in a photon gas is given by:
V ~ω 3
1
.
u(ω) = ~ωn(ω) = 2 3
π c exp(~ω/kB T ) − 1
This also describes the black body radiation spectrum.
Photons and phonons
34
3.3 Einstein’s model
Photons and phonons
35
Einstein’s model for solids
The Simple Harmonic Oscillator is one of the early attempts to
understand the heat capacity of solids using quantum
mechanics.
It was proposed by Einstein. The idea is to think of - to model
- a solid as atoms vibrating independently of one another other.
Glazer and Wark (2001)
Each atom sits in its own potential well. It could undergo
simple harmonic oscillation in all 3 directions x, y and z.
Vibration in any one direction is independent of vibration in the
other two directions.
Photons and phonons
36
The 1-D case
From quantum mechanics, we know that the energy for a 1-D
simple harmonic oscillator is
εn = (n + 1/2)hν.
where n is an integer, and ν is the frequency.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html
Photons and phonons
37
The equipartition theorem
We have seen that at high temperature, the heat capacity
approaches 3N kB .
This can be understood this using the ”equipartition theorem,”
which states that:
”The mean value of each independent quadratic term in the
energy is equal to 1
2 kB T .”
In our case, ”quadratic term” refers to the kinetic energy
1 mv 2 , or the potential energy of the harmonic oscillator, 1 kx2 .
2
2
We can prove this by finding, for example, the mean kinetic
energy in the Boltzmann distribution:
R 1
2 exp(− 1 mv 2 /k T )dv
mv
1
B
2
2
= kB T
R
1
2
2
exp(− 2 mv /kB T )dv
Photons and phonons
38
The Dulong and Petit Law
The equipartition theorem can help us the understand the
Dulong and Petit Law - that the heat capacity is 3N kB at high
temperature.
There are N atoms.
Each atom vibrates in 3 directions.
In each direction, there is kinetic energy and potential energy.
Each type of energy has a mean value of 1
2 kB T .
The total is U = N × 3 × 2 × 1
2 kB T = 3N kB T .
Therefore, the heat capacity is C = U/T = 3N kB .
Photons and phonons
39
At low temperature
At very low temperature, the atoms are mostly in the ground
state, with some in the first excited state. Only these two levels
are important, so its behaviour is similar to a spin 1/2 salt.
Suppose there are N atoms, with n in the excited state. The
total energy is approximately
hν
3hν
+n
.
2
2
n is small, so N − n is close to N and n is given by the
Boltzmann factor:
hν
3hν
U ≈N
+ exp(−hν/kB T )
.
2
2
The heat capacity than has an exponential form:
U = (N − n)
C=
Photons and phonons
dU
hν
3hν
=
exp(−hν/k
T
)
.
B
2
dT
kB T
2
40
The heat capacity
We can sketch the graph for the heat capacity predicted by
Einstein’s model. It looks like this:
As we have just shown, the heat capacity approaches 3N kB at
high T .
This agrees with the empirical Law of Dulong and Petit, who
measured many solids at room temperature in 1819.
Photons and phonons
41
Compare with experiment
It also appears to agree with experiment at lower temperature.
A. Einstein, Ann. Physik, vol. 22, p. 186 (1907)
However, when the temperature gets too low, it falls off
exponentially.
This does not agree with experiments, which show that C ∝ T 3.
This is because the effects of co-ordinated movements among
atoms become important. A model using phonons is needed.
Photons and phonons
42
3.4 Useful Integrals
Photons and phonons
43
Useful Integrals
For Debye model, to find heat capacity at low temperature;
and for black body radiation, to find Stefan’s law:
Z ∞
0
π4
x3
dx =
.
ex − 1
15
For liquid helium-4, later in the course:to find condensation temperature of Bose Einstein condensate:
Z ∞ 1/2
x
dx = 2.315
0 ex − 1
to the find heat capacity:
Z ∞ 3/2
x
dx = 1.783
x
0 e −1
Photons and phonons
44