3Quantum gases

II Statistical Physics 3.3 Phonons and the Debye model
One could still reasonably imagine a “gas of photons” as a gas. In this section,
we are going to study solids. It turns out, this works.
In the first example sheet, we studied a model of a solid, by treating it
as some harmonic oscillators. We saw it was good at high temperatures, but
rubbish at low temperatures. Here we are going to use a better model.
If we have a crystal lattice, then they vibrate, and these give sound waves in
the crystal. In quantum mechanics, we know that light is made out of particles,
called photons. Similarly, in quantum mechanics, sound is made up of “particles”
called phonons. Similar to the case of photons, the energy of a phonon of
frequency ω is
E = ~ω,
and the frequency of a phonon is described by a dispersion relation involving
the wavevector k.
ω = ω(k).
In general, this function is very complicated. We learn how to compute these in
the IID Applications of Quantum Mechanics course, but we will not do that.
Suppose the spacing in the crystal is
a
. If
|k|a
1, then the dispersion
relation becomes linear, and we have
ω |k|c
s
,
where
c
s
is the speed of sound in the solid. This is all very similar to photons,
except we replace c with c
s
.
However, there is some important difference compared to photons. While
photons have two polarizations, phonons have 3. Two of these are transverse,
while the third is longitudinal. As one may know from the IID Waves course,
they travel at different speeds, which we write as
c
T
and
c
L
for transverse and
longitudinal respectively.
We now want to count the number of phonons, and obtain the density of
states. Repeating what we did before, we find
g(ω) dω =
V ω
2
2π
2
c
3
T
+
1
c
3
L
dω.
This is valid for |k|a 1. Alternatively, we need ωa/c
s
1.
It is convenient to write this as
g(ω) dω =
3V ω
2
2π
2
¯c
3
s
dω, ()
where
3
¯c
3
s
=
2
c
3
T
+
1
c
3
L
defines ¯c
s
as the average speed.
The Debye model ignores the
|k|a
1 assumption, and supposes it is also
valid for ωa/c
3
1.
There is another difference between photons and phonons for phonons,
ω
cannot be arbitrarily large. We know high
ω
corresponds to small wavelength,
but if the wavelength gets too short, it is less than the separation of the atoms,
which doesn’t make sense. So we have a minimum possible wavelength set by the
atomic spacing. Consequently, there is a maximum frequency
ω
D
(
D
for
Debye
).
We would expect the minimum wavelength to be
a
V
N
1/3
. So we
expect
ω
D
¯c
s
N
V
1/3
.
Here we are assuming that the different speeds of sound are of similar orders of
magnitude.
Is this actually true? Given a cut-off
ω
0
, the total number of 1-phonon states
is
Z
ω
0
0
g(ω) dω =
V ω
2
0
2π
2
¯c
3
s
.
This is the number of different ways the lattice can vibrate. In other words, the
number of normal modes of the lattice. Thus, to find
ω
0
, it suffices to find the
number of normal modes, and equate it with the above quantity.
Claim. There are in fact 3N normal modes.
Proof. We consider the big vector
X =
x
1
x
2
.
.
.
x
N
,
where each x
i
is the position of the ith atom. Then the Lagrangian is
L =
1
2
˙
X
2
V (X),
where V is the interaction potential that keeps the solid a solid.
We suppose V is minimized at some particular X = X
0
. We let
δX = X X
0
.
Then we can write
L =
1
2
˙
X
2
V
0
1
2
δX
T
VδX + ··· ,
where
V
is the Hessian of
V
at
X
0
, which is a symmetric positive definite matrix
since we are at a minimum. The equation of motion is then, to first order,
¨
X = VδX.
We assume that X = Re(e
t
Q) for some Q and ω, and then this reduces to
VQ =
2
Q.
This is an eigenvalue equation for
V
. Since
V
is a 3
n ×
3
n
symmetric matrix,
it has 3
N
independent eigenvectors, and so we are done, and these are the 3
N
normal modes.
If we wanted to, we can diagonalize
V
, and then the system becomes 3
N
independent harmonic oscillators.
If we accept that there are 3N normal modes, then this tells us
V ω
2
0
2π
2
¯c
3
3
= 3N.
Using this, we can determine
ω
0
=
6π
2
N
V
1/3
¯c
s
,
which is of the form we predicted above by handwaving. If we live in a frequency
higher than this, then we are exciting the highest frequency phonons. We also
define the Debye temperature
T
0
=
~ω
0
k
.
How high is this temperature? This depends only on
N, V
and
¯c
s
, and these
are not hard to find. For example
T
0
100 K
, since lead is a soft
substance and speed is slow. On the other hand,
T
0
2000 K
for diamond, since
it is hard.
We can compute partition functions for these, entirely analogous to that
of photons. The only important difference is that we have that cutoff
ω ω
0
.
What we get is
log Z =
Z
ω
0
0
dω g(ω) log
1 e
β~ω
.
From the partition function, we can get the energy
E =
log Z
β
V
=
Z
ω
0
0
dω ~ωg(ω)
e
β~ω
1
=
3V ~
2π
2
¯c
3
s
Z
ω
0
0
dω ω
3
e
β~ω
1
.
Again we set
x
=
β~ω
. Then the limit becomes
x
=
T
0
/T
. So the integral
becomes
E =
3V (kT )
4
2π
3
(~¯c
s
)
3
Z
T
0
/T
0
dx x
3
e
x
1
.
This is the same as the case of a photon, but the integral has a limit related to
the temperature. This is integral not something we can investigate easily, but
we can easily analyze what happens at extreme temperatures.
If T T
0
, then x takes a very small range, and we can Taylor expand
Z
T
0
/T
0
dx x
3
e
x
1
=
Z
T
0
/T
0
dx (x
2
+ ···) =
1
3
T
0
T
3
.
So we find that E T, and
C
V
=
E
T
V
=
V k
4
T
3
0
2π
2
(~¯c
s
)
3
= 3Nk.
This agrees reasonably well with experiment, and is called the Dulong–Petit
law. This is also predicted by the Einstein’s model. This is essentially just
a consequence of the equipartition of energy, as there are 3
N
degrees of
freedom, and we have kT for each.
If
T T
0
, then we might as well replace the upper limit of the integral by
infinity, and we have
Z
T
0
/T
0
dx x
3
e
x
1
Z
0
dx x
3
e
x
1
=
π
4
15
.
So we have
E T
4
, which is exactly like photons. If we work out the heat
capacity, then we find
C
V
=
2π
2
V k
4
5(~¯c
s
)
3
T
3
= Nk
12π
4
5
T
T
0
3
.
Remarkably, this
T
3
behaviour also agrees with experiment for many
substances, unlike the Einstein model.
This is pattern is observed for most solids. There is one important exception,
which is for metals. In metals, we have electrons which are free to move in the
lattice. In this case, they can also be considered to form a gas, and they are
important at low temperatures.
Note that in the model, we made the assumption
ω |k|c
s
. This is valid
for
ω ω
D
. Because of this, we would expect the Debye model to work at low
temperatures
T T
0
. At high temperature, we saw it also happens to works,
but as we said, this is just equipartition of energy with 3
N
oscillators, and any
model that involves 3N harmonic oscillators should give the same prediction.