1Fundamentals of statistical mechanics

II Statistical Physics

1.3 The canonical ensemble
So far, we have been using the microcanonical ensemble. The underlying as-
sumption is that our system is totally isolated, and we know what the energy of
the system is. However, in real life, this is most likely not the case. Even if we
produce a sealed box of gas, and try to do experiments with it, the system is
not isolated. It can exchange heat with the environment.
On the other hand, there is one thing that is fixed the temperature.
The box is in thermal equilibrium with the environment. If we assume the
environment is “large”, then we can assume that the environment is not really
affected by the box, and so the box is forced to have the same temperature as
the environment.
Let’s try to study this property. Consider a system
S
interacting with a
much larger system
R
. We call this
R
a heat reservoir. Since
R
is assumed to
be large, the energy of
S
is negligible to
R
, and we will assume
R
always has
a fixed temperature
T
. Then in this set up, the systems can exchange energy
without changing T .
As before, we let
|ni
be a basis of microstates with energy
E
n
. We suppose
we fix a total energy
E
total
, and we want to find the total number of microstates
of the combined system with this total energy. To do so, we fix some state
|ni
of
S
, and ask how many states of
S
+
R
there are for which
S
is in
|ni
. We then
later sum over all |ni.
By definition, we can write this as
R
(E
total
E
n
) = exp
k
1
S
R
(E
total
E
n
)
.
By assumption, we know
R
is a much larger system than
S
. So we only get
significant contributions when
E
n
E
total
. In these cases, we can Taylor expand
to write
R
(E
total
E
n
) = exp
k
1
S
R
(E
total
) k
1
S
R
E
V
E
n
.
But we know what
S
R
E
is it is just T
1
. So we finally get
R
(E
total
E
n
) = e
k
1
S
R
(E
total
)
e
βE
n
,
where we define
Notation (β).
β =
1
kT
.
Note that while we derived this this formula under the assumption that
E
n
is small, it is effectively still valid when
E
n
is large, because both sides are very
tiny, and even if they are very tiny in different ways, it doesn’t matter when we
add over all states.
Now we can write the total number of microstates of S + R as
Ω(E
total
) =
X
n
R
(E
total
E
n
) = e
k
1
S
R
(E
total
)
X
n
e
βE
n
.
Note that we are summing over all states, not energy.
We now use the fundamental assumption of statistical mechanics that all
states of
S
+
R
are equally likely. Then we know the probability that
S
is in
state |ni is
p(n) =
R
(E
total
E
n
)
Ω(E
total
)
=
e
βE
n
P
k
e
βE
k
.
This is called the Boltzmann distribution for the canonical ensemble. Note that
at the end, all the details have dropped out apart form the temperature. This
describes the energy distribution of a system with fixed temperature.
Note that if
E
n
kT
=
1
β
, then the exponential is small. So only states
with
E
n
kT
have significant probability. In particular, as
T
0, we have
β , and so only the ground state can be occupied.
We now define an important quantity.
Definition (Partition function). The partition function is
Z =
X
n
e
βE
n
.
It turns out most of the interesting things we are interested in can be expressed
in terms of
Z
and its derivatives. Thus, to understand a general system, what
we will do is to compute the partition function and express it in some familiar
form. Then we can use standard calculus to obtain quantities we are interested
in. To begin with, we have
p(n) =
e
βE
n
Z
.
Proposition. For two non-interacting systems, we have Z(β) = Z
1
(β)Z
2
(β).
Proof. Since the systems are not interacting, we have
Z =
X
n,m
e
β(E
(1)
n
+E
(2)
n
)
=
X
n
e
βE
(1)
n
!
X
n
e
βE
(2)
n
!
= Z
1
Z
2
.
Note that in general, energy is not fixed, but we can compute the average
value:
hEi =
X
n
p(n)E
n
=
X
E
n
e
βE
n
Z
=
β
log Z.
This partial derivative is taken with all
E
i
fixed. Of course, in the real world,
we don’t get to directly change the energy eigenstates and see what happens.
However, they do depend on some “external” parameters, such as the volume
V
,
the magnetic field
B
etc. So when we take this derivative, we have to keep all
those parameters fixed.
We look at the simple case where
V
is the only parameter we can vary. Then
Z = Z(β, V ). We can rewrite the previous formula as
hEi =
β
log Z
V
.
This gives us the average, but we also want to know the variance of
E
. We have
E
2
= h(E hEi)
2
i = hE
2
i hEi
2
.
On the first example sheet, we calculate that this is in fact
E
2
=
2
β
2
log Z
V
=
hEi
β
V
.
We can now convert
β
-derivatives to
T
-derivatives using the chain rule. Then
we get
E
2
= kT
2
hEi
T
V
= kT
2
C
V
.
From this, we can learn something important. We would expect
hEi N
, the
number of particles of the system. But we also know C
V
N. So
E
hEi
1
N
.
Therefore, the fluctuations are negligible if
N
is large enough. This is called the
thermodynamic limit
N
. In this limit, we can ignore the fluctuations in
energy. So we expect the microcanonical ensemble and the canonical ensemble to
give the same result. And for all practical purposes,
N
10
23
is a large number.
Because of that, we are often going to just write E instead of hEi.
Example. Suppose we had particles with
E
= ε, E
= 0.
So for one particle, we have
Z
1
=
X
n
e
βE
n
= 1 + e
βε
= 2e
βε/2
cosh
βε
2
.
If we have
N
non-interacting systems, then since the partition function is
multiplicative, we have
Z = Z
N
1
= 2
n
e
βεN/2
cosh
N
βε
2
.
From the partition function, we can compute
hEi =
d log Z
dβ
=
Nε
2
1 tanh
βε
2
.
We can check that this agrees with the value we computed with the microcanon-
ical ensemble (where we wrote the result using exponentials directly), but the
calculation is much easier.
Entropy
When we first began our journey to statistical physics, the starting point of
everything was the entropy. When we derived the canonical ensemble, we used
the entropy of the everything, including that of the reservoir. However, we are
not really interested in the reservoir, so we need to come up with an alternative
definition of the entropy.
We can motivate our new definition as follows. We use our previous picture
of an ensemble. We have
W
1 many worlds, and our probability distribution
says there are
W p
(
n
) copies of the world living in state
|ni
. We can ask what is
the number of ways of picking a state for each copy of the world so as to reach
this distribution.
We apply the Boltzmann definition of entropy to this counting:
S = k log
This time, is given by
Ω =
W !
Q
n
(W p(n))!
.
We can use Stirling’s approximation, and find that
S
ensemble
= kW
X
n
p(n) log p(n).
This suggests that we should define the entropy of a single copy as follows:
Definition
(Gibbs entropy)
.
The Gibbs entropy of a probability distribution
p(n) is
S = k
X
n
p(n) log p(n).
If the density operator is given by
ˆρ =
X
n
p(n) |nihn|,
then we have
S = Tr(ˆρ log ˆρ).
We now check that this definition makes sense, in that when we have a micro-
canonical ensemble, we do get what we expect.
Example. In the microcanonical ensemble, we have
p(n) =
(
1
Ω(E)
E E
n
E + δE
0 otherwise
Then we have
S = k
X
n:EE
n
E+δE
1
Ω(E)
log
1
Ω(E)
= kΩ(E) ·
1
Ω(E)
log
1
Ω(E)
= k log Ω(E).
So the Gibbs entropy reduces to the Boltzmann entropy.
How about the canonical ensemble?
Example. In the canonical ensemble, we have
p(n) =
e
βE
n
Z
.
Plugging this into the definition, we find that
S = k
X
n
p(n) log
e
βE
n
Z
= k
X
n
p(n)(βE
n
log Z)
= kβhEi + k log Z,
using the fact that
P
p(n) = 1.
Using the formula of the expected energy, we find that this is in fact
S = k
T
(T log Z)
V
.
So again, if we want to compute the entropy, it suffices to find a nice closed
form of Z.
Maximizing entropy
It turns out we can reach the canonical ensemble in a different way. The second
law of thermodynamics suggests we should always seek to maximize entropy. Now
if we take the optimization problem of “maximizing entropy”, what probability
distribution will we end up with?
The answer depends on what constraints we put on the optimization problem.
We can try to maximize
S
Gibbs
over all probability distributions such that
p
(
n
) = 0 unless
E E
n
E
+
δE
. Of course, we also have the constraint
P
p(n) = 1. Then we can use a Lagrange multiplier α and extremize
k
1
S
Gibbs
+ α
X
n
p(n) 1
!
,
Differentiating with respect to p(n) and solving, we get
p(n) = e
α1
.
In particular, this is independent of
n
. So all microstates with energy in this
range are equally likely, and this gives the microcanonical ensemble.
What about the canonical ensemble? It turns out this is obtained by max-
imizing the entropy over all
p
(
n
) such that
hEi
is fixed. The computation is
equally straightforward, and is done on the first example sheet.