PHYC - 505: Statistical Mechanics
Midterm Exam 1
Solutions
Thursday, March 6, 2014
11:00 am - 12:15 pm
You may use a one-sided 8-1/2 × 11 cheat sheet. The total points for this exam are 30.
10 pts. Calculate (do not merely state) the partition function of a harmonic oscillator of mass m and frequency
ω (i) classically, and (ii) quantum mechanically. From your result also evaluate the entropy of N such
non-interacting oscillators as a function of temperature.
Classical Case:
For the classical case, we have that the Hamiltonian of a one-dimensional harmonic oscillator of mass
m and frequency ω is written as
p2
1
H=
+ mω 2 x2
2m 2
yielding that the energy, E of such an oscillator is merely
E=
p2
1
+ mω 2 x2
2m 2
The partition function, remembering that for Gaussian integrals we have that
r
Z ∞
π
−ax2
e
dx =
a
−∞
can then be calculated, in the classical regime, as
Z
Z ∞
1 ∞
Z1 =
dx
dpe−βE
h −∞
−∞
Z
Z
2
p
1
−β 2m
+ 12 mω 2 x2
=
dx dpe
h
Z
Z
β
β
2 2
2
1
=
dxe− 2 mω x
dpe− 2m p
h
s
s
1
π
π
=
β
β
2
h 2 mω
2m
2π
hβω
kB T
=
~ω
=
where β = 1/kB T . In order to find the entropy for N such non-interacting oscillators as a function of
temperature, we must first define the partition function for the system of N distinguishable harmonic
oscillators, which is
N
kB T
Z = Z1N =
~ω
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Now we know that the Helmholtz free energy of the system is given as
F = −kB T ln Z
Plugging in our partition function we get that
F = −N kB T ln
kB T
~ω
In order to find the entropy, we recall the relation
S=−
∂F
∂T
We then find that the entropy of N non-interacting oscillators as a function of temperature is calculated
as
S=−
∂F
∂T
kB T
= N kB ln
+ N kB
~ω
kB T
+1
= N kB ln
~ω
Quantum Mechanical Case:
For the quantum mechanical case, we have that the Hamiltonian of a one-dimensional harmonic oscillator of mass m and frequency ω is written as
1
Ĥ = n̂ +
~ω
2
yielding that the energy, E of such an oscillator is merely
1
E = n+
~ω
2
where n takes on integer values from 0 to ∞. Remembering that for x < 1,
∞
X
xn =
n=0
1
1−x
is a geometric series that converges, the partition function, can then be calculated, in the classical
regime, as
Z1 =
=
∞
X
n=0
∞
X
e−βE
e−β (n+ 2 )~ω
1
n=0
= e−β~ω/2
∞
X
e−β~ω
n
n=0
−β~ω/2
=
e
1 − e−β~ω
where β = 1/kB T . In order to find the entropy for N such non-interacting oscillators as a function of
temperature, we must, again, first define the partition function for the system of N -oscillator system,
which is
−β~ω/2 N
e
Z = Z1N =
1 − e−β~ω
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Now we know that the Helmholtz free energy of the system is given as
F = −kB T ln Z
Plugging in our partition function we get that
F =
N ~ω
+ N kB T ln 1 − e−β~ω
2
In order to find the entropy, we recall the relation
S=−
∂F
∂T
We then find that the entropy of N non-interacting oscillators as a function of temperature is calculated
as
S=−
∂F
∂T
~ωe−β~ω
= −N kB ln 1 − e
+ N kB T
1 − e−β~ω
β~ωe−β~ω
−β~ω
= N kB − ln 1 − e
+
1 − e−β~ω
β~ω
−β~ω
= N kB β~ω
− ln 1 − e
e
−1
−β~ω
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1
kB T 2
10 pts A 2-dimensional polarizable material consists of a large number N of non-interacting electric dipoles
each of moment µ, which can be aligned within the plane in ONLY 4 different ways with respect to
an applied electric field E: two of them parallel to the field (directed along and opposite to the field,
respectively) and two of them perpendicular. Neglect the kinetic energy of the dipoles (consider the
entire energy to be given by the interaction of the dipoles with the field) and calculate (i) the electric
polarization (total dipole moment in the direction of the field) and (ii) the specific heat of the system
at temperature T .
For this situation, we know that the energy of an electric dipole with dipole moment µ in an electric
field E is given as
~ = −µE cos θ
E = −~
µ·E
We know that for the 2-dimensional case that the partition function for one of these dipoles is written
as
Z 2π
Z1 =
g(θ)e−βE dθ
0
where g(θ) is the “weight of each state allowable” in the system in a torus of radius θ and θ + dθ. We
know from the configurations given that the only allowed orientations for each dipole are θ = 0, π2 , π,
and 3π
2 , meaning that our g(θ) would have to be
π
3π
g(θ) = δ(θ) + δ θ −
+ δ (θ − π) + δ θ −
2
2
Plugging this into our partition function relation we have that
Z 2π
Z1 =
g(θ)e−βE dθ
0
Z 2π 3π
π
eβµE cos θ dθ
=
+ δ (θ − π) + δ θ −
δ(θ) + δ θ −
2
2
0
= eβµE(1) + βµE(0) + eβµE(−1) + βµE(0)
= eβµE + 1 + e−βµE + 1
= 2 + 2 cosh (βµE)
βµE
= 4 cosh2
2
For N of these particles we have that the partition function for the system of N of these electric dipoles
is
βµE
Z = Z1N = 4N cosh2N
2
The electric polarization, that is, the total dipole moment in the direction of the field is calculated
using the relation
∂
1 ∂Z
P =
ln Z =
∂(βE)
Z ∂(βE)
as
1 ∂Z
1
βµE
∂
2N
N
P =
=
4 cosh
Z ∂(βE)
∂(βE)
2
4N cosh2N βµE
2
1
βµE
βµE µ
2N cosh2N −1
=
sinh
2N βµE
2
2
2
cosh
2
sinh βµE
2
= Nµ
βµE
cosh 2
βµE
= N µ tanh
2
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The specific heat of the system, that is, Cv /N , is calculated by using the relation for the heat capacity
dhEi
d
∂
Cv =
=
−
ln Z
dT
dT
∂β
So for this system we have that
Cv = −
d
dT
1 ∂Z
Z ∂β

∂
d 
1
=−
cosh2N
dT cosh2N βµE ∂β
2


βµE
N
µE
sinh
2
d 

=−
βµE
dT
cosh 2
d
βµE
= −N µE
tanh
dT
2
2
βµE
N kB µE
2
sech
=
2
kB T
2

βµE 
2
Yielding that the specific heat is
kB
Cv
=
N
2
µE
kB T
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2
sech
2
βµE
2
10 pts Consider a 2-dimensional solid in which there are wave motions having the dispersion relation ω = Aq r
between the frequency ω and the wave number q, where A and r are constants. Show in detail that the
specific heat of such a solid at low temperatures goes as T s where T is the temperature, and obtain
the relation between s and r.
Here we want to begin with the Einstein heat capacity for one quantum harmonic oscillator of frequency
ω, which is defined as,
eβ~ω
2
CEinstein (ω) = kB (β~ω)
2,
(eβ~ω − 1)
where our ω is defined as above. Since we know that interactions between the atoms in a solid allows
us to treat the solid as several decoupled harmonic oscillators, each with different frequency, we use
this to calculate the Dybye heat capacity,
Z ωmax
CDebye =
CEinstein (ω)g(ω)dω
0
where g(ω) is the density of modes and ωmax is the maximum frequency of oscillation. In two dimensions
g(ω)dω ∝ d2 q = 2πqdq⊥
which gives us that
g(ω) ∝
2πq
dω
dq⊥
Plugging in ω = Aq r to our expression above, we have that,
ω 1/r
ω 1/r
2π A
2π A
2π ω 2/r−1
g(ω) ∝
=
=
(r−1)/r
rAq r−1
rA A
rA ω
A
∝ω
2/r−1
Plugging this into our expression for the Debye heat capacity we have
Z ωmax
CDebye =
CEinstein (ω)g(ω)dω
0
Z ωmax
eβ~ω
2
2/r−1
dω
∝
kB (β~ω)
2ω
(eβ~ω − 1)
0
2/r
Z ωmax
1
eβ~ω
2
2/r−1
∝
kB (β~ω)
(β~ω)
d (β~ω)
2
β~ω
β~
(e
− 1)
0
2/r Z ωmax
1
eβ~ω
2/r+1
d (β~ω)
∝
2 (β~ω)
β~
(eβ~ω − 1)
0
Now letting x = β~ω, we have that
CDebye ∝
1
β~
2/r Z
xmax
0
ex
(ex − 1)
2/r+1
dx
2x
Now for low temperatures, that is, ~ω >> kB T or x = β~ω >> 1 we can approximate the above
−2
integral to be a constant as xmax → ∞ and (ex − 1) ≈ e−2x
2/r Z ∞
1
e−x x2/r+1 dx
CDebye ∝
β~
0
2/r 2
1
∝
Γ
+2
β~
r
∝ T 2/r
= Ts
So we see that at low temperature the specific heat of such a solid goes as T s , where s = 2r .
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