Tutorial 4
Problem 1. Show that when a system is in thermal and diffusive equilibrium with a reservoir,
the average number of particles in the system is
N̄ =
kT ∂Z
,
Z ∂µ
(1)
where the partial derivative is taken at fixed temperature and volume. Show also that the mean
square number of particles is,
(kT )2 ∂ 2 Z
N¯2 =
.
Z ∂µ2
(2)
Use these results to show that the standard deviation of N is,
q
σN = kT (∂ N̄ /∂µ),
(3)
in analogy with problem 6.18. Finally apply this formula to an ideal gas, to obtain an expression
for σN in terms of N̄ . Discuss your results briefly.
Answer. We start with the Grand canonical partition function,
Z = Σs e−(E(s)−µ N (s))/kT
(4)
Then
N̄ = Σs N (s)
N¯2 = Σs N 2 (s)
e−(E(s)−µ N (s))/kT
Z
e−(E(s)−µ N (s))/kT
Z
=
=
kT ∂Z
Z ∂µ
(5)
(kT )2 ∂ 2 Z
.
Z ∂ 2µ
(6)
So,
s
σN =
(kT )2
∂ 2Z
Z
∂µ2
−
kT ∂Z
Z ∂µ
s
=
(kT )2
∂ 2Z
Z
∂µ2
+
∂Z ∂
∂µ ∂µ
(kT )2
Z
s
=
kT
∂ N̄
.
∂µ
(7)
Let’s Now consider ideal gas with the number of particles is N = N̄ . From eqn 6.93, we have,
V Zint
µ = −kT ln
.
(8)
N̄ vQ
Now one can calculate the σN for ideal gas:
−1
p
∂ N̄
N̄
∂µ
=
=
⇒ σN = N̄ .
∂µ
kT
∂ N̄
(9)
This means that the deviation of particles is same as the average of particles and in this case
the distribution is called Poisson distribution.
Problem 2. Suppose you have a ”box” in which each particle may occupy any of 10 singleparticle states. For simplicity, assume that each of these states has energy zero.
• What is the partition function of this system if the box contains only one particle?
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Ans. Since the states has energy zero, each Boltzmann factor equals to 1 and it means
that the partition function Z is same as the total number of states (Ω). So, in this case,
Z1 = 10.
(10)
• What is the partition function of this system if the box contains two distinguishable
particles?
Ans.
If there are two distinguishable particles,
Z2 = Z12 = 100.
(11)
• What is the partition function if the box contains two identical bosons?
Ans.
There are two cases, one is that the particles are different states and another is
that the particles are same state. The number of possible states are 45 (=10 C 2)for the
first case and 10 for the second case. Then, Z3 = 10C2 + 10 = 55
• What is the partition function if the box contains two identical fermions?
Ans. Both fermions cannot sit in the same place due to pauli exclusion principle, therefore we only have one case to consider here as compared to the previous one. Therefore
we have,
Z4 = 10C2 = 45.
(12)
• What would be the partition function of this system according to equation 7.16?
Ans.
If there are two indistinguishable non interacting particles,
Z = Z12 /2! = 50.
(13)
ZN
This shows that the formula for N particle partition function ZN = N1! is only an
approximation, and that it actually matters what type of indistinguishable particles
(Bosons/Fermions) they actually are. (Compare the answer with the previous answers.)
• What is the probability of finding both particles in the same single-particle state, for the
three cases of distinguishable particles, identical bosons, and identical fermions?
Ans.
Consider the case that the both particles in the same state, there are 10 possible states for distinguishable particles, also 10 possible states for bosons, and 0 state
for fermions. Then, the probability to find both particle in same state is 1/10 for distinguishable particles, 2/11 for identical bosons and 0 for identical fermions
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Problem 3. Consider a system of five particles, inside a container where the allowed energy
levels are non degenerate and evenly spaced. For instance, the particles could be trapped in
a one-dimensional harmonic oscillator potential. In this problem you will consider the allowed
states for this system, depending on whether the particles are identical fermions, identical
bosons, or distinguishable particles.
• Describe the ground state of this system, for each of these three cases.
Let s1 , s2 , s3 , s4 , s5 , ... etc be the first five energy states of the system in the order of
increasing energy. The ground state of distinguishable particles and bosons is then all
the five particles in the state s1 . for fermions, considering the Pauli exclusion principle,
the ground state is the one with the first five states states s1 to s5 singly occupied.
• Suppose that the system has one unit of energy (above the ground state). Describe the
allowed states of the system, for each of the three cases. How many possible system states
are there in each case?
For fermions, there is one possible state, that is with state s6 occupied along with states
from s1 to s4 . For distinguishable particles and bosons, the first excited state is the state
with 4 particles in s1 and one particle in s2 . For Distinguishable particles, this state is
five fold degenerate.
• Repeat part (b) for two units of energy and for three units of energy.
• Suppose that the temperature of this system is low, so that the total energy is low (though
not necessarily zero). In what way will the behavior of the bosonic system differ from
that of the system of distinguishable particles? Discuss.
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Answer. Bosonic particles are more likely to be found in the ground state compared to
distinguishable particles for which the degeneracy is higher comparatively.
Problem 4. For a system obeying Boltzmann statistics, we know what µ is from Chapter
6. Suppose, though, that you knew the distribution function(equation 7.31) but didn’t know
µ. You could still determine µ by requiring that the total number of particles, summed over
all single-particle states, equal N. Carry out this calculation, to rederive the formula µ =
−kT ln (Z1 /N ). (This is normally how µ is determined in quantum statistics, although the
math is usually more difficult.)
Answer.
Z
N =
g()dn̄
(14)
n̄ = e(−µ)
3/2
Z ∞
√
π 8mL2
−β(−µ)
d e
, where g0 =
⇒ N = g0
4
h2
0
Z ∞
√
= g0 eβµ
d e−β using change of variables, β = x,
0
Z ∞
√
βµ
3/2
= g0 e (kT )
x dx e−x
0
βµ
= g0 e (kT )
3/2
Γ(3/2)
√
βµ
3/2 π
= g0 e (kT )
2
3/2
v [2πmkT ]
V
βµ
⇒N =
e ⇒ µ = −kT ln
.
h3
N vQ
Problem 5. Imagine that there exists a third type of particle, which can share a singleparticle state with one other particle of the same type but no more. Thus the number of
these particles in any state can be 0, 1, or 2. Derive the distribution function for the average
occupancy of a state by particles of this type, and plot the occupancy as a function of the
state’s energy, for several different temperatures.
Answer. We start with the probability of a state being occupied by n particles is given by,
P(n) =
1 −n(−µ)/kT
e
,
Z
(15)
where the partition function in this case when n=0,1,2 is given by,
Z = 1 + e−(−µ)/kT + e−2(−µ)/kT
(16)
the average occupancy is obtained as,
n̄ =
0 × 1 + 1 × +e−(−µ)/kT + 2 × e−2(−µ)/kT
e−(−µ)/kT + 2 × e−2(−µ)/kT
=
Z
1 + e−(−µ)/kT + e−2(−µ)/kT
(17)
Here is a plot of these distributions for µ = 2.Notice that that plot becomes a step function as
in the case of fermions as the temperature goes to zero (at = µ).
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