Introduction to Thermodynamics and Statistical Physics phần 10 pptx

5. Exam Winter 2010 A
5.1 Problems
1. Two identical non-interacting particles each having mass M are confined
in a one dimensional parabolic poten tial given by
V (x)=
1
2
Mω
2
x
2
, (5.1)
where the angular frequency ω is a constant.
a) Calculate the canonical partition function of the system Z
c,B
for the
case where the particles are Bosons.
b) Calculate the canonical partition func tion of t he system Z
c,F
for the
case where the particles are Fermions.
2. Consider a one dimensional gas con taining N non-interacting electrons
mo ving along the x direction. The e lectrons are con fined to a section of
length L. A t zero temperature τ =0calculatetheratioU/ε
F
between
the total energy of the system U and the Fermi energy ε
F
.
3. Consider an ideal classical gas at temperature τ.Thesetofinternal

eigenstates of each p article in the gas, when a magnetic field H is applied,
contains 2 states having energies ε
−
= −µ
0
H and ε
+
= µ
0
H,wherethe
magnetic moment µ
0
is a constant. Calculate the magnetization of the
system, which is defined b y
M = −
µ
∂F
∂H
¶
τ
, (5.2)
where F is the Helmholtz free energy.
4. (Note: replace this with Ex. 3.9 in the lecture notes) Consider an ideal gas
made of N electrons in the extreme relativistic limit. The gas is contained
in a box having a cube shape with a volume V = L
3
. In the extreme
relativistic limit the dispersion relation ε (k)ismodified: the energy ε of
a single particle quantum state having a wavefunction ψ given by
ψ (x, y, z)=

µ
2
L
¶
3/2
sin (k
x
x)sin(k
y
y)sin(k
z
z) , (5.3)
Chapter5. ExamWinter2010A
is given by
ε (k)=~kc , (5.4)
where c is the speed of ligh t and where k =
q
k
2
x
+ k
2
y
+ k
2
z
(contrary
to the non-r elativistic case where it is given by ε (k)=~
2
k

2
/2M). The
system is in thermal equilibrium at zero temperature τ =0.Calculate
the ratio p/U between the pressure p and the total energy of the system
U.
5. Consider a mixture of two classical ideal gases, consisting of N
A
particles
of type A and N
B
particles of type B. T he heat capacities c
p,A
and c
V,A
(c
p,B
and c
V,B
)atconstantpressureandatconstantvolumerespectively
of gas A (B) are assumed to be temperature independent. The volume of
the mixture is initially V
1
and the pressure is initially p
1
. The mixture
undergoes an adiabatic (slow) and isentropic (at a constant en t ropy)
process leading to a final volume V
2
. Calculate the final pressure p
2

.
5.2 Solutions
1. The single pa rticle eigen en ergies are given by

n
= ~ω
µ
n +
1
2
¶
, (5.5)
where n =0, 1, 2, ···.
a) For Bosons
Z
c,B
=
1
2
∞
X
n=0
∞
X
m=0
exp [−β (
n
+ 
m
)] +

1
2
∞
X
n=0
exp (−2β
n
)
=
1
2
Ã
∞
X
n=0
exp (−β
n
)
!
2
+
1
2
∞
X
n=0
exp (−2β
n
)
=

exp (−β~ω)
2(1− exp (−β~ω))
2
+
exp (−β~ω)
2(1− exp (−2β~ω))
.
(5.6)
Note that the av erage energy U
B
is given by
U
B
= −
∂ log Z
c,B
∂β
= ~ω
1+2e
−2β~ω
+ e
−β~ω
1 −e
−2β~ω
. (5.7)
b) For Fermions
Z
c,F
=
1

2
∞
X
n=0
∞
X
m=0
exp [−β (
n
+ 
m
)] −
1
2
∞
X
n=0
exp (−2β
n
)
=
exp (−β~ω)
2(1− exp (−β~ω))
2
−
exp (−β~ω)
2(1− exp (−2β~ω))
.
(5.8)
Eyal Buks Thermodynamics and Statistical Physics 148

5.2. Solutions
Note that for this case the a verage energy U
F
is given by
U
F
= −
∂ log Z
c,F
∂β
= ~ω
2+e
−2β~ω
+ e
−β~ω
1 −e
−2β~ω
. (5.9)
2. Th e orbital eigenen ergies are given by
ε
n
=
~
2
2m
³
π
L
´
2

n
2
, (5.10)
where n =1, 2, 3, ···. The grandcanonical par tition functio n of the gas
is given by
Z
gc
=
Y
n
ζ
n
, (5.11)
where
ζ
n
=
Y
l
(1 + λ exp (−βε
n
)exp(−βE
l
)) (5.12)
is the orbital grandcanonical Fermionic partition function where,
λ =exp(βµ)=e
−η
, (5.13)
is the fugacity, β =1/τ and {E
l

} are the eigenenergies of a particle due
to internal degrees of freedom. For electrons, in the absen ce of magnetic
field both spin states have the same energy, which is taken to be zero.
Thus, log Z
gc
can be written as
log Z
gc
=
∞
X
n=1
log ζ
n
=2
∞
X
n=1
log (1 + λ exp(−βε
n
))
' 2
∞
Z
0
dn log
µ
1+λ exp
µ
−β

~
2
2m
³
π
L
´
2
n
2
¶¶
.
(5.14)
By employing the variable transformation
ε =
~
2
2m
³
π
L
´
2
n
2
, (5.15)
one has
log Z
gc
=

1
2
∞
Z
0
dεD(ε)log(1+λ exp (−βε)) , (5.16)
Eyal Buks Thermodynamics and Statistical Physics 149
Chapter5. ExamWinter2010A
where
D (ε)=
(
2L
π
q
2m
~
2
ε
−1/2
ε ≥ 0
0 ε<0
(5.17)
is the 1D density of states. Using Eqs. (1.80) and (1.94) for the energy
U and the number of particles N ,namelyusing
U = −
µ
∂ log Z
gc
∂β
¶

η
, (5.18)
N = λ
∂ log Z
gc
∂λ
, (5.19)
one finds that
U =
∞
Z
−∞
dεD(ε) εf
FD
(ε) , (5.20)
N =
∞
Z
−∞
dεD(ε) f
FD
(ε) , (5.21)
where f
FD
is the Fermi-Dirac distribution function [see Eq. (2.35)]
f
FD
()=
1
exp [β ( − µ)] + 1

. (5.22)
At zero temperature, where µ = ε
F
one has
U =
D (ε
F
)
ε
−1/2
F
ε
F
Z
0
dεε
1/2
=
2D (ε
F
)
3
ε
2
F
, (5.23)
N =
D (ε
F
)

ε
−1/2
F
ε
F
Z
0
dεε
−1/2
=2D (ε
F
) ε
F
, (5.24)
thus
U
ε
F
=
N
3
. (5.25)
3. The Helmholtz free energy is given by
F = Nτ
µ
log
n
n
Q
− log Z

int
− 1
¶
, (5.26)
where
Z
int
=exp(βµ
0
H)+exp(−βµ
0
H)=2cosh(βµ
0
H)(5.27)
Eyal Buks Thermodynamics and Statistical Physics 150
5.2. Solutions
is the internal partition function. Thus the magnetization is given b y
M = −
µ
∂F
∂H
¶
τ
= Nµ
0
tanh (βµ
0
H) . (5.28)
4. The grandcanonical partition function of the gas is giv en by
Z

gc
=
Y
n
ζ
n
, (5.29)
where
ζ
n
=
Y
l
(1 + λ exp (−βε
n
)exp(−βE
l
)) (5.30)
is a grandcanonical Fermionic partition function of an orbital having
energy ε
n
given by
ε
n
=
π~cn
L
, (5.31)
where n =
q

n
2
x
+ n
2
y
+ n
2
z
, n
x
,n
y
,n
z
=1, 2, 3, ···,
λ =exp(βµ)=e
−η
(5.32)
is the fugacity, β =1/τ and {E
l
} are the eigenenergies of a particle due
to internal degrees of freedom. For electrons, in the absen ce of magnetic
field both spin states have the same energy, which is taken to be zero.
Thus, log Z
gc
can be written as
log Z
gc
=

X
l
∞
X
n
x
=1
∞
X
n
y
=1
∞
X
n
z
=1
log (1 + λ exp (−βε
n
)exp(−βE
l
)) . (5.33)
For a macroscopic system the sum over n can be approximately replaced
by an integral
∞
X
n
x
=0
∞

X
n
y
=0
∞
X
n
z
=0
→
4π
8
∞
Z
0
dnn
2
, (5.34)
thus, one has
log Z
gc
=2
4π
8
∞
Z
0
dnn
2
log

µ
1+λ exp
µ
−β
π~cn
L
¶¶
. (5.35)
By employing the variable transformation
Eyal Buks Thermodynamics and Statistical Physics 151
Chapter5. ExamWinter2010A
ε =
π~cn
L
. (5.36)
one has
log Z
gc
=
∞
Z
0
dε
Vε
2
π
2
~
3
c

3
log (1 + λ exp (−βε)) . (5.37)
The energy U and the number of particles N are given b y
U = −
µ
∂ log Z
gc
∂β
¶
η
=
∞
Z
0
dε
Vε
3
π
2
~
3
c
3
f
FD
() , (5.38)
N = λ
∂ log Z
gc
∂λ

=
∞
Z
0
dε
Vε
2
π
2
~
3
c
3
f
FD
() , (5.39)
where f
FD
is the Fermi-Dirac distribution function [see Eq. (2.35)]
f
FD
()=
1
exp [β ( − µ)] + 1
. (5.40)
At zero temperature
U =
ε
F
Z

0
dε
Vε
3
π
2
~
3
c
3
=
V
π
2
~
3
c
3
ε
4
F
4
, (5.41)
N =
ε
F
Z
0
dε
Vε

2
π
2
~
3
c
3
=
V
π
2
~
3
c
3
ε
3
F
3
, (5.42)
and therefore
U =
3N
4
ε
F
. (5 .43)
The energy U can be expressed as a function of V and N as
U =
(3N)

4/3
¡
π
2
~
3
c
3
¢
1/3
V
−1/3
4
.
At zero temperature the Helmholtz free energy F equals the energy U,
thus the pressure p is given by
p = −
µ
∂F
∂V
¶
τ,N
= −
µ
∂U
∂V
¶
τ,N
=
1

3
¡
3N
V
¢
4/3
¡
π
2
~
3
c
3
¢
1/3
4
, (5.44)
thus
p
U
=
1
3V
. (5.45)
Eyal Buks Thermodynamics and Statistical Physics 152
5.2. Solutions
5. First, consider the case of an ideal gas made of a unique type of particles.
Recall that the entropy σ, c
V
and c

p
are given by [see Eqs. (2.87), (2.88)
and (2.89)]
σ = N
µ
5
2
+log
n
Q
n
+
∂ (τ log Z
int
)
∂τ
¶
, (5.46)
c
V
= N
µ
3
2
+ h
int
¶
, (5.47)
c
p

= c
V
+ N, (5.48)
where n = N/V is the density,
n
Q
=
µ
Mτ
2π~
2
¶
3/2
(5.49)
is the quantum density, M is the mass of a particle in the gas, and
h
int
= τ
∂
2
(τ log Z
int
)
∂τ
2
=
c
V
N
−

3
2
. (5.50)
The requirement that h
int
is temperature independent leads to
∂ (τ lo g Z
int
)
∂τ
= g
int
+ h
int
log
τ
τ
0
, (5.51)
where both g
int
and τ
0
are constants. Using this notation, the change in
entrop y due to a change in V from V
1
to V
2
and a change in τ from τ
1

to τ
2
is given by
∆σ = σ
2
− σ
1
= N
Ã
log
V
2
τ
3/2
2
V
1
τ
3/2
1
+
µ
c
V
N
−
3
2
¶
log

τ
2
τ
1
!
= N log
Ã
V
2
V
1
µ
τ
2
τ
1
¶
c
V
N
!
.
(5.52)
Thus the total change in the entropy of the mixture is given by
∆σ = ∆σ
A
+ ∆σ
B
= N
A

log
Ã
V
2
V
1
µ
τ
2
τ
1
¶
c
V,A
N
A
!
+ N
B
log
Ã
V
2
V
1
µ
τ
2
τ
1

¶
c
V,B
N
B
!
(5.53)
=(N
A
+ N
B
)log


µ
V
2
V
1
¶µ
τ
2
τ
1
¶
c
V,A
+c
V,B
N

A
+N
B


, (5.54)
and the requirement ∆σ =0leadsto
Eyal Buks Thermodynamics and Statistical Physics 153
Chapter5. ExamWinter2010A
µ
V
2
V
1
¶µ
τ
2
τ
1
¶
c
V,A
+c
V,B
N
A
+N
B
=1. (5.55)
Alternatively, by employing the equation of state

pV = Nτ , (5.56)
this can be rewritten as
µ
V
2
V
1
¶
c
V,A
+c
V,B
N
A
+N
B
+1
µ
p
2
p
1
¶
c
V,A
+c
V,B
N
A
+N

B
=1, (5.57)
or with the help of Eq. (5.48) as
p
2
= p
1
µ
V
2
V
1
¶
−
c
V,A
+c
V,B
N
A
+N
B
+1
c
V,A
+c
V,B
N
A
+N

B
= p
1
µ
V
1
V
2
¶
c
p,A
+c
p,B
c
V,A
+c
V,B
.
(5.58)
Eyal Buks Thermodynamics and Statistical Physics 154
6. Exam Winter 2010 B
6.1 Problems
1. Consider two particles, both having the same mass m,movinginaone-
dimensional potential with coordinates x
1
and x
2
respectively. The po-
tential energy is given by
V (x

1
,x
2
)=
mω
2
x
2
1
2
+
mω
2
x
2
2
2
+ mΩ
2
(x
1
− x
2
)
2
, (6.1)
where the angular frequen cies ω and Ω are real constan ts Assume that
the temperature τ of the system is sufficiently high to allow treating it
classically. Calculate the following average values
a)


x
2
1
®
for the case Ω =0.
b)

x
2
1
®
, however without assuming that Ω =0.
c)
D
(x
1
− x
2
)
2
E
, again without assuming that Ω =0.
2. Consider an ideal classica l gas containing N identical particles having
each mass M in the extreme relativistic limit.Thegasiscontainedin
avesselhavingacubeshapewithavolumeV = L
3
. In the extreme
relativistic limit the dispersion relation ε (k)ismodified: the energy ε of
a single particle quantum state having a wavefunction ψ given by

ψ (x, y, z)=
µ
2
L
¶
3/2
sin (k
x
x)sin(k
y
y)sin(k
z
z) , (6.2)
is given by
ε (k)=~kc , (6.3)
where c is the speed of ligh t and where k =
q
k
2
x
+ k
2
y
+ k
2
z
(contrary
to the non-r elativistic case where it is given by ε (k)=~
2
k

2
/2M). The
system is in thermal equilibrium at tem perature τ.Calculate:
a) the total energy U of the system.
b) the pressure p.
Chapter6. ExamWinter2010B
3. Consider a system made of two localized spin 1/2 particles whose energy
is given by
ε
σ
1
,σ
2
= −µ
0
H (σ
1
+ σ
2
)+Jσ
1
σ
2
, (6.4)
where both σ
1
and σ
2
can take one of two possible values σ
n

= ±1(n =
1, 2). While H is the externally applied magnetic field, J is the coupling
constant bet ween both spins. The system is in thermal equilibrium at
temperature τ . Calculate the m a gnetic susceptibility
χ = lim
H→0
∂M
∂H
, (6.5)
where
M = −
µ
∂F
∂H
¶
τ
(6.6)
is the magnetization of the system, and where F is the Helmholtz free
energy.
4. Consider a one dimensional gas con taining N non-interacting electrons
mo ving along the x direction. Th e electrons are confined b y a potential
given by
V (x)=
1
2
mω
2
x
2
, (6.7)

where m is the electron mass and where ω is the angular frequency of
oscillations. Calculate the chemical potent ial µ
a) in the limit of zero temperature τ =0.
b) in the limit of high tem peratures τ À ~ω.
5. The state equation of a given matter is
p =
Aτ
n
V
, (6.8)
where p, V and τ are the pressure, volume and temperature, respectively,
and A and n are both constants. Calculate the difference c
p
−c
V
between
the heat capacities at constant pressure and at constant volume.
6.2 Solutions
1. It is convenient to employ the coordinate transformation
x
+
=
x
1
+ x
2
√
2
, (6.9)
x

−
=
x
1
− x
2
√
2
. (6.10)
Eyal Buks Thermodynamics and Statistical Physics 156
6.2. Solutions
The inverse transformation is given by
x
1
=
x
+
+ x
−
√
2
, (6.11)
x
2
=
x
+
− x
−
√

2
. (6.12)
The following holds
x
2
1
+ x
2
2
= x
2
+
+ x
2
−
, (6.13)
and
˙x
2
1
+ ˙x
2
2
= ˙x
2
+
+ ˙x
2
−
. (6.14)

Thus, the kinetic energy T of the system is given by
T =
m
¡
˙x
2
1
+ ˙x
2
2
¢
2
=
m
¡
˙x
2
+
+ ˙x
2
−
¢
2
, (6.15)
and the poten tial energy V is given by
V (x
1
,x
2
)=

mω
2
x
2
1
2
+
mω
2
x
2
2
2
+ mΩ
2
(x
1
− x
2
)
2
=
mω
2
x
2
+
2
+
m

¡
ω
2
+4Ω
2
¢
x
2
−
2
.
(6.16)
The equipartition theorem yields
mω
2

x
2
+
®
2
=
m
¡
ω
2
+4Ω
2
¢
x

2
−
®
2
=
τ
2
, (6.17)
thus
D
(x
1
+ x
2
)
2
E
=
2τ
mω
2
, (6.18)
and
D
(x
1
− x
2
)
2

E
=
2τ
m (ω
2
+4Ω
2
)
. (6.19)
Furthermore, since by symmetry hx
+
x
−
i =0onehas

x
2
1
®
=
1
2
D
(x
+
+ x
−
)
2
E

=
1
2
¡
x
2
+
®
+

x
2
−
®¢
=
τ
mω
2
Ã
1 −
1
2
4Ω
2
ω
2
1+
4Ω
2
ω

2
!
.
(6.20)
Eyal Buks Thermodynamics and Statistical Physics 157
Chapter6. ExamWinter2010B
2. The k vector is restricted due to boundary conditions to the values
k =
πn
L
, (6.21)
where
n =(n
x
,n
y
,n
z
) , (6.22)
and n
x
,n
y
,n
z
=1, 2, 3, ···. The single particle partition function is given
by
Z
1
=

∞
X
n
x
=1
∞
X
n
y
=1
∞
X
n
z
=1
exp
µ
−
ε (k)
τ
¶
. (6.23)
Approximating the discrete sum by a con t inuous integral according to
∞
X
n
x
=0
∞
X

n
y
=0
∞
X
n
z
=0
→
4π
8
∞
Z
0
dnn
2
, (6.24)
one has
Z
1
=
4π
8
∞
Z
0
dnn
2
exp
µ

−
n~πc
Lτ
¶
=
4Vτ
3
8π
2
~
3
c
3
∞
Z
0
dxx
2
exp (−x)
|
{z }
2
=
Vτ
3
π
2
~
3
c

3
.
(6.25)
In the classical limit the grandcanonical partition function Z
gc
is given
by [see Eq. (2.44)]
log Z
gc
= λZ
1
, (6.26)
where λ =exp(βµ) is the fugacity. In terms of the Lagrange multipliers
η = −µ/τ and β =1/τ the last result can be rewritten as
log Z
gc
= e
−η
V
π
2
~
3
c
3
β
3
. (6.27)
a) The a v erage energy U and average number of particle N are calcu-
lated using Eqs. (1.80) and (1.81) respectively

Eyal Buks Thermodynamics and Statistical Physics 158
6.2. Solutions
U = −
µ
∂ log Z
gc
∂β
¶
η
=
3
β
log Z
gc
, (6.28)
N = −
µ
∂ log Z
gc
∂η
¶
β
=logZ
gc
, (6.29)
thus
U =3Nτ , (6.30)
and
η =log
µ

Vτ
3
π
2
N~
3
c
3
¶
. (6.31)
b) The en tropy is evaluate using Eq. (1.86)
σ =logZ
gc
+ βU + ηN
= N (1 + 3 + η)
= N
·
4+log
µ
Vτ
3
π
2
N~
3
c
3
¶¸
,
(6.32)

and the Helmholtz free energy by the definition (1.116)
F = U − τσ = −Nτ
·
1+log
µ
Vτ
3
π
2
N~
3
c
3
¶¸
. (6.33)
Thus the pressure p is given by
p = −
µ
∂F
∂V
¶
τ,N
=
Nτ
V
. (6.34)
3. The partition function is given b y
Z =
X
σ

1
,σ
2
=±1
exp (−βε
σ
1
,σ
2
)
=exp(−βJ)[exp(−2βµ
0
H)+exp(2βµ
0
H)] + 2 exp (βJ) ,
(6.35)
where β =1/τ . The free energy is given b y
F = −τ log Z, (6.36)
thus the magnetization is given by
M = −
µ
∂F
∂H
¶
τ
=
2µ
0
exp (−βJ)[−exp (−2βµ
0

H)+exp(2βµ
0
H)]
exp (−βJ)[exp(−2βµ
0
H)+exp(2βµ
0
H)] + 2 exp (βJ)
,
(6.37)
Eyal Buks Thermodynamics and Statistical Physics 159
Chapter6. ExamWinter2010B
and the magnetic susceptibility is given by
χ =
4βµ
2
0
1+e
2βJ
. (6.38)
Note that in the high temperature limit βJ ¿ 1
χ '
2µ
2
0
τ + J
. (6.39)
4. The orbital eigenenergies in this case are given by
ε
n

= ~ω
µ
n +
1
2
¶
, (6.40)
where n =0, 1, 2, ···. The grandcanonical par tition functio n of the gas
is given by
Z
gc
=
Y
n
ζ
n
, (6.41)
where
ζ
n
=
Y
l
(1 + λ exp (−βε
n
)exp(−βE
l
)) (6.42)
is the orbital grandcanonical Fermionic partition function where,
λ =exp(βµ)=e

−η
(6.43)
is the fugacity, β =1/τ and {E
l
} are the eigenenergies of a particle due
to internal degrees of freedom. For electrons, in the absen ce of magnetic
field both spin states have the same energy, which is taken to be zero.
Thus, log Z
gc
can be written as
log Z
gc
=
∞
X
n=0
log ζ
n
=2
∞
X
n=0
log (1 + λ exp (−βε
n
)) .
(6.44)
The num ber of particles N is given by
N = λ
∂ log Z
gc

∂λ
=2
∞
X
n=0
f
FD
(ε
n
) , (6.45)
where f
FD
is the Fermi-Dirac distribution function
f
FD
(ε)=
1
exp [β (ε − µ)] + 1
. (6.46)
Eyal Buks Thermodynamics and Statistical Physics 160
6.2. Solutions
a) At zero temperature the chemical potential µ is the Fermi energy ε
F
,
and the Fermi-Dirac distribution f unction becomes a step function,
th us with the help of Eq. (6.45) one finds that
N =
2ε
F
~ω

, (6.47)
thus
µ = ε
F
=
N~ω
2
. (6.48)
b) Using the appro ximation
f
FD
(ε) ' exp [−β (ε − µ)] , (6.49)
for the the limit of high temperatures and approximating the sum by
an integral one has
N =2
∞
X
n=0
exp
·
−β
µ
~ω
µ
n +
1
2
¶
− µ
¶¸

=2exp
µ
β
µ
µ −
~ω
2
¶¶
∞
Z
0
dn exp (−β~ωn)
=
2exp
¡
β
¡
µ −
~ω
2
¢¢
β~ω
,
(6.50)
thus
µ = τ
µ
log
µ
Nβ~ω

2
¶
+
β~ω
2
¶
' τ log
µ
Nβ~ω
2
¶
.
(6.51)
5. Using Eq. (2.239), which is given by
c
p
− c
V
= τ
µ
∂p
∂τ
¶
V,N
µ
∂V
∂τ
¶
p,N
, (6 .52)

one finds that
c
p
− c
V
=
A
2
τ
pV
n
2
τ
2(n−1)
= n
2
Aτ
n−1
. (6.53)
Eyal Buks Thermodynamics and Statistical Physics 161