7. equilibrium statistical mechanics - ncku
TRANSCRIPT
7. Equilibrium Statistical Mechanics
7.A. Introduction
7.B. The Microcanonical Ensemble
7.C. Einstein's Fluctuation Theory
7.D. The Canonical Ensemble
7.E. Heat Capacity of A Debye Solid
7.F. Order-Disorder Transitions
7.G. The Grand Canonical Ensemble
7.H. Ideal Quantum Gases
7.A. Introduction
A closed and isolated system tends to a stationary state that is characterised by a few
time-independent state variables. Such a state is said to be in thermodynamical
equilibrium. According to the 1st law, its total (internal) energy is fixed. According
to the 2nd law, its entropy is at a maximum since all entropy-increasing spontaneous
processes have ceased.
For a system with 3N degrees of freedom, its states with fixed internal energy of, say,
U E , form a 6 1N hyper-surface in the phase space. The phase space
trajectory of a closed and isolated system is restricted on this constant energy surface.
Consider now an infinite number of such systems, all identical and prepared in
equilibrium with U E . In short, we have an equilibrium ensemble of closed
isolated systems.
According to the postulate of equal probability, we assign an equal probability to
each state point on the constant energy surface so that the probability density is given
by
1
,
0
N EE
X for NH E
otherwise
X(7.1)
where ,N N NX q p is the 6N-dimensional phase space vector. E is called
the structure function and denotes the area of the constant energy surface. Note
that as shown in chapter 6, this postulate is always satisfied for ergodic systems.
The entropy is defined by Gibbs as
lnN N N NBS k d C X X X (7.2)
where C is a constant inserted to give the proper dimensions.
For quantum systems, the Gibbs entropy (7.2) is generalized to
lnBS k Tr (7.3)
where is the density operator and the trace is over any complete orthonormal set
of basis states.
7.B. The Microcanonical Ensemble
7.B.1. Quantum Version
7.B.2. Exercise 7.1 (Einstein Solid)
7.B.3. Classical Version
7.B.4. Exercise 7.2 (Ideal Gas)
7.B.1. Quantum Version
Consider a closed isolated system of volume V , particle number N, and energy E.
Let ,E n , with 1, ,n N E , be a complete set of orthonormal eigenstates of the
hamiltonian H with eigenvalue E, i.e.,
, ,H E n E E n
1
, ,N E
n
E n E n I
, , mnE m E n
where I
is the identity operator. The probability of finding the system in state
,E n is
, ,nP E n E n . The Gibbs entropy (7.3) can be written in terms of
nP as follows. First, we write
lnBS k Tr
1
, ln ,N E
Bn
k E n E n
, 1
, , , ln ,N E
Bn m
k E n E m E m E n
(7.4a)
Using
, , , ,nm nm nE n E m E n E n P
, ln , ln , , lnnm nm nE n E m E n E n P (7.4b)
eq(7.4a) becomes
1
lnN E
B n nn
S k P P
(7.4)
Proof of (7.4b)
An analytic function f of the operator is defined by
0 0
1
!
mm
mm
d ff
m d
where is obtained by treating as an ordinary variable. Therefore,
0 0
1, , , ,
!
mm
mm
d fE n f E n E n E n
m d
0 0
1
!
mm
nmm
d fP
m d
nf P
Maximization Of S
The equilibrium values of nP are those that maximize the entropy subject to the
constraint
1
1N E
nn
Tr P
(a)
This is equivalent to the variational problem 1 0S Tr S Tr
where α is a Lagrange multiplier. With the help of (7.4), we have
1 1
ln ln 1 0N E N E
B n n n B n nn n
k P P P k P P
ln 1 0B nk P
exp 1nB
P constk
Normalization (a) then requires
1
nPN E
(7.7)
which is simply the postulate of equal probability. A collection of closed and
isolated systems distributed according to (7.7) is called a microcanonical ensemble.
The entropy (7.4) becomes
1
1 1ln
N E
Bn
S kN E N E
lnBk N E (7.8)
With the identification of E as the thermodynamic internal energy U, all other
thermodynamic quantities can be obtained through the usual thermodynamic relations.
7.B.2. Exercise 7.1 (Einstein Solid)
An Einstein solid is a lattice of N sites, each of which is endowed with 3 independent
quantum harmonic oscillators of the same frequency ω. The total energy is3
1
1
2
N
ii
E n
3
1
3
2
N
ii
Nn
3
2M N
where in is the number of quanta on the ith oscillator and3
1
N
ii
M n
the total
number of quanta present.
(a) What is the total number of microscopic states for a given E?
(b) Find the entropy ,S T N .
(c) Find the heat capacity NC .
Answer
(a)
The problem is to find the number of ways W to put3
2
EM N
identical
quanta into 3N distinguishable sites. Now, a given configuration can be represented
as a line of M quanta separated into 3N partitions by 3N1 walls. [Note: 2 walls at
adjacent positions represent a site with no quanta.] Thus, all possible configurations
can be generated by permutations of the combined M+3N1 quanta and walls
"objects". [Note that the 3N partitions, or sites, are distinguishable by virtue of the
order of their positions in the line.] Since the quanta and walls themselves are
indistinguishable, we have
3 1 !
! 3 1 !
M NW
M N
(1)
(b)
Using (7.8), we have
3 1 !ln ln
! 3 1 !B B
M NS k W k
M N
(2)
For large M and N, we can use the Stirling formula
ln ! ln ln NN N N N N N to write
3 1
3 1
3 1ln
3 1
M N
B NM
M NS k
M N
3
3
3ln
3
M N
B NM
M Nk
M N
33 3ln
3
M M N
B M
N M Nk
M N
3
ln 13 3
M M N
B
M Mk
N N
(3)
Now, setting
3
2dU TdS dN dE dM dN
we have
1
N
S
T E
1
N
S
M
(4a)
Using
ln ln ln 1xd dx x x x
dx dx
we have
3 1ln 1 ln 1
3 3 313
BN
S M M M Nk
MM N N NN
3ln 1B
Nk
M
(4)
so that (4a) becomes
3ln 1
B
N
k T M
3
exp 1B
N
M k T
exp 1
1
3 exp 1
M
N
exp
1 exp
(5)
so that
ln ln 1 exp3
M
N
exp1
3 exp 1
M
N
1
1 exp
ln 1 ln 1 exp3
M
N
3 exp
ln ln 1 exp3 1 exp
NMM
N
33 ln 1 ln 1 exp
3 1 exp
M NM N
N
Putting these into (3) gives
exp
3 ln 1 exp1 expBS Nk
(6)
(c)
The internal energy is
13
3 2
MU E N
N
exp 13
1 exp 2N
1 1
3exp 1 2
N
(7)
Using
exp expT T
2 expBk
the heat capacity becomes
NN N
S UC T
T T
2
2
exp3
exp 1BNk
2
2
exp3
1 expBNk
2
22
exp3
1 expB
N
k T
(8)
7.B.3. Classical Version
Consider a closed isolated system of volume V and fixed particle number N.
Let its energy be constrainted to the shell E E E with 0E .
The phase space volume of the energy shell is , , , ,E E V N E V N E , where
, ,E V N is the structure function (area of the energy surface).
The equilibrium probability density N X corresponds to the maximum of
lnN
N N N NB
E H E E
S k d C
X
X X X (7.2)
subject to the normalization constraint
1N
N N
E H E E
d
X
X X (7.9)
With the help of the Lagrange multiplier 0 , this extremum problem becomes
0ln 0N
N NB
E H E E
d k C
X
X (7.10a)
Using
ln ln 1N NC C
we have
0ln 1 0N
N NB
E H E E
d k C
X
X (7.10)
which, since is arbitrary, can be satisfied only if
0ln 1 0NBk C
0exp 1N
B
C Kk
(7.10b)
From (7.9), we have , , 1EK E V N , so that
1
, ,
0
NE E V N
X for
E H E E
otherwise
(7.11)
Eq(7.2) thus becomes
1 1ln lnN E
B E B NE E
S k C kC
or
, ,, , ln E
B N
E V NS E V N k
C (7.12)
The constant NC cannot be determined classically.
Quantum mechanically, the uncertainty principle assigns a phase space volume of h to
each state so that we have3N NC h for distinguishable particles
and3!N NC N h for indistinguishable particles
where h is the Planck’s constant. The insertion the factor !N is called the correct
Boltzmann counting. It was originally introduced by hand to solve the Gibbs
paradox: in effect, it removes the entropy of mixing of indistinguishable particles.
In a practical calculation, it is often easier to work with the volume , ,E V N
instead of the shell , ,E E V N . Towards this end, we divide intoE
Eshells
of thickness E each. Hence,
/
1
, , , ,E E
E ii
E V N E V N
(7.13)
where iE i E and /E E
EE E E
E
. Obviously, the volume of any shell
cannot be larger than the whole volume, i.e., , , , ,E iE V N E V N , and the
outermost shell has the largest volume, i.e., , , , ,E i EE V N E V N . Hence
, , , , , ,E E
EE V N E V N E V N
E
(7.14)
Taking the logarithm gives
ln ln ln lnE E
E
E
(7.15)
Now, ln N ,E
NE so that as N , (7.15) implies ln lnE .
Hence, (7.12) becomes
, ,, , lnB N
E V NS E V N k
C
(7.16)
7.B.4. Exercise 7.2 (Ideal Gas)
Find the entropy and equation of state for an ideal gas.
Answer
Quoting the result from Exercise 6.3, we have
3 / 22
31
2
NNV mEE
N
(1)
For very large N and V, the entropy is given by (7.16). Using the indistinguishable
particles value of 3! NNC N h , we have
3
ln!B N
ES k
N h
3 / 2
3
2ln
3! 1
2
NN
BN
V mEk
N h N
(2)
3/ 2
2
2 3ln ln ! 1
2B B
mEk N V k N N
h
Now, for large N, we have
ln ! lnN N N N lnN
Ne
(Stirling formula)
1 !N
NN n
e
so that
3 3 3 3ln ! 1 ln ln
2 2 2 2
N NN N N N N N
5 5 3 3ln ln
2 2 2 2N N
3/ 25/ 23 5
ln2 2
N N
and
3/ 2 3/ 25/ 2
2
2 3 5ln ln
2 2B
mES k N V N
h
3/ 2
2
4 5ln
3 2B
V mEk N
N h N
(3)
Thus, the entropy per particle is
3/ 2
2
4 5ln
3 2B
mes k v
h
(4)
where the lower case letters denote quantities per particle.
Note that the independence of (4) of N is the result of the N! factor in NC .
Now, (3) is the fundamental equation for the ideal gas. We can rewriting it as
3/ 2
0 5/ 2lnB
VUS S k N
N
(5)
where
3/ 2
0 2
4 5ln
3 2B
mS k N
h
For fixed N, we have
3
2B
dV dUdS k N
V U
so that
2
3V B
U UT
S k N
3
2 BU Nk T
2
3S
U UP
V V
BNk T
V
which is the equation of state.
7.C. Einstein's Fluctuation Theory
7.C.1. General Discussion
7.C.2. Fluid Systems
7.C.1. General Discussion
Consider a closed and isolated ergodic system with energy within shell ,E E E .
According to (7.8), the entropy is
lnBS k E (7.17)
where E is the number of microscopic states in shell ,E E E .
Now, subdivide the system into n cells and denote the value of a state variable A in
cell i by iA . The probability of finding the system in the macrostate 1, , , nE A A
is therefore
1
1
, , ,, , , n
n
E A AP E A A
E
(7.18)
where 1, , , nE A A is the number of microstates satisfying 1, , , nE A A .
Similarly,
1 1, , , ln , , ,n B nS E A A k E A A (7.19)
so that
1 1
1 1, , , exp , , ,n n
B
P E A A S E A AE k
(7.20)
According to the 2nd law, S is a maximum when the system is in an equilibrium state
0 01, , , nE A A . The Taylor expansion of S about this state is
0 01 1
, 1
1, , , , , ,
2
n
n n ij i ji j
S E A A S E A A g
(7.22)
where the 1st order terms vanish since the expansion is about an extremum and
0i i iA A (7.21)
2
0
0ij jii j
Sg g
A A
[S0 is a maximum] (7.23)
Putting (7.22) into (7.20) gives
1, 1
1, , , exp
2
n
n ij i ji jB
P P E A A C gk
α
1exp
2T
B
Ck
α Gα (7.24a)
where 1, ,Tn α , T
ijg G G , and
0 01
1 1exp , , , n
B
C S E A AE k
From the normalization condition, we have,
11 nd d P
α 1
1exp
2T
nB
C d dk
α Gα
2
1
1exp
2
n
i i ii B
C dk
where α Oβ and Tidiag O GO
Γ. Hence
1
21
nB
i i
kC
2
det
n
BkC
G
Hence
det 1exp
22T
nBB
Pkk
Gα α Gα (7.24)
which is valid only for small fluctuations. We now define a "characteristic function"
by the Laplace transform
exp TI D P
h
α α h α
(7.24a)
where
1 nD d d
α
Thus,
det 1exp
22T T
nBB
I Dkk
Gh
α α Gα h α
Using T T T hα h Oβ δ β
, where Tδ O h , we have,
2
1
1 1
2 2
nT T
i i i iiB Bk k
α Gα h α
2
2
1
1
2 2
nB B
i i i ii B i i
k k
k
11 1
2 2T T
BB
kk
μ Gμ δ Γ δ
and1 1 1T T T T δ Γ δ h OΓ O h h G h
so that
11 det 1exp exp
2 22T T
B nBB
I k Dkk
Gh h G h
μ μ Gμ
11exp
2T
Bk
h G h (7.26)
This is useful for calculating moments since from (7.24a), we have
1 11 0
m
m mm
ID P
h h
h
hα α
(7.27a)
Now, with 11
2T
BM k h G h , we have,
i i
I MI
h h
h
h
2 2
i j i j i j
I M M MI
h h h h h h
hh
where
1 11
2 B j jij jiji
Mk h h
h G G
1B jij
j
k h G 1B i
k G h
2
1B ij
i j
Mk
h h
G
so that (7.27a) gives
1 0m for odd m
1i j B ij
k G (7.27)
As with Ex.4.9, all moments of higher order can be expressed as linear combinations
of products of the 2nd moments.
7.C.2. Fluid Systems
7.C.2.1. Densities
7.C.2.2. Fluctuations
7.C.2.1. Densities
We shall express all extensive variables by upper case letters, say, X, and the
corresponding densities by lower case letters, say, x. Thus, X xV . One
exception to this is the mass density which we write as M V . Using
dX Vdx xdV (1)
the combined 1st and 2nd law
dU TdS PdV dM (2)
becomes
Vdu udV V Tds d Ts P dV
Since this is valid for arbitrary V and dV, we must have
du Tds d (3)
u Ts P (4)
where (3) is the combined law and (4) is the fundamental equation with densities as
independent variables. Combining (3) and (4), we get the Gibbs-Duhem equation
0sdT dP d (5)
Similarly, using (1) and collecting terms, we have
T S P V M V s T s T P V
V s T (6)
where (5) was used to get the last equality.
7.C.2.2. Fluctuations
Consider a closed and isolated box of fluid subdivided into cells. The fluctuations
of the entropy is given by (2.171) as [see Fig.7.1 for notations],
10
1
2T i i i i iii
S T S P V MT
(7.28)
According to (7.24), we have
310
det 1exp
22T i i i i ii
iBB
P S T S P V Mk Tk
G
(7.29)
where G is to be determined through (7.24) once the independent variables are
decided. By eq(6) of section 7.C.2.1., we have
T S P V M V T s (a)
whereS
sV
andM
V are densities. Next, we choose T and as independent
variables so that
T
s ss T
T
T
c sT
T
(b)
T
TT
T T
sT
(c)
where the Maxwell relation
T
s
T
can be gleaned from the diagram
g
u
T
a f
s
Putting (b,c) into (a) gives
2 2
0 00
i i i i ii
T
cV T s V T
T
where, accurate to order 2T , we've replaced iT with 0T .
Eq(7.29) thus becomes
2 202
10 0
det, exp
22i i i i
iBB T
cVP T T
k T Tk
G
which, when comparing with
det 1exp
22T
nBB
Pkk
Gα α Gα (7.24)
suggests
1 2, , α 1 1, , , ,T T
02
0
0
0
0
ij
T
V c
T
Vg
T
for
i j odd
i j even
otherwise
(7.31)
so that
2
1
det iii
g
G
0 02
0 0 T
V c V
T T
and
0 0
2 20 0
1,
2i i
B T
V c VP T
T Tk
2 20
10 0
exp2 i i
iB T
cVT
k T T
(7.32)
The 2nd moments of the fluctuations are given by
0 02 2
0 0
1
2i j i j
B T
V c Vd d
T Tk
221exp
2i j ii i jj jB
g gk
so that
0i j i jT T [odd integral vanishes] (7.33,36)
2 0
0
1
2i iB T
Vd
k T
2 20
0
exp2i i
B T
V
k T
1
0
0
B
T
k T
V
(7.34)
2 02
0
1
2i iB
V cT d T
k T
2 202
0
exp2i i
B
V cT T
k T
20
0
Bk T
V c (7.35)
where we've used the Gaussian integral formula [see, e.g., Reif, Appendix.4]
2 23
1exp
2dx x ax
a
7.D. Canonical Ensemble
7.D.1. Probability Density Operator
7.D.2. Systems Of Indistinguishable Particles
7.D.3. Systems Of Distinguishable Particles
7.D.1. Probability Density Operator
Consider a closed but not isolated system that can exchange energy but not matter
with its surroundings. A collection of such identically prepared systems is called a
canonical ensemble. As will be shown below, the corresponding probability
distribution describes a system kept at constant temperature.
Once again, the probability density operator is determined by maximizing the
entropy S. Since the number of particles is fixed, we have the normalization
constraint 1NTr (7.37)
where N is the (fixed) number of particles. Furthermore, the free exchange of energy
is expected to stabilize, at equilibrium, to some constant average value E , i.e.,
N NTr H E U (7.38)
Thus, the variational problem is 0 0N E N NS Tr Tr H
00 lnN B E NTr k H
0ln 1N B E NTr k H
which can be satisfied only if
0ln 1 0B E Nk H (7.40)
0
1exp 1 E N
B
Hk
(7.41)
The constraints (7.37-8) thus become
0exp 1 exp 1EN N
B B
Tr Hk k
(a)
0exp 1 exp EN N N
B B
Tr H H Uk k
(b)
Eq(a) gives the partition function
0exp exp 1EN N N
B B
Z Tr Hk k
(7.42)
so that
0 1 lnB Nk Z (c)
while (b) becomes
1exp E
N N NN B
U Tr H HZ k
Now, taking the average of (7.40) gives 00 ln 1N B E NTr k H
0B ES k U
lnB N ES k Z U
Comparing with the fundamental equation for the Helmholtz free energy
A U TS we have
1E T
and lnB NA k T Z (7.44)
so that
(c) 0 1BB
Ak
k T
(7.42) expNB
AZ
k T
exp N
NB
HTr
k T
(7.45)
(7.41)
exp ln NN
B
HZ
k T
1exp N
N B
H
Z k T
(7.46)
which is the probability density operator for the canonical ensemble.
Once NZ and hence A is known, all other thermodynamic quantities can be
calculated. For example, for , ,A A T X N , we have
XN
AS
T
TN
AY
X
'
TX
A
N
Furthermore,
U A TS XN
AA T
T
XN
AA
XN
A
(e)
Now, since the canonical ensemble only specifies a fixed average E U , we
expect fluctuations in E about this value. The corresponding variance can be
calculated as follows.
Starting with the normalization
exp 1N NTr A H (7.47)
we can differentiate with respect to to get
exp 0N N N
XN
ATr A H A H
exp 0N N N
XN
Tr H A A H
Doing it again gives
2
2
2exp 0N N N
XNXN
Tr A H A A H
Using (e) gives
2
exp 0N N N
XN
UTr H E A H
20
XN
UE E
i.e.,22
XN
UE E
XN
dT U
d T
2B XNk T C (7.49)
22 2
B XNE E k T C
E E
1
N (7.50)
since both U and XNC are proportional to N. Thus, in the thermodynamic limit
( ,N V ), the relative deviation vanishes, i.e., the canonical ensemble is equivalent
to the microcanonical ensemble.
7.D.2. Systems Of Indistinguishable Particles
As shown in Appendix B, the states of a system of indistinguishable particles must be
properly symmetrized with respect to particle interchanges, namely, symmetric (+) for
bosons and antisymmetric () for fermions. The trace of thus becomes
1
1 1
11
!N
N N Nk k
Tr k k k kN
(7.51)
where momentum eigenstates are assumed to take the direct product form
1 2a b l a b l Nk k k k k k (7.52)
which says particles 1,2,…,N have momenta , , ,a b lk k k , respectively. Thus,
1 1, , 1 , ,P
N NP
k k P k k (7.53)
where the sum is over all possible permutations of state indices.
7.D.2a. Exercise 7.3
7.D.2b. Semi-Classical Limit
7.D.2c. Exercise 7.4
7.D.2a. Exercise 7.3
Compute the partition function 3Z T for an ideal gas of 3 identical particles in a
cubic volume 3V L . For convenience, neglect the spin degrees of freedom.
What approximations can be made for a high T and low density gas?
Answer
For a particle in the box with hamiltonian 2
1 2H
m
p
, the wave function is
sin sin sinx y zA k x k y k z k x x k (1)
which vanishes at the walls at , , 0,x y z L provided
a ak nL
where , ,a x y z and 1,2, ,an
Note: k x and k x are linearly dependent so that only positive an need be
counted.
The 1-particle partition function is therefore
2 2
1 exp2
Z Tm
k
k
2 2
2 2 22
1 1 1
exp2
x y z
x y zn n n
n n nmL
(3)
Using a ak nL
, we have, in the thermodynamic limit,
3 2
2 2 21 3
0 0 0
exp2x y z x y z
LZ T dk dk dk k k k
m
3 22 2 2
3 exp22
x y z x y z
Ldk dk dk k k k
m
3
33 2
2
2
L m
322
mV
3
22Bmk T
V
32
2 Bmk TV
h
3T
V
(4)
where
2T
B
h
mk T
(5)
is the thermal wavelength, which is a measure of the particle’s coherence length.
For 3 identical particles, the hamiltonian is
2 2 2
3 1 2 3
1
2H
m p p p
with wave function [see Appendix B]
//1 2 3 1 2 3, , , , , ,
a b c
s as aa b c k k k x x x x x x k k k
where
/ 1, , , ,
3! !
s a
a b c a b cn
k
k
k k k k k k
, , , ,P
a b c a b cP
P k k k k k k
where nk is the occupation number of state k. Hence, for a b c ,
, , , , , , , ,a b c a b c b c a c a b
k k k k k k k k k k k k
, , , , , ,b a c c b a a c b k k k k k k k k k (7)
/ 1, , , ,
3!
s a
a b c a b c
k k k k k k (7')
2 , , , , , ,, ,
0
a a b a b a b a aa a b
k k k k k k k k kk k k (7a)
1, , , ,
2 3
s
a a b a a b
k k k k k k (7a')
6 , ,, ,
0a a a
a a a
k k kk k k (7b)
1, , , ,
6s
a a a a a a
k k k k k k (7b')
The partition function is then, by eq(B.30),
3 3
1, , , ,
3!a b c
a b c a b cZ T k k k
k k k k k k (6)
where
3 1 1 1exp 1 2 3H H H
2 2 2
1 2 3exp2m
p p p
so that for a b c ,
2
2 2 23, , , , exp
2a b c a b c a b cm
k k k k k k k k k
2
2 2 23, , , , exp
2a b c a b c a b cm
k k k k k k k k k
2
2 23, , , , exp 2
2a a b a a b a bm
k k k k k k k k
2
2 2
3
2exp 2, , , , 2
0
a ba a b a a b m
k kk k k k k k
2
23, , , , exp 3
2a a a a a a am
k k k k k k k
2
2
3
6exp 3, , , , 2
0
aa a a a a a m
kk k k k k k
In view of eqs(7,a,b), the sum in (6) must be splitted into 3 parts according to
a b c a b c a b c a b c a c b a b c
f f
k k k k k k k k k k k k k k k k k k
so that
3 3
1, , , ,
3!a b c
a b c a b cZ T
k k k
k k k k k k
3 33 , , , , , , , ,
a b a
a a b a a b a a a a a a
k k k
k k k k k k k k k k k k
Thus, for bosons,
2
2 2 23
1exp
3! 2a b c
a b cZ Tm
k k k
k k k
2 2
2 2 26 exp 2 6 exp 32 2
a b a
a b am m
k k k
k k k
For fermions,
2
2 2 23
1exp
3! 2a b c
a b cZ Tm
k k k
k k k
Now,
2
21exp 3
2 3a
a
TZ
m
k
k
Usinga b a b a b
k k k k k k
, we have
2 2 2
2 2 2 2 2exp 2 exp 2 exp 32 2 2
a b a b a
a b a b am m m
k k k k k
k k k k k
1 1 12 3
T TZ Z T Z
Usinga b c a b c a b c a b c a c b a b c
k k k k k k k k k k k k k k k k k k
, we have
2 2
2 2 2 2 2 2exp exp2 2
a b c a b c
a b c a b cm m
k k k k k k
k k k k k k
2 2
2 2 23 exp 2 exp 32 2
a b a
a b am m
k k k
k k k
3
1 1 1 1 132 3 3
T T TZ T Z Z T Z Z
3
1 1 1 13 22 3
T TZ T Z Z T Z
Hence, for fermions, we have
3
3 1 1 1 1
13 2
3! 2 3
T TZ T Z T Z Z T Z
For bosons, we have
3
3 1 1 1 1
13 2
3! 2 3
T TZ T Z T Z Z T Z
1 1 1 16 62 3 3
T T TZ Z T Z Z
3
1 1 1 1
13 2
3! 2 3
T TZ T Z Z T Z
Both cases can be combined to give
3
3 1 1 1 1
13 2
3! 2 3
T TZ T Z T Z Z T Z
Using (5), we have
3
3 3 3/ 2 3 3 3/ 2 3
13 2
3! 2 3T T T T
V V V VZ T
3 23 3
3 3/ 2 3/ 2
1 3 21
3! 2 3T T
T
V
V V
(8)
In the semiclassical limit (high T and large V),3T
V
is small so that
3
3 3
1
3! T
VZ T
(9)
which implies
1
!
N
N NT
VZ T
N
(10)
for a N particle system. The factor N! arises from the quantum indistinguishability
and resolves the Gibb's paradox.
7.D.2b. Semi-Classical Limit
As shown in Ex.7.3, NZ should be caculated using properly symmetrized basis states.
However, as compared to the results obtained from unsymmetrized bases, the most
significant correction is just the factor N!, which can be added by hand. Other
corrections (exchange and correlation effects) are proportional to powers of 3T
N
V
3Tn
3/ 2
n
T . Thus, they become negligible in the semi-classical limit of high T
and small n. Hence,
1
1 1, ,
1, , , ,
!N
HN N N
k k
Z T k k e k kN
for high T, low n (7.54)
The non-interacting molecules in a gas can possess internal degrees of freedom. For
example, the hamiltonian of a typical molecule may be written as
2
2 rot vib el nuclH H H H Hm
p
(7.55a)
where the successive terms on the right denote contributions from the translational,
rotational, vibrational, electronic, and nucleonic degrees of freedom, respectively.
The implicit, and usually valid, assumption of (7.55a) is of course that the various
degrees of freedom are decoupled. Thus, for a gas of N molecules contained in fixed
volume V, we have
2
1
1, exp
! 2
N
N N rot vibi
iZ T V Tr H i H i
N m
p
el nuclH i H i (7.55)
If the various degrees of freedom are truly decoupled, the corresponding terms in H
will commute with one another. Now, since
, 0A B A B A Be e e
we can write
1
,!N N tr N rot N vib N el N nuclZ T V Z Z Z Z Z
N
1 1 1 1 1
1
!
N
tr rot vib el nuclZ Z Z Z ZN
(7.56)
For a semi-classical gas with no internal degrees of freedom, we have, from Ex.7.3,
3 3
1,
!
N N
NT T
V eVZ T V
N N
(7.57)
where the last equality made use of the Stirling formula !N
NN
e valid for large
N. The Helmholtz free energy is
lnB NA k T Z 3lnB
T
eVk TN
N
3/ 2
2
21 ln B
B
V mk TNk T
N h
(7.58)
so that
VN
AS
T
3/ 2
2
2 31 ln
2B
B B
V mk TNk Nk T
N h T
3/ 2
2
5 2ln
2B
B
V mk TNk
N h
(7.59)
which is just the Sackur-Tetrode equation first introduced in Ex.2.3.
7.D.2c. Exercise 7.4
Consider a cubic box of volume 3V L containing an ideal gas of N identical atoms,
each with spin 1/2 and magnetic moment . A magnetic B is applied to the system.(a) Find NZ .
(b) Find U and VNC .
(c) Find M.
Answer
(a) ZN
Following (7.56), we write
1 1
1
!
N N
N tr magZ Z ZN
(1)
where, as shown in Ex.7.3,
1 3trT
VZ
with
2T
B
h
mk T
The magnetic energy is
1
2E s s B with 1s
so that
11
1exp
2mags
Z s B
12cosh
2B
(2)
Hence, (1) becomes
3
1, 2cosh
! 2
NN
NT B
V BZ T V
N k T
(3)
32cosh
2
NN
T B
eV B
N k T
(b) U, CVN
From the definition (7.45)
expN N NZ Tr H
we have
expNN N N
ZTr H H
N N NZ Tr H NZ E NZ U
so that
1 N
N VN
ZU
Z
ln N
VN
Z
Note: as will be shown in (c), ,U U S B so that it is actually an "enthalpy".
Using
3ln ln 2 ln ln cosh
2NT B
V BZ N e
N k T
2T h
m
1
2 2
h
m
2T
we have
sinh / 23
2 cosh / 2 2T
T
B BU N
B
3tanh
2 2 2B
B BN k T
3tanh
2 2 2BB B
B BNk T
k T k T
(4)
and
VNVNB
UC
T
3tanh
2 2 2BB B
B BNk
k T k T
22 2
tanh sech2 2 2 2 2B
B B B B B
B B B B BNk T
k T k T k T k T k T
2
23sech
2 2 2BB B
B BNk
k T k T
(5)
(c) M
Analogous to the Helmholtz free energy (7.58), the magnetic free energy of the
system is given by (see note below)
, lnB NT k T Z B
with
d SdT d M B
so that
ln NB
TN TN
ZM k T
B B
tanh2 2B
B B
BNk T
k T k T
1tanh
2 2 B
BN
k T
(6)
Note:
Consider the non-magnetic case 0H H where E U . This can be interpreted
as the magnetic case 0H H m B with 0B . Hence, 0H H corresponds to
0
,E U S
B
M since U depends only on extensive variables. Therefore, the
hamiltonian 0H H m B averages to E U m B U M B
,H S B , which is an "enthalpy". The corresponding canonical ensemble thus
gives rise to a "Gibbs energy" , lnB NT k T Z B .
7.D.3. Systems Of Distinguishable Particles
For distiguishable particles, there is no need for symmetrization. Hence,
1
1 1, ,
, , , ,N
N N Nk k
Tr k k k k
(7.60)
Exercise 7.5 Einstein Solid
Use the canonical ensemble to calculate U and C for an Einstein solid (see Ex.7.1)
Answer
From Ex.7.1, we have
3
1
1
2
N
ii
H n
(1)
With eigenstates
i i i in n n n where 0,1,2,in
the partition function becomes
3
1
1exp
2
N
N N ii
Z T Tr n
(2)
1 3
3
1 3 1 3, , 1
1, , exp , ,
2N
N
N i Nn n i
n n n n n
3
1 0
1exp
2i
N
i i ii n
n n n
3
1 0
1exp
2i
N
ii n
n
3
1 0
1exp exp
2i
N
ii n
n
3
1
1 1exp
2 1 exp
N
i
3
1 1exp
2 1 exp
N
(3)
The Helmholtz free energy is
, lnB NA T N k T Z T
13 ln 1 exp
2BNk T
13 ln 1 exp
2 BB
N k Tk T
(4)
The entropy is
N
AS
T
2 exp
3 ln 1 exp
1 exp
B BB B
B
B
k T k TN k k T
k T
k T
exp
3 ln 1 exp
1 exp
BB
B B
B
k TNk
k T k T
k T
(5)
The internal energy is (see 7.D.1)
N
AU
13 ln 1 exp
2N
exp13
2 1 expN
exp13
2 1 expN
1 1
32 exp 1
N
(6)
The heat capacity is
NN
UC
T
2
2
exp3
exp 1Bk T
N
2
2
exp3
exp 1B
B
Nkk T
2
2
exp3
1 expB
B
Nkk T
(7)
7.E. Heat Capacity Of A Debye Solid
Consider a simple crystalline solid with one atom at each lattice site.
Since the size and shape of the solid are fixed macroscopically, the motion of the
atoms must be restricted to oscillations about their equilibrium positions.
The degrees of freedom of such vibrations is 3N , where N is the number of atoms.
If the amplitudes of the vibrations are small enough, the oscillations become harmonic,
i.e., the potential energy is quadratic in atomic displacements. Thus, the atomic
vibrations of a solid can be approximated as a set of 3N coupled harmonic oscillators.
By means of a transformation into the so-called normal coordinates, the vibrations
can be de-coupled into independent normal modes.
Typical measured values of the heat capacity of monatomic solids are shown in Fig.
7.2. At high temperatures, VC approaches a constant value of 6 cal/K mole, in
agreement with the classical theory. For low temperatures, VC drops as 3T , which
can only be explained in terms of quantum theory.
The Debye theory is a quantum theory of harmonic oscillations in a continuum.
According to classical elastic theory, there are 2 types of waves governed by the wave
equations
2
22 2
1, 0T
T
tc t
u r 0T u transverse, doubly degenerate
2
22 2
1, 0L
L
tc t
u r 0L u longitudinal
where c is the phase velocity. The propagating modes thus obey linear dispersion
i ic k ,i L T
and are called sound waves.
Let the solid be a rectangular lattice of sides , ,x y zL L L . The normal modes are
standing waves which vanishes at the surfaces. This means the allowable wave
vectors are
i ii
k nL
where , ,i x y z and 1,2, ,i in N (a)
where ii
i
LN
a , with ia being the lattice spacing, is the number of sites in the ith
direction. Thus, the total number of sites is x y zN N N N . With each site having 3
degrees of freedom, the total degrees of freedom is 3N.
In the quantized version of the theory, the hamiltonian is
1 2
,, ,
1
2i ii L T T
H n
kk
k (7.62)
where the sum over k involves N modes for each branch i. The partition function is
1 2
,, ,
1exp
2N N i ii L T T
Z T Tr n
kk
k
,
,0
1exp
2i
i ii n
n
k
kk
k
1 1
exp2 1 expi
i i
k
kk
11
2sinh2
ii
k k
(7.63)
Hence,
1ln ln 2sinh
2N ii
Z T
k
k
ln NE Z
1cosh
121 2sinh2
i
ii
i
k
kk
k
1 1exp exp
12 21 1 2exp exp2 2
i i
ii
i i
k
k kk
k k
exp 1 1
2exp 1i
ii i
k
kk
k
1 1
2 exp 1 ii i
k
kk
(7.64)
,
1
2 i ii
n
kk
k
where the average occupation of the mode ,ik
,
1
exp 1ii
n
k k(7.65)
is called the Planck's formula.
The sum over k can be approximated by an integral
d k
k k
where the density of states in k-space, k , can be calculated from eq(a) as
3 3x y zL L L V
k
so that
33
0ik
Vd k
k
where the condition 0ik from eq(a) restricts the integration to the 1st quadrant.
Using kc
, we have
33
0ik
Vf d k f
k
222
Vdk k f
22 32
Vd f
c
(b)
where we've used the fact that integration of the angular part over the 1st quadrant
gives
1
1
8 2st Q
d d
Summing over the branches gives
22 32i i i i i i
i i i
Vf d f
c
k
For the special case that if f , we have
23 2
1
2ii i i
Vf d f
c
k
22 3
3
2
Vd f
c
where
3 3 3 3
3 1 2 1
i i T Lc c c c
To restrict the total number of modes to 3N, we must introduce a cutoff (Debye)
frequency D so that
22 3
0
33
2
DVN d
c
3
2 32DV
c
(7.68)
1/ 326
D D
Nc c k
V
(7.69)
Defining the density of states in -space by
i
d g k
we have
22 3
3
2
Vg
c
2
3
9
D
N
(7.70)
Eq(7.64) thus becomes
0
1
2
D
E d g n
(7.71)
where
1
exp 1n
With the help of (7.70), we get
33
0
9 1
2
D
D
NE d n
4 3
30
9
8 exp 1
D
D
D
Nd
(7.72)
The heat capacity is therefore
4
23 20
exp9
exp 1
D
ND B
NC d
k T
52 4
23 20
9
1
Dx xB
xD B
N k T x edx
k T e
where
B
xk T
3 4
20
91
Dx xB
B xD
k T x eNk dx
e
3 4
20
91
Dx x
B xD
T x eNk dx
T e
For low T, we have Dx . Using
4 4
20
4
151
x
x
x edx
e
we get
3412
5N BD
TC Nk
T
(7.74)
which is the famous Debye 3T rule.
7.F. Order-Disorder Transitions
The transition from a disordered state to an ordered one can be studied using methods
of equilibrium statistics.
Consider a lattice of N spin 1/2 objects with magnetic moment . Interactions
between objects further than nearest neighbors are assumed negligible. In the
presence of an applied magnetic field B, the hamiltonian of the system is
, 1
N
ij i j ii j i
H s s B s
(7.75)
where 1is indicates the orientation (up/down) of the spin at site i along the z-axis,
ij is the interaction energy, and ,i j denotes a sum over nearest neighbors with
each pair of sites counted once. This is known as the Ising model.
If 0ij , the lowest energy for 0B happens when all spins are parallel, i.e., they
are all up or all down, both cases being equally probable but can be chosen by an
infinitesimal B. This corresponds to ferromagnetism if ij ; otherwise, the
system is ferrimagnetic. Analogously, the case 0ij corresponds to anti-ferro or
anti-ferri- magnetism.
The partition function is
, 1
expN
N ij i j iall config i j i
Z T s s B s
(7.76)
where "all config" denotes the 2N possible spin configurations, i.e.,
1 , , Nall config s s
with all 1is
In contrast to the magnetic interaction which tends to align the spins in an orderly
fashion, the presence of thermal energy Bk T tends to randomize the spin orientations
and increases the entropy or degree of disordered. These competing forces then
leads to an order- disorder phase transition.
7.F.1. Exact Solution For A 1-D Lattice
7.F.2. Mean Field Theory For A d-D Lattice
7.F.1. Exact Solution For A 1-D Lattice
Consider a 1-D lattice of N sites with ij and impose the periodic boundary
conditions so that i N is s . For a given spin configuration, the total energy is
11 1
N N
i i i ii i
E s s s B s
(7.77)
The partition function is therefore,
1
11 1 1 1
, expN
N N
N i i is s i i
Z T B s s B s
Under the periodic boundary condition, we can write
1
1 11 2 1 1
1
2
N N N N
i i i i ii i i i
s s s s s
where we've used 1 1Ns s . Hence,
1
1 11 1 1 1
1, exp
2N
N N
N i i i is s i i
Z T B s s B s s
(7.78)
To proceed, we introduce a transfer-matrix P with matrix elements
1 1 1
1exp
2i i i i i is s s s B s s
P (7.80)
Using for each site the bases,
11
0
01
1
P is found to be
1 1 1 1
1 1 1 1
P PP
P P
exp exp
exp exp
B
B
(7.79)
Putting (7.80) into (7.78) gives
1
1 2 2 3 1 11 1
,N
N N N Ns s
Z T B s s s s s s s s
P P P P
Using the completeness relation
1
1 01 0 0 1
0 1i
i is
s s
1 01
0 1
we have
1
1 11
,N
Ns
Z T B s s
P NTr P i
i
(7.81a)
where i are the eigenvalues of the 22 matrix NP . Since P is symmetric, it can
be diagonalized by an orthogonal transformation, i.e., Tidiag O PO , where i
are the eigenvalues of P and T T O O OO 1 . Thus,
T N T T TO P O O POO PO O PO N
idiag Nidiag
i.e., Ni i . Solving the secular equation
exp expdet 0
exp exp
B
B
we get
exp exp exp 2 0B B
or 2 2 exp cosh 2sinh 2 0B
with roots
2exp cosh exp 2 cosh 2sinh 2B B
2exp cosh cosh 2exp 2 sinh 2B B
(7.82)
so that (7.81a) becomes
, N NNZ T B 1
N
N
(7.81)
In the thermodynamic limit, the Gibbs free energy per site is
1, lim ,N
Ng T B G T B
N 1
lim ln ,B NN
k T Z T BN
1lim ln 1
N
NB
Nk T
N
lnBk T
where we've used 1
. Hence,
2, ln cosh cosh 2exp 2 sinh 2Bg T B k T B B
(7.84)
The order parameter is
T
gs
B
2
2
sinh coshsinh
cosh 2exp 2 sinh 2
cosh cosh 2exp 2 sinh 2B
B BB
Bk T
B B
2
sinh
cosh 2exp 2 sinh 2
B
B
7.F.2. Mean Field Theory For A d-D Lattice
Consider the hamiltonian of a d-D spin lattice of N sites and ij
, 1
N
i j ii j i
H s s B s
(7.75)
1 . . 1
1
2
N N
i j ii j n n of i i
s s B s
where n.n. stands for "nearest neighbors" and the 1/2 factor removes the double
counting of each pair of sites in the double sum. Note that periodic boundary
conditions are assumed to remove any spurious surface effects.
7.F.2a. Mean Field
7.F.2b. Heat Capacity
7.F.2c. Magnetic Susceptibility
7.F.2a. Mean Field
In the mean field approximation, one sets
. .j
j n n of i
s s
where is the number of nearest neighbors at each site and is s is the average
spin at a site. Eq(7.75) thus becomes
1 1
1
2
N N
i ii i
H s s B s
1
N
eff ii
B s
(7.86)
where the effective, or mean, field
1
2effB B s
is to be determined self-consistently.
Now, the partition function for (7.86) is
1
expN
N N eff ii
Z Tr B s
1
expN
eff ii
Tr B s
1
expN
effs
B s
2cosh
N
effB (7.87)
with a Gibbs free energy per site in the thermodynamic limit
1, lim lnB N
Ng T B k T Z
N ln 2coshB effk T B (7.88)
The probability of having spin is at site i is
1
1expi eff iP s B s
Z
exp
2cosheff i
eff
B s
B
(7.89)
so that
1s
s P s s
1
1exp
2cosh effseff
s B sB
sinh
cosheff
eff
B
B
tanh effB
1tanh
2B s
(7.91)
which is the self-consistent equation for s .
7.F.2b. Heat Capacity
For 0B , we have
1tanh
2s s
tanh
2 B
sk T
tanh s (7.92)
where
2C
B
T
k T T
and2C
B
Tk
(7.91a)
Eq(7.92) can be solved graphically as shown in Fig.7.6. By inspection, we have
0
0s
s
for1
1
or C
C
T T
T T
(7.92a)
Thus, CT is the critical temperature of the ferromagnetic phase transition. If
viewed as an order- disorder transition, s stands for the order parameter. Typical
temperature dependence of s is shown in Fig.7.7. Note that (7.92a) predicts a
finite CT for all d and is thus in disagreement with the exact result for 1d . In
fact, mean field theories typically overestimate CT for 3d . [see Chapter 8]
Eq(7.92a) gives rise to partition function of
12cosh
2
N
NZ s
0
2
2cosh2
N
N
B
s
k T
for C
C
T T
T T
and a Gibbs free energy per site of [see (7.88)],
0
ln 2
,0ln 2cosh
B
CB
k T
g T Tk T s
T
for C
C
T T
T T
(7.93)
The internal energy is, from (7.86),
ln NU Z
1
1
2
N
ii
H s s
21
2N s
Hence, the heat capacity is
NN
UC
T
N
sN s
T
2B
N
sNk s
Using (7.92), we have
2 1 1sech
2 2N N
s ss s
(7.96)
2
2
1 1sech
2 2
1 11 sech
2 2
s s
s
2 12cosh
2
s
s
(7.97)
so that
2
2 12cosh
2
N B
sC Nk
s
2
2
12
cosh
CB
C C
TNk s
T TT sT T
(7.98)
As T approaches CT from below, i.e., CT T or 1CT
T , eq(7.92) simplifies
to
31
3s s s
Neglecting 0s solution, we have
2 20 3
31s s
where 1CT
T
Thus, (7.98) becomes
231
23
3 1lim 2 lim 1
3cosh 1
CN B
T TC Nk
12
1 12 lim 3
1cosh 3
BNk
Now,
2
2 1 3 1cosh 3 1
2
1
1 3
2 1 1cosh 3 1 3
13
so that
1
1lim 2 lim 3
1 3CN B
T TC Nk
3 BNk
Now, for CT T , we have 0s and 0NC . Therefore, NC is as shown in
Fig.7.8.
7.F.2c. Magnetic Susceptibility
We now turn to the calculation of the magnetic susceptibility
TN
TN
MB
B
TN
sN
B
(7.99)
From (7.91), we get
2 1 1sech
2 2TN TN
s sB s
B B
(7.100)
2
2
1sech
21 1
1 sech2 2
B s
B s
2 1 1cosh
2 2B s
(7.101)
Hence
2
2 1 1cosh
2 2
TN
NB
B s
(7.102)
so that
2
2
01 1
cosh2 2
TN
N
s
2
2cosh C C
NT T
sT T
2
2
1
cosh
C
C C B C
N TT T T k TsT T
2
2
2
cosh
C
C C
N TT T TsT T
(7.103)
7.G. Grand Canonical Ensemble
An open system allows exchange of both heat and matter with its surrounding.
Both of its energy and particle number fluctuate about their equilibrium values.
A collection of such identical systems is called a grand canonical ensemble.
At equilibrium, the Gibbs entropy
lnBS k Tr is maximized subject to the
constraints 1Tr (7.104)
Tr H E (7.105)
and Tr N N (7.106)
With the help of the Lagrange multipliers 0 , N and E , we have
0ln 0B E NTr k H N
0ln 1B E NTr k H N (7.107)
0ln 1 0B E Nk H N (7.108)
i.e., 0exp 1 N E
B B B
N Hk k k
1exp N E
G B B
N HZ k k
(7.108a)
where the grand partition function is defined as
0exp 1GB
Zk
exp N E
B B
Tr N Hk k
(7.109)
Now, 7.108Tr gives
0 0B E NS k E N (7.110)
i.e., ln 0B G E NS k Z E N
Comparing with the fundamental equation for the grand potential :
'U TS N (2.104)
or1 1 '
0S U NT T T
we have
lnB Gk T Z (7.111)
1E T
(7.111a)
'N T
(7.111b)
so that (7.108a,9) becomes 1
exp 'G
H NZ
exp 'H N (7.113)
exp exp 'GZ Tr H N (7.112)
Eq(7.112) is the fundamental equation for an open system from which all other
thermodynamical properties can be obtained. For example, as discussed in section
(2.F.5), , , 'T X and
'd SdT YdX N d
Thus,
'X
ST
'T
YX
'XT
N
7.G.1. Exercise 7.6
7.G.2. Fluctuations
7.G.1. Exercise 7.6
Consider the photons in equilibrium inside a cubic box with volume 3V L andtemperature T at the walls. The photon energies are i ick , where ki is the
wavevector of the ith standing wave. Compute pressure P of this photon gas.
Answer
The allowed photon states are standing waves that vanish at the walls. Thus,
, ,x y zn n nL
k with 1,2,jn for , ,j x y z
so that
2 2 2x y z
cck n n n
L
(3)
and
k
33
, , 0x y zk k k
Ld k
33
2
Ld k
Since there are 2 transverse modes for each k, the sum over modes is
1,2
,f
k
k 3
3
1,2
,2
Ld k f
k
When f is independent of polarization, we have
1,2
2f f
k k
k k 3
322
Ld k f
k
Since ' 0 , the grand partition function is
expZ T Tr H , 0 ,
expn
n
k
k kk k
, 0
expn
n
k
k kk
,
1
1 exp
k k
2
1
1 exp
k k(1)
which gives a grand potential
, , lnBT V k T Z T ,
ln 1 expBk T
kk
(2)
3
32 ln 1 exp2B
Lk T d k
k
Using
3 24d k f k dk k f k 23
4d f
c c
we have
3
2
0
ln 1 expB
Lk T d
c
(2a)
Now,
2
0
ln 1 expI d
3
0
1ln 1 exp
3d
3 3
0 0
exp1 1ln 1 exp
3 3 1 expd
(2b)
The 1st term involves the function
ln 1 xf x x e
In particular
21lim 0
2x x
xf x e e
where we've used
21ln 1
2x x x
and, by repeated application of the L'Hospital rule,
'lim lim
'x a x a
f x f
g x g
to get1 !
lim lim lim lim 0n n
n xx x xx x x x
x nx nx e
e e e
Also
0
1
0 lim ln!
nn
xn
xf x
n
1
02
lim ln 1!
nn
xn
xx x
n
1
02
lim ln ln 1!
nn
xn
xx x x
n
2
0
1 1lim ln ln 1
2 3!xx x x x x
0
lim lnx
x x
10
lnlimx
x
x
1
20limx
x
x
0lim 0x
x
Hence, (2b) becomes
3
0
exp1
3 1 expI d
3
30
1
13
x
x
edx x
e
(2c)
Now,
0
1
1n
n xJ dx x
e
0 1
xn
x
edx x
e
00
n x m x
m
dx x e e
1
0 0
m xn
m
dx x e
10 0
1
1n y
nm
dy y em
10
1!
1n
m
nm
11
1!
nm
nm
! 1n n
where is the Reimann zeta function. Thus, (2c) becomes
33
1
3I J
3
13! 4
3
3
13! 4
3
4
3
13!
903
4
345
Hence, eq(2a) is
3
B
Lk T I
c
3 4
345
B
Lk T
c
3 2
4
45B
Lk T
c
The pressure is
PV
3 24 1
45Bk Tc
41
3T (6)
where3 21
15Bkc
is the Stefan- Boltzmann constant.
7.G.2. Fluctuations
The fixed quantities specified in a grand canonical ensemble are
T, ', U E, and NThus, it is of interest to find the fluctuations in E and N. Since the derivation for the
variance in E is similar to that for the canonical ensemble (see section 7.D.1), we shall
consider only the case for N here.
Combining the normalization (7.104) with (7.113) gives
exp ' 1Tr H N (7.114)
where , , 'T X . Thus, 7.114'
gives
exp ' 0'
TX
Tr N H N
0'
TX
N
or '
TX
N
(7.114a)
while 2
27.114
'
gives
2
22
2exp ' 0
' 'TXTX
Tr N H N
2
2
20
' 'TXTX
N
2 22
2
12
' ' 'TX TX TX
N N
22
2
1
'TX
N
[(7.114a) used]
Hence, the deviation is
222
2
1
'TX
N N
'B
TX
Nk T
(7.115)
The fractional deviation is
221
'B
TX
N N Nk T
N NN
1/ 2N
(7.116)
which vanishes as N . Hence, the grand canonical ensemble approaches the
canonical ensemble in the thermodynamic limit. In fact, all 3 ensembles give the
same macroscopic (average) physical quantities in the thermodynamic limit.
We now attempt to relate the deviation (7.115) to directly measurable quantities for
the case of a PVT system. To this end, we start with
'
'
'TN
TV
T
V
NVN
(a)
Using the diagram
' P
V N
(b)
we have
'
TN TV
P
V N
TP
TN
V
NV
P
so that (a) becomes
'
1''
TV
T TN
NVN V
'
TN
T TP
V
PV V
N N
(c)
Now, the diagram (b) also gives
'
TP TN
V
N P
1
TN
G
N P
V
N (d)
where we've used 'G N and 'dG SdT VdP dN .
Also from (b), we have
1
' 'T T
V N
N V
1
'TV
P
'
TVP
Since ' ' ,T P , we have
' ' '
TV T TN
V
P P P N
where (d) was used. Hence (c) becomes
2
' T
TV
N NV
V
where1
TTN
V
V P
so that finally, with the thermodynamic N identified with the statistical N , eq(7.115)
becomes2
22B T
NN N k T
V (7.117)
7.H. Ideal Quantum Gases
In quantum mechanics, the uncertainty principle implies that identical particles must
be indistinguishable since there is no way to ascertain their identities when the
distance between them is smaller thanp
, where p is the uncertainty in their
relative momentum. As was shown in the spin-statistics theorem of quantum field
theory, indistinguishable particles in 3-D space must obey Bose-Einstein (Fermi-Dirac)
statistics if they have integral (half-integral) spins.
[Note: recent studies in the non-integral quantum hall effects as well as high TC
superconductors led to the possible existence in 2-D systems of anyons that can have
arbitrary value of spins.]
In general, quantum effects are most noticeable when the system is near its ground
state, i.e., low T for macroscopic systems.
For an ideal gas of N identical particles, the hamiltonian is
1
1
N
Ni
H H i
where 1H i is the 1-particle hamiltonian of the ith particle, i.e.,
2
1
1
2H i i
m p
where ip is the momentum of the ith particle. If the particles are confined in a
rectangular volume x y zV L L L with periodic boundary conditions, the allowable
eigenvalues for p
are
p k 2 2 2, ,x y z
x y z
l l lL L L
(7.119)
where 0, 1,l for , ,x y z . Those for 1H are
2 2
2m k
k(7.120)
where zs is the spin component along the z-axis.
In the n-representation, we have
NH n
k kk
with N n
kk
where n k is the number of particles in state k . The grand partition function is
, exp 'NZ T V Tr H N (7.118)
exp 'Tr n
k kk
exp 'n
n
k
k kk
exp 'n
n
k
k kk
(7.118a)
For spin independent hamiltonians, this simplifies to
2 1
, exp '
s
n
Z T V n
k
k kk
For bosons, there is no restrictions on n k so that
2 1
0
, , ' exp '
s
BEn
Z T V n
k
k kk
(7.121)
For fermions, 0,1n k , so that
2 1
1
0
, , ' exp '
s
FDn
Z T V n
k
k kk
(7.122)
7.H.1. Bose-Einstein Ideal Gases
7.H.2. Fermi-Dirac Ideal Gases
7.H.1. Bose-Einstein Ideal Gases
7.H.1a. Basics
7.H.1b. Integration
7.H.1c. Pressure
7.H.1d. Number
7.H.1e. Thermodynamic Limit
7.H.1f. Heat Capacity
7.H.1g. High T limit
7.H.1h. Exercise 7.7
7.H.1a. Basics
For 0-spin free bosons,
0
, , ' exp 'BEn
Z T V n
k
k kk
0
exp 'n
n
k
k kk
(7.121)
where , , ,n n nk 0 k so that 0 0 0n n n
k 0 k
.
Doing the summation, we get
1
, , '1 exp 'BEZ T V
k k
(7.124)
which gives a grand potential
, , ' ln , , 'BE B BET V k T Z T V
ln 1 exp 'Bk T kk
(7.125)
The average number of particles is
'BE
TV
N
exp '
1 exp 'Bk T
k
k k
1
exp ' 1
k k
n kk
(7.126)
where the nk is called the occupation number of state k and
1
exp ' 1n
k
k exp
z
z
k
(7.127)
where exp 'z is the fugacity.
[ Note: meaningful usage of the fugacity as defined abopve requires the implicit
assumption of min 0 k . A more general definition is minexp 'z .]
Now, the lowest value of k is 0 0 with 0,0,0k and
1
exp ' 1n
0 1
z
z
(7.128)
Since the number of particles cannot be negative, we must have
1exp ' 1
z or ' 0
which implies
1 0z and ' 0
Thus, adding particles to the gas actually decreases its internal energy. The case
' 0 means that N is no longer fixed. This applies to photons, phonons, or
any bosons that serve as "carrier" of interactions.
In the classical limit valid for small , we expect (7.127) to become the Boltzmann
distribution expn k k . This can be achieved if expz k for all k,
or, since 0 k , if 1z . Thus, the limit exp ' 0z corresponds to
1 and ' . Incidentally, this means that, carriers of interactions, with
' 0 , have no classical limit.
Of particular interest is the case 1z for which n 0 . This means the
ground state is macroscopically occupied and we have a Bose- Einstein
condensation. Since the ground state plays a prominent role, we expect the
situation applies when 1 and ' 0 . For example, as a helium liquid
becomes a superfluid, its chemical potential drops from a finite value to zero.
7.H.1b. Integration
For a macroscopic system, we can replace the summation over states with integrals,
i.e.,
3 3
3 32 2
V Vd k d p
k
However, the possibility of a Bose-Einstein condensation means that the state k 0
requires special attention. Thus,
33
'2
Vf d p f
k
k p
where 32
V
is the volume in p-space occupied by the ground state, and '
is the p-space volume with taken out. Since f f 0 within , we have
33
'2
Vf f d p f
k
k 0 p
33
'2
Vf d p f
0 p
If f is isotropic, i.e., f f p , then
0
23
40
2 p
Vf f dp p f p
k
k
(7.130)
where the "radius" 0p of the ground state is given approximately by
0
2p
L
with 1/ 3L V 1/ 3
x y zL L L
As an example, (7.126) becomes
0
23
2
1 4 1
exp ' 1 12 exp ' 12
p
VN dp p
pm
0
23
2
411 2 exp
2p
z V zdp p
z p zm
(7.131)
Similarly, (7.125) becomes
, , ' ln 1 exp 'BE BT V k T
0
2 23
4 1ln 1 exp '
22B
p
Vk T dp p p
m
0
2 23
4 1ln 1 ln 1 exp
22B B
p
Vk T z k T dp p z p
m
(7.132)
Setting
2x p
m
2 B
p
mk T
2Tp
where the thermal wavelength is given by
22T
Bmk T
2 B
h
mk T (7.135)
eq(7.130) becomes
0
3
23
4 2 20
2 T Tx
Vf f dx x f x
k
k
0
2
3
4 20
TT x
Vf dx x f x
where
0 02
Tx p
T
L
Thus, eq(7.132) becomes
, , ' ln 1BE BT V k T z 0
2 2
3
4ln 1 expB
T x
Vk T dx x z x
(7.133)
Similarly, eq(7.131) gives
0
2
23
4
1 expT x
z V zN dx x
z x z
(7.134)
7.H.1c. Pressure
The pressure of the gas can be obtained from (2.122) and (7.133) as
BEPV
0
2 2
3
4ln 1 ln 1 expB
B
T x
k Tz k T dx x z x
V
(7.136a)
Consider now the integral
2
0
ln 1 expnnI z dx x z x
2
1 0
expm
n
m
zdx x mx
m
1 / 2
1
1 1
2 2
mn
m
z nm
m
3 / 21
1 1
2 2
m
nm
n z
m
3 / 2
1 1
2 2 n
ng z
(7.137a)
where we've used the Gaussian integral formula
1 / 22
0
1 1exp
2 2nn n
dx x mx m
and the definition
1
m
m
zg z
m
(7.137)
Hence
2 5/ 21
1 3
2 2
m
m
zI z
m
5/ 214
m
m
z
m
5/ 24g z
From (7.137), we see that, (cf. Fig.7.10),
5/ 2 0 0g
5/ 2 5/ 21
1 51 1.342
2m
gm
(7.142)
where is the Riemann zeta function. Next,
2
00
lim ln 1 expa
nn
ai z dx x z x
00
lim ln 1 0a
n
adx x z
Hence, (7.136) becomes
2 23
4ln 1B
B
T
k TP z k T I z i z
V
5/ 23
1ln 1B
BT
k Tz k T g z
V (7.136)
Note that since 2 0i z , the contribution of the ground state would have been lost
without the special treatment of (7.130).
7.H.1d. Number
The average particle density of the gas can be obtained from (7.134) as
nN
V
0
2
23
1 4
1 expT x
z zdx x
V z x z
(7.139a)
Consider now the integral
20 exp
nn
zK z dx x
x z
(7.139b)
Using
2
2
2
expln 1 exp
1 exp
xz x
z z x
2
1
exp x z
we have, from (7.137a) that
n
n
dI zK z z
dz
3 / 2
1
1 1
2 2
m
nm
n d zz
dz m
1 / 21
1 1
2 2
m
nm
n z
m
1 / 2
1 1
2 2 n
ng
(7.139c)
Another, perhaps more common, form of (7.139b) can be obtained by putting 2y xin (7.139b) so that
12
0
1
2
n
n y
zK z dy y
e z
12
10
1 1
2 1
n
ydy y
z e
(7.139d)
Now, from (7.139c), we have
2 3/ 24K z g z
Note that, (cf. Fig.7.10),
3/ 2 0 0g
3/ 2 3/ 21
1 31 2.612
2m
gm
(7.142)
where is the Riemann zeta function. Next,
200
limexp
an
na
zk z dx x
x z
00
lim 01
an
a
zdx x
z
Hence, (7.139a) becomes
2 23
1 4
1 T
zn K z k z
V z
3/ 23
1 1
1 T
zg z
V z
(7.139)
which can be used to obtain '.
7.H.1e. Thermodynamic Limit
We now consider the thermodynamic limits,
,N V withN
nV (a)
of the expressions
5/ 23
1ln 1B
BT
k TP z k T g z
V (7.136)
3/ 23
1 1
1 T
zn g z
V z
(7.139)
Obviously, for 1z ,
5/ 23
1B
T
P k T g z
(b)
3/ 23
1
T
n g z
(c)
Hence, we need only consider the case 1z , which, as discussed in section 7.H.1a,
corresponds to the Bose- Einstein condensation limit with ' 0 and usually
1 . For 1z , the 1st the term in (7.139), i.e.,
01
1lim
1z
zn T
V z
(d)
denotes the density of particles occupying the ground state when the system is at
temperature T. If n is kept constant, we have
010
10 lim
1zT
zn n
V z
(e)
Combining (d) and (c) turns (7.139) into
3/ 23
0 3/ 23
1
11
T
T
g z
n
n T g
for1
1
z
z
(7.145)
As can be seen in Fig.7.10, 3/ 2g z is a monotonically increasing function of z with
a maximum at 1z . Thus, if we keep reducing T while keeping n constant,
3/ 23
1
CT
g z
will become smaller than n below a critical temperature CT given
by
3/ 23
11
CT
n g
3/ 2
3/ 221
2B Cmk T
g
2 / 32
3/ 2
2
1CB
nT
mk g
2 / 322
2.612B
n
mk
(7.148)
For CT T , we have 3/ 23
11
T
n g
so that 0 0n and we have a new phase
characterized by a macroscopic occupation of the ground state. The fraction of
particles in the condensate is
0n TT
n
3/ 23
11
T
g
n
3
31 CT
T
3/ 2
1C
T
T
(7.149)
Condensation can also occur if we keep increasing n while keeping T constant.
the critical per particle volume is given by
1C
C
v Tn
3
3/ 2 1T
g
(7.147)
Note that also serves as the order parameter of the phase transition. The case0 1 thus indicates the coexistence of the "normal" and "condensed" phases. A
plot of vs T is shown in Fig.7.13.
Turning now to the pressure as given in (7.136), we begin by writing (d) as
00
1 1limx
xn T
V x
0
1limx Vx
so that
0x Vn T or 01z Vn T
Thus
01
1 1 1lim ln 1 lim lnV Vz
zV V Vn T
1 1
lim ln 0V V V
where we've used0
lim ln 0a
a a
. If this is combined with the fact that for 1z ,
1lim ln 1 0V
zV
the 1st term in eq(7.136) can be dropped so that
5/ 23
5/ 23
1
11
BT
BT
k T g z
P
k T g
for1
1
z
z
(7.144)
At the critical point, CT T , and 1z , so that
5/ 23
11
C
C B CT
P k T g
Using (7.147) and (7.148) gives
2 / 3
2
5/ 23/ 2 3/ 2
2 11
1 1C BB C C
nP k g
mk g v T g
5/ 3
2
5/ 23/ 2
2 11
1C C
gm v T g
(7.150)
where we've used 1C Cv T
n . For CT T , (7.144,7) gives
5/ 23
11B
T
P k T g
5/ 2
3/ 2
11
1BC
gk T
v T g (7.150a)
irregardless of the fraction of particles in the condensed phase. Thus, treating
(7.150a) as a function P P v where Cv v T gives the coexistence curve in
the P-v plane [see dotted line in Fig.7.14]. For a given T, the region to the left of the
coexistence curve is the coexistence region where P is a constant given by (7.150a).
7.H.1f. Heat Capacity
Since the independent variables for a grand canonical ensemble is , , 'T X , the
entropy per unit volume may be written as
0'
limV
T
S Ss
V V
Using the diagramP T
S V
, we have
'V
Ps
T
.
Since,
5/ 23
5/ 23
1
11
BT
BT
k T g z
P
k T g
for1
1
z
z
(7.144)
where exp 'z and22
TBmk T
, we have
2'
'
V B
z z
T k T
' 2T T
VT T
and
5/ 2 5/ 22'
'
V B
z dg z g z
T k T dz
3/ 22
'
B
g zk T
where we've used (7.140). Thus, for 1z ,
5/ 2 5/ 2 3/ 23 4 3 2
1 3 1 '
2T
B B BT T T B
s k g z k T g z k T g zT k T
5/ 2 3/ 23 3
5 1 '
2 BT T
k g z g zT
5/ 23
5 1 '
2 BT
k g z nT
[see (7.145)]
5/ 23
5 1ln
2 B BT
k g z k n z
(7.151a)
Setting 1z , we have
5/ 23
5 11
2 BT
s k g
for 1z (7.151b)
so that 3/ 2s T as 0T , in agreement with the 3rd law.
Now, the heat capacity at constant density is
nn
sc T
T
Now, from (7.145), we see that for 1z ,
3/ 23/ 24 3
3 10 T
n nT T
dg z zg z
T dz T
3/ 2 1/ 23 3
3 10
2 nT T
zg z g z
T z T
so that
3/ 2
1/ 2
3
2n
g zzz
T T g z
(7.152)
Hence, for 1z ,
5/ 24
5 3
2 2T
n BT
c T k g zT
3/ 2 3/ 23
1/ 2
5 1 1 3
2 2B BT
g z g zT k k n z
z z T g z
3/ 25/ 23
1/ 2
15 1 9
4 4B BT
g zk g z k n
g z (7.153a)
where (7.145) was used. For the case 1z , eq(7.151b) gives
5/ 23
15 11
4n BT
c k g
(7.153b)
Note that
1/ 2 1/ 21
11
m
gm
so that nc is continuous at CT as shown in Fig.7.15.
7.H.1g. High T Limit
In the high T limit, 0z so that
2 25/ 2
0
4ln 1 expg z dx x z x
(7.137)
becomes
2 25/ 2
0
4expg z z dx x x
4 1 3
2 2z z
Now, by definition,
1
m
m
zg z
m
so that
111
m
m
dg z zz g z
dz m
Therefore, we have,
5/ 2 3/ 2 1/ 2g z g z g z z as 0z
Hence, (7.145) becomes
3/ 2
3 2exp '
2B
T
z mk Tn
(7.154)
while (7.144) simplifies to
3B
B BT
Nk TzP n k T k T
V (7.155)
Finally, the heat capacity (7.153a) becomes
3
15 9
4 4n B BT
zc k k n
3
2 Bk n3
2 B
Nk
V (7.156)
7.H.1h. Exercise 7.7
Compute the variance 2N N for 0T .
Answer
From the definition
exp 'N Tr N Tr N H N
we have
1exp '
' 'TVTV
NTr N N H N
2
'TV
N N
22N N 2N N (1)
Now, for CT T , we have 1z so that (7.139) gives
3/ 231 T
z VN g z
z
(2)
Hence, with
'TV
zz
and
1
dg zz g z
dz
eq(2) becomes
1/ 22 3
1 1
' 1 1 TTV
N z Vz g z
z z
2
3/ 2 3/ 2 1/ 23 3 3T T T
V V VN g z N g z g z
2
1/ 23T
VN N g z
2
N N (3)
where we've dropped the 3/ 23
1
T
g z
terms since they vanishes as 0T .
7.H.2. Fermi-Dirac Ideal Gases
7.H.2a. Basics
7.H.2b. Integration
7.H.2c. Pressure
7.H.2d. Number
7.H.2e. Low Temperature Limit
7.H.2f. Heat Capacity
7.H.2g. Exercise 7.8
7.H.2h. Exercise 7.9
7.H.2a. Basics
For s-spin free fermions,
1
0
, , ' exp 's s
FDs sn
Z T V n
k
k kk
1
0
exp 's
s n
n
k
k kk
where zs and , , ,n n n k 0 k so that
1 1 1
0 0 0n n n
k 0 k
. Also,
we’ve assumed a spin independent hamiltonian. Doing the summation gives
, , ' 1 exp 's
FDs
Z T V
kk
2 11 exp '
s
k
k
(7.158)
which gives a grand potential
, , ' ln , , 'FD B FDT V k T Z T V
2 1 ln 1 exp 'Bk T s kk
(7.159)
The average number of particles is
'FD
TV
N
exp '2 1
1 exp 'Bk T s
k
k k
2 1
exp ' 1
s
k k
exp ' 1
g
k k
n kk
(7.160)
where 2 1g s and nk is called the occupation number of state l and
exp ' 1
gn
k
k
1 exp 1
g
z k exp
gz
z
k
(7.161)
where exp 'z is the fugacity.
[ Note: meaningful usage of fugacity requires the implicit assumption of min 0 k ].
Now, (7.161) has no singularity if all quantities are real. Hence, we can have
' k , which corresponds to 0 n g k . [see Fig.7.16].
At T 0 or , we have
0
gn
k for'
'
k
k
(7.161a)
which is a step function with discontinuity at0
' FT
. This zero temperature
chemical potential F is also called the Fermi energy. The magnitude of the
corresponding momentum 2F Fp m is called the Fermi momentum. A
picturesque way to describe (7.161a) is to say that the fermions are in a Fermi seawith a Fermi surface at F . This is a helpful reminder that only particles near the
surface are easily excited [see Fig.7.16].
The fact that 0F means that adding particles to the gas is "difficult" since it
increases the internal energy of the gas.
In the classical limit valid for small , we expect (7.161) to become the Boltzmann
distribution expn k k . This can be achieved if expz k for all k,
or, since 0 k , if 1z . Thus, the limit exp ' 0z corresponds to
1 and ' .
7.H.2b. Integration
For a macroscopic system, we can replace the summation over states with integrals,
i.e.,
3 3
3 32 2
V Vd k d p
k
If f is isotropic, i.e., f f p , then
23
0
4
2
Vf dp p f p
k
k
(7.162)
In practice, it is sometimes more convenient to work with energies. To this end, we
define the density of states by
3
32
Vd k d
3
2
V dSd
k
k
where Sk is a k space surface on which the energy has a constant value .
For a free particle,
2 2
2m k
k k k
2
m
k 2k
m
23 2
2
V md k
k
2 22
Vmk
2 2 2
2
2
Vm m
so that
3/ 2
2 2
2
4
V md
k
(7.162a)
As an example, (7.160) becomes
2
320
4
12 exp ' 12
V gN dp p
pm
2
320
4
2 exp2
V gzdp p
p zm
(7.164)
3/ 2
2 20
2
4 exp ' 1
V m gd
3/ 2
2 20
2
4 exp
V m gzd
z
(7.164a)
Similarly, (7.159) becomes
2 23
0
4 1, , ' ln 1 exp '
22FD B
VT V k T g dp p p
m
2 2
30
4ln 1 exp
22B
Vk T g dp p z p
m
(7.163)
3/ 2
2 20
2ln 1 exp
4B
V mk T g d z
(7.163a)
Setting
2x p
m
2 B
p
mk T
2Tp
2y x where the thermal wavelength is given by
22T
Bmk T
2 B
h
mk T (7.135)
eq(7.162,a) becomes
3
23
0
4 2 2
2 T T
Vf dx x f x
k
k
23
0
4
T
Vdx x f x
3/ 2
2 20
2
4B
B
V mk Tdy y f yk T
30
2
T
Vdy y f y
(7.162b)
Note that
2
0 0
1
2dx x f x dy y f y
(7.162c)
Thus, eq(7.163) becomes
2 2
30
4, , ' ln 1 expFD B
T
VT V k T g dx x z x
(7.163a)
3
0
2ln 1 expB
T
Vk T g dy y z y
(7.163b)
Similarly, eq(7.164) gives
2
230
4
expT
V zN g dx x
x z
(7.164a)
30
2
expT
V zg dy y
y z
(7.164b)
7.H.2c. Pressure
The pressure of the gas can be obtained from (2.122) and (7.163a) as
FDPV
2 2
30
4ln 1 expB
T
k T g dx x z x
(7.165a)
1/ 2
30
2ln 1 expB
T
k T g dy y z y
(7.165b)
Consider now the integral
2
0
ln 1 expnnI z dx x z x
1 / 2
0
1ln 1 exp
2ndy y z y
1 2
1 0
expm
m n
m
zdx x mx
m
1 1 / 2
1
1 1
2 2
mm n
m
z nm
m
1
3 / 21
1 1
2 2
mm
nm
n z
m
3 / 2
1 1
2 2 n
nf z
(7.165b)
where
1
1
mm
m
zf z
m
(7.166)
Hence,
2 5/ 24I z f z
while (7.165a) becomes
5/ 23
1B
T
P k T g f z
(7.165)
7.H.2d. Number
The average particle density of the gas can be obtained from (7.164a) as
nN
V
2
230
4
expT
zg dx x
x z
(7.164b)
30
2
expT
zg dy y
y z
(7.164c)
Note that n is a constant since both N and V are kept constant in the grand
canonical ensemble. Consider now the integral
20 exp
nn
zK z dx x
x z
1 / 2
0
1
2 expn z
dy yy z
Using
2
2
2
expln 1 exp
1 exp
xz x
z z x
2
1
exp x z
we have, from (7.165b) that
n
n
dI zK z z
dz
1
1 / 21
1 1
2 2
mm
nm
n z
m
1 / 2
1 1
2 2 n
nf z
so that
1
0
1
expn
n
zf z dy y
n y z
(7.164d)
or, with ' ,
1
0
1 1
exp 1n
nf dy yn y
(7.164e)
which is the more familiar definition of the Fermi integral nf . Hence
2 3/ 24K z f z
Hence, (7.164b) becomes
3/ 23T
gn f z
(7.167)
or3
2 33/ 2 3/ 2
1 1
2 3T n z z z
g
(7.167a)
Now, the 1st few coefficients in the inversion of a series
0 01
n
nn
y y a x x
0 01
n
nn
x x b y y
are given by (see Arfken )
11
1b
a 2
2 31
ab
a 2
3 2 1 351
12b a a a
a
Applying these to (7.167a) gives
2 33 3 3
3/ 2 2 3/ 2
1 1 1
2 2 3T T Tz n n n
g g g
(7.169)
which gives the T dependence of z and hence '.
Since2
1/ 22T
B
Tmk T
, we have
0z as T z as 0T
Since exp 'z , this means
' as T ' 0 as 0T
7.H.2e. Low Temperature Limit
We now turn to the evaluation of the Fermi integral
1
0
1 1
exp 1n
nf dy yn y
(7.164e)
in the limit of low T or 1 . Thus, 1exp 1n y y
is essentially a
step function so that its derivative
2
exp1
exp 1
ydnn n
dy y
vanishes everywhere except for y . To take advantage of this, we integrate
(7.164e) by part to get
0
1 1
exp 1n
nf dyn n y
200
exp1 1
exp 1 exp 1
n n yy dy y
n n y y
20
exp1
exp 1
n ydy y
n n y
2
exp1
exp 1
n tdt t
n n t
[ t y ]
20
exp1 !
! ! exp 1
nn m m
m
tnv dt t
n n m n m t
Note that the sum over m has no upper limit if n is not an integer. For 1 , we
can replace the lower limit of the integral by so that
2
0
1 ! exp1
! ! exp 1
nn m m
nm
n tf v dt t
n m n m t
0
1 !1
! !
nn m
mm
nv I
n m n m
0
1
! !
nn m
mm
v Im n m
where
2
exp
exp 1
mm
tI dt t
t
(7.173)
Now, taking t t , we have
2
exp
exp 1
m
m
tI d t t
t
2
exp
exp 1
m m tdt t
t
2
exp
1 exp
m m tdt t
t
m
mI
Hence, 0mI for m odd. For m even, we have
2
0
2 exp 1 expmmI dt t t t
0 0
2 1 exp 1j m
j
j dt t j t
0 0
12 exp
1
j mm
j
dt t tj
0
12 !
1
j
mj
mj
1
12 !
j
mj
mj
while from (7.173), we have 0 1I . Now,
1 1 1
1 12
2
n
mm mn n nn n n
1
1 1
1 12 m
m mn nn n
11 2 m m
so that
12 ! 1 2 mmI m m [ m even ]
Putting everything together, we have
2
21
1 1
! 2 ! 2 !n n m
n mm
f v In m n m
2 1 2
1
1 12 1 2 2
! 2 !n n m m
m
v mn n m
(7.172a)
with
2
26
4
490
6
6945
For example, (7.167) becomes
3
3/ 2T n f
g
11 2
3/ 2 1/ 2 13 1! 2 ! 1 2
2 2 6
Using
3 5 3 3 3 1 1 3 1!
2 2 2 2 2 2 2 2 2
1 1
!2 2
we have
3 3/ 2
3/ 2 1/ 24' '
63T n
g
(7.174)
which, for 0T , becomes
3/ 22
3/ 21 2 4'
3n
g m
so that
2 / 32
0
32'
4T
n
m g
2 / 322 6
2
n
m g
F (7.175)
Thus, (7.174) can be written as
2
3/ 2 3/ 2 1/ 22' '8F
so that
2 / 322
1/ 2
3/ 2' 1 '8
BF
F
k T
221/ 2
3/ 21 '12
BF
F
k T
22
1/ 2
3/ 2112
BF F
F
k T
22
112
BF
F
k T
(7.176)
7.H.2f. Heat Capacity
The internal energy is given by
U H n
k kk
2 1s n k kk
3/ 2
30
2
expBT
V zgk T dy y
y z
5/ 23
2 5
2BT
Vgk T f
where we've used (7.162b), (7.164e), Bk Ty , and exp ' expz .
From (7.172a), we have
1 1
5/ 2 1/ 2 15/ 2
5 1! 2 ! 1 2 2
2 2f
2
5/ 2 1/ 25/ 2
5 2
2 5 4f
25/ 2 1/ 2
3
2 2
5 4BT
VU gk T
Using3 3/ 2
3/ 2 1/ 24
63T n
g
(7.174)
we have
13/ 23/ 2 1/ 2
3
4
63T
gn
123/ 2 23
14 8
n
23/ 2 23
14 8
n
so that
2 23/ 2 2 5/ 2 1/ 23 2
12 8 5 4BU Vk T n
23/ 2 5/ 2 1/ 23 2
2 5 5BVk T n
213
5 2Bk T N
[ N V n ]
223'
5 2 'Bk T
N
Using
22
' 112
BF
F
k T
(7.176)
we have
12 222 2 23
1 15 12 2 12
BB BF
F F F
k Tk T k TU N
223 51
5 12B
FF
k TN
(7.177)
so that
VV N
UC
T
2
21
2B
FF
kN T
2
2B
BF
k TN k
(7.178)
7.H.2g. Exercise 7.8
Compute 2N N for 0T .
Answer
From
, exp 'Z T V Tr H N
we have
exp ''
TV
ZTr N H N
Z N
2
222
exp ''
TV
ZTr N H N
2 2Z N
so that
22
2 2
1
'TV
ZN
Z
2
1
'TV
Z NZ
1
' 'TV TV
Z NN Z
Z
2
'B
TV
NN k T
Therefore
2 22N N N N 'B
TV
Nk T
(1)
For 0T , eq(7.174) simplifies to
3
3/ 24'
3T
N
V g
(2)
3/ 2
3
4'
3T
gVN
3/ 2
2
4 '
23
gV m
(3)
Hence
3/ 2
2
2'
' 2TV
N gV m
(4)
Now, (3) gives
2 / 322 3
'4
Nm gV
so that (4) becomes
1/ 33/ 2 2
2
2 2 3
' 2 4TV
N gV mN
m gV
1/ 32
2 4
3
4
m gV N
V
(5)
Hence, by (1), we have
1/ 32
2
2 4
3
4B
m gN N k TV N
V
(6)
7.H.2h. Exercise 7.9
See section 7.H.2b.
Summary
Bose-Einstein Gases
5/ 23
1ln 1B
BT
k TP z k T g z
V
3/ 23
1 1
1 T
zn g z
V z
2 / 32
3/ 2
2
1CB
nT
mk g
0
1Cv T
n n T
3
3/ 2 1T
g