Quantum Statistics of Ideal Gas

Kumneger T.

kumneger.tadele@astu.edu.et
Adama Science and Technology University

July 2, 2021

Kumneger T. (ASTU) July 2, 2021 1 / 37

Identical particles and symmetry requirements

Consider a system of gas consisting of N identical particles enclosed

in a container of volume V .
Suppose the collective coordinates of th ith particle is represented by
Qi and its quantum state by Si .
The state of the system of gas is then described by the set of
quantum numbers
S1 , S2 , ..., SN (1)
The system of gas in this state can be described by the following
wave function
ψ = ψS1 ...SN (Q1 , ..., QN ) (2)

Kumneger T. (ASTU) July 2, 2021 2 / 37

If the particles are distinguishable and any number of particles are
allowed to be in the same state S, then the particles are said to obey
Maxwell-Boltzman statistics. This is called classical description of a
system and it doesn’t impose symmetry requirements on the wave
function when two particles are interchanged.
However, the quantum-mechanical description, where identical
particles are considered to be indistinguishable, impose symmetry
requirements on the wave function during interchanging two particles,
i.e., interchanging two identical particles doesn’t lead the whole
system to a new state.
If each particle in the system has integral spin, the wave function of
the system ψ must be symmetric under interchanging two particles.

ψ(...Qi ...Qj ...) = ψ(...Qj ...Qi ...) (3)

Kumneger T. (ASTU) July 2, 2021 3 / 37

Particles satisfying this symmetry condition are said to obey
Bose-Enstein statistics and they are called bosons.
On the other hand if each particle in the system has half integral spin,
then the wave function of the system satisfy antisymmetric condition
during interchanging two particles
ψ(...Qi ...Qj ...) = −ψ(...Qj ...Qi ...) (4)
Particles satisfying this antisymmetric condition are said to obey
Fermi-Dirac statistics and they are called fermions.
If two particles, say i and j, of the same state S are interchanged, the
wave function of the system should remain the same. At the same
time if the particles have half-integral, spin condition (4) must be
satisfied. This leads to the conclusion that for a system containing
two particles in the same state, the wave function should vanish.
ψ=0 (5)
Thus in the Fermi-Dirac case there exists no state of the whole gas
for which two or more particles are in the same single particle state.
This is called Pauli exclusion principle.
Kumneger T. (ASTU) July 2, 2021 4 / 37
Formulation of the statistical problem

Consider a system of gas consisting of N weakly interacting identical

particles enclosed in a container of volume V .
Suppose the system is in equilibrium at a temperature T .
Let nr particles are in state r characterised by energy r .
The energy of the whole system which is supposed to be in state R is
then given by
X
ER = n1 1 + n2 2 + ... = nr r (6)
r

The partition function of the system can be given by

X X
Z= exp −βER = exp −β(n1 1 + n2 2 + ...) (7)
R R

Kumneger T. (ASTU) July 2, 2021 5 / 37

The mean number of particles in a particular state r is then
P
R nr exp −β(n1 1 + n2 2 + ...)
n¯r = P
R exp −β(n1 1 + n2 2 + ...)
1X 1 ∂
= − exp −β(n1 1 + n2 2 + ...)
Z β ∂r
R
1 1 ∂ X
= (− ) exp −β(n1 1 + n2 2 + ...)
Z β ∂r
R
1 1 ∂Z
= −
β Z ∂r
1 ∂ ln Z
= − (8)
β ∂r

Kumneger T. (ASTU) July 2, 2021 6 / 37

 P 2
R nr exp −β(n1 1 + n2 2 + ...)
n¯r2 = P
R exp −β(n1 1 + n2 2 + ...)
1 1 ∂ 1 ∂ X
= (− )(− ) exp −β(n1 1 + n2 2 + ...)
Z β ∂r β ∂r
R
1 1 ∂2Z
=
β 2 Z 2 ∂2r
1 ∂ 1 ∂Z 1 ∂Z 2
= 2
[ ( ) + 2( ) ]
β ∂r Z ∂r Z ∂r
1 ∂ ∂ ln Z 1 ∂Z 2
= 2
[ +( ) ]
β ∂r ∂r Z ∂r
1 ∂ ∂ ln Z
= [ + β 2 n¯r 2 ]
β 2 ∂r ∂r
1 ∂ 2 ln Z
= + n¯r 2 (9)
β 2 ∂2r

Kumneger T. (ASTU) July 2, 2021 7 / 37

 2
¯r )2 = (nr −¯ ¯ n)2r = n¯2 − n¯r 2 = 1 ∂ ln Z
=⇒ (∆n (10)
r
β 2 ∂2r

If the particles under consideration obey Maxwell-Boltzman statistics,

partition function of the system can be obtained by eq. (6) after
summing over all possible values of nr , (nr = 0, 1, 2, ...) for each r
provided that total number of particles in the system is constant
X
nr = N (11)
r

Since the particles are considered to be distinguishable, in addition to

specifying the number of particles in each state, it is necessary to
specify which particular particle is in which state.
In the case of Bose-Einstein statistics, the summation in eq. (6)
should be taken over all possible values of nr for each r , and the total
number of particles must be fixed. However, since the particles are
considered to be indistinguishable, specifying the number of particles
in each state is sufficient.
Kumneger T. (ASTU) July 2, 2021 8 / 37
A special case of Bose-Einstein statistics where total number of
particles in the system is not constant, i.e., where the restriction in
eq. (10) is lifted, is called Photon statistics.
In Fermi-Dirac statistics the summation should be over the two
possible values of nr , (nr = 0, 1), for each r , since no state can be
occupied by more than one particle and the total number of the
particles must be fixed.
The Quantum Distribution Function
Consider a system of gas containing N particles.
Let the lowest energy level of a single particle is denoted by o
For a system of particles that obey BE statistics, where there is no
restriction on the number of particles in any state, the lowest energy
level of the whole system can be obtained by placing all the particles
in the lowest energy level, and hence for the lowest energy level of the
whole system one can write

Eo = No (12)

Kumneger T. (ASTU) July 2, 2021 9 / 37

In the case of FD statistics however, where we are not allowed to
have more than one particle in any state, the lowest energy level of
the whole system can only obtained by placing one particle in each
consecutive states of increasing energy starting from the lowest
energy level o . Hence even at the lowest energy level of the system
there are particles which have very high energy as compared to o .
For a system maintained at absolute temperature T , the mean
number of particles in a particular state r is
P
n ,n ,... nr exp −β(n1 1 + n2 2 + ...)
n¯r = P1 2 (13)
n1 ,n2 ,... exp −β(n1 1 + n2 2 + ...)

nr exp −βnr r rn1 ,n2 ,... nr exp −β(n1 1 + n2 2 + ...)

P P
nr
n¯r = P Pr (14)
nr exp −βnr r n1 ,n2 ,... exp −β(n1 1 + n2 2 + ...)

Kumneger T. (ASTU) July 2, 2021 10 / 37

Photon Statistics
The summation in eq. (10) should be over all possible values of nr for
each r . Since
P there is no restriction on the total number of particles,
the sums r in the numerator and denuminator of eq. (10) are the
same. Hence one can write
P
nr nr exp −βnr r
n¯r = P
nr exp −βnr r
∂ P
−1 ∂r nr exp −βnr r −1 ∂ X
= P = ln exp −βnr r
β nr exp −βnr r β ∂r n r

−1 ∂ 1 1 ∂
= ln = ln 1 − exp −βr
β ∂r 1 − exp −βr β ∂r
1 β exp −βr exp −βr 1
= = =
β 1 − exp −βr 1 − exp −βr exp βr − 1
(15)

This is called Planck distribution.

Kumneger T. (ASTU) July 2, 2021 11 / 37
FD Statistics
In this case nr only has two values, nr = o, 1, for each r and there is
restriction
P on the total number of particles where it is supposed to be
fixed r nr = N. Let
r
X
Zr (N 0 ) = exp −β(n1 1 + n2 2 + ...) (16)
n1 ,n2 ,...

where N 0 is supposed to be the number of particles distributed over

the remaining states (excluding particular state r )
r
X
nr = N 0 (17)
s
Thus eq. (10) becomes
0 ∗ Zr (N) + exp −βr Zr (N − 1)
n¯r =
Zr (N) + exp −βr Zr (N − 1)
exp −βr Zr (N − 1)
=
Zr (N) + exp −βr Zr (N − 1)
Kumneger T. (ASTU) July 2, 2021 12 / 37
 1
n¯r =
[ ZrZ(N−1)
r (N)
] exp βr +1

For small ∆N, ∆N  N, one can write

∂ ln Zr (N)
ln Zr (N − ∆N) = ln Zr (N) − ∆N − ....
∂N
∂ ln Zr (N)
= ln Zr (N) − ∆N
∂N
∂ ln Zr (N)
= ln Zr (N) − αr ∆N; αr = (18)
∂N
=⇒ Zr (N − ∆N) = Zr (N) exp −αr ∆N (19)
with an approximation that αr is independent of state r , such that

∂ ln Z (N)
αr = α = (20)
∂N

Kumneger T. (ASTU) July 2, 2021 13 / 37

Thus one can obtain

Zr (N − ∆N) = Zr (N) exp −α∆N (21)

=⇒ Zr (N − 1) = Zr (N) exp −α (22)

The mean number of particles in a given state r is then
1 1
n¯r = Zr (N)
= (23)
[ Zr (N) exp (α + βr ) + 1
exp −α ] exp βr +1

This is called Fermi-Dirac distribution.

Kumneger T. (ASTU) July 2, 2021 14 / 37

Helmholtz free energy is given by
F = Ē − TS
= Ē − TkB (ln Z + β Ē ); S = kB (ln Z + β Ē )
= Ē − kB T ln Z − Ē
= −kB T ln Z
F
=⇒ ln Z = −
kB T
∂ ln Z 1 ∂F µ
=⇒ = − =−
∂N kB T ∂N kB T
∂ ln Z µ
=⇒ α = =− (24)
∂N kB T
The mean number of particles in a particular state r is then given by
1
n¯r = (25)
exp β(r − µ) + 1
At higher energy level, when r is large enough, n¯r → 0.
For small r , n¯r → 1.
=⇒ 0 ≤ n¯r ≤ 1. (26)
Kumneger T. (ASTU) July 2, 2021 15 / 37
BE Statistics
The summation in eq. (10) is range over all possible values of
nr , (nr = 0, 1, 2, ...), for each r and the total number of particles is
restricted to be constant. Applying the concept of eq. (12) in eq.
(10) yeilds
0 + exp −βr Zr (N − 1) + 2 exp −2βr Zr (N − 2) + ...
n¯r =
Zr (N) + exp −βr Zr (N − 1) + 2 exp −2βr Zr (N − 2) + ...
(27)
But we have

Zr (N − 1) = Zr (N) exp −α, Zr (N − 2) = Zr (N) exp −2α, ... (28)

Zr (N)[0 + exp −(βr + α) + 2 exp −2(βr + α) + ...]

=⇒ n¯r =
Z (N)[1 + exp −(βr + α) + exp −2(βr + α) + ...]
Pr
nr nr exp −nr (βr + α)
= P
nr exp −nr (βr + α)

Kumneger T. (ASTU) July 2, 2021 16 / 37

 − β1 ∂∂ r
P
nr exp −nr (βr + α)
= P
exp −nr (βr + α)
nr
1 ∂ X
= − ln exp −nr (βr + α)
β ∂r n r

1 ∂ 1
= − ln
β ∂r 1 − exp −(α + βr )
1 ∂
= ln 1 − exp −(α + βr )
β ∂r
1 β exp −(α + βr )
=
β 1 − exp −(α + βr )
exp −(α + βr ) 1
= = (29)
1 − exp −(α + βr ) exp (α + βr ) − 1

This is called Bose-Einstein distribution.

Kumneger T. (ASTU) July 2, 2021 17 / 37

Maxwell-Boltzman Statistics
X
Z= exp −β(n1 1 + n2 2 + ...) (30)
R

where the sum ranges over all states R of the system, or over all
possible values of nr .
Since particles are distinguishable, for a total number of N molecules
and for given values of n1 , n2 the number of possible ways of placing
particles in the given states becomes
N!
(31)
n1 !n2 !n3 !...
Each of these arrangements yields distinct state for the system.
Hence partition function of the system becomes
X N!
Z = exp −β(n1 1 + n2 2 + ...)
n !n !n !...
n ,n ,... 1 2 3
1 2

Kumneger T. (ASTU) July 2, 2021 18 / 37

 X N!
= (exp −β1 )n1 (exp −β2 )n2 ...
n1 ,n2
n !n !n
,... 1 2 3
!...
= (exp −β1 + exp −β2 + ...)N
X
= ( exp −βr )N
r
X
=⇒ ln Z = N ln exp −βr (32)
r

Kumneger T. (ASTU) July 2, 2021 19 / 37

 The mean number of particles n¯r in state r is then
1 ∂ ln Z 1 ∂ X
n¯r = − =− N ln exp −βr
β ∂r β ∂r r
N ∂ X
= − ln exp −βr
β ∂r r
N −β exp −βr β exp −βr
= − P =P (33)
β r exp −βr r exp −βr

Photon Statistics
We have the following relation for the partition function
X
Z= exp −β(n1 1 + n2 2 + ...) (34)
R

where the summation is over all possible states R of the whole

system, or equivalently over all values of nr = 0, 1, 2, ... for each r .

Kumneger T. (ASTU) July 2, 2021 20 / 37

 X
=⇒ Z = exp −β(n1 1 exp −β(n2 2 ...
n1 ,n2 ,...
X X
= exp −β(n1 1 exp −β(n2 2 ...
n1 n2
1 1
= ( )( )...
1 − exp −β1 1 − exp −β2
X 1
=
r
1 − exp −βr
X 1 X 1
=⇒ ln Z = ln =− ln (35)
r
1 − exp −βr r
1 − exp −βr
(36)

The mean number of particles in a particular state r with the

corresponding energy r is, by eq. (7),
1 ∂ ln Z 1 ∂ X
n¯r = − = ln (1 − exp −βr )
β ∂r β ∂r r
Kumneger T. (ASTU) July 2, 2021 21 / 37
 1 β exp −βr 1
= = (37)
β 1 − exp −βr exp βr − 1

The dispersion in nr is then

¯r )2 = 1 ∂ 2 ln Z 1 ∂ 1 ∂ ln Z
(∆n =
β 2 ∂2r β ∂r β ∂r
1 ∂ 1 ∂ n¯r
= (−n¯r ) = −
β ∂r β ∂r
1 −β exp βr exp βr
= − 2
=
β (exp βr − 1) (exp βr − 1)2
(exp βr − 1) + 1 exp βr − 1 1
= 2
= 2
+
(exp βr − 1) (exp βr − 1) (exp βr − 1)2
1 1
= +( )2
exp βr − 1 exp βr − 1
= n¯r + n¯r 2 = n¯r (1 + n¯r ) (38)

Kumneger T. (ASTU) July 2, 2021 22 / 37

Bose-EinsteinStatistics
The summation in the partition function
X
Z= exp −β(n1 1 + n2 2 + ...) (39)
R

should be over all possible values of nr for each r provided that the
values of nr satisfy the restrictive condition
X
nr = N (40)
r

This restriction implies that Z = Z (N). Hence for a system of N 0

particles the partition function is Z = Z (N 0 ). This function is a
rapidly increasing function of N 0 , and hence multiplying it by a rapidly
decreasing function exp −αN 0 yields a function Z (N 0 ) exp −αN 0
which usually has a very sharp maximum.
The proper choice of the positive parameter α will made the sharp
maximum to occur at N 0 = N. Thus one can write
Kumneger T. (ASTU) July 2, 2021 23 / 37
 X
Z (N 0 ) exp −αN 0 = Z (N) exp −αN∆∗ N 0 (41)
N0

where Z (N) exp −αN is the maximum value of the summand while
∆∗ N 0  N is the width of the maximum.
Let X
Ż = Z (N 0 ) exp −αN 0 (42)
N0
Thus one can write

Ż = Z (N) exp −αN∆∗ N 0 (43)

ln Ż = ln Z (N) − αN + ln ∆∗ N 0 (44)
But ln ∆∗ N 0 is negligible as compared to other terms which are of
order N.

=⇒ ln Ż = ln Z (N) − αN
=⇒ ln Z (N) = αN + ln Ż (45)

Kumneger T. (ASTU) July 2, 2021 24 / 37

The sum in the grand partition function Ż extends over all possible
values with out any restriction, hence the partition function can be
rewritten as

Kumneger T. (ASTU) July 2, 2021 25 / 37

Suppose the parameter α is chosen properly so that the function
Z (N 0 ) exp −αN 0 has its maximum at N 0 = N.

ln(Z (N 0 ) exp −αN 0 ) = ln Z (N 0 ) − αN 0

∂ ∂
=⇒ 0
ln(Z (N 0 ) exp −αN 0 )|N 0 =N = [ln Z (N 0 ) − αN 0 ]|N 0 =N = 0
∂N ∂N 0
∂ ln Z (N 0 )
=⇒ |N 0 =N − α = 0
∂N 0
∂ ln Z (N)
=⇒ −α = 0 (46)
∂N
Thus we obtain

(47)

Kumneger T. (ASTU) July 2, 2021 26 / 37

Hence eq. (50) becomes

(48)

Similarly from eq. (47) one obtains

(49)

∂ ln Z (N)
=⇒ =0 (50)
∂α
Substituting eq. (49) in eq. (54) yields
X exp −(βr + α)
N− = 0
r
1 − exp −(βr + α)
X 1
=⇒ = N (51)
r
exp (βr + α) − 1

Kumneger T. (ASTU) July 2, 2021 27 / 37

The mean number of particles in state r is then
1 ∂ ln Z 1 ∂ X
n¯r = − =− [αN − ln(1 − exp −(βr + α)]
β ∂r β ∂r r
1 β exp −(βr + α) ∂ ln Z ∂α
= − [− + ]
β 1 − exp −(βr + α) ∂α ∂r
exp −(βr + α) 1
= = (52)
1 − exp −(βr + α) exp (βr + α) − 1

And hence it proves that

X
n¯r = N. (53)
r

Kumneger T. (ASTU) July 2, 2021 28 / 37

Fermi-Dirac Statistics

The summation in the partition function

Then the grand partition function becomes

(54)

Kumneger T. (ASTU) July 2, 2021 29 / 37

We have
X
= αN + ln(1 + exp −(α + βr ))
r
∂ ln Z X exp −(α + βr )
=⇒ = N− =0
∂α r
1 + exp −(α + βr )
X exp −(α + βr )
=⇒ N = (55)
r
1 + exp −(α + βr )

The mean number of particles in state r is

1 ∂ ln Z 1 −β exp −(α + βr ) ∂ ln Z ∂α

n¯r = − =− [ + ]
β ∂r β 1 + exp −(α + βr ) ∂α ∂r
1
=⇒ n¯r = (56)
exp (α + βr ) + 1

Kumneger T. (ASTU) July 2, 2021 30 / 37

Quantum statistics in the classical limit
The Quantum statistics description of a system of ideal gas can be
summuraised by
1
n¯r = (57)
exp (α + βr ) ± 1
where the ± sign in the denuminator represents FD/BE statistics and
α is supposed to be obtained from the relation
X X 1
n¯r = =N (58)
r r
exp (α + βr ) ± 1
We want to evaluate α under some limiting cases
Consider a system of gas where the concentration is sufficiently low.
Thus, eq. (3.64) can only be satisfied when each term in the sum over
all states is sufficiently low, n¯r  1 or exp (α + βr )  1 for all r .
Similarly for a system consisting of fixed number N of particles, if the
temperature is sufficiently large so that βr  α, the parameter α
must be large enough to prevent the sum from exceeding N, then one
can have exp (α + βr )  1 or n¯r  1.
Kumneger T. (ASTU) July 2, 2021 31 / 37
Under these conditions eq. (3.63) becomes
n¯r ∼
= exp −(α + βr ) (59)
The parameter α can be determined from
X X
exp −(α + βr ) = exp −α exp −βr = N
r r
X
=⇒ exp −α = N( exp −βr )−1
r
exp −βr
=⇒ n¯r ∼
= exp −α exp −βr = NP (60)
r exp −βr
Therefore, in the classical limit of sufficiently low concentration or
sufficiently high temperature the Quantum statistics, FD and BE
statistics reduces to MB statistics.
In the Quantum statistics the partition function satisfy the relation
X
ln Z = αN ± ln(1 ± exp −α − βr )
r
∼
X
= αN ± ± exp −α − βr
r
Kumneger T. (ASTU)
X July 2, 2021 32 / 37
 X
exp −α = N( exp −βr )−1
r
X
=⇒ −α = ln N − ln exp −βr
r
X
=⇒ α = − ln N + ln exp −βr
r
X
=⇒ ln Z = −N ln N + N ln exp −βr + N
r
X
= N ln exp −βr − ln N!; ln N! = N ln
r
X
= ln ZMB − ln N!; ZMB = ( exp −βr )
r
ZMB
=⇒ Z =
N!

Kumneger T. (ASTU) July 2, 2021 33 / 37

Evaluation of partition function

Consider a system of monatomic ideal gas. Suppose the system is in

the classical limit, i.e., it has sufficiently low concentration or it is at
sufficiently high temperature. The partition function of the system is
then given by
X
ln Z = N(− ln N + ln exp −βr + 1)
r
= N(− ln N + ln ξ + 1) (69)
P
where ξ = ln r exp −βr and the sum is over all possible states of a
single particle.

Kumneger T. (ASTU) July 2, 2021 34 / 37

To evaluate this sum we need to know the energy of a single particle
corresponding to the possible states.
A system of single noninteracting particle of mass m, position vector
r, and momentum p confined in a container of volume V can be
described by a wave function ψ(r, t) of the plane wave form

ψ(r, t) = A exp i(k· r − ωt) = ψ(r) exp −iωt (70)

which propagates in the direction of the wave vector k.

The energy of this particle is then given by

p2 ~2 k2
= = (71)
2m 2m
where momentum p of the particle given by de Broglie relation

p = ~k
~2 2
=⇒  = (k + ky2 + kz2 ) (72)
2m x
Kumneger T. (ASTU) July 2, 2021 35 / 37
The wave function in eq. (3.76) is assumed to satisfy the periodic
boundary condition, provided that dimensions of the container are
large enough compared to the de Broglie wavelength of the particle.
ψ(x + Lx , y , z) = ψ(x, y , z)
ψ(x, y + Ly , z) = ψ(x, y , z)
ψ(x, y , z + Lz ) = ψ(x, y , z) (73)
where Lx , Ly , Lz are the dimensions of the container.
ψ(r) = exp i(k· r = exp i(kx x + ky y + kz z) (74)
Hence to satisfy the periodic boundary condition, eq. (3.79), one
must have
kx (x + Lx ) = kx x + 2πnx
2πnx 2πny 2πnz
=⇒ kx = , ky = , kz = (75)
Lx Ly Lz
where nx , ny , nz are integers.
Thus eq. (3.78) can be rewritten as
Kumneger T. (ASTU) July 2, 2021 36 / 37
 ~2 n2 ny2 n2
 = (2π)2 [ x2 + 2 + z2 ]
2m Lx Ly Lz
X X −β~2 2
=⇒ ξ = exp −βr = exp [ (kx + ky2 + kz2 )]
r
2m
kx ,ky ,kz
X −β~2 X −β~2 2 X −β~2 2
= ( exp ( kx2 ))( exp ( ky ))( exp ( k ))
2m 2m 2m z
kx ky kz
(76)

Successive terms in eq. (3.82) correspond to a very small change like

∆kx = 2π Lx , hence a small range between kx andkx + dkx contains
Lx
∆nx = 2π dkx terms which have nearly the same magnitude.
Thus the sums in eq. (3.82) can be replaced by integrals.
∞ Z ∞
X −β~2 2 ∼ −β~2 2 Lx
=⇒ exp kx = exp k dkx
2m −∞ 2m x 2π
kx =−∞
Lx ∞ −β~2 2
Z
Kumneger T. (ASTU) July 2, 2021 37 / 37