Quantum 

Statistics (Chapter 10 McQuarrie)
• half integral spin – Fermi‐Dirac                              electron, proton
• integral spin – Bose‐Einstein                                   deuteron, photon

Composite particles
odd # of fermions – acts as fermion
even # of fermions – acts as a boson

( )
±1
Ξ (V , T , λ ) = ∏ 1 ± λ e − βε k
, λ =e βμ
k

λ e − βε λ e − βε
N =∑
k k

− βε
; nk =
(1) k 1 ± λ e 1 ± λ e − βε
k k

upper – FD
λε k e − βε
(2) E = ∑
k
lower ‐ BE
− βε
k 1 ± λe
k

(3) pV = ± kT ∑ n (1 ± λ e βε )
k
− k

solve (1) for λ, and substitute into (2) and (3)
 In general, can’t solve analytically for λ

N (high T and
λ=
If λ small,                           classical statistics
q low densities)

Even if particles are non‐interacting, quantum effects cause
deviations from pV = NkT

Weakly degenerate ideal Fermi‐Dirac gas

λ e − βε
N =∑ (
pV = kT ∑ n 1 + λ e − βε k )
k

− βε
k 1 + λe
k
k

h2
εk =
8mV 2/3 ( n 2
x + n 2
y + n z ),
2
nx , n y , nz = 1, 2,...
 Counting states for translational problem

h2
2 ( x
ε= n 2 + n y2 + mz2 )
8ma

8ma 2ε
degeneracy = # ways                  can be written as 
2 nx2 + n y2 + nz2
h

Take sphere of radius  8ma 2
ε
=R
h2

8ma 2ε
n +n +n =
2
x
2
y
2
z = R 2

h2

for large R treat ε, R as continuous

# states with energy ≤ ε

3/ 2
1 4π R 3
π ⎛ 8ma ε ⎞ 2
Φ= = ⎜ ⎟
8 3 6 ⎝ h2 ⎠
 # between ε+εΔε, such that Δε/ε << 1

ω = Φ ( ε + Δε ) − Φ ( ε )
3/ 2
π ⎛ 8ma 2 ⎞ ⎡( ε + Δε )3/ 2 − ε 3/ 2 ⎤
⎜ ⎟ ⎣ ⎦
6 ⎝ h2 ⎠
3/ 2
π ⎛ 8ma 2 ⎞
= ⎜ ⎟ ε Δε + ...
4 ⎝ h2 ⎠

3kT
ε=
If                  , T = 300° K, m = 10‐22 g, a = 10 cm, and Δε = 0.01ε
2
ω ∼ 1028

for N particle system, the degeneracy is much larger: ω ~ 10
1023

3/2
⎛ 2m ⎞
ω = 2π ⎜ 2 ⎟ V ε Δε
⎝h ⎠
3/2
⎛ 2m ⎞
∑ → 2π ⎜⎝ h2 ⎟⎠ V ∫ ε dε
 ∞ λ εe dε
3/2 − βε
⎛ 2m ⎞
N = 2π ⎜ 2 ⎟ V ∫ using the density
⎝h ⎠ 0 1 + λ e − βε
3/2
of states from 
⎛ 2m ⎞
pV = 2π kT ⎜ 2 ⎟ V ∫ ε 1/2 n (1 + λ e − βε )d ε
∞
pages 10‐11
⎝h ⎠ 0

Expand in powers of λ and integrate

1 ∞ ( −1) λ
+1
N 1 ⎡λ λ2 ⎤
ρ= = 3∑ = − +
Λ 3 ⎢⎣ 1 23/2 ⎥
A ...
V Λ =1 3/2
⎦

( −1) λ
+1
p 1 1 ⎡ λ2 ⎤
B = 3∑ = λ − +
Λ 3 ⎢⎣ ⎥
...
kT Λ =1 5/2
25/2 ⎦

λ = λ (ρ)
solve                      using  A

B p
plug into              to get            as a function of  ρ
kT
 Write  λ = a0 + a1ρ + a2 ρ 2 + ...

⎡ ( ) ( ) ⎤
2 3
a + a ρ + a ρ 2
+ ... a + a ρ + a ρ 2
+ ...
ρ = 3 ( a0 + a1ρ + a2 ρ + ...) +
1 ⎢ ⎥
+
2 0 1 2 0 1 2

Λ ⎢ 2 3/2
33/2
⎥
⎣ ⎦

1 ⎡⎛ a02 ⎞ ⎛ 2a0 a1 3a0 a1 ⎞ ⎤

ρ = 3 ⎢⎜ a0 + 3/2 + ... ⎟ + ⎜ a1 + 3/2 + 3/2 + ... ⎟ ρ + ( a2 + ...) ρ 2 + ...⎥
Λ ⎣⎝ 2 ⎠ ⎝ 2 3 ⎠ ⎦

⇒ a0 = 0
λ = a1 ρ + a2 ρ 2 + a3 ρ 3 + …
⇒ a1 = Λ 3

Λ6 2
a12 = Λ ρ + 3/ 2 ρ + …
3

⇒ a2 − 3/2 = 0 2
2

( ρΛ3 )
2
⎛1 1 ⎞
+ ⎜ − 3/ 2 ⎟ ( ρΛ 3 ) + ...
3
⇒ λ = ρΛ 3 +
23/ 2 ⎝4 3 ⎠
 p 1 ⎡λ λ2 λ3 ⎤
= 3 ⎢ − 5/ 2 + 5 / 2 − ...⎥
kT Λ ⎣ 1 2 3 ⎦

1 ⎢ ( a1ρ + a2 ρ + ...) ( a1ρ + a2 ρ + ...)

⎡ 2 2 2
⎤
p
= 3 − + ...⎥
kT Λ ⎢ 2 5/2
⎥
⎣ ⎦
( a ρ + ...)
3

+ 1
35/2
p 1 ⎡Λ ρ ⎛
3
a ⎞ 2 6
2
⎤
= 3⎢ + ⎜ a2 − ⎟ ρ Λ
1
+ ...⎥ a12 a12 ⎛ 1 1 ⎞ p 1 ⎡ 3 Λ6 ⎤
kT Λ ⎣ 1 ⎝ 2 ⎠ 5/2
⎦ − 5/2 = ⎜ 3/2 − 5/2 ⎟ Λ 6 = ⎢ Λ ρ + 3/2 ⎥
⎝2 2 ⎠ kT Λ 3 ⎣
3/2
2 2 2 ⎦
1 ⎛1 1⎞ 6 1
= ⎜ − ⎟Λ = Λ6
p Λ3 2 ⎛ 1 2 ⎞ 6 3 2 ⎝2 4⎠
= ρ + 5/2 ρ + ⎜ − 5/2 ⎟ Λ ρ ... 4 2
kT 2 ⎝8 3 ⎠

= ρ + B2 ρ 2 + B3 ρ 3 + ...
Virial coefficients
2nd virial 3rd virial reflect deviations away
coeff coeff from ideality 

B2 is  +, thus increases pressure beyond that for an ideal classical gas
Λ = thermal de Broglie wavelength

quantum effects < as de Broglie λ <

Λ 3
Actually it is            that is a measure of quantum effects
V

( −1) λ
+1
3 VkT
E=
2 Λ3
∑
=1
5/2

3 ⎛ Λ3 ⎞
= NkT ⎜1 + 5/2 ρ + ... ⎟
2 ⎝ 2 ⎠

get expansion for μ from λ = eμ/kT

and S from G = μN=E – TS + pV

Of course, above approach only valid if quantum corrections are small
 Now consider the strongly degenerate Fermi‐Dirac gas

A model for the electrons in a metal

λ e − βε k
1
nK = =
1 + λ e − βε 1 + e β (ε
k k −μ )
since ε is essentially continuous
1
f (ε ) = β (ε − μ )
= prob. a state is occupied
1+ e 1

T =0 ε < μ0
states with               are occupied f(ε)
μ0 = μ ε > μ0
states with               are unoccupied
0 μ0 ε

from 1‐35
3/2
⎛ 2m ⎞
ω ( ε ) d ε = 4π ⎜ 2 ⎟
V ε dε (includes factor of 2 for spin)
⎝h ⎠
3/2
⎛ 2m ⎞ μ0
N = 4π ⎜ 2 ⎟ V ∫ ε d ε = # valence e −
⎝h ⎠ 0

2/3 2/3
8π ⎛ 2m ⎞
3/2
h2 ⎛ 3 ⎞ ⎛ N ⎞
= ⎜ 2 ⎟ V μ0
3/2 μ0 = ⎜ ⎟ ⎜ ⎟
3 ⎝h ⎠ 2m ⎝ 8π ⎠ ⎝ V ⎠
 at T = 0, the levels are double occupied
ε up to μ0

μ0 a finite T, the boundary is smeared out
i.e., some electrons are excited leaving “holes”

Even at room T
f (ε ) = 1 ε < μ0
f (ε ) = 0 ε > μ0
is a good approximation

μ0 / k = TF = Fermi T , typically a few thousand degrees
 3/2
⎛ 2m ⎞ μ0 3
E0 = 4π ⎜ 2 ⎟ V ∫ ε 3/2 d ε = N μ0 T =0 K
⎝h ⎠ 0 5

ZPE of FD gas

only a very small fraction of the e‐ are excited, so
contribution to heat capacity is ~ 0 
3
equipartition theorem would lead us to expect          for each electron
k
2

( )
3/2
⎛ 2m ⎞ μ0
p = 4π kT ⎜ 2 ⎟ ∫ ε n 1 + e β ( μ −ε ) d ε
0

⎝h ⎠ 0

ignore the “1”
⎛ 2 m ⎞ μ0
p0 = 4π ⎜ 2 ⎟ ∫ ε 1/ 2 ( μ0 − ε ) d ε
⎝h ⎠ 0
2 μ0
= N
5 V zero‐point pressure on the order of (106 atm)

S0 = 0 only one way to occupy levels at T = 0 K
 It can be show that

μ π2 2 1
= 1 − η + ..., η=
μ0 12 βμ0

μ ~ μ0 for temperatures for which a 
metal is solid

⎛μ ⎞
5 Note at T =0 K, μ = μo
⎡ 5 2 2 ⎤
E = E0 ⎜ ⎟ 1
⎢⎣ 8+ π η + ...⎥⎦
⎝ μ0 ⎠

π 2 NkT π 2 ⎛T ⎞
Cv = = Nk ⎜ ⎟
2 μ0 / k 2 ⎜T ⎟ ~ 10‐4 T cal/deg‐mol
⎝ f ⎠
weakly degenerate ideal Bose‐Einstein gas

λ e − βε
N =∑
k

1 − λ e − βε k
0 ≤ λ < e βε 0 , otherwise get 0 in denominator
(
pV = − kT ∏ n 1 − λ e − βε k )
k

λ e − βε λ e− βε
+∑
0 k

N= − βε − βε
1 − λe k ≠0 1 − λ e
0 k

redefine ε0 to be zero

λ λε 1/ 2 e − βε
3/ 2
N ⎛ 2m ⎞ ∞
ρ= =
V V (1 − λ )
+ 2π ⎜ 2 ⎟
⎝h ⎠
∫ε >0 1 − λ e− βε dε , (0 ≤ λ < 1)
 3/ 2
⎛ 2m ⎞
ε n (1 − λ e− βε ) d ε
p 1 ∞
=− n (1 − λ ) − 2π ⎜ 2 ⎟ ∫ε ε
kT V ⎝h ⎠ > 0

if λ << 1 can ignore 1/V terms

1
3 3/2 ( )
ρ= g λ
Λ ∞
λ
gn = ∑ n
p 1
= 3 g5/2 ( λ ) =1
kT Λ

effective interaction between
p Λ3
= 1 − 5/ 2 ρ + ... ideal bosons is attractive
ρ kT 2

3 ⎛ Λ3 ⎞
E = NkT ⎜1 − 5 / 2 ρ + ... ⎟
2 ⎝ 2 ⎠
 Strongly degenerate ideal Bose‐Einstein gas

T < T0 condensation into ground state

Behavior seen for CV
He‐4 Nk
T0
T
From Wikepedia: A Bose–Einstein condensate (BEC) is a state of matter of a dilute gas of
weakly interacting bosons confined in an external potential and cooled to T near to absolute
zero. Under such conditions, a large fraction of the bosons occupy the lowest quantum state of
the external potential, and quantum effects become apparent on a macroscopic scale. This
state of matter was predicted by Bose and Einstein in 1924–25.

The first such condensate was produced by Cornell and Wieman in 1995 at the Univ. of
Colorado NIST-JILA lab, using a gas of Rb atoms cooled to 170 (nK) [2]. Cornell, Wieman,
and Ketterle (MIT) received the 2001 Nobel Prize in Physics.

Note: the fact that Rb atoms act as bosons is due to interplay of electronic and nuclear spins.
 From Wikepedia

A Bose–Einstein condensate (BEC) is a state of matter of a dilute gas of weakly interacting

bosons confined in an external potential and cooled to T near to absolute zero. Under such
conditions, a large fraction of the bosons occupy the lowest quantum state of the external
potential, and quantum effects become apparent on a macroscopic scale. This state of matter
was predicted by Bose and Einstein in 1924–25.

The first such condensate was produced by Cornell and Wieman in 1995 at the Univ. of
Colorado NIST-JILA lab, using a gas of Rb atoms cooled to 170 (nK) [2]. Cornell, Wieman,
and Ketterle (MIT) received the 2001 Nobel Prize in Physics.

Note: the fact that Rb atoms act as bosons is due to interplay of electronic and nuclear spins.
Ideal gas of photons

photons
mass = 0 cavity
ang mom ħ
cavity emits/absorbs photons
N  not fixed

Assume harmonic electromagnetic waves

2π
E ( x, t ) = sin ( x − ct ) = sin(kx − ωt )
λ
ε = hυ = ω ; p = h/λ = k

Consider black-body radiation to be due to standing waves

φ ( x, t ) = 2sin kx cos ωt

Fix at 0,L → k = nπ/L

ε = ħck
k2 = (π/L)2(nx2 + ny2 + nz2)

E = ∑ nk ε k
k

1
Q = ∏ (∑ e − βε k n ) =∏
k n k (1 − e − βε k )
π 2V (kT ) 4
E=
15( c)3

Can be used to derive the Stephan-Boltzmann law

Can also show that the chemical potential = 0 (follows from the fact
that the number of particles is not conserved)
So could have used the Bose-Einstein formulas with λ = 1
Density matrices
All the expressions described above were derived assume there are no
interactions between particles

Q = ∑e
−β Ej

1
∑
−β Ej
M= M j e , M j = q.m. expectation value of operator Mˆ
Q j
Hψ j = E jψ j
−β Ej
e − β Hψ j = e ψj
Q = ∑e = ∑ ψ j e− β H ψ j
−β Ej

j j

Q = Tr (e − β H )

The trace is independent of the basis

 ϕ j = ∑ a jnψ n
n

Q is the same when evaluated over the ϕ j

ˆ −β H )
Tr ( Me
M=
Tr (e − β H )
e− β H
ρ=
Tr (e − β H )
M = Tr ( Mˆ ρ )

− 2
H=
2m
∑∇
l
2
l + U (r1 ,… , rN )

∑
i
pk i rk
u ( p1 ,… , rN ) = e = e.f. of momentum operator
∑ pk i rk
i
ϕ j (r1 ,… rN ) = ∫ Aj ( p1 , …, pN )e dp1 ,… , dpN
∑ pk i rk
i
1 −
Aj ( p1 , …, pN ) =
(2π )3 N ∫ ϕ (r ,… r
j 1 N )e dr1 ,… drN Inverse Fourier transform
 − ∑ pk i rk ' ∑ pk i rk
i i
1
Q = 3 N ∫ ϕ j *(r1 , … rN )ϕ j (r1 ', … rN ')e −β H
e e
h Note we have not
* dp1 ,… , dpN dr1 ,… drN dr1 ',… drN ' included the
symmetry of the
− ∑ pk i rk ' ∑ pk i rk
i i
1
Q = 3N ∫ e −β H
e e p1 ,… drN wavefunction
h which is why the
Now adopt a strategy due to Kirkwood N! is missing.

∑ pk i rk ∑ pk i rk
i i
−β H −β H
=e w( p1 ,… rN , β ) = F ( p1 ,… rN , β )
QM Cl
e e e
1
Q = 3N ∫ e − β H Cl
w( p1 ,… rN , β ) p1 ,… drN
h Contains the
∂F quantum corrections
= − H QM F It is easier to work with
∂β
this diff.eq.
w = ∑ wl l

l There is a w1 term but it

does not contribute to Q
w0 = 1
1 β2 2 β3 1 p4
w2 = − { ∇ U − [(∇U ) + ( p i∇) U ] +
2 2
( p i∇U ) 2 }
2m 2 3 m 2m