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5.62 Physical Chemistry II

Spring 2008

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 5.62 Lecture #8: Boltzmann, Fermi-Dirac, and
Bose–Einstein Statistics
THE DIRECT APPROACH TO THE PROBLEM OF INDISTINGUISHABILITY

We could have approached the problem of indistinguishability by treating

particles as indistinguishable fermions or bosons at the outset. QM tells us

1. All particles are indistinguishable

2. All particles are either fermions or bosons. The odd/even symmetry of a

particle's wavefunction with respect to exchange is determined by whether
the particle is a fermion or boson.

FERMION — a particle which obeys Fermi-Dirac statistics;

many-particle wavefunction is antisymmetric (changes sign) with respect to
exchange of any pair of identical particles: P12ψ = –ψ

1/2 integer spin

e–, proton, 3He

Single state occupation number: ni = 0 or ni = 1, no other possibilities!

BOSON — a particle which obeys Bose-Einstein statistics;

many-particle wavefunction is symmetric (does not change sign) with respect to
exchange of any pair of identical particles

4He, H2, photons integer spin

ni = any number, without restriction

What kind of particle is 6Li, 7Li, D, D+?

We are going to figure out how to write
t
ω ( n i ,g i )
Ω ({n i }) = ∏
i=1 N!
where we are considering level i with energy ει and degeneracy gi. Previously we had
considered Ω({ni}) for non-degenerate states rather than gi-fold degenerate εi levels.
5.62 Spring 2008 Lecture 8, Page 2

Let us play with some simple examples before generalizing results for each type of
system.

3 degenerate states: A, B, C (states could be x, y, z directions for particle in

cube)
2 particles: 1, 2
A B C
1 2
1 2
2 1
2 1
1 2
2 1
1,2
1,2
1,2

If particles are distinguishable and there are no restrictions on occupancy, total #

of distinguishable arrangements is 32. Note that this is different from the
degeneracy of a particular set of occupation numbers for non-degenerate states,
N!
.

∏ n i
!
For each degenerate level occupied by particles, we have a factor:
degeneracy of atomic
particles
state

ω(n,g) = gn
to correct for particle indistinguishability. We divide by N!

t t

∏ ω(n i ,g i ) ∏g n i

i
Ω B ({n i }) = i=1
= i=1
N! N!

32 gn
For our case 2! = 4.5 which is not an integer so can be only an approximation
n!
to the correct total # of ways.

Now go to F–D system

occupation # is 0 or 1, indistinguishable particles, therefore g ≥ n.

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5.62 Spring 2008 Lecture 8, Page 3

ωFD(n,g) = put excitation first in any of g states, 2nd in any of g–1, then
⎡ g! ⎤ 1 divide by n! for indistinguishability of particles. Finally, divide
⎢ (g − n)! ⎥ n! by (g – n)! for indistinguishability of “holes”.
⎣ ⎦

t t
⎡
gi ! ⎤

∏ω FD (g i , n i ) ∏ ⎢⎣ (g ⎥
i − n i )!n! ⎦
ΩFD ({n i }) = i =1
=
i =1

N! N!

A B C

X
X g = 3, n = 2
X X 3!
X X ωFD = 2!1! = 3

Now for B–E

what is the combinatorial factor?

n particles
g distinguishable states → g–1 indistinguishable partitions

arrange n indist. particles and g–1 indist. partitions in all possible orders

(n +g–1)!
ωBE(N,g) = n!(g–1)!

t t
(n i + g − 1)!
∏ ω BE (n i ,g i ) ∏ n i !(g − 1)!
ΩBE ({n i }) = i=1
= i=1

N! N!

(2+3–1)! 4!
for our case 2!(3–1)! = 2!2! = 6

A B C
XX
XX
XX
X X
X X
X X

for our example ωFD < ωB < ωBE

3 4.5 6

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5.62 Spring 2008 Lecture 8, Page 4

always true — compare ω(ni,gi) factors in Ω({ni}) term by term. This means that n i for a
degenerate level is always largest for BE and smallest for FD. Why?

Before starting the general, rigorously derivable result for n i for BE, corrected
Boltzmann, and FD, we need to derive a relationship between µ and q for corrected
Boltzmann statistics.

A = –kT ln Q = –kT ln (qN/N!) = –NkT ln q + kT ln N!

for large N, ln N! = N ln N – N. This is Stirling’s approximation. Very, very useful.

A = −NkTln q + NkTln N − NkT

⎛ ∂A ⎞
µ≡⎜ = −kTln q + kTln N + NkT (1 N ) − kT
⎝ ∂N ⎟⎠ T,V
= −kTln q + kTln N
= −kTln ( q / N )
µ
− = ln(q / N)
kT
e − µ kT = q / N
q = Ne – µ /kT which is a very convenient form.

The probability of finding one particle out of N in level εi is

n i e − εi kT which are the standard definition and

Pi = = statisitical mechanical values of Pi
N q

n i = Ne − εi kT
q

replace q

1
n i = Ne − εi kT
( Ne − µ kT ) =
kT ( εi − µ)
e
This is the corrected Boltzmann result for n i . Notice that when εi < µ, n i > 1 which
violates the assumption upon which the validity of corrected Boltzmann depends. Note
also, that when T → 0, the only occupied levels are those where εi ≤ µ. When εi > µ and
T = 0, n i = 0. Note further, from the derived T dependence of µ

µ = –kT ln (q/N)

lim µ = 0
T→0

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5.62 Spring 2008 Lecture 8, Page 5

Thus, as T → 0, the only occupied level in εi = 0 which has occupancy n ε = 0 = 1.

It is clear that we need to replace corrected Boltzmann statistics by BE or FD as T → 0

and whenever niB becomes comparable to 1.

Assert (derived for Grand Canonical Ensemble, where µ, V, and T are held constant, pp.

431-439 of Hill)
+1 is FD
1
n i = ( εi − µ) kT −1 is BE
e ±1
no 1 is B
This equation obeys the expectation
niFD < niB < niBE
In the limit of T → 0
εi = µ εi < µ εi > µ
ni → ∞
BE < 0 (impossible) 0
ni → 2
FD 1 1 0
niB → 1 ∞ (illegal) 0

First, let's make sure “exact” result for n i is correctly normalized to total number of
particles.

1
N = F ∑ ni = F ∑
i i e ( εi − µ )/kT ± 1

normalization correction factor

so
N
F= normalization factor
1
∑
i e( ε i − µ ) /kT ± 1
N
ni = Fn i =
1
(e ( ± 1)∑
ε i −µ ) kT
(ε j − µ) kT
j e ±1
normalized “exact” result
approximate result being
checked for normalization

Now make the Boltzmann approximation e( j )  1 for all j

ε −µ kT

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5.62 Spring 2008 Lecture 8, Page 6

The factors of ±1 are small and can be neglected, thus:

N
ni = =
e(εi −µ) kT ∑ e ( j )
− ε −µ kT

Ne −εi kT
Ne −εi kT
ni = −ε j kT
=
∑e q
j

CORRECTLY REDUCES TO BOLTZMANN STATISTICS RESULT!

— ε −µ
PLOT the FD and BE distribution functions for ni . ni vs. i
kT
Note that:

* n i cannot be larger than 1 for FD

* n i goes to ∞ when εi = µ for BE

* when n i ≈ 1, we are no longer allowed to use corrected Boltzmann.

(εi–µ)/kT —
For large εi – µ and consequently large e or ni << 1,

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5.62 Spring 2008 Lecture 8, Page 7

niFD = niBE

and when these occupation numbers are equal

—
ni << 1 !

Most atoms and molecules at "ordinary" temperatures are in Boltzmann regime where
—
ni << 1 or q >> N. In this case, there is no difference between FD and BE statistics.
Doesn't matter if the molecule is a fermion or boson. So the Boltzmann statistics we have
developed is valid over a wide range of molecules and conditions.

EXCEPTIONS: 4He and H2 are BOSONS.

Must be treated as such at T close to 0K.
3He is FERMION.
Must be treated as such at T close to 0K.

Notice that the exceptions are lighter atoms and molecules at low T. That's because as
you make particle less massive, the spacings in energy between the particle's states get
—
larger, leading to fewer available states. If fewer states are available, ni goes up, and
eventually the difference between FD and BE statistics becomes discernible.

εi =
h2
8ma 2 (
n 2x + n 2y + n 2z )

ε3 ε7

ε5
ε2
ε ε3
ε

ε1 ε1

LIGHT PARTICLE HEAVY PARTICLE

— —
* ni is larger * ni is smaller
* may have to use
FD or BE statistics

At "normal" temperatures > ~20K, can treat 3He, 4He, H2 with Boltzmann
statistics.

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5.62 Spring 2008 Lecture 8, Page 8

So temperature plays a role here too. We'll talk about this next time.

BUT ELECTRONS always have to be treated as FERMIONS for all "normal"

—
temperatures (<3000K), because their ni ’s  1.

The valence electrons of the Au atoms which make up a gold crystal are
delocalized throughout the crystal. These electrons can be thought of as an
electron gas contained within the crystal. This is called a free electron model
where the energy levels of electrons are particle-in-box energy levels. The
average number of electrons in each electron state is given by the Fermi-Dirac
distribution function

1
ni = ( ε i − µ ) kT
F.D.

e +1

At T = 0

We fill each state with 1 e– in order of increasing energy. The energy at which we run
out of e–s is εi = µ. In Solid State Physics language

µ ≡ εF FERMI ENERGY

which is the maximum energy that an e– can have at T = 0.

AT T > 0K

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5.62 Spring 2008 Lecture 8, Page 9

Electrons move from occupied to unoccupied states as T is increased. Must move to

—
unoccupied states because ni > 1 not allowed. This is the origin of conductivity in
metals. [More on this in second half of course.]

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5.62 Spring 2008 Lecture 8, Page 10

Non-Lecture
Alternative Derivation of ni for B-E and F-D Particles

(this is actually a sum over sets of

Canonical p. f. Q = ∑ Ω ({ni }) e−βni εi occupation numbers, {ni}, and over
{ni }
states, i)
=∑ ∏ ⎡⎣ω ( ni ) e−βni εi ⎤⎦
{ni } i

ω(ni) is the number of distinguishable ways of arranging ni particles in the εi single-

particle energy level.

The form of ω(n) depends on the particle statistics.

If it were possible to evaluate this sum, then we could determine ni from

⎛ ∂lnQ ⎞ −kT ∂Q
ni ≡ −kT ⎜ ⎟ =
⎝ ∂εi ⎠V ,T , N Q ∂εi
kT ⎛ −ni ⎞
=−
Q
∑ ⎜
⎝ kT ⎠
⎟Ω ({ni }) e−βni εi
{ni }

∑ ni Ω ({ni }) e−βni εi
= { i} ≡ ni .
n

Now we will evaluate Q approximately by finding the single set of occupation numbers
{ni} that gives the maximum term in the sum over occupation numbers that defines Q.
The approximation is to set Q equal to the value of this maximum term in the sum. There
remains a sum over states, i.

This approximation can only be valid if the maximum term in the sum is vastly larger
than any term corresponding to a different set of occupation numbers.

This is a common and useful approximation in statistical mechanics.

Find the maximum term in the sum, call it QM and assume QM ≈ Q

A = −kT lnQM

−βA = lnQM = ∑′ {ln ⎡⎣ω (n )⎤⎦ − βε n },

i i i
i
we have kept only the single set of occupation numbers that gives the maximum
term in the sum that defines Q.

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5.62 Spring 2008 Lecture 8, Page 11

The prime on the ∑

′ implies the constraint
i

N= ∑ ni
i

Use Lagrange multipliers to impose this constraint so that the constrained sum can be
replaced by an unconstrained sum,

⎛ ⎞
−βA = lnQM = ∑ {ln ⎡⎣ω (ni )⎤⎦ − βεi ni } + λ ⎜∑ ni − N ⎟
⎝ i ⎠
i  
unconstrained =0
sum
⎛ ∂A ⎞ value to be
−β ⎜ ⎟ = −λ
chosen
⎝ ∂N ⎠V ,T

⎛ ∂A ⎞ µ
but ⎜ ⎟ = µ Thus λ = βµ = .
⎝ ∂N ⎠V ,T kT
Insert the derived specific value for λ,

−βA = ∑ {ln ⎡⎣ω (n )⎤⎦ − β (ε

i i − µ) ni } − Nβµ ,
i

rearrange

{
Nβµ − βA = ∑ ln ⎡⎣ω(n i ) − β ( ε i − µ ) n i ⎤⎦
i
}
and use the thermodynamic identity

G = Nµ = A + pV

β ( Nµ − A) = βpV = lnQ + Nβµ = ∑ {ln ⎡⎣ω (n )⎤⎦ − β (ε

i i − µ) ni }

Now we choose particular forms for ω(ni) and insert them into the above equation for
βpV. Note that we have not yet addressed the approximation of Q by QM.

A. Corrected Boltzmann

gi is the degeneracy of the single-particle εi energy level, and ni is the number of particles
in the assembly in the εi energy level.

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5.62 Spring 2008 Lecture 8, Page 12

gn
ωB =
n!

ln ω B = n ln g − n ln n + n

βpV = ∑ {ni ln gi − ni ln ni + ni − β (
εi − µ) ni }
i

to obtain the maximum term in this sum, we take a derivative wrt each ni. For each ni

⎛ ∂ ⎞
⎜ (βpV ) ⎟ = {ln ( gi ) − ln ni −1+1 − β (εi − µ)} = 0, thus
⎝ ∂ni ⎠V ,T , N

niB = gi e−βεi eβµ

∂2 (βpV )
(note = –(1/ni) < 0 which assures that the extremum is a maximum and not a

∂ni2
minimum).

Since q = Ne–βµ, we get the standard corrected Boltzmann result when we replace eβµ by

N/q,

niB gi e−βεi
= ,
N q

moreover, when we replace the original sum over sets of {ni} that defines βpV by the
specific set {ni} that gives the maximum term in the sum, we get

⎧ g ⎫
βpV = ∑ ⎨ni ln i + ni − β ( ε i − µ) ni ⎬
i ⎩ ni ⎭
gi
= eβ ( εi − µ)
ni
ni ln ( gi ni ) = niβ ( ε i − µ) .
Thus

βpV = ∑ ni = N
i
which is the ideal gas law.

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5.62 Spring 2008 Lecture 8, Page 13

B. Fermi-Dirac

g!
ω FD ( g, n ) =
n!( g − n )!

βpV = ∑ {ln ⎡⎣ω (n ) − β (ε

i i − µ) ni ⎤⎦}
i

βpV = ∑ {gi ln gi − gi − ni ln ni + ni − (gi − ni )ln(gi − ni ) + ( gi − ni ) − β (εi − µ) ni }

i

−gi + ni + ( gi − ni ) = 0

∂
The extremum term in the sum over sets of {ni} is obtained from taking and setting
dn i
result = 0.

⎛ ∂ (βpV ) ⎞
⎜ ⎟ = {ln ( gi − ni ) − ln ni − β (εi − µ)} = 0 for all i.
⎝ ∂ni ⎠V ,T , N

gi − ni
ln = β (εi − µ)
ni
gi − ni gi
= −1 = eβ (εi − µ)
ni ni

gi
= eβ (εi − µ) +1

ni

Thus
1
niFD = gi β (εi − µ)
,
e +1

replacing the sum in Q by its maximum term, QM

βpV = ∑ {gi ln gi − ni ln ni − gi ln ( gi − ni ) + ni ln ( gi − ni ) − β (εi − µ) ni }

i

⎧ g − ni ⎫
= ∑ ⎨ gi ln gi − gi ln ( gi − ni ) + ni ln i − β (εi − µ) ni ⎬
i ⎩ ni ⎭

but, for the term in the sum over sets of {ni} that has the maximum value
g −n
ln i i = β (εi − µ)
ni
and the last two terms in { } cancel. We obtain

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5.62 Spring 2008 Lecture 8, Page 14

gi − ni ⎛ n⎞
βpV = −∑ gi ln = −∑ gi ln ⎜1 − i ⎟
i gi i ⎝ gi ⎠
⎛ 1 ⎞
= −∑ gi ln ⎜1 − β (ε − µ) ⎟
i
⎝ e i +1 ⎠
⎛ eβ (εi − µ) +1 −1 ⎞
= −∑ gi ln ⎜ β (εi − µ) ⎟
i
⎝ e +1 ⎠
⎛ eβ (εi − µ) +1 ⎞
= ∑ gi ln ⎜ β (ε − µ) ⎟ = ∑ gi ln [1 + eβ ( µ−εi ) ]
i
⎝ e i ⎠ i
which is not the ideal gas law. However, at high ε,

ln [1+ eβ ( µ−ε) ] → eβ ( µ−ε)

because, for x  1
1
ln (1+ x ) = ln1+ x +…
1+ 0
≈x
β ( µ−ε)
and, for x = e when ε  µ
eβ( µ−ε)  1 .
Thus
βpV = ∑ gi eβ ( µ−εi ) = ∑ gi eβµe−βεi
i i

e −βµ
= (q N )
Ngi e−βεi q
βpV = ∑ =N =N
i q q
which is the ideal gas law!

C. Bose-Einstein

ω BE ( ni ) =
( gi + ni −1)!
ni !( gi −1)!
Thus
ln ω BE ( ni ) = ∑ {( gi + ni −1) ln ( gi + ni −1) − ( gi + ni −1) − ni ln ni + ni − ( gi −1) ln ( gi −1) + gi −1}
i

=∑ {( gi + ni −1) ln ( gi + ni −1) − ni lni ni − ( gi −1) ln ( gi −1)}

βpV = lnQ + Nβµ = ∑ {ln ω BE (ni ) − β (εi − µ) ni } .

{ni }

The set of occupation numbers that gives the largest contribution to βpV is obtained from

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5.62 Spring 2008 Lecture 8, Page 15

⎛ ∂(βpV ) ⎞ g + n −1 n
⎜ ⎟ = 0 = ln ( gi + ni −1) + i i − ln ni − i − β (εi − µ)
⎝ ∂ni ⎠V ,T , N gi + ni −1 ni
thus
gi + ni −1
ln = β (εi − µ)
ni
gi −1
+1 = eβ (εi − µ)
ni
1
niBE = ( gi −1) β (εi − µ)
e −1
Since gi  1 , we obtain the usual result
g
niBE = i
β (εi − µ)
e −1
and, when εi = µ, the denominator vanishes and ni can be very large. This is the Bose-
BE

Einstein condensation. If εi < µ, the nonsense result of a negative occupation number is

obtained.

D. Accuracy of the Maximum Term Approximation

(see Hill, Appendix II, pp. 478-480).

′
Q≡∑ ∏ ⎡⎣ω ( ni ) e−βεi ni ⎤⎦ ≡ ∑ t ({ni })
{ni } ni {ni }

where t({ni}) is a typical term in the sum and ∑′ imposes the implicit constraint

∑n i =N.
i
Expand the value of the typical term t({ni}) in the sum as a power series in deviations
from the special set of occupation numbers {n̂i } that give the maximum value of

∏ ⎡⎣ω ( ni ) e−βεi ni ⎤⎦ ,
ni

t M = t ({n̂i }) .
∂ ⎡
⎣ω ( ni ) e i i ⎤⎦ for all ni to find the
−βε n
We have already used the requirement that 0 =
∂ni
value of tM and the set {nˆi } .

Thus
2 ∂ ln ω ( n
1
2
ˆi )
∑
ln t ({ni }) = ln t ({nˆi }) +
2 i
δn i
∂ni2
.

The first nonzero term in the expansion involves the second derivatives because all of the
first derivatives were required to be zero (condition for the maximum term)

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5.62 Spring 2008 Lecture 8, Page 16

∂ln t ({n̂i })
=0
∂ni
and the e−βεi ni term is “used up” in the extremum condition. Thus the true value of Q is
given in terms of the value of the maximum term in the sum over {ni} as
⎛
⎡
1
⎤
⎞

2 ∂ ln ω ( n̂i )
2
⎜ ′
Q = QM ⎜1+ ∑
exp ⎢ ∑
δni ⎟
⎥ +…⎟
⎝ {δni } ⎣
2
i ∂ni2
⎦
⎠

but, since we showed that the set {n̂i } maximizes t({ni}), all of the second derivatives
must be negative. This means the factor multiplying QM is (1 + e–x) where x > 0 hence
Q ≈ QM. To make this argument stronger we need to compute these second derivatives
and also realize that the exp[ ] contains many additive negative terms, thus e–x → 0.

The derivation of the second derivatives listed below is left as an exercise for you:

∂2 ln ω ( n̂i )
Statistics
∂ni2
Boltzmann -1/ni
gi
Fermi-Dirac −
ni ( gi − ni )
gi −1
Bose-Einstein −
ni ( gi + ni −1)

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