td sm lecture notes · 2017-09-24 · chapter1. theprincipleoflargestuncertainty δσ= ∇¯σ ⊥...

Eyal Buks

Introduction toThermodynamics and StatisticalPhysics (114016) - Lecture Notes

September 24, 2017

Technion

Preface

to be written...

Contents

1. The Principle of Largest Uncertainty . . . . . . . . . . . . . . . . . . . . . 11.1 Entropy in Information Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Example - Two States System . . . . . . . . . . . . . . . . . . . . . 11.1.2 Smallest and Largest Entropy . . . . . . . . . . . . . . . . . . . . . . 21.1.3 The composition property . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.4 Alternative Definition of Entropy . . . . . . . . . . . . . . . . . . . 8

1.2 Largest Uncertainty Estimator . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.1 Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.2 The Free Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3 The Principle of Largest Uncertainty in Statistical Mechanics 141.3.1 Microcanonical Distribution . . . . . . . . . . . . . . . . . . . . . . . 141.3.2 Canonical Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3.3 Grandcanonical Distribution . . . . . . . . . . . . . . . . . . . . . . 161.3.4 Temperature and Chemical Potential . . . . . . . . . . . . . . . 17

1.4 Time Evolution of Entropy of an Isolated System . . . . . . . . . . . 181.5 Thermal Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.5.1 Externally Applied Potential Energy . . . . . . . . . . . . . . . . 201.6 Free Entropy and Free Energies . . . . . . . . . . . . . . . . . . . . . . . . . . 211.7 Problems Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.8 Solutions Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2. Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.1 A Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.2 Gibbs Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.3 Fermions and Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.3.1 Fermi-Dirac Distribution . . . . . . . . . . . . . . . . . . . . . . . . . 532.3.2 Bose-Einstein Distribution . . . . . . . . . . . . . . . . . . . . . . . . 542.3.3 Classical Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.4 Ideal Gas in the Classical Limit . . . . . . . . . . . . . . . . . . . . . . . . . . 552.4.1 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572.4.2 Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582.4.3 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592.4.4 Internal Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . 59

2.5 Processes in Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Contents

2.5.1 Isothermal Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642.5.2 Isobaric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642.5.3 Isochoric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652.5.4 Isentropic Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

2.6 Carnot Heat Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.7 Limits Imposed Upon the Efficiency . . . . . . . . . . . . . . . . . . . . . . 682.8 Problems Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 732.9 Solutions Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

3. Bosonic and Fermionic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 993.1 Electromagnetic Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

3.1.1 Electromagnetic Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . 993.1.2 Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.1.3 Cube Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.1.4 Average Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043.1.5 Stefan-Boltzmann Radiation Law . . . . . . . . . . . . . . . . . . . 105

3.2 Phonons in Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.2.1 One Dimensional Example . . . . . . . . . . . . . . . . . . . . . . . . . 1073.2.2 The 3D Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

3.3 Fermi Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1123.3.1 Orbital Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . 1123.3.2 Partition Function of the Gas . . . . . . . . . . . . . . . . . . . . . . 1123.3.3 Energy and Number of Particles . . . . . . . . . . . . . . . . . . . . 1143.3.4 Example: Electrons in Metal . . . . . . . . . . . . . . . . . . . . . . . 114

3.4 Semiconductor Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1163.5 Problems Set 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1173.6 Solutions Set 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4. Classical Limit of Statistical Mechanics . . . . . . . . . . . . . . . . . . . 1294.1 Classical Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

4.1.1 Hamilton-Jacobi Equations . . . . . . . . . . . . . . . . . . . . . . . . 1304.1.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1304.1.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

4.2 Density Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1324.2.1 Equipartition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 1324.2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

4.3 Nyquist Noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1344.4 Thermal Equilibrium From Stochastic Processes . . . . . . . . . . . . 138

4.4.1 Langevin Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1384.4.2 The Smoluchowski-Chapman-Kolmogorov Relation . . . 1394.4.3 The Fokker-Planck Equation . . . . . . . . . . . . . . . . . . . . . . . 1394.4.4 The Potential Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 1414.4.5 Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1424.4.6 Fokker-Planck Equation in One Dimension . . . . . . . . . . 1434.4.7 Ornstein—Uhlenbeck Process in One Dimension . . . . . . . 145

Eyal Buks Thermodynamics and Statistical Physics 6

Contents

4.5 Problems Set 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1474.6 Solutions Set 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

5. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1595.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1595.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175


1. The Principle of Largest Uncertainty

In this chapter we discuss relations between information theory and statisticalmechanics. We show that the canonical and grand canonical distributionscan be obtained from Shannon’s principle of maximum uncertainty [1, 2, 3].Moreover, the time evolution of the entropy of an isolated system and the Htheorem are discussed.

1.1 Entropy in Information Theory

The possible states of a given system are denoted as em, wherem = 1, 2, 3, · · · ,and the probability that state em is occupied is denoted by pm. The normal-ization condition reads

m

pm = 1 . (1.1)

For a given probability distribution pm the entropy is defined as

σ = −

m

pm log pm . (1.2)

Below we show that this quantity characterizes the uncertainty in the knowl-edge of the state of the system.

1.1.1 Example - Two States System

Consider a system which can occupy either state e1 with probability p, orstate e2 with probability 1− p, where 0 ≤ p ≤ 1. The entropy is given by

σ = −p log p− (1− p) log (1− p) . (1.3)

Chapter 1. The Principle of Largest Uncertainty

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.2 0.4 0.6 0.8 1x

−p log p− (1− p) log (1− p)As expected, the entropy vanishes at p = 0 and p = 1, since in both casesthere is no uncertainty in what is the state which is occupied by the system.The largest uncertainty is obtained at p = 0.5, for which σ = log 2 = 0.69.

1.1.2 Smallest and Largest Entropy

Smallest value. The term −p log p in the range 0 ≤ p ≤ 1 is plotted inthe figure below. Note that the value of −p log p in the limit p → 0 can becalculated using L’Hospital’s rule

limp→0

(−p log p) = limp→0

−d log pdp

ddp

1p

= 0 . (1.4)

From this figure, which shows that −p log p ≥ 0 in the range 0 ≤ p ≤ 1, it iseasy to infer that the smallest possible value of the entropy is zero. Moreover,since −p log p = 0 iff p = 0 or p = 1, it is evident that σ = 0 iff the systemoccupies one of the states with probability one and all the other states withprobability zero. In this case there is no uncertainty in what is the state whichis occupied by the system.


1.1. Entropy in Information Theory

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

-p*log(p)

0.2 0.4 0.6 0.8 1p

−p log p

Largest value. We seek a maximum point of the entropy σ with respect toall probability distributions pm which satisfy the normalization condition.This constrain, which is given by Eq. (1.1), is expressed as

0 = g0 (p) =

m

pm − 1 , (1.5)

where p denotes the vector of probabilities

p = (p1, p2, · · · ) . (1.6)

A small change in σ (denoted as δσ) due to a small change in p (denoted asδp = (δp1, δp2, · · · )) can be expressed as

δσ =

m

∂σ

∂pmδpm , (1.7)

or in terms of the gradient of σ (denoted as ∇σ) as

δσ = ∇σ · δp . (1.8)

In addition the variables (p1, p2, · · · ) are subjected to the constrain (1.5).Similarly to Eq. (1.8) we have

δg0 = ∇g0 · δp . (1.9)

Both vectors ∇σ and δp can be decomposed as

∇σ =∇σ +

∇σ⊥ , (1.10)

δp = (δp) + (δp)⊥ , (1.11)

where∇σ and (δp) are parallel to ∇g0, and where

∇σ⊥ and (δp)⊥ are

orthogonal to ∇g0. Using this notation Eq. (1.8) can be expressed as



δσ =∇σ · (δp) +

∇σ⊥ · (δp)⊥ . (1.12)

Given that the constrain g0 (p) = 0 is satisfied at a given point p, one hasg0 (p+ δp) = 0 to first order in δp provided that δp, is orthogonal to ∇g0,namely, provided that (δp) = 0. Thus, a stationary (maximum or minimum

or saddle point) point of σ occurs iff for every small change δp, which isorthogonal to ∇g0 (namely, δp · ∇g0 = 0) one has 0 = δσ = ∇σ · δp. Ascan be seen from Eq. (1.12), this condition is fulfilled only when

∇σ⊥ = 0,

namely only when the vectors ∇σ and ∇g0 are parallel to each other. In otherwords, only when

∇σ = ξ0∇g0 , (1.13)

where ξ0 is a constant. This constant is called Lagrange multiplier . UsingEqs. (1.2) and (1.5) the condition (1.13) is expressed as

log pm + 1 = ξ0 . (1.14)

Let M be the number of available states. From Eq. (1.14) we find that allprobabilities are equal. Thus using Eq. (1.5), one finds that

p1 = p2 = · · · =1

M. (1.15)

After finding this stationary point it is necessary to determine whether itis a maximum or minimum or saddle point. To do this we expand σ to secondorder in δp

σ (p+ δp) = expδp · ∇

σ (p)

=

1 + δp · ∇+

δp · ∇

2

2!+ · · ·

σ (p)

= σ (p) + δp · ∇σ +

δp · ∇

2

2!σ + · · ·

= σ (p) +

m

∂σ

∂pmδpm +

1

2

m,m′

δpmδpm′

∂2σ

∂pm∂pm′

+ · · ·

(1.16)

Using Eq. (1.2) one finds that

∂2σ

∂pm∂pm′

= − 1

pmδm,m′ . (1.17)

Since the probabilities pm are non-negative one concludes that any stationarypoint of σ is a local maximum point. Moreover, since only a single stationarypoint was found, one concludes that the entropy σ obtains its largest value,which is denoted as Λ (M), and which is given by



Λ (M) = σ

1

M,1

M, · · · , 1

M

= logM , (1.18)

for the probability distribution given by Eq. (1.15). For this probability dis-tribution that maximizes σ, as expected, the state which is occupied by thesystem is most uncertain.

1.1.3 The composition property

The composition property is best illustrated using an example.

Example - A Three States System. A system can occupy one of thestates e1, e2 or e3 with probabilities p1, p2 and p3 respectively. The uncer-tainty associated with this probability distribution can be estimated in twoways, directly and indirectly. Directly, it is simply given by the definition ofentropy in Eq. (1.2)

σ (p1, p2, p3) = −p1 log p1 − p2 log p2 − p3 log p3 . (1.19)

Alternatively [see Fig. 1.1], the uncertainty can be decomposed as follows:(a) the system can either occupy state e1 with probability p1, or not occupystate e1 with probability 1 − p1; (b) given that the system does not occupystate e1, it can either occupy state e2 with probability p2/ (1− p1) or occupystate e3 with probability p3/ (1− p1). Assuming that uncertainty (entropy)is additive, the total uncertainty (entropy) is given by

σi = σ (p1, 1− p1) + (1− p1)σ

p2

1− p1,

p31− p1

. (1.20)

The factor (1− p1) in the second term is included since the uncertainty asso-ciated with distinction between states e2 and e3 contributes only when statee1 is not occupied, an event which occurs with probability 1− p1. Using thedefinition (1.2) and the normalization condition

p1 + p2 + p3 = 1 , (1.21)

one finds

σi = −p1 log p1 − (1− p1) log (1− p1)

+ (1− p1)

− p21− p1

logp2

1− p1− p3

1− p1log

p31− p1

= −p1 log p1 − p2 log p2 − p3 log p3

− (1− p1 − p2 − p3) log (1− p1)

= σ (p1, p2, p3) ,

(1.22)

that is, for this example the entropy satisfies the decomposition property.



1p 11 p−

1e 32 ,ee

1

2

1 p

p

− 1

3

1 p

p

−

2e 3e

1p 11 p−

1e 32 ,ee

1

2

1 p

p

− 1

3

1 p

p

−

2e 3e

Fig. 1.1. The composition property - three states system.

The general case. The composition property in the general case can bedefined as follows. Consider a system which can occupy one of the statese1, e2, · · · , eM0

with probabilities q1, q2, · · · , qM0respectively. This set of

states is grouped as follows. The first group includes the first M1 statese1, e2, · · · , eM1; the second group includes the nextM2 states eM1+1, eM1+2, · · · , eM1+M2,etc., whereM1+M2+ · · · =M0. The probability that one of the states in thefirst group is occupied is p1 = q1+q2+· · ·+qM1 , the probability that one of thestates in the second group is occupied is p2 = qM1+1+ qM1+2+ · · ·+qM1+M2 ,etc., where

p1 + p2 + · · · = 1 . (1.23)

The composition property requires that the following holds [see Fig. 1.2]

σ (q1, q2, · · · , qM0) = σ (p1, p2, · · · )

+ p1σ

q1p1,q2p1, · · · , qM1

p1

+ p2σ

qM1+1

p2,qM1+2

p2, · · · , qM1+M2

p2

+ · · ·

(1.24)

Using the definition (1.2) the following holds

σ (p1, p2, · · · ) = −p1 log p1 − p2 log p2 − · · · , (1.25)



1p

1e

1,...,, 21 Meee

1

1

p

q

2e1Me

…

211,...,1 MMM ee

++

2p

…1

2

p

q

1

1

p

qM

11+Me2

1

p

qM

21 +Me21 MMe

+

…2

21

p

qM +

2

21

p

q MM +

1...211 Mqqqp +++=

211...12 MMM qqp

++++=

1p

1e

1,...,, 21 Meee

1

1

p

q

2e1Me

…

211,...,1 MMM ee

++

2p

…1

2

p

q

1

1

p

qM

11+Me2

1

p

qM

21 +Me21 MMe

+

…2

21

p

qM +

2

21

p

q MM +

1...211 Mqqqp +++=

211...12 MMM qqp

++++=

Fig. 1.2. The composition property - the general case.

p1σ

q1p1,q2p1, · · · , qM1

p1

= p1

−q1p1

logq1p1− q2p1

logq2p1− · · · − qM1

p1log

qM1

p1

= −q1 log q1 − q2 log q2 − · · · − qM1 log qM1

+ p1 log p1 ,

(1.26)

p2σ

qM1+1

p2,qM1+2

p2, · · · , qM1+M2

p2

= −qM1+1 log qM1+1 − qM1+2 log qM1+2

− · · · − qM1+M2 log qM1+M2

+ p2 log p2 ,

(1.27)

etc., thus it is evident that condition (1.24) is indeed satisfied.



1.1.4 Alternative Definition of Entropy

Following Shannon [1, 2], the entropy function σ (p1, p2, · · · , pN) can be al-ternatively defined as follows:

1. σ (p1, p2, · · · , pN) is a continuous function of its arguments p1, p2, · · · , pN .2. If all probabilities are equal, namely if p1 = p2 = · · · = pN = 1/N , then

the quantity Λ (N) = σ (1/N, 1/N, · · · , 1/N) is a monotonic increasingfunction of N .

3. The function σ (p1, p2, · · · , pN) satisfies the composition property givenby Eq. (1.24).

Exercise 1.1.1. Show that the above definition leads to the entropy givenby Eq. (1.2) up to multiplication by a positive constant.

Solution 1.1.1. The 1st property allows approximating the probabilitiesp1, p2, · · · , pN using rational numbers, namely p1 = M1/M0, p2 = M2/M0,etc., where M1,M2, · · · are integers and M0 = M1 +M2 + · · ·+MN . Usingthe composition property (1.24) one finds

Λ (M0) = σ (p1, p2, · · · , pN) + p1Λ (M1) + p2Λ (M2) + · · · (1.28)

In particular, consider the case were M1 = M2 = · · · = MN = K. For thiscase one finds

Λ (NK) = Λ (N) + Λ (K) . (1.29)

Taking K = N = 1 yields

Λ (1) = 0 . (1.30)

Taking N = 1 + x yields

Λ (K +Kx)− Λ (K)

Kx=

1

K

Λ (1 + x)

x. (1.31)

Taking the limit x→ 0 yields

dΛ

dK=

C

K, (1.32)

where

C = limx→0

Λ (1 + x)

x. (1.33)

Integrating Eq. (1.32) and using the initial condition (1.30) yields

Λ (K) = C logK . (1.34)


1.2. Largest Uncertainty Estimator

Moreover, the second property requires that C > 0. Choosing C = 1 andusing Eq. (1.28) yields

σ (p1, p2, · · · , pN) = Λ (M0)− p1Λ (M1)− p2Λ (M2)− · · ·

= −p1 logM1

M0− p2 log

M2

M0− · · · − pM log

MN

M0

= −p1 log p1 − p2 log p2 − · · · − pN log pN ,

(1.35)

in agreement with the definition (1.2).

1.2 Largest Uncertainty Estimator

As before, the possible states of a given system are denoted as em, wherem = 1, 2, 3, · · · , and the probability that state em is occupied is denoted bypm. Let Xl (l = 1, 2, · · · , L) be a set of variables characterizing the system(e.g., energy, number of particles, etc.). Let Xl (m) be the value which thevariable Xl takes when the system is in state em. Consider the case wherethe expectation values of the variables Xl are given

Xl =

m

pmXl (m) , (1.36)

where l = 1, 2, · · · , L. However, the probability distribution pm is not given.Clearly, in the general case the knowledge of X1 , X2 , · · · , XL is not

sufficient to obtain the probability distribution because there are in generalmany different possibilities for choosing a probability distribution which isconsistent with the contrarians (1.36) and the normalization condition (1.1).For each such probability distribution the entropy can be calculated accordingto the definition (1.2). The probability distribution pm, which is consistentwith these conditions, and has the largest possible entropy is called the largestuncertainty estimator (LUE).

The LUE is found by seeking a stationary point of the entropy σ withrespect to all probability distributions pm which satisfy the normalizationconstrain (1.5) in addition to the constrains (1.36), which can be expressedas

0 = gl (p) =

m

pmXl (m)− Xl , (1.37)

where l = 1, 2, · · ·L. To first order one has

δσ = ∇σ · δp , (1.38a)

δgl = ∇gl · δp , (1.38b)



where l = 0, 1, 2, · · ·L. A stationary point of σ occurs iff for every smallchange δp, which is orthogonal to all vectors ∇g0, ∇g1, ∇g2, · · · , ∇gL onehas

0 = δσ = ∇σ · δp . (1.39)

This condition is fulfilled only when the vector ∇σ belongs to the subspacespanned by the vectors

∇g0, ∇g1, ∇g2, · · · , ∇gL

[see also the discussion

below Eq. (1.12) above]. In other words, only when

∇σ = ξ0∇g0 + ξ1∇g1 + ξ2∇g2 + · · ·+ ξL∇gL , (1.40)

where the numbers ξ0, ξ1, · · · , ξL, which are called Lagrange multipliers, areconstants. Using Eqs. (1.2), (1.5) and (1.37) the condition (1.40) can beexpressed as

− log pm − 1 = ξ0 +L

l=1

ξlXl (m) . (1.41)

From Eq. (1.41) one obtains

pm = exp (−1− ξ0) exp

−L

l=1

ξlXl (m)

. (1.42)

The Lagrange multipliers ξ0, ξ1, · · · , ξL can be determined from Eqs. (1.5)and (1.37)

1 =

m

pm = exp (−1− ξ0)

m

exp

−L

l=1

ξlXl (m)

, (1.43)

Xl =

mpmXl (m)

= exp (−1− ξ0)

m

exp

−L

l=1

ξlXl (m)

Xl (m) .

(1.44)

Using Eqs. (1.42) and (1.43) one finds

pm =

exp

−

L

l=1

ξlXl (m)

mexp

−

L

l=1

ξlXl (m)

. (1.45)

In terms of the partition function Z, which is defined as



Z =

m

exp

−L

l=1

ξlXl (m)

, (1.46)

one finds

pm =1

Z exp

−L

l=1

ξlXl (m)

. (1.47)

Using the same arguments as in section 1.1.2 above [see Eq. (1.16)] it is easy toshow that at the stationary point that occurs for the probability distributiongiven by Eq. (1.47) the entropy obtains its largest value.

1.2.1 Useful Relations

The expectation value Xl can be expressed as

Xl =

m

pmXl (m)

=1

Z

mexp

−L

l=1

ξlXl (m)

Xl (m)

= − 1

Z∂Z∂ξl

= −∂ logZ∂ξl

.

(1.48)

Similarly, X2l

can be expressed as

X2l

=

m

pmX2l (m)

=1

Z

m

exp

−L

l=1

ξlXl (m)

X2l (m)

=1

Z∂2Z∂ξ2l

.

(1.49)

Using Eqs. (1.48) and (1.49) one finds that the variance of the variable Xl isgiven by

(∆Xl)

2=(Xl − Xl)2

=

1

Z∂2Z∂ξ2l

−1

Z∂Z∂ξl

2. (1.50)

However, using the following identity



∂2 logZ∂ξ2l

=∂

∂ξl

1

Z∂Z∂ξl

=1

Z∂2Z∂ξ2l

−

1

Z∂Z∂ξl

2, (1.51)

one finds(∆Xl)

2=∂2 logZ∂ξ2l

. (1.52)

Note that the above results Eqs. (1.48) and (1.52) are valid only when Z isexpressed as a function of the the Lagrange multipliers, namely

Z = Z (ξ1, ξ2, · · · , ξL) . (1.53)

Using the definition of entropy (1.2) and Eq. (1.47) one finds

σ = −

m

pm log pm

= −

m

pm log

1

Z exp

−L

l=1

ξlXl (m)

=

m

pm

logZ +L

l=1

ξlXl (m)

= logZ +L

l=1

ξl

m

pmXl (m) ,

(1.54)

thus

σ = logZ +L

l=1

ξl Xl . (1.55)

Using the above relations one can also evaluate the partial derivative ofthe entropy σ when it is expressed as a function of the expectation values,namely

σ = σ (X1 , X2 , · · · , XL) . (1.56)

Using Eq. (1.55) one has

∂σ

∂ Xl=∂ logZ∂ Xl

+L

l′=1

Xl′∂ξl′

∂ Xl+

L

l′=1

ξl′∂ Xl′∂ Xl

=∂ logZ∂ Xl

+L

l′=1

Xl′∂ξl′

∂ Xl+ ξl

=L

l′=1

∂ logZ∂ξl′

∂ξl′

∂ Xl+

L

l′=1

Xl′∂ξl′

∂ Xl+ ξl ,

(1.57)



thus using Eq. (1.48) one finds

∂σ

∂ Xl= ξl . (1.58)

1.2.2 The Free Entropy

The free entropy σF is defined as the term logZ in Eq. (1.54)

σF = logZ

= σ −L

l=1

ξl

m

pmXl (m)

= −

m

pm log pm −L

l=1

ξl

m

pmXl (m) .

(1.59)

The free entropy is commonly expressed as a function of the Lagrange mul-tipliers

σF = σF (ξ1, ξ2, · · · , ξL) . (1.60)

We have seen above that the LUE maximizes σ for given values of expectationvalues X1 , X2 , · · · , XL. We show below that a similar result can beobtained for the free energy σF with respect to given values of the Lagrangemultipliers.

Claim. The LUE maximizes σF for given values of the Lagrange multipliersξ1, ξ2, · · · , ξL.

Proof. As before, the normalization condition is expressed as

0 = g0 (p) =

m

pm − 1 . (1.61)

At a stationary point of σF, as we have seen previously, the following holds

∇σF = η∇g0 , (1.62)

where η is a Lagrange multiplier. Thus

− (log pm + 1)−L

l=1

ξlXl (m) = η , (1.63)

or

pm = exp (−η − 1) exp

−L

l=1

ξlXl (m)

. (1.64)



Table 1.1. The microcanonical, canonical and grandcanonical distributions.

energy number of particles

microcanonical distribution constrained U (m) = U constrained N (m) = N

canonical distribution average is given U constrained N (m) = N

grandcanonical distribution average is given U average is given N

This result is the same as the one given by Eq. (1.42). Taking into accountthe normalization condition (1.61) one obtains the same expression for pm asthe one given by Eq. (1.47). Namely, the stationary point of σF correspondsto the LUE probability distribution. Since

∂2σF∂pm∂pm′

= − 1

pmδm,m′ < 0 , (1.65)

one concludes that this stationary point is a maximum point [see Eq. (1.16)].

1.3 The Principle of Largest Uncertainty in Statistical

Mechanics

The energy and number of particles of state em are denoted by U (m) andN (m) respectively. The probability that state em is occupied is denoted aspm. We consider below three cases (see table 1.1). In the first case (micro-canonical distribution) the system is isolated and its total energy U and num-ber of particles N are constrained , that is for all accessible states U (m) = Uand N (m) = N . In the second case (canonical distribution) the system isallowed to exchange energy with the environment, and we assume that itsaverage energy U is given. However, its number of particles is constrained, that is N (m) = N . In the third case (grandcanonical distribution) the sys-tem is allowed to exchange both energy and particles with the environment,and we assume that both the average energy U and the average numberof particles N are given. However, in all cases, the probability distributionpm is not given.

According to the principle of largest uncertainty in statistical mechanicsthe LUE is employed to estimate the probability distribution pm, namely,we will seek a probability distribution which is consistent with the normal-ization condition (1.1) and with the given expectation values (energy, in thesecond case, and both energy and number of particles, in the third case),which maximizes the entropy.

1.3.1 Microcanonical Distribution

In this case no expectation values are given. Thus we seek a probabilitydistribution which is consistent with the normalization condition (1.1), and


1.3. The Principle of Largest Uncertainty in Statistical Mechanics

which maximizes the entropy. The desired probability distribution is

p1 = p2 = · · · = 1/M , (1.66)

whereM is the number of accessible states of the system [see also Eq. (1.18)].Using Eq. (1.2) the entropy for this case is given by

σ = logM . (1.67)

1.3.2 Canonical Distribution

Using Eq. (1.47) one finds that the probability distribution is given by

pm =1

Zcexp (−βU (m)) , (1.68)

where β is the Lagrange multiplier associated with the given expectationvalue U, and the partition function is given by

Zc =

m

exp (−βU (m)) . (1.69)

The term exp (−βU (m)) is called Boltzmann factor.Moreover, Eq. (1.48) yields

U = −∂ logZc∂β

, (1.70)

Eq. (1.52) yields

(∆U)2

=∂2 logZc∂β2

, (1.71)

and Eq. (1.55) yields

σ = logZc + β U . (1.72)

Using Eq. (1.58) one can expressed the Lagrange multiplier β as

β =∂σ

∂U. (1.73a)

The temperature τ = 1/β is defined as

1

τ= β . (1.74)

Exercise 1.3.1. Consider a system that can be in one of two states having

energies ±ε/2. Calculate the average energy U and the variance(∆U)2

in thermal equilibrium at temperature τ .



Solution: The partition function is given by Eq. (1.69)

Zc = exp

βε

2

+ exp

−βε

2

= 2cosh

βε

2

, (1.75)

thus using Eqs. (1.70) and (1.71) one finds

U = −ε2tanh

βε

2

, (1.76)

and

(∆U)2

=ε2

2 1

cosh2 βε2

, (1.77)

where β = 1/τ .

-1

-0.8

-0.6

-0.4

-0.2

0

-tanh(1/x)

1 2 3 4 5x

1.3.3 Grandcanonical Distribution

Using Eq. (1.47) one finds that the probability distribution is given by

pm =1

Zgcexp (−βU (m)− ηN (m)) , (1.78)

where β and η are the Lagrange multipliers associated with the given expec-tation values U and N respectively, and the partition function is givenby

Zgc =

m

exp (−βU (m)− ηN (m)) . (1.79)

The term exp (−βU (m)− ηN (m)) is called Gibbs factor.Moreover, Eq. (1.48) yields


1.3. The Principle of Largest Uncertainty in Statistical Mechanics

U = −∂ logZgc

∂β

η

, (1.80)

N = −∂ logZgc

∂η

β

(1.81)

Eq. (1.52) yields

(∆U)2

=

∂2 logZgc

∂β2

η

, (1.82)

(∆N)

2=

∂2 logZgc

∂η2

β

, (1.83)

and Eq. (1.55) yields

σ = logZgc + β U+ η N . (1.84)

1.3.4 Temperature and Chemical Potential

Probability distributions in statistical mechanics of macroscopic parametersare typically extremely sharp and narrow. Consequently, in many cases nodistinction is made between a parameter and its expectation value. That is,the expression for the entropy in Eq. (1.72) can be rewritten as

σ = logZc + βU , (1.85)

and the one in Eq. (1.84) as

σ = logZgc + βU + ηN . (1.86)

Using Eq. (1.58) one can expressed the Lagrange multipliers β and η as

β =

∂σ

∂U

N

, (1.87)

η =

∂σ

∂N

U

. (1.88)

The chemical potential µ is defined as

µ = −τη . (1.89)

In the definition (1.2) the entropy σ is dimensionless. Historically, the entropywas defined as

S = kBσ , (1.90)

where



kB = 1.38× 10−23 JK−1 (1.91)

is the Boltzmann constant. Moreover, the historical definition of the temper-ature is

T =τ

kB. (1.92)

When the grandcanonical partition function is expressed in terms of βand µ (instead of in terms of β and η), it is convenient to rewrite Eqs. (1.80)and (1.81) as (see homework exercises 14 of chapter 1).

U = −∂ logZgc

∂β

µ

+ τµ

∂ logZgc

∂µ

β

, (1.93)

N = λ∂ logZgc

∂λ, (1.94)

where λ is the fugacity , which is defined by

λ = exp (βµ) = e−η . (1.95)

1.4 Time Evolution of Entropy of an Isolated System

Consider a perturbation which results in transitions between the states of anisolated system. Let Γrs denotes the resulting rate of transition from state rto state s. The probability that state s is occupied is denoted as ps.

The following theorem (H theorem) states that if for every pair of statesr and s

Γrs = Γsr , (1.96)

then

dσ

dt≥ 0 . (1.97)

Moreover, equality holds iff ps = pr for all pairs of states for which Γsr = 0.To prove this theorem we express the rate of change in the probability ps

in terms of these transition rates

dprdt

=

s

psΓsr −

s

prΓrs . (1.98)

The first term represents the transitions to state r, whereas the second onerepresents transitions from state r. Using property (1.96) one finds

dprdt

=

s

Γsr (ps − pr) . (1.99)


1.5. Thermal Equilibrium

The last result and the definition (1.2) allows calculating the rate of changeof entropy

dσ

dt= − d

dt

r

pr log pr

= −

r

dprdt

(log pr + 1)

= −

r

s

Γsr (ps − pr) (log pr + 1) .

(1.100)

One the other hand, using Eq. (1.96) and exchanging the summation indicesallow rewriting the last result as

dσ

dt=

r

s

Γsr (ps − pr) (log ps + 1) . (1.101)

Thus, using both expressions (1.100) and (1.101) yields

dσ

dt=

1

2

r

s

Γsr (ps − pr) (log ps − log pr) . (1.102)

In general, since logx is a monotonic increasing function

(ps − pr) (log ps − log pr) ≥ 0 , (1.103)

and equality holds iff ps = pr. Thus, in general

dσ

dt≥ 0 , (1.104)

and equality holds iff ps = pr holds for all pairs is states satisfying Γsr =0. When σ becomes time independent the system is said to be in thermalequilibrium. In thermal equilibrium, when all accessible states have the sameprobability, one finds using the definition (1.2)

σ = logM , (1.105)

where M is the number of accessible states of the system.Note that the rates Γrs, which can be calculated using quantum mechan-

ics, indeed satisfy the property (1.96) for the case of an isolated system.

1.5 Thermal Equilibrium

Consider two isolated systems denoted as S1 and S2. Let σ1 = σ1 (U1, N1) andσ2 = σ2 (U2,N2) be the entropy of the first and second system respectively



and let σ = σ1+σ2 be the total entropy. The systems are brought to contactand now both energy and particles can be exchanged between the systems.Let δU be an infinitesimal energy, and let δN be an infinitesimal number ofparticles, which are transferred from system 1 to system 2. The correspondingchange in the total entropy is given by

δσ = −∂σ1∂U1

N1

δU +

∂σ2∂U2

N2

δU

−∂σ1∂N1

U1

δN +

∂σ2∂N2

U2

δN

=

− 1

τ1+

1

τ2

δU −

−µ1τ1

+µ2τ2

δN .

(1.106)

The change δσ in the total entropy is obtained by removing a constrain.Thus, at the end of this process more states are accessible, and therefore,according to the principle of largest uncertainty it is expected that

δσ ≥ 0 . (1.107)

For the case where no particles can be exchanged (δN = 0) this implies thatenergy flows from the system of higher temperature to the system of lowertemperature. Another important case is the case where τ1 = τ2, for whichwe conclude that particles flow from the system of higher chemical potentialto the system of lower chemical potential.

In thermal equilibrium the entropy of the total system obtains its largestpossible value. This occurs when

τ1 = τ2 (1.108)

and

µ1 = µ2 . (1.109)

1.5.1 Externally Applied Potential Energy

In the presence of externally applied potential energy µex the total chemicalpotential µtot is given by

µtot = µint + µex , (1.110)

where µint is the internal chemical potential . For example, for particles havingcharge q in the presence of electric potential V one has

µex = qV , (1.111)


1.7. Problems Set 1

whereas, for particles having mass m in a constant gravitational field g onehas

µex =mgz , (1.112)

where z is the height. The thermal equilibrium relation (1.109) is generalizedin the presence of externally applied potential energy as

µtot,1 = µtot,2 . (1.113)

1.6 Free Entropy and Free Energies

The free entropy [see Eq. (1.59)] for the canonical distribution is given by[see Eq. (1.85)]

σF,c = σ − βU , (1.114)

whereas for the grandcanonical case it is given by [see Eq. (1.86)]

σF,gc = σ − βU − ηN . (1.115)

We define below two corresponding free energies, the canonical free energy(known also as the Helmholtz free energy )

F = −τσF,c = U − τσ , (1.116)

and the grandcanonical free energy

G = −τσF,gc = U − τσ + τηN = U − τσ − µN .

In section 1.2.2 above we have shown that the LUE maximizes σF forgiven values of the Lagrange multipliers ξ1, ξ2, · · · , ξL. This principle can beimplemented to show that:

• In equilibrium at a given temperature τ the Helmholtz free energy obtainsits smallest possible value.

• In equilibrium at a given temperature τ and chemical potential µ the grand-canonical free energy obtains its smallest possible value.

Our main results are summarized in table 1.2 below

1.7 Problems Set 1

Note: Problems 1-6 are taken from the book by Reif, chapter 1.



Table 1.2. Summary of main results.

general

micro

−canonical

(M states)

canonical grandcanonical

given

expectation

values

Xl where

l = 1, 2, ..., LU U , N

partition

function

Z =

m

e−

L

l=1ξlXl(m)

Zc =

m

e−βU(m)Zgc =

m

e−βU(m)−ηN(m)

pm

pm =

1Ze−

L

l=1ξlXl(m)

pm = 1M

pm =1Zce−βU(m)

pm =1Zgc

e−βU(m)−ηN(m)

Xl Xl = −∂ logZ∂ξl

U = − ∂ logZc∂β

U = −∂ logZgc

∂β

η

N = −∂ logZgc

∂η

β

(∆Xl)

2 (∆Xl)

2 = ∂2 logZ

∂ξ2l

(∆U)2

= ∂2 logZc

∂β2

(∆U)2

=∂2 logZgc

∂β2

η (∆N)2

=∂2 logZgc

∂η2

β

σσ =

logZ +L

l=1

ξl Xlσ = logM

σ =

logZc + β U

σ =

logZgc + β U+ η N

Lagrange

multipliersξl =

∂σ

∂Xl

Xnn =l

β = ∂σ∂U

β =∂σ∂U

N

η =∂σ∂N

U

min /max

principle

max

σF (ξ1, ξ2, ..., ξL)

σF = σ −L

l=1

ξl Xl

max σmin F (τ )

F = U − τσ

min G (τ, µ)

G = U − τσ − µN

1. A drunk starts out from a lamppost in the middle of a street, taking stepsof equal length either to the right or to the left with equal probability.What is the probability that the man will again be at the lamppost aftertaking N steps

a) if N is even?b) if N is odd?

2. In the game of Russian roulette, one inserts a single cartridge into thedrum of a revolver, leaving the other five chambers of the drum empty.One then spins the drum, aims at one’s head, and pulls the trigger.

a) What is the probability of being still alive after playing the game Ntimes?


1.7. Problems Set 1

b) What is the probability of surviving (N − 1) turns in this game andthen being shot the Nth time one pulls the trigger?

c) What is the mean number of times a player gets the opportunity ofpulling the trigger in this macabre game?

3. Consider the random walk problem with p = q and let m = n1 − n2, de-note the net displacement to the right. After a total of N steps, calculatethe following mean values: m,

m2, m3, and

m4. Hint: Calculate

the moment generating function.4. The probability W (n), that an event characterized by a probability p

occurs n times in N trials was shown to be given by the binomial distri-bution

W (n) =N !

n! (N − n)!pn (1− p)N−n . (1.117)

Consider a situation where the probability p is small (p << 1) and whereone is interested in the case n << N . (Note that if N is large, W (n)becomes very small if n → N because of the smallness of the factor pn

when p << 1. Hence W (n) is indeed only appreciable when n << N .)Several approximations can then be made to reduce Eq. (1.117) to simplerform.

a) Using the result ln(1− p) ≃ −p, show that (1− p)N−n ≃ e−Np.b) Show that N !/(N − n)! ≃ Nn.c) Hence show that Eq. (1.117) reduces to

W (n) =λn

n!e−λ ,

where λ ≡ Np is the mean number of events. This distribution iscalled the "Poisson distribution."

5. Consider the Poisson distribution of the preceding problem.

a) Show that it is properly normalized in the sense that∞

n=0

W (n) = 1.

(The sum can be extended to infinity to an excellent approximation,since W (n) is negligibly small when n N .)

b) Use the Poisson distribution to calculate n.c) Use the Poisson distribution to calculate

(∆n)2

=(n− n)2

.

6. A molecule in a gas moves equal distances l between collisions with equalprobability in any direction. After a total of N such displacements, whatis the mean square displacement

R2of the molecule from its starting

point ?7. A multiple choice test contains 20 problems. The correct answer for each

problem has to be chosen out of 5 options. Show that the probability topass the test (namely to have at least 11 correct answers) using guessingonly, is 5.6× 10−4.



8. Consider two objects traveling in the xy plane. Object A starts from thepoint (0, 0) and object B starts from the point (N,N), where N is aninteger. At each step both objects A and B simultaneously make a singlemove of length unity. Object A makes either a move to the right (x, y)→(x+ 1, y) with probability 1/2 or an upward move (x, y) → (x, y + 1)with probability 1/2. On the other hand, object B makes either a moveto the left (x, y) → (x− 1, y) with probability 1/2 or a downward move(x, y) → (x, y − 1) with probability 1/2. What is the probability thatobjects A and B meet along the way in the limit N →∞?

9. Consider A dice having 6 faces. All faces have equal probability of out-come. Initially, n faces are colored white and 6−n faces are colored black,where n ∈ 0, 1, 2, · · · , 6. Each time the outcome is white (black) oneblack (white) face is turned into a white (black) face before the next roll.The process continues until all faces have the same color. What is theprobability pn that all faces will become white?

10. Alice, Bob and other N −2 people are randomly seated at a round table.What is the probability pC that Alice and Bob will be seated next toeach other? What is the probability pR that Alice and Bob will be seatednext to each other for the case where the group is randomly seated in arow.

11. Write a computer function returning the value 1 with probability p andthe value 0 with probability 1− p for any given 0 ≤ p ≤ 1. The functioncan use another given function, which returns the value 1 with probability1/2 and the value 0 with probability 1/2. Make sure the running time isfinite.

12. Consider a system of N spins. Each spin can be in one of two possiblestates: in state ’up’ the magnetic moment of each spin is +m, and instate ’down’ it is −m. Let N+ (N−) be the number of spins in state ’up’(’down’), where N = N++N−. The total magnetic moment of the systemis given by

M = m (N+ −N−) . (1.118)

Assume that the probability that the system occupies any of its 2N pos-sible states is equal. Moreover, assume that N ≫ 1. Let f (M) be theprobability distribution of the random variable M (that is, M is consid-ered in this approach as a continuous random variable). Use the Stirling’sformula

N ! = (2πN)1/2NN exp

−N +

1

2N+ · · ·

(1.119)

to show that

f (M) =1

m√2πN

exp

− M2

2m2N

. (1.120)

Use this result to evaluate the expectation value and the variance of M .


1.7. Problems Set 1

13. Consider a one dimensional random walk. The probabilities of transitingto the right and left are p and q = 1 − p respectively. The step size forboth cases is a.

a) Show that the average displacement X after N steps is given by

X = aN (2p− 1) = aN (p− q) . (1.121)

b) Show that the variance(X − X)2

is given by

(X − X)2

= 4a2Npq . (1.122)

14. A classical harmonic oscillator of massm, and spring constant k oscillateswith amplitude a. Show that the probability density function f(x), wheref(x)dx is the probability that the mass would be found in the intervaldx at x, is given by

f(x) =1

π√a2 − x2

. (1.123)

15. A coin having probability p = 2/3 of coming up heads is flipped 6 times.Show that the entropy of the outcome of this experiment is σ = 3.8191(use log in natural base in the definition of the entropy).

16. A fair coin is flipped until the first head occurs. Let X denote the numberof flips required.

a) Find the entropy σ. In this exercise use log in base 2 in the definitionof the entropy, namely σ = −i pi log2 pi.

b) A random variable X is drawn according to this distribution. Findan “efficient” sequence of yes-no questions of the form, “Is X con-tained in the set S?” Compare σ to the expected number of questionsrequired to determine X.

17. In general the notation

∂z

∂x

y

is used to denote the partial derivative of z with respect to x, where thevariable y is kept constant. That is, to correctly calculate this derivativethe variable z has to be expressed as a function of x and y, namely,z = z (x, y).

a) Show that:

∂z

∂x

y

= −

∂y∂x

z∂y∂z

x

. (1.124)



b) Show that:

∂z

∂x

w

=

∂z

∂x

y

+

∂z

∂y

x

∂y

∂x

w

. (1.125)

18. Let Zgc be a grandcanonical partition function.

a) Show that:

U = −∂ logZgc

∂β

µ

+ τµ

∂ logZgc

∂µ

β

. (1.126)

where τ is the temperature, β = 1/τ , and µ is the chemical potential.b) Show that:

N = λ∂ logZgc

∂λ, (1.127)

where

λ = exp (βµ) (1.128)

is the fugacity.

19. Consider an array on N distinguishable two-level (binary) systems. Thetwo-level energies of each system are ±ε/2. Show that the temperatureτ of the system is given by

τ =ε

2 tanh−1−2U

Nε

, (1.129)

where U is the average total energy of the array. Note that the tem-perature can become negative if U > 0. Why a negative temperatureis possible for this system ?

20. Consider an array of N distinguishable quantum harmonic oscillators inthermal equilibrium at temperature τ . The resonance frequency of alloscillators is ω. The quantum energy levels of each quantum oscillatoris given by

εn = ℏω

n+

1

2

, (1.130)

where n = 0, 1, 2, · · · is integer.a) Show that the average energy of the system is given by

U = Nℏω

2coth

βℏω

2, (1.131)

where β = 1/τ .


1.7. Problems Set 1

b) Show that the variance of the energy of the system is given by

(∆U)2

=

Nℏω2

2

sinh 2 βℏω2

. (1.132)

21. Consider a lattice containing N non-interacting atoms. Each atom has 3non-degenerate energy levels E1 = −ε, E2 = 0, E3 = ε. The system is atthermal equilibrium at temperature τ .

a) Show that the average energy of the system is

U = −2Nε sinh (βε)

1 + 2 coshβε, (1.133)

where β = 1/τ .b) Show the variance of the energy of the system is given by

(U − U)2

= 2Nε2

cosh (βε) + 2

[1 + 2 cosh (βε)]2. (1.134)

22. Consider a one dimensional chain containing N ≫ 1 sections (see figure).Each section can be in one of two possible sates. In the first one thesection contributes a length a to the total length of the chain, whereasin the other state the section has no contribution to the total length ofthe chain. The total length of the chain in Nα, and the tension appliedto the end points of the chain is F . The system is in thermal equilibriumat temperature τ .

a) Show that α is given by

α =a

2

1 + tanh

Fa

2τ

. (1.135)

b) Show that in the limit of high temperature the spring constant isgiven approximately by

k ≃ 4τ

Na2. (1.136)

NαNα

23. A long elastic molecule can be modelled as a linear chain of N links. Thestate of each link is characterized by two quantum numbers l and n. Thelength of a link is either l = a or l = b. The vibrational state of a link



is modelled as a harmonic oscillator whose angular frequency is ωa for alink of length a and ωb for a link of length b. Thus, the energy of a linkis

En,l =

ℏωa

n+ 1

2

for l = a

ℏωbn+ 1

2

for l = b

, (1.137)

n = 0, 1, 2, · · ·

The chain is held under a tension F . Show that the mean length L ofthe chain in the limit of high temperature T is given by

L = Naωb + bωaωb + ωa

+NFωbωa (a− b)

2

(ωb + ωa)2 β +O

β2, (1.138)

where β = 1/τ .24. The elasticity of a rubber band can be described in terms of a one-

dimensional model of N polymer molecules linked together end-to-end.The angle between successive links is equally likely to be 0 or 180. Thelength of each polymer is d and the total length is L. The system is inthermal equilibrium at temperature τ . Show that the force f required tomaintain a length L is given by

f =τ

dtanh−1

L

Nd. (1.139)

25. Consider a system which has two single particle states both of the sameenergy. When both states are unoccupied, the energy of the system iszero; when one state or the other is occupied by one particle, the energyis ε. We suppose that the energy of the system is much higher (infinitelyhigher) when both states are occupied. Show that in thermal equilibriumat temperature τ the average number of particles in the level is

N = 2

2 + exp [β (ε− µ)], (1.140)

where µ is the chemical potential and β = 1/τ .26. Consider an array of N two-level particles. Each one can be in one of two

states, having energy E1 and E2 respectively. The numbers of particlesin states 1 and 2 are n1 and n2 respectively, where N = n1+n2 (assumen1 ≫ 1 and n2 ≫ 1). Consider an energy exchange with a reservoir attemperature τ leading to population changes n2 → n2−1 and n1 → n1+1.

a) Calculate the entropy change of the two-level system, (∆σ)2LS.b) Calculate the entropy change of the reservoir, (∆σ)R.c) What can be said about the relation between (∆σ)2LS and (∆σ)R in

thermal equilibrium? Use your answer to express the ration n2/n1 asa function of E1, E2 and τ .


1.7. Problems Set 1

27. Consider a lattice containing N sites of one type, which is denoted as A,and the same number of sites of another type, which is denoted as B. Thelattice is occupied by N atoms. The number of atoms occupying sites oftype A is denoted as NA, whereas the number of atoms occupying atomsof type B is denoted as NB, where NA +NB = N . Let ε be the energynecessary to remove an atom from a lattice site of type A to a latticesite of type B. The system is in thermal equilibrium at temperature τ .Assume that N,NA, NB ≫ 1.

a) Calculate the entropy σ.b) Calculate the average number NB of atoms occupying sites of type

B.

28. Consider a microcanonical ensemble of N quantum harmonic oscillatorsin thermal equilibrium at temperature τ . The resonance frequency of alloscillators is ω. The quantum energy levels of each quantum oscillator isgiven by

εn = ℏω

n+

1

2

, (1.141)

where n = 0, 1, 2, · · · is integer. The total energy E of the system is givenby

E = ℏω

m+

N

2

, (1.142)

where

m =N

l=1

nl , (1.143)

and nl is state number of oscillator l.

a) Calculate the number of states g (N,m) of the system with totalenergy ℏω (m+N/2).

b) Use this result to calculate the entropy σ of the system with totalenergy ℏω (m+N/2). Approximate the result by assuming thatN ≫1 and m≫ 1.

c) Use this result to calculate (in the same limit of N ≫ 1 and m≫ 1)the average energy of the system U as a function of the temperatureτ .

29. The energy of a donor level in a semiconductor is −ε when occupiedby an electron (and the energy is zero otherwise). A donor level can beeither occupied by a spin up electron or a spin down electron, however, itcannot be simultaneously occupied by two electrons. Express the averageoccupation of a donor state Nd as a function of ε and the chemicalpotential µ.



1.8 Solutions Set 1

1. Final answers:

a) N !

(N2 )!(N2 )!

12

N

b) 0

2. Final answers:

a)56

N

b)56

N−1 16

c) In general

∞

N=0

NxN−1 =d

dx

∞

N=0

xN =d

dx

1

1− x=

1

(1− x)2,

thus

N =1

6

∞

N=0

N

5

6

N−1=

1

6

11− 5

6

2 = 6 .

3. Let W (m) be the probability for for taking n1 steps to the right andn2 = N − n1 steps to the left, where m = n1 − n2, and N = n1 + n2.Using

n1 =N +m

2,

n2 =N −m

2,

one finds

W (m) =N !

N+m2

!N−m2

!pN+m2 q

N−m2 .

It is convenient to employ the moment generating function, defined as

φ (t) = etm.

In general, the following holds

φ (t) =∞

k=0

tk

k!

mk,

thus from the kth derivative of φ (t) one can calculate the kth momentof m

mk= φ(k) (0) .


1.8. Solutions Set 1

Using W (m) one finds

φ (t) =N

m=−NW (m) etm

=N

m=−N

N !N+m2

!N−m2

!pN+m2 q

N−m2 etm ,

or using the summation variable

n1 =N +m

2,

one has

φ (t) =N

n1=0

N !

n1! (N − n1)!pn1qN−n1et(2n1−N)

= e−tNN

n1=0

N !

n1! (N − n1)!

pe2t

n1qN−n1

= e−tNpe2t + q

N.

Using p = q = 1/2

φ (t) =

et + e−t

2

N= (cosh t)N .

Thus using the expansion

(cosh t)N = 1 +1

2!Nt2 +

1

4![N + 3N (N − 1)] t4 +O

t5,

one finds

m = 0 , m2= N ,

m3= 0 ,

m4= N (3N − 2) .

4. Using the binomial distribution

W (n) =N !

n! (N − n)!pn (1− p)N−n

=N (N − 1) (N − 1)× · · · (N − n+ 1)

n!pn (1− p)N−n

∼= (Np)n

n!exp (−Np) .



5.

a)∞

n=0

W (n) = e−λ∞

n=0

λn

n! = 1

b) As in Ex. 1-6, it is convenient to use the moment generating function

φ (t) = etn=

∞

n=0

etnW (n) = e−λ∞

n=0

λnetn

n!= e−λ

∞

n=0

(λet)n

n!= exp

λet − 1

.

Using the expansion

expλet − 1

= 1 + λt+

1

2λ (1 + λ) t2 +O

t3,

one finds

n = λ .

c) Using the same expansion one finds n2= λ (1 + λ) ,

thus(∆n)2

= n2− n2 = λ (1 + λ)− λ2 = λ .

6.

R2=

N

n=1

rn

2

=N

n=1

r2n

=l2

+

n =mrn · rm

=0

= Nl2

7.

20

n=11

20!

n! (20− n)!0.2n × 0.820−n = 5.6× 10−4 . (1.144)

8. Let σAn = 1 (σBn = 1) if object A (B) makes a move to the right (left)at step n, and σAn = 0 (σBn = 0) if object A (B) makes an upward(downward) move at step n. The location (xAm, yAm) of object A andthe location (xBm, yBm) of object B after m steps is given by

(xAm, yAm) = (SAm,m− SAm) , (1.145)

(xBm, yBm) = (N − SBm,N −m+ SBm) , (1.146)

where

SAm =m

n=1

σAn , (1.147)

SBm =m

n=1

σBn . (1.148)



A meeting occurs if for some m

SAm = N − SBm , (1.149)

m− SAm = N −m+ SBm , (1.150)

i.e. if

SAm + SBm = N = 2m−N . (1.151)

Thus, a meeting is possible only after m = N steps, and it occurs ifSAm + SBm = N . Therefore, the probability is given by

pN =

2NN

22N=

(2N)!

(N !2N)2 . (1.152)

With the help of the Stirling’s formula (1.119) one finds that

limN→∞

pN =1√Nπ

. (1.153)

9. The following holds

p0 = 0 ,

p1 =5

6p0 +

1

6p2 ,

p2 =4

6p1 +

2

6p3 ,

p3 =3

6p2 +

3

6p4 ,

p4 =2

6p3 +

4

6p5 ,

p5 =1

6p4 +

5

6p6 ,

p6 = 1 ,

and thus

p1 =1

32, p2 =

3

16, p3 =

1

2, p4 =

13

16, p5 =

31

32.

10. For the case of a round table (and N > 2)

pC =N × 2× (N − 2)!

N !=

2

N − 1, (1.154)

and for the case of a row

pR =(2 + (N − 2)× 2)× (N − 2)!

N !=

2

N. (1.155)



11. Let the binary representation of p be given by

p =∞

m=1

σm

1

2

m, (1.156)

where σm ∈ 0, 1. Let Σm be a sequence of random variables generatedby the given computer function (i.e. Σm = 1 with probability 1/2 andΣm = 0 with probability 1/2). The proposed function has a while looprunning over integer values of the variablem starting from the valuem =1. At each iteration the random variable Σm is generated and comparedwith σm. If σm = Σm the value of m is increased by 1, i.e. m→ m+ 1,and the loop continues. If σm = Σm the program stops and the value 1 isreturned if σm > Σm and the value 0 is returned if σm < Σm. Note thatthe probability that the program will never stop vanishes even when p isirrational and/or the number of nonzero binary digits σm is infinite.

12. Using

N+ +N− = N , (1.157a)

N+ −N− =M

m, (1.157b)

one has

N+ =N

2

1 +

M

mN

, (1.158a)

N− =N

2

1− M

mN

, (1.158b)

or

N+ =N

2(1 + x) , (1.159a)

N− =N

2(1− x) , (1.159b)

where

x =M

mN.

The number of states having total magnetization M is given by

Ω (M) =N !

N+!N−!=

N !N2 (1 + x)

!N2 (1− x)

!. (1.160)

Since all states have equal probability one has

f (M) =Ω (M)

2N. (1.161)



Taking the natural logarithm of Stirling’s formula one finds

logN ! = N logN −N +O

1

N

, (1.162)

thus in the limit N ≫ 1 one has

log f = − log 2N +N logN −N

−N

2(1 + x)

log

N

2(1 + x)

+

N

2(1 + x)

−N

2(1− x)

log

N

2(1− x)

+

N

2(1− x)

= −N log 2 +N logN

−N

2(1 + x)

log

N

2(1 + x)

−N

2(1− x)

log

N

2(1− x)

=

−N

2

−2 log N

2+ (1 + x) log

N

2(1 + x)

+ (1− x) log

N

2(1− x)

=

−N

2

−2 log N

2+ (1 + x)

log

N

2+ log (1 + x)

+ (1− x)

log

N

2+ log (1− x)

=

−N

2

log1− x2

+ x log

1 + x

1− x

.

(1.163)

The function log f (x) has a sharp peak near x = 0, thus we can approx-imate it by assuming x≪ 1. To lowest order

log1− x2

+ x log

1 + x

1− x= x2 +O

x3, (1.164)

thus

f (M) = A exp

− M2

2m2N

, (1.165)

where A is a normalization constant, which is determined by requiringthat

1 =

∞

−∞

f (M) dM . (1.166)

Using the identity

∞

−∞

exp−ay2

dy =

π

a, (1.167)



one finds

1

A=

∞

−∞

exp

− M2

2m2N

dM =m

√2πN , (1.168)

thus

f (M) =1

m√2πN

exp

− M2

2m2N

, (1.169)

The expectation value is giving by

M =∞

−∞

Mf (M) dM = 0 , (1.170)

and the variance is given by

(M − M)2

= M2=

∞

−∞

M2f (M) dM =m2N . (1.171)

13. The probability to have n steps to the right is given by

W (n) =N !

n! (N − n)!pnqN−n . (1.172)

a)

n =N

n=0

N !n

n! (N − n)!pnqN−n (1.173)

= p∂

∂p

N

n=0

N !

n! (N − n)!pnqN−n

= p∂

∂p(p+ q)N = pN (p+ q)N−1 = pN .

Since

X = an− a (N − n) = a (2n−N) (1.174)

we find

X = aN (2p− 1) = aN (p− q) . (1.175)

b)



n2=

N

n=0

N !n2


=N

n=0

N !n (n− 1)

n! (N − n)!pnqN−n +

N

n=0

N !n


= p2∂2

∂p2

N

n=0

N !

n! (N − n)!pnqN−n + n

= p2∂2

∂p2(p+ q)

N+ n = p2N (N − 1) + pN .

Thus(n− n)2

= p2N (N − 1) + pN − p2N2 = Npq , (1.176)

and(X − X)2

= 4a2Npq . (1.177)

14. The total energy is given by

E =kx2

2+mx2

2=ka2

2, (1.178)

where a is the amplitude of oscillations. The time period T is given by

T = 2

a

−a

dx

x= 2

m

k

a

−a

dx√a2 − x2

= 2π

m

k, (1.179)

thus

f(x) =2

T |x| =1

π√a2 − x2

. (1.180)

15. The six experiments are independent, thus

σ = 6×−2

3ln

2

3− 1

3ln

1

3

= 3.8191 .

16. The random variable X obtains the value n with probability pn = qn,where n = 1, 2, 3, · · · , and q = 1/2.

a) The entropy is given by

σ = −∞

n=1

pn log2 pn = −∞

n=1

qn log2 qn =

∞

n=1

nqn .

This can be rewritten as

σ = q∂

∂q

∞

n=1

qn = q∂

∂q

1

1− q− 1

=

q

(1− q)2= 2 .



b) A series of questions is of the form: "Was a head obtained in the 1sttime ?", "Was a head obtained in the 2nd time ?", etc. The expectednumber of questioned required to find X is

1

2× 1 +

1

4× 2 +

1

8× 3 + · · · = 2 ,

which is exactly the entropy σ.

17.

a) Consider an infinitesimal change in the variable z = z (x, y)

δz =

∂z

∂x

y

δx+

∂z

∂y

x

δy . (1.181)

For a process for which z is a constant δz = 0, thus

0 =

∂z

∂x

y

(δx)z +

∂z

∂y

x

(δy)z . (1.182)

Dividing by (δx)z yields

∂z

∂x

y

= −∂z

∂y

x

(δy)z(δx)z

= −∂z

∂y

x

∂y

∂x

z

= −

∂y∂x

z∂y∂z

x

.

(1.183)

b) Consider a process for which the variable w is kept constant. Aninfinitesimal change in the variable z = z (x, y) is expressed as

(δz)w =

∂z

∂x

y

(δx)w +

∂z

∂y

x

(δy)w . (1.184)

Dividing by (δx)w yields

(δz)w(δx)w

=

∂z

∂x

y

+

∂z

∂y

x

(δy)w(δx)w

. (1.185)

or∂z

∂x

w

=

∂z

∂x

y

+

∂z

∂y

x

∂y

∂x

w

. (1.186)



18. We have found in class that

U = −∂ logZgc

∂β

η

, (1.187)

N = −∂ logZgc

∂η

β

. (1.188)

a) Using relation (1.125) one has

U = −∂ logZgc

∂β

η

= −∂ logZgc

∂β

µ

−∂ logZgc

∂µ

β

∂µ

∂β

η

= −∂ logZgc

∂β

µ

− η

β2

∂ logZgc

∂µ

β

= −∂ logZgc

∂β

µ

+ τµ

∂ logZgc

∂µ

β

.

(1.189)

b) Using Eq. (1.188) one has

N = −∂ logZgc

∂η

β

= −∂µ

∂η

β

∂ logZgc

∂µ

β

= τ

∂ logZgc

∂µ

β

,

(1.190)

or in terms of the fugacity λ, which is defined by

λ = exp (βµ) = e−η , (1.191)

one has

N = τ

∂ logZgc

∂µ

β

= τ∂λ

∂µ

∂ logZgc∂λ

= λ∂ logZgc

∂λ.

(1.192)



19. The canonical partition function is given by

Zc = ZN1 , (1.193)

where

Z1 = exp

βε

2

+ exp

−βε

2

= 2cosh

βε

2

. (1.194)

Thus


= −N ∂ logZ1∂β

= −Nε2

tanhβε

2, (1.195)

and

τ =ε

2 tanh−1−2U

Nε

. (1.196)

The negative temperature is originated by our assumption that the en-ergy of a single magnet has an upper bound. In reality this is never thecase.


Zc = ZN1 , (1.197)

where

Z1 = exp

−βℏω

2

∞

n=0

exp (−βℏωn) (1.198)

=exp

−βℏω

2

1− exp (−βℏω) =1

2 sinh βℏω2

.

a)


= −N ∂ logZ1∂β

=Nℏω

2coth

βℏω

2(1.199)

b)

(∆U)2

=∂2 logZc∂β2

= N∂2 logZ1∂β2

=Nℏω2

2

sinh 2 βℏω2

(1.200)


Zc = [exp (βε) + 1 + exp (−βε)]N = [1 + 2 cosh (βε)]N , (1.201)

where β = 1/τ .



a) Thus the average energy is


= −2Nε sinh (βε)

1 + 2 coshβε, (1.202)

b) and the variance is

(U − U)2

=∂2 logZc

∂β2= −∂ U

∂β= 2Nε2

cosh (βε) + 2

[1 + 2 cosh (βε)]2.

(1.203)

22. Each section can be in one of two possible sates with corresponding en-ergies 0 and −Fa.a) By definition, α is the mean length of each segment, which is given

by

α =a exp (Faβ)

1 + exp (Faβ)=a

2

1 + tanh

Faβ

2

, (1.204)

where β = 1/τ .b) At high temperature Faβ ≪ 1 the length of the chain L = Nα is

given by

L =Na

2

1 + tanh

Faβ

2

≃ Na

2

1 +

Faβ

2

, (1.205)

or

F = k

L− Na

2

, (1.206)

where the spring constant k is given by

k =4τ

Na2. (1.207)

23. The average length of a single link is given by

l =a exp (βFa)

∞

n=0

exp−βℏωa

n+ 1

2

+ b exp (βFb)

∞

n=0

exp−βℏωb

n+ 1

2

exp (βFa)∞

n=0

exp−βℏωa

n+ 1

2

+ exp (βFb)

∞

n=0

exp−βℏωb

n+ 1

2

(1.208)

=

a exp(βFa)

2 sinh βℏωa2

+ b exp(βFb)

2 sinhβℏωb2

exp(βFa)

2 sinh βℏωa2

+ exp(βFb)

2 sinhβℏωb2

.



To first order in β

l = aωb + bωaωb + ωa

+Fωbωa (a− b)

2

(ωb + ωa)2 β +O

β2. (1.209)

The average total length is L = n l .24. The length L is given by

L = N l ,

where l is the average contribution of a single molecule to the totallength, which can be either +d or −d. The probability of each possibilityis determined by the Boltzmann factor. The energy change due to flippingof one link from 0 to 180 is 2fd, therefore

l = deβfd − e−βfd

eβfd + e−βfd= d tanh (βfd) ,

where β = 1/τ . Thus

L = Nd tanh (βfd) ,

or

f =τ

dtanh−1

L

Nd.

25. The grand partition function Zgc is given by

Zgc = 1 + 2exp [β (µ− ε)] , (1.210)

thus

N = 1

β

∂ logZgc

∂µ=

2

2 + exp [β (ε− µ)]. (1.211)

26.

a)

(∆σ)2LS = logN !

(n2 − 1)! (n1 + 1)!− log

N !

n2!n1!

= logn2

n1 + 1≃ log

n2n1

.

(1.212)

b)

(∆σ)R =E2 −E1

τ(1.213)



c) For a small change near thermal equilibrium one expects (∆σ)2LS +(∆σ)R = 0, thus

n2n1

= exp

−E2 −E1

τ

. (1.214)

27. The number of ways to select NB occupied sites of type B out of N sitesis N !/n! (N − n)!. Similarly the number of ways to select NB empty sitesof type A out of N sites is N !/n! (N − n)!.

a) Thus

σ = log

N !

NB! (N −NB)!

2≃ 2 [N logN −NB logNB − (N −NB) log (N −NB)] .

(1.215)

b) The energy of the system is given by U = NBε. Thus, the Helmholtzfree energy is given by

F = U−τσ = U−2τ

N logN − U

εlog

U

ε−N − U

ε

log

N − U

ε

.

(1.216)

At thermal equilibrium (∂F/∂U)τ = 0, thus

0 =

∂F

∂U

τ

= 1 +2τ

ε

log

U

ε− log

N − U

ε

, (1.217)

or

N −NB

NB= exp

ε

2τ

, (1.218)

therefore

NB =N

1 + expε2τ

.

Alternatively, one can calculate the chemical potential from the re-quirement

1 =NA

N+NB

N, (1.219)

where

NA

N=

exp (βµ)

1 + exp (βµ), (1.220a)

NB

N=

exp (βµ− βε)

1 + exp (βµ− βε), (1.220b)



which is satisfied when

µ =ε

2, (1.221)

thus

NB =N

1 + expε2τ

. (1.222)

28. In general,

g (N,m) = # os ways to distribute m identical balls in N boxes

Moreover

# os ways to distribute m identical balls in N boxes= # os ways to arrange m identical balls and N − 1 identical partitions in a line

…

…

…

…a) Therefore

g (N,m) =(N − 1 +m)!

(N − 1)!m!. (1.223)

b) The entropy is given by

σ = log(N − 1 +m)!

(N − 1)!m!≃ (N +m) log (N +m)−N logN −m logm ,

(1.224)

or in terms of the total energy E = ℏω (m+N/2)

σ =

N +

E

ℏω− N

2

log

N +

E

ℏω− N

2

−N logN −E

ℏω− N

2

log

E

ℏω− N

2

.

(1.225)



c) The temperature τ is given by

1

τ=

∂σ

∂E

=1− ln 2ℏω

2E+Nℏω

ℏω+

ln 2ℏω2E−Nℏω − 1

ℏω

=1

ℏωln

2E +Nℏω

2E −Nℏω

.

(1.226)

In the thermodynamical limit (N ≫ 1, m≫ 1) the energy E and itsaverage U are indistinguishable, thus

exp

ℏω

τ

=

2U +Nℏω

2U −Nℏω, (1.227)

or

U =Nℏω

2coth

ℏω

2τ. (1.228)

29. The grand canonical partition function is given by

ζ = 1 + 2λ exp (βε) , (1.229)

where λ = exp (βµ) is the fugacity, thus

Nd = λ∂ log ζ

∂λ=

2λeβε

1 + 2λeβε=

1

1 + 12e−β(ε+µ) . (1.230)


2. Ideal Gas

In this chapter we study some basic properties of ideal gas of massive identicalparticles. We start by considering a single particle in a box. We then discussthe statistical properties of an ensemble of identical indistinguishable particlesand introduce the concepts of Fermions and Bosons. In the rest of this chapterwe mainly focus on the classical limit. For this case we derive expressions forthe pressure, heat capacity, energy and entropy and discuss how internaldegrees of freedom may modify these results. In the last part of this chapterwe discuss an example of an heat engine based on ideal gas (Carnot heatengine). We show that the efficiency of such a heat engine, which employs areversible process, obtains the largest possible value that is allowed by thesecond law of thermodynamics.

2.1 A Particle in a Box

Consider a particle having massM in a box. For simplicity the box is assumedto have a cube shape with a volume V = L3. The corresponding potentialenergy is given by

V (x, y, z) =

0 0 ≤ x, y, z ≤ L∞ else

. (2.1)

The quantum eigenstates and eigenenergies are determined by requiring thatthe wavefunction ψ (x, y, z) satisfies the Schrödinger equation

− 2

2M

∂2ψ

∂x2+∂2ψ

∂y2+∂2ψ

∂z2

+ V ψ = Eψ . (2.2)

In addition, we require that the wavefunction ψ vanishes on the surfaces ofthe box. The normalized solutions are given by

ψnx,ny,nz (x, y, z) =

2

L

3/2sin

nxπx

Lsin

nyπy

Lsin

nzπz

L, (2.3)

where

nx, ny, nz = 1, 2, 3, · · · (2.4)

Chapter 2. Ideal Gas

The corresponding eigenenergies are given by

εnx,ny,nz =2

2M

πL

2 n2x + n2y + n2z

. (2.5)

For simplicity we consider the case where the particle doesn’t have any in-ternal degree of freedom (such as spin). Later we will release this assumptionand generalize the results for particles having internal degrees of freedom.

The partition function is given by

Z1 =∞

nx=1

∞

ny=1

∞

nz=1

exp−εnx,ny,nz

τ

=∞

nx=1

∞

ny=1

∞

nz=1

exp−α2

n2x + n2y + n2z

,

(2.6)

where

α2 =2π2

2ML2τ. (2.7)

The following relation can be employed to estimate the dimensionless para-meter α

α2 =7.9× 10−17

Mmp

Lcm

2 τ300K

, (2.8)

where mp is the proton mass. As can be seen from the last result, it is oftenthe case that α2 ≪ 1. In this limit the sum can be approximated by anintegral

∞

nx=1

exp−α2n2x

≃

∞

0

exp−α2n2x

dnx . (2.9)

By changing the integration variable x = αnx one finds

∞

0

exp−α2n2x

dnx =

1

α

∞

0

exp−x2

dx =

√π

2α, (2.10)

thus

Z1 =

√π

2α

3=

ML2τ

2π2

3/2= nQV , (2.11)

where we have introduced the quantum density


2.1. A Particle in a Box

nQ =

Mτ

2π2

3/2. (2.12)

The partition function (2.11) together with Eq. (1.70) allows evaluatingthe average energy (recall that β = 1/τ)

ε = −∂ logZ1∂β

= −∂ log

ML2

2π2β

3/2

∂β

= −∂ log β−3/2

∂β

=3

2

∂ log β

∂β

=3τ

2.

(2.13)

This result can be written as

ε = dτ

2, (2.14)

where d = 3 is the number of degrees of freedom of the particle. As we will seelater, this is an example of the equipartition theorem of statistical mechanics.Similarly, the energy variance can be evaluated using Eq. (1.71)

(∆ε)

2=∂2 logZ1

∂β2

= −∂ ε∂β

= − ∂

∂β

3

2β

=3

2β2

=3τ2

2.

(2.15)

Thus, using Eq. (2.13) the standard deviation is given by

(∆ε)2

=

2

3ε . (2.16)

What is the physical meaning of the quantum density? The de Brogliewavelength λ of a particle having mass M and velocity v is given by



λ =h

Mv. (2.17)

For a particle having energy equals to the average energy ε = 3τ/2 one has

Mv2

2=

3τ

2, (2.18)

thus in this case the de-Broglie wavelength, which is denoted as λT (thethermal wavelength)

λT =h√3Mτ

, (2.19)

and therefore one has (recall that = h/2π)

nQ =

πh2

2Mτ

3/2

=

1

λT

2π

3

3

. (2.20)

Thus the quantum density is inversely proportional to the thermal wavelengthcubed.

2.2 Gibbs Paradox

In the previous section we have studied the case of a single particle. Let usnow consider the case where the box is occupied by N particles of the sametype. For simplicity, we consider the case where the density n = N/V issufficiently small to safely allowing to neglect any interaction between theparticles. In this case the gas is said to be ideal.

Definition 2.2.1. Ideal gas is an ensemble of non-interacting identical par-

ticles.

What is the partition function of the ideal gas? Recall that for the singleparticle case we have found that the partition function is given by [see Eq.(2.6)]

Z1 =

n

exp (−βεn) . (2.21)

In this expression Z1 is obtained by summing over all single particle orbitalstates, which are denoted by the vector of quantum numbers n = (nx, ny, nz).These states are called orbitals .

Since the total number of particles N is constrained we need to calculatethe canonical partition function. For the case of distinguishable particles onemay argue that the canonical partition function is given by


2.2. Gibbs Paradox

Zc ?= ZN

1 =

n

exp (−βεn)N

. (2.22)

However, as was demonstrated by Gibbs in his famous paradox, this answeris wrong. To see this, we employ Eqs. (1.72) and (2.11) and assume that thepartition function is given by Eq. (2.22) above, thus

σ − βU = logZc ?= logZN

1 = N log (nQV ) , (2.23)

or in terms of the gas density

n =N

V, (2.24)

we find

σ − βU?= N log

NnQn

. (2.25)

What is wrong with this result? It suggests that the quantity σ − βU is notsimply proportional to the size of the system. In other words, for a given nand a given nQ, σ−βU is not proportional to N . As we will see below, such abehavior may lead to a violation of the second law of thermodynamics. To seethis consider a box containing N identical particles having volume V . Whathappens when we divide the box into two sections by introducing a partition?Let the number of particles in the first (second) section be N1 (N2) whereasthe volume in the first (second) section be V1 (V2). The following hold

N = N1 +N2 , (2.26)

V = V1 + V2 . (2.27)

The density in each section is expected to be the same as the density in thebox before the partition was introduced

n =N

V=N1

V1=N2

V2. (2.28)

Now we use Eq. (2.25) to evaluate the change in entropy ∆σ due to theprocess of dividing the box. Since no energy is required to add (or to remove)the partition one has

∆σ = σtot − σ1 − σ2?= N log

NnQn

−N1 log

N1

nQn

−N2 log

N2

nQn

= N logN −N1 logN1 −N2 logN2 .

(2.29)

Using the Stirling’s formula (1.162)

logN ! ≃ N logN −N , (2.30)



…

orbital 1

N1=0

orbital 2

N2=1

orbital 3

N3=2

orbital 4

N4=0

3 1

2

…

orbital 1

N1=0

orbital 2

N2=1

orbital 3

N3=2

orbital 4

N4=0

33 11

22

Fig. 2.1. A figure describing an example term in the expansion (2.22) for a gascontaining N = 3 identical particles.

one finds

∆σ ≃ logN !

N1!N2!> 0 . (2.31)

Thus we came to the conclusion that the process of dividing the box leadsto reduction in the total entropy! This paradoxical result violates the secondlaw of thermodynamics. According to this law we expect no change in theentropy since the process of dividing the box is a reversible one.

What is wrong with the partition function given by Eq. (2.22)? Ex-panding this partition function yields a sum of terms each having the form

exp

−β

nNnεn

, where Nn is the number of particles occupying orbital n.

Let g (N1, N2, · · · ) be the number of terms in such an expansion associatedwith a given set of occupation numbers N1, N2, · · · . Since the partitionfunction (2.22) treats the particles as being distinguishable, g (N1, N2, · · · )may in general be larger than unity. In fact, it is easy to see that

g (N1,N2, · · · ) =N !

N1!N2!× · · ·. (2.32)

For example, consider the state that is described by Fig. 2.1 below for agas containing N = 3 particles. The expansion (2.22) contains 3!/1!/2! =3 terms having the same occupation numbers (Nn = 1 if n = 2,Nn = 2if n = 3, and Nn = 0 for all other values of n). However, for identicalparticles these 3 states are indistinguishable. Therefore, only a single termin the partition function should represent such a configuration. In general,the partition function should include a single term only for each given set ofoccupation numbers N1, N2, · · · .

2.3 Fermions and Bosons

As we saw in the previous section the canonical partition function given byEq. (2.22) is incorrect. For indistinguishable particles each set of orbital oc-


2.3. Fermions and Bosons

cupation numbers N1,N2, · · · should be counted only once. In this sectionwe take another approach and instead of evaluating the canonical partitionfunction of the system we consider the grandcanonical partition function.This is done by considering each orbital as a subsystem and by evaluatingits grandcanonical partition function, which we denote below as ζ. To do thiscorrectly, however, it is important to take into account the exclusion rulesimposes by quantum mechanics upon the possible values of the occupationnumbers Nn.

The particles in nature are divided into two type: Fermion and Bosons.While Fermions have half integer spin Bosons have integer spin. According toquantummechanics the orbital occupation numbersNn can take the followingvalues:

• For Fermions: Nn = 0 or 1• For Bosons: Nn can be any integer.

These rules are employed below to evaluate the grandcanonical partitionfunction of an orbital.

2.3.1 Fermi-Dirac Distribution

In this case the occupation number can take only two possible values: 0 or1. Thus, by Eq. (1.79) the grandcanonical partition function of an orbitalhaving energy is ε is given by

ζ = 1 + λ exp (−βε) , (2.33)

where

λ = exp (βµ) (2.34)

is the fugacity [see Eq. (1.95)]. The average occupation of the orbital, whichis denoted by fFD (ε) = N (ε), is found using Eq. (1.94)

fFD (ε) = λ∂ log ζ

∂λ

=λ exp (−βε)

1 + λ exp (−βε)

=1

exp [β (ε− µ)] + 1.

(2.35)

The function fFD (ε) is called the Fermi-Dirac function .



2.3.2 Bose-Einstein Distribution

In this case the occupation number can take any integer value. Thus, by Eq.(1.79) the grandcanonical partition function of an orbital having energy is εis given by

ζ =∞

N=0

λN exp (−Nβε)

=∞

N=0

[λ exp (−βε)]N

=1

1− λ exp (−βε) .

(2.36)

The average occupation of the orbital, which is denoted by fBE (ε) = N (ε),is found using Eq. (1.94)

fBE (ε) = λ∂ log ζ

∂λ

= λexp (−βε)

1− λ exp (−βε)

=1

exp [β (ε− µ)]− 1.

(2.37)

The function fBE (ε) is called the Bose-Einstein function .

2.3.3 Classical Limit

The classical limit occurs when

exp [β (ε− µ)]≫ 1 . (2.38)

As can be seen from Eqs. (2.35) and (2.37) the following holds

fFD (ε) ≃ fBE (ε) ≃ exp [β (µ− ε)]≪ 1 , (2.39)

and

ζ ≃ 1 + λ exp (−βε) . (2.40)

Thus the classical limit corresponds to the case where the average occupationof an orbital is close to zero, namely the orbital is on average almost empty.The main results of the above discussed cases (Fermi-Dirac distribution, Bose-Einstein distribution and the classical limit) are summarized in table 2.1below.


2.4. Ideal Gas in the Classical Limit

Table 2.1. Fermi-Dirac, Bose-Einstein and classical distributions.

orbital partition function average occupation

Fermions 1 + λ exp (−βε) 1exp[β(ε−µ)]+1

Bosons 11−λ exp(−βε)

1exp[β(ε−µ)]−1

classical limit 1 + λ exp (−βε) exp [β (µ− ε)]

0

0.5

1

1.5

2

-2 2 4 6 8 10 12 14 16 18 20energy

2.4 Ideal Gas in the Classical Limit

The rest of this chapter is devoted to the classical limit. The grandcanonicalpartition function ζn of orbital n having energy εn is given by Eq. (2.40)above. The grandcanonical partition function of the entire system Zgc isfound by multiplying ζn of all orbitals

Zgc =

n

(1 + λ exp (−βεn)) . (2.41)

Each term in the expansion of the above expression represents a set of or-bital occupation numbers, where each occupation number can take one of thepossible values: 0 or 1. We exploit the fact that in the classical limit

λ exp (−βε)≪ 1 (2.42)

and employ the first order expansion

log (1 + x) = x+Ox2

(2.43)

to obtain



logZgc =

n

log (1 + λ exp (−βεn))

≃ λ

n

exp (−βεn)

= λZ1 ,

(2.44)

where

Z1 = V

Mτ

2π2

3/2(2.45)

[see Eq. (2.11)] is the single particle partition function. In terms of the La-grange multipliers η = −µ/τ and β = 1/τ the last result can be rewrittenas

logZgc = e−ηV

M

2π2β

3/2. (2.46)

The average energy and average number of particle are calculated using Eqs.(1.80) and (1.81) respectively

U = −∂ logZgc

∂β

η

=3

2βlogZgc , (2.47)

N = −∂ logZgc

∂η

β

= logZgc . (2.48)

In what follows, to simplify the notation we remove the diagonal bracketsand denote U and N by U and N respectively. As was already pointedout earlier, probability distributions in statistical mechanics of macroscopicparameters are typically extremely sharp and narrow. Consequently, in manycases no distinction is made between a parameter and its expectation value.Using this simplified notation and employing Eqs. (2.47) and (2.48) one findsthat

U =3Nτ

2. (2.49)

Namely, the total energy is N ε, where ε is the average single particleenergy that is given by Eq. (2.13).

The entropy is evaluate using Eq. (1.86)

σ = logZgc + βU + ηN

= N

1 +

3

2− µ

τ

= N

5

2− µβ

.

(2.50)



Furthermore, using Eqs. (2.44), (2.48), (2.11) and (1.95) one finds that

µβ = logn

nQ, (2.51)

where n = N/V is the density. This allows expressing the entropy as

σ = N

5

2+ log

nQn

. (2.52)

Using the definition (1.116) and Eqs. (2.49) and (2.52) one finds that theHelmholtz free energy is given by

F = Nτ

log

n

nQ− 1

. (2.53)

2.4.1 Pressure

The pressure p is defined by

p = −∂F

∂V

τ,N

. (2.54)

Using Eq. (2.53) and keeping in mind that n = N/V one finds

p =Nτ

V. (2.55)

The pressure represents the force per unit area acting on the walls of the boxcontaining the gas due to collisions between the particles and the walls. Tosee that this is indeed the case consider a gas of N particles contained in abox having cube shape and volume V = L3. One of the walls is chosen tolie on the x = 0 plane. Consider and elastic collision between this wall anda particle having momentum p = (px, py, pz). After the collision py and pzremain unchanged, however, px becomes −px. Thus each collision results intransferring 2 |px| momentum in the x direction from the particle to the wall.The rate at which a particle collides with the wall x = 0 is |px| /2mL. Thusthe pressure acting on the wall due to a single particle is

pressure = forcearea

=

rate of

momentum change

area

=2 |px| × |px|

2mL

L2

=p2xmV

.

(2.56)



The average energy of a particle is given by Eq. (2.13)

3τ

2= ε =

p2x + p2y + p2z

2m, (2.57)

thus one finds that p2x

m= τ . (2.58)

Using this result and Eq. (2.56) one finds that the pressure due to a singleparticle is p = τ/V , thus the total pressure is

p =Nτ

V. (2.59)

2.4.2 Useful Relations

In this section we derive some useful relations between thermodynamicalquantities.

Claim. The following holds

p = −∂U

∂V

σ,N

Proof. Using the definition (2.54) and recalling that F = U − τσ one finds

−p =∂F

∂V

τ,N

=

∂U

∂V

τ,N

−∂ (τσ)

∂V

τ,N

. (2.60)

Using identity (1.125), which is given by

∂z

∂x

w

=

∂z

∂x

y

+

∂z

∂y

x

∂y

∂x

w

, (2.61)

one finds∂U

∂V

τ,N

=

∂U

∂V

σ,N

+

∂U

∂σ

V,N τ

∂σ

∂V

τ,N

, (2.62)

thus

−p =∂U

∂V

σ,N

+ τ

∂σ

∂V

τ,N

− τ

∂σ

∂V

τ,N

=

∂U

∂V

σ,N

. (2.63)



In a similar way the following relations can be obtained

p = τ

∂σ

∂V

U,N

= −∂U

∂V

σ,N

= −∂F

∂V

τ,N

, (2.64)

µ = −τ∂σ

∂N

U,V

=

∂U

∂N

σ,V

=

∂F

∂N

τ,V

. (2.65)

Another useful relation is given below.

Claim. The following holds∂σ

∂V

τ

=

∂p

∂τ

V

. (2.66)

Proof. See problem 2 of set 2.

2.4.3 Heat Capacity

The heat capacity at constant volume is defined by

cV = τ

∂σ

∂τ

V

, (2.67)

whereas the heat capacity at constant pressure is defined by

cp = τ

∂σ

∂τ

p

. (2.68)

Using Eq. (2.52) and recalling that nQ ∝ τ3/2 one finds

cV = τ3N

2τ=

3N

2. (2.69)

Claim. The following holds

cp =5N

2. (2.70)


2.4.4 Internal Degrees of Freedom

In this section we generalize our results for the case where the particles in thegas have internal degrees of freedom. Taking into account internal degrees offreedom the grandcanonical partition function of an orbital having orbitalenergy εn for the case of Fermions becomes

ζFD,n =

l

(1 + λ exp (−βεn) exp (−βEl)) , (2.71)



where El are the eigenenergies of a particle due to internal degrees offreedom, and where λ = exp (βµ) and β = 1/τ . As is required by the Pauliexclusion principle, no more than one Fermion can occupy a given internaleigenstate. Similarly, for the Bosonic case, where each state can be occupiedby any integer number of Bosons, one has

ζBE,n =

l

∞

m=0

λm exp (−βmεn) exp (−βmEl)

. (2.72)

In the classical limit the average occupation of an orbital is close to zero.In this limit, namely when

λ exp (−βεn)≪ 1 , (2.73)

[see Eq. (2.38)] the following holds

ζFD,n ≃ ζBE,n ≃ ζn , (2.74)

where

ζn = 1 + λ exp (−βεn)Zint , (2.75)

and where

Zint =

l

exp (−βEl) , (2.76)

is the internal partition function.Using Eq. (1.94) one finds that the average occupation of the orbital fn

in the classical limit is given by

fn = λ∂ log ζn∂λ

=λZint exp (−βεn)

1 + λZint exp (−βεn)≃ λZint exp (−βεn) .

(2.77)

The total grandcanonical partition function is given by

Zgc =

n

ζn , (2.78)

thus



logZgc =

n

log ζn

=

n

log [1 + λZint exp (−βεn)]

≃ λZint

n

exp (−βεn)

= λZintZ1 ,

(2.79)

where we have used the fact that in the classical limit λZint exp (−βεn)≪ 1.Furthermore, using Eq. (2.11) and recalling that η = −µ/τ one finds

logZgc = e−ηZintV

M

2π2β

3/2. (2.80)

This result together with Eqs. (1.80) and (1.81) yield

U = −∂ logZgc

∂β

η

=3

2βlogZgc + El logZgc , (2.81)

N = −∂ logZgc

∂η

β

= logZgc , (2.82)

where

El =

l

El exp (−βEl)

l

exp (−βEl)= −∂ logZint

∂β. (2.83)

Claim. The following hold

µ = τ

log

n

nQ− logZint

, (2.84)

U = N

3τ

2− ∂ logZint

∂β

, (2.85)

F = Nτ

log

n

nQ− logZint − 1

, (2.86)

σ = N

5

2+ log

nQn

+∂ (τ logZint)

∂τ

, (2.87)

cV = N

3

2+ τ

∂2 (τ logZint)

∂τ2

, (2.88)

cp = cV +N . (2.89)




2.5 Processes in Ideal Gas

The state of an ideal gas is characterized by extensive parameters (by def-inition, parameters that are proportional to the system size) such as U , V ,N and σ and by intensive parameters (parameters that are independent onthe system size) such as τ , µ and p. In this section we discuss some exam-ples of processes that occur by externally changing some of these parameters.We will use these processes in the next section to demonstrate how one canconstruct a heat engine based on an ideal gas.

In general, the entropy is commonly expresses as a function of the energy,volume and number of particles σ = σ (U, V,N). A small change in σ isexpressed in terms of the partial derivatives

dσ =

∂σ

∂U

V,N

dU +

∂σ

∂V

U,N

dV +

∂σ

∂N

U,V

dN . (2.90)

Using Eqs. (1.87), (2.64) and (2.65)

dσ =1

τdU +

p

τdV − µ

τdN , (2.91)

or

dU = τdσ − pdV + µdN . (2.92)

This relation expresses the change in the energy of the system dU in termsof

dQ = τdσ heat added to the systemdW = pdV work done by the system

µdN energy change due to added particles

For processes that keep the number of particles unchanged

dN = 0 ,

one has

dU = dQ− dW . (2.93)

Integrating this relation for the general case (not necessarily an infinitesimalprocess) yields

∆U = Q−W , (2.94)

We discuss below some specific examples for processes for which dN = 0.The initial values of the pressure, volume and temperature are denoted asp1, V1 and τ1 respectively, whereas the final values are denoted as p2, V2and τ2 respectively. In all these processes we assume that the gas remains in


2.5. Processes in Ideal Gas

V

pisobaric

isothermalisentropic

iso

cho

ric

V

pisobaric

isothermalisentropic

iso

cho

ric

Fig. 2.2. Four processes for which dN = 0.

thermal equilibrium throughout the entire process. This can be achieved byvarying the external parameters at a rate that is sufficiently slow to allow thesystem to remain very close to thermal equilibrium at any moment duringthe process. The four example to be analyzed below are (see fig. 2.2):

• Isothermal process - temperature is constant• Isobaric process - pressure is constant• Isochoric process - volume is constant• Isentropic process - entropy is constant

Note that in general, using the definition of the heat capacity at constantvolume given by Eq. (2.67) together with Eq. (1.87) one finds that

cV =

∂U

∂τ

N,V

. (2.95)

Furthermore, as can be seen from Eq. (2.85), the energy U of an ideal gasin the classical limit is independent on the volume V (it can be expressed asa function of τ and N only). Thus we conclude that for processes for whichdN = 0 the change in energy dU can be expressed as

dU = cV dτ . (2.96)



2.5.1 Isothermal Process

Since τ is constant one finds using Eq. (2.96) that ∆U = 0. Integrating therelation dW = pdV and using Eq. (2.55) one finds

Q = W =

V2

V1

pdV

= Nτ

V2

V1

dV

V

= Nτ logV2V1

.

(2.97)

2.5.2 Isobaric Process

Integrating the relation dW = pdV for this case where the pressure is con-stant yields

W =

V2

V1

pdV = p (V2 − V1) . (2.98)

The change in energy ∆U can be found by integrating Eq. (2.96)

∆U =

τ2

τ1

cV dτ . (2.99)

The heat added to the system Q can be found using Eq. (2.94)

Q = W +∆U

= p (V2 − V1) +

τ2

τ1

cV dτ .

(2.100)

Note that if the temperature dependence of cV can be ignored to a goodapproximation one has

∆U = cV (τ2 − τ1) . (2.101)


2.5. Processes in Ideal Gas

2.5.3 Isochoric Process

In this case the volume is constant, thus W = 0. By integrating Eq. (2.96)one finds that

Q = ∆U =

τ2

τ1

cV dτ . (2.102)

Also in this case, if the temperature dependence of cV can be ignored to agood approximation one has

Q = ∆U = cV (τ2 − τ1) .

2.5.4 Isentropic Process

In this case the entropy is constant, thus dQ = τdσ = 0, and thereforedU = −dW , thus using the relation dW = pdV and Eq. (2.96) one has

cV dτ = −pdV , (2.103)

or using Eq. (2.55)

cVdτ

τ= −N dV

V. (2.104)

This relation can be rewritten using Eq. (2.89) as

dτ

τ= (1− γ)

dV

V. (2.105)

where

γ =cpcV

. (2.106)

The last result can be easily integrated if the temperature dependence of thefactor γ can be ignored to a good approximation. For that case one has

logτ2τ1

= log

V2V1

1−γ. (2.107)

Thus

τ1Vγ−11 = τ2V

γ−12 , (2.108)

or using Eq. (2.55)

p1Vγ1 = p2V

γ2 . (2.109)



τh τl

Qh Ql

W

τh τl

Qh Ql

W

Fig. 2.3. Carnot heat engine.

In other words both quantities τV γ−1 and pV γ remain unchanged duringthis process. Using the last result allows integrating the relation dW = pdV

−∆U = W =

V2

V1

pdV

= p1Vγ1

V2

V1

V −γdV

= p1Vγ1

V −γ+11 − V −γ+12

γ − 1

=p2V2 − p1V1

1− γ

=N (τ2 − τ1)

1− γ

= −cV (τ2 − τ1) .

(2.110)

2.6 Carnot Heat Engine

In this section we discuss an example of a heat engine proposed by Carnotthat is based on an ideal classical gas. Each cycle is made of four steps (seeFigs. 2.3 and 2.4)

1. Isothermal expansion at temperature τh (a→ b)2. Isentropic expansion from temperature τh to τ l (b→ c)3. Isothermal compression at temperature τ l (c→ d)4. Isentropic compression from temperature τ l to τh (d→ a)

All four steps are assumed to be sufficiently slow to maintain the gas inthermal equilibrium throughout the entire cycle. The engine exchanges heat


2.6. Carnot Heat Engine

V

p

a

b

cd

Va VbVd Vc

pa

pb

pd

pc

τh

τl V

p

a

b

cd

Va VbVd Vc

pa

pb

pd

pc

τh

τl

Fig. 2.4. A cycle of Carnot heat engine.

with the environment during both isothermal processes. Using Eq. (2.97) onefinds that the heat extracted from the hot reservoir Qh at temperature τhduring step 1 (a→ b) is given by

Qh = Nτh logVbVa

, (2.111)

and the heat extracted from the cold thermal reservoir Ql at temperature τ lduring step 3 (c→ d) is given by

Ql = Nτ l logVdVc

, (2.112)

where Vn is the volume at point n ∈ a, b, c, d. Note that Qh > 0 sincethe system undergoes expansion in step 1 whereas Ql < 0 since the systemundergoes compression during step 3. Both thermal reservoirs are assumedto be very large systems that can exchange heat with the engine withoutchanging their temperature. No heat is exchanged during the isentropic steps2 and 4 (since dQ = τdσ).

The total work done by the system per cycle is given by

W = Wab +Wcd +Wbc +Wda

= Nτh logVbVa

+Nτ l logVdVc

+N (τ l − τh)

1− γ+N (τh − τ l)

1− γ

= N

τh log

VbVa

+ τ l logVdVc

,

(2.113)



where the work in both isothermal processesWab andWcd is calculated usingEq. (2.97), whereas the work in both isentropic processes Wbc and Wda iscalculate using Eq. (2.110). Note that the following holds

W = Qh +Ql . (2.114)

This is expected in view of Eq. (2.94) since the gas returns after a full cycleto its initial state and therefore ∆U = 0.

The efficiency of the heat engine is defined as the ratio between the workdone by the system and the heat extracted from the hot reservoir per cycle

η =W

Qh= 1 +

Ql

Qh. (2.115)

Using Eqs. (2.111) and (2.113) one finds

η = 1 +τ l log

VdVc

τh logVbVa

. (2.116)

Employing Eq. (2.108) for both isentropic processes yields

τhVγ−1b = τ lV

γ−1c , (2.117)

τhVγ−1a = τ lV

γ−1d , (2.118)

thus by dividing these equations one finds

V γ−1b

V γ−1a

=V γ−1c

V γ−1d

, (2.119)

or

VbVa

=VcVd

. (2.120)

Using this result one finds that the efficiency of Carnot heat engine ηC isgiven by

ηC = 1− τ lτh

. (2.121)

2.7 Limits Imposed Upon the Efficiency

Is it possible to construct a heat engine that operates between the same heatreservoirs at temperatures τh and τ l that will have efficiency larger thanthe value given by Eq. (2.121)? As we will see below the answer is no. Thisconclusion is obtained by noticing that the total entropy remains unchangedin each of the four steps that constructs the Carnot’s cycle. Consequently,


2.7. Limits Imposed Upon the Efficiency

τ

Q

W

Mτ

Q

W

M

Fig. 2.5. An idle heat engine (Perpetuum Mobile of the second kind).

the entire process is reversible, namely, by varying the external parametersin the opposite direction, the process can be reversed.

We consider below a general model of a heat engine. In a continuos oper-ation the heat engine repeats a basic cycle one after another. We make thefollowing assumptions:

• At the end of each cycle the heat engine returns to the same macroscopicstate that it was in initially (otherwise, continuous operation is impossible).

• The workW done per cycle by the heat engine does not change the entropyof the environment (this is the case when, for example, the work is usedto lift a weight - a process that only changes the center of mass of theweight, and therefore causes no entropy change).

Figure (2.5) shows an ideal heat engine that fully transforms the heat Qextracted from a thermal reservoir into work W , namely Q = W . Such anidle engine has a unity efficiency η = 1. Is it possible to realized such anidle engine? Such a process does not violate the law of energy conservation(first law of thermodynamics). However, as we will see below it violates thesecond law of thermodynamics. Note also that the opposite process, namelya process that transforms work into heat without losses is possible, as can beseen from the example seen in Fig. (2.6). In this system the weigh normallygoes down and consequently the blender rotates and heats the liquid in thecontainer. In principle, the opposite process at which the weigh goes up andthe liquid cools down doesn’t violate the law of energy conservation, however,it violates the second law (Perpetuum Mobile of the second kind), as we willsee below.

To show that the idle heat engine shown in Fig. (2.5) can not be realizedwe employ the second law and require that the total change in entropy ∆σper cycle is non-negative

∆σ ≥ 0 . (2.122)

The only change in entropy per cycle is due to the heat that is subtractedfrom the heat bath



Fig. 2.6. Transforming work into heat.

∆σ = −Qτ, (2.123)

thus since Q =W (energy conservation) we find that

W

τ≤ 0 . (2.124)

Namely, the work done by the heat engine is non-positive W ≤ 0. This resultis known as Kelvin’s principle.

Kelvin’s principle: In a cycle process, it is impossible to extract heatfrom a heat reservoir and convert it all into work.

As we will see below, Kelvin’s principle is equivalent to Clausius’s principlethat states:

Clausius’s principle: It is impossible that at the end of a cycle process,heat has been transferred from a colder to a hotter thermal reservoirs withoutapplying any work in the process.

A refrigerator and an air conditioner (in cooling mode) are examples ofsystems that transfer heat from a colder to a hotter thermal reservoirs. Ac-cording to Clausius’s principle such systems require that work is consumedfor their operation.

Theorem 2.7.1. Kelvin’s principle is equivalent to Clausius’s principle.

Proof. Assume that Clausius’s principle does not hold. Thus the systemshown in Fig. 2.7(a) that transfers heat Q0 > 0 from a cold thermal reser-voir at temperature τ l to a hotter one at temperature τh > τ l is possible.In Fig. 2.7(b) a heat engine is added that extracts heat Q > Q0 from thehot thermal reservoir, delivers heat Q0 to the cold one, and performs workW = Q −Q0. The combination of both systems extracts heat Q−Q0 fromthe hot thermal reservoir and converts it all into work, in contradiction withKelvin’s principle.

Assume that Kelvin’s principle does not hold. Thus the system shown inFig. 2.8(a) that extracts heat Q0 from a thermal reservoir at temperature τhand converts it all into work is possible. In Fig. 2.8(b) a refrigerator is added


2.7. Limits Imposed Upon the Efficiency

τh τl

-Q0 Q0

M

(a)

τh τl

Q -Q0

W=Q-Q0

M

-Q0 Q0

M

(b)

τh τl

-Q0 Q0

M

(a)

τh τl

Q -Q0

W=Q-Q0

M

-Q0 Q0

M

(b)

Fig. 2.7. The assumption that Clausius’s principle does not hold.

that employs the work W = Q0 to remove heat Q from a colder thermalreservoir at temperature τ l < τh and to deliver heat Q0 + Q to the hotthermal reservoir. The combination of both systems transfers heat Q froma colder to a hotter thermal reservoirs without consuming any work in theprocess, in contradiction with Clausius’s principle.

As we have seen unity efficiency is impossible. What is the largest possibleefficiency of an heat engine?

Theorem 2.7.2. The efficiency η of a heat engine operating between a hotterand colder heat reservoirs at temperature τh and τ l respectively can not exceedthe value

ηC = 1− τ lτh

. (2.125)

Proof. A heat engine (labeled as ’I’) is seen in Fig. 2.9. A Carnot heat engineoperated in the reverse direction (labeled as ’C’) is added. Here we exploitthe fact the Carnot’s cycle is reversible. The efficiency η of the heat engine’I’ is given by

ηI =W

Q′h, (2.126)

whereas the efficiency of the reversed Carnot heat engine ’C’ is given by Eq.(2.121)



τh τl

Q0

M

(a)

W= Q0

τh τl

Q0

M

(b)

W= Q0

Q

M-Q0- Q

τh τl

Q0

M

(a)

W= Q0

τh τl

Q0

M

(b)

W= Q0

Q

M-Q0- Q

Fig. 2.8. The assumption that Kelvin’s principle does not hold.

τh τl

Qh

C

W

Q’l

IQ’h

Ql

τh τl

Qh

C

W

Q’l

IQ’h

Ql

Fig. 2.9. Limit imposed upon engine efficiency.

ηC =W

Qh= 1− τ l

τh. (2.127)

For the combined system, the Clausius’s principle requires that

Q′h −Qh > 0 . (2.128)

thus

ηI ≤ ηC = 1− τ lτh

.

The same argument that was employed in the proof above can be used todeduce the following corollary:


2.8. Problems Set 2

Corollary 2.7.1. All reversible heat engines operating between a hotter heat

reservoir and a colder one at temperatures τh and τ l respectively have thesame efficiency.

Note that a similar bound is imposed upon the efficiency of refrigerators(see problem 16 of set 2).

2.8 Problems Set 2

1. The heat capacity at constant pressure is defined as

Cp = τ

∂σ

∂τ

p

. (2.129)

Calculate Cp of an classical ideal gas having no internal degrees of free-dom.

2. Show that∂σ

∂V

τ

=

∂p

∂τ

V

, (2.130)

where σ is entropy, V is volume, and p is pressure.3. Consider a classical ideal gas having internal partition function Zint.

a) Show that the chemical potential µ is given by

µ = τ

log

n

nQ− logZint

, (2.131)

where τ is the temperature, n = N/V , V is the volume, and nQ isthe quantum density.

b) Show that the energy U is related to the number of particles by thefollowing relation:

U = N

3τ

2− ∂ logZint

∂β

, (2.132)

where β = bl/τ .c) Show that the Helmholtz free energy is given by

F = Nτ

log

n

nQ− logZint − 1

. (2.133)

d) Show that the entropy is given by

σ = N

5

2+ log

nQn

+∂ (τ logZint)

∂τ

. (2.134)



e) Show that the heat capacity at constant volume is given by

cV = N

3

2+ τ

∂2 (τ logZint)

∂τ2

. (2.135)

f) Show that the heat capacity at constant pressure is given by

cp = cV +N . (2.136)

4. The heat capacity c of a body having entropy σ is given by

c = τ∂σ

∂τ, (2.137)

where τ is the temperature. Show that

c =

(∆U)2

τ2, (2.138)

where U is the energy of the body and where ∆U = U − U.5. Consider an ideal classical gas made of diatomic molecules. The internal

vibrational degree of freedom is described using a model of a one dimen-sional harmonic oscillator with angular frequency ω. That is, the eigenenergies associated with the internal degree of freedom are given by

εn =

n+

1

2

ω , (2.139)

where n = 0, 1, 2, · · · . The system is in thermal equilibrium at tempera-ture τ , which is assumed to be much larger than ω. Calculate the heatcapacities cV and cp.

6. A thermally isolated container is divided into two chambers, the firstcontaining NA particles of classical ideal gas of type A, and the secondone contains NB particles of classical ideal gas of type B. Both gases haveno internal degrees of freedom. The volume of first chamber is VA, andthe volume of the second one is VB. Both gases are initially in thermalequilibrium at temperature τ . An opening is made in the wall separatingthe two chambers, allowing thus mixing of the two gases. Calculate thechange in entropy during the process of mixing.

7. Consider an ideal gas of N molecules in a vessel of volume V . Showthat the probability pn to find n molecules in a small volume v (namely,v << V ) contained in the vessel is given by

pn =λn

n!e−λ , (2.140)

where λ = Nv/V .


2.8. Problems Set 2

8. A lattice contains N sites, each is occupied by a single atom. The set ofeigenstates of each atom, when a magnetic field H is applied, contains2 states having energies ε− = −µ0H and ε+ = µ0H, where the mag-netic moment µ0 is a constant. The system is in thermal equilibrium attemperature τ .

a) Calculate the magnetization of the system, which is defined by

M = −∂F

∂H

τ

, (2.141)

where F is the Helmholtz free energy.b) Calculate the heat capacity

C = τ

∂σ

∂τ

H

. (2.142)

where σ is the entropy of the system.c) Consider the case where initially the magnetic field is H1 and the

temperature is τ1. The magnetic field is then varied slowly in anisentropic process from H1 to H2. Calculate the final temperature ofthe system τ2.

9. A lattice contains N sites, each occupied by a single atom. The set ofeigenstates of each atom, when a magnetic field H is applied, contains 3states with energies

ε−1 = −∆− µ0H ,

ε0 = 0 ,

ε1 = −∆+ µ0H ,

where the magnetic moment µ0 is a constant. The system is in thermalequilibrium at temperature τ . Calculate the magnetic susceptibility

χ = limH→0

M

H, (2.143)

where

M = −∂F

∂H

τ

, (2.144)

is the magnetization of the system, and where F is the Helmholtz freeenergy.

10. A lattice contains N sites, each occupied by a single atom. The setof eigenstates of each atom, when a magnetic field H is applied, con-tains 2J + 1 states with energies εm = −mµH, where J is integer,m = −J,−J + 1, · · ·J − 1, J , and the magnetic moment µ is a constant.The system is in thermal equilibrium at temperature τ .



a) Calculate the free energy F of the system.b) Show that the average magnetization, which is defined as

M = −∂F

∂H

τ

, (2.145)

is given by

M =Nµ

2

(2J + 1) coth

(2J + 1)

µH

2τ

− coth

µH

2τ

. (2.146)

11. Assume the earth’s atmosphere is pure nitrogen in thermodynamic equi-librium at a temperature of 300 K. Calculate the height above sea levelat which the density of the atmosphere is one-half its sea-level value(answer: 12.6 km).

12. Consider a box containing an ideal classical gas made of atoms of massMhaving no internal degrees of freedom at pressure p and temperature τ .The walls of the box have N0 absorbing sites, each of which can absorb 0,1, or 2 atoms of the gas. The energy of an unoccupied site and the energyof a site occupying one atom is zero. The energy of a site occupying twoatoms is ε. Show that the mean number of absorbed atoms is given by

Na = N0λ+ 2λ2e−βε

1 + λ+ λ2e−βε, (2.147)

where β = 1/τ and

λ =

M

2πℏ2

−3/2τ−5/2p . (2.148)

13. An ideal gas containing N atoms is in equilibrium at temperature τ .The internal degrees of freedom have two energy levels, the first onehas energy zero and degeneracy g1, and the second one energy ε anddegeneracy g2. Show that the heat capacities at constant volume and atconstant pressure are given by

cV = N

3

2+ ετ

2 g1g2 exp− ε

τ

g1 + g2 exp

− ε

τ

2

!

, (2.149)

cp = N

5

2+ ετ

2 g1g2 exp− ε

τ

g1 + g2 exp

− ε

τ

2

!

. (2.150)

14. A classical gas is described by the following equation of state

p (V − b) = Nτ , (2.151)

where p is the pressure, V is the volume, τ is the temperature, N is thenumber of particles and b is a constant.


2.8. Problems Set 2

a) Calculate the difference cp − cV between the heat capacities at con-stant pressure and at constant volume.

b) Consider an isentropic expansion of the gas from volume V1 andtemperature τ1 to volume V2 and temperature τ2. The number ofparticles N is kept constant. Assume that cV is independent on tem-perature. Calculate the work W done by the gas during this process.


p+

a

V 2

(V − b) = Nτ , (2.152)

where p is the pressure, V is the volume, τ is the temperature, and aand b are constants. Calculate the difference cp − cV between the heatcapacities at constant pressure and at constant volume.


p+

a

V 2

(V − b) = Nτ , (2.153)

where p is the pressure, V is the volume, τ is the temperature, and a andb are constants. The gas undergoes a reversible isothermal expansion at afixed temperature τ0 from volume V1 to volume V2. Show that the workW done by the gas in this process, and the heat Q which is supplied tothe gas during this process are given by

W = Nτ0 logV2 − b

V1 − b− a

V2 − V1V2V1

, (2.154)

Q = ∆U +W = Nτ0 logV2 − b

V1 − b. (2.155)

17. The energy of a classical ideal gas having no internal degrees of freedomis denoted as E, the deviation from the average value U = E as ∆E =E − U . The gas, which contains N particles and has volume is V , is inthermal equilibrium at temperature τ .

a) Calculate(∆E)2

.

b) Calculate(∆E)3

.

18. A body having a constant heat capacity C and a temperature τa is putinto contact with a thermal bath at temperature τb. Show that the totalchange in entropy after equilibrium is establishes is given by

∆σ = C

τaτ b− 1− log

τaτb

. (2.156)

Use this result to show that ∆σ ≥ 0.19. The figure below shows a cycle of an engine made of an ideal gas. In

the first step a → b the volume is constant V2, the second one b → c is



V

p

1p

2pa

b

c

2V 1VV

p

1p

2pa

b

c

2V 1V

Fig. 2.10.

an isentropic process, and in the third one the pressure is constant p2.Assume that the heat capacities CV and Cp are temperature independent.Show that the efficiency of this engine is given by

η = 1− γp2 (V1 − V2)

V2 (p1 − p2), (2.157)

where γ = Cp/Cv.20. Consider a refrigerator consuming workW per cycle to extract heat from

a cold thermal bath at temperature τ l to another thermal bath at highertemperature τh. LetQl be the heat extracted from the cold bath per cycleand −Qh the heat delivered to the hot one per cycle. The coefficient ofrefrigerator performance is defined as

γ =Ql

W. (2.158)

Show that the second law of thermodynamics imposes an upper boundon γ

γ ≤ τ lτh − τ l

. (2.159)

21. A room air conditioner operates as a Carnot cycle refrigerator betweenan outside temperature τh and a room at a lower temperature τ l. Theroom gains heat from the outdoors at a rate A (τh − τ l); this heat isremoved by the air conditioner. The power supplied to the cooling unitis P . Calculate the steady state temperature of the room.

22. The state equation of a given matter is


2.8. Problems Set 2

p =Aτ3

V, (2.160)

where p, V and τ are the pressure, volume and temperature, respectively,A is a constant. The internal energy of the matter is written as

U = Bτn logV

V0+ f (τ) , (2.161)

where B and V0 are constants, f (τ) only depends on the temperature.Find B and n.

23. An ideal classical gas is made of N identical molecules each having massM . The volume of the gas is V and the temperature is τ . The energyspectrum due to internal degrees of freedom of each molecule has a groundstate, which is nondegenerate state (singlet state), and a first excitedenergy state, which has degeneracy 3 (triplet state). The energy gapbetween the ground state and the first excited state is ∆ and all otherstates have a much higher energy. Calculate:

a) the heat capacity at constant volume CV .b) the heat capacity at constant pressure Cp.

24. Two identical bodies have internal energy U = Cτ , with a constant heatcapacity C. The initial temperature of the first body is τ1 and that of thesecond one is τ2. The two bodies are used to produce work by connectingthem to a reversible heat engine and bringing them to a common finaltemperature τf .

a) Calculate τf .b) Calculate the total work W , which is delivered by the process.

25. An ideal classical gas having no internal degrees of freedom is containedin a vessel having two parts separated by a partition. Each part containsthe same number of molecules, however, while the pressure in the firstone is p1, the pressure in the second one is p2. The system is initially inthermal equilibrium at temperature τ . Calculate the change of entropycaused by a fast removal of the partition.

26. A classical ideal gas contains N particles having massM and no internaldegrees of freedom is in a vessel of volume V at temperature τ . Expressthe canonical partition function Zc as a function of N , M , V and τ .

27. Consider an engine working in a reversible cycle and using an ideal clas-sical gas as the working substance. The cycle consists of two processes atconstant pressure (a→ b and c→ d), joined by two isentropic processes(b→ c and d→ a), as show in Fig. 2.11. Assume that the heat capacitiesCV and Cp are temperature independent. Calculate the efficiency of thisengine.

28. Consider an engine working in a cycle and using an ideal classical gasas the working substance. The cycle consists of two isochoric processes(constant volume) a→b at volume V1 and c→d at volume V2, joined by



V

p

1p

2p

a b

d c

V

p

1p

2p

a b

d c

Fig. 2.11.

V

p

V1 V2

a

b

c

d V

p

V1 V2

a

b

c

d

Fig. 2.12.

two isentropic processes (constant entropy) b→c and d→a, as shown inFig. 2.12. Assume that the heat capacities CV and Cp are temperatureindependent. Calculate the efficiency η of this engine.



29. Consider two vessels A and B each containing ideal classical gas of par-ticles having no internal degrees of freedom . The pressure and numberof particles in both vessels are p and N respectively, and the tempera-ture is τA in vessel A and τB in vessel B. The two vessels are broughtinto thermal contact. No heat is exchanged with the environment duringthis process. Moreover, the pressure is kept constant at the value p inboth vessels during this process. Find the change in the total entropy∆σ = σfinal − σinitial.

2.9 Solutions Set 2

1. The entropy is given by

σ = N

log

"Mτ

2πℏ2

3/2V

N

#

+5

2

!

, (2.162)

or using pV = Nτ

σ = N

log

"M

2πℏ2

3/2τ5/2

p

#

+5

2

!

, (2.163)

thus

Cp = τ

∂σ

∂τ

p

=5

2N . (2.164)

2. Since

∂2F

∂V ∂τ=

∂2F

∂τ∂V, (2.165)

where F is Helmholtz free energy, one has

∂

∂V

∂F

∂τ

V

τ

=

∂

∂τ

∂F

∂V

τ

V

. (2.166)

By definition∂F

∂V

τ

= −p .

Moreover, using F = U − τσ one finds∂F

∂τ

V

=

∂U

∂τ

V

− τ

∂σ

∂τ

V

− σ

=

∂U

∂τ

V

−∂U

∂σ

V

∂σ

∂τ

V

− σ

= −σ ,(2.167)



thus∂σ

∂V

τ

=

∂p

∂τ

V

. (2.168)

3. We have found in class the following relations

nQ =

Mτ

2π2

3/2, (2.169)

η = −µτ, (2.170)

logZgc = e−ηZintV nQ , (2.171)

U =

3τ

2− ∂ logZint

∂β

logZgc , (2.172)

N = logZgc . (2.173)

a) Using Eqs. (2.171) and (2.173) one finds

logn

nQZint=µ

τ, (2.174)

thus

µ = τ

log

n

nQ− logZint

. (2.175)

b) Using Eqs. (2.172) and (2.173)

U = N

3τ

2− ∂ logZint

∂β

. (2.176)

c) Using the relations

F = U − τσ , (2.177)

σ = logZgc + βU + ηN , (2.178)

one finds

F = U − τσ (2.179)

= Nτ (−η − 1) (2.180)

= Nτµτ− 1

(2.181)

= Nτ

log

n

nQ− logZint − 1

. (2.182)

(2.183)



d) Using the relation

σ = −∂F

∂τ

V

, (2.184)

one finds

σ = −∂F

∂τ

V

= N

−

∂τ log n

nQ

∂τ

V

+∂ (τ logZint)

∂τ+ 1

= N

−τ

∂log n

nQ

∂τ

V

− logn

nQ+∂ (τ logZint)

∂τ+ 1

= N

3

2− log

n

nQ+∂ (τ logZint)

∂τ+ 1

= N

5

2+ log

nQn

+∂ (τ logZint)

∂τ

.

(2.185)

e) By definition

cV = τ

∂σ

∂τ

V

= N

3

2+ τ

∂2 (τ logZint)

∂τ2

.

(2.186)

f) The following holds

cp = τ

∂σ

∂τ

p

(2.187)

= τ

∂σ

∂τ

V

+ τ

∂σ

∂V

τ

∂V

∂τ

p

= cV + τ

∂σ

∂V

τ

∂V

∂τ

p

.

Using V p = Nτ and Eq. (2.185) one finds

cp = cV + τN

V

N

p= cV +N . (2.188)



4. With the help of Eqs. (1.87), (1.70) and (1.71) together with the followingrelation

∂

∂τ= − 1

τ2∂

∂β, (2.189)

one finds that

c = τ∂σ

∂τ=∂U

∂τ= − 1

τ2∂U

∂β=

(∆U)2

τ2. (2.190)

5. The internal partition function is given by

Zint =1

2 sinh ω2τ

≃ τ

ω, (2.191)

thus using Eqs. (2.186) and (2.188)

cV = N

3

2+ τ

∂2τ log τ

ω

∂τ2

=5N

2, (2.192)

cp =7N

2. (2.193)

6. Energy conservation requires that the temperature of the mixture willremain τ . The entropy of an ideal gas of density n, which contains Nparticles, is given by

σ (N,n) = N

log

nQn

+5

2

thus the change in entropy is given by

∆σ = σmix − σA − σB

= σ

NA,

NA

VA + VB

+ σ

NB,

NB

VA + VB

− σ

NA,

NA

VA

− σ

NB,

NB

VB

= NA logVA + VB

VA+NB log

VA + VBVB

7. The probability to find a molecule in the volume v is given by p = v/V .Thus, pn is given by

pn =N !

n! (N − n)!pn (1− p)N−n . (2.194)

Using the solution of problem 4 of set 1

pn =λn

n!e−λ ,

where λ = Nv/V .



8. The partition function of a single atom is given by

Z1 = exp (µ0Hβ) + exp (−µ0Hβ) = 2 cosh (µ0Hβ) ,

where β = 1/τ , thus the partition function of the entire system is

Z = (2 cosh (µ0Hβ))N ,

a) The free energy is given by

F = −τ logZ = −Nτ log (2 cosh (µ0Hβ)) . (2.195)

The magnetization is given by

M = −∂F

∂H

τ

= Nµ0 tanh (µ0Hβ) . (2.196)

b) The energy U is given by

U = −∂ logZ∂β

= −Nµ0H tanh (µ0Hβ) , (2.197)

thus

C = τ

∂σ

∂τ

H

=

∂U

∂τ

H

= −Nµ0H∂ tanh µ0H

τ

∂τ

H

= N

µ0H

τ

1

cosh µ0Hτ

2

.

(2.198)c) The entropy σ, which is given by

σ = β (U − F )

= N

log

2 cosh

µ0H

τ

− µ0H

τtanh

µ0H

τ

,

(2.199)and which remains constant, is a function of the ratio H/τ , therefore

τ2 = τ1H2

H1. (2.200)




Z =1

m=−1exp (−βεm)

= 1 + 2 exp (β∆) cosh (βµ0H) ,

(2.201)

where β = 1/τ . The free energy is given by

F = −Nτ logZ , (2.202)

thus the magnetization is given by

M = −∂F

∂H

τ

=2Nµ0 exp (β∆) sinh (βµ0H)

1 + 2 exp (β∆) cosh (βµ0H).

(2.203)

and the magnetic susceptibility is given by

χ =Nµ20

τ1 + 1

2 exp (−β∆) , (2.204)


Z =J

m=−Jexp (mµHβ) ,

where β = 1/τ . By multiplying by a factor sinh (µHβ/2) one finds

sinh

µHβ

2

Z =

1

2

exp

µHβ

2

− exp

−µHβ

2

J

m=−Jexp (mµHβ)

=1

2

exp

J +

1

2

µHβ

− exp

−J +

1

2

µHβ

,

(2.205)

thus

Z =sinh

J + 1

2

µHβ

sinhµHβ2

. (2.206)

a) The free energy is given by

F = −Nτ logZ = −Nτ log

sinhJ + 1

2

µHβ

sinhµHβ2

. (2.207)



b) The magnetization is given by

M = −∂F

∂H

τ

=Nµ

2

(2J + 1) coth

(2J + 1)

µH

2τ

− coth

µH

2τ

.

(2.208)

11. The internal chemical potential µg is given by Eq. (2.131). In thermalequilibrium the total chemical potential

µtot = µg +mgz , (2.209)

(m is the mass of the each diatomic molecule N2, g is the gravity acceler-ation constant, and z is the height) is z independent. Thus, the densityn as a function of height above see level z can be expressed as

n (z) = n (0) exp

−mgzkBT

. (2.210)

The condition n (z) = 0.5× n (0) yields

z =kBT ln 2

mg=

1.3806568× 10−23 JK−1 × 300K× ln 2

14× 1.6605402× 10−27 kg× 9.8ms−2= 12.6 km .

(2.211)

12. The Helmholtz free energy of an ideal gas of N particles is given by

F = −τN log

"Mτ

2πℏ2

3/2V

#

+ τN logN − τN , (2.212)

thus the chemical potential is

µ =

∂F

∂N

τ,V

= −τ log

Mτ

2πℏ2

3/2V

+ τ logN , (2.213)

and the pressure is

p = −∂F

∂V

τ,V

=Nτ

V. (2.214)

Using these results the fugacity λ = exp (βµ) can be expressed in termsof p

λ = eβµ =

Mτ

2πℏ2

−3/2N

V=

M

2πℏ2

−3/2τ−5/2p . (2.215)

At equilibrium the fugacity of the gas and that of the system of absorb-ing sites is the same. The grand canonical partition function of a singleabsorption site is given by



Z = 1 + eβµ + eβ(2µ−ε) , (2.216)

or in terms of the fugacity λ = exp (βµ)

Z = 1 + λ+ λ2e−βε . (2.217)

Thus

Na = N0λ∂ logZ∂λ

= N0λ+ 2λ2e−βε

1 + λ+ λ2e−βε, (2.218)

where λ is given by Eq. (2.215).13. The internal partition function is given by

Zint = g1 + g2 exp− ετ

. (2.219)

Using Eq. (2.135)

cV =3

2N +Nτ

∂2

∂τ2(τ logZint)

V

= N

3

2+ ετ

2 g1g2 exp− ε

τ

g1 + g2 exp

− ε

τ

2

!

.

(2.220)

Using Eq. (2.136)

cp = N

5

2+ ετ

2 g1g2 exp− ε

τ

g1 + g2 exp

− ε

τ

2

!

. (2.221)

14. Using Maxwell’s relation

∂σ

∂V

τ,N

=

∂p

∂τ

V,N

, (2.222)

and the equation of state one finds that

∂σ

∂V

τ,N

=N

V − b. (2.223)

a) Using the definitions

cV = τ

∂σ

∂τ

V,N

, (2.224)

cp = τ

∂σ

∂τ

p,N

, (2.225)



and the general identity

∂z

∂x

α

=

∂z

∂x

y

+

∂z

∂y

x

∂y

∂x

α

, (2.226)

one finds

cp − cV = τ

∂σ

∂V

τ,N

∂V

∂τ

p,N

, (2.227)

or with the help of Eq. (2.223) and the equation of state

cp − cV = NNτ

p (V − b)= N . (2.228)

b) Using the identity

∂U

∂V

τ,N

=

∂U

∂V

σ,N

+

∂U

∂σ

V,N

∂σ

∂V

τ,N

, (2.229)

together with Eq. (2.223) yields

∂U

∂V

τ,N

= −p+ Nτ

V − b= 0 . (2.230)

Thus, the energy U is independent on the volume V (it can be ex-pressed as a function of τ and N only), and therefore for processesfor which dN = 0 the change in energy dU can be expressed as

dU = cV dτ . (2.231)

For an isentropic process no heat is exchanged, and therefore dW =−dU , thus since cV is independent on temperature one has

W = −∆U = −cV (τ2 − τ1) . (2.232)

15. Using the definitions

cV = τ

∂σ

∂τ

V,N

, (2.233)

cp = τ

∂σ

∂τ

p,N

, (2.234)

and the general identity

∂z

∂x

α

=

∂z

∂x

y

+

∂z

∂y

x

∂y

∂x

α

, (2.235)



one finds

cp − cV = τ

∂σ

∂V

τ,N

∂V

∂τ

p,N

. (2.236)

Using Maxwell’s relation∂σ

∂V

τ,N

=

∂p

∂τ

V,N

, (2.237)

and the equation of statep+

a

V 2

(V − b) = Nτ , (2.238)

one finds

cp − cV = τ

∂p

∂τ

V,N

∂V

∂τ

p,N

=τ N(V−b)

−aV+2ab+pV 3

NV 3

=N

1 + −2aV+2abV 3(p+ a

V 2)

,

(2.239)

or

cp − cV =N

1− 2a(1− bV )

2

V Nτ

. (2.240)

16. The work W is given by

W =

V2

V1

pdV . (2.241)

Usingp+

a

V 2

(V − b) = Nτ , (2.242)

one finds

W =

V2

V1

Nτ0V − b

− a

V 2

dV = Nτ0 log

V2 − b

V1 − b− a

V2 − V1V2V1

. (2.243)

Using the identity

∂U

∂V

τ,N

=

∂U

∂V

σ,N

+

∂U

∂σ

V,N

∂σ

∂V

τ,N

= −p+ τ

∂σ

∂V

τ,N

,



(2.244)

and Maxwell’s relation∂σ

∂V

τ,N

=

∂p

∂τ

V,N

, (2.245)

one finds∂U

∂V

τ,N

= τ

∂p

∂τ

V,N

− p . (2.246)

In the present case

∂U

∂V

τ,N

=Nτ

V − b− p =

a

V 2, (2.247)

thus

∆U =

V2

V1

∂U

∂V

τ,N

dV = a

V2

V1

dV

V 2= a

V2 − V1V2V1

, (2.248)

and

Q = ∆U +W = Nτ0 logV2 − b

V1 − b. (2.249)

17. In general the following holds

En = 1

Zgc

−∂

nZgc∂βn

η

, (2.250)

and

U = −∂ logZgc

∂β

η

. (2.251)

Thus the variance is given by

(∆E)2

= E2− E2

=1

Zgc

∂2Zgc∂β2

η

− 1

Z2gc

∂Zgc

∂β

2

η

=

∂2 logZgc

∂β2

η

.

(2.252)

Furthermore, the following holds:



(∆E)3

= E3 − 3E2U + 3EU2 − U3

= E3− 3U

E2+ 2U3

= −"

1

Zgc

∂3Zgc∂β3

η

− 3

Z2gc

∂Zgc∂β

η

∂2Zgc∂β2

η

+2

Z3gc

∂Zgc∂β

3

η

#

= − ∂

∂β

"1

Zgc

∂2Zgc∂β2

η

− 1

Z2gc

∂Zgc∂β

2

η

#

= −∂3 logZgc

∂β3

η

.

(2.253)

For classical gas having no internal degrees of freedom one has

N = logZgc = e−ηV

M

2π2β

3/2, (2.254)

thus

U = −∂ logZgc

∂β

η

=3Nτ

2. (2.255)

a) Using Eq. (2.252)

(∆E)2

= − ∂

∂β

3N

2β=

3

2

N

β2=

2U2

3N. (2.256)

b) Using Eq. (2.253)

(∆E)3

= − ∂

∂β

3N

2β2=

3N

β3=

8U3

9N2. (2.257)

18. The entropy change of the body is

∆σ1 = C

τb

τa

dτ

τ= C log

τ bτa

, (2.258)

and that of the bath is

∆σ2 =∆Q

τ b=C (τa − τ b)

τ b, (2.259)

thus

∆σ = C

τaτ b− 1− log

τaτb

. (2.260)

The function f (x) = x − 1 − logx in the range 0 < x < ∞ satisfyf (x) ≥ 0, where f (x) > 0 unless x = 1.



19. The efficiency is given by

η = 1+Ql

Qh= 1+

Qca

Qab= 1+

Cp (τa − τ c)

Cv (τ b − τa)= 1−γ p2 (V1 − V2)

V2 (p1 − p2), (2.261)

where γ = Cp/CV .20. Energy conservation requires thatW = Ql+Qh. Consider a Carnot heat

engine operating between the same thermal baths producing workW percycle. The Carnot engine consumes heat Q′h from the hot bath per cycleand delivered −Q′l heat to the cold one per cycle, where W = Q′l +Q′hand

ηc =W

Q′h= 1− τ l

τh. (2.262)

According to Clausius principle

Ql +Q′l ≤ 0 , (2.263)

thus

γ =Ql

W≤ −Q

′l

W=Q′h −W

W=

τhτh − τ l

− 1 =τ l

τh − τ l. (2.264)

21. Using Eq. (2.264)

A (τh − τ l)

P=

τ lτh − τ l

, (2.265)

thus

τ2l − 2τ l

τh +

P

2A

+ τ2h = 0 , (2.266)

or

τ l = τh +P

2A±

,τh +

P

2A

2− τ2h . (2.267)

The solution for which τ l ≤ τh is

τ l = τh +P

2A−

,τh +

P

2A

2− τ2h . (2.268)

22. Using Eq. (2.244)

∂U

∂V

τ,N

= τ

∂p

∂τ

V,N

− p , (2.269)



thus

Bτn

V=

3Aτ3

V− p =

2Aτ3

V, (2.270)

therefore

B = 2A , (2.271)

n = 3 . (2.272)

23. In general the following holds

CV = N

3

2+ τ

∂2 (τ logZint)

∂τ2

, (2.273)

Cp = cv +N , (2.274)

where in our case

Zint = 1 + 3exp

−∆τ

, (2.275)

thus

a) CV is given by

CV = N

3

2+

3∆τ

2e−

∆τ

1 + 3e−

∆τ

2

, (2.276)

b) and Cp is given by

Cp = N

5

2+

3∆τ

2e−

∆τ

1 + 3e−

∆τ

2

, (2.277)

24. Consider an infinitesimal change in the temperatures of both bodiesdτ1 and dτ2. The total change in entropy associated with the reversibleprocess employed by the heat engine vanishes, thus

0 = dσ = dσ1 + dσ2 =dQ1

τ1+

dQ2

τ2= C

dτ1τ1

+dτ2τ2

. (2.278)

a) Thus, by integration the equation

dτ1τ1

= −dτ2τ2

, (2.279)

one finds



τf

τ1

dτ1τ1

= − τf

τ2

dτ2τ2

, (2.280)

or

logτfτ1

= logτ2τf

, (2.281)

thus

τf =√τ1τ2. (2.282)

b) Employing energy conservation law yields

W = ∆U1+∆U2 = C (τ1 − τf )+C (τ2 − τf ) = C (√τ1 −

√τ2)

2.

(2.283)

25. Energy conservation requires that the temperature of the mixture willremain τ . The entropy of an ideal gas of density n, which contains Nparticles, is given by

σ (N,n) = N

log

nQn

+5

2

, (2.284)

where

nQ =

Mτ

2π2

3/2, (2.285)

n =N

V. (2.286)

Using the relation

pV = Nτ , (2.287)

one finds that the final pressure of the gas after the partition has beenremoved and the system has reached thermal equilibrium is given by

pfinal =2p1p2p1 + p2

.

Thus, the change in entropy is given by

∆σ = σfinal − σ1 − σ2

= 2N

log

(p1 + p2) τnQ2p1p2

+5

2

−N

log

τnQp1

+5

2

−N

log

τnQp2

+5

2

= N log(p1 + p2)

2

4p1p2.

(2.288)



26. Using

σ = logZc + βU = logZgc + βU + ηN

one finds

logZc = logZgc + ηN . (2.289)

The following holds for classical ideal gas having no internal degrees offreedom

logZgc = N

η = −βµ = lognQV

N,

(2.290)

where

nQ =

Mτ

2π2

3/2, (2.291)

thus

logZc = N

1 + log

nQV

N

= N log (nQV ) +N −N logN

≃ N log (nQV )− logN !

= log(nQV )N

N !,

(2.292)

or

Zc =1

N !

Mτ

2π2

3/2V

N

. (2.293)

27. The efficiency is defined as η =W/Qh, whereW is the total work, andQh

is the heat extracted from the heat bath at higher temperature. Energyconservation requires that W = Qh +Ql, where Ql is the heat extractedfrom the heat bath at lower temperature, thus η = 1 + Ql/Qh. In thepresent case Qh is associated with process a→ b, while Ql is associatedwith process c → d. In both isentropic processes (b → c and d → a) noheat is exchanged. Thus

η = 1 +Ql

Qh= 1 +

Cp (τd − τ c)

Cp (τ b − τa). (2.294)



Using pV = Nτ yields

η = 1 +τd − τcτ b − τa

= 1 +p2 (Vd − Vc)

p1 (Vb − Va). (2.295)

Along the isentropic process pV γ is constant, where γ = Cp/Cv, thus

η = 1 +p2p1

p1p2

1γ

(Va − Vb)

Vb − Va= 1−

p2p1

γ−1γ

. (2.296)

28. No heat is exchanged in the isentropic processes, thus the efficiency isgiven by

η = 1 +Ql

Qh

= 1 +Qc→d

Qa→b

= 1 +cV (τd − τc)

cV (τb − τa).

(2.297)

Since τV γ−1 remains unchanged in an isentropic process, where

γ =cpcV

, (2.298)

one finds that

τbVγ−11 = τ cV

γ−12 , (2.299)

τdVγ−12 = τaV

γ−11 , (2.300)

or

τ cτb

=τdτa

=

V2V1

1−γ, (2.301)

thus

η = 1−V2V1

1−γ. (2.302)

29. Let VA1 = NτA/p (VB1 = NτB/p) be the initial volume of vessel A (B)and let VA2 (VB2) be the final volume of vessel A (B). In terms of thefinal temperature of both vessels, which is denoted as τ f , one has

VA2 = VB2 =Nτ fp

. (2.303)



The entropy of an ideal gas of density n = N/V , which contains Nparticles, is given by

σ = N

log

nQn

+5

2

, (2.304)

where

nQ =

Mτ

2π2

3/2, (2.305)

or as a function of τ and p

σ = N

log

M2π2

3/2τ5/2

p+

5

2

. (2.306)

Thus the change in entropy is given by

∆σ = σfinal − σinitial

=5N

2log

τ2fτAτB

.

(2.307)

In general, for an isobaric process the following holds

Q =W +∆U = p (V2 − V1) + cV (τ2 − τ1) , (2.308)

where Q is the heat that was added to the gas, W the work done by thegas and ∆U the change in internal energy of the gas. Using the equationof state pV = Nτ this can be written as

Q = (N + cV ) (τ2 − τ1) . (2.309)

Since no heat is exchanged with the environment during this process thefollowing holds

QA +QB = 0 ,

where

QA = (N + cV ) (τ f − τA) , (2.310)

QB = (N + cV ) (τ f − τB) , (2.311)

thus

τ f =τA + τB

2, (2.312)

and therefore

∆σ =5N

2log

(τA + τB)2

4τAτB. (2.313)


3. Bosonic and Fermionic Systems

In the first part of this chapter we study two Bosonic systems, namely photonsand phonons. A photon is the quanta of electromagnetic waves whereas aphonon is the quanta of acoustic waves. In the second part we study twoFermionic systems, namely electrons in metals and electrons and holes insemiconductors.

3.1 Electromagnetic Radiation

In this section we study an electromagnetic cavity in thermal equilibrium.

3.1.1 Electromagnetic Cavity

Consider an empty volume surrounded by conductive walls having infiniteconductivity. The Maxwell’s equations in SI units are given by

∇×H = ǫ0∂E

∂t, (3.1)

∇×E = −µ0∂H

∂t, (3.2)

∇ ·E = 0 , (3.3)

and

∇ ·H = 0 , (3.4)

where ǫ0 = 8.85× 10−12 Fm−1 and µ0 = 1.26× 10−6NA−2 are the permit-tivity and permeability respectively of free space, and the following holds

ǫ0µ0 =1

c2, (3.5)

where c = 2.99× 108ms−1 is the speed of light in vacuum.In the Coulomb gauge, where the vector potential A is chosen such that

∇ ·A = 0 , (3.6)

Chapter 3. Bosonic and Fermionic Systems

the scalar potential φ vanishes in the absence of sources (charge and current),and consequently both fields E and H can be expressed in terms of A onlyas

E = −∂A∂t

, (3.7)

and

µ0H =∇×A . (3.8)

The gauge condition (3.6) and Eqs. (3.7) and (3.8) guarantee thatMaxwell’s equations (3.2), (3.3), and (3.4) are satisfied

∇×E = −∂ (∇×A)

∂t= −µ0

∂H

∂t, (3.9)

∇ ·E = −∂ (∇ ·A)

∂t= 0 , (3.10)

∇ ·H =1

µ0∇ · (∇×A) = 0 , (3.11)

where in the last equation the general vector identity ∇ · (∇×A) = 0 hasbeen employed. Substituting Eqs. (3.7) and (3.8) into the only remainingnontrivial equation, namely into Eq. (3.1), leads to

∇× (∇×A) = − 1

c2∂2A

∂t2. (3.12)

Using the vector identity

∇× (∇×A) = ∇ (∇ ·A)−∇2A , (3.13)

and the gauge condition (3.6) one finds that

∇2A =1

c2∂2A

∂t2. (3.14)

Consider a solution in the form

A = q (t)u (r) , (3.15)

where q (t) is independent on position r and u (r) is independent on time t.The gauge condition (3.6) leads to

∇ · u = 0 . (3.16)

From Eq. (3.14) one finds that

q∇2u =1

c2ud2q

dt2. (3.17)


3.1. Electromagnetic Radiation

Multiplying by an arbitrary unit vector n leads to

∇2u

· n

u · n =1

c2q

d2q

dt2. (3.18)

The left hand side of Eq. (3.18) is a function of r only while the right handside is a function of t only. Therefore, both should equal a constant, which isdenoted as −κ2, thus

∇2u+κ2u = 0 , (3.19)

and

d2q

dt2+ω2κq = 0 , (3.20)

where

ωκ = cκ . (3.21)

Equation (3.19) should be solved with the boundary conditions of a per-fectly conductive surface. Namely, on the surface S enclosing the cavity wehave H · s = 0 and E× s = 0, where s is a unit vector normal to the surface.To satisfy the boundary condition for E we require that u be normal to thesurface, namely, u = s (u · s) on S. This condition guarantees also that theboundary condition for H is satisfied. To see this we calculate the integral ofthe normal component of H over some arbitrary portion S′ of S. Using Eq.(3.8) and Stoke’s’ theorem one finds that

S′(H · s) dS =

q

µ0

S′[(∇× u) · s] dS

=q

µ0

/

C

u·dl ,

(3.22)

where the close curve C encloses the surface S′. Thus, since u is normal tothe surface one finds that the integral along the close curve C vanishes, andtherefore

S′(H · s) dS = 0 . (3.23)

Since S′ is arbitrary we conclude that H · s =0 on S.Each solution of Eq. (3.19) that satisfies the boundary conditions is called

an eigen mode. As can be seen from Eq. (3.20), the dynamics of a modeamplitude q is the same as the dynamics of an harmonic oscillator havingangular frequency ωκ = cκ.



3.1.2 Partition Function

What is the partition function of a mode having eigen angular frequency ωκ?We have seen that the mode amplitude has the dynamics of an harmonicoscillator having angular frequency ωκ. Thus, the quantum eigenenergies ofthe mode are

εs = sωκ , (3.24)

where s = 0, 1, 2, · · · is an integer1 . When the mode is in the eigenstate havingenergy εs the mode is said to occupy s photons. The canonical partitionfunction of the mode is found using Eq. (1.69)

Zκ =∞

s=0

exp (−sβωκ)

=1

1− exp (−βωκ).

(3.25)

Note the similarity between this result and the orbital partition function ζof Bosons given by Eq. (2.36). The average energy is found using

εκ = −∂ logZκ∂β

=ωκ

eβωκ − 1.

(3.26)

The partition function of the entire system is given by

Z =

κ

Zκ , (3.27)

and the average total energy by


=

κ

εκ . (3.28)

3.1.3 Cube Cavity

For simplicity, consider the case of a cavity shaped as a cube of volumeV = L3. We seek solutions of Eq. (3.19) satisfying the boundary condition

1 In Eq. (3.24) above the ground state energy was taken to be zero. Note thatby taking instead εs = (s+ 1/2) ωκ , one obtains Zκ = 1/2 sinh (βℏωκ /2) andεn = (ωκ /2) coth (βℏωκ /2). In some cases the offset energy term ωκ /2 isvery important (e.g., the Casimir force), however, in what follows we disregardit.



that the tangential component of u vanishes on the walls. Consider a solutionhaving the form

ux =

8

Vax cos (kxx) sin (kyy) sin (kzz) , (3.29)

uy =

8

Vay sin (kxx) cos (kyy) sin (kzz) , (3.30)

uz =

8

Vaz sin (kxx) sin (kyy) cos (kzz) . (3.31)

While the boundary condition on the walls x = 0, y = 0, and z = 0 isguaranteed to be satisfied, the boundary condition on the walls x = L, y = L,and z = L yields

kx =nxπ

L, (3.32)

ky =nyπ

L, (3.33)

kz =nzπ

L, (3.34)

where nx, ny and nz are integers. This solution clearly satisfies Eq. (3.19)where the eigen value κ is given by

κ =0k2x + k2y + k2z . (3.35)

Alternatively, using the notation

n = (nx, ny, nz) , (3.36)

one has

κ =π

Ln , (3.37)

where

n =0n2x + n2y + n2z . (3.38)

Using Eq. (3.21) one finds that the angular frequency of a mode characterizedby the vector of integers n is given by

ωn =πc

Ln . (3.39)

In addition to Eq. (3.19) and the boundary condition, each solution has tosatisfy also the transversality condition ∇·u = 0 (3.16), which in the presentcase reads

n · a = 0 , (3.40)



where

a = (ax, ay, az) . (3.41)

Thus, for each set of integers nx, ny, nz there are two orthogonal modes(polarizations), unless nx = 0 or ny = 0 or nz = 0. In the latter case, only asingle solution exists.

3.1.4 Average Energy

The average energy U of the system is found using Eqs. (3.26), (3.28) and(3.39)

U =

n

ωneβωn − 1

= 2τ∞

nx=0

∞

ny=0

∞

nz=0

αn

eαn − 1,

(3.42)

where the dimensionless parameter α is given by

α =βπc

L. (3.43)

The following relation can be employed to estimate the dimensionless para-meter α

α =2.4× 10−3

Lcm

τ300K

. (3.44)

In the limit where

α≪ 1 (3.45)

the sum can be approximated by the integral

U ≃ 2τ4π

8

∞

0

dn n2αn

eαn − 1. (3.46)

Employing the integration variable transformation [see Eq. (3.39)]

n =L

πcω , (3.47)

allows expressing the energy per unit volume U/V as

U

V=

∞

0

dω uω , (3.48)



where

uω =

c3π2ω3

eβω − 1. (3.49)

This result is know as Plank’s radiation law. The factor uω represents thespectral distribution of the radiation. The peak in uω is obtained at βω0 =2.82. In terms of the wavelength λ0 = 2πc/ω0 one has

λ0µm

= 5.1

T

1000K

−1. (3.50)

0.2

0.4

0.6

0.8

1

1.2

1.4

0 2 4 6 8 10x

The function x3/ (ex − 1).The total energy is found by integrating Eq. (3.48) and by employing thevariable transformation x = βω

U

V=

τ4

c3π23

∞

0

x3dx

ex − 1

π4

15

=π2τ4

153c3.

(3.51)

3.1.5 Stefan-Boltzmann Radiation Law

Consider a small hole having area dA drilled into the conductive wall of anelectromagnetic (EM) cavity. What is the rate of energy radiation emittedfrom the hole? We employ below a kinetic approach to answer this question.Consider radiation emitted in a time interval dt in the direction of the unitvector u. Let θ be the angle between u and the normal to the surface of thehole. Photons emitted during that time interval dt in the direction u camefrom the region in the cavity that is indicated in Fig. 3.1, which has volume



θ

dAcosθ

cdt

u

θ

dAcosθ

cdt

u

Fig. 3.1. Radiation emitted through a small hole in the cavity wall.

Vθ = dA cos θ × cdt . (3.52)

The average energy in that region can be found using Eq. (3.51). Integratingover all possible directions yields the total rate of energy radiation emittedfrom the hole per unit area

J =1

dAdt

1

4π

π/2

0

dθ sin θ

2π

0

dϕU

VVθ

=π2τ4

153c21

4π

π/2

0

dθ sin θ cos θ

2π

0

dϕ

1/4

=π2τ4

603c2

(3.53)

In terms of the historical definition of temperature T = τ/kB [see Eq. (1.92)]one has

J = σBT4 , (3.54)

where σB, which is given by

σB =π2k4B603c2

= 5.67× 10−8W

m2K4 , (3.55)

is the Stefan-Boltzmann constant.


3.2. Phonons in Solids

m mω2 m mω2 m mω2 m mω2 mm mω2 m mω2 m mω2 m mω2 m

Fig. 3.2. 1D lattice.

3.2 Phonons in Solids

In this section we study elastic waves in solids. We start with a one-dimensional example, and then generalize some of the results for the caseof a 3D lattice.

3.2.1 One Dimensional Example

Consider the 1D lattice shown in Fig. 3.2 below, which contains N ’atoms’having mass m each that are attached to each other by springs having springconstant mω2. The lattice spacing is a. The atoms are allowed to move inone dimension along the array axis. In problem 5 of set 3 one finds that thenormal mode angular eigen-frequencies are given by

ωn = ω12 (1− cos kna) = 2ω

2222sinkna

2

2222 , (3.56)

where a is the lattice spacing,

kn =2πn

aN, (3.57)

and n is integer ranging from −N/2 to N/2.

0

0.2

0.4

0.6

0.8

-1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1x

The function |sin (πx/2)|.



What is the partition function of an eigen-mode having eigen angular fre-quency ωn? The mode amplitude has the dynamics of an harmonic oscillatorhaving angular frequency ωn. Thus, as we had in the previous section, wherewe have discussed EM modes, the quantum eigenenergies of the mode are

εs = sωn , (3.58)

where s = 0, 1, 2, · · · is an integer. When the mode is in an eigenstate havingenergy εs the mode is said to occupy s phonons. The canonical partitionfunction of the mode is found using Eq. (1.69)

Zκ =∞

s=0

exp (−sβωκ)

=1

1− exp (−βωκ).

(3.59)

Similarly to the EM case, the average total energy is given by

U =

N/2

n=−N/2

ℏωnexp (βℏωn)− 1

, (3.60)

where β = 1/τ , and the total heat capacity is given by

CV =∂U

∂τ=

N/2

n=−N/2

(βℏωn)2 exp (βℏωn)

[exp (βℏωn)− 1]2. (3.61)

High Temperature Limit. In the high temperature limit βℏω ≪ 1


[exp (βℏωn)− 1]2≃ 1 , (3.62)

therefore

CV = N . (3.63)

Low Temperature Limit. In the low temperature limit βℏω ≫ 1 themain contribution to the sum in Eq. (3.61) comes from terms for which|n| N/βℏω. Thus, to a good approximation the dispersion relation canbe approximated by

ωn = 2ω

2222sinkna

2

2222 ≃ 2ω

2222kna

2

2222 = ω2π

N|n| . (3.64)

Moreover, in the limit N ≫ 1 the sum in Eq. (3.61) can be approximatedby an integral, and to a good approximation the upper limit N/2 can besubstituted by infinity, thus



CV =

N/2

n=−N/2


[exp (βℏωn)− 1]2

≃ 2∞

n=0

βℏω 2π

N n2

expβℏω 2π

N n

exp

βℏω 2π

N n− 12

≃ 2

∞

0

dn

βℏω 2π

N n2

expβℏω 2π

N n

exp

βℏω 2π

N n− 12

=N

π

τ

ℏω

∞

0

dxx2 exp (x)

(exp (x)− 1)2

π2/3

=Nπ

3

τ

ℏω.

(3.65)

3.2.2 The 3D Case

The case of a 3D lattice is similar to the case of EM cavity that we havestudied in the previous section. However, there are 3 important distinctions:

1. The number of modes of a lattice containing N atoms that can move in3D is finite, 3N instead of infinity as in the EM case.

2. For any given vector k there are 3, instead of only 2, orthogonal modes(polarizations).

3. Dispersion: contrary to the EM case, the dispersion relation (namely, thefunction ω (k)) is in general nonlinear.

Due to distinctions 1 and 2, the sum over all modes is substituted by anintegral according to

∞

nx=0

∞

ny=0

∞

nz=0

→ 3

84π

nD

0

dn n2 , (3.66)

where the factor of 3 replaces the factor of 2 we had in the EM case. Moreover,the upper limit is nD instead of infinity, where nD is determined from therequirement

3

84π

nD

0

dn n2 = 3N , (3.67)

thus

nD =

6N

π

1/3. (3.68)



Similarly to the EM case, the average total energy is given by

U =

n

ℏωnexp (βℏωn)− 1

(3.69)

=3π

2

nD

0

dn n2ℏωn

exp (βℏωn)− 1.

(3.70)

To proceed with the calculation the dispersion relation ωn (kn) is needed.Here we assume for simplicity that dispersion can be disregarded to a goodapproximation, and consequently the dispersion relation can be assumed tobe linear

ωn = vkn , (3.71)

where v is the sound velocity. The wave vector kn is related to n =0n2x + n2y + n2z by

kn =πn

L, (3.72)

where L = V 1/3 and V is the volume. In this approximation one finds usingthe variable transformation

x =βℏvπn

L, (3.73)

that

U =3π

2

nD

0

dn n2ℏvπnL

expβℏvπn

L

− 1

=3V τ4

2ℏ3v3π2

xD

0

dxx3

expx− 1.

(3.74)

where

xD =βℏvπnD

L=βℏvπ

6Nπ

1/3

L.

Alternatively, in terms of the Debye temperature, which is defined as

Θ = ℏv

6π2N

V

1/3, (3.75)

one has



xD =Θ

τ, (3.76)

and

U = 9Nτ τΘ

3xD

0

dxx3

expx− 1. (3.77)

As an example Θ/kB = 88K for Pb, while Θ/kB = 1860K for diamond.Below we calculate the heat capacity CV = ∂U/∂τ in two limits.

High Temperature Limit. In the high temperature limit xD = Θ/τ ≪ 1,thus

U = 9Nτ τΘ

3xD

0

dxx3

expx− 1

≃ 9Nτ τΘ

3xD

0

dx x2

= 9Nτ τΘ

3 x3D3

= 3Nτ ,

(3.78)

and therefore

CV =∂U

∂τ= 3N . (3.79)

Note that in this limit the average energy of each mode is τ and consequentlyU = 3Nτ . This result demonstrates the equal partition theorem of classicalstatistical mechanics that will be discussed in the next chapter.

Low Temperature Limit. In the low temperature limit xD = Θ/τ ≫ 1,thus

U = 9Nτ τΘ

3xD

0

dxx3

expx− 1

≃ 9Nτ τΘ

3∞

0

dxx3

expx− 1

π4/15

=3π4

5Nτ τΘ

3,

(3.80)



and therefore

CV =∂U

∂τ=

12π4

5N τΘ

3. (3.81)

Note that Eq. (3.80) together with Eq. (3.75) yield

U

V=

3

2

π2τ4

15ℏ3v3. (3.82)

Note the similarity between this result and Eq. (3.51) for the EM case.

3.3 Fermi Gas

In this section we study an ideal gas of Fermions of mass m. While only theclassical limit was considered in chapter 2, here we consider the more generalcase.

3.3.1 Orbital Partition Function

Consider an orbital having energy εn. Disregarding internal degrees of free-dom, its grandcanonical Fermionic partition function is given by [see Eq.(2.33)]

ζn= 1 + λ exp (−βεn) , (3.83)

where

λ = exp (βµ) = e−η , (3.84)

is the fugacity and β = 1/τ . Taking into account internal degrees of freedomthe grandcanonical Fermionic partition function becomes

ζn=

l

(1 + λ exp (−βεn) exp (−βEl)) , (3.85)

where El are the eigenenergies of a particle due to internal degrees offreedom [see Eq. (2.71)]. As is required by the Pauli exclusion principle, nomore than one Fermion can occupy a given internal eigenstate and a givenorbital.

3.3.2 Partition Function of the Gas

The grandcanonical partition function of the gas is given by

Zgc =

n

ζn. (3.86)


3.3. Fermi Gas

As we have seen in chapter 2, the orbital eigenenergies of a particle of massm in a box are given by Eq. (2.5)

εn =2

2m

πL

2n2 , (3.87)

where

n = (nx, ny, nz) , (3.88)

n =0n2x + n2y + n2z , (3.89)

nx, ny, nz = 1, 2, 3, · · · , (3.90)

and L3 = V is the volume of the box. Thus, logZgc can be written as

logZgc =∞

nx=1

∞

ny=1

∞

nz=1

log ζn. (3.91)

Alternatively, using the notation

α2 =β2π2

2mL2, (3.92)

and Eq. (3.85) one has

logZgc =

l

∞

nx=1

∞

ny=1

∞

nz=1

log1 + λ exp

−α2n2

exp (−βEl)

. (3.93)

For a macroscopic system α ≪ 1, and consequently the sum over n canbe approximately replaced by an integral

∞

nx=0

∞

ny=0

∞

nz=0

→ 1

84π

∞

0

dn n2 , (3.94)

thus, one has

logZgc =π

2

l

∞

0

dn n2 log1 + λ exp

−α2n2

exp (−βEl)

. (3.95)

This can be further simplified by employing the variable transformation

βε = α2n2 . (3.96)

The following holds



√ε

2

β

α2

3/2dε = n2dn .

Thus, by introducing the density of states

D (ε) =

V2π2

2m2

3/2ε1/2 ε ≥ 0

0 ε < 0, (3.97)

one has

logZgc =1

2

l

∞

−∞

dε D (ε) log (1 + λ exp (−β (ε+El))) . (3.98)

3.3.3 Energy and Number of Particles

Using Eqs. (1.80) and (1.94) for the energy U and the number of particlesN , namely using

U = −∂ logZgc

∂β

η

, (3.99)

N = λ∂ logZgc

∂λ, (3.100)

one finds that

U =1

2

l

∞

−∞

dε D (ε) (ε+El) fFD (ε+El) , (3.101)

N =1

2

l

∞

−∞

dε D (ε) fFD (ε+El) , (3.102)

where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)]

fFD (ǫ) =1

exp [β (ǫ− µ)] + 1. (3.103)

3.3.4 Example: Electrons in Metal

Electrons are Fermions having spin 1/2. The spin degree of freedom gives riseto two orthogonal eigenstates having energies E+ and E− respectively. In theabsent of any external magnetic field these states are degenerate, namelyE+ = E−. For simplicity we take E+ = E− = 0. Thus, Eqs. (3.101) and(3.102) become


3.3. Fermi Gas

U =

∞

−∞

dε D (ε) εfFD (ε) , (3.104)

N =

∞

−∞

dε D (ε) fFD (ε) , (3.105)

Typically for metals at room temperature or below the following holds τ ≪ µ.Thus, it is convenient to employ the following theorem (Sommerfeld expan-sion) to evaluate these integrals.

Theorem 3.3.1. Let g (ε) be a function that vanishes in the limit ε→−∞,and that diverges no more rapidly than some power of ε as ε → ∞. Then,the following holds

∞

−∞

dε g (ε) fFD (ε)

=

µ

−∞

dε g (ε) +π2g′ (µ)

6β2+O

1

βµ

4.


With the help of this theorem the number of particles N to second orderin τ is given by

N =

µ

−∞

dε D (ε) +π2τ2D′ (µ)

6. (3.106)

Moreover, at low temperatures, the chemical potential is expected to be closethe the Fermi energy εF, which is defined by

εF = limτ→0

µ . (3.107)

Thus, to lowest order in µ− εF one has

µ

−∞

dε D (ε) =

εF

−∞

dε D (ε) + (µ− εF)D (εF) +O (µ− εF)2 , (3.108)

and therefore

N = N0 + (µ− εF)D (εF) +π2τ2D′ (εF)

6, (3.109)

where



N0 =

εF

−∞

dε D (ε) , (3.110)

is the number of electrons at zero temperature. The number of electrons Nin metals is expected to be temperature independent, namely N = N0 andconsequently

µ = εF −π2D′ (εF)

6β2D (εF). (3.111)

Similarly, the energy U at low temperatures is given approximately by

U =

∞

−∞

dε D (ε) εfFD (ε)

=

εF

−∞

dε D (ε) ε

U0

+ (µ− εF)D (εF) εF +π2τ2

6(D′ (εF) εF +D (εF))

= U0 −π2τ2D′ (εF)

6D (εF)D (εF) εF +

π2τ2

6(D′ (εF) εF +D (εF))

= U0 +π2τ2

6D (εF) ,

(3.112)

where

U0 =

εF

−∞

dε D (ε) ε . (3.113)

From this result one finds that the electronic heat capacity is given by

CV =∂U

∂τ=π2τ

3D (εF) . (3.114)

Comparing this result with Eq. (3.81) for the phonons heat capacity, whichis proportional to τ3 at low temperatures, suggests that typically, while theelectronic contribution is the dominant one at very low temperatures, athigher temperatures the phonons’ contribution becomes dominant.

3.4 Semiconductor Statistics

To be written...


3.5. Problems Set 3

3.5 Problems Set 3

1. Calculate the average number of photonsN in equilibrium at temperatureτ in a cavity of volume V . Use this result to estimate the number ofphotons in the universe assuming it to be a spherical cavity of radius1026m and at temperature τ = kB × 3K.

2. Write a relation between the temperature of the surface of a planet and itsdistance from the Sun, on the assumption that as a black body in thermalequilibrium, it reradiates as much thermal radiation, as it receives fromthe Sun. Assume also, that the surface of the planet is at constant temper-ature over the day-night cycle. Use TSun = 5800K; RSun = 6.96× 108m;and the Mars-Sun distance of DM−S = 2.28 × 1011m and calculate thetemperature of Mars surface.

3. Calculate the Helmholtz free energy F of photon gas having total energyU and volume V and use your result to show that the pressure is givenby

p =U

3V. (3.115)

4. Consider a photon gas initially at temperature τ1 and volume V1. The gasis adiabatically compressed from volume V1 to volume V2 in an isentropicprocess. Calculate the final temperature τ2 and final pressure p2.

5. Consider a one-dimensional lattice of N identical point particles of massm, interacting via nearest-neighbor spring-like forces with spring constantmω2 (see Fig. 3.2). Denote the lattice spacing by a. Show that the normalmode eigen-frequencies are given by

ωn = ω12 (1− coskna) , (3.116)

where kn = 2πn/aN , and n is integer ranging from−N/2 toN/2 (assumeN ≫ 1).

6. Consider an orbital with energy ε in an ideal gas. The system is in thermalequilibrium at temperature τ and chemical potential µ.

a) Show that the probability that the orbital is occupied by n particlesis given by

pF (n) =exp [n (µ− ε)β]

1 + exp [(µ− ε)β], (3.117)

for the case of Fermions, where n ∈ 0, 1, and by

pB (n) = 1− exp [(µ− ε)β] exp [n (µ− ε) β] , (3.118)

where n ∈ 0, 1, 2, · · · , for the case of Bosons.



b) Show that the variance (∆n)2=(n− n)2

is given by

(∆n)2F = nF (1− nF ) , (3.119)

for the case of Fermions, and by

(∆n)2B = nB (1 + nB) , (3.120)

for the case of Bosons.

7. Let g (ε) be a function that vanishes in the limit ε → −∞, and thatdiverges no more rapidly than some power of ε as ε→∞. Show that thefollowing holds (Sommerfeld expansion)

I =

∞

−∞

dε g (ε) fFD (ε)

=

µ

−∞

dε g (ε) +π2g′ (µ)

6β2+O

1

βµ

4.

8. Consider a metal at zero temperature having Fermi energy εF, numberof electrons N and volume V .

a) Calculate the mean energy of electrons.b) Calculate the ratio α of the mean-square-speed of electrons to the

square of the mean speed

α =

v2

v2. (3.121)

c) Calculate the pressure exerted by an electron gas at zero tempera-ture.

9. For electrons with energy ε ≫ mc2 (relativistic fermi gas), the energyis given by ε = pc. Find the fermi energy of this gas and show that theground state energy is

E (T = 0) =3

4NεF (3.122)

10. A gas of two dimensional electrons is free to move in a plane. The massof each electron is me, the density (number of electrons per unit area) isn, and the temperature is τ . Show that the chemical potential µ is givenby

µ = τ log

exp

nπ2

meτ

− 1

. (3.123)



3.6 Solutions Set 3

1. The density of states of the photon gas is given by

dg =V ε2

π2ℏ3c3dε . (3.124)

Thus

N =V

π2ℏ3c3

∞

0

ε2

eε/τ − 1dε

= V τℏc

3α , (3.125)

where

α =1

π2

∞

0

x2

ex − 1dx . (3.126)

The number α is calculated numerically

α = 0.24359 . (3.127)

For the universe

N =4π

3

1026m

3

1.3806568× 10−23 JK−13K

1.05457266× 10−34 J s2.99792458× 108ms−1

3× 0.24359

≃ 2.29× 1087 . (3.128)

2. The energy emitted by the Sun is

ESun = 4πR2SunσBT

4Sun , (3.129)

and the energy emitted by a planet is

Eplanet = 4πR2planetσBT

4planet . (3.130)

The fraction of Sun energy that planet receives is

πR2planet

4πD2M−S

ESun , (3.131)

and this equals to the energy it reradiates. Therefore

πR2planet

4πD2ESun = Eplanet ,



thus

Tplanet =

RSun

2DTSun ,

and for Mars

TMars =

6.96× 108m

2× 2.28× 1011m5800K = 226K .

3. The partition function is given by

Z =

n

∞

s=0

exp (sβℏωn) =

n

1

1− exp (−βℏωn),

thus the free energy is given by

F = −τ logZ = τ

n

log [1− exp (−βℏωn)] .

Transforming the sum over modes into integral yields

F = τπ

∞

0

dnn2 log [1− exp (−βℏωn)] (3.132)

= τπ

∞

0

dnn2 log

1− exp

−βℏπcn

L

,

or, by integrating by parts

F = −1

3

ℏπ2c

L

∞

0

dnn3

expβℏπcnL

− 1

= −1

3U ,

where

U =π2τ4V

15ℏ3c3. (3.133)

Thus

p = −∂F

∂V

τ

=U

3V. (3.134)

4. Using the expression for Helmholtz free energy, which was derived in theprevious problem,

F = −U3

= −π2τ4V

45ℏ3c3,

one finds that the entropy is given by



σ = −∂F

∂τ

V

=4π2τ3V

45ℏ3c3.

Thus, for an isentropic process, for which σ is a constant, one has

τ2 = τ1

V1V2

1/3.

Using again the previous problem, the pressure p is given by

p =U

3V,

thus

p =π2τ4

45ℏ3c3,

and

p2 =π2τ4145ℏ3c3

p1

V1V2

4/3.

5. Let u (na) be the displacement of point particle number n. The equationsof motion are given by

mu (na) = −mω2 2u (na)− u [(n− 1) a]− u [(n+ 1) a] . (3.135)

Consider a solution of the form

u (na, t) = ei(kna−ωnt) . (3.136)

Periodic boundary condition requires that

eikNa = 1 , (3.137)

thus

kn =2πn

aN. (3.138)

Substituting in Eq. 3.135 yields

−mω2nu (na) = −mω22u (na)− u (na) e−ika − u (na) eika

, (3.139)

or

ωn = ω12 (1− coskna) = 2ω

2222sinkna

2

2222 . (3.140)



6. In general using Gibbs factor

p (n) =exp [n (µ− ε)β]

n′

exp [n′ (µ− ε)β], (3.141)

where β = 1/τ , one finds for Fermions

pF (n) =exp [n (µ− ε)β]

1 + exp [(µ− ε)β], (3.142)

where n ∈ 0, 1, and for Bosons

pB (n) =exp [n (µ− ε)β]

∞

n′=0

exp [n′ (µ− ε)β]

= 1− exp [(µ− ε)β] exp [n (µ− ε)β] ,

(3.143)

where n ∈ 0, 1, 2, · · · . The expectation value of n in general is givenby

n =

n′

n′p (n′) =

n′

n′ exp [n (µ− ε)β]

n′

exp [n′ (µ− ε)β], (3.144)

thus for Fermions

nF =1

exp [(ε− µ)β] + 1, (3.145)

and for Bosons

nB = 1− exp [(µ− ε)β]∞

n′=0

n′ exp [n′ (µ− ε)β]

= 1− exp [(µ− ε)β] exp [(µ− ε)β]

(1− exp [(µ− ε)β])2

=1

exp [(ε− µ)β]− 1.

(3.146)

In general, the following holds

τ

∂ n∂µ

τ

=

n′′

(n′′)2 exp [n (µ− ε)β]

n′

exp [n′ (µ− ε)β]−

n′

n′ exp [n′ (µ− ε)β]

n′

exp [n′ (µ− ε)β]

2

= n2− n2 =

(n− n)2

.

(3.147)



Thus

(∆n)2F =

exp [(ε− µ)β]

(exp [(ε− µ) β] + 1)2= nF (1− nF ) , (3.148)

(∆n)2B =exp [(ε− µ)β]

(exp [(ε− µ) β]− 1)2= nB (1 + nB) . (3.149)

7. Let

G (ε) =

ε

−∞

dε′ g (ε′) . (3.150)

Integration by parts yields

I =

∞

−∞

dε g (ε) fFD (ε)

= [G (ε) fFD (ε)|∞−∞ =0

+

∞

−∞

dε G (ε)

−∂fFD

∂ε

,

(3.151)

where the following holds

−∂fFD

∂ε

=

βeβ(ε−µ)eβ(ε−µ) + 1

2 =β

4 cosh2β2 (ε− µ)

. (3.152)

Using the Taylor expansion of G (ε) about ε− µ, which has the form

G (ε) =∞

n=0

G(n) (µ)

n!(ε− µ)n , (3.153)

yields

I =∞

n=0

G(n) (µ)

n!

∞

−∞

β (ε− µ)n dε

4 cosh2β2 (ε− µ)

. (3.154)

Employing the variable transformation

x = β (ε− µ) , (3.155)

and exploiting the fact that (−∂fFD/∂ε) is an even function of ε−µ leadsto



I =∞

n=0

G(2n) (µ)

(2n)!β2n

∞

−∞

x2ndx

4 cosh2 x2

. (3.156)

With the help of the identities

∞

−∞

dx

4 cosh2 x2

= 1 , (3.157)

∞

−∞

x2dx

4 cosh2 x2

=π2

3, (3.158)

one finds

I = G (µ) +π2G(2) (µ)

6β2+O

1

βµ

4

=

µ

−∞

dε g (ε) +π2g′ (µ)

6β2+O

1

βµ

4.

(3.159)

8. In general, at zero temperature the average of the energy ε to the powern is given by

εn =

εF3

0

dε D (ε) εn

εF3

0

dε D (ε)

, (3.160)

where D (ε) is the density of states

D (ε) =V

2π2

2m

2

3/2ε1/2 , (3.161)

thus

εn = εnF2n3 + 1

. (3.162)

a) Using Eq. (3.162) one finds that

ε = 3εF5

. (3.163)

b) The speed v is related to the energy by

v =

2ε

m, (3.164)



thus

α =ε

ε1/2

2 =

εF23+14

ε−1/2F2312+1

52 =16

15. (3.165)

c) The number of electrons N is given by

N =

εF

0

dε D (ε) =D (εF)

ε1/2F

εF

0

dε ε1/2 =D (εF)

ε1/2F

2

3ε3/2F ,

thus

εF =2

2m

3π2N

V

2/3, (3.166)

and therefore

U =3N

5

2

2m

3π2N

V

2/3. (3.167)

Moreover, at zero temperature the Helmholtz free energy F = U −τσ = U , thus the pressure is given by

p = −∂F

∂V

τ,N

= −∂U

∂V

τ,N

=3N

5

2

2m

3π2N

V

2/32

3V

=2NεF5V

.

(3.168)

9. The energy of the particles are

εn,± = pc (3.169)

where p = k, and k = πnL , n =

0n2x + n2y + n2z and ni = 1, 2, · · · .

Therefore

N = 2× 1

8× 4

3πn3F =

π

3n3F (3.170)

nF =

3N

π

1/3

εF =cπ

L

3N

π

1/3= cπ

3N

πV

1/3



The energy is given by

E (T = 0) =

εF

0

εg (ε) dε =L3

π23c3

εF

0

ε3dε = (3.171)

=L3

π23c3× ε4F

4=

1

4

L3

π23c3ε3F × εF =

=1

4

L3

π23c33π23c3N

L3εF =

3

4NεF

10. The energy of an electron having a wave function proportional to exp (ikxx) exp (ikyy)is

2

2me

k2x + k2y

. (3.172)

For periodic boundary conditions one has

kx =2πnxLx

, (3.173)

ky =2πnyLy

, (3.174)

where the sample is of area LxLy, and nx and ny are both integers. Thenumber of states having energy smaller than E′ is given by (includingboth spin directions)

2π2meE′

2

LxLy4π2

, (3.175)

thus, the density of state per unit area is given by

D (E) =

me

π2 E > 00 E < 0

. (3.176)

Using Fermi-Dirac function

f (E) =1

1 + exp [β (E − µ)], (3.177)

where β = 1/τ , the density n is given by

n =

∞

−∞D (E) f (E) dE

=me

π2

∞

0

dE

1 + exp [β (E − µ)]

=meτ

π2log1 + eβµ

,

(3.178)



thus

µ = τ log

exp

nπ2

meτ

− 1

. (3.179)


4. Classical Limit of Statistical Mechanics

In this chapter we discuss the classical limit of statistical mechanics. We dis-cuss Hamilton’s formalism, define the Hamiltonian and present the Hamilton-Jacobi equations of motion. The density function in thermal equilibrium isused to prove the equipartition theorem. This theorem is then employed toanalyze an electrical circuit in thermal equilibrium, and to calculate voltagenoise across a resistor (Nyquist noise formula).

4.1 Classical Hamiltonian

In this section we briefly review Hamilton’s formalism, which is analogous toNewton’s laws of classical mechanics . Consider a classical system having ddegrees of freedom. The system is described using the vector of coordinates

q = (q1, q2, · · · , qd) . (4.1)

Let E be the total energy of the system. For simplicity we restrict the dis-cussion to a special case where E is a sum of two terms

E = T + V ,

where T depends only on velocities, namely T = T ·q, and where V depends

only on coordinates, namely V = V (q). The notation overdot is used toexpress time derivative, namely

·q =

dq1dt

,dq2dt

, · · · , dqddt

. (4.2)

We refer to the first term T as kinetic energy and to the second one V aspotential energy.

The canonical conjugate momentum pi of the coordinate qi is defined as

pi =∂T

∂qi. (4.3)

The classical Hamiltonian H of the system is expressed as a function of thevector of coordinates q and as a function of the vector of canonical conjugatemomentum variables

Chapter 4. Classical Limit of Statistical Mechanics

p = (p1, p2, · · · , pd) , (4.4)

namely

H = H (q, p) , (4.5)

and it is defined by

H =d

i=1

piqi − T + V . (4.6)

4.1.1 Hamilton-Jacobi Equations

The equations of motion of the system are given by

qi =∂H∂pi

(4.7)

pi = −∂H∂qi

, (4.8)

where i = 1, 2, · · · d.

4.1.2 Example

q

V(q)

m

q

V(q)

m

Consider a particle having mass m in a one dimensional potential V (q).The kinetic energy is given by T = mq2/2, thus the canonical conjugate mo-mentum is given by [see Eq. (4.3)] p = mq. Thus for this example the canon-ical conjugate momentum equals the mechanical momentum. Note, however,that this is not necessarily always the case. Using the definition (4.6) onefinds that the Hamiltonian is given by


4.1. Classical Hamiltonian

H = mq2 − mq2

2+ V (q)

=p2

2m+ V (q) .

(4.9)

Hamilton-Jacobi equations (4.7) and (4.8) read

q =p

m(4.10)

p = −∂V∂q

. (4.11)

The second equation, which can be rewritten as

mq = −∂V∂q

, (4.12)

expresses Newton’s second law.

4.1.3 Example

L CL C

Consider a capacitor having capacitance C connected in parallel to aninductor having inductance L. Let q be the charge stored in the capacitor.The kinetic energy in this case T = Lq2/2 is the energy stored in the inductor,and the potential energy V = q2/2C is the energy stored in the capacitor.The canonical conjugate momentum is given by [see Eq. (4.3)] p = Lq, andthe Hamiltonian (4.6) is given by

H =p2

2L+

q2

2C. (4.13)

Hamilton-Jacobi equations (4.7) and (4.8) read

q =p

L(4.14)

p = − q

C. (4.15)

The second equation, which can be rewritten as

Lq +q

C= 0 , (4.16)

expresses the requirement that the voltage across the capacitor is the sameas the one across the inductor.



4.2 Density Function

Consider a classical system in thermal equilibrium. The density functionρ (q, p) is the probability distribution to find the system in the point (q, p).The following theorem is given without a proof. Let H (q, p) be an Hamil-tonian of a system, and assume that H has the following form

H =d

i=1

Aip2i + V (q) , (4.17)

where Ai are constants. Then in the classical limit, namely in the limit wherePlank’s constant approaches zero h→ 0, the density function is given by

ρ (q, p) = N exp (−βH (q, p)) , (4.18)

where

N =13

dq3dp exp (−βH (q, p))

(4.19)

is a normalization constant, β = 1/τ , and τ is the temperature. The notation3dq indicates integration over all coordinates, namely

3dq =

3dq13dq1 ·

· · · ·3dqd, and similarly

3dp =

3dp1

3dp1 · · · · ·

3dpd.

Let A (q, p) be a variable which depends on the coordinates q and theircanonical conjugate momentum variables p. Using the above theorem theaverage value of A can be calculates as:

A (q, p) =

dq

dp A (q, p) ρ (q, p)

=

3dq3dp A (q, p) exp (−βH (q, p))3dq3dp exp (−βH (q, p))

.

(4.20)

4.2.1 Equipartition Theorem

Assume that the Hamiltonian has the following form

H = Biq2i + H , (4.21)

where Bi is a constant and where H is independent of qi. Then the followingholds

Biq

2i

=τ

2. (4.22)

Similarly, assume that the Hamiltonian has the following form


4.2. Density Function

H = Aip2i + H , (4.23)

where Ai is a constant and where H is independent of pi. Then the followingholds

Aip

2i

=τ

2. (4.24)

To prove the theorem for the first case we use Eq. (4.20)

Biq

2i

=

3dq3dp Biq

2i exp (−βH (q, p))3

dq3dp exp (−βH (q, p))

=

3dqi Biq

2i exp

−βBiq

2i

3dqi exp (−βBiq2i )

= − ∂

∂βlog

dqi exp

−βBiq

2i

= − ∂

∂βlog

π

βBi

=1

2β.

(4.25)

The proof for the second case is similar.

4.2.2 Example

Here we calculate the average energy of an harmonic oscillator using both,classical and quantum approaches. Consider a particle having mass m in aone dimensional parabolic potential given by V (q) = (1/2) kq2, where k isthe spring constant. The kinetic energy is given by p2/2m, where p is thecanonical momentum variable conjugate to q. The Hamiltonian is given by

H =p2

2m+kq2

2. (4.26)

In the classical limit the average energy of the system can be easily calculatedusing the equipartition theorem

U = H = τ . (4.27)

In the quantum treatment the system has energy levels given by

Es = sω ,

where s = 0, 1, 2, · · · , and where ω =1k/m is the angular resonance fre-

quency. The partition function is given by



Z =∞

s=0

exp (−sβω) = 1

1− exp (−βω) , (4.28)

thus the average energy U is given by


=ω

eβω − 1. (4.29)

Using the expansion

U = β−1 +O (β) , (4.30)

one finds that in the limit of high temperatures, namely when βω ≪ 1, thequantum result [Eq. (4.30)] coincides with the classical limit [Eq. (4.27)].

4.3 Nyquist Noise

Here we employ the equipartition theorem in order to evaluate voltage noiseacross a resistor. Consider the circuit shown in the figure below, which consistsof a capacitor having capacitance C, an inductor having inductance L, and aresistor having resistance R, all serially connected. The system is assumed tobe in thermal equilibrium at temperature τ . To model the effect of thermalfluctuations we add a fictitious voltage source, which produces a randomfluctuating voltage V (t). Let q (t) be the charge stored in the capacitor attime t. The classical equation of motion, which is given by

q

C+ Lq +Rq = V (t) , (4.31)

represents Kirchhoff’s voltage law.

R

L

CV(t) ~

R

L

CV(t) ~

Fig. 4.1.


4.3. Nyquist Noise

Consider a sampling of the fluctuating function q (t) in the time interval(−T/2, T/2), namely

qT (t) =

q(t) −T/2 < t < T/20 else

. (4.32)

The energy stored in the capacitor is given by q2/2C. Using the equipartitiontheorem one finds

q2

2C=τ

2, (4.33)

where q2is obtained by averaging q2 (t), namely

q2≡ lim

T→∞1

T

+∞

−∞dt q2T (t) . (4.34)

Introducing the Fourier transform:

qT (t) =1√2π

∞

−∞dω qT (ω)e

−iωt, (4.35)

one finds

q2= lim

T→∞1

T

+∞

−∞dt

1√2π

∞

−∞dω qT (ω)e

−iωt 1√2π

∞

−∞dω′ qT (ω

′)e−iω′t

=1

2π

∞

−∞dω qT (ω)

∞

−∞dω′ qT (ω

′) limT→∞

1

T

+∞

−∞dte−i(ω+ω

′)t

2πδ(ω+ω′)

= limT→∞

1

T

∞

−∞dω qT (ω)qT (−ω) .

(4.36)

Moreover, using the fact that q(t) is real one finds

q2= lim

T→∞1

T

∞

−∞dω |qT (ω)|2 . (4.37)

In terms of the power spectrum Sq(ω) of q (t), which is defined as

Sq(ω) = limT→∞

1

T|qT (ω)|2 , (4.38)

one finds

q2=

∞

−∞dω Sq(ω) . (4.39)



Taking the Fourier transform of Eq. (4.31) yields

1

C− iωR− Lω2

q(ω) = V (ω) , (4.40)

where V (ω) is the Fourier transform of V (t), namely

V (t) =1√2π

∞

−∞dω V (ω)e−iωt . (4.41)

In terms of the resonance frequency

ω0 =

1

LC, (4.42)

one has

Lω20 − ω2

− iωR

q(ω) = V (ω) . (4.43)

Taking the absolute value squared yields

Sq(ω) =SV (ω)

L2 (ω20 − ω2)2+ ω2R2

, (4.44)

where SV (ω) is the power spectrum of V (t). Integrating the last result yields

∞

−∞dω Sq(ω) =

∞

−∞dω

SV (ω)

L2 (ω20 − ω2)2+ ω2R2

=1

L2

∞

−∞dω

SV (ω)

(ω0 + ω)2 (ω0 − ω)2 + ω2R2

L2

.

(4.45)

The integrand has a peak near ω0, having a width ≃ R/2L. The Qualityfactor Q is defined as

ω0Q

=R

2L. (4.46)

Assuming SV (ω) is a smooth function near ω0 on the scale ω0/Q, and as-suming Q≫ 1 yield


4.3. Nyquist Noise

∞

−∞dω Sq(ω) ≃

SV (ω0)

L2

∞

−∞

dω

(ω0 + ω)2(ω0 − ω)

2+2ωω0Q

2

≃ SV (ω0)

4ω40L2

∞

−∞

dωω0−ωω0

2+1Q

2

=SV (ω0)

4ω30L2

∞

−∞

dx

x2 +1Q

2

πQ

=SV (ω0)πQ

4ω30L2

.

(4.47)

On the other hand, using Eqs. (4.33) and (4.39) one finds

∞

−∞dω Sq(ω) =

q2= Cτ , (4.48)

therefore

SV (ω0) =4Cω30L

2

πQτ , (4.49)

or using Eqs. (4.42) and (4.46)

SV (ω0) =2Rτ

π. (4.50)

Thus, Eq. (4.44) can be rewritten as

Sq(ω) =2Rτ

π

1

L2 (ω20 − ω2)2+ ω2R2

. (4.51)

Note that the spectral density of V given by Eq. (4.50) is frequency in-dependent. Consider a measurement of the fluctuating voltage V (t) in afrequency band having width ∆f . Using the relation

V 2=

∞

−∞dω SV (ω) , (4.52)

one finds that the variance in such a measurement V 2∆f

is given by

V 2∆f

= 4Rτ∆f . (4.53)

The last result is the Nyquist’s noise formula.



4.4 Thermal Equilibrium From Stochastic Processes

This section demonstrates that under appropriate conditions a stochasticprocess can lead to thermal equilibrium in steady state.

4.4.1 Langevin Equation

Consider the Langevin equation

x = A (x, t) + q (t) , (4.54)

where x is a vector of coordinates that depends on the time t, overdot denotestime derivative, the vector A (x, t) is a deterministic function of x and t, andthe vector q (t) represents random noise that satisfies

q (t) = 0 , (4.55)

and

qi (t) qj (t′) = gijδ (t− t′) . (4.56)

Let

δx = x (t+ δt)− x (t) . (4.57)

To first order in δt one finds by integrating Eq. (4.54) that

(δx)i = Ai (x, t) δt+

t+δt

t

dt′ qi (t′) +O

(δt)2

. (4.58)

With the help of Eqs. (4.55) and (4.56) one finds that

(δx)i = Ai (x, t) δt+O(δt)2

, (4.59)

and(δx)i (δx)j

= Ai (x, t)Aj (x, t) (δt)2

+

t+δt

t

dt′ t+δt

t

dt′′ qi (t′) qj (t′′)+ · · ·

= Ai (x, t)Aj (x, t) (δt)2 + gijδt+ · · · ,

(4.60)

thus to first order in δt(δx)i (δx)j

= gijδt+O

(δt)2

. (4.61)

In a similar way one can show that all higher order moments (e.g. third ordermoments (δx)i′ (δx)i′′ (δx)i′′′) vanish to first order in δt.


4.4. Thermal Equilibrium From Stochastic Processes

4.4.2 The Smoluchowski-Chapman-Kolmogorov Relation

Let p1 (x, t) be the probability density to find the system at point x at timet, let p2 (x

′′, t′′;x′, t′) be the probability density to find the system at point x′

at time t′ and at point x′′ at time t′′, and similarly let p3 (x′′′, t′′′;x′′, t′′;x′, t′)be the probability density to find the system at point x′ at time t′, at pointx′′ at time t′′ and at point x′′′ at time t′′′. The following holds

p2 (x3, t3;x1, t1) =

dx2 p3 (x3, t3;x2, t2;x1, t1) . (4.62)

Let P (x, t|x′, t′) be the conditional probability density to find the system atpoint x at time t given that it was (or will be) at point x′ at time t′. Thefollowing holds

p2 (x3, t3;x1, t1) = P (x3, t3|x1, t1) p1 (x1, t1) . (4.63)

Moreover, by assuming that t1 ≤ t2 ≤ t3 and by assuming the case of aMarkov process, i.e. the case where the future (t3) depends on the present(t2), but not on the past (t1), one finds that

p3 (x3, t3;x2, t2;x1, t1)

= P (x3, t3|x2, t2)P (x2, t2|x1, t1) p1 (x1, t1) .

With the help of these relations Eq. (4.62) becomes

P (x3, t3|x1, t1) p1 (x1, t1)

=

dx2 P (x3, t3|x2, t2)P (x2, t2|x1, t1) p1 (x1, t1) ,

(4.64)

thus by dividing by p1 (x1, t1) one finds that

P (x3, t3|x1, t1) =

dx2 P (x3, t3|x2, t2)P (x2, t2|x1, t1) . (4.65)

4.4.3 The Fokker-Planck Equation

Equation (4.65) can be written as

P (x, t+ δt|x0, t0) =

dx′ P (x, t+ δt|x′, t′)P (x′, t′|x0, t0) . (4.66)

On the other hand

P (x, t+ δt|x′, t′) = δ (x− δx− x′) , (4.67)



where

δx = x (t+ δt)− x (t) . (4.68)

For a general scalar function F of x′ the following holds

F (x′ + δx)

= expδx ·∇′F

= F (x′) + (δx)idF

dx′i+

(δx)i (δx)j2!

d2F

dx′idx′j

+ · · · ,

(4.69)

thus

δ (x− δx− x′) = δ (x− x′)

+ (δx)idδ (x− x′)

dx′i

+(δx)i (δx)j

2!

d2δ (x− x′)dx′idx

′j

+ · · · .

(4.70)

Inserting this result into Eq. (4.66)

P (x, t+ δt|x0, t0)

=

dx′ δ (x− δx− x′)P (x′, t′|x0, t0)

=

dx′ δ (x− x′)P (x′, t′|x0, t0)

+

dx′ (δx)i

dδ (x− x′)dx′i

P (x′, t′|x0, t0)

+1

2

dx′

(δx)i (δx)j


′j

P (x′, t′|x0, t0)

+ · · · ,(4.71)

employing Eqs. (4.59) and (4.61) (recall that higher order moments vanishto first order in δt), dividing by δt, taking the limit δt→ 0

∂P∂t

=

dx′ Ai (x

′, t)dδ (x− x′)

dx′iP (x′, t′|x0, t0)

+1

2

dx′ gij


′j

P (x′, t′|x0, t0) ,

(4.72)



and finally integrating by parts and assuming that P → 0 in the limit x →±∞ yield

∂P∂t

= − d

dx′i(AiP) +

1

2

d2

dx′idx′j

(gijP) . (4.73)

This result, which is known as the Fokker-Planck equation, can also be writtenas

∂P∂t

+∇ · J = 0 , (4.74)

where the probability current density J is given by

Ji = AiP −1

2

d

dx′j(gijP) . (4.75)

In steady state the Fokker-Planck equation (4.74) becomes

∇ · J = 0 . (4.76)

4.4.4 The Potential Condition

Consider the case whereA can be expressed in terms of a scalar ’Hamiltonian’H as (the potential condition)

A (x, t) = −∇H . (4.77)

Moreover, for simplicity assume that

gij = 2τ δij , (4.78)

where the ’temperature’ τ is a constant. For this case one has

Ji = −PdHdx′i

− τ dPdx′i

, (4.79)

thus

J = −P∇H− τ∇P= −P∇ (H+ τ logP) .

(4.80)

Substituting a solution having the form

P = Ne−Hτ (4.81)

yields



J = −τNe−Hτ ∇ (logN) = −τe−H

τ ∇N , (4.82)

and thus

∇ · J = [∇H · ∇N − τ∇ · (∇N)] e−Hτ . (4.83)

In terms of N the Fokker-Planck equation (4.74) becomes

∂N

∂t+ [∇H · ∇N − τ∇ · (∇N)] = 0 . (4.84)

In steady state Eq. (4.84) becomes

∇H · ∇N = τ∇ · (∇N) . (4.85)

This equation can be solved by choosing N to be a constant, which can bedetermined by the normalization condition. In terms of the partition functionZ, where Z = 1/N , the steady state solution is expressed as

P =1

Ze−

Hτ , (4.86)

where

Z =

dx′ P . (4.87)

4.4.5 Free Energy

The free energy functional is defined by

F (P) = U − τσ , (4.88)

where U is the energy

U =

dx′ HP , (4.89)

and σ is the entropy

σ = −

dx′ P logP . (4.90)

The distribution P is constrained to satisfy the normalization constrain

G (P) =

dx′ P − 1 = 0 . (4.91)

Minimizing F (P) under the constrain (4.91) is done by introducing the La-grange multiplier η



δ (F (P) + ηG (P))

=

dx′ [H+ τ (1 + logP) + η] δP .

(4.92)

At a stationary point the following holds

H+ τ (1 + logP) + η = 0 , (4.93)

thus

P = e−ητ−1e−

Hτ , (4.94)

where the constant e−ητ−1 is determined by the normalization condition,

i.e. the obtained distribution is identical to the steady state solution of theFokker-Planck equation (4.86).

4.4.6 Fokker-Planck Equation in One Dimension

In one dimension the Fokker-Planck equation (4.74) becomes [see Eq. (4.79)]

∂P∂t

=∂

∂x

P ∂H∂x

+ τ∂P∂x

, (4.95)

or

∂P∂t

= LP , (4.96)

where the operator L is given by

L =∂

∂x

∂H∂x

+ τ∂2

∂x2. (4.97)

It is convenient to define the operator L, which is given by

L = eH2τ Le− H

2τ . (4.98)

The following holds



L = eH2τ

∂

∂x

∂H∂x

e−H2τ + τe

H2τ

∂2

∂x2e−

H2τ

= eH2τ

∂

∂x

∂H∂x

e−H2τ

+∂H∂x

∂

∂x

+ τeH2τ

∂2

∂x2e−

H2τ

+ 2τe

H2τ∂e−

H2τ

∂x

∂

∂x+ τ

∂2

∂x2

=∂2H∂x2

− 1

2τ

∂H∂x

2+∂H∂x

∂

∂x

− 1

2

∂2H∂x2

+1

4τ

∂H∂x

2− ∂H

∂x

∂

∂x+ τ

∂2

∂x2

=1

2

∂2H∂x2

− 1

4τ

∂H∂x

2+ τ

∂2

∂x2,

(4.99)

thus

L = τ∂2

∂x2− V , (4.100)

where the potential V is given by

V =1

4τ

∂H∂x

2− 1

2

∂2H∂x2

. (4.101)

Note that while the operator L is not Hermitian, the operator L is since

L† = e−H2τ L†e H

2τ

= e−H2τ L†eHτ e− H

2τ

= e−H2τ e

Hτ Le−H

2τ

= eH2τ Le−H

2τ = L .

(4.102)

In terms of L Eq. (4.96) becomes (it is assume that H is time independent)

∂P∂t

= LP , (4.103)

where

P = eH2τ Pe−H

2τ . (4.104)

Let ψn (x) be a set of eigenvectors of L

Lψn = λnψn . (4.105)



The following holds [see Eq. (4.98)]

Lϕn = λnϕn , (4.106)

where

ϕn = e−H2τ ψn . (4.107)

The conditional probability distribution P (x, t|x′, t′) is given by [see Eq.(4.96)]

P (x, t|x′, t′) = eL(t−t′)δ (x− x′) . (4.108)

With the help of the closure relation (recall that L is Hermitian)

δ (x− x′) =

n

ψ∗n (x′)ψn (x)

= eH(x)+H(x′)

2τ

n

ϕ∗n (x′)ϕn (x)

= eH(x′)τ

n

ϕ∗n (x′)ϕn (x) ,

(4.109)

one finds that

P (x, t|x′, t′) = eL(x)(t−t′)e

H(x′)τ

n

ϕ∗n (x′)ϕn (x)

= eH(x′)τ

n

ϕ∗n (x′) eL(x)(t−t

′)ϕn (x)

= eH(x′)τ

n

ϕ∗n (x′) eλn(t−t

′)ϕn (x)

= eH(x′)τ

n

e−H(x′)2τ ψ∗n (x

′) eλn(t−t′)e−

H(x)2τ ψn (x) ,

(4.110)

thus

P (x, t|x′, t′) = eH(x′)−H(x)

2τ

n

ψ∗n (x′)ψn (x) e

λn(t−t′) . (4.111)

4.4.7 Ornstein—Uhlenbeck Process in One Dimension

Consider the following Langevin equation



x+ Γx = q (t) , (4.112)

where x can take any real value, Γ is a positive constant and where the realnoise term q (t) satisfies q (t) = 0 and

q (t) q (t′) = 2τδ (t− t′) , (4.113)

where τ is positive.For this case [see Eq. (4.77)]

H (x) =Γx2

2, (4.114)

the Fokker-Planck equation for the conditional probability distribution P isgiven by [see Eq. (4.74)]

∂P∂t

=∂

∂x

PΓx+ τ

∂P∂x

, (4.115)

and the operator L is given by [see Eq. (4.100)]

L = −Γ2

−

∂

∂

xx0

2

+

x

x0

2− 1

,

where

x0 =

2τ

Γ. (4.116)

The eigenvectors of L are given by [see Eq. (4.131)]

ψn (x) =e− 12

xx0

2

Hn

xx0

π1/4x1/20

√2nn!

, (4.117)

and the corresponding eigenvalues by

λn = −Γn+

1

2− 1

2

, (4.118)

where n = 0, 1, 2, · · · .Using these results one finds that P is given by [see Eq. (4.111)]

P (x, t|x′, t′) = e−xx0

2

n

Hn

x′

x0

Hn

xx0

e−Γn(t−t

′)

√πx02nn!

. (4.119)


4.5. Problems Set 4

With the help of the general identity (4.138) one finds that the followingholds

αe−X2

√π

∞

n=0

α2

nHn (Y )Hn (X)

n!=

exp

−Y−Xα−1√α−2−1

2

1π (α−2 − 1)

, (4.120)

thus Eq. (4.119) becomes

P (x, t|x′, t′) =exp

−x−x′e−Γ(t−t′)

δ

2

√πδ

, (4.121)

where

δ =0x201− e−2Γ (t−t′)

=

,2τ1− e−2Γ (t−t′)

Γ. (4.122)

4.5 Problems Set 4

1. A gas at temperature τ emits a spectral line at wavelength λ0. Thewidth of the observed spectral line is broadened due to motion of themolecules (this is called Doppler broadening). Show that the relationbetween spectral line intensity I and wavelength is given by

I (λ) ∝ exp

"

−mc2 (λ− λ0)

2

2λ20τ

#

, (4.123)

where c is velocity of fight, and m is mass of a molecule.2. The circuit seen in the figure below, which contains a resistor R, ca-

pacitor C, and an inductor L, is at thermal equilibrium at tempera-ture τ . Calculate the average value

I2, where I is the current in the

inductor.

R C LR C L



3. Consider a random real signal q(t) varying in time. Let qT (t) be a sam-pling of the signal q(t) in the time interval (−T/2, T/2), namely

qT (t) =

q(t) −T/2 < t < T/20 else

. (4.124)

The Fourier transform is given by

qT (t) =1√2π

∞

−∞dω qT (ω)e

−iωt , (4.125)

and the power spectrum is given by

Sq(ω) = limT→∞

1

T|qT (ω)|2 .

a) Show that

q2≡ lim

T→∞1

T

+∞

−∞dt q2T (t) =

∞

−∞dω Sq(ω) . (4.126)

b) Wiener-Khinchine Theorem - show that the correlation functionof the random signal q(t) is given by

q (t) q (t+ t′) ≡ limT→∞

1

T

+∞

−∞dt qT (t) qT (t+ t′) =

∞

−∞dω eiωt

′

Sq(ω) .

(4.127)

4. Consider a resonator made of a capacitor C, an inductor L, and a resistorR connected in series, as was done in class. Let I (t) be the current in thecircuit. Using the results obtained in class calculate the spectral densitySI (ω) of I at thermal equilibrium. Show that in the limit of high qualityfactor, namely when

Q =2

R

L

C≫ 1 , (4.128)

your result is consistent with the equipartition theorem applied for theenergy stored in the inductor.

5. A classical system is described using a set of coordinates q1, q2, · · · , qNand the corresponding canonically conjugate variables p1, p2, · · · , pN.The Hamiltonian of the system is given by

H =N

n=1

Anpsn +Bnq

tn , (4.129)

where An and Bn are positive constants and s and t are even positiveintegers. Show that the average energy of the system in equilibrium attemperature τ is given by


4.5. Problems Set 4

U = Nτ

1

s+

1

t

, (4.130)

6. A small hole of area A is made in the walls of a vessel of volume V con-taining a classical ideal gas of N particles of massM each in equilibriumat temperature τ . Calculate the number of particles dN , which escapethrough the opening during the infinitesimal time interval dt.

7. Consider an ideal gas of Fermions having massM and having no internaldegrees of freedom at temperature τ . The velocity of a particle is denoted

as v =0v2x + v2y + v2z . Calculate the quantity

v41

v

5

(the symbol denoted averaging) in the:

a) classical limit (high temperatures).b) zero temperature.

8. Consider an ideal gas of N molecules, each of mass M , contained in acentrifuge of radius R and length L rotating with angular velocity ωabout its axis. Neglect the effect of gravity. The system is in equilibriumat temperature τ = 1/β. Calculate the particle density n (r) as a functionof the radial distance from the axis r ( where 0 ≤ r ≤ R) .

9. Consider an ideal classical gas of particles having massM and having no

internal degrees of freedom at temperature τ . Let v =0v2x + v2y + v2z .

be the velocity of a particle. Calculate

a) vb)1v2

10. A mixture of two classical ideal gases, consisting of N1 and N2 particlesof mass M1 and M2, respectively, is enclosed in a cylindrical vessel ofheight h and area of bottom and top side S. The vessel is placed in agravitational field having acceleration g. The system is in thermal equi-librium at temperature τ . Find the pressure exerted on the upper wall ofthe cylinder.

11. The Hermite polynomial Hn (X) of order n is defined by

Hn (X) = exp

X2

2

X − d

dX

nexp

−X

2

2

. (4.131)

For some low values of n the Hermite polynomials are given by

H0 (X) = 1 , (4.132)

H1 (X) = 2X , (4.133)

H2 (X) = 4X2 − 2 , (4.134)

H3 (X) = 8X3 − 12X , (4.135)

H4 (X) = 16X4 − 48X2 + 12 . (4.136)



Show that

exp2Xt− t2

=

∞

n=0

Hn (X)tn

n!. (4.137)

12. Show that

∞

n=0

α2

nHn (X)Hn (Y )

n!=

exp

α(2XY−αX2−αY 2)

1−α2

√1− α2

. (4.138)

4.6 Solutions Set 4

1. Let λ be the wavelength measured by an observer, and let λ0 be thewavelength of the emitted light in the reference frame where the moleculeis at rest. Let vx be the velocity of the molecule in the direction of thelight ray from the molecule to the observer. Due to Doppler effect

λ = λ0 (1 + vx/c) . (4.139)

The probability distribution f (vx) is proportional to

f (vx) ∝ exp

−mv

2x

2τ

, (4.140)

thus using

vx =c (λ− λ0)

λ0, (4.141)

the probability distribution I (λ) is proportional to

I (λ) ∝ exp

"

−mc2 (λ− λ0)

2

2λ20τ

#

. (4.142)

2. The energy stored in the inductor is UL = LI2/2. Using the equipartitiontheorem UL = τ/2, thus

I2=

τ

L. (4.143)

3.

a)



q2= lim

T→∞1

T

+∞

−∞dt

1√2π

∞

−∞dω qT (ω)e

−iωt 1√2π

∞

−∞dω′ qT (ω

′)e−iω′t

=1

2π

∞

−∞dω qT (ω)

∞

−∞dω′ qT (ω

′) limT→∞

1

T

+∞

−∞dt e−i(ω+ω

′)t

2πδ(ω+ω′)

= limT→∞

1

T

∞

−∞dω qT (ω)qT (−ω) .

(4.144)

Since q(t) is real one has qT (−ω) = q∗T (ω), thus

q2=

∞

−∞dω Sq(ω) . (4.145)

b)

q (t) q (t+ t′) ≡ limT→∞

1

T

+∞

−∞dt qT (t) qT (t+ t′)

=1

2πlimT→∞

1

T

+∞

−∞dt

∞

−∞dω qT (ω)e

−iωt ∞

−∞dω′ qT (ω

′)e−iω′(t+t′)

=1

2πlimT→∞

1

T

∞

−∞dω e−iω

′t′qT (ω)

∞

−∞dω′ qT (ω

′)

+∞

−∞dt e−i(ω+ω

′)t

2πδ(ω+ω′)

= limT→∞

1

T

∞

−∞dω eiωt

′

qT (ω)qT (−ω)

=

∞

−∞dω eiωt

′

Sf (ω) .

(4.146)

4. Using I (ω) = −iωq (ω) and q2= Cτ one finds for the case Q≫ 1

I2=

∞

−∞dω SI(ω) ≃ ω20

∞

−∞dω Sq(ω) = ω20

q2=

τ

L, (4.147)

in agreement with the equipartition theorem for the energy stored in theinductor LI2/2.

5. Calculate for example



Bnq

tn

=

3∞−∞Bnq

tn exp (−βBnq

tn) dqn3∞

−∞ exp (−βBnqtxn) dqn

=

3∞0 Bnqtn exp (−βBnqtn) dqn3∞

0 exp (−βBnqtxn) dqn

= − d

dβlog

∞

0

exp−βBnq

tn

dqn .

(4.148)

where β = 1/τ . Changing integration variable

x = βBnqtn, (4.149)

dx = tβBnqt−1n dqn , (4.150)

leads to

Bnq

tn

= − d

dβlog

(βBn)

− 1t t−1

∞

0

x1t−1e−xdx

=τ

t. (4.151)

Thus

U =N

n=1

Anpsn+

Bnq

tn

= Nτ

1

s+

1

t

. (4.152)

6. Let f (v) be the probability distribution of velocity v of particles in thegas. The vector u is expressed in spherical coordinates, where the z axisis chosen in the direction of the normal outward direction

v = v (sin θ cosϕ, sin θ sinϕ, cos θ) . (4.153)

By symmetry, f (v) is independent of θ and ϕ. The number dN is calcu-lated by integrating over all possible values of the velocity of the leavingparticles (note that θ can be only in the range 0 ≤ θ ≤ π/2)

dN =

∞

0

dv

1

0

d (cos θ)

2π

0

dϕv2v (dt)A cos θN

Vf (v) . (4.154)

Note that v (dt)A cos θ represents the volume of a cylinder from whichparticles of velocity v can escape during the time interval dt. Since f (v)is normalized

1 =

∞

0

dv

1

0

d (cos θ)

2π

0

dϕv2f (v) = 4π

∞

0

dvv2f (v) , (4.155)

thus

dN

dt=πNA

V

∞

0

dvv2vf (v) =NA v4V

. (4.156)



In the classical limit

f (v) ∝ exp

−Mv2

2τ

, (4.157)

thus, by changing the integration variable

x =Mv2

2τ(4.158)

one finds

v =3∞0

dvv3 exp−Mv2

2τ

3∞0

dvv2 exp−Mv2

2τ

=

2τ

M

1/2 3∞0 dxx exp (−x)

3∞0 dxx1/2 exp (−x)

=

8τ

πM

1/2,

(4.159)

and

dN

dt=NA

4V

8τ

πM

1/2. (4.160)

7. The probability that an orbital having energy ε is occupied is given by

fF (ε) =1

1 + exp [(ε− µ)β], (4.161)

where β = 1/τ and µ is the chemical potential. The velocity v of such anorbital is related to the energy ε by

ε =Mv2

2. (4.162)

The 3D density of state per unit volume is given by

g (ε) =1

2π2

2M

2

3/2ε1/2 . (4.163)

Thus



v41

v

5=

∞3

0

dε g (ε) fF (ε) v

∞3

0

dε g (ε) fF (ε)

∞3

0

dε g (ε) fF (ε) 1v

∞3

0

dε g (ε) fF (ε)

=

∞3

0

dε εfF (ε)

∞3

0

dε fF (ε)

∞3

0

dε ε1/2fF (ε)

2 .

(4.164)

a) In the classical limit

fF (ε) ∝ exp (−βε) , (4.165)

thus using the identities

∞

0

dε εn exp (−βε) = Γ (n)β−nn

β

∞

0

dε exp (−βε) = 1

β

Γ (1) = 1

Γ

1

2

=√π

one finds

v41

v

5=

∞3

0

dε ε exp (−βε)∞3

0

dε exp (−βε)

∞3

0

dε ε1/2 exp (−βε)2

=Γ (1)β−1 1β

1β

Γ12

β−1/2 1

2β

2

=4

π.

(4.166)

b) Using the identity

εF

0

dε εn =εn+1F

n+ 1



one finds

v41

v

5=

εF3

0

dε ε

εF3

0

dε

εF3

0

dε ε1/22 .

=12ε

2FεF

23ε

32

F

2

=9

8.

(4.167)

8. The effect of rotation is the same as an additional external field withpotential energy given by

U (r) = −1

2Mω2r2 , (4.168)

thus

n (r) = A exp [−βU (r)] = A exp

βMω2

2r2, (4.169)

where the normalization constant A is found from the condition

N = 2πL

R

0

n (r) rdr

= 2πLA

R

0

exp

βMω2

2r2rdr

=2πLA

βMω2

exp

βMω2

2R2

− 1

,

(4.170)

thus

n (r) =NβMω2

2πL8exp

βMω2

2 R2− 19 exp

β

2Mω2r2

. (4.171)

9. In the classical limit the probability distribution of the velocity vector vsatisfies

f (v) ∝ exp

−Mv2

2τ

, (4.172)

where v = |v|.



a) By changing the integration variable

x =Mv2

2τ(4.173)

one finds

v =3∞0

dvv3 exp−Mv2

2τ

3∞0 dvv2 exp

−Mv2

2τ

(4.174)

=

2τ

M

1/2 3∞0 dxx exp (−x)

3∞0 dxx1/2 exp (−x)

=

8τ

πM

1/2.

(4.175)

b) Similarly

v2=

3∞0

dv v4 exp−Mv2

2τ

3∞0

dv v2 exp−Mv2

2τ

=2τ

M

3∞0 dx x3/2 exp (−x)3∞0

dx x1/2 exp (−x)

=2τ

M

3

2,

(4.176)

thus

1v2 =

3τ

M

1/2=

3π

8

1/2v . (4.177)

10. For each gas the density is given by

nl (z) = nl (0) exp (−βMlgz) ,

where l ∈ 1, 2, 0 ≤ z ≤ h and the normalization constant is found fromthe requirement

S

h

0

n (z) dz = Nl , (4.178)

therefore

nl (0) =Nl

Sh3

0

exp (−βMlgz) dz

=βMlgNl

S (1− e−βMlgh). (4.179)



Using the equation of state p = nτ , where n = N/V is the density, onefinds that the pressure on the upper wall of the cylinder is given by

p = (n1 (h) + n2 (h)) τ

=

M1N1

exp (βM1gh)− 1+

M2N2

exp (βM2gh)− 1

g

S.

(4.180)

11. The relation (4.137), which is a Taylor expansion of the function f (t) =exp

2Xt− t2

around the point t = 0, implies that

Hn (X) =dn

dtnexp

2Xt− t2

2222t=0

. (4.181)

The identity 2Xt− t2 = X2 − (X − t)2 yields

Hn (X) = expX2 dn

dtnexp

− (X − t)2

2222t=0

. (4.182)

Moreover, using the relation

d

dtexp

− (X − t)2

= − d

dXexp

− (X − t)2

, (4.183)

one finds that

Hn (X) = expX2(−1)n dn

dXnexp

− (X − t)2

2222t=0

= expX2(−1)n dn

dXnexp

−X2

.

(4.184)

Note that for an arbitrary function g (X) the following holds

− expX2 d

dXexp

−X2

g =

2X − d

dX

g , (4.185)

and

exp

X2

2

X − d

dX

exp

−X

2

2

g =

2X − d

dX

g , (4.186)

thus

Hn (X) = exp

X2

2

X − d

dX

nexp

−X

2

2

. (4.187)



12. With the help of Eq. (4.184) and the general identity

∞

−∞

exp−ax2 + bx+ c

dx =

π

ae144ca+b2

a , (4.188)

according to which the following holds (for the case a = 1, b = 2iX andc = 0)

exp−X2

=

1√π

∞

−∞

exp−x2 + 2iXx

dx , (4.189)

one finds that

Hn (X) =exp

X2

√π

− d

dX

n ∞

−∞

exp−x2 + 2iXx

dx

=1√π

∞

−∞

(−2ix)n expX2 − x2 + 2iXx

dx

=1√π

∞

−∞

(−2ix)n e(X+ix)2dx ,

(4.190)

thus the following holds [see Eq. (4.188)]

∞

n=0

α2

nHn (X)Hn (Y )

n!

=1

π

∞

−∞

dx

∞

−∞

dy e(X+ix)2e(Y+iy)2∞

n=0

(−2αxy)nn!

e−2αxy

=1

π

∞

−∞

dx e(X+ix)2∞

−∞

dy e(Y+iy)2

e−2αxy

√πeαx(αx−2iY )

=1√π

∞

−∞

dx e−(1−α2)x2+2i(X−Y α)x+X2

=

exp

α(2XY−αX2−αY 2)

1−α2

√1− α2

.

(4.191)


5. Exercises

5.1 Problems

1. Two identical non-interacting particles each having massM are confinedin a one dimensional parabolic potential given by

V (x) =1

2Mω2x2 , (5.1)

where the angular frequency ω is a constant.

a) Calculate the canonical partition function of the system Zc,B for thecase where the particles are Bosons.

b) Calculate the canonical partition function of the system Zc,F for thecase where the particles are Fermions.

2. Consider a one dimensional gas containing N non-interacting electronsmoving along the x direction. The electrons are confined to a section oflength L. At zero temperature τ = 0 calculate the ratio U/εF betweenthe total energy of the system U and the Fermi energy εF.

3. Consider an ideal classical gas at temperature τ . The set of internaleigenstates of each particle in the gas, when a magnetic fieldH is applied,contains 2 states having energies ε− = −µ0H and ε+ = µ0H, where themagnetic moment µ0 is a constant. Calculate the magnetization of thesystem, which is defined by

M = −∂F

∂H

τ

, (5.2)

where F is the Helmholtz free energy.4. (Note: replace this with Ex. 3.9 in the lecture notes) Consider an ideal gas

made of N electrons in the extreme relativistic limit. The gas is containedin a box having a cube shape with a volume V = L3. In the extremerelativistic limit the dispersion relation ε (k) is modified: the energy ε ofa single particle quantum state having a wavefunction ψ given by

ψ (x, y, z) =

2

L

3/2sin (kxx) sin (kyy) sin (kzz) , (5.3)

Chapter 5. Exercises

is given by

ε (k) = kc , (5.4)

where c is the speed of light and where k =0k2x + k2y + k2z (contrary

to the non-relativistic case where it is given by ε (k) = 2k2/2M). The

system is in thermal equilibrium at zero temperature τ = 0. Calculatethe ratio p/U between the pressure p and the total energy of the systemU .

5. Consider a mixture of two classical ideal gases, consisting of NA particlesof type A and NB particles of type B. The heat capacities cp,A and cV,A(cp,B and cV,B) at constant pressure and at constant volume respectivelyof gas A (B) are assumed to be temperature independent. The volume ofthe mixture is initially V1 and the pressure is initially p1. The mixtureundergoes an adiabatic (slow) and isentropic (at a constant entropy)process leading to a final volume V2. Calculate the final pressure p2.

6. Consider two particles, both having the same mass m, moving in a one-dimensional potential with coordinates x1 and x2 respectively. The po-tential energy is given by

V (x1, x2) =mω2x21

2+mω2x22

2+mΩ2 (x1 − x2)

2 , (5.5)

where the angular frequencies ω and Ω are real constants Assume thatthe temperature τ of the system is sufficiently high to allow treating itclassically. Calculate the following average values

a) x21for the case Ω = 0.

b) x21, however without assuming that Ω = 0.

c)(x1 − x2)

2, again without assuming that Ω = 0.

7. Consider an ideal classical gas containing N identical particles havingeach mass M in the extreme relativistic limit. The gas is contained ina vessel having a cube shape with a volume V = L3. In the extremerelativistic limit the dispersion relation ε (k) is modified: the energy ε ofa single particle quantum state having a wavefunction ψ given by

ψ (x, y, z) =

2

L

3/2sin (kxx) sin (kyy) sin (kzz) , (5.6)

is given by

ε (k) = kc , (5.7)

where c is the speed of light and where k =0k2x + k2y + k2z (contrary

to the non-relativistic case where it is given by ε (k) = 2k2/2M). The

system is in thermal equilibrium at temperature τ . Calculate:


5.2. Solutions

a) the total energy U of the system.b) the pressure p.

8. Consider a system made of two localized spin 1/2 particles whose energyis given by

εσ1,σ2 = −µ0H (σ1 + σ2) + Jσ1σ2 , (5.8)

where both σ1 and σ2 can take one of two possible values σn = ±1 (n =1, 2). While H is the externally applied magnetic field, J is the couplingconstant between both spins. The system is in thermal equilibrium attemperature τ . Calculate the magnetic susceptibility

χ = limH→0

∂M

∂H, (5.9)

where

M = −∂F

∂H

τ

(5.10)

is the magnetization of the system, and where F is the Helmholtz freeenergy.

9. Consider a one dimensional gas containing N non-interacting electronsmoving along the x direction. The electrons are confined by a potentialgiven by

V (x) =1

2mω2x2 , (5.11)

where m is the electron mass and where ω is the angular frequency ofoscillations. Calculate the chemical potential µ

a) in the limit of zero temperature τ = 0.b) in the limit of high temperatures τ ≫ ω.

10. The state equation of a given matter is

p =Aτn

V, (5.12)

where p, V and τ are the pressure, volume and temperature, respectively,and A and n are both constants. Calculate the difference cp−cV betweenthe heat capacities at constant pressure and at constant volume.

5.2 Solutions

1. The single particle eigen energies are given by

ǫn = ω

n+

1

2

, (5.13)

where n = 0, 1, 2, · · · .



a) For Bosons

Zc,B =1

2

∞

n=0

∞

m=0

exp [−β (ǫn + ǫm)] +1

2

∞

n=0

exp (−2βǫn)

=1

2

∞

n=0

exp (−βǫn)2

+1

2

∞

n=0

exp (−2βǫn)

=exp (−βω)

2 (1− exp (−βω))2+

exp (−βω)2 (1− exp (−2βω)) .

(5.14)Note that the average energy UB is given by

UB = −∂ logZc,B∂β

= ω1 + 2e−2βω + e−βω

1− e−2βω. (5.15)

b) For Fermions

Zc,F =1

2

∞

n=0

∞

m=0

exp [−β (ǫn + ǫm)]− 1

2

∞

n=0

exp (−2βǫn)

=exp (−βω)

2 (1− exp (−βω))2− exp (−βω)

2 (1− exp (−2βω)) .

(5.16)Note that for this case the average energy UF is given by

UF = −∂ logZc,F∂β

= ω2 + e−2βω + e−βω

1− e−2βω. (5.17)

2. The orbital eigenenergies are given by

εn =2

2m

πL

2n2 , (5.18)

where n = 1, 2, 3, · · · . The grandcanonical partition function of the gasis given by

Zgc =

n

ζn , (5.19)

where

ζn =

l

(1 + λ exp (−βεn) exp (−βEl)) (5.20)

is the orbital grandcanonical Fermionic partition function where,

λ = exp (βµ) = e−η , (5.21)

is the fugacity, β = 1/τ and El are the eigenenergies of a particle dueto internal degrees of freedom. For electrons, in the absence of magnetic


5.2. Solutions

field both spin states have the same energy, which is taken to be zero.Thus, logZgc can be written as

logZgc =∞

n=1

log ζn

= 2∞

n=1

log (1 + λ exp (−βεn))

≃ 2

∞

0

dn log

1 + λ exp

−β

2

2m

πL

2n2

.

(5.22)

By employing the variable transformation

ε =2

2m

πL

2n2 , (5.23)

one has

logZgc =1

2

∞

0

dε D (ε) log (1 + λ exp (−βε)) , (5.24)

where

D (ε) =

2Lπ

02m2ε−1/2 ε ≥ 0

0 ε < 0(5.25)

is the 1D density of states. Using Eqs. (1.80) and (1.94) for the energyU and the number of particles N , namely using

U = −∂ logZgc

∂β

η

, (5.26)

N = λ∂ logZgc

∂λ, (5.27)

one finds that

U =

∞

−∞

dε D (ε) εfFD (ε) , (5.28)

N =

∞

−∞

dε D (ε) fFD (ε) , (5.29)


fFD (ǫ) =1

exp [β (ǫ− µ)] + 1. (5.30)



At zero temperature, where µ = εF one has

U =D (εF)

ε−1/2F

εF

0

dε ε1/2 =2D (εF)

3ε2F , (5.31)

N =D (εF)

ε−1/2F

εF

0

dε ε−1/2 = 2D (εF) εF , (5.32)

thus

U

εF=N

3. (5.33)

3. The Helmholtz free energy is given by

F = Nτ

log

n

nQ− logZint − 1

, (5.34)

where

Zint = exp (βµ0H) + exp (−βµ0H) = 2 cosh (βµ0H) (5.35)

is the internal partition function. Thus the magnetization is given by

M = −∂F

∂H

τ

= Nµ0 tanh (βµ0H) . (5.36)

4. The grandcanonical partition function of the gas is given by

Zgc =

n

ζn, (5.37)

where

ζn=

l

(1 + λ exp (−βεn) exp (−βEl)) (5.38)

is a grandcanonical Fermionic partition function of an orbital havingenergy εn given by

εn =πcn

L, (5.39)

where n =0n2x + n2y + n2z, nx, ny, nz = 1, 2, 3, · · · ,

λ = exp (βµ) = e−η (5.40)

is the fugacity, β = 1/τ and El are the eigenenergies of a particle dueto internal degrees of freedom. For electrons, in the absence of magnetic


5.2. Solutions

field both spin states have the same energy, which is taken to be zero.Thus, logZgc can be written as

logZgc =

l

∞

nx=1

∞

ny=1

∞

nz=1

log (1 + λ exp (−βεn) exp (−βEl)) . (5.41)

For a macroscopic system the sum over n can be approximately replacedby an integral

∞

nx=0

∞

ny=0

∞

nz=0

→ 4π

8

∞

0

dn n2 , (5.42)

thus, one has

logZgc = 24π

8

∞

0

dn n2 log

1 + λ exp

−β πcn

L

. (5.43)

By employing the variable transformation

ε =πcn

L. (5.44)

one has

logZgc =∞

0

dεV ε2

π23c3log (1 + λ exp (−βε)) . (5.45)

The energy U and the number of particles N are given by

U = −∂ logZgc

∂β

η

=

∞

0

dεV ε3

π23c3fFD (ǫ) , (5.46)

N = λ∂ logZgc

∂λ=

∞

0

dεV ε2

π23c3fFD (ǫ) , (5.47)


fFD (ǫ) =1

exp [β (ǫ− µ)] + 1. (5.48)

At zero temperature

U =

εF

0

dεV ε3

π23c3=

V

π23c3ε4F4, (5.49)

N =

εF

0

dεV ε2

π23c3=

V

π23c3ε3F3, (5.50)



and therefore

U =3N

4εF . (5.51)

The energy U can be expressed as a function of V and N as

U =(3N)4/3

π23c3

1/3V −1/3

4.

At zero temperature the Helmholtz free energy F equals the energy U ,thus the pressure p is given by

p = −∂F

∂V

τ,N

= −∂U

∂V

τ,N

=1

3

3NV

4/3 π23c3

1/3

4, (5.52)

thus

p

U=

1

3V. (5.53)

5. First, consider the case of an ideal gas made of a unique type of particles.Recall that the entropy σ, cV and cp are given by [see Eqs. (2.87), (2.88)and (2.89)]

σ = N

5

2+ log

nQn

+∂ (τ logZint)

∂τ

, (5.54)

cV = N

3

2+ hint

, (5.55)

cp = cV +N , (5.56)

where n = N/V is the density,

nQ =

Mτ

2π2

3/2(5.57)

is the quantum density, M is the mass of a particle in the gas, and

hint = τ∂2 (τ logZint)

∂τ2=cVN− 3

2. (5.58)

The requirement that hint is temperature independent leads to

∂ (τ logZint)

∂τ= gint + hint log

τ

τ0, (5.59)

where both gint and τ0 are constants. Using this notation, the change inentropy due to a change in V from V1 to V2 and a change in τ from τ1to τ2 is given by


5.2. Solutions

∆σ = σ2 − σ1 = N

logV2τ

3/22

V1τ3/21

+

cVN− 3

2

log

τ2τ1

= N log

V2V1

τ2τ1

cVN

.

(5.60)

Thus the total change in the entropy of the mixture is given by

∆σ = ∆σA +∆σB

= NA log

V2V1

τ2τ1

cV,ANA

+NB log

V2V1

τ2τ1

cV,BNB

(5.61)

= (NA +NB) log

V2V1

τ2τ1

cV,A+cV,BNA+NB

, (5.62)

and the requirement ∆σ = 0 leads to

V2V1

τ2τ1

cV,A+cV,BNA+NB

= 1 . (5.63)

Alternatively, by employing the equation of state

pV = Nτ , (5.64)

this can be rewritten as

V2V1

cV,A+cV,BNA+NB

+1p2p1

cV,A+cV,BNA+NB

= 1 , (5.65)

or with the help of Eq. (5.56) as

p2 = p1

V2V1

−cV,A+cV,BNA+NB

+1

cV,A+cV,BNA+NB

= p1

V1V2

cp,A+cp,BcV,A+cV,B

.

(5.66)

6. It is convenient to employ the coordinate transformation

x+ =x1 + x2√

2, (5.67)

x− =x1 − x2√

2. (5.68)

The inverse transformation is given by



x1 =x+ + x−√

2, (5.69)

x2 =x+ − x−√

2. (5.70)

The following holds

x21 + x22 = x2+ + x2− , (5.71)

and

x21 + x22 = x2+ + x2− . (5.72)

Thus, the kinetic energy T of the system is given by

T =mx21 + x22

2=mx2+ + x2−

2, (5.73)

and the potential energy V is given by

V (x1, x2) =mω2x21

2+mω2x22

2+mΩ2 (x1 − x2)

2

=mω2x2+

2+mω2 + 4Ω2

x2−

2.

(5.74)

The equipartition theorem yields

mω2 x2+

2=mω2 + 4Ω2

x2−

2=τ

2, (5.75)

thus(x1 + x2)

2=

2τ

mω2, (5.76)

and(x1 − x2)

2=

2τ

m (ω2 + 4Ω2). (5.77)

Furthermore, since by symmetry x+x− = 0 one has

x21=

1

2

(x+ + x−)

2

=1

2

x2++ x2−

=τ

mω2

1− 1

2

4Ω2

ω2

1 + 4Ω2

ω2

.

(5.78)


5.2. Solutions

7. The k vector is restricted due to boundary conditions to the values

k =πn

L, (5.79)

where

n = (nx, ny, nz) , (5.80)

and nx, ny, nz = 1, 2, 3, · · · . The single particle partition function is givenby

Z1 =∞

nx=1

∞

ny=1

∞

nz=1

exp

−ε (k)

τ

. (5.81)

Approximating the discrete sum by a continuous integral according to

∞

nx=0

∞

ny=0

∞

nz=0

→ 4π

8

∞

0

dn n2 , (5.82)

one has

Z1 =4π

8

∞

0

dn n2 exp

−nπc

Lτ

=4V τ3

8π23c3

∞

0

dx x2 exp (−x)

2

=V τ3

π23c3.

(5.83)

In the classical limit the grandcanonical partition function Zgc is givenby [see Eq. (2.44)]

logZgc = λZ1 , (5.84)

where λ = exp (βµ) is the fugacity. In terms of the Lagrange multipliersη = −µ/τ and β = 1/τ the last result can be rewritten as

logZgc = e−ηV

π23c3β3. (5.85)

a) The average energy U and average number of particle N are calcu-lated using Eqs. (1.80) and (1.81) respectively



U = −∂ logZgc

∂β

η

=3

βlogZgc , (5.86)

N = −∂ logZgc

∂η

β

= logZgc , (5.87)

thus

U = 3Nτ , (5.88)

and

η = log

V τ3

π2N3c3

. (5.89)

b) The entropy is evaluate using Eq. (1.86)σ = logZgc + βU + ηN

= N (1 + 3 + η)

= N

4 + log

V τ3

π2N3c3

,

(5.90)and the Helmholtz free energy by the definition (1.116)

F = U − τσ = −Nτ1 + log

V τ3

π2N3c3

. (5.91)

Thus the pressure p is given by

p = −∂F

∂V

τ,N

=Nτ

V. (5.92)

8. The partition function is given by

Z =

σ1,σ2=±1exp (−βεσ1,σ2)

= exp (−βJ) [exp (−2βµ0H) + exp (2βµ0H)] + 2 exp (βJ) ,

(5.93)

where β = 1/τ . The free energy is given by

F = −τ logZ , (5.94)

thus the magnetization is given by

M = −∂F

∂H

τ

=2µ0 exp (−βJ) [− exp (−2βµ0H) + exp (2βµ0H)]

exp (−βJ) [exp (−2βµ0H) + exp (2βµ0H)] + 2 exp (βJ),

(5.95)


5.2. Solutions

and the magnetic susceptibility is given by

χ =4βµ20

1 + e2βJ. (5.96)

Note that in the high temperature limit βJ ≪ 1

χ ≃ 2µ20τ + J

. (5.97)

9. The orbital eigenenergies in this case are given by

εn = ω

n+

1

2

, (5.98)

where n = 0, 1, 2, · · · . The grandcanonical partition function of the gasis given by

Zgc =

n

ζn , (5.99)

where

ζn =

l

(1 + λ exp (−βεn) exp (−βEl)) (5.100)

is the orbital grandcanonical Fermionic partition function where,

λ = exp (βµ) = e−η (5.101)

is the fugacity, β = 1/τ and El are the eigenenergies of a particle dueto internal degrees of freedom. For electrons, in the absence of magneticfield both spin states have the same energy, which is taken to be zero.Thus, logZgc can be written as

logZgc =∞

n=0

log ζn

= 2∞

n=0

log (1 + λ exp (−βεn)) .

(5.102)

The number of particles N is given by

N = λ∂ logZgc

∂λ= 2

∞

n=0

fFD (εn) , (5.103)

where fFD is the Fermi-Dirac distribution function

fFD (ε) =1

exp [β (ε− µ)] + 1. (5.104)



a) At zero temperature the chemical potential µ is the Fermi energy εF,and the Fermi-Dirac distribution function becomes a step function,thus with the help of Eq. (5.103) one finds that

N =2εFω

, (5.105)

thus

µ = εF =Nω

2. (5.106)

b) Using the approximation

fFD (ε) ≃ exp [−β (ε− µ)] , (5.107)

for the the limit of high temperatures and approximating the sum byan integral one has

N = 2∞

n=0

exp

−βω

n+

1

2

− µ

= 2 exp

β

µ− ω

2

∞

0

dn exp (−βωn)

=2 exp

βµ− ω

2

βω,

(5.108)thus

µ = τ

log

Nβω

2

+βω

2

≃ τ log

Nβω

2

.

(5.109)

10. Using Eq. (2.239), which is given by

cp − cV = τ

∂p

∂τ

V,N

∂V

∂τ

p,N

, (5.110)

one finds that

cp − cV =A2τ

pVn2τ2(n−1) = n2Aτn−1 . (5.111)


References

1. C. E. Shannon, Bell System Tech. J. 27, 379 (1948).2. C. E. Shannon, Bell System Tech. J. 27, 623 (1948).3. E. T. Jaynes, Phys. rev. Lett. 106, 620 (1957).

Index

Bose-Einstein distribution, 54Bose-Einstein function, 54Boson, 52

canonical conjugate momentum, 129canonical distribution, 15Carnot, 66cavity, 99chemical potential, 17, 20classical limit, 54Clausius’s principle, 70composition property, 5Coulomb gauge, 99

Debye temperature, 110density function, 132density of states, 114

efficiency, 68, 71electromagnetic radiation, 99entropy, 1equipartition theorem, 132

Fermi energy, 115Fermi-Dirac distribution, 53Fermi-Dirac function, 53Fermion, 52Fokker-Planck equation, 139free energy, 142fugacity, 18, 53

grandcanonical distribution, 16

H theorem, 18Hamilton-Jacobi equations, 130Hamiltonian, 129heat capacity, 59Helmholtz free energy, 21, 57

ideal gas, 47

information theory, 1internal degrees of freedom, 59isentropic process, 65isobaric process, 64isochoric process, 65isothermal process, 64

Kelvin’s principle, 70

Lagrange multiplier, 4Langevin equation, 138largest uncertainty estimator, 9

Maxwell’s equations, 99microcanonical distribution, 14

Nyquist noise, 134

orbitals, 50Ornstein—Uhlenbeck process, 145

partition function, 10permeability, 99permittivity, 99phonon, 107photon, 102Plank’s radiation law, 105pressure, 57

quantum density, 48

Shannon, 1Sommerfeld expansion, 115Stefan-Boltzmann constant, 106Stefan-Boltzmann radiation law, 105

temperature, 15, 17thermal equilibrium, 19

vector potential, 99

td sm lecture notes · 2017-09-24 · chapter1. theprincipleoflargestuncertainty δσ= ∇¯σ ⊥...

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