Kpn Stat Mech

182
K. P. N. Murthy Statistical Mechanics January 29, 2014 Springer

description

a textbook on statistical mechanics

Transcript of Kpn Stat Mech

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K. P. N. Murthy

Statistical Mechanics

January 29, 2014

Springer

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Contents

1 Why should we study statistical mechanics ? . . . . . . . . . . . . . . 11.1 Micro-Macro connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 S = kB ln Ω(E, V,N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 S = −kB∑

i pi ln(pi) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.2 Q =

∑i Ei dpi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.3 W =∑

i pi dEi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.4 F = −kB T lnQ(T, V,N) . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.5 σ2

E = kB T 2 CV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.6 σ2N =

〈N〉2kBTV

kT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.7 Micro world : Determinism and time-reversal invariance 31.3 Macro world : Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Micro-Macro synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 What am I going to teach ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.6 Syllabus prescribed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.7 Books on statistical mechanics and thermodynamics . . . . . . . . . 61.8 Extra reading : Books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.9 Extra reading : Papers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Experiment, outcomes, events, probabilities and ensemble . 132.1 Toss a coin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Roll a die . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 Pick randomly a molecule from this room. . . . . . . . . . . . . . . . . . . 132.4 Sample space and events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.6 Rules of probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.7 Random variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.8 Maxwell’s mischief : ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.9 Calculation of probabilities from an ensemble . . . . . . . . . . . . . . . 172.10 Construction of ensemble from probabilities . . . . . . . . . . . . . . . . . 17

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2.11 Counting of the number of elements in events of the samplespace : Coin tossing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.12 Gibbs ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.13 Why should a Gibbs ensemble be of large size ? . . . . . . . . . . . . . 19

2.13.1 Stirling and his Approximation to Large Factorials . . . . 202.13.2 N ! = NN exp(−N)

√2πN . . . . . . . . . . . . . . . . . . . . . . . . . 21

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Binomial, Poisson, and Gaussian . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.1 Binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.1.1 Moment generating function . . . . . . . . . . . . . . . . . . . . . . . . 273.1.2 Poisson distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.1.3 Binomial → Poisson a la Feller . . . . . . . . . . . . . . . . . . . . . . 303.1.4 Poisson process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.1.5 Easy method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.1.6 Easier method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1.7 Characteristic function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1.8 Cumulant generating function . . . . . . . . . . . . . . . . . . . . . . . 343.1.9 Sum of identically distributed independent random

variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.1.10 Poisson → Gaussian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4 Isolated system: Micro canonical ensemble . . . . . . . . . . . . . . . . . 414.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Configurational entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.3 Ideal gas law : Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.4 Boltzmann entropy and Clausius’ entropy are the same . . . . . . . 444.5 Some issues on extensitivity of entropy . . . . . . . . . . . . . . . . . . . . . 454.6 Boltzmann counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.7 Heaviside Theta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.8 Dirac delta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.9 Area of a circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.10 Volume of an N -dimensional sphere . . . . . . . . . . . . . . . . . . . . . . . . 494.11 Classical counting of micro states . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.11.1 Counting of the volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.12 Density of states : g(E) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.12.1 A sphere lives on its outer shell : Power law can beintriguing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.13 Entropy of an isolated system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.14 Properties of an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.15 Quantum counting of micro states . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.15.1 Energy eigenvalues : Integer Number of Half Wavelengths in L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.16 Chemical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

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4.16.1 Toy model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.16.2 Chemical potential of an ideal gas . . . . . . . . . . . . . . . . . . . 59

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5 Closed system : Canonical ensemble . . . . . . . . . . . . . . . . . . . . . . . 635.1 What is a closed system ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.2 Toy model a la H B Callen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.3 Canonical partition function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.4 Canonical partition function :

Method of most probable distribution . . . . . . . . . . . . . . . . . . . . . . 665.5 Lagrange’s method of undetermined multipliers . . . . . . . . . . . . . 705.6 Generalisation to a function of N variables . . . . . . . . . . . . . . . . . 725.7 Derivation of Boltzmann weight . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.8 Canonical partition function : Transform of density of states . . 755.9 Canonical partition function and Helmholtz free energy . . . . . . 765.10 Canonical ensemble and entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.11 Free energy to entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.12 Energy fluctuations and heat capacity . . . . . . . . . . . . . . . . . . . . . . 805.13 Canonical partition function for an ideal gas . . . . . . . . . . . . . . . . 81

5.13.1 Easy method: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.13.2 Easier method : Transform of density of (energy) states 82

5.14 Microscopic interpretation of heat and work . . . . . . . . . . . . . . . . 845.15 Work in statistical mechanics : W =

∑i pi dEi . . . . . . . . . . . . . . 84

5.16 Heat in statistical mechanics : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6 Grand canonical ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.1 Grand canonical partition function and grand potential . . . . . . 956.2 Euler formula in the context of homogeneous function . . . . . . . . 976.3 PV = kBT lnQ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 986.4 Gibbs-Duhem relation : dµ = −sdT + vdP . . . . . . . . . . . . . . . . . . 986.5 Grand canonical ensemble : Number fluctuations . . . . . . . . . . . . 996.6 Number fluctuations and isothermal compressibility . . . . . . . . . . 1006.7 Alternate derivation of the relation : σ2

N/〈N〉2 = kB T kT /V . . 102Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

7 Quantum Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1097.1 Occupation number representation of a micro state . . . . . . . . . . 1097.2 Open system and grand canonical partition function . . . . . . . . . 1107.3 Fermi-Dirac Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1127.4 Bose-Einstein Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1127.5 Classical Distinguishable Particles . . . . . . . . . . . . . . . . . . . . . . . . . 1137.6 Maxwell-Boltzmann Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

7.6.1 QMB(T, V,N) → QMB(T, V, µ) . . . . . . . . . . . . . . . . . . . . 1147.6.2 QMB(T, V, µ) → QMB(T, V,N) . . . . . . . . . . . . . . . . . . . . 115

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7.7 Grand canonical partition function, grand potential,and thermodynamic properties of an open system. . . . . . . . . . . . 115

7.8 Expressions for 〈N〉 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1177.8.1 Maxwell-Boltzmann Statistics . . . . . . . . . . . . . . . . . . . . . . . 1177.8.2 Bose-Einstein Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1177.8.3 Fermi-Dirac Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.8.4 Study of a system with fixed N employing grand

canonical formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.9 All the three statistics are the same at high temperature

and/or low densities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.9.1 Easy Method : ρΛ3 → 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.9.2 Easier Method : λ→ 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.9.3 Easiest Method Ω(n1, n2, · · · ) = 1 . . . . . . . . . . . . . . . . . . . 122

7.10 Statistics of Occupation Number - Mean . . . . . . . . . . . . . . . . . . . 1237.10.1 Ideal Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.10.2 Ideal Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1247.10.3 Classical Indistinguishable Ideal Particles . . . . . . . . . . . . . 124

7.11 Some Remarks on 〈nk〉 from the three statistics . . . . . . . . . . . . . 1257.11.1 Fermi-Dirac Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1257.11.2 Bose-Einstein Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1267.11.3 Maxwell-Boltzmann Statistics . . . . . . . . . . . . . . . . . . . . . . . 1267.11.4 At high T and/or low ρ all statistics give the same 〈nk〉 126

7.12 Statistics of Occupation Number - Fluctuations . . . . . . . . . . . . . 1277.12.1 Fermi-Dirac Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1287.12.2 Bose-Einstein Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1297.12.3 Maxwell-Boltzmann statistics . . . . . . . . . . . . . . . . . . . . . . . 132

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

8 Bose Einstein Condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1358.1 Some Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

8.1.1 〈N〉 =∑k〈nk〉 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1368.1.2

∑k(·) →

∫(·)dǫ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

8.1.3 g3/2(λ) versus λ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1408.1.4 Graphical inversion to determine fugacity . . . . . . . . . . . . . 1418.1.5 Treatment of the Singular Behaviour . . . . . . . . . . . . . . . . . 1418.1.6 Bose-Einstein Condensation Temperature . . . . . . . . . . . . . 1478.1.7 Grand Potential for Bosons . . . . . . . . . . . . . . . . . . . . . . . . . 1478.1.8 Average Energy of Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . 1488.1.9 Specific Heat Capacity of Bosons . . . . . . . . . . . . . . . . . . . . 1508.1.10 Mechanism of Bose-Einstein Condensation . . . . . . . . . . . 154

9 Statistical Mechanics of Harmonic Oscillators . . . . . . . . . . . . . . 1579.1 Classical Harmonic Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1579.2 Helmholtz Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1589.3 Thermodynamic Properties of the Oscillator System . . . . . . . . . 159

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9.4 Quantum Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1609.5 Specific Heat of a Crystalline Solid . . . . . . . . . . . . . . . . . . . . . . . . 1639.6 Einstein Theory of Specific Heat of Crystals . . . . . . . . . . . . . . . . 1669.7 Debye theory of Specific Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

9.7.1 Bernoulli Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

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List of Figures

3.1 Binomial distribution : B(n) =N !

n!(N − n)!pn(1 − p)N−n with

N = 10; B(n) versus n; depicted as sticks; (Left) p = 0.5;(Right) p = .35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.2 Poisson distribution : P (n) =µn

n!exp(−µ) with mean µ;

P (n) versus n; depicted as sticks; Gaussian distribution :

G(x) =1

σ√2π

exp

[(x− µ)2

2σ2

]with mean µ and variance

σ2 = µ : continuous line.(Left) µ = 1.5; (Right) µ = 9.5. For large µ Poisson andGaussian coincide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.1 Two ways of keeping a particle in a box divided into two equalparts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.2 Four ways of keeping two distinguishable particles in a boxdivided into two equal halves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

7.1 Average occupation number of a quantum state underBose-Einstein, Fermi-Dirac, and Maxwell-Boltzmann statistics . 127

8.1 g3/2(λ) versus λ. Graphical inversion to determine fugacity . . . . 1408.2 Singular part of 〈N〉 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1438.3 ρΛ3 versus λ. The singular part [Λ3/V ][λ/(1− λ)] (the bottom

most curve), the regular part g3/2(λ) (the middle curve) , andthe total (ρΛ3) are plotted. For this plot we have taken Λ3/Vas 0.05 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

8.4 Fugacity λ versus ρΛ3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1458.5 Ground state occupation as a function of temperature . . . . . . . . . 1468.6 Heat capacity in the neighbourhood of Bose - Einstein

condensation temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

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List of Tables

3.1 Probabilities calculated from Binomial distribution :B(n;N = 10, p = .1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.1 Micro states of of two particles with total energy 2ǫ . . . . . . . . . . . 584.2 Micro states of three particles with total energy 2ǫ . . . . . . . . . . . . 59

5.1 Micro states of three dice with the constraint that they add tosix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5.2 A micro state with occupation number representation (2, 3, 4) . . 685.3 A few micro states with the same occupation number

representation of (2, 3, 4) There are 1260 micro states with thesame occupation number representation . . . . . . . . . . . . . . . . . . . . . 68

7.1 Terms in the restricted sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

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1

Why should we study statistical mechanics ?

A quick answer : because it is one of the core courses, like Classical Mechan-ics, QuantumMechanics, Electrodynamics, and Mathematical Physics, in yourpost graduate curriculum. Fortunately, you have only one course in statisti-cal mechanics, unlike quantum mechanics, electrodynamics and mathematicalphysics !

1.1 Micro-Macro connections

On a more serious note, statistical mechanics provides a theoretical bridgethat takes you from the micro world of Newton, Schrodinger, and Maxwell,etc. to the macro world of thermodynamics. It attempts to derive the empiricalthermodynamics, especially the Second law - the law of ever-increasing entropy- from the microscopic laws of classical and quantum mechanics.

• When do we call something, a macroscopic object ?• What do we call something, a microscopic constituent of a macroscopicobject ?

The answer depends crucially on the object and the properties under study.For example,

• if we are interested in the properties like density, pressure, temperatureetc. of a cup of coffee, then the molecules of coffee are the microscopicconstituents; the cup of coffee is the macroscopic object.

• in another context, an atom can be considered a macroscopic object; theelectrons, protons and neutrons form its microscopic constituents.

• A polymer is a macroscopic object; the monomers are its microscopicconstituents.

• A society is a macroscopic object; men, women, children and monkeys areits microscopic constituents.

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2 1 Why should we study statistical mechanics ?

1.2 S = kB ln Ω(E, V,N)

This is the first and the most important link, between the microscopic and themacroscopic worlds; it was proposed by Boltzmann1. S stands for entropy andbelongs to the macro world described by thermodynamics. Ω is the numberof micro states of a macroscopic system2. kB is the Boltzmann constant 3

that establishes correspondence of the statistical entropy of Boltzmann to thethermodynamic entropy of Clausius.

1.2.1 S = −kB

∑i pi ln(pi)

I would call this Boltzmann-Gibbs-Shannon entropy. The sum is over all themicro states of a macroscopic system; the micro states are labelled by i. Theprobability of a micro state i is denoted by pi.

An interesting question : We resort to probabilistic description to hideour ignorance or to reconcile with our inability to keep track of the innumer-able micro states through which an equilibrium macroscopic system would gothrough, dictated by Newton’s equations of motion and initial conditions.

In thermodynamics, entropy is a property of a system. However in statis-tical mechanics entropy is defined in terms of the probabilities of the microstates. Does it imply that entropy is not only determined by then system butalso also by the ignorance or the inability of the observer ? Looks paradoxical?

1.2.2 Q =∑

i Ei dpi

This equation provides a microscopic description of heat. The sum runs overall the micro states of the macroscopic system. Ei is the energy of the systemwhen it is in micro state i. The probability that the system can be found inmicro state i given by pi. We need to impose an additional constraint that∑

i dpi is zero to ensure that the total probability is unity.

1.2.3 W =∑

i pi dEi

This equation defines work in the vocabulary of the micro world; the sum runsover all micro states of the system.

1 engraved on the tomb of Ludwig Eduard Boltzmann (1844-1906) in Zentralfried-hof, Vienna.

2 For example a set of six numbers, three positions and three momenta specify asingle particle. A set of 6N numbers specify a macroscopic system of N particles.The string labels a micro state.

3 kB = 1.381 × 10−23 Joules (Kelvin)−1. We have kB = R/A where R = 8.314Joules (Kelvin)−1 is the universal gas constant and A = 6.022× 1023 (mole)−1 isthe Avagadro number.

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1.3 Macro world : Thermodynamics 3

1.2.4 F = −kB T lnQ(T, V,N)

Helmholtz free energy F (T, V,N), defined in thermodynamics, is related tothe canonical partition function Q(T, V,N) of statistical mechanics. This isanother important micro-macro connection.

1.2.5 σ2E = kB T 2 CV

This relation connects the thermodynamic specific heat at constant volume(CV ) to the fluctuations (σ2

E) of statistical energy of a closed system.

1.2.6 σ2N =

〈N〉2kBT

VkT

This relation shows that fluctuations in the number of particles in an opensystem is proportional to the measurable thermodynamic property kT , calledthe isothermal compressibility defined as,

kT = − 1

V

(∂V

∂P

)

T

We shall see of several such micro-macro connections in the course of studyof Statistical Mechanics.

We can say the aim of statistical mechanics is to synthesise the macroworld from the micro world. This is not an easy task. Why ? Let us spend alittle bit of time on this question.

1.2.7 Micro world : Determinism and time-reversal invariance

The principal character of the micro world is determinism and time-reversalinvariance.

Determinism implies that the entire past and the entire future is frozen inthe present.

The solar system is a good example. If you know the positions and mo-menta of all the planets now, then the Newtonian machinery is adequate totell you where the planets shall be a thousand years from now and where theywere some ten thousand years ago.

Microscopic laws do not distinguish the future from the past. The equa-tions are invariant under transformation of t→ −t.

1.3 Macro world : Thermodynamics

On the other hand the macro world obeys the laws of thermodynamics4.

4 The laws of thermodynamics can be stated, in a lighter vein as follows.

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4 1 Why should we study statistical mechanics ?

• The zeroth law that tells of thermal equilibrium, provides a basis for thethermodynamic property, we call temperature. It is the starting point forthe game of thermodynamics.

• The first law that articulates in a smart way, the law of conservationof energy; it provides a basis for the thermodynamic property called theinternal energy. You can put in energy into a system by heat or by work.

• The second law that tells, come what may, the entropy increases; itprovides a basis for the thermodynamic property called entropy. An enginecan draw energy from the surroundings by work and deliver the sameamount or energy by heat. On the other hand if the machine draws energyfrom the surroundings by heat, then the energy it can deliver by work isinvariably less. The second law is a statement of this basic assymetry.

• The third law that tells that entropy vanishes at absolute zero. We cansay that the third law provides the basis for absolute zero temperature onentropy scale. The third law is also about the unattainability of absolutezero. You can go as close as you desire but you can never reach it.

Of these the second law is tricky. It breaks the time-symmetry presentin the microscopic descriptors. Macroscopic behaviour is not time-reversalinvariant. There is a definite direction of time - the direction of increasingentropy.

1.4 Micro-Macro synthesis

How do we comprehend the time asymmetric macroscopic behaviour emergingfrom the time symmetric microscopic laws ?

Let us make life simpler by attributing two aims to statistical mechanics.The first is to provide a machinery for calculating the thermodynamic prop-erties of a system on the basis of the properties of its microscopic constituentse.g. atoms and molecules, and their mutual interactions. Statistical Mechanicshas been eminently successful in this enterprise. This is precisely why we arestudying this subject.

The second aim is to derive the Second law of thermodynamics. StatisticalMechanics has not yet had any spectacular success on this count. However,some recent developments in non linear dynamics and chaos, have shown thereis indeed an unpredictability in some (nonlinear) deterministic system; wenow know that determinism does not necessarily imply predictability. This

Zeroth Law : You can play.First Law : You can’t win.Second Law : You can’t even win even.Third Law : You can’t quit.

The above or some variations of the above are usually attributed to Ginsberg andC P Snow.

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1.5 What am I going to teach ? 5

statement, perhaps, provides the raison d’etre for the ’statistics’ in statisticalmechanics.

In these lectures I shall not address the second issue - concerning the emer-gence of time asymmetry observed in macroscopic phenomena. I shall leavethis question to the philosophers and/or better equipped theoretical physi-cists. Instead we shall concentrate on how to derive the macroscopic propertiesfrom the properties of its microscopic constituents and their interactions.

1.5 What am I going to teach ?

I shall tell you of the elementary principles of statistical mechanics. I shall beas pedagogical as possible. Stop me when you do not understand.

I shall cover topics in

• micro canonical ensemble that provides a description of isolated system;• canonical ensemble, useful in the study of closed system;• grand canonical ensemble that describes open system.

I shall discuss equilibrium fluctuations of

• energy in canonical ensemble and relate them to heat capacity;• number of molecules in open system and relate them to isothemal com-pressibility.

Within the frame work of grand canonical ensemble I shall discuss Bose-Einstein, Fermi-Dirac and Maxwell Boltzmann statistics.

I shall deal with ideal gas, and classical and quantum harmonic oscillators.I shall discuss Bose Einstein condensation in some details with an emphasison the mechanism.

While on quantum harmonic oscillators, I shall discuss statistical mechan-ics of

• phonons emerging due to quantization of displacement field in a crystallinelattice, and

• photons arising due to quantization of electromagnetic field.

Maxwell’s demon is a mischievous and interesting entity. The idea is sim-ple. For every thermodynamic property we have, in statistical mechanics, acorresponding statistical quantity. We take the average of the statistical quan-tity over a suitable ensemble - micro canonical, for isolated system, canonicalfor closed system and grand canonical for open systems - and relate it to itsthermodynamic counter part. Thus entropy is a statistical entity. It is a merenumber. It corresponds to the number of micro states accessible to a macro-scopic system, consistent with the constraints. If entropy is statistical thenthe second law which states that entropy invariable increases in a process,should be a statistical statement. If so, there is a non-zero probability thatthe second law would be violated. Maxwell constructed a demon that violates

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6 1 Why should we study statistical mechanics ?

the second law ... a demon that extracts work from an equilibrium system. Iftime permits, I shall discuss Maxwell’s demon and its later incarnations.

Toward the end, and again if time permits, I shall discuss some recentdevelopments in thermodynamics - work fluctuation theorems and second lawviolations.

1.6 Syllabus prescribed

Course No. : PY454Course Title : Statistical MechanicsBasic Statistical ideas :Probability concepts, states of classical and quantum systems.Isolated systems :Micro canonical ensemble, statistical entropy, most probable state. Sys-tems in thermal and diffusive contact. Conditions for equilibrium. Canon-ical and grand canonical ensemble, and partition functions.Thermodynamics :Extensive and intensive variables, laws of thermodynamics, various ther-modynamic potentials and their connection to partition functions.Ideal Fermi and Bose gases :Distribution functions, classical limit. Electron gas in a metal. Black bodyradiation. Debye theory. Bose Einstein Condensation.Phase Transitions :Elementary ideas about phase transitions of different kinds. Examples ofsome phase transitions.Recommended books :

1. Thermal Physics, C. Kittel2. Statistical Physics, L. D. Landau and E. M. Lifshitz3. Problems in Thermodynamics and Statistical Physics, P. T. Landsberg

(Ed.)4. Introduction to Statistical Mechanics, F. Reif

I shall assume you are all comfortable with calculus; I shall also assume youhave a nodding acquaintance with thermodynamics. If you have difficulty atany time about thermodynamics, stop me. I shall explain the relevant portions.

Let me end this section by giving a list of some books and articles onthermodynamics and statistical mechanics.

1.7 Books on statistical mechanics and thermodynamics

• R K Pathria, Statistical Mechanics, Second Edition, Butterworth-Heinemann(1996)

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1.7 Books on statistical mechanics and thermodynamics 7

A popular book in statistical mechanics. Starts with a beautiful historicalaccount of the subject. Contains a very large collection of interesting andnon-trivial problems.

• Donald A McQuarrie, Statistical Mechanics, Harper & Row (1976)A beautiful book with emphasis on applications. Contains excellent prob-lems at the end of each chapter; suitable for self-study.

• R Balescu, Equilibrium and Non-Equilibrium Statistical Mechanics, Wi-ley (1975)An insightful book with emphasis on concepts. The issues on irreversibilityare explained beautifully.

• David Goodstein, States of MatterDelightful and entertraining. You are reminded of Feynman’s writing whenyou read this book. The discussion on dimensional analysis is excellent.This book is a must in your bookshelf.

• Debashish Chowdhury and Dietrich Stauffer, Principles of Equilib-rium Statistical Mechanics, Wiley-VCH (2000)An easy to read and enjoyable book. Contains applications to severalfields - condensed matter physics, materials science, polymers, solid statephysics, and astrophysics.

• F Rief, Fundamentals of statistical and thermal physics, McGraw-Hill(1965)One of the best text books on statistical thermodynamics. Rief developsthermal physics entirely in the vocabulary of statistical mechanics. As aresult after reading this book, you will get an uneasy feeling that thermo-dynamics has been relegated to the status of an uninteresting appendixto statistical mechanics. My recommendation : read this book for learningstatistical - thermodynamics; then read Callen or Van Ness (probably fora second time) for thermodynamics. Then you will certainly fall in lovewith both statistical mechanics and thermodynamics, separately!

• Palash B Pal, An Introductory Course of Statistical Mechanics, Narosa(2008)A book with a broad perspective; emphasis on relativistic systems.

• D Chandler, Introduction to Modern Statistical Mechanics, Oxford Uni-versity Press, New York (1987)A book that connects neatly the traditional to modern methods of teach-ing statistical mechanics; gives an excellent and simple introduction torenormalization groups. A great book for the teachers also.

• Claude Garrod, Statistical Mechanics and Thermodynamics, OxfordUniversity Press (1995)A good book at introductory level; neatly organized; pedagogic; nice prob-lems and exercises.

• Kerson Huang, Statistical Mechanics, Second Edition, Wiley India(2011)A whole generation of physicists has learnt statistical mechanics from thisbook. It is perhaps one of a very few books that take kinetic theory and

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8 1 Why should we study statistical mechanics ?

Boltzmann transport equation as a starting point. Historically, statisticalmechanics originated from Boltzmann transport equation. After all, it isBoltzmann transport - with its collision term obeying stosszahlansatz (col-lision number assumption or also known as molecular chaos) that estab-lishes thermal equilibrium - an important point that is either overlookedor not adequately emphasized in most of the text books on statisticalmechanics.The book contains three parts : the first on thermodynamics; the sec-ond on statistical mechanics; and the third on special topics in statisticalmechanics.Do not learn thermodynamics from this book. You will lose interest. Theother two parts are excellent - especially the third on special topics.I would recommend, retain your copy of the first edition of this book.Huang has omitted in his second edition, several interesting discussionspresent in the first edition.

• Kerson Huang, Introduction to Statistical Physics, Second Edition, CRCPress (2009).I think Huang has made an hurried attempt to ’modernize’ his earlierclassic : ”Statistical Mechanics”. I do not recommend this book to studentstaking their first course in statistical mechanics. However a discerningteacher will find this book very useful.

• J W Gibbs, Elementary Principles in Statistical Mechanics, Schribner,New York (1902)A seminal work in statistical mechanics. It feels good to read statisticalmechanics in the words of one of its masters. The book looks at canonicalformalism in a purely logical fashion and justifies it because of its simplic-ity and proximity to thermodynamics ! I will not recommend this book fora first reading. Learn statistical mechanics well and then read this book.

• Avijit Lahiri, Statistical Mechanics : An Elementary Outline, RevisedEdition, Universities Press (2008)A neat and well planned book. Focuses on bridging the micro world de-scribed by quantum mechanics to the macro world of thermodynamics.Concepts like mixed states, reduced states etc. provide the basic elementsin the development of the subject. The book contains a good collection ofworked examples and problems.

• Francis W Sears, and Gerhard L Salinger, Thermodynamics, KineticTheory, and Statistical Mechanics, Narosa (1974)Provides a balanced treatment of thermodynamics, kinetic theory andstatistical mechanics. Contains nice set of problems. Verbose in severalplaces that tests your patience.

• Joon Chang Lee, Thermal physics - Entropy and Free Energies, WorldScientific (2002)Joon Chang Lee presents statistical thermodynamics in an unorthodoxand distinctly original style. The presentation is so simple and so beautiful

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1.7 Books on statistical mechanics and thermodynamics 9

that you do not notice that the book is written in awful English and atseveral places, flawed.

• James P Sethna, Entropy, Order Parameters, and Complexity, Claren-don Press, Oxford (2008).James Sethna covers an astonishingly wide range of modern applications;a book, useful not only to physicists, but also to biologists, engineers, andsociologists. I find exercises and footnotes very interesting; often moreinteresting than the main text! However thermodynamics gets bruised.Is entropy a property of the system or a property of the (knowledge orignorance) of the fellow, observing the system ?

• C Kittel, and H Kromer, Thermal physics, W H Freeman (1980)A good book; somewhat terse. I liked the parts dealing with entropy,temperature, chemical potential, and Boltzmann weight; contains a goodcollection of examples.

• Daniel V Schrhroder, An Introduction to Thermal Physics, Pearson(2000).Schroder has excellent writing skills. The book reads well. Contains plentyof examples. Somewhat idiosyncratic.

• M Glazer, and J Wark, Statistical Mechanics : A Survival Guide, OxfordUniversity Press (2010) This book gives a neat introduction to statisti-cal mechanics; well organized; contains a good collection of worked-outproblems; a thin book and hence does not threaten you !

• H C Van Ness, Understanding Thermodynamics, Dover (1969).This is an awesome book; easy to read and very insightful. In particular,I enjoyed reading the first chapter on the first law of thermodynamics,the second on reversibility, and the fifth and sixth on the second law. Myonly complaint is that Van Ness employs British Thermal Units. Anotherminor point : Van Ness takes the work done by the system as positive andthat done on the system as negative. Engineers always do this. Physicistsand chemists employ the opposite convention. For them the sign coincideswith the sign of change of internal energy caused by the work process.When the system does work, its internal energy decreases; hence the workdone by the system is negative. When work is done on the system itsinternal energy increases; hence work done on the system is positive.

• H B Callen, Thermodynamics, John Wiley (1960).A standard textbook. This book has influenced generations of teachersand students alike, all over the world. Callen is a house hold name in thecommunity of physicists. The book avoids all the pitfalls in the historicaldevelopment of thermodynamics by introducing a postulational formula-tion.

• H B Callen, Thermodynamics and an Introduction to thermostatistics,Second Edition, Wiley, India (2005).Another classic from H B Callen. He has introduced statistical mechanicswithout undermining the inner strength of thermodynamics. In fact, thestatistical mechanics he presents, enhances the beauty of thermodynamics.

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10 1 Why should we study statistical mechanics ?

The simple toy problem with a red die (the closed system) and two whitedice (the heat reservoir), and the restricting sum to a fixed number (con-servation of total energy) explains beautifully the canonical formalism.The pre-gas model introduced for explaining grand canonical ensemble ofFermions and Bosons is simply superb. I also enjoyed the discussions onthe subtle mechanism underlying Bose condensation. I can go on listingseveral such examples. The book is full of beautiful insights.A relatively inexpensive, Wiley student edition of the book is available inthe Indian market. Buy your copy now !

• Gabriel Weinreich, Fundamental Thermodynamics, Addison Wesley(1968).Weinreich’s is original; he has a distinctive style. Perhaps you will feeluneasy when you read his book for the first time. But very soon, you willget used to Weireich’s idiosyncracy; and you would love this book 5.

• C B P Finn, Thermal Physics, Nelson Thornes (2001).Beautiful; concise; develops thermodynamics from first principles. Finnbrings out the elegance and power of thermodynamics.

• Max Planck, Treatise on Thermodynamics, Third revised edition, Dover;first published in the year 1897. Translated from the seventh Germanedition (1922).A carefully scripted master piece; emphasises chemical equilibrium. I donot think any body can explain irreversibility as clearly as Planck does.If you think third law of thermodynamics is irrelevant, then read the lastchapter. You may change your opinion.

• E Fermi, Thermodynamics, Dover (1936)A great book from a great master; concise; the first four chapters (on ther-modynamic systems, first law, the Second law, and entropy) are superb.I also enjoyed the parts covering Clapeyron and van der Waal equations.

• J S Dugdale, Entropy and its physical meaning, Taylor and Francis(1998).An amazing book. Dugdale de-mystifies entropy. This book is not justabout entropy alone, as the name would suggest. It teaches you thermody-namics and statistical mechanics. A book that cleverly avoids unnecessaryrigour.

• M W Zamansky, and R H Dittman, Heat and Thermodynamics, anintermediate textbook, Sixth edition, McGraw-Hill (1981)A good and dependable book for a first course in thermodynamics. I amnot very excited about the problems given in the book. Most of them areroutine and requires uninteresting algebraic manipulations.

• R Shanthini, Thermodynamics for the Beginners, Science EducationUnit, University of Peredeniya (2009)

5 Thanks to H S Mani, now at Chennai Mathematical Institute, for bringing myattention to this book and for presenting me with a copy.

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1.9 Extra reading : Papers 11

Student-friendly. Shanthini has anticipated several questions that wouldarise in the minds of the students when they learn thermodynamics forthe first time. The book has a good collection of worked out examples. Abit heavy on heat engines.

• Dilip Kondepudi and Ilya Prigogine,Modern Thermodynamics : Fromheat engines to Dissipative Structures, John Wiley (1998)Classical, statistical, and non equilibrium thermodynamics are woven intoa single fabric. Usually chemists present thermodynamics in a dry fashion.This book is an exception; it tells us learning thermodynamics can be fun.Contains lots of interesting tit-bits on history. Deserves a better coverdesign; the present cover looks cheap.

1.8 Extra reading : Books

• Nicolaus Sadi Carnot, Reflexions sur la puissance motrice du feu et surles machines propres a developer cette puissance, Paris (1824); for Englishtranslation see Sadi carnot, Reflections on the motive power of fire andon machines fitted to develop that power, in J Kestin (Ed.) The secondlaw of thermodynamics, Dowden, Hutchinson and Ross, Stroudsburg, PA(1976)p.16

• J Kestin (Ed.), The second law of thermodynamics, Dowden, Hutchinsonand Ross (1976)

• P Atkin, The Second Law, W H Freeman and Co. (1984)• G Venkataraman, A hot story, Universities Press (1992)• Michael Guillen, An unprofitable experience : Rudolf Clausius and thesecond law of thermodynamics p.165, in Five Equations that Changed theWorld, Hyperion (1995)

• P Atkins, Four Laws that drive the Universe, Oxford university Press(2007).

• Christopher J T Lewis, Heat and Thermodynamics : A Historical Per-spective, First Indian Edition, Pentagon Press (2009)

• S G Brush, Kinetic theory Vol. 1 : The nature of gases and of heat,Pergamon (1965) Vol. 2 : Irreversible Processes, Pergamon (1966)

• S G Brush, The kind of motion we call heat, Book 1 : Physics and theAtomists Book 2 : Statistical Physics and Irreversible Processes, NorthHolland Pub. (1976)

• I Prigogine, From Being to Becoming, Freeman, San Francisci (1980)• K P N Murthy, Excursions in thermodynamics and statistical mechan-ics, Universities Press (2009)

1.9 Extra reading : Papers

• K K Darrow, The concept of entropy, American Journal of Physics 12,183 (1944).

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12 1 Why should we study statistical mechanics ?

• M C Mackay, The dynamical origin of increasing entropy, Rev. Mod.Phys. 61, 981 (1989).

• T Rothman, The evolution of entropy, pp.75-108, in Science a la mode: physical fashions and fictions Princeton University Press (1989)

• Ralph Baierlein, Entropy and the second law : A pedagogical alternative,American Journal of Physics 62, 15 (1994)

• G. Cook, and R H Dickerson, Understanding the chemical potential,American Journal of Physics 63, 738 (1995).

• K. P. N. Murthy, Ludwig Boltzmann, Transport Equation and the Sec-ond Law, arXiv: cond-mat/0601566 (1996)

• Daniel F Styer, Insight into entropy, American Journal of Physics 68,1090 (2000)

• B J Cherayil, Entropy and the direction of natural change, Resonance6, 82 (2001)

• J K Bhattacharya, Entropy a la Boltzmann, Resonance 6, 19 (2001)• J Srinivasan, Sadi Carnot and the second law of thermodynamics, Res-onance 6 42 (2001)

• D C Shoepf, A statistical development of entropy for introductory physicscourse, American Journal of Physics 70, 128 (2002).

• K P N Murthy, Josiah Willard Gibbs and his Ensembles, Resonance12, 12 (2007).

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2

Experiment, outcomes, events, probabilitiesand ensemble

2.1 Toss a coin

You get either ”Heads” or ”Tails”. The experiment has two outcomes.Consider tossing of two coins. Or equivalently consider tossing a coin twice.

There are four outcomes : HH, HT, TH, TT . An outcome is an orderedpair. Each entry in the pair is drawn from the set H, T .

We can consider, in general, tossing of N coins. There are 2N outcomes.Each outcome is an ordered string of size N with entries from the set (H,T ).

2.2 Roll a die

You get one of the six outcomes :

• ;

;

;

• •

• •

;

• •

• •

;

• • •

• • •

Consider throwing of N dice. There are then 6N outcomes. Each outcomeis an ordered string of N entries drawn from the basic set of six elements givenabove.

2.3 Pick randomly a molecule from this room

In classical mechanics, a molecule is completely specified by giving its threeposition and three momentum coordinates. An ordered set of six numbers

q1, q2, q3, p1, p2, p3is an outcome of the experiment. A point in the six dimensional phase spacerepresents an outcome.

We impose certain constraints e.g. the molecule is always confined to thisroom. Then all possible strings of six numbers, consistent with the constrains,are the outcomes of the experiment.

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14 2 Experiment, outcomes, events, probabilities and ensemble

2.4 Sample space and events

The set of all outcomes of an experiment is called the sample space. Let usdenote it by the symbol Ω.

• Ω = H, T for the toss of a single coin.• Ω = HH,HT, TH, TT for the toss of two coins.

A subset of Ω is called an event. Let A denote an event. A is a subset ofΩ underlying an experiment. When we carry out the experiment and if theoutcome belongs to A, then we say the event A has occurred.

Consider tossing of two coins. Let an event A be described by the statementthat the first toss is H . Then A consists of the following elements: HH, HT .

The event corresponding to the roll of an even number in the game of dice,is the subset

;

• •

• •

;

• • •

• • •

.

2.5 Probabilities

Probability is defined for an event.What is the probability of ”H” in a toss of a coin ?One-half. This would be your immediate response. The logic is simple.

There are two outcomes : ”Heads” and ”Tails”. We have no reason to believewhy should the coin prefer ”Heads” over ”Tails” or vice versa. Hence we sayboth outcomes are equally probable.

What is the probability of having at least one ”H” in a toss of two coins ?The event corresponding this statement is HH,HT, TH and contains threeelements. The sample size contains four elements. The required probability isthus 3/4. All the four outcomes are equally probable 1.

Thus, if all the outcomes are equally probable, then the probability of anevent is the number of elements in that event divided by the total number ofelements in the sample space. For e.g., the event A of rolling an even numberin a game of dice, P (A) = 3/6 = 0.5.

The outcome can be a continuum. For example, the angle of scattering ofa neutron is a real number between zero and π. We then define an interval(θ1, θ2) where 0 ≤ θi ≤ π : i = 1, 2 as an event. A measurable subset of asample space is an event.

1 Physicists have a name for this. They call it the axiom (or hypothesis or assump-tion) of Ergodicity. Strictly ergodicity is not an assumption; it signifies absenceof an assumption.

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2.7 Random variable 15

2.6 Rules of probability

The probability p that you assign to an event, should be obey the followingrules.

p ≥ 0

p(A ∪ B) = p(A) + p(B)− p(A ∩B)

p(φ) = 0

p(Ω) = 1 (2.1)

In the above φ denotes a null event and Ω is a sure event.How does one assign probability to an event ?Actually, this question does not bother the mathematicians. It is the physi-

cists who should worry about this2.

2.7 Random variable

The next important concept in probability theory is random variable, x =X(ω) where ω denotes an outcome and x a real number. Random variableis a way of stamping an outcome with a number : Real number, for a realrandom variable; integer, for an integer random variable; complex number,for a complex random variable3. Thus the random variable x = X(ω) is a setfunction.

Consider a continuous random variable x = X(ω). We define a probabilitydensity function f(x) by

f(x)dx = P (ω|x ≤ X(ω) ≤ x+ dx) (2.2)

In other words f(x)dx is the probability of the event (measurable subset) thatcontains all the outcomes to which we have attached a real number betweenx and x+ dx.

2 Maxwell and Boltzmann attached probabilities to events in some way; we gotMaxwell-Boltzmann statistics.

Fermi and Dirac had their way of assigning probabilities to Fermions e.g. elec-trons, occupying quantum states. We got Fermi-Dirac statistics.

Bose and Einstein came up with their scheme of assigning probabilities toBosons populating quantum states; we got Bose-Einstein statistics.

3 In fact, we stamped dots on the faces of die; this is equivalent to implementingthe idea of a random variable : attach a number between one and six to eachoutcome.

For a coin, we stamped ”Heads” on one side and ”Tails” on the other. This isin the spirit of defining a random variable, except that we have stamped figures;for the random variable, however, we should stamp numbers.

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16 2 Experiment, outcomes, events, probabilities and ensemble

Now consider a continuous random variable defined between a to b witha < b. We define a quantity called the ”average” of the random variable xas

µ =

∫ b

a

dx x f(x).

µ is also called the mean, expectation, first moment etc.Consider a discrete random variable n, taking values from say 0 to N .

Let P (n) define the discrete probability. We define the average of the randomvariable as

µ =

N∑

n=0

n P (n).

But then, we are accustomed to calculating the average in a differentway. For example I am interested in knowing the average marks obtainedby the students in a class, in the last mid-semester examination. How do Icalculate it ? I take the marks obtained by each of you, sum them up anddivide by the total number of students. That is it. I do not need notionslike probabilities, probability density, sum over the product of the randomvariable and the corresponding probability, integration of the product of thecontinuous random variable and its probability density function etc.

Historically, before Boltzmann and Maxwell, physicists had no use forprobability theory in their work. Newton’s equations are deterministic. Thereis nothing chancy about a Newtonian trajectory. We do not need probabilisticdescription in the study of electrodynamics described by Maxwell equations;nor do we need probability to comprehend and work with Einstein’s relativity- special or general.

However mathematicians had developed the theory of probability as animportant and sophisticated branch of mathematics4.

It was Ludwig Eduard Boltzmann who brought, for the first time, theidea of probability into physical sciences; he was championing the cause ofkinetic theory of heat and of matter. Boltzmann transport equation is thefirst ever equation written for describing the time evolution of a probabilitydistribution.

2.8 Maxwell’s mischief : ensemble

However, Maxwell, had a poor opinion about a physicist’s ability to com-prehend mathematicians’ writings on probability theory, in general, and themeaning of average as an integral over a probability density, in particular.

After all, if you ask a physicist to calculate the average age of a studentin the class, he’ll simply add the ages of all the students and divide by thenumber of students.

4 perhaps for assisting the kings in their gambling.

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2.11 Counting of the number of elements in events of the sample space : Coin tossing 17

To be consistent with this practice, Maxwell proposed the notion of anensemble of outcomes of an experiment (or an ensemble of realisations of arandom variable). Let us call it Maxwell ensemble 5.

Consider a collection of a certain number of independent realisations ofthe toss of a single coin. We call this collection a Maxwell ensemble if it itobeys certain conditions, see below.

Let N denote the number of elements in the ensemble. Let nH denotethe number ”Heads” and nT number of ’Tails”. We have nH + nT = N .If nH = Np, and hence nT = Nq, then we say the collection of outcomesconstitutes a Maxwell ensemble.

Thus an ensemble holds information about the outcomes of the experimentconstituting the sample space; it also holds information about the probabilityof each outcome. The elements of an ensemble are drawn from the samplespace; however each element occurs in an ensemble as often as to reflect cor-rectly its probability.

2.9 Calculation of probabilities from an ensemble

Suppose we are given the following ensemble : H,T,H,H, T . By looking atthe ensemble, we can say the sample space contains two outcomes H, T . Wealso find that the outcome H occurs thrice and T occurs twice. We concludethat the probability of H is 3/5 and that of T is 2/5.

2.10 Construction of ensemble from probabilities

We can also do the reverse. Given the outcomes and their probabilities, wecan construct an ensemble. Let ni denote the number of times an outcomei occurs in an ensemble. Let N denote the total number of elements of theensemble. Choose ni such that ni/N equals pi; note that we have assumedthat pi is already known.

2.11 Counting of the number of elements in events ofthe sample space : Coin tossing

Consider tossing of N identical coins or tossing of a single coin N times. Letus say the coin is fair. In other words P (H) = P (T ) = 1/2.

Let Ω(N) denote the set of all possible outcomes of the experiment. Anoutcome is thus a string N entries, each entry being ”H” or ”T”. The number

5 Later we shall generalise the notion of Maxwell ensemble and talk of ensembleas a collection identical copies of a macroscopic system. We shall call it a Gibbsensemble.

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18 2 Experiment, outcomes, events, probabilities and ensemble

of elements of this setΩ(N) is denoted by the symbol, Ω(N). We have Ω(N) =2N .

Let Ω(n;N) denote a subset of Ω(N), containing only those outcomeswith n ’Heads’ (and hence (N − n) ’Tails’). How many outcomes are there inthe set Ω(n;N) ?

Let Ω(n;N) denote the number of elements in the event Ω(n;N). I shalltell you how to count the number of elements of this set6.

Take one outcome belonging to Ω(n;N). There will be n ’Heads’ in thatoutcome. Imagine for a moment that all these ’Heads’ are distinguishable. Ifyou like, you can label them as H1, H2, · · · , Hn. Permute all the ’Heads’and produce n! new configurations. From each of these new configurations,produce (N−n)! configurations by carrying out the permutations of the (N−n) ’Tails’. Thus from one outcome belonging to the set Ω(n;N), we haveproduced n!× (N −n) new configurations. Repeat the above for each element

of the set Ω(n;N), and produce Ω(n;N) × n! × (N − n)! configurations. Amoment of thought will tell you that this number should be the same as N !7.We thus have,

Ω(n;N) n! (N − n)! = N ! . (2.3)

It follows, then,

Ω(n;N) =N !

n!(N − n)!(2.4)

These are called the binomial coefficients.Show that

N∑

n=0

Ω(n;N) = Ω(N) = 2N . (2.5)

The binomial coefficients add to 2N :

N∑

n=1

N !

n!(N − n)!= 2N (2.6)

6 I remember I have seen this method described in a book on Quantum Mechanicsby Gassiarowicz. Check it out

7 if you do not get the hang of it, then work it out explicitly for four coins withtwo ”Heads” and two ”Tails”. Show explicitly that

Ω(n = 2, N = 4)× 2!× 2! = 4!

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2.13 Why should a Gibbs ensemble be of large size ? 19

2.12 Gibbs ensemble

Following Gibbs, we can think of an ensemble as consisting of large numberof identical mental copies of a macroscopic system. 8. All the members of anensemble are in the same macro state9. However they can be in different microstates. Let the micro states of the system under consideration, be indexedby i = 1, 2, · · · . The number of elements of the ensemble in micro state jdivided by the size of the ensemble is equal to the the probability of the systemto be in micro state j. It is intuitively clear that the size of the ensemble shouldbe large (→ ∞) so that it can capture exactly the probabilities of differentmicro states of the system10. Let me elaborate on this issue, see below.

2.13 Why should a Gibbs ensemble be of large size ?

What is the value of n for which Ω(n;N) is maximum11 ?

It is readily shown that for n = N/2 the value of Ω(n;N) is maximum.

Let us denote this number by the symbol Ω⋆(N). We have, Ω⋆(N) = Ω(n =N/2, N).

Thus we have

Ω(N) =

N∑

n=0

N !

n! (N − n)!= 2N (2.7)

8 For example the given coin is a system. Let p denote the probability of ”Heads”and q = 1−p the probability of ”tails”. The coin can be in a micro state ”Heads”or in a micro state ”Tails”.

9 This means the values of p and q are the same for all the coins belonging to theensemble.

10 If you want to estimate the probability of Heads in the toss of a single coinexperimentally then you have to toss a large number of identical coins. Largerthe size of the ensemble more (statistically) accurate is your estimate .

11 you can find this in several ways. Just guess it. I am sure you would have guessedthe answer as N/2. We know that the binomial coefficient is largest when n = N/2if N is even, or when n equals the two integers closest to N/2 for N odd. That isit.

If you are more sophisticated, take the derivative of Ω(n;N) with respect to nand set it zero; solve the resulting equation to get the value of N for which thefunction is an extremum.

You may find it useful to take the derivative of logarithm of Ω(n;N); employStirling approximation for the factorials : ln(m!) = m ln(m) − m for large m.Stirling approximation to large factorials is described in the next section.

You can also employ any other pet method of yours to show that for n = N/2

the function Ω(n;N) is maximum.Take the second derivative and show that the extremum is a maximum.

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20 2 Experiment, outcomes, events, probabilities and ensemble

Ω⋆(N) = Ω(n = N/2;N) =N !

(N/2)! (N/2)!(2.8)

Let us evaluate Ω⋆ for large values of N . We employ Stirling approxima-tion12 :

N ! = NN exp(−N)√2πN (2.9)

We have

Ω⋆(N) = Ω(n = N/2;N) =NN exp(−N)

√2πN

[(N/2)(N/2) exp(−N/2)

√2π(N/2)

]2

= 2N√2√πN

(2.10)

Let us evaluate the natural logarithm of both the quantities under discussion.Let

SG = ln Ω(N) = N ln 2 (2.11)

SB = ln Ω⋆(N) = N ln 2− (1/2) lnN + (1/2) ln(2/π) (2.12)

∼N→∞ N ln 2− (1/2) lnN (2.13)

Thus SB is less than SG by a negligibly small amount. SB and SG are both ofthe order of N ; SG − SB is of the order of logarithm of N , which is negligiblefor large N . For example take a typical value for N = 1023. We have SG =0.69× 1023 and SB = 0.69× 1023 − 24.48.

Note that only when N is large, we have SB equals SG. It is preciselybecause of this, we want the number of elements to be large, while constructinga Gibbs ensemble. We should ensure that all the micro states of the systemare present in the ensemble in proportions, consistent with their probabilities.For example I can simply toss N independent fair coins just once and if Nis large then I am assured that there shall be N/2 ± ǫ ’Heads’ and N/2 ∓ ǫ’Tails’, where ǫ is negligibly small : of the order of

√N .

2.13.1 Stirling and his Approximation to Large Factorials

N ! = NN exp(−N)

Stirling’s formula is an approximation to large factorials. James Stirling (1692-1770) was a Scottish mathematician. We have,

12 Stirling approximation is described in the next section

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2.13 Why should a Gibbs ensemble be of large size ? 21

N ! = N × (N − 1)× · · · × 3× 2× 1

ln N ! = ln 1 + ln 2 + ln 3 + ·+ ln N

=

N∑

k=1

ln(k)

≈∫ N

1

ln x dx

= (x ln x− x)∣∣N1

= N ln N −N − 1

≈ N lnN −N (2.14)

Thus, for large N we have

n! ≈ NN exp(−N).

Stirling formula is explained well in the following references.

• Daniel V. Schroeder, An Introduction to Thermal Physics, Addison Wes-ley (2000)

• F. Rief, Fundamentals of Statistical and Thermal Physics, McGraw Hill(1965)

2.13.2 N ! = NN exp(−N)√2πN

To show this, proceed as follows.

Γ (N + 1) = N ! =

∫ ∞

0

dx xN exp(−x),

=

∫ ∞

0

dx exp [N ln(x) − x] ,

=

∫ ∞

0

dx exp [F (x)] , (2.15)

where, F (x) = N ln(x) − x. Determine the value of x, say x = x⋆, at which

F (x) is extremum. Note that x⋆ is solution of the equation,dF

dx= 0. We see

immediately that x⋆ = N . Thus, F (x) is extremum at x = N .

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22 2 Experiment, outcomes, events, probabilities and ensemble

First check whether the extremum is maximum or minimum. I leave thisto you, as an exercise. The result : F (x) is maximum at x = x⋆.

Carry out the Taylor expansion of F (x) around x⋆ and keep terms up to thesecond derivative. Substitute it in the integral and carry out the integrationto get the Stirling approximation.

Problems

2.1. Consider a coin with probability of ”Heads” given by 0.3. The experimentconsists of tossing the coin once. Write down a possible ensemble of realisationsof the experiment.

2.2. Consider a p-coin; i.e. a coin for which p is the probability of ”Heads”.Consider an experiment of tossing the p-coin independently twice. Write downa possible ensemble of realisations, for the following cases.

(a) p = 1/2(b) p = 1/4

2.3. Let x = X(ω) denote a continuous random variable defined in the range0 ≤ x ≤ 1 with a uniform probability density function. Find the mean andvariance of x.

2.4. Let x = X(ω) denote a continuous random variable defined in the range0 ≤ x ≤ +∞, with an exponential probability density : f(x) = exp(−x). LetMn denote the n-th moment of the random variable x. It is defined as

Mn =

∫ ∞

0

xn exp(−x)dx.

Show that M0 = 1; M1 = 1; and M2 = 2. Obtain an expression for Mn.

2.5. Betrand’s paradoxSelect a chord randomly in a circle. What is the probability that the chordlength is greater than the side of an inscribed equilateral triangle?

2.6. Throw two fair independent dice. Let (n1, n2) denote the result of theexperiment : n1 is the result of the first die and n2 that of the second die.Define a random variable

n = maximum(n1, n2).

What is the probability distribution of the random variable n ? Determinethe mean and variance of n.

2.7. Calculate YE , the exact value of ln N ! =∑N

k=1 ln k and the Stirlingapproximation YS = N ln N −N for N = 2, 3, · · · 100 . Plot the results in thesame graph. Plot also the relative error ǫ = |YE − YI |/YE as a function of N .

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2.13 Why should a Gibbs ensemble be of large size ? 23

2.8. Calculate YE , the exact value of ln N ! =∑N

k=1 ln k and the improvedStirling approximation YSI = N ln N − N + 1

2 ln(2πN) for N = 2, 3, · · ·100. Plot the results in the same graph. Plot also the relative error ǫ = |YE −YSI |/YE as a function of n.

Page 36: Kpn Stat Mech
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3

Binomial, Poisson, and Gaussian

3.1 Binomial distribution

Consider a system consisting of one coin. It has two micro states : ’H’ and’T’. The probability for the system to be in micro state ’H’ is p and that inmicro state ’T’ is q = 1− p.

Consider the case with p = 0.6 and hence q = 1 − p = 0.4. A possibleensemble of systems that contain exact information about the micro states ofthe system and of their probabilities is

T, H, H H, T, H, H T, H, T

Notice that the ensemble contains ten systems (or mental copies of the sys-tem). Six systems are in micro state ’H’ and four are in micro state ’T’.

As I said earlier, we usually take the size of the ensemble to be arbitrarilylarge. Let N denote the size of the ensemble.

Imagine, we attempt to construct the ensemble by actually carrying outthe experiment of tossing identical coins or by tossing the same coin severaltimes independently. What is the probability that in the experiment thereshall be n ’Heads’ and hence (N-n) ’Tails’ ? Let us denote this by the symbolB(n). It is readily seen

B(n) =N !

n! (N − n)!pn qN−n (3.1)

Figure (??) depicts Binomial distribution for N = 10, p = 0.5 and 0.35. Whatis average value of n ? The average is also called the mean, the first moment,the expectation value etc.. Denote it denoted by the symbol M1 or 〈n〉. It isgiven by

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26 3 Binomial, Poisson, and Gaussian

0 5 100

0.05

0.1

0.15

0.2

0.25

n

B(n)

0 5 100

0.05

0.1

0.15

0.2

0.25

0.3

n

B(n)

Fig. 3.1. Binomial distribution : B(n) =N !

n!(N − n)!pn(1− p)N−n with N = 10;

B(n) versus n; depicted as sticks; (Left) p = 0.5; (Right) p = .35

M1 = 〈n〉 =N∑

n=0

n B(n;N)

=N∑

n=0

nN !

n! (N − n)!pn qN−n

=

N∑

n=0

Np(N − 1)!

(n− 1)! [(N − 1)− (n− 1)]!pn−1 q(N−1)−(n−1)

= Np

N∑

n=0

(N − 1)!

n![(N − 1)− n]!pn q(N−1)−n

= NpN−1∑

n=0

B(n;N − 1)

= Np (3.2)

Thus the first moment (or the average) of the random variable n is Np.We can define higher moments. The k-th moment is defined as

Mk = 〈nk〉 =N∑

n=0

nk B(n) (3.3)

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3.1 Binomial distribution 27

The next important property of the random variable is variance. It is definedas,

σ2n =

N∑

n=0

(n−M1)2 B(n)

=

N∑

n=0

n2B(n)−M21

= M2 −M21 (3.4)

Let me now describe a smart way of generating the moments of a randomvariable.

3.1.1 Moment generating function

Let B(n) denote the probability that n coins are in micro state ”Heads” inan ensemble of N coins. We have shown that

B(n) =N !

n!(N − n)!pnqN−n (3.5)

The moment generating function is defined as

B(z) =

N∑

n=0

zn B(n), (3.6)

The first thing we notice is that B(z = 1) = 1. This guarantees that the prob-ability distribution B(n) is normalized. The moment generating function islike a discrete transform of the probability distribution function. We transformthe variable n to z.

Let us now take the first derivative of the moment generating functionwith respect to z. We have,

dB

dz= B′(z) =

N∑

n=0

n zn−1 B(n)

zB′(z) =N∑

n=0

n zn B(n) (3.7)

. Substitute in the above z = 1. We get,

B′(z = 1) = 〈n〉 (3.8)

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28 3 Binomial, Poisson, and Gaussian

Thus the first derivative of B evaluated at z = 1 generates the first moment.Now take the second derivative of B(z) to get

d2B

dz2=

N∑

n=0

n(n− 1)zn−2B(n)

z2d2B

dz2=

N∑

n=0

zn n(n− 1) B(n) (3.9)

Substitute in the above z = 1 and get,

d2B

dz2

∣∣∣∣z=1

= 〈n(n− 1)〉 (3.10)

For the Binomial random variable, we can derive the moment generatingfunction :

B(z) =

N∑

n=0

znB(n)

=

N∑

n=0

N !

n! (N − n)!(zp)n qN−n

= (q + zp)N (3.11)

When N is large, it is clumsy to calculate quantities employing Binomialdistribution. Consider the following situation.

I have N molecules of air in this room of volume V . The molecules aredistributed uniformly in the room. In other words the number density, denotedby ρ is same at all points in the room. Consider now an imaginary small volumev < V completely contained in the room. Consider an experiment of choosingrandomly an air molecule from this room. The probability that the moleculeshall be in the small volume is p = v/V ; the probability that it shall be outside the small volume is q = 1 − (v/V ). There are only two possibilities. Wecan use Binomial distribution to calculate the probability for n molecules tobe present in v.

Consider first the problem with V = 10M3, v = 6M3 and N = 10. Thevalue of p for the Binomial distribution is 0.6. The probability of finding nmolecules in v is then,

B(n;N = 10) =10!

n!(10− n)!(0.1)n(0.9)10−n (3.12)

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3.1 Binomial distribution 29

n B(n; 10) n B(n; 10)

0 0.0001 6 0.25081 0.0016 7 0.21502 0.0106 8 0.12093 0.0425 9 0.04034 0.1115 10 0.00605 0.2007 − −

Table 3.1. Probabilities calculated from Binomial distribution : B(n;N = 10, p =.1)

The table below gives the probabilities calculated from the Binomial distri-bution.

Consider the same problem with v = 10−3 M3 and N = 105. We havep = 10−4 and Np = 10. Immediately we recognize that Binomial distributionis not appropriate for this problem. Calculation of the probability of findingn molecules in v involves evaluation of 100000!.

What is the right distribution for this problem and problems of this kind ?To answer this question, consider what happens to the Binomial distributionin the limit of N → ∞, p→ 0, and Np = µ, a constant1. Note that

Np = Nv/V = ρv = constant.

We shall show below that in this limit, Binomial goes over to Poissondistribution.

3.1.2 Poisson distribution

We start with

B(z) = (q + zp)N (3.13)

We can write the above as

B(z) = qN (1 + zp/q)N

= (1− p)N (1 + zp/q)N (3.14)

1 Note that for a physicist, large is infinity and small is zero.

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30 3 Binomial, Poisson, and Gaussian

(N→∞p→0 Np = µ

)B(z) ∼ exp(−Np) exp(zNp/q)

= exp(−µ) exp(µz)

= P (z) (3.15)

Thus in the limit N → ∞, p → 0 and Np = µ, we find B(z) → P (z), givenby

P (z) = exp[−µ(1− z)] (3.16)

The coefficient of zn in the power series expansion of P (z) gives P (n),

P (n) =µn

n!exp(−µ) (3.17)

The above is called the Poisson distribution2. Thus in the limit of N → ∞,p → 0, Np = µ, the Binomial distribution goes over to Poisson distribution.Figure (??) depicts Poisson distribution for µ = 1.5 and 9.5.

3.1.3 Binomial → Poisson a la Feller

Following Feller3, we have

B(n;N)

B(n− 1;N)=

N ! pn qN−n

n! (N − n)!

(n− 1)! (N − n+ 1)!

N ! pn−1 qN−n+1

=p (N − n+ 1)!

n q

=N p− p (n− 1)

n q

(N→∞p→ 0 Np=µ

)∼ µ

n(3.18)

Start with

2 We shall come across Poisson distribution in the context of Maxwell-Boltzmannstatistics. Let nk denote the number of ’indistinguishable classical’ particles in asingle-particle state k. The random variable nk is Poisson-distributed.

3 William Feller, An Introduction to PROBABILITY : Theory and its Applications,Third Edition Volume 1, Wiley Student Edition (1968)p.153

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3.1 Binomial distribution 31

B(n = 0;N) = qN

= (1− p)N

∼p→0 exp(−Np)

= P (n = 0;µ) = exp(−µ) (3.19)

We get

P (n = 1;N) = µ exp(−µ) (3.20)

P (n = 2;N) =µ2

2!exp(−µ) (3.21)

P (n = 3;N) =µ3

3!exp(−µ) (3.22)

Finally prove by induction

P (n;N) =µn

n!exp(−µ) (3.23)

3.1.4 Poisson process

Consider dots on the time axis recording the arrival time of a neutron in yourdetector, the time at which a car passes by you, the time at which an atomof radioactive substance decays etc. Let ∆t be a small time interval. We take∆t to be adequately small such that where ever you place the interval on thetime axis, there shall be either no point or only one point it it. Let

p = λ∆t,

be the probability that a point is present in ∆t and q = 1−p be the probabilitythat a point is not present in the interval. λ is a constant characteristic of thePoisson process4. Note that ∆t must be chosen such that

∆t < < λ−1.

Let P (n, t) denote the probability that there are n points between time 0 andtime t. Note that the origin of the time axis - i.e. the time at which you startthe experiment, is arbitrary. The results do not depend on time origin. Hence

4 in radioactive decay, λ is called the decay constant. Problem: Find how λ and thehalf - life of the radioactive substance are related to each other?

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32 3 Binomial, Poisson, and Gaussian

we can interpret P (n, t) as the probability that your get n when you countover a duration of t. Show that,

P (n, t) = P (n, t−∆t)[1 − λ∆t] + P (n− 1, t−∆t)λ∆t (3.24)

The above is called a Master equation. Rewrite the Master equation as

P (n, t)− P (n, t−∆t)

∆t= −λ [P (n, t−∆t)− P (n− 1, t−∆t)] . (3.25)

Take the limit ∆t → 0 and derive an equation differential in t and differencein n, given by

∂P (n, t)

∂t= −λ [P (n, t)− P (n− 1, t)] (3.26)

The above equation can be solved by an easy method and an easier method.

3.1.5 Easy method

Write down the differential equation for n = 0 and solve it to get,

P (0, t) = exp(−λt) (3.27)

where we have taken the intial condition as

P (n, t = 0) = δn,0 =

0 n 6= 0

1 n = 0(3.28)

Write down the equation for n = 1. In the resulting equation substitute forP (0, t) and solve the resulting differential equation and get,

P (1, t) = λt exp(−λt) (3.29)

In the same way proceed for n = 2, 3, and show that

P (2, t) =(λt)2

2!exp(−λt) (3.30)

P (3, t) =(λt)3

3!exp(−λt) (3.31)

From the pattern that emerges we can conclude

P (n, t) =(λt)n

n!exp(−λt) (3.32)

More rigorously, employing the method of induction, we can prove thatthe above is indeed the solution of the difference - differential equation.

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3.1 Binomial distribution 33

3.1.6 Easier method

Employing the generating function method, show that

∂P (z; t)

∂t= −λ(1− z)P (z; t) (3.33)

Solve the above and show

P (z, t) = P (z, t = 0) exp [−λ (1− z) t] (3.34)

Show that

P (z, t = 0) =

∞∑

n=1

znP (n.t = 0)

=

∞∑

n=1

znδn,0

= 1 (3.35)

and hence

P (z, t) = exp [−λ (1− z) t] (3.36)

The next item in the agenda is on Gaussian distribution. It is a continuousdistribution defined for −∞ ≤ x ≤ +∞. Before we take up the task ofobtaining Gaussian from Poisson (in the limit µ∞), let us learn a few relevantand important things about continuous distribution.

3.1.7 Characteristic function

Let x = X(ω) be a continuous random variable, and f(x) its probabilitydensity function. The Fourier transform of f(x) is called the characteristicfunction of the random variable x = X(ω):

φX(k) =

∫ +∞

−∞

dx exp(−ikx) f(x) (3.37)

Taylor expanding the exponential in the above, we get

φX(k) =

∞∑

n=0

(−ik)nn!

∫ ∞

−∞

dx xn f(x)

=

∞∑

n=0

(−ik)nn!

Mn (3.38)

Thus the characteristic function generates the moments.

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34 3 Binomial, Poisson, and Gaussian

3.1.8 Cumulant generating function

The logarithm of the characteristic function is called the cumulant generatingfunction.

ψX(k) = lnφX(k) (3.39)

Let us write the above as,

ψX(k) = ln

(1 +

∞∑

n=1

(−ik)nn!

Mn

)

= ln(1 + ξ)

=∞∑

n=1

(−1)n+1

nξn

=

∞∑

n=1

(−1)n+1

n!

(∞∑

m=1

(−ik)mm!

Mm

)n

(3.40)

We now express ψX(k) as a power series in k as follows

ψX(k) =

∞∑

n=1

(−ik)nn!

ζn (3.41)

where ζn is called the n-th cumulant.From the above equations we can find the relation between moments and

cumulants.

3.1.9 Sum of identically distributed independent random variable

Let us consider the sum of N independent and identically distributed randomvariable, Xi : i = 1, 2, · · · , N. We thus have, Y ′ =

∑Ni=1Xi. Let us

consider scaled random variable Y = Y ′/N and enquire about its distributionin the limit N → ∞. We have,

φY (k) =

∫ ∞

−∞

dy exp(−iky)f(y)

=

∫ ∞

−∞

dx1

∫ ∞

−∞

dx2 · · ·∫ ∞

−∞

dxN

exp

[−ik

(x1 + x2 + · · ·xN

N

)]f(x1, x2, · · ·xN ) (3.42)

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3.1 Binomial distribution 35

The random variables are independent. Hence

f(x1, x2, · · ·xN ) = f(x1)f(x2) · · · f(xN ).

We have,

φY (k) =

∫ ∞

−∞

dx1 exp(−ikx1/N)f(x1)

∫ ∞

−∞

dx2 exp(−ikx2/N)f(x2) · · ·

· · ·∫ ∞

−∞

dxN exp(−ikxN/N)f(xN )

=

[∫ ∞

−∞

dx exp(−ikx/N)f(x)

]N

= [φX(k → k/N)]N

= exp [N lnφX(k → k/N)]

= exp

[N

∞∑

n=1

(−ik)nn!

ζnNn

]

= exp

[∞∑

n=1

(−ik)nn!

ζnNn−1

]

= exp

[−ikµ− k2

2!

σ2

N+O(1/N2)

]

∼N→∞ exp

[−ikµ− k2

2!

σ2

N

](3.43)

Thus the characteristic function of Y , in the limit N → ∞ is exp(−ikµ −(k2/2!)σ2/N). We will show below, that this is the characteristic function ofGaussian random variable with mean µ and variance σ2/N .

Thus the sum of N independent and identically distributed random vari-ables (with finite variance) tends to have a Gaussian distribution for large N .This is called the central limit theorem.

3.1.10 Poisson → Gaussian

Start with the moment generating function of the Poisson random variable:

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36 3 Binomial, Poisson, and Gaussian

P (z;µ) = exp[−µ(1− z)]. (3.44)

In the above substitute z = exp(−ik) and get,

P (k;µ) = exp [−µ 1− exp(−ik)] (3.45)

Substitute the power series expansion of the exponential function and get,

P (k;µ) = exp

[∞∑

n=1

(−ik)nn!

µ

](3.46)

We recognise the above as the cumulant expansion of a distribution for whichall the cumulants are the same µ. For large value µ it is adequate to consideronly small values of k. Hence we retain only terms upto quadratic in k. Thusfor k small, we have,

P (k) = exp

[−ikµ− k2

2!µ

](3.47)

The above is the Fourier transform or the characteristic function of a Gaussianrandom variable with mean as µ and variance also as µ.

−5 0 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 5 10 15 200

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Fig. 3.2. Poisson distribution : P (n) =µn

n!exp(−µ) with mean µ; P (n) versus n;

depicted as sticks; Gaussian distribution : G(x) =1

σ√2π

exp

[(x− µ)2

2σ2

]with mean

µ and variance σ2 = µ : continuous line.(Left) µ = 1.5; (Right) µ = 9.5. For large µ Poisson and Gaussian coincide

Thus in the limit µ → ∞, Gaussian distribution with mean and varianceboth equal to µ is a good approximation to Poisson distribution with meanµ, see figure above.

Page 49: Kpn Stat Mech

3.1 Binomial distribution 37

Problems

3.1. Derive an expression for 〈n(n − 1)〉 - called the first factorial moment.Find the variance of the random variable n.

3.2. Consider a system to two coins each with P (H) = p = 0.6 and P (T ) =q = 1− p = 0.4.

(a) Write down the micro states of the experiment.(b) Write down a possible ensemble of micro states describing the probabilities

of the micro states.

3.3. Consider the moment generating function of the binomial random vari-able given by B(z) = (q+ zp)N . Let 〈nk〉 denote the k− the moment of n. Bytaking the derivatives with respect to z calculate the first four moments of n. Let σ2 = 〈n2〉 − 〈n〉2 denote the variance of n. Calculate σ2. Relative fluc-tuations of n are given by σ/〈n〉. How does this quantity vary with increaseof N?

3.4. Consider a coin for which the probability of Heads is p and the probabilityof Tails is q = (1 − p). The experiment consists of tossing the coin until youget Heads for the first time. The experiment stops once you get Heads. Let ndenote the number of tosses in an experiment.

1. What is the sample space or micro-state space underlying this experi-ment ?

2. What is the discrete probability density function of n ?3. From the probability density function calculate the mean and variance ofn

4. Derive an expression for the moment generating function/partition func-tion of the random variable n

5. From the moment generating function calculate the mean and variance ofthe random variale n.

3.5. Consider a random walker starting from origin of a one dimensional lat-tice. He tosses a coin; if ’Heads’ he takes a step toward right; if ’Tails’ he stepsto the left. Let p be the probability for ’Heads’ and q = 1−p be the probabilityfor ’Tails’. Let P (m,n) denote the probability that the random walker is atm after n steps. Derive an expression for P (m,n). Let nR be the number ofright jumps and nL be the number of left jumps. We have n = nR + nL andm = nR − nL. We have P (nR, nL;n) = [n!/(nR!nL!)]p

nRqnL . etc.

3.6. Consider the problem with V = 10 M3; v = 10−3 M3 and N = 105. Theprobability of finding n molecules in v is given by the Poisson distributionwith µ = Np = 10. Plot the Poisson distribution.

3.7. Show that the mean and variance of a Poisson random variable are thesame : M1 =

∑∞n=0 n P (n;µ) = µ; M2 =

∑∞n=0 n2 P (n;µ) =?; σ2 =

M2 −M21 = µ

Page 50: Kpn Stat Mech

38 3 Binomial, Poisson, and Gaussian

3.8. Show that, ζ1 = M1; ζ2 = M2 −M21 ; ζ3 = M3 − 3M2M1 + 2M3

1 ; ζ4 =M4 − 4M3M1 − 3M2

2 + 12M2M21 − 6M4

1 .

3.9. Let x = X(ω) be an exponential random variable. It is defined for x ≥ 0;the probability density function is given by

f(x) = exp(−x).

Show that the characteristic function is given by

φ(k) =1

1 + ik.

Show that the n-th moment is given by

Mn = Γ (n+ 1) = n!

3.10. The characteristic function of the Gaussian random variable is formallyexpressed as,

φ(k) =1

σ√2π

∫ ∞

−∞

dx exp

[− (x− µ)2

2σ2− ikx

]

Carry out the above integral and show that,

φ(k) = exp

[−ikµ− k2

2!σ2

]

3.11. Let

Y =1

N

N∑

i=1

Xi.

where Xi : i = 2, N are independent and identically distributed exponen-tial random variables.

An exponential random variable is defined for x ≥ 0 and its probabilitydensity function is given by exp(−x). The n-th cumulant is (n−1)!. In particu-lar its mean is unity and its variance is also unity. Show that the characteristicfunction of Y is given by

φY (k) = exp

[−N ln

(1 +

ik

N

)]

We have,

− ln(1 − x) =∞∑

n=1

xn

n.

Substitute the above in φY (k) and show that

φY (k) = exp

∞∑

n=1

(−ik)nn!

(n− 1)!

Nn−1(3.48)

Page 51: Kpn Stat Mech

3.1 Binomial distribution 39

Demonstrate that the terms with n ≥ 3 can be neglected when N → ∞. Theresulting characteristic function is that of a Gaussian with mean unity andvariance (1/N).

3.12. There are N ideal gas molecules in a big room of volume V = 10 M3.The molecules are in equilibrium. There is a box of volume v < V , con-tained completely inside the room. The walls of the box are permeable andconducting. In other words the box can exchange energy and matter with thesurroundings.

Let ν be a random variable defined as the number of molecules in the box.Let P (n) denote the probability that the random variable ν takes a value n.Let µ = Nv/V . Define, nL = floor0.99× µ, nU = ceil1.01× µ, where, floorxis called the floor of x and it denotes the largest integer less than or equal tox; e.g. floor9.8 = 9; floorπ = 3; etc., and ceilx is called ceil of x and it denotesthe smallest integer greater that or equal to µ; e.g. ceil9.8 = 10; ceilπ = 4; etc..Calculate the probability that the random variable ν takes a value betweennL and nU , for the following three cases.

(i) v = 6 cubic m ; N = 10(ii) v = 10 cubic cm. ; N = 105

(iii) v = 10 cubic mm. ; N = 1014.

3.13. Start with a Poisson distribution of mean µ : P (n) = µn exp(−µ/n!.Employ Stirling approximation for n! (n! = nn exp(−n)) and write P (n) for-mally as P (n) = exp[−F (n)], where, F (n) = n ln(n) − n − n lnµ + µ. Solvethe equation

dF

dn= 0

and show that F is an extremum at n = µ. Then expand F (n) around µ as aTaylor series retaining terms upto second derivatives and and show that P (n)is a Gaussian with mean µ and variance µ. This is valid only for large n : thePoisson distribution peaks at n = µ only when µ is large.

Plot the Poisson distribution for various values of µ = .1; .5; 2.0; 10. Plotalso Gaussian distribution with mean =variance =µ. Demonstrate that thePoisson and Gaussian coincide for large µ.

Page 52: Kpn Stat Mech
Page 53: Kpn Stat Mech

4

Isolated system: Micro canonical ensemble

4.1 Preliminaries

We are going to study an isolated system of N particles confined to a volumeV . The particle do not interact with each other. We will count the number ofmicro states, denoted by the symbol Ω, of the system. This will be in generala function of energy E, volume V and the number of particles N . We shalldo the counting for both classical and quantum particles. Before we addressthe full problem, we shall consider a simpler problem of counting the microstates taking into account only the spatial coordinates neglecting completelythe momentum coordinates. Despite this simplification, we shall discover thatstatistical mechanics helps you derive the ideal gas law1.

1 I must tell you of a beautiful derivation of the ideal gas law by DanielBernoulli(1700-1782). It goes as follows. Bernoulli imagined air to be made oftis billiard balls all the time in motion, colliding with each other and with thewalls of the container. When a billiard ball bounces off the wall, it transmits acertain momentum to the wall and Bernoulli imagined it as pressure. It makessense. First consider air contained in a cube of side one meter. There is a certainamount of pressure felt by the wall. Now imagine the cube length to be doubledwith out changing the speeds of the molecule. In modern language this assump-tion is the same as keeping the temperature constant. The momentum transferredper collision remains the same. However since each billiard ball molecule has totravel twice the distance between two collision the force on the wall should besmaller by an factor of two. Also pressure is force per unit area. The ares of theside of the cube is four times more now. Hence the pressure should be less by afurther factor of four. Taking into account both these factors, we find the pressureshould be eight times less. We also find the volume of cube is now eight timesmore. Bernoulli concluded that the product of pressure and volume must be aconstant when there is no change in the molecular speeds - a brilliant argumentbased on simple scaling ideas.

Page 54: Kpn Stat Mech

42 4 Isolated system: Micro canonical ensemble

4.2 Configurational entropy

Consider placing a single particle in a volume V divided into two equal halves.Let ǫ = V/2. There are two ways, see figure below.

Fig. 4.1. Two ways of keeping a particle in a box divided into two equal parts.

Ω(V,N = 1, ǫ = V/2) =V

ǫ= 2 (4.1)

S = kB ln Ω = kB ln(2) (4.2)

Now consider two distinguishable particles into these two cells each ofvolume ǫ = V/2, see figure below.

Fig. 4.2. Four ways of keeping two distinguishable particles in a box divided intotwo equal halves.

We then have

Ω(V,N = 2, ǫ = V/2) =

(V

ǫ

)2

= 4 (4.3)

S = kB ln Ω = 2kB ln(2) (4.4)

For N particles we have,

Page 55: Kpn Stat Mech

4.3 Ideal gas law : Derivation 43

Ω(V,N, ǫ = V/2) =

(V

ǫ

)N

= 2N (4.5)

S = kB ln Ω = NkB ln(2) (4.6)

Let us now divide the volume equally into V/ǫ parts and count the numberof ways or organizing N (distinguishable) particles. We find

Ω(V,N) =

(V

ǫ

)N

(4.7)

S = kB ln Ω

= NkB ln(V/ǫ)

= NkB lnV −NkB ln ǫ (4.8)

We will discover later that the above formula captures the volume dependenceof entropy quite accurately.

4.3 Ideal gas law : Derivation

Differentiate S given by Eq. (??), with respect V . We get,

(∂S

∂V

)

E,N

=NkBV

(4.9)

From thermodynamics2 we have

2 In thermodynamics we start with U ≡ U(S, V ) for a given quantity of say anideal gas. This internal can change in a (quasi static) reversible process either byheat, TdS, or by work, −PdV . Hence we have the first law of thermodynamicsdu = TdS − Pdv. We have then, by definition

T =

(∂U

∂S

)

V

P = −(∂U

∂V

)

S

Let us start with S ≡ S(U, V ), which is natural for statistical mechanics. Wehave,

Page 56: Kpn Stat Mech

44 4 Isolated system: Micro canonical ensemble

(∂S

∂V

)

E,N

=P

T(4.10)

Thus we see from Eq. (??) and Eq. (??)

PV = NkBT (4.11)

4.4 Boltzmann entropy and Clausius’ entropy are thesame

From Eq. (??), we have,

dS =NkBV

dV (4.12)

Employing the equation of state : PV = NkBT , which we have derived, wecan rewrite the above as

dS =PdV

T(4.13)

Consider an isothermal process in an ideal gas. We have dU = 0. This impliesTdS = PdV . When the system absorbs a certain quantity of heat q isother-mally and reversibly, we have q = TdS = PdV . Equating PdV to q in Eq.(??), we get

dS =q

T(4.14)

which shows that Boltzmann entropy and thermodynamic entropy are thesame.

dS =

(∂S

∂U

)

V

+

(∂S

∂V

)

U

To express the partial derivatives in the above in terms of T and P , we rearrangethe terms in the first law equation as,

dS =1

TdU +

P

TdV

Equating the pre-factors of dU and dV in the above two equation, we get,

(∂S

∂U

)

V

=1

T

(∂S

∂V

)

U

=P

T

Page 57: Kpn Stat Mech

4.6 Boltzmann counting 45

4.5 Some issues on extensitivity of entropy

The expression for entropy given below,

S(V,N) = NkB lnV −NkB ln ǫ

is not extensive. If I double the value of V and of N , I expect S to be doubled.It does not. Mathematically S is extensive if it is a first order homogeneousfunction V and N . In other words we should have S(λV, λN) = λS(V,N).The above expression for entropy does not satisfy this rule3.

4.6 Boltzmann counting

To restore extensivity of entropy, Boltzmann introduced an ad-hoc notion ofindistinguishable particles. N ! permutations of the particles, should all becounted as one micro state since they are ’indistinguishable’4. Hence,

Ω(V,N) =1

N !

(V

ǫ

)N

S(V,N) = kB ln Ω(V,N) (4.15)

= NkB ln

(V

N

)+NkB −NkB ln ǫ (4.16)

S(λV, λN) = λS(V,N)

Time has come for us to count the micro states of an isolated systemof N non interacting point particles confined to a volume V , taking intoconsiderations the positions and the momenta of all the particles.

Each particle for its specification requires six numbers : three positions andthree momenta. The entire system can be specified by a string of 6N numbers.In a 6N dimensional phase space the system is specified by a point. The phasespace point is all the time moving. We would be interested determining theregion of the phase space accessible to the system when it is in equilibrium.

3 This is called Gibbs’ paradox. More precisely Gibbs formulated the paradox interms of entropy of mixing of like and unlike gases. We shall see these in detailslater when we consider closed system described by canonical ensembles.

4 The remedy suggested by Boltzmann is only temporary. Non extensivity of en-tropy points to a deeper malady in the statistical mechanics based on classicalformalism. For the correct resolution of the non-extensivity-paradox we have towait for the arrival of Quantum Mechanics. We shall see of these issues in detailslater when we consider quantum statistics.

Page 58: Kpn Stat Mech

46 4 Isolated system: Micro canonical ensemble

If we are able to count the phase space volume, then we can employ the firstmicro-macro connection proposed by Boltzmann and get an expression forentropy as a function of energy, volume and the number of particles.

The system is isolated. It does not transact energy or matter with thesurroundings. Hence its energy remains a constant. The potential energy iszero since the particles do not interact with each other. The kinetic energy isgiven by

E =3N∑

i=1

p2i2m

(4.17)

The system is thus confined to the surface of a 3N dimensional sphere. Weneed a formula for the volume of an hyper-sphere in 3N dimensions. To thisend we need to know of Heaviside5 theta function and Dirac6 delta function.

4.7 Heaviside Theta function

Define a function

f(x; ǫ) =

0 for −∞ ≤ x ≤ − ǫ

2

(1

ǫ

)x+

1

2for − ǫ

2≤ x ≤ +

ǫ

2

1 for +ǫ

2≤ x ≤ +∞

(4.18)

where ǫ > 0.Define

Θ(x) =lim.ǫ→ 0 f(x; ǫ).

Θ(x) is called the step function, Heaviside step function, unit step functionor theta function. It is given by,

Θ(x) =

0 for −∞ ≤ x < 0

1 for 0 < x ≤ +∞(4.19)

5 Oliver Heaviside(1850-1925)6 Paul Adrien Maurice Dirac(1902-1984)

Page 59: Kpn Stat Mech

4.9 Area of a circle 47

4.8 Dirac delta function

Start with the function f(x; ǫ) defined by Eq. (??). Take the derivative of thefunction. We find that the derivative is 1/ǫ, when −ǫ/2 < x + ǫ/2 and zerootherwise, Define,

g(x; ǫ) =df

dx=

0 for −∞ ≤ x < −ǫ/2

1

ǫfor −ǫ/2 < x + ǫ/2

0 for +ǫ/2 < x ≤ +∞

(4.20)

The Dirac-delta function is defined as,

δ(x) = limitǫ→0 g(x; ǫ) (4.21)

Consider the following integral.

I =

∫ +∞

−∞

dx g(x; ǫ) (4.22)

We find that the integral is the same for all values of ǫ. This gives us animportant property of the Dirac-delta cunction:

∫ +∞

−∞

dx δ(x) = 1 (4.23)

4.9 Area of a circle

Let us demonstrate how to use the theta function and delta function to derivean expression for a circle of radius R. Let us denote the area of a circle bythe symbol V2(R) - the ’volume’ of a two dimensional ’sphere’ of radius R. Alittle thought will tell you,

V2(R) =

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 Θ

(R2 −

2∑

i=1

x2i

)(4.24)

Letyi = xi/R for i = 1, 2.

Then,

V2(R) = R2

∫ +∞

−∞

dy1

∫ +∞

−∞

dy2 Θ

(R2(1 −

2∑

i=1

y2i

)(4.25)

We have

Page 60: Kpn Stat Mech

48 4 Isolated system: Micro canonical ensemble

Θ(λx) = θ(x) ∀ λ > 0.

Therefore,

V2(R) = R2

∫ +∞

−∞

dy1

∫ +∞

−∞

dy2 Θ

(1−

2∑

i=1

y2i

)(4.26)

= R2 V2(R = 1) (4.27)

We can now write Eq. (??) as

V2(R = 1)R2 =

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 Θ

(R2 −

2∑

i=1

x2i

)(4.28)

Now differentiate both sides of the above equation with respect to the variableR. We have already seen that the derivative of a Theta function is the Dirac-delta function. Therefore

V2(R = 1)2R = 2R

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 δ

(R2 −

2∑

i=1

x2i

)(4.29)

Now multiply both sides of the above equation by exp(−R2)dR and integrateover the variable R from 0 to ∞. We get,

V2(R = 1)

∫ ∞

0

exp(−R2)2RdR =

∫ ∞

0

exp(−R2) 2RdR

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 δ

(R2 −

2∑

i=1

x2i

)(4.30)

V2(R = 1)

∫ ∞

0

dt exp(−t) =∫ ∞

−∞

dx1

∫ ∞

−∞

dx2 exp(−x21 − x22) (4.31)

V2(R = 1)× 1 =

[∫ +∞

−∞

dx exp(−x2)]2

(4.32)

V2(R = 1) =

[2

∫ ∞

0

dx exp(−x2)]2

=

[∫ ∞

0

dx x−1/2 exp(−x)]2

(4.33)

V2(R = 1) =

[∫ ∞

0

x(1/2)−1 exp(−x)dx]2

= [Γ (1/2)]2= π (4.34)

Thus V2(R) = V2(R = 1)×R2 = πR2, a result we are all familiar with.

Page 61: Kpn Stat Mech

4.10 Volume of an N-dimensional sphere 49

4.10 Volume of an N -dimensional sphere

The volume of an N - dimensional sphere of radius R is formally given by theintegral,

VN (R) =

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 · · ·∫ +∞

−∞

dxN Θ

(R2 −

N∑

i=1

x2i

)(4.35)

Change the coordinate system from

xi : i = 1, N to yi = xi/R : i = 1, N.

dxi = Rdyi ∀ i = 1, N

Θ

(R2

[1−

N∑

i=1

y2i

])= Θ

(1−

N∑

i=1

y2i

)

We have,

VN (R) = RN

∫ +∞

−∞

dy1

∫ +∞

−∞

dy2 · · ·∫ +∞

−∞

dyN Θ

(1−

N∑

i=1

y2i

)(4.36)

= VN (R = 1)RN (4.37)

where VN (R = 1) is the volume of an N - dimensional sphere of radius unity.To find the volume of N -dimensional sphere of radius R, we proceed as

follows.

VN (R = 1)RN =

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 · · ·∫ +∞

−∞

dxN Θ

(R2 −

N∑

i=1

x2i

)(4.38)

Differentiate both sides of the above expression with respect to R and get,

NVN (R = 1)RN−1 =

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 · · ·

· · ·∫ +∞

−∞

dxN δ

(R2 −

N∑

i=1

x2i

)2R (4.39)

Now, multiply both sides by exp(−R2)dR and integrate over R from 0 to ∞.The Left Hand Side:

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50 4 Isolated system: Micro canonical ensemble

LHS = NVN (R = 1)

∫ ∞

0

dR exp(−R2)RN−1 (4.40)

Let x = R2; then dx = 2RdR. This give

dR =1

2

dx

x1/2.

We get,

LHS = VN (R = 1)N

2

∫ ∞

0

xN2−1 exp(−x)dx

= VN (R = 1)N

(N

2

)

= Γ

(N

2+ 1

)VN (R = 1) (4.41)

The Right Hand Side :

RHS =

∫ ∞

0

dR exp(−R2)

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 · · ·

· · ·∫ +∞

−∞

dxN δ

(R2 −

N∑

i=1

x2i

)2R (4.42)

t = R2

dt = 2RdR

RHS =

∫ ∞

0

dt exp(−t)∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 · · ·∫ +∞

−∞

dxN δ

(t−

N∑

i=1

x2i

)

=

∫ +∞

−∞

dx1

∫ +∞

−∞

dx2 · · ·∫ +∞

−∞

dxN exp[−(x21 + x22 + · · ·x2N )

]

=

[∫ ∞

−∞

dx exp(−x2)]N

= πN/2 (4.43)

Page 63: Kpn Stat Mech

4.12 Density of states : g(E) 51

Thus we get

VN (R = 1) =πN/2

Γ(N2 + 1

) (4.44)

VN (R) =πN/2

Γ(N2 + 1

)RN (4.45)

4.11 Classical counting of micro states

Consider an isolated system of N non-interacting point particles. Each par-ticle requires 3 position coordinates and 3 momentum coordinates for for itsspecification. A string of 6N numbers denotes a micro state of the system.Let us first find the volume of the phase space accessible to the system. Theintegral over 3N spatial coordinates gives V N . We have

E =

3N∑

i=1

p2i2m

The volume of the phase space of the system with energy ≤ E is the volumeof a 3N dimensional sphere of radius

√2mE.

4.11.1 Counting of the volume

Let us measure the volume of the phase space in units of h3N , where h isPlanck constant. We have

∆x∆px ≥ h.

Thus h3N is the volume of a ”minimum uncertainty” cube. Thus we have

Ω(E, V,N) =V N

h3N(2πmE)3N/2

Γ(3N2 + 1

) (4.46)

4.12 Density of states : g(E)

Let g(E) denote the density of (energy) states. g(E)dE gives the number ofmicro states with energy between E and E + dE. In other words,

Ω(E) =

∫ E

0

g(E′)dE′ (4.47)

From the above, we find

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52 4 Isolated system: Micro canonical ensemble

g(E, V,N) =

(∂Ω(E, V,N)

∂E

)

V,N

(4.48)

Let us take the partial derivative of Ω(E, V,N) with respect to E and get,

g(E, V,N) =V N

h3N(2πm)3N/2

Γ (3N2 + 1)

3N

2E(3N/2)−1 (4.49)

Let us substitute N = 1 in the above and get the single particle density ofstates, g(E, V ) as,

g(E, V ) =V

h3π

4(8m)3/2E1/2 (4.50)

4.12.1 A sphere lives on its outer shell : Power law can beintriguing

In the limit of N → ∞, the volume of a thin outer shell tends to the volumeof the whole sphere. This intriguing behaviour is a consequence of the powerlaw behaviour.

VN (R)− VN (R−∆R)

VN (R)=RN − (R−∆R)N

RN(4.51)

= 1−(1− ∆R

R

)N

(4.52)

= 1 for N → ∞ (4.53)

Hence in the limit of N → ∞ the number of micro states with energy lessthan or equal to E is nearly the same as the number of micro states withenergy between E −∆E and E.

4.13 Entropy of an isolated system

From Eq. (??) we see that

S(E, V,N) = NkB

[lnV +

3

2ln

(E

N

)+

3

2ln

(4πm

3h2

)+

3

2

]

We find that the above expression for entropy is not extensive. In other wordsS is not an extensive function of V 7 : S(λE, λV, λN) 6= λS(E, V,N). We

7 Note that S is extensive in E and N .

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4.14 Properties of an ideal gas 53

shall follow Boltzmann’s prescription and divide Ω(E, V,N), see Eq. (??), byN !.

Ω(E, V,N) =V N

h3N1

N !

(2πmE)3N/2

Γ(3N2 + 1

) (4.54)

The corresponding entropy is then,

S(E, V,N) = NkB

[ln

(V

N

)+

3

2ln

(E

N

)+

3

2ln

(4πm

3h2

)+

5

2

](4.55)

4.14 Properties of an ideal gas

The temperature of an ideal gas, as a function of E, V , and N , is given by

(∂S

∂E

)

V,N

=1

T=

3NkB2E

(4.56)

T =2E

3NkB(4.57)

The energy of the system is thus given by,

E = 3N

(1

2kBT

)(4.58)

The above is called equi-partition theorem. Each quadratic term in the Hamil-tonian carries an energy of kBT/2

8.The pressure of an isolated system of ideal gas as a function of E, V , and

N , is given by,

(∂S

∂V

)

E,N

=P

T=NkBV

(4.59)

P =NkBT

V(4.60)

Substituting in the above T as a function of E, V , and N , see Eq. (??), weget,

8 For an ideal gas

H =3N∑

i=1

p2i2m

.

There are 3N quadratic terms in the Hamiltonian.

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54 4 Isolated system: Micro canonical ensemble

P =2E

3V(4.61)

An expression for the chemical potential as a function of E, V , and N isderived as follows.

(∂S

∂N

)

E,V

= −µ

T(4.62)

Therefore,

µ = −NkBT ln

(V

N

)− 3

2NkBT ln

(4πmE

3Nh2

)(4.63)

Substituting in the above the expression for T from Eq. (??), we get,

µ = −2E

3ln

(V

N

)− 2E ln

(4πmE

3Nh2

)(4.64)

In the above expression for the micro canonical chemical potential, let ussubstitute

E = 3NkBT/2

and express chemical potential in terms of T , V and N . We get,

µ = −NkBT ln

(V

N

)− 3

2NkBT ln

(2πmkBT

h2

)(4.65)

= −NkBT ln

(V

N

)+ 3NkBT ln(Λ) (4.66)

where Λ is the thermal or quantum wavelength9 given by,

Λ =h√

2πmkBT(4.67)

Let the number density be denoted by the symbol ρ. We have thus ρ =N/V . We can write the chemical potential in a compact form

µ = NkBT ln(ρΛ3) (4.68)

Let me end this section by saying that the micro canonical ensemble formalismleads to the ideal gas law: See the expression for P given in Eq.(??). We have,

PV = NkBT (4.69)

9 we shall see about it later

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4.15 Quantum counting of micro states 55

4.15 Quantum counting of micro states

We have done classical counting of micro states and showed that for an isolatedsystem of a single particle confined to a volume V , the number of micro stateswith energy less than ǫ, is given by

Ω(E, V ) =V

h3(2πmǫ)

3/2

Γ ((3/2) + 1)(4.70)

We have obtained the above by substituting N = 1 in Eq. (??).Now let us do quantum counting of micro states 10.

10 Some blah .... blah ... on Quantum Mechanics

A fundamental entity in quantummechanics is the wave function ψ(q, t), whereq is the position vector. The wave function is given a physical interpretation that

ψ⋆(q, t)ψ(q, t)dq

gives the probability of finding the system in an elemental volume dq around thepoint q at time t. Since the system has to be somewhere, for, otherwise we wouldnot be interested in it, we have the normalization,

∫ψ⋆(q, t)ψ(q, t)dq = 1,

where the integral is taken over the entire coordinate space - each of the x, y andz coordinates extending from −∞ to +∞.

A central problem in quantum mechanics is the calculation of ψ(q, t) for thesystem of interest. We shall be interested in the time independent wave functionψ(q) describing a stationary states of the system.How do we get ψ(q) ?Schrodinger gave a prescription : Solve the equation Hψ(q) = Eψ(q), with ap-propriate boundary conditions.We call this the time independent Schrodinger equation.H is the Hamiltonian operator

H = − ~2

2m2 +U(q)

The first operator on the right is kinetic energy and the second the potentialenergy.E in the Schrodinger equation is a scalar ... a real number... called energy. It

is an eigenvalue of the Hamiltonian operator : we call it energy eigenvalue.The Schrodinger equation is a partial differential equation. Once we impose

boundary condition on the solution, then only certain discrete energies are per-mitted. We call these energy eigenvalues.

Energy Eigenvalue by Solving Schrodinger Equation Once we specifyboundary conditions, then a knowledge of the Hamiltonian is sufficient to deter-mine its eigenvalues and the corresponding eigenfunctions. There will be usuallyseveral eigenvalues and corresponding eigenfunctions for a given system.

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56 4 Isolated system: Micro canonical ensemble

4.15.1 Energy eigenvalues : Integer Number of Half Wave lengthsin L

Let me tell you how to obtain the energy eigenvalues without invokingSchrodinger equation. Consider a particle confined to a one dimensional boxof length L. We recognise the segment L must contain integral number of halfwave lengths - so that the wave function vanishes at the boundaries of the onedimensional box. In other words,

L = n× λ

2: n = 1, 2, , · · · (4.71)

λ =2L

n: n = 1, 2, , · · · (4.72)

Substitute the above in the de Broglie relation

p =h

λ=

h

2Ln : n = 1, 2, · · · .

This yields

ǫn =p2

2m=

h2

8mL2n2 : n = 1, 2, · · · .

Consider a particle in an L × L × L cube - a three dimensional infinite well.The energy of the system is given by

ǫnx,ny,nz=

h2

8mL2(n2

x + n2y + n2

z)

where nx = 1, 2, · · · , , ny = 1, 2, · · · , and nz = 1, 2 · · · , .The ground state is (nx, ny, nz) = (1, 1, 1); it is non degenerate; the energy

eigenvalue is

ǫ1,1,1 =3h2

8mL2(4.73)

For a single particle in a one dimensional infinite well

H = − ~2

2m

∂2

∂x2

Solve the one dimensional Schrodinger equation with the boundary condi-tion : the wave function vanishes at the boundaries. Show that the energyeigenvalues are given by

ǫn =h2

8mL2n2; n = 1, 2, · · ·

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4.15 Quantum counting of micro states 57

The first excited state is three-fold degenerate. The corresponding energyeigenvalue is

ǫ2,1,1 = ǫ1,2,1 = ǫ1,1,2 =3h2

4mL2. (4.74)

We start with,

ǫ =h2

8mL2(n2

x + n2y + n2

z)

We can write the above as,

n2x + n2

y + n2z =

8mL2ǫ

h2= R2

(nx, ny, nz) represents a lattice point in the three dimensional space. Theequation n2

x + n2y + n2

z = R2 says we need to count the number of latticepoints that are at a distance R from the origin. It is the same as the numberof lattice points that are present on the surface of a sphere of radius R in thepositive quadrant; note that the x, y, and z coordinates of the lattice pointsare all positive. It is difficult to count the number of lattice points lying onthe surface of a sphere. Instead we count the number of points contained ina thin spherical shell. To calculate this quantity we first count the number ofpoints inside a sphere of radius

R =

(8mL2ǫ

h2

)1/2

and take one-eighth of it. Let us denote this number by Ω(ǫ). We have,

Ω(ǫ) =1

8

4

3π R3 =

π

6

(8mL2ǫ

h2

)3/2

(4.75)

We recognize V = L3 and write the above equation as

Ω(ǫ, V ) =V

h3π

6(8mǫ)3/2 =

V

h34π

3(2mǫ)3/2

=V

h34π

3

(2πmǫ)3/2

π3/2=V

h3(2πmǫ)3/2

(3/2)(1/2)√π

=V

h3(2πmǫ)3/2

(3/2)(1/2)Γ (1/2)=V

h3(2πmǫ)3/2

Γ ((3/2) + 1)(4.76)

The above is exactly the one we obtained by classical counting, see Eq. (??)Notice that in quantum counting of micro states, the term h3 comes naturally,while in classical counting it is hand-put11.

11 We wanted to count the phase space volume. We took h3 as the volume of asix-dimensional cube. We considered the six-dimensional phase space (of a single

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58 4 Isolated system: Micro canonical ensemble

The density of (energy) states is obtained by differentiating Ω(ǫ, V ) withrespect to the variable ǫ. We get

g(ǫ, V ) =V

h3π

4(8m)3/2ǫ1/2 (4.77)

The important point is that the density of energy states is proportional toǫ1/2.

4.16 Chemical Potential

4.16.1 Toy model

Consider an isolated system of two identical, distinguishable and non-interactingparticles occupying non-degenerate energy levels 0, ǫ, 2ǫ, 3ǫ, · · · , such that

the total energy of the system is 2ǫ. Let Ω(E = 2ǫ,N = 2) denote the numberof micro states of the two-particle system with total energy E = 2ǫ. For anisolated system, since all the micro states are equally probable the micro statespace is a good candidate for the micro canonical ensemble.

We label the two particles as A and B. The micro states with total energy2ǫ are given below.

0 ǫ 2 ǫ

A − B

B − A

− A,B −

Table 4.1. Micro states of of two particles with total energy 2ǫ

We find thatΩ(E = 2ǫ,N = 2) = 3.

The entropy of the two-particle system with energy E = 2ǫ is given by

S(E = 2ǫ,N = 2) = kB ln Ω(E = 2ǫ,N = 2).

Now add a particle, labelled C, such that the energy of the system doesnot change. In other words, the three-particle system has energy 2ǫ which isthe same as that of the two particle system. Let Ω(E = 2ǫ,N = 3) denotethe number of micro states of the three-particle system with a total energy ofE = 2ǫ. The table below gives the micro states.

particle) as filled with non-overlapping exhaustive set of such tiny cubes. We haveto do all these because of Boltzmann ! He told us that entropy is logarithm ofnumber of micro states. We need to count the number of micro states.

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4.16 Chemical Potential 59

0 ǫ 2ǫ

A,B − C

B,C − A

C,A − B

A B,C −B C,A −C A,B −

Table 4.2. Micro states of three particles with total energy 2ǫ

We findΩ(E = 2ǫ,N = 3) = 6.

The entropy is given by

S(E = 2ǫ,N = 3) = kB ln Ω(E = 2ǫ,N = 3).

We find S(E = 2ǫ,N = 3) > S(E = 2ǫ,N = 2). Note that µ =

(∂U

∂N

)

S,V

.

In other words, µ is the change in energy of the system when one particleis added in such a way that the entropy and volume of the system remainunchanged. To achieve this we must remove ǫ amount of energy from thethree particle system. In other words we demand S(E = ǫ,N = 3) = S(E =2ǫ,N = 2).

Thus µ = −ǫ. This problem that the chemical potential of a system ofnon-interacting particles is negative.

4.16.2 Chemical potential of an ideal gas

The number of micro states of a classical system ofN non-interacting particlesconfined to a volume V is given by,

Ω(E, V,N) =V N

h3N1

N !

(2πmE)3N/2

Γ (3N2 + 1)

The entropy of the system defined as

S(E, V,N) = kB ln Ω(E, V,N)

is given by,

S(E, V,N) = NkB

[lnV

N+

3

2ln

(E

N

)+

3

2ln

(4πm

3h2

)+

5

2

]. (4.78)

The chemical potential µ is given by

µ(E, V,N) = −T(∂S

∂N

)

E,V

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60 4 Isolated system: Micro canonical ensemble

Problems

4.1. Consider an isolated system of N non-interacting point particles occupy-ing two states of energies −ǫ and +ǫ. The energy of the system is E. Define

x =E

Nǫ. Show that the entropy of the system is given by

S(x) = −NkB2

[(1 + x) ln

(1 + x

2

)+ (1− x) ln

(1− x

2

)]

Also show that

1

T=kB2ǫ

ln

(1− x

1 + x

)

4.2. Sketch the function f(x; ǫ) for ǫ = 2, 1, 1/2, 1/4; sketch also the functionin the limit of ǫ→ 0.

4.3. Sketch g(x; ǫ) for ǫ = 2, 1, 1/2, 1/4. How does it look in the limit ǫ→ 0?

4.4. Define the Dirac-delta function centered at x0 as δ(x− x0). To this end,start with g(x; ǫ, x0) and show that in the limit ǫ → 0 it defines a thetafunction with a step at x = x0 6= 0; the theta function is denoted by Θ(x−x0).Take the derivative of g(x; ǫ, x0) and show that in the limit ǫ→ 0 we get theDirac-delta function centered at x0. It is denoted by δ(x− x0). Show that

∫ +∞

−∞

dx δ(x− x0) = 1

∫ +∞

−∞

dx ρ(x)δ(x − x0) = ρ(x0)

4.5. Show that Γ (1/2) =√π

4.6. The volume of a (three dimensional) sphere of radius R is denoted byV3(R). Show that

V3(R) =4

3π R3

4.7. Substitute N = 1, 2, and 3 in the expression above for VN (R) and showthat you recover the standard results V1(R) = 2R, V2(R) = πR2, and V3(R) =4

3πR3.

4.8. Consider a particle in a two dimensional infinite well. In other words,consider particle in a square of area A. Carry out ”quantum counting” andshow that the density of (energy) states is

g(ǫ, A) =A

h22πm

Notice that the density of (energy) states is independent of ǫ.

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4.16 Chemical Potential 61

4.9. Consider a particle in a one dimensional infinite well. In other wordsconsider a particle confined to a line segment of length L. Carry out ”quantumcounting” and show that the density of (energy) states is

g(ǫ, L) =L

h(2m)

1/2ǫ−1/2

4.10. Carry out ”classical counting” of the number of micro states of an iso-lated system of N particles confined to an area A, with energy E. Since theparticles are confined to a plane we require 2N position and 2N momentumcoordinates to specify a micro state. Calculate Ω(E,A,N) the number of mi-cro states with energy less than or equal to E. Substitute N = 1. Differentiatethe resulting expression with respect to energy and verify whether you get thesame answer as of Problem No. 4.7

4.11. Carry out ”classical counting” of the number of micro state of an iso-lated system of N particles confined to a length L with energy E. Since theparticles are confined to a line we require N position and N momentum coor-dinates to specify a micro state. Calculate Ω(E,L,N) - the number of microstates with energy less than or equal to E. Substitute N = 1. Differentiate theresulting expression with respect to E and verify whether you get the sameanswer as of Problem No. 4.9

4.12. Show that we must remove ǫ of energy from the system to restore en-tropy to its original value. In other words show that Ω(E = ǫ,N = 3) =

Ω(E = 2ǫ,N = 2) = 3

4.13. Derive an expression for µ as a function of E, V and N . Substitute inthe expression12

E = 3N

(1

2kBT

)

and show that µ as a function of T , V and N is given by13 ,

µ(T, V,N) = −kBT ln

(V

N

)− 3

2kBT ln

(2πmkBT

h2

).

Now add a particle to the system such that the energy change from E to E+ηand the entropy changes from S(E, V,N)to S(E + η, V,N + 1).

To derive an expression for S(E + η, V,N + 1) replace in Eq. (??): E byE + η and N by N + 1.

Show that for η = µ we get S(E + η, V,N + 1) = S(E, V,N). In thederivation, assume N to be large and the magnitude of η to be small comparedto total energy i.e |η| << E.

12 Equipartition theorem : Every quadratic term in the Hamiltonian carries an en-ergy of kBT/2.

13 Read G Cook and R H Dickerson, Understanding the chemical potential, AmericanJournal of Physics 63(8), 737 (1995)

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5

Closed system : Canonical ensemble

5.1 What is a closed system ?

A closed system is one which does not exchange material with the surround-ings. However, it does exchange energy. It is in thermal contact with thesurroundings. We idealize the surroundings of a closed system as a ”heatbath”1.

Thus, a system in thermal equilibrium, is characterized by T , V and N .The system is not isolated. Hence its micro states are not all equi-probable.

5.2 Toy model a la H B Callen

Let us illustrate this by considering a toy problem proposed by H B Callen2.Consider a fair red die representing the system and two fair white dice rep-resenting the surroundings. Let P (k) denote the probability that the red dieshows up k given the three dice together add to 6. Note if we do not impose thecondition (the three die add to six), the six micro states are equally probablefor the red die. However because of the condition imposed the micro statesare not equally probable. We find there are 10 micro states with the propertythat the three dice add to 6. These are listed below.

Of these, there are four micro states for which the system die shows up1; therefore P (1) = 0.4. Similarly we can calculate the other probabilities :

1 A ”heat bath” is one which transacts energy with the system, but its temperaturedoes not change. Example: Keep a cup of hot coffee at temperature say 60 Cin a room; let the room be at temperature 30 C. The coffee cools to 30C. Thetemperature of the room does not increase. May be, I should say the temperatureof the room increases by an extremely small amount which for all practical pur-poses can be considered as zero. We idealize and say the room is a heat bath andits temperature does not change when it transacts energy with the cup of coffee.

2 H B Callen, Thermodynamics and and Itroduction to Thermostatistics, WileyStudent Edition (2005)

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64 5 Closed system : Canonical ensemble

W R W

1 1 4

2 1 3

3 1 2

4 1 1

1 2 3

2 2 2

3 2 1

1 3 2

2 3 1

1 4 1

Table 5.1. Micro states of three dice with the constraint that they add to six

P (2) = 0.3; P (3) = 0.2; P (4) = 0.1; P (5) = P (6) = 0. The important pointis that the micro states of the system are not equi probable.

In other words, if the system interacts thermally with the surroundingsthen the probability differs from one micro state to the other.

What is the probability of a micro state in a closed system ? Let us calcu-late the probability in the next section by a fairly straight forward procedureinvolving Taylor expansion of S(E). I learnt of this first, from the book ofBalescu3.

5.3 Canonical partition function

Consider the system, its boundary and the bath - all the three together - asan isolated system. We know for an isolated system, all the micro states areequally probable. Let E denote the energy of the isolated system. It remainsa constant.

Now consider a particular micro state of the system. Let us label it as C.Let its energy be E(C). Note that E(C) < < E . When the closed system is in

its micro state C, the surroundings can be in any one of Ω(E − E(C)) microstates of the isolated system4.

3 R Balescu, Equilibrium and nonequilibrium statistical mechanics, Wiley (1975).4 I am considering that a micro state of the isolated system can be thought of asa simple juxtaposition of the micro state of closed system and the micro state of

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5.3 Canonical partition function 65

For the isolated system, all micro states are equally probable. Thus we cansay that the probability of finding the closed system in its micro state C isgiven by

P (C) =Ω(E − E(C))

Ωt

where we have denoted the total number of micro states of the isolated systemas Ωt.

We have S(E − E(C)) = kB ln Ω(E − E(C)). Therefore

Ω(E − E(C) = exp

[1

kBS(E − E(C))

]

Also since E(C) < < E , we can Taylor expand S(E−E(C)) around E retainingonly the first two terms. We get,

S(E − E(C)) = S(E)− E(C)

(∂S

∂E

)

E=E

= S(E)− 1

TE(C)

Substituting the above in the expression for P (C), we get,

P (C) =exp [S(E)/kB]

Ωt

exp

[−E(C)

kBT

]

= α exp[−βE(C)]

where α is a constant and β = 1/(kBT ). We can evaluate α completely interms of the properties of the closed system by the normalization conditionfor the probabilities, see below.

C

P (C) = 1

α∑

C

exp [−βE(C)] = 1

Q(T, V,N) =1

α=∑

C

exp [−βE(C)]

the surroundings. The system and the surroundings interact at the boundariesand hence there shall exist micro states of the isolated system which can not beneatly viewed as the system micro state juxtaposed with the surroundings microstate. Such micro states are so few in number we shall ignore them.

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66 5 Closed system : Canonical ensemble

where Q(T, V,N) is called the canonical partition function.Thus we have,

Q(T, V,N) =∑

C

exp [−βE(C)]

=∑

E

Ω(E) exp(−βE)

where Ω(E) is the degeneracy of the eigenvalue E. In other words Ω(E) isthe number of micro states of the equilibrium closed system with energy E.

If energy is a continuous variable the we consider g(E), the density of(energy) states. Then g(E)dE denotes the number of micro states with energybetween E and E + dE. Canonical partition function can be expressed as anintegral

Q(T, V,N) =

∫ +∞

0

dE g(E) exp[−βE) (5.1)

The above is just a transform of the g(E) → Q(T ) where we have transformedenergy E in favour of T - the temperature5. Physically the transform meansthat We are going from a micro canonical description with independent vari-ables E, (V and N) to a canonical description with independent variablesT , (V , and N). In other words we are going from an isolated system withenergy as an independent variable to a closed system with temperature as anindependent variable.

5.4 Canonical partition function :Method of most probable distribution

Let us now derive an expression for the canonical partition function employingan easier method - called the method of most probable distribution.

Consider an isolated system. For convenience we imagine it as a big cube.It contains molecules moving around here and there, hitting against each otherand hitting against the wall. The isolated system is in equilibrium. Rememberthat an isolated system left to itself will eventually reach a state of equilibriumwhence all its macroscopic properties are unchanging with time; also the valueof a macroscopic property shall be the same at any region in the system. Lettemperature be T . Note that the temperature is determined by the isolatedsystem; it is not determined by you. The system attains that temperature forwhich its entropy is maximum.

5 If we take β = ik, then we have one sided Fourier transform. If we take β = sthen we have Laplace transform. etc.

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5.4 Canonical partition function : Method of most probable distribution 67

Let us imagine that the isolated system represented by a big cube is dividedinto a set of small cubes of equal volumes by means of imaginary walls. Eachcube represents a macroscopic part of the isolated system.

Each small cube is, in its own right, a macroscopic system with a volumeV . Since the the walls of a small cube permits molecules and energy to moveacross, the number of molecules in a cube, is not fixed. It shall fluctuate aroundsome mean value; the fluctuations, however, are extremely small. The aboveobservations hold good for energy also. Let A denote the number of cubescontained in the big cube.

The isolated system - the big cube, has a certain amount energy say E andcertain number of molecules N and a certain volume V and these quantitiesare constants.

You can immediately see that what we have is a grand canonical ensembleof open systems - each cube represents an open system. Each cube is a mem-ber of a grand canonical ensemble. All the members are identical as far astheir macroscopic properties are concerned. This is to say the volume V , thetemperature T and chemical potential µ are all the same for all the members.

Now, let us imagine that the walls are made impermeable to movementof molecules across. A cube can not exchange matter with its neighbouringcubes. Let us also assume that each cube contains exactly N molecules. En-ergy in a cube is however not fixed. Energy can flow from one cube to itsneighbouring cubes. This constitutes a canonical ensemble6.

Aim :To find the probability for the closed system to be in its micro state i.

First we list down all the micro states of the equilibrium closed system. Letus denote the micro states as 1, 2, · · · . Note that the macroscopic propertiesT , V , and N are the same for all the micro states. In fact the system switchesfrom one micro state to another all the time. Let Ei denote the energy of thesystem when it is in micro state i. The energy can vary from one micro stateto another. However the fluctuations of energy from its mean value are verysmall for an equilibrium macroscopic system.

To each cube, we can attach an index i. The index i denotes the microstate of the closed system with fixed T , V and N . An ordered set of A indicesuniquely specifies a micro state of the isolated system.

6 In books, canonical ensemble is constructed by taking a system with a fixed valueof V and N and assembling a large number of them in such a way that eachis in thermal contact with its neighbours. Usually these are called metal copiesof the system. The system and its mental copies are then isolated. The isolatedsystem of a large number of mental copies of the system plus the system placedsomewhere in the middle is called a canonical ensemble. Note : all the mentalcopies are identical macroscopically in the sense they all have the same value ofT , V and N . Also other macroscopic properties defined as averages of a stochasticvariable e.g. energy are also the same for all the mental copies. But they mightdiffer one from the other in their microscopic properties

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68 5 Closed system : Canonical ensemble

Let us take an example. Let the micro states of the closed system bedenoted by the indices 1,2,3. There are only three micro states. Let usrepresent the isolated system by a big square and construct nine small squares,each of which represents a member of the ensemble. Each square is attachedwith an index which can be 1, 2 or 3. Thus we have a micro state of theisolated system represented by

3 1 2

2 3 3

2 3 1

Table 5.2. A micro state with occupation number representation (2, 3, 4)

In the above micro state, there are two squares with index 1, three withindex 2 and four with index 3. Let a1 = 2, a2 = 3, a3 = 4 be the occupationnumber representation of the microstate. There are several micro states havingthe same occupation number representation. I haver given below a few of them.

1 3 3

2 1 2

3 2 3

2 3 3

3 1 2

3 2 1

1 2 1

2 3 2

2 3 2

1 1 2

2 2 3

3 3 3

1 2 3

1 2 3

2 3 3

Table 5.3. A few micro states with the same occupation number representation of(2, 3, 4) There are 1260 micro states with the same occupation number representation

Notice that all the micro states given above have the same occupationnumber string 2, 3, 4. How many micro states are there with this occupationnumber string ? We have

Ω(2, 3, 4) =9!

2!3!4!= 1260

I am not going to list all the 1260 of the microstates belonging to the occu-pation number string 2, 3, 4

Let me generalize and say that a string (of occupation numbers) is denotedby the symbol a = a1, a2, · · · , where a1 + a2 + · · · = A. We also have anadditional constraint namely a1E1 + a2E2 + · · · = E .

Let Ω(a) = Ω(a1, a2, · · · ) denote the number of micro states of theisolated system belonging to the string a. For a given string, we can definethe probability for the closed system to be in its micro state indexed by i as

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5.4 Canonical partition function : Method of most probable distribution 69

pi(a) =ai(a)

A (5.2)

Note, the string a = a1, a2 · · · obeys the following constraints.

A∑

i=1

ai(a) = A ∀ strings a (5.3)

A∑

i=1

ai(a)Ei = E ∀ strings a (5.4)

Note that the value of pi varies from one string to another. It is reasonableto obtain the average value of pi over all possible strings a. We have

Pi =∑

a

(ai(a)

A

)P(a) (5.5)

where P(a) is the number of micro states of the isolated system in the string adivided by the total number of micro states of the isolated system : All microstates of an isolated system are equally probable. We have,

P(a) =Ω(a)

∑a Ω(a)

(5.6)

where,

Ω(a) =A!

a1!a2! · · ·(5.7)

In the above we have used the simple notation ai = ai(a) ∀ i = 1, 2, · · · .Let us take a look at Ω(a) for various strings a. For large A the num-

ber Ω(a) will be overwhelmingly large for a particular string, which we shalldenote as a⋆. Thus we can write

Pi =

∑a

(ai(a)A

)Ω(a)

∑a Ω(a)

(5.8)

By taking A → ∞ we can always ensure7 ai(a) → ∞ ∀ i. In this limit,

7 the size of the ensemble is arbitrarily large. It should be large enough so thateven a micro state of smallest probability is present in the ensemble.

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70 5 Closed system : Canonical ensemble

Pi =ai(a

⋆)

AΩ(a⋆)

Ω(a⋆)(5.9)

=ai(a

⋆)

A (5.10)

=a⋆iA (5.11)

Thus the problem reduces to finding that string a⋆ for which Ω(a) is a maxi-mum. Ofcourse there are two constraints on the string. They are

j

aj(a) = A ∀ a (5.12)

j

aj(a)Ej = E ∀ a (5.13)

We need to find the maximum (or minimum) of a function of a many variableunder one or several constraints on the variables. In the above example thereare two constraints. We shall tackle this problem employing the Lagrange’smethod of undetermined multipliers.

To this we turn our attention, below.

5.5 Lagrange’s method of undetermined multipliers

Let me pose the problem through a simple example.A mountain be described by h(x, y) where h is a function of the variable

x and y. h is the elevation of the mountain at a point (x, y) on the plane.I want to find out (x⋆, y⋆) at which h is maximum.We write

dh =∂h

∂xdx+

∂h

∂ydy = 0 (5.14)

If dx and dy are independent then dh = 0 if and only if

∂h

∂x= 0 (5.15)

∂h

∂y= 0 (5.16)

We have two equations and two unknowns. In principle we can solve the abovetwo equations and obtain (x⋆, y⋆) at which h is maximum.

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5.5 Lagrange’s method of undetermined multipliers 71

Now imagine there is a road on the mountain which does not necessarilypass through the peak of the mountain. If you are travelling on the road, thenwhat is the highest point you will pass through ? In the equation

dh =∂h

∂xdx+

∂h

∂ydy = 0 (5.17)

the infinitesimals dx and dy are not independent. You can choose only oneof them independently. The other is detemined by the constraint which saysthat you have to be on the road.

Let the projection of the mountain-road on the plane be described by thecurve

g(x, y) = 0.

This gives us a constraint

∂g

∂xdx+

∂g

∂ydy = 0 (5.18)

From the above we get,

dy = −

(∂g

∂x

)

(∂g

∂y

)dx (5.19)

We then have,

dh =∂h

∂xdx+

∂h

∂ydy = 0 (5.20)

=∂h

∂xdx+

∂h

∂y

(∂g∂x

)

(∂g∂y

)

dx (5.21)

=

∂h∂x

(∂h∂y

)

(∂g∂y

)

∂g

∂x

dx = 0 (5.22)

In the above dx is an arbitrary non-zero infinitesimal. Hence the above equalityholds good if and only if the terms inside the square bracket is zero. We have,

∂h

∂x− λ

∂g

∂x= 0 (5.23)

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72 5 Closed system : Canonical ensemble

where we have set,

λ =

(∂h∂y

)

(∂g∂y

) (5.24)

We have a similar equation involving the partial derivative with respect tothe varible y, which follows from the definition of the Lagrange undeterminedmultipler λ.

Thus we have two independent equations

∂h

∂x− λ

∂g

∂x= 0 (5.25)

∂h

∂y− λ

∂g

∂y= 0 (5.26)

We can solve and and get x⋆ ≡ x⋆(λ) and y⋆ = y⋆(λ). The value of x and y atwhich h(x, y) is maximum under constraint g(x, y) = 0 can be found in termsof the unknown Lagrange multiplier λ.

Of course we can determine the value of λ by substituting the solution(x⋆(λ), y⋆(λ)) in the constraint equation : g(x⋆(λ), y⋆(λ)) = 0.

5.6 Generalisation to a function of N variables

Let f(x1, x2, · · ·xN ) be a function of N variables. The aim is to maximize funder one constraint g(x1, x2, · · · , xN ) = 0.

We start with

df =N∑

i=1

∂f

∂xidxi = 0 (5.27)

for maximum. In the set dx1, dx2, · · · dxµ, · · · dxN, not all are independent.They are related by the constraint

N∑

i=1

∂g

∂xidxi = 0 (5.28)

We pick up one of the variable, say xµ and write

dxµ = −N∑

i=1,i6=µ

∂g∂xi

∂g∂xµ

dxi (5.29)

Substitute the above in the expression for df . We get,

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5.6 Generalisation to a function of N variables 73

N∑

i=1;i6=µ

[∂h

∂xi− λ

∂g

∂xi

]dxi = 0 (5.30)

where

λ =

(∂h∂xµ

)

(∂g∂xµ

) (5.31)

There are only N − 1 values of dxi. We have eliminated dxµ. Instead wehave the undetermined multiplier λ. Since dxi : i = 1, N and i 6= µ are allindependent of each other we can set each term in the sum to zero. Therefore

∂h

∂xi− λ

∂g

∂xi= 0 ∀ i 6= µ (5.32)

From the definition of λ we get

∂h

∂xµ− λ

∂g

∂xµ= 0 (5.33)

Thus we have a set of N equations

∂h

∂xi− λ

∂g

∂xi= 0 ∀ i = 1, N (5.34)

There areN equations andN unknowns. In principle we can solve the equationand get

x⋆i ≡ x⋆i (λ) ∀ i = 1, N,

where the function h is maximum under constraint

g(x1, x2, · · ·xN ) = 0.

The value of the undetermined multiplier λ can be obtained by substitutingthe solution in the constraint equation.

If we have more than one constraints we introduce separate Lagrangemultipliers for each constraint. Let there be m ≤ N constraints. Let theseconstraints be given by

gi(x1, x2, · · ·xN ) = 0 ∀i = 1,m.

We introduce m number of Lagrange multipliers, λi : i = 1,m and write

∂f

∂xi− λ1

∂g1∂xi

− λ2∂g2∂xi

· · · − λm∂gm∂xi

= 0 ∀ i = 1, N

where the m ≤ N .

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74 5 Closed system : Canonical ensemble

5.7 Derivation of Boltzmann weight

Let us return to our problem of finding Pi - the probability that a closedequilibrium system (with macroscopic properties T, V,N) will be found init micro state i with energy Ei. Employing the method of most probabledistribution, we have found that,

Pi =a⋆iA

where A is the number of elements of the canonical ensemble and a⋆j = aj(a⋆).

a⋆ is that string for which Ω(a) is maximum, under two constraints.

Ω(a) =A!

a1!a2! · · ·

The two constraints are∑

j

aj(a) = A

j

aj(a)Ej = E

For convenience we extremize ln Ω(a).

ln Ω(a1, a2, · · · ) = lnA!−∑

j

ln aj ln aj +∑

j

aj

We introduce two Lagrange multipliers α and β and write

∂ ln Ω(a1, a2, · · · )∂ai

− α∂(a1 + a2 + · · · − A)

∂ai− β

∂(a1E1 + a2E2 + · · · − E)∂ai

= 0

Let a⋆j denote the solution of the above equation. We get,

− ln a⋆j − α− 1− βEj = 0 ∀ j = 1, 2, · · ·

The above can be written in a convenient form

a⋆j = γ exp(−βEj)

where γ = exp(−α− 1).We thus have,

Pj = η exp(−βEj)

where η = γ/A.

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5.8 Canonical partition function : Transform of density of states 75

Thus we get the probability that a closed system shall be found in itsmicro state j in terms of the constants η which can be expressed as a functionof the Lagrange multiplier α and β which is the Lagrange multiplier for theconstraint on the total energy of the isolated system.

The task now is to evaluate the constants η and β.The constant η can be evaluated by imposing the normalization condition

:∑

j Pj = 1. The closed system has to be in one of its micro state with unitprobability. Thus we have,

Pj =1

Qexp(−βEj)

Q(β, V,N) =∑

j

exp(−βEj)

We call Q the canonical partition function.What is the nature of the Lagrange multiplier β ? On physical ground we

identify

β =1

kBT.

I shall refer you to

• Donald AMcQuairrie, Statistical Mechanics, Harper and Row (1976)pp.35-44

for this.I shall return to this issue of identifying a physical quantity for the La-

grange multiplier β at a later time when I teach you a bit more of thermody-namics. For the present, assume that the Lagrange multiplier β = 1/[kBT ].

5.8 Canonical partition function : Transform of densityof states

We start with

Q(β, V,N) =∑

i

exp[−βEi(V,N)]

where, β = 1/[kBT ], and the sum runs over all the micro states of a closed sys-

tem at temperature T , volume V and number of particles N . Let Ω(E, V,N)denote the density of (energy) states. In other words

Ω(E, V,N)dE

is the number of micro states having energy between E and E + dE. Thecanonical partition function can be written as an integral over energy,

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76 5 Closed system : Canonical ensemble

Q(β, V,N) =

∫ ∞

0

dE Ω(E, V,N) exp [−βE(V,N)]

The canonical partition function is a ’transform’ of the density of states. The”variable” energy is transformed to the ”variable” temperature.

The density of states is a steeply increasing function of E. The exponentialfunction exp(−βE) decays with E for any fine value of β. The decay is steeperat higher value of β or equivalently at lower temperatures. The product shallbe, in general, sharply peaked at a value of E determined by β.

When β is small (or temperature is large) the integrand would peak at alarge value of E. When β is high (at low temperatures) it would peak at alow value of E.

5.9 Canonical partition function and Helmholtz freeenergy

The thermodynamic energy U of a closed system is the statistical energy Eaveraged over a canonical ensemble. A closed system will invariably be foundwith an energy U = 〈E〉 but for extremely small fluctuations around U ; thesefluctuations are proportional to the inverse of the square root of the numberof molecules.

Hence we can replace the integral over E by the value of the integrand,evaluated atE = 〈E〉 = U . We get,

Q = Ω(E = U) exp(−βU)

lnQ = ln Ω(U)− βU

−kBT lnQ = U − TkB ln Ω

= U − TS

We identify the right hand side of the above as (Helmholtz) free energy :

F (T, V,N) = U(T, V,N)− TS(T, V,N).

Thus we get a relation between (the microscopic description enshrined in)the canonical partition function (of statistical mechanics) and (the macro-scopic description given in terms of) (Helmholtz) free energy (of thermody-namics) :

F (T, V,N) = −kBT lnQ(T, V,N)

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5.10 Canonical ensemble and entropy 77

Statistical mechanics aims to connect the micro world (of say atoms andmolecules) to the macro world (of solids and liquids). In other words it helpsyou calculate the macroscopic properties of a system say a solid, in terms ofthe properties of its microscopic constituents (atoms and molecules) and theirinteractions.

Boltzmann started the game of statistical mechanics by first proposinga micro - macro connection for an isolated system, in the famous formulaengraved on his tomb:

S = kB ln Ω

You will come across several micro-macro connection during this course onstatistical mechanics. The formula

F (T, V,N) = −kBT lnQ(T, V,N),

provides a micro - macro connection for a closed system.

5.10 Canonical ensemble and entropy

Consider the expression

−∑

i

pi ln pi

in the context of canonical ensemble.In an earlier lecture we talked of Ω(a1, a2. · · · ) which gives the number of

micro states of an isolated system having a macroscopic property described bya1, a2, · · · . The isolated system is constructed as follows : Assemble a largenumber of closed systems. Each closed system is in thermal contact with itsneighbouring closed systems. Let η = i = 1, 2, · · · denote the set of all microstates of a closed system. Let A denote the number of closed systems in theassembly. The entire assembly is isolated. We have thus an isolated system.

We describe a micro state of the isolated system by specifying the index iof each closed system in the assembly. The index comes from the set η defiedearlier. The micro states of the isolated system are all equally probable; wegroup them and denote a group by specifying the string a1, a2, · · · where aiis the number of closed systems having the index i. We have

Ω(a1, a2, · · · ) =A!

a1!a2! · · ·

The aim is to find a⋆i : i = 1, 2, · · · that maximizes Ω. For convenience

we maximize ln Ω. Let us consider the limit ai → ∞ ∀ i. Also consider thevariables pi = ai/A. Then

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78 5 Closed system : Canonical ensemble

ln Ω = A lnA−A−∑

i

ai ln ai +∑

i

ai

=∑

i

ai lnA−∑

ai ln ai

= −∑

ai ln(aiA)

= −A∑

i

pi ln pi

ln Ω

A = −∑

i

pi ln pi

The above is the entropy of one of the A number of closed systems constitutingthe the isolated assembly. Thus, −∑i pi ln pi provides a natural formula forthe entropy of a system whose micro states are not equi-probable.

Physicists would prefer to measure entropy in units of Joules per Kelvin.For, that is what they have learnt from Claussius, who defined

dS =q

T,

where dS is the entropy gained by a system when it receives an energy ofq Joules by heat in a reversible process at constant temperature, T Kelvin.Hence we define,

S = −kB∑

i

pi ln pi.

We call this Boltzmann-Gibbs-Shannon entropy. This expression for entropyis natural for a closed system described by a canonical ensemble.

5.11 Free energy to entropy

We start with the thermodynamic relation

F (T, V,N) = U − TS

We recognize that F (T, V,N) is the Legendre transform of the fundamentaleqution U(S, V,N) where we transform the variable S in favour of T definedas

T =

(∂U

∂S

)

V,N

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5.11 Free energy to entropy 79

Read the chapter on Legendre transform in my notes on thermodynamics. Ifrequired, I shall discuss these issues in one of the extra classes. Remind methen.

We re-write the expression for F as,

− S

kB= β(F − U)

We make use of the following,

Q =∑

i

exp(−βEi)

pi =1

Qexp(−βEi)

F = −kBT lnQ

U = 〈E〉 = 1

Q

i

Ei exp(−βEi)

and write,

− S

kB= β

[−kBT lnQ− β

1

Q

i

Ei exp(−βEi)

]

= − lnQ

[1

Q

i

exp(−βEi)

]− β

1

Q

i

Ei exp(−βEi)

=∑

i

[1

Qexp(−βEi)

− lnQ− Eiβ

]

=∑

i

[1

Qexp(−βEi)

ln

exp(−βEi)

Q

]

=∑

i

pi ln pi

from which we get the Boltzmann-Gibbs-Shannon entropy,

S = −kB∑

i

pi ln pi.

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80 5 Closed system : Canonical ensemble

5.12 Energy fluctuations and heat capacity

The average energy of a system is formally given by

〈E〉 =∑

i

Ei pi (5.35)

where pi is the probability of the micro state i and Ei is the energy of thesystem when in micro state i. For a closed system

pi =1

Qexp(−βEi) (5.36)

where Q(T, V,N) is the (canonical) partition function given by

Q =∑

i

exp(−βEi). (5.37)

We have,

〈E〉 =∑

iEi exp(−βEi)∑i exp(−βEi)

(5.38)

=1

Q

i

Ei exp(−βEi) (5.39)

= − 1

Q

∂Q

∂β(5.40)

= −∂ lnQ∂β

(5.41)

We identify 〈E〉 with the internal energy, usually denoted by the symbol U inthermodynamics.

We have,

U = − 1

Q

∂Q

∂β(5.42)

(∂U

∂β

)

V

= − 1

Q

∂2Q

∂β2+

(1

Q

∂Q

∂β

)2

(5.43)

= −[〈E2〉 − 〈E〉2

](5.44)

= −σ2E (5.45)

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5.13 Canonical partition function for an ideal gas 81

Now write

∂U

∂β=∂U

∂T× ∂T

∂β(5.46)

= CV (−kBT 2) (5.47)

We get the relation between the fluctuations of energy of an equilibrium sys-tem and the reversible heat required to raise the temperature of the systemby one degree Kelvin :

σ2E = kBT

2CV . (5.48)

The left hand side of the above equation represents the fluctuations ofenergy when the system is in equilibrium. The right hand side is about howthe system would respond when you heat it8. Note CV is the amount ofreversible heat you have to supply to the system at constant volume to raiseits temperature by one degree Kelvin. The equilibrium fluctuations in energyare related to the linear response; i.e. the response of the system to smallperturbation9.

5.13 Canonical partition function for an ideal gas

I shall derive an expression for the canonical partition function of an ideal gasof N molecules confined to a volume V and at temperature T . I shall do thederivation by an easy method and an easier method.

5.13.1 Easy method:

Formally the partition function is given by,

Q(T, V,N) =V N

N !

1

h3N

∫ +∞

−∞

dp1

∫ +∞

−∞

dp2 · · ·

· · ·∫ +∞

−∞

dp3N exp

[− β

2m

(p21 + p22 · · · + p23N

)]

=V N

N !

1

h3N

[∫ +∞

−∞

dp exp

(−1

2

p2

mkBT

)]3N

Consider the integral

8 Notice that σ2 is expressed in units of Joule2 . The quantity kB T 2 is expressed inunits of Joule-Kelvin. CV is in Joule/Kelvin. Thus kB T 2 CV has units of Joule2.

9 first order perturbation.

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82 5 Closed system : Canonical ensemble

I =

∫ +∞

−∞

dp exp

(−1

2

p2

mkBT

)

since the integrand is an even function, we can write the above as

I = 2

∫ +∞

0

dp exp

(−1

2

p2

mkBT

)

Let

x =p2

2mkBT

Therefore,

dx =p

mkBTdp

dp =

√mkBT

2

1

x1/2dx

The integral can now expressed as

I =√2mkBT

∫ ∞

0

dx x−1/2exp(−x)

=√2mkBT

∫ ∞

0

dx x(1/2)−1exp(−x)

=√2mkBT Γ

(1

2

)

=√2πmkBT since Γ (1/2) =

√π

The canonical partition function is thus given by,

=V N

N !

1

h3N(2πmkBT )

3N/2

5.13.2 Easier method : Transform of density of (energy) states

We first derive an expression for the density of (energy) states, denoted byg(E) from micro canonical ensemble. g(E)dE is the number of micro statesof an isolated system with energy between E and E + dE. Formally, we have

Page 95: Kpn Stat Mech

5.13 Canonical partition function for an ideal gas 83

g(E) =∂Ω

∂E

Ω =V N

N !

1

h3N(2πmE)3N/2

Γ (3N2 + 1)

Therefore the density of (energy) states is given by

g(E) =∂Ω

∂E=V N

N !

1

h3N(2πm)3N/2

Γ (3N2 + 1)

3N

2E

3N2

−1

=V N

N !

1

h3N(2πm)3N/2

Γ (3N2 )E

3N2

−1

where we have made use of the relation

Γ

(3N

2+ 1

)=

(3N

2

(3N

2

)

The partition function is obtained as a ”transform” of the density of stateswhere the variable E transformed to the variable β.

Q(β, V,N) =V N

N !

1

h3N(2πm)3N/2

Γ(3N2

)∫ ∞

0

dE exp(−β E) E3N2

−1

Consider the integral

I =

∫ ∞

0

dE exp(−βE)E3N2

−1

Let,

x = βE then dx = βdE

I =1

β3N/2

∫ ∞

0

dx x3N2

−1 exp(−x)

=Γ (3N2 )

β3N/2

Substituting the above in the expression for the partition function we get,

Q(T, V,N) =V N

N !

1

h3N(2πmkBT )

3N/2

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84 5 Closed system : Canonical ensemble

5.14 Microscopic interpretation of heat and work

The thermodynamic energy U is identified with statistical energy E averagedover a suitable Gibbs ensemble of micro states. We use micro canonical en-semble for isolated system, canonical ensemble for closed system, and grandcanonical ensemble for open system. Let pi : i = 1, 2, denote formally an en-semble. pi is the probability of a micro state of the system under consideration.

For example if the system is isolated, then all micro states are equallyprobable. We simply count the number of micro states of the system; let ussay there are Ω micro states; then pi = 1/Ω.

For a closed system, pi = (1/Q) exp(−βEi).For an open system pi = (1/Q exp(−βEi + βµNi)We can write, in general,

U =∑

i

pi Ei (5.49)

The above equation suggests that the internal energy of a closed system canbe changed by two ways.

1. change Ei : i = 1, 2, · · · keeping pi : i = 1, 2, · · · the same. Thiswe call as work.

2. change pi i = 1, 2, keeping Ei : i = 1, 2, the same. The changesin pi should be done in such way that

∑i pi = 1. This we call as heat.

Thus we have,

dU =∑

i

pidEi +

⋆∑

i

Eidpi

where the super script ⋆ in the second sum should remind us that all dpis arenot independent and that they should add up to zero.

In the first sum we change Ei by dEi ∀ i keeping pi ∀ i unchanged.In the second sum we change pi by dpi ∀ i keeping Ei unchanged for all i

and ensuring∑

i dpi = 0.

5.15 Work in statistical mechanics : W =∑

ipi dEi

Show that d W =∑

i pi dEi

We start with

i

pi dEi =∑

i

pi∂Ei

∂VdV =

∂ (∑

i piEi)

∂VdV

=∂〈E〉∂V

dV =∂U

∂VdV = −PdV = d W (5.50)

Page 97: Kpn Stat Mech

5.16 Heat in statistical mechanics : 85

5.16 Heat in statistical mechanics :

Show that d q =∑⋆

i Ei dpiWe start with

S = −kB∑

i

pi ln pi (5.51)

dS = −kB⋆∑

i

[1 + ln pi] dpi (5.52)

TdS = d q = −kBT⋆∑

i

ln pi dpi

= −kBT⋆∑

i

[−βEi − lnQ] dpi

=

⋆∑

i

Eidpi (5.53)

Problems

5.1. Find the probability of the micro states of the red die under the conditionthat the three die add to 12.

5.2. Maximise A(x1, x2) = x1 × x2 under constraint x1 + x2 = 10.We have

∂A

∂x1− λ

∂g

∂x1= 0

∂A

∂x2− λ

∂g

∂x2= 0

For the given problem we have,

x2 − λ = 0

x1 − λ = 0

The constraint x1 + x2 = 10 gives λ = 5. Thus for x1 = x2 = 5 the functionf is maximum under constraint x1 + x2 = 10. For a fixed perimeter, the areaof a rectangle is maximum only if the sides are same.

5.3. Maximise x3y5 under the constraint x+ y = 8. Answer: x = 3; y = 5.

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86 5 Closed system : Canonical ensemble

5.4. Let (x, y, x) be a point on the surface of a sphere x2 + y2 + z2 = 1. Let(2, 1,−2) be a point. Let D(x, y, z) denote the distance between the point(x, y, z) on the sphere and the point (2, 1,−2). Employing Lagrange’s methodof undetermined multiplier, find the maximum and minimum value of D.Answer: 4 and 2

5.5. Practice Problems employing the method of Lagrange Unde-termined Multipliers

5.6. Start with S = −kB∑

i pi ln pi, where the sum runs over all the microstates of the closed system. Treat S as a function of the variables p1, p2, · · · .There are two constraints on the variables :

∑i pi = 1;

∑i piEi = 〈E〉 = U,

where U is constant. Employing Lagrange method of undetermined multipli-ers, maximize the entropy S. Show that pi : i = 1, 2, · · · that maximize Sare given by

pi =exp(−βEi)∑i exp(−βEi)

where β is the Lagrange multiplier associated with the second constraint.

5.7. A rectangle is inscribed in an ellipse whose major and minor axes areof lengths a and b respectively. The major axis is along the X axis and theminor axis is along the Y axis. The centre of the ellipse is at the origin. Thecentre of the inscribed rectangle is also at origin. Find the length and breathof the rectangle with largest possible area. Employ the method of Lagrangemultiplier. If you have a circle of radius R with centre at origin instead ofellipse what would be the length and breath of the inscribed rectangle withlargest possible area ?

5.8. Consider a right circular cylinder of volume 2π cubic meter. Employingthe method of Lagrangemultiplier find out what height and radius will providethe minimum total surface area for the cylinder ?

5.9. Let the sample space associated with the experiment of throwing a dicebe denoted by

Ω = ω1, ω2, ω3, ω4, ω5, ω6.There are six outcomes. Consider throwing of N fair and independent dice. Astring of ωs denote an outcome of this experiment. The length of the stringis N . The number of outcomes is 6N . Let ni denote the number of times theoutcome ωi occurs. n = (n1, n2, n3, n4, n5, n6) denote a macro state of thesystem. The number of outcomes associated with a given macro state n isgiven by multinomial distribution, see below.

Ω(n1, n2, n3, n4, n5, n6) =N !∏6i=1 ni!

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5.16 Heat in statistical mechanics : 87

Find n for which Ω(n) is maximum. Employ Lagrange method of undeter-mined multiplier. The constraint is

6∑

i=1

ni = N

5.10. Consider energy levels En = n, where n = 0, 1, 2, · · · . The degeneracyof the n-th energy level is n+ 1. Noninteracting distinguishable particles aredistributed over the energy levels. The system is in equilibrium at temperatureT . Calculate

1. the canonical partition function2. the thermodynamic energy defined as U = 〈E〉3. the entropy S.

We have,

− S

kB=∑

i

pi ln pi

In the above the right hand side can be interpreted as 〈ln p〉, where the angularbracket denotes an average over pi : i = 1, 2, · · · . Consider a closed systemfor which pi = exp(−βEi)/Q. Show that −S/kB = −βU − lnQ. From thisdeduce that F = −kBT lnQ.

5.11. See S. B. Cahn, G. D. Mahan, and B. E. Nadgorny, A Guide to PhysicsProblems Part 2: Thermodynamics, Statistical Physics, and Quantum Me-chanics, Plenum ((1997) problem No. 4.45 page 24Consider a system composed of a very large number N of distinguishable par-ticles at rest. The particles do not interact with each other. Each particle hasonly two non-degenerate energy levels: 0 and ǫ > 0. Let E denote the totalenergy of the system. Note that E is a random variable; it varies, in general,from one micro state of the system to another. Let ξ = E/N denote energyper particle.

1. Assume that the system is not necessarily in thermal equilibrium. Whatis the maximum possible value of ξ ?

2. Let the system be in thermal equilibrium at temperature T . The canonicalensemble average of E is the the thermodynamic energy, denoted by U . i.e.U = 〈E〉, where 〈·〉 denote an average over a canonical ensemble of microstates 10. Let ζ = U/N denote the (thermodynamic, equilibrium) energyper particle. Derive an expression for ζ as a function of temperature.

10 Note that it is meaningful to call U as thermodynamic energy only when theaverage of energy is calculated for N → ∞; only in this limit the average energywill be unchanging with time. Fluctuations around the average value, defined asthe standard deviation (i.e. square-root of the variance) of energy divided by themean energy will be of the order of 1/

√N ; this goes to zero only in the limit of

N → ∞.

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88 5 Closed system : Canonical ensemble

3. Find the value of ζ in the limit T → 0 and in the limit T → ∞.4. What is the maximum possible value that ζ can take ?

5.12. Consider a system of N distinguishable non-interacting particles each ofwhich can be in states designated as 1 and 2. Energy of state 1 is ǫ1 = −ǫ andthat of state 2 is ǫ2 = +ǫ. Let the number of particles in states 1 and 2 be N1

andN2 respectively.We haveN = N1+N2 andE = N1ǫ1+N2ǫ2 = (2N2−N)ǫ.

(i) Evaluate canonical partition function Q(T, V,N). Do not forget the de-

generacy factor, Ω which gives the number of ways we can organize N1

particles in state 1 and N2 particles in state 2.(ii) Let q(T, V ) be the single-particle partition function. How Q(T, V,N) and

q(t, V ) are related ?(iii) Calculate and sketch heat capacity CV of the system.

5.13. Consider a system of two non-interacting particles in thermal equilib-rium at temperature T = 1/[kBβ]. Each of the particles can occupy any ofthe three quantum states. The energies of the quantum states are

−ǫ, 0 and + ǫ.

Obtain canonical partition function of the system for particles obeying

(i) classical statistics and are distinguishable(ii) Maxwell-Boltzmann statistics and are ‘indistinguishable ’ .

For each of the above two cases calculate average energy of the system.

5.14. A zipper has N links. Each link can be in any one of the two states

(a) a closed state with zero energy(b) an open state with energy ǫ > 0.

The zipper can be unzipped from top to bottom. A link can be open if andonly if all the links above it are also open. In other words, if we number thelinks as 1, 2, · · · , N from top to bottom, then link k can be open if and onlyif all the links from 1 to k − 1 are also open.

(i) Derive an expression for a canonical partition function(ii) Let n denote the number of open links. Derive an expression for the av-

erage number, 〈n〉 of open links. Employ canonical ensemble for carryingout the averaging process.

(iii) Show that at low temperatures (kBT << 1), the average 〈n〉 is indepen-dent of N .

5.15. The expression for the average energy which corresponds to thermody-namic energy is given by

〈E〉 = U = −∂ ln Q

∂β

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5.16 Heat in statistical mechanics : 89

Show that for an ideal gas the energy is given by

〈E〉 = U = 3NkBT

2

consistent with equi-partition theorem which says that each degree of freedom(each quadratic term in the Hamiltonian) carries an energy of kBT/2.

5.16. Helmholtz free energy and the canonical partition function are related:

F (T, V,N) = −kBT ln Q(T, V,N)

Derive an expression for the free energy of an ideal gas ofN molecules confinedto a volume V at temperature T .

5.17. From thermodynamics we know that

dF = −PdV − SdT + µdN.

Consider the expression for the free energy of an ideal gas, see last problem.Take the partial derivative of F (T, V,N), with respect to V and show that itleads to ideal gas law PV = NkBT .

5.18. R K Pathria, Statistical Mechanics Second Edition Butterworth andHeinemann (1996) p. 87; Problem 3.32The quantum states available to a given physical system are

(i) a group of g1 equally likely states with a common energy ǫ1.(ii) a group of g2 equally likely states with a common energy ǫ2 6= ǫ1.

Show that the entropy of the system is given by,

S = −kB[p1 ln

(p1g1

)+ p2 ln

(p2g2

)](5.54)

where p1 and p2 are, respectively, the probabilities of the system being in astate belonging to group 1 or to group 2. Also p1 + p2 = 1. Assuming that p1and p2 are given by canonical distribution show that,

S = kB

[ln g1 + ln

1 +

(g2g1

)exp(−x)

+ (5.55)

x

1 +

(g1g2

)exp(x)

](5.56)

where x = β(ǫ2 − ǫ1) assumed positive.

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90 5 Closed system : Canonical ensemble

5.19. Adiabatic process and Canonical ensembleA M Glazer and J S Wark, Statistical Mechanics : A survival guide Oxford(2001) sec. 4.4; pages : 52-53

We have earlier discussed the canonical partition function of an ideal gas.We have

Q(T, V,N) =1

h3NV N

N !(2πmkBT )

3N/2(5.57)

From the partition function derive an expression for the entropy. The expres-sion for entropy is called the Sackur-Tetrode equation. Consider a process inwhich there is no change in entropy : ∆S = 0. The volume and temperatureof the gas changes during such an isentropic (iso-entropic process; constantentropy process etc.) process. Show that TV 2/3 is a constant during such aprocess. Combined with the equation of state for an ideal gas : PV is a con-stant at constant N and T , this gives the formula for an adiabatic process :PV 5/3 is a constant.

5.20. Adiabatic Process : A microscopic viewA M Glazer and J S Wark, Statistical Mechanics : A survival guide Oxford

(2001) sec. 4.5; pages : 53-56Change in internal energy,

U =∑

i

piǫi,

can be brought about by two processes :

1. Work :w =

i

pidǫi

change ǫi : i = 1, 2, · · · keeping pi : i = 1, 2, · · · same2. Heat :

q =∑

i

ǫidpi

Change pi : i = 1, 2, · · · keeping ǫi : i = 1, 2, · · · the same. Also∑i pi = 1.

Adiabatic process is one in which energy transacted by heat is zero. Considera particle of mass m in a cube of side L. The energy eigenvalues are given by

ǫn1,n2,n3=

h2

8mL2(n2

1 + n22 + n2

3) ni = 1, 2, 3, · · · for i = 1, 2, 3

We can change L keeping pi : i = 1, 2, · · · the same. From these consid-erations show that for an adiabatic process PV 5/3 is a constant for a monoatomic ideal gas.

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6

Grand canonical ensemble

An open system is one which exchanges energy and matter with its surround-ings. The surroundings act as a heat bath as well as a particle (or material)bath.

• A heat bath transacts energy with the system. The temperature of theheat bath does not change because of the transaction of energy.

• A material (or particle) bath transacts matter (or particles) with the sys-tem. The chemical potential of the material bath does not change becauseof the transaction matter.

The system is in thermal equilibrium with the surroundings1. The systemis also in diffusional equilibrium with the surroundings2.

The system can be described by its temperature3, T , volume, V and chem-ical potential4, µ. Notice that the temperature, T , chemical potential, µ, andvolume V are independent properties of an open system.

Since the system is not isolated, its micro states are not equi-probable.Aim : To calculate the probability of a micro state of the open system.Let us take the open system, its boundary and surroundings and construct

an isolated system. We are interested in constructing an isolated system be-cause, we want to start with the only assumption we make in statistical me-chanics : all the micro states of an isolated system are equally probable. Wehave called this the ergodic hypothesis.

1 equality of temperature signals thermal equilibrium.2 equality of chemical potential ensures diffusional or material equilibrium. Ofcourse, equality of pressure shows mechanical equilibrium.

3

T =

(∂U

∂S

)

V,N

4

µ =

(∂U

∂N

)

S,V

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92 6 Grand canonical ensemble

Let E denote the total energy of the isolated system.

E >> E,

where E is the (average or typical) energy of the open system.

N >> N,

where N is the (average or typical) number of particles in the open system. Vis the total volume of the isolated system. An isolated system is characterizedby E , N , and V and are held at constant values.

Our aim is to describe the open system in terms of its own micro states.Let c be a micro state of the open system. Let E(c) be the energy of the opensystem when it is in its micro state c. Let N(c) be the number of particles inthe open system when it is in its micro state c.

When the open system is in its micro state c the surrounding can be inany one of the innumerable micro states (of the isolated system)5 such thatthe energy of the surroundings is E −E(c) and the number of particles in thesurroundings is N −N(c).

LetΩ(E − E(c),V ,N −N(c)),

denote the number of micro states of the isolated system such that the energyof the surrounding is E − E(c) and number of particles in the surroundingsis N − N(c) and the open system is in its micro state c. The volume of thesurroundings is a constant V − V .

Following Boltzmann we define a statistical entropy as

S(E − E(c), V − V,N −N(c)) = kB ln Ω(E − E(c), V − V, N −N(c)).(6.1)

SinceE(c) << E ,

andN(c) << N ,

we can Taylor-expand S retaining only the first two terms. We have

5 The picture I have is the following. I am visualizing a micro state of the isolatedsystem as consisting of two parts. One part holds the signature of the open system;the other holds the signature of the surroundings. For example a string of positionsand momenta of all the particles in the isolated system defines a micro state. Thisstring consists of two parts. The first part contains the string of positions andmomenta of all the particle in the open system and the second part contains thepositions and momenta of all the particles in the surroundings. Since the systemis open the length system-string is a fluctuating quantity and so is the lengthof bath-string. However the string of the isolated system is of fixed length. I amneglecting those micro states of the isolated system which hold the signature ofthe interaction between the system and the surroundings at the boundaries.

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6 Grand canonical ensemble 93

S

(E − E(c),V − V,N −N(c)

)= S

(E ,V ,N

)(6.2)

−E(c)

(∂S

∂E

)

V,N

∣∣∣∣E,N

(6.3)

−N(c)

(∂S

∂N

)

E,V

∣∣∣∣E,N

(6.4)

From the first law of thermodynamics we have, for a reversible process,

dE = Tds− PdV + µdN (6.5)

dS =1

TdE +

P

TdV − µ

TdN (6.6)

We have

S ≡ S(E, V,N) (6.7)

dS =

(∂S

∂E

)

V,N

dE +

(∂S

∂V

)

E,N

dV +

(∂S

∂N

)

E,V

dN (6.8)

Comparing the coefficients of dE, dV and dN , we get,(∂S

∂E

)

V,N

=1

T(6.9)

(∂S

∂V

)

E,N

=P

T(6.10)

(∂S

∂N

)

E,V

= −µ

T(6.11)

Therefore,

S

(E − E(c),N −N(c)

)= S

(E ,N

)− 1

TE(c) +

µ

TN(c) (6.12)

The probability of the micro state c is given by

P (c) =

Ω

(E − E(c),V ,N −N(c)

)

ΩTotal

(6.13)

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94 6 Grand canonical ensemble

We are able to write the above because of the postulate of ergodicity : Allmicro states of an isolated system are equally probable. We have,

P (c) =1

ΩTotal

exp

[1

kBS

(E − E(c),N −N(c)

)](6.14)

=1

ΩTotal

exp

[S(E ,N )

kB− E(c)

kBT+µN(c)

kBT

](6.15)

= α exp[−β (E(c) − µN(c)] (6.16)

where the constant α can be determined by the normalization condition,

c

P (c) = 1,

where the sum runs over all the micro states of the open system. We have,

P (c) =1

Q exp[−βE(c)− µN(c)] (6.17)

where the grand canonical partition function is given by

Q(T, V, µ) =∑

c

exp [−β E(c)− µN(c)] (6.18)

Let λ = βµ; then we can write the grand canonical partition function as,

Q(T, V, λ) =∑

c

λN(c) exp[−βE(c)] (6.19)

Collect those micro states of a grand canonical ensemble with a fixed valueof N . Then these micro states constitute a canonical ensemble described thecanonical partition function, Q(T, V,N), see also the footnote6. Thus we canwrite the grand canonical partition function as,

Q(T, V, λ) =∑

N

λNQ(T, V,N) (6.20)

Grand canonical ensemble is an extremely useful ensemble. The reasonis that the constraint of constant N required for calculating the canonicalensemble is often mathematically awkward7.

6 We can further collect all those micro states of the canonical ensemble with afixed energy. Then these micro states constitute a micro canonical ensemble.

7 We shall experience this while trying to evaluate the canonical partition functionfor Fermions and Bosons. We will not be able to carry out the sum over occupation

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6.1 Grand canonical partition function and grand potential 95

6.1 Grand canonical partition function and grandpotential

• Entropy or energy is the thermodynamic counter part of the statisticalmechanical micro canonical ensemble. We have

S(E, V,N) = kB ln Ω(E, V,N).

• Helmholtz free energy is the thermodynamic counter part of the statisticalmechanical canonical ensemble. We have

F (T, V,N) = −kBT lnQ(T, V,N).

What is the thermodynamic counter part of the grand canonical ensemble?

Let us call it, the ”grand” potential and denote it by the symbol G. It is afunction of temperature, volume and chemical potential. G(T, V, µ), obtainedfrom U(S, V,N) by Legendre transform of S → T and N → µ. We have,

G(T, V, µ) = U(S, V,N)− TS − µN (6.21)

T =

(∂U

∂S

)

V,N

(6.22)

µ =

(∂U

∂N

)

S,V

(6.23)

Some authors e.g. Donald A McQuairrie, Statistical Mechanics, Harperand Row (1976) would like to identify PV as the thermodynamic counterpart of the grand canonical ensemble. The correspondence is

G = −PV = −kBT lnQ.

First let us establish the above relation between the grand potential G andgrand canonical partition function Q.

numbers because of the constraint that they add up to a constant N . Hence weshall multiply the restricted sum by λN and sum over all possible values of N .This would remove the restriction and we shall express the partition function assum over (micro states) of product (over occupation numbers). We shall interpretλ as fugacity. The chemical potential and fugacity are related : λ = exp(βµ).All these can be viewed as mathematical tricks. The language of grand canonicalensemble gives a physical meaning to these mathematical tricks.

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96 6 Grand canonical ensemble

G = −kBT ln Q

We follow the same method we employed for establishing the connection be-tween Helmholtz free energy and canonical partition function. We have,

Q(T, V, λ) =∑

i

exp(Ei − µNi) (6.24)

where the sum runs over all the microstates i of the open system. Ei is theenergy of the micro state i, and Ni is the number of particles in the systemwhen it is in its micro state i.

We replace the sum over micro states by sum over energy and number ofparticles. Let g(E,N) denote the density of states. We have then,

Q(T, V, µ) =

∫dE

∫dNg(E,N) exp[−β(E − µN)] (6.25)

The contribution to the integrals come overwhelmingly from a single term at〈E〉, and 〈N〉. We then get,

Q(T, V, µ) = g(〈E〉, 〈N〉) exp[−β(〈E〉 − µ〈N〉)] (6.26)

Let us denote 〈E〉 by E and 〈N〉 by N . Taking logarithm on both sides,

lnQ = ln g(E,N)− β(E − µN) (6.27)

−kBT lnQ = −T [kB ln g(E,N)] + E − µN (6.28)

= E − TS − µN (6.29)

= G (6.30)

Alternately, we start with

Q =∑

N ′

λN′

Q(T, V,N ′) (6.31)

In principle, the number of particles in an equilibrium open system is nota fixed number. It fluctuates from one micro state to another. However thefluctuations are very small; it can be shown that the relative fluctuations areinversely proportional to the size of the system. In the above expression for Q,only one term contributes overwhelmingly to the sum over N ′. Let the valueof N ′ for which the λN

Q(T, V,N ′) is maximum be N . Hence the sum overN ′ can be replaced by a single entry with N ′ = N .

Q(T, V, µ) = λNQ(T, V,N) (6.32)

lnQ(T, V, µ) = βµN + lnQ(T, V,N) =µN

kBT+ lnQ(T, V,N) (6.33)

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6.2 Euler formula in the context of homogeneous function 97

kBT lnQ = µN + kBT lnQ(T, V,N) (6.34)

Refer to notes on Canonical ensemble. We have shown that

F (T, V,N) = −kBT lnQ(T, V,N).

Therefore we can write the above equation as,

kBT lnQ = µN − F (T, V,N) (6.35)

−kBT lQ = F − µN (6.36)

= U − TS − µN (6.37)

= G (6.38)

Recall the discussions on Legendre Transform.We start with U ≡ U(S, V,N).Transform S in favour of the slope T (partial derivative of U with respect toS). We get the ”intercept” F (T, V,N) as U − TS. Let us do one more trans-form : N → µ. The partial derivative of U with respect to N is the chemicalpotential µ. We get the intercept G(T, V, µ) - the grand potential. We haveG(T, V, µ) = U − TS − µN . Thus we have,

G(T, V, µ) = −kBT lnQ(T, V, µ) (6.39)

Our next task is to show that G(T, V, µ) = −PV . To this end, let me tellyou of a beautiful formula proposed by Euler, in the context of homogeneousfunction.

6.2 Euler formula in the context of homogeneousfunction

U is a homogeneous function of S, V and N . U is an extensive property; so areS, V and N . In the words of mathematicians, U is a first order homogeneousfunction of S, V , and N . This means,

U(λS, λV, λN) = λU(S, V,N) (6.40)

where λ is a constant. Euler’s trick consists of differentiating both sides of theabove equation with respect to λ. We get,

∂U

∂(λS)

∂λS

∂λ+

∂U

∂(λV )

∂λV

∂λ+

∂U

∂(λN)

∂λN

∂λ= U(S, V,N) (6.41)

S∂U

∂(λS)+ V

∂U

∂(λV )+N

∂U

∂(λN)= U(S, V,N) (6.42)

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98 6 Grand canonical ensemble

The above is true for any value of λ. In particular it is true for λ = 1.Substitute in the above λ = 1 and get,

S∂U

∂S+ V

∂U

∂V+N

∂U

∂N= U(S, V,N) (6.43)

TS − PV + µN = U (6.44)

6.3 PV = kBT lnQWe proceed as follows. From Eq. (??) we have

−PV = U − TS − µN (6.45)

The RHS of the above is grand potential. Hence,

−PV = G(T, V, µ) (6.46)

= −kBT lnQ(T, V, µ) (6.47)

PV = kBT lnQ(T, V, µ) (6.48)

6.4 Gibbs-Duhem relation : dµ = −sdT + vdP

Now that we are on the Euler’s formula, let me digress a little bit and see ifthe equations we have derived, can be used to establish a relation amongstthe intensive properties T , P and µ of the system.

Derivation from U(S, V,N)

To this end we proceed as follows.

U = TS − PV + µN (6.49)

dU = TdS − PdV + µdN + SdT − V dP +Ndµ (6.50)

From the first law of thermodynamics, we have dU = TdS − PdV + µdN .Hence,

Ndµ+ SdT − V dP = 0 (6.51)

dµ = − S

NdT +

V

NdP (6.52)

= −sdT + vdP (6.53)

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6.5 Grand canonical ensemble : Number fluctuations 99

where s is the specific entropy - entropy per particle and v is specific volume- volume per particle.

6.5 Grand canonical ensemble : Number fluctuations

In an open system, the energy E and the number of molecules N are randomvariables. Energy fluctuates when the system goes from micro state to another.The number of molecules fluctuates from one micro state to the other. Let usnow derive an expression for the variance of N :

σ2 = 〈N2〉 − 〈N〉2.

To this end, we start with

Q(T, V, µ) =∑

c

exp

[− βE(c)− µN(c)

](6.54)

In the above

c denotes a micro state of the open systemE(c) denotes the energy of the open system when in micro state cN(c) denotes the number of particles of the open when in micro state c

Let us now take the partial derivative of all the terms in the above equation,with respect to the variable µ, keeping the temperature and volume constant.We have,

(∂Q∂µ

)

T,V

=∑

c

βN(c) exp

[− βE(c)− µN(c)

]= β〈N〉Q(T, V, µ)(6.55)

(∂2Q∂µ2

)

T,V

= β

[〈N〉

(∂Q∂µ

)

T,V

+Q(∂〈N〉∂µ

)

T,V

](6.56)

The left hand side of the above equation equals 8 β2〈N2〉Q.Substituting this in the above, we get,

8

(∂Q∂µ

)

T,V

=∑

c

βN(c) exp

[− βE(c)− µN(c)

]

(∂2Q∂µ2

)

T,V

=∑

c

β2[N(c)]2 exp

[− βE(c)− µN(c)

]

= β2〈N2〉Q.

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100 6 Grand canonical ensemble

β2〈N2〉Q = β2〈N〉2Q+ βQ(∂〈N〉∂µ

)

T,V

(6.57)

σ2 = 〈N2〉 − 〈N〉2 = kBT

(∂〈N〉∂µ

)

T,V

(6.58)

In the above, we have an example of fluctuation-dissipation theorem. We havethe Gibbs free energy. G(T, P,N) = µN .

6.6 Number fluctuations and isothermal compressibility

Let us express the number fluctuations in terms of experimentally measurableproperties 9 of the open system. Let us define

v =V

〈N〉 .

It is called specific volume. It is the volume per particle. We have,

(∂〈N〉∂µ

)

T,V

=

(∂(V/v)

∂µ

)

T,V

(6.59)

= − V

v2

(∂v

∂µ

)

T,V

(6.60)

= −〈N〉2V

(∂v

∂µ

)

T,V

(6.61)

In the above we can express,

9 We shall show that,

(∂〈N〉∂µ

)

T,V

=〈N〉2V

kT

where kT denotes isothermal compressibility - an experimentally measurable prop-erty. Isothermal compressibility is defined as

kT = − 1

V

(∂V

∂P

)

T

Page 113: Kpn Stat Mech

6.6 Number fluctuations and isothermal compressibility 101

(∂v

∂µ

)

T,V

=

(∂v

∂P

)

T,V

(∂P

∂µ

)

T,V

(6.62)

Employing Gibbs-Duhem relation, we find 10,

(∂P

∂µ

)

T

=〈N〉V

(6.64)

Therefore,

(∂v

∂µ

)

T,V

=〈N〉V

(∂v

∂P

)

T,V

=1

v

(∂v

∂P

)

T,V

= −kT (6.65)

Finally we get,

(∂〈N〉∂µ

)

T,V

=〈N〉2V

kT (6.66)

σ2 = kBT

(∂〈N〉∂µ

)

T,V

(6.67)

= kBT〈N〉2V

kT (6.68)

σ2

〈N〉2 =kBT

VkT (6.69)

Thus the fluctuations in the number of molecules of an open system is directlyproportional to the isothermal compressibility. Hence we expect isothermalcompressibility to be positive 11.

The number fluctuations are small in the thermodynamic limit; they areof the order of inverse of the square-root of the number of particles in thesystem.

Thus equilibrium fluctuations are related to an appropriate susceptibilitywhich measures the response of the system to a small external perturba-tion. When heated, the system responds by raising its temperature. Energy

10 Gibbs - Duhem relation reads as 〈N〉dµ = V dP −SdT. At constant temperature,we have,〈N〉dµ = V dP,

(∂P

∂µ

)

T

=〈N〉V

(6.63)

11 The relative fluctuations of energy in a canonical ensemble is proportional to heatcapacity at constant volume. Hence we expect heat capacity to be positive .

Page 114: Kpn Stat Mech

102 6 Grand canonical ensemble

absorbed by heat divided by the increase in temperature is called the heatcapacity. The volume does not change during this process. We saw that theequilibrium energy fluctuations are proportional heat capacity.

Thus, the susceptibility is heat capacity for energy fluctuations; it isisothermal compressibility for the fluctuations of number density. These arespecial cases of a more general principle called Fluctuation-Dissipation theo-rem enunciated by Albert Einstein in the context of Brownian motion. If timepermits, I shall speak on Brownian motion in one of the extra classes.

However, close to first order phase transition, the isothermal compressibil-ity diverges. The fluctuations in the number of particles are large; they are ofthe order of the system size. The pressure-volume phase diagram has a flatregion very near first order phase transition temperature, see discussions onvan der Waal gas and Maxwell’s construction.

6.7 Alternate derivation of the relation :σ2

N/〈N〉2 = kB T kT/V

In my last lecture I had derived a relation is between the number fluctuationand isothermal compressibility. I had followed closely R K Pathria, StatisticalMechanics, Second Edition, Butterworth and Henemann (1996), I told youthat I was not very excited about Pathria’s derivation and that I would tryand produce a better one; or at least an alternate derivation. Here it is.

Start with a fluctuation-dissipation relation derived in the last lecture12,

σ2 = kBT

(∂〈N〉∂µ

)

V,T

(6.70)

We consider the reciprocal,

(∂µ

∂〈N〉

)

V,T

(6.71)

We observe that µ is a function13 of T and P . We are keeping T a constant.Hence µ can change only when P changes.

Gibbs-Duhem relation tells us,

〈N〉dµ = V dP − SdT. (6.72)

When T is a constant, we have dT = 0. This gives us

12 relating number fluctuations to response to small changes in the chemical poten-tial.

13 Gibbs and Duhem told us that the three intensive properties T , P , and µ are notall independent. Only two of them are independent. The third is automaticallyfixed by the other two.

Page 115: Kpn Stat Mech

6.7 Alternate derivation of the relation : σ2

N/〈N〉2 = kB T kT /V 103

dµ =V

〈N〉dP (6.73)

(∂µ

∂〈N〉

)

T,V

=V

〈N〉

(∂P

∂〈N〉

)

T,V

(6.74)

(∂µ

∂P

)

T,V

=V

〈N〉 (6.75)

Let us now define

ρ =〈N〉V

,

which denotes the particle density : number of particles per unit volume. Wehave,

(∂P

∂〈N〉

)

V,T

=

(∂P

∂(ρV )

)

V,T

(6.76)

=1

V

(∂P

∂ρ

)

V,T

(6.77)

The density can be changed either by changing 〈N〉 and/or V . Here we changeρ infinitesimally, keeping V constant. As a result the pressure changes in-finitesimally. Let me repeat : both these changes happen at constant V andT .

As far as P is concerned, it does not care whether ρ has changed by changeof 〈N〉 or of V . Hence it is legitimate to write

(∂P

∂ρ

)

V,T

=

(∂P

∂(〈N〉/V )

)

〈N〉,T

(6.78)

= − V 2

〈N〉

(∂P

∂V

)

〈N〉,T

(6.79)

Thus we get,(

∂µ

∂〈N〉

)

V,T

=V

〈N〉

(∂P

∂〈N〉

)

V,T

(6.80)

=1

〈N〉

(∂P

∂ρ

)

V,T

(6.81)

= − V 2

〈N〉2(∂P

∂V

)

〈N〉,T

(6.82)

Page 116: Kpn Stat Mech

104 6 Grand canonical ensemble

Take a reciprocal of the above and get,(∂〈N〉∂µ

)

V,T

= −〈N〉2V 2

(∂V

∂P

)

〈N〉,T

(6.83)

Then we get,

σ2N = kBT

[−〈N〉2

V 2

](∂V

∂P

)

T,〈N〉

(6.84)

= kBT〈N〉2V

kT (6.85)

Problems

6.1. Start with Helmholtz free energy : F ≡ F (T, V,N). F is an extensivethermodynamic variable. F is a first order homogeneous function of the ex-tensive thermodynamic variables V,N . Note that F also depends on the inten-sive variable T . Therefore we can write, F (T, λV λN) = λF (T, V,N). EmployEuler’s formula and derive Gibbs-Duhem relation.

6.2. Start with enthalpy: H ≡ H(S, P,N). H is an extensive thermodynamicvariable. H is a first order homogeneous function of the extensive thermody-namic variables S,N . Note that H also depends on the intensive variable P .Therefore we can write H(λS, P, λN) = λH(S, P,N). Employ Euler’s formulaand show that H = TS + µN . Also derive Gibbs-Duhem relation.

6.3. Start with Gibbs free energy G ≡ G(T, P,N). Employ Euler’s formulaand show that G = µN. Derive Gibbs-Duhem relation.

6.4. Let f be a second order homogeneous function of the variable x1, x2, · · ·xN . By this we mean

f(λx1, λx2, · · ·λxN ) = λ2 f(x1, x2, · · ·xN ).

Show that

2 f(x1, x2. · · ·xN ) =N∑

i=1

xi∂f

∂xi

6.5. Let f be an n-th order homogeneous function of the variables x1, x2, · · ·xN .:

f(λx1, λx2, · · · , λxN ) = λnf(x1, x2, · · ·xN ).

Derive the Euler relation

nf(x1, x2, · · ·xN ) =

N∑

i=1

xi∂f

∂xi

Page 117: Kpn Stat Mech

6.7 Alternate derivation of the relation : σ2

N/〈N〉2 = kB T kT /V 105

6.6. For a grand canonical ensemble show that,

S = kB lnQ+ kBT

(∂ lnQ∂T

)

V,µ

N = kBT

(∂ lnQ∂µ

)

V,T

P = kBT

(∂lnQ∂V

)

µ,T

= kBT1

VlnQ

6.7. Start with the Boltzmann-Gibbs-Shannon entropy

S(p1, p2, · · · ) = −kB∑

i

pi ln pi,

where pj is the probability of the micro state i of the open system.Show that the pj : j = 1, 2, · · · for which the entropy is maximum

under the three constraints,

i

pi = 1;

i

piǫi = 〈E〉 = U ;

i

piNi = 〈N〉

are given by,

pi =1

Q exp[−βEi − γNi)]

where the β and γ are the Lagrange multipliers for the second and thirdconstraints respectively. The first constraint gives rise to the grand canonicalpartition function Q.

The energy of the system when in micro state i is Ei; the number ofparticles in the system when in micro state i is Ni.

6.8. Start with

Q(β, V, µ) =∑

c

exp [−β E(c)− µN(c)]

where the sum runs over all the micro states of an open system. The averagenumber of particles is given by

〈N〉 = 1

Q∑

c

N(c) exp [−β E(c)− µN(c)]

Page 118: Kpn Stat Mech

106 6 Grand canonical ensemble

Show that the variance of N is given by

σ2 = 〈N2〉 − 〈N〉2 = kBT

(∂〈N〉∂µ

)

T,V

6.9. Donald A McQuarrie, Statistical Mechanics, Harper and Row (1976) page: 65; Problem : 3-4.

Show that the pressure of an open system, in a canonical ensemble, is givenby

P = kBT

(∂ lnQ∂V

)

µ,V

(6.86)

Use Euler’s theorem for first order homogeneous functions and show that

P =kBT

VlnQ(T, V, µ).

6.10. R K Pathria, Statistical Mechanics, Second Edition, Butterworth andHenemann (1996) page : 102; Problem : 4.4

The probability that an open system, has N particles is given by,

P (N) =exp(βµN)Q(T, V,N)

Q(T, V, µ)

Verify the above statement. Consider a system of classical, indistinguishable,ideal gas. Show that N has a Poisson distribution. Calculate the variance ofN employing the general formula

σ2 = kBT

(∂〈N〉∂µ

)

T,V

Show that the result agrees with the one obtained from the Poisson distribu-tion.

6.11. Donald A McQuarrie, Statistical Mechanics, Harper and Row (1976)page : 67; Problem : 3-22.

Show that the fluctuations of energy in a grand canonical ensemble is

σ2E = kBT

2CV +

(∂〈E〉∂〈N〉

)

T,V

σ2N

Page 119: Kpn Stat Mech

6.7 Alternate derivation of the relation : σ2

N/〈N〉2 = kB T kT /V 107

6.12. Donald A McQuarrie, Statistical Mechanics, Harper and Row (1976)page : 67; Problem : 3-26.

Show that(

∂µ

∂〈N〉

)

V,T

= − V 2

〈N〉2(∂P

∂V

)

〈N〉,T

Hint : Use Gibbs-Duhem relation :

〈N〉dµ− V dP + SdT = 0.

Use the chain rule:(∂µ

∂P

)

T,V

=

(∂µ

∂〈N〉

)

T,V

(∂〈N〉∂P

)

T,V

Use the cyclic product rule of partial derivatives with variables 〈N〉, P, V .

(∂〈N〉∂P

)

T,V

(∂P

∂V

)

T,〈N〉

(∂V

∂〈N〉

)

T,P

= −1

Page 120: Kpn Stat Mech
Page 121: Kpn Stat Mech

7

Quantum Statistics

7.1 Occupation number representation of a micro state

Let1, 2, · · ·

denote the single-particle quantum states. Let

ǫi : i = 1, 2, · · ·

denote the corresponding energies. Notice that there can be several quantumstates with the same energy.

We haveN non-interacting particles occupying the single particle quantumstates. A micro state of the macroscopic system of N particles in a volume Vis uniquely specified by a string of numbers

n1, n2, · · · ,

where ni is the number of particles in the single- particle quantum state i.Thus, a string of occupation numbers uniquely describes a micro state of

the quantum system. Note that such a string should obey the constraint :

n1 + n2 + · · · = N.

The energy of a micro state (specified by a string of occupation numbers)is

n1ǫ1 + n2ǫ2 + · · · .The canonical partition function can now be written as

Q(T, V,N) =

⋆∑

n1,n2,···

exp [−β(n1ǫ1 + n2ǫ2 + · · · )] (7.1)

where the sum runs over all possible strings of occupation numbers (i.e. microstates) obeying the constraint

Page 122: Kpn Stat Mech

110 7 Quantum Statistics

i

ni = N.

To remind us of this constraint I have put a star over the summation sign.

7.2 Open system and grand canonical partition function

The presence of the constraint renders evaluation of the sum a difficult task.A way out is to transform the variable N in favour of λ. Do not confuse thisλ here with the one I had mentioned while discussing quantum or thermal orde Broglie wave length. λ here is the fugacity and is related to the chemicalpotential µ; the relation is λ = exp(βµ). The transform defines the grandcanonical partition function 1 see below.

Q(β, V, λ) =∞∑

N=0

λN Q(β, V,N) (7.2)

This provides a description of an open system with its natural variable :temperature, volume and chemical potential. The system can exchange energyas well as matter with the environment. Thus energy E, and the number ofparticles N , of the system would be fluctuating around the means 〈E〉 and〈N〉 respectively, where the angular bracket denotes averaging over a grandcanonical ensemble of the open system.

Thus we have,

Q(T, V, µ) =

∞∑

N=0

λN⋆∑

n1,n2,···

exp [−β(n1ǫ1 + n2ǫ2 + · · · )] (7.3)

=∞∑

N=0

⋆∑

n1,n2,···

λn1+n2+··· exp [−β(n1ǫ1 + n2ǫ2 + · · · )] (7.4)

=

∞∑

N=0

⋆∑

n1,n2,···

exp [−β(ǫ1 − µ)n1 − β(ǫ2 − µ)n2 + · · · )] (7.5)

1 we have already seen that a partition function is a transform. We start withdensity of states Ω(E) and transform the variable E in favour of β = 1/[kBT ] toget canonical partition function:

Q(β, V,N) =

∫∞

0

dEΩ(E,V,N) exp(−βE(V,N)).

I shall discuss formally the grand canonical ensemble in full glory in some laterlectures. For the present it is sufficient to consider the grand canonical partitionfunction as a transform of the canonical partition function with the variable Ntransformed to fugacity λ or chemical potential µ = kBT ln(λ).

Page 123: Kpn Stat Mech

7.2 Open system and grand canonical partition function 111

=∞∑

N=0

⋆∑

n1,n2,···

[exp −β (ǫ1 − µ)]n1 ×

[exp −β (ǫ2 − µ)]n2 × · · · (7.6)

=∞∑

N=0

⋆∑

n1,n2,···

xn1

1 × xn2

2 × · · · (7.7)

where we have introduced in the last line a short hand notation

xi = λ exp(−βǫi) = exp[−β(ǫi − µ)]. (7.8)

We have a restricted sum over strings of occupation numbers. The restric-tion is that the occupation numbers constituting a string should add to N .We then take a sum over N from 0 to ∞. which removes the restriction. Toappreciate this, see Donald A McQuairrie, Statistical Mechanics, Harper andRow (1976)p.77;Problem:4-6. I have worked out this problem below.

Consider first the sum over restricted sums :

I1 =∞∑

N=0

⋆∑

n1,n2

xn1

1 xn2

2 (7.9)

where the star over the summation sign reminds us of the restriction n1+n2 =N . Also let us assume ni can be 0, 1 or 2, for i = 1, 2. Thus we have We canwrite down I1 as,

I1 = 1 + x1 + x2 + x21 + x22 + x1x2 + x21x2 + x1x22 + x21x

22 (7.10)

Now consider the product over the unrestricted sums.

I2 =

2∏

i=1

2∑

ni=0

xni

i (7.11)

=

2∏

i=1

(1 + xi + x2i ) (7.12)

= (1 + x1 + x21)× (1 + x2 + x22) (7.13)

= 1 + x2 + x22 + x1 + x1x2 + x1x22 + x21 + x21x2 + x21x

22 (7.14)

= I1 (7.15)

We can now write the grand canonical partition function of N quantum par-ticles occupying single-particle quantum states determined by volume V .

Q(T, V, µ) =∏

i

all∑

ni=0

xni

i =∏

i

all∑

ni=0

[λ exp(−βǫi)]ni (7.16)

=∏

i

all∑

ni=0

[exp −β (ǫi − µ)

]ni(7.17)

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112 7 Quantum Statistics

N n1 n2 xn1

1xn2

2

0 0 0 1

1 1 0 x1

0 1 x2

2 2 0 x2

1

0 2 x2

2

1 1 x1x2

3 3 0 −2 1 x2

1x2

1 2 x1x2

2

0 3 −

4 − − −2 2 x2

1x2

2

− − −

5 − − −

Table 7.1. Terms in the restricted sum

7.3 Fermi-Dirac Statistics

We have ni = 0, 1 ∀ i. This is a consequence of Pauli exclusion principle. Nosingle quantum state can have more than one Fermion.

QFD(T, V, µ) =∏

i

[1 + exp[−β(ǫi − µ)

](7.18)

7.4 Bose-Einstein Statistics

We have ni = 0, 1, · · ·∞ ∀ i. Let exp[−β(ǫi − µ)] < 1 ∀ i. This condition ismet if ǫi > µ ∀ i. Then we have

Page 125: Kpn Stat Mech

7.6 Maxwell-Boltzmann Statistics 113

QBE =∏

i

1

1− exp[−β(ǫi − µ)](ǫi > µ ∀ i) (7.19)

7.5 Classical Distinguishable Particles

If the particles are classical and distinguishable, then the string n1, n2, · · · will not uniquely specify a micro state. The string will have a set of classicalmicro states associated with it. The number of micro states associated withthe string is given by,

Ω(n1, n2, · · · ) =N !

n1!n2! · · ·(7.20)

We then have

QCS(T, V, µ) =

∞∑

N=0

λN⋆∑

n1,n2m···

N !

n1!n2! · · ·[exp(−βǫ1)]n1 ×

[exp(−βǫ2)]n2 × · · ·

=∞∑

N=0

λN

[∑

i

exp(−βǫi)]N

=∞∑

N=0

λN [Q1(T, V )]N (7.21)

where Q1(T, V ) is the single-particle partition function.We have already seen that the partition function for classical distinguish-

able non-interacting point particles leads to an entropy which is not exten-sive. This is called Gibbs’ paradox. To take care of this Boltzmann askedus to divide the number of micro states by N !, saying that the particles areindistinguishable.

The non-extensitivity of entropy indicates a deep flaw in classical formula-tion of statistical mechanics. Classical particles do not simply exist in nature.We have only quantum particles : Bosons and Fermions. It is quantum me-chanics that would set the things right eventually. But quantum mechanicshad not arrived yet, in the times of Boltzmann.

Boltzmann corrected the (non-extensive entropy) flaw by introducing astrange notion of indistinguishability of particles. Genius he was, Boltzmannwas almost on the mark. Division by N ! arises naturally in quantum mechan-ical formulation, because of the symmetry of the wave function

In the derivation of grand canonical partition function, we shall divide theclassical degeneracy by N !, as recommended by Boltzmann, see below, andcall the resulting statistics as Maxwell-Boltzmann statistics.

7.6 Maxwell-Boltzmann Statistics

Formally we have,

Page 126: Kpn Stat Mech

114 7 Quantum Statistics

QMB =∞∑

N=0

λN⋆∑

n1,n2,···

1

n1!n2! · · ·exp[−β(n1ǫ1 + n2ǫ2 + · · · )]

=

∞∑

N=0

⋆∑

n1,n2,···

[λ exp(−βǫ1)]n1

n1!× [λ exp(−βǫ2)]n2

n2!× · · ·

=

∞∑

N=0

⋆∑

n1,n2,···

[exp(−β(ǫ1 − µ)]n1

n1!× [exp(−β(ǫ2 − µ)]n2

n2!× · · ·

=

∞∑

N=0

⋆∑

n1,n2,···

xn1

1

n1!

xn2

2

n2!· · ·

=

(∞∑

n1=0

xn1

1

n1!

)(∞∑

n2=0

xn2

2

n2!

)· · ·

= exp(x1) exp(x2) · · ·

=∏

i

exp(xi)

=∏

i

exp[λ exp(−βǫi)] =∏

i

exp[exp−β(ǫi − µ)] (7.22)

We can also express the grand canonical partition function for classical indis-tinguishable ideal gas as,

QMB = exp(x1) exp(x2) · · ·

= exp

[∑

i

xi

]

= exp

[∑

i

exp[−β(ǫi − µ)]

]

= exp[λ∑

i

exp(−βǫi)]

= exp[λQ1(T, V )]

7.6.1 QMB(T, V,N) → QMB(T, V, µ)

We could have obtained the above in a simple manner, by recognizing that

QMB(T, V,N) =[QMB(T, V,N = 1)]N

N !

Page 127: Kpn Stat Mech

7.7 Grand canonical partition function, grand potential,and thermodynamic properties of an open system 115

Then we have,

=

∞∑

N=0

λN[QMB(T, V,N = 1)]N

N !(7.23)

= exp[λQMB(T, V,N = 1)] (7.24)

7.6.2 QMB(T, V, µ) → QMB(T, V,N)

We start with QMB(T, V, µ) = exp[λQMB(T, V,N = 1)]. Taylor expandingthe exponential,

QMB(T, V, µ) =∞∑

N=0

λNQN

MB(T, V,N = 1)

N !

The coefficient of λN is the canonical partition function2 and hence is givenby

QMB(T, V,N) =QMB(T, V,N = 1)N

N !

7.7 Grand canonical partition function, grand potential,and thermodynamic properties of an open system

We can write the grand canonical partition function for the MB,BE, and FDstatistics as

Q =∏

i

exp[exp−β(ǫi − µ)] Maxwell− Boltzmann

1

1− exp [−β(ǫi − µ)]BoseEinstein

1 + exp −β(ǫi − µ) Fermi−Dirac

(7.25)

The connection to thermodynamics is established by the expression for grandpotential denoted by the symbol G(T, V, µ). We have,

G(T, V, µ) = −kBT lnQ(T, V, µ) (7.26)

2 Recall that

QMB(T, V, λ) =∞∑

N=0

λNQMB(T, V,N)

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116 7 Quantum Statistics

Recall from thermodynamics that the G is obtained as a Legendre trans-form of U(S, V,N) : S → T ; N → µ; and U → G.

G(T, V, µ) = U − TS − µN (7.27)

T =

(∂U

∂S

)

V,N

(7.28)

µ =

(∂U

∂N

)

S,V

(7.29)

From the above, we get,

dG = −PdV − SdT −Ndµ (7.30)

It follows,

P (T, V, µ) = −(∂G∂V

)

T,µ

(7.31)

S(T, V, µ) = −(∂G∂T

)

V,µ

(7.32)

N(T, V, µ) = −(∂G∂µ

)

T,µ

(7.33)

If we have an open system of particles obeying Maxwell-Boltzmann, Bose-Einstein, or Fermi-Dirac statistics at temperature T and chemical potentialµ, in a volume V , then the above forrmulae help us calculate the pressure,entropy and average number of particles in the system. In fact, in the lastsection on grand ganonical ensemble, we have derived formal expressions forthe mean and fluctuations of the number of particles in an open system; wehave related the fluctuations to isothermal compressibility - an experimentallymeasurable property.

The grand potential for the three statistics is given by,

G(T, V, µ) = −kBT lnQ (7.34)

=

−kBT∑

i exp[−β(ǫi − µ)] Maxwell− Boltzmann

kBT∑

i ln [1− exp −β(ǫi − µ)] Bose− Einstein

−kBT∑

i ln [1 + exp −β(ǫi − µ)] Fermi−Dirac

(7.35)

Page 129: Kpn Stat Mech

7.8 Expressions for 〈N〉 117

7.8 Expressions for 〈N〉

7.8.1 Maxwell-Boltzmann Statistics

Q(T, V, µ) =∏

i

exp [exp −β (ǫi − µ) (7.36)

G(T, V, µ) = −kBT lnQ= −kBT

i

exp [−β(ǫi − µ)] (7.37)

(∂G∂µ

)

T,V

= −∑

i

exp[−β(ǫi − µ) (7.38)

〈N〉 =∑

i

exp[−β(ǫi − µ)]

= λ∑

i

exp(−βǫi)

= λQ1(T, V, µ) (7.39)

In the above Q1 is the single-particle canonical partition function. The aboveresult on 〈N〉 is consistent with formal result from grand canonical formalism,see below.

Q(T, V, µ) =

∞∑

N=0

Q(T, V,N) (7.40)

=

∞∑

N=0

λN ,QN

1

N !(7.41)

=

∞∑

N=0

exp(βµN)QN

1

N !(7.42)

= λQ1 (7.43)

G(T, V, µ) = −kBT exp(βµ)Q1 (7.44)

〈N〉 = −(∂G∂µ

)

T,V

= λQ1(T, V ) (7.45)

7.8.2 Bose-Einstein Statistics

Q(T, V, µ) =∏

i

1

1− exp[−β(ǫi − µ)](7.46)

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118 7 Quantum Statistics

lnQ = −∑

i

ln [1− exp −β(ǫi − µ)] (7.47)

〈N〉 = −(∂G∂µ

)

T,V

=∑

i

exp[−β(ǫi − µ)]

1− exp[−β(ǫi − µ)](7.48)

7.8.3 Fermi-Dirac Statistics

Q =∏

i

1 + exp[−β(ǫi − µ)] (7.49)

G = −kBT lnQ = −kBT∑

i

ln [1 + exp−β(ǫi − µ)] (7.50)

〈N〉 = −(∂G∂µ

)

T,V

=∑

i

exp[−β(ǫi − µ)]

1 + exp[−β(ǫi − µ)](7.51)

7.8.4 Study of a system with fixed N employing grand canonicalformalism

In my next lecture, I shall derive an expression for the mean of occupationnumber, nk, of a single particle quantum state. Then we can express 〈N〉 =∑

k〈nk〉. These results on 〈N〉 shall be the same as the one derived in the lastthree subsections.

Grand canonical formalism is not the best suited for studying ideal gas- classical or quantum. Recall, we introduced an adhoc-indistinquishability-factor of N !, suggested by Boltzmann, to restore the extensivity of entropywhile studying closed system. In an open system the number of particlesfluctuate and the issue of non-extensitivity becomes more awkward.

We shall adopt the following strategy. We shall employ grand canonicalformalism, but with a fixed N . The price we have to pay is that in suchan approach the chemical potential is no longer an independent property. Itbecomes a function of temperature. Let me explain.

Let us say we keep µ fixed and change the temperature3. The statisticsof the number of particles, in particular the mean number of particles in thesystem changes when temperature changes. To restore the value of 〈N〉 tothe fixed value N , we change the chemical potential. In what follows, weshall investigate the behaviour of particles under Maxwell-Boltzmann, Bose-Einstein, and Fermi-Dirac statistics employing grand canonical formalism but

3 This is permitted in grand canonical formalism since T and µ are independentproperties of the open system. T is determined by the heat bath and µ is deter-mined by particle bath.

Page 131: Kpn Stat Mech

7.9 All the three statistics are the same at high temperature and/or low densities 119

with a fixed N ; This implies µ is not any more an independent property; it isa function of T .

I must say there is nothing unphysical about this strategy. We are studyinga physical system enclosed by a non-permeable wall - a wall that does notpermit particle exchange. The chemical potential µ, is a well defined propertyof the system. It is just that µ is not any more under our control4. The systemautomatically selects the value of µ depending on the temperature.

7.9 All the three statistics are the same at hightemperature and/or low densities

I shall show below that three statistics are the same at high temperaturesand / or low densities by an easy method, an easier method and perhaps theeasiest method.

7.9.1 Easy Method : ρΛ3 → 0

For ease of notation let,

ηi =ǫi − µ

kBT(7.52)

and write the grand canonical partition function under Maxwell-Boltzmann,Bose-Einstein, and Fermi-Dirac statistics as

Q =∏

i

exp[exp(−ηi)] Maxwell− Boltzmann

1

1− exp(−ηi)Bose− Einstein

1 + exp(−ηi) Fermi−Dirac

(7.53)

Let us make an estimate of ηi in the limit of high temperatures and lowdensities.

We had shown earlier that the canonical partition function for an ideal gascan be written as, see notes,

Q(T, V,N) =V N

N !

1

Λ3N(7.54)

where the thermal wave length is given by

4 µ is not determined by us externally by adjusting the particle bath.

Page 132: Kpn Stat Mech

120 7 Quantum Statistics

Λ =h√

2πmkBT(7.55)

The free energy is given by

F (T, V,N) = −kBT lnQ(T, V,N) (7.56)

= −kBT [−3N ln Λ+N ln V −N ln N +N ] (7.57)

When you take the partial derivative of the free energy with respect to Nkeeping temperature and volume constant, you get the chemical potential5.Therefore,

µ =

(∂F

∂N

)

T,V

= kBT [3 ln Λ− ln V + 1 + ln N − 1] (7.62)

= kBT[lnΛ3 + ln(N/V )

](7.63)

= kBT ln(ρΛ3) (7.64)

µ

kBT= ln(ρΛ3) (7.65)

where ρ = N/V is the number density, i.e. number of particles per unitvolume. Thus we get,

ηi =ǫikBT

− ln(ρΛ3) (7.66)

We have earlier shown that classical limit obtains when ρΛ3 → 0. Thus inthis limit, ηi → ∞ or exp(−ηi) → 0. Let xi = exp(−ηi). Let us express thegrand canonical partition function for the three statistics in terms of the smallparameter xi.

5 In thermodynamics we have

F = U − TS (7.58)

dF = dU − TdS − SdT (7.59)

From the first law of thermodynamics we have dU = TdS −PdV + µdN . Substi-tuting this in the expression for dF above we get,

dF = −PdV + µdN − SdT (7.60)

Therefore,

µ =

(∂F

∂N

)

T,V

(7.61)

Page 133: Kpn Stat Mech

7.9 All the three statistics are the same at high temperature and/or low densities 121

Q =∏

i

exp(xi) Maxwell− Boltzmann

1

1− xiBose− Einstein

1 + xi Fermi−Dirac

(7.67)

In the limit xi → 0 we have exp(±xi) = 1 ± xi and (1 − xi)−1 = 1 + xi.

Therefore in the limit xi → 0, we have

Q =∏

i

exp(xi)∼

xi→0 1 + xi Maxwell− Boltzmann

1

1− xi∼

xi→0 1 + xi Bose− Einstein

1 + xi = 1 + xi Fermi−Dirac

(7.68)

For all the three statistics, the grand canonical partition function take thesame expression. Bosons, Fermions and classical indistinguishable particlesbehave the same way when ρΛ3 → 0.

When do we get ρΛ3 → 0 ?Note that Λ is inversely proportional to square-root of the temperature.

Λ =h√

2πmkBT

Hence Λ→ 0, when T → ∞. For a fixed temperature (Λ is constant), ρΛ3 → 0when ρ→ 0. For a fixed ρ, when T → ∞ ( the same as Λ→ 0), then ρΛ3 → 0.

• Classical behaviour obtains at low densities and/or high tempera-tures.

• Quantum effects manifest only at low temperatures and/or high densities.

7.9.2 Easier Method : λ → 0

Another simple way to show that the three statistics are identical in the limitof high temperatures and low densities is to recognise (see below) that ρΛ3 → 0implies λ→ 0. Here λ = exp(βµ), is the fugacity. Let us show this first.

We have shown that

µ

kBT= ln(ρΛ3) (7.69)

Therefore the fugacity is given by

Page 134: Kpn Stat Mech

122 7 Quantum Statistics

λ = exp

kBT

)= exp(ln[ρΛ3]) = ρΛ3 (7.70)

Thus ρΛ3 → 0 implies λ→ 0.In the limit λ→ 0 we have, for Maxwell-Boltzmann statistics,

QMB =∏

i

exp[λ exp(−βǫi)] (7.71)

∼λ→0

i

(1 + λ exp(−βǫi)) (7.72)

In the limit λ→ 0, for Bose-Einstein statistics, we have,

QBE =∏

i

1

1− λ exp(−βǫi)(7.73)

∼λ→0

i

[1 + λ exp(−βǫi)] (7.74)

For Fermi-Dirac statistics, we have exatly,

QFD =∏

i

[1 + λ exp(−βǫi)] (7.75)

Thus in the limit of hight temperatures and low densities Maxwell Boltzmannstatistics and Bose Einstein statistics go over to Fermi - Dirac statistics.

7.9.3 Easiest Method Ω(n1, n2, · · · ) = 1

We could have shown easily that in the limit of high temperature and lowdensity, the three statistics are identical by considering the degeneracy factor

Ω =

1 Bose− Einstein and Fermi−Dirac statistics

1

n1!n2! · · ·Maxwell− Boltzmann statistics

(7.76)

When the temperature is high the number of quantum states that becomeavailable for occupation is very large; When the density is low the numberof particles in the system is low. Thus we have a very few number of parti-cles occupying a very large of quantum states. In other words the number ofquantum states is very large compared to the number of particles.

Page 135: Kpn Stat Mech

7.10 Statistics of Occupation Number - Mean 123

Hence the number of micro states with ni = 0 or 1 ∀ i are overwhelminglylarge compared to the those with ni ≥ 2. In any case, in Fermi-Dirac statisticsni is always 0 or 1.

In other words, when particles are a few in number(number density islow) and the accessible quantum levels are large in number (temperature ishigh), then micro states with two or more particles in one or more quantumstates are very rare. In other words bunching of particles is very rare at lowdensities and high temperatures. Almost always every quantum state is eitherunoccupied or occupied by one particle. Very rarely will you find two or moreparticles in a single quantum state. Hence the degeneracy factor is unity athigh temperatures and/or low densities. Hence all the three statistics areidentical in the limit of high temperatures and/or low densities.

7.10 Statistics of Occupation Number - Mean

Let us consider the single particle quantum state k with energy ǫk. Let nk

be the number of particles occupying the state k. nk is called the occupationnumber. It is a random variable. Here we shall calculate the statistics of therandom variable nk.

Let 〈nk〉 denote the average occupation number; the averaging is done overa grand canonical ensemble of micro states. Formally we have for Bosons andFermions,

〈nk〉 =

i6=k

all∑

n=0

xni

all∑

n=0

nxnk

i6=k

all∑

n=0

xni

all∑

n=0

xnk

=

all∑

n=0

nxnk

all∑

n=0

xnk

7.10.1 Ideal Fermions

For Fermions, n = 0, 1. Hence

〈nk〉FD =xk

1 + xk

=1

x−1k + 1

=1

exp[β(ǫk − µ)] + 1

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124 7 Quantum Statistics

7.10.2 Ideal Bosons

For Bosons n = 0, 1, 2, · · · ,∞. In other words, a single particle quantum statecan hold any number of particles. In fact, we shall see later, that at low enoughtemperature all the particles would condense into the lowest energy state, i.e.the ground state. We call this Bose-Einstein condensation.

We carry out the summation6, in the numerator and and the denominatorof the expression for 〈nk〉, see above, analytically and get,

〈nk〉BE =xk

(1− xk)2(1− xk)

=xk

1− xk=

1

x−1k − 1

=1

exp[β(ǫk − µ)]− 1

7.10.3 Classical Indistinguishable Ideal Particles

〈nk〉MB =

i6=k

∞∑

n=0

xnin!

(

∞∑

n=0

nxnkn!

)

i6=k

∞∑

n=0

xnin!

(

∞∑

n=0

xnkn!

)

=

∞∑

n=0

nxnkn!

∞∑

n=0

xnkn!

In the above the summation in the numerator and the denominator are eval-uated analytically as follows. We start with the definition,

6 Consider

S(x) = 1 + x+ x2 + · · · = 1

1− x

dS

dx= 1 + 2x+ 3x2 + 4x3 + · · · = 1

(1− x)2

xdS

dx= x+ 2x2 + 3x3 + · · · = x

(1− x)2

Page 137: Kpn Stat Mech

7.11 Some Remarks on 〈nk〉 from the three statistics 125

∞∑

n=0

xn

n!= exp(x)

Differentiate both sides of the above equation with respect to x. You get

∞∑

n=0

nxn−1

n!= exp(x)

1

x

∞∑

n=0

nxn

n!= exp(x)

∞∑

n=0

nxn

n!= x exp(x)

Therefore,

〈nk〉MB =xk exp(xk)

exp(xk)= xk = exp[−β(ǫk − µ)] =

1

exp[β(ǫk − µ)]

7.11 Some Remarks on 〈nk〉 from the three statistics

We can write for all the three statistics,

〈nk〉 =1

exp[β(ǫk − µ)] + awhere a =

+ 1 Fermi−Dirac

0 Maxwell− Boltzmann

− 1 Bose− Einstein

Variation of 〈nk〉 with energy is shown in the figure, see next page. Note that

the x axis isǫk − µ

kBTand the y axis is 〈nk〉.

7.11.1 Fermi-Dirac Statistics

We see that for Fermi-Dirac statistics the occupation number never exceedsunity. When ǫk − µ is negative and |ǫk − µ| is large the value of 〈nk〉 tends tounity. For ǫ = µ and T 6= 0, we have 〈nk〉 = 1/2.

Page 138: Kpn Stat Mech

126 7 Quantum Statistics

It is also clear that at high temperature the chemical potential of a Fermionmust be negative, because its behaviour coincides with that of Bosons 7 andclassical indistinguishable particles8.

7.11.2 Bose-Einstein Statistics

For Bose-Einstein statistics we must have

ǫk > µ ∀ k.

In particular the lowest value of energy, say ǫ0 corresponding to the groundstate of the macroscopic system, must be greater than µ. Let us take ǫ0 = 0without loss of generality. Hence for Bosons µ must be negative. Also µ is afunction of temperature. As we lower T the chemical potential increases andat T = TBE , µ becomes zero. TBE is called the Bose-Einstein temperature. Atthis temperature the occupancy of the ground state becomes infinitely high.This leads to the phenomenon of Bose-Einstein Condensation.

7.11.3 Maxwell-Boltzmann Statistics

For the indistinguishable classical particles 〈nk〉 takes the familiar exponentialdecay form,

〈nk〉 = exp[−β(ǫk − µ)].

7.11.4 At high T and/or low ρ all statistics give the same 〈nk〉

Whenǫk − µ

kBT→ ∞,

all the three statistics coincide. We have already seen that at high tempera-tures classical behaviour obtains. Then, the only way

ǫk − µ

kBT

can become large at high temperature (note in the expression T is in thedenominator) is when µ is negative and its magnitude also should increase withincrease of temperature. Thus for all the statistics at high temperature thechemical potential µ is negative and its magnitude must be large. Essentiallyµ

kBT must be negative and its magnitude must be large at high temperature,when classical behaviour obtains. But the we know

7 for Bosons the chemical potential is negative at all temperature, and zero at zerotemperature and at temperatures less than a critical temperature called Bose-Einstein condensation temperature.

8 for classical indistinguishable the chemical potential is negative at high temper-ature, positive at low temperatures and zero at zero temperature.

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7.12 Statistics of Occupation Number - Fluctuations 127

−3 −2 −1 0 1 2 3 4 5−1

−0.5

0

0.5

1

1.5

2

ǫk −µ

kBT

〈nk〉

F ermi−Dirac →

←− Maxwell − Boltzmann

← Bose − Einstein

Fig. 7.1. Average occupation number of a quantum state under Bose-Einstein,Fermi-Dirac, and Maxwell-Boltzmann statistics

µ

kBT= ln(ρΛ3).

This means that ρΛ3 << 1 for classical behaviour to emerge9. This is incomplete agreement with our earlier surmise that classical behaviour obtainsat low ρ and/or high T .

Hence all the approaches are consistent with each other and all the issuesfall in place.

In this lecture we saw of the first order statistics - the mean/average/expectation/firstmoment - of the occupation number in Fermi-Dirac, Bose-Einstein, andMaxwell-Boltzmann statistics. In the next lecture I shall tell you of the secondorder statistics - the fluctuations of the occupation number.

7.12 Statistics of Occupation Number - Fluctuations

In the last lecture, I introduced a random variable nk, which denotes thenumber of particles in the quantum state k of energy ǫk. We call nk a random

9 Note that ln(x) = 0 for x = 1 and is negative for x < 1. As x goes from 1 to 0,the quantity ln(x) goes from 0 to −∞.

Page 140: Kpn Stat Mech

128 7 Quantum Statistics

variable because it takes values that are, in general, different for different microstates. For Bose-Einstein and Fermi-Dirac statistics, a string of occupationnumbers specifies completely a micro state. We found that for Bosons andFermions, the average value of nk can be expressed as,

〈nk〉 =

all∑

n=0

nxnk

all∑

n=0

xnk

where,

xk = exp[−β(ǫk − µ)]

Formally

〈nk〉 =all∑

n=0

nP (n)

In the above P (n) ≡ P (nk = n) is the probability that the random variablenk takes a value n. Comparing the above with the first equation, we find

P (n) ≡ P (nk = n) =1

all∑

m=0

xmk

xnk

where the denominator ensures that the total probability is normalized tounity.

Here, I shall work out explicitly P (n) for Fermi-Dirac, Bose-Einstein, andMaxwell-Boltzmann statistics. We shall calculate the variance of the randomvariable nk; we shall derive the relative fluctuations : the standard deviationdivided by the mean, for the three statistics.

7.12.1 Fermi-Dirac Statistics

In Fermi-Dirac statistics the random variable nk can take only two values :n = 0 and n = 1. Thus,

P (n) =

1

1 + xkfor n = 0

xk1 + xk

for n = 1

We have

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7.12 Statistics of Occupation Number - Fluctuations 129

〈nk〉 =1∑

n=0

nP (n) =xk

1 + xk,

consistent with the result obtained earlier.For convenience of notation let us denote the mean of the random variable

nk by the symbol ζ. We have,

ζ = 〈nk〉 =xk

1 + xk.

We thus have,

P (n) =

1− ζ for n = 0

ζ for n = 1

Thus Fermi-Dirac statistics defines a quantum coin with ζ and 1 − ζ as theprobabilities of ”Heads” (n = 0) and ”Tails” (n = 1) respectively.

Formally,

〈nk〉 =1∑

n=0

nP (n) = ζ

〈n2k〉 =

1∑

n=0

n2P (n) = ζ

σ2 = 〈n2k〉 − 〈nk〉2 = ζ(1− ζ)

The relative fluctuations of the random variable nk is defined as the standarddeviation σ divided by the mean ζ. Let us denote the relative fluctuation bythe symbol η. For Fermi-Dirac statistics we have,

ηFD =σ

ζ=

√1

ζ− 1

7.12.2 Bose-Einstein Statistics

For Bosons,

P (nk = n) = P (n) = xnk (1− xk)

from which it follows,

ζ =

∞∑

n=0

nP (n) =xk

1− xk

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130 7 Quantum Statistics

consistent with the result obtained earlier. Inverting the above, we get,

xk =ζ

1 + ζ

Then the probability distribution of the random variable nk can be writtenin a convenient form

P (n) =ζn

(1 + ζ)n+1

The distribution is geometric, with a constant common ratio ζ/(ζ + 1). Wehave come across geometric distribution earlier10.

Calculation of variance is best done by first deriving an expression for themoment generating function given by,

P (z) =

∞∑

n=0

znP (n)

=1

1 + ζ

∞∑

n=0

zn(

ζ

1 + ζ

)n

=1

1 + ζ

1(1− ζ z

1 + ζ

)

=1

1 + ζ(1− z)

Let us now differentiate P (z) with respect to z and in the resulting expressionset z = 1. We shall get 〈nk〉, see below.

∂P

∂z=

ζ

(1 + ζ(1 − z))2

∂P

∂z

∣∣∣∣z=1

= ζ

10 see Problem No. 7 (Problem set : 2, page 4. Let me recollect : The simplestproblem in which geometric distribution arises is in coin tossing. Take a p-coin;ı.e. a coin for which the probability of ”Heads” is p and that of ”Tails” is q =1 − p. Toss the coin until the side ”Heads” appears. The number of tosses is arandom variable with a geometric distribution P (n) = qn−1p. We can write thisdistribution in terms of ζ = 〈n〉 = 1/p and get P (n) = (ζ − 1)n−1/ζn.

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7.12 Statistics of Occupation Number - Fluctuations 131

Differentiating twice with respect to z and setting z = 1 in the resultingexpression shall yield the factorial moment 〈nk(nk − 1)〉, see below.

∂2P

∂z2=

2ζ2

[1 + ζ(1 − z)]3

∂2P

∂z2

∣∣∣∣z=1

= 〈nk(nk − 1)〉 = 2ζ2

〈n2k〉 = 2ζ2 + ζ

σ2 = 〈n2k〉 − 〈nk〉2 = ζ2 + ζ

ηBE =σ

ζ=

√1

ζ+ 1

For doing the problem , You will need the following tricks.

S(x) =∞∑

n=0

xn = 1 + x+ x2 + x3 + · · · = 1

1− x

dS

dx=

∞∑

n=1

nxn−1 = 1 + 2x+ 3x2 + 4x3 + · · · = 1

(1− x)2

xdS

dx=

∞∑

n=1

nxn = x+ 2x2 + 3x3 + 4x4 + · · · = x

(1 − x)2

d

dx

(xdS

dx

)=

∞∑

n=1

n2xn−1 = 1 + 22x+ 32x2 + 42x3 + · · · = 2x

(1− x)3+

1

(1− x)2

xd

dx

(xdS

dx

)=

∞∑

n=1

n2xn = x+ 22x2 + 32x3 + 42x4 + · · · = 2x2

(1 − x)3+

x

(1− x)2

You can employ the above trick to derive power series for ln(1 ± x), seebelow.

∫dx

1− x= − ln(1− x) = x+

x2

2+x3

3+x4

4· · ·

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132 7 Quantum Statistics

7.12.3 Maxwell-Boltzmann statistics

For Maxwell-Boltzmann statistics we have,

〈nk〉 = xk

P (nk = n) ≡ P (n) =xnkn!

1

exp(xk)

=ζn

n!exp(−ζ)

The random variable nk has Poisson distribution. The variance equals themean. Thus the relative standard deviation is given by

ηMB =σ

ζ=

1√ζ

We can now write the relative fluctuations for the three statistics in onesingle formula as,

η =

√1

ζ− a with a =

+1 for Fermi−Dirac Statistics

0 for Maxwell− Boltzmann Statistics

−1 for Bose− Einstein Statistics

Let us look at the ratio,

r =P (n)

P (n− 1).

For the Maxwell-Boltzmann statistics, r = ζ/n. The ratio r is inversely pro-portional to n. This is the normal behaviour; inverse dependence of r on n iswhat we should expect, when the events are uncorrelated. Recall the discus-sions we had on Poisson process : Problem 22, Assignment 6.

On the other hand, for Bose-Einstein statistics, the ratio is given by

r =P (n)

P (n− 1)=

ζ

ζ + 1

r is independent of n. This means, a new particle will get into any of thequantum states, with equal probability irrespective of how abundantly orhow sparsely that particular quantum state is already populated. An emptyquantum state has the same probability of acquiring an extra particle as anabundantly populated quantum state.

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7.12 Statistics of Occupation Number - Fluctuations 133

Thus, compared to classical particles obeying Maxwell-Boltzmann statis-tics, Bosons exhibit a tendency to bunch together. By nature, Bosons liketo be together. Note that this ”bunching-tendency” is not due to interactionbetween Bosons. We are considering ideal Bosons. This bunching is purely aquantum mechanical effect; it arises due to symmetry property of the wavefunction.

For Fermions, the situation is quite the opposite. There is what we maycall an aversion to bunching; call it anti-bunching if you like. No Fermionwould like to have another Fermion in its quantum state. A Fermion behaveslike a dog in the manger.

Problems

7.1. Consider Fermi-Dirac statistics at T = 0. Show how the graph of

〈nk〉 versusǫk − µ

kBT

will look like. In particular investigate the region around ǫk = µ.

7.2. See R K Pathria, Statistical Mechanics - Second Edition, Buttteworthand Heinemann (1996)p.152;Problem:6.1Show for ideal Bosons in thermal equilibrium the entropy is given by

SFD = kB∑

i

〈ni + 1〉 ln〈ni + 1〉 − kB∑

i

〈ni〉 ln〈ni〉

7.3. See R K Pathria, Statistical Mechanics - Second Edition, Buttteworthand Heinemann (1996)p.152;Problem:6.1Show that for ideal Fermions in thermal equilibrium, the entropy is given by

SMB = kB∑

i

−〈1− ni〉 ln〈1− ni〉 − kB∑

i

〈ni〉 ln〈ni〉

7.4. See R K Pathria, Statistical Mechanics - Second Edition, Buttteworthand Heinemann (1996)p.152;Problem:6.3The occupation number nk is random variable. nk is the number of particlesin quantum state k.

In Fermi-Dirac statistics nk can either be zero or one. In Bose-Einsteinstatistics nk can take any value from zero to infinity.

Now imagine an intermediate statistics in which nk can take any valuebetween zero and say L. Let us call the particles that obey such a statisticsas ”anyons”. Show that

〈nk〉 =1

exp[β(ǫk − µ)]− 1− L+ 1

exp[β(ǫk − µ)(L+ 1)]− 1

Page 146: Kpn Stat Mech

134 7 Quantum Statistics

7.5. Start with

P (n) =ζn

(1 + ζ)n+1

for Bose-Einstein statistics and show that

〈nk〉 =∞∑

n=0

nP (n) = ζ

7.6. For Bose-Einstein statistics, calculate the second moment, 〈n2k〉, the hard

way

〈n2k〉 =

1

1 + ζ

∞∑

n=0

n2

1 + ζ

)n

Calculate the relative fluctuations and show that your results agree with theones obtained employing generating function technique

7.7. Show that the chemical potential of Bosons is negative at all temper-atures. At best it can become zero at very low temperatures. Consider theexpression

〈nk〉 =1

exp[β(ǫk − µ)]− 1

Take the lowest energy quantum state to be of energy zero. i.e. ǫ0 = 0. Showµ positive is unphysical.

Page 147: Kpn Stat Mech

8

Bose Einstein Condensation

8.1 Some Preliminaries

For Bosons we found that the grand canonical partition funtion is given by,

Q(T, V, µ) =∏

i

1

1− exp[−β(ǫi − µ)](8.1)

The correspondence with thermodynamics is established by the expression forgrand potential denoted by the symbol G(T, V, µ). We have,

G(T, V, µ) = −kBT lnQ(T, V, µ) (8.2)

= kBT ln [1 + exp −β(ǫi − µ)] (8.3)

Recall, from thermodynamics, that the G is obtained by Legendre trans-form of U(S, V,N) : S → T ; N → µ; and U → G.

G(T, V, µ) = U − TS − µN (8.4)

T =

(∂U

∂S

)

V,N

(8.5)

µ =

(∂U

∂N

)

S,V

(8.6)

From the above, we get,

dG = −PdV − SdT −Ndµ (8.7)

It follows,

Page 148: Kpn Stat Mech

136 8 Bose Einstein Condensation

P (T, V, µ) = −(∂G∂V

)

T,µ

(8.8)

S(T, V, µ) = −(∂G∂T

)

V,µ

(8.9)

N(T, V, µ) = −(∂G∂µ

)

T,µ

(8.10)

If we have an open system of Bosons at temperature T and chemical potentialµ, in a volume V , then the above forrmulae help us calculate the pressure,entropy and number of Bosons in the system. In fact we have calculated thefluctuations in the number of particles in the open system and related it toisothermal compressibility - an experimentally measurable property.

The grand potential of the Bosonic system is given by,

G(T, V, µ) = −kBT lnQ (8.11)

(8.12)

8.1.1 〈N〉 =∑

k〈nk〉

For Bosons, we found that the average occupancy of a (single-particle) quan-tum state k, is given by,

〈nk〉 =λ exp(−βǫk)

1− λ exp(−βǫk)

where λ is fugacity. We have

λ = exp(βµ).

In the above µ is the chemical potential and equals the energy change due toaddition of a single particle under constant entropy and volume :

µ =

(∂U

∂N

)

S,V

.

The average number of particles in the open system is then

〈N〉 =∑

k

〈nk〉

=∑

k

λ exp(−βǫk)1− λ exp(−βǫk)

(8.13)

Page 149: Kpn Stat Mech

8.1 Some Preliminaries 137

We shall study Bosonic system with a fixed number of Bosons. We shallemploy grand canonical ensemble formalism in the study. This means µ isnot an independent property of the system which we can control through aparticle bath. The system chooses the appropriate value of µ and not we. Inother words µ is a function of temperature.

8.1.2∑

k(·) →∫(·)dǫ

Let us now convert the sum over quantum states to an integral over energy.To this end we need an expression for the number of quantum states in in-finitesimal intervals dǫ centered at ǫ. Let us denote this quantity by g(ǫ)dǫ.We call g(ǫ) the density of (energy) states. Thus we have,

N =

∫ ∞

0

λ exp(−βǫ)1− λ exp(−βǫ)g(ǫ)dǫ (8.14)

We need an expression for the density of states. We have done this exerciseearlier. In fact we have carried out classical counting and quantum countingand found both lead to the same result. The density of states is given by,

g(ǫ) = V 2π

(2m

h2

)3/2

ǫ1/2 (8.15)

We then have,

N = V 2π

(2m

h2

)3/2 ∫ ∞

0

λ exp(−βǫ)1− λ exp(−βǫ) ǫ

1/2dǫ (8.16)

We note that 0 ≤ λ < 1. This suggests that the integrand in the above canbe expanded in powers of λ. To this end we write

1

1− λ exp(−βǫ) =

∞∑

k=0

λk exp(−kβǫ) (8.17)

This gives us

λ exp(−βǫ)1− λ exp(−βǫ) =

∞∑

k=0

λk+1 exp[−β(k + 1)ǫ] (8.18)

=

∞∑

k=1

λk exp[−βkǫ] (8.19)

Substituting the above in the integral we get,

Page 150: Kpn Stat Mech

138 8 Bose Einstein Condensation

N = V 2π

(2m

h2

)3/2 ∞∑

k=1

λk∫ ∞

0

exp(−kβǫ)ǫ1/2dǫ (8.20)

= V 2π

(2m

h2

)3/2 ∞∑

k=1

λk∫ ∞

0

exp(−kβǫ)(kβǫ)1/2d(kβǫ)β3/2k3/2

(8.21)

= V 2π

(2mkBT

h2

)3/2 ∞∑

k=1

λk

k3/2

∫ ∞

0

exp(−x)x1/2dx (8.22)

= V 2π

(2mkBT

h2

)3/2 ∞∑

k=1

λk

k3/2Γ (3/2) (8.23)

= V 2π

(2mkBT

h2

)3/21

2Γ (1/2)

∞∑

k=1

λk

k3/2(8.24)

= V 2π

(2mkBT

h2

)3/2 √π

2

∞∑

k=1

λk

k3/2(8.25)

= V

(2πmkBT

h2

)3/2 ∞∑

k=1

λk

k3/2(8.26)

(8.27)

In an earlier lecture, we defined a thermal wave length denoted by the symbolΛ. This is the de Broglie wave length associated with a particle having thermalenergy of kBT . It is also called quantum wavelength. It is given by, see earliernotes,

Λ =h√

2πmkBT(8.28)

The sum over k, in the expression for 〈N〉, is usually denoted by the symbolg3/2(λ):

g3/2(λ) =

∞∑

k=1

λk

k3/2(8.29)

= λ+λ2

2√2+

λ3

3√3+ · · · (8.30)

Page 151: Kpn Stat Mech

8.1 Some Preliminaries 139

Thus we get,

N =V

Λ3g3/2(λ) (8.31)

We can write the above as,

NΛ3

V= g3/2(λ) (8.32)

It is easily verified that at high temperature we get results consistent withMaxwell Boltzmann statistics1

1 The fugacity λ is small at high temperature. For small λ we can replace g3/2(λ)by λ. We get

N =V

Λ3λ (8.33)

This result is consistent with Maxwell-Boltzmann statistics, see below.For Maxwell-Boltzmann statistics

〈nk〉 = λ exp(−βǫk) (8.34)

N =∑

k

〈nk〉 = λ∑

k

exp(−βǫk) = λ 2πV

(2m

h2

)3/2 ∫ ∞

0

dǫ ǫ1/2 exp(−βǫ)(8.35)

= λ 2πV

(2m

h2

)3/2 ∫ ∞

0

(βǫ)1/2 exp(−βǫ) d(βǫ)β3/2

(8.36)

= λ 2πV

(2mkBT

h2

)3/2

Γ (3/2) (8.37)

= λ 2πV

(2mkBT

h2

)3/21

2Γ (1/2) (8.38)

= λ 2πV

(2mkBT

h2

)3/2

1

2

√π (8.39)

= λ V

(2πmkBT

h2

)3/2

(8.40)

=V

Λ3λ (8.41)

Page 152: Kpn Stat Mech

140 8 Bose Einstein Condensation

8.1.3 g3/2(λ) versus λ

I have plotted g3/2(λ) as a function of λ in the figure below. Note that the

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

3.5

4

g3/2(λ = 1) = ζ(1) = 2.612

λ

g3/2(λ)→

Fig. 8.1. g3/2(λ) versus λ. Graphical inversion to determine fugacity

infinite series converges only for 0 ≤ λ ≤ 1. We have

g3/2(λ = 1) =

∞∑

k=1

1

k3/2(8.42)

= ζ(3/2) (8.43)

= 2.612 (8.44)

where ζ(n) is the Riemann zeta function, defined as,

ζ(n) =

∞∑

k=1

1

kn. (8.45)

Page 153: Kpn Stat Mech

8.1 Some Preliminaries 141

We note that the value of . g3/2 does not exceed 2.612.

8.1.4 Graphical inversion to determine fugacity

Let us say we know the values of N , V and temperature T of a system. Thenwe can find the value of fugacity by graphical inversion : Draw a line parallelto the x axis at y = NΛ3/V and read off the value of λ at which the line cutsthe curve g3/2(λ).

Once we get the fugacity, we can determine all other thermodynamic prop-erties of the open system employing the formalism of grand canonical ensem-ble.

So far so good.But then we realise that the above graphical inversion scheme does not

permit evaluation of the fugacity of a system with NΛ3/V greater than 2.612.This is absurd.There must be something wrong with what we have done.

8.1.5 Treatment of the Singular Behaviour

We realise that when NΛ3/V approaches 2.612, the fugacity λ approachesunity; the chemical potential µ approaches zero2. We have already seen thatat µ = 0 the occupancy of the ground state diverges. The singular behaviourof the ground state occupancy was completely lost when we replaced the sumover quantum states by an integral over energy : the weighting function is thedensity of states, given by, g(ǫ) ∼ ǫ1/2. The density of states vanishes at zeroenergy. Hence we must take care of the singular behaviour separately.

We have,

N =∑

k

λ exp(−βǫk)1− λ exp(−βǫk)

(8.46)

1− λ+∑

k

λ exp(−βǫk1− λ exp(−βǫk)

(8.47)

In the above, we have separated the ground state occupancy and the occu-pancy of all the excited states. Let 〈N0〉 denote the ground state occupancy.It is given by the first term,

N0 =λ

1− λ(8.48)

The occupancy of all the excited states is given by the second term, wherethe sum is taken only over the indices k representing the excited states. LetNe denote the occupancy of excited states. It is given by,

2 The chemical potential approaches the energy of the ground state. With out lossof generality, we can set the ground state at zero energy.

Page 154: Kpn Stat Mech

142 8 Bose Einstein Condensation

Ne =∑

k

λ exp(−βǫk1− λ exp(−βǫk)

(8.49)

In the above, the sum over k can be replaced by an integral over energy. Inthe integral over energy, we can still keep the lower limit of integration aszero, since the density of states giving weight factors for occupancy of statesis zero at zero energy. Accordingly we write

N = 〈N0〉+ 〈Ne〉 (8.50)

1− λ+V

Λ3g3/2(λ) (8.51)

We thus have,

NΛ3

V=Λ3

V

λ

1− λ+ g3/2(λ) (8.52)

Let us define the number of density - number of particles per unit volume,denoted by the symbol ρ. It is given by

ρ =N

V(8.53)

The function λ/(1 − λ) diverges at λ = 1, as you can see from the figurebelow.

Hence the relevant curve for carrying out graphical inversion should be theone that depicts the sum of the singular part (that takes care of the occupancyof the ground state) and the regular part (that takes care of the ocupancyof the excited states). For a value of Λ3/V = .05 we have plotted both thecurves and their sum in the figure below. Thus for any value of ρΛ3 we cannow determine the fugacity by graphical inversion.

We carry out such an exercise and obtain the values of λ for various valuesof ρΛ3 and the figure below depicts the results.

It is clear that when ρΛ3 > 2.612, the fugacity λ is close unity. How closecan it approach unity ?

Let us postulate3

λ = 1− a

N.

where a is a number. To determine a we proceed as follows.

3 We have reasons to postulate λ = 1− a

N. This is related to the mechanism underly-

ing Bose-Einstein condensation we shall discuss the details later. In fact, followingDonald A McQuarrie, Statistical Mechanics, Harper and Row (1976)p.173 we can

make a postulate λ = 1− a

V. This should also lead to the same conclusions.

Page 155: Kpn Stat Mech

8.1 Some Preliminaries 143

0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

60

70

80

90

100

λ

λ

1−λ→

Fig. 8.2. Singular part of 〈N〉

We have,

λ

1− λ=N

a− 1 (8.54)

≈ N

aif N >> a (8.55)

We start with,

ρΛ3 =Λ3

V

λ

1− λ+ g3/2(λ) (8.56)

Sunstitute λ = 1− a/N in the above and get4,

ρΛ3 =ρΛ3

a+ g3/2(1) (8.57)

4 g3/2(1− a/N) ≈ g3/2(1),

Page 156: Kpn Stat Mech

144 8 Bose Einstein Condensation

0 0.2 0.4 0.6 0.8 10

1

2

3

4

5

6

7

8

λ

g3/2(λ = 1) = ζ(1) = 2.612

Λ3

V

λ

1−λ+ g3/2(λ)→

Fig. 8.3. ρΛ3 versus λ. The singular part [Λ3/V ][λ/(1−λ)] (the bottom most curve),the regular part g3/2(λ) (the middle curve) , and the total (ρΛ3) are plotted. For thisplot we have taken Λ3/V as 0.05

Thus we get,

a =ρΛ3

ρΛ3 − g3/2(1)(8.58)

The point ρΛ3 = g3/2(1) = 2.612 is a special point indeed. What is thephysical significance of this point ? To answer this question, consider thequantity ρΛ3 as a function of temperature with ρ kept at a constant value.The temperature dependence of this quantity is shown below.

ρΛ3 = ρ

(h√

2πmkBT

)3

(8.59)

At high temperature for which ρΛ3 < g3/2(1) = 2.612, we can determinethe value of λ from the equation g3/2(λ) = ρΛ3 by graphical or numericalinversion.

Page 157: Kpn Stat Mech

8.1 Some Preliminaries 145

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ρΛ3

2.612

λ

ρΛ3 =Λ3

V

λ

1−λ+ g3/2(λ)

Λ3

V= .05

Fig. 8.4. Fugacity λ versus ρΛ3

At low temperatures for which ρΛ3 > 2.612, we have λ = 1 − a/〈N〉where

a =ρΛ3

ρΛ3 − g3/2(1)(8.60)

The quantity λ/(1 − λ) is the average number of particles in the groundstate. At temperatures for which ρΛ3 > 2.612, we have,

N0 =λ

1− λ(8.61)

=N

a(8.62)

N0

N=

1

a(8.63)

= 1− 1

ρΛ3g3/2(1) (8.64)

We can write the above in a more suggestive form by defining a temperatureTBEC by

Page 158: Kpn Stat Mech

146 8 Bose Einstein Condensation

ρΛ3BEC = g3/2(1) (8.65)

Therefore,

N0

N=

1

a(8.66)

= 1− ρΛ3BEC

ρΛ3(8.67)

= 1−(ΛBEC

Λ

)3

(8.68)

= 1−( √

T√TBEC

)3

(8.69)

= 1−(

T

TBEC

)3/2

fer T < TBEC (8.70)

We have depicted the above behaviour of the fractional number of particlesin the ground state as a function of temperature in the figure below.

0 0.5 1 1.50

0.5

1

1.5

T

TBEC

N0

N

N0

N= 1−

(

T

TBEC

)3/2

Fig. 8.5. Ground state occupation as a function of temperature

Page 159: Kpn Stat Mech

8.1 Some Preliminaries 147

8.1.6 Bose-Einstein Condensation Temperature

Thus we can define the temperature at which Bose-Einstein condensationtakes place as,

N

V

(h√

2πmkBT

)3

= 2.612 (8.71)

kBTBEC =h2

2πm

(N

2.612 V

)2/3

(8.72)

At T = TBEC , Bose- Einstein condensation sets in and the ground stateoccupancy becomes anomalously larger and larger as temperature decreasesfurther.

8.1.7 Grand Potential for Bosons

The grand potential for Bosons is given by

G(T, V, µ) = −kBT lnQ(T, V, µ) (8.73)

= −kBT ln

[∏

k

1

(1− λ exp(−βǫk)

](8.74)

= kBT∑

k

ln[1− λ exp(−βǫk)] (8.75)

Now we shall be careful and separate the singular part and regular part toget,

G = kBT ln(1− λ) + kBT∑

k

ln[1− λ exp(−βǫk)] (8.76)

We have

ln[1− λ exp(−βǫk)] = −∞∑

k=1

λk exp(−kβǫk) (8.77)

Substitute the above in the expression for G. Convert the sum of k by anintegral over ǫ by the prescription below :

k

(·) −→ V 2π

(2m

h2

)3/2 ∫ ∞

0

(·) ǫ1/2 dǫ, (8.78)

We proceed as follows:

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148 8 Bose Einstein Condensation

G = kBT ln(1− λ) + kBT∑

k

ln[1− λ exp(−βǫk)] (8.79)

= kBT ln(1− λ)− kBT V 2π

(2m

h2

)3/2 ∞∑

k=1

λk∫ ∞

0

dǫ ǫ1/2 exp(−kβǫ)(8.80)

= kBT ln(1− λ)− kBT V 2π

(2m

h2

)3/2 ∞∑

k=1

λk ×∫ ∞

0

(kβǫ)1/2 exp(−kβǫ)k3/2β3/2

d(kβǫ)(8.81)

= kBT ln(1− λ)− kBT V 2π

(2mkBT

h2

)3/2

Γ (3/2)

∞∑

k=1

λk

k3/2(8.82)

= kBT ln(1− λ)− kBT V

(2πmkBT

h2

)3/2 ∞∑

k=1

λk

k3/2(8.83)

= kBT ln(1− λ)− kBTV

Λ3g3/2(λ) (8.84)

Thus we have,

G(T, V, λ) = kBT ln(1− λ)− V

Λ3kBT g3/2(λ) (8.85)

8.1.8 Average Energy of Bosons

An open system is described by a grand canonical partition function. It isformally given by,

Q(β, V, µ) =∑

i

exp[−β(Ei − µNi)] (8.86)

In the above Ei is the energy of the open system when in micro state i; Ni isthe number of particles in the open system when in micro state i. Let γ = βµ.Then we get,

Q(β, V, µ) =∑

i

exp(−βEi) exp(+γNi) (8.87)

We differentiate Q with respect to the variable β, keeping γ constant. We get

Page 161: Kpn Stat Mech

8.1 Some Preliminaries 149

∂Q∂β

= −∑

i

ǫi exp[−βǫi + γNi) (8.88)

− 1

Q∂Q∂β

= 〈E〉 = U (8.89)

For Bosons, we have,

Q =∏

i

1

1− λ exp(−βǫi)(8.90)

lnQ = −∑

i

ln[1− λ exp(−βǫi)] (8.91)

U =∑

i

ǫiλ exp(−βǫi)1− λ exp(−βǫi)

(8.92)

Let us now go to continuum limit by converting the sum over micro states byan integral over energy and get,

U =3

2V kBT

1

Λ3g5/2(λ) (8.93)

Let us now investigate the energy of the system at T > TBEC . Whentemperature is high, the number of Bosons in the ground state is negiligiblysmall. Hence the total energy of the system is the same as the one given above.

For temperatures less that TBEC, the ground state gets populated anoma-lously. The Bosons in the ground state do not contribute to the energy. Hencefor T < TBEC, we have

U =3

2V kBT

1

Λ3g5/2(1)

(1− N0

N

)(8.94)

=3

2V kBT

1

Λ3g5/2(1)

(T

TBEC

)3/2

(8.95)

Thus we have,

U =

3

2kBT

V

Λ3g5/2(λ) for T > TBEC

3

2kBT

V

Λ3g5/2(1)

(T

TBEC

)3/2

for T < TBEC

(8.96)

We can cast the above in a suggestive form, see below.

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150 8 Bose Einstein Condensation

We have,

N = N0 +Ne (8.97)

N0 =λ

1− λ(8.98)

Ne =V

Λ3g3/2(λ) (8.99)

For T > TBEC, we have Ne = N . Therefore,

V

Λ3=

N

g3/2(λ)(8.100)

substituting the above in the expression for U we get

U =

[3

2〈N〉kBT

]g5/2(λ)

g3/2(λ)for T > TBEC

[3

2〈N〉kBT

]g5/2(1)

g3/2(1)

(T

TBEC

)3/2

for T < TBEC

(8.101)

8.1.9 Specific Heat Capacity of Bosons

CV

NkB

for T > TBEC

Let us consider first the case with T > TBEC. We have

U

NkB=

3

2Tg5/2(λ)

g3/2(λ)(8.102)

1

NkB

∂U

∂T=

CV

NkB=

∂T

[3T

2

g5/2(λ)

g3/2(λ)

](8.103)

To carry out the derivative in the above, we need the following :

∂T

[g3/2(λ)

]= − 3

2Tg3/2(λ)

First Relation:

∂T

[g3/2(λ)

]= − 3

2Tg3/2(λ)

Page 163: Kpn Stat Mech

8.1 Some Preliminaries 151

Proof : We start with ρΛ3 = g3/2(λ). Therefore,

g3/2(λ) = ρΛ3 (8.104)

∂T[g3/2(λ)] = 3 ρΛ2 ∂Λ

∂T(8.105)

= 3 ρΛ3 ∂

∂T

(h√

2πmkBT

)(8.106)

= −3

2ρΛ2 h√

2πmkB

1

T 3/2(8.107)

= − 3

2TρΛ2 h√

2πmkBT(8.108)

= − 3

2TρΛ3 (8.109)

= − 3

2Tg3/2(λ) (8.110)

—————————— ————————————————————Q.E.D

∂λ[gn/2(λ)] =

1

λg(n/2)−1(λ)

Second Relation

∂λ[gn/2(λ)] =

1

λg(n/2)−1(λ)

Proof : We have by definition, gn/2(λ) =∞∑

k=1

λk

kn/2. Therefore,

Page 164: Kpn Stat Mech

152 8 Bose Einstein Condensation

∂λ[gn/2(λ)] =

∂λ

[∞∑

k=1

λk

kn/2

](8.111)

=

∞∑

k=1

kλk−1

kn/2(8.112)

=1

λ

∞∑

k=1

λk

k(n/2)−1(8.113)

=1

λg(n/2)−1(λ) (8.114)

——————————– ——————————–——————————– Q.E.D

1

λ

dT= − 3

2T

g3/2(λ)

g1/2(λ)

Third Relation

1

λ

dT= − 3

2T

g3/2(λ)

g1/2(λ)

Proof :We proceed as follows :

∂T[g3/2(λ)] =

∂λ[g3/2(λ)]

[dλ

dT

](8.115)

− 3

2Tg3/2(λ) =

1

λg1/2(λ)

dT(8.116)

From the above we get,

1

λ

dT= − 3

2T

g3/2(λ)

g1/2(λ)(8.117)

——————————- ——————————————————–Q.E.D

We have,

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8.1 Some Preliminaries 153

CV

NkB=

∂T

[3T

2

g5/2(λ)

g3/2(λ)

]

=3

2

g5/2(λ)

g3/2(λ)+

3T

2

∂T

[g5/2(λ)

g3/2(λ)

]

=3

2

g5/2(λ)

g3/2(λ)− 3T

2

[g5/2(λ)

g23/2(λ)

∂g3/2(λ)

∂T− 1

g3/2(λ)

∂g5/2(λ)

∂λ

dT

]

=3

2

g5/2(λ)

g3/2(λ)− 3T

2

[g5/2(λ)

g23/2(λ)

− 3

2Tg3/2(λ)

− 1

g3/2(λ)

1

λg3/2(λ)

dT

]

=3

2

g5/2(λ)

g3/2(λ)− 3T

2

[g5/2(λ)

g23/2(λ)

− 3

2Tg3/2(λ)

− 1

λ

dT

]

=3

2

g5/2(λ)

g3/2(λ)− 3T

2

[g5/2(λ)

g23/2(λ)

− 3

2Tg3/2(λ)

+

3

2T

g3/2(λ)

g1/2(λ)

]

=3

2

g5/2(λ)

g3/2(λ)+

9

4

g5/2(λ)

g3/2(λ)− 9

4

g3/2(λ)

g1/2(λ)

=15

4

g5/2(λ)

g3/2(λ)− 9

4

g3/2(λ)

g1/2(λ)(8.118)

CV

NkB

for T < TBEC

Now, let us consider the case with T < TBEC. We have,

U

NkB=

3

2Tg5/2(1)

g3/2(1)

(T

TBEC

)3/2

(8.119)

1

NkBCV =

3

2

g5/2(1)

g3/2(1)

5

2

(T

TBEC

)3/2

(8.120)

Thus we have,

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154 8 Bose Einstein Condensation

1

NkBCV =

15

4

g5/2(λ)

g3/2(λ)− 9

4

g3/2(λ)

g1/2(λ)for T > TBEC

15

4

g5/2(1)

g3/2(1)

(T

TBEC

)3/2

for T < TBEC

(8.121)

The specific heat is plotted against temperature in the figure below. The

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

T

TBEC

CV

NkB

Classical : 3NkB/2

Fig. 8.6. Heat capacity in the neighbourhood of Bose - Einstein condensation tem-perature

cusp in the heat capacity at T = TBEC is the signature of the Bose-Einsteincondensation. Asymptotically T → ∞, the heat capacity tends to the classicalvalue consistent with equi-partition.

8.1.10 Mechanism of Bose-Einstein Condensation

Let the ground state be of energy ǫ0 ≥ 0. For example consider particle in athree dimensional box of length L. The ground state is

(nx, ny, nx) = (1, 1, 1).

The ground state energy is

ǫ1,1,1 = ǫ0 =3h2

8mL2

.

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8.1 Some Preliminaries 155

The chemical potential is always less than or equal to ǫ0. As temperaturedecreases, the chemical potential increases and comes closer and closer to theground state energy ǫ0 ≥ 0. Let us estimate how close µ can get to ǫ0. In otherwords, we want to estimate the smallest possible value of (ǫ0 − µ)/[kBT ]. Tothis end, consider the expression for the average number of Bosons in theground state. Let us denote this by N0. It is given by,

N0 =1

exp

[ǫ0 − µ

kBT

]− 1

(8.122)

As temperature goes to zero, the chemical potential goes toward the

ground state energy. For a non-zero value of T , whenǫ0 − µ

kBTis small, we

can write

exp

(ǫ0 − µ

kBT

)= 1 +

ǫ0 − µ

kBT(8.123)

Substituting this in the expression for N0, given above, we get,

N0 =kBT

ǫ0 − µ(T )(8.124)

N0 goes to zero5 as T → ∞. At high temperature, the ground state occupancyis extremely small, as indeed it should.

Therefore we have,

ǫ0 − µ

kBT=

1

N0(8.125)

The largest value thatN0 can take isN , i.e. when all the particles condenseinto the ground state. In other words, the smallest value that 1/N0 can take is1/N . Hence (ǫ0−µ)/[kBT ] can not be smaller than 1/N . The smallest possiblevalue it can take is 1/N - inverse of the average number of particles in theentire system.

ǫ0 − µ

kBT≥ 1

N(8.126)

Thus, measured in units of kBT , the chemical potential shall always be lessthat the ground state energy at any non-zero temperature. At best, the quan-tity (ǫ − µ), expressed in units of thermal energy (kBT ), can only be of theorder of 1/N .

Therefore the chemical potential can never take a value close to any ofthe excited states, since all of them invariably lie above the ground state. In

5 For large T , the numerator is large; but the denominator is also large. Note thatµ(T ) is negative and large for large T . In fact the denominator goes to infinityfaster than the numerator.

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156 8 Bose Einstein Condensation

a sense, the ground state forbids the chemical potential to come close to anyenergy level other than the ground state energy. It sort of guards all the excitedstates from a close visit of µ. As T → 0, the number of Bosons in the groundstate increases.

This precisely is the subtle mechanism underlying Bose-Einsteincondensation.

Page 169: Kpn Stat Mech

9

Statistical Mechanics of Harmonic Oscillators

9.1 Classical Harmonic Oscillators

Consider a closed system of 3N harmonic oscillators at temperature T . Theoscillators do not interact with each other and are distinguishable. Let usderive an expression for the single-oscillator partition function.

The energy of an harmonic oscillator is given by

E =p2

2m+

1

2mω2q2 (9.1)

where q is the distance between the current position of the harmonic oscillatorand its mean position and p its momentum. ω is the characteristic frequencyof the oscillator and m its mass.

We have,

Q1(T ) =1

h

∫ +∞

−∞

dq

∫ +∞

−∞

dp exp

[−β(p2

2m+

1

2mω2q2

)](9.2)

We can write the above in a convenient way as a product of two Gaussianintegrals, one over dq and the other over dp, as

Q1(T ) =1

h

∫ +∞

−∞

dq exp

−1

2

q2(√

kBT

mω2

)2

× (9.3)

∫ +∞

−∞

dp exp

[−1

2

p2(√mkBT

)3

](9.4)

Let σ1 and σ2 denote the standard deviations of of the two zero-mean Gaussiandistributions. These are given by,

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158 9 Statistical Mechanics of Harmonic Oscillators

σ1 =

√kBT

mω2(9.5)

σ2 =√mkBT (9.6)

σ1σ2 =kBT

ω(9.7)

We have normalization identity for a Gaussian∫ +∞

−∞

dx exp

[−1

2

x2

σ2

]= σ

√2π (9.8)

Therefore,

Q1(T ) =2π

hσ1σ2 =

kBT

~ω(9.9)

If all the oscillators are identical i.e. they all have the same characteristicfrequency of oscillations, then

Q3N (T ) =

(kBT

)3N

(9.10)

See footnote1 where we have considered 3N harmonic oscillators with 3Ncharacteristic frequencies.

9.2 Helmholtz Free Energy

The free energy of a system of 3N non-interacting, identical classical harmonicoscillators is given by

F (T, V,N) = −3NkBT ln

(kBT

)= 3NkBT ln

(~ω

kBT

)(9.12)

See footnote2 where I have expressed the free energy as an integral over thedistribution of frequencies of the harmonic oscillator for N large.

Once we know of free energy, we can employ the machinery of thermo-dynamics and get expressions for all other thermodynamic properties of thesystem, see below.

1 On the other hand if the oscillators have distinct characteristic frequency ωi :i = 1, 2, · · · , 3N, then

Q3N (T ) =3N∏

i=1

kBT

~ωi(9.11)

2 If the oscillators have different frequencies then

Page 171: Kpn Stat Mech

9.3 Thermodynamic Properties of the Oscillator System 159

9.3 Thermodynamic Properties of the Oscillator System

F (T, V,N) = U − TS (9.16)

dF = dU − TdS − SdT (9.17)

= −SdT − PdV + µdN (9.18)

Thus for a system of identical, non-interacting classical harmonic oscillators

P = −(∂F

∂V

)

T,N

(9.19)

= 0 why? (9.20)

µ =

(∂F

∂N

)

T,V

(9.21)

= kBT ln

(~ω

kBT

)(9.22)

S = −(∂F

∂T

)

V,N

(9.23)

= NkB

[ln

(kBT

)+ 1

](9.24)

We also have,

F (T, V,N) = −kBT3N∑

i=1

ln

(kBT

~ωi

)(9.13)

If N is large we can define g(ω)dω as the number of harmonic oscillators withfrequencies in an interval dω around ω. The sum can be replaced by an integral,

F (T ) = −kBT∫

0

ln

(kBT

)g(ω)dω (9.14)

We have the normalization∫

0

g(ω)dω = 3N (9.15)

Page 172: Kpn Stat Mech

160 9 Statistical Mechanics of Harmonic Oscillators

U = −(∂ lnQ

∂β

)

V,N

(9.25)

= 3NkBT, (9.26)

consistent with equipartition theorem which says each quadratic term in theHamiltonian carries kBT/2 of energy. The Hamiltonian of a single harmonicoscillator has two quadratic terms - one in position q and the other in mo-mentum p.

We also find that the results are consistent with the Dulong and Petit’slaw which says that the heat capacity at constant volume is independent oftemperature:

CV =

(∂U

∂T

)

V

(9.27)

= 3NkB = 3nR (9.28)

CV

n= 3R = 6 calories (mole)−1 (Kelvin)−1 (9.29)

More importantly, the heat capacity is the same for all the materials; it de-pends only on the number of molecules or the number of moles of the substanceand not on what the substance is. The heat capacity per mole is approximately6 calories per Kelvin.

9.4 Quantum Harmonic Oscillator

Now let us consider quantum harmonic oscillators. The energy eigenvalues ofa single one dimensional harmonic oscillator is given by

ǫn =

(n+

1

2

)~ω : n = 0, 1, 2, · · · (9.30)

The canonical partition function for a single (quantum) harmonic oscillator isthen,

Q1(β) = exp(−β~ω/2)∞∑

n=0

[exp(−β~ω)]n (9.31)

=exp(−β~ω/2)1− exp(−β~ω) (9.32)

The partition function of a collection of 3N non-interacting quantum harmonicoscillators is then given by

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9.4 Quantum Harmonic Oscillator 161

QN (T ) =exp(−3Nβ~ω/2)

[1− exp(−β~ω)]3N(9.33)

See footnote3 for an expression of the partition function for 3N independentharmonic oscillators with different frequencies.

The free energy is given by,

F (T, V,N) = −kBT lnQ3N (T ) (9.35)

= 3N

[1

2~ω + kBT ln 1− exp(−β~ω)

](9.36)

See footnote4 for an expression for free energy for 3N independent harmonicoscillators with different frequencies.

We can obtain the thermodynamic properties of the system from the freeenergy. We get,

3 If the harmonic oscillators are all of different frequencies, the partition functionis given by

Q(T ) =3N∏

i=1

exp(−β~ωi/2)

1− exp(−β~ωi)(9.34)

4 For 3N independent harmonic oscillators with different frequencies we have

F =3N∑

i=1

[~ωi

2+ kBT ln 1− exp(−β~ωi)

](9.37)

=

∫∞

0

[~ω

2+ kBT ln 1− exp(−β~ω)

]g(ω) (9.38)

∫∞

0

dω g(ω) = 3N (9.39)

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162 9 Statistical Mechanics of Harmonic Oscillators

µ =

(∂F

∂N

)

T,V

(9.40)

=1

2~ω + kBT ln [1− exp(−β~ω)] (9.41)

P = −(∂F

∂V

)

T,N

(9.42)

= 0 (9.43)

S = −(∂F

∂T

)

V,N

(9.44)

= 3NkB

[β~ω

exp(β~ω)− 1− ln 1− exp(−β~ω)

](9.45)

U = −∂ lnQ∂β

(9.46)

= 3N

[~ω

2+

exp(β~ω)− 1

](9.47)

The expression for U tells that the equipartition theorem is the first victim ofquantum mechanics : Quantum harmonic oscillators do not obey equipartitiontheorem. The average energy per oscillator is higher than the classical valueof kBT . Only for T → ∞, we have kBT >> ~ω, the ”quantum” resultscoincide with the ”classical” results.

The heat capacity at constant volume is given by

CV =

(∂U

∂T

)

V,N

(9.48)

=3N

kB

(~ω

T

)2exp[β~ω]

(exp[β~ω]− 1)2 (9.49)

The second victim of quantum mechanics is the law of Dulong and Petit. Theheat capacity depends on temperature and on the oscillator frequency. Theheat capacity per mole will change from substance to substance because ofits dependence on the oscillator frequency. Only in the limit of T → ∞ (thesame as β → 0), do we get the classical results.

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9.5 Specific Heat of a Crystalline Solid 163

The temperature dependence of heat capacity is an important ”quantum”outcome. In fact we find that the heat capacity goes to zero exponentially asT → 0. However experiments suggest that the fall is algebraic rather thanexponential. The heat capacity goes to zero as T 3. This is called T 3 law.

9.5 Specific Heat of a Crystalline Solid

In the above we studied the behaviour of a collection of independent identicalharmonic oscillators in a canonical ensemble. We shall see below how sucha study is helpful toward understanding of the behaviour of specific heat ofcrystalline solid as a function of temperature.

A crystal is a collection of say N atoms organized in a regular lattice. Letx1, x2, · · · , x3N specify the 3N positions of these atoms. For example wecan consider a periodic array of atoms arranged at regular intervals along thethree mutually perpendicular directions, constituting a three dimensional cu-bic structure. We can think of other structures like face-centred cubic (FCC),body centred cubic (BCC) lattices.

An atom vibrates around its lattice location say (xi, xi+1, xi+2). It doesnot make large excursions away from its lattice location. We must bear inmind that the atoms are not independently bound to their lattice position.They are mutually bound5.

Consider a total of N atoms organized in a three dimensional lattice. Eachatom executes small oscillations about its mean position. In the process ofoscillations each atom pulls or pushes its neighbours; these neighbours in turnpull or push their neighbours and so on. The disturbance propagates in thecrystal. We can set up equations of motion for the three coordinates of eachof the atoms. We shall have 3N coupled equations. Consider the Hamiltonianof a solid of N atoms with position coordinates are x1, x2, · · ·x3N. Whenthe system of atoms is in its lowest energy, the coordinates are x1, x2, · · · x3N .Let V (x1, x2, · · ·x3N ) denote the potential energy. We express the potentialenergy under harmonic approximation, as

5 To appreciate the above statement, consider a class room wherein the chairsare already arranged with constant spacing along the length and breadth of theclass room. The students occupy these chairs and form a regular structure. Thiscorresponds to a situation wherein each student is bound independently to hischair.

Now consider a situation wherein the students are mutually bound to eachother. Let us say that the students interact with each other in the following way :Each is required to keep an arm’s length from his four neighbours. If the distancebetween two neighbouring students is less, they are pushed outward; if more,they are pulled inward. Such mutual interactions lead to the student organizingthemselves in a two dimensional regular array

I shall leave it to you to visualize how such mutual nearest neighbour interac-tions can give rise to three dimensional arrays.

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164 9 Statistical Mechanics of Harmonic Oscillators

V (x1, x2, · · ·x3N ) = V (x1, x2, · · · x3N ) + (9.50)

3N∑

i=1

(∂V

∂xi

)

x1,x2,··· ,x3N

(xi − xi) + (9.51)

3N∑

i=1

3N∑

j=1

1

2

(∂2V

∂xi∂xj

)

x1,x2,···x3N

(xi − xi)(xj − xj)(9.52)

The first term gives the minimum energy of the solid when all its atoms arein their equilibrium positions. We can denote this energy by V0.

The second set of terms involving the first order partial derivatives ofthe potential are all identically zero by definition : V has a minimum atxi = xi ∀i = 1, 3N

The third set of terms involving second order partial derivatives describeharmonic vibrations. We neglect the terms involving higher order derivativesand this is justified if only small oscillations are present in the crystalline .

Thus under harmonic approximations we can write the Hamiltonian as,

H = V0 +3N∑

i=1

1

2

(dξidt

)2

+3N∑

i=1

3N∑

j=1

αi,jξiξj (9.53)

where

ξ = xi − xi (9.54)

αi,j =1

2

(∂2V

∂xi∂xj

)

x1,x2,··· ,x3N

(9.55)

We shall now introduce a linear transformation from the coordinates ξi :i = 1, 3N to the normal coordinates qi : i = 1, 3N. We choose the lineartransformation matrix such that the Hamiltonian does not contain any crossterms in the q coordinates.

H = V0 +3N∑

i=1

1

2m(q2 + ω2

i q2i

)(9.56)

where ωi : i = 1, 3N are the characteristic frequencies of the normal modesof the system. These frequencies are determined by the nature of the potentialenergy function V (x1, x2, · · ·x3N . Thus the energy of the solid can be consid-ered as arising out of a set of 3N one dimensional, non interacting, harmonicoscillators whose characteristic frequencies are determined by the nature ofthe atoms of the crystalline solid, the nature of their mutual interaction, thenature of the lattice structure etc..

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9.5 Specific Heat of a Crystalline Solid 165

Thus we can describe the system in terms of independent harmonic oscilla-tors by defining a normal coordinate system, in which the equations of motionare decoupled. If there are N atoms in the crystals there are 3N degrees offreedom. Three of the degrees of freedom are associated with the translationof the whole crystal; and three with rotation. Thus, there are strictly 3N − 6normal mode oscillations. If N is of the order of 1025 or so, it doesn’t matterif the number of normal modes is 3N − 6 and not 3N .

We can write the canonical partition function as

Q =

3N∏

i=1

exp(−β~ωi/2)

1− exp(−β~ωi)(9.57)

There are 3N normal frequencies. We can imagine them to be continuouslydistributed. Let g(ω)dω denote the number of normal frequencies between ωand ω + dω. The function g(ω) obeys the normalization

∫ ∞

0

g(ω)dω = 3N (9.58)

We have,

− lnQ =

3N∑

i=1

[β~ωi

2+ ln 1− exp(−β~ωi)

](9.59)

=

∫ ∞

0

[β~ω

2+ ln 1− exp(−β~ω)

]g(ω)dω (9.60)

The problem reduces to finding the function g(ω). Once we know g(ω), wecan calculate the thermodynamic properties of the crystal. In particular wecan calculate the internal energy U and heat capacity, see below.

U =

∫ ∞

0

[~ω

2+

~ω exp(−β~ω)1− exp(−β~ω)

]g(ω)dω (9.61)

=

∫ ∞

0

[~ω

2+

exp(β~ω)− 1

]g(ω)dω (9.62)

CV = kB

∫ ∞

0

(β~ω)2 exp(β~ω)

[exp(β~ω)− 1]2g(ω)dω (9.63)

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166 9 Statistical Mechanics of Harmonic Oscillators

The problem of determining the function g(ω) is a non-trivial task. It is pre-cisely here that the difficulties lie. However, there are two well known ap-proximations to g(ω). One of them is due to Einstein and the other due toDebye.

9.6 Einstein Theory of Specific Heat of Crystals

Einstein assumed all the 3N harmonic oscillators to have the same frequency.In other words,

g(ω) = 3Nδ(ω − ωE) (9.64)

where ωE is the Einstein frequency or the frequency of the Einstein oscillator.The Einstein formula for the heat capacity is then given by

CV = 3NkB

(~ω

kBT

)2exp(~ωE/[kBT ])

(exp(~ωE/[kBT ])− 1)2(9.65)

Let us define

ΘE =~ωE

kB(9.66)

and call ΘE as Einstein temperature. Verify that this quantity has the unitof temperature. In terms of Einstein temperature we have,

CV = 3NkB

(ΘE

T

)2exp(ΘE/T )

[exp(ΘE/T )− 1]2 (9.67)

9.1 Show that in the limit of T → ∞, the heat capacityof the Einstein solid tends to the value 3NkB = 3R =6 cal (mole)−1K−1 predicted by Dulong and Petit.

9.2 Show that in the low temperature limit,

CV∼

T→0 3NkB

(ΘE

T

)2

exp(−ΘE/T ) (9.68)

Experiments suggest T 3 decay of CV with temperature. In the next class Ishall discuss Debye’s theory of heat capacity. We will find that Debye’s theorygives the T 3 law.

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9.7 Debye theory of Specific Heat 167

9.7 Debye theory of Specific Heat

Debye assumed a continuous spectrum of frequencies, cut off at an upper limitωD. Let us call it Debye frequency. Debye assumed based on an earlier workof Rayleigh, that g(ω) = αω2, where the proportionality constant depends onthe speed of propagation of the normal mode, its nature6, its degeneracy7.From the normalization condition,

α

∫ ωD

0

ω2dω = 3N (9.69)

we get, α = 9N/ω3D. Thus we have

g(ω) =

9N

ω3D

ω2 for ω ≤ ωD

0 for ω > ωD

(9.70)

Let us now calculate CV under Debye’s theory. We start with

CV (T ) = kB

∫ ∞

0

(β~ω)2 exp(β~ω)

[exp(β~ω)− 1]2g(ω)dω (9.71)

Let

x = β~ω (9.72)

ΘD =~ωD

kB(9.73)

ΘD is called the Debye temperature. Then we have,

CV = (3NkB)× 3

(T

ΘD

)3 ∫ ΘD/T

0

x4 exp(x)

[exp(x) − 1]2dx (9.74)

Let us consider the integral in the above expression and write,

I =

∫ Θ/T

0

x4 exp(x)

(exp(x) − 1)2dx (9.75)

Integrating by parts8 we get,

6 transverse or longitudinal7 transverse mode is doubly degenerate and longitudinal mode is non-degenerate,etc..

8 Take u(x) = x4 and dv(x) = exp(x)dx/[exp(x)− 1]2

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168 9 Statistical Mechanics of Harmonic Oscillators

I =1

exp(ΘD/T )− 1

(ΘD

T

)4

+ 4

∫ ΘD/T

0

x3

exp(x) − 1dx (9.76)

The expression for heat capacity is then,

CV = (3NkB)

[−3

(ΘD

T

)1

exp(ΘD/T )− 1+ 12

(T

ΘD

)3 ∫ ΘD/T

0

x3

exp(x) − 1dx

](9.77)

Let us now consider the behaviour CV in the limit of T → ∞. we haveT >> ΘD. We can set exp(ΘD/T ) ≈ 1 + (ΘD/T ); also in the integral wecan set exp(x) = 1 + x. Then we get,

CV = 3NkB

[−3 + 12

(T

ΘD

)3 ∫ ΘD/T

0

x2dx

](9.78)

= 3NkB(−3 + 4) (9.79)

= 3NkB (9.80)

In the low temperature limit we have T << ΘD. We start with,

CV = (3NkB)

[−3

(ΘD

T

)1

exp(ΘD/T )− 1+ 12

(T

ΘD

)3 ∫ ΘD/T

0

x3

exp(x) − 1dx

](9.81)

In the limit T → 0, the first term inside the square bracket goes to zero likeexp(−ΘD/T ). The upper limit of the integral in the second term inside thesquare bracket can be set to ∞. From standard integral tables9, we have,

∫ ∞

0

x3

exp(x) − 1dx =

π4

15(9.82)

Thus we have in the low temperature limit,

CV∼

T→0

12π4

5NkB

(T

ΘD

)3

(9.83)

Riemann Zeta Function

In an earlier class, we came across an integral,

∫ ∞

0

x3

exp(x) − 1dx (9.84)

9 The integral equals Γ (4)ζ(4), where Γ (·) is the gamma function and ζ(·) is theRiemann zeta function. Γ (4) = 3! = 6 and ζ(4) = π4/90. See e.g. G B Arfkenand H J Weber, Mathematical Methods for Physicists, Fourth Edition, AcademicPress, INC, Prism Books PVT LTD (1995).

Page 181: Kpn Stat Mech

9.7 Debye theory of Specific Heat 169

The value of the integral is π4/15.This is a particular case of a more general result based on Riemann zeta

function,

∫ ∞

0

xp

exp(x)− 1dx = Γ (p+ 1)ζ(p+ 1) (9.85)

where Γ (·) is the usual gamma function defined as

Γ (z) =

∫ ∞

0

xz−1 exp(−x)dx for Real(z) > 0, (9.86)

and ζ(·) is the Riemann zeta function, see below. Note ζ(2) = π2/6 andζ(4) = π4/90, etc..

Riemann zeta function is defined as

ζ(p) =

∞∑

n=1

n−p (9.87)

Take f(x) = x−p and then

∫ ∞

1

x−pdx =

x−p+1

−p+ 1

∣∣∣∣∞

1

for p 6= 1

lnx

∣∣∣∣∞

1

for p = 1

(9.88)

The integral and hence the series is divergent for p ≤ 1 and convergent forp > 1.

9.7.1 Bernoulli Numbers

Bernoulli numbers Bn are defined by the series

x

exp(x)− 1=

∞∑

n=0

xn

n!Bn (9.89)

which converges for |x| < 2π.By differentiating the power series repeatedly and then setting x = 0, we

obtain

Bn =

[dn

dxn

(x

exp(x) − 1

)]

x=0

. (9.90)

Page 182: Kpn Stat Mech

170 9 Statistical Mechanics of Harmonic Oscillators

9.3 Show that B0 = 1; B1 = −1/2; B2 = 1/6; B4 = −1/30;B6 = 1/42; B2n+1 = 0 ∀ n ≥ 1.

Euler showed that,

B2n = (−1)n−1 2(2n)!

(2π)2n

∞∑

p=1

p−2n, n = 1, 2, 3, · · · (9.91)

= (−1)n−1 2(2n)!

(2π)2nζ(2n), n = 1, 2, 3, · · · (9.92)

9.4 Employing the above relation between Bernoulli numbersand Riemann zeta function, show that

ζ(2) =π2

6ζ(4) =

π4

90(9.93)

ζ(6) =π6

945ζ(8) =

π8

9450(9.94)