1

CHEM-GA 2600: Statistical Mechanics

Professor M. E. Tuckerman

Office: 1166E Waverly

Phone: 8-8471

E-mail: mark.tuckerman@nyu.edu CHEM-GA 2600: Statistical Mechanics Fall, 2016, Location: 429 Waverly Professor M.

E. Tuckerman Office: 1166E Waverly Phone: 998-8471 E-mail:

mark.tuckerman@nyu.edu

Books The main text for the course is Statistical Mechanics: Theory and Molecular

Simulation by M. E. Tuckerman, published by Oxford University Press.

In addition, I have compiled a list of other excellent books on the subject (listed below),

which can provide alternative explanations and supplementary information. R. K. Pathria,

Statistical Mechanics. Oxford, New York D.A. McQuarrie, Statistical Thermodynamics

K. Huang, Statistical Mechanics, 2nd edition W. Greiner, L. Neise and H. St¨ocker,

Thermodynamics and Statistical Mechanics R.P. Feynman, Statistical Mechanics, A set

of lectures D. Chandler, Introduction to Modern Statistical Mechanics M. Kardar,

Statistical Physics of Particles When needed, we will cover some basic principles from

quantum mechanics, wherein, we will draw from the following references: R. Shankar,

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Principles of Quantum Mechanics R. P. Feynman and A. R. Hibbs, Quantum Mechanics

and Path Integrals For further reading and background to the mathematics used in this

course, I highly recommend: G. Arfken, Mathematical Methods for Physicists M. E.

Starzak, Mathematical Methods in Chemistry and Physics M. Boas, Mathematical

Methods in the Physical Sciences Material will also be drawn from current literature in

statistical mechanics.

1 Course Outline

I. Classical mechanics (Chapter 1, Sections 1-3, 6)

A. Newton’s laws of motion

B. Phase spaces and the Lagrangian and Hamiltonian formulations

II. Foundations of classical statistical mechanics (Chapter 2)

A. The ensemble concept

B. Liouville’s Theorem

C. The Liouville equation

D. Equilibrium solutions

III. The microcanonical ensemble (Chapter 3, Sections 1-10)

A. Basic thermodynamic relations

B. Phase space distribution and the partition function

C. Applications: Free particles and harmonic oscillators

D. Introduction to molecular dynamics calculations i. Integrators from Taylor expansions

ii. Integrators from the Liouville operator iii. Trajectories and the calculation of

observables: Ergodicity

IV. The canonical ensemble (Chapter 4, Sections 1-5, 8, 9, 11)

A. Basic thermodynamic relations

B. Phase space distribution and the partition function

C. Energy fluctuations

D. Applications: Free particles, harmonic oscillators, and the Gaussian random chain

model E. Canonical molecular dynamics i. Generalized Liouville theorem and non-

Hamiltonian algorithms ii. Nos´e-Hoover and Nos´e-Hoover chains iii. The isokinetic

ensemble and isokinetic molecular dynamics

V. The isothermal-isobaric ensemble (Chapter 5, Sections 1-10)

A. Basic thermodynamic relations

B. Phase space distribution and the partition function

C. Virial theorems and volume fluctuations

D. Applications: Free particles and periodic potentials

E. Isothermal-isobaric molecular dynamics i. Isotropic volume fluctuations ii.

Anisotropic cell fluctuations 2

VI. The grand canonical ensemble (Chapter 6 plus literature)

A. Basic thermodynamic relations

B. Phase space distribution and the partition function

C. Particle number fluctuations

D. Applications: Free particles and harmonic oscillators D. Adaptive resolution molecular

dynamics

VII. Distribution functions and liquid structure (Chapter 4, Sections 6, 7)

A. Spatial distribution functions, the radial distribution function, and the Virial equation

of state

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B. Diffraction experiments and radial distribution functions

C. Thermodynamics from radial distribution functions

D. Perturbation theory and the Van der Waals equation

VIII. Rare-event sampling and free energy calculations (Chapter 8, Sections 1-3, 6-11)

A. Free-energy perturbation theory

B. Adiabatic switching and thermodynic integration

C. Adiabatic free energy dynamics and metadynamics

D. Collective variables, reactions, and conformational changes

E. The “blue moon” ensemble

F. Umbrella sampling

G. Adiabatic free energy dynamics and metadynamics

IX. Review of the basic postulates of quantum mechanics (Chapter 9)

A. Hilbert space and state vectors

B. Operators, eigenvalues and observables

C. Measurement

D. Microscopic dynamics and the Schr¨odinger equation

E. Representations: Momentum and coordinate basis wave functions

F. The Heisenberg picture G. Spin and wave functions for identical particles.

X. Foundations of quantum statistical mechanics (Chapter 10)

A. Quantum ensembles and the density matrix

B. Properties of the density matrix

C. Time evolution of the density matrix and the quantum Liouville equation

D. Equilibrium solutions of the Liouville equation

E. Density matrix for the basic quantum ensembles 1. Canonical ensemble 2. Isothermal-

isobaric ensemble 3. Grand canonical ensemble 3

XI. Introduction to path integrals in quantum mechanics and quantum statistical

mechanics (Chapter 12)

A. Trotter theorem and derivation of the path integral for the canonical density matrix

B. Path integrals for the time-evolution operator

C. Continuous and discrete path integrals: 1. Functional integration 2. Most probable path

and derivation of classical mechanics

D. Thermodynamics from path integrals

E. Example applications: 1. The free particle 2. The harmonic oscillator

F. Numerical evaluation of path integrals

XII. Time-dependent processes: Classical case (Chapter 13, Sections 1-4)

A. Perturbative solution of the Liouville equation

B. Linear response, Green-Kubo theory, and transport properties

C. Classical time-correlation functions

XIII. Time-dependent processes: Quantum case (Chapter 14)

A. Theory of spectroscopy

B. Quantum linear response theory

C. Quantum time correlation functions i. Exact formulations in energy and path integral

representations ii. Imaginary time approximations

XIV. The Langevin and generalized Langevin equations (Chapter 15, Sections 1-5)

A. The harmonic bath model.

B. Definitions of friction and random forces.

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C. Exactly solvable cases.

D. Numerical integration and ensembles.

E. Modern uses. Grading basis Homework:............20%

Midterm:...............40%

Final:....................40%

Web resources Lecture material can be found on the course web page:

http://www.nyu.edu/classes/tuckerman/stat.mech

http://www.nyu.edu/classes/tuckerman/stat.mechII

MARK TUCKERMAN’S LECTURES ON STATISTICAL

MECHANICS

Lecture notes

Lecture 1 -- Classical microstates, Newtonian, Lagrangian and

Hamiltonian mechanics, ensemble concept.

Lecture 2 -- Liouville's Theorem, non-Hamiltonian systems, the

microcanonical ensemble

Lecture 3 -- Thermal equilibrium; the arrow of time.

Lecture 4 -- Classical virial theorem; Legendre transforms; the canonical

ensemble.

Lecture 5 -- Estimators, energy fluctuations, the isothermal-isobaric

ensemble

Lecture 6 -- The classical ideal gas

Lecture 7 -- The grand canonical ensemble

Lecture 8 -- Structure and distribution functions in classical liquids and

gases

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Lecture 9 -- Distribution functions and perturbation theory

Lecture 10 -- Review of the postulates of quantum mechanics

Lecture 11 -- Basic principles of quantum statistical mechanics

Lecture 12 -- The quantum ideal gases -- general formulation

Lecture 13 -- The ideal fermion gas

Lecture 14 -- The ideal boson gas

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Reeditor: Associate Prof., PhD Nguyen Quang Hoc

Finished day: 26th July, 2017

LECTURE 1: CLASSICAL MICROSTATES, NEWTONIAN,

LAGRANGIAN, HAMILTONIAN MECHANICS, ENSEMBLE

CONCEPT

Defining statistical mechanics: Statistical Mechanics provies the connection

between microscopic motion of individual atoms of matter and macroscopically

observable properties such as temperature, pressure, entropy, free energy, heat

capacity, chemical potential, viscosity, spectra, reaction rates, etc.

Why do we need Statistical Mechanics:

- Statistical Mechanics provides the microscopic basis for thermodynamics,

which, otherwise, is just a phenomenological theory.

- Microscopic basis allows calculation of a wide variety of properties not dealt

with in thermodynamics, such as structural properties, using distribution

functions, and dynamical properties - spectra, rate constants, etc., using time

correlation functions.

- Because a statistical mechanical formulation of a problem begins with a

detailed microscopic description, microscopic trajectories can, in principle and

in practice, be generated providing a window into the microscopic world. This

window often provides a means of connecting certain macroscopic properties

with particular modes of motion in the complex dance of the individual atoms

that compose a system, and this, in turn, allows for interpretation of

experimental data and an elucidation of the mechanisms of energy and mass

transfer in a system.

Content of Lecture 1

1.1.The microscopic laws of motion

1.2. The ensemble concept (heuristic definition)

1.3. The Lagrangian formulation of classical mechanics

1.4. The Hamiltonian formulation of classical mechanics

1.5. Phase space

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o Classical microscopic states or microstates and ensembles

1.6. Phase space distribution functions and Liouville's theorem

1.1.The microscopic laws of motion

Consider a system of N classical particles. The particles are confined to a

particular region of space by a ``container'' of volume V. The particles have a

finite kinetic energy and are therefore in constant motion, driven by the forces

they exert on each other (and any external forces which may be present). At a

given instant in time t, the Cartesian positions of the particles are 1( ),..., ( ),Nr t r t

The time evolution of the positions of the particles is then given by Newton's

second law of motion:

2

12,...i

i i i i N

d rm m r F r r

dt (1.1)

where 1,..., NF F are the forces on each of the N particles due to all the other

particles in the system. Newton's equations of motion constitute a set of

3N coupled second order differential equations. In order to solve these, it is

necessary to specify a set of appropriate initial conditions on the coordinates

and their first time derivaties 1 1(0),..., (0), (0),..., (0) .N Nr r r r Then, the solution of

Newton's equations gives the complete set of coordinates and velocities for all

time t.

1.2. The ensemble concept (heuristic

definition)

For a typical macroscopic system, the total number of particles 23~10 .N Since

an essentially infinite amount of precision is needed in order to specify the

initial conditions (due to exponentially rapid growth of errors in this

specification), the amount of information required to specify a trajectory is

essentially infinite. Even if we contented ourselves with quadrupole precision,

however, the amount of memory needed to hold just one phase space point

would be about 128 bytes = 7 22 ~10 bytes for each number or 2 23 1710 6 10 ~10 Gbytes. The largest computers we have today have perhaps 310 Gbytes of memory, so we are off by 14 orders of magnitude just to specify 1

point in phase space.

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Do we need all this detail? (Yes and No).

Yes - There are plenty of chemically interesting phenomena for which we

really would like to know how individual atoms are moving as a process

occurs. Experimental techniques such as ultrafast laser spectroscopy can

resolve short time scale phenomena and, thus, obtain important insights

into such motions. From a theoretical point of view, although we cannot

follow 2310 particles, there is some hope that we could follow the motion

of a system containing 410 or 510 particles, which might capture most of

the features of true macroscopic matter. Thus, by solving Newton's

equations of motion numerically on a computer, we have a kind of

window into the microscopic world. This is the basis of what are known

as molecular dynamics calculations.

No - Intuitively, we would expect that if we were to follow the evolution of a

large number of systems all described by the same set of forces but

having starting from different initial conditions, these systems would

have essentially the same macroscopic characteristics, e.g. the same

temperature, pressure, etc. even if the microscopic detailed evolution of

each system in time would be very different. This idea suggests that the

microscopic details are largely unimportant.

Since, from the point of view of macroscopic properties, precise microscopic

details are largely unimportant, we might imagine employing a construct

known as the ensemble concept in which a large number of systems with

different microscopic characteristics but similar macroscopic characteristics is

used to ``wash out'' the microscopic details via an averaging procecure. This is

an idea developed by individuals such as Gibbs, Maxwell, and Boltzmann.

Ensemble: Consider a large number of systems each described by the same set

of microscopic forces and sharing some common macroscopic property (e.g.

the same total energy). Each system is assumed to evolve under the

microscopic laws of motion from a different initial condition so that the time

evolution of each system will be different from all the others. Such a collection

of systems is called an ensemble. The ensemble concept then states that

macroscopic observables can be calculated by performing averages over the

systems in the ensemble. For many properties, such as temperature and

pressure, which are time-independent, the fact that the systems are evolving in

time will not affect their values, and we may perform averages at a particular

instant in time. Thus, let A denote a macroscopic property and let a denote a

microscopic function that is used to compute A. An example of A would be the

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temperature, and a would be the kinetic energy (a microscopic function of

velocities). Then, A is obtained by calculating the value of a in each system of

the ensemble and performing an average over all systems in the ensemble:

1

1A a (1.2)

where is the total number of members in the ensemble and a is the value

of a in the th system.

The questions that naturally arise are:

1. How do we construct an ensemble?

2. How do we perform averages over an ensemble?

3. How many systems will an ensemble contain?

4. How do we distinguish time-independent from time-dependent properties in

the ensemble picture?

Answering these questions will be our main objective in this course.

1.3. The Lagrangian formulation of

classical mechanics

In order to begin to make a connection between the microscopic and

macroscopic worlds, we need to better understand the microscopic world and

the laws that govern it. We will begin placing Newton's laws of motion in a

formal framework which will be heavily used in our study of classical

statistical mechanics.

First, we begin by restricting our discussion to systems for which the forces are

purely conservative. Such forces are derivable from a potential energy

function 1,..., NU r r by differentiation:

i

i

UF

r (1.3)

It is clear that such forces cannot contain dissipative or friction terms. An

important property of systems whose forces are conservative is that they

conserve the total energy

10

2

1

1

1,...,

2

N

i i N

i

E K U m r U r r (1.4)

To see this, simply differentiate the energy with respect to time:

1 1 1 1

0N N N N

i i i i i i i i

i i i ii

dE Um r r r r F F r

dt r (1.5)

where, the facts that /i i ir F m (Newton's law) and i

i

UF

r (conservative force

definition) have been used. This is known as the law of conservation of energy.

For conservative systems, there is an elegant formulation of classical

mechanics known as the Lagrangian formulation. The Lagrangian function, L,

for a system is defined to be the difference between the kinetic and potential

energies expressed as a function of positions and velocities. In order to make

the nomenclature more compact, we shall introduce a shorthand for the

complete set of positions in an N-particle system: 1,..., Nr r r and for the

velocities: 1,..., Nr r r . Then, the Lagrangian is defined as follows:

2

1

1

1, ,...,

2

N

i i N

i

L r r K U m r U r r (1.6)

In terms of the Lagrangian, the classical equations of motion are given by the

so called Euler-Lagrange equation:

0ii

d L L

dt rr (1.7)

The equations that result from application of the Euler-Lagrange equation to a

particular Lagrangian are known as the equations of motion. The solution of the

equations of motion for a given initial condition is known as a trajectory of the

system. The Euler-Lagrange equation results from what is known as an action

principle.

The Euler-Lagrange formulation is completely equivalent to Newton's second

law. In order to see this, note that

,i i i

ii i

L L Um r F

rr r (1.8)

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Therefore,

0i i i

ii

d L Lm r F

dt rr (1.9)

which is just Newton's equation of motion.

An important property of the Lagrangian formulation is that it can be used to

obtain the equations of motion of a system in any set of coordinates, not just

the standard Cartesian coordinates, via the Euler-Lagrange equation (see

problem set #1).

1.4. The Hamiltonian formulation of

classical mechanics

The Lagrangian formulation of mechanics will be useful when we study the

Feynman path integral. For our purposes now, the Lagrangian formulation is an

important springboard from which to develop another useful formulation of

classical mechanics known as the Hamiltonian formulation. The Hamiltonian

of a system is defined to be the sum of the kinetic and potential energies

expressed as a function of positions and their conjugate momenta. What are

conjugate momenta?

Recall from elementary physics that momentum of a particle, ip , is defined in

terms of its velocity ir by

i i ip m r (1.10)

In fact, the more general definition of conjugate momentum, valid for any set

of coordinates, is given in terms of the Lagrangian:

i

i

Lp

r (1.11)

Note that these two definitions are equivalent for Cartesian variables.

In terms of Cartesian momenta, the kinetic energy is given by

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2

1 2

Ni

i i

pK

m (1.12)

Then, the Hamiltonian, which is defined to be the sum, K+U, expressed as a

function of positions and momenta, will be given by

2

1

1

, ,...,2

Ni

N

i i

pH p r U r r

m (1.13)

Where 1,..., Np p p . In terms of the Hamiltonian, the equations of motion of a

system are given by Hamilton's equations:

,i i

i i

H Hr p

p r (1.14)

The solution of Hamilton's equations of motion will yield a trajectory in terms

of positions and momenta as functions of time. Again, Hamilton's equations

can be easily shown to be equivalent to Newton's equations, and, like the

Lagrangian formulation, Hamilton's equations can be used to determine the

equations of motion of a system in any set of coordinates.

The Hamiltonian and Lagrangian formulations possess an interesting

connection. The Hamiltonian can be directly obtained from the Lagrangian by a

transformation known as a Legendre transform. We will say more about

Legendre transforms in a later lecture. For now, note that the connection is

given by

1

, ,N

i i

i

H p r p r L r r (1.15)

which, when the fact that /i i ir p m is used, becomes

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1 1

1 1 1

1, ,..., ,...,

2 2

N N Ni i i

i i N N

i i ii i i

p p pH p r p m U r r U r r

m m m (1.16)

Because a system described by conservative forces conserves the total energy,

it follows that Hamilton's equations of motion conserve the total Hamiltonian.

Hamilton's equations of motion conserve the Hamiltonian

( ), ( ) (0), (0)H p t r t H p r E (1.17)

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Proof: const =0dH

Hdt

1 1

0N N

i i

i ii i i i i i

dH H H H H H Hr p

dt r p r p p r (1.18)

This, then, provides another expression of the law of conservation of energy.

1.5. Phase space

We construct a cartesian space in which each of the 6N coordinates and

momenta is assigned to one of 6N mutually orthogonal axes. Phase space is,

therefore, a 6N dimensional space. A point in this space is specified by giving a

particular set of values for the 6N coordinates and momenta. Denote such a

point by

1 1,..., , ,...,N Nx p p r r

(1.19)

x is a 6N dimensional vector. Thus, the time evolution or trajectory of a system

as specified by Hamilton's equations of motion, can be expressed by giving the

phase space vector x as a function of time.

The law of conservation of energy, expressed as a condition on the phase space

vector:

( ) = constH x t E (1.20)

defines a 6N-1 dimensional hypersurface in phase space on which the trajectory

must remain.

Classical microscopic states or microstates and ensembles

A microscopic state or microstate of a classical system is a specification of the

complete set of positions and momenta of the system at any given time. In the

language of phase space vectors, it is a specification of the complete phase

space vector of a system at any instant in time. For a conservative system, any

valid microstate must lie on the constant energy hypersurface H x E . Hence,

specifying a microstate of a classical system is equivalent to specifying a point

on the constant energy hypersurface.

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The concept of classical microstates now allows us to give a more formal

definition of an ensemble.

An ensemble is a collection of systems sharing one or more macroscopic

characteristics but each being in a unique microstate. The complete ensemble is

specified by giving all systems or microstates consistent with the common

macroscopic characteristics of the ensemble.

The idea of ensemble averaging can also be expressed in terms of an average

over all such microstates (which comprise the ensemble). A given macroscopic

property, A, and its microscopic function ( )a a x , which is a function of the

positions and momenta of a system, i.e. the phase space vector, are related by

ensemble

1

1A a a x (1.21)

where x is the microstate of the th member of the ensemble.

However, recall the original problem of determining the microscopic detailed

motion of each individual particle in a system. In reality, measurements are

made only on a single system and all the microscopic detailed motion is

present. However, what one observes is still an average, but it is an average

over time of the detailed motion, an average that also washes out the

microscopic details. Thus, the time average and the ensemble average should

be equivalent, i.e.

ensemble

0

1lim ( )

T

TA a dta x t

T (1.22)

This statement is known as the ergodic hypothesis. A system that is ergodic is

one for which, given an infinite amount of time, it will visit all possible

microscopic states available to it (for Hamiltonian dynamics, this means it will

visit all points on the constant energy hypersurface). No one has yet been able

to prove that a particular system is truly ergodic, hence the above statement

cannot be more than a supposition. However, it states that if a system is

ergodic, then the ensemble average of a property A x can be equated to a time

average of the property over an ergodic trajectory.

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1.6. Phase space distribution functions

and Liouville's theorem

Given an ensemble with many members, each member having a different phase

space vector x corresponding to a different microstate, we need a way of

describing how the phase space vectors of the members in the ensemble will be

distributed in the phase space. That is, if we choose to observe one particular

member in the ensemble, what is the probability that its phase space vector will

be in a small volume dx around a point x in the phase space at time t. This

probability will be denoted

,f x t dx (1.22)

where ,f x t is known as the phase space probability density or phase space

distribution function. It's properties are as follows:

, 0,f x t (1.23)

,dx f x t Number of members in the ensemble (1.24)

Liouville's Theorem: The total number of systems in the ensemble is a

constant. What restrictions does this place on ,f x t ? For a given volume in

phase space, this condition requires that the rate of decrease of the number of

systems from this region is equal to the flux of systems into the volume. Let

n be the unit normal vector to the surface of this region.

Figure 1

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The flux through the small surface area element, dS is just ,n xf x t dS . Then

the total flux out of volume is obtained by integrating this over the entire

surface that encloses :

, ,xdSn xf x t dx xf x t (1.25)

which follows from the divergence theorem. x is the 6N dimensional gradient

on the phase space

1 1

1 1

,..., , ,..., ,..., , ,...,N Nx p p r r

N Np p r r (1.26)

On the other hand, the rate of decrease in the number of systems out of the

volume is

, ,d

dxf x t dx f x tdt t

(1.27)

Equating these two quantities gives

, ,xdx xf x t dx f x tt

(1.28)

But this result must hold for any arbitrary choice of the volume , which we

may also allow to shrink to 0 so that the result holds locally, and we obtain the

local result:

, , 0xf x t xf x tt

(1.29)

But

, , ,x x xxf x t x f x t f x t x (1.30)

This equation resembles an equation for a ``hydrodynamic'' flow in the phase

space, with ,f x t playing the role of a density. The quantity x x , being the

divergence of a velocity field, is known as the phase space compressibility, and

it does not, for a general dynamical system, vanish. Let us see what the phase

space compressibility for a Hamiltonian system is:

17

1

i i

N

x p i r i

i

x p r (1.31)

However, by Hamilton's equations:

, i ii r i pp H r H (1.32)

Thus, the compressibility is given by

1

0i i i i

N

x p r r p

i

x H H (1.33)

Thus, Hamiltonian systems are incompressible in the phase space, and the

equation for ,f x t becomes

, , 0x

dff x t x f x t

t dt (1.34)

which is Liouville's equation, and it implies that ,f x t is a conserved quantity

when x is identified as the phase space vector of a particular Hamiltonian

system. That is, ,f x t will be conserved along a particular trajectory of a

Hamiltonian system. However, if we view x is a fixed spatial label in the phase

space, then the Liouville equation specifies how a phase space distribution

function ,f x t evolves in time from an initial distribution ,0f x .

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LECTURE 2: LIOUVILLE’S THEOREM AND THE

MICROCANONICAL ENSEMBLE

Content of Lecture 2:

2.1. The Liouville operator and the Poisson bracket

2.2. Preservation of phase space volume and Liouville's theorem

2.3. Equilibrium ensembles

2.1. The Liouville operator and the

Poisson bracket

From the last lecture, we saw that Liouville's equation could be cast in the form

0x

fxf

t (2.1)

The Liouville equation is the foundation on which statistical mechanics rests.

Define an operator

ˆxiL x (2.2)

known as the Liouville operator. 1i , the i is there as a matter of convention

and has the effect of making L̂ a Hermitian operator. Then Liouville's equation

can be written

ˆ 0f

iLft

(2.3)

The Liouville operator also be expressed as

1

ˆ ...,N

i i i i i

H HiL H

p r r p (2.4)

where ,A B is known as the Poisson bracket between A x and B x :

19

1

,N

i i i i i

A B A BA B

r p p r (2.5)

Thus, the Liouville equation can be written as

, 0f

f Ht

(2.6)

The Liouville equation is a partial differential equation for the phase space

probability distribution function. Thus, it specifies a general class of

functions ,f x t that satisfy it. In order to obtain a specific solution requires

more input information, such as an initial condition on f, a boundary condition

on f, and other control variables that characterize the ensemble.

2.2. Preservation of phase space volume

and Liouville's theorem

Consider a phase space volume element 0dx at t = 0, containing a small

collection of initial conditions on a set of trajectories. The trajectories evolve in

time according to Hamilton's equations of motion, and at a time t later will be

located in a new volume element tdx as shown in the figure below:

Figure 2

How is 0dx related to tdx ? To answer this, consider a trajectory starting from a

phase space vector 0x in 0dx and having a phase space vector tx at

time t in tdx . Since the solution of Hamilton's equations depends on the choice

of initial conditions, tx depends on 0x :

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0 1 1

1 1

1

0 0

p (0),..., p (0), r (0),..., r (0) ,

p (t),..., p (t), r (t),..., r (t) ,

x x ,..., .

N N

t N N

i i n

t t

x

x

x x

(2.7)

Thus, the phase space vector components can be viewed as a coordinate

transformation on the phase space from t=0 to time t. The phase space volume

element then transforms according to

0 0;t tdx J x x dx (2.8)

where 0;tJ x x is the Jacobian of the transformation:

1

0 1

0 0

...;

...

n

t t

t n

x xJ x x

x x (2.9)

where n=6N. The precise form of the Jacobian can be determined as will be

demonstrated below.

The Jacobian is the determinant of a matrix M

Tr ln

0; dettJ x x e (2.10)

whose matrix elements are

0

i

tij j

xM

x (2.11)

Taking the time derivative of the Jacobian, we therefore have

1 Trln 1

1 1

Trn n

ji

ij

i j

ddJ de J

dt dt dt (2.12)

The matrices 1 and d

dt can be seen to be given by

1 0

0

,i j

ji tij j i

t

dx xM

x dt x (2.13)

21

Substituting into the expression for dJ

dt gives

0 0

, 1 , , 10 0

i j i j kn nt t t

j i j k ii j i j kt t t

x x x x xdJJ J

dt x x x x x (2.14)

where the chain rule has been introduced for the derivative 0

j

t

i

x

x. The sum

over i can now be performed:

1 10

1 1 10

i kn n nt

ij ki ki ij kjj ii i it

x x

x x (2.15)

Thus,

, 1 1

j jn nt t

jk xk jj k jt t

x xdJJ J J x

dt x x (2.16)

The initial condition on this differential equation is 0 0(0) ; 1.J J x x

Moreover, for a Hamiltonian system 0.x x This says that dJ

dt0 and J(0)=1.

Thus, 0; 1.tJ x x If this is true, then the phase space volume element

transforms according to

0 tdx dx (2.17)

which is another conservation law. This conservation law states that the phase

space volume occupied by a collection of systems evolving according to

Hamilton's equations of motion will be preserved in time. This is one statement

of Liouville's theorem.

Combining this with the fact that 0df

dt, we have a conservation law for the

phase space probability:

0 0,0 ,t tf x dx f x t dx (2.18)

which is an equivalent statement of Liouville's theorem.

22

2.3. Equilibrium ensembles

An equilibrium ensemble is one for which there is no explicit time-dependence

in the phase space distribution function, 0f

t. In this case, Liouville's

equation reduces to

, 0f H (2.19)

which implies that f x must be a pure function of the Hamiltonian

f x F H x (2.20)

The specific form that F H x has depends on the specific details of the

ensemble.

The integral over the phase space distribution function plays a special role in

statistical mechanics:

dxF H x (2.21)

It is known as the partition function and is equal to the number of members in

the ensemble. That is, it is equal to the number of microstates that all give rise

to a given set of macroscopic observables. Thus, it is the quantity from which

all thermodynamic properties are derived.

If a measurement of a macroscopic observable A x is made, then the value

obtained will be the ensemble average:

dxA x F H x

AdxF H x

(2.22)

Eqs. (2.21) and (2.22) are the central results of ensemble theory, since they

determine all thermodynamic and other observable quantities.

23

LECTURE 3: MICROCANONICAL ENSEMBLE AND THERMAL

EQUILIBRIUM

Content of Lecture 3:

3.1. Introduction to the Microcanonical Ensemble

3.2. Microcanonical ensemble: conditions for thermal equilibrium

3.1. Introduction to the Microcanonical

Ensemble

The microcanonical ensemble is built upon the so called postulate of equal a

priori probabilities:

24

3.1.1. Postulate of equal a priori probabilities: For an isolated macroscopic

system in equilibrium, all microscopic states corresponding to the same set of

macroscopic observables are equally probable.

3.1.2. Basic definitions and thermodynamics

Consider a thought experiment in which N particles are placed in a container of

volume V and allowed to evolve according to Hamilton's equations of motion.

The total energy E H x is conserved. Moreover, the number of

particles N and volume V are considered to be fixed. This constitutes a set of

three thermodynamic variables N, V, E that characterize the ensemble and can

be varied to alter the conditions of the experiment.

The evolution of this system in time generates a trajectory that samples the

constant energy hypersurface H x E . All points on this surface correspond

to the same set of macroscopic observables. Thus, by the postulate of equal a

priori probabilities, the corresponding ensemble, called the microcanonical

ensemble, should have a distribution function F H x that reflects the fact

that all points on the constant energy hypersurface are equally probable. Such a

distribution function need only reflect the fact that energy is conserved and can

be written as

F H x H x E (3.1)

where x is the Dirac delta function. The delta function has the property that

x a f x dx f a (3.2)

for any function f x .

Averaging over the microcanonical distribution function is equivalent to

computing the time average in our thought experiment. The microcanonical

partition function , ,N V E is given by

, , NN V E C dx H x E (3.3)

In Cartesian coordinates, this is equivalent to

25

( ), , ,N N

ND V

N V E C d p d r H p r E

(3.4)

Where NC is a constant of proportionality. It is given by

0

3!N N

EC

N h (3.5)

Here h is a constant with units Energy Time, and 0E is a constant having units

of energy. The extra factor of 0E is needed because the function has units of

inverse energy. Such a constant has no effect at all on any properties. Thus,

, ,N V E is dimensionless. The origin of NC is quantum mechanical in nature

(h turns out to be Planck's constant) and must be put into the classical

expression by hand. Later, we will explore the effects of this constant on

thermodynamic properties of the ideal gas.

The microcanonical partition function measures the number of microstates

available to a system which evolves on the constant energy hypersurface.

Boltzmann identified this quantity as the entropy, S of the system, which, for

the microcanonical ensemble is a natural function of N, V and E:

, ,S S N V E (3.6)

Thus, Boltzmann's relation between , ,N V E , the number of microstates

and S(N,V,E) is

, , ln , ,S N V E k N V E (3.7)

where k is the Boltzmann's constant. The importance of Boltzmann's relation is

that it establishes a connection between the thermodynamic properties of a

system and its microscopic details.

Recall the standard thermodynamic definition of entropy:

dQ

ST

(3.8)

where an amount of heat dQ is assumed to be absorbed reversibly, i.e., along a

thermodynamic path, by the system. The first law of thermodynamics states

that the energy E of the system is given by the sum of the heat absorbed by the

system and the work done on the system in a thermodynamic process:

26

E Q W (3.9)

If the thermodynamic transformation of the system is carried reversibly, i.e.,

along a thermodynamic path, then the first law will be valid for

the differential change in energy dE due to absorption of a differential amount

of heat dQ and a differential amount of work dW done on the system:

dE dQ dW (3.10)

The work done on the system can be in the form of compression/expansion

work at constant pressure P, leading to a change dV in the volume and/or the

insertion/deletion of particles from the system at constant chemical potential ,

leading to a change dN in the particle number. Thus, in general

dW PdV dN (3.11)

The above relation for the work is true only for a one-component system. If

there are M types of particles present, then the second term must be generalized

according to 1

M

k k

k

dN . Then, using the fact that dQ = TdS, we have

dE TdS PdV dN (3.12)

or

dE P

dS dV dNT T T

(3.13)

But since S = S(N,V,E) is a natural function of N, V, and E, the

differential, dS is also given by

, , ,N V N E E V

S S SdS dE dV dN

E V N (3.14)

Comparing these two expressions, we see that

,

1,

N V

S

E T (3.15)

,

,N E

S P

V T (3.16)

27

,E V

S

N T (3.17)

Finally, using Boltzmann's relation between the entropy S and the partition

function , we obtain a prescription for obtaining the thermodynamic

properties of the system starting from a microscopic, particle-based description

of the system:

,

1 ln,

N V

kT E

(3.18)

,

ln,

N E

Pk

T V (3.19)

,

ln

E V

kT N

(3.20)

Of course, the ultimate test of Boltzmann's relation between entropy and the

partition function is that the above relations correctly generate the known

thermodynamic properties of a given system, e.g. the equation of state. We will

soon see several examples in which this is, indeed, the case.

3.2. Microcanonical ensemble: conditions

for thermal equilibrium

Consider bringing two systems into thermal contact. By thermal contact, we

mean that the systems can only exchange heat. Thus, they do not exchange

particles, and there is no potential coupling between the systems. In this case, if

system 1 has a phase space vector 1x and system 2 has a phase space vector 2x ,

then the total Hamiltonian can be written as

1 1 2 2H x H x H x (3.21)

Furthermore, let system 1 have 1N particles in a volume 1V and system 2

have 2N particles in a volume 2V . The total particle number N and

volume V are 1 2N N N and 1 2V V V . The entropy of each system is given

by

28

1 1 1 1 1 1 1 1, , ln , , ,S N V E k N V E (3.22)

2 2 2 2 2 2 2 2, , ln , ,S N V E k N V E (3.23)

The partition functions are given by

11 1 1 1 1 1 1 1, , ,NN V E C dx H x E (3.24)

22 2 2 2 2 2 2 2, , ,NN V E C dx H x E (3.25)

1 1 2 2 1 1 1 1 2 2 2 2, , , , , ,NN V E C dx H x H x E N V E N V E (3.26)

However, it can be shown that the total partition function can be written as

1 1 1 2 1

0

E

E C dE E E E (3.27)

where C is an overall constant independent of the energy. Note that the

dependence of the partition functions on the volume and particle number has

been suppressed for clarity.

Now imagine expressing the integral over energies in the above expression as a

Riemann sum:

( ) ( )

1 1 2 1

1

Pi i

i

E C E E E (3.28)

where is the small energy interval (which we will allow to go to 0) and

/P E . The reason for writing the integral this way is to make use of a

powerful theorem on sums with large numbers of terms.

Consider a sum of the form

1

P

i

i

a (3.29)

Where 0ia for all ia . Let maxa be the largest of all the ia 's. Clearly, then

max

1

,P

i

i

a a (3.30)

29

max

1

P

i

i

Pa a (3.31)

Thus, we have the inequality

max maxa Pa (3.32)

Or

max maxln ln ln lna a P (3.33)

This gives upper and lower bounds on the value of ln a . Now suppose that

maxln lna P . Then the above inequality implies that

maxln ln a (3.34)

This would be the case, for example, if max ~ Pa e . In this case, the value of the

sum is given to a very good approximation by the value of its maximum term.

Why should this theorem apply to the sum expression for ( )E ? Consider the

case of a system of free particles 2

1,

2

N i

ii

pH

m i.e., no potential. Then the

expression for the partition function is

2

( )1

( ) ~ ~2

NN N Ni

D Vi i

pE d r d p E V

m (3.35)

since the particle integrations are restricted only the volume of the container.

Thus, the terms in the sum vary exponentially with N. But the number of terms

in the sum P also varies like N since /P E and ~E N since E is extensive.

Thus, the terms in the sum under consideration obey the conditions for the

application of the theorem.

Let the maximum term in the sum be characterized by energies

1E and 2 1E E E . Then, according to the above analysis,

1 1 2 1( ) ln ( ) ln ln ln lnS E k E k k E E E k P k C (3.36)

30

Since / , ln ln lnP E P E . But ~E N , while 1ln ~ N . Since lnN N ,

the above expression becomes, to a good approximation

1 1 2 1( ) ln ln constS E k E E E N (3.37)

where ln N is a small infinity of ln .N Thus, apart from constants, the

entropy is approximately additive:

1 1 2 2 1 1 2 2( ) ln ln ln constS E k E k E S E S E N (3.38)

Finally, in order to compute the temperature of each system, we make a small

variation in the energy 1 1,E dE . But since

1 2 1 2, .E E E dE dE Also, this

variation is made such that the total entropy S and energy E remain constant.

Thus, we obtain 1 2

1 1

0 ,dS dS

dE dE 1 2

1 2

0dS dS

dE dE and

1 2

1 10

T Tfrom which it is

clear that 1 2T T , the expected condition for thermal equilibrium.

It is important to point out that the entropy S(N,V,E) defined via the

microcanonical partition function is not the only entropy that satisfies the

properties of additivity and equality of temperatures at thermal equilibrium.

Consider an ensemble defined by the condition that the Hamiltonian H x is

less than a certain energy E. This is known as the uniform ensemble and its

partition function, denoted , ,N V E is defined by

, ,H x E

N V E C dx C dx E H x (3.39)

where ( )x is the Heaviside step function. Clearly, it is related to the

microcanonical partition function by

, , , ,N V E N V EE

(3.40)

Although we will not prove it, the entropy , ,S N V E defined from the uniform

ensemble partition function via

, , ln , ,S N V E k N V E (3.41)

31

is also approximately additive and will yield the condition 1 2T T for two

systems in thermal contact. In fact, it differs from S(N,V,E) by a constant of

order so that one can also define the thermodynamics in terms of

, ,S N V E . In particular, the temperature is given by

,,

1 ln

N VN V

Sk

T E E (3.42)

LECTURE 4: CLASSICAL VIRIAL THEOREM, LEGENDRE

TRANSFORMS, THE CANONICAL RESEMBLE

Content of Lecture 4:

4.1. The classical virial theorem (microcanonical derivation)

4.2. Legendre transforms

4.3. The canonical ensemble

o Basic Thermodynamics

o The partition function

o Relation between canonical and microcanonical ensembles

o Classical Virial Theorem (canonical ensemble derivation)

4.1. The classical virial theorem

(microcanonical derivation)

32

Consider a system with Hamiltonian H x . Let ix and jx be specific

components of the phase space vector. The classical virial theorem states that

i ij

j

Hx kT

x (4.1)

where the average is taken with respect to a microcanonical ensemble.

To prove the theorem, start with the definition of the average:

( )

i i

j j

H C Hx dxx E H x

x E x (4.2)

where the fact that ( ) ( )x x has been used. Also, the N and V dependence of

the partition function have been suppressed. Note that the above average can be

written as

( )

i i

j j

H C Hx dxx E H x

x E E x

( )

i i

j jH x E

H C Hx dxx

x E E x

( )

i

jH x E

H ECdxx

E E x (4.3)

However, writing

i i ij

j j

H Ex x H E H E

x x (4.4)

allows the average to be expressed as

( )

i i ij

j jH x E

H Cx dx x H E H E

x E E x

33

( )

i j ij

H E H x E

Cx H E dS dx E H x

E E (4.5)

The first integral in the brackets is obtained by integrating the total derivative

with respect to jx over the phase space variable

jx . This leaves an integral that

must be performed over all other variables at the boundary of phase space

where H = E, as indicated by the surface element jdS . But the integrand

involves the factor H - E, so this integral will vanish. This leaves:

( )( ) ( ) ( )

ij ij ij

i

j H x E H x E

C CHx dx E H x dx E

x E E E E (4.6)

Where ( )E is the partition function of the uniform ensemble. Recalling that

( ) ( )E EE

(4.7)

we have

( ) 1 1

( ) ln ( )i ij ij ij ij

j

H Ex k kT

E Ex S

E E E

(4.8)

which proves the theorem.

Example: i ix p and i = j. The virial theorem says that

,i

i

Hp kT

p (4.9)

2

,i

i

pkT

m (4.10)

2

2 2

i

i

p kT

m (4.11)

Thus, at equilibrium, the kinetic energy of each particle must be kT/2. By

summing both sides over all the particles, we obtain a well know result

34

23 3

2

1 1

1 3

2 2 2

N Ni

i i

i ii

pm v NkT

m (4.12)

4.2. Legendre transforms

The microcanonical ensemble involved the thermodynamic

variables N, V and E as its variables. However, it is often convenient and

desirable to work with other thermodynamic variables as the control variables.

Legendre transforms provide a means by which one can determine how the

energy functions for different sets of thermodynamic variables are related. The

general theory is given below for functions of a single variable.

Consider a function f(x) and its derivative

( ) ( ).df

y f x g xdx

(4.13)

The equation y = g(x) defines a variable transformation from x to y. Is there a

unique description of the function f(x) in terms of the variable y? That is, does

there exist a function ( )y that is equivalent to f(x)?

Given a point 0x , can one determine the value of the function 0f x given only

0f x ? No, for the reason that the function 0f x c for any constant c will

have the same value of 0f x as shown in the figure below.

35

Figure 4.1

However, the value 0f x can be determined uniquely if we specify the slope

of the line tangent to f at 0x , i.e., 0f x and the y-intercept, 0b x of this line.

Then, using the equation for the line, we have

0 0 0 0f x x f x b x (4.14)

This relation must hold for any general x:

f x xf x b x (4.15)

Note that f x is the variable y, and 1( )x g y , where 1g is the functional

inverse of g, i.e., 1g g x x . Solving for 1( )b x b g y gives

1 1 1 ( )b g y f g y yg y y (4.16)

where ( )y is known as the Legendre transform of f(x). In shorthand notation,

one writes

( ) ( )y f x xy (4.17)

36

however, it must be kept in mind that x is a function of y.

4.3. The canonical ensemble

Content of section 4.3

4.3.1. Basic Thermodynamics

4.3.2. The partition function

4.3.3. Relation between canonical and microcanonical ensembles

4.3.4. Classical Virial Theorem (canonical ensemble derivation)

4.3.1. Basic Thermodynamics

In the microcanonical ensemble, the entropy S is a natural function of N,

V and E, i.e., S = S(N,V,E). This can be inverted to give the energy as a

function of N, V, and S, i.e., E = E(N,V,S). Consider using Legendre

transformation to change from S to T using the fact that

,N V

ET

S (4.18)

The Legendre transform E of E(N,V,S) is

, , , ,S( ) , ,S( )E

E N V T E N V T S E N V T TSS

(4.19)

The quantity , ,E N V T is called the Hemlholtz free energy and is given the

symbol A(N,V,T). It is the fundamental energy in the canonical ensemble.

The differential of A is

, , ,V N T N V T

A A AdA dT dV dN

T V N (4.20)

However, from A = E - TS, we have

37

dA dE TdS SdT (4.21)

From the first law, dE is given by

dE TdS PdV dN (4.22)

Thus,

dA PdV SdT dN (4.23)

Comparing the two expressions, we see that the thermodynamic relations are

,V N

AS

T, (4.24)

,

,T N

AP

V (4.25)

,V T

A

N (4.26)

4.3.2. The partition function

Consider two systems 1 and 2 in thermal contact such that

2 1 2 1 1 2 1 2, ; ; ;N N E E N N N E E E

1 2dim dimx x (4.27)

and the total Hamiltonian is just 1 1 2 2H x H x H x

Since system 2 is infinitely large compared to system 1, it acts as an infinite

heat reservoir that keeps system 1 at a constant temperature T without gaining

or losing an appreciable amount of heat, itself. Thus, system 1 is maintained at

canonical conditions, N, V, T.

The full partition function , ,N V E for the combined system is the

microcanonical partition function

1 2 1 1 2 2, ,N V E dx H x E dx dx H x H x E (4.28)

38

Now, we define the distribution function, 1f x of the phase space variables of

system 1 as

1 2 1 1 2 2f x dx H x H x E (4.29)

Taking the natural log of both sides, we have

1 2 1 1 2 2ln lnf x dx H x H x E (4.30)

Since 2 1E E , it follows that 2 2 1 1H x H x , and we may expand the

above expression about 1 0H . To linear order, the expression becomes

1 11 2 2 2 1 1 2 1 1 2 2

01 1

ln ln lnH x

f x dx H x E H x dx H x H x EH x

2 2 2 1 1 2 2 2ln lndx H x E H x dx H x E

E (4.31)

where, in the last line, the differentiation with respect to 1H is replaced by

differentiation with respect to E. Note that

22 2 2

( )ln ,

S Edx H x E

k

22 2 2

( ) 1ln

S Edx H x E

E E k kT (4.32)

where T is the common temperature of the two systems. Using these two facts,

we obtain

2 1 1

1ln ,S E H x

f xk kT

2 1 1

1 exp expS E H x

f xk kT

(4.33)

Thus, the distribution function of the canonical ensemble is

expH x

f xkT

(4.34)

39

The prefactor 2exp

S E

k is an irrelevant constant that can be disregarded as it

will not affect any physical properties.

The normalization of the distribution function is the integral:

exp , ,H x

dx N V TkT

(4.35)

where , ,N V T is the canonical partition function. It is convenient to define

an inverse temperature 1

kT. , ,N V T is the canonical partition function.

As in the microcanonical case, we add in the ad hoc quantum corrections to the

classical result to give

3

1, , exp

! NN V T dx H x

N h (4.36)

The thermodynamic relations are thus,

Hemlholtz free energy:

1

, , ln , ,A N V T N V T (4.37)

To see that this must be the definition of A(N,V,T), recall the definition

of A:

A E TS H x TS (4.38)

But we saw that

,N V

AS

T (4.39)

Substituting this in gives

A

A H x TT

(4.40)

or, noting that

40

2

1A A A

T T kT (4.41)

it follows that

A

A H x (4.42)

This is a simple differential equation that can be solved for A. We will

show that the solution is

1

ln ( )A (4.43)

Note that

1 1

lnA

A H x (4.44)

Substituting in gives, therefore

A H x A H x A (4.45)

so this form of A satisfies the differential equation.

Other thermodynamics follow:

Average energy:

1

ln , ,H x

NE H x C dxH x e N V T (4.46)

Pressure:

, ,

ln , ,

N T N T

N V TAP kT

V V (4.47)

Entropy:

2

1A A AS

T T kT

2 1 lnln , , lnk N V T k k

ln lnE

k E k kT

(4.48)

41

Heat capacity at constant volume:

2

2

,

, ,V

N V

E EC k N V T

T T (4.49)

4.3.3. Relation between canonical and microcanonical ensembles

We saw that the E(N,V,S) and A(N,V,T) could be related by a Legendre

transformation. The partition functions , ,N V E and , ,N V T can be

related by a Laplace transform. Recall that the Laplace transform ( )f of a

function f(x) is given by

0

( ) ( )xf dxe f x (4.50)

Let us compute the Laplace transform of , ,N V E with respect to E:

0

, , E

NN V C dEe dx H x E (4.51)

Using the -function to do the integral over E:

, ,H x

NN V C dxe (4.52)

By identifying , we see that the Laplace transform of the microcanonical

partition function gives the canonical partition function , ,N V T .

4.3.4. Classical Virial Theorem (canonical ensemble derivation)

Again, let ix and jx be specific components of the phase space vector

1 3 1 3,..., , ,...,N Nx p p q q . Consider the canonical average

i

j

Hx

x (4.53)

given by

1 1 1H x H x

i N i N i

j j j

H Hx C dxx e C dxx e

x x x (4.54)

42

But

H x H x H x H x H xi

i i i ij

j j j j

xx e x e e x e e

x x x x (4.55)

Thus,

1 H x ij H x

i N i N

j j

Hx C dx x e C dxe

x x

1 H x

N j i ij

j

C dx dx x e kTx

j

H x

i ijx

dx x e kT (4.56)

Several cases exist for the surface termH x

ix e :

1. i ix p a momentum variable. Then, since 2~ iH p , He evaluated

at ip clearly vanishes.

2. ~i ix q and U as iq , thus representing a bound system. Then, He also vanishes at iq .

3. ~i ix q and 0U as iq , representing an unbound system. Then the

exponential tends to 1 both at iq , hence the surface term vanishes.

4. i ix q and the system is periodic, as in a solid. Then, the system will be

represented by some supercell to which periodic boundary conditions can be

applied, and the coordinates will take on the same value at the boundaries.

Thus, H and He will take on the same value at the boundaries and the surface

term will vanish.

5. i ix q , and the particles experience elastic collisions with the walls of the

container. Then there is an infinite potential at the walls so that U at the

boundary and He at the boundary.

Thus, we have the result

i ij

j

Hx kT

x (4.57)

43

The above cases cover many but not all situations, in particular, the case of a

system confined within a volume V with reflecting boundaries. Then, surface

contributions actually give rise to an observable pressure (to be discussed in

more detail in the next lecture).

LECTURE 5: ESTIMATORS, ENERGY FLUCTUATIONS, THE

ISOTHERMAL-ISOBARIC ENSEMBLE

5.1. Temperature and pressure estimators

5.2. Energy fluctuations in the canonical ensemble

5.3. Isothermal-Isobaric ensemble

o Basic Thermodynamics

o The partition function and relation to thermodynamics

o Pressure and work virial theorems

5.1. Temperature and pressure estimators

44

From the classical virial theorem

i ij

j

Hx kT

x (5.1)

we arrived at the equipartition theorem:

2

1

3

2 2

Ni

i i

pNkT

m (5.2)

where 1,..., Np p are the N Cartesian momenta of the N particles in a system.

This says that the microscopic function of the N momenta that corresponds to

the temperature, a macroscopic observable of the system, is given by

2

1

1

,...,2

Ni

N

i i

pK p p

m (5.3)

The ensemble average of K can be related directly to the temperature

1 1 3

2 2,..., ,...,

3 3NT K p p K p p

Nk nR (5.4)

1,..., NK p p is known as an estimator (a term taken over from the Monte Carlo

literature) for the temperature. An estimator is some function of the phase space

coordinates, i.e., a function of microscopic states, whose ensemble average

gives rise to a physical observable.

An estimator for the pressure can be derived as well, starting from the basic

thermodynamic relation:

, ,

ln , ,

N T N T

N V TAP kT

V V (5.5)

With

,, ,

H x H p rN N

N N

V

N V T C dxe C d p d re (5.6)

The volume dependence of the partition function is contained in the limits of

integration, since the range of integration for the coordinates is determined by

45

the size of the physical container. For example, if the system is confined within

a cubic box of volume 3V L , with L the length of a side, then the range of

each q integration will be from 0 to L. If a change of variables is made to

/i is q L , then the range of each s integration will be from 0 to 1. The

coordinates is are known as scaled coordinates. For containers of a more

general shape, a more general transformation is

1/3

i is V r (5.7)

In order to preserve the phase space volume element, however, we need to

ensure that the transformation is a canonical one. Thus, the corresponding

momentum transformation is

1/3

i iV p (5.8)

With this coordinate/momentum transformation, the phase space volume

element transforms as

N N N Nd pd r d d s (5.9)

Thus, the volume element remains the same as required. With this

transformation, the Hamiltonian becomes

2 2/3 2

1/3 1/3

1 1

1 1

,..., ,...,2 2

N Ni i

N N

i ii i

p VH U r r U V s V s

m m (5.10)

and the canonical partition function becomes

2/3 2

1/3 1/3

1

1

, , exp ,...,2

NN N i

N N

i i

VN V T C d d s U V s V s

m (5.11)

Thus, the pressure can now be calculated by explicit differentiation with respect

to the volume, V:

2

5/3 2/3

1/31 1

1 2

3 2 3

N NN N Hi

N i

i ii i

kT UP kT C d d s V V s e

V m V s

2

,

1 13 2 3

N NH p rN N i

N i

i ii i

pkT UC d p d r r e

V m V r

46

2

1

1

3 2

Ni

i i

i i

p Hr F

V m V (5.12)

Thus, the pressure estimator is

2

1 1

1

1,..., , ,...,

3 2

Ni

N N i i

i i

pp p r r x r F

V m (5.13)

and the pressure is given by

P x (5.14)

For periodic systems, such as solids and currently used models of liquids, an

absolute Cartesian coordinate iq is ill-defined. Thus, the virial part of the

pressure estimator i ii

q F must be rewritten in a form appropriate for periodic

systems. This can be done by recognizing that the force iF is obtained as a sum

of contributions ijF , which is the force on particle i due to particle j. Then, the

classical virial becomes

1 1

1 1

2 2

N N

i i i ij i ij j ji i ij j ij

i i j i i j i j i j i j i j i j

r F r F r F r F r F r F

, , , ,

1 1

2 2i j ij ij ij

i j i j i j i j

r r F r F (5.15)

where ijr is now a relative coordinate.

ijr must be computed consistent with

periodic boundary conditions, i.e., the relative coordinate is defined with

respect to the closest periodic image of particle j with respect to particle i. This

gives rise to surface contributions, which lead to a nonzero pressure, as

expected.

5.2. Energy fluctuations in the canonical

ensemble

47

In the canonical ensemble, the total energy is not conserved. ( constH x ).

What are the fluctuations in the energy? The energy fluctuations are given by

the root mean square deviation of the Hamiltonian from its average H :

2 22E H H H H (5.16)

where

ln , , ,H N V T (5.17)

2 2

2 2

2 2

1H x H xN NC CH dxH x e dx e

2 2 22 2 2

2 2 2 2

1 1 lnln ln ln (5.18)

Therefore

2

22

2lnH H (5.19)

But

2

2

2ln VkT C (5.20)

Thus,

2

VE kT C (5.21)

Therefore, the relative energy fluctuation is given by

2

VkT CE

E E (5.22)

Now consider what happens when the system is taken to be very large. In fact,

we will define a formal limit called the thermodynamic limit, in which

N and V such that N/V remains constant.

48

Since VC and E are both extensive variables, ~VC N and ~E N ,

1

~ 0E

E N khi N (5.23)

But /E E would be exactly 0 in the microcanonical ensemble. Thus, in the

thermodynamic limit, the canonical and microcanonical ensembles are

equivalent, since the energy fluctuations become vanishingly small.

5.3. Isothermal-Isobaric ensemble

Content of Section 5.3

5.3.1. Basic Thermodynamics

5.3.2. The partition function and relation to thermodynamics

5.3.3. Pressure and work virial theorems

5.3.1. Basic Thermodynamics

The Helmholtz free energy A(N,V,T) is a natural function of N, V and T. The

isothermal-isobaric ensemble is generated by transforming the volume V in

favor of the pressure P so that the natural variables are N, P and T (which are

conditions under which many experiments are performed – “standard

temperature and pressure,'' for example).

Performing a Legendre transformation of the Helmholtz free energy

, , , ( ), ( )A

A N P T A N V P T V PV

(5.24)

But

A

PV

(5.25)

Thus,

, , , ( ), , ,A N P T A N V P T PV G N P T (5.26)

49

where G(N,P,T) is the Gibbs free energy.

The differential of G is

, , ,N T N P P T

G G GdG dP dT dN

P T N (5.27)

But from G = A + PV, we have

dG dA PdV VdP (5.28)

But dA SdT PdV dN , thus

dG SdT VdP dN (5.29)

Equating the two expressions for dG, we see that

,

,N T

GV

P (5.30)

,

,N P

GS

T (5.31)

,P T

G

N (5.32)

5.3.2. The partition function and relation to thermodynamics

In principle, we should derive the isothermal-isobaric partition function by

coupling our system to an infinite thermal reservoir as was done for the

canonical ensemble and also subject the system to the action of a movable

piston under the influence of an external pressure P. In this case, both the

temperature of the system and its pressure will be controlled, and the energy

and volume will fluctuate accordingly.

However, we saw that the transformation from E to T between the

microcanonical and canonical ensembles turned into a Laplace transform

relation between the partition functions. The same result holds for the

transformation from V to T. The relevant ``energy'' quantity to transform is the

work done by the system against the external pressure P in changing its volume

from V = 0 to V, which will be PV. Thus, the isothermal-isobaric partition

50

function can be expressed in terms of the canonical partition function by the

Laplace transform:

0 0

1, , , ,PVN P T dVe N V T

V (5.33)

where 0V is a constant that has units of volume. Thus,

3

0 0

1, ,

!

H x PV

NN P T dV dxe

V N h (5.34)

The Gibbs free energy is related to the partition function by

1

, , ln , ,G N P T N P T (5.35)

This can be shown in a manner similar to that used to prove the 1/ lnA .

The differential equation to start with is

G

G A PV A PP

(5.36)

Other thermodynamic relations follow:

Volume:

,

ln , ,

N T

N P TV kT

P (5.37)

Enthalpy:

ln , ,H H x PV N P T (5.38)

Heat capacity at constant pressure

2

2

2

,

ln , ,P

N P

HC k N P T

T (5.39)

Entropy:

,

ln , ,N P

G HS k N P T

T T (5.40)

The fluctuations in the enthalpy H are given, in analogy with the canonical

ensemble, by

51

2

PH kT C (5.41)

so that

2

PkT CH

H H (5.42)

so that, since PC and H are both extensive, / ~ 1/H H N which vanish in the

thermodynamic limit.

5.3.3. Pressure and work virial theorems

As noted earlier, the quantity /H V is a measure of the instantaneous value

of the internal pressure intP . Let us look at the ensemble average of this quantity

int

0

1 H xPV

N

HP C dVe dx e

V

0 0

1 1, ,

H xPV PV

NC dVe dxkT e dVe kT N V TV V

(5.43)

Doing the volume integration by parts gives

int 0

0

1 1, , , ,PV PVP e kT N V T dVkT e N V T

V

0

1, ,PVP dVe N V T P (5.44)

This result is known as the pressure virial theorem. It illustrates that the average

of the quantity /H V gives the fixed pressure P that defines the ensemble.

Another important result comes from considering the ensemble average

int

0

1, ,PVP V dVe kTV N V T

V (5.45)

Once again, integrating by parts with respect to the volume yields

52

int 0

0

1 1, , , ,PV PVP V e kTV N V T dVkT Ve N V T

V

0 0

1( ) ( ) ( )PV PVkT dVe V P dVe V V kT P V (5.46)

Or

int ( )P V kT P V (5.47)

This result is known as the work virial theorem. It expresses the fact that

equipartitioning of energy also applies to the volume degrees of freedom, since

the volume is now a fluctuating quantity.

LECTURE 6: THE CLASSICAL IDEAL GAS

Content of Lecture 6

6.1. The ideal gas

o Microcanonical ensemble treatment

o Relation to thermodynamic entropy

o Canonical ensemble treatment

o The Gibbs paradox

o Isothermal-isobaric ensemble

6.1. The ideal gas

53

The ideal gas is simplest thermodynamic system that lends itself to analytical

solution in all the ensembles we have studied thus far. It's importance lies in the

fact that, for many ``real'' systems, the ideal gas behavior is the zeroth order

approximation to these systems. Thus, deviations from ideal behavior become

manifestly clear. It is therefore important that we establish what this ideal

behaviour is. We will begin with the microcanonical ensemble treatment.

6.1.1. Microcanonical ensemble treatment

Consider a system of N particles in a cubic box of volume 3V L .

The particles are assumed not to interact with each other. Thus, the

Hamiltonian in Cartesian coordinates may be taken to be

2

1 2

Ni

i

pH

m (6.1)

where we are assuming that all particles are of the same type.

The microcanonical partition function is

2

0

31

, ,! 2

NN N i

Ni

E pN V E d pd r E

N h m (6.2)

Since the Hamiltonian is independent of the coordinates, the 3N coordinate

integrations can be done straightforwardly. The range of each one is 0 to L.

Thus, these integrations give an overall factor 3N NL V :

2

0

31

, ,! 2

NN N i

Ni

E pN V E V d p E

N h m (6.3)

To do the momentum integrals, we first change variables to

3 /2

, 22

NN Nii

pp d p m d p

m (6.4)

Substitution into the partition function gives

3 /2 20

31

, , 2!

NNN N

iNi

EN V E V m d p p E

N h (6.5)

54

The 3N dimensional integral can now be seen to an integration over the surface

of a sphere defined by the equation

2

1

N

i

i

p E (6.6)

Therefore, it proves useful to transform to 3N dimensional spherical

coordinates, 1 3 1, ,..., NR , where

2 2

1

N

i

i

R p (6.7)

Then

3 1 3 1N N Nd p d dR R (6.8)

where 3 1Nd is the 3N-1 solid angle integral over the 3N - 1 angles. The

partition function now becomes:

3 /2 3 1 3 1 20

3, , 2

!

NN N N

N

EN V E V m d dR R R E

N h (6.9)

At this point, we use an identity of -functions:

2 2 1

2x a x a x a

a (6.10)

to write

3 /2 3 1 3 10

3

1, , 2

! 2

NN N N

N

EN V E V m d dR R R E R E

N h E

3 /2 3 /2 1 3 10

3

12

! 2

NN N N

N

EV m E d

N h (6.11)

We will also make use of the fact that N is large, so that we may take

3 1 3N N . Using the general formula for an n dimensional solid angle

integral:

55

1 /2

2

1

2

n

ndn

(6.12)

Where ( )x is the Gamma function:

1

0

( ) x tx dtt e (6.13)

which satisfies

1/22 1 !!1

( ) 1 !, 2 2n

nn n n (6.14)

Thus, the solid angle integral is

3 /2

3 1 2

3

2

NNd

N (6.15)

and the partition function finally becomes

3/20

3

1 1, , 2

3!

2

NE V

N V E mENN E h

(6.16)

The entropy S(N,V,E) is given by

3/2

3

0

3, , ln , , ln 2 ln ! ln ln (6.17)

2

V N ES N V E k N V E Nk mE k N k k

h E

Note that the term 0ln / lnk E E N since E N , which is negligibly small

compared to the term proportional to N and ln !N . Thus, we can neglect it.

Now, we can simplify ln !N using Stirling's approximation

ln ! lnN N N N (6.18)

which is valid for N very large. Also, note that

3 3 3

1 ! !2 2 2

N N N (6.19)

56

so that

3 3 3 3 3

ln ln ! ln2 2 2 2 2

N N N N N (6.20)

Substituting these approximations into the expression for the entropy, we

obtain

3/2

3

4 3, , ln , , ln ln !

3 2

V mES N V E k N V E Nk Nk k N

h N (6.21)

We could also simplify the ln !N using Stirling's approximation, however, let us

keep it as it is for now, since, as we remember from our past treatment of the

microcanonical ensemble, this factor was included in the partition function in

an ad hoc manner, in order to account for the indistinguishability of the

particles. We will want to explore the effect of removing this term. Without it,

the entropy is the purely classical entropy

3/2

3

4 3, , ln , , ln

3 2

V mES N V E k N V E Nk Nk

h N (6.22)

Other thermodynamic quantities can be easily obtained. For example, the

temperature is

1 ln 3

2

N

kT E E (6.23)

or

3

2E NkT (6.24)

which is the result we obtained from our analysis of the classical virial theorem.

The pressure is given by

ln N

P kTV V

(6.25)

or

PV NkT (6.26)

57

which is the famous ideal gas law. This is actually the equation of state of the

ideal gas, as it expresses the pressure as a function of the volume and

temperature. It can also be written as

P N

kT V (6.27)

where is the constant density of the gas.

One often expresses the equation of state graphically. For the ideal gas, if we

plot P vs. V, for different values of T, we obtain the following plot:

Figure 6.1

The different curves shown are called the isotherms, since they

represent P vs. V for fixed temperature.

The heat capacity of the ideal gas follows from the expression for the energy

,

3

2V

N V

EC Nk

T (6.28)

From our previous analysis of the virial theorem, we can conclude that each

kinetic mode contributes k/2 to the heat capacity.

6.1.2. Relation to thermodynamic entropy

In thermodynamics, the change in entropy in a reversible process which

transforms the system from state 1 to state 2 is

2

rev

1

dQS

T (6.29)

58

Where revdQ is the heat absorbed in the process. We can now ask if the entropy

obtained starting from the microscopic description agrees with the standard

thermodynamic definition. We will consider two types of processes as

described below:

6.1.2.1. Isothermal expansion/compression of the system from volume, 1V to

2V

In an isothermal process, the temperature, T, does not change. Thus, the

entropy relation can be integrated immediately to yield

2

revrev

1

1 QS dQ

T T (6.30)

where revQ is the heat absorbed as the state changes from 1 to 2. Now, from the

first law of thermodynamics, the change in total internal energy of the system is

rev revE Q W (6.31)

where revW is the work done on the system. Since, for the ideal gas,

3 3

, 2 2

E NkT E Nk T (6.32)

and 0, 0T E and

rev revQ W (6.33)

The expansion/compression of the system gives rise to a change in pressure

such that rev ( )dW P V dV , where P(V) = NkT/V, is given by the equation of

state (ideal gas law). Thus, the total work done on the system is

2

1

2rev

1

ln

V

V

VNkTW dV NkT

V V (6.34)

Thus,

2rev

1

lnV

Q NkTV

(6.35)

59

and

2

1

lnV

S NkTV

(6.36)

If we now use the statistical definition of entropy

3/2

3

4 3ln ln !

3 2

V mES Nk Nk k N

h N (6.37)

the change in entropy is

3/2 3/2

2 1

3 3

4 3 4 3ln ln ! ln ln !

3 2 3 2

V VmE mES Nk Nk k N Nk Nk k N

h N h N

22 1

1

ln ln lnV

Nk cV Nk cV NkV

(6.38)

where 3/231/ 4 / 3c h mE N . Thus, we see that the two agree exactly.

6.1.2.2. Isochoric heating/cooling from temperature 1T to 2T

In an isochoric process, the volume remains constant. Hence,

rev 0W (6.39)

and, from the first law,

rev rev, Q E dQ dE (6.40)

However, for the ideal gas

2 21 1 2 2

1 1

3 3 3 3, , , ,

2 2 2 2

E TE NkT dE NkdT E NkT E NkT

E T (6.41)

Thus, the change in entropy is

2 2

1 1

rev 2

1

3 3ln

2 2

T T

T T

dQ TdTS Nk Nk

T T T (6.42)

60

From the statistical definition:

3/2 3/2

2 1

3 3

4 43 3ln ln ! ln ln !

3 2 3 2

mE mEV VS Nk Nk k N Nk Nk k N

h N h N

3/2

3/2 3/2 2 2 22 1

1 1 1

3 3ln ln ln ln ln

2 2

E E TNk aE Nk aE Nk Nk Nk

E E T (6.43)

which agrees exactly with the thermodynamic entropy change.

These two examples illustrate that the statistical approach agrees exactly with

the standard thermodynamic definition of entropy.

6.1.3. Canonical ensemble treatment

The canonical partition function for the ideal gas is much easier to evaluate

than the microcanonical partition function. Recall the expression for the

canonical partition function:

2

3 1

1, , exp

! 2

NN N i

N i

pN V T d pd r

N h m (6.44)

Note that this can be expressed as

2

3

1, , exp

! 2

N

N

N

pN V T V dp

N h m (6.45)

since the Hamiltonian is completely separable. Evaluating the Gaussian integral

gives us the final result immediately:

3/2

3

1 2, ,

!

N

V mN V T

N h (6.46)

The expressions for the energy

ln , ,E N V T (6.47)

and pressure

61

ln , ,N V T

P kTV

(6.48)

gives rise to the results E = 3NkT/2 and P V= NkT just as for the

microcanonical ensemble. Note also that the entropy S(N,V,T) given by

ln , ,E

S k N V TT

(6.49)

becomes

3/2

3

2 3, , ln ln !

2

V mS N V T Nk Nk k N

h (6.50)

which reduces to the microcanonical expression exactly if we use the fact that 3

2

N

E:

3/2

3

4 3, , ln , , ln ln !

3 2

V mES N V E k N V E Nk Nk k N

h N (6.51)

Thus, the canonical and microcanonical ensembles gives rise to exactly the

same thermodynamics!! Let us now look more carefully at the expression for

the entropy.

6.1.4. The Gibbs paradox

Consider an ideal gas of N particles in a container with a volume V. A partition

separates the container into two sections with volumes 1V and 2V ,

respectively, such that 1 2V V V . Also, there are 1N particles in the volume

1V and 2N particles in the volume 2V . It is assumed that the number density is

the same throughout the system

1 2

1 2

N N

V V (6.52)

62

Figure 6.2

If the partition is now removed, what should happen to the total entropy? Since

the particles are identical, the total entropy should not increase as the partition

is removed because the two states cannot be differentiated due to the

indistinguishability of the particles. Let us analyze this thought experiment

using the classical expression entropy derived above (i.e., we leave off

the ln !N term).

The entropies 1S and 2S before the partition is removed are

1 1 1 1

3ln ,

2S N V N k (6.53)

2 2 2 2

3ln

2S N V N k (6.54)

and the total entropy is 1 2S S S .

After the partition is removed, the total entropy is

1 2 1 2 1 2

3ln ,

2S N N V V N N k (6.55)

Thus, the difference (f i iS S S S is initial entropy and fS is final entropy) is

63

1 2 1 2 1 1 2 2 1 2

1 2

ln ln ln ln ln 0V V

S N N k V V N k V N k V N k N kV V

(6.56)

This contradicts our predicted result that 0S . Therefore, the classical

expression must not be quite right.

Let us now restore the ln !N . Using the Stirling approximation

ln ! lnN N N N , the entropy can be written as

3/2

3

2 5

2

N mS Nk Nk

Nh (6.57)

which is known as the Sackur-Tetrode equation. Using this expression for the

entropy, the difference now becomes

1 2 1 21 2 1 2 1 2

1 2 1 2 1 2

ln ln ln ln lnV V V V V V

S N N k N k N k N k N kN N N N V V

1 21 2 1 2

1 2 1 2

ln ln ln lnN NN N V V

N k N k N k N kN N N V N V

(6.58)

However, since the density 1 1 2 2/ / /N V N V N V is constant, the terms

appearing in the log are all 1 and, therefore, vanish. Hence, the change in

entropy, 0S as expected. Thus, it seems that the 1/N! term is absolutely

necessary to resolve the paradox. This means that only a correct quantum

mechanical treatment of the ideal gas gives rise to a consistent entropy.

6.1.5. Isothermal-isobaric ensemble

Finally, let us look at the ideal gas in the isothermal-isobaric ensemble. The

partition function is given by

3 /2

3

0 00 0

1 2 1. , , ,

!

N

PV PV N

N

vmN P T dVe N V T dVe V

V V N h (6.59)

We change variables to

v PV (6.60)

so that

64

3 /2 3/2

13 3

0 00

2 1 1 1 2 1 1. , (6.61)

!

NN

v N

NN

vm vmN P T dve v

V N h h P PVP

The enthalpy is given by

ln , ,H N P T

(6.62)

which, for the ideal gas, becomes

3 5 1

12 2

N NH kT N kT kT (6.63)

But recall that the enthalpy is H E P V , where E is the average energy.

Since E = 3NkT/2, we have

1P V N kT (6.64)

This result can also be derived by considering the average volume, given by

ln 1N

V kT kTP P

(6.65)

But recall the work virial theorem, which states that

intP V P V kT (6.66)

Where intP is the instantaneous internal pressure obtained from int /P H V .

Thus, the appropriate form of the ideal gas law is

intP V NkT (6.67)

65

LECTURE 7: THE GRAND CANONICAL ENSEMBLE

Content of Lecture 7

7.1. The grand canonical ensemble

o Thermodynamics

o Partition function

o Ideal gas

o Particle number fluctuations

7.1. The grand canonical ensemble

In the grand canonical ensemble, the control variables are the chemical

potential , the volume V and the temperature T. The total particle number N is

therefore allowed to fluctuate. It is therefore related to the canonical ensemble

66

by a Legendre transformation with respect to the particle number N. Its utility

lies in the fact that it closely represents the conditions under which experiments

are often performed and, as we shall see, it gives direct access to the equation

of state.

7.1.1. Thermodynamics

In the canonical ensemble, the Helmholtz free energy A(N,V,T) is a natural

function of N, V and T. As usual, we perform a Legendre transformation to

eliminate N in favor of /A N :

,

, , , , , ,V T

AA V T A N V T N A N V T N

N (7.1)

It turns out that the free energy , ,A V T is the quantity PV . We shall derive

this result below in the context of the partition function. Thus,

, ,PV A N V T N (7.2)

To motivate the fact that PV is the proper free energy of the grand canonical

ensemble from thermodynamic considerations, we need to introduce a

mathematical theorem, known as Euler's theorem:

Euler's Theorem: Let 1,..., Nf x x be a function such that

1 1,..., ,...,n

N Nf x x f x x (7.3)

Then f is said to be a homogeneous function of degree n. For example, the

function 2( ) 3f x x is a homogeneous function of degree 2, 2 3, ,f x y z xy z is a homogeneous function of degree 3,

however, ( , ) xyf x y e xy is not a homogeneous function. Euler's

Theorem states that, for a homogeneous function f,

1

1

,...,N

N i

i i

fnf x x x

x (7.4)

Proof: To prove Euler's theorem, simply differentiate the homogeneity

condition with respect to lambda:

67

1

1 1 1

1

,..., ,..., , ,...,N

n n

N N i N

i i

d d ff x x f x x x n f x x

d d x (7.5)

Then, setting 1, we have

1

1

,...,N

i N

i i

fx nf x x

x

which is exactly Euler's theorem.

Now, in thermodynamics, extensive thermodynamic functions are

homogeneous functions of degree 1. Thus, to see how Euler's theorem applies

in thermodynamics, consider the familiar example of the Gibbs free energy:

, ,G G N P T (7.6)

The extensive dependence of G is on N, so, being a homogeneous function of

degree 1, it should satisfy

, , , ,G N P T G N P T (7.7)

Applying Euler's theorem, we thus have

, ,G

G N P T N NN

(7.8)

or, for a multicomponent system,

j j

j

G N (7.9)

But, since

G E TS PV (7.10)

it can be seen that G N is consistent with the first law of thermodynamics.

Now, for the Legendre transformed free energy in the grand canonical

ensemble, the thermodynamics are

dA dA dN Nd PdV SdT Nd (7.11)

68

But, since

, , ,

, , ,

V T T V

A A AA A V T dA d dV dT

V T (7.12)

the thermodynamics will be given by

,

,

V T

AN (7.13)

,

,

T

AP

V (7.14)

,V

AS

T (7.15)

Since, A is a homogeneous function of degree 1, and its extensive argument

is V, it should satisfy

, , , ,A V T A V T (7.16)

Thus, applying Euler's theorem,

, ,A

A V T V PVV

(7.17)

and since

A A N E TS N (7.18)

the assignment A PV is consistent with the first law of thermodynamics. It

is customary to work with PV, rather than PV , so PV is the natural free

energy in the grand canonical ensemble, and, unlike the other ensembles, it is

not given a special name or symbol!

7.1.2. Partition function

Consider two canonical systems, 1 and 2, with particle numbers 1N and 2N ,

volumes 1V and 2V and at temperature T. The systems are in chemical contact,

69

meaning that they can exchange particles. Furthermore, we assume that

2 1N N and 2 1V V so that system 2 is a particle reservoir. The total particle

number and volume are

1 2 1 2, V V V N N N (7.19)

The total Hamiltonian ,H x N is

1 1 1 2 2 2, , ,H x N H x N H x N (7.20)

If the systems could not exchange particles, then the canonical partition

function for the whole system would be

1 1 1 2 2 23

1, , exp , ,

! NN V T dx H x N H x N

N h

1 21 1 1 2 2 2

! !, , , ,

!

N NN V T N V T

N (7.21)

Where

1

1 1 1 1 1 1 13

1

1, , exp , ,

!N

N V T dx H x NN h

(7.22)

2

2 2 2 2 2 2 23

2

1, , exp ,

!N

N V T dx H x NN h

(7.23)

However, 1N and 2N are not fixed, therefore, in order to sum over all

microstates, we need to sum over all values that 1N can take on subject to the

constraint 1 2N N N . Thus, we can write the canonical partition function for

the whole system as

1

1 21 1 1 1 2 2 2

0

! !, , , , , , ,

!

N

N

N NN V T f N N N V T N V T

N (7.24)

where 1,f N N is a function that weights each value of 1N for a given N.

Thus, f(0,N) is the number of configurations with 0 particles in

1V and N particles in 2V ; f(1,N) is the number of configurations with 1

particles in 1V and 1N particles in 2V ; etc.

70

Determining the values of 1,f N N amounts to a problem of counting the

number of ways we can put N identical objects into 2 baskets. Thus,

0, 1,f N (7.25)

!

1, ,1! 1 !

Nf N N

N (7.26)

1 !

2, ,2 2! 2 !

N N Nf N

N (7.27)

etc.

or generally,

1

1 1 1 2

! !,

! ! ! !

N Nf N N

N N N N N (7.28)

which is clearly a classical degeneracy factor. If we were doing a purely

classical treatment of the grand canonical ensemble, then this factor would

appear in the sum for , ,N V T , however, we always include the ad

hoc quantum correction 1/N! in the expression for the canonical partition

function, and we see that these quantum factors will exactly cancel the classical

degeneracy factor, leading to the following expression:

1

1 1 1 2 2 2

0

, , , , , ,N

N

N V T N V T N V T (7.29)

which expresses the fact that, in reality, the various configurations are not

distinguishable from each other, and so each one should count with equal

weighting. Now, the distribution function x is given by

,

3

1

!,, ,

H x N

Ne

N hx NN V T

(7.30)

which is chosen so that

, 1dx x N (7.31)

However, recognizing that 2N N , we can obtain the distribution for

1 1 1,x N immediately, by integrating over the phase space of system 2:

71

1 1 1 2 2 2

1 2

, ,

1 1 1 23 3

1 2

1 1 1,

, , ! !

H x N H x N

N Nx N e dx e

N V T N h N h (7.32)

where the 13

1

1

!N

N h prefactor has been introduced so that

1

1 1 1

0

, 1N

N

dx x N (7.33)

and amounts to the usual ad hoc quantum correction factor that must be

multiplied by the distribution function for each ensemble to account for the

identical nature of the particles. Thus, we see that the distribution function

becomes

1 1 1

1

,2 2 2

1 1 1 3

1

, , 1,

, , !

H x N

N

N V Tx N e

N V T N h (7.34)

Recall that the Hemlholtz free energy is given by

1

lnA (7.35)

Thus,

, ,, , ,

A N V TN V T e (7.36)

2 2 1 1, , , ,

2 2 2, ,A N V T A N N V V T

N V T e e (7.37)

Or

1 1, , , ,2 2 2, ,

, ,

A N N V V T A N V TN V Te

N V T (7.38)

But since 1N N and 1V V , we may expand:

1 1 1 1 1 1, , , , ... , , ...(7.39)A A

A N N V V T A N V T N V A N V T N PVN V

Therefore the distribution function becomes

72

1 1 1 1 1 11 1 1

1 1 1

, ,

1 1 1 3 3

1 1

1 1 1,

! !

H x N H x NN PV N

N N PVx N e e e e e

N h N h e (7.40)

Dropping the ``1'' subscript, we have

,

3

1 1,

!

H x NN

PV Nx N e e

e N h (7.41)

We require that ,x N be normalized:

,

30 0

1 1, 1, 1

!

H x NN

PV NN N

dx x N e dxee N h

(7.42)

Now, we define the grand canonical partition function

,

30

1, ,

!

H x NN

NN

V T e dxeN h

(7.43)

Then, the normalization condition clearly requires that

, , , ln , ,PV PVV T e V T

kT (7.44)

Therefore PV is the free energy of the grand canonical ensemble, and the

entropy S , ,V T is given by

S,,

, , ln , , ln , ,VV

PVV T k V T k V T

T (7.45)

We now introduce the fugacity defined to be

e (7.46)

Then, the grand canonical partition function can be written as

,

30 0

1, , , ,

!

H x NN N N

NN N

V T e dxe N V TN h

(7.47)

which allows us to view the grand canonical partition function as a function of

the thermodynamic variables , V, and T.

73

Other thermodynamic quantities follow straightforwardly:

Energy:

,

30 ,

ln , ,1, ,

!

NH x N

NN V

V TE H x N dxH x N e

N h (7.48)

Average particle number:

,

ln , ,

V T

V TN kT (7.49)

This can also be expressed using the fugacity by noting that

(7.50)

Thus,

ln , ,N V T (7.51)

7.1.3. Ideal gas

Recall the canonical partition function expression for the ideal gas:

3/2

3

1 2, ,

!

N

V mN V T

N h (7.52)

Define the thermal wavelength as

1/22

2

h

m (7.53)

which has a quantum mechanical meaning as the width of the free particle

distribution function. Here it serves as a useful parameter, since the canonical

partition can be expressed as

3

1, ,

!

NV

N V TN

(7.54)

The grand canonical partition function follows directly from , ,N V T :

74

3 3

0

1, , exp

!

N

N

V VV T

N (7.55)

Thus, the free energy is

3

lnPV V

kT (7.56)

In order to obtain the equation of state, we first compute the average particle

number

3

lnV

N (7.57)

Thus, eliminating in favor of N in the equation of state gives

PV N kT (7.58)

as expected. Similarly, the average energy is given by

4

,

ln 3 3

2V

VE N kT (7.59)

where the fugacity has been eliminated in favor of the average particle number.

Finally, the entropy

S3

,

ln , , 5, , ln , , ln (7.60)

2V

V T VV T k V T k N k N k

N

which is the Sackur-Tetrode equation derived in the context of the canonical

and microcanonical ensembles.

7.1.4. Particle number fluctuations

In the grand canonical ensemble, the particle number N is not constant. It is,

therefore, instructive to calculate the fluctuation in this quantity. As usual, this

is defined to be

22N N N (7.61)

75

Note that

2

2

20 0

1 1ln , , , , , ,N N

N N

V T N N V T N N V T

22N N (7.62)

Thus,

2 2

2 2

2 2ln , , ln , ,

PN V T kT V T kTV (7.63)

In order to calculate this derivative, it is useful to introduce the Helmholtz free

energy per particle defined as follows:

1

, , ,a v T A N V TN

(7.64)

where / 1/v V N is the volume per particle.

The chemical potential is defined by

, ,A a v a

a v T N a v T vN v N v

(7.65)

Similarly, the pressure is given by

A a v a

P NV v V v

(7.66)

Also,

2

2

av

v v (7.67)

Therefore,

12 2

2 2

1P P v a av

v v v v (7.68)

and

76

12 2

2 2 2 3

1 1

/

P P v av

v v v v P v (7.69)

But recall the definition of the isothermal compressibility:

1 1

/T

V

V P v p v (7.70)

Thus,

2

2 2

TP

v (7.71)

And

TN kTN

v (7.72)

and the relative fluctuation is given by

1 1

~ 0TN kTN

N N v N khi N (7.73)

Therefore, in the thermodynamic limit, the particle number fluctuations vanish,

and the grand canonical ensemble is equivalent to the canonical ensemble.

77

LECTURE 8: DISTRIBUTION FUNCTIONS AND CORRELATION

FUNCTIONS IN CLASSICAL LIQUIDS AND GASES

So far, we have developed the classical theory of ensembles and applied it to

the ideal gas, for which there was no potential of interaction between the

particles: U=0. We were able to derive all the thermodynamics for this system

using the various ensembles available to us, and, in particular, we could

compute the equation of state. Now, we wish to consider the general case that

the potential is not zero. Of course, all of the interesting physics and chemistry

of real systems results from the specific interactions between the particles. Real

systems can exhibit spatial structure, undergo phase transitions, undergo

chemical changes, exhibit interesting dynamics, basically, a wide variety of

rich behavior.

Content of Lecture 8

8.1. General distribution functions and correlation functions 8.2. General correlation functions

8.3. The pair correlation function

8.4. Thermodynamic quantities in terms of g(r)

78

8.1. General distribution functions and correlation functions

We begin by considering a general N-particle system with Hamiltonian

23

1

1

,...,2

Ni

N

i

pH U r r

m (8.1)

For simplicity, we consider the case that all the particles are of the same type.

Having established the equivalence of the ensembles in the thermodynamic

limit, we are free to choose the ensemble that is the most convenient on in

which to work. Thus, we choose to work in the canonical ensemble, for which

the partition function is

23

3 3

131

1, , exp ,...,

! 2

NN N i

NNi

pN V T d pd r U r r

N h m (8.2)

The 3N integrations over momentum variables can be done straighforwardly,

giving

1 13 3

1, , ... exp ,...,

! !

NN NN N

ZN V T dr dr U r r

N N (8.3)

where 2 / 2h m is the thermal wavelength and the quantity NZ is

known as the configurational partition function

1 1... exp ,...,N N NZ dr dr U r r (8.4)

The quantity

1 ( )

1 1 1

exp ,...,... ,..., ...

N N

N N N

N

U r rdr dr P r r dr dr

Z (8.5)

represents the probability that particle 1 will be found in a volume element

1dr at the point 1r , particle 2 will be found in a volume element 2dr at the point

2r ,..., particle N will be found in a volume element Ndr at the point Nr . To

obtain the probability associated with some number n, irrespective of the

locations of the remaining 1,...,n N particles, we simply integrate this

expression over the particles with indices 1,...,n N :

79

( )

1 1 1 1 1

1,..., ... ... exp ,..., ...N

N N n N N n

N

P r r dr dr dr dr U r r dr drZ

(8.6)

The probability that any particle will be found in the volume element 1dr at the

point 1r and any particle will be found in the volume element 2dr at the point 2r

,...,any particle will be found in the volume element ndr at the point nr is

defined to be

( ) ( )

1 1 1 1

!,..., ... ,..., ...

!

n n

n n n n

Nr r dr dr P r r dr dr

N n (8.7)

which comes about since the first particle can be chosen in N ways, the second

chosen in 1N ways, etc.

Consider the special case of n = 1. Then, by the above formula,

(1)

1 2 1

1 !... exp ,...,

1 !N N

N

Nr dr dr U r r

Z N

2 1... exp ,...,N N

N

Ndr dr U r r

Z (8.8)

Thus, if we integrate over all 1r , we find that

(1)

1 1

1 Ndr r

V V (8.9)

Thus, (1) actually counts the number of particles likely to be found, on

average, in the volume element 1dr at the point 1r . Thus, integrating over the

available volume, one finds, not surprisingly, all the particles in the system.

8.2. General correlation functions

A general correlation function can be defined in terms of the probability

distribution function ( )

1,...,n

nr r according to

( ) ( )

1 1 1 1

1 !,..., ,..., ... exp ,...,

!

nn n

n n n N Nn n

N

V Ng r r r r dr dr U r r

Z N N n (8.10)

80

Another useful way to write the correlation function is

( )

1 1 1 1 1

!,..., ... exp ,..., ...

!

nn

n N N n nn

N

V Ng r r dr dr U r r r r r r

Z N N n

1

1 ,...,

!

!N

n n

i iniN r r

V Nr r

Z N N n (8.11)

i.e., the general n-particle correlation function can be expressed as an ensemble

average of the product of functions, with the integration being taken over

the variables 1,..., Nr r .

8.3. The pair correlation function

Of particular importance is the case n = 2, or the correlation

function (2)

1 2,g r r known as the pair correlation function. The explicit

expression for (2)

1 2,g r r is

2

(2)

1 2 1 1 2 22

!,

2 !

V Ng r r r r r r

N N

1

2

3 1 1 1 2 22 ,...,

1 1... exp ,...,

NN N r r

N

V N N Ndr dr U r r r r r r

NZ(8.12)

In general, for homogeneous systems in equilibrium, there are no special points

in space, so that (2)g should depend only on the relative position of the particles

or the difference 1 2r r . In this case, it proves useful to introduce the change of

variables

1 2 1 2 1 2

1 1 1, , ,

2 2 2r r r R r r r R r r R r (8.13)

Then, we obtain a new function (2)g , a function of r and R :

2

(2)

3 3

1 1 1, ... exp , , ,...,

2 2N N

N

V Ng r R dr dr U R r R r r r

NZ

81

1

1 22

,...,

1 1 1

2 2Nr r

N NR r r R r r (8.14)

In general, we are only interested in the dependence on r . Thus, we integrate

this expression over R and obtain a new correlation function g r defined by

(2)

3 3

11 1 1, ... exp , , ,...,

2 2N N

N

V Ng r dRg r R dRdr dr U R r R r r r

V NZ

3 3

1 1 1... exp , , ,...,

2 2N N

N

NdRdr dr U R r R r r r

Z

(8.15)

For an isotropic system such as a liquid or gas, where there is no preferred

direction in space, only the maginitude or , r r r is of relevance. Thus, we

seek a choice of coordinates that involves r explicitly. The spherical-polar

coordinates of the vector r is the most natural choice. If ( , , )r x y z then the

spherical polar coordinates are

2sin cos , sin sin , cos , sinx r y r z r dr r drd d (8.16)

where and are the polar and azimuthal angles, respectively. Also, note

that

r rn (8.17)

where

sin cos ,sin sin ,cosn (8.18)

Thus, the function g(r) that depends only on the distance r between two

particles is defined to be

1

( ) sin4

g r d d g r

3 3

1 1 1sin ... exp , , ,...,

4 2 2N N

N

Nd d dRdr dr U R r R r r r

Z

82

3, , , , ,...,

1

4Nr R r r

r rN

rr (8.19)

Integrating g(r) over the radial dependence, one finds that

2

0

4 ( ) 1drr g r N N (8.20)

The function g(r) is important for many reasons. It tells us about the structure

of complex, isotropic systems, as we will see below, it determines the

thermodynamic quantities at the level of the pair potential approximation, and it

can be measured in neutron and X-ray diffraction experiments. In such

experiments, one observes the scattering of neutrons or X-rays from a particular

sample. If a detector is placed at an angle from the wave-vector direction of

an incident beam of particles, then the intensity ( )I that one observes is

proportional to the structure factor

2

,

1 1( ) ~ mm n

ik r rik r

m m n

I e e S kN N

(8.21)

where k is the vector difference in the wave vector between the incident and

scattered neutrons or X-rays (since neutrons and X-rays are quantum

mechanical particles, they must be represented by plane waves of the form

exp ik r ). By computing the ensemble average, one finds that S k S k

and S(k) is given by

0

4( ) 1 sin ( )S k drr kr g r

k (8.22)

Thus, if one can measure S(k), g(r) can be determined by Fourier

transformation.

8.4. Thermodynamic quantities in terms of g(r)

In the canonical ensemble, the average energy is given by

ln , , , ln , , ln 2 ln ln !NE N V N V Z N N (8.23)

83

Therefore,

3 1 N

N

ZNE

Z (8.24)

Since

1/22

,2 2

h

m (8.25)

Thus,

1 1 1

3 1 3... ,..., exp ,...,

2 2N N N

N

E NkT dr dr U r r U r r NkT UZ

(8.26)

In order to compute the average energy, therefore, one needs to be able to

compute the average of the potential U . In general, this is a nontrivial task,

however, let us work out the average for the case of a pairwise-

additive potential of the form

1 pair 1

, ,

1,..., ,...,

2N i j N

i j i j

U r r u r r U r r (8.27)

i.e., U is a sum of terms that depend only the distance between two particles at

a time. This form turns out to be an excellent approximation in many

cases. U therefore contains 1N N total terms, and U becomes

1 pair 1

, ,

1... exp ,...,

2N i j N

i j i jN

U dr dr u r r U r rZ

1 pair 1

1... exp ,...,

2N i j N

N

N Ndr dr u r r U r r

Z (8.28)

The second line follows from the fact that all terms in the first line are the exact

same integral, just with the labels changed. Thus,

1 2 1 2 3 pair 1

11... exp ,...,

2N i j N

N

N NU drdr u r r dr dr u r r U r r

Z

84

2

(2) (2)

1 2 1 2 1 2 1 2 1 2 1 22

1, ,

2 2

Ndrdr u r r r r drdr u r r g r r

V (8.29)

Once again, we change variables to 1 2r r r and 1 2 / 2R r r . Thus, we find

that

2 2

(2) (2)

2 2, ,

2 2

N NU drdRu r g r R dru r dRg r R

V V

2 2

2

0

( ) 4 ( ) ( )2 2

N Ndru r g r dr r u r g r

V V (8.30)

Therefore, the average energy becomes

2

0

34 ( ) ( )

2 2

NE NkT drr u r g r (8.31)

Thus, we have an expression for E in terms of a simple integral over the pair

potential form and the radial distribution function. It also makes explicit the

deviation from ``ideal gas'' behavior, where E = 3NkT/2.

By a similar procedure, we can develop an equation for the pressure P in terms

of g(r). Recall that the pressure is given by

1 ln 1 N

N

ZP

V Z (8.32)

The volume dependence can be made explicit by changing variables of

integration in NZ to

1/3

i is V r (8.33)

Using these variables, NZ becomes

1/3 1/3

1 1... exp ,...,N

N N NZ V ds ds U V s V s (8.34)

Carrying out the volume derivative gives

1/3 1/3

1 1

1

1... exp ,...,

3

NNN

N N i N

i i

Z N UZ V ds ds r U V s V s

V V V r

85

1 1

1

1... exp ,...,

3

N

N N i i N

i

NZ dr dr r F U r r

V V (8.35)

Thus,

1

1

3

NN

i i

iN

Z Nr F

Z V V V (8.36)

Let us consider, once again, a pair potential. We showed in an earlier lecture

that

1 1 1,

N N N

i i i ij

i i j j i

r F r F (8.37)

where ijF is the force on particle i due to particle j. By interchanging

the i and j summations in the above expression, we obtain

1 , , , ,

1

2

N

i i i ij j ji

i i j i j i j i j

r F r F r F (8.38)

However, by Newton's third law, the force on particle i due to particle j is equal

and opposite to the force on particle j due to particle i:

ij jiF F (8.39)

Thus,

1 , , , , , , , ,

1 1 1

2 2 2

N

i i i ij j ij i j ij ij ij

i i j i j i j i j i j i j i j i j

r F r F r F r r F r F (8.40)

The ensemble average of this quantity is

1 pair 1

1 , , , ,

... exp ,..., (8.41)3 6 6

N

i i ij ij N ij ij N

i i j i j i j i jN

r F r F dr dr r F U r rV V VZ

As before, all integrals are exactly the same, so that

1 12 12 pair 1

1

1... exp ,...,

3 6

N

i i N N

i N

N Nr F dr dr r F U r r

V VZ

86

1 2 12 12 3 pair 1

1... exp ,...,

6N N

N

N Ndrdr r F dr dr U r r

V Z

2

(2) (2)

1 2 12 12 1 2 1 2 12 12 1 23, ,

6 6

Ndrdr r F r r drdr r F g r r

V V (8.42)

Then, for a pair potential, we have

pair 1 2 1212 1 2 12

12 1 2 12

U r r rF u r r u r

r r r r (8.43)

where /u r du dr , and 12 12r r . Substituting this into the ensemble average

gives

2

(2)

1 2 12 12 1 231

,3 6

N

i i

i

Nr F drdr u r r g r r

V V (8.44)

As in the case of the average energy, we change variables at this point

to 1 2r r r and 1 2 / 2R r r . This gives

2

(2)

31

,3 6

N

i i

i

Nr F drdRu r rg r R

V V

2 2

3

2 2

0

4 ( )6 6

N Ndru r rg r dr r u r g r

V V (8.45)

Therefore, the pressure becomes

2

3

0

4 ( ) ( )6

Pdr r u r g r

kT kT (8.46)

which again gives a simple expression for the pressure in terms only of the

derivative of the pair potential form and the radial distribution function. It also

shows explicitly how the equation of state differs from the that of the ideal gas

/ ( )P kT .

From the definition of g(r) it can be seen that it depends on the density and

temperature T: ( ) ; ,g r g r T . Note, however, that the equation of state,

derived above, has the general form

87

2PB

kT (8.47)

which looks like the first few terms in an expansion about ideal gas behavior.

This suggests that it may be possible to develop a general expansion in all

powers of the density about ideal gas behavior. Consider representing

; ,g r T as such a power series:

0

; , ;j

j

j

g r T g r T (8.48)

Substituting this into the equation of state derived above, we obtain

2

2

0

( ) j

j

j

PB T

kT (8.49)

This is known as the virial equation of state, and the coefficients 2( )jB T are

given by

3

2

0

1( ) 4 ( ) ;

6j jB T dr r u r g r T

kT (8.50)

are known as the virial coefficients. The coefficient 2 ( )B T is of particular

interest, as it gives the leading order deviation from ideal gas behavior. It is

known as the second virial coefficient. In the low density

limit, 0; , ;g r T g r T and 2 ( )B T is directly related to the radial distribution

function.

88

LECTURE 9: DISTRIBUTION FUNCTIONS AND

PERTURBATION THEORY

Content of Lecture 9

9.1. Distribution functions and perturbation theory

o General formulation

o Derivation of the Van der Waals equation

9.1. Distribution functions and

perturbation theory

89

9.1.1. General formulation

Recall the expression for the configurational partition function:

1 1... exp ,...,N N NZ dr dr U r r (9.1)

Suppose that the potential U can be written as a sum of two contributions

1 0 1 1 1,..., ,..., ,...,N N NU r r U r r U r r (9.2)

where 1U is, in some sense, small compared to 0U . An extra bonus can be had

if the partition function for 0U can be evaluated analytically.

Let

(0)

1 0 1... exp ,...,N N NZ dr dr U r r (9.3)

Then, we may express NZ as

(0)

1 0 1 1 1(0)... exp ,..., exp ,...,N

N N N N

N

ZZ dr dr U r r U r r

Z

(0)

1 10

exp ,...,N NZ U r r (9.4)

where 0

... means average with respect to 0U only. If 1U is small, then the

average can be expanded in powers of 1U :

1

2 32 3

1 1 1 10 0 0 000

1 ...2! 3! !

k

U k

k

e U U U Uk

(9.5)

The free energy is given by

1

(0)

3 3 0

1 1 1, , ln ln ln

! !

UN N

N N

Z ZA N V T e

N N (9.6)

Separating A into two contributions, we have

(0) (1), , , , , ,A N V T A N V T A N V T (9.7)

90

where (0)A is independent of 1U and is given by

(0)

(0)

3

1, , ln

!

N

N

ZA N V T

N (9.8)

and

1(1)

1 000

1 1, , ln ln

!

k

U k

k

A N V T e Uk

(9.9)

We wish to develop an expansion for (1)A of the general form

1

(1)

0 !

k

k

k

Ak

(9.10)

where k are a set of expansion coefficients that are determined by the

condition that such an expansion be consistent with 1 00

ln!

k

k

k

Uk

.

Using the fact that

1

1

ln 1 ( 1)!

kk

k

xx

k (9.11)

we have that

1 10 00 0

1 1ln ln 1

! !

k k

k k

k k

U Uk k

1

1 01 1

1 11

! !

kl

k l

k l

Uk l

(9.12)

Equating this expansion to the proposed expansion for (1)A , we obtain

1

1 01 1 1

11

! ! !

kl

k kl k

k l k

Uk l k

(9.13)

91

This must be solved for each of the undetermined parameters k , which can be

done by equating like powers of on both sides of the equation. Thus, from

the 1 term, we find, from the right side:

Right Side: 1

1! (9.14)

and from the left side, the l = 1 and k = 1 term contributes:

Left Side: 1 0

1!

U (9.15)

from which it can be easily seen that

1 1 0U (9.16)

Likewise, from the 2 term,

Right Side: 2

22!

(9.17)

and from the left side, we see that the l = 1, k = 2 and l = 2 , k = 1 terms

contribute:

Left Side: 2

22

1 1 002!U U (9.18)

Thus,

22

2 1 1 00U U (9.19)

For 3 , the right sides gives:

Right Side: 3

33!

(9.20)

the left side contributes the l = 1, k = 3, k = 2, l = 2 and l = 3,k = 1 terms:

Left Side: 3

33 2

1 1 1 10 00 02 3

3!U U U U (9.21)

92

Thus,

33 2

3 1 1 1 10 00 02 3U U U U (9.22)

Now, the free energy, up to the third order term is given by

2(0)

1 2 3 ...2 6

A A

(0) 22 32 3 2

1 1 1 1 1 1 13 0 0 0 00 0 0

1ln 3 2 ...(9.23)

! 2 6

N

N

ZU U U U U U U

N h

In order to evaluate 1 0U , suppose that 1U is given by a pair potential

1 1 1

1,...,

2N i j

i j

U r r u r r (9.24)

Then,

1 1 1 0 1(0)0

1 1... exp ,...,

2N i j N

N

U dr dr u r r U r rZ

1 2 1 1 2 3 0 1(0)

1... exp ,...,

2N N

N

N Ndrdr u r r dr dr U r r

Z

2 2

(2) 2

1 2 1 1 2 0 1 2 1 02

0

, 42 2

N Vdr dr u r r g r r r u r g r dr

V (9.25)

The free energy is therefore given by

(0)22 2 2

1 0 1 13 000

1 1, , ln 4 ( ) ...

! 2 2

N

N

ZA N V T V r u r g r dr U U

N h(9.26)

9.1.2. Derivation of the Van der Waals equation

As a specific example of the application of perturbation theory, we consider the

Van der Waals equation of state. Let 0U be given by a pair potential:

0 1 0

1,...,

2N i j

i j

U r r u r r (9.27)

93

with

0

0 ( )

( )

ru r

r (9.28)

This potential is known as the hard sphere potential. In the low-density limit,

the radial distribution function can be shown to be given correctly by

0 0~ expg r u r or

0

1 ( )

0 ( )

rg r r

r (9.29)

1u r is taken to be some arbitrary attractive potential, whose specific form is

not particularly important. Then, the full potential 0 1( ) ( ) ( )u r u r u r might look

like:

Figure 9.1

Now, the first term in (1)A is

(1) 2 2 2 2

1 0 1

0 0

1 14 4

2 2A V r u r g r V r u r r

2 2

12 V r u r dr aN (9.30)

where

2

12 0a drr u r (9.31)

is a number that depends on and the specific form of 1( )u r .

94

Since the potential 0u r is a hard sphere potential, (0)

NZ can be determined

analytically. If were 0, then 0u would describe an ideal gas and

(0) N

NZ V (9.32)

However, because two particles may not approach each other closer than a

distance between their centers, there is some excluded volume:

Figure 9.2

If we consider two hard spheres at closest contact and draw the smallest

imaginary sphere that contains both particles, then we find this latter sphere has

a radius :

95

Figure 9.3

Hence the excluded volume for these two particles is

34

3 (9.33)

and hence the excluded volume per particle is just half of this:

32

3b (9.34)

Therefore Nb is the total excluded volume, and we find that, in the low density

limit, the partition function is given approximately by

(0) N

NZ V Nb (9.35)

Thus, the free energy is

2

3

1, , ln

!

N

N

N Nb aNA N V T

N V (9.36)

If we now use this free energy to compute the pressure from

96

,N T

AP

V (9.37)

we find that

2 2

2 1

P N aN a

kT V Nb kTV B kT (9.38)

This is the well know Van der Waals equation of state. In the very low density

limit, we may assume that 1b , hence

1

11

bb

(9.39)

Thus,

2P ab

kT kT (9.40)

from which we can approximate the second virial coefficient:

3 2

2 1

2 2( ) ( )

3

aB T b drr u r

kT kT (9.41)

A plot of the isotherms of the Van der Waals equation of state is shown below:

Figure 9.4

The red and blue isotherms appear similar to those of an ideal gas, i.e., there is

a monotonic decrease of pressure with increasing volume. The black isotherm

97

exhibits an unusual feature not present in any of the ideal-gas isotherms - a

small region where the curve is essentially horizontal (flat) with no curvature.

At this point, there is no change in pressure as the volume changes. Below this

isotherm, the Van der Waals starts to exhibit unphysical behavior. The green

isotherm has a region where the pressure decreases with decreasing volume,

behavior that is not expected on physical grounds. What is observed

experimentally, in fact, is that at a certain pressure, there is a dramatic,

discontinuous change in the volume. This dramatic jump in volume signifies

that a phase transition has occurred, in this case, a change from a gaseous to a

liquid state. The dotted line shows this jump in the volume. Thus, the small flat

neighborhood along the black isotherm becomes larger on isotherms below this

one. The black isotherm just represents a boundary between those isotherms

along which no such phase transition occurs and those that exhibit phase

transitions in the form of discontinuous changes in the volume. For this reason,

the black isotherm is called the critical isotherm, and the point at which the

isotherm is flat and has zero curvature is called a critical point.

A critical point is a point at which

2

20, 0

P P

V V (9.42)

Using these two conditions, we can solve for the critical volume cV and critical

temperature cT :

8

3 , 27

c c

aV Nb kT

b (9.43)

and the critical pressure cP is therefore

227

c

aP

b (9.44)

Using these values for the critical pressure, temperature and volume, we can

show that the isothermal compressibility, given by

1

T

V

V P (9.45)

diverges as the critical point is approached. To see this, note that

98

2

2 2 3 3 2 3 2

2 2 1 8~

2 27 4 27 4 27c

c

V V

P NkT aN kT a akT T T

V N b N b Nb Nb Nb b (9.46)

Thus,

1

~T cT T (9.47)

It is observed that at a critical point, T diverges, generally, as cT T . To

determine the heat capacity, note that

2

, ,

3

8, , exp

!

N

A N V T

N

V Nb aNN V T e

N V (9.48)

so that

2

ln , , ln ln ! 3 lnaN

E N V T N V Nb N NV

2 2 23 3 3

2 2

N aN N aN aNNkT

V V V (9.49)

Then, since

V

V

EC

T (9.50)

it follows that

03

~2

V cC Nk T T (9.51)

The heat capacity is observed to diverge as cT T . Exponents such as and

are known as critial exponents.

Finally, one other exponent we can easily determine is related to how the

pressure depends on density near the critical point. The exponent is called ,

and it is observed that

99

~ const c

PC

kT (9.52)

What does our theory predict for ? To determine we expand the equation

of state about the critical density and temperature:

2 32 3

2 3

1 1 1...

2 6c c c

cc c c

c c c c

PP P P P

kT kT kT kT kT

22 2

2

2 2

1 1

2c

c

cc c

c c c

P P V P V P V

kT kT V kT V V

3

3

3

1...

6c

c

c

P

kT (9.53)

The second and third terms vanish by the conditions of the critical point. The

third derivative term can be worked out straightforwardly and does not vanish.

Rather

3

2

3

1 2430

8c

c

Pb

kT (9.54)

Thus, we see that, by the above expansion, 3 .

The behavior of these quantities near the critical temperature determine three

critical exponents. To summarize the results, the Van der Waals theory predicts

that

0, 1, 3 (9.55)

The determination of critical exponents such as these is an active area in

statistical mechanical research. The reason for this is that such exponents can

be grouped into universality classes- groups of systems with exactly the same

sets of critical exponents. The existence of universality classes means that very

different kinds of systems exhibit essentially the same behavior at a critical

point, a fact that makes the characterization of phase transitions via critical

exponents quite general. The values obtained above for , and are known

as the mean-field exponents and shows that the Van der Waals theory is really a

mean field theory. These exponents do not agree terribly well with

experimental values ( 0.1, 1.35, 4.2 ). However, the simplicty of

100

mean-field theory and its ability to give, at least, qualitative results, makes it,

nevertheless, useful. To illustrate universality classes, it can be shown that,

within mean field theory, the Van der Waals gas/liquid and a magnetic system

composed of spins at particular lattice sites, which composes the so called Ising

model, have exactly the same mean field theory exponents, despite the

completely different nature of these two systems.

Another problem with the Van der Waals theory that is readily apparent is the

fact that it predicts / 0P V for certain values of the density . Such

behavior is unstable. Possible ways of improving the approximations used are

the following:

1. Improve the approximation to 0 ( )g r .

2. Choose a better zeroth order potential 0 1,..., NU r r .

3. Go to a higher order in perturbation theory.

Barker and Henderson have shown that going to second order in perturbation

theory (see discussion in McQuarrie's book), yields well converged results for a

square well fluid, compared to ``exact'' results from a molecular dynamics

simulation.

101

LECTURE 10; REVIEW OF THE POSTULATES OF QUANTUM

MECHANICS

Content of Lecture 10

10.1. The fundamental postulates of quantum mechanics

o The physical state of a quantum system

o Physical Observables

o Measurement

o Time evolution of the state vector

o The Heisenberg uncertainty principle

o The Heisenberg picture

10.1. The fundamental postulates of

quantum mechanics

The fundamental postulates of quantum mechanics concern the following

questions:

1. How is the physical state of a system described?

2. How are physical observables represented?

3. What are the results of measurements on quantum mechanical systems?

4. How does the physical state of a system evolve in time?

5. The uncertainty principle.

10.1.1. The physical state of a quantum system

The physical state of a quantum system is represented by a vector denoted

( )t (10.1)

which is a column vector, whose components are probability amplitudes for

different states in which the system might be found if a measurement were

made on it.

102

A probability amplitude is a complex number, the square modulus of which

gives the corresponding probability

2

P (10.2)

The number of components of ( )t is equal to the number of possible states

in which the system might observed. The space that contains ( )t is called a

Hilbert space . The dimension of is also equal to the number of states in

which the system might be observed. It could be finite or infinite (countable or

not).

( )t must be a unit vector. This means that the inner product:

( ) ( ) 1t t (10.3)

In the above, if the vector ( )t , known as a Dirac ``ket'' vector, is given by

the column

1

2

( ) .

.

.

t (10.4)

then the vector ( )t , known as a Dirac ``bra'' vector, is given by

* *

1 2( ) . . .t (10.5)

so that the inner product becomes

2

( ) ( ) 1i

i

t t (10.6)

We can understand the meaning of this by noting that i , the components of

the state vector, are probability amplitudes, and 2

i are the corresponding

probabilities. The above condition then implies that the sum of all the

probabilities of being in the various possible states is 1, which we know must

be true for probabilities.

103

10.1.2. Physical Observables

Physical observables are represented by linear, hermitian operators that act on

the vectors of the Hilbert space. If A is such an operator, and is an arbitrary

vector in the Hilbert space, then A might act on to produce a vector ,

which we express as

A (10.7)

Since is representable as a column vector, A is representable as a matrix

with components

11 12 13

21 22 23

...

...

. . . ...

A A A

A A A A (10.8)

The condition that A must be hermitian means that

A A (10.9)

or

*

ij jiA A (10.10)

10.1.3. Measurement

The result of a measurement of the observable A must yield one of the

eigenvalues of A. Thus, we see why A is required to be a hermitian operator:

Hermitian operators have real eigenvalues. If we denote the set of eigenvalues

of A by ia , then each of the eigenvalues ia satisfies an eigenvalue equation

i i iA a a a (10.11)

where ia is the corresponding eigenvector. Since the operator A is hermitian

and ia is therefore real, we have also the left eigenvalue equation

i i ia A a a (10.12)

104

The probability amplitude that a measurement of A will yield the

eigenvalue ia is obtained by taking the inner product of the corresponding

eigenvector ia with the state vector ( )t , ( )ia t . Thus, the probability

that the value ia is obtained is given by

2

( )ia iP a t (10.13)

Another useful and important property of hermitian operators is that their

eigenvectors form a complete orthonormal basis of the Hilbert space, when the

eigenvalue spectrum is non-degenerate. That is, they are linearly independent,

span the space, satisfy the orthonormality condition

i j ija a (10.14)

and thus any arbitrary vector can be expanded as a linear combination of

these vectors:

i i

i

c a (10.15)

By multiplying both sides of this equation by ja and using the orthonormality

condition, it can be seen that the expansion coefficients are

i ic a (10.16)

The eigenvectors also satisfy a closure relation:

2

i

i

I c (10.17)

where I is the identity operator.

Averaging over many individual measurements of A gives rise to an average

value or expectation value for the observable A, which we denote A and is

given by

( ) ( )A t A t (10.18)

105

That this is true can be seen by expanding the state vector ( )t in the

eigenvectors of A:

( ) ( )i i

i

t t a (10.19)

where i are the amplitudes for obtaining the eigenvalue ia upon measuring A,

i.e., ( )i ia t . Introducing this expansion into the expectation value

expression gives

2* *

, ,

( ) ( ) ( ) ( ) ( ) ( )i j i i i j ij i i

i j i j i

A t t t a A a t t a t (10.20)

The interpretation of the above result is that the expectation value of A is the

sum over possible outcomes of a measurement of A weighted by the probability

that each result is obtained. Since 22

( )i ia t is this probability, the

equivalence of the expressions can be seen.

Two observables are said to be compatible if AB = BA. If this is true, then the

observables can be diagonalized simultaneously to yield the same set of

eigenvectors. To see this, consider the action of BA on an

eigenvector ia of A. i i iBA a a B a . But if this must equal iAB a , then the

only way this can be true is if iB a yields a vector proportional to ia which

means it must also be an eigenvector of B. The condition AB = BA can be

expressed as

0, , 0AB BA A B (10.21)

where, in the second line, the quantity ,A B AB BA is know as the

commutator between A and B. If , 0A B , then A and B are said to commute

with each other. That they can be simultaneously diagonalized implies that one

can simultaneously predict the observables A and B with the same

measurement.

As we have seen, classical observables are functions of position x and

momentum p (for a one-particle system). Quantum analogs of classical

observables are, therefore, functions of the operators X and P corresponding to

position and momentum. Like other observables X and P are linear hermitian

106

operators. The corresponding eigenvalues x and p and eigenvectors x and

p satisfy the equations

,X x x x P p p p (10.22)

which, in general, could constitute a continuous spectrum of eigenvalues and

eigenvectors. The operators X and P are not compatible. In accordance with the

Heisenberg uncertainty principle (to be discussed below), the commutator

between X and P is given by

,X P i I (10.23)

and that the inner product between eigenvectors of X and P is

/1

2

ipxx p e (10.24)

Since, in general, the eigenvalues and eigenvectors of X and P form a

continuous spectrum, we write the orthonormality and closure relations for the

eigenvectors as:

, , ,x x x x p px p p dx x

2 2

, ,dp p I dx x I dp p (10.25)

The probability that a measurement of the operator X will yield an

eigenvalue x in a region dx about some point is

2

, ( )P x t dx x t dx (10.26)

The object ( )x t is best represented by a continuous function ,x t often

referred to as the wave function. It is a representation of the inner product

between eigenvectors of X with the state vector. To determine the action of the

operator X on the state vector in the basis set of the operator X, we compute

(t) ,x X x x t (10.27)

The action of P on the state vector in the basis of the X operator is

consequential of the incompatibility of X and P and is given by

107

(t) ,x P x ti x

(10.28)

Thus, in general, for any observable A(X,P), its action on the state vector

represented in the basis of X is

( , ) (t) , ,x A X P A x x t

i x (10.29)

10.1.4. Time evolution of the state vector

The time evolution of the state vector is prescribed by the Schrödinger equation

( ) ( )i t H tt

(10.30)

where H is the Hamiltonian operator. This equation can be solved, in principle,

yielding

/( ) (0)iHtt e (10.31)

where (0) is the initial state vector. The operator

/( ) iHtU t e (10.32)

is the time evolution operator or quantum propagator. Let us introduce the

eigenvalues and eigenvectors of the Hamiltonian H that satisfy

i i iH E E E (10.33)

The eigenvectors for an orthonormal basis on the Hilbert space and therefore,

the state vector can be expanded in them according to

( ) ( )i i

i

t c t E (10.34)

where, of course, ( ) ( )i ic t E t , which is the amplitude for obtaining the

value iE at time t if a measurement of H is performed. Using this expansion, it

is straightforward to show that the time evolution of the state vector can be

written as an expansion:

108

//( ) (0) (0)iiE tiHt

i ii

t e e E E (10.35)

Thus, we need to compute all the initial amplitudes for obtaining the different

eigenvalues iE of H, apply to each the factor /iiE t

ie E and then sum over all

the eigenstates to obtain the state vector at time t.

If the Hamiltonian is obtained from a classical Hamiltonian H(x,p), then, using

the formula from the previous section for the action of an arbitrary

operator A(X,P) on the state vector in the coordinate basis, we can recast the

Schrödiner equation as a partial differential equation. By multiplying both sides

of the Schrödinger equation by x , we obtain

, ( ) ( ) ,x H X P t i x tt

, , ,H x x t i x ti x t

(10.36)

If the classical Hamiltonian takes the form

2

,2

pH x p U x

m (10.37)

then the Schrödinger equation becomes

2 2

2( ) , ,

2U x x t i x t

m x t (10.38)

which is known as the Schrödinger wave equation or the time-

dependent Schrödinger equation.

In a similar manner, the eigenvalue equation for H can be expressed as a

differential equation by projecting it into the X basis:

, , ( ) ( ),i i i i i ix H E E x E H x x E xi x

2 2

2( )

2i i iU x x E x

m x (10.39)

109

where i ix x E is an eigenfunction of the Hamiltonian.

10.1.5. The Heisenberg uncertainty principle

Because the operators X and P are not compatible, , 0X P , there is no

measurement that can precisely determine both X and P simultaneously. Hence,

there must be an uncertainty relation between them that specifies how uncertain

we are about one quantity given a definite precision in the measurement of the

other. Presumably, if one can be determined with infinite precision, then there

will be an infinite uncertainty in the other. Recall that we had defined the

uncertainty in a quantity by

22A A A (10.40)

Thus, for X and P, we have

2 22 2, x X X p P P (10.41)

These quantities can be expressed explicitly in terms of the wave function

,x t using the fact that

*( ) ( ) ( ) ( ) ( , ) ,X t X t dx t x x X t dx x t x t (10.42)

and

2 2 * 2( ) , ,X t X t x t x x t (10.43)

Similarly,

*( ) ( ) ( ) ( ) ( , ) ,P t P t dx t x x P t dx x t x ti x

(10.44)

and

2

2 2 * 2

2( ) , ,P t P t x t x t

x (10.45)

Then, the Heisenberg uncertainty principle states that

110

x p (10.46)

which essentially states that the greater certainty with which a measurement

of X or P can be made, the greater will be the uncertainty in the other.

10.1.6. The Heisenberg picture

In all of the above, notice that we have formulated the postulates of quantum

mechanics such that the state vector ( )t evolves in time but the operators

corresponding to observables are taken to be stationary. This formulation of

quantum mechanics is known as the Schrödinger picture. However, there is

another, completely equivalent, picture in which the state vector remains

stationary and the operators evolve in time. This picture is known as

the Heisenberg picture. This particular picture will prove particularly useful to

us when we consider quantum time correlation functions.

The Heisenberg picture specifies an evolution equation for any operator A,

known as the Heisenberg equation. It states that the time evolution of A is given

by

1

,dA

A Hdt i

(10.47)

While this evolution equation must be regarded as a postulate, it has a very

immediate connection to classical mechanics. Recall that any function of the

phase space variables A(x, p) evolves according to

,dA

A Hdt

(10.48)

where ...,... is the Poisson bracket. The suggestion is that in the classical limit

( small), the commutator goes over to the Poisson bracket. The Heisenberg

equation can be solved in principle giving

/ /( ) ( ) ( )iHt iHtA t e Ae U t AU t (10.49)

where A is the corresponding operator in the Schrödinger picture. Thus, the

expectation value of A at any time t is computed from

( ) ( )A t A t (10.50)

111

where is the stationary state vector.

Let's look at the Heisenberg equations for the operators X and P. If H is given

by

2

( )2

PH U X

m (10.51)

then Heisenberg's equations for X and P are

1 1

, , ,dX P dP U

X H P Hdt i m dt i X

(10.52)

Thus, Heisenberg's equations for the operators X and P are just Hamilton's

equations cast in operator form. Despite their innocent appearance, the solution

of such equations, even for a one-particle system, is highly nontrivial and has

been the subject of a considerable amount of research in physics and

mathematics.

Note that any operator that satisfies [A(t), H] =0 will not evolve in time. Such

operators are known as constants of the motion. The Heisenberg picture shows

explicitly that such operators do not evolve in time. However, there is an

analog with the Schrödinger picture: Operators that commute with the

Hamiltonian will have associated probabilities for obtaining different

eigenvalues that do not evolve in time. For example, consider the Hamiltonian,

itself, which it trivially a constant of the motion. According to the evolution

equation of the state vector in the Schrödinger picture,

/( ) (0)iHt

i ii

t e E (10.53)

the amplitude for obtaining an energy eigenvalue jE at time t upon

measuring H will be

// /( ) (0) (0) (0)ji i

iE tiE t iE t

j j i ij i jii i

E t e E E e E e E (10.54)

Thus, the squared modulus of both sides yields the probability for obtaining jE ,

which is

2 2

( ) (0)j jE t E (10.55)

112

Thus, the probabilities do not evolve in time. Since any operator that commutes

with H can be diagonalized simultaneously with H and will have the same set

of eigenvectors, the above arguments will hold for any such operator.

113

LECTURE 11: BASIC PRINCIPLES OF QUANTUM STATISTICAL

MECHANICS

Content of Lecture 11

11.1. Principles of quantum statistical mechanics

o The density matrix and density operator

o Time evolution of the density operator

o The quantum equilibrium ensembles

The microcanonical ensemble

The canonical ensemble

Isothermal-isobaric and grand canonical ensembles

o A simple example - the quantum harmonic oscillator

11.1. Principles of quantum statistical

mechanics

The problem of quantum statistical mechanics is the quantum mechanical

treatment of an N-particle system. Suppose the corresponding N-particle

classical system has Cartesian coordinates

1 3,..., Nq q (11.1)

and momenta

1 3,..., Np p (11.2)

and Hamiltonian

23

1 3

1

,...,2

Ni

N

i i

pH U q q

m (11.3)

Then, as we have seen, the quantum mechanical problem consists of

determining the state vector ( )t from the Schrödinger equation

( ) ( )H t i tt

(11.4)

114

Denoting the corresponding operators, 1 3,..., NQ Q and 1 3,..., NP P , we note that

these operators satisfy the commutation relations:

, , 0, ,i j i j i j ijQ Q P P Q P i I (11.5)

and the many-particle coordinate eigenstate 1 3... Nq q is a tensor product of the

individual eigenstate 1 3,..., Nq q :

1 3 1 3... ...N Nq q q q (11.6)

The Schrödinger equation can be cast as a partial differential equation by

multiplying both sides by 1 3... Nq q :

1 3 1 3... ( ) ... ( ) ,N Nq q H t i q q t

t (11.7)

2 23

1 3 1 3 1 321

,..., ,..., , ,..., ,2

N

N N N

i i i

U q q q q t i q q tm q t

(11.8)

where the many-particle wave function is 1 3 1 3,..., , ... ( )N Nq q t q q t .

Similarly, the expectation value of an operator 1 3 1 3,..., , ,...,N NA Q Q P P is given

by

*

1 3 1 3 1 3 1 3

1 3

... ,..., ,..., , ,..., ,...,N N N N

N

A dq dq q q A q q q qi q i q

(11.9)

11.1.1. The density matrix and density operator

In general, the many-body wave function 1 3,..., ,Nq q t is far too large to

calculate for a macroscopic system. If we wish to represent it on a grid with just

10 points along each coordinate direction, then for 2310N , we would need

231010

total points, which is clearly enormous.

We wish, therefore, to use the concept of ensembles in order to express

expectation values of observables A without requiring direct computation of

115

the wavefunction. Let us, therefore, introduce an ensemble of systems, with a

total of Z members, and each having a state vector ( ) 1,..., Z .

Furthermore, introduce an orthonormal set of vectors k ( k j kj

) and

expand the state vector for each member of the ensemble in this orthonormal

set:

( ) ( )

k k

k

C (11.10)

The expectation value of an observable, averaged over the ensemble of systems

is given by the average of the expectation value of the observable computed

with respect to each member of the ensemble:

( ) ( )

1

1 Z

A AZ

(11.11)

Substituting in the expansion for ( ) , we obtain

( )* ( ) ( ) ( )*

, 1

1 1Z

k l k l l k k l

k l

A C C A C C AZ Z

(11.12)

Let us define a matrix

( ) ( )*

1

Z

lk l kC C (11.13)

and a similar matrix

( ) ( )*

1

1 Z

lk l kC CZ

(11.14)

Thus, lk is a sum over the ensemble members of a product of expansion

coefficients, while lk is an average over the ensemble of this product. Also,

let kl k lA A . Then, the expectation value can be written as follows:

,

1 1 1Tr Trlk kl kk

k l k

A A A A AZ Z Z

(11.15)

116

where and A represent the matrices with elements lk and klA in the basis of

vectors k

. The matrix lk is known as the density matrix. There is an

abstract operator corresponding to this matrix that is basis-independent. It can

be seen that the operator

( ) ( )

1

Z

(11.16)

and similarly

( ) ( )

1

1 Z

Z (11.17)

have matrix elements lk when evaluated in the basis set of vectors k

( ) ( ) ( ) ( )*

1 1

Z Z

l k l k l k lkC C (11.18)

Note that is a hermitian operator

(11.19)

so that its eigenvectors form a complete orthonormal set of vectors that span

the Hilbert space. If k and k represent the eigenvalues and eigenvectors of

the operator , respectively, then several important properties they must satisfy

can be deduced.

Firstly, let A be the identity operator I. Then, since 1I , it follows that

1

1 Tr Tr k

kZ (11.20)

Thus, the eigenvalues of must sum to 1. Next, let A be a projector onto an

eigenstate of , k kk

A P . Then

Trk k kP (11.21)

But, since can be expressed as

117

k k k

k

(11.22)

and the trace, being basis set independent, can be therefore be evaluated in the

basis of eigenvectors of , the expectation value becomes

,

k j i i i k k j i ij ik kj k

j i i j

P (11.23)

However,

2

( ) ( ) ( )

1 1

1 10

Z Z

k k k kPZ Z

(11.24)

Thus, 0k . Combining these two results, we see that, since

1kk and 0k , 0 1k , so that k satisfy the properties of

probabilities.

With this in mind, we can develop a physical meaning for the density matrix.

Let us now consider the expectation value of a projector ii ai

a onto one of

the eigenstates of the operator A. The expectation value of this operator is given

by

2

( ) ( ) ( ) ( ) ( )

1 1 1

1 1 1i i

Z Z Z

a a i iia a

Z Z Z (11.25)

But 2

( ) ( )

ii aa P is just probability that a measurement of the operator A in

the th member of the ensemble will yield the result ia . Thus,

( )

1

1i i

P

a aPZ

(11.26)

or the expectation value of iais just the ensemble averaged probability of

obtaining the value ia in each member of the ensemble. However, note that the

expectation value of iacan also be written as

,

Tr Tri ia a k k i i l k k i i lk k

k k l

a a a a

118

2

,lk kl k i i k i k

k l k

a a a (11.27)

Equating the two expressions gives

2( )

1

1i

Z

a k i k

k

P aZ

(11.28)

The interpretation of this equation is that the ensemble averaged probability of

obtaining the value ia if A is measured is equal to the probability of obtaining

the value ia in a measurement of A if the state of the system under

consideration were the state k , weighted by the average probability k that

the system in the ensemble is in that state. Therefore, the density

operator (or ) plays the same role in quantum systems that the phase

space distribution function f plays in classical systems.

11.1.2. Time evolution of the density operator

The time evolution of the operator can be predicted directly from the

Schrödinger equation. Since ( )t is given by

( 0 ( 0

1

( ) ( ) ( )Z

t t t (11.29)

the time derivative is given by

( 0 ( 0 ( 0 ( 0

1

( ) ( ) ( ) ( )Z

t t t tt t t

( 0 ( 0 ( 0 ( 0

1

1( ) ( ) ( ) ( )

Z

H t t t t Hi

1 1

,H H Hi i

(11.30)

where the second line follows from the fact that the Schrödinger equation for

the bra state vector ( 0 ( )t is

( 0 ( 0( ) ( )i t t Ht

(11.31)

119

Note that the equation of motion for ( )t differs from the usual Heisenberg

equation by a minus sign! Since ( )t is constructed from state vectors, it is not

an observable like other hermitian operators, so there is no reason to expect that

its time evolution will be the same. The general solution to its equation of

motion is

/ /( ) e (0)e ( ) (0) ( )iHt iHtt U t U t (11.32)

The equation of motion for ( )t can be cast into a quantum Liouville equation

by introducing an operator

1

...,iL Hi

(11.33)

In term of iL, it can be seen that ( )t satisfies

, ( ) (0)iLtiL t et

(11.34)

What kind of operator is iL? It acts on an operator and returns another operator.

Thus, it is not an operator in the ordinary sense, but is known as

a superoperator or tetradic operator (see S. Mukamel, Principles of Nonlinear

Optical Spectroscopy, Oxford University Press, New York (1995)).

Defining the evolution equation for this way, we have a perfect analogy

between the density matrix and the state vector. The two equations of motion

are

( ) ( ) , ( ) ( )i

t H t t iL tt t

(11.35)

We also have an analogy with the evolution of the classical phase space

distribution ,f t , which satisfies

f

iLft

(11.36)

with ...,iL H being the classical Liouville operator. Again, we see that the

limit of a commutator is the classical Poisson bracket.

120

11.1.3. The quantum equilibrium ensembles

At equilibrium, the density operator does not evolve in time; thus, / 0t .

Thus, from the equation of motion, if this holds, then , 0H , and ( )t is a

constant of the motion. This means that it can be simultaneously diagonalized

with the Hamiltonian and can be expressed as a pure function of the

Hamiltonian

( )f H (11.37)

Therefore, the eigenstates of , the vectors, we called k are the

eigenvectors iE of the Hamiltonian, and we can write H and as

, i i i i i i

i i

H E E E f E E E (11.38)

The choice of the function f determines the ensemble.

11.1.3.1. The microcanonical ensemble

Although we will have practically no occasion to use the quantum

microcanonical ensemble (we relied on it more heavily in classical statistical

mechanics), for completeness, we define it here. The function f, for this

ensemble, is

i i if E E E E E E E (11.39)

where ( )x is the Heaviside step function. This says that if E E is 1 if

iE E E E and 0 otherwise. The partition function for the ensemble is

Tr , since the trace of is the number of members in the ensemble:

, , Tr i i

i

N V E E E E E E (11.40)

The thermodynamics that are derived from this partition function are exactly

the same as they are in the classical case

,

1 ln, , ln , , ,

N V

S N V E k N V E kT E

(11.41)

121

etc.

11.1.3.2. The canonical ensemble

In analogy to the classical canonical ensemble, the quantum canonical

ensemble is defined by

, iEH

ie f E e (11.42)

Thus, the quantum canonical partition function is given by

, , Tr iEH

i

N V T e e (11.43)

and the thermodynamics derived from it are the same as in the classical case:

1

, , ln , , ,A N V T N V T (11.44)

, , ln , , ,E N V T N V T (11.45)

1

, , ln , ,P N V T N V TV

(11.46)

etc. Note that the expectation value of an observable A is

1

Tr e HA A (11.47)

Evaluating the trace in the basis of eigenvectors of H (and of ), we obtain

1 1

iEH

i i i i

i i

A E Ae E e E A E (11.48)

The quantum canonical ensemble will be particularly useful to us in many

things to come.

11.1.3.3. Isothermal-isobaric and grand canonical ensembles

Also useful are the isothermal-isobaric and grand canonical ensembles, which

are defined just as they are for the classical cases:

122

0 0

, , , , Tr ,H PVPVN P T dVe N V T dV e (11.49)

0 0

, , , , TrH NN

N N

V T e N V T e (11.50)

11.1.4. A simple example - the quantum harmonic oscillator

As a simple example of the trace procedure, let us consider the quantum

harmonic oscillator. The Hamiltonian is given by

2

2 21

2 2

PH m X

m (11.51)

and the eigenvalues of H are

1

, 0,1,2,...2

nE n n (11.52)

Thus, the canonical partition function is

1/2 /2

0 0

nn

n n

e e e (11.53)

This is a geometric series, which can be summed analytically, giving

/2

/2 /2

1( )

1

e

e e e (11.54)

The thermodynamics derived from it as as follows:

1. The free energy

1 1

ln ( ) ln 12

A e (11.55)

2.The average energy

1

ln2 1 2

eE H n

e (11.56)

123

3.The entropy

ln ln 11

E eS k k e

T T e (11.57)

Now consider the classical expressions. Recall that the partition function is

given by

2

2 21 1( ) exp

2 2

pdpdx m x

h m

1/2 1/2

2

1 2 2 2 1m

h m h (11.58)

Thus, the classical free energy is

cl

1lnA (11.59)

In the classical limit, we may take to be small. Thus, the quantum expression

for A becomes, approximately, in this limit:

Q

1ln

2A (11.60)

and we see that

Q cl

2A A (11.61)

The residual / 2 (which truly vanishes when 0 ) is known as the

quantum zero point energy. It is a pure quantum effect and is present because

the lowest energy quantum mechanically is not E = 0 but the ground state

energy / 2E .

124

LECTURE 12: THE QUANTUM IDEAL GASES – GENERAL

FORMULATION

Content of Lecture 12

12.1. Introduction to spin

12.2. Solution of the N-particle eigenvalue problem

12.3. An ideal gas of distinguishable quantum particles

12.4. General formulation for fermions and bosons

12.1. Introduction to spin

The path integral formulation of quantum statistical mechanics is particularly

useful for situations in which particle spin statistics can be largely ignored. In

the quantum ideal gases, we have a situation in which the spin statistics

determine all of the interesting behavior! The fully quantum treatment of the

ideal gas will be the subject of the next several lectures.

The spin degree of freedom of a particle is a purely quantum mechanical aspect

(with no classical analog). In quantum mechanics, spin is analogous to an

angular momentum. It is described by a Hermitian operator , ,x y zS S S S ,

where the components satisfy angular momentum type commutation relations:

, , , , ,x y z y z x z x yS S i S S S i S S S i S (12.1)

The spin operators for a spin-s particle are represented by 2 1 2 1s s

matrices (which define different representations of the group SU(2)). For

example, for a spin-1/2 particle, such as an electron, the three spin operators are

0 1 0 1 0

, , 1 0 0 0 12 2 2

x y z

iS S S

i (12.2)

which can be shown to satisfy the above commutation relations. Since the three

components of spin to do not commute, we choose, by convention, to work in a

basis in which zS is diagonal. Thus, there will be (2s + 1) eigenvalues given by

,...,s s . In the example of the spin-1/2 particle, we see that the allowed spin

125

eigenvalues (denoted m) are / 2m and / 2m . The corresponding

eigenstates are just

1/2 1/2

1 0/ 2 , / 2

0 1m m (12.3)

which are denoted the ``spin-up'' and ``spin-down'' states, respectively. Note

that the operator 2 2 2 2

x y zS S S S is also diagonal so that the spin-up and spin-

down eigenstates of zS are also eigenstate of 2S , both having the eigenvalue 2( 1)s s . Thus, given a Hamiltonian H for a system, if H is independent of

spin, then the eigenstates of H must also be eigenstates of 2S and zS since all

three can be simultaneously diagonalized.

What happens in quantum mechanics when we have systems of identical

particles of a given type of spin? Consider the simple example of a system of

two identical spin-1/2 particles. Suppose we perform a measurement which is

able to determine that one of the particles has an zS eigenvalue of am and the

other bm such that a bm m . Is the state vector of the total system just after

this measurement

;a bm m or ;b am m (12.4)

where, in the first state, particles 1 and 2 have zS eigenvalues am and bm ,

respectively, and, in the second state, it is the reverse of this? The answer is that

neither state is the correct state vector since the measurement is not able to

assign the particular spin states of each particle. In fact, the two

state ;a bm m and ;b am m are not physically equivalent states. Two states

and can only be physicall equivalent if there is a complex

number such that

(12.5)

and there is no such number connecting ;a bm m and ;b am m . However, it is

possible to construct a new state vector ;a bm m such that ;b am m is

physically equivalent to ;a bm m . Let

; ; ;a b a b b am m C m m C m m (12.6)

126

If we require that

; ;a b b am m m m (12.7)

then

; ; ; ;a b b a b a a bC m m C m m C m m C m m (12.8)

from which we see that

, C C C C (12.9)

or

2C C (12.10)

from which 1 and C C . This gives us two possible physical states of

the system

; ; ; ,S a b a b b am m m m m m (12.11)

; ; ;A a b a b b am m m m m m (12.12)

which are symmetric and antisymmetric, respectively, with respect to an

exchange of the particle spin eigenvalues. The analog in ordinary one-

dimensional quantum mechanics would be the case of two identical particles

moving along the x axis. If a measurement performed on the system determined

that a particle was at position x = a and the other was at x = b, then the state of

the system after the measurement would be one of the two following

possibilities:

, ,S a b a b b a (12.13)

, A a b a b b a (12.14)

The standard postulates of quantum mechanics now need to supplemented by

an additional postulate that allows us to determine which of the two possible

physical states a system will assume. The new postulate states the following: In

nature, particles are of two possible types - those that are always found in

symmetric (S) states and those that are always found in antisymmetric (A)

127

states. The former of these are known as bosons and the latter are known as

fermions. Moreover, fermions possess only half-integer spin, s = 1/2, 3/2,

5/2,..., while bosons possess only integer spin, s = 0, 1, 2,....

Suppose a system is composed of N identical fermions or bosons with

coordinate labels 1,..., Nr r and spin labels 1,..., Ns s . Let us define, for each

particle, a combined label ,i i ix r s . Then, for a given permutation P(1),...,P(N)

of the particle indices 1,..,N, the wave function will be totally symmetric if the

particles are bosons:

B 1 B (1) ( ),..., ,...,N P P Nx x x x (12.15)

For fermions, as a result of the Pauli exclusion principle, the wave function is

antisymmetric with respect to an exchange of any two particles in the systems.

Therefore, in creating the given permutation, the wave function will pick up a

factor of 1 for each exchange of two particles that is performed:

F 1 F (1) ( ),..., 1 ,...,exN

N P P Nx x x x (12.16)

where exN is the total number of exchanges of two particles required in order to

achieve the permutation P(1),...,P(N). An N-particle bosonic or fermionic state

can be created from a state 1,..., Nx x which is not properly symmetrized but

which, nevertheless, is an eigenfunction of the Hamiltonian

H E (12.17)

Noting that there will be N! possible permutations of the N particle labels in

an N-particle state, the bosonic state B 1,..., Nx x is created from 1,..., Nx x

according to

!

B 1 1

1

1,..., ,...,

!

N

N Nx x P x xN

(12.18)

where P creates 1 of the N! possible permutations of the indices. The

fermionic state is created from

!

( )

F 1 1

1

1,..., 1 ,...,

!

ex

NN

N Nx x P x xN

(12.19)

128

where exN is the number of exchanges needed to create permutation .

This simple difference in the symmetry of the wavefunction leads to stark

contrasts in the properties of fermonic and bosonic systems. With these

quantum mechanical rules in mind, let us work out what these properties are.

12.2. Solution of the N-particle eigenvalue

problem

The Hamiltonian for an ideal gas of N particles is

2

1 2

Ni

i

PH

m (12.20)

The eigenvalue problem for the Hamiltonian is in the form of the time-

independent Schrödinger equation for the (unsymmetrized) eigenfunctions

2

2

1 1

1

,..., ,...,2

N

i N N

i

x x E x xm

(12.21)

First, we notice that the equation is completely separable in the N particle

coordinate/spin labels 1,..., Nx x , meaning that the Hamiltonian is of the form

2

1

, 2

Ni

i i

i

PH h h

m (12.22)

Note, further, that H is independent of spin, hence, the eigenfunctions must also

be eigenfunctions of 2S and zS . Therefore, the solution can be written as

1 1 ,..., 1

1

,...,N N i i

N

m m N m i

i

x x x (12.23)

where i im ix is a single particle wave function characterized by a set of

spatial quantum numbers i and zS eigenvalues im . The spatial quantum

numbers i are chose to characterized the spatial part of the eigenfunctions in

terms of appropriately chosen observables that commute with the Hamiltonian.

Note that each single-particle function i im ix can be further decomposed into

a product of a spatial function i ir and a spin eigenfunction

im is , where

129

m m mss s (12.24)

Substituting this ansatz in to the wave equation yields a single-particle wave

equation for each single particle function:

2

2

2 i i ii i ir rm

(12.25)

Here, i is a single particle eigenvalue, and the N-particle eigenvalue is,

therefore, given by

1 ,...,

1N i

N

i

E (12.26)

We will solve the single-particle wave equation in a cubic box of side L for

single particle wave functions that satisfy periodic boundary conditions:

, , , , , , , ,i i i ii i i i i i i i i i i ix y z x L y z x y L z x y z L (12.27)

Note that the momentum operator i commutes with the corresponding single-

particle Hamiltonian

, 0i ih (12.28)

This means that the the momentum eigenvalue ip is a good number for

characterizing the single particle states i ip

. In fact, the solutions of the

single-particle wave equation are of the form

expi

i ip i

ip rr C (12.29)

provided that the single particle eigenvalues are given by

2

2i

ip

p

m (12.30)

The constant C is an overall normalization constant on the single-particle states

to ensure that 2

3 1i id r r .

130

Now, we apply the periodic boundary condition. Consider the boundary

condition in the x-direction. The condition

, , , ,i ip i i i p i i ix y z x L y z (12.31)

leads to

exp expi ix i x iip x ip x L (12.32)

or

1 exp cos sini i ix x xip L p L p Li (12.33)

which will be satisfied if

2i

i

x

x

p Ln (12.34)

where ixn is an integer, 0, 1, 2,...Thus, the momentum

ixp can take on only

discrete values, i.e., it is quantized, and is given by

2

i ix xp nL

(12.35)

Applying the boundary conditions in y and z leads to the conditions

2

,i iy yp n

L (12.36)

2

i iz zp nL

(12.37)

Thus, the momentum vector ip can be written generally as

2

i ip nL

(12.38)

131

where in is a vector of integers , ,i i ii x y zn n n n . This vector of integers can be

used in place of ip to characterize the single-particle eigenvalues and wave

functions. The single-particle energy eigenvalues will be given by

2 2 2

2

2

2

2i

in i

pn

m mL (12.39)

and the single-particle eigenfunctions are given by

2

expi

i in i

in rr C

L (12.40)

Finally, the normalization constant C is determined by the condition

2 23

0 0 0

2 2exp exp 1,

i

L L L

i i i ii n i i i i

in r in rd r r C dx dy dz

2 2 3

30 0 0

1 11,

L L L

i i iC dx dy dz C L CVL

(12.41)

Therefore, the complete solution for the single-particle eigenvalues and

eigenfunctions is

2 2

2

2

21 2exp

i i i i i

i ii i m n m i m n ii

in rx n x s n

L mLV (12.42)

and the total energy eigenvalues are given by

1

2 22

,..., 21

2N

N

n n i

i

E nmL

(12.43)

Another way to formulate the solution of the eigenvalue problem is to consider

the single particle eigenvalue and eigenfunction for a given vector of integers n

2 2

2

2

21 2exp , i i

n n

in rr n

L mLV (12.44)

and ask how many particles in the N-particle system occupy this state. Let this

number be nmf . nmf is called an occupation number and it tells just how many

particles occupy the state characterized by a vector of integers n . Since there

132

are an infinite number of possible choices for n , there is an infinite number of

occupation numbers. However, they must satisfy the obvious restriction

nm

m n

f N (12.45)

where

x y zn n n n

(12.46)

And

s

m m s

(12.47)

runs over the 2s + 1 possible values of m for a spin-s particle. These occupation

numbers can be used to characterize the total energy eigenvalues of the system.

The total energy eigenvalue will be given by

nm

n nmfm n

E f (12.48)

12.3. An ideal gas of distinguishable

quantum particles

As an illustration of the use of occupation numbers in the evaluation of the

quantum partition function, let us consider the simple case of Boltzmann

statistics (ignoring spin statistics or treating the particles as distinguishable).

The canonical partition function , ,N V T can be expressed as a sum over the

quantum numbers 1,..., Nn n for each particle:

,...,1 1 2

1 2 1 2

, , ... ... ...n n nn nN N

N N

E

n n n n n n

N V T e e e e

1 2

1 2

... n nn n N N

N

N

n n n n

e e e e (12.49)

In terms of occupation numbers, the partition is

133

, ,n nn

f

f

N V T g f e (12.50)

where g f is a factor that tells how many different physical states can be

represented by a given set of occupation numbers f . For Boltzmann

particles, exchanging the momentum labels of two particles leads to a different

physical state but leaves the occupation numbers unchanged. In fact the

counting problem is merely one of determining how many different ways

can N particles be placed in the different physical states. This is just

!

!nn

Ng f

f (12.51)

For example, if there are just two states, then the occupation numbers must

be 1N and 2N where 1 2N N N . The above formula gives

1 2

1 2 1 1

! !,

! ! ! !

N Ng N N

N N N N N (12.52)

which is the expected binomial coefficient.

The partition function therefore becomes

!

, , n nf

f nnn

NN V T e

f (12.53)

which is just the multinomial expansion for

, , n

N

n

N V T e (12.54)

Again, if there were two states, then the partition function would be

1 2 1 1 2 2

1 2 1 2, , 1 2

!

! !

NN N

N N N N N

Ne e e e

N N (12.55)

using the binomial theorem.

Therefore, we just need to be able to evaluate the sum

134

22 2 22 /

nn mL

n n

e e (12.56)

But we are interested in the thermodynamic limit, where L . In this limit,

the spacing between the single-particle energy levels becomes quite small, and

the discrete sum over n can, to a very good approximation, be replaced by an

integral over a continuous variable:

2 22 2 2 2 2 22 / 2 /3n mL n mL

n

e d ne (12.57)

Since the single-particle eigenvalues only depend on the magnitude of n , this

becomes

22 2 2

3/2

2 /3 2

2 3

0

42

n mL m Vd nn e V (12.58)

where is the thermal deBroglie wavelength.

Hence,

3

, ,

NV

N V T (12.59)

which is just the classical result. Therefore, we see that an ideal gas of

distinguishable particles, even when treated fully quantum mechanically, will

have precisely the same properties as a classical ideal gas. Clearly, all of the

quantum effects are contained in the particle spin statistics. In the next few

lectures we will see just how profound an effect the spin statistics can have on

the equilibrium properties.

12.4. General formulation for fermions

and bosons

For systems of identical femions and identical bosons, an exchange of particles

does not change the physical state. Therefore the factor nmg f is just 1 for

both of kinds of systems. Moreover, the occupation number of a state

characterized by n for a system of identical bosons can be any number between

0 and N:

135

n 0,1,...,mf N (12.60)

For fermions, the Pauli exclusion principle forbids two identical particles from

occupying the same quantum state. This restricts the occupation numbers to be

either 0 or 1:

n 0,1mf (12.61)

Given these possibilities for the occupation numbers, the canonical partition

function can be formulated:

n nn n n

n n n

, ,mm

m m

f f

f f m

N V T e e (12.62)

Note that the sum over occupation numbers must be performed subject to the

restriction

n

n

m

m

f N (12.63)

a condition that makes the evaluation of , ,N V T extremely difficult.

Therefore, it seems that the canonical ensemble is not the best choice for

carrying out the calculation. No worry, there are other ensembles from which to

choose, and of these, it turns out that the grand canonical ensemble is

significantly easier to work with. Recall that in the grand canonical ensemble,

, V and T are the control variables and the partition function is given by

n n

n0 0 n

, , , , m

m

fN N

N N f m

V T N V T e e

(12.64)

Note that the inner sum over occupation numbers is still subject to the

restriction nn mm

f N . However, there is a final sum over all possible

values that N, the number that restricts the sum over occupation numbers, can

take on. Therefore, if we let the sum over occupation numbers be unrestricted,

then they could sum to any value they liked. This would be equivalent to

performing an unrestricted sum over occupation numbers without performing

the final sum over N, since in the course of summing, unrestricted, over

occupation numbers, we would obtain every possible value of N as required by

the final sum over N. This is the main advantage of using this ensemble for

bosonic and fermonic systems. Thus, the grand canonical partition function

becomes

136

n n

n n

, , m

m

f

f m

V T e

(12.65)

Note also that the sum of products is just

3 3 3 31 1 2 2 1 1 2 2

1 2 3 1 2 3

... ... ...f ff f f f

f f f f f f

e e e e e e

n n

nn

m

m

f

fm

e (12.66)

For bosons, each individual sum is just the sum of a geometric series. Hence,

n

n

1, ,

1m

V Te

(12.67)

whereas, for fermions, each individual sum contains only two terms

corresponding to n 0f and n 1f . Thus, for fermions:

n

n

, , 1m

V T e

(12.68)

Note that the summands are independent of the quantum number m so that we

may perform the product over m values trivially with the result

n

n

1, ,

1

g

V Te

(12.69)

for bosons and

n

n

, , 1

g

V T e

(12.70)

for fermions, where g = 2s + 1 is the number of eigenstates of Sz (also known

as the spin degeneracy).

At this point, let us recall the procedure for calculating the equation of state in

the grand canonical ensemble. The free energy in this ensemble is PV/kT given

by

137

ln , ,PV

V TkT

(12.71)

and the average particle number is given by

ln , ,N V T (12.72)

The fugacity must be eliminated in favor of N using the second equation

and substituted into the first equation to yield the equation of state. Recall that,

for the classical ideal gas,

3

, , exp ,V

V T (12.73)

3

,PV V

kT (12.74)

3

lnV

N (12.75)

Eliminating in favor N is trivial in this case, leading to the classical ideal

gas equation

PV N kT (12.76)

For the ideal gas of identical fermions, the equations one must solve are

n n

nn

ln , , ln 1 ln 1 ,

g

PVV T e g e

kT (12.77)

n

nn

ln1

eN g

e (12.78)

and for bosons, they are

n

nnn

1ln , , ln ln 1 ,

1

g

PVV T g e

kT e (12.79)

138

n

nn

ln1

eN g

e (12.80)

It is not difficult to see that the problem of solving for in terms of N is

highly non-trivial for both systems. The next two lectures will be devoted to

just this problem and exploring the rich behavior that the quantum ideal gases

exhibit.

139

LECTURE 13: THE IDEAL FERMION GAS

Content of Lecture 13

13.1. The fermiobn quantum ideal gas: Introduction

13.2. The high temperature, low density limit

13.3. The high density, low temperature limit

13.1. The fermion quantum ideal gas:

Introduction

For an ideal gas of fermions, we had shown that the problem of determining the

equation of state was one of solving two equations

ln 1 ,n

n

PVg e

kT (13.1)

n

nn 1

eN g

e (13.2)

where the second of these must be solved for in terms of and substituted

into the first to obtain P as a function of .

As we did in the Boltzmann case, let us consider the thermodynamic limit

L so that the spacing between energy levels becomes small. Then the

sums can be replaced by integrals over the continuous variable n . For the

pressure, this replacement give rise to

22 2 2

n2 /3 3ln 1 ln 1

n mLPVg d n e g d n e

kT

22 2 22 /2

0

4 ln 1n mL

g dnn e (13.3)

Change variables to

2 2

2

2x n

mL (13.4)

140

Then,

2 2

3/2

2 2

2 2 3

0 0

44 ln 1 ln 1

2

x xPV m ggV dxx e dxx e

kT (13.5)

The remaining integral can be evaluated by expanding the log in a power series

and integrating the series term by term:

2 2

1

1

1ln 1 ,

l l

x lx

l

e el

2

1 1

2

3 5/231 10

1 14l ll l

lx

l l

PV g Vgdxx e

kT l l (13.6)

By the same technique, the average particle number N can be shown to be

equal to

1

3 3/21

1l l

l

VgN

l (13.7)

Multipling both over these equations on both sides by 3 /V gives

13

5/21

1,

l l

l

P

gkT l

13

3/21

1l l

lg l (13.8)

Although exact solution of these equations analytically is intractable, we will

consider their solutions in two interesting limits: The high temperature, low

density limit and its counterpart, the low temperature, high density limit.

13.2. The high temperature, low density

limit

Since ( ) , in the low density limit, the fugacity can be expanded in the

form

141

2 3

1 2 3 ...a a a (13.9)

Writing out the first few terms in the pressure and density equations, we have

3 2 3 4

5/2 5/2 5/2...,

2 3 4

P

gkT

3 2 3 4

5/2 5/2 5/2...,

2 3 4g (13.10)

Substituting the expansion for into the density equation gives

3

22 3 2 3

1 2 3 1 2 33/2

1... ...

2a a a a a a

g

3

2 3

1 2 33/2

1... ...

3a a a (13.11)

This equation can now be solved perturbatively, equating like powers of on

both sides. For example, working only to first order in , yields:

3 3 3

1 1 a ag g g

(13.12)

When this is substituted into the pressure equation, and only first order terms in

the density are kept, we find

3 3

NP P

gkT g kT V (13.13)

which is just the classical ideal gas equation. Working, now, to second order in

, we have, from the density equation

3 3 6 2

2

2 3/2 2

1

2a

g g g (13.14)

or

6

2 3/2 22a

g (13.15)

142

Thus,

3 6

2

3/2 22g g (13.16)

and the equation of state becomes

3

2

5/22

P

kT g (13.17)

From this, we can read off the second virial coefficient

3 3

2 5/20.1768 0

2B T

g g (13.18)

It is particularly interesting to note that there is a nonzero second virial

coefficient in spite of the fact that there are no interactions among the particles.

The implication is that there is an ``effective'' interaction among the particles as

a result of the fermionic spin statistics. Moreover, this effective interaction is

such that is tends to increase the pressure above the classical ideal gas result (

2 0B T ). Thus, the effective interaction is repulsive in nature. This is a

consequence of the Pauli exclusion principle: The particle energies must be

distributed among the available levels in such a way that no two particles can

occupy the same quantum state, thus giving rise to an ``effective'' repulsion

between them.

If we look at the third order correction to the pressure, we find that

9

3 3/2 3

1 1,

4 3a

g

3 6 9

2 3

3/2 2 3/2 3

1 1,

2 4 3g g g

3 6

2 3

5/2 2 5.2

1 2

2 8 3

P

kT g g (13.19)

so that 3 0B T . Thus, one must go out to third order in the density expansion

to find a contribution that tends to decrease the pressure.

143

13.3. The high density, low temperature

limit

Recall that the density equation could be expressed as an integral

2

23

10

4

1x

g x dx

e (13.20)

which lead to an expansion in powers of . It is also possible to develop an

expansion in powers of ln /mn kT . This is accomplished by letting

lnkT

(13.21)

and developing an expansion in powers of . In order to see how this is done,

consider making a change of variables in the integral 2y x , x y ,

/ 2dx dy y . Then

3

0

2

1y

ydyg

e (13.22)

Integrate by parts using

2

1 1, ,

1 1

y

y yu du e dy

e e

1/2 3/22,

3dv y dy v y (13.23)

so that

3/2

3

2

0

4

3 1

y

y

g y e dy

e (13.24)

If we now expand 3/2y about y :

144

23/2 3/2 1/2 1/23 3

...2 8

y y y (13.25)

substitute this expansion into the integral and perform the resulting integrals

over y, we find

2

3/2 1/23 4ln ln ... 1/

83

gO (13.26)

where the fact that / ( ) 1kT has been used owing to the low temperature.

Since we are in the high density limit, is expected to be large as well so

that the series, whose error goes as powers of 1/ will converge. As

0,T and only one term in the above expansion survives:

3/22

3/23 2 4 4ln

3 3

g g

mkT kT (13.27)

Solving for gives

2/32 2

0

6

2F

m g (13.28)

which is independent of T. The special value of the chemical potential

0 0T is known as the Fermi energy. To see what its physical meaning

is, consider the expression for the average number of particles:

n

nn 1m

eN

e (13.29)

However, recall that

n

n

m

m

f N (13.30)

for a specific number of particles. Averaging both sides gives

n

n

m

m

N f (13.31)

145

Comparing these two expressions, we see that the average occupation number

of a given state with quantum number n and m is

n

n nn

1

1 1m

ef

e e (13.32)

As 0,T and n 0e if n 0 and n 0 0e if n 0 . Thus,

at T = 0, we have the result

n F

n F n

n F

0

1 mf (13.33)

A plot of the occupation average occupation number vs n at T = 0 is shown in

the plot below:

Figure 13.1

Thus, at T = 0, the particles will exactly fill up all of the energy levels up to an

energy value F above which no energy levels will be occupied. As T is

increased, the probability of an excitation above the Fermi energy becomes

nonzero, and the average occupation (shown for several different values of )

appears as follows:

146

Figure 13.2

Thus, there is a finite probability that some of the levels just above the Fermi

energy will become occupied as T is raised slightly above T = 0. At T = 0, the

highest occupied energy eigenvalue must satisfy

n F ,

2 2

2

F2

2n ,

mL

2 2

2 2 2

F2

2x y zn n n

mL (13.34)

This defines a spherical surface in n space, which is known as the Fermi

Surface. Note that the Fermi surface is only a sphere for the ideal gas. For

systems in which interactions are included, the Fermi surface can be a much

more complicated surface, and studying the properties of this surface is a task

that occupies the time of many a solid-state physicist.

13.3.1. Zero-temperature thermodynamics

In order to derive an expression for the average particle number, recall that

n F n F n

n n n

m

m m

N f g (13.35)

In the thermodynamic limit, we may take the sum over to an integration:

147

2

F n

0

4N g dnn (13.36)

But

2 2

2

n 2

2n

mL (13.37)

Therefore, it proves useful to change variables of integration from n to n ,

using the above relation:

1/22

1/2

n2 2,

2

mLn

1/22

1/2

n n2 2

1

2 2

mLdn d (13.38)

Thus,

3/22

2 1/2

F n n n F n2 2

0 0

4 22

mLN g dnn d

F

3/2 3/221/2 3/2

F2 2 2

0

42

2 3 2

mL g mg d V (13.39)

In order to derive an expression for the average energy, recall that the energy

eigenvalues were given by

n

n n

n

mfm

E f (13.40)

Therefore, the average energy is given by

n n

n

m

m

H E f (13.41)

At T = 0, this becomes

3 2

F n n F n n F n n

n 0

n 4E g g d g dnn (13.42)

148

If the same change of variables is made, one finds that

3/22

3/2

n n F n2 2

0

14

2 2

mLE g d

F3/2 3/2

3/2 5/2

n n F2 2 2 2

0

42

2 5 2

m g mg V d (13.43)

Thus, the average energy can be seen to be related to the average particle

number by

F

3

5E N (13.44)

which is clearly not 0 (as it would be classically).

Note that the pressure can be obtained simply in the following way: Recognize

that

1

3 5/21

1ln , ,

l l

l

PV VgV T

kT l

(13.45)

The energy is given by

,

ln , ,V

E V T (13.46)

Thus,

1

3 5/21

13

2

l l

l

VgE

l

(13.47)

Comparing these two equations for the energy and pressure shows that

3 2

2 3

EE PV P

V (13.48)

Note, that just like the energy, the pressure at T = 0 is not zero. The T = 0

values of both the energy and pressure are:

149

F

3,

5E N (13.49)

F

2

5

NP

V (13.50)

These are referred to as the zero-point energy and pressure and are purely

quantum mechanical in nature. The fact that the pressure does not vanish at T =

0 is again a consequence of the Pauli exclusion principle and the effective

repulsive interaction that also showed up in the low density, high temperature

limit. Using the expansion for 3 , we can derive the thermodynamics in this

limit.

13.3.2. Thermodynamics at low temperature

Finite temperature thermal corrections can be obtained by starting with the

expansion derived earlier: Note that

2

3/2 1/23 4ln ln ...

83

g

23/2 1/2 3/22 24 4... 1 ...

8 83 3

g g kT

kT kT kT (13.51)

The term proportional to 2T is a small thermal correction to the T = 0 limit. As

such, it is small and we can replace the appearing there with 0 F to the

same order in T:

23/2 23 F

F

41 ...

83

g kT

kT (13.52)

Solving this, now, for (which is equivalent to solving for ) gives

2/3 23 2

F2/32

2 F

F

3 11 ...

4 12

18

kTkT

gkT

(13.53)

where the second line is obtained by expanding 2/3

1/ 1 x about x = 0.

150

In order to obtain the thermal corrections, one must expand the average

occupation number formula about the 0 F value using the expansion

obtained above for and the do the integrals. The result is simply

2

2

F

F

3 51 ...

5 12

kTE N (13.54)

The thermal correction is necessary in order to obtain the heat capacity at

constant volume, which is given by

V

V

EC

T (13.55)

Using the above expression for the energy, one finds

2

F2

VC kT

N k (13.56)

From the thermally-corrected expression for the energy, the pressure can be

obtained immediately:

2

2

F

F

2 51 ...

5 12

kTP (13.57)

151

LECTURE 14: THE IDEAL BOSON GAS

Content of Lecture 14

14.1. The ideal boson gas: Introduction

14.2. Low density, small limit

14.3. The high density, 1 limit

14.1. The ideal boson gas: Introduction

For the bosonic ideal gas, one must solve the equations

n

n

ln 1 ,PV

g ekT

(14.1)

n

nn 1

eN g

e (14.2)

in order to obtain the equation of state. Examination of these equations,

however, shows an immediate problem: The term n 0,0,0 is divergent both

for the pressure and the average particle number. These terms need to be treated

carefully, and so we split them off from the rest of the sum, giving:

n

'

n

ln 1 ln 1 ,PV

g e gkT

(14.1)

n

n

'

n 1 1

eN g g

e (14.2)

where '

n

means that the n 0,0,0 term is excluded. With these divergent

terms split off, the thermodynamic limit can be taken and the remaining sums

converted to integrals as was done in the fermion case. Thus, for the pressure,

we find

nn ln 1 ln 1PV

g d e gkT

152

22 2 22 n /2

0

4 . ln 1 ln 1mL

g dn n e g

22

3

0

4. ln 1 ln 1xVg

dx x e g (14.3)

where the change of variables

2 2

2

2x n

mL (14.4)

has been made. Using the expansion

2 2

1

ln 1l

x lx

l

e el

(14.5)

the pressure equation becomes

3 3

5/21

ln 1l

l

P

gkT l V (14.6)

and by a similar procedure, the average particle number becomes

3 3

3/21 1

l

lg l V (14.7)

In this equation, the term that has been split off represents the average

occupation of the ground ( n 0,0,0 ) energy state:

0

1f (14.8)

Since 0f must be greater than or equal to 0, it can be seen that there are

restrictions on the allowed values of . Firstly, since e , must be a

positive number. However, in order that the average occupation of the ground

state be positive,

0 1 (14.9)

from which it follows that

153

0 (14.10)

The fact that as 1 causes 0f to diverge will have interesting

consequences to be discussed below. However, let us first consider the low

density limit with 1.

14.2. Low density, small limit

In a manner completely analogous to what was done for the fermion case, the

low density limit can be treated by perturbation theory. Note that if is not

close to 1, then the divergent terms, which have a 3 /V prefactor

accompanying them, will vanish in the thermodynamic limit. Thus, for the

proceeding analysis, these terms can be neglected.

As before, we assume the fugacity can be expanded as

2 3

1 2 3 ...a a a (14.11)

Then the equation for the density becomes

3

22 3 2 3

1 2 3 1 2 33/2

1... ...

2a a a a a a

g

3

2 3

1 2 33/2

1...

3a a a (14.12)

By equating like powers of on both sides, the coefficients 1 2 3, , ,...a a a can be

determined as they were for the fermion gas. Working to first order in gives

3 3

1 , ag g

(14.13)

and the equation of state is

P

kT (14.14)

which is just the classical ideal gas equation. To second order, however, we

find

154

6 3 6

2

2 3/2 2 3/2 2,

2 2a

g g g (14.15)

and the second order equation of state becomes

3

2

5/22

P

kT g (14.16)

The second virial coefficient can be read off and is given by

3 3

2 5/2

1 0.17680

2B T

g g (14.17)

Interestingly, in contrast to the fermionic system, the pressure is actually

decreased from its classical value as a result of bosonic spin statistics. Thus, it

appears that there is an ``effective attraction'' between the particles. This fact is

not entirely unexpected, given that any number of bosons can occupy the same

quantum state.

14.3. The high density, 1 limit

The 1 limit is the limit of maximum chemical potential, which is expected

at high density. However, since 0 , maximum chemical potential will be the

limit 0. In this limit, the full problem, including the divergent terms, must

be solved:

3 3

5/21

ln 1 ,l

l

P

gkT l V

3 3

3/21 1

l

lg l V (14.18)

We will need to refer to these two sums often in this section, so let us define

them to be

3/2 3/21

,l

l

gl

155

5/2 5/21

l

l

gl

(14.19)

Thus, the problem becomes one of solving

3 3

5/2 ln 1 ,P

ggkT V

3 3

3/21

gg V

(14.20)

We examine, first the density equation. The second term will diverge at 1.

It is instructive to ask what is the behavior of the first term 3/2g at 1. In

fact 3/2 1g is nothing but a Riemann zeta-function:

3/2 3/2

1

11 3 / 2

l

g Rl

(14.21)

In general, a Riemann zeta-function R(n) is given by

1

1n

l

R nl

(14.22)

and the values of this function are given in many standard math tables. The

particles value of R(3/2) is approximately 2.612... Moreover, from the form of

3/2g , it is clear that, since 1, 3/2 1g is the maximum value of 3/2g . A

plot of 3/2g is given below:

156

Figure 14.1

The figure also indicates that the derivative 3/2g diverges at 1 despite the

fact that the value of the function is finite. Note that, since 1

3/2 5/2g g (14.23)

It is possible to solve the density equation for by noting that unless is very

close to 1, the divergent term will still vanish in the thermodynamic limit as a

result of its 3 /V prefactor. How close to 1 must it be for this term to

dominate? It can only be different from 1 by an amount on the order of 1/V.

Thus, let us take to be of the form

1a

V (14.24)

where a is a positive constant. Substituting this ansatz into the equation for the

density gives

3 3

3/2

1 /1 /

/

a Vg a V

g V a V (14.25)

Since 3/2g does not change its value much if is displaced just a little from

1, we can replace the first term by R(3/2). Then,

3 3

3/2

1 /1

/

a Vg

g V a V (14.26)

157

can be solved for a to yield

3

3

3 / 2

a

Rg

(14.27)

where we have neglected a term 3 /V , which vanishes in the thermodynamic

limit. Since a must be positive, this solution is only valid for 3 / 3 / 2g R .

For 3 / 3 / 2 , g R will be different from 1 by more than an amount 1/V so

in this regime, the / 1 term can be neglected, leaving the problem of

solving 3

3/2/ g g . Therefore, the solution for can be expressed as

3 3

3

3 3

3/2

/1 3 / 2 ,

3 / 2

root of 3 / 2

VR

gR

g

g Rg g

(14.28)

which, in the thermodynamic limit, becomes

3

3 3

3/2

1 3 / 2 ,

root of 3 / 2

Rg

g Rg g

(14.29)

A plot of vs. 3 3/ /v V N is shown below:

158

Figure 14.2

Clearly, point R(3/2) is special, as undergoes a transition there to a constant

value of 1.

Recall that the occupation of the ground state is

0

1f (14.30)

Thus, for 1 /a V , this becomes

3

0 33 / 2

V Vf R

a g (14.31)

for 3 / 3 / 2g R . At 3 / 3 / 2g R the occupation of the ground state

becomes 0. To what temperature does this correspond? We can find this out by

solving

2/33/23 2 2

0

0

2 23 / 2 , 3 / 2 ,

3 / 2R R kT

g g mkT gR m (14.32)

so that for temperatures less than 0T the occupation of the ground state becomes

0 3 31 3 / 2 1 3 / 2

NV g gf R R

g g

159

3/23/2

2

0

3 / 21 1 ,

2

N NgR mkT T

g g T

3/2

0

0

11

f T

N g T (14.33)

Thus, at T = 0

0

Nf

g (14.34)

which is equivalent to

n 0,0,0 ,m

Nf

g (14.35)

If we sum both sides over m, this gives

n 0,0,0 ,

,m

m m

Nf

g

0f N (14.36)

where 0f indicates that the spin degeneracy has been summed over. For

3

0 , / 3 / 2T T g R and is not within 1/V of 1. This means that

/ 1 is finite and

0 10

1

f

N N (14.37)

as N . Therefore, we have, for the occupation of the ground state:

3/2

0 00

0

1 /

0

T T T Tf

N T T (14.38)

which is shown in the figure below:

160

Figure 14.3

The occupation of the ground state undergoes a transition from a finite value to

0 at 0T T and for all higher temperatures, remains 0. Now,

0 /f N represents the probability that a particle will be found in the ground

state. It also represents the fraction of the total number of particles that will be

found in the ground state. For 0T T , this number is very close to 1, and at T =

0, it becomes exactly 1, implying that at T = 0 all particles will be found in the

ground state. This is a phenomenon known as Bose-Einstein condensation. The

occupation number of the ground state as a function of temperature is shown in

the plot below:

Note that there is also a critical density corresponding to this temperature. This

will be given by the solution of

3

3 / 2Rg

(14.39)

which can be solved to yield

3/2

003 2

3 / 23 / 2

2

gR mkTgR (14.40)

and the occupation number, expressed in terms of the density is

0 0 0

0

1 /

0

f

N (14.41)

161

The term in the pressure equation

3

ln 1V

(14.42)

becomes, for very close to 1

3 ln

ln ~V V

V a V (14.43)

which clearly vanishes in the thermodynamic limit, since ~V N . This allows

to deduce the equation of state as

3

5/2 0

3

5/2 0

(1) /

( ) /

gP

gkT g (14.44)

where in the above equation comes from the actual solution of 3

3/2/ g g . What is particularly interesting to note about the equation of

state is that the pressure is independent of the density for 0 . Isotherms of

the ideal Bose gas are shown below:

Figure 14.4

Here, 0v corresponds to the critical density 0 . As a function of temperature,

we see that 5/2~P T , which is quite different from the classical ideal gas. This is

also in contrast to the fermion ideal gas, where as 0T the pressure remains

finite. For the Boson gas, as 0T the pressure vanishes, in keeping with the

notion of an ``effective'' attraction between the particles.

162

Other thermodynamic quantities can be determined in a similar manner. The

energy can be obtained from E = 3PV/2 straightforwardly:

5/2 0 03

5/2 0 03

3(1) ,

2

3( ) ,

2

kTVg T T

EkTV

g T T

(14.45)

and the heat capacity at constant volume from

V

V

EC

T (14.46)

which gives

5/203

5/2 3/203

1/2

(1)15

4

( ) ( )15 9

4 4 ( )

V

gT T

C

g gN kT T

g

(14.47)

A plot of the heat capacity exhibits a cusp at 0T T :

Figure 14.5

Experiments carried out on liquid He , which has been observed to undergo

Bose-Einstein condensation at around T = 2.18 K, have measured an actual

discontinuity in the heat capacity at the transition temperature, suggesting that

163

Bose-Einstein condensation is a phase transition known as the transition. The

experimental heat capacity is shown roughly below:

Figure 14.6

By contrast, the ideal Bose gas undergoes a first order phase transition.

However, using the mass and density of liquid He in the expression for

0T given above, one would predict that 0T is about 3.14 K, which is not far off

the experimental transition temperature of 2.18 K for real liquid helium.

For completeness, other thermodynamic properties of the ideal Bose gas are

given as follows: The entropy is

5/2 03

5/2 03

5 1(1)

2

5 1( ) ln

2

g T TS

N kg T T

(14.48)

The Gibbs free energy is given by

0

0

0

ln

T TG

N k T T (14.49)

It is clear from the analysis of this and the fermion ideal gas that quantum

statistics give rise to an enormously rich behavior, even when there are no

particle interactions!

END

164