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The document is a textbook titled 'Group Theory for Physicists' by Pichai Ramadevi and Varun Dubey, focusing on the application of group theory in physics, covering discrete and Lie groups. It includes essential concepts such as molecular symmetry, representations of finite groups, and real-life applications in various fields, supported by solved problems and exercises. The authors aim to provide a comprehensive understanding of group theory to undergraduate students and researchers in physics and related disciplines.

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Ramadevi Dubey

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Group Theory for Physicists

Group theory helps readers in understanding energy spectrum and the degeneracy of
systems possessing discrete symmetry and continuous symmetry. This text covers two
essential aspects of group theory, namely discrete groups and Lie groups. Important
concepts including permutation groups, point groups, and irreducible representation
related to discrete groups are discussed with the aid of solved problems. Topics such
as the matrix exponential, the circle group, tensor products, angular momentum
algebra, and the Lorentz group are explained to help readers in understanding the
quark model and the corresponding bound states of quarks. Real life applications
including molecular vibration, level splitting perturbation, and the orthogonal group
are also covered. Application-oriented solved problems and exercises are interspersed
throughout the text to reinforce understanding of the key concepts.
Pichai Ramadevi is Professor of Physics at the Indian Institute of Technology Bombay,
India. She has taught courses including general relativity, group theory methods,
particle physics, quantum mechanics and quantum physics. She has published more
than 50 papers in journals of national and international repute. Her research interests
include string theory, knot theory, quantum field theory, and supersymmetry.
Varun Dubey is a researcher at the International Center for Theoretical Sciences,
Bangalore, India. His research interests include probability theory, stochastics in
quantum dynamics, and geometric quantization.
Group Theory for Physicists
with Applications

Pichai Ramadevi
Varun Dubey
University Printing House, Cambridge CB2 8BS, United Kingdom
One Liberty Plaza, 20th Floor, New York, NY 10006, USA
477 Williamstown Road, Port Melbourne, VIC 3207, Australia
314 to 321, 3rd Floor, Plot No.3, Splendor Forum, Jasola District Centre, New Delhi 110025, India
79 Anson Road, #06–04/06, Singapore 079906

Cambridge University Press is part of the University of Cambridge.

It furthers the University’s mission by disseminating knowledge in the pursuit of
education, learning and research at the highest international levels of excellence.

www.cambridge.org
Information on this title: www.cambridge.org/9781108429474
© Pichai Ramadevi and Varun Dubey 2019
This publication is in copyright. Subject to statutory exception
and to the provisions of relevant collective licensing agreements,
no reproduction of any part may take place without the written
permission of Cambridge University Press.
First published 2019
Printed in India
A catalogue record for this publication is available from the British Library
Library of Congress Cataloging-in-Publication Data
Names: Ramadevi, Pichai, author. | Dubey, Varun, author.
Title: Group theory for physicists : with applications / Pichai Ramadevi,
Varun Dubey.
Description: Cambridge ; New York, NY : Cambridge University Press, 2019. |
Includes bibliographical references and index.
Identifiers: LCCN 2019016363 | ISBN 9781108429474 (alk. paper)
Subjects: LCSH: Group theory–Textbooks. | Mathematical physics–Textbooks.
Classification: LCC QC20.7.G76 R354 2019 | DDC 512/.2–dc23
LC record available at https://lccn.loc.gov/2019016363
ISBN 978-1-108-42947-4 Hardback
ISBN 978-1-108-45427-8 Paperback
Cambridge University Press has no responsibility for the persistence or accuracy
of URLs for external or third-party internet websites referred to in this publication,
and does not guarantee that any content on such websites is, or will remain,
accurate or appropriate.
Contents

List of Figures ix
List of Tables xi
Preface xiii

1 Introduction 1
1.1 Definition of a Group 1
1.2 Subgroups 4
1.3 Conjugacy Classes 5
1.4 Further Examples of Groups 6
1.5 Homomorphism of Groups 8
1.6 The Symmetric Group 9
1.7 Direct and Semi-direct Products 14
Exercises 16

2 Molecular Symmetry 18
2.1 Elements of Molecular Symmetry 18
2.2 The Symmetry Group of a Molecule 20
2.3 Symmetry Point Groups 23
Exercises 28

3 Representations of Finite Groups 29

3.1 Vector Spaces 29
3.2 Group Action on Vector Spaces 36
3.3 Reducible and Irreducible Representations 38
3.4 Irreducible Representations of Point Groups 41
3.5 The Regular Representation 44
3.6 Tensor Product of Representations 45
vi Contents

3.7 Decomposition of Reducible Representations 46

3.8 Irreducible Representations of Direct Products 52
3.9 Induced Representations 54
Exercises 57

4 Elementary Applications 59
4.1 General Considerations 59
4.2 Level Splitting under Perturbation 62
4.3 Selection Rules 64
4.4 Molecular Vibrations 71
Exercises 85

5 Lie Groups and Lie Algebras 87

5.1 The Circle Group U(1) 87
5.2 The Matrix Exponential 89
5.2.1 Generators of the Lie group 89
5.2.2 Convergence property of matrix exponentials 90
5.3 Finite Dimensional Lie Algebras 92
5.3.1 su(2) algebra 99
5.4 Semi-simple Lie Algebras 101
5.5 Lie Algebra of a Lie Group 103
5.6 Examples of Lie Groups 105
5.7 Compact Simple Lie Algebras 107
5.7.1 su(3) algebra 110
5.7.2 Cartan matrix 113
5.7.3 Fundamental weights 113
Exercises 114

6 Further Applications 115

6.1 Continuous Symmetry and Constant of Motion 117
6.1.1 Translational symmetry in three-dimensional space 117
6.2 Tensor Product Rule for SU (2) Irreducible Representations 119
6.2.1 Digression on the Young tableau approach 119
6.2.2 SU (2) Clebsch–Gordan matrix 124
6.2.3 Wigner–Eckart theorem 128
6.3 Elementary Particles in Nuclear and Particle Physics 130
6.3.1 Isospin symmetry 131
6.4 Quark Model 132
6.4.1 SU (3) group approach quark model 133
6.4.2 Antiparticles 136
6.4.3 Symmetry breaking from SU (3) → SU (2) 137
Contents vii

6.5 Non-compact Groups 139

6.5.1 The Lorentz group 139
6.5.2 Poincare group 143
6.5.3 Scale invariance 143
6.5.4 Conformal group 143
6.6 Dynamical Symmetry in Hydrogen Atom 144
6.6.1 Lie algebra symmetry 145
6.6.2 Energy levels of hydrogen atom 146
Exercises 148

Appendix A: Maschke’s Theorem 151

Appendix B: Schur’s Lemma 153
Bibliography 156
Index 157
Figures

2.1.1 Roto-Reflection Symmetry. The figure shows the action of Cn σh and σh Cn

on an atom A of some molecule 20
2.1.2 Inversion Symmetry. The central dark sphere is the point of symmetry of
this hypothetical molecule 20
2.2.1 Equivalent and Non-equivalent Reflection Planes 22
2.2.2 Symmetry Axes of a Cube 22
2.2.3 Bilateral Axis 23
2.3.1 Stereographic Projections 24
2.3.2 σd U2 Transformation in Dnd Group 26
2.3.3 Symmetry Axes of the Tetrahedron 27
2.3.4 Hypothetical Molecule 28
3.1.1 R2 35
4.4.1 Normal Modes (Non-linear Tri-atomic Molecule) 75
4.4.2 Hypothetical Equilateral Molecule 78
5.1.1 The Circle Group U(1) 88
5.3.1 Group Manifold of the SO(3) (doubly-connected space) 95
5.7.1 Dynkin Diagrams 110
5.7.2 su(3) Weight Diagram for Defining Representation 111
5.7.3 su(3) Root Diagram 112
6.4.1 Quark Model 133
Cambridge University Press
978-1-108-42947-4 — Group Theory for Physicists
Pichai Ramadevi , Varun Dubey
Frontmatter
More Information

Tables

1.1 Klein-4 Group V 3

1.2 Symmetric Group S(3) 3

© in this web service Cambridge University Press www.cambridge.org

Preface

This text grew out of the group theory lectures taught by the first author to
undergraduate students at IIT Bombay. While most students attending the course
were majoring in physics, there were also students from electrical, mechanical
and aerospace engineering streams. The continued interest of students in the topic
motivated us to compile the contents into a textbook suitable for a course pitched
at the undergraduate level. We also hope that the text will be useful to researchers
who wish to familiarize themselves with the basic principles and typical examples of
applications of group theory in physics.
In our day-to-day life as well as in the laboratories, we observe patterns with
symmetry. These could be in the symmetric shape of the wings of a butterfly, the
arrangement of atoms in a molecule, the shape of nuclei as inferred by advanced
experimental techniques, among many others. Group theory is a mathematical
formulation of such symmetries. While group theory is of interest in mathematics,
the prevalence of symmetries in the physical world enable it to find applications
in several other disciplines. For instance, the synthesis of molecules in chemistry,
crystallographic structures in physics, elasticity properties in continuum mechanics,
etc., require the basics of the symmetry principles.
Understanding complex physical systems in nature by solving complicated
mathematical equations can be a daunting task. Group theory offers an elegant and
powerful framework to extract certain details about such complex systems. For
example, the underlying group symmetry can assert whether a given scattering or
decay process is allowed or forbidden without performing any explicit computation.
This book employs solved examples as well as a variety of exercises (sometimes drawn
from sources listed in the Bibliography) to help the reader appreciate the role of group
theory in explaining certain experimental observations. We have tried to balance
between the formal mathematical aspects and the applications so that the student
can effortlessly absorb the group theory arguments behind ad hoc postulates and
xiv Preface

selection rules. The contents and presentation style enable undergraduate students to
connect with the physics they have learned, or will be learning in courses like quantum
mechanics, continuum mechanics, atomic and molecular physics, condensed matter
physics, nuclear physics, and particle physics in their undergraduate curriculum.
A comprehensive understanding of the concepts in this book will build a strong
foundation, preparing the students for further study in diverse areas such as
theoretical condensed matter physics or high energy physics, theory including string
theory. The first four chapters address discrete group concepts and their applications.
The introductory chapter contains definitions and examples, including permutation
group. Chapter 2 contains discussion on molecular symmetry, broadly known as point
groups. After an initial review of vector spaces and group action on vector spaces,
Chapter 3 elaborates on reducible and irreducible representation, character tables,
tensor products and their decomposition. This provides the necessary background to
determine the vibrational modes of molecules, splitting of degenerate energy level due
to impurity or defect which breaks the symmetry of the system partially or completely,
and selection rules for transition between energy levels. These applications are
elaborated in Chapter 4.
Continuous groups (also known as Lie groups) are discussed in Chapter 5 and
Chapter 6. Some of the concepts and notations elaborated in the context of discrete
groups are useful while studying systems possessing continuous group symmetry.
The key highlights of these two chapters are: (i) Young diagram presentation of
irreducible representations for both permutation group as well as special unitary
groups which clarifies concepts like tensor products and the decomposition of
irreducible representations; (ii) a warm-up on angular momentum algebra which
naturally facilitates the description of formal aspects of special unitary group where
we introduce weight vectors, root vectors and the Dynkin diagrams; (iii) orthogonal
groups and their applications to the energy spectrum of hydrogen atom.
The task of structuring this book in print form would not have been possible
without the help of our colleagues and students. We are grateful to Deepak and
Himanshu for their timely help in fixing LYX syntax errors which we faced while
compiling the chapters. We would like to thank Vivek for helping us with the
drawings for some of the figures. We would also like to thank Adiba for helping us
in typesetting two sections in LYX package, as well as for proof reading the contents
of this book. Our thanks are due also to Himanshu, Aman, Gurbir, Anish, Amihay,
Ayaz, Saswati, Zodin, Lata, and Abhishek for their help and valuable comments
during various stages in the progress of the book. Besides these students, we greatly
appreciate Urjit, Uma, Punit, Pradeep, and Vikram for their comments. Finally, we
would like to thank our family members for their encouragement.
We envision that the book will be most suited to teachers offering a one-semester
group theory course to undergraduate students. It is also of relevance to researchers
embarking on a study of this field. We wish all the readers enjoy their journey through
the contents.

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1
Introduction

We observe objects in nature to have some pattern or symmetry. In fact, the symmetries
inherent in any physical system play a very crucial role in the study of such
systems. Group theory is a branch of mathematics that facilitates classification of
these symmetries. Hence learning the group theory tools will prove useful to studying
applications in physics. Readers will particularly appreciate the power and elegance
of the group theoretical techniques in reproducing the experimental observations.
Before delving into its applications, it is important to understand the concept of an
abstract group from a purely mathematical standpoint. In this chapter, we present the
formal definition of a group and also the notations which will be followed in the rest
of the book.

1.1 Definition of a Group

Definition 1. A group G is a set on which is defined a binary operation called the
product having the following properties:
1. Closure: For all a and b in G, the product ab is in G. Here a and b need not be
distinct.
2. Associativity: ( ab)c = a(bc) for all a, b and c in G.
3. Existence of Identity: There exists a unique e in G such that ae = ea = a for all a
in G. e is called the identity element of the group.
4. Existence of Inverse: For every a in G there exists a unique b in G such that
ab = ba = e. b is called the inverse of a and is conventionally denoted by a−1 .
2 Group Theory for Physicists

In addition to the above mentioned axioms, if it is also true that ab = ba for all a and b
in G, then G is said to be an abelian group. It must be noted that the property of being
abelian is special in the sense that not all groups need be abelian. In any group, it is
trivial to prove the following statements:
ax = bx ⇒ a = b (1.1.1)
xa = xb ⇒ a = b
( ab)−1 = b−1 a−1 .
Due to the obvious simplicity of the definition, many familiar sets in mathematics are
indeed seen to be examples of groups.

Example 1. The set of all integers Z is a group if the group product is taken to be the
usual addition of integers. This group is clearly abelian and has an infinite number of
elements.

Example 2. The set of all complex numbers C is a group under addition of complex
numbers. This group again is abelian and infinite.
Example 3. The set C − {0} is an infinite abelian group under the usual multiplication
of complex numbers.
Example 4. The set of all 2 × 2 matrices with complex entries is an infinite abelian
group under matrix addition.
Example 5. The set of all invertible 2 × 2 matrices with complex entries is an infinite
non-abelian group under matrix multiplication.
A group G that contains a finite number of elements is called a finite group. The
number of elements in a finite group G is called the order of the group and is denoted
by | G |. For any element a in a group and a positive integer n, an represents aa...a where
there are n factors in the product. Similarly a−n represents ( a−1 )n . Before looking at
some examples of finite groups, the following definition may be noted.

Definition 2. A subset S of elements of a group G is said to generate G if every

element of G can be expressed as a finite product of finite powers of elements (or
their inverses) of S in some order. If the set S is finite then the group G is said to
be finitely generated. The elements of the minimal set S that generates a group G are
called the generators of the group and S itself is called the generating set. A group whose
generating set contains a single element is said to be a cyclic group.
A group is completely specified if its generators are known along with all the
relations that exist between them. It is important to realize that a group may have
more than one generating set. In case of finite groups though, the number of elements
in all possible generating sets is the same.

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Introduction 3

Table 1.1 Klein-4 Group V

V e a b ab
e e a b ab
a a e ab b
b b ab e a
ab ab b a e

Example 6. The set {e} along with the relation e2 = e generates the trivial group {e}
called the identity group of order 1. Henceforth, this group would be represented by
the symbol E.
Example 7. The set { a} along with the relation a2 = e generates the group {e, a} of
order 2.
Example 8. The set { a} along with the relation an = e (n being a positive integer)
generates the cyclic group {e, a, a2 , . . . , an−1 } of order n. Henceforth, this group
would be represented by the symbol Cn .
Example 9. The set { a, b} along with the relations a2 = b2 = e and ab = ba generates
the abelian group V = {e, a, b, ab}. The group V is called the Klein-4 group. It is
useful to depict this group in the form of a multiplication table showing all possible
products of various group elements as in Table 1.1. Such a table can be drawn for all
finite groups.
Example 10. A slightly less trivial example is that of a finite group generated by the
set { a, b} where a and b satisfy the relations a2 = b3 = e and ab = b2 a. The generated
group S(3) = {e, a, b, b2 , ab, ab2 } is of order 6 and is non-abelian (Table 1.2). Any
product involving a finite number of a’s and b’s can be reduced to one of the elements
of S(3) by use of the relations on a and b. The notation S(3) will be clarified later.

Table 1.2 Symmetric Group S(3)

S(3) e a b b2 ab ab2
e e a b b2 ab ab2
a a e ab ab2 b b2
b b ab2 b2 e a ab
b2 b2 ab e b ab2 a
ab ab b2 ab2 a e b
ab2 ab2 b a ab b2 e

1.2 Subgroups
Definition 3. A subset H of a group G is called a subgroup if H is a group under
the product operation defined on G. The identity group E and the group G are both
subsets of the group G and are called the trivial subgroups of G. Other subgroups of G
are called non-trivial subgroups.
For a subset H of a group G to be a subgroup, H must satisfy all the axioms stated
in the Definition 1. If H is closed under the group product and every element of H
has its inverse in H, then other axioms are automatically satisfied since H is merely a
subset of the group G. If G is a finite group, then the condition on H to be a subgroup
of G is even simpler: H would then just have to be closed under the group product.
Given a finite group G, it is easy to find cyclic subgroups of G. For example, if a
is an element of G, all positive integral powers of a are also elements of G. G being a
finite group implies that there are only finitely many integral powers of a which are
distinct elements of G. This could happen only if there was some minimum positive
integer n for which an = e, so that further powers of a would merely be repetitions
of elements in the set H = {e, a, . . . , an−1 }. But then we have generated a cyclic
subgroup H of G. The order of the subgroup generated by a is also called the order of
the element a. Evidently, the order of every element of a finite group is finite.

Example 11. S(3) has {e, a}, { e, ab}, {e, ab2 }, {e, b, b2 } as its cyclic subgroups.
Two subsets of a group G are said to be equal if they contain the same elements. If
A and B are two subsets of G, then AB is the set of all elements of G which are equal
to the product of an element of A with an element of B in that order. It is worth noting
that AB and BA need not be equal sets. Suppose that H is a subgroup of G. If a is any
element of G, Ha denotes a subset of G containing elements of the form ha where h
runs through all the elements of H. Then Ha is called a left coset of the subgroup H.
In the same way, a right coset aH can be defined. From hereon, by a coset we mean a
left coset. It is evident that H = He, and therefore H is one of the cosets of H. Every
element of G belongs to some coset of H because the coset Ha definitely contains a
which could be any arbitrary element of G. Thus each and every coset of H contains
every possible element of G. It is possible that two distinct elements a, b of G may
belong to the same coset of H. This can happen only if there is some h in H such that
ha = b, or in other words, ab−1 is in H. Also, two different cosets of H are disjoint.
Suppose the two cosets had a common element a, then both the cosets must be equal
to the coset Ha indicating that the intersection between two different cosets must be
a null set. Hence for the finite group G, all the cosets of H contain same number
of elements and are disjoint whose union is equal to G. This proves the important
Lagrange’s theorem which states that the order of every subgroup H of a finite group
G divides the order of G. Lagrange’s theorem does not imply that a subset of a finite
group is definitely a subgroup if the number of elements in the subset divides the
group order.

A group G may have several subgroups of various orders. Let H be a non-trivial

subgroup of G. If a is an element of G, then aHa−1 is a set which contains elements of
form aha−1 for every h in H. It can be verified that aHa−1 is also a subgroup of G. By
choosing a different a each time, we can generate all possible subgroups of the form
aHa−1 . Such subgroups are called conjugate subgroups. It is possible that one may get
the same subgroup for different choices of a. If it turns out that for every choice of
a, aHa−1 = H, then H is said to be a normal subgroup or invariant subgroup. Normal
subgroups are special in the sense that they are invariant under conjugation by all the
elements of the group, i.e., if k is an element of a normal subgroup K, then for all a
in G, aka−1 is again an element of K. Further aKa−1 = K ⇒ aK = Ka, i.e., the left
and right cosets of a normal subgroup are identical. The trivial subgroups of G are
clearly normal. If G is abelian then every subgroup of G is normal. In case G has no
non-trivial normal subgroups then G is said to be a simple group. Groups of prime
order are examples of simple groups.

Example 12. Consider the group S(3). With the notation Ha = {e, a} for the cyclic
subgroup generated by a and using the relations on the generators a and b, the
following can be verified

aHa a−1 = Ha ; ( ab) Ha ( ab)−1 = Hab2 ; ( ab2 ) Ha ( ab2 )−1 = Hab .

Thus Ha , Hab , Hab2 are conjugate subgroups. Also, Hb can be seen to be a normal
subgroup of S(3) and hence S(3) is not a simple group. In fact Hb is the only
non-trivial normal subgroup of S(3).
The non-trivial normal subgroups of a group play an important role in the study
of the group’s structure. If K is a normal subgroup of the group G, then the set of
cosets of K is also a group, called the factor group of G with K. The factor group is
denoted as G/K. Let Ka and Kb be two cosets of K. Defining the product of the
two cosets (Ka)(Kb) to be the set that contains elements of G which are equal to the
product of an element of Ka with an element of Kb in that order. Then the elements
of (Ka)(Kb) have the form k1 ak2 b = k1 ak2 a−1 ab = k1 k3 ab = k4 ab. When k1 takes
all values in K, k4 also takes all values in K. Thus (Ka)(Kb) = Kab and we have
closure in the set of cosets of K under the defined product. Associativity follows from
(KaKb)Kc = KabKc = K ( ab)c = Ka(bc) = KaKbc = Ka(KbKc). The coset K serves as
the identity in G/K and (Ka)−1 = Ka−1 . This proves G/K is a group.

1.3 Conjugacy Classes

Suppose a is an element of a group G. The set of all elements of G which are
equal to gag−1 for some choice of g in G is called the conjugacy class of the
element a. The elements of a conjugacy class are said to be conjugate elements of G.
Conjugate elements, even though distinct, have important common properties. One

such property is that if a and b are conjugate then both must have same order. For
instance, if a has order n and b = gag−1 for some g, then

bm = ( gag−1 )( gag−1 ) . . . ( gag−1 ) = gam g−1 . (1.3.1)

| {z }
m factors

The smallest positive integer m for which gam g−1 is equal to e is clearly n. Hence b
has the same order as a. The above expression also proves that if a and b are conjugate
then so are their equal powers. Additionally, if a is conjugate to b( a = gbg−1 ) and b
is conjugate to c(b = hch−1 ), then a = ghch−1 g−1 = ( gh)c( gh)−1 . It follows that a is
conjugate to c. The property of being conjugate is for this reason transitive.
The conjugate elements of G belong to the same conjugacy class. As every element
of G is in its own conjugacy class ( a = eae−1 ), the whole group G can be divided into
several conjugacy classes. It is important that two different conjugacy classes cannot
have a common element. For if there was a common element, then that element would
be conjugate to all the elements in both the conjugacy classes. By the transitivity
property, it follows that the elements in the two classes would be conjugate and
therefore must belong to the one and same class. The conjugacy classes of a group
decompose the group into mutually exclusive sets. In case of an abelian group, the
conjugacy classes consist of the individual group elements.

Example 13. Consider S(3). Since geg−1 = e for all g in S(3), {e} forms a conjugacy
class. Also bab−1 = bab2 = ab4 = ab and (b2 ) a(b2 )−1 = b2 ab = b4 a = ba = ab2 , thus
{ a, ab, ab2 } is a conjugacy class. aba−1 = aba = b2 · a2 = b2 , and it follows {b, b2 } is
a conjugacy class. We note finally that

S(3) = {e} ∪ { a, ab, ab2 } ∪ {b, b2 }.

1.4 Further Examples of Groups

Two important groups are considered here in light of the concepts introduced in
previous sections. They are namely: the Quaternion Group and the Dihedral Group.
The quaternion group is a group structure on algebraic entities called quaternions.
Quaternions were first considered in connection to classical mechanics. They have
deep significance as regards mathematical constructs in various physical theories. We
do not intend to explore this significance, but we would like to understand what
quaternions are in principle. The dihedral group is of importance in the study of
symmetries of a regular polygon. We will often encounter the dihedral group in later
chapters.

The Quaternion Group Q

The quaternion group Q has 8 elements. Explicitly, the group Q = {e, s, i, j, k, si, sj, sk }.
Various elements satisfy the relationships

s2 = e; i2 = j2 = k2 = ijk = s.

The symbols i, j, k can be regarded as imaginary units, e as the real unit and s as
negative e. The defining relations specify the group completely. The relation i2 = ijk
implies i = jk. Likewise, j = ki and k = ij. One may note the cyclic nature of these
equalities. Furthermore, i2 = j2 ⇒ ( ji )(ij) = e. Because ij = k and (sk )k = e, it follows
ij = sji. Likewise, jk = skj and ki = sik. The identity element e obviously commutes
with all the elements of Q, but so does the element s. For example, i = jk ⇒ si = sjk
and ijk = s ⇒ sjk = is. It follows that si = is. In a similar fashion it may be shown s
commutes with j and k and hence with all the elements of Q. To sum up, we may note
the relations that follow from the defining relationships.
i = jk, j = ki, k = ij,
ij = sji, jk = skj, ki = sik,
si = is, sj = js, sk = ks.
The non-trivial subgroups of Q can be of order 2 or 4. The subgroup of order 2 is
{e, s}. The three subgroups of order 4 are {e, s, i, si }, {e, s, j, sj} and {e, s, k, sk}.
Each of the order-4 subgroups is isomorphic to the cyclic group C4 . It can be verified
that all the subgroups of Q are normal in Q. The conjugacy classes of Q can be found
from the relationships given above. Since e and s commute with all the elements, they
are the only elements in their respective classes. Since jij−1 = sijj−1 = si, it follows i
and si are conjugate. Similarly, j and sj are conjugate, and also k and sk are conjugate.
The decomposition of Q in conjugacy classes is.

Q = {e} ∪ {s} ∪ {i, si } ∪ { j, sj} ∪ {k, sk }.

The Dihedral Group Dn (n ≥ 3)

The dihedral group Dn has 2n elements. It can be generated by the symbols r and s
satisfying the relationships as under

r n = s2 = e, sr = r −1 s.

Explicitly, Dn = {e, s, r, r2 , . . . r n−1 , sr, sr2 , . . . sr n−1 }. Any product of a finite

number of r’s and s’s can be reduced to one of the group elements using the defining
relationships. Consider

r k s = rr . . . }r
| {z s = rr . . . }r sr −1 = sr −k = sr n−k ,
| {z
k factors k −1 factors

and likewise in other cases. The subgroup {e, s} is the smallest non-trivial subgroup.
The cyclic subgroup generated by r, {e, r, r2 , . . . r n−1 } is a normal subgroup of Dn . If a

positive integer k is a divisor of n then r k will generate a cyclic subgroup of Dn . In

order to calculate the conjugacy classes, the following relationships are useful:
sr k s = r −k ,
r −k sr k = sr2k ,
(sr t )r k (sr t )−1 = sr t r k r −t s = r −k .
In the case when n is 2p + 1, there are p + 2 classes. The decomposition of D2p+1 into
classes is
D2p+1 = {e} ∪ {r, r2p } ∪ {r2 , r2p−1 } ∪ . . . {r p , r p+1 }∪ (1.4.1)
| {z }
p classes

∪{s, sr, sr2 , . . . sr2p }.

In the case when n is 2p, there are p + 3 classes. The decomposition of D2p is seen to be

D2p = {e} ∪ {r p } ∪ {r, r2p−1 } ∪ . . . {r p−1 , r p+1 } ∪ (1.4.2)

| {z }
p−1 classes

∪{s, sr2 , . . . sr2p−2 } ∪ {sr, sr3 , . . . sr2p−1 }.

1.5 Homomorphism of Groups

A homomorphism between two groups is a function from one to the other that
preserves products. More specifically, a function ϕ from a group G to a group T is
a homomorphism if for all a, b in G

ϕ( ab) = ϕ( a) ϕ(b). (1.5.1)

Here the notation ϕ( a) stands for the member of T which is the image of a under the
function ϕ. If a and b are both chosen to be identity element e of G then ϕ(e) = [ ϕ(e)]2
and it follows that the identity of G is mapped to the identity of T. Also, if b = a−1 ,
then ϕ(e) = ϕ( a) ϕ( a−1 ) and it follows that ϕ( a−1 ) = [ ϕ( a)]−1 . In other words, under
a homomorphism, identity is mapped to identity and inverse is mapped to inverse.
The subset K of G which contains all the elements which are mapped to the identity
of T is called the kernel of the homomorphism ϕ. It can be shown that the kernel
K is a normal subgroup of G. In fact all the normal subgroups of G would cause a
homomorphism of G into some group. It can be also shown that all elements of a
coset of K are mapped to the same element of T.

Example 14. Let C2 = { E, A} where C2 is the cyclic group of order 2 in our notation
and E here is the identity element of C2 . Consider a function ϕ from S(3) to C2
defined as

ϕ(e) = ϕ(b) = ϕ(b2 ) = E,

ϕ( a) = ϕ( ab) = ϕ( ab2 ) = A.

Then ϕ is a homomorphism. The kernel of this homomorphism is the normal subgroup

Hb .
When the kernel of a homomorphism is simply the trivial group E, then the
homomorphism is one to one. In this case the two groups are said to be isomorphic.
Isomorphic groups are essentially the same group and differ merely in the manner of
labelling of their elements. In the study of point groups, it will be seen that some of
them are isomorphic groups. In case of the group S(3), the factor group S(3)/Hb is
isomorphic to the group C2 . In notation,

C2 ∼
= S(3)/Hb . (1.5.2)

Suppose now that ϕ1 is a homomorphism from G0 into G1 and that ϕ2 is a

homomorphism from G1 into G2 . It is then possible to compose ϕ1 and ϕ2 together
to give a homomorphism ψ from G0 into G2 . For a ∈ G0 , define ψ so that

ψ( a) = ϕ2 ( ϕ1 ( a)).

It is easy to show that with the above definition, ψ is a homomorphism from G0 into G2 .

1.6 The Symmetric Group

Consider a set of n distinct letters {1, 2, . . . n} arranged in that order. Any
rearrangement of this set of elements is called a permutation. For example, if n = 3,
then all the permutations are

1 2 3 1 2 3
π1 = ; π4 =
1 2 3 3 2 1

1 2 3 1 2 3
π2 = ; π5 =
2 1 3 2 3 1

1 2 3 1 2 3
π3 = ; π6 = .
1 3 2 3 1 2
A permutation that leaves the positions of all the letters unchanged is called
the identity permutation. In the example above, π1 is the identity permutation.
Remember that the exchange amongst columns in the above elements denote the same
permutation operation. That is,

1 2 3 2 1 3
π5 = = .
2 3 1 3 2 1

A product operation may be defined in the set of permutations so that the product of
two permutations is another permutation. The operation of forming the product of
two permutations can be most easily understood by considering the concrete case of
n = 3. Consider the product π2 π5 . π2 takes 1 to 2 and π5 takes 2 to 3, thus π2 π5 takes
1 to 3. In this manner the action of π2 π5 can be ascertained on all the letters

1 2 3 2 1 3 1 2 3
π2 π5 = = = π4 ,
2 1 3 3 2 1 3 2 1

1 2 3 2 3 1 1 2 3
π5 π2 = = = π3 .
2 3 1 1 3 2 1 3 2

Similarly, one may verify that the product so defined is associative. Also (π2 )−1 =
π2 , (π3 )−1 = π3 , (π4 )−1 = π4 , (π5 )−1 = (π5 )2 = π6 , (π6 )−1 = (π6 )2 = π5 . Thus
the set of permutations on three letters is a group. In fact, this is the same group
as S(3) if we identify π1 with e, π2 with a, π5 with b, π6 with b2 , π4 with ab and
π3 with ab2 . Since S(3) is the group of all permutations on 3 letters, it is called the
symmetric group of degree 3. In the general case of permutations on n letters, S(n)
is called the symmetric group of degree n. The order of S(n) is n!, the total number
of permutations of n letters. Subgroups of S(n) are called permutation groups. The
very important Cayley’s Theorem states that every group is isomorphic to a permutation
group which is embedded in some symmetric group. In particular, if G is a group of
order n, then it is isomorphic to a subgroup of S(n). In order to see this, label the
elements of G so that G = { g1 (= e), g2 , . . . gn }. Let g be some element of G. If every
element of G is multiplied with g from right, then we have a permutation π g induced
on the letters { g1 , g2 , . . . gn } which can be written as
!
g1 g2 · · · gn
πg = .
g1 g g2 g · · · g n g

Another element h of G induces a permutation πh so that the product π g πh is given by

! !
g1 g2 · · · gn g1 g2 · · · gn
π g πh =
g1 g g2 g · · · g n g g1 h g2 h · · · g n h
!
g1 g2 · · · gn
⇒ π g πh =
g1 gh g2 gh · · · gn gh
⇒ π g πh = π gh .
This proves that the mapping g → π g is a homomorphism (Equation 1.5.1). Because
two distinct elements of G induce distinct permutations, it follows that the group
G and the set of permutations {π g1 , π g2 , . . . π gn } are isomorphic. This proves the
Cayley’s Theorem for finite groups.

A permutation can also be represented as a product of disjoint permutation cycles.

Consider the following permutation on 7 letters.

1 2 3 4 5 6 7
π= .
5 7 3 4 6 1 2

π permutes the letters 1, 5, 6 among themselves, and also 2, 7 among themselves while
leaving 3 and 4 in their places. This can also be represented as
π π π π π π π
1 → 5 → 6 → 1, 2 → 7 → 2, 3 → 3, 4 → 4.

Here (1, 5, 6) is a three-cycle, (2, 7) is a two-cycle, (3) and (4) are one-cycles. It is now
simpler to write π = (1, 5, 6)(2, 7)(3)(4). Since elements in one-cycles are unmoved by
the permutation, it is redundant to mention them [for this reason it is conventional to
represent the identity permutation simply by the empty cycle ()]. The decomposition
of π in disjoint cycles is therefore

π = (1, 5, 6)(2, 7).

The cycle decomposition is particularly useful in carrying out calculations with

permutations. Consider another permutation

1 2 3 4 5 6 7
σ= = (3, 6)(2, 4, 5, 7).
1 4 6 5 7 3 2

The products πσ and σπ can be easily read from the cycle decompositions of the two
to be

πσ = (1, 7, 4, 5, 3, 6); σπ = (1, 5, 2, 4, 6, 3).

It may be noted that π (product of a two-cycle and a three-cycle) and σ (product of a

two-cycle and a four-cycle) have different cycle structure while σπ and πσ both has
the same six-cycle structure. Further, (1, 5, 2, 4, 6, 3), (5, 2, 4, 6, 3, 1) and (3, 1, 5, 2, 4, 6)
represent the same permutations, i.e., positions in a cycle can be changed cyclically
without affecting the permutation.
The inverse of a permutation can be obtained by inspection. For example, π is
a product of a three-cycle and a two-cycle. Since both the cycles are disjoint, π −1
would be a product of the inverses of the two cycles. A two-cycle like (2, 7) is called
a transposition. It is obvious that the inverse of a transposition is the transposition
itself. As for the three-cycle (1, 5, 6), its inverse should take 5 to 1, 1 to 6 and 6 to 5,
i.e., (1, 5, 6)−1 = (1, 6, 5). Thus π −1 = (1, 6, 5)(2, 7), and calculating similarly, σ−1 =
(3, 6)(2, 7, 5, 4).
Since a transposition interchanges positions of only two letters, it is the most
elementary type of permutation. It is natural to expect that all permutations can be

built up from transpositions. It can be checked readily that the k-cycle (1, 2, .., k) is
equivalent to a product of k − 1 transpositions

(1, 2, . . . , k) = (1, 2)(1, 3) . . . (1, k).

As any permutation is a product of disjoint cycles, it can be expressed as a product

of transpositions. Depending on whether an odd or an even number of transpositions
are involved in the product, the permutation is called odd or even correspondingly. No
permutation is both even and odd.
In the symmetric group S(n), it is evident that not all elements would have similar
cycle structure. Any permutation in S(n) can be written as a product of disjoint ik
k-cycles, where k ranges from 1 to n as in

(1) (1) (1) (2) (1) (2)
a1 . . . ai b1 , b1 . . . bi , bi ...
1 2 2
| {z }| {z }
i1 1−cycles i2 2−cycles

(1) (2) (n) (1) (2) (n)
. . . t1 , t1 , . . . t1 . . . tin , tin , . . . tin
| {z }
in n−cycles

It is clear that Σnk=1 kik = n, where some of the ik ’s can be 0. Using elementary
combinatorics, it can be proved that the total number of such permutations is given by
the following expression:
n
1
n! ∏ . (1.6.1)
i
k =1 k
! ( k )ik

The significance of having identical cycle structure is that all such elements are
conjugate in S(n) and hence belong to the same conjugacy class. Suppose that σ and
π are permutations on same set of letters. Let them be represented by:
σ = ( a11 . . . a1s1 )( a21 . . . a2s2 ) . . . ( at1 . . . atst ),
!
a11 · · · a1s1 a21 · · · a2s2 · · · at1 · · · atst
π = .
b11 · · · b1s1 b21 · · · b2s2 · · · bt1 · · · btst

π −1
Consider the action of π −1 σπ on b11 : b11 → a11 → a12 → b12 , which means that
σ π

π −1 σπ sends b11 to b12 and so on. In effect,

π −1 σπ = (b11 . . . b1s1 )(b21 . . . b2s2 ) . . . (bt1 . . . btst )

and the conclusion is that conjugate elements have the same cycle structure. From
the above it is also clear that given two permutations in S(n) having identical cycle
structures (the RHS of σ and π −1 σπ), one can always find a permutation in S(n)(the
RHS of π) so that the first two permutations are related via conjugation by the third.

The conjugacy classes of S(n) are distinguished by the cycle structure of elements in
the class. Expression 1.6.1 gives the number of permutations in a conjugacy class of a
specified cycle structure.
The number of conjugacy classes in S(n) is the number of ways in which n can
be expressed as a sum of positive integers. For example, if n = 5 then there are 7
partitions of n, namely 1 +1 + 1 + 1 + 1, 1 + 2 + 2, 3 + 2, 1 + 4, 1+1+1+2, 1+1+3 and 5.
S(5) therefore has 7 conjugacy classes. These conjugacy classes may be conveniently
represented by Young diagrams. A Young diagram is an arrangement of a certain
number of boxes in left justified rows such that the number of boxes in a row is at
least as large as the number of boxes in the row below it. Below are listed the Young
diagrams for conjugacy classes of S(5).
1. The Identity class contains the identity permutation as its only member. The
identity is simply the product of 5 one-cycles.

2. Product of 2 two-cycles and 1 one-cycle

3. Product of 1 three-cycle and 1 two-cycle

4. Product of 1 four-cycle and 1 one-cycle

5. Product of 1 two-cycle and 3 one-cycles

6. Product of 1 three-cycle and 2 one-cycles

7. A five-cycle

In the symmetric group S(n), the set of all even permutations forms a group
called the alternating group U(n). That U(n) is a group follows from the fact that
product of two even permutations is an even permutation. The conjugates of an even
permutation would have identical cycle structure, and therefore would all be even. It
follows that U(n) is a normal subgroup of S(n). This further implies that the factor
S( n )
group is the set of cosets of U(n). Suppose now
U( n )

S( n )
= {U(n), U(n)(1, 2), O1 , . . .}.
U( n )

Here U(n)(1, 2) is a coset of U(n) obtained by multiplying all elements of U(n)

with the transposition (1, 2) and O1 is another coset of U(n) containing only odd
permutations. Then O1 U(n)(1, 2) is also a coset of U(n). O1 U(n)(1, 2) can contain only
even permutations. Because the only coset of U(n) containing even permutations is
U(n) itself, we have O1 U(n)(1, 2) = U(n). U(n) is the identity of the factor group and
it follows that O1 = [U(n)(1, 2)]−1 But [U(n)(1, 2)]−1 = U(n)(1, 2) and we have the
result that there are only two cosets of U(n), one being U(n) and the other being the
n!
set of all odd permutations. Thus the order of the group U(n) is .
2

1.7 Direct and Semi-direct Products

Two groups G1 and G2 can be used to form a larger group G by a construction called
the direct product of groups. The direct product of G1 and G2 is denoted as G = G1 × G2 .
Every element of G should be uniquely expressible as a product of an element of G1
and an element of G2 . In forming the direct product, G1 × G2 is not distinguished from
G2 × G1 . Consequently, every element of G1 should commute with every element of
G2 . With these properties, it can be seen that G1 and G2 are normal subgroups of G.
The only element common to G1 and G2 is the identity, for if g1 is in G1 and g2 is in G2 ,
then g1 g2 g1−1 g2−1 = e because of commutation. Because G1 is normal g1 ( g2 g1−1 g2−1 )
is an element of G1 , and because G2 is normal ( g1 g2 g1−1 ) g2−1 is an element of G2 , and
both are equal to the identity element. It is clear that | G | = | G1 | × | G2 |.

Example 15. Consider the groups C2 = {e, a} and C3 = {e, b, b2 }. From the
prescription above, the direct product

C2 × C3 = {e, a, b, b2 , ab, ab2 }.

It may now be noted that the element ab generates the group C2 × C3 , and order of ab
is 6. In fact C2 × C3 is isomorphic to C6 . In other words

C6 = C2 × C3 .

In the general case, a direct product may be constructed for any number of groups, the
groups themselves being finite or infinite. We do not present this construction here.
Finally it may be noted that a finite group G may be expressed as a direct product of
its normal subgroups K1 , K2 , ., Ks under the following conditions:
1. G = K1 K2 . . . Ks .
2. Every element of G is uniquely expressible as a product of elements from each of
the normal subgroups Ki .
In a group G, let K be a non-trivial normal subgroup. Let T be another non-trivial
subgroup of G such that the identity is the only common element in K and T. Due to
normality of K, we have KT = TK. Suppose all the group elements gi ∈ G can be
written as product k i t j where k i ∈ K and ti ∈ T. That is, G = KT. In this circumstance,
G is said to be a semi-direct product of K by T. The notation for the same is

G = K o T.

Notice the positioning of the groups K and T about o. For t1 and t2 in T, Kt1 and
Kt2 are cosets of K. These two cosets will be same only if t1 t2−1 is in K. Because t1 t2−1
is a member of both K and T, from aforementioned conditions on K and T it follows
that t1 = t2 . The conclusion is that distinct elements of T are in distinct cosets of K
and therefore can be chosen as the coset representatives. Additionally Kti , with ti in T
exhaust all the cosets of K in G since G = KT. If the subgroup T is also normal in G
then G is the direct product K × T.

Example 16. It was noted in Section 1.6 that the alternating group U(n) is a normal
subgroup of S(n). For n ≥ 3, let T be the group of order 2 generated by a transposition
such as (1, 2). Then

S(n) = U(n) × T.

Exercises
1. For the following sets S and binary operation ∗ on S, determine which of the group
axioms stated in Definition 1 are satisfied
1) S = N, m ∗ n = max{m, n} for m, n in N .
2) S = Z, m ∗ n = m + n + 1 for m, n in Z.
3) S is the set of n × n invertible real matrices (n ≥ 2), A ∗ B = [ AB + BA].

2. Prove the statements in Equations 1.1.1.

3. Show that a subset H of a finite group G is a subgroup of G if H is closed under the
group product.
4. If H is a subgroup of G and a is any arbitrary element of G, then show that aHa−1
is a subgroup of G.
5. Write the multiplication table of D3 and determine the conjugacy classes.
6. If K is a normal subgroup of the group G and A is any subset of elements of G, show
that KA = AK.
7. For subsets A and B of a group G, let AB represent the subset of G containing
elements ab such that a is in A and b is in B. If A and B are subgroups of G, and at
least one of them is normal in G, show that every finite product of elements of A and
B is an element of AB.
8. With reference to Equation 1.5.1, show that the kernel K of the homomorphism φ is
a subgroup of G. Further show that K is a normal subgroup of G.
9. In a homomorphism φ, is it always true that the orders of the element a and its image
φ( a) are equal? What if φ is an isomorphism?
10. Find an isomorphism between a set of positive reals (a group under ordinary
multiplication) and all the reals (a group under ordinary addition).
11. Of the groups C4 , D2 and the Kelin-4 group V , which are the isomorphic groups? Are
the groups D4 and Q isomorphic?
12. Consider the group D4 . List all its conjugacy classes and its subgroups. Are all the
non-trivial subgroups of Dn cyclic?
13. Show that all possible products generated by the Pauli matrices iσ1 and iσ2 form a
group. Write the multiplication table of this group and determine the conjugacy classes.
This group is isomorphic to which one of the groups considered in the chapter?

0 1 0 −i
σ1 = , σ2 = .
1 0 i 0

14. Show that the composition of homomorphisms defined in Section 1.5 is again a
homomorphism.

15. Consider the direct product C2 × C6 and state whether it is isomorphic to the cyclic
group C12 .
16. For the symmetric group S(6), determine the number of conjugacy classes and
the number of permutations belonging to each class. Prepare a Young diagram
representation of the classes as illustrated for S(5) in the text.
17. Prove that the number of elements belonging to a conjugacy class in S(n) is given by
Expression 1.6.1.
18. Show that the alternating group U(n) is generated by the set of all the three-cycles on
n letters.
19. Consider the group S(4). Determine its conjugacy classes and the number of elements
in each class. Repeat the same exercise for U(4). Do the elements appearing in a
conjugacy class in S(4) appear in the same class in U(4)?
20. In the group S(4), consider the subgroups

V = {(), (1, 2), (3, 4), (1, 2)(3, 4)}

and VN = {(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. Show that VN is normal
in S(4) while V is not. Realize S(4) as a semi-direct product of VN with another
subgroup of S(4).
21. Verify (1, 2, 3, . . . a, a + 1, . . . n) = ( a, a + 1, . . . n)(1, 2, 3, . . . a).
22. The generators of the symmetric group S(n) are (1, 2), (1, 2, . . . n). Verify this
statement for S(4).
23. Let Pi = (i, i + 1) denote transposition element of S(n) involving neighbouring
objects. Show that

Pi Pi+1 Pi = Pi+1 Pi Pi ; Pi Pj = Pj Pi for |i − j| ≥ 2.

Incidentally, these relations are similar to the defining relations of generators bi0 s of
braid group B(n) which emerge as natural symmetry of exchange of neighbouring
objects in a two-dimensional plane with bi2 6= Pi2 = I.

In this chapter, we will focus on the symmetries possessed by molecules. By

symmetrical transformation of an object, we mean any rigid geometric transformation
under which the object remains invariant. The three types of rigid transformations
are translations, reflections and rotations. Translations are relevant to the study of
symmetries of crystals which is not considered here. For objects of finite size, e.g. a
molecule, a translation of the object as a whole definitely does not change its shape and
size, but the final position of the object does not exactly overlap its initial position. For
this reason, a translation is not an element of symmetry of a molecule. However,
rotations and reflections or any combinations of the two are elements of symmetry of
a molecule. A rotation is a proper transformation because it preserves handedness. On
the other hand, reflection is an improper transformation. We will present symmetry
operations and their resemblance to finite group elements discussed in the previous
chapter.

2.1 Elements of Molecular Symmetry

All the possible symmetry transformations of a molecule can be described in terms
of some basic transformations. Of these transformations, the simplest is the identity
transformation E, which leaves the molecule untouched. This transformation should
remind the reader of the identity element of a group.
A molecule may have an axis of symmetry about which, if the molecule is rotated
by a certain minimum angle (say 2π n ), then it appears just as before the rotation. This
transformation is symbolically represented as Cn , and the axis is a n-fold axis. It is
obvious that rotation by the same angle for n consecutive times would bring the

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Molecular Symmetry 19

molecule to the exact same configuration as it was before the rotations were carried
out. The order of the transformation Cn is n.

Cnn = E. (2.1.1)

Because Cnk Cnn−k = E, Cnn−k may be regarded as the inverse operation of Cnk . It
may so happen that a molecule has more than one axis of symmetry. In such a case
the axis with maximum n may be called the principal axis of symmetry, though this
nomenclature is not absolutely necessary. It is conventional to identify the principal
axis with the z-axis of the coordinate system. In the special case when there is a
two-fold axis of symmetry perpendicular to the principal axis, the two-fold axis would
be symbolically represented by U2 .
A certain plane through a molecule may exist such that reflection of the molecule in
it leaves the molecule indistinguishable from the original. Such a plane is called a plane
of symmetry A molecule may have more than one such plane. In case the plane contains
the principal axis, it is the vertical plane and the transformation corresponding to
reflection through this plane is denoted by σv . Reflections in a plane of symmetry
perpendicular to the principal axis are denoted by σh (horizontal). Occasionally, as
in case of symmetries of a cube, diagonal planes of symmetry exist, denoted by σd .
Order of all the reflection operations is clearly 2.

σv2 = σh2 = σd2 = E. (2.1.2)

A molecule may have a roto-reflection symmetry. This happens if there is a σh plane

perpendicular to a Cn axis. A roto-reflection transformation, Sn , can be visualized as a
Cn rotation followed by a σh reflection.

Sn = Cn σh = σh Cn . (2.1.3)

It is obvious from Figure 2.1.1 that the order in which Cn and σh are carried out is
immaterial. If n is an odd integer, then n consecutive roto-reflections on the molecule
would merely be equivalent to a reflection in the horizontal plane, for Snn = (Cn σh )n =
Cnn σhn = σh . It follows that Sn is a truly distinct element of symmetry only if n is an
even integer, the order of Sn in this case being n.
Lastly, a molecule may have inversion symmetry, Ci . Such a symmetry occurs if the
molecule has a point of symmetry. The property of such a point is that all the atoms
of the molecule lie on lines passing through the point and for every atom lying on
some line, there exists another identical atom at the same distance behind the point
of symmetry as the first atom is in front of it. Clearly Ci2 = E. Figure 2.1.2 illustrates
inversion symmetry.

2 p /n
A
A.C n

A.s h As h C n = AC n s h

Figure 2.1.1 Roto-Reflection Symmetry.

The figure shows the action of Cn σh and σh Cn on an
atom A of some molecule

Figure 2.1.2 Inversion Symmetry.

The central dark sphere is the point of symmetry of this
hypothetical molecule

2.2 The Symmetry Group of a Molecule

A molecule may possess one or more of the symmetry elements depending upon
its structure. Once the symmetry elements of the molecule are identified, one
may compose them (in other words, form their products) in various ways and

generate all possible symmetry transformations of the molecule. The manner of

composition of symmetries is to perform each transformation in the prescribed
sequence. For example, if a molecule has a five-fold symmetry axis and a horizontal
plane of symmetry, then the operation C53 σh would mean 3 consecutive rotations
of 72◦ about the principal axis followed by a reflection in the horizontal plane. As
each elementary transformation leaves the molecule invariant, it is clear that
any transformations composed of elementary transformations are also symmetry
transformations. This shows that the set of symmetry transformations is closed under
product operation. Equivalently, these transformations imply that all the symmetry
axes and symmetry planes of a molecule must coincide at one point.
Note that every elementary symmetry transformation or composed transformation
on a molecule permutes all the identical atoms in the molecule (bijection of space
points). It is evident that the set of all elementary as well as composed symmetry
transformations form a group. The set contains identity element E which corresponds
to no symmetry operation. For every elementary symmetry operation A, we can show
that their inverse operation A−1 is in the set. If C and B are elementary symmetry
operations, then for the symmetry operation CB, the inverse can be defined to be
B−1 C −1 . Such a group is an example of a point group as there is at least one point
in the molecule which remains fixed in all the symmetry transformations.
The number of elements in a point group is finite except for the special case of
linear molecules. This follows from the Cayley’s Theorem since the point group of
a non-linear molecule will be isomorphic to a subgroup of some symmetric group
S(n). Just like any abstract group, a point group can be decomposed into its conjugacy
classes. The conjugacy classes are of importance in the study of group representations
as will be seen in the next chapter. It is therefore necessary to figure out the conjugacy
classes of a given point group. Having identified the various symmetry axes (planes),
those axes (planes) are labeled as equivalent which can be brought to coincide in
direction by a symmetry transformation of the molecule. Symmetry axes of different
orders cannot be equivalent. Consider the ammonia molecule shown in Figure 2.2.1.
All the σv planes are equivalent as the alternate planes can be made to coincide with
00 0 C3 C3 C3
each other after rotation by an angle of 2π 3 about the C3 axis ( σv → σv → σv → σv ).
However, the σv planes in the water molecule are not equivalent. In the Figure 2.2.2
are shown the symmetry axes of a cube. There are 3 C4 axes, 4 C3 axes and 6 C2 axes.
It is easy to verify that all the C4 axes are equivalent, so are all the C3 axes as well as
all the C2 axes.

C3 C2

N
O

H H

sv H sv

s v¢
H

s v¢
s v² H
(a) NH 3 (C 3v symmetry) (b) H 2 O (C 2v symmetry)

Figure 2.2.1 Equivalent and Non-equivalent Reflection Planes

C3 C4

Figure 2.2.2 Symmetry Axes of a Cube

Equal rotations about equivalent axes are conjugate transformations in the

symmetry group of a molecule. In order to see this, suppose A1 and A2 are two
equivalent symmetry axes (Cn ) of a molecule. Then there exists some symmetry
transformation S of the molecule which carries A1 to A2 . Then the transformation

SCnk S−1 carries A1 to A2 , performs a Cnk rotation about A2 and bring A2 axis back to
A1 . This is clearly equivalent to performing a Cnk rotation about A1 itself. If an axis
Cn lies in a symmetry plane σ, then it is called the bilateral axis. For such an axis Cnk
and Cn−k are conjugate. With reference to Figure 2.2.3, the following equalities can be
noted:
Aσ = A, AσCn−k = A0 ,
⇒ AσCn−k σ−1 = AσCn−k σ = A00 = ACnk
⇒ σCn−k σ−1 = Cnk .
By a similar construction it is easy to see that an axis is bilateral also when there is a
U2 axis (two-fold axis) perpendicular to it (see Exercise 1).
Cn

A²

A
A¢ C –k
n

Figure 2.2.3 Bilateral Axis

2.3 Symmetry Point Groups

Finite point groups that describe the symmetries of molecules are considered here.
The standard Schoenflies notation is used to designate the various groups. The graphical
depiction of the groups is done by means of stereographic projections. Consider a unit
sphere which is divided by the plane of the diagram into two equal parts. In order to
form the stereographic projections, we project on the plane of the diagram any general
point of the sphere and all its transformed positions obtained by application of the
various symmetry transformations of the concerned group. Projections of the points of
the upper hemisphere are indicated with a + sign while those of the lower hemisphere
are indicated with . Further details on stereographic projections are provided as we
consider specific groups.

Cn Group. A molecule of Cn symmetry has an n-fold symmetry axis as its only element
of symmetry. This group is clearly cyclic, has n elements and each conjugacy class
consists of a single element. The special case of n = 1, i.e., C1 corresponds to no
symmetry. Figure 2.3.1(a) shows the stereographic projection for a C3 group. Note
that the diagram plane cuts the unit sphere along the dotted circle. The central triangle
indicates that the axis of symmetry (perpendicular to the diagram plane) is three-fold.
A + is taken into the next in the anticlockwise direction upon a C3 operation, and then
to the second next upon a C32 operation.

S2n Group. This is the symmetry group of a molecule having a rotary reflection axis
of order 2n. It is a cyclic group as is evident from the nature of a rotary reflection
operation. Thus every element of S2n is the unique member of its conjugacy class. In
the Figure 2.3.1(b) is shown the projection diagram for S6 . The 6–order rotary reflection
axis is indicated by a hollow hexagon. Note that S62 = C6 σh C6 σh = (C6 )2 (σh )2 = C3 ,
implying the presence of a three-fold axis of symmetry which is indicated by a solid
triangle in the diagram.

+ + + + +

+ +
+ + + + + + + +

(a) C 3 (b) S6 (c) C 3h (d) C 3v

+ + + +
+

+
+
+ + +
+ + + +
+

(e) D 3 (f ) D 3h (g) D3d

Figure 2.3.1 Stereographic Projections

Cnh Group. If to an n-fold axis is added a horizontal plane of symmetry σh , then

the Cnh group is obtained. This group has n elements of the Cn group and in addition,
further n elements of the type Cnk σh . It is easy to check that the reflection in a horizontal
plane and rotation about the principal axis are commuting operations. Therefore Cnh
is a cyclic group. Consequently every element of Cnh is the unique member of its
conjugacy class. In the special case of n = 1, i.e., C1h contains only the elements E and
σh ; this group is designated Cs . In the stereographic projection for the C3h group, the
solid circle indicates a horizontal symmetry plane.

The aforementioned groups Cn , S2n and Cnh are all cyclic groups. For odd n =
2p + 1 the groups S2p+1 and C(2p+1)h are the same. As abstract groups, it may be
noted that two cyclic groups of same order are isomorphic.

Cnv Group. The Cnv group is obtained by adding a σv plane to a Cn axis.

Consequently, a total of n vertical reflection planes appear corresponding to operations
σv , σv Cn1 , σv Cn2 , . . . σv Cnn−1 . The total number of elements in the Cnv group is therefore
2n. The number of conjugacy classes depends upon whether n is odd or even. For
n = 2p, the vertical planes divide into two sets of equivalent planes. This gives two
equivalent classes:
n o n o
2 2p−2 1 2p−1
σv , σv C2p , ... σv C2p & σC2p , ... σv C2p ,
k and C 2p−k
containing p reflections each. The C2p axis is bilateral and thus C2p 2p are
conjugate. This further gives p − 1 classes, each containing two rotations and one
p
class containing the rotation C2p = C2 . Then, there is one class containing the identity
E. Thus for even n = 2p, Cnv has p + 3 equivalence classes. In a similar manner it
may be seen that for odd n = 2p + 1, all the reflection planes are equivalent and the
number of equivalence classes is p + 2. In the stereographic projection for C3v , the
solid diametrical lines indicate the vertical reflection planes (Figure 2.3.1(d)).

Dn Group. The Dn group is obtained by adding a U2 axis to the principal Cn

axis. As a result a total of n such U2 axes appear corresponding to operations
U2 , U2 Cn1 , . . . U2 Cnn−1 . The total number of elements in Dn is therefore 2n. The
decomposition of the group Dn in conjugacy classes is similar to that of Cnv group
for both the even and odd n. In fact, in may be noted that Dn and Cnv are both
isomorphic to the abstract dihedral group discussed in Section 1.4. The stereographic
projection for D3 shows the U2 axes marked as the double arrowed dotted diametrical
lines (Figure 2.3.1(e)).

Dnh Group. To the group Dn , if a plane σh is added as an element of symmetry, then

n vertical planes, σv appear automatically. Besides the 2n elements of group Dn , the
group Dnh contains nσv reflections as well as n rotary reflections σh Cnk . Each σv plane
contains the principal axis and a U2 axis. Since σh commutes with all the other group
elements, Dnh = Dn × Cs . The number of classes in Dnh is twice the number in Dn
enumerated as follows: Half of them are same as those of the group Dn , while the
remainder are obtained by multiplying these by σh . Note that the reflections in σv
planes belong to the same class if n is odd or forms two classes if n is even. Also, the
rotary reflection elements σh Cnk and σh Cnn−k are conjugate pairwise.

Dnd Group. If a diagonal plane σd bisecting the angle between two adjacent U2 axes of
the Dn is added as a symmetry element, then a further n − 1 such planes appear. The
resultant group is the Dnd group containing 4n elements. To the 2n elements of the Dn
group are added n reflections in the diagonal planes and further n transformations of
the type σd U2 . The angle between a U2 axis and the adjacent σd planes is clearly π/2n.

Figure 2.3.2 illustrates the action of the transformation σd U2 which is nothing but a
rotary reflection. That is, σd U2 = S2n . All the diagonal reflection planes do contain the
Cn axis and hence the principal axis is bilateral. For n = 2p, the group D2p,d has 2p + 3
conjugacy classes: E, C2 , ( p − 1) classes of conjugate rotations about principal axis, a
single class of 2pU2 rotations, a single class of 2p reflections σd and p classes of two
2k +1 −2k−1
rotary reflection transformations S4p and S4p where k can take values 0 to p − 1.
For odd n = 2p + 1, inversion is an element of the group and D2p+1,d = D2p+1 × Ci .

s d U2 A

U2 A
sd

Figure 2.3.2 σd U2 Transformation in Dnd Group

T Group. The group T consists of all the rotational symmetries of a regular tetrahedron
(Figure 2.3.3). A regular tetrahedron is a flat sided solid with 4 triangular faces, all
the faces being same. Through each vertex of the tetrahedron, passes a C3 axis. All
these axes are equivalent. Additionally there are C2 axes going through midpoints of
opposite edges of the tetrahedron. These axes are also equivalent. Hence, the classes
of the group T are: E, 4 C3 rotations, 4 C32 rotations and 3 C2 rotations.

Td Group. This is the full symmetry group of the tetrahedron. Apart from the proper
transformations of the group T, Td contains the improper transformations as well.
These transformations are obtained by adding 6 diagonal reflection planes. One such
plane is shown in Figure 2.3.3. As is evident from the diagram, the plane contains the
C3 and C2 axes. Consequently, the C3 axes are bilateral as they contain the mirror
planes. Another important feature is that the C2 axes now become S4 axes. This
can be seen by rotating the tetrahedron by π/2 about the C2 axis and reflecting in
a perpendicular plane that is midway through the C2 axis. Also S4 axes are bilateral
as they are all contained in diagonal planes. Thus the class structure is: E, all 8 C3
rotations, all 6 σd reflections, 3 C2 (S42 ) rotations, 6 ( S4 and S43 are conjugate) rotary
reflections.

Th Group. To the group Td , if an inversion operation is added, the Th group is obtained,

Th = Ci × Td . Consequently the number of conjugacy classes in Th is twice as many as
in Td (see Exercise 2).
O Group. The O group or the octahedral group is the group of all the rotational
symmetries of the cube (Figure 2.2.2). The 4 body diagonals are the C3 axes. Then
there are 3 C4 axes and 6 C2 axes. Thus the total number of elements of this group is
equal to 24 (4 × 2 + 3 × 3 + 6 + 1). All the C4 axes are equivalent, and so are all the
C3 axes and the C2 axes. The C4 axes are bilateral as the C2 axes are perpendicular to
them. For the same reason C3 axes are also bilateral. The class structure is: E, 6 C2
rotations, 8 rotations of the type C3 and C32 6 rotations of the type C4 and C43 and 3
C2 (= C42 ) rotations.
C3
C2

Figure 2.3.3 Symmetry Axes of the Tetrahedron

Oh Group. Oh is the full octahedral group and contains all the possible symmetry
transformations of the cube. These include the rotational symmetries contained in the
O group as well as those symmetries arising from the fact that the center of the cube
is a point of symmetry. Consequently, it can be stated that Oh = Ci × O.Oh has 48
elements and twice as many conjugacy classes as the group O.
Y and Yh Groups. These are the symmetries of a regular icosahedron. An icosahedron
is a convex polyhedron with twenty triangular faces arranged so that at each vertex
of the icosahedron 5 triangular faces meet. The group Y is the group of 60 rotational
symmetries of the icosahedron and Yh = Ci × Y. These symmetries occur rather rarely
in molecules.
Example 17. Consider the group C3v , the symmetry group of the ammonia molecule
(Figure 2.2.1). Label the hydrogen atoms by the numbers 1, 2 and 3. Now a C3 rotation
sends the atom 1 to atom 2, atom 2 to atom 3 and the atom 3 to atom 1. Thus it is
correct to represent the C3 operation by a 3–cycle (1, 2, 3). Now consider the σv plane

that contains the nitrogen atom and the hydrogen atom 3. A reflection in this plane
switches the positions of 1 and 2, therefore the operation σv can be represented by the
transposition (1, 2). With this notation, the following equalities are easily verified:

C3 ≡ (1, 2, 3), σv ≡ (1, 2), C32 ≡ (1, 3, 2),

σv C3 ≡ (1, 3), σv C32 ≡ (2, 3).
This way the permutation representation of the group C3v is obtained. In fact, it must be
noted that the group C3v is isomorphic to the group S (3).

Exercises
1. Show that an axis of symmetry Cn is bilateral if there exists a U2 axis perpendicular
to Cn . Find the conjugate rotations about Cn for both cases of n being even and odd.
2. Show that the inversion operation commutes with all other symmetry elements. If G is
a point group then show that the group G × Ci contains twice as many classes as G;
to each class A of G there correspond two classes A and Ci A in the group G × Ci .
3. Draw the stereographic projection for the group D4h . Determine all the conjugacy
classes.
4. Find the permutation representation for the Tetrahedral Group T . Identify a
permutation group that is isomorphic to T .
5. Determine the permutation representation for the group Td . Show that C3v is a
subgroup of Td .
6. What is the symmetry group of the hypothetical molecule shown in Figure 2.3.4?
7. What could be the Schoenflies notation for designating the symmetry group of the O2
molecule? What could it be for the CO molecule? Are these groups finite?

Figure 2.3.4 Hypothetical Molecule

In the previous chapter, we observed that group multiplication tables of elements

of point groups like C3v and T also apply to the corresponding elements of an
appropriate permutation group. In fact, the one to one and onto correspondence
between these groups were referred to as isomorphism. Such alternative descriptions
of the multiplication table of an abstract group G can also be called as representations
denoted by Γ( g) for every element g ∈ G.
One of the extensively studied approaches towards generating representations
of a group G is through the familiar matrices. One associates with each element
g ∈ G a matrix Γ( g) such that the group multiplication table is obeyed by the
matrices under matrix multiplication. Such representations are particularly useful as
they are convenient for performing calculations. We present the necessary aspects of
the theory of representations combined with examples. In this chapter our emphasis is
on the statements of theorems and the results obtained from their application. Readers
interested in detailed proofs are advised to refer to the many available standard texts
on group theory. For continuity and clarity, we will now begin with vectors and vector
spaces which leads to the construction of matrices acting on such vectors.

3.1 Vector Spaces

An abstract vector space is an important mathematical structure which is frequently
encountered in physics. The elements of a vector space are called vectors. The set
of vectors forms an abelian group under the addition operation. The identity of
this abelian structure is called the null vector, denoted usually by 0. Apart from the

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30 Group Theory for Physicists

operation of addition, the vector space is also closed under the operation of scalar
multiplication. A scalar, for the purposes of this chapter, is simply a complex number.
It is natural to assume that for any vector x in a vector space V, the complex number 1
times x is same as x, i.e., 1x = x. For scalars α, β and vectors x, y : (α + β) x = αx + βx
as well as α( x + y) = αx + αy. Consequently 0x = 0. The same symbol is used to
represent the scalar 0 and the null vector as there is no chance of confusion here. Also,
if x, y are any two vectors and α, β are any two complex numbers then the linear
combination αx + βy is also a vector in V. A linear combination of any finite number
of vectors can be formed in this way. It may be noted that the scalars themselves need
not be members of V. As described, V is a complex vector space since all the scalars
belong to the set of complex numbers.
A set of non-null vectors { xi }in=1 in V is said to be a linearly independent set if the only
possible linear combination of these vectors that is equal to 0 is the one in which all the
multiplicative scalars are themselves 0. The set of all possible linear combinations of
the linearly independent vectors { xi }in=1 is the linear span of the linearly independent
set. Denote the linear span of { xi }in=1 by L({ xi }in=1 ). Clearly L({ xi }in=1 ) is itself a
vector space contained in V, and hence is called a subspace of V. When L({ xi }in=1 )
is equal to V, then V is said to be of dimension n and { xi }in=1 are said to constitute a
basis for V. Note that L(S) is defined for any set S of vectors but here the set { xi }in=1 is
assumed to be linearly independent.
Any general vector x in V can be uniquely expressed as a linear combination of the
n n
basis vectors. For, if the same vector x could be expressed as ∑ αi xi as well as ∑ β i xi
i =1 i =1
where at least one αi 6= β i , then
n
∑ ( αi − β i ) xi = 0
i =1
and it would follow that the basis vectors are not linearly independent. The scalars αi
are called the components of the vector x in the basis { xi }in=1 .
From here onwards, it would be assumed that V is of finite dimension, though
infinite dimensional vector spaces are quite relevant as well. As the number of
vectors in V is infinite (unless of course V contains only the null vector), it may
be possible to choose a different set of linearly independent vectors {y j }m j=1 as a
basis for V. For the concept of dimension of V to be well defined, it is necessary
that both the basis sets contain same number of elements, i.e., m = n. In order
to see this, suppose { xi }in=1 and {y j }m j=1 are both basis sets and m > n. Since
x1 can be expressed as a linear combination of y0j s, some yk can be expressed as
a linear combination of x1 and the remaining y0j s. This would mean that the set
B1 = {y1 , . . . , yk−1 , x1 , yk+1 , . . . ym } is a basis set. x2 can be expressed as a linear
combination of vectors in B1 , and because x2 and x1 are linearly independent, it would
follow that some yl can be expressed as a linear combination of x2 and the remaining
elements of B1 . Now yl can be replaced by x2 and a new basis set is obtained to
be B2 = {y1 , . . . , yk−1 , x1 , yk+1 , . . . , yl −1 , x2 , yl +1 , . . . , ym }. Continuing in this

manner, the basis set Bn would contain all the { xi }in=1 and some y0j s. But because
{ xi }in=1 is itself a complete basis for V, Bn would no longer be linearly independent
which is a contradiction. Thus the conclusion is m ≤ n. By the same argument, it
follows the n ≤ m and hence m = n.
The subspace {0} and the subspace V are the trivial subspaces of the vector space V.
All other subspaces are non-trivial. The dimension of a subspace of V is clearly less
than or equal to the dimension of V. If there exist subspaces W1 and W2 of V such that
every vector v in V can be uniquely expressed as the sum w1 + w2 where wi ∈ Wi , then
V is said to be a direct sum of the subspaces W1 and W2 .

V = W1 ⊕ W2 . (3.1.1)

A Linear Operator on a vector space V is a mapping T of V into V such that for all x
and y in V : T (αx + βy) = αTx + βTy. A consequence of this definition is that T0 = 0.
Suppose { xi }in=1 is a basis set for V and x is any element of V. By linearity of T it
follows that
!
n n
Tx = T ∑ αj xj = ∑ α j Tx j ,
j =1 j =1

i.e., the action of T on any vector is determined completely if the action of T is known
on the basis set. Suppose now that

n
Tx j = ∑ tij xi ,
i =1

where tij are some scalars. Combining the above two

!
n n
Tx = ∑ ∑ tij α j xi .
i =1 j =1

Thus the action of T on the vector x = (α1 , . . . α j , . . . αn ) is to transform it into the

vector Tx = ( β 1 , . . . , β i , . . . β n ) whose component β i = (∑nj=1 tij α j ). This is most
conveniently represented as the matrix product
    
β1 t11 ··· t1n α1
 ..   .. .. .. ..  .
 . =

. . .  . 
βn tn1 ··· tnn αn
The matrix {tij } is identified with the operator T. It may be noted that the jth column of
the operator matrix contains the components of the vector Tx j . However, the matrix
associated with a linear operator is not unique in the sense that the same operator
would be represented by a different matrix for a different choice of basis vectors. If T
is such that Tx = Ty ⇒ x = y, then T is a one to one map of V onto V. In such a case

the matrix of T is an invertible n × n matrix. It is a fact that the set of all invertible
square n × n matrices form a non-abelian group under matrix multiplication.
It is sometimes necessary to relate the matrix representations of a linear operator
T in two different basis sets { xi }in=1 and {yi }in=1 of a vector space V. Suppose Tx
and Ty are the matrices of the operator T in the two bases. The basis vectors { xi }in=1
themselves must be linearly dependent on the vectors {yi }in=1 by relations of the form

n
xi = ∑ s ji y j ,
j =1

where {s ji } are some scalars. The scalars s ji can be thought of as entries of a matrix S.
It is clear that S is an invertible matrix, for { xi }in=1 and {yi }in=1 form a complete basis.
If x = ∑in=1 αi xi is any general vector in V, then the representation of x in the basis
{yi }in=1 can now be obtained as
! !
n n n n
x= ∑ αi ∑ s ji y j = ∑ ∑ s ji αi yj.
i =1 j =1 j =1 i =1

It follows that Sx is the representation of x in {yi }in=1 basis. Because a change of basis
should not change the end result of a linear transformation STx x = Ty Sx. In matrix
form, this condition can be written as a similarity transformation

Tx = S−1 Ty S. (3.1.2)

It may as well be pointed out that the traces (sum of diagonal elements) of matrices
that are related by a similarity transformation are equal.
Given an invertible linear operator T on a vector space V, the Eigenvalue Equation
of the operator is

Tx = λx.

The non-null vectors x satisfying the above equation are called the eigenvectors of T and
the scalars λ, the corresponding eigenvalues. An important example of an eigenvalue
equation is the Schrödinger equation Hψ = Eψ for calculating the stationary states of
a closed quantum mechanical system. An operator T which has the same eigenvalue
for all x ∈ V is called a scaling transformation.
A vector space V is a normed space if for every vector x in V is defined a
non-negative real number called the norm (or length) of x denoted by k x k such that
the following properties are satisfied:
1. k x + yk ≤ k x k + kyk for all x and y in V.
2. kαx k = |α|k x k for all x in V and all scalars α.
3. k x k = 0 if and only if x = 0.

A vector of unit norm is called a unit vector.

A vector space V is an inner product space (or a unitary space) if for all x, y in V is
defined a complex number ( x, y), called their inner product such that the following
properties are satisfied:

1. ( x, y) = (y, x ) where the overbar means complex conjugation.

2. ( x, αy + βz) = α( x, y) + β( x, z) for all scalars α and β.
3. ( x, x ) ≥ 0 for all x in V.
4. ( x, x ) = 0 if and only if x = 0.

Notice that it follows from Property 1 and Property 2 that (αx, y) = α( x, y) and also
that (0, x ) = ( x, 0) = 0. Two non-null vectors x and y are said to be orthogonal if
( x, y) = 0. Clearly if ( x, y) = 0 then (y, x ) = 0. It can be proven that if x and y are
orthogonal then they are linearly independent.
p A unitary space is naturally a normed space if for a vector x, the norm k x k =
( x, x ). It is easy to check that Property 2 and Property 3 for the norm are satisfied
with this definition. In order to verify Property 1, one may note

k x + y k2 = ( x + y, x + y) = ( x, x ) + (y, y) + 2Re ( x, y)

⇒ k x + yk2 ≤ k x k2 + kyk2 + 2|( x, y)|.

It follows from above that k x + yk ≤ k x k + kyk will be true if the following, well
known Cauchy–Schwartz inequality is satisfied:

|( x, y)| ≤ k x kkyk. (3.1.3)

The Cauchy–Schwartz inequality can be proved as follows. For some scalar α and
vectors x and y,

k x + αyk2 = k x k2 + 2Re [α( x, y)] + |α|2 kyk2 ≥ 0.

Since the above inequality is true for any arbitrary scalar α, it should as well be true in
(y, x )
the special case α = − ||y||2 (assuming y 6= 0, otherwise the inequality is trivially true).
On substituting this value of α, inequality 3.1.3 follows.
The basis set for a unitary space can be chosen in such a way that all the basis
vectors are of unit length and are mutually orthogonal. Such a basis is called an
orthonormal basis. Suppose that { xi }in=1 is a basis for the unitary space V, which is
not necessarily orthonormal. From this basis one may construct an orthonormal basis
{ei }in=1 by the Gram–Schmidt orthogonalisation process. It can be checked that with the
following definitions for ei0 s, the basis {ei }in=1 is indeed orthonormal.

x1 x 2 − ( e1 , x 2 ) e1 x 3 − ( e1 , x 3 ) e1 − ( e2 , x 3 ) e2
e1 = , e = , e = , ... (3.1.4)
k x1 k 2 k x 2 − ( e1 , x 2 ) e1 k 3 k x 3 − ( e1 , x 3 ) e1 − ( e2 , ) e2 k
(ei , e j ) = δij ,

where δij is the Kronecker delta which assumes the value 1 if i = j and 0 otherwise.
Suppose that T is a linear operator on a unitary space V and {ei }in=1 an orthonormal
basis of V. It is then possible to define another operator T † related to T, called the
adjoint of T by the following relation:

tij† = t ji . (3.1.5)

Thus the entries of the adjoint T † are complex conjugates of the corresponding entries
of the transpose of the operator T. For this reason the terms adjoint and transpose
conjugate are used interchangeably. The following chains of equalities follow from the
definition of the adjoint.
!
n n n
(ei , Te j ) = ei , ∑ tkj ek = ∑ tkj (ei , ek ) = ∑ tkj δik = tij .
k =1 k =1 k =1
!
n n n
( T † ei , e j ) = ∑ t†ki ek , e j = ∑ t†ki (ek , ej ) = ∑ t†ki δkj = t†ji = tij .
k =1 k =1 k =1

Consequently, (ei , Te j ) = ( T † ei , e j ). In general, for any two vectors x and y

( x, Ty) = ( T † x, y). (3.1.6)

An operator is self-adjoint if it is equal to its adjoint. In other words ( x, Ty) = ( Tx, y)

for all x and y in V. Self-adjoint operators are indeed special as their eigenvalues are
real. Let T be self-adjoint and x be an eigenvector with eigenvalue λ. Then
( x, Tx ) = ( Tx, x )
⇒ ( x, λx ) = (λx, x )

⇒ λ( x, x ) = λ( x, x )

⇒ (λ − λ)k x k2 = 0.
As x is not a null vector it follows λ is real. Self-adjoint operators are particularly
important for this reason. In fact, operators corresponding to physical quantities in
quantum mechanics are necessarily self-adjoint.

The notion of a unitary operator is an important one in physical applications. Such

an operator is an invertible linear operator with additional conditions. If U is a unitary
operator then its inverse operator is the adjoint of U.

UU † = U † U = 1. (3.1.7)

Some of the important properties of unitary operators are worth noting. The norm of
a vector remains invariant under a unitary transformation, for kUx k2 = (Ux, Ux ) =
(U † Ux, x ) = ( x, x ) = k x k2 . If x is an eigenvector of U with eigenvalue λ, then it
follows from the same chain of equalities that |λ|2 = 1, i.e., the eigenvalues of a unitary
operator are of unit modulus. The rows of a unitary matrix are orthonormal, and so
are the columns. The product of two unitary operators (matrix multiplication) is again
a unitary operator. Also, the unit matrix is indeed a unitary operator. These properties
show that the set of all unitary operators on the vector space V forms a group under
matrix multiplication. If V has dimension n, then this group is designated as U (n),
also referred to as the unitary group of degree n. The unitary group and its subgroups
will be studied in Chapter 5, Lie Groups.
y

Tq a

j q
a
i x

Figure 3.1.1 R2

The vector space V that is considered here is of the finite dimension. For
applications in quantum mechanics, the specialized vector space called Hilbert space
is sufficient. The Hilbert space is a unitary space with the additional condition of
completeness. The completeness condition will not be of concern in this text, though
it may be implicitly in use. Let it be said that the notions of linear transformations,
inverse transformations, eigenvalue equations, adjoints, self-adjoint transformations
and unitary operators continue to remain relevant whether the Hilbert space under
consideration is finitely dimensional or infinitely dimensional. Also, all the numbered
formulae in this section continue to remain valid irrespective.

Example 18. Consider the two-dimensional Euclidean plane endowed with a

coordinate system consisting of mutually perpendicular x-y axes. Then the set of
radius vectors of all points in the plane is a two-dimensional real vector space R2 .
A real vector space is one whose multiplicative scalars are chosen from the set of
real numbers instead of complex numbers. The reader should be familiar with the
parallelogram law of vector addition and scaling of vectors via multiplication by real
numbers. The unit vectors i and j (Figure 3.1.1) together form an orthonormal basis.
The linear transformation Rθ rotates any general radius vector a by an angle of θ
about the origin without changing the length of a. If k ak is the length of a then in the
chosen basis
a = k ak(i cos α + j sin α),
Rθ a = k ak(i cos(α + θ ) + j sin(α + θ )).
On eliminating α, the transformation matrix is obtained to be

cosθ −sinθ

Rθ = .
sinθ cosθ

It can be verified that indeed Rθ (αa + βb) = αRθ a + βRθ b for any two vectors a and b
in R2 and reals α and β. The inverse transformation of Rθ is R−θ which is also equal to
the transpose of Rθ as is evident from the matrix form of Rθ . A transformation whose
inverse is the transpose of the transformation is called an orthogonal transformation.
R2 is an inner product space if the inner product of a and b, ( a, b) = a.b, where
the right hand side is the usual dot product of vectors one learns about in elementary
vector algebra. With this definition it follows that ( R−θ a, b) = ( a, Rθ b). Hence R−θ is
the adjoint of Rθ in the sense of Equation 3.1.6.
Example 19. All homogeneous polynomials of degree two in the variables x and y
form a three-dimensional vector space. The basis set for this vector space can be chosen
to be { x2 , y2 , xy}. A general element of this vector space is αx2 + βy2 + γxy where
α, β, γ are some complex numbers. If the basis set is chosen to be { x2 + y2 , x2 −
α+ β α− β
y2 , xy}, then the same general element can be written as 2 ( x2 + y2 ) + 2 ( x2 −
y2 ) + γxy.

3.2 Group Action on Vector Spaces

Given an abstract group G and a vector space V, it is often possible to regard the
elements of the group as invertible linear operators on the vector space. Once this is
done, every element of G can be associated with the matrix of the operator in a fixed
basis for V. Let Γ be a function that maps an element g of G to the matrix Γ( g). Then
the totality of matrices Γ( g) is called a representation of G if Γ is a homomorphism, i.e.,

Γ ( g1 g2 ) = Γ ( g1 ) Γ ( g2 ) . (3.2.1)

Notice that the product on the right hand side is a matrix product, while that on the
left hand side is the product defined in the group. If the dimension of V is n then the
representation Γ is said to be of degree n. For a given group element g, the entries of
Γ( g) are easily ascertained from the fact that g acts as an invertible linear operator on
V. Thus the jth column of Γ( g) are the coordinates of the transformation of the basis
vector x j of V by the group element g in accordance with the map

n
gx j → ∑ [Γ( g)]ij xi .
i =1

A representation of Γ is said to be faithful if Γ( g1 ) 6= Γ( g2 ) for g1 6= g2 .

Example 20. Consider the point group C3v and the three-dimensional real vector
space R3 spanned by the unit vectors {i, j, k} (using the conventional notation).
The symmetry axis of C3v points along k. The objective is to generate a matrix
representation of C3v by letting its elements operate on-R3 . Under the C3 operation
(rotation by 2π3 in the counterclockwise sense about the z-axis), the unit vectors are
clearly transformed as follows:
√
1 3
i → − i+ j + 0k,
2 2
√
3 1
j → − i − j + 0k,
2 2
k → 0i + 0j + k.

It is therefore plausible to make following associations

 √   √ 
−
√2
1
− 23 0 −√12 − 23 0
2
C3 ≡  3
− 12 0  , C3 ≡  − 23 − 12 0 .
   
2
0 0 1 0 0 1
Now consider the σv plane to be the z-x plane. Then a σv reflection leaves i and k
invariant and reverses j.
 
1 0 0
σv ≡  0 −1 0  .
0 0 1
The matrix forms of C3 σv and C32 σv can be found from above matrices. The identity
element is the 3 × 3 identity matrix.
 √ 
1 3
 
1 0 0 −
√2 2 0
E ≡  0 1 0  , C3 σv ≡  3 1
0 ,
 
2 2
0 0 1 0 0 1

 √ 
3
−1 − 0
 √23 2
C32 σv ≡  − 1
0 .

2 2
0 0 1

If Γ( g) is the matrix corresponding to group element g in some given basis of V, then

under a change of basis effected by an invertible matrix S, the matrix of g becomes
S−1 Γ( g)S in accordance with Equation 3.1.2.

Definition 4. Let Γ be a representation of degree n of a group G. Then the character of

an element g of G under the representation Γ is the trace of the matrix Γ( g).
n
χΓ ( g) = trΓ( g) = ∑ [Γ( g)]ii . (3.2.2)
i =1

The symbol χΓ ( g) would from now onwards stand for the character of the group
element g under the representation Γ.
A consequence of this definition is that if Γ and Θ are two different representations
of G related by a similarity transformation, then the character of any element of G
has the same value in both the representations. Such representations are equivalent
representations and are fundamentally not distinguished from each other. A further
consequence is that all the group elements which belong to the same conjugacy class
have the same character. For if g1 = ag2 a−1 , then Γ( g1 ) = Γ( a)Γ( g2 )Γ( a)−1 , thereby
Γ( g1 ) and Γ( g2 ) have the same character. The group character is therefore a class
function. In the example of the matrix representation of group C3v , χ( E) = 3, χ(C3 ) =
χ(C32 ) = 0 and χ(σv ) = χ(C3 σv ) = χ(C32 σv ) = 1.

3.3 Reducible and Irreducible Representations

In a representation Γ of a group G over a vector space V, it may so happen that some
non-trivial subspace V1 of V remains invariant under the group action. By this it is
meant that the group elements transform V1 onto V1 , however the vectors outside
of V1 are transformed to vectors outside V1 . When this happens, then Γ is also a
representation of G on V1 and in the context of V, a subrepresentation of G. With
an appropriate choice and ordering of basis vectors, the representing matrix for any
general element g of G takes the form

Θ( g) Π( g)

Γ( g) = ,
0 4( g)

where Θ( g), Π( g) and 4( g) are fixed size matrix blocks whose entries vary with
g. The bottom left block has 0 as all entries. A representation such as Γ which
has a subrepresentation is called a reducible representation. A representation with no
subrepresentations is called an irreducible representation.

A reducible representation is completely reducible if it can be decomposed into

irreducible representations. If Γ were a completely reducible representation then for
every g
Γ ( g) 0
 1 
0
Γ( g) =  0
 .. .

. 0
0 0 Γk ( g)

In other words, the representation matrices of all the group elements can be
simultaneously put in a block diagonal form of identical structure, with each block
being an irreducible subrepresentation. Γ is then a direct sum of the irreducible
representations.
k
Γ= Γi .
M
(3.3.1)
i =1

It is also obvious from the above decomposition that

k
χΓ ( g) = ∑ χ Γ ( g ).
i
(3.3.2)
i =1

It is a theorem that all reducible representations of a finite group obtained via group
action on complex or real vector spaces are completely reducible. The interested reader
is encouraged to study Maschke’s Theorem (see Appendix A) in this regard. It may as
well be noted that every representation of a finite group over a real or a complex
vector space is equivalent to a unitary representation, i.e., the matrices of all the group
elements are unitary.
It is fairly obvious from the foregoing that the irreducible representations are the
building blocks for all possible representations of a finite group. Therefore a study
of the irreducible representations of a group is essential. To begin with, a group
has the trivial irreducible representation of degree 1 which is generated when every
group element maps the unit vector of a one-dimensional vector space to itself. This
representation is also called the unit representation and is designated by the symbol A.
The character of every group element in this representation is equal to 1. It is clear that

∑ χ A ( g ) = | G |, (3.3.3)
geG

where summation is over all the elements of the group.

We may recall that the degree of a representation is the dimensionality of the vector
space on which the group acts. It may be possible for a group to have non-equivalent
irreducible representations of the same degree, and these representations would be
treated as distinct. It is a theorem that the degree of an irreducible representation
divides the order of the group. Even the number of distinct irreducible representations

of a finite group is finite and is equal to the number of conjugacy classes in the
group. It has already been noted that group elements in a conjugacy class have the
same character under a given representation. Suppose that the finite group G has r
conjugacy classes. If Γ1 , Γ2 , . . . Γr are distinct irreducible representations of a group
G whose degrees are `1 , `2 , . . . `r respectively then

r
∑ `2i = |G|. (3.3.4)
i =1

The criteria that G has at least one representation of degree 1, that each ì divides | G |
and that together the ì ’s satisfy the above equation put stringent constraints on the
possible values of ì0 s. In most cases, these conditions are sufficient to ascertain the
values of ì0 s.
Let Γ and Θ be two irreducible unitary representations of a finite group G. Then the
following orthogonality relation holds (read Schur’s Lemma in this connection which
we have briefly discussed in Appendix B) between the entries of the matrices of group
elements under the two representations.

|G|
∑ [Γ( g)]ik [Θ( g)]lm = δ δ δ .
`Γ ΓΘ il km
(3.3.5)
geG

The group characters also follow an orthogonality relation given by

∑ χΓ ( g).χΘ ( g) = |G|δΓΘ . (3.3.6)

geG

As the identity or unit representation A is always an irreducible representation of any

group, it follows from Equation 3.3.6 that for any irreducible representation Γ other
than A

∑ χΓ ( g) = 0. (3.3.7)
geG

If Γ = Θ in Equation 3.3.6, then we have

∑ |χΓ ( g)|2 = |G|. (3.3.8)

geG

Another fact to bear in mind is that the character of the identity element of a group
is always equal to the degree of the representation. If c1 , c2 . . . cr represent the classes
of G and ni the number of elements in the conjugacy class ci , then Equation 3.3.6 may
once again be written as
r
n
∑ |Gi| χΓ (ci )χΘ (ci ) = δΓΘ . (3.3.9)
i =1

As the number of distinct irreducible representations is also equal to number of classes

th
q a r × r matrix whose k row contains that characters of the
in G, one may construct
classes multiplied by |nGi| in the kth representation. Then the orthogonality of group
characters is equivalent to stating that the row vectors of this matrix are orthonormal.
Since the columns of such a matrix would also be orthonormal, it immediately follows
that
r
n
∑ |Gi| χΓl (ci )χΓl (c j ) = δij . (3.3.10)
l =1

3.4 Irreducible Representations of Point Groups

In physical applications of the representation theory, it is often only the group
characters that are of relevance as against the actual matrices corresponding to the
group elements. Equations 3.3.6, 3.3.7 and 3.3.8 prove very useful in the calculation
of these characters. The characters of the group classes under various representations
are arranged to create a character table for the group. A typical character table for
representations of a finite group G formally looks like

G n 1 g1 ... n r gr basis
Γ1 Γ
χ 1 (C1 ) ... Γ
χ 1 (Cr )
.. ..
. .
Γr χΓr (C1 ) χΓr (Cr )

where the symbols have their usual meanings. ni gi in the top row states that there
are ni group elements in the conjugacy class ci , and gi is a representative element of
ci . The columns to the right of the character entries contain the basis states of the
corresponding representations.
While working with point groups, it is conventional to adopt the Mulliken symbols
for naming irreducible representations. The principal symmetry axis is taken along
the z-axis. Representations of degree 1 are identified by symbols A and B, degree 2
representations are identified by the symbol E (not to be confused with the identity
element) and degree 3 representations by the symbols F or T. Given a symmetry
operation g in the group G, the basis vectors of a representation Γ get transformed
by Γ( g). For certain operations, it may happen that the basis vectors are transformed
onto themselves, in which case the representation is symmetric. If on the other
hand, the basis vectors are reversed, then the representation is antisymmetric. In
case of representations A, the base functions are symmetric with respect to rotation
about principal axis, while they are antisymmetric with respect to rotations about
principal axis in case of representations B. For representations which are symmetric or
antisymmetric with respect to reflections σh , their symbols are marked with one prime
or two primes respectively. Representations which are symmetric or antisymmetric

with respect to inversions are marked with the letter g (gerade) or u (ungerade)
respectively in the subscript to the representation symbol. Following are some
examples which illustrate how character tables may be constructed in simple cases.

Example 21. Consider the group C2h (Chapter 2, Section 2.3). The number of classes
in C2h is 4. Since 12 + 12 + 12 + 12 = 4, all the irreducible representations are of degree
1. The character of identity element of C2h is 1 in all the representations. However,
since the order of every other element of C2h is 2, and that each representation is of
degree 1, it follows that the character of every element can be either +1 or −1. The
character table follows:

C2h E C2 σh I
Ag 1 1 1 1 Rz x2 ; y2 ; z2 ; xy
Au 1 1 −1 −1 z
Bg 1 −1 −1 1 R x ; Ry xz; yz
Bu 1 −1 1 −1 x; y

Particular attention should be paid to the names of the irreducible representations.

Observe that the character of the inversion operation (I) is +1 in the gerade
representations and −1 in the ungerade ones. Similarly the character of C2 operation
is +1 in A representations and −1 in B representations. Two choice of bases have been
shown, linear bases in second last column and quadratic bases in the last column. In
case of the linear bases, x, y, z are the polar vectors along the axes and R x , Ry , Rz
are the axial vectors along the axes. The reflection and inversion properties of these
vectors are well known from elementary vector algebra and it is left for the reader to
verify that indeed the group operations produce the indicated effects: For example,
C2 R x = − R x but C2 Rz = Rz and so on. In case of quadratic bases it is the signs of
the expressions that get affected: For example, C2 x2 = x2 but C2 xz = − xz and so on.
Notice that C2 (a rotation by π about z-axis) changes the sign of x but leaves that of z
unchanged.

Example 22. Consider the group D2d . This is a group of 8 elements. The 5 conjugacy
classes in D2d are: one class containing E, one class containing C2 rotation, one
containing two rotary reflection transformations S4 , one class containing 2 reflections
in the diagonal planes σd and finally one class containing two rotations about the
U2 axes. There are therefore 5 irreducible representations of D2d . From the equation
12 + 12 + 12 + 12 + 22 = 8, it follows that 4 of these representations are of degree 1 and
one representation is of degree 2.

D2d E C2 2S4 2U2 2σd

A1 1 1 1 1 1 x 2 + y2 , z2
A2 1 1 1 −1 −1 Rz
B1 1 1 −1 1 −1 x 2 − y2
B2 1 1 −1 1 1 z xy
( x, y);
E 2 −2 0 0 0 ( R x , Ry ) xz, yz

In the unit representation A1 , all the group characters are 1 as usual. In any of the
representations, the group characters satisfy Equations 3.3.7 and 3.3.8. From these it
follows that
χ( E) + χ(C2 ) + 2χ(S4 ) + 2χ(U2 ) + 2χ(σd ) = 0,

|χ( E)|2 + |χ(C2 )|2 + 2|χ(S4 )|2 + 2|χ(U2 )|2 + 2|χ(σd )|2 = 8.
The character of identity is 1 in all the other representations of degree 1. Since
the orders of C2 , U2 and σd are two, their character values in a one-dimensional
representation can be +1 or −1. The second of the above equations then forces the
|χ(S4 )| = 1 and the first of the above equations forces it to be real. The only
possible combinations of characters satisfying the above equations for representations
of degree 1 are then as shown in the character table. In the representation of degree 2
(i.e., the representation E), the character of identity is clearly 2. The characters of other
elements in this representation can now be ascertained easily using orthogonality of
the columns of the character table (Equation 3.3.10). For example, the columns of the
group elements E and C2 are orthogonal.

χA1 ( E)χA1 (C2 ) + χA2 ( E)χA2 (C2 ) + χB1 ( E)χB1 (C2 )+

+χB2 ( E)χB2 (C2 ) + χE ( E)χE (C2 ) = 0.

It follows then that χE (C2 ) = −2. In the same manner the characters of other elements
in the representation E can be easily checked to be 0. Lastly, it may be noted that the
principal axis of symmetry in the group is the S4 axis. In accordance with this fact, the
symmetric representations A1 and A2 are the ones in which the χ(S4 ) = 1, while in
the antisymmetric representations B1 and B2 , χ(S4 ) = −1.
Example 23. Consider the cyclic group C3 . Since there are three conjugacy classes in
C3 , there are 3 irreducible representations of C3 . All the representations are of degree 1,
as follows from the equation 12 + 12 + 12 = 3. In the unit representation, all elements
have character 1. In a representation of degree 1 (not the unit representation), let the
character of rotation C3 be ω.

C3 E C3 C32
A 1 1 1 z; Rz x 2 + y2 ; z2
(
2πi 2πi
E 1 e 3 e− 3 ( x, y); ( x2 − y2 , xy);
2πi 2πi
1 e− 3 e 3 ( R x , Ry ) (yz, xz)

Suppose the basis vector for this representation is ψ. The action of C3 rotation on ψ is
in accordance with C3 ψ = ωψ. Then

C32 ψ = C3 (C3 ψ) = C3 (ωψ) = ωC3 ψ = ω 2 ψ.

Thus the character of C32 in the same representation is ω 2 . It follows from Equation
3.3.9 that 1 + ω + ω 2 = 0. Hence ω is the complex cube root of unity. These
representations are shown in the last two rows of the character table. However, instead
of identifying them as two separate 1 degree representations, these representations are
combined into a single 2 degree representation, identified as E in the first column. The
reason for doing so will be made clear when applications are considered.

Example 24. Consider the group C3v . The group is non-abelian with 6 members in 3
conjugacy classes. Since 12 + 12 + 22 = 6, C3v has three irreducible representations,
two of which are of degree 1, and one is of degree 2. The calculations in this case are
straightforward and the reader can easily verify that indeed the following characters
are obtained:

C3v E 2C3 3σv

A1 1 1 1 z x 2 + y2 ; z2

A2 1 1 −1 Rz
E 2 −1 0 ( x, y); ( R x Ry ) xz; yz

3.5 The Regular Representation

Consider a finite group G of order n whose elements are g1 (= E), g2 , . . . gn . Let
{ xi }in=1 be a linearly independent set of unit vectors spanning an n-dimensional vector
space. The group elements can be put in a one to one correspondence with the basis
vectors clearly by associating gi with xi . Then the action of the group element g on
this vector space can be formally defined by xk = xi g if gk = gi g. If g = E, clearly
xk = xi , i.e., action of the identity element fixes every basis vector. For this reason,
E can be associated with the n × n identity matrix. For any other group element g,
right multiplication of G with g is a mere permutation of elements of G in which
none of the elements of G are fixed. Correspondingly, g transforms any given unit
vector in the basis to another distinct unit vector in the basis. It follows that the matrix

representation of g would be a n × n matrix in which each row and each column

would contain exactly one non-zero entry which will be equal to 1. This particular
representation of G is called the right regular representation, which we denote by the
symbol Γ R . It is clear the degree of Γ R is n, χΓ ( E) = n and χΓ ( g) = 0 for g 6= E. From
R R

orthogonality relations, it follows that Γ cannot be an irreducible representation.

The immediate task is to decompose Γ R as a direct sum of the irreducible

representations of G. In order to do this, assume more generally that Γ is some
reducible representation of G. Suppose now that
r
Γ= mi Γi ,
M

i =1
where mi is the multiplicity with which the irreducible representation Γi appears in Γ
and r is the number of distinct irreducible representations of G. It follows from above
that for any group element g

r
χΓ ( g) = ∑ mi χ Γ ( g )
i

i =1
r
⇒ χΓ ( g ) χΓk ( g ) = ∑ mi χ Γ ( g ) χ Γ ( g )
i k

i =1
r
∑ χΓ ( g)χΓ ( g) = ∑ m i ∑ χ Γ ( g ) χ Γ ( g ).
k i k
⇒
geG i =1 geG

Using Equation 3.3.6, it now easily follows that

1
∑ χ Γ ( g ) χ Γ ( g ).
k
mk =
|G| geG

The equation above can be used to obtain the multiplicity mk with which any
irreducible representation Γk appears in a given reducible representation Γ. In the
specific case of Γ = Γ R , it was earlier noted that χΓ ( E) = n and χΓ ( g) = 0
R R

for g 6= E. Since χΓ ( E) = `k , one has mk = `k from the above equation. Each

irreducible representation of G appears as many times as its degree in the right

regular representation of G. Additionally, since χΓ ( E) = ∑ri=1 mi χΓ ( E), Equation
R i

3.3.4 follows as a consequence. Finally it may be noted that the trivial representation
(of degree 1) occurs once in Γ R .

3.6 Tensor Product of Representations

It is possible to generate reducible representations of a group from its irreducible
ones. This can be done by forming what is called the tensor product of the

irreducible representations. Let G be a finite group which has Γ1 and Γ2 as irreducible

representations. Γ1 is a representation of G on a unitary space V spanned by the basis
{ xi }in=1 1 whereas Γ2 is a representation of G on a unitary space W spanned by the basis
{yi }in=2 1 . The tensor product of V and W, denoted by V ⊗ W, is another vector space
n ,n
spanned by the basis { xi ⊗ y j }i=1 1,j2=1 . The dimension of V ⊗ W is clearly equal to the
product of the dimensions of V and W. Once the product space has been defined, it
remains to construct a representation Γ of G on V ⊗ W. The representation Γ is called
the tensor product of the representations Γ1 and Γ2 and symbolically Γ = Γ1 ⊗ Γ2 . For
completing the definition, Γ( g) should be simply related to Γ1 ( g) and Γ2 ( g) for all g
in G. This essentially requires the specification of action of g on every basis vector of
V ⊗ W. For the basis vector xi ⊗ y j in V ⊗ W let

g ( xi ⊗ y j ) → g ( xi ) ⊗ g ( y j )
! !
n1 n2
g ( xi ⊗ y j ) → ∑ [Γ1 ( g)]ki xk ⊗ ∑ [Γ2 ( g)]lj yl
k =1 l =1

n1 n2
g ( xi ⊗ y j ) → ∑ ∑ [Γ1 ( g)]ki [Γ2 ( g)]lj xk ⊗ yl . (3.6.1)
k =1 l =1

The above mapping defines the action of g on a basis state xi ⊗ y j of V ⊗ W. It remains

to be shown that with such a definition, Γ is indeed a representation of G on V ⊗ W, or
in other words, Equation 3.2.1 is satisfied. The rows and columns of the matrix Γ( g)
can be indexed by the basis states xi ⊗ y j . Equation 3.2.1 is true if

[Γ( g1 g2 )] xu ⊗yv , xs ⊗yi = [Γ( g1 )Γ( g2 )] xu ⊗yv ,xs ⊗yi .

The verification of the above equality is trivial after noticing

[Γ( g)] xu ⊗yv , xs ⊗yi = [Γ1 ( g)]us [Γ2 ( g)]vi , (3.6.2)

which follows from the mapping defined in 3.6.1. The character of g can be easily
obtained by summing the diagonal elements. A direct consequence of the above is
that

χ Γ = Γ1 ⊗ Γ2 ( g ) = χ Γ1 ( g ) χ Γ2 ( g ). (3.6.3)

3.7 Decomposition of Reducible Representations

A reducible representation Γ of a group G over a unitary space V can be expressed
as a direct sum of the irreducible representations Γα of G in which each irreducible
representation appears with a certain multiplicity mα .

r
Γ= mα Γα .
M
(3.7.1)
α =1

In Section 3.5, the formula for mα was obtained which is restated below for emphasis.

1
∑ χΓ ( g)χΓ
α
mα = ( g ). (3.7.2)
|G| g∈ G

It may be desirable sometimes to not only know the multiplicities of the various
irreducible representations in Γ but also the subspace of V which is left invariant by
the irreducible representation Γα (i.e., the subspace of V on which Γα is an irreducible
representation of G).
Let V be a unitary space of which U and W are proper subspaces such that
V = U ⊕ W. Then any vector v in V can be expressed uniquely as a sum of vectors u
and w which belong to U and W respectively. The projection operator PU is a linear
operator on V with values in the subspace U such that

PU v = u.

Similarly PW may be defined so that PW v = w. Since the projection operator is

a mapping onto a proper subspace, it is not invertible. The projection operator
essentially projects a vector in V on a subspace (for which the operator is defined). As
a trivial example, the projection operators that project out the x, y and z components
of a vector in R3 are
     
1 0 0 0 0 0 0 0 0
Px =  0 0 0  ; Py =  0 1 0  ; Pz =  0 0 0  .
0 0 0 0 0 0 0 0 1

The problem of finding the subspace of V which is left invariant by the irreducible
representation Γα (Equation 3.7.1) reduces to constructing the project operator PΓα . In
the orthogonality relation (Equation 3.3.5)

|G|
∑ [Γα ( g)]ik [Γβ ( g)]lm = δ α βδ δ ,
`Γα Γ Γ il km
geG

if i is set equal to k and summation over all i is performed, then the relation reduces to

`Γα
|G| ∑ χα ( g)[Γβ ( g)]lm = δΓα Γβ δlm .
geG

It is evident that when Γα and Γ β are the same representations then the right hand
side of the above is the identity matrix of order `α × `α and in other cases the right

hand side is the null matrix. This important property motivates the following
definition of the projection operator PΓα on the reducible representation Γ.

`Γα
∑ χΓ ( g ) Γ ( g ).
α
PΓα = (3.7.3)
|G| geG

In the direct sum in Equation 3.7.1, upon action of PΓα only those blocks in the
matrix Γ( g) will be non-zero which correspond to subspaces on which Γα gives a
representation of G.

Example 25. Consider the group C3v . In an earlier example (Section 3.2, Example
20), the explicit form of the matrices for various group elements was calculated
by considering the group’s action on R3 . The k vector was left invariant under
all transformations and in fact, the representation generated was of degree 2. For
reference, the representing matrices of various group elements of C3v in the irreducible
representation E are given below:
√ !
3
− 12 −

1 0 2
E( E) = ; E (C3 ) = √ ;
0 1 3
− 21
2
√ !
3
− 12

1 0

E C32 = √ 2
; E(σv ) = ;
− 3
− 21 0 −1
2
√ ! √ !
3 3
− 21 − 12 −
E (C3 σv ) = √ 2
; E C32 σv = √ 2
.
3 1 3 1
2 2 − 2 2

Let Γ = E ⊗ E be the tensor product of the irreducible representation of degree 2

with itself. From Equation 3.6.3, it follows that χΓ ( E) = 4, χΓ (C3 ) = χΓ (C32 ) = 1
and χΓ (σv ) = χΓ (σv C3 ) = χΓ (σv C32 ) = 0. It is now possible to decompose Γ. The
multiplicity of various irreducible representations in Γ can be calculated by application
of Equation 3.7.2.
1
Multiplicity of A1 = [1 × 4 × 1 + 2 × 1 × 1 + 3 × 0 × 1] = 1.
6
1
Multiplicity of A2 = [1 × 4 × 1 + 2 × 1 × 1 + 3 × 0 × 0] = 1.
6
1
Multiplicity of E = [1 × 4 × 2 + 2 × 1 × −1 + 3 × 0 × 0] = 1.
6
It may be noted that Γ = A1 ⊕ A2 ⊕ E. Each of the irreducible representations of
C3v occur only once in Γ. From the matrices of the representation E, the matrices of
the representation Γ can be calculated by using Equation 3.6.2. The basis for the tensor
product space in this case can be taken as i1 ⊗ i2 , i1 ⊗ j2 , j1 ⊗ i2 and j1 ⊗ j2 where

{i1 , j1 } is the basis of one Γ E and {i2 , j2 } is the basis for the other. The rows and
columns of the matrices in Γ representation are indexed in the order of the basis states
of the tensor product space as mentioned in the previous line. With this convention,
following matrices in Γ representation are obtained:
 
1 0 0 0
 0 1 0 0 
Γ( E) =   0 0 1 0 ,


0 0 0 1
 √ √ 
1 3 3 3
 4√ 4 4 4
√ 
 − 3 1
− 34 3
 

4 4 4
Γ(C3 ) =  √ ,
 
√
3 3
 − 4

− 34 1
4 4


 √ √ 
3 3 3 1
4 − 4 − 4 4
 √ √ 
1 3 3 3
4 − 4 − 4 4
 √ √ 
3 1
− 34 − 43
 
 
2 4 4
Γ(C3 ) =  ,
 
√ √
3

 4 − 34 1
4 − 43 

 √ √ 
3 3 3
4 4 4 − 14
 
1 0 0 0
 0 −1 0 0 
Γ(σv ) =  ,
 0 0 −1 0 
0 0 0 1
 √ √ 
1 3 3 3
4 − 4 − 4 4
 √ √ 
 − 3 − 14 3 3
 

4 4 4
Γ(C3 σv ) =  √ ,
 
√
3 3 3
 − 4

4 − 14 4


 √ √ 
3 3 3 1
4 4 4 4
 √ √ 
1 3 3 3
4 4 4 4
 √ √ 
3
− 14 3
− 43
 
 
2 4 4
Γ(C3 σv ) =  .
 
√ √
3

 4
3
4 − 14 − 43 

 √ √ 
3 3 3 1
4 − 4 − 4 4

Finally the projection operators for the irreducible representations in Γ may be

calculated from these matrices, the characters of irreducible representations and
Equation 3.7.3. For example, all the characters in A1 are unit, and Equation 3.7.3 takes
the form
1h i
PA1 = Γ( E) + Γ(C3 ) + Γ(C32 ) + Γ(σv ) + Γ(C3 σv ) + Γ(C32 σv ) .
6
In a similar fashion, the projection operators for other irreducible representations may
be obtained. The results follow:
 1
0 0 − 12
 1 1
 
2 0 0 2 2
 0 0 0 0   0 1 1
0 
   
2 2
PA1 =  
 , PE = 
 
1 1
,
 0 0 0 0   0 2 2 0


1 1 1 1
2 0 0 2 −2 0 0 2
 
0 0 0 0
 0 12 − 12 0 
 
PA2 =  1 1
.
 0 −2 0 
 
2
0 0 0 0

By letting these projection operators act on any general vector in the tensor product
space spanned by i1 ⊗ i2 , i1 ⊗ j2 , j1 ⊗ i2 and j1 ⊗ j2 , the subspace on which the
corresponding representation exists is easily found. Let v = α(i1 ⊗ i2 ) + β(i1 ⊗ j2 ) +
Γ( j1 ⊗ i2 ) + δ( j1 ⊗ j2 ) be a typical vector. Then
 1
0 0 − 21
    
2 α α−δ
 0 1 1  β+γ 
0   β 
     
PE v =  2 2   = 1 
1 1   Γ  2 +
 0 0 β γ
 
2 2     
− 1
0 0 2
1 δ
2
−(α − δ)

α−δ β+γ
⇒ PE = (i1 ⊗ i2 − j1 ⊗ j2 ) + (i1 ⊗ j2 + j1 ⊗ i2 ).
2 2
Thus PE projects onto the two-dimensional subspace spanned by (i1 ⊗ i2 − j1 ⊗ j2 ) and
(i1 ⊗ j2 + j1 ⊗ i2 ). Likewise, one may show that PA1 projects onto the one-dimensional
subspace spanned by

(i1 ⊗ i2 + j1 ⊗ j2 )

and PA2 projects onto the one-dimensional subspace spanned by

(i1 ⊗ j2 − j1 ⊗ i2 ).

The preceding development illustrated the method by which a reducible

representation can be decomposed into irreducible parts. In the remainder of this
section, decomposition of tensor product of two irreducible representations is
considered. Let Γ1 and Γ2 be two irreducible representations of a group G and Γ1 ⊗ Γ2
be their tensor product. The multiplicity of the unit representation of G in Γ1 ⊗ Γ2 may
be calculated by using Equations 3.6.3 and 3.7.2. Since every group element has a
unit character in the unit representation, the multiplicity of the unit representation is
given by

1 1
∑ χA ( g ) χ Γ ⊗ Γ ( g ) = | G | ∑ χ Γ χ Γ .
1 2 1 2
|G| geG geG

The right hand side of the above equation is equal to 1 if Γ1 and Γ2 are conjugate
representations and 0 in all other cases (Equation 3.3.6). If the characters of the
representation are real, then the unit representation is present only in the tensor
product of the representation with itself. For now suppose Γ is an irreducible
representation of G on a unitary space V which is spanned by the basis { xi }in=1 . Then
the tensor product space V ⊗ V is of dimension n2 and is spanned by basis states
{ xi ⊗ x j }in,n
=1,j=1 . Note that xi ⊗ x j is distinguished from x j ⊗ xi . Apart from this
particular choice of basis, another choice is sometimes useful. Consider the subspaces
(V ⊗ V )σ and (V ⊗ V )α (read symmetric and antisymmetric subspaces respectively).
The symmetric space (V ⊗ V )σ is spanned by basis states xi ⊗ x j + x j ⊗ xi when i 6= j
n ( n + 1)
and the states xi ⊗ xi . Clearly the dimension of (V ⊗ V )σ is . In a similar
2
manner, the antisymmetric subspace (V ⊗ V )α spanned by basis states xi ⊗ x j − x j ⊗ xi
n ( n − 1)
for i 6= j is of dimension . Note that the dimensions of the symmetric and the
2
antisymmetric subspaces add up to n2 as they should. Let g be some transformation
in G. Consider the action of g on a basis state in the antisymmetric subspace. In
accordance with Equation 3.6.1, one has

g xi ⊗ x j − x j ⊗ xi → g ( xi ) ⊗ g x j − g x j ⊗ g ( xi )
! !
n n
∑ [Γ( g)]ki xk ∑ [Γ( g)]lj xl

g xi ⊗ x j − x j ⊗ xi → ⊗
k =1 l =1
! !
n n
− ∑ [Γ( g)]lj xl ⊗ ∑ [Γ( g)]ki xk
l =1 k =1
n
∑ [Γ( g)]ki [Γ( g)]lj ( xk ⊗ xl − xl ⊗ xk )

g xi ⊗ x j − x j ⊗ xi →
k,l =1

1 n
2 k,l∑
Γ Γ Γ Γ

g xi ⊗ x j − x j ⊗ xi → [ ( g )] ki [ ( g )] lj − [ ( g )] li [ ( g )] kj ( xk ⊗ xl − xl ⊗ xk )
=1

As is evident from the above map, the antisymmetric subspace gives a

subrepresentation (Γ ⊗ Γ)α of the group G. The characters χ2α of the antisymmetric
representation can be easily obtained from the above map to be
1 n
χ2α ( g) = ∑
2 i,j=1
[Γ( g)]ii [Γ( g)] jj − [Γ( g)] ji [Γ( g)]ij

1
(χ( g))2 − χ g2 .
⇒ χ2α ( g) = (3.7.4)
2
The reader will do well to convince themselves that the symmetric subspace also
gives a subrepresentation (Γ ⊗ Γ)σ of G. The characters χ2σ ( g) of the symmetric
representation can be shown to be

1
χ2σ ( g) = (χ( g))2 + χ g2 . (3.7.5)
2
Finally, it may be noted that in the decomposition of the representation Γ ⊗ Γ as
(Γ ⊗ Γ)σ ⊕ (Γ ⊗ Γ)α , the symmetric as well as the antisymmetric component may be a
reducible representation of Γ.

3.8 Irreducible Representations of Direct Products

As has been noted previously, it is sometimes possible that a group is isomorphic to
a direct product of two of its subgroups (e.g. Dnh = Dn × Cs ). The tensor product
described in Section 3.6 allows the construction of all the irreducible representations
of the direct product from the irreducible representations of the component groups.
Let G be isomorphic to the the direct product H × K of two groups H and K. Then any
element g of G can be expressed uniquely as

g = hk,

where h is in H and k is in K. It is also known that H and K are normal subgroups

in G and elements of H commute with elements of K. Let it be assumed that h1 and
h2 are conjugate in H and similarly for k1 and k2 in K. Then h1 k1 = hh2 h−1 kk2 k−1
for some h in H and some k in K. Because of the commutation it follows that
h1 k1 = (hk )h2 k2 (hk )−1 , i.e., h1 k1 is conjugate to h2 k2 and by the same reasoning
h1 k1 , h1 k2 , h2 k1 and h2 k2 are all conjugates in G. A conjugacy class of G consists
of all elements of the form hk in which h’s come from one conjugacy class of H
and the k’s come from one conjugacy class of K. It follows that the number of
conjugacy classes in G is equal to the product of the number of conjugacy classes
in H and the number of conjugacy classes in K. For this reason the number of
irreducible representations of G is equal to the number of irreducible representations
of H times the number of irreducible representations of K. It is natural to expect that
an irreducible representation of G could be obtained by forming a tensor product of
the an irreducible representation of H with an irreducible representation of K. If Γ H is

n
an irreducible representation of H on a unitary space V spanned by the basis { xi }i=1 1
whereas ΓK is an irreducible representation of K on a unitary space W spanned by the
basis {yi }in=2 1 , then a representation of G on V ⊗ W is obtained from the map

(hk)( xi ⊗ y j ) → h( xi ) ⊗ k(y j ),

where hk is an element of G. Proceeding as before, with maps such as above, it is not

very difficult to show that the representations so obtained indeed satisfy Equations
3.3.6, 3.3.7, 3.3.8 and 3.3.9. Additionally, in analogy with Equation 3.6.3, in the current
case one has

χΓ H ⊗ΓK (hk) = χΓ H (h)χΓK (k). (3.8.1)

Example 26. The group D2h is a direct product of D2 and Cs . The character tables for
D2 and Cs follow:

D2 E C2 (z) C2 (y) C2 ( x )
A1 1 1 1 1 x 2 , y2 , z2
B1 1 1 −1 −1 z, Rz xy
B2 1 −1 1 −1 y, Ry zx
B3 1 −1 −1 1 x, R x yz

CS E σh
A0 1 1 x, y, Rz x 2 , y2 , z2
A00 1 −1 z, R x , Ry yz, xz

It is evident now that D2h has 8 conjugacy classes and 8 irreducible representations.
Apart from all the conjugacy classes in D2 , there are other classes in D2h . For example,
C2 (z) in D2 multiplied with σh in Cs gives a class containing a single operation of
inversion in D2h , denoted by I. Similarly, multiplication of C2 (y) with σh is a class
containing the single element corresponding to reflection in the xz plane, and so
on. The characters are various classes in D2h and are obtained from those of D2 and Cs
by use of Equation 3.8.1. The names of various irreducible representations of D2h , in
accordance with the discussion in Section 3.4, are as mentioned in the character table:

D2h E C2 (z) C2 (y) C2 ( x ) I σxy σxz σyz

Ag 1 1 1 1 1 1 1 1 x 2 , y2 , z2
B1g 1 1 −1 −1 1 1 −1 −1 Rz xy
B2g 1 −1 1 −1 1 −1 1 −1 Ry zx
B3g 1 −1 −1 1 1 −1 −1 1 Rx yz .
Au 1 1 1 1 −1 −1 −1 −1
B1u 1 1 −1 −1 −1 −1 1 1 z
B2u 1 −1 1 −1 −1 1 −1 1 y
B3u 1 −1 −1 1 −1 1 1 −1 x

3.9 Induced Representations

Let G be a group and Γ a representation of G on a vector space V. For a subgroup
H of G, the matrices Γ(h) for all h ∈ H together form a representation of H on V.
A representation of H obtained by restricting a representation of G to H is denoted
by Γ∗ in this section. The reverse problem of obtaining a representation of G from
a representation Θ of H is considered here. Such a representation of G is said to be
induced by the representation Θ of H and is denoted by Θ∗ in this section.
The first step in the construction of Θ∗ is to realize a representation of G on the
set of cosets of H in G. Let { xi }in=1 be a set of representatives, one from each of the
distinct cosets of H in G. Then one has G = ∪ xi H. If g ∈ G, then gxi H is again a coset
and therefore equal to xk H for some xk . Because { xi }in=1 are representatives of distinct
cosets, gxi H 6= gx j H unless i = j. In other words, left multiplication by g permutes
the cosets of H among themselves. Every g ∈ G in this way can be identified with a
permutation τg in S(n), the symmetric group on the n letters { xi }in=1 . It follows from
τg1 g2 ( xi H ) = g1 g2 ( x i H ) = g1 ( g2 x i H )

⇒ τg1 g2 ( xi H ) = τg1 τg2 ( xi H )

that τ is a representation of G into S(n), i.e., τ : G → S(n) is a homomorphism. One
notices that the regular representation described in the Section 3.5 corresponds to the
case when H is the trivial subgroup { E} of G.

Example 27. Consider the group S(3) of Example 10 in Chapter 1. The subgroup
H = {e, a} has 3 cosets eH = {e, a}, bH = {b, ba} and b2 H = {b2 , b2 a}. Consider
the action of the element ab ∈ S(3) on the cosets. Evidently

ab : eH → b2 H, ab : bH → bH, ab : b2 H → eH.

In matrix form, the above action can be written as

 
0 0 1
τab =  0 1 0  .
1 0 0

The representation for other elements can be similarly worked out.

     
1 0 0 1 0 0 0 0 1
τe =  0 1 0  , τa =  0 0 1  , τb =  1 0 0  ,
0 0 1 0 1 0 0 1 0
   
0 1 0 0 1 0
τb2 =  0 0 1  , τab2 =  1 0 0  .
1 0 0 0 0 1

As previously, let Θ be a representation of the subgroup H over a vector space W of

dimension m. Θ is extended to Θ∗ by τ action of G on the coset space of H. The τ
representation of G consists of permutation matrices. Such matrices have a single 1
as the only non-zero entry in each row and each column as is evident from the above
example. For g ∈ G, the (i, j)0 th entry in τg is equal to 1 if and only if xi−1 gx j ∈ H.
Define matrices gij as

Θ( xi−1 gx j ) if xi−1 gx j ∈ H
(
ij
g = (3.9.1)
0m × m otherwise.

Here 0m×m is the m × m matrix all of whose entries are equal to 0. The matrix Θ∗ ( g)
in the induced representation is now given in terms of matrix blocks gij as

g11 g1n
 
···
Θ∗ ( g) =  ... .. ..
. (3.9.2)
 
. .
gn1 ··· g nn

Evidently, the order of Θ∗ ( g) is mn × mn. Because of the form of τ, only one matrix
block is non-zero in any row or any column of matrix blocks in Θ∗ ( g). It remains to
be verified that Θ∗ is a representation of G. For g1 , g2 ∈ G the (i, j)0 th matrix block of
the product matrix Θ∗ ( g1 )Θ∗ ( g2 ) is given by
n
[Θ∗ ( g1 )Θ∗ ( g2 )]ij = ∑ ( g1 )ik ( g2 )kj .
k =1

The matrix in the right hand side of the above can be non-zero only when ( g1 )ik and
( g2 )kj are non-zero for the same k = k0 . This happens when both xi−1 g1 xk0 and xk−1 g2 x j 0

are in H. Then ( g1 )ik0 = Θ( xi−1 g1 xk0 ), ( g2 )k0 j = Θ( xk−1 g2 x j ) and one has
0
∗ ∗
[Θ ( g1 )Θ ( g2 )] ij
= Θ( xi−1 g1 xk0 )Θ( xk−1 g2 x j )
0

⇒ [Θ∗ ( g1 )Θ∗ ( g2 )]ij = [Θ∗ ( g1 g2 )]ij . (3.9.3)

Also note that g2 x j ∈ xk H for some xk and if ∈ H, then both xi−1 g1 g2 x j xi−1 g1 xk
−1
and xk g2 x j are in H. Hence the corresponding non-zero blocks of Θ∗ ( g1 )Θ∗ ( g2 ) and
Θ∗ ( g1 g2 ) are equal. It is similarly shown that the same holds for the zero blocks and
Equation 3.9.3 holds generally. It follows that Θ∗ is indeed a representation of G.

Example 28. Continuing with the previous example, a representation of H of degree

2 is

1 0 0 1
Θ(e) = ; Θ( a) = .
0 1 1 0

The reader should follow the above description to obtain Θ∗ . A couple of representing
matrices are
   
0 0 0 0 0 1 0 0 1 0 0 0
 0 0 0 0 1 0   0 0 0 1 0 0 
   
0 0 0 1 0 0   0 0 0 0 1 0 
Θ∗ ( ab) =  Θ ∗ 2
 
 0 0 1 0 0 0  ; ( b ) =  0 0 0 0 0 1 .
 
   
 0 1 0 0 0 0   1 0 0 0 0 0 
1 0 0 0 0 0 0 1 0 0 0 0

If χ is the character of Θ and χ∗ that of Θ∗ , then it follows from Equation 3.9.2 that
χ∗ ( g) = ∑in=1 tr gii . By use of Equation 3.9.1, one has

n
χ∗ ( g) = ∑ χ( xi−1 gxi ). (3.9.4)
i =1, xi−1 gxi ∈ H

For xi−1 gxi ∈ H, one also has χ( xi−1 gxi ) = χ(h−1 xi−1 gxi h) ∀ h ∈ H. Employing this
fact, the reader can easily establish

1
| H | t∈G,t∑
χ∗ ( g) = χ(t−1 gt). (3.9.5)
−1 gt ∈ H

Suppose now that Γ is an irreducible representation of G and Θ an irreducible

representation of H, a subgroup of G. In general, Γ∗ and Θ∗ are reducible
representations of H and G respectively. The multiplicity m∗ of Θ in Γ∗ is given by

1
χ Γ ( h ) χ Θ ( h −1 ),
| H | h∑
m∗ =
∈H

where use has been made of the facts Γ∗ (h) = Γ(h) ∀ h ∈ H and χΘ (h) = χΘ (h−1 ).
The multiplicity m∗ of Γ in Θ∗ is given by
1 ∗
m∗ =
|G| ∑ χΘ ( g −1 ) χ Γ ( g )
g∈ G
 
n
1
⇒ m∗ =
|G| ∑ χΓ ( g)  ∑ χΘ ( xi−1 g−1 xi ) .
g∈ G i =1,xi−1 g−1 xi ∈ H

Rearranging the summations and using the fact that χΓ ( g) = χΓ ( xi−1 gxi ), one has

1 n
χΓ xi−1 gxi χΘ xi−1 g−1 xi .
| G | i∑ ∑
m∗ =
=1 g∈ G, xi−1 g−1 xi ∈ H

The conjugation xi−1 Gxi is an automorphism of G. The second summation is therefore

over exactly all the elements of H. Since there are | G |/| H | representatives, it follows
that

1 |G|
χ Γ ( h ) χ Θ ( h −1 ) = m ∗ .
| G | | H | h∑
m∗ = (3.9.6)
∈H

The above result is called the Frobenius reciprocity theorem.

Assuming again H to be a subgroup of G, let Θ1 and Θ2 be representations of H
over vector spaces of W1 and W2 . Let the degrees of these representations be n1 and
n2 . Then in the representation Θ1 ⊕ Θ2 over W1 ⊕ W2 , one has for h ∈ H

Θ1 ( h )
!
0n1 × n2
(Θ1 ⊕ Θ2 )(h) = .
0n2 × n1 Θ2 ( h )

The matrices in the representation (Θ1 ⊕ Θ2 )∗ would have blocks such as above in
places where 1 appears in the permutation representation of G on the coset space of H
and 0(n1 +n2 )×(n1 +n2 ) blocks elsewhere. Such a matrix can be written as a sum of two
matrices which operate on complimentary space isomorphic to W1 and W2 . Therefore,
one has

(Θ1 ⊕ Θ2 )∗ = Θ1∗ ⊕ Θ2∗ . (3.9.7)

Exercises
1. Show that all invertible n × n matrices with complex entries form a group. Also show
that the set of matrices whose determinant is of unit modulus is a subgroup.
2. Show that two matrices related by a similarity transformation have the same trace.

3. Using the orthogonality relations, write the character table of the group C4 and indicate
the irreducible representations using Mulliken’s symbols. Write down the matrix form
of the regular representation of C4 .
4. Show that the characters of a finite cyclic group in any irreducible unitary representation
are unimodular.
5. In a certain representation Γ of C3v , it is known that χΓ ( E) = 7, χΓ (C3 ) = 1 and
χΓ (σv ) = −3. Determine the multiplicities of various irreducible representations of
C3v in Γ.
6. Obtain the maximal point group symmetry for a particle in a two-dimensional box.
Write down the character table for this group along with the basis elements (consider
the action of group elements on the coordinate space). Further figure out which
stationary states of the particle give the irreducible representations of the group.
7. Construct the character table of the group C4v . Determine the irreducible
representations present in the tensor product E ⊗ E where E is the two-dimensional
irreducible representation of C4v .
8. For the symmetric group S(3), determine the characters of all the irreducible
representations. Give a complete reducible 3 × 3 representation Γ of S(3) on a
three-dimensional vector space V spanned by (1, 0, 0), (0, 1, 0) and (0, 0, 1).
Determine the multiplicity of all the irreducible representations in Γ. Using the
projection operator on Γ, project out the subspaces of V on which Γ is irreducible.
9. Prove Equation 3.7.5.
10. Construct the character table for the point group D3d . Consider D3d as a direct product
of two of its subgroups.
11. Complete the proof of Equation 3.9.5.
12. Let G be a finite group and H a subgroup of G. Show that the regular representation
of H induces the regular representation G.
13. If H is a subgroup of G, then show that any irreducible representation of G is a
component in the representation induced by some irreducible representation of H . The
result of the previous exercise is useful.
14. In the notation of the last section of the chapter, show that Γ ⊗ Θ∗ = (Γ∗ ⊗ Θ)∗ .

In this chapter, we will focus on solving problems in physics using the group
theory representations extensively discussed in the previous chapter. The character
tables for any discrete group will be the primary tool to apply in these physical
problems. Particular attention will be on atomic and nuclear systems obeying the
postulates of quantum mechanics.
For instance, a physical system possessing a discrete group symmetry like C4v may
get disturbed by impurities or defects. Such a disturbance could result in a system
with a residual symmetry like C2 . In fact, we will see how these effects are reflected
in energy spectrum observed experimentally which can be validated using the group
representation theory tools.
We hope the readers will appreciate how the formal mathematical steps provide
meaning to results observed in the physical systems. We plan to present several
examples in atomic, nuclear and particle physics which will highlight the elegance
of group theory tools.

4.1 General Considerations

We will briefly review the salient features of quantum mechanics which describe the
symmetries possessed by microscopic systems. In fact, this review will be useful for
validating these quantum mechanical results applying group theory tools.
For a closed quantum mechanical system, the eigenstates ψ of the system are
solutions of the time independent Schrödinger’s equation:

Hψ = Eψ, (4.1.1)

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60 Group Theory for Physicists

where H denotes Hamiltonian of the system and E is the energy when the system is
in the stationary state ψ. The Hamiltonian H of the system is a self-adjoint operator
that acts on the Hilbert space of system states. Equation 4.1.1 essentially states that the
action of the Hamiltonian on a stationary state leaves the state invariant apart from a
constant factor of E. The eigenstate ψ can be always taken to be a unit vector in the
Hilbert space. All vectors which are complex multiples of the same unit vector are
assumed to be physically equivalent. In this sense, ψ represents that normalized state
of the system. In the notation of Section 3.1, this fact is represented by
(ψ, ψ) = 1. (4.1.2)
Let O be a unitary operator which commutes with H. By the fact that O and H
commute, it is implied that the order in which O and H act on a state ψ is immaterial,
i.e., HO ψ = OHψ. If ψ is a stationary state of the system satisfying Equation 4.1.1,
then one has
H(Oψ) = EOψ.
In other words, if ψ is a stationary state of the system, then so is Oψ. In general, Oψ
would differ from ψ not merely by a constant complex multiple. In this circumstance,
the energy level E is said to be degenerate.
There can be a set of unitary operators {O1 , O2 , · · · } commuting with H. By a
similar application of these operators to the Schrödinger equation 4.1.1, we deduce
energy E to be same for states {Oi ψ} which may be linearly independent to ψ. Suppose
this set of unitary operators form a group G. Then we interpret such a quantum
mechanical system to possess the group symmetry G.
In Section 3.2, it was noted how the elements of a group G may act on a
vector space and give a representation of G on that vector space. Suppose that the
group G is a symmetrical group of the system under consideration. Such symmetry
transformations would leave the form of the Schrödinger equation invariant. We
denote the representation of unitary operator Oi as Γ( gi ), corresponding to any group
element gi ∈ G, acting on the Hilbert space. Particularly the elements of the set {Γ( gi )}
commutes with the Hamiltonian of the system possessing group symmetry G. The
energy of a stationary state cannot change due to action of Γ( gi ). Thus the subspace of
degenerate states corresponding to some energy level E gives a representation of G. As
this subrepresentation is unitary, it is completely reducible. It is then possible to choose
the degenerate eigenstates corresponding to an energy level E so that they transform
according to irreducible representations of G. If a particular irreducible representation
occurs only once in the representation of G on the full Hilbert space, then all the basis
states of this representation are also all the degenerate eigenstates corresponding to
some energy E. It may so happen that the degenerate levels of another energy E0
give a reducible representation of G. This happens if the Hamiltonian of the system
has a higher order symmetry than G, a circumstance we will assume does not occur
in the subsequent discussion. To sum up, it would be assumed that the degenerate
states corresponding to an energy level can be so chosen that they give an irreducible
representation of the symmetry group of the system.

Example 29. For a one-dimensional harmonic oscillator given by the Hamiltonian

}2 d2 1
H=− 2
+ mω 2 ,
2m dx 2
determine the group symmetry and deduce whether the energy eigenstates are
degenerate or non-degenerate.
Observe that the inversion transformation g : x → − x, where g ∈ Ci , commutes
with H. Hence Ci is the group symmetry. Recall that the character table of abelian
groups contain only one-dimensional irreducible representations. Hence, the energy
eigenstates of the harmonic oscillator possessing Ci group symmetry must be
non-degenerate.
We have illustrated the power of the group theory tool to deduce the
non-degenerate nature of energy eigenstates without solving the Schrödinger
equation. Further, the stationary states

ψ ( x ) ∝ Γ ( g ) ψ ( x ),

where Γ( g) denotes the representation of the group element acting on the Hibert space.
By using the projection method (see Section.3.7), the basis states of group Ci can be
shown to be either

odd : ψ0 ( x ) = −ψ0 (− x ) or even : ψe ( x ) = ψe (− x ).

Thus group theory arguments have justified that the stationary states are
non-degenerate states which are either ψ0 ( x ) or ψe ( x ). However, to determine the
explicit form of stationary states and the energy eigenvalues, we have to solve the
Schrödinger equation.
We will now look at another system with degenerate energy levels.
Example 30. Consider a particle of mass m moving in two dimensions subjected to
potential V( x, y) which is zero inside a square region − a ≤ ( x, y) ≤ a and ∞
elsewhere. This is the familiar example in quantum mechanics known as particle
in a square box whose energy eigenstates are either non-degenerate or two-fold
degenerate. Can we explain the degenerate energy level using the discrete symmetry
possessed by the system ?
The Hamiltonian describing such a system is

}2
2
∂2

∂
H=− + + V(x, y),
2m ∂x2 ∂y2
whose stationary states can be determined by solving the Schrodinger Equation
4.1.1. We quote the result which is extensively discussed in any quantum mechanics
textbook. The energy eigenstates and eigenvalues are

1 n πx n πy π 2 }2
2
ψ(n1 ,n2 ) ( x, y) = sin 1 sin ; E(n1 ,n2 ) = n21 + n22 .
a 2a 2a 8ma2

Clearly, ψ(n1 ,n2 ) ( x, y) and ψ(n2 ,n1 ) ( x, y) are linearly independent states sharing the
same energy eigenvalues.
The rotation operation by angle π/2 about z-axis, which transforms ( x, y) →
(y, − x ), as well as mirror reflection σv commutes with the Hamiltonian. Hence, the
group symmetry possessed by this system is C4v . We observe the character table of C4v
has one-dimensional as well as two-dimensional irreducible representations. Hence,
we can claim that the energy states are either non-degenerate or two-fold degenerate.
It is appropriate to mention that besides the group symmetry accounting for
degenerate energy levels, there could be accidental degeneracy. For example, the
stationary states with (n1 , n2 ) = (5,5), (7, 1), (1, 7) share the same energy eigenvalues
where the accidental degenerate state ψ(5,5) ( x, y) cannot be deduced using group
theory tools.
Suppose the square region, where potential is V ( x, y) = 0, gets slightly modified
to a rectangular region due to mild disturbance. We know from quantum mechanics
that the two-fold degenerate levels in a square box split into two non-degenerate
levels in such a rectangular box of arbitrary length and breath (we neglect accidental
degeneracy which can arise due to suitable ratio of the sides of the rectangle). From
the group theory point of view, we can say that the two-fold degenerate energy levels
show splitting due to disturbance. In the following section, we will understand how
disturbance or perturbation can split the degenerate energy levels using group theory
tools.

4.2 Level Splitting under Perturbation

For a system described by H possessing symmetry group G, Equation 4.1.1 will give
the stationary states. Let the degenerate eigenstates {ψi }in=1 give an irreducible unitary
representation Γ of the symmetry group G of the system. The characters of G in Γ are
of course assumed known. On application of a perturbation H1 to the system, the
Schrödinger equation takes the form

(H + H1 )ψ = Eψ.
If the perturbation is assumed to be small, the stationary states and the energy levels of
the perturbed system are expected to differ only slightly from those of the unperturbed
system. It is known from perturbation theory methods in quantum mechanics that a
small perturbation usually lifts the degeneracy in a set of degenerate levels such as
{ψi }in=1 . While a complete calculation of the perturbed levels is made possible only
through application of perturbation methods to the known form of H1 , group theory
allows one to deduce the extent to which the degeneracy gets lifted from the form of
H1 in some cases.
Suppose K is the maximal group of spatial transformations which commutes with
the perturbation. If K is a subgroup of G then the perturbation H1 has a lower
order symmetry than the unperturbed Hamiltonian H. Consequently, the complete
Hamiltonian of the perturbed system also has its symmetry group as K. Since Γ is a
representation of G, it is also a representation of the subgroup K. As Γ is unitary, it is

now possible to completely reduce Γ as a direct sum of irreducible representations of

K. These irreducible representations of K correspond to different energy eigenvalues
of the perturbed system in accordance with our assumption. It is immediate that
Equations 3.7.1 and 3.7.2 are directly applicable in this case.
Using the above arguments, we can now revisit Example 30 where V ( x, y) = 0
in a rectangular region due to perturbation. The group symmetry possessed by the
particle in a rectangular box system is no longer G = C4v but K = C2v which is
abelian. Hence, any two-dimensional irreducible representation Γ ∈ C4v will become
the direct sum of irreducible representation: ⊕Γα ∈ C2v which are one-dimensional
representations. Therefore all the two-fold degenerate levels of the square box will
split into non-degenerate levels (ignoring accidental degeneracy). We will elaborate
another example which will illustrate the splitting of energy levels using the group
theory tools.
Example 31. Let the unperturbed Hamiltonian of a system have Td symmetry
(Chapter 2, Section 2.3). Suppose a small perturbation breaks the symmetry to the
subgroup C3v . For reference, character tables for Td and C3v are given below. Let us
determine the splitting of energy level corresponding to a representation of degree 3
(say F1 ) in Td .
C3v E 2C3 3σv
Td E 8C2 3C2 6σd 6S4
A1 1 1 1
A1 1 1 1 1 1
A2 1 1 −1
A2 1 1 1 −1 −1
E 2 −1 0
E 2 −1 2 0 0
F1 3 0 −1 1 −1
F2 3 0 −1 −1 1
Some observations are in order before calculating the split. The permutation
representation of Td is the symmetric group S(4). Similarly, the permutation
representation of C3v is the symmetric group S(3). Since S(3) is a subgroup of S(4),
it follows that C3v is a subgroup of Td . It may be recalled here that S(4) = VN o S(3).
In the point group Td , the normal subgroup corresponding to VN (the Klein-4 normal
subgroup in S(4)) consists of the identity transformation along with the three C2
rotations which interchange pairs of vertices of the tetrahedron. In the isomorphism
Td ∼
= C3v ,
VN
the 8C3 class in Td is mapped to 2C3 class in C3v , the classes 6σd and 6S4 in Td are both
mapped to the class 3σv in C3v while VN being the kernel of the homomorphism is
mapped to the identity class in C3v . The characters of the representations of C3v can
be extended in this case to the characters of representations of the same degrees in Td
by noting the fact that all representations of the kernel VN are of degree 1 since VN is
an abelian group. For example, consider the representation E. The characters of the
identity class and the 3C2 class in Td are both equal to 2, the character of the identity

class of C3v (to which both the classes of Td are mapped in the homomorphism).
Similarly one may obtain the characters of other representations of Td from those of
C3v . However, since C3v has no irreducible representations of degree 3, the characters
of the representations F1 and F2 of Td would have to be obtained via application of the
orthogonality relations on the characters.
Returning to the splitting of the degeneracy of F1 under a C3v perturbation, the
basis states of the representation F1 of Td give a reducible representation of C3v . In
this reducible representation, the characters of the various classes of C3v can be read
directly from the characters of F1 by considering the homomorphism mentioned above
and orthogonality relations:

C3v E 2C3 3σv

.
Γ 3 0 1

Now apply Equation 3.7.2 (which we note once again)

1
∑ χΓ ( g)χΓ
α
mα = ( g ).
|G| geG

In the above, Γ is the reducible representation of C3v effected by the basis states of the
F1 representation of Td and Γα is any of the irreducible representations of C3v . Upon
carrying out the routine calculations, it is found that

Γ = A1 ⊕ E.

The above indicates that the perturbation splits the triply degenerate energy level
of the original Hamiltonian into two levels, one singlet and the other one doubly
degenerate.
Stationary states are the stable energy levels in which the system stays as long
as there is no external interactions. Experimentalists in the laboratory measure the
frequencies {νi }0 s of spectral lines emitted whenever an atom undergoes a transition
from one energy level to another energy level due to interactions. The observed
frequencies are characteristic of an atom and also depend on the nature of interaction.
This brings us to topic of selection rules where group theory tools can deduce whether
an atomic transition is allowed or disallowed due to such interactions.

4.3 Selection Rules

In quantum mechanics, physical quantities are represented by self-adjoint operators
acting on the Hilbert space of system states. The Hamiltonian of the system is itself
an example of such an operator which is useful to determine the energy eigenvalues
Ei and their corresponding stationary states ψi solving Schrödinger equation. The
stationary states (indexed by a complete set of quantum numbers) form a basis for the
Hilbert space. There are other linear operators corresponding to physical quantities
like position, momentum, angular momentum and so forth, which can be expressed

in a matrix form in this basis. Let O represent a self-adjoint operator corresponding to

a physical quantity. The matrix of O in the basis of stationary states is then given by
Z
∗
Omn = ψm O ψn = (ψm , O ψn ), (4.3.1)

where {ψi } are the stationary states indexed by a complete set of quantum numbers.
The integration in the equation above is carried out over the complete configuration
space of the system. The non-zero matrix element Omn implies that the transition from
energy level En to another energy level Em due to interaction represented by operator
O is allowed. Instead of working this matrix element by integrating Equation 4.3.1,
we would like to exploit the power of group theory to deduce whether Omn is zero or
non-zero.
Suppose the system possesses group symmetry G corresponding to spatial
transformation. We expect the above matrix element Omn to be unchanged when any
group element g ∈ G acts on the states ψm , ψn as well as on the operator O . In fact,
one can deduce which of the matrix elements Omn vanishes by applying the principal
ideas of group theory as follows.
For the system described by Equation 4.1.1 having G as its non-trivial symmetry
group, we know from Section 4.1 that the degenerate states of same energy belong to
an irreducible representation of G. Consider the integral
Z
ψiα

where ψiα is a basis state in some irreducible representation Γα of the G. The action
of any group element on ψiα transforms into a linear combination of the basis states
of same energy to which ψiα belongs. However, a mere spatial transformation which
brings the
R system to an identical configuration should not change the value of the
integral ψiα . In other words it is expected that

Z d Z
ψiα = ∑ [Γ α
( g)]ki ψkα ,
k =1

where d is the degree of the representation Γα . Also, the above must be true for any
transformation g in G. Adding the expressions for all transformations g in G, one has
!Z
d
1
Z
ψiα =
|G| ∑ ∑ [Γα ( g)]ki ψkα .
k =1 g∈ G

If Γα is not the trivial representation, then it is easy to verify using Equation 3.3.5
that (∑ g∈G [Γα ( g)]ki ) vanishes. Therefore, if ψiα is a basis state in some non-trivial
irreducible unitary representation Γα of the G then
Z
ψiα = 0. (4.3.2)

f
Returning to Equation 4.3.1, let ψm and ψni be eigenstates in two distinct irreducible
representations Γ f and Γi respectively, of the symmetry group G of the system.
The operator O is in general some tensor (for example, a dipole moment or a
quadrupole moment) which has a certain number of independent components. It
will be assumed that these independent components themselves are transformed into
a linear combination of each other under the spatial transformations effected by the
elements of G. It is then possible to treat those independent components of O as a
basis for irreducible representation of G. For instance, see Example 32 for electric
dipole moment components. Formally for any operator O , its representation of G is
denoted as ΓO . By suitable linear combinations, it is always possible to choose the
eigenfunctions of the stationary states to be real and ψm in Equation 4.3.1 can be taken
f∗
as real. Then the states ψm O i would be transformed by G according to the states
of the tensor product Γ f ⊗ ΓO ⊗ Γi . The representation Γ f ⊗ ΓO ⊗ Γi can obviously be
decomposed into a direct sum of irreducible representations of G and the multiplicities
of these irreducible representations can be found by using Equation 3.7.2. RThe
evaluation of Omn reduces to the calculation of the sum of integrals of the form ψkα
where the ψkα ’s transform in accordance with some irreducible representation of G.
That all such integrals vanish has already been established unless of course some of
the ψkα ’s transform in accordance with the trivial representation. Therefore, the matrix
elements of O vanish unless the trivial representation is present in the decomposition
of Γ j ⊗ ΓO ⊗ Γi . Alternatively, when the tensor product ΓO ⊗ Γi is decomposed into
irreducible components, then these irreducible components are the possible values of
Γ f for which the transition elements may be non-zero. This is a very convenient way
to decide which matrix elements vanish as against a full-fledged evaluation of the
integral. The exact values of the non-vanishing matrix elements would of course have
to be calculated by integration. The implementation of this formalism is detailed in
Examples 32 and 33.
The aforementioned method for calculation of Omn works when ψm and ψn belong
to different energy levels of the system. In order to consider transition elements which
are diagonal with respect to energy, we make a small digression to recall an important
fact related to the solution of the time dependent Schrödinger equation

∂
i Ψ = HΨ.
∂t
Let the Hamiltonian H of the system described by the time dependent wave function
Ψ be a time independent real operator. If the state Ψ of the system develops in
accordance with the above equation, then the conjugates of quantities on both sides of
the equation should be always equal, i.e.,

∂
−i Ψ = HΨ
∂t
∂
⇒i Ψ = HΨ. (4.3.3)
∂(−t)

Equation 4.3.3 is identical to the time dependent Schrödinger equation. In fact, the
above equation states that under time reversal, the system that was described by state
Ψ would become Ψ. In this sense Ψ is the time reversed state of the system. Under time
reversal, various physical quantities related to the system may behave differently (as
a simple example, for a particle moving in space-time, under time reversal, assuming
there are no spatial transformations involved, the coordinates of the particle remain
the same no matter how its velocity vector reverses in direction). The physical
quantity O itself may remain invariant or change sign upon time reversal. This
behaviour is important in determining the selection rules for transitions among states
that are diagonal with respect to energy. Let the closed system under consideration
be in a stationary state whose basis states transform according to the irreducible
representation Γ. If the degree of the representation Γ is d, then the energy level is
d-fold degenerate with {ψ1 , · · · , ψd } serving as a basis. It is of course assumed that
all the basis states are real. If ψ is the state of the system belonging to irreducible
representation Γ, then in general ψ is a linear combination of the basis states:

d
ψ= ∑ cm ψm .
m =1

The average value of the quantity O in this state is obtained according to the usual
rules of quantum mechanics. If hOi represents the average value then
Z d
hOi = ψO ψ = ∑ cm cn Omn .
m,n=1

In the time reversed state ψ, the average value of O would be obtained to be

Z d
hOirev = ψO ψ = ∑ cm cn Onm .
m,n=1

If the physical quantity represented by the operator O was such that it remained
invariant under time reversal, then its average value would remain unchanged under
such a transformation. The value hOi and hOirev would then be equal. As the
coefficients ci can be complex, it follows that

Omn = Onm
Z Z
⇒ ψm O ψn = ψn O ψm .

The transition matrix elements would therefore be non-zero when ΓO is present in

(Γ ⊗ Γ)σ , the symmetric component of the tensor product. For a physical quantity

that changes its sign under time reversal, the matrix elements would be non-zero if the
antisymmetric component (Γ ⊗ Γ)α contained ΓO . This detour on time reversal action
leading to the complex conjugation of wavefunction is generally known as antilinear
property within the linear algebra context.

Example 32. Consider the character table for C4v .

C4v E 2C4 C2 2σv 2σd

A1 1 1 1 1 1 z x 2 + y2 ; z2
A2 1 1 1 −1 −1 Rz
B1 1 −1 1 1 −1 x 2 − y2
B2 1 −1 1 −1 1 xy
E 2 0 −2 0 0 ( x, y); ( R x Ry ) xz; yz

The electric dipole moment p for a system of charges is defined to be ∑ qi ri . For a single
charge, the x-component of the dipole moment vector ( p x ) clearly has transformation
properties of the x coordinate which belongs to the E representation of C4v . Hence
Γ px ∈ ΓE . Consider a transition from an initial state corresponding E representation
to a final state corresponding to the A2 representation due to x-component dipole
moment interaction. The matrix element of p x corresponding to this transition would
be non-zero only if the tensor product A2 ⊗ E contained the representation E (the
representation transforming p x ). It can be observed from the table given above that in
fact A2 ⊗ E = E. Thus hA2 |q x̂ |Ei is non-zero and the dipole transition is possible. This
selection rule of non-zero matrix element will also be valid for the py operator because
the y-component dipole moment belongs to the irreducible representation E as well.
However, pz belongs to trivial representation and hence hA2 |qẑ|Ei = 0.
Example 33. Consider the electric quadrupole moment tensor defined as

Qik = e(3xi xk − r2 δik ), (4.3.4)

where the indices i and k take values 1, 2 and 3. In the coordinate notation used in the
text, x1 = x, x2 = y, x3 = z and r2 = x2 + y2 + z2 . The tensor is clearly symmetric.
Explicitly, the nine components of the quadrupole tensor for unit charge e are

Q xx ∼ 3x2 − r2 , Qyy ∼ 3y2 − r2 , Qzz ∼ 3z2 − r2 ,

Q xx + Qyy + Qzz = 0,

( Q xy = Qyx ) ∼ 3xy, ( Q xz = Qzx ) ∼ 3xz, ( Qyz = Qzy ) ∼ 3yz.

From above relationships, it follows that Qik has 5 independent quantities. It is

our purpose to study quadrupole moment selection rules in a system having the
octahedral (O) symmetry. The character table for O symmetry group follows:

O E 8C3 3C2 6C4 6C2

A1 1 1 1 1 1 x 2 + y2 + z2
A2 1 1 1 −1 −1
E 2 −1 2 0 0 0 (2z2 − x2 − y2 , x2 − y2 )
T1 3 0 −1 1 −1 ( x, y, z)
( R x , Ry , Rz )
T2 3 0 −1 −1 1 ( xy, xz, yz)

Comparing the bases of the representation T2 , it is clear that the triplet ( Q xy , Q xz , Qyz )
transforms according to the T2 representation of O. Also notice that Qzz ∼ 2z2 − x2 −
y2 and Q xx − Qyy ∼ x2 − y2 . Therefore pair ( Qzz , Q xx − Qyy ) transforms according to
the E representation of O.
Let us first find which matrix elements of ( Q xy , Q xz , Qyz ) are non zero for
transitions between states of different representations. For this purpose one needs
to find which irreducible representations are present in the tensor product of T2
representation with various other irreducible representations of the group. This is
accomplished, as has already been shown through previous examples, by use of
Equation 3.7.2. We will leave it to the readers to verify the following:

T2 ⊗ A1 = T2 ,

T2 ⊗ A2 = T1 ,

T2 ⊗ E = T1 ⊕ T2 ,

T2 ⊗ T1 = A2 ⊕ E ⊕ T1 ⊕ T2 ,

T2 ⊗ T2 = A1 ⊕ E ⊕ T1 ⊕ T2 .

From the above decomposition, it is obvious that Q xy , Q xz , Qyz have non-zero matrix
elements for transitions

A1 ↔ T2 ; A2 ↔ T1 ; E ↔ T1 , T2 ; T1 ↔ T2 .

To calculate the non-zero matrix elements for Q xx , Qyy , Qzz , proceeding as above, the
following decompositions of the tensor product of the E representation with various
other representations of the group are obtained.

E ⊗ A1 = E,

E ⊗ A2 = E,

E ⊗ E = A1 ⊕ A2 ⊕ E,

E ⊗ T1 = T1 ⊕ T2 ,

E ⊗ T2 = T1 ⊕ T2 .

Once again it follows from the above decomposition that Q xx , Qyy , Qzz have non zero
matrix elements for transitions

E ↔ A1 , A2 ; T1 ↔ T2 .

Till now, the transition matrix elements between states of different representations
have been considered. In order to consider matrix elements that are diagonal with
respect to energy, the first thing to note is that electric quadrupole moment tensor
remains invariant under time reversal. This follows directly from the form of Qik
which is itself defined in terms of quantities that remain unaffected under time
reversal. Therefore the symmetric component of the tensor product of an irreducible
representation with itself needs to be considered. As an example, consider the
symmetric component of the tensor product T2 ⊗ T2 , and let this component be
represented (T2 ⊗ T2 )σ (in the notation of Chapter 3, Section 3.7). The class characters
in the representation (T2 ⊗ T2 )σ are easily found using Equation 3.7.5.

χ2σ ( E) = 6, χ2σ (8C3 ) = 0, χ2σ (3C2 ) = 2, χ2σ (6C4 ) = 0, χ2σ (6C2 ) = 2.

From the above character values, the decomposition of (T2 ⊗ T2 )σ is obtained to be

(T2 ⊗ T2 )σ = A1 ⊕ E ⊕ T2 .

In a similar manner, the symmetric components of the tensor products of the other
irreducible representations may be obtained.

(A1 ⊗ A1 )σ = A1 ,
(A2 ⊗ A2 )σ = A1 ,

(E ⊗ E)σ = A1 ⊕ E,

(T1 ⊗ T1 )σ = A1 ⊕ E ⊕ T2 ,

(T2 ⊗ T2 )σ = A1 ⊕ E ⊕ T2 .

The matrix elements of Q xx , Qyy , Qzz that are diagonal with respect to energy
are non-zero for transitions in E, T1 and T2 . Similarly, the matrix elements of
Q xy , Q xz , Qyz that are diagonal with respect to energy are non-zero for transitions
in T1 and T2 .
The above two examples indicate the power of group theory concepts in deducing
selection rules. We have clearly demonstrated that the group representation theory can
determine whether any matrix elements of self-adjoint operators between stationary

states is zero or non-zero. In the following section, we will exploit group theory tools to
determine the independent vibrational modes of molecules possessing discrete group
symmetry. These are well known as normal modes which characterize the molecules.

4.4 Molecular Vibrations

We will first briefly review the concepts of normal modes of vibrations of simple
systems which are well known in classical mechanics. Then we will elaborate the
elegant group theory approach in reproducing those normal modes of molecules.
Consider the classical problem of two masses m1 and m2 connected by a Hooke’s
spring of stiffness κ . It is well known that a small disturbance/displacement leads to
oscillations/vibration of the system with a unique frequency ω given by
r
κ
ω= .
µ
In the above expression for ω, µ = mm1+mm22 is the reduced mass of the system.

1
When the number of participating masses and springs is increased, the number of
degrees of freedom also increases and as a result the oscillatory motion of the system
becomes more complicated. Consider a system consisting of n mass points. Such a
system of mass points has in general 3n degrees of freedom of movement. Of these, 3
degrees of freedom correspond to an overall motion of the system in the three space
directions. Additionally, the system will have rotational degrees of freedom. If the
distribution of the mass points in the equilibrium state is in three-dimensional space,
then the system has 3 rotational degrees of freedom. All the other degrees correspond
to the vibrational motion and therefore such a system has a total of 3n − 6 vibrational
degrees of freedom. In case the mass points were distributed along a straight line,
the rotational degree of freedom associated with rotation about the axis of the system
would have to be discounted since here it is assumed that the masses have negligible
physical dimensions. For such an arrangement of mass points, there will be then a
total of 3n − 5 vibrational degrees of freedom.
Let us recall the basic theory of small oscillations from classical mechanics.
Consider a system with s vibrational degrees of freedom. Let xi (i = 1, 2, · · · , s)
denote small excursions of the mass points from their mean positions. Let x be a s × 1
column vector such that {x}i1 = xi . In general, the kinetic energy T of the system is
a quadratic function of ẋi (time derivative of xi ). The most general form of T is then
given by
1 s
2∑
T= m jk ẋ j ẋk ,
j, k
where m jk are symmetric constants (m jk = mkj ). It is expected that the system
will exhibit oscillatory motion when disturbed from its mean position. In this
circumstance, the potential energy of the system can be taken to be minimum (and
equal to 0) in the undisturbed state. When the excursions xi are small, then the
potential energy U may be written as,

1 s
2∑
U= κ jk x j xk ,
j, k
2
where κ jk (= ( ∂x∂∂x U ) x j=0 ,xk=0 ) are some other symmetric constants. Let M and K be
j k
symmetric matrices such that {M} jk = m jk and {K} jk = κ jk . The Lagrangian L of the
system can therefore be written as
1 s 1 s
L = T−U = ∑
2 j,k
m jk ẋ j ẋk − ∑ κ jk x j xk
2 j,k
1h T i
⇒L= ẋ Mẋ − xT Kx . (4.4.1)
2
We note the Euler–Lagrange equation for convenience here.

d ∂ ∂
L = L.
dt ∂ ẋi ∂xi
The equations of motion are obtained by substituting the form of L in the
Euler–Lagrange equation. It may be noted that

∂ 1 s s
L = ∑ (mik + mki ) ẋk = ∑ mik ẋk ,
∂ ẋi 2 k =1 k =1

∂
because of symmetry of mik . In a similar manner ∂xi L can be obtained. The equations
of motion in their final form are
s
∑ (mik ẍk + κik xk ) = 0; i = (1, 2 · · · , s). (4.4.2)
k =1

Equation 4.4.2 admits simple harmonic solutions of the form xk = ak exp(iωt).

Substituting for xk in the above system of equations, we obtain a homogeneous system
of equations in complex quantities ak .
s
∑ −mik ω 2 + κik ak = 0; i = (1, 2 · · · , s). (4.4.3)
k =1

The above system has non-trivial solutions for complex quantities ak if the determinant
of the coefficient matrix vanishes.

κ11 − ω 2 m11 . . . κ1n − ω 2 m1n

.. .. .. = 0. (4.4.4)
. . .
2
κn1 − ω mn1 . . . κnn − ω mnn2

Equation 4.4.4 is called the secular equation while the determinant in the left hand side
is called the secular determinant. Upon explicitly expanding the secular determinant,
a polynomial of degree n in ω 2 is obtained. All the values of ω 2 obtained from
solving this polynomial equation can be shown to be positive, but we do not prove
this here. The positive roots of the various values of ω 2 are called the eigenfrequencies
of the system. For a system with s vibrational degrees of motion, there are s possible
eigenfrequencies. It may so happen that some of the eigenfrequencies coincide, in
which case they are called degenerate. Suppose that the system has no degenerate
frequencies. Upon substituting each of the values of ω, say ωl in Equation 4.4.3, the
values for ak are obtained up to the same constant complex multiplier cl for all ak . Let
(l )
this value of ak be denoted by akl cl so that xk = akl cl exp(iωl t) where akl are all real.
Noting that Equation 4.4.2 is homogeneous and linear, the general solution may now
be obtained as a linear combination of solutions corresponding to each eigenfrequency.
s
xk = < ∑ akl [cl exp(iωl t)], (4.4.5)
l =1

where < indicates the real part. With the definition ηl = <cl exp(iωl t), and
representing the excursion xk ’s and ηl ’s as s × 1 column vectors x and η respectively,
the above equation may be written as a matrix equation

x = Aη, (4.4.6)

where A is the coefficient matrix such that [A]kl = akl . As the xk ’s are independent
and so are ηl ’s, the above equation can be inverted to express ηl in terms of xk ’s. This
allows us to choose a different set of generalized coordinates, namely ηl to describe the
motion of the system. The important property which makes this choice of generalized
coordinates more convenient is that

η̈l + ωl2 ηl = 0; l = (1, 2, · · · , s).

Thus the description of the motion of the system is much simpler in terms of
the coordinates ηl . These coordinates are called the normal coordinates. A normal
coordinate is essentially a linear combination of the original coordinates that oscillates
at a unique eigenfrequency. The motion of the system corresponding to a normal
coordinate is called a normal mode. It may be noted that if some particular
eigenfrequency ωl was degenerate with a multiplicity νl , then there would be νl
normal coordinates associated with that frequency. More generally

η̈l p + ωl2 ηl p = 0, ∑ νl = s, (4.4.7)

where the index l ranges over the number of distinct eigen frequencies and the index p
ranges from 1 to νl . Note that the range of p varies with l. Applying the transformation

in Equation 4.4.6, it is now possible to express the Lagrangian and the total energy H
of the vibrating system in a matrix form.

1h T T i
L= η̇ A MA η̇ − η T AT KA η , (4.4.8)
2
1h T T i
H= η̇ A MA η̇ + η T AT KA η .
2
The important consequence of the transformation to normal coordinates is that the
symmetric matrices M and K are simultaneously diagonalized, whereby leading to an
expression of the energy of the vibrating system as a sum of energies associated with
each of the normal vibrational modes. As a result, one has
" #
vl vl
1
2 ∑ ∑ ml,p η̇l,p + ∑ ωl ∑ ml,p ηl,p ,
2 2 2
H=
l p =1 l p =1

where ml,p are positive constants. It is conventional to redefine the normal coordinates
√
as Θl,p = ml p ηl,p so that the total energy can be written simply as
" #
vl vl
1
H=
2 ∑∑ Θ̇2l,p +∑ ωl2 ∑ Θ2l,p . (4.4.9)
l p =1 l p =1

We will apply the classical mechanics approach on a triatomic molecule example and
obtain the normal modes.

Example 34. (Landau and Lifshitz, Mechanics) Consider a non-linear triatomic

molecule such as the one shown in Figure 4.4.1(a). The three mass point m, M and m
are nothing but the nuclei of the atoms constituting the molecule. The size of the nuclei
is negligible compared to the size of the molecule itself. As the three mass points will
always lie in a plane, we need to consider motions that occur only in the x-z plane. For
motions restricted to the plane, there are a total of 6 degrees of freedom, out of which 2
are translational and 1 rotational (about an axis perpendicular to the plane of motion).
This leaves a total of 6 − 3 = 3 vibrational degrees of freedom. Let the displacements
of the atoms of mass m from their mean positions be ( x1 , z1 ) and ( x2 , z2 ) and that of
M be (X, Z). The overall translation of the molecule may be eliminated by imposing
the condition that the center of mass of the molecule does not get displaced.

m( x1 + x2 ) + MX = 0,

m(z1 + z2 ) + MZ = 0. (4.4.10)

As is obvious from the Figure 4.4.1(a), the location of the center of mass is (0, 2ml cos α
2m+ M )
where l is the bond length between M and m in the undisturbed state and α is the

bond angle. The rotational motion of the molecule may be eliminated by equating
the angular momentum of the molecule (with respect to the center of mass of the
molecule) to zero. In case of small oscillations, the instantaneous positions of the
atoms from the center of mass can always be approximated by their positions with
respect to the center of mass in the undisturbed state.

2ml cos α
− k̂ × M Żk̂ + Ẋ î +
2m + M

Ml cos α
l sin αî + k̂ × m ẋ1 î + ż1 k̂ +
2m + M

Ml cos α
−l sin αî + k̂ × m ż2 k̂ + ẋ2 î = 0.
2m + M

Simplifying the above by use of the first of Equation 4.4.10, the angular motion is
eliminated completely by the condition

(z1 − z2 ) sin α − ( x1 + x2 ) cos α = 0. (4.4.11)

M
X
x
l 2a
Z x1
m x2 m

z2 z1
z
(a) (b)

Figure 4.4.1 Normal Modes (Non-linear Tri-atomic Molecule)

As long as displacements of the masses satisfy Equations 4.4.10 and 4.4.11, the motion
will be purely vibrational. As the molecule vibrates, it will generally get deformed.
There are essentially two types of deformations possible in the current case:

(1) The bond length l changes for either or both the bonds
(2) The bond angle 2α changes.

With both such deformations, we can associate springs of stiffness κl and κα . It is

easily seen that the associated deformations are

δl1 = ( x1 − X ) sin α + (z1 − Z ) cos α,

δl2 = ( X − x2 ) sin α + (z2 − Z ) cos α, (4.4.12)

by considering the components of the displacements of atoms along the bond lengths.
Likewise, considering the displacements perpendicular to the bond length, the change
in the bond angle is obtained to be

lδ(2α) = ( x1 − X ) cos α + ( Z − z1 ) sin α

−( x2 − X ) cos α − (z2 − Z ) sin α. (4.4.13)

It is possible to eliminate three of the variables from Equations 4.4.10 and 4.4.11. A
convenient change of variable is effected by the transformations

q1 = x1 + x2 ,

q2 = x1 − x2 ,

q3 = z1 + z2 ,

q1 cot α = z1 − z2 . (4.4.14)

Note that the last equation in the above is a consequence of the first three and Equation
4.4.11. The coordinates q1 , q2 and q3 are three coordinates corresponding to the
three vibrational degrees of freedom of the molecule. The above equations can be
inverted to obtain z1 , z2 , Z, x1 , x2 , X in terms of qi ’s. Upon carrying out the relevant
substitutions, one has

q1 2m 1 q2 q3 2m
δl1 = + sin α + sin α + 1+ cos α,
2 M sin2 α 2 2 M

q1 2m 1 q2 q3 2m
δl2 = − + sin α + sin α + 1+ cos α,
2 M sin2 α 2 2 M

2m
lδ(2α) = q2 cos α − q3 1+ sin α. (4.4.15)
M

The potential energy U of the deformed molecule is calculated from U = κl 2

2 [( δl1 ) +
(lδ(2α))2
(δl2 +)2 ] 2
κα
. The reader is advised to carry out rather straightforward
calculations leading to the following form of the Lagrangian.
2
m 2m 1 2 κl 2m 1
L= + q̇1 − + q21 sin2 α+
4 M sin2 α 4 M sin2 α 2
m 2 h iq
2
q̇2 − κl sin2 α + 2κα cos2 α +
4 4

2m 2 h i q2

m 2m 2
1+ q̇3 − 1 + κl cos2 α + 2κα sin2 α 3 +
4 M M 4

2m q q
1+ [κl − 2κα ] 2 3 sin α cos α. (4.4.16)
M 2

Finally, the equations of motion obtained from the above form of the Lagrangian are

2m 2
mq̈1 + 1 + sin α κl q1 = 0,
M
h i
mq̈2 + κl sin2 α + 2κα cos2 α q2 +

2m
+ 1+ [κl − 2κα ] (sin α cos α)q3 = 0,
M

2m h i
mq̈3 + 1 + κl cos2 α + 2κα sin2 α q3 +
M

+[κl − 2κα ](sin α cos α)q2 = 0. (4.4.17)

r is clear that the coordinate q1 = x1 + x2 oscillates normally with a frequency

It
κl 2m
[1 + sin2 α]. When only this particular mode is excited in the molecule, then
m M
q2 = q3 = 0, or in other words x1 = x2 and z1 = −z2 . During such an oscillation,
the central atom of mass M does not get displaced in the z-direction. This mode is
depicted in Figure 4.4.1 (b).
The other coordinates q2 and q3 are coupled. The secular equation for the
eigenfrequencies of the normal modes associated with q2 and q3 is

2m
[κl sin2 α + 2κα cos2 α] − ω 2 m

1+ M [κl − 2κα ] × (sin α cos α)
= 0.
2m
κl cos2 α + 2κα sin2 α − ω 2 m

[κl − 2κα ] × (sin α cos α)ν2 1+ M

The solution of the secular equation gives 2 more eigenfrequencies corresponding to

distinct normal modes. These modes would in general be some linear combinations
of q2 and q3 . When only one of these modes is excited then q1 = 0, i.e., x1 = − x2 . The
central atom of mass M cannot get displaced in the x-direction now. Figure 4.4.1 (c),
(d) depicts the two modes.
It will be interesting to obtain all the three non-degenerate normal modes depicted
in Figure 4.4.1 (b), (c), (d) using group theory. Clearly, this molecule has C2v symmetry.
We revisit this triatomic molecule (see Example 36) where we reproduce these normal
modes by exploiting the group symmetry. We will elaborate another example using
the conventional classical mechanics procedure to emphasize that the methodology is
tedious but systematic. Then, the readers will be able to appreciate the results when
we discuss elegant group theory approaches towards deriving normal modes.

Example 35. Consider a hypothetical equilateral molecule of three identical atoms

shown in Figure 4.4.2. The Lagrangian of the system is given by
mh 2 i κh i
L= ẋ1 + ẏ21 + ẋ22 + ẏ22 + ẋ32 + ẏ23 − 2
∆l12 2
+ ∆l23 2
+ ∆l13 ,
2 2
y1

y2 y3

x2 x3

Figure 4.4.2 Hypothetical Equilateral Molecule

where ∆l12 is the extension in the bond between nuclei 1 and 2 etc. The extensions are
easily calculated to be

x1 − x2 √ y1 − y2
∆l12 = + 3 ,
2 2
x − x1 √ y1 − y3
∆l13 = 3 + 3 ,
2 2
∆l23 = x3 − x2 .

The conditions for zero momentum of the molecule and zero angular momentum are
given by

x1 + x2 + x3 = 0,

y1 + y2 + y3 = 0,
x2 + x3 √ y3 − y2
− x1 + + 3 = 0.
2 2
With a choice of q1 = x1 , q2 = x3 − x2 and q3 = y1 − y3 and from above equations,
three coordinates can be eliminated. From here the Lagrangian can be expressed
completely in terms of the three vibrational coordinates q1 , q2 , q3 and their time
derivatives. The intervening steps are left as an exercise for the reader who shall
eventually obtain the following equations of motion:

3κ
q̈1 + q = 0,
2m 1
√
3κ 9κ 3κ
q1 + q̈2 + q2 + q3 = 0,
4m 4m 2m
√ √
3 3κ 3 3κ 9κ
q + q2 + q̈3 + q3 = 0.
8m 1 8m 4m
q
3κ
The secular equation of this system gives a doubly degenerate mode of frequency 2m
q
and another mode of frequency 3mκ .
The technique illustrated in Examples 34 and 35 for calculating eigenfrequencies
of molecular vibrations is generalizable for molecules with more atoms. A thorough
calculation as performed in the previous examples is desirable for all these polyatomic
molecules. However, the computation becomes very tedious. In many situations,
it is not the exact values of these eigenfrequencies that are needed but rather a
classification of the normal modes. In the case of the non-linear triatomic molecule,
this classification could have been done without actually going through the full
calculations. With reference to Figure 4.4.1, one may note that the three normal modes
of the motion are quite obvious from mere inspection of the structure of the molecule.
The symmetry of the molecule allows one to guess the simplest possible ways in which
the molecule can be vibrating. While such intuitive guesswork might work in the
simplest of cases, a more formal approach is necessary when one needs to classify the
normal modes of polyatomic molecules. Group theory provides an elegant solution to
this classification problem.
In a vibrating molecule, the atomic nuclei do not remain in their mean positions. If
the nuclei were assumed fixed in their mean positions, then a given molecule would

possess a certain point group symmetry. For small vibrations of the molecule, the
symmetry is amply retained so as to be useful in classification of the normal modes.
If the molecule is vibrating in some specific normal mode, upon a symmetry
transformation the frequency of the oscillation should remain unchanged but the
coordinate designations change for various nuclei. The vibration in the transformed
state of the molecule can at best be a linear combination of the degenerate normal
modes associated with this particular frequency. The degenerate normal coordinates
of the molecule are transformed into linear combinations of each other upon action
of the symmetry group and therefore they give a representation of the symmetry
group. That this representation is always irreducible is ensured by the fact that the
v
quadratic form ∑ pl=1 Θ2l,p (Equation 4.4.9) remains invariant under every symmetry
transformation. For, if the representation of the symmetry group given by the normal
coordinates of a degenerate frequency were reducible, then the cancellation of the
v
cross terms in the transformed value of ∑ pl=1 Θ2l,p could no more be guaranteed. The
conclusion may now be stated that the degenerate normal modes of vibration give an
irreducible representation of the symmetry group of the molecule.
If the normal coordinates are known at the outset, it would be a simple matter to
calculate the symmetry group action on them and thereby construct a representation
of the same. Such a representation is called the full vibrational representation of
the symmetry group of the molecule. The vibrational representation is a reducible
representation. From the characters of the vibrational representation it is possible to
decompose the representation into irreducible parts using a projection operator. Since
the knowledge of the normal coordinates is not assumed to begin with, the characters
of the vibrational representation would have to be calculated by other means. In
Equation 4.4.6, it was noted that the transformation A is an invertible transformation.
It follows that the representation of the symmetry group over the normal coordinates
and the representation of the symmetry group over any other basis for the vibrational
degrees of freedom are equivalent, and hence have the same characters. Once a basis
for the vibrational degrees of freedom is obtained, the symmetry group action on the
basis yields a representation of the symmetry group. The characters of the group
elements can be easily calculated from the representing matrices.

Example 36. The non-linear triatomic molecule is considered here again. The
symmetry group of the molecule is clearly C2v . With reference to Figure 4.4.1(a), the
C2 axis is along the z-axis, the σv (σv0 ) planes are the x-z plane and the y-z plane. To
each of the molecules is attached a basis of its excursions from the mean position,
namely ( x1 , z1 ), ( x2 , z2 ) and (X, Z). Upon eliminating the translational and rotational
motion of the molecule it was shown in the previous example that coordinates q1 =
x1 + x2 , q2 = x1 − x2 and q3 = z1 + z2 correspond to the three vibrational degrees
of freedom, though not all of them are the normal coordinates. Consider the basis set
(q1 , q2 , q3 ) and let the symmetry group C2v act on this basis. The representation of
C2v on this basis is the full vibrational representation ΓV . The reader should have no
difficulty in verifying that the representing matrices for various group elements are

   
1 0 0 −1 0 0
ΓV ( E) =  0 1 0  , ΓV (C2 ) =  0 1 0 ,
0 0 1 0 0 1
   
1 0 0 −1 0 0
ΓV (σv ) =  0 1 0  , ΓV (σv0 ) =  0 1 0 .
0 0 1 0 0 1

Note that the action of C2 is such that x1 7→ − x2 , x2 7→ − x1 , z1 7→ z1 and z2 7→ z2 ;

consequently q1 7→ −q1 , q2 7→ q2 and q3 7→ q3 . Likewise the whole group action
can be worked out. The character table for C2v that includes the characters for the
representation ΓV is

C2v E C2 σv σv0
A1 1 1 1 1 z
A2 1 1 −1 −1 Rz
.
B1 1 −1 1 −1 x, Ry
B2 1 −1 −1 1 y, R x
ΓV 3 1 3 1 ( q1 , q2 , q3 )

From the character table, it is easily seen that ΓV = A1 ⊕ A1 ⊕ B1 . Since all the
irreducible representations present in ΓV are of degree 1, all the eigenfrequencies are
non-degenerate. Two of the eigenfrequencies transform according to the symmetric
representation A1 , the ones depicted in Figures 4.4.1(c) and (d). Then there is one
frequency corresponding to the antisymmetrical representation B1 of Figure 4.4.1(b).

Instead of generating the vibrational representation of the molecule, it is often more
convenient to generate the representation on a basis consisting of excursions. We do
this in the following example.

Example 37. Continuing from the previous example, let the excursions be
labeled ( x1 , z1 ), ( x2 , z2 ) and (X, Z). Consider the group action on the basis
( x1 , z1 , x2 , z2 , X, Z ). In this case, identify the degree 6 reducible representation
as Γ M . The following representing matrices are obtained for Γ M :
 
1 0 0 0 0 0

 0 1 0 0 0 0 

 
M 0 0 1 0 0 0
Γ ( E) = 
 
,

 0 0 0 1 0 0 

0 0 0 0 1 0
 
 
0 0 0 0 0 1

 
0 0 −1 0 0 0

 0 0 0 1 0 0 

−1
 
0 0 0 0 0
Γ M (C2 ) = 
 
,

 0 1 0 0 0 0 

0 0 0 0 −1 0
 
 
0 0 0 0 0 1
 
1 0 0 0 0 0

 0 1 0 0 0 0 

 
M 0 0 1 0 0 0
Γ (σv ) = 
 
,

 0 0 0 1 0 0 

0 0 0 0 1 0
 
 
0 0 0 0 0 1
 
0 0 −1 0 0 0

 0 0 0 1 0 0 

−1
 
M 0 0 0 0 0 0
Γ (σv ) = 
 
.

 0 1 0 0 0 0 

0 0 0 0 −1 0
 
 
0 0 0 0 0 1

As for the characters of Γ M , one has

C2v E C2 σv σv0
.
ΓM 6 0 6 0

Upon decomposition, it is found that ΓM = 3A1 ⊕ 3B1 . Now consider the

representations of pure translations and rotations that keep the molecule in the x-z
plane. Translations in z and x directions correspond to the representations A1 and B1
respectively, while the rotation about the y-axis corresponds to another instance of the
irreducible representation B1 . The vibrational representation ΓV is therefore equal to
ΓM − (A1 ⊕ 2B1 ) which is same as A1 ⊕ A1 ⊕ B1 , as was found before for vibrational
representation.
Thus having determined the irreducible representations representing the normal
modes of the triatomic molecule possessing C2v symmetry, we can extract the three
non-degenerate normal coordinates in the following manner: Construct the projection
operators PA1 and subtract the translation operator Tz denoting the translation along
the z direction which belongs to A1 basis. The explicit matrix form of the two operators
in the excursion basis is

   
2 0 −2 0 0 0 0 0 0 0 0 0

 0 2 0 2 0 0 


 0 1 0 1 0 1 

−2
   
M 1  0 2 0 0 0  , TzM = 1 
  0 0 0 0 0 0 
PA = = .
1 4 
 0 2 0 2 0 0 
 3 0 1 0 1 0 1 

0 0 0 0 0 0 0 0 0 0 0 0
   
   
0 0 0 0 0 4 0 1 0 1 0 1

In fact, the rank of PA1 − Tz is two whose non-trivial eigenvectors

   
−1 0

 0 


 −1/2 

   

 1 
 = x2 − x1 ,

 0  = Z − z 1+ z 2 ,


 0 


 −1/2 
 2
0 0
   
   
0 1

which are proportional to q2 and q3 in the center of the mass frame.

The projection operator eigenvectors are not in general the frequency eigenbasis.
However, this group theory approach will assert that the frequency basis (normal
modes) belonging to the A1 irreducible representation will involve only linear
superposition of q2 and q3 . Similarly, we can obtain the projection operator for
irreducible representation PB1 and subtract translation operator Tx . The rank of
PB1 − Tx is again two whose the non-trivial eigenvectors are q1 = x1 + x2 (vibrational
mode) and z1 − z2 (rotational mode) consistent with the classical mechanics approach
discussed in Example 34.
The above two examples illustrate how the representation of the symmetry group
of a molecule on the basis of excursions can be utilized to infer the normal modes of
the vibrations of the molecule. Also, the matrix representation in the excursion basis
is useful to infer the normal coordinates. However, the characters of the vibrational
representation can be inferred directly in all cases without actually determining the
representing matrices.
Consider a molecule with point group symmetry G in a coordinate system oriented
in the usual way so that z-axis is the symmetry axis of the molecule. With each
nucleus of the molecule as an origin, construct coordinate systems of excursions such
that their axes are parallel to the original coordinate system (Figure 4.4.1(a)). We
consider the group action on the excursions. There are 3n coordinates in general for
an n-atomic molecule. If a symmetry transformation displaces a nucleus into another
identical one, then the representing matrix of the transformation cannot have diagonal
elements corresponding to coordinates of the first nucleus. Thus the character of the
transformation will depend on the action of the transformation on the coordinates of
the nuclei which are unmoved from their positions by the transformation. Suppose the
symmetry transformation Cθ is a rotation of the molecule about z-axis by an angle θ.

Let the molecule whose excursion coordinates are ( x a , y a , z a ) remain unmoved by the
Cθ transformation. The excursion coordinates are transformed due to rotation as

x a 7→ x a cos θ + y a sin θ,
y a 7→ − x a sin θ + y a cos θ,
za → za .

The contribution to the character of the rotation Cθ from one such nucleus is (1 +
2 cos θ ) If there are NC nuclei which are unmoved, then the total contribution to the
character is NC (1 + 2 cos θ ). However, in the process of calculating the character,
one has to keep track of the translational motion of the molecule as a whole as
well as its rotation about the symmetry axis. Under pure rotation, both the vectors
corresponding to small displacements of the center of mass as well as small angular
displacements behave as polar vectors and consequently contribute 2 (1 + 2 cos θ ) to
the character of Cθ . Therefore the character of Cθ in the vibrational representation ΓV
is given by

χΓ (Cθ ) = ( NC − 2)(1 + 2 cos θ ).

V
(4.4.18)

For calculating the character of the rotary reflection transformation Sθ , it may be noted
first of all that under such a transformation the coordinates x a and y a transform as
indicated above, however z a 7→ −z a . If there are NS nuclei which are unmoved by Sθ ,
then the total contribution to the character of Sθ is NS (−1 + 2 cos θ ). The polar vector
corresponding to the overall translational motion transforms in the same manner
and has a contribution of (−1 + 2 cos θ ). The vector corresponding to small angular
displacement no more transforms as a polar vector since there is reflection involved
in Sθ . Under reflection, the angular displacement transforms are an axial vector. For
such a vector, it is well known that the components of the vector parallel to the plane
of reflection reverse direction while the component of the vector perpendicular to the
plane remains unchanged. After carrying out a rotation by θ, the components of the
angular displacement transform as above. Since the plane of reflection is the x-y plane
after reflection in this plane the final transformations are given by

x a 7→ − x a cos θ − y a sin θ,
y a 7→ x a sin θ − y a cos θ,
z a 7→ z a .

The contribution to the character from the transformation of a small angular

displacement of the molecule by Sθ is therefore (1–2 cos θ ). Adding this contribution
to the contribution from the translational motion, and subtracting from the overall, the
character of Sθ in the vibrational representation ΓV is obtained to be

χΓ (Sθ ) = NS (−1 + 2 cos θ ).

V
(4.4.19)

It is also clear that NS can be either 0 or 1. A reflection in the x-y plane σ is simply a
rotary reflection transformation with θ = 0. Its character is therefore

χΓ = Nσ ,
V
(4.4.20)

where Nσ is the number of the nuclei in the reflection plane. The same formula holds
for σv reflections. For inversions, take θ = π in Equation 4.4.19.

Example 38. Consider the methane molecule (CH4 ). In the equilibrium state of the
molecule, the hydrogen atoms are at the vertices of a regular tetrahedron and the
carbon is at the geometrical center. Clearly this molecule has Td point group symmetry.
The reader may refer to the character table for the Td group given in Section 4.2.The
molecule has 9 vibrational degrees of freedom. The character of the identity element E
is therefore 9. On a C3 axis, there are two nuclei (one hydrogen and the central carbon)
which are therefore unmoved by C3 rotations. In accordance with Equation 4.4.18, the
character of C3 rotations is zero. The C2 and the S4 axes leave only the central carbon
unmoved and the respective characters are 1 and −1 (Equation 4.4.19). The characters
of σd reflections are 3 as the σd planes contain two hydrogen atoms and the central
carbon. To summarize

Td E 8C3 3C2 6S4 6σd

.
ΓV 9 0 1 −1 3

Decomposing the vibrational representation, one has

ΓV = A1 ⊕ E ⊕ F2 ⊕ F2 .

There are a total of 4 distinct vibrational frequencies,1 two of which (F2 ) are triply
degenerate, one is doubly degenerate (E) and one is non-degenerate (A1 ).

Exercises
1. Suppose we confine a particle to move in two dimensions where the potential is
U ( x, y) = 0 in a region enclosing an equilateral triangle and zero elsewhere.
Purely from group theory, deduce whether the stationary states are non-degenerate
or degenerate.
2. Suppose the C3v symmetry of a molecule breaks down to a subgroup Cs = { E, σv }
due to external perturbation. How does the degree 2 degenerate level of C3v split with
respect to the Cs group.

1 https://www.youtube.com/watch?v=3RqEIr8NtMI& index=29& list=PLEB476ECA0DA9481C.

3. Determine the irreducible representations corresponding to the normal modes of the

ammonia molecule.
4. Write down the decompositions of all possible tensor products of two irreducible
representations of C3v . Show that the electric and magnetic dipole transitions between
states of representations A1 and A2 are forbidden.
5. From Equation 4.4.19, calculate the character of the inversion operation I .
6. For the symmetry group D3d , find all the allowed electric and magnetic dipole
transitions. The character table for D3d is produced below for reference.

D3d E 2C3 3C2 I 2S6 3σd

A1g 1 1 1 1 1 1 x 2 + y2 , z2
A2g 1 1 −1 1 1 −1 Rz
Eg 2 −1 0 2 −1 0 ( R x Ry ) ( x2 − y2 , xy) .
A1u 1 1 1 −1 −1 −1
A2u 1 1 −1 −1 −1 1 z
Eu 2 −1 0 −2 1 0 ( x, y)

7. Find the allowed electric quadrupole transitions for the symmetry group D3d .
8. Exploiting the C3v symmetry of an equilateral molecule made of three identical
nuclei, determine the degree of degeneracy of the eigenfrequencies. Calculate the
eigenfrequencies of small oscillations of the molecule by utilizing the symmetry of the
normal modes. Verify that the frequencies are the same as those obtained in Example
35. The bonds connecting the nuclei, all have same spring constant κ .
9. Using discrete group symmetry, analyze the normal modes of a square system of four
identical masses. Identical springs connect two masses along the edges of the square.
The mass points are constrained to move in a plane.

The previous chapters dealt with some systems whose symmetry could be described
by finite groups. Such systems possess discrete symmetry. One does not need to go too
far afield to find an example of a system whose symmetry is not discrete. Consider a
sphere which appears the same when viewed from all orientations. If the sphere were
to be rotated by any angle whatsoever about an axis that passes through the sphere’s
center, it would still appear the same. Any diametrical plane is also a reflection plane
of symmetry. Therefore, there are an infinite number of symmetrical transformations
of the sphere. Apart from this fact, it is possible to develop the notion of closeness
between certain symmetry transformations. If Cα is the symmetry transformation
corresponding to a rotation by α about the axis C passing through the sphere’s center,
then Cα+e is also a symmetry transformation for the sphere where e could be made
arbitrarily small in magnitude. This is an instance of continuous symmetry. The theory
of Lie groups is the suitable tool for study of such symmetries. It is the purpose of this
chapter to introduce the basic concepts of the theory of Lie groups and their associated
Lie algebras so as to be able to quickly apply these for problem solving.

5.1 The Circle Group U(1)

Consider the group of complex numbers of unit modulus with the group operation
being the usual multiplication. All such numbers lie on the unit circle centered at
the origin in the complex plane (Figure 5.1.1). This group is called the U(1) group.
Apart from satisfying the group axioms, U(1) has a richer structure owing to the fact
that it is a subset of the complex plane. For any two complex numbers z1 and z2 ,

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88 Group Theory for Physicists

their product z1 z2 is uniquely defined. Additionally, for small variations in z1 or z2

or both, the value of the product varies by a small amount. More precisely, the map
(z1 , z2 ) → z1 z2 is continuous. The inversion operation z → 1z for z 6= 0 can be shown
to be a continuous map. As U(1) is a subset of the complex plane, these properties
are inherited in U(1). The product and inversion operations in U(1) are continuous
operations and this makes U(1) a continuous group.

z –1 = 1/z

Figure 5.1.1 The Circle Group U(1)

Another important feature of U(1) is that there is a homomorphism from the

additive group of real numbers onto U(1). This homomorphism is clearly given by
the map

t → exp(iωt)

where ω is any non-zero real constant, and t varies over the real line R. Thus U(1) is
a one-parameter group in the sense that all the group elements are parametrized using
a single real parameter t. It may be noticed that such a parametrization is not unique
and depends on the choice of the value of ω. The identity of U(1) is prametrized by
all integral multiples of 2π/ω. For group elements z1 and z2 parametrized by t1 and
t2 , their product is parametrized by t1 + t2 . If z(t) = z1 z2 and t is assumed to be
functionally related to t1 and t2 such as t = f (t1 , t2 ), then it is clear that in case of
U(1) this functional dependence is given by

f ( t1 , t2 ) = t1 + t2 .

Likewise, if the parameters of z and z−1 are assumed to be related by the function g(t),
then in case of U(1)

g(t) = −t.

Both f and g are not merely continuous but are in fact analytic functions (i.e., they can
be expanded as a Taylor Series about any point in their domains of definition). U(1)
is an almost trivial example of a class of groups called Lie groups.

5.2 The Matrix Exponential

It is clear from the above discussion that any given parametrization exp(iωt) of U(1)
is a one-dimensional unitary representation of the group U(1). The generalisation of
such a parametrization and exponentiation for other Lie group elements involving
higher dimensional matrices will be
s
g({t a }) = exp(i ∑ Xa ta ) ≡ exp(Ω) where g({t a = 0}) = I, (5.2.1)
a =1

and t a ’s are the parameters. Note that the number of parameters depends on the
group properties of the Lie group. For instance, a set of 2 × 2 unitary matrices forms a
group U (2). Though there are 8 real entries (4 complex entries) in these matrices, the
unitarity property imposes four real constraints. Hence, the number of independent
real parameters to describe these matrices reduces from 8 to 4. Associated with every
real parameter, there is a corresponding matrix Xa . Hence there will be four linearly
independent Xa ’s for U (2) group. The role of Xa ’s is to take group elements away
from identity I and hence they are called generators of the Lie group.

5.2.1 Generators of the Lie group

Suppose u is a real parameter and Ω ≡ i ∑ a Xa t a a square matrix. Let f (u) = exp(Ωu)
be a matrix valued function of u. Following the usual rules of differentiation of
functions of a real variable, one has

d
f (u) = exp(Ωu)Ω = Ω exp(Ωu). (5.2.2)
du
Such a differentiation cannot determine each of the generators separately. In order to
determine the generators, we need to take the parameters δt a ’s to be infinitesimal so
that differentiation with respect to δt a is meaningful; leading to:
!
∂
X a = −i exp i ∑ δtb Xb |δta =0 . (5.2.3)
∂δt a b

Using the group multiplication law, g({t a }) g({ub }) = g({vc (t a , ub )}), it is a

straightforward exercise (as discussed in Section 5.5) to show that the generators X a ’s
satisfy

c
Xa Xb − Xb Xa = iCab Xc (5.2.4)
c are known as structure constants and the above relation is called
where coefficients Cab
Lie algebra g whose formal definitions and properties are elaborated in Section 5.3.

5.2.2 Convergence property of matrix exponentials

It is not at all clear whether the exponential of the square matrix which can also be
formally written as the power series expansion

∞
Ωk
exp Ω = ∑ k! , (5.2.5)
k =0

is a convergent series. In the case of U(1), Ω = [iωt] is a 1 ×1 matrix, and the

above expansion is same as that of exp(iωt). However, for a general square matrix
Ω, Equation 5.2.5 is meaningful only if the infinite series in its right hand side
converges. Noting that all the summands are matrices of same order, the critierion
for convergence can be set simply to be that all the series of corresponding matrix
entries in the summands converge. We will now illustrate this convergence property
through a simple example.

Example 39. Consider the matrix

0 −θ
Ω ≡ iωθ = .
θ 0

It is easily seen that

(−1)s θ 2s

2s 0
Ω =
0 (−1)s θ 2s
s = {0, 1, 2, . . .}.
(−1)s+1 θ 2s+1

0
Ω2s+1 =
(−1)s θ 2s+1 0

With the definition exp Ω ≡ exp iωθ as in Equation 5.2.5, one has

cosθ −sinθ
exp Ω ≡ exp iωθ = .
sinθ cosθ

The reader should immediately recognize exp Ω as the familiar rotation matrix Rθ
about an axis perpendicular to the x-y plane.

The inverse of such a matrix will be given by R− 1 T

θ ≡ R−θ = Rθ implying that the
matrix Rθ denotes an orthogonal matrix with the corresponding Ω matrix being the

antisymmetric matrix Ω T = −Ω. Note that the equivalent expression in terms of the
ω matrix will imply that the matrices exp(iωθ ) are unitary with ω being self-adjoint
as illustrated below.
The adjoint of a linear transformation was defined in Section 3.1. If Ω ≡ iωθ
is a square antisymmetric matrix, then it follows from the definition of adjoint that
[(iω )† ] jk = [(iω )kj ] = −i [ωkj ]. Further, if ω itself was self-adjoint then

[(iω )† ] jk = −iω jk

⇒ (iω )† = −iω.

Assuming that the matrix exponential exp(iωθ ) converges for some self-adjoint ω,
upon taking the adjoint of Equation 5.2.5 one has

∞ ∞
θk θk
[exp(iωθ )]† = ∑ k!
[(iω )k ]† = ∑ (−iω )k
k!
k =0 k =0

⇒ [exp(iωθ )]† = exp(−iωθ ).

From the definition of the matrix exponential, it can be shown that

exp(−iωθ ) exp(iωθ ) = exp(iωθ ) exp(−iωθ ) = I. (5.2.6)

In other words, exp(iωθ ) is a unitary matrix if ω is self-adjoint. This property has

several applications in quantum mechanics as will be seen in the next chapter. It is
important to bear in mind that for two square matrices Ω1 and Ω2 of same order, the
equality exp Ω1 exp Ω2 = exp(Ω1 + Ω2 ) holds only if the Ω1 and Ω2 commute. Such
a property is obeyed by the set of rotation matrices {Rθ } which are orthogonal 2 × 2
matrices. Incidentally, these 2 × 2 matrices, involving four real entries, are dependent
only on one parameter θ. This is because the orthogonality property gives three
constraints reducing four real entries of 2 × 2 matrices to one. Hence this set Rθ forms
a group SO(2) where the letter O denotes the matrices are orthogonal and S implies
that their determinants are unity. Like U (1), the one parameter group SO(2) is also
abelian group:

R θ1 R θ2 = R θ2 R θ1 (5.2.7)

The commutator of Ω1 and Ω2 , denoted by the symbol [Ω1 , Ω2 ], is defined as

[ Ω1 , Ω2 ] = Ω1 Ω2 − Ω2 Ω1 . (5.2.8)

Thus the condition for exp Ω1 exp Ω2 = exp(Ω1 + Ω2 ) to be true is that [Ω1 , Ω2 ] = 0.
The following properties of the commutator are directly obtained from its definition.
(1) [λΩ1 + µΩ2 , Ω3 ] = λ[Ω1 , Ω3 ] + µ[Ω2 , Ω3 ] for any complex λ and µ.

(2) [Ω1 , Ω2 ] = −[Ω2 , Ω1 ].

(3) [Ω1 , [Ω2 , Ω3 ]] + [Ω2 , [Ω3 , Ω1 ]] + [Ω3 , [Ω1 , Ω2 ]] = 0.
The last equality is called the Jacobi identity. All the three aforementioned properties
of the commutator are also properties of the Poisson bracket if Ωi were dynamical
quantities of a Hamiltonian describing classical system. The generators of a Lie group
(5.2.4) indeed have such non-trivial commutators. We will formally introduce the
algebra obeyed by the generators Xa ’s in the following section.

5.3 Finite Dimensional Lie Algebras

Definition 5. A Lie Algebra g is a vector space on which is defined a binary operation
[ , ] having following properties:
(1) For all x and y in g, [ x, y] is in g.
(2) For all x, y and z in g, and scalars λ and µ, [λx + µy, z] = λ[ x, z] + µ[y, z].
(3) [ x, y] = −[y, x ].
(4) [ x, [y, z]] + [y, [z, x ]] + [z, [ x, y]] = 0. [Jacobi identity]
If the scalars are chosen from the set of real numbers, then the Lie algebra is called a real
Lie algebra. In case the scalars come from the set of complex numbers, one has a complex
Lie algebra. The properties of the composition [ , ] are identical to the properties of the
commutator discussed in the last section. The composition [ x, y] of x, y ∈ g would
be refered to as the Lie bracket of x and y. A Lie algebra g in which the Lie bracket
vanishes identically over g is an abelian algebra. Being a vector space, g has a basis. It
will be assumed throughout that g is finite dimensional. If { xi }in=1 is a basis for g then
for any xs and xt in the basis it must be true that
n
[ xs , xt ] = ∑ ckst xk (5.3.1)
k =1

since [ xs , xt ] is after all yet another vector in g. The coefficients ckst are called the
structure constants of the Lie algebra. The obvious relation [ xs , xt ] + [ xt , xs ] = 0 for
commutator bracket puts the following restriction on the structure constants:
n
∑ (ckst + ckts )xk = 0.
k =1

From linear independence of { xi }in=1 , it follows that ckst = −ckts . In a Lie algebra, it can
be shown further that
n
∑ [clsk cktu + cltk ckus + cluk ckst ] = 0
k =1

The following definitions of Lie algebra and the properties of the structure functions
are essential and will henceforth be assumed for the rest of the chapter. We will now
present two familiar examples where the structure constants and Lie algebra appear.

Example 40. The cross product of vectors in the three-dimensional space R3 turns
into a three-dimensional Lie algebra. If the basis vectors are taken to be the standard
i, j and k, then all the four properties of Definition 5 are easily seen to be satisfied.
The non-zero structure constants are
j
cijk = cki = cijk = +1

j
cikj = cik = ckji = −1,

and the other structure constants are all equal to zero.

Example 41. In quantum mechanics, the operators of position x̂ and the corresponding
component of momentum p̂ x of a particle do not commute and the identities

[ x̂, p̂ x ] = [ŷ, p̂y ] = [ẑ, p̂z ] = i} (5.3.2)

are well known. For simplicity, let us set } (proportional to Planck constant) to be
1. The position components and the momentum components commute amongst
themselves. Also, position components commute with any orthogonal momentum
component. The components of the orbital angular momentum operators `ˆ = r̂ × p̂ of
the particle

`ˆ x = ŷ p̂z − ẑ p̂y

lˆy = ẑ p̂ x − x̂ p̂z

lˆz = x̂ p̂y − ŷ p̂ x ,

generate a three-dimensional Lie algebra. Using the position-momentum algebra

(5.3.2) and the definitions of the commutator, the readers can verify that the Lie algebra
amongst the angular momentum components become:

[lˆx , lˆy ] = i lˆz

[lˆy , lˆz ] = i lˆx (5.3.3)

[lˆz , lˆx ] = i lˆy

We will now focus on rotation operations in three-dimensional space and apply the
above algebra. We can consider three independent rotations depending on the choice

of perpendicular axis. Let R(θ ≡ θ n̂) denote 3 × 3 matrix representation of proper

rotations θ about n̂ axis, acting on any position vector ~
r1 as follows:

r1 → ~
R(θ) : ~ r2 ,

where rotations keeps the norm of vectors ~r 1 .~

r1 = ~
r 2 .~
r2 same. This implies that the
rotation R(θ ) must obey the orthogonal property

R(θ) R T (θ) = I.

The determinant of these proper rotations matrices must be one. Depending on the
axis of rotation, there will be an appropriate Ω ≡ −iL.θ in the exponential map:
R(θ ) = exp(−iθ.L). Just like the collection of one parameter elements {exp(iωt)}
of U (1) group and {exp(iωθ )} of SO(2) group, the set of orthogonal 3 × 3 matrices
R(θ) = exp(Ω) = exp(−iθ.L), involving three parameters θ and three generators L,
forms a group SO(3). We leave it to the readers to verify that orthogonal 3 × 3 matrices
will require only three independent entries. Unlike the SO(2) group (5.2.7) where the
elements commute, the SO(3) group elements do not obey

exp(iΩ1) exp iΩ2 6= exp[i (Ω1+ Ω2 )]

and hence the group is non-abelian. For infinitesimal angle δθ, the exponential map
of SO(3) elements can be approximated as

R(δθ) = I − iδθ n̂.L, (5.3.4)

where we have denoted the magnitude of the rotation angle about an axis n̂
Example 42. It is straightforward to check that the infinitesimal rotations of
magnitude δθ about x-axis R(δθ î) and the infinitesimal rotations of same magnitude
δθ about y-axis R(δθ ĵ) obey the following relation:

R(δθ î)R(δθ ĵ) − R(δθ ĵ)R(δθ î) = R(δθ 2 k̂) − I. (5.3.5)

Using the definition (5.3.4) in the above equation, verify that the 3 × 3 matrices
L x , Ly , Lz obey angular momentum algebra (5.3.3):

[ L x , Ly ] = iLz .

Using R(δθ î) ≡ I − iδθL x and R(δθ ĵ) ≡ I − iδθLy , the LHS (5.3.5) simplifies to
(−i )2 δθ 2 [ L x , Ly ] whereas the RHS (5.3.5) gives −iδθ 2 Lz leading to the conventional
angular momentum algebra (5.3.3).
Thus L x , Ly , Lz , which are the generators of the rotation group SO(3), constitute
the Lie algebra g of the rotation group SO(3) denoted as g ≡ so(3) which is isomorphic

to the angular momentum algebra (5.3.3). Incidentally, the range of the magnitude of
the parameter θ = θ n̂ must be −π ≤ θ < π. Hence the parameter space describing
the group SO(3) will be a solid sphere of radius π. The groups with such bounded
parameters are referred to as compact groups.
Every point inside the solid sphere corresponds to a distinct group element of
SO(3). However, the diametrically opposite points on the boundary of the solid
sphere of radius π will denote the same group element ( R[π n̂] = R[π (−n̂)]. Hence,
this identification of diametrically opposite points allows two topologically distinct
closed curves inside the sphere as shown in Figure 5.3.1. In the literature, such a
parameter space is called doubly-connected space.

m
K¢
p

O n

Figure 5.3.1 Group Manifold of the SO(3) (doubly-connected

space)

Definition 6. If g is a Lie algebra and h is a subset of it such that the elements of h form
a Lie algebra under the Lie bracket of g, then h is called a subalgebra. The subalgebra h
is an invariant subalgebra if [ g, h] ∈ h for all g ∈ g and h ∈ h.

Example 43. Suppose g ∈ g and h ∈ h where h is an invariant subalgebra of g. Show

that the set of elements {exp(ih)} forms an invariant subgroup H ⊂ G where the set
{exp(ig)} belongs to Lie group G. In other words, show that

exp(ig) exp(ih) exp(−ig) = exp(ih1 )

where h1 ∈ h.
Using the well-known properties of matrices and the fact that h ∈ h, we can deduce

1
exp(ig)h exp(−ig) = h + i [ g, h] − [ g, [ g, h]] + . . . = h1 ,
2!

as [ g, h], [ g, [ g, h]], . . . ∈ h. Hence, for any exponential form of h in LHS of the above
equation, we will get RHS as exp(ih1 ).
In Chapter 3, linear representations of finite groups were studied in some length.
Analogously, representations of a Lie algebra may be defined. To begin with, let
V be a complex (or real) vector space of dimension n. Let gl(V ) be the set of all
linear operators on V. For T1 , T2 ∈ gl(V ) and scalars α, β one may define a linear
combination αT1 + βT2 to be the linear operator which acts on V in accordance with

(αT1 + βT2 ) x = αT1 ( x ) + βT2 ( x ),

for all x in V. With the above definition, gl(V ) itself is a vector space. If { xi }in=1
is a basis for V, then define linear operators Xij which act on the basis of V such that
Xij ( xk ) = δik x j . Any linear operator in gl(V ) may be expressed as a linear combination
of Xij ’s. It is left to the reader to verify that Xij are in fact linearly independent. It
follows that the dimension of the space gl(V ) is n2 . Apart from being a vector space,
gl(V ) has an additional important structure. Two linear operators can be composed
according to the usual rules of function composition, ( T1 T2 )( x ) = T1 ( T2 ( x )). Linearity
of T1 and T2 immediately leads to linearity of T1 T2 so that T1 T2 is also in gl(V ). One can
now employ the vector space structure of gl(V ) along with this property to define the
Lie bracket [ T1 , T2 ] = T1 T2 − T2 T1 . The conclusion that, for a given finite dimensional
vector space V, the space gl(V ) of all linear operators on V can be turned into Lie
algebra is obvious.

Example 44. It is trivial that the set of real numbers R is a one-dimensional real vector
space. The linear operators are the real numbers themselves and the commutator
identically vanishes. The Lie algebra is abelian.
Example 45. Consider the two-dimensional complex vector space. In any basis, the
linear operators on this space can be represented by 2 × 2 matrices with complex
entries. The Lie algebra of this four-dimensional complex vector space of linear
operators is denoted gl(2, C). gl(2, C) is clearly non-abelian which is spanned by
four 2 × 2 matrices such as

1 0 0 1
E1 = , E2 = ,
0 0 0 0

0 0 0 0
E3 = , E4 = .
1 0 0 1

Example 46. There are several important subalgebras of gl(2, C). For instance,
consider matrices of the form

a z
X= ,
z∗ − a

where a is a real number and z a complex number. The important properties of such
matrices are that they are traceless and Hermitian. The complex subspace of gl(2, C)
spanned by matrices of this type is a subalgebra denoted by sl(2, C). If { Xi }in=1 are
traceless Hermitian matrices and {αi }in=1 complex numbers, then X = ∑in=1 αi Xi is in
sl(2, C). Similarly, for Y = ∑m j=1 β j Yj in sl(2, C), by linearity of the Lie bracket, one
has

[ X, Y ] = ∑ αi β j [ Xi , Yj ] ≡ Z.
i,j

The reader can verify that the brackets [ Xi , Yj ] are all equal to the imaginary unit times
Zij , where Zij is again a traceless Hermitian matrix. Then the above equality shows
that [ X, Y ] ∈ sl(2, C). Hence, sl(2, C) is indeed a subalgebra of gl(2, C). Since
sl(2, C) is a proper subspace of gl(2, C), its dimension is less than 4. Notice that a
traceless Hermitian matrix is completely specified if the 3 independent quantities a, z
and z∗ are known. Therefore the dimension of sl(2, C) must be 3. This can be seen
more explicitly by considering the matrices

0 1 0 −i 1 0
σx = , σy = , σz = .
1 0 i 0 0 −1

σx , σy and σz are traceless Hermitian matrices which are also linearly independent.
Their linear span is definitely a subset of sl(2, C). It follows that their linear span is
all of sl(2, C). The matrices σx , σy and σz are the familiar Pauli matrices.
Complex numbers z in C are pairs of independent real numbers (<(z), =(z)).
Similarly, an n-tuple of complex numbers is a 2n-tuple of real numbers. Thus a
complex vector space of dimension n can be regarded as a real vector space of
dimension 2n. In the above discussion, gl(2, C) was regarded as a complex vector
space of dimension 4. As a real vector space, gl(2, C) has dimension 8 spanned by the
basis consisting of E1 , E2 , E3 and E4 and the matrices

i 0 0 i
,
0 0 0 0

0 0 0 0
, .
i 0 0 i

Likewise, sl(2, C) is a six-dimensional real Lie algebra spanned by a basis consisting

of σx , σy , σz and the matrices

0 i 0 1 i 0
Sx = , Sy = , Sz = .
i 0 −1 0 0 −i

The following Lie brackets are easily computed from the above representations of the
basis elements of sl(2, C).

[σx , σy ] = 2iσz [σy , σz ] = 2iσx [σz , σx ] = 2iσy (5.3.6)

[Sx , Sy ] = −2Sz [Sy , Sz ] = −2Sx [Sz , Sx ] = −2Sy .

The lower set of 3 equations is completely equivalent to the upper 3 equations since
Sx = iσx , etc. The three-dimensional real subalgebra of sl(2, C) spanned by Sx , Sy and
Sz is called the su(2) algebra. Notice that Sx , Sy and Sz are skew-Hermitian (S†x = −Sx
etc. ) traceless matrices.

Definition 7. Let g be a Lie algebra and gl(V ) the Lie algebra of operators on the
vector space V. A linear map π from g into gl(V ) is said to be a representation of g on
the vector space V if

π ([ x, y]) = [π ( x ), π (y)] (5.3.7)

for all x, y in g. The dimension of V is the degree of the representation π.

Since a Lie algebra g is also a vector space, the above definition of representation
allows one to represent g over itself. For some x ∈ g, consider a mapping adx from g
into g defined by adx (y) = [ x, y] for all y ∈ g. It is clear that adx is a linear operator
on the vector space g since the Lie bracket is linear in y by definition. Therefore adx ∈
gl(g) and one has a mapping ad from g into gl(g) defined as ad ( x ) = adx for all x in
g. This mapping is a representation if ad ([ x, y]) = [ad( x ), ad(y) ]. Let z ∈ g so that
one has

ad[ x,y] (z) = [[ x, y], z] = [ x, [y, z]] + [y, [z, x ]]

⇒ ad[ x,y] (z) = [ x, [y, z]] − [y, [ x, z]]

⇒ ad[ x,y] = adx ady − ady adx (5.3.8)

and one has shown that ad is indeed a representation. Such a representation of a Lie
algebra is called an adjoint representation. From Equation 5.3.1 it immediately follows
that the representing matrices of the bases { xs }in=1 of g are given by

[adxs ]kt = ckst . (5.3.9)

Example 47. The su(2) algebra was described in the previous example (Equation
5.3.6). If one takes Sx /2, Sy /2 and Sz /2 as the basis for the algebra then in the adjoint
representation, the matrices are easily found to be

   
0 0 0 0 0 −1
Sx /2 =  0 0 1  , Sy /2 =  0 0 0  ,
0 −1 0 1 0 0
 
0 1 0
Sz /2 =  −1 0 0  .
0 0 0

5.3.1 su(2) algebra

In quantum mechanics, we define generators of su(2) Lie algebra as Ji ’s obeying

[ Ji , Jj ] = ieijk Jk .

The representation for Ji = 21 σi is known as the fundamental matrix representation

which √act on a two-dimensional complex vector space V. Using the redefinition ( J1 ±
iJ2 )/ 2 = J± where J± are known as raising and lowering operators, the algebra
becomes

[ J+ , J− ] = J3 ; [ J3 , J± ] = ± J± . (5.3.10)

In fact such a redefinition naturally appears in the formal discussion of Cartan

classification of semi-simple Lie algebras. We will briefly present such ladder
operators in Section 5.7.
As the su(2) algebra has no abelian subalgebra, it is always possible to choose one
of the generators to have a diagonal matrix representation. Following the convention
in many books, we take the representation of J3 as diagonal matrix and work with its
eigenvectors as basis {| j, mi} ∈ V. That is,

J3 | j, mi = m| j, mi,

where m can be either integers or half-odd integers. We refer to maximum eigenvalue

of J3 as j and the corresponding highest value state is | j, ji.
Applying J± on the states | j, mi we observe

J3 { J± | jmi} = J± { J3 | j, mi} ± J± | j, mi = (m ± 1){ J± | j, mi}.

From this relation, it is clear that the J3 eigenvalue of states J± | jmi is m ± 1. For highest
value state | j, ji, we cannot get J3 eigenvalue as j + 1 for J+ | j, ji which implies

J+ | j, ji = 0.

The other eigenbasis of J3 can be obtained by applying lowering operator J− on the

highest value state | j, ji:

J− | j, ji = Nj,j | j, j − 1i.

Taking inner product of the above state with its dual state h j, j| J+ , we get

h j, j| J+ J− | j, ji = | Nj,j |2 = h j, j| J3 + J− J+ | j, ji = h j, j| J3 | j, ji = j,
p
implying Nj,j = j. We can now determine

1 1 p
J+ | j, j − 1i = J+ J− | j, ji = { J3 + J− J+ }| j, ji = j| j, ji = Nj,j | j, ji.
Nj,j Nj,j

This application of J− can be continued to obtain the basis states | j, j − 2i, | j, j − 3i, . . .
belonging to the vector space V:

J− | j, j − r i = Nj,j−r | j, j − r − 1i ; J+ | j, j − r − 1i = Nj,j−r | j, j − r i.

Using the su(2) algebra involving ladder operators, it is easy to deduce

2 2
Nj,j −r = h j, j − r | J+ J− | j, j − r i = Nj,j−r +1 + j − r,

giving a recursion relation amongst the coefficients whose solution turns out to be

1
q
Nj,m = √ ( j + m)( j − m + 1).
2

As the vector space is finite dimensional, we will also have a lowest eigenvalue state
| j, j − ki such that

J− | j, j − ki = 0.

Clearly, Nj,m = 0 when m = j + 1 (highest value state | j, ji) and m = j − k = − j (lowest

value state | j, j − ki) which means k = 2j. The highest value j can be also determined
by acting J.J = J32 + J+ J− + J− J+ on| j, mi:

J.J | j, mi = ( J32 + J+ J− + J− J+ )| j, mi = j( j + 1)| j, mi.

Notice that J.J commutes with the all the three su(2) generators.

[ J.J, J3 ] = [ J.J, J+ ] = [ J.J, J3 ] = 0.

Such quantities are called Casimir operators.

We will see that the above discussion on su(2) algebra and their action on
basis states of finite dimensional vector space V will be useful when we formally
introduce root vectors and study su(2) subalgebras constituting compact semi-simple
Lie algebras in Section 5.7.

By exponential mapping, the special unitary group SU (2) group elements will be

g(θ) = exp(iθ.J ),

whose parameters are similar to that of group SO(3) but there is a subtle difference.
For the fundamental representation J = σ/2, using the properties of the Pauli
matrices, these group elements can be written as

g(θ n̂) = cos(θ/2)I + i n̂.σ sin(θ/2),

which implies g(2π ) 6= g(0), suggesting that the parameter space will not be a solid
sphere of radius π. Instead, the SU (2) parameter space is a solid sphere of radius
2π. Further, all points on the boundary of the solid sphere are identified making
this SU (2) parameter space simply connected. In other words, there will be only one
class of closed curves in the parameter space. This subtle difference in the parameter
space also indicates that the group SO(3) (Figure 5.3.1) is not isomorphic to group
SU (2). That is, g(θ n̂) and g((θ + π )n̂) are mapped to R(θ n̂) ∈ SO(3) giving two to
one mapping (homomorphism). In the literature, SU (2) is refered to as a double cover
of SO(3) because of the above properties.

5.4 Semi-simple Lie Algebras

The explicit construction of the adjoint representation of a Lie algebra allows for
introduction of the Killing form on it. The advantage of the Killing form lies in that
it bears some resemblance to the concept of inner product on an inner product space.
One should bear in mind that this resemblance is exact only for a certain type of Lie
algebras. It is very convenient that such Lie algebras turn out to be quite useful in
physical applications.
Let g be a Lie algebra and ad its adjoint representation. For x, y ∈ g, the Killing
form κ ( x, y) is given by

κ ( x, y) = −tr(ad( x ) ad(y)), (5.4.1)

where tr stands for trace of the matrix products. Evidently κ ( x, y) = κ (y, x ). The
linearity of the map ad gives κ (αx + βy, z) = ακ ( x, z) + βκ (y, z) for scalars α, β and
x, y, z ∈ g. Killing form is therefore a symmetric bilinear form. If either x or y is 0,
then κ ( x, y) = 0.
Consider the set g⊥ = { x ∈ g |κ ( x, y) = 0 ∀ y ∈ g}. g⊥ is clearly a subspace of g as
follows from linearity of κ. The notation seems to suggest that g⊥ consists of vectors
orthogonal to the linear space g. However, this orthogonality is with respect to the
Killing form (5.4.1) and not in the usual sense of orthogonality in an inner product
space. It very well might be that g⊥ is not merely the null space and g⊥ ⊂ g. When
this is true, the Killing form κ is said to be degenerate.

Definition 8. A Lie algebra whose Killing form is non-degenerate is called semi-simple.

Let h be an invariant subalgebra (Definition 6) of a semi-simple Lie algebra
g. h⊥ = { x ∈ g |κ ( x, y) = 0 ∀ y ∈ h} is the orthogonal subspace of h in g. Also
suppose g ∈ g, h ∈ h and x ∈ h⊥ so that [ g, h] ∈ h and κ ( x, h) = 0. Then one has
κ ([ x, g], h) = tr (ad([ x, g])ad(h)) and from the property of trace and Equation 5.3.8,
κ ([ x, g], h) = κ ( x, [ g, h]) = 0. It follows that [ x, g] ∈ h⊥ and since g is arbitrary in
g, h⊥ is an invariant subagebra in g. The reader can easily verify that h ∩ h⊥ is also an
invariant subalgebra in g. Since κ ( x, y) = 0 for x, y ∈ h⊥ ∩ h, it follows that for any
g ∈ g, κ ( g, [ x, y]) = κ ( x, [y, g]) = 0 requires h⊥ ∩ h is an abelian subalgebra because
κ is non-degenerate for g.
A basis { xi }ik=1 for the subspace h ∩ h⊥ can be chosen which can be extended to
a complete basis { xi }ik=+1l for g. { x j }kj=
+l
k+1 is then a basis for the complement of the
subspace h ∩ h⊥ in g. As h ∩ h⊥ is an abelian subalgebra, it immediately follows that
the matrix of the operator ad ( g) ad ( x ) is of the form

0k × k Bk×l
0l × k 0l × l

for any x ∈ h ∩ h⊥ and g ∈ g. The size of the various 0-blocks and the matrix B
has been indicated by subscripts. From the above it follows that tr (ad( g)ad( x )) =
κ ( g, x ) = 0 for all g ∈ g. Because κ is non-degenerate by assumption, the only possible
conclusion is x = 0 and one has h ∩ h⊥ = {0}. Hence, for semi-simple Lie algebra, it
can be shown that g = h ⊕ h⊥ .
Suppose {hi }im=1 is a basis of the subspace h. Any x ∈ h can be expressed
x = ∑im=1 ci hi where ci are real or complex (depending upon whether g is a real or
complex vector space). Thus every x ∈ h corresponds to an m-tuple of numbers. Let
the collection of all such m-tuples be called F m . Here F stands for the field R or C. With
component wise addition and scalar multiplication of the m-tuples, F m is a vector
space which has the same dimension as the subspace h. Consider a map T (Section
3.1) from g into F m such that for g ∈ g,

T ( g) = (κ ( g, h1 ), . . . , κ ( g, hm )).

T is then a linear transformation which is also a group homomorphism (Section 1.5)

between the abelian group structures on g and F m . The kernel of this homomorphism,
Ker( T ) consists of g ∈ h such that κ ( g, hi ) = 0 for all i ∈ {1, . . . , m}. In other words,
Ker( T ) = h⊥ . Therefore the quotient group g/h⊥ is isomorphic to the image of g in F m
under the mapping T. The reader can easily infer that the group g/h⊥ is a vector space
of dimension equal to dim g–dim h⊥ . Since g/h⊥ is a subspace in F m , it’s dimension
can be no more than m. Thus one has the inequality

dim g ≤ dim h + dim h⊥ .

Since the only common subspace of h and h⊥ is the null space, dim g = dim h + dim h⊥
and one has g = h ⊕ h⊥ . Also, the restriction of κ to h is non-degenerate and h
is semi-simple. For emphasis, we sum up the arguments of this paragraph as the
following:

Theorem. If h is an invariant subalgebra of a finitely dimensional semi-simple Lie algebra

g and h⊥ is the orthogonal complement of h with respect to the Killing form on g, then g =
h ⊕ h⊥ . Additionally, h is semi-simple and h⊥ is an invariant subalgebra.

It is clear from the above that a semi-simple Lie algebra can have no non-trivial
invariant abelian subalgebras. Also note the following

Definition 9. A simple Lie algebra that has no non-trivial invariant subalgebras is a

semi-simple Lie algebra.
It is an obvious consequence of the above theorem that a semi-simple Lie algebra
is a direct sum of simple Lie algebras.

5.5 Lie Algebra of a Lie Group

Definition 10. A Lie group is a group with the structure of an analytic manifold in
which group multiplication and inversion are analytic functions of their arguments.
The precise concept of an analytic manifold will not be made here. The analytic
nature of the group product will be central in subsequent development. The fact
that a Lie group is an analytic manifold leads to the consequence that any small
neighborhood of the identity of the group has the structure of a vector space. In
the example of the circle group, this vector space is one-dimensional and in the
exponential representation exp(iωt), it is spanned by the basis [ω ]. Generally, the
number of parameters needed to span the local vector space near the group identity
may be more than one. It is assumed that this number is always finite. Let G be a
Lie group whose elements in the vicinity of identity can be parametrized by n real
numbers tα† . Then G is said to be of dimension n. The identity itself is parametrized
so that each of tα is set to zero. For two group elements g1 and g2 parametrized by the
sets of real quantities t1α and t2α respectively, the parameters tα of their product g = g1 g2
are analytic functions of t1α and t2α . Suppose

t α = f α ( t1 , t2 ),

where t1 and t2 represent the complete sets of n parameters of g1 and g2 respectively.

Letting t1 = 0 or t2 = 0 to be parameters for the identity element, the condition on f α
obtained is

tα = f α (t, 0) = f α (0, t).

† The superscript α should not be confused as an exponent.

With this condition, and the fact that the functions f α are analytic functions of t1 and
t2 , a Taylor series for f α may be written as

1
2∑
α γ δ
tα = f α (t1 , t2 ) = t1α + t2α + f γδ t1 t2 + . . . (5.5.1)
γ.δ

α are derivatives of f α at t = t = 0. If Γ is a unitary representation of G

where f γδ 1 2
(assuming such a representation exists) and g an element of G close to identity then
the representing matrix Γ( g) of g is an analytic function of the parameters tα for g.

1
Γ ( g ) = I + i ∑ t α Xα −
2∑
tα t β Xαβ + . . . (5.5.2)
α α,β

where Xα , Xα X β etc are matrices of the same order and do not depend on the value
of the real parameters tα . Suppose g = g1 g2 , one has Γ( g1 )Γ( g2 ) = Γ( g). The
parameters of g are given by Equation 5.5.1. Similar expressions for Γ( g1 ) and Γ( g2 )
in combination with Γ( g1 )Γ( g2 ) = Γ( g) give
" #
1
I + i ∑ t1α Xα − ∑ t1α t1 Xαβ + · · · · · · ×
β

α 2 α,β
" #
1
I + i ∑ t2α Xα − ∑ t2α t2 Xαβ + · · · =
β

α 2 α,β
" #
1
I + i ∑ t1 + t2 + ∑ f γ,δ t1 t2 . . . Xα −
α α α γ δ
α 2 γ,δ
" #
1 1
2∑
t1 + t2 + ∑ f γ,δ t1 t2 . . . ×
α α α γ δ
α,β
2 γ,δ
" #
1
t1 + t2 + ∑ f γ,δ t1 t2 . . . Xαβ + . . .
β β β γ δ
2 γ,δ

Expanding the above equation, the reader should observe that the leading terms
remaining after cancellations are quadratic in $. For reference, an intermediate step
is
i 1
− ∑ t1α t2 Xα Xβ = ∑ t1 t2δ f γ,δ Xα − ∑[t1α t2 + t2α t1 ] Xαβ .
β γ α β β

α,β
2 α,γ,δ 2 α,β

Here one may notice that the Xαβ = X βα since these are coefficients in the Taylor
expansion Equation. 5.5.2. After relabelling the indices in the first term in the right
hand side of above and using the fact that tiα are independent, one has

i
∑
γ
− Xα X β = f αβ Xγ − Xαβ . (5.5.3)
2 γ

Interchanging α and β in above equation and subtracting gives

i
∑
γ γ
[ Xα , X β ] = ( f βα − f αβ ) Xγ , (5.5.4)
2 γ

where the [ Xα , X β ] stands for the commutator of Xα and X β . The reader will
immediately recognize the above equation as the commutation relation in a Lie algebra
(Equation 5.3.1) with the structure constants cαβ = 12 ∑γ ( f βα − f αβ ). The structure
γ γ γ

γ
constant cαβ is clearly antisymmetric with respect to interchange of indices α and β.
In the other direction, if the structure constant of this Lie algebra are known, then
Equation 5.5.3 allows us to calculate Xαβ and consequently, the representing matrix
for any group element near the identity up to a precision of second order by use of
Equation 5.5.2.
A more general result is that in fact, all the matrices in the infinite series in the right
hand side of Equation 5.5.2 are known from the commutations in Equation 5.5.4. This
result will not be proven here. An immediate consequence is that knowledge of Xα
and its commutations are sufficient to calculate Γ( g) for all g in a finite neighborhood
of the identity. For this reason all the constituents of Xα are called the generators of
the group and the vector space spanned by generators is the Lie algebra g associated
with the group G. A Lie group whose Lie algebra is real is called a real Lie group and
similarly for complex Lie group corresponds to Lie algebra which is complex. In the
following section, we will discuss examples of Lie groups.

5.6 Examples of Lie Groups

Important examples of Lie groups are the general linear groups and some of their
subgroups. These are Lie groups which can be faithfully represented by matrices of
finite size, and so are called matrix Lie groups.
The general linear group of degree n is the group of invertible n × n matrices with
the group operation being the usual multiplication of matrices. The general linear
group of degree n whose representing matrices have only real entries is denoted by
GL(n, R). In the complex case the notation is GL(n, C). The group GL(n, R) is itself
a subgroup of GL(n, C). Regarded as Lie groups on their own, there is an important
difference between GL(n, R) and GL(n, C). GL(n, R) consists of invertible matrices
whose determinants are non vanishing. Therefore these determinants can be either
positive valued or negative valued.
For M1 and M2 in GL(n, R) such that det M1 > 0 and det M2 < 0, there
cannot then exist a continuous real parameter smoothly connecting M1 to M2 . On the
otherhand, GL(n, C) does not suffer from this discontinuity as it is always possible to
find a path in the complex plane. GL(n, C) is therefore an example of a connected Lie

group. GL(n, R) can be decomposed into two components. One component consists
of matrices with positive determinant while the other of matrices with negative
determinant. Each component is connected in itself but disconnected from the other.
The identity matrix of GL(n, R) lies in the positive component.
Special linear groups are subgroups of general linear groups. The corresponding
notations are SL(n, R) and SL(n, C). These groups consist of matrices whose
determinants equal unity. As before, SL(n, R) is a subgroup of SL(n, C). Both the
groups are connected Lie groups.
The orthogonal group O(n) is a subgroup of GL(n, R). The defining condition for a
matrix M = [mij ]n×n to be in O(n) is that

MT M = MMT = I. (5.6.1)

Upon explicitly writing out the product MT M, the above condition reduces to
∑in=1 mij mik = δjk . In other words, the columns of any matrix in O(n) are orthonormal.
The same can be inferred about the rows. From the defining Equation 5.6.1 it
also follows that the determinant of the matrices in O(n) can be either +1 or −1.
The subgroup of O(n) consisting of matrices whose determinants equal unity is
called the special orthogonal group SO(n). The group SO(n) essentially consists of
transformations that preserve handedness (rotations), while in O(n) there are also
improper transformations like reflections which do not preserve handedness.
The generalized orthogonal group O(m, n) is another subgroup of GL(m + n, R).
Define g to be a (m + n) × (m + n) matrix in which the the first m entries in the
principal diagonal are +1, the last n entries in the principal diagonal are −1 while
all the other entries are 0. A real matrix M is in O(m, n) if and only if

MT gM = MgMT = g. (5.6.2)

It is obvious from the above definition that det M = ±1. The subgroup of O(m, n)
which contains only matrices of unit determinant is the generalized special orthogonal
group SO(m, n). In fact, the Lorentz group is the group SO(1, 3). As this group is the
symmetry group of physical systems respecting the laws of special theory of relativity,
we will elaborate the representations of Lorentz group in the next chapter.
The symplectic group Sp(2n, R) is another subgroup of SL(2n, R). The defining
equation of a symplectic matrix M2n×2n ∈ Sp(2n, R) is

MT JM = J,

where
" #
0n × n In×n
J= .
− In×n 0n×n

Symplectic groups find applications in the transformation properties of the position

and velocity variables describing classical systems. Now turning to GL(n, C), the
subgroup consisting of such matrices M for which

M† M = MM† = I (5.6.3)

is called the unitary group U (n). The circle group U (1) was the first example of this
type. It is obvious from the above definition that determinants of matrices in U (n)
are unimodular complex numbers. The subgroup of U (n) containing matrices whose
determinants equal unity is call the special unitary group SU (n). The group SU (2)
plays an important role in description of particle spin in quantum mechanics. Hence
their applications in quantum mechanics and particle physics will be discussed in the
following chapter.
We will now discuss the classification of semi-simple Lie algebras known in the
literature as Cartan classification.

5.7 Compact Simple Lie Algebras

In Section 5.5, we came across the fact that the Lie algebra of a Lie group can reveal
the structure of the group in a finite neighborhood of the identity element. Here, we
will focus on the broad classification of simple Lie algebras.
The works of Killing and Cartan led to the classification of simple Lie algebras
g to be of four infinite types and five exceptional types. We will now present the
definitions that govern operations followed by the n basis elements of Lie algebra g
without proofs.
Let ` independent elements { Hi } ∈ g obey abelian subalgebra:

[ Hi , Hj ] = 0; i, j = 1, . . . `

which is known in the literature as Cartan subalgebra. The number of commuting

elements (`) of the algebra g is called the rank of the algebra.
The remaining n − ` elements E±α ∈ g are even in number. Note that the subscript
±α are `-component vectors which are referred to as positive and negative root vectors
of the algebra. Hence, there will be n − ` non-zero root vectors associated with the Lie
algebra g. The algebra involving E±α will yield

`
[ Eα , E−α ] = ∑ αi Hi ,
i =1

where αi refers to the i-th component of the root vector α. If α and β are two different
root vectors of g then

[ Eα , Eβ ] = Nα,β Eα+ β ,

where Nα,β = −N β,α 6= 0 if and only if α + β is also a root vector of g. Further, for
complete definition of the algebra g, we need the following commutator as well:

[ Hi , Eα ] = αi Eα ; [ Hi , E−α ] = −αi E−α .

Note that the above commutator bracket resembles ladder operators and J3 of su(2)
algebra (5.3.10). As su(2) algebra has only one diagonal generator, the rank of the su(2)
algebra is ` = 1 leading to one-component root vectors (numbers). The comparison
of the formal g algebra with su(2) implies that there are two non-zero roots ±α = ±1
and hence the corresponding generators E±1 ≡ J± ·
For a general Lie algebra g, we can construct many su(2) subalgebras in the
following way:

(α) (α)
J± = |α|−1 E±α ; J3 = |α|−2 α.H.

Similar to the su(2) eigenbasis | jmi of J3 with half integer values m, we have eigenbasis
(α)
|Λ, µi ∈ V of J3 where Λ is the highest weight vector and µ denotes ` component
weight vector with the following properties:

(α) α.µ (α)

J3 |Λ, µi = |Λ, µi ; J± |Λ, µi ∝ |Λ, µ ± αi
| α |2
(α)
where the eigenvalues of J3 are half integers. Hence,

2α.µ
= integer.
| α |2

This integer can be determined by finding p, q such that

(α) (α)
( J+ ) p+1 |Λ, µi = 0 ; ( J− )q+1 |Λ, µi = 0, (5.7.1)

(α)
implying that the eigenvalue of J3 is maximum (+ j) for state |Λ, µ + pαi and
minimum (− j) for state |Λ, µ − qαi. That is,

α.(µ + pα) α.(µ − qα)

2
=j=− ,
|α| | α |2

which implies

2α.µ
= ( q − p ). (5.7.2)
| α |2

Comparing with the su(2) ladder operations, we can say highest weight state |Λ, Λi is
defined as
(α)
J+ |Λ, µ = Λi = 0 ∀ α.

Suppose we choose the vector space V = g (adjoint representation), then the

(α)
eigenbasis of the J3 can be represented by {| Hi i}, {| E± β i}. In fact, for adjoint
(α) ( β)
representation, the weight vector of states are root vectors. Applying J3 , J3 on such
states will give the following eigenvalue equations:

(α) α.β ( β) α.β

J3 |Λ, E+ β i = |Λ, E+ β i ; J3 |Λ, E+α i = |Λ, E+α i.
| α |2 | β |2

Using the two equations (Equation 5.7.1 and 5.7.2) for the above states, we can infer:

2α.β 2β.α
2
= ( q − p ), = ( q 0 − p 0 ),
|α| | β |2

suggesting that the angle between two root vectors must obey:

(q − p)(q0 − p0 )
cos2 θα,β = ,
4

where p, q, p0 , q0 are integers. This forces the non-trivial angle between two root
vectors to assume values of 900 , 600 or 1200 , 450 or 1350 , 300 or 1500 .
If the rank of the Lie algebra g is `, then there are ` simple roots. The remaining
positive and negative roots can be obtained from linear combination of the simple
roots.There is a neat concise way of classifying simple Lie algebras g using connected
graphs called Dynkin diagrams. Suppose we denote the ` simple root vectors by `
filled circles. We connect these ` filled circles by single bond or double bond or triple
bond depending on the angle between any two simple roots α, β. In fact,

4 cos2 θα,β = m,

where m is the number of bonds connecting the simple roots α, β. For instance, the
Lie algebra A` ≡ su(` + 1) has ` simple roots of equal length (α(1) , α(2) , . . . α(`) ) and
has a single bond connecting adjacent simple roots as shown in Figure 5.7.1. That is,

4 cos2 θα(i) ,α(i±1) = 1.

Besides the A` infinite series Dynkin diagram depicting unitary algebras su(` + 1)
of arbitrary ranks, we have other connected graphs corresponding to orthogonal
algebras so (n) which belong to infinite series Dynkin diagrams B`≥3 for n = 2` + 1

and D`≥4 for n = 2` as shown in Figure 5.7.1. Note that the last two simple roots in
B` are connected by double bond and the orientation of the arrow indicates that the
length of the root vectors obey |α(i) | > |α(`) | for all i < `. The Dynkin diagram of the
symplectic algebra C`≥2 ≡ sp(2`) has a reversed orientation implying |α(i) | < |α(`) |
for all i < `. Besides these connected graphs of arbitrary ranks (see Figure 5.7.1(a)),
we can have finite rank Dynkin diagrams which are known in the literature as five
exceptional Lie algebras (as shown in Figure 5.7.1(b)). Basically, these four infinite
series and five exceptional series are the only allowed connected graphs for linearly
independent simple roots which can represent simple Lie algebras g.

G2
su( + 1) =

so(2 + 1) = ³3

Usp(2 ) = ³2

so(2 ) = ³4
E8

(a) Infinite series (b) Exceptional series

Figure 5.7.1 Dynkin Diagrams

With the formal definition of simple Lie algebras where we introduced root and
weight vectors, we will now extensively discuss A`=2 ≡ su(3) algebra.

5.7.1 su(3) algebra

Just like the 2 × 2 σ matrices are the fundamental representation of su(2) acting
on a two-dimensional complex vector space V, we have 3 ×3 Gell-Mann matrices
{λ a }’s which are referred to as the defining representation of su(3) acting on
three-dimensional complex vector space V. These Gell-Mann matrices are Hermitian
and traceless. Hence, there will be 8 λ a ’s denoting the basis elements of su(3) algebra
because hermiticity and traceless conditions reduce the nine complex entries of these
3 × 3 matrices to eight independent real entries. The explicit form of the Gell-Mann
matrices are
     
0 1 0 0 −i 0 1 0 0
λ1 =  1 0 0 , λ2 =  i 0 0 , λ3 =  0 −1 0 
0 0 0 0 0 0 0 0 0

     
0 0 1 0 0 −i 0 0 0
λ4 =  0 0 0 , λ5 =  0 0 0 , λ6 =  0 0 1 
1 0 0 i 0 0 0 1 0
   
0 0 0 1 0 0
1
λ7 =  0 0 −i , λ8 = √  0 1 0 .
0 i 0 3 0 0 −2

Note that there are two diagonal matrices λ3 , λ8 which constitute the Cartan
subalgebra of rank ` = 2. Hence the three basis states
    
1 0 0
|Λµ1 i =  0 , |Λµ2 i =  1 , |Λµ3 i =  0  .
0 0 1

of the fundamental three-dimensional complex vector space V are simultaneous

eigenstates of both H1 = λ3 /2 and H2 = λ8 /2. We have indicated both the
eigenvalues in the two-dimensional H1 − H2 diagram (Figure 5.7.2) for the three basis
states. Hence, we can equivalently represent the basis state as two-component weight
vectors:
√ √ √
|Λµ1 i ≡ (1/2, 3/6); |Λµ2 i ≡ (−1/2, 3/6); |Λµ3 i ≡ (0, − 3/3).

The weight vectors are called negative weight vectors if the first non-zero component
is negative. In the above set of weight vectors, µ1 is positive weight vector whereas
µ2 , µ3 are negative weight vectors.

Ö 1 , Ö3
– 1, 3
2 6 2 6

Ö3
0, –
3

Figure 5.7.2 su(3) Weight Diagram for Defining Representation

The raising and lowering operators will take one basis state to another basis state.
The explicit form of such operators will involve the remaining six Gell-Mann matrices
in the following way:

1 1
E±α(3) = √ (λ1 ± iλ2 ) ; E±α(1) = √ (λ4 ± iλ5 );
2 2 2 2
1
E∓α(2) = √ (λ6 ± iλ7 ),
2 2

where

E−α(3) |Λ, µ1 i ∝ |Λ, µ1 − α(3) i = |Λ, µ2 i ; E+α(i) |Λ, µ1 i = |Λ, µ1 + α(i) i = 0 ∀ i, (5.7.3)

implying that the weight vector |Λ, µ1 i is the highest weight state (that is., Λ = µ1 and
the root vector

α(3) = µ1 − µ2 = (1, 0).

In the similar fashion, you can check that

√ √
α(1) = µ1 − µ3 = (1/2, 3/2), α(2) = µ3 − µ2 = (1/2, − 3/2). (5.7.4)

Note that α(3) = (1, 0) = α(1) + α(2) indicating that the two simple roots of the su(3) ≡
A2 algebra are α(1) , α(2) whereas α(3) is not a simple root vector. These non-zero root
vectors can be plotted in the H1 − H2 graph (see Figure 5.7.3 where the right side of
the graph has positive roots and left side of the graph has negative roots) which are
the non-zero weight vector states of the adjoint representation which also represent
the raising and lowering operators of su(3) algebra. The origin of the graph are the
null root vectors denoting the Cartan subalgebra operators H1 , H2 .

– a1 a1

– a1 – a2 a1 + a2

– a2 a2

Figure 5.7.3 su(3) Root Diagram

5.7.2 Cartan matrix

For a Lie algebra of rank `, there will be ` simple roots. One can construct the Cartan
matrix

2α(i) .α( j)
Aij =
| α (i ) |2

from the Dynkin diagram. For A2 ≡ su(3), the Cartan matrix elements using the
simple root vectors (5.7.4):

2 −1
.
−1 2

5.7.3 Fundamental weights

We will have ` fundamental highest weight vectors µ(i) ’s such that

2α(i) .µ( j)
= δij . (5.7.5)
| α (i ) |2

Example 48. For su(3) algebra, determine the two fundamental highest weight
vectors.
Let us take the 2-component fundamental highest weight vector as (u, √ v), Using
the orthogonality property (5.7.5) with simple roots (5.7.4), we get u = ± 3v. The
normalized form of the two fundamental highest weight vectors are
√ √
µ(1) = (1/2, 3/6) ; µ(2) = (1/2, − 3/6).

Recall µ(1) (5.7.3) is the highest weight vector of the defining representation which we
discussed using Gell-Mann matrices. For arbitrary representation of su(3), the highest
weight vector will be

Λ ≡ µ = aµ(1) + bµ(2)

where a, b are positive integers. For example, the highest weight vector of the adjoint
representation is α(1) + α(2) = µ(1) + µ(2) = (1, 0).
Armed with the formal aspects of Lie groups and Lie algebras discussed in this
chapter, we will focus on their applications in the following chapter. Also, the tensor
product of irreducible representations of su(2) and su(3) algebra will be discussed in
detail.

Exercises
1. We have discussed that the group U (2) has four independent parameters. Suppose we
impose that the determinant of these unitary matrices to be +1, then we call the group
as SU (2). What will be the number of independent parameters for SU (2) group.
2. Determine the number of independent parameters (equal to number of generators) for
a general U ( N ) group and SU ( N ) group.
3. For a Lie algebra g in which the structure constants are ckst as in Equation 5.3.1, show
that
n
∑ [clsk cktu + cltk ckus + cluk ckst ] = 0
k =1

[Hint: Jacobi identity]

4. Let g be a semi-simple Lie algebra and h an invariant subalgebra in g. Show that
(h⊥ )⊥ = h.
5. For vectors x and y in Rn , define the product of x and y as
n
x.y = ∑ xi yi .
i =1
Show that under a transformation M in O(n), the value of the product is invariant, i.e.,
x.y= Mx.My.
6. For vectors x and y in Rm+n , define the product of x and y as
m n
x.y = ∑ xi yi − ∑ x m+ j y m+ j .
i =1 j =1
Show that under a transformation M in O(m, n), the value of the product is invariant.
7. For vectors x and y in an n-dimensional complex vector space, define the product of
x and y as
n
xy = ∑ xi yi
i =1
where xi represents the complex conjugate of xi . Show that under the transformations
in U (n), the value of the product is invariant.
8. Given a Lie algebra g having two simple roots
√
α(1) = (0, 1), α(2) = ( 3/2, −3/2),

work out the Cartan matrix and the corresponding Dynkin diagram. Also determine
their fundamental highest weight vectors.

With the formal aspects of Lie algebra and Lie groups discussed in the previous
chapter, it is now possible to apply the power of continuous symmetry to physics
of many simple and complex systems. This chapter will focus upon the applications
of Lie algebra and groups to physical systems possessing orthogonal group or
unitary group symmetry. For completeness, note that the symplectic group sp(2n)
is nothing but the canonical transformation of 2n phase space variables (position and
velocity variables). Readers are advised to look up canonical transformations in any
classical mechanics textbooks such as Goldstein for further appreciation of the natural
emergence of symplectic group symmetry in classical systems.
Many classical systems in three-dimensional space respect spherical symmetry.
Such systems remain unchanged under proper rotations. Noether’s theorem states
that ‘associated with any continuous group symmetry, there is always a constant
of motion’. In Section 6.1, we will briefly discuss classical systems possessing
translational symmetry and show that the constant of motion is linear momentum p.
We have extensively discussed, in the previous chapter, the rotation group SO(3)
and its Lie algebra so(3) whose generators are the three components of orbital angular
momentum L. Thus the generators of the group are nothing but constants of motion
in the classical system respecting rotational symmetry. The rotational invariance of
quantum mechanical systems leads to a natural emergence of group SU (2) extensively
presented in the previous chapter. Particularly, the Lie algebra generators are the
components of the total angular momentum operator Ĵ = L̂ + Ŝ incorporating the
intrinsic spin Ŝ in accordance with the famous Stern–Gerlach experiment. Any basic
course on the quantum mechanics of rotational invariant systems is based on the
theory of angular momentum algebra which is isomorphic to su(2).

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116 Group Theory for Physicists

In the previous chapter, we have discussed the actions of Lie algebra generators
E±α , associated with a root vector α, on the highest weight vector Λ leading to weight
vectors Λ ± α. Through su(2) as an example (Chapter 5, Section 5.3.1), we showed
that angular momentum corresponds to the highest weights in the Lie group context
and the the range of z–components of angular momentum are the weights. In Section
6.2, we will construct the tensor product of two or more angular momentum states
using the Young tableau approach. In quantum mechanics literature, they are known
as addition of angular momentum. In Subsection 6.2.1, we have a small digression on
Young Tableau approach for both symmetry group of degree n and SU ( N ) group.
Then, we will discuss selection rules for the quantum mechanical transition from initial
state to final state due to interactions. The important theorem known as Wigner–Eckart
theorem will be dealt with proof. Through examples, readers will come to appreciate
the power and elegance of SU (2) group theory tools in validating experimentally
observed allowed and forbidden transitions in quantum systems possessing rotational
symmetry. We also remind the readers to compare the selection rules due to discrete
symmetry discussed in Chapter 4 to the Wigner–Eckart theorem rules discussed here.
This theorem is applicable to many other systems possesing SU (2) symmetry.
For instance, theoretical validation of (i) almost same mass mp ≈ mn of proton and
neutron (ii) experimentally observed strong interaction processes (scattering and
decay) involving such elementary particles (broadly classified as baryons and mesons)
are deduced invoking SU (2) Lie group symmetry. For baryons and mesons, SU (2)
represents rotational symmetry in an abstract internal space called isospin space whose
algebra is same as su(2). We discuss these features in elementary particle physics in
Section 6.3.
Interestingly, these elementary particles with same angular momentum J but
with different isospin I and charge Q can be plotted on a two-dimensional diagram
which appears identical to the su(3) weight diagrams discussed in the previous
chapter. Historically, Gell-Mann introduced a simple model called the quark model to
theoretically predict the baryons and mesons observed in the laboratory. According to
the quark model, particles like protons, neutrons and other baryons must be composed
of three quarks. Assuming that there are three flavors of quarks: u, d, s (up, down
and strange), Gell-Mann applied the unitary group SU (3) and tensor product of
quarks to account for baryons observed in experiments. Interestingly, he predicted
that the SU (3) symmetry requires the presence of a baryon Ω− which was detected
experimentally three years later. This is one of the instances where an abstract theory
was validated by the experiment after some years. We discuss the quark model in
Section 6.4. Similar to the symmetry breaking discussed in the discrete group context,
we will briefly present symmetry breaking in the context of continuous groups. The
SU (3) quark model can be generalized to SU ( N ) assuming quarks come in N flavors
using Young diagrams and their tensor product discussed in Subsection 6.2.1.
Besides compact groups, there exist non-compact groups whose Lie algebra can
be systematically deduced. In Section 6.5, we will elaborate on one such group called
Lorentz group : SO(3, 1). One of the reasons for focussing on this Lorentz group is
because it is the symmetry possessed by relativistic particles in the 3 + 1 dimensional

Minkowski space-time. We will briefly discuss Poincare algebra and Poincare groups
which serve as a underlying symmetry of many physical systems. Many systems near
criticality (near phase transition point) are invariant under scale transformation. The
group symmetry possessed by such systems are known as conformal groups. We will
give an overview of the conformal group symmetry and their generators. Finally, the
readers can see through the exercise problem, the resemblence between the Lorentz
algebra so(3, 1) and so(4) associated with rotations in four-dimensional space.
Even though four-dimensional space is not physical, we will show in Section 6.6
that the abstract group SO(4) has an elegant way of reproducing familiar hydrogen
atom energy levels without solving the Schrödinger equation. Besides angular
momentum L, the hydrogen atom has another constant of motion called Runge–Lenz
vector M. The abstract group SO(4) generators and their algebra turns out to be
isomorphic to the algebra involving L and M. Unlike angular momentum L which can
be attributed to geometrical transformation of rotations, M has no such geometrical
interpretation. Such a symmetry which has no geometrical interpretation is known as
dynamical symmetry.

6.1 Continuous Symmetry and Constant of Motion

We will focus on the familiar translation symmetry to formally understand the
corresponding constant of motion (Noether theorem). Further, we will also examine
representations of these translation group elements and deduce the generators of such
transformation. This detailed description will show that the constants of motion act
as generators of the corresponding group transformation.

6.1.1 Translational symmetry in three-dimensional space

At the start, we will take the simplest symmetry of a system under the space
translation operation. Consider a free particle moving in our three-dimensional space.
The classical Hamiltonian H = p2 /(2m) is unchanged under translation x → x + a
for any constant shift vector a. For such a translationally invariant classical system, it
is easy to show that the Hamilton’s equation of motion involving Poisson bracket {, }
(see Classical Mechanics by Goldstein) is

dp
= { H, p} = 0,
dt
which implies linear momentum p is a constant of motion for translationally invariant
systems.
In quantum mechanics, the free particle wavefunction ψ(x) under translation
operation T̂(a) transforms as follows:

T̂(a)ψ(x) = ψ 0 (x),

where wavefunctions obey ψ(x) = ψ0 (x + a) for translation invariant systems. This

implies ψ0 (x) = ψ(x − a). Substituting this functional form in the above equation and
writing as a Taylor series expansion, we can deduce the form of the operator from

T̂(a)ψ(x) = exp(−a.∇)ψ(x).

The differential operator is related to the well-known momentum operator, p̂ = −i}∇,

leading to the explicit form of the translation operator as

−ia.p̂

T̂(a) = exp .
}

For an infinitesimal translation (a small), for the above operator the expansion up to
O(a) will be

T̂(a) ≈ I − ia.p̂/} + O(a2 ).

In fact, as discussed in the previous chapter on Lie groups, the deviation from
identity I operator generates the infinitesimal translation operation. Thus for spatial
translations, linear momentum p̂ is the generator and a ∈ R3 is the parameter in
three-dimensional space. We have explicitly shown that the constant of motion is
indeed the linear momentum generating translational symmetry. This methodology
can be applied to time translations as well where the generator turns out to be the
Hamiltonian H and the corresponding time translation parameter τ ∈ R belongs
to a real line. We have already seen rotational symmetry groups SO(3) and their
Lie algebra so(3) whose generators are the orbital angular momentum L in the
three-dimensional space. For completeness, rotational symmetry and corresponding
constant of motion is presented through the following example.

Example 49. For a rotationally invariant system described by the Hamiltonian

p2 k
H= + ,
2m r2
where r is the radial coordinate and constant k is a dimensionful constant. Show that
dL/dt = 0, confirming that the generators L are constants of motion of such rotational
invariant systems.
The Poisson bracket

∂H ∂Li ∂H ∂Li k 1
{ H, Li } = ∑ − = ∑ ∈ilj x x
j l − p p = 0.
j
∂x j ∂p j ∂p j ∂x j j,l
r3 m j l

Hence the Poisson equation of motion for dL/dt is zero indicating that the angular
momentum is a constant of motion for rotational invariant systems.

In the context of discrete groups, the tensor product of two or more polar vectors
and their decomposition as binary basis, tertiary basis and so on, were dealt with
using character tables and projection operators. Interestingly, there is an equivalent
systematic procedure to deal with the tensor product of vectors and higher rank
tensors, leading to irreducible tensors of SO(3). As the algebra so(3) is same as
su(2) algebra, the construction of a tensor product in the SU (2) context turns out to
be applicable to SO(3) as well. Hence in the following section, we will elaborate on
this tensor product aspect using the irreducible angular momentum states | j, mi ∈ V j
(vector space whose dimension is d j = 2j + 1 as m ∈ [− j, − j + 1, . . . j]) and the action
of su(2) generators J± , Jz discussed in the previous chapter. This is the main core of the
addition of angular momentum and the construction of irreducible states in quantum
mechanics.

6.2 Tensor Product Rule for SU (2) Irreducible Representations

Let us take the tensor product of two vector spaces V j1 ⊗ V j2 whose decomposition
into irreducible components can be formally written as

∑ Nj1 ,j2 V j ,
j
V j1 ⊗ V j2 = (6.2.1)
j

j j
where it can be shown that Nj ,j = 1 when | j1 − j2 | ≤ j ≤ j1 + j2 and Nj ,j = 0 for the
1 2 1 2
other j’s. Basically, the proof for the allowed range of j and minimum value of j = jmin
is based on taking maximum j = jmax = j1 + j2 and constraining the dimension of the
vector space to obey

jmax
∑
j
d j1 .d j2 = Nj dj.
1 ,j2
j= jmin

There is an equivalent elegant Young tableau approach of obtaining such a tensor

product decomposition. We present this diagrammatic method where we can directly
determine d j as well as visualize decomposition.

6.2.1 Digression on the Young tableau approach

Symmetric Group S(n)
Recall the definition of the Young tableau diagram which we reviewed in the context
of cycle structure of the elements of the symmetric group S(n). We can also use the
set of integers λ1 ≥ λ2 . . ., associated with a Young diagram Y to denote an irreducible
representation of S(n). Here λ1 denotes the number of boxes in first row, λ2 being the
number of boxes in second row and so on such that ∑ λi = n. Note that n entries
i

in these boxes must be distinct for a permutation group. The diagram implies totally
symmetric property along any row and totally antisymmetric along any column.
For instance, an irreducible representation Y ∈ S(5) denoted by

1 2 3
4 5

with integer entries inside the boxes indicating that first row boxes are totally
symmetric under exchange. Similarly, the boxes on the second row are also totally
symmetric. The exchanges of boxes within the first column as well as exchange of
boxes within the second column are each totally antisymmetric. Note that there is no
symmetry between boxes numbered 3 and boxes numbered 5 belonging to a different
row and a different column.
The dimension dY of the irreducible representation Y can be determined by
counting the possible options of putting integer entries on the boxes such that they
are increasing along their row or their column. Such a prescription incorporates the
symmetric or antisymmetric or no-symmetry nature of the given Young diagram Y.
The following example will illustrate the prescription giving dimension dY .

Example 50. Determine the dimensions of the irreducible representation Y ∈ S(3)

whose Young tableau is

We can place integers in the boxes following the above mentioned prescription

1 2 1 3
, .
3 2

Thus there are only two possibilities indicating that the dimension dY = 2.
This procedure can be done for any Y ∈ S(n) but the method may become tedious
for large n. There is an alternative formula giving the dimensions dY :

dY = n!/ ∏ hi , (6.2.2)
i ∈Y

where hi denotes the hook length of a box i in the Young diagram Y. Basically, hi counts
the number of boxes to the right of the box i along the same row plus the number of
boxes below the box i along the column plus one (for the box i).
In the above example, the three hook lengths associated with the three boxes of Y
will be h1 = 3, h2 = 1, h3 = 1 resulting in dY = 3!/(3 × 1 × 1) = 2 in agreement with
Example 50.
The irreducible representations Y of S(n) has a close resemblance to the irreducible
representation of SU (n) but there are also differences. This naturally leads us to
discuss SU ( N ) group derived from the Young diagram approach.

SU ( N ) Young tableau
We will now briefly review the Young diagrams Y depicting irreducible SU ( N )
representations. Here N denotes the number of possible states which we can place in
every box of Y. Remember that the entry in each box of a Young diagram can be any of
the N values (with or without repetition) such that the properties of symmetric nature
along the horizontal direction and antisymmetric nature along the vertical direction
are maintained.
Suppose we try to place these states in a vertical column of N boxes, we have only
one possibility as the states along a column must be antisymmetric. Hence, one or
more vertical columns of N boxes denote a singlet or trivial representation. Unlike
the ∑ λi = n constraint in the Young diagram Y of the permutation group S(n),
i
there is no such restriction on the total number of boxes of Y denoting non-trivial
irreducible representation of SU ( N ). However, the number of rows must not exceed
N − 1 for non-trivial irreducible representation. The dimension dY for any SU ( N )
Young diagram can be determined through combinatorial consideration of symmetric,
antisymmetric and mixed symmetry properties of N variables.
Example 51. Obtain the dimension of the irreducible representation Y ∈ SU (3) whose

Young diagram is

The SU ( N = 3) rule here is to put three states u, d, s in the boxes such that
repetition along the row is allowed but forbidden along the column. Hence the
possible states are

u u u u d d d d s s s s u d u s
, , , , , , ,
d s u s u d s d

giving dimension dY = 8.
Alternatively, there is a formula involving hook number

dY = NrY / ∏ hi , (6.2.3)
i ∈Y
where the numerator NrY is calculated as follows: place N, N + 1, along the boxes in
the first row and decreasing those integers in steps of one along the vertical columns.
Then NrY is the product of all the integers in the boxes of Y.
3 4
The integer entries for the above example will be , giving NrY = 3 × 4 × 2 =
2
24 and its hook length as discussed in Example 50 is 3. Hence, we get dimension
dY = 24/3 = 8 which is in agreement with that obtained in Example 51.

Example 52. Deduce the Young diagram Y ∈ SU (2) corresponding to the irreducible
representation V j whose dimension is 2j + 1.
Recall that the non-trivial irreducible representation of SU (2) can be shown by
a single row Young diagram a1 a2 . . . am where ai ’s can be either u

a1
or d. Note that diagrams with one or more columns of two boxes are trivial
b1
one-dimensional representations because if a1 is chosen as u then b1 is necessarily d.
Hence, for an irreducible representation Y = a1 a2 . . . am of SU (2), the
dimension can be calculated using formula (6.2.3). Placing integers N = 2 in the first
box and increasing them one by one along the row, we can determine the dimension
to be

dY = NrY / ∏ hi = (2 × 3 × . . . m + 1)/(m)! = m + 1.
i ∈Y

This implies that the irreducible representation V j , corresponding to angular

momentum j, whose dimension is 2j + 1 can be presented as the following Young
tableau: a1 a2 . . . a2j .
This has been a quick overview of Young tableau presentation of irreducible
representation of SU ( N ). We can now understand the diagrammatic construction of
tensor product Y1 ⊗ Y2 of two irreducible representations and their decomposition ⊕Y.

Highest weight associated with SU ( N ) Young diagrams

In the previous chapter, we deduced the two SU (3) fundamental weight vectors from
the two simple root vectors. The fundamental weight vector µ(1) is the highest weight
state
of representation and the fundamental weight µ(2) vector is the highest weight

state of representation . Any arbitrary representation of SU (3) will have a1

single boxes and a2 double vertical boxes. For example, the Young diagram

has a1 = 5 and a2 = 2. The highest weight vector of such

a SU (3) representation will be Λ = a1 µ(1) + a2 µ(2) . The Young diagram can be

equivalently denoted as ( a1 , a2 ).
This procedure can be generalized to SU ( N ) group as follows: there will be
N − 1 component fundamental weight vectors µ(i) ’s where i = 1, 2, . . . N − 1 whose

corresponding Young diagrams will be , , , . . . respectively. Further arbitrary

SU ( N ) representation can be written as ( a1 , a2 , . . . , a N −1 ) whose equivalent Young

diagram will have a1 single boxes, a2 double vertical boxes and so on, and the
corresponding highest weight will be

Λ = a 1 µ (1) + a 2 µ (2) . . . + a N −1 µ ( N −1) .

Diagrammatic understanding of tensor product

Let us explain this aspect through the following simple example:

a1 a2 b1 b2 b3 a1 a2 b1 b2
⊗ = a1 a2 b1 b2 b3 ⊕ ⊕
b3
| {z } | {z }
Y1 Y2
(6.2.4)
a1 a2 b1
b2 b3

where we have placed ai ’s in the boxes of Y1 and bi ’s in the boxes of Y2 . This will keep
track of decomposition possibilities ⊕α Yα as depicted above on the right hand side.
Note that the symmetric nature amongst boxes with entries ai ’s and amongst boxes
with entries bi ’s are maintained in the irreducible representations Y. Also observe that
the boxes with entries ai can be symmetric or antisymmetric with respect to boxes with
entries bi in the irreducible representations Y.

Example 53. Work out the dimensions for the above Young diagrams Y1 , Y2 , Y ∈
SU (4) and verify

dY1 dY2 = ∑ dY .
Y ∈Y1 ×Y2

The dimensions of representations of SU (4)(6.2.3):

d = (4 × 5)/(1 × 2) = 10,
4 5
d = (4 × 5 × 6)/(1 × 2 × 3) = 20.
4 5 6

Hence LHS: dY1 dY2 = 200. In the similar fashion, we can work out

d = 56,
4 5 6 7 8
d = (3 × 4 . . . 7)/(1 × 5 × 3 × 2 × 1) = 84 and
4 5 6 7
3
d = 1440/(2 × 4 × 3 × 1) = 60.
4 5 6
3 4

Thus RHS: ∑Y ∈Y1 ×y2 dY = 56 + 84 + 60 = 200.

The tensor product (6.2.4) and SU ( N ) decomposition can be further simplified for
the SU (2) group as vertical column(s) with two boxes are trivial:

a1 a2 ⊗ b1 b2 b3 = a1 a2 b1 b2 b3 ⊕ a2 b1 b2 ⊕ b1
(6.2.5)

which agrees with the tensor product Equation 6.2.1 V 1 ⊗ V 3/2 = ⊕5/2 j
j=1/2 V where V
j

denotes single row Young tableau with 2j boxes. Just like the basis states | j, mi belong
to V j , we could also have rank k irreducible tensors O(k, q) whose transformation
properties are similar to the state |k, qi. For example, vector A~ belongs to rank 1
irreducible tensor. Tensor product of two vectors A~ and ~B whose decomposition will
be exactly like Equation 6.2.1:

~ ⊗ ~B = ⊕2 O(k, q).
A k =0

Comparing the form of spherical harmonics Ym`=1 (θ, φ) with the components of
~ can be rewritten in the spherical rank 1 tensor form as
position vector, the vector A

A x ± iAy
A(1, 0) = Az , A(1, ±1) = ∓ √ . (6.2.6)
2

We had elaborated the projector method of obtaining binary and tertiary basis from
tensor product in the discrete group context. In similar fashion, the states | j, mi
belonging to the irreducible space V j ∈ V j1 ⊗ V j2 must be rewritable as the linear
combination of tensor product states | j1 , m1 i| j2 , m2 i. The projection method for tensor
product of SU (2) representations is the focus of the following section.

6.2.2 SU (2) Clebsch–Gordan matrix

The decomposition into irreducible component states | j, mi ∈ V j from the tensor
j,m
product of states | j1 , m1 i ⊗ | j2 , m2 i involves a matrix Cj ,m ;j ,m called Clebsch–Gordan
1 1 2 2
(CG) coefficient matrix:

∑
j,m
| j, mi = Cj
1 ,m1 ;j2 ,m2
| j1 , m1 i ⊗ | j2 , m2 i, (6.2.7)
m1 m2

where CG coefficients are non-zero if and only if | j1− j2 | ≤ j ≤ j1+ j2 and m = m1 + m2 .

These CG coefficients are available in the form of tables. We can also obtain these
coefficients by assuming CG coefficient of the maximum angular momentum state,
j + j ,j + j
| jmax = j1 + j2 , m = j1 + j2 i (also called as highest weight state), as Cj1,j ;j2 ,j1 2 = 1, i.e., 1 1 2 2

| j1 + j2 , j1 + j2 i = | j1 , j1 i| j2 , j2 i. (6.2.8)

Recall the action of the su(2) generators J ±, Jz on the states discussed in the previous
chapter:
q
J± | j, mi = } ( j ∓ m)( j ± m + 1)| j, m ± 1i; Jz | j, mi = }m| j, mi. (6.2.9)

Incidentally, su(2) generators on the tensor product state | j1 , m1 i| j2 , m2 i can be

deduced from the tensor product of the infinitesimal group elements : (I +
(1) (2)
∑3a=1 Ja δθ a ) ⊗ (I + ∑3a=1 Ja δθ a ) where the superscript on the generators keeps
track of the angular momentum. Keeping up to order O(δθ 2) , the tensor product
simplifies to
" ! !#
3 3
∑ ∑
(1) (2)
I+ Ja θ a ⊗I+I⊗ Ja θ a + O(δθ 2 ),
a =1 a =1

indicating that the generators acting on tensor product state must be

[ J ⊗ I + I ⊗ J ] | j1 , m1 i ⊗ | j2 , m2 i.

Applying these su(2) generators on both the LHS and RHS of Equation 6.2.8 leads to
the following:
p
J− | j1 + j2 , j1 + j2 i = 2j1 + 2j2 | j1 + j2 , j1 + j2 − 1i. (6.2.10)

The action of these generators on the tensor product state of RHS of Equation 6.2.8
will be
p p
{ J− | j1 , j1 i}| j2 , j2 i + | j1 , j1 i{ J− | j2 , j2 i} = 2j1 | j1 , j1 − 1i| j2 , j2 i + 2j2 | j1 , j1 i| j2 , j2 − 1i.
(6.2.11)

From the two equations given above (Equations 6.2.10 and 6.2.11), we obtain
s s
j1 j2
| j1 + j2 , j1 + j2 − 1i = | j , j − 1i| j2 , j2 i + | j , j i| j2 , j2 − 1i.
j1 + j2 1 1 j1 + j2 1 1
(6.2.12)

Comparing the above equation with Equation 6.2.7, we can read off CG coefficients
s s
j + j ,j + j −1 j2 j + j ,j1 + j2 −1 j1
Cj1,j ;j2 ,j1 −21 = , Cj1,j −2 1;j = . (6.2.13)
1 1 2 2 j1 + j2 1 1 2 ,j2 j1 + j2

This procedure can be similarly continued acting J− on Equation 6.2.12.

1/2,1/2 1/2,1/2
Example 54. Work out the CG coefficients C1/2,1/2;1,0 , C1/2,−1/2;1,1 . From Equation
6.2.12, the state |3/2, 1/2i is
r r
1 2
|3/2, 1/2i = |1/2, −1/2i|1, 1i + |1/2, 1/2i|1, 0i.
3 3
1/2,1/2
We need to write the state |1/2, 1/2i to obtain the CG coefficients C1/2,1/2;1,0 ,
1/2,1/2
C1/2, −1/2;1,1 . This state must be orthogonal to |3/2, 1/2i. Hence
r r
1 2
|1/2, 1/2i = |1/2, −1/2i|1, 1i − |1/2, 1/2i|1, 0i,
3 3

which leads to the CG coefficients:

r r
1/2,1/2 2 1/2,1/2 1
C1/2,1/2;1,0 = − , C1/2,−1/2;1,1 = .
3 3

The CG method, which we described for the tensor product of any two states, is also
applicable to the tensor product of two irreducible rank tensors O(k1 , q1 ), O(k2 , q2 )
or the tensor product of O(k, q) with state | j1 , m1 i. Similar to the action of su(2)
generators on states | j, mi in Equation 6.2.9, their action on the irreducible tensors
can be shown to satisfy:

[ Jz , O(k, q)] = }q O(k, q), (6.2.14)

q
[ J± , O(k, q)] = } k(k + 1) − (q ± 1)O(k, q ± 1). (6.2.15)

~ under
Example 55. Using the familiar transformation of rank one tensor A
infinitesimal rotation R by angle δθ about axis n̂ to the quantum mechanical operator
transformation

~ 0 = RA
A ~ =A
~ + δθ n̂ × A ~ R† ,
~ = UR AU

iδθ ~
prove Equations 6.2.14 and 6.2.15 where UR = 1 + } n̂. J .
Equating O(δθ ) terms in the above equation, we get

i ~ ~ ~
[n̂. J, A] = n̂ × A,
}
~ For instance, the above equation for A x turns
which simplifies for components of A.
out to be

[ A x , Jx ] = 0 ; [ A x , Jy ] = ih̄A x ; [ A x , Jz ] = −i} Ay ,

which implies the following compact form

[ Ji , A j ] = eijk i} Ak ,

leading to obtaining the algebra for A(1, q)(6.2.6):

q
[ Jz , A(1, q)] = }qA(1, q) ; [ J ±, A(1, q)] = } 2 − q(q ± 1) A(1, q ± 1).

Example 56. Construct the rank 2 tensor component T (2, 1) using two rank one
tensors A(1, q1 ) and B(1, q2 ).
Following the Clebsch–Gordan procedure for states, we start with the maximum q
tensor as

T (2, 2) = A(1, 1) B(1, 1),

and applying J− on both LHS as well as RHS will give

1
T (2, 1) = √ [ A(1, 1) B(1, 0) + A(1, 0) B(1, 1)].
2

Even though we confined ourselves to the construction of SU (2)CG matrix and

angular momentum states, the method can be extended to SU ( N ) starting with the
highest weight vector state |Λ, Λi and applying su( N ) generators E±α , Hi associated
with simple root vector α.
Recall our discussion on highest weight vector states of the irreducible

representation of SU ( N ). For the Young diagram depicted: ∈ SU (3),

the highest weight vector will be

Λ = 3µ(1) + 2µ(2) ,

where µ(1) , µ(2) are the two fundamental weights of SU (3) discussed in the previous
chapter (see Example 48).
Similar to the SU (2) CG construction, we can take the tensor product of two SU (3)
irreducible representations of highest weights |Λ1 , Λ1 i ⊗ |Λ2 , Λ2 i to give highest
weight vector Λ = Λ1 + Λ2 . The other weight vector states are obtained by applying
E−α(1) , E−α(2) , E−α(3) on both sides as follows:

E−α(i) |Λ, Λi ∝ |Λ, Λ − α(i) i = [ E−α(i) |Λ1 , Λ1 i]|Λ2 , Λ2 i + |Λ1 , Λ1 i[ E−α(i) |Λ2 , Λ2 i].

The proportionality factor can be deduced by the action of the corresponding su(2)
(α) (α)
subalgebra generators J3 , J− on the highest weight states:

(α) (Λ.α)
J3 |Λ, Λi = |Λ, Λi,
| α |2

[(Λ + Λ).α][(Λ − Λ).α + 1]

p
(α)
J− |Λ, Λi = |Λ, Λ − αi.
|α|

For the lower weight states |Λ, Λ − m(α)(1) − nα(2) i where m, n are integers, the
(α) (α)
J3 , J± operator action will be

(α)
Λ − mα(1) − nα(2) .α
J3 |Λ, Λ − mα(1) − nα(2) i = |Λ, Λ − mα(1) − nα(2) i, (6.2.16)
| α |2
1
q
(α)
J∓ |Λ, Λ − mα(1) − nα(2) i = [{Λ ± (Λ − mα(1) − nα(2) )}.α]
|α|
q
[{Λ ∓ (Λ − mα(1) − nα(2) )}.α + 1]|Λ, Λ − mα(1) − nα(2) ∓ αi.

This CG construction of tensor product of weight vector states will be useful for
determining the states of hadrons which are bound states of fundamental quarks.
These hadronic states in the continuous group context are similar to the binary or
tertiary basis discussed in the discrete group context. We will illustrate in Section 6.4.1
for a bound state when we discuss hadrons in particle physics.
In the following subsection, we will look at the selection rules for transition from
the initial state to the final state due to interactions respecting SU (2) symmetry. In
Chapter 4, we had deduced that the matrix elements in the discrete symmetry
situation would be non-zero if the tensor product of the irreducible representations
corresponding to basis states and operators turned out to be trivial representation. In
the same fashion, the matrix elements h βj f , m f |O(k, q)|αji , mi i between initial state
α with angular momentum ji and final state β angular momentum j f due to
interactions, like electric dipole moment interaction or quadrapole moment interaction
and suchlike, denoted by the irreducible rank k tensor operator O(k, q) can be deduced
to be vanishing or non-vanishing. This is the content of Wigner–Eckart theorem which
we will present with proof.

6.2.3 Wigner–Eckart theorem

The theorem states that the matrix elements of the irreducible tensor operator between
two angular momentum states consists of product of two terms

j ,m
h βj f , m f |O(k, q)|αji , mi i = h βj f ||O(k)||αji iCj f,m ;k,q
f
,
i i

where the first factor is called reduced matrix element which depends on the quantum
states |αji i, | βj f i and the rank of the tensor operator whereas the second factor is
purely geometrical given by the CG coefficient. Even though we need experimental
data to determine the reduced matrix element, we can at least rule out (using CG
coefficients) whether the process is allowed or forbidden in any system with rotational
spherical symmetry: SU (2).
From Equation 6.2.14 we know that

h βj f , m f |[ jz , O(k, q)] − }qO(k, q)|αji , mi i = 0,

leading to the constraint m f − mi = q for non-zero h βj f , m f |O(k, q)|αji , mi i when we

expand the commutator. This is one of the properties of the CG coefficient. Similarly,
we can impose the following constraint using Equation 6.2.15:
q
h βj f , m f |[ j±, O(k, q)] − } (k ∓ q) − (k ± q + 1)O(k, q ± 1)|αji , mi i = 0.

Again expanding the commutator, the equation resembles the recursion relation
obeyed by the Clebsch–Gordan coefficients. Thus we conclude that

j ,m
h βj f , m f |O(k, q)|αji , mi i ∝ Cj f,m ;k,q
f
,
i i

and the proportionality constant will be independent of m f , mi , q proving the

Wigner–Eckart Theorem.
Example 57. Given that the diagonal matrix element of hαj, m0 = j| Q(2, 0)|αj, m =
ji = K, determine the matrix element of

hαj, m0 | Q(2, −2)|αj, m = ji.

Using Wigner–Eckart theorem, hαj, m0 = j| Q(2, 0)|αj, m = ji = K = hαj|| Q(2)||αji

j,j
Cj,j;2,0 . This determines the reduced matrix element in terms of K and Clebsch–Gordan
coefficient:
j,j
hαj|| Q(2)||αji = K/Cj,j;2,0 .

Once we know the reduced matrix element then

j,m0 = j−2

j,j
hαj, m0 | Q(2, −2)||αj, m = ji = KCj,j;2,−2 /Cj,j;2,0 .

Thus the theorem is useful in determining the matrix elements of other components
of Q(2, q) from the experimental result for one component.

6.3 Elementary Particles in Nuclear and Particle Physics

In this section, we will focus on the applications of SU (2) and SU (3) symmetries in
nuclear and particle physics. The periodic table of elementary particles is broadly
classified as hadrons and leptons. The electromagnetic and nuclear interactions amongst
these particles are mediated by new particles which are referred to as force carriers. For
instance, photon γ is the mediator of electromagnetic interaction. Similarly, there are
eight gluons gi ’s which are mediators of strong nuclear interaction and three vector
bosons W ±, Z mediating weak nuclear interaction. In order to give masses to these
particles, the Higgs particle was theoretically proposed and recently discovered in the
Large Hadron Collider experiments at CERN, Geneva.
Hadrons are further subclassified as baryons and mesons. The protons and neutrons
inside the atomic nucleus are examples of baryons. Surprisingly, the nuclear magnetic
moment of the proton and neutron are proportional to nuclear magneton but the
proportionality constant is a real number and not an integer. Hence the protons and
neutrons cannot be elementary but must have a structure. The other massive baryons
like 4, Σ, . . . appear in the collision experiments involving high energy proton beams.
Similar to the Rutherford scattering experiment which suggested a massive nucleus
inside the atom, deep inelastic scattering experiments of baryons indicated that they
are bound states of three quarks. Unlike electrons, protons and neutrons, free quarks
are not seen in nature.
In the language of group theory, we will say that baryons can be obtained by
taking the tensor product of three fundamental representations associated with the
quarks. Unlike baryons, mesons are the bound states of quark and antiquark pairs.
For example, pions π and ρ particle are mesons.
Murray Gell-Mann was influenced by the pattern diagrams emerging from plotting
baryons and mesons based on their mass m and charge Q for a fixed angular
momentum J (some patterns resemble an eight-fold path). Importantly, he proposed
the quark model to reproduce the observed patterns. The readers can actually visualize
that these pattern diagrams are nothing but the weight diagrams discussed in Chapter
5 for SU (3) Lie group.
From the experimentally observed strong nuclear scattering or decay process of
hadrons, theoreticians postulated that there must be an additional conservation law
(isospin conservation) besides linear momentum, energy and angular momentum
conservation laws. Similar to Wigner–Eckart theorem, discussed in the context of
su(2) angular momentum states and operators, there will be selection rules for such
scattering or decay processes obeying isospin conservation.
We will briefly discuss selection rules of strong scattering and decay processes
involving hadrons in the following subsection and then elaborate the Gell-Mann quark
model, in Section 6.4, to understand the theoretical construction of observed baryons
and mesons and the pattern diagrams.

6.3.1 Isospin symmetry

There is a quantity called isospin ~I in nuclear and particle physics whose Lie algebra

[ Ii , Ij ] = ieijk Ik

is exactly similar to angular momentum algebra. Interestingly any decay or scattering

process involving hadrons A → B + C must conserve isospin. In other words, isospin
symmetry is respected in such strong interaction processes. The tensor operators
describing such scattering/decay processes are scalar operators S(0, 0). We can now
study allowed and forbidden processes using Wigner–Eckart theorem.
The isospin I and the z-component of isospin Iz for few baryons and mesons which
are needed to determine Clebsch–Gordan coefficients are as follows:

delta particles 4: I = 3/2, Iz = +3/2, 1/2, −1/2, −3/2 for 4++ , 4+ , 40 , 4− ,

nucleons N: I = 1/2, Iz = 1/2, Iz = −1/2 for p and n,
pi meson π: I = 1, Iz = +1, 0, −1 for π + , π 0 and π − ,
rho meson ρ: I = 1, Iz = +1, 0, −1 for ρ+ , ρ0 and ρ− .

With this data, let us attempt to figure out the allowed and forbidden processes of
strongly interacting baryons and mesons.

Example 58. Show that the the ratio of the decay rate between the two strong decay
processes respecting isospin symmetry is

Γ[4+ → pπ 0 ] |h pπ 0 |S(0, 0)|4+ i|2

= = 2.
Γ[4+ → nπ + ] |hnπ + |S(0, 0)|4+ i|2

Using the Wigner–Eckart theorem, the decay process 4+ → pπ 0 is given by the matrix
element
3/2,1/2
h pπ 0 |S(0, 0)|4+ i ∝ h pπ 0 |3/2, 1/2i = C1/2,1/2;1,0 , (6.3.1)

where | pπ 0 i ≡ |1/2, 1/2; 1, 0i = |1/2, 1/2i|1, 0i. Similarly the decay amplitude of
4+ → nπ + is given by
3/2,1/2
hnπ + |S(0, 0)|4+ i ∝ hnπ + |3/2, 1/2i = C1/2, −1/2;1,1 . (6.3.2)

The Clebsch–Gordan decomposition of the state |3/2, 1/2i obtained from the tensor
product of I1 = 1/2 ⊗ I2 = 1 will be
r r
2 1
|3/2, 1/2i = |1/2, 1/2; 1, 0i + |1/2, −1/2; 1, 1i .
3| {z } 3| {z }
| p,π 0 i |nπ + i

Using the above CG coefficients, the ratio of Equations 6.3.1 and 6.3.2 gives
q 2
2
Γ[4+ → pπ 0 ] 3
= q 2 = 2.
Γ[4+ → nπ + ]

1
3

6.4 Quark Model

The angular momentum of the baryons are either J = 3/2 or J = 1/2. Further, the
observed magnetic moments of protons are not integral multiples of nuclear magneton
µ N = e}/(2m p ) indicating the baryons are not elementary particles but composites
made up of n number of quarks which are spin 1/2 particles.
The angular momentum state of these spin 1/2 quarks belong to the
two-dimensional space V 1/2 . Baryonic states belong to the irreducible spaces obtained
from the tensor product of these n quarks:

V
|
1/2
⊗ V 1/2 1/2
{z ⊗ . . . V } = V
n/2
⊕ V n/2−1 ⊕ . . .
n

The angular momentum of the irreducible representations, denoting baryonic states,

must have either spin 3/2 or spin 1/2. Equating highest spin n/2 of the irreducible
space V n/2 to spin 3/2, we deduce that the baryons must be made up of n = 3 quarks.
Even though we have accounted for the observed angular momentum of baryons,
we still need a radical idea to explain the reason for eight baryons including nucleons
to have J = 1/2 and the rest of the baryons including 4 particles to have J = 3/2.
Figure 6.4.l(b) shows the plots of J = 1/2 baryons resembling eight-fold path diagram
and Figure 6.4.l(a) for J = 3/2 baryons. Note that the dotted lines indicate the line
of same charge Q and the horizontal lines have same isospin I and decreasing Iz
components in steps of one from right to left. Interestingly, Gell-Mann’s mathematical
formulations used Lie groups to produce the observed plots. This is famously known
as Gell-Mann’s quark model for which he received the Nobel Prize in the year 1969. We
will explain the rescaling of the vertical axis and the definition of Y in the next
subsection.

Ö3 H2 = Y
2

D– D0 D+ D++
ddd dud uud uuu

H 1 = Iz
dds uds uus
S* – S* 0 S* +
Q = +2

sds sus
X* – X* 0 Q = +1

sss Q=0
W–

Q = –1

(a) Decimet diagram of spin J = 3/2 baryons

2
H2 = Y
Ö3
n p

S– S0 S+ H 1 = Iz
L
Q = +1

X– X0
Q=0
Q = –1

(b) Octet diagram of spin J = 1/2 baryons

Figure 6.4.1 Quark Model

6.4.1 SU (3) group approach quark model

Gell-Mann assumed that the quarks come in three flavors u, d, s (up, down, strange
quarks). In the flavor space, (u, d) are states having isospin I = 1/2 and have the
z-component of isospin as +1/2, −1/2 respectively, whereas s has isospin I = 0.

We can denote the representation of the quark in flavor space as a Young diagram
∈ SU (3). Its dimension is d = 3 indicating that we can place any of the
three-flavors in the single box which agrees with the dimension in formula (6.2.3)
for group SU ( N = 3). We have already studied the tensor product of SU ( N )
representations and their irreducible decomposition through the Young diagram
approach. In the present context, we can construct baryon states from the tensor
product of three quarks in flavor space as follows:
!

⊗ ⊗ = ⊗ ⊕ = (6.4.1)

⊕ ⊕ ⊕

The baryons belonging to irreducible representation refers to all possible

totally symmetric three quark states with and without repetition of the three flavors.
Hence, the total number of possible states in the above representation will be 10 with
the following possible entries in the boxes : uuu; ddd; sss; uud; uus; ddu; dds; ssu; ssd;
uds. Using the dimension formula (6.2.3), d = 10. We would like to emphasize
that the pattern of the ten possible states indeed coincides with the observed pattern
diagram of baryons with angular momentum J = 3/2. When Gell-Mann proposed
the SU (3) quark model deducing the decimet states, the baryon Ω− had not been
discovered by experimentalists. They discovered Ω− three years after the quark model
hypothesis, demonstrating the power of group theory prediction.
√ that the highest weight of the decimet representation is Λ = 3µ ++
Note (1) =

(3/2, 3 3/6). In the decimet diagram 6.4.1a, this highest weight state is the 4
baryon whose Iz = 3/2 and charge Q = 2. The first component of the weight vector
is the eigenvalue of Cartan generator H1 which matches with the Iz = 3/2 whereas
the second component of the weight vector is the eigenvalue of Cartan generator H2 .
Using the Gell-Mann–Nishijima formula Q = Iz + ( B + S)/2, the following relation to
H2 can be deduced:

2
√ H2 = 2( Q − Iz ) ≡ Y = ( B + S),
3

where Y is referred to as hypercharge whose value can be deduced using the charge
Q and z-component of isospin of the particles. Equivalently, all baryons are assigned
baryon number B = +1. If the baryons have strange quarks as constituents, these
baryons have a non-zero strangeness S quantum number. The 4 particles have B = 1,
S = 0 whereas Σ∗+ has B = 1, S = −1. That is, for every s quark, we associate S = −1.
In terms of B, S, the hypercharge Y = B + S can be obtained. Using the lowering

operators E−α(i) on the highest weight vector state |Λ, Λ = 3µ(1) ≡ 4++ i, we can
obtain the remaining nine baryons in the deciment diagram. Hence,√the deciment
diagram is indeed a weight diagram with the H1 ≡ Iz and H2 = 3Y/2 for the
irreducible representation ∈ SU (3). Following the relation of H1 and H2
to isospin and hypercharge, the SU (3) defining representation weight diagram states
discussed in Chapter 5 can be equivalently interpreted as |µ1 ≡ ui, |µ2 ≡ di, |µ3 ≡ si.
We leave the readers to compute the dimensions of the other irreducible
representation of SU (3) in the tensor product ⊗ ⊗ and verify that the
dimension of Equation 6.4.1 is

3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1.

Note that there are two eight-dimensional representations from the group theory
tensor product whereas there is only one octet weight diagram for baryons. The
weight vector states of each irreducible representation can be obtained using the
CG method and shown that they are orthonormal to each other. In fact, the
experimental value of the magnetic moment data can be reproduced only for a suitable
linear combination of the two eight-dimensional representation. This experimentally
stringent requirement justifies one octet diagram of baryons. Interested readers can
refer to any elementary particle physics textbook for details.
Let us now work out a simple example to determine SU (3) CG coefficients.

Example 59. For a di-quark bound state, determine the six weight vector states
associated with the Young diagram ∈ SU (3) using the CG construction.
The highest weight vector state is |2µ(1) , 2µ(1) i ≡ | u u i = |uui. We can now
( α (1) )
apply the lowering operators J− on the highest weight state. Before we do that, we
know that

(2α(1) .2µ(1) )/|α(1) |2 = q = 2.

Thus there will be two weight vector states

|2µ(1) , 2µ(1) − α(1) i, |2µ(1) , 2µ(1) − 2α(1) i.

Applying the lowering operator (6.2.16):

( α (1) ) √
J− |2µ(1) , 2µ(1) i = 2|2µ(1) , 2µ(1) − α(1) i = |usi + |sui.

Thus we get the weight vector state

1
|2µ(1) , 2µ(1) − α(1) i = √ (|usi + |sui) ≡ | u s i.
2

Further operating the same lowering operator, we can show that

|2µ(1) , 2µ(1) − 2α(1) i = | s s i ≡ |ssi.

( α (2) )
Remember that J− |2µ(1) , 2µ(1) i = 0 as α(2) .µ(1) = 0. However, we can operate the
lowering operator associated with the positive root α(3) = α(1) + α(2) to derive the
following two states:

1
|2µ(1) , 2µ(1) − α(1) − α(2) i = √ (|udi + |dui) ≡ | u d i;
2

|2µ(1) , 2µ(1) − 2α(1) − 2α(2) i = | d d i ≡ |ddi.

( α (2) )
On the state | d d i, we cannot apply J− as the dot product of the root vector
with the weight vector being 2α .(2µ − 2α ) − 2α(2) )/|α(2) |2 = (q − p) = −2.
( 2 ) ( 1 ) ( 1

Hence, we will try raising the operator twice on the state. Applying once gives

1
|2µ(1) , 2µ(1) − 2α(1) − α(2) i = √ (|dsi + |sdi) ≡ | d s i,
2

and applying the raising operator on the | d s i will give the state | s s i
already obtained. Thus we have derived the six weight vector states and presented
the CG coefficient decomposition of the di-quark bound state by applying the su(3)
raising and lowering operators as briefly mentioned in Section 6.2.2. In the similar
way, we can write the highest weight vector state |3µ(1) , 3µ(1) i ≡ = |uuui
and determine the weight states corresponding to the remaining nine baryons in the
decimet diagram using the CG construction method.

6.4.2 Antiparticles
Antiparticles of quarks are called antiquarks. Just as for particles, we can have
decimet diagrams and octet diagrams for antibaryons. The values of Iz , Y, Q, S, B of
antiparticles are opposite in sign with respect to their corresponding particles. We will
now describe the Young diagram for the antiparticle multiplet of the SU ( N ) group.
Suppose a particle multiplet is (a1 , a2 , . . . a N −1 ) where a1 , a2 , . . . a N −1 are the
number of single box, number of double vertical boxes, · · · number of columns of
N − 1 vertical boxes in the corresponding Young diagram presentation. Then the
antiparticle multiplet will be given by (a N −1 , a N −2 ; . . . a2 , a1 ). For example, SU (3)
decimet baryon multiplet denoted by (a1 = 3, a2 = 0). The corresponding

antibaryon decimet is (a2 = 0, a1 = 3) whose Young diagram is . Similarly,

SU (3) antiquark representation is (0, 1) whose Young diagram is . The highest

weight vector of the antiquark state is µ(2) . If we change the signs of xi ’s and yi ’s of the
points (xi , yi ) in the SU (3) fundamental representation weight diagram in the H1 , H2
plane (see Figure 5.7.2) denoting u, d, s quarks, we get the weight diagram of SU (3)
antiquarks. The readers can verify that the highest weight vector of the antiquark
multiplet is µ(2) (as determined in Example 48). Using these Young diagrams of the
antiparticle multiplet, we can study mesons which are bound states of quarks and
antiquarks.

Mesons
In group theory language, we say that the SU (3) mesons belong to irreducible
representations obtained from tensor products of quark and antiquark representation:

⊗ = ⊕ (6.4.2)

where the first irreducible representation (vertical 3 boxes) is a trivial SU (3)

representation whose isospin ~I, Iz , hypercharge Y are zero. The second irreducible
representation whose dimension is 8 is called octet meson multiplet. Notice that the
meson bound state multiplet is its own antimeson multiplet. That is, some mesons are
their own antimesons and some mesons get mapped to another set of mesons within
the octet multiplet.
During the 1974 - 1984 period, new mesons and baryons were discovered which
required the introduction of more flavors for quarks besides u, d, s. As of today, there
are actually six flavors of quarks indirectly observed through scattering experiments.
They are called up (u), down (d), strange (s), charm (c), beauty (b) and top (t) quarks.
We could extend Gell-Mann’s quark model proposal by including these flavors as well.
That is, the Lie group will be SU ( N ) where N denotes the number of quark flavors.
We have given exercise problems to use the Young diagram tensor product tools to
understand such new baryons and meson bound states.
We will now discuss in the following subsection as to how the SU (3) irreducible
representations like decimet and octet will break if there is a perturbation that breaks
the symmetry to a lower rank group. In Chapter 4, we discussed (in the discrete
group context) how some degenerate energy levels split into non-degenerate states.
In similar fashion, we would like to understand how the ten-dimensional decimet
belonging to SU (3) breaks to a lower rank group symmetry by adding perturbations.

6.4.3 Symmetry breaking from SU (3) → SU (2)

In the absence of perturbation, the system described by the Hamiltonian H0 possesses
SU (3) symmetry if and only if the Hamiltonian commutes with all the eight generators

of su(3). Suppose we add a perturbation term to the Hamiltonian H1 = k ∑3i=1 λ2i

where k is a constant (|k| < 1) and λ{i=1,2,3} are the first three Gell-Mann matrices,
then the total Hamiltonian

[ H = H0 + H1 , λi ] = 0,

where i = 1, 2, 3. Note that the first three Gell-Mann matrices satisfy su(2) algebra.
Suppose the system is the quark system described by defining representation, we see
that

(λ1 + iλ2 )|Λ, µ1 ≡ ui ∝ |Λ, µ2 ≡ di; (λ1 − iλ2 )|Λ, µ2 ≡ di ∝ |Λ, µ1 ≡ ui,

whereas the state |Λ, µ3 i ≡ |si remains invariant under the action of λ{i=1,2,3} which
means that the state transforms as a one-dimensional representation of SU (2). That is,
the three-dimensional defining representation under the perturbation H1 breaks as

3 → 2 + 1.

The three-dimensional multiplet becomes

( u d s ) = ( u d ) ⊕ ( s ).

Example 60. Suppose the perturbation breaks the SU (3) symmetry to SU (2) such that
s quarks behave like a trivial (singlet) representation of SU (2). How will the SU (3)
baryons in the decimet multiplet break under such a perturbation?
Recall, that in the decimet weight diagram ∆ particles are made of non-strange
quarks. Hence they transform like ∈ SU (2) which is four-dimensional. The
baryons ∑∗ have one strange quark and hence will transform as ∈ SU (2) as one
∗
of the boxes with s quark behaves trivial. Similarly Ξ has two s quarks which will
transform as ∈ SU (2). The Ω− has three s quarks and hence will transform as a
singlet representation. Therefore

10 → 4 + 3 + 2 + 1

is the way in which decimet breaks under the perturbation which reduces the SU (3)
symmetry to SU (2) symmetry.
So far, we have discussed applications in quantum mechanics, particle and nuclear
physics in an elaborate fashion. Particularly, we have examined the tensor product
of SU (2), SU (3) representations which is generalizable for SU ( N ) groups and their
relevance to selection rules and elementary particle multiplets. In the following
section, we will discuss the non-compact group symmetry which forms the backbone
of quantum theories in 3+1 dimensional flat space-time (also known as Minkowski
space-time).

6.5 Non-compact Groups

We will discuss the non-compact group symmetries like Lorentz group, Poincare
groups, conformal groups in this section.

6.5.1 The Lorentz group

The Lorentz group SO(3, 1) is a set of {Λ} 4 × 4 matrices with real matrix elements
and determinant +1 satisfying

Λ T ηΛ = η,

where η is a 4 × 4 diagonal matrix (also known as Minkowski metric):

 
1 0 0 0
 0 −1 0 0 
η=
 0 0 −1 0  .


0 0 0 −1

Using the above condition, we can deduce that the number of independent matrix
elements is 6. That is, there must be six Lie algebra generators. These 4 × 4 matrices
Λ act on the four-dimensional space-time coordinates x µ = ( x0 , ~x ) = (ct, x, y, z).
Alternatively, the group is a set of transformations

ct0
   
ct
 x0   x
x µ → x 0µ = Λν x ; ≡ 
 y0  = Λν  y
µ ν  µ 



z0 z

such that
  
1 0 0 0 cdt
 0 −1 0   dx  =
0   
(cdt dx dy dz) 
 0 0 −1 0   dy 
0 0 0 −1 dz

cdt0
  
1 0 0 0
 0 −1 0 0 
(cdt0 dx 0 dy0 dz0 )    dx 
0  
 0 0 −1 0   dy0  .
0 0 0 −1 dz0

The familiar rotation group SO(3) in the three-dimensional space ( x, y, z) is a

subgroup SO(3, 1). Besides the rotation operation, we can also consider the following
operation called Lorentz transformation by boosting along the x-direction:

0
t − vcx2x x − vx t
t = q 2
; x0 = q 2
; y0 = y; z0 = z, (6.5.1)
1 − vc2x 1 − vc2x

where v x is the x-component boost velocity which serves as one of the parameters of
the Lorentz group. Thus, the parameters corresponding to three rotations θ x , θy , θz
about the x, y, z axes plus the three boost parameters v x , vy , vz corresponding to
Lorentz transformation along x, y, z directions combine to give six independent
parameters. We can determine the matrix form of the corresponding generators
associated with the infinitesimal transformations.

Generators of the Lorentz group

The generators of rotation for the Lorentz group along the three coordinate axes
( x, y, z) are the same as the generators of the SO(3) group that forms a closed algebra
of angular momentum operators L as discussed in Chapter 5. Instead of our earlier
notion of rotation about axis, we will state the rotation operation in a plane so that the
we can generalize the operation to four-dimensional Minkowski space-time as well
as rotations in any higher dimensional space. For example, consider an infinitesimal
rotation about z-axis. We will denote the rotation parameter as δφxy (as a rotation
about z-axis which mixes x and y components):

R(δθz ) ≡ R(δφxy ), (6.5.2)

   
cos δφxy − sin δφxy 0 0 −1 0
R(δφxy ) =  sin δφxy cos δφxy 0  = I +  1 0 0  δφxy = I + L xy δφxy ,
0 0 1 0 0 0
 
0 −1 0
where L xy =  1 0 0  is the generator for rotation in the x-y plane and δφxy
0 0 0
is the corresponding parameter for this transformation. The matrix form of these
generators must be 4 × 4 in the four-dimensional Minkowski space-time. That is we
should add an extra first row and a first column associated with the time coordinate
with 0 as entries:
 
0 0 0 0
 0 0 −1 0 
L xy = 
 0 1 0 0 .
 (6.5.3)
0 0 0 0

Similarly, the generators for rotation in the y-z plane (about the x-axis) and z-x plane
(about the y-axis) respectively are
   
0 0 0 0 0 0 0 0
 0 0 0 0   0 0 0 1 
Lyz = ; Lzx =  (6.5.4)
 0 0 0 −1   0 0 0 0 
0 0 1 0 0 −1 0 0

These three rotation generators in Equations 6.5.3 and 6.5.4 satisfy the commutation
relation

[ L̂i , L̂ j ] = eijk L̂k , (6.5.5)

1
where L̂ p = e pqr Lqr , with the p, q, r ∈ ( x, y, z) and e pqr is the Levi–Civita
2
antisymmetric tensor of rank 3 where

exyz = eyzx = ezxy = −eyxz = −exzy = −ezyx = +1,

and zero otherwise. Similar to the matrix form of rotation generators, we will
determine the matrix form of boost generators by rewriting the Lorentz transformation
vx 1
(6.5.1) operation using the notation β = ;γ = p
c 1 − β2

ct0
    
γ − βγ 0 0 ct
 x 0   − βγ γ 0 0   x 
 0 =  . (6.5.6)
 y   0 0 1 0  y 
z0 0 0 0 1 z

Implementing the following change of variables:

γ = cosh φxt and βγ = sinh φxt ⇒ β = tanh φxt , (6.5.7)

where the quantity φxt is known in literature as ‘rapidity’ which is related to velocity
v x as φxt = tanh−1 β. The subscript xt on the rapidity parameter keeps track of the
boost along x-direction which mixes the x, t coordinates. We can now determine the
boost generator from the infinitesimal transformation:
 
γ − βγ 0 0
 − βγ γ 0 0 
Λ(δφxt ) = 
 0
 = I + Kxt δφxt , (6.5.8)
0 1 0 
0 0 0 1

where Kxt is the generator for Lorentz boost along x-direction and δφxt is the
parameter corresponding to the transformation
   
cosh δφxt − sinh δφxt 0 0 0 −1 0 0
 −sinhδφxt coshδφxt 0 0  −1 0 0 0 
Λ(δφxt ) =   = I+

 δφ .
 0 0 1 0   0 0 0 0  xt
0 0 0 1 0 0 0 0
(6.5.9)
So, now, comparing Equations 6.5.8 and 6.5.9, we get the Lorentz boost generator
along x-direction:
 
0 −1 0 0
 −1 0 0 0 
Kxt =
 0
. (6.5.10)
0 0 0 
0 0 0 0

Similarly, the generators for boost along the y-direction and z-direction respectively
are
   
0 0 −1 0 0 0 0 −1
 0 0 0 0   0 0 0 0 
Kyt =  −1 0 0 0  ; Kzt =  0 0 0
  . (6.5.11)
0 
0 0 0 0 −1 0 0 0

Rewriting K̂i = Kit where i denotes the x, y, z coordinates, it is simple to verify that the
matrices in Equations 6.5.10, 6.5.11, 6.5.3 and 6.5.4 satisfy the following commutation
relations:

[K̂i , K̂ j ] = −eijk L̂k , (6.5.12)

[ L̂i , K̂ j ] = eijk K̂k . (6.5.13)

So, the six generators of the Lorentz group ( L xy , Lyz , Lzx , Kxt , Kyt , Kzt ) satisfy
the three commutation relations given in Equations 6.5.5, 6.5.12 and 6.5.13. The Lie
algebra of the Lorentz group is based on these three commutation relationships of the
generators.
The rotation group is compact, i.e., the parameter space for rotation, consisting of
angles, is bound between 0 and 2π (0 ≤ φxy , φyz , φzx ≤ 2π ); while the boost operation
is non-compact, i.e., the parameter space for boost transformations ranges from −∞ to
∞(−∞ ≤ φxt , φyt , φzt ≤ ∞), although the boost velocity is constrained by the special
theory of relativity by −c ≤ v x , vy , vz ≤ c.
Thus the Lorentz group SO(3, 1) is a non-compact group whose Lie algebra
involves three rotation generators as well as three boost generators.

6.5.2 Poincare group

Besides the Lorentz group symmetry, physical systems can respect space translational
and the time translational symmetry. The generators of the space-time translation in
relativistic notation is four momentum vector: pµ = ( E/c, ~p). Recall that the linear
momentum ~p corresponds to space translation generator and energy E corresponds
to the time translational generator. The readers can extend the algebra by including
E, p x , py , pz along with the six Lorentz generators and thereby construct Poincare
algebra.

6.5.3 Scale invariance

All physical systems near their phase transition point are called critical systems. For
example, near liquid-vapor phase transition, water molecules of all possible sizes will
be present. In other words, the systems at criticality possess scale symmetry: x 0µ =
exp(α) x µ .

Example 61. Determine the generator D for the scale transformation:

x 0µ = exp(α) x µ .

For infinitesimal scale transformation any function

∂
f x 0µ = f ( x µ + αx µ ) = f ( x µ ) + αx µ µ f ( x µ ),

∂x
implying that the generator is

∂
D = xµ .
∂x µ

6.5.4 Conformal group

In addition to the scaling symmetry and Poincare symmetry, there is a symmetry
called special conformal transformation which is defined as follows:

xµ x 00µ
x µ → x 0µ = 2
→ x 00µ = x 0µ + bµ → x 000µ = 00 2 .
|x| |x |

In words, it is an inversion followed by a translation by vector bµ and again another

inversion. There will be four generators associated with the four parameters. The
algebra of E, p x , py , pz , L̂i , K̂i , D plus the four special conformal transformation
generators form an algebra called conformal algebra. These symmetries find
applications in quantum field theories called conformal field theories, the conformal
group symmetries. Interested readers can pursue these areas for research.

6.6 Dynamical Symmetry in Hydrogen Atom

The resemblance between the Kepler problem of planetary motion and the charged
particle in Coulombic potential motivates us to understand symmetries possessed by
such systems. We can formally write the Hamiltonian as

~p2 κ
H= − , (6.6.1)
2µ r

where µ is the reduced mass [ Mm/( M + m)], κ = GMm (for gravitational potential)
and κ = Ze2 (for Coulombic potential where Z is the atomic number). These systems
possess rotational symmetry and hence the angular momentum ~L = ~r × ~p (generators
of rotations in three-dimensional space) is a conserved quantity. Further, for such a
central force potential satisfying the inverse-square law, there is one another conserved
quantity known as the Runge–Lenz vector:

~
~ = ~p × L − κ r̂.
M (6.6.2)
µ r

Unlike the geometrical interpretation for rotation generators, we do not have a

~ represents
geometrical description for the Runge–Lenz vector. Hence we say that M
~
generators of dynamical (hidden) symmetry. Classically, M satisfies the following
relations:

~L · M ~ 2 = 2H~L2 + κ 2 ,
~ =0 ; M (6.6.3)
µ

where H is the Hamltonian of the system.

~ will be
The quantum mechanical operator M̂ of the classical Runge–Lenz vector M
defined as follows:
1 κ
M̂ = ( p̂ × L̂ − L̂ × p̂) − r̂, (6.6.4)
2µ r

as we require Hermitian operators corresponding to observables in quantum

mechanics. This operator (6.6.4) commutes with the Hamiltonian, [ M̂, H ] = 0, and
hence is a conserved quantity. The relations in Equation 6.6.3 become

2H 2
L̂ · M̂ = M̂ · L̂ = 0 ; M̂2 = ( L̂ + }2 ) + κ 2 . (6.6.5)
µ

We will now present the Lie algebra determined by working out the commutator
brackets amongst the components of M̂ as well as the commutator bracket between
M̂ and L̂.

6.6.1 Lie algebra symmetry

The commutation relations satisfied by the components Mx , My , Mz are as follows:

[ Mi , Li ] = 0, (6.6.6)

[ Mi , L j ] = i}eijk Mk , (6.6.7)

2i}
[ Mi , M j ] = − e HLk , (6.6.8)
µ ijk

where i, j, k = x, y, z = 1, 2, 3.
Replacing the Hamiltonian operator H by its corresponding energy eigenvalue E,
and rescaling M as

M̂ → M̂0 = a M̂,

the commutation relation (6.6.8) becomes

1 0 0 2i}
[ M , My ] = − E Lz . (6.6.9)
a2 x µ

So, the Mi ’s and Li ’s as generators constitute a closed SO(4) algebra if

−2
µ 1/2
a2 E =1⇒a= − , (6.6.10)
µ 2E

µ 1/2
M̂0 = − M̂. (6.6.11)
2E
Using this rescaling factor, Equation 6.6.8 becomes

[ Mi0 , M0j ] = i}eijk Lk . (6.6.12)

Comparing with the exercise problem (14), these generators can be mapped to SO(4)
group generators as follows:

1
M̂i0 ≡ Li4 ; L̂i ≡ e L ,
2 ijk jk
In other words, we have to add a fourth fictitious coordinate ω such that the following
operation defines SO(4) group operation:

x0
   
x
 y0  y 
 0  = eΣi< j θij Lij 

. (6.6.13)
 z   z 
ω 0 ω

Thus the SO(4) group symmetry respected by the hydrogen atom Hamiltonian has
no geometric interpretation in the three-dimensional physical space. Hence the SO(4)
symmetry is referred to as the dynamical symmetry respected by the hydrogen atom.
In quantum mechanics textbooks, the hydrogen atom energy spectrum En =
−13.6eV/n2 is determined by solving the time independent Schrödinger equation.
In the following subsection, our aim is to highlight the power of SO(4) dynamical
symmetry. By exploiting this dynamical symmetry, we obtain the energy levels
of the hydrogen atom (1/n2 dependence of energy spectrum) without solving the
Schrödinger equation.

6.6.2 Energy levels of hydrogen atom

The SO(4) group (generated by operators L̂ and M̂0 ) can be decomposed into two
independent SU(2) groups. That is,

SO(4) ≡ SU (2) × SU (2).

The generators of the two SU (2) groups are

1 1
Î = ( L̂ + M̂0 ); K = ( L̂ − M̂0 ), (6.6.14)
2 2
which satisfies the following commutation relations:

[ Ii , Ij ] = i}eijk Ik ,

[Ki , K j ] = i}eijk Kk ,

[ Î, K̂ ] = 0.

We observe that the algebra of the operators involving Î and K̂ form two independent
su(2) algebra which commutes with the Hamiltonian:

[ Î, H ] = 0; [K̂, H ] = 0.

Hence we can construct states | E, i, iz , k, k z i which are simultaneous eigenstates of

Îz , K̂z , H where E is the energy eigenvalue, i, k denote highest weights and iz , k z are
the weights obtained by the operation of Îz , K̂z operators:

Iz | E, i, iz , k, k z i = iz }| E, i, iz , k, k z i,

Kz | E, i, iz , k, k z i = k z }| E, i, iz , k, k z i.

The highest weight can be obtained by the action of Î. Î = Iz2 + ( I+ I− + I− I+ ) and
K̂.K̂ = Kz2 + (K+ K− + K− K+ ) on the states as follows:

Î. Î | E, i, iz , k, k z i = i (i + 1)}2 | E, i, iz , k, k z i,

K̂.K̂ | E, i, iz , k, k z i = k(k + 1)}2 | E, ; i, iz , k, k z i (6.6.15)

We now construct two new operators,

C1 = Î. Î + K̂.K̂ ; C2 = Î. Î − K̂.K̂. (6.6.16)

Using Equations 6.6.5 and 6.6.14, we can confirm that

C2 = Î. Î − K̂.K̂ = 0 ⇒ Î. Î = K̂.K̂. (6.6.17)

The above condition forces that the highest weights i, k cannot be independent but
equal. Further incorporating this condition, the action of C1 on the states will be

C1 | E, i, iz , k, k z i = 2i (i + 1)}2 | E, i, iz , k, k z i. (6.6.18)

Substituting Equations 6.6.5, 6.6.11, 6.6.14, and 6.6.16, the form of the operator C1 is:

1 2 }2 µκ 2
C1 = ( L̂ + M̂02 ) = − − . (6.6.19)
2 2 4H
Comparing the above equation with Equation 6.6.18, we deduce:

C1 | E, i, iz , k, k z i = 2i (i + 1)}2 | E, i, iz , k, k z i

}2 µκ 2

= − − | E, i, iz , k, k z i.
2 4E

which implies

}2 µκ 2
2i (i + 1)h̄2 = − − , (6.6.20)
2 4E

indicating that the energy eigenvalues are

−µκ 2
Ei = , (6.6.21)
2}2 (2i + 1)2

where we have put a subscript on the energy E to keep track of the highest weight i.
Recall that the highest weights of su(2) algebra could be half-odd integers or integers.
Hence,

2i + 1 = n,

where n is always an integer. Substituting the value of κ for hydrogen atom, the
energy eigenvalues (6.6.21) in terms of n are exactly same as that obtained from the
Schrödinger equation:

µe4
En = − , (6.6.22)
2}2 n2
indicating that the integer n can be referred to as the principal quantum number.
The energy levels in the hydrogen atom are degenerate with degeneracy of n2 .
Interestingly, using the SU (2) × SU (2) symmetry arguments we can reproduce the
degeneracy of the energy levels. The physical orbital angular momentum operator
L̂ = Î + K̂ involves addition of two su(2) generators.
Applying the CG construction method, we can obtain eigenstates |``z i of L̂z from
the simultaneous eigenstates of Iz , Kz In fact, we can show that the the allowed highest
weight {`} corresponding to L̂ generators will range from i + k, i + k − 1, . . . |i − k|.
Substituting i = k (6.6.17) will give ` = 2i, 2i − 1, . . . 0. In terms of n, the range
of ` = 0, 1, . . . n − 1. For each `, the states |``z i allow 2` + 1 states. The energy
eigenvalues (6.6.22) are independent of both ` and `z accounting for degeneracy as
n −1
degeneracy = ∑ (2` + 1) = n2 .
`=0

Exercises
1 Quadrapole moment tensor Q(2, q) is derivable from the tensor product of ~r ⊗~r =
⊕2q=−2 Q(2, q). Show that Q(2, 0) = 2z2 − r2 and Q(2, 2) − Q(2, −2) = x2 − y2 .
2 Consider two spin 1/2 particles (proton and neutron) described by the Hamiltonian
H = kŝn .ŝ p where k is a constant. Find the symmetry possessed by the system and the
corresponding conserved quantity. Determine the energy eigenstates and eigenvalues.
3 If S(k, q) and T (k, q) are two irreducuble tensor operators of rank k, prove that ω =
k
∑ (−1)q T (k, q)S(k, −q) is a scalar operator.
q=−k

4 A deuteron has spin 1. Use the Wigner–Eckart theorem to find the ratios of the
expectation values of the electric quadrupole moment operator Q(2, 0) for the three
orientations of deutron (m = 0, +1, −1).
5 Using the Clebsch–Gordan coefficient,
s
j(2j − 1)
h j1 = j, j2 = 2, m1 = j, m2 = 0| J = j, m = ji = ,
( j + 1)(2j + 3)

verify the following statement for quadrupole moment tensor: The static 2k pole
moment of a charge distribution has zero expectation value in any state with angular
momentum j < (k/2).
6 Show that the strong decay process ρ0 → π 0 π 0 is forbidden.
7 The highest weight vector state for decimet irreducible representation belonging to
SU (3) group is

|3µ(1) , 3µ(1) i ≡ | i = |uuui.

Determine the weight vector states corresponding to the remaining nine baryons in the
decimet diagram using the CG construction method on the tensor product of three
fundamental quarks ∈ SU (3).

8 Suppose an irreducible representation ∈ SU (3) denotes a hypothetical

particle multiplet. What will be the Young diagram depicting the corresponding
antiparticle multiplet?
9 Suppose we rewrite the Lorentz group SO(3, 1) generators as Mi = L̂i + iK̂i and
Ni = L̂i − iK̂i . Note that the Mi , Ni are complex conjugates of each other. Write
the algebra satisfied by these Mi ’s and Ni ’s.
10 Suppose we are given a six-dimensional multiplet belonging to a irreducible
representation of a unitary group of rank less than 3. How do we check whether
the six states |i i belong to spin 5/2 of SU (2) ( ) or symmetric tensor
of rank 2 representation of SU (3)? Explain.
11 Assume quarks occur in four flavors. For this case, determine the irreducible
representations and their dimensions for baryons and mesons.
12 For quarks belonging to the fundamental representation of SU ( N ), determine the
dimensions of the meson multiplets.
13 Determine the irreducible representations and their dimensions for di-baryon bound
states obtained from baryons belonging to the octet multiplet of SU (3).
14 SO(4) denotes the group of orthogonal matrices in four-dimensional space with
determinant 1. (i) Using the orthogonality property, show that the number of
independent parameters is 6. (ii) We know that φk (k = 1, 2, 3) denotes rotation

about the k-axis in three dimensions. Similarly, we can give geometrical meaning by
choosing the six parameters as φij (i > j, i = 1, 2, 3, 4) to denote rotation in the i- j
plane. Let Lij = ri p j − r j pi be the generators of rotation about i- j plane. Show that
the generators obey a closed algebra.

Let V be a finite dimensional vector space (real or complex) and T a linear operator on
V. The kernel of the transformation T is the subspace K of V on which T vanishes. The
image of T is also a subspace of V, let this subspace be R. One seeks the condition on
the operator T so that any given vector v ∈ V can be expressed uniquely as v = k + r
for some k ∈ K and r ∈ R. Notice that v = ( I − T )v + Tv and Tv ∈ R. If this
representation of v is unique, then ( I − T )v ∈ K. It follows that T ( I − T )v = 0 for all
v ∈ V and one has the condition

T2 = T (A.0.1)

if V = K ⊕ R. Conversely, suppose the above condition holds for T. If u ∈ K ∩ R, then

Tu = 0 and also there exists a w ∈ V such that Tw = u. Then

u = Tw = TTw = Tu = 0.

It also follows from here that v = ( I − T )v + Tv is the desired unique representation

of any given v ∈ V, i.e., V = K ⊕ R. Any operator T that satisfies the condition A.0.1 is
called a projection operator. The reader can easily construct an operator on R2 which
has a non-trivial kernel but is not a projection operator.
Now let Γ1 and Γ2 be representations of a group G on vector spaces V and W
respectively. A linear transformation T from V to W is said to be G-linear if

T ◦ Γ1 ( g ) = Γ2 ( g ) ◦ T

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152 Appendices

for all g ∈ G. K be the kernel of T and L = T (V ) be the subspace of W. If l ∈ L, then

there exists a v ∈ V such that l = Tv. For any g ∈ G, Γ2 ( g)l = Γ2 ( g) ◦ Tv = T ◦ Γ1 ( g)v,
i.e., Γ2 ( g)l ∈ L. Hence, Γ2 is a subrepresentation of G on L. A similar argument shows
that Γ1 is a subrepresentation of G on K.

THEOREM. Let Γ be a representation of G on a finite dimensional vector space V ( Real or

Complex). If Γ is a subrepresentation on a subspace L of V, then Γ is a subrepresentation on
another subspace K of V such that V = K ⊕ L.

PROOF. A basis of L in V can always be extended to a basis of V. Then the subspace K 0

spanned by basis vectors required for the extension is such that V = K 0 ⊕ L. For every
v ∈ V, one has v = k + l uniquely for some k ∈ K 0 and l ∈ L. Define the projection
operator P on V such that Pv = l. Then K 0 is the kernel of P and L is the image of P. If
P was G-linear, then the theorem follows from the discussion prior to the statement of
the theorem. However P need not be G-linear. Consider the operator T defined as

1
T=
|G| ∑ Γ( g−1 ) PΓ( g).
g∈ G

By construction, T is G-linear (see also the discussion of Equation B.0.2). Also, for
every l ∈ L, because L is invariant under Γ( g) for all g ∈ G and P is projection onto
L, Tl = l. This shows that T a map of V into V whose image is the subspace L. For
these same reasons, T 2 = T and it follows that T is a projection. Let the kernel of T be
K, so that one has by the property of projection operator T

V =K⊕L

and Γ is a subrepresentation of G over the subspace K.

The Maschke’s theorem states that every representation of a finite group over a
finite dimensional vector space (real or complex) is completely reducible. This is a
direct consequence of the above result.

Schur’s Lemma is stated and proved in the following. Some of its consequences,
as regards irreducible representations and their characters are also developed. It is
assumed that the vector spaces are finite dimensional over real or complex numbers
and the groups are finite. The reader should be familiar with the notation and
terminology introduced in Sections 3.1 to 3.3.
LEMMA. (Schur) Let Γ1 and Γ2 be irreducible representations of a group G over vector spaces
V1 and V2 . If T is a linear transformation of V1 into V2 such that

TΓ1 ( g) = Γ2 ( g) T (B.0.1)

for all g in G, then either T is an isomorphism of spaces V1 and V2 or the kernel of T is V1 . In the
case T is an isomorphism with Γ1 ( g) = Γ2 ( g) for all g in G, T is an scaling transformation.
PROOF. Let the image of T in V2 be L. If L is a non-trivial proper subspace of V2 , then
the condition implies that Γ2 is a representation on L. This is not possible since Γ2 is
irreducible over V2 . Then L is either the null vector space or all of V2 . Similarly, the
kernel K of T is either the null vector space or all of V1 .
In the case K is the null vector space and L = V2 , T is evidently an invertible
linear transformation satisfying TΓ1 ( g) T −1 = Γ2 ( g) for all g in G, i.e., Γ1 and Γ2 are
equivalent. T may now be regarded as an invertible linear operator on the vector space
V (= V1 ). There then exists a scalar λ 6= 0 and a vector v 6= 0 in V such that

Tv = λv.

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154 Appendices

Assuming Γ1 = Γ2 = Γ, the linear span L({Γ( g)v} g∈G ) is a representation subspace

of V, and by irreducibility of Γ, L({Γ( g)v} g∈G ) = V. Any vector v0 ∈ V, can be
expressed in the form v0 = ∑ α g Γ( g)v for some choice of scalars α g . Then one has
g∈ G

!
Tv0 = T ∑ α g Γ( g)v = ∑ αg TΓ( g)v = ∑ αg Γ( g)Tv
g∈ G g∈ G g∈ G

⇒ Tv0 = λv0 .

In the general case, a transformation T need not satisfy the condition (B.0.1).
In order to successfully apply the Lemma, it is desirable to obtain a suitable
transformation TS from a given T so that TS fits the condition. Consider the following
expression for TS :

1
TS =
|G| ∑ Γ2 ( g−1 )TΓ1 ( g), (B.0.2)
g∈ G

where Γ1 and Γ2 are irreducible representations of the the group G on vector spaces V1
and V2 , while T is a linear transformation from V1 into V2 . Clearly, TS is also a linear
transformation from V1 into V2 . The form of TS suggests an averaging of T over the
group. For h ∈ G,

1
TS Γ1 (h) =
|G| ∑ Γ2 ( g−1 )TΓ1 ( g)Γ1 (h)
g∈ G

1
⇒ TS Γ1 (h) =
|G| ∑ Γ2 (h)Γ2 (h−1 )Γ2 ( g−1 )TΓ1 ( g)Γ1 (h)
g∈ G

" #
1
⇒ TS Γ1 (h) = Γ2 (h)
|G| ∑ Γ2 (( gh) −1
) TΓ1 ( gh)
g∈ G

⇒ TS Γ1 (h) = Γ2 (h) TS

which is exactly the condition one needs for application of the Lemma. In Equation
B.0.2, if Γ1 = Γ2 = Γ, then TS is a transformation of scale λ. If the degree of
representation Γ is `Γ , then upon equating the trace of both sides of Equation B.0.2,
one obtains

tr ( T )
λ= . (B.0.3)
`Γ

In this case, TS is diagonal and each entry in the diagonal is equal to λ. Upon explicitly
writing, say the element ( TS )ii from Equation B.0.2

`
Γ
∑ j= 1 Tjj 1
`Γ
=
|G| ∑ [Γ( g)]ik Tkm [Γ( g−1 )]mi (B.0.3)
g∈ G

and because the matrix T is an arbitrary linear transformation, one can equate
coefficients of Tkm on both sides of above to obtain

|G|
∑ [Γ( g)]ik [Γ( g−1 )]ml = δ δ .
`Γ il km
g∈ G

Similar calculation for Γ1 6= Γ2 readily leads to

∑ [Γ1 ( g)]ik [Γ2 ( g−1 )]ml = 0.

g∈ G

If it is further assumed that the representations are unitary, and noting that in such a
case [Γ( g−1 )]ml = [Γ( g)]lm , the above two expressions combined give Equation 3.3.5

|G|
∑ [Γ( g)]ik [Θ( g)]lm = δ δ δ .
`Γ ΓΘ il km
geG

[1] Georgi, Howard. 1999. Lie Algebras in Particle Physics: From Isospin to Unified
Theories (Frontiers in Physics). Boulder, Colorado: Westview Press.
[2] Hamermesh, M. 1962. Group Theory and Its Applications to Physical Problems.
Reading, MA: Addison-Wesley.
[3] Joshi, A. W. 1973. Elements of Group Theory for Physicists. Delhi: New Age
International (P) Ltd. Pub.
[4] Landau, L. D., and E. M. Lifshitz. 1965. Quantum Mechanics: Non-Relativistic
Theory. Oxford: Pergamon Press. (See Chapter 13: Polyatomic Molecules, where
molecular vibrations are discussed.)
[5] Ma, Zhong-Qi, and Xiao-Yan Gu. 2004. Problems and Solutions in Group Theory for
Physicists. Singapore: World Scientific Publishing.
[6] O’Raifeartaigh, Lochlainn. 1986. Group Structures of Gauge Theories. N. York:
Cambridge University Press.
[7] Ramond, Pierre. 2010. Group Theory: A Physicist’s Survey. N. York: Cambridge
University Press.
[8] Schiff, L. I. 1955. Quantum Mechanics. N. York: McGraw-Hill International
Editions. (See Chapter 7: Symmetry in Quantum Mechanics.)
[9] Serre, Jean-Pierre. 1977. Linear Representations of Finite Groups. N. York:
Springer-Verlag.
[10] Sury, B. 2004. Group Theory: Selected Problems. Hyderabad: Universities Press.
[11] Tung, Wu-Ki. 1985. Group Theory in Physics. Singapore: World Scientific
Publishing.
[12] Zee, A. 2106. Group Theory in a Nutshell for Physicists. Princeton: Princeton
University Press.

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Index

abelian group 2 conjugate subgroup 5

adjoint 34 coset 4
adjoint representation 98 cyclic group 2
alternating group 14
angular momentum algebra 93–94 degenerate states 60
addition of angular momentum dihedral group 6
116, 119 direct product 14
antisymmetric subspace 51 double cover 101
doubly connected space 95
dynamical symmetry 117, 144
baryons 116
Dynkin diagrams 109
bilateral axis 23

eigenfrequencies 73
Cartan eigenvalue equation 32
matrix 113 electric dipole moment 66, 68
subalgebra 107 exceptional Lie algebra 110
Cauchy–Schwartz inequality 33
Cayley’s Theorem 10
character factor group 5
of the group element 38 Frobenius reciprocity theorem 57
table 41 fundamental weights 113
circle group U(1) 87
Clebsch–Gordan coefficient 124 generating set 2
commutator 91 generators 2, 89
compact groups 95 constants of motion 117
conformal group 117, 143 gerade 41
conjugacy class 5 Gram–Schmidt Orthogonalisation 33

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158 Index

Group 1 Lie groups

continuous 88 general Linear groups 106
direct product of groups 14 orthogonal groups 106
order of the group 2 generalized orthogonal group 106
properties of the group 1 Lorentz group 116
representations 21 symplectic group 106
unitary group 107
highest weight vector 108 special linear groups 106
homomorphism 8 special orthogonal group 106
kernel of homormorphism 8 special unitary groups 107
hydrogen atom 117, 144 linear operator 31, 96
Runge–Lenz vector 117, 144 linearly independent set 30
hypercharge 134 linear span of 30
Lorentz transformation 139
induced representation 54
inner product space 33 mesons 116
invariant subalgebra 102 Mulliken symbols 41
irreducible representation 38
of direct product groups 52 negative roots 109
irreducible tensor 124 Noether theorem 117
isomorphic groups 9 non-abelian 3
isospin 116, 131 norm 32
normal mode 73
Jacobi identity 92 normal subgroup 5

Killing form 101 one-parameter group 88

degenerate Killing form 101 orthogonal 33
nondegenerate Killing form 102
Klein-4 group 3 particle physics 130
antiquarks 136
Lagrange theorem 4 force carriers 130
Lagrangian 72 gluons 130
Euler–Lagrange equation 72 photons 130
level splitting 62 vector bosons 130
Lie algebra 92 hadrons 130
Cartan subalgebra 107 baryons 130
complex 92 mesons 130, 137
real 92 leptons 130
semi-simple Lie algebra 101 quarks 132
simple Lie algebra 103 permutation cycles 11
compact simple Lie algebra 107 two-cycle, transposition 11
structure constants 92 permutation groups 10

Poincare algebra 117 permutation group 10

Poincare group 117, 143 symmetric subspace 51
point groups 9, 21 symmetry
Poisson bracket 92 axis of 18
positive roots 112 equivalent 21
projection operator 47 principal 19
continuous 87
quadrupole moment 66, 68 inversion 19
quark model 116, 130, 132 plane of 19
quaternion group 6 equivalent 21
point of 19
rank 107 roto-reflection 19
reducible representation 38 transformation 18
completely reducible 39
multiplicity of irreducible components tensor product 45
45–46 of irreducible representations 46
regular representation 44–45 time reversal 67
root vectors 107 antilinear property 68
rotational symmetry 118 translational symmetry 117
angular momentum conservation 118 space translation 117
linear momentum conservation 117
scalar multiplication 30 time translation 118
scaling transformation 32 Hamiltonian 118
Schoenflies notation 23
secular equation 73
secular determinant 73 ungerade 42
selection rules 64, 116 unitary group 107
self-adjoint operator 34, 60 unitary operator 35
semi-direct product 15
semi-simple Lie algebras 100 vector space
similarity transformation 32 complex 30
simple group 5 real 36
simple Lie algebras 103 vibrational degrees of freedom 71
simple roots 109
simply connected 101
special unitary group 101 weight vectors 110
stereographic projections 23 Wigner–Eckart Theorem 116, 128
structure constant 92, 105
subalgebra 95 Young diagrams 13
invariant subalgebra 95 dimension of irreducible representation
subgroups 4 121
subrepresentation 38 Young tableau method 116, 119
symmetric group 9–10 symmetric group 119

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Immediate Download Group Theory and Computation N.S. Narasimha Sastry Ebooks 2024
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The Geometry of Möbius Transformations: John Olsen
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Math 113 Homework 4 Solutions: Due July 11, 2011
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Introduction To Modern Algebra (GROUP)
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