(2001) d.d.vvedensky - group theory (lecture note)

Group Theory

First Term 2001 Office Hours:D. D. Vvedensky ([email protected]) Tu 2-3, Fr 11-12 (Blackett 807)

1. Introductiona. Symmetry in physicsb. Discrete and continuous symmetriesc. Symmetry in quantum mechanics

2. Mathematical Background for Discrete Groupsa. Groupsb. Subgroupsc. Cosetsd. Conjugacy classes

3. Representations of Groupsa. Reducible and irreducible representationsb. Schur’s lemmas and the Great Orthogonality Theoremc. Character tablesd. Direct products and their decomposition

4. Physical Applications of Discrete Groupsa. The group of the Hamiltonianb. Eigenfunctions and irreducible representationsc. Bloch’s theoremd. Selection rules

5. Continuous Groups, Lie Groups, and Lie Algebrasa. Linear transformation groupsb. Infinitesimal generatorsc. Algebra of infinitesimal generators

6. Irreducible Representations of SO(2) and SO(3)a. Orthogonality relations and the density functionb. Basis functions for irreducible representations of SO(2)c. Spherical Harmonics and characters for SO(3)

7. Unitary Groupsa. Unitary groups and particle physicsb. Basis states for SU(N)c. Multiparticle states and direct productsd. Young tableaux

Recommended Books

Basic course material:

H.F. Jones,Groups, Representations and Physics(Adam Hilger, Bristol, 1998)M. Tinkham,Group Theory and Quantum Mechanics(McGraw–Hill, 1964)

Related and advanced treatments:

R. Hermann,Lie Groups for Physicists(Benjamin, 1966)D.B. Lichtenberg,Unitary Symmetry and Elementary Particles(Academic, 1978)

Chapter 1

Introduction

As far as I can see, all a priori statements in physics have their originin symmetry.

—Hermann Weyl1

1.1 Symmetry in Physics

Symmetry is a fundamental human concern, as evidenced by its pres-ence in the artifacts of virtually all cultures. Symmetric objects areaesthetically appealing to the human mind and, in fact, the Greekwork symmetros was meant originally to convey the notion of “well-proportioned” or “harmonious.” This fascination with symmetry firstfound its rational expression around 400 B.C. in the Platonic solids andcontinues to this day unabated in many branches of science.

1.1.1 What is a Symmetry?

An object is said to be symmetric, or to have a symmetry, if there isa transformation, such as a rotation or reflection, whereby the objectlooks the same after the transformation as it did before the transforma-tion. In Fig. 1.3, we show an equilateral triangle, a square, and a circle.The triangle is indistinguishable after rotations of 1

3π and 2

3π around

its geometric center, or symmetry axis. The square is indistinguishable

1In Symmetry (Princeton University Press, 1952)

1

2 Introduction

after rotations of 12π, π, and 3

2π, and the circle is indistinguishable after

all rotations around their symmetry axes. These transformations aresaid to be symmetry transformations of the corresponding object, whichare said to be invariant under such transformations. The more symme-try transformations that an object admits, the more “symmetric” it issaid to be. One this basis, the circle is “more symmetric” than thesquare which, in turn, is more symmetric than the triangle. Anotherproperty of the symmetry transformations of the objects in Fig. 1.3that is central to this course is that those for the triangle and squareare discrete, i.e., the rotation angles have only discrete values, whilethose for the circle are continuous.

(a) (b) (c)

Figure 1.1: An equilateral triangle (a), square (b) and circle (c). These ob-jects are invariant to particular rotations about axes that are perpendicularto their plane and pass through their geometric centers (indicated by dots).

1.1.2 Symmetry in Physical Laws

In the physical sciences, symmetry is of fundamental because there aretransformations which leave the laws of physics invariant. Such trans-formations involve changing the variables within a physical law suchthat the equations describing the law retain their form when expressedin terms of the new variables. The relationship between symmetryand physical laws began with Newton, whose equations of motion werefound to be the same in different frames of reference related by Galileantransformations. Symmetry was also the guiding principle that en-abled Lorentz and Poincare to derive the transformations, now knownas Lorentz transformations, which leave Maxwell’s equations invari-ant. The incompatibility between the Lorentz invariance of Maxwell’s

Introduction 3

equations and the Galilean invariance of Newtonian mechanics was, ofcourse, resolved by Einstein’s special theory of relativity.

As an example of a symmetry in a physical law, consider the prop-agation of an impulse at the speed of light c. This is governed by thewave equation, which is obtained from Maxwell’s equations:

1

c2

∂2u

∂t2=∂2u

∂x2+∂2u

∂y2+∂2u

∂z2. (1.1)

The Lorentz transformation of space-time coordinates corresponding toa velocity v = (v, 0, 0) is

x′ = γ(x− vt), y′ = y, z′ = z, t′ = γ(t− v

c2x), (1.2)

where γ = (1−v2/c2)−1/2. When expressed in terms of the transformedcoordinates (x′, y′, z′, t′), the wave equation (1.1) is found to retain itsform under this transformation:

1

c2

∂2u′

∂t′2=∂2u′

∂x′2+∂2u′

∂y′2+∂2u′

∂z′2. (1.3)

This implies that the wave propagates in the same way with the samevelocity in two inertial frames that are in uniform motion with respectto one another. The Lorentz transformation is thus a symmetry trans-formation of the wave equation (1.1) and this equation is said to becovariant with respect to these transformations. In general, symmetrytransformations of physical laws involve the space-time coordinates,which are sometimes called geometrical symmetries , and/or internal co-ordinates, such as spin, which are called internal symmetries .

1.1.3 Noether’s Theorem

Identifying appropriate symmetry transformations is one of the centralthemes of modern physics since their mathematical expression affectsthe structure and predictions of physical theories. Work by both math-ematicians and physicists, culminating with Emmy Noether, led to thedemonstration that there was a deep relationship between symmetryand conservation laws. This is now known as Noether’s Theorem:

4 Introduction

Noether’s Theorem. The covariance of the equations of motion withrespect to a continuous transformation with n parameters implies theexistence of n quantities, or constants of motion, i.e., conservation laws.

In classical mechanics, the conservation of linear momentum resultsfrom the translational covariance of Newton’s equations of motion, i.e.,covariance with respect to transformations of the form r′ = r + a, forany vector a. The conservation of angular momentum similarly resultsfrom rotational covariance, i.e., covariance with respect rotations inspace: r′ = Rr, where R is a 3 × 3 rotation matrix. Finally, the con-servation of energy results from the covariance of Newton’s equationsto translations in time, i.e., transformations of the form t′ = t+ τ .

1.1.4 Symmetry and Quantum Mechanics

The advent of quantum mechanics and later quantum field theory fos-tered entirely new avenues for investigating the consequences of sym-metry. London and Weyl introduced a type of transformation known asa gauge transformation into quantum theory, with total electric chargeas the conserved quantity. In the early 1960s, Gell–Mann and Ne’emanproposed the unitary symmetry SU(3) for the strong interactions. Thisled to the proposal by Gell–Mann and Zweig of a new, deeper, levelof quanta, “quarks,” to account for this symmetry. Heisenberg, Gold-stone and Nambu suggested that the ground state (i.e., the vacuum)of relativistic quantum field theory may not have the full global sym-metry of the Hamiltonian, and that massless excitations (Goldstonebosons) accompany this “spontaneous symmetry breaking.” Higgs andothers found that for spontaneously broken gauge symmetries there areno Goldstone bosons, but instead massive vector mesons. This is nowknown as the Higgs phenomenon and its verification verification hasbeen the subject of extensive experimental effort.

Another aspect of symmetry, also due to the quantum mechanicalnature of matter, arises from the arrangement of atoms in moleculesand solids. The symmetry of atomic arrangements, whether in a sim-ple diatomic molecule or a complex crystalline material such as a high-temperature superconductor, affects many aspects of their electronic

Introduction 5

and vibrational properties and especially their response to externalthermal, mechanical, and electromagnetic perturbations. The trans-formation properties of wavefunctions in quantum mechanics are anexample of what is known as Representation Theory , which was devel-oped by the mathematicians Frobenius and Schur near the turn of the20th century. This inspired a huge effort by physicists and chemiststo determine the physical consequences of the symmetries of wavefunc-tions which continues to this day. Notable examples include Bloch’swork on wavefunctions in periodic potentials, which forms the basis ofthe quantum theory of solids, Pauling’s work on the chemical interpre-tation orbital symmetries, and Woodward and Hoffman’s work on howthe conservation of orbital symmetry determines the course of chemicalreactions. Recent scientific advances that highlight the prominent rolethat symmetry maintains in condensed-matter physics is the discov-ery of quasicrystals, which have rotational symmetries (e.g., fivefold, asshown in Fig. 1.2) which are incompatible with the translational sym-metry of ordinary crystals and are thus sometimes called aperiodic, andthe C60 form of carbon, known as “Buckminsterfullerene,” or “Buck-yballs”, a name derived from its resemblance to structures (geodesicdomes) proposed by R. Buckminster Fuller as an alternative to conven-tional architecture.

Figure 1.2: A section of a Penrose tile, which has a fivefold rotationalsymmetry, but no translational symmetry. This two-dimensional structureshares a number of features with quasicrystals.

6 Introduction

1.2 Examples from Quantum Mechanics

1.2.1 One-Dimensional Systems

To appreciate how symmetry enters into the description of quantum me-chanical systems, we consider the time-independent Schrodinger equa-tion for the one-dimensional motion of a particle of mass m bound bya potential V (x): [

− h2

2m

d2

dx2+ V (x)

]ϕ(x) = Eϕ(x) , (1.4)

where h = h/2π, h is Planck’s constant, ϕ is the wavefunction, and Eis the energy eigenvalue. By writing this equation as Hϕ = Eϕ, weidentify the coordinate representation of the Hamiltonian operator as

H = − h2

2m

d2

dx2+ V (x) . (1.5)

In the following discussion, we will utilize the fact that the energy eigen-values of one-dimensional quantum mechanical problems such as thatin (1.4) are nondegenerate, i.e., each energy eigenvalue is associatedwith one and only one eigenfunction.2

Suppose that the potential in (1.4) is an even function of x. Themathematical expression of this fact is the invariance of this potentialunder the inversion transformation x→ −x:

V (−x) = V (x) . (1.6)

Examples of such potentials are the symmetric square well and theharmonic oscillator (Fig. 1.3), but the particular form of the potentialis unimportant for this discussion. The kinetic energy term in (1.4) isalso invariant under the same inversion transformation as the potential,since

d2

d(−x)2=

d2

dx2(1.7)

2This follows directly from the fact that this equation, together with appropri-ate boundary conditions, constitute a Sturm–Liouville problem. Other well-knownproperties of solutions of Schrodinger’s equation (real eigenvalues, discrete eigen-values for bound states, and orthogonality of eigenfunctions) also follow from theSturm–Liouville theory.

Introduction 7

x

E

Figure 1.3: The first four eigenfunctions of the Schrodinger equation (1.4)for an infinite square-well potential, V (x) = 0 for |x| ≤ L and V (x) → ∞for |x| > L (left), and a harmonic oscillator potential, V (x) = 1

2kx2, where

k is the spring constant of the oscillator (right). The abscissa is the spatialposition x and the ordinate is the energy E, with the vertical displacement ofeach eigenfunction given by its energy. The origins are indicated by brokenlines.

Thus, the Hamiltonian operator in (1.5) is itself invariant under inver-sion, i.e., inversion is a symmetry transformation of this Hamiltonian.We now use this property of H to change variables from x to −x in(1.4) and thereby obtain the Schrodinger equation for ϕ(−x):[

− h2

2m

d2

dx2+ V (x)

]ϕ(−x) = Eϕ(−x) (1.8)

Since E is nondegenerate, there can be only one eigenfunction associ-ated with this eigenvalue, so the ϕ(−x) cannot be linearly independentof ϕ(x). The only possibility is that ϕ(−x) is proportional to ϕ(x):

ϕ(−x) = Aϕ(x) (1.9)

where A is a constant. Changing x to −x in this equation,

ϕ(x) = Aϕ(−x) (1.10)

and then using (1.9) to replace ϕ(−x), yields

ϕ(x) = A2ϕ(x) (1.11)

8 Introduction

This requires that A2 = 1, i.e., A = 1 or A = −1. Combining this resultwith (1.9) shows that the eigenfunctions ϕ of (1.4) must be either even

ϕ(−x) = ϕ(x) (1.12)

or oddϕ(−x) = −ϕ(x) (1.13)

under inversion. As we know from the solutions of Schrodinger’s equa-tion for square-well potentials and the harmonic oscillator (Fig. 1.3),both even and odd eigenfunctions are indeed obtained. Thus, not alleigenfunctions have the symmetry of the Hamiltonian, although theground state usually does.3 Nevertheless, the symmetry (1.6) does pro-vide a classification of the eigenfunctions according to their parity underinversion. This is a completely general result which forms one of thecentral themes of this course.

1.2.2 Symmetries and Quantum Numbers

The example discussed in the preceding section showed how symmetryenters explicitly into the solution of Schrodinger’s equation. In fact, wecan build on our discussion in Sec. 1.1.2, and especially Noether’s theo-rem, to establish a general relationship between continuous symmetriesand quantum numbers.

Consider the time-dependent Schrodinger equation for a free particleof mass m in one dimension:

ih∂ϕ

∂t= − h2

2m

∂2ϕ

∂x2. (1.14)

The solutions to this equation are plane waves:

ϕ(x, t) = ei(kx−ωt) , (1.15)

where k and ω are related to the momentum and energy by p = hkand E = hω. In other words, the quantum numbers k and ω of the

3A notable exception to this is the phenomenon of spontaneous symmetry-breaking discussed in Sec. 1.1, where the symmetry of the equations of motionand the boundary conditions is not present in the observed solution for the groundstate.

Introduction 9

solutions to Eq. (1.14) correspond to the momentum and energy which,because of the time- and space-translational covariance of this equation,correspond to conserved quantities. Thus, quantum numbers are asso-ciated with the symmetries of the system. Similarly, for systems withrotational symmetry, such the hydrogen atom or, indeed, any atom,the appropriate quantum numbers are the energy and the angular mo-mentum, the latter producing two quantum numbers, as required byNoether’s theorem, because the transformations have two degrees offreedom.

1.2.3 Matrix Elements and Selection Rules

One of the most important uses of symmetry is to identify the matrixelements of an operator which are required to vanish. Continuing withthe example in the preceding section, we consider the matrix elementsof an operator H′ whose position representation H′(x) has a definiteparity. The matrix elements of this operator are given by

H′ij =∫ϕi(x)H′(x)ϕj(x) dx (1.16)

where the range of integration is symmetric about the origin. If H′has even parity, i.e., if H′(−x) = H′(x), as in (1.6), then these matrixelements are nonvanishing only if ϕi(x) and ϕj(x) are both even or bothodd, since only in these cases is the integrand an even function of x.This is called a selection rule , since the symmetry of H′(x) determines,or selects, which matrix elements are nonvanishing.

Suppose now that H′(x) has odd parity, i.e., H′(−x) = −H′(x).The matrix elements in (1.16) now vanishes if ϕi(x) and ϕj(x) areboth even or both odd, since these choices render the integrand anodd function of x. In other words, the selection rule now states thatonly eigenfunctions of opposite parity are coupled by such an operator.Notice, however, that the use of symmetry only identifies which matrixelements must vanish; it provides no information about the magnitudeof the nonvanishing matrix elements.

Suppose that

H′(x) = Ax (1.17)

10 Introduction

where A is a constant, i.e., H′(x) is proportional to the coordinate x.Such operators arise in the quantum theory of transitions induced byan electromagnetic field.4 H′(x) clearly has odd parity, so the matrixelements (1.16) are nonvanishing only if ϕi(x) and ϕj(x) have oppositeparity. But, if

H′(x) = − h2

2m

d2

dx2(1.18)

which is the coordinate representation of the kinetic energy operator,then the matrix elements (1.16) are nonvanishing only if ϕi(x) andϕj(x) have the same parity.

Selection rules are especially useful if there are broken symmetries .For example, the Hamiltonian of an atom, which is the sum of thekinetic energies of the electrons and their Coulomb potentials, is in-variant under all rotations. But when an atom is placed in an electricor magnetic field, the Hamiltonian acquires an additional term whichis not invariant under all rotations, since the field now defines a pre-ferred direction. These are the Stark and Zeeman effects, respectively.A similar situation is encountered in quantum field theory when, be-ginning with a Lagrangian that is invariant under certain symmetryoperations, a term is added which does not have this invariance. Ifthe symmetry-breaking terms in these cases are small, then selectionrules enter into the perturbative calculation around the solutions of thesymmetric theory.

1.3 Summary

The notion of symmetry implicit in all of the examples cited in thischapter is endowed with the algebraic structure of “groups.” This is atopic in mathematics that had its beginnings as a formal subject onlyin the late 19th century. For some time, the only group that was knowand whose properties were studied were permutation groups. Cauchyplayed a major part in developing the theory of permutations, but itwas the English mathematician Cayley who first formulated the notionof an abstract group and used this to identify matrices and quaternions

4E. Merzbacher, Quantum Mechanics 2nd edn. (Wiley, New York, 1970), Ch. 18.

Introduction 11

as groups. In a later paper, Cayley showed that every finite group couldbe represented in terms of permutations, a result that we will prove inthis course. The fact that geometric transformations, as discussed inthis chapter, and permutations, share the same algebraic structure ispart of the richness of the subject and is rooted in its history as anadjunct to the study of algebraic solutions of equations. In the nextchapter, we discuss the basic properties of groups that form the basisof this course.

12 Introduction

Chapter 2

Elements of Abstract GroupTheory

Mathematics is a game played according to certain simple rules withmeaningless marks on paper.

—David Hilbert1

The importance of symmetry in physics, and for quantum mechanicsin particular, was discussed in the preceding chapter. In this chapter,we begin our development of the algebraic structure which enables usto formalize what we mean by “symmetry” by introducing the notionof a group and some related concepts. In the following chapters we willexplore the consequences of this algebraic structure for applications tophysics.

2.1 Groups: Definitions and Examples

The motivation for introducing an algebraic structure to describe sym-metry in physical problems is based on transformations. But the def-inition of a group is based on a much more abstract notion of what a“transformation” entails. Accordingly, we first set out the conditions

1As quoted in, N. Rose, Mathematical Maxims and Minims (Rome Press,Raleigh, North Carolina, 1988).

13

14 Elements of Abstract Group Theory

that an abstract group must satisfy and then consider both abstractand concrete examples.

Definition. A group G is a set of elements a, b, c, . . . together witha binary composition law, called multiplication, which has the followingproperties:

1. Closure. The composition of any two elements a and b in G, calledthe product and written ab, is itself an element c of G: ab = c.

2. Associativity. The composition law is associative, i.e., for anyelements a, b, and c in G, (ab)c = a(bc).

3. Identity. There exists an element, called the unit or identity anddenoted by e, such that ae = ea = a for every element a in G.

4. Inverses. Every element a in G has an inverse, denoted by a−1,which is also in G, such that a−1a = aa−1 = e.

The closure property ensures that the binary composition law doesnot generate any elements outside of G. Associativity implies thatthe computation of an n-fold product does not depend on how theelements are grouped together.2 For example, the product abc is un-ambiguous because the two interpretations allowed by the existence ofa binary composition rule, (ab)c and a(bc), are equal. As will be shownin Sec. 2.3, the left and right identities are equal and unique, as arethe left and right inverses of each element. Thus we can replace theexistence of an identity and inverses in the definition of a group withthe more “minimal” statements:

3′. Identity. There exists a unique element, called the unit or identityand denoted by e, such that ae = a for every element a in G.

4′. Inverses. Every element a in G has a unique inverse, denoted bya−1, which is also in G, such that a−1a = e.

2In abstract algebra (the theory of calculation), binary composition can be asso-ciative or non-associative. The most important non-associative algebras in physicsare Lie algebras, which will be discussed later in this course.

Elements of Abstract Group Theory 15

The terms “multiplication,” “product,” and “unit” used in this def-inition are not meant to imply that the composition law corresponds toordinary multiplication. The multiplication of two elements is only anabstract rule for combining an ordered pair of two group elements toobtain a third group element. The difference from ordinary multiplica-tion becomes even more apparent from the fact that the compositionlaw need not be commutative, i.e., the product ab need not equal ba fordistinct group elements a and b. If a group does have a commutativecomposition law, it is said to be commutative or Abelian .

Despite the somewhat abstract tone of these comments, a moment’sreflection leads to the realization that the structure of groups is ideallysuited to the description of symmetry in physical systems. The groupelements often correspond to coordinate transformations of either ge-ometrical objects or of equations of motion, with the composition lawcorresponding to matrix multiplication or the usual composition lawof functions,3 so the associativity property is guaranteed.4 If two op-erations each correspond to symmetry operations, then their productclearly must as well. The identity corresponds to performing no trans-formation at all and the inverse of each transformation correspondsto performing the transformation in reverse, which must exist for thetransformation to be well-defined (cf. Example 2.4).

Example 2.1. Consider the set of integers,

. . . ,−3,−2,−1, 0, 1, 2, 3, . . .

with the composition rule being ordinary addition. The sum of any twointegers is an integer, thus ensuring closure, addition is an associativeoperation, 0 is the identity, and the inverse of n is −n, which is clearlyan integer. Thus, the integers form a group under addition. This groupis denoted by Z (derived from the German word Zahlen for integers).

3For two functions f(x) and g(x), the application of f , followed by the applicationof g is g[f(x)], and the application of g followed by the application of f is f [g(x)].

4The associativity of linear operations in general, and matrices in particular,is discussed by Wigner in Group Theory (Academic, New York, 1959), along withother group properties.


Since the order in which two integers are added is immaterial, Z is anAbelian group.

Example 2.2. The importance of the composition law for determiningwhether a set of elements forms a group can be seen by again consideringthe integers, but now with ordinary multiplication as the compositionrule. The product of any two integers is again an integer, multiplicationis associative, the unit is 1, but the inverse of n is 1/n, which is not aninteger if n 6= 1. Hence, the integers with ordinary multiplication donot form a group.

Example 2.3. Consider the elements 1,−1 under ordinary multipli-cation. This set is clearly closed under multiplication and associativityis manifestly satisfied. The unit element is 1 and each element is itsown inverse. Hence, the set 1,−1 is a two-element group under mul-tiplication.

Example 2.4. Consider the set of 2×2 matrices with real entries(a b

c d

), (2.1)

such that the determinant, ad − bc, is non-zero. The composition lawis the usual rule for matrix multiplication:(

a1 b1

c1 d1

)(a2 b2

c2 d2

)=

(a1a2 + b1c2 a1b2 + b1d2

c1a2 + d1c2 c1b2 + d1d2

).

To determine if this set of matrices forms a group, we must first showthat the product of two matrices with non-zero determinant is also amatrix with non-zero determinant. This follows from that fact thatfor any pair of 2×2 matrices A and B, their determinants, denotedby det(A) and det(B), satisfy det(AB) = (detA)(detB). Associativitycan be verified with a straightforward, but laborious, calculation. Theidentity is (

1 0

0 1

)


and the inverse of (2.1) is

1

ad− bc

(d −b−c a

),

which explains the requirement that ad−bc 6= 0. This group is denotedby GL(2,R), for general linear group of 2×2 matrices with real entries.Note that the elements of this group form a continuous set, so GL(2,R)is a continuous group.

2.2 Permutation Groups

A permutation of n objects is a rearrangement of those objects. Whencombined with the usual rule for function composition for successivepermutations (see below), these permutations are endowed with thestructure of a group, which is denoted by Sn. At one time, permuta-tion groups were the only groups studied by mathematicians and theymaintain a special status in the subject through Cayley’s theorem , whichestablishes a relationship between Sn and every group with n elements.In this section, we will examine the structure of S3, both as an abstractgroup and as the symmetry group of an equilateral triangle.

The group S3 is the set of all permutations of three distinguishableobjects, where each element corresponds to a particular permutationof the three objects from a given reference order. Since the first objectcan be put into any one of three positions, the second object into eitherof two positions, and the last object only into the remaining position,there are 3× 2× 1 = 6 elements in the set. These are listed below:

e =

(1 2 3

1 2 3

)a =

(1 2 3

2 1 3

)b =

(1 2 3

1 3 2

)

c =

(1 2 3

3 2 1

)d =

(1 2 3

3 1 2

)f =

(1 2 3

2 3 1

)

In this notation, the top line represents the initial, or reference, orderof the objects and the bottom line represents the effect of the per-mutation. The composition law corresponds to performing successive


permutations and is carried out by rearranging the objects according tothe first permutation and then using this as the reference order to rear-range the objects according to the second permutation. As an example,consider the product ad, where we will use the convention that opera-tions are performed from right to left, i.e., permutation d is performedfirst, followed by permutation a. Element d permutes the referenceorder (1, 2, 3) into (3, 1, 2). Element a then permutes this by puttingthe first object in the second position, the second object into the firstposition, and leaves the third object in position three, i.e.,

a =

(1 2 3

2 1 3

)=

(3 1 2

1 3 2

).

Notice that it is only the permutation of the distinct objects, not theirlabelling, which is important for specifying the permutation. Hence,

ad =

(1 2 3

1 3 2

)= b ,

An analogous procedure shows that

da =

(2 1 3

3 2 1

)(1 2 3

2 1 3

)=

(1 2 3

3 2 1

)= c ,

which shows that the composition law is not commutative, so S3 is anon-Abelian group.

A geometric realization of S3 can be established by considering thesymmetry transformations of an equilateral triangle (Fig. 2.1). Theelements a, b, and c correspond to reflections through lines which in-tersect the vertices at 3, 1, and 2, respectively, and d and f correspondto clockwise rotations of this triangle by 2

3π and 4

3π radians, respec-

tively. The effects of each of these transformations on the positionsof the vertices of the triangle is identical with the corresponding ele-ment of S3. Thus, there is a one-to-one correspondence between thesetransformations and the elements of S3. Moreover, this correspondenceis preserved by the composition laws in the two groups. Consider forexample, the products ad and da calculated above for S3. For the equi-lateral triangle, the product ad corresponds to a rotation followed by


3

2

1c

3

1

2d

2

3

1f

1

2

3e

2

1

3a

1

3

2b

Figure 2.1: The symmetry transformations of an equilateral triangle labelledby the corresponding elements of S3. The lines in the diagram correspondingto the identity are lines through which reflections of transformations a, b andc are taken. The transformations d and f are rotations.

a reflection. Thus, beginning with the standard order shown for theidentity the successive application of these transformations is shownbelow:

1

2

3

1

23

3

21

d a

By comparing with Fig. 2.1, we see that the result of these transfor-mations is equivalent to the transformation b. Similarly, one can showthat da = c and, in fact, that all the products in S3 are identical tothose of the symmetry transformations of the equilateral triangle. Twosuch groups that have the same algebraic structure are said to be iso-morphic to one another and are, to all intents and purposes, identical.This highlights the fact that it is the algebraic structure of the group


which is important, not any particular realization of the group. Furtherdiscussion of this point will be taken up in the next chapter.

2.3 Elementary Properties of Groups

The examples in the preceding section showed that all groups are en-dowed with several general properties. In this section, we deduce someadditional properties which, although evident in particular examples,can be shown generally to follow from the properties of abstract groups.

Theorem 2.1. (Uniqueness of the identity) The identity element ina group G is unique.

Proof. Suppose there are two identity elements e and e′ in G. Then,according to the definition of a group, we must have that

ae = a

and

e′a = a

for all a in G. Setting a = e′ in the first of these equations and a = ein the second shows that

e′ = e′e = e ,

so e = e′.

This theorem enables us to speak of the identity e of a group. Thenotation e is derived from the German word Einheit for unity.

Another property common to all groups is the cancellation of com-mon factors within equations. This property owes its existence to theassociativity of the group composition rule.


Theorem 2.2. (Cancellation) In a group G, the left and right cancel-lation laws hold, i.e., ab = ac implies b = c and ba = ca implies b = c.

Proof. Suppose that ab = ac. Let a−1 be an inverse of a. Then, byleft-multiplying by this inverse,

a−1(ab) = a−1(ac)

and invoking associativity,

(a−1a)b = (a−1a)c ,

we obtain

eb = ec ,

so b = c. Similarly, beginning with ba = ca and right-multiplying bya−1 shows that b = c in this case also.

Notice that the proof of this theorem does not require the inverseof a group element to be unique; only the existence of an inverse wasrequired. In fact, the cancellation theorem can be used to prove thatinverses are, indeed, unique.

Theorem 2.3. (Uniqueness of inverses) For each element a in agroup G, there is a unique element b in G such that ab = ba = e.

Proof. Suppose that there are two inverses b and c of a. Then ab = eand ac = e. Thus, ab = ac, so by the Cancellation Theorem, b = c.

As in the case of the identity of a group, we may now speak of theinverse of every element in a group, which we denote by a−1. As wasdiscussed in Sec. 2.1, this notation is borrowed from ordinary multipli-cation, as are most other notations for the group composition rule. Forexample, the n-fold product of a group element g with itself is denoted


by gn. Similarly gngm = gn+m, which conforms to the usual rule of ex-ponents for real numbers. However, there are some notable exceptions.For two group elements a and b, the equality of (ab)n and anbn doesnot generally hold. As the examples in Sec. 2.1 demonstrated, as longas this notation is interpreted in the context of the appropriate groupcomposition rule, no confusion should arise.

2.4 Discrete and Continuous Groups

Groups are divided into two general categories: discrete and continuous.The basis definitions apply to both types of group, but the discussionof a number of properties depends sensitively on the discrete or con-tinuous nature of the group. In this course, we will focus our attentionon discrete groups first, to establish a conceptual base, and considercontinuous groups later in the course.

2.4.1 Finite Groups

One of the most fundamental properties of a group G is number ofelements contained in the group. This is termed the order of G and isdenoted by |G|. The group Z of integers under addition, has infiniteorder and the order of S3, the group of permutations of three objects,is 6. We will be concerned initially with finite groups which, apartfrom their applicability to a range of physical problems, have a numberinteresting arithmetic properties.

Finite groups also have properties which are not shared by eitherinfinite or continuous groups. For example, if an element g of a finitegroup G is multiplied by itself enough times, the unit e is eventually re-covered. Clearly, multiplying any element g by itself a number of timesgreater than |G| must eventually lead to a recurrence of the product,since the number of distinct products is bounded from above by |G|.To show this explicitly, we denote a recurring product by a and write

a = gp = gq ,

where p = q + n. Then, by using the associativity of the composition


law, gq+n = gqgn = gngq, so

gp = gqgn = gngq = gq ,

and, from the definition of the identity and its uniqueness, we concludethat

gn = e .

Thus, the set of elements g, g2, g3, . . . represents a recurring sequence.The order of an element g, denoted by |g|, is the smallest value of ksuch that gk = e. The period of such an element g is the collection ofelements e, g, g2, . . . , gk−1.

Example 2.5. Using S3 as an example, |a| = |b| = |c| = 2 and|d| = |f | = 3. The corresponding periods are e, a, e, b, e, c,and e, d, f = d2.

Theorem 2.4. (Rearrangement Theorem) If e, g1, g2, . . . , gn arethe elements of a group G, and if gk is an arbitrary group element,then the set of elements

Ggk = egk, g1gk, g2gk, . . . , gngk

contains each group element once and only once.

Proof. The set Ggk contains |G| elements. Suppose two elements ofGgk are equal: gigk = gjgk. By the Cancellation Theorem, we must havethat gi = gj. Hence, each group element appears once and only once inGgk, so the sets G and Ggk are identical apart from a rearrangementof the order of the elements if gk is not the identity.


2.4.2 Multiplication Tables

One application of this theorem is in the representation of the compo-sition law for a finite group as a multiplication table . Such a table is asquare array with the rows and columns labelled by the elements of thegroup and the entries corresponding to the products, i.e., the elementgij in the ith row and jth column is the product of the element gi la-belling that row and the element gj labelling that column: gij = gigj.To see how the construction of multiplication tables proceeds by utiliz-ing only the abstract group properties, consider the simplest nontrivialgroup, that with two distinct elements e, a. We clearly must have theproducts e2 = e and ea = ae = a. The Rearrangement Theorem thenrequires that a2 = e. The multiplication table for this group is shownbelow:

e ae e aa a e

Note that the entries of this table are symmetric about the main diag-onal, which implies that this group is Abelian.

Now consider the group with three distinct elements: e, a, b. Theonly products which we must determine explicitly are ab, ba, a2, and b2

since all other products involve the unit e. The product ab cannot equala or b, since that would imply that either b = e or a = e, respectively.Thus, ab = e. The Cancellation Theorem then requires that a2 = e,b2 = a, and ba = e. The multiplication table for this group is shownbelow:

e a be e a ba a b eb b e a

Because the entries of this table are symmetric about the main diagonal,this group is also Abelian. Our procedure shows that every groupwith two or three elements must have the multiplication tables just


calculated, i.e., the algebraic structures of group with two and threeelements are unique! Thus, we can speak of the group with two elementsand the group with three elements. A similar procedure for groupswith four elements e, a, b, c yields two distinct multiplication tables(Problem Set 2). As a final example, the multiplication table for S3 isshown below:

e a b c d fe e a b c d fa a e d f b cb b f e d c ac c d f e a bd d c a b f ef f b c a e d

As is immediately evident from this table, S3 not Abelian.

2.5 Subgroups and Cosets

If, from a group G, we select a subset H of elements which themselvesform a group under the same composition law, H is said to be a sub-group of G. According to this definition, the unit element e forms asubgroup of G, and G is a subgroup of itself. These are termed impropersubgroups. The determination of proper subgroups is one of the cen-tral concerns of group theory. In physical applications, subgroups arisein the description of symmetry-breaking, where a term is added to aHamiltonian or a Lagrangian which lowers the symmetry to a subgroupof the original symmetry operations.

Example 2.6. The group S3 has a number of proper subgroups: e, a,e, b, e, c, and e, d, f. The identification of these subgroups ismost easily carried out by referring to the symmetry operations of anequilateral triangle (Fig. 2.1).


If H = e, h1, h2, . . . , hr is a subgroup of a group G, and g is anelement of G, then the set

Hg = eg, h1g, h2g, . . . , hrg

is a right coset of H. Similarly, the set

gH = ge, gh1, gh2, . . . , ghr

is a left coset of H. A coset need not be a subgroup; it will be a subgrouponly if g is an element of H.

Theorem 2.5. Two cosets of a subgroup either contain exactly thesame elements or else have no common elements.

Proof. These cosets either have no common elements or have atleast one common element. We will show that if there is a single incommon, then all elements are common to both subgroups. Let Hg1

and Hg2 be two right cosets. If one common element of these cosets ishig1 = hjg2, then

g2g−11 = h−1

j hi

so g2g−11 is in H. But also contained in H are the elements

Hg2g−11 = eg2g

−11 , h1g2g

−11 , h2g2g

−11 , . . . , hrg2g

−11

since, according to the Rearrangement Theorem, each element of Happears once and only once in this sequence. Therefore, the elementsof Hg1 are identical to those of

(Hg2g−11 )g1 = Hg2(g−1

1 g1) = Hg2

so these two cosets have only common elements.

Example 2.7. Consider again the group S3 and its subgroupH = e, a(Example 2.6). The right cosets of this subgroup are

e, ae = e, a, e, aa = a, e, e, ab = b, d

e, ac = c, f, e, ad = d, b, e, af = f, c


We see that there are three distinct right cosets of e, a,

e, a, b, d, c, f

of which only the first is a subgroup (why?). Similarly, there are threeleft cosets of e, a:

e, a, c, d, b, f

Notice that the left and right cosets are not the same.

Theorem 2.6 (Lagrange’s theorem). The order of a subgroup H ofa finite group G is a divisor of the order of G, i.e., |H| divides |G|.

Proof. Cosets either have all elements in common or they are dis-tinct (Theorem 2.5). This fact, combined with the Rearrangement The-orem, means that every element of the group must appear in exactlyone distinct coset. Thus, since each coset clearly has the same numberof elements, the number of distinct cosets, which is called the indexof the subgroup, multiplied by the number of elements in the coset, isequal to the order of the group. Hence, since the order of the coset andthe subgroup are equal, the order of the group divided by the order ofthe subgroup is equal to the number of distinct cosets, i.e., an integer.

Example 2.8. The subgroup e, a of S3 is of order 2 and the subgroupe, d, f is of order 3. Both 2 and 3 are divisors of |S3| = 6.

Lagrange’s theorem identifies the allowable orders of the subgroupsof a given group. But the converse of Lagrange’s theorem is not gener-ally valid, i.e., the orders of the subgroups of a group G need not spanthe divisors of G.


2.6 The Quotient Group

2.6.1 Conjugacy Classes

Two elements a and b of a group G are said to be conjugate if there isan element g in the group, called the conjugating element, such thata = gbg−1. Conjugation is an example of what is called an equivalencerelation , which is denoted by “≡,” and is defined by three conditions:

1. a ≡ a (reflexive).

2. If a ≡ b, then b ≡ a (symmetric).

3. If a ≡ b and b ≡ c, then a ≡ c (transitive).

To show that conjugacy corresponds to an equivalence relation weconsider each of these conditions in turn. By choosing g = e as theconjugating element, we have that a = eae−1 = a, so a ≡ a. If a ≡ b,then a = gbg−1, which we can rewrite as

g−1ag = g−1a(g−1)−1 = b

so b ≡ a, with g−1 as the conjugating element. Finally, to show tran-sitivity, the relations a ≡ b and b ≡ c imply that there are elements g1

and g2 such that b = g1ag−11 and c = g2bg

−12 . Hence,

c = g2bg−12 = g2g1ag

−11 g−1

2 = (g2g1)a(g2g1)−1

so c is conjugate to a with the conjugating element g1g2. Thus, conju-gation fulfills the three conditions of an equivalency class.

One important consequence of equivalence is that it permits theassembly of classes , i.e., sets of equivalent quantities. In particular,a conjugacy class is the totality of elements which can be obtainedfrom a given group element by conjugation. Group elements in thesame conjugacy class have several common properties. For example,all elements of the same class have the same order. To see this, webegin with the definition of the order n of an element a as the smallestinteger such that an = e. An arbitrary conjugate b of a is b = gag−1.Hence,

bn = (gag−1)(gag−1) · · · (gag−1)︸︷︷︸n factors

= gang−1 = geg−1 = e


so b has the same order as a.

Example 2.9. The group S3 has three classes: e, a, b, c, and d, f.As we discussed in Example 2.5, the order of a, b, and c is two, and theorder of d and f is 3. The order of the unit element is 1 and is alwaysin a class by itself. Notice that each class corresponds to a distinct kindof symmetry operation on an equilateral triangle. The operations a, b,and c correspond to reflections, while d and f correspond to rotations.In terms of operations in S3, the elements d and f correspond to cyclicpermutations of the reference order, e.g., 1 → 2, 2 → 3, and 3 → 1,while a, b, and c correspond to permutations which are not cyclic.

2.6.2 Self-Conjugate Subgroups

A subgroup H of G is self-conjugate if the elements gHg−1 are identicalwith those of H for all elements g of G. The terms invariant subgroupand normal subgroup are also used. A group with no self-conjugateproper subgroups is called simple . If gHg−1 = H for all g in G, thengiven an element h1 in H, for any a, we can find an element h2 in H suchthat ah1a

−1 = h2, which implies that ah1 = h2a, or that aH = Ha.This last equality yields another definition of a self-conjugate subgroupas one whose left and right cosets are equal. From the definition of self-conjugacy and of classes, we can furthermore conclude that a subgroupH of G is self-conjugate if and only if it contains elements of G incomplete classes, i.e., H contains either all or none of the elements ofclasses of G.

The cosets of a self-conjugate subgroup are themselves endowed witha group structure, with multiplication corresponding to an element-by-element composition of two cosets and discounting duplicate products.We show first that the multiplication of the elements of two right cosetsof a conjugate subgroup yields another right coset. Let H be a self-conjugate subgroup of G and consider the two right cosets Ha and Hb.Then, the multiplication of Ha and Hb produces products of the form

hiahjb = hi(ahj)b


The product ahj can be written as hka for some hk in H, since H isassumed to be self-conjugate. Thus, we have

hi(ahj)b = hi(hka)b = (hihk)(ab)

which is clearly an element of a right coset of H.

Example 2.10. Consider the subgroup e, d, f of S3. Right-multiplyingthis subgroup by each element of S3 yields the right cosets of this sub-group:

e, d, fe = e, d, f, e, d, fa = a, c, b, e, d, fb = b, a, c

e, d, fc = c, b, a, e, d, fd = d, f, e, e, d, ff = f, e, d

Similarly, left-multiplying by each element of S3 produces the left cosetsof this subgroup:

ee, d, f = e, d, f, ae, d, f = a, b, c, be, d, f = b, c, a

ce, d, f = c, a, b, de, d, f = d, f, e, fe, d, f = f, e, d

Thus, since the right and left cosets of e, d, f are the same, theseelements form a self-conjugate subgroup of S3 whose distinct cosetsare e, d, f and a, b, c. Multiplying these subgroups together andneglecting duplicate elements yields

e, d, fe, d, f = e, d, f, e, d, fa, b, c = a, b, c

a, b, ce, d, f = a, b, c, a, b, ca, b, c = e, d, f

The quotient group (also called the factor group ) of a self-conjugatesubgroup is the collection of cosets, each being considered an element.The order of the factor group is equal to the index of the self-conjugatesubgroup. With the notation used above, the quotient group is denotedby G/H.


Example 2.11. The cosets of the self-conjugate subgroup e, d, f ofS3 are e, d, f and a, b, c, so the order of the factor group is two. Ifwe use the notation

E = e, d, f, A = a, b, c (2.2)

for the elements of the factor group, we can use the results of Example2.8 to construct the multiplication table for this group (shown below)from which see that E is the identity of the factor group, and E and

E AE E AA A E

A are their own inverses. Note that this multiplication table has theidentical structure as the two-element group e, a discussed in Sec. 2.4.

2.7 Summary

In this chapter, we have covered only the most basic properties ofgroups. One of the remarkable aspects of this subject, already evidentin some of the discussion here, is that the four properties that define agroup, have such an enormous implication for the properties of groups,quite apart from their implications for physical applications, which willbe explored throughout this course. A comprehensive discussion of themathematical theory of groups, including many wider issues in bothpure and applied mathematics, may be found in the book by Gallian.5

5J.A. Gallian, Contemporary Abstract Algebra 4th edn. (Houghton Mifflin,Boston, 1998).

Chapter 3

Representations of Groups

How can it be that mathematics, being after all a product of humanthought which is independent of experience, is so admirably appropriateto the objects of reality?

—Albert Einstein

The structure of abstract groups developed in Chapter 2 forms thebasis for the application of group theory to physical problems. Typi-cally in such applications, the group elements correspond to symmetryoperations which are carried out on spatial coordinates. When theseoperations are represented as linear transformations with respect to acoordinate system, the resulting matrices, together with the usual rulefor matrix multiplication, form a group that is equivalent to the groupof symmetry operations in a sense to be made precise later in this chap-ter. In essence, these matrices form what is called a representation ofthe symmetry group with each element corresponding to a particularmatrix.

For applications to quantum mechanics, as we have seen in Sec-tion 1.2, the symmetry operations are performed on the Hamiltonian,whose invariance properties determine the symmetry group. The wave-functions, which do not all share the symmetry of the Hamiltonian,will be seen to determine the representations of the symmetry group inthe sense described above. These representations will, in turn, providea classification scheme for the eigenfunctions of the Hamiltonian, in

33

34 Representations of Groups

analogous fashion to the identification of even and odd eigenfunctionsin Section 1.2. The strength of the group-theoretic formalism that wewill develop is that this procedure can be carried out in a systematicfashion for a Hamiltonian having any symmetry without undue com-putational effort.

In this chapter, we will set out the basic definitions that enable us toconstruct a mathematical definition of what we mean by a representa-tion and discuss the basic types of representation. In the next chapterwe will develop a number of remarkable properties of representationsthat lie at the heart of applications of discrete group theory to quantummechanics.

3.1 Homomorphisms and Isomorphisms

Consider two finite groups G and G′ with elements e, a, b, . . . ande′, a′, b′, . . . and which need not be of the same order. Suppose thereis a mapping φ between the elements of G and G′ which preserves theircomposition rules, i.e., if a′ = φ(a) and b′ = φ(b), then

φ(ab) = φ(a)φ(b) = a′b′

If the order of the two groups is the same, then this mapping is said tobe an isomorphism and the two groups are isomorphic to one another.This is denoted by G ≈ G′. If the order of the two groups is notthe same, then the mapping is a homomorphism and the two groupsare said to be homomorphic . Thus, an isomorphism is a one-to-onecorrespondence between two groups, while a homomorphism is a many-to-one correspondence. An isomorphism preserves the structure of theoriginal group, but a homomorphism causes some of the structure of theoriginal group to be lost. Both properties are reflected in the behaviorof multiplication tables under these mappings. Homomorphisms andisomorphisms are not limited to finite groups nor even to groups withdiscrete elements.

Example 3.1. We saw in Sec. 2.2 that S3 is isomorphic to the planarsymmetry operations of an equilateral triangle, since there is a one-to-one correspondence between the elements of the two groups and they

Representations of Groups 35

have the same multiplication table. On the other hand, consider thecorrespondence between the elements of S3 and the elements of thequotient group of S3 discussed in Examples 2.9 and 2.11):

e, d, f 7→ E, a, b, c 7→ A (3.1)

i.e. the mapping φ is defined by

φ(e) = E φ(d) = E φ(f) = Eφ(a) = A φ(b) = A φ(c) = A

(3.2)

This is a homomorphism because three elements of S3 correspond to asingle element of the quotient group. To see that this mapping preservesmultiplication, we rearrange the multiplication table of S3 (Example2.2) as follows:

e d f a b ce e d f a b cd d f e c a bf f e d b c aa a b c e d fb b c a f e dc c a b d f e

7→E A

E E AA A E

where the mapping of the multiplication table onto the elements E ,Ais precisely that of a two-element group (cf. Example 2.9). This homo-morphism clearly causes some of the structure of the original group tobe lost. For example, S3 is non-Abelian group, but the two-elementgroup is Abelian.

3.2 Representations

A representation of dimension n of an abstract group G is a homomor-phism or isomorphism between the elements of G and the group ofnonsingular n × n matrices (i.e. n × n matrices with non-zero deter-minant) with complex entries and with ordinary matrix multiplicationas the composition law (Example 2.4). An isomorphic representation


x

y

Figure 3.1: The coordinate system used to generate a two-dimensional rep-resentation of the symmetry group of the equilateral triangle. The origin ofthe coordinate system coincides with the geometric center of the triangle.

is called a faithful representation and a homomorphic representation iscalled an unfaithful representation.

According to this definition, if elements a and b of G are assignedmatrices D(a) and D(b), then D(a)D(b) = D(ab). The nonsingularnature of the matrices is required because inverses must be containedin the set (Example 2.4). Representations can also be comprised ofnumbers; the dimensionality of such representations is unity.

Example 3.2. Consider the following matrix representation of S3 basedon the correspondence with planar symmetry operations of an equilat-eral triangle:

e =

(1 0

0 1

), a = 1

2

(1 −

√3

−√

3 −1

), b = 1

2

(1√

3√

3 −1

)

c =

(−1 0

0 1

), d = 1

2

(−1 −√

3√

3 −1

), f = 1

2

( −1√

3

−√

3 −1

)(3.3)

These matrices were generated by regarding each of the symmetry op-erations as a linear transformation in the coordinate system shown inFig. 3.1. Matrices a, b, and c correspond to reflections, so their deter-minant is −1, while matrices d and f correspond to rotations, so theirdeterminant is 1. These matrices form a faithful representation of S3.


Consider now the following mapping between the elements of S3

and the set 1,−1:

e, f 7→ 1, a, b, c 7→ −1 (3.4)

This is essentially a mapping between the elements of S3 and the de-terminant of their matrix representation discussed above. Thus, theidentity matrix e and the rotations d and f have determinants of 1,while the reflections a, b, and c have determinants of −1. The physicalinterpretation of this homomorphism is therefore as a mapping fromthe individual elements of S3 to their parity, i.e., whether they changethe orientation of the coordinate system (−1) or not (1).1 Since thedeterminant provides less information about a transformation than itsmatrix representation, it is clear that some information about the groupstructure of S3 is not preserved by this homomorphism.

Finally, we note that the mapping of all elements to unity,

e, a, b, c, d, f 7→ 1

is a representation of any group, though clearly an unfaithful one. Thisis called the identical representation . In the present case, the identicalrepresentation corresponds to a mapping from the group element tothe absolute value of the determinant. Since all of the transformationspreserve the lengths of vectors, any product of these transformationsdoes so as well.

Representations of groups are important in quantum mechanics forseveral reasons. First, the eigenfunctions of a Hamiltonian will trans-form under the symmetry operations of that Hamiltonian according toa particular representation of that group. Second, quantum mechanicaloperators are often written in terms of their matrix elements, so it isconvenient to write symmetry operations in the same kind of matrixrepresentation. Moreover, the evaluation of these matrix elements maysometimes be simplified by identifying the appropriate selection rules

1In terms of the operations of S3, even parity corresponds to an even numberof pairwise interchanges, while odd parity corresponds to an odd number of suchinterchanges.


(Section 1.2). Finally, the algebra of matrices is generally simpler tocarry out than abstract symmetry operations. Thus, in the next section,we discuss some of the important properties of matrix representationsof groups.

3.3 Reducible and Irreducible Represen-

tations

The definition of a representation provides for considerable flexibilityin constructing matrix representations, which is manifested in severalways, but also indicates that representations are not unique. We con-sider some examples.

Given a matrix representationD(e), D(a), D(b), . . .

of an abstract group with elements e, a, b, . . ., we can obtain a new setof matrices which also form a representation by performing a transfor-mation known variously as a similarity , equivalence , or canonical trans-formation (cf. Sec. 2.6):

BD(e)B−1, BD(a)B−1, BD(b)B−1, . . .

(3.5)

Such transformations arise quite naturally, for example, in carrying outa change of basis for a set of matrices. Thus, suppose one begins withthe matrix equation b = Aa relating two vectors a and b through atransformation A. If we now wish to express this equation in anotherbasis which is obtained from the original basis by applying a transfor-mation B, we can write

Bb = BAa = BAB−1Ba

so in the new basis, our original equation becomes

b′ = A′a′

where b′ = Bb, a′ = Ba, and A′ = BAB−1. A similarity transfor-mation can therefore be interpreted as a sequence of transformations


involving first a transformation to the original basis (B−1), then per-forming the transformation A, and finally transforming back to thenew basis (B). Referring back to the discussion in Section 2.6 on con-jugacy classes, we see that group elements in the same conjugacy classrepresent the same type of transformation (e.g., reflection or rotation)which can be transformed into one another by particular symmetryoperations.

Suppose we have representations of dimensions m and n. We canconstruct a representation of dimensionm+n by forming block-diagonalmatrices:[

D(e) 0

0 D′(e)

],

[D(a) 0

0 D′(a)

],

[D(b) 0

0 D′(b)

], . . .

(3.6)

where D(e), D(a), D(b), . . . is an n-dimensional representation andD′(e), D′(a), D′(b), . . . an m-dimensional representation of the groupG, and the symbol 0 is an n×m or an m×n zero matrix, as required byits position in the supermatrix. Each of the m+n-dimensional matricesformed in this manner is called a direct sum of the n- and m-dimensionalcomponent matrices. The direct sum is denoted by “⊕” to distinguishit from the ordinary addition of two matrices. Thus, we can write therepresentation in (3.6) as

D(e)⊕D′(e), D(a)⊕D′(a), D(b)⊕D′(b), · · ·

The two representations that form this direct sum can be either distinctor identical and, of course, the block-diagonal form can be continuedindefinitely simply by incorporating additional representations in diag-onal blocks. However, in all such constructions, we are not actuallygenerating anything intrinsically new; we are simply reproducing theproperties of known representations. Thus, although representationsare a convenient way of associating matrices with group elements, thefreedom we have in constructing representations, exemplified in (3.5)and (3.6), does not readily demonstrate that these matrices embodyany intrinsic characteristics of the group they represent. Accordingly,we now describe a way of classifying equivalent representations andthen introduce a refinement of our definition of a representation.


To overcome the problem of nonuniqueness posed by representationsthat are related by similarity transformations we consider the sum ofthe diagonal elements of an n × n matrix A, called the trace of A andby “tr”:

tr(A) =n∑i=1

Aii

The utility of the trace stems from its invariance under similarity trans-formations, i.e.,

tr(A) = tr(BAB−1)

The importance of this invariance, the proof of which is discussed inProblem Set 4, is that, although there is an infinite variety of represen-tations related by similarity transformations, each such representationhas the same set of traces associated with each of its elements.

But working with the trace alone does not alleviate the nonunique-ness of representations posed by (3.6). To address this issue, we in-troduce the concept of an irreducible representation. Representationssuch as those in (3.6) are termed reducible because they are the directsum of two (or more) representations. We could, of course, performa similarity transformation to obtain a representation that is not inblock form, but the representation so obtained is still deemed to bereducible because it was obtained from matrices which originally werein block form. Based on these considerations, we define reducible andirreducible representations as follows:

Definition. If the same similarity transformation brings all of the ma-trices of a representation into the same block form (by which we meanmatrices of the same dimension in the same positions), then this repre-sentation is said to be reducible . Otherwise, the representation is saidto be irreducible .

Thus, irreducible representations cannot be expressed in terms ofrepresentations of lower dimensionality. One-dimensional representa-tions are, by definition, always irreducible. Determining the irreducible


representations of groups is one of the central issues to be covered inthe following chapters.

Example 3.4. All of the representations of S3 discussed in Example3.2 are irreducible. This is clear for the identical representation and forthe representation in (3.4), since they are composed of numbers. Butwe can use these representations to construct the following manifestlyreducible representation of S3:

e =

(1 0

0 1

), a =

(1 0

0 −1

), b =

(1 0

0 −1

)

c =

(1 0

0 −1

), d =

(1 0

0 1

), f =

(1 0

0 1

)

The representation in (3.3) is irreducible. There is no similaritytransformation that will bring all of the matrices into block-diagonalform which, for the case here, means simple diagonalization. The easiestway to see this is from the point of view of the commutability of twomatrices. If two matrices can be brought into diagonal form by the samesimilarity transformation, then they commute. As diagonal matrices,they certainly commute, so they must also commute in their originalform. But a glance at the multiplication table for these matrices (recallthat they are a faithful representation of S3) in Example 2.2 shows thatthey do not all commute. Hence, they cannot all be simultaneouslydiagonalized, so this representation is irreducible.

3.4 Unitary Representations

Representations of groups are useful because of orthogonality theoremswhich we will prove in the next chapter. As background to that dis-cussion, we will prove in this section an important result about theunitarity of representations. But we first review some general proper-ties of matrices.

We begin by considering the transformation of an n × n matrix Awith entries Aij, i, j = 1, 2, . . . , n, under the action of various opera-


tions. The complex conjugate of A, denoted by A∗, has entries whichare the complex conjugates of the corresponding entries of A:

(A∗)ij = (Aij)∗ (3.7)

The transpose of A, denoted by At, has its rows and columns inter-changed with respect to those of A:

(At)ij = Aji (3.8)

When applied to vectors, the transpose transforms a row vector into acolumn vector and vice versa. The transpose of a product of matricesA,B,C, . . . is

(ABC · · ·)t = · · ·C tB tA t (3.9)

i.e., the order of matrix multiplication is reversed. This can be proveneasily from the definition (3.8). Finally, the adjoint or Hermitian conju-gate of A, denoted by A†, is the transposed conjugate of A, i.e.

(A†)ij = (Aji)∗ (3.10)

In common with the transpose, the application of the Hermitian conju-gate to a product of matrices A,B,C, . . . can be expressed as a productof Hermitian conjugates of the individual matrices, but with the orderreversed:

(ABC · · ·)† = · · ·C †B †A † (3.11)

3.4.1 Hermitian and Orthogonal Matrices

A matrix A is Hermitian if

A† = A (3.12)

Hermitian matrices and Hermitian operators are familiar from quan-tum mechanics, where their properties of having real eigenvalues andorthogonal eigenvectors are of fundamental importance. A matrix A isorthogonal if its transpose is its inverse:

AtA = AAt = I (3.13)


where I is the n× n unit matrix. In terms of matrix components, thiscondition reads

n∑k=1

akiakj =n∑k=1

aikajk = δij (3.14)

where

δij =

0, if i 6= j;

1, if i = j(3.15)

is the Kronecker delta . Thus, the rows of an orthogonal matrix aremutually orthogonal, as are the columns. The consequences of theorthogonality of a transformation matrix can be seen by examining theeffect of applying an orthogonal matrix A to two n-dimensional vectorsu and v, yielding vectors u′ and v′:

u′ = Au, v′ = Av

We now take the scalar, or ‘dot,’ product between u′ and v′:

(u′,v′) = (u′)tv′ = (Au)tAv = utAtAv = utv = (u,v) (3.16)

where we have used (3.11) and the fact that A is orthogonal. Thisshows that the relative orientations and the lengths of vectors are pre-served by orthogonal transformations. Such transformations are eitherrigid rotations, which preserve the “handedness” (i.e., left or right)of a coordinate system, and are called proper rotations, or reflections,which reverse the “handedness” of a coordinate system, and are calledimproper “rotations.”

3.4.2 Unitary Matrices

A third type of matrix, called unitary , has the property that

A†A = AA† = I (3.17)

By writing this condition in terms of matrix components,

n∑k=1

a∗kiakj =n∑k=1

aika∗jk = δij (3.18)


we see that, in common with orthogonal matrices, the rows and columnsof a unitary matrix are orthogonal, but with respect to a different scalarproduct. For two vectors u and v, this scalar product is defined as

(u,v) = u†v (3.19)

This generalizes the familiar dot product to complex vectors in n di-mensions. We can now show, by proceeding as above, that unitarytransformations leave the scalar product invariant:

(u′,v′) = (u′)†v′ = (Au)†Av = u†A†Av = u†v = (u,v) (3.20)

The property of unitarity, when applied to operators, is of immenseimportance in quantum mechanics because it enables changes of basesto be performed while preserving the orthogonality of bases and, thus,the overlap between wavefunctions. In this sense, unitary matricesare associated with proper and improper “rotations,” in analogy withorthogonal matrices.

3.4.3 Diagonalization of Hermitian Matrices∗

Let H be an n× n Hermitian matrix. The eigenvalue equation for thismatrix is

Ha = λa (3.21)

By writing this equation as

(H − λI)a = 0 (3.22)

the eigenvalue equation in (3.21) has nontrivial solutions for a if andonly if the determinant of the matrix of coefficients in (3.22) vanishes:

det(H − λI) = 0 (3.23)

The expansion of the determinant leads to an nth-order polynomial inλ whose solution yields the n (not necessarily distinct) eigenvalues ofH: λ1, λ2, . . . , λn.

We now show that the eigenvectors of H which correspond to dis-tinct eigenvalues are orthogonal. Consider the eigenvalue equations for


two eigenvectors a and b corresponding to distinct eigenvalues λ andµ, respectively:

Ha = λa (3.24)

Hb = µb (3.25)

We now take the scalar product between b and (3.24) and that between(3.25) and a:

b†(Ha) = λb†a (3.26)

(b†H†)a = µb†a (3.27)

Subtracting (3.27) from (3.26), and using the fact that H is Hermitianyields

(λ− µ)b†a = b†Ha− b†H†a = 0 (3.28)

which, since λ 6= µ, implies that b†a = 0, i.e., that a and b are or-thogonal. If λ and µ are not distinct, we must use a Gram–Schmidtprocedure to explicitly construct an orthogonal set of eigenvectors as-sociated with the degenerate eigenvalue. Thus, the eigenvectors of aHermitian matrix can always be chosen to form an orthogonal set.

Consider the matrix U whose columns are the eigenvectors of H:

U = (a1,a2, . . . ,an)

We can then write (3.21) in a form that subsumes all the eigenvectorsof H as follows:

HU = UD (3.29)

where D is the diagonal matrix whose entries are the eigenvalues of H:

D =

λ1 0 · · · 0

0 λ2 · · · 0

......

. . ....

0 0 · · · λd


Since the rows of U are composed of the (orthogonal) eigenvectors ofH, it has the property that (cf. 3.18)

U †U = UU † = I

i.e., U−1 = U †, so U is unitary. Hence, we can rewrite (3.29) as

U−1HU = U †HU = D

We have proven the following theorem:

Theorem 3.1. Any Hermitian matrix can be diagonalized by an ap-propriate unitary transformation.

This theorem will be used in the next section to prove an importantresult concerning the existence of unitary group representations.

3.4.4 Transformation to Unitary Representations

We have seen in Section 3.3 that there is considerable flexibility inconstructing group representations. In this section, we take a first stepin restricting this freedom by showing that any representation can beexpressed entirely in terms of unitary matrices. Quite apart from theconvenient properties of unitary matrices discussed in Section 3.4.2,this theorem allows to think of group representations as proper andimproper complex “rotations.”

Theorem 3.2. Every representation can be brought into unitary formby a similarity transformation.

Proof. Let A1, A2, . . . , A|G| be a d-dimensional representation ofa group G, i.e., the Aα are a set of |G| d×d matrices with nonvanishingdeterminants. From these matrices we form a matrix H given by thesum

H =|G|∑α=1

AαA†α


This matrix is Hermitian because, using the property (3.11),

H† =∑α

(AαA†α)† =

∑α

AαA†α = H

According to Theorem 3.1, any Hermitian matrix can be diagonalizedby some unitary transformation U . Denoting the diagonalized form ofH by D, we have D = U †HU , which enables us to write D as

D =∑α

U †AαA†αU =

∑α

(U †AαU)(U †A†αU) =∑α

(U †AαU)(U †AαU)†

By introducing the notation Aα = U †AαU , we can write the last equa-tion in a more concise form as

D =∑α

AαA†α (3.30)

The diagonal elements of D are real, because

Dkk =∑α

∑j

(Aα)kj(A†α)jk

=∑α

∑j

(Aα)kj(Aα)∗kj

=∑α

∑j

∣∣∣(Aα)kj

∣∣∣2for k = 1, 2, . . . , d, and positive, because the summation over j includesa diagonal element of the identity, which is a d × d unit matrix, andhence is equal to unity. Thus, the diagonal matrix D1/2,

D1/2 =

D1/211 0 · · · 0

0 D1/222 · · · 0

......

. . ....

0 0 · · · D1/2dd

and D−1/2, which is given by an analogous expression, both have posi-tive entries.


We now form the matrices

Bα = D−1/2AαD1/2

from which we obtain the corresponding Hermitian conjugates:

B†α = (D−1/2AαD1/2)† = D1/2A†αD

−1/2

We will now demonstrate that the Bα are unitary by first showing thatthe product BαB

†α is equal to the identity matrix. The product BαB

†α

is given by

BαB†α =

(D−1/2AαD

1/2)(D1/2A†αD

−1/2)

= D−1/2AαDA†αD−1/2

The definition of D in (3.30) and the associativity of matrix multipli-cation allow us to write this expression as

BαB†α = D−1/2

∑j

AαAβA†βA†αD−1/2

= D−1/2∑j

(AαAβ)(AαAβ)†D−1/2

Since the Aα are a representation of G, then so are the Aα (Problem 3,Problem Set 4). Hence, the product AαAβ is another matrix Aγ in thisrepresentation. Moreover, according to the Rearrangement Theorem,the sum over all β means that the set of Aγ obtained from these prod-ucts contains the matrix corresponding to each group element once andonly once. Thus,

BαB†α = D−1/2

∑γ

AγA†γ︸︷︷︸

D

D−1/2 = I

where I is the d× d unit matrix. This result can also be used to showthat B†αBα = I. Thus, the Bα, which are obtained from the originalrepresentation by a similarity transformation,

Bα = D−1/2U−1AαUD1/2 = (UD1/2)−1Aα(UD1/2)

form a unitary representation of G. Hence, without any loss of gener-ality, we may always assume that a representation is unitary.


3.5 Summary

The main concepts introduced in this chapter are faithful and unfaithfulrepresentations, based on isomorphic and homomorphic mappings, re-spectively, reducible and irreducible representations, and the fact thatwe may confine ourselves to unitary representations of groups. In thenext chapter we will focus on irreducible representations, both faithfuland unfaithful, since these cannot be decomposed into representationsof lower dimension and are, therefore, “intrinsic” to a symmetry group,since all reducible representations will be shown to be composed ofdirect sums of irreducible representations. Irreducible representationsoccupy a special place in group theory because they can be classified fora given symmetry group solely according to their traces and dimension.

Chapter 4

Properties of IrreducibleRepresentations

Algebra is generous; she often gives more than is asked of her.—Jean d’Alembert

We have seen in the preceding chapter that a reducible representa-tion can, through a similarity transformation, be brought into block-diagonal form wherein each block is an irreducible representation. Thus,irreducible representations are the basic components from which allrepresentations can be constructed. But the identification of whether arepresentation is reducible or irreducible is a time-consuming task if itrelies solely on methods of linear algebra.1 In this chapter, we lay thefoundation for a more systematic approach to this question by derivingthe fundamental theorem of representation theory, called the Great Or-thogonality Theorem. The utility of this theorem, and its central rolein the applications of group theory to physical problems, stem from thefact that it leads to simple criteria for determining irreducibility andprovides a direct way of identifying the number of inequivalent repre-sentations for a given group. This theorem is based on two lemmas ofSchur, which are the subjects of the first two sections of this chapter.

1K. Hoffman and R. Kunze, Linear Algebra 2nd edn (Prentice–Hall, EnglewoodCliffs, New Jersey, 1971), Ch. 6,7.

51

52 Properties of Irreducible Representations

4.1 Schur’s First Lemma

Schur’s two lemmas are concerned with the properties of matrices thatcommute with all of the matrices of a irreducible representations. Thefirst lemma addresses the properties of matrices which commute witha given irreducible representation:

Theorem 4.1 (Schur’s First Lemma) . A non-zero matrix which com-mutes with all of the matrices of an irreducible representation is aconstant multiple of the unit matrix.

Proof. Let A1, A2, . . . , A|G| be the matrices of a d-dimensionalirreducible representation of a group G, i.e., the Aα are d× d matriceswhich cannot all be brought into block-diagonal form by the same sim-ilarity transformation. According to Theorem 3.2, we can take thesematrices to be unitary without any loss of generality. Suppose there isa matrix M that commutes with all of the Aα:

MAα = AαM (4.1)

for α = 1, 2, . . . , |G|. By taking the adjoint of each of these equations,we obtain

A†αM† = M †A†α . (4.2)

Since the Aα are unitary, A†α = A−1α , so multiplying (4.2) from the left

and right by Aα yields

M †Aα = AαM† , (4.3)

which demonstrates that, if M commutes with every matrix of a repre-sentation, then so does M †. Therefore, given the commutation relationsin (4.1) and (4.3) any linear combination of M and M † also commutewith these matrices:

(aM + bM †)Aα = Aα(aM + bM †) ,

where a and b are any complex constants. In particular, the linearcombinations

H1 = M +M †, H2 = i(M −M †)

Properties of Irreducible Representations 53

yield Hermitian matrices: Hi = H†i for i = 1, 2. We will now showthat a Hermitian matrix which commutes with all the matrices of anirreducible representation is a constant multiple of the unit matrix. Itthen follows that M is also such a matrix, since

M = 12(H1 − iH2) (4.4)

The commutation between a general Hermitian matrix H and theAα is expressed as

HAα = AαH . (4.5)

Since H is Hermitian, there is a unitary matrix U which transforms Hinto a diagonal matrix D (Theorem 3.1):

D = U−1HU .

We now perform the same similarity transformation on (4.5):

U−1HAiU = U−1HUU−1AiU

= U−1AiHU = U−1AiUU−1HU

By defining Aα = U−1AαU , the transformed commutation relation (4.5)reads

DAα = AαD . (4.6)

Using the fact that D is a diagonal matrix, i.e., that its matrix elementsare Dij = Diiδij, where δij is the Kronecker delta, the (m,n)th matrixelement of the left-hand side of this equation is

(DAα)mn =∑k

Dmk(Aα)kn =∑k

Dmmδmk(Aα)kn = Dmm(Aα)mn .

Similarly, the corresponding matrix element on the right-hand side is

(AαD)mn =∑k

(Aα)mkDkn =∑k

(Aα)mkDnnδkn = (Aα)mnDnn .

Thus, after a simple rearrangement, the (m,n)th matrix element of(4.6) is

(Aα)mn(Dmm −Dnn) = 0 . (4.7)


There are three cases that we must consider to understand the impli-cations of this equation.

Case I. Suppose that all of the diagonal elements of D are dis-tinct: Dmm 6= Dnn if m 6= n. Then, (4.7) implies that

(Aα)mn = 0, m 6= n ,

i.e., the off-diagonal elements of Aα must vanish, these are diagonal ma-trices and, therefore, according to the discussion in Section 3.3, theyform a reducible representation composed of d one-dimensional rep-resentations. Since the Ai are obtained from the Ai by a similaritytransformation, the Ai themselves form a reducible representation.

Case II. If all of the diagonal elements of D are equal, i.e. Dmm =Dnn for all m and n, then D is proportional to the unit matrix. The(Aα)mn are not required to vanish for any m and n. Thus, only thiscase is consistent with the requirement that the Aα form an irreduciblerepresentation. If D is proportional to the unit matrix, then so isH = UDU−1 and, according to (4.4), the matrix M is as well.

Case III. Suppose that the first p diagonal entries of D are equal,but the remaining entries are distinct from these and from each other:D11 = D22 = · · · = Dpp, Dmm 6= Dnn otherwise. The (Aα)mn mustvanish for any pair of unequal diagonal entries. These correspond tothe cases where both m and n lie in the range 1, 2, . . . , p and where mand n are equal and both greater than p, so all the Ai all have thefollowing general form:

Ai =

(B1 0

0 B2

),

where B1 is a p×p matrix and B2 is a (p−d)× (p−d) diagonal matrix.Thus, the Ai are block diagonal matrices and are, therefore, reducible.

We have shown that if a matrix that not a multiple of the unitmatrix commutes with all of the matrices of a representation, thenthat representation is necessarily reducible (Cases I and III). Thus, ifa non-zero matrix commutes with all of the matrices of an irreduciblerepresentation (Case III), that matrix must be a multiple of the unitmatrix. This proves Schur’s lemma.


4.2 Schur’s Second Lemma

Schur’s first lemma is concerned with the commutation of a matrix witha given irreducible representation. The second lemma generalizes thisto the case of commutation with two distinct irreducible representationswhich may have different dimensionalities. Its statement is as follows:

Theorem 4.2 (Schur’s Second Lemma). Let A1, A2, . . . , A|G| andA′1, A′2, . . . , A′|G| be two irreducible representations of a group G ofdimensionalities d and d′, respectively. If there is a matrix M such that

MAα = A′αM

for α = 1, 2, . . . , |G|, then if d = d′, either M = 0 or the two represen-tations differ by a similarity transformation. If d 6= d′, then M = 0.

Proof. Given the commutation relation between M and the twoirreducible representations,

MAα = A′αM , (4.8)

we begin by taking the adjoint:

A†αM† = M †A′†α . (4.9)

Since, according to Theorem 3.2, the Aα may be assumed to be unitary,A†α = A−1

α , so (4.9) becomes

A−1α M † = M †A′−1

α . (4.10)

By multiplying this equation from the left by M ,

MA−1α M † = MM †A′−1

α ,

and utilizing the commutation relation (4.8) to write

MA−1α = A′−1

α M ,

we obtain

A′−1α MM † = MM †A′−1

α .


Thus, the d′ × d′ matrix MM † commutes with all the matrices of anirreducible representation. According to Schur’s First Lemma, MM †

must therefore be a constant multiple of the unit matrix,

MM † = cI , (4.11)

where c is a constant. We now consider individual cases.

Case I. d = d′. If c 6= 0, Eq. (4.11) implies that2

M−1 =1

cM † .

Thus, we can rearrange (4.8) as

Aα = M−1A′αM ,

so our two representations are related by a similarity transformationand are, therefore, equivalent.

If c = 0, then MM † = 0. The (i, j)th matrix element of this productis

(MM †)ij =∑k

Mik(M†)kj =

∑k

MikM∗jk = 0 .

By setting i = j, we obtain∑k

MikM∗ik =

∑k

|Mik|2 = 0 ,

which implies that Mik = 0 for all i and k, i.e., that M is the zeromatrix. This completes the first part of the proof.

Case II. d 6= d′. We take d < d′. Then M is a rectangular matrixwith d columns and d′ rows:

M =

M11 · · · M1d

M21 · · · M2d

.... . .

...

Md′1 · · · Md′d

.

2By multiplying (4.10) from the right by M and following analogous steps asabove, one can show that M†M = cI, so that the matrix c−1M† is both the leftand right inverse of M .


We can make a d′ × d′ matrix N from M by adding d′ − d columns ofzeros:

N =

M11 · · · M1d 0 · · · 0

M21 · · · M2d 0 · · · 0

.... . .

......

. . ....

Md′1 · · · Md′d 0 · · · 0

≡ (M, 0) .

Taking the adjoint of this matrix yields

N † =

M11 M∗21 · · · M∗

d′1

M∗12 M∗

22 · · · M∗d′2

......

. . ....

M∗1d M∗

2d · · · M∗d′d

0 0 · · · 0...

.... . .

...

0 0 · · · 0

=

(M †

0

).

Note that this construction maintains the product MM †:

NN † = (M, 0)

(M †

0

)= MM † = cI .

The determinant of N is clearly zero. Thus,

det(NN †) = det(N) det(N †) = cd′= 0

so c = 0, which means that MM † = 0. Proceeding as in Case I, weconclude that this implies M = 0. This completes the second part ofthe proof.

4.3 The Great Orthogonality Theorem

Schur’s lemmas provide restrictions on the form of matrices which com-mute with all of the matrices of irreducible representations. But the


group property enables the construction of many matrices which sat-isfy the relations in Schur’s First and Second Lemmas. The interplaybetween these two facts provides the basis for proving the Great Or-thogonality Theorem. The statement of this theorem is as follows:

Theorem 4.3 (Great Orthogonality Theorem). Let A1, A2, . . . , A|G|and A′1, A′2, . . . , A′|G| be two inequivalent irreducible representationsof a group G with elements g1, g2, . . . , g|G| and which have dimen-sionalities d and d′, respectively. The matrices Aα and A′α in the tworepresentations correspond to the element gα in G. Then∑

α

(Aα)∗ij(A′α)i′j′ = 0

for all matrix elements. For the elements of a single unitary irreduciblerepresentation, we have

∑α

(Aα)∗ij(Aα)i′j′ =|G|dδi,i′δj,j′ ,

where d is the dimension of the representation.

Proof. Consider the matrix

M =∑α

A′αXA−1α , (4.12)

where X is an arbitrary matrix with d′ rows and d columns, so that Mis a d′ × d′ matrix. We will show that for any matrix X, M satisfies acommutation relation of the type discussed in Schur’s Lemmas.

We now multiply M from the left by the matrix A′β correspondingto some matrix in the the “primed” representation:

A′βM =∑α

A′βA′αXA

−1α

=∑α

A′βA′αXA

−1α A−1

β Aβ

=∑α

A′βA′αX(AβAα)−1Aβ . (4.13)


Since the Aα and A′α form representations of G, the products AαAβand A′αA

′β yield matrices Aγ and A′γ, respectively, both corresponding

to the same element in G because representations preserve the groupcomposition rule. Hence, by the Rearrangement Theorem (Theorem2.1), we can write the summation over α on the right-hand side of thisequation as ∑

α

A′βA′αX(AβAα)−1 =

∑γ

A′γXA−1γ = M .

Substituting this result into (4.13) yields

A′βM = MAβ . (4.14)

Depending on the nature of the two representations, this is precisely thesituation addressed by Schur’s First and Second Lemmas. We considerthe cases of equivalent and inequivalent representations separately.

Case I. d 6= d′ or, if d = d′, the representations are inequivalent (i.e.,not related by a similarity transformation). Schur’s Second Lemmathen implies that M must be the zero matrix, i.e., that each matrixelement of M is zero. From the definition (4.12), we see that thisrequires

Mii′ =∑α

∑jj′

(A′α)ijXjj′(A−1α )j′i′ = 0 . (4.15)

By writing this sum as (note that because all sums are finite, theirorder can be changed at will)

∑jj′Xjj′

[∑α

(Aα)′ij(A−1α )j′i′

]= 0 , (4.16)

we see that, since X is arbitrary, each of its entries may be variedarbitrarily and independently without affecting the vanishing of thesum. The only way to ensure this is to require that the coefficients ofthe Xjj′ vanish: ∑

α

(A′α)ij(A−1α )j′i′ = 0 .


For unitary representations, (A−1α )j′i′ = (Aα)∗i′j′ , so this equation re-

duces to ∑α

(A′α)ij(Aα)∗i′j′ = 0 ,

which proves the first part of the theorem.

Case II. d = d′ and the representations are equivalent. Accordingto Schur’s First Lemma, M = cI, so,

cI =∑α

AαXA−1α . (4.17)

Taking the trace of both sides of this equation,

tr(cI)︸︷︷︸cd

= tr(∑

α

AαXA−1α

)=∑α

tr(AαXA−1α ) =

∑α

tr(X)︸︷︷︸|G| tr(X)

,

yields an expression for c:

c =|G|d

tr(X) .

Substituting this into Eq. (4.17) and expressing the resulting equationin terms of matrix elements, yields

∑jj′Xjj′

[∑α

(Aα)ij(A−1α )j′i′

]=|G|dδi,i′

∑j

Xjj ,

or, after a simple rearrangement,

∑jj′Xjj′

[∑α

(Aα)ij(A−1α )j′i′ −

|G|dδi,i′δj,j′

]= 0 .

This equation must remain valid under any independent variation ofthe matrix elements of X. Thus, we must require that the coefficientof Xjj′ vanishes identically:

∑α

(Aα)ij(A−1α )j′i′ =

|G|dδi,i′δj,j′ .


Since the representation is unitary, this is equivalent to

∑α

(Aα)ij(Aα)∗i′j′ =|G|dδi,i′δj,j′ .

This proves the second part of the theorem.

4.4 Some Immediate Consequences of the

Great Orthogonality Theorem

The Great Orthogonality Theorem establishes a relation between ma-trix elements of the irreducible representations of a group. Suppose wedenote the αth matrix in the kth irreducible representation by Akα andthe (i, j)th element of this matrix by (Akα)ij. We can then combine thetwo statements of the Great Orthogonality Theorem as

∑α

(Akα)ij(Ak′α )∗i′j′ =

|G|dδi,i′δj,j′δk,k′ (4.18)

This expression helps us to understand the motivation for the name“Orthogonality Theorem” by inviting us to consider the matrix ele-ments of irreducible representations as entries in |G|-component vec-tors, i.e., vectors in a space of dimensionality |G|:

V kij =

[(Ak1)ij, (A

k2)ij, . . . , (A

k|G|)ij

]According to the statement of the Great Orthogonality Theorem, twosuch vectors are orthogonal if they differ in any one of the indices i, j,or k, since (4.18) requires that

V kij · V k′

i′j′ =|G|dδi,i′δj,j′δk,k′

But, in a |G|-dimensional space there are at most |G| mutually or-thogonal vectors. To see the consequences of this, suppose we haveirreducible representations of dimensionalities d1, d2, . . ., where the dk


are positive integers. For the k representations, there are dk choicesfor each of i and j, i.e., there are d2

k matrix elements in each matrixof the representation. Summing over all irreducible representations, weobtain the inequality ∑

k

d2k ≤ |G| (4.19)

Thus, the order of the group acts as an upper bound both for thenumber and the dimensionalities of the irreducible representations. Inparticular, a finite group can have only a finite number of irreduciblerepresentations. We will see later that the equality in (4.19) alwaysholds.

Example 4.1. For the group S3, we have that |G| = 6 and we havealready identified two one-dimensional irreducible representations andone two-dimensional irreducible representation (Example 3.2). Thus,using (4.19), we have ∑

k

d2k = 12 + 12 + 22 = 6

so the Great Orthogonality Theorem tells us that there are no addi-tional distinct irreducible representations.

For the two element group, we have found two one-dimensional rep-resentations, 1, 1 and 1,−1 (Example 3.3). According to the in-equality in (4.19), ∑

k

d2k = 1 + 1 = 2

so these are the only two irreducible representations of this group.

4.5 Summary

The central result of this chapter is the statement and proof of theGreat Orthogonality Theorem. Essentially all of the applications inthe next several chapters are consequences of this theorem. The impor-tant advance provided this theorem is that it provides an orthogonality


relation between the entries of the matrices of the irreducible repre-sentations of a group. While this can be used to test whether a givenrepresentation is reducible or irreducible (Problem Set 6), its main rolewill be in a somewhat “reduced” form, such as that used in Sec. 4.4 toplace bounds on the number of irreducible representations of a finitegroup. One of the most important aspects of the Great OrthogonalityTheorem for applications to physical problems is in the constructionof “character tables,” i.e., tables of traces of matrices of an irreduciblerepresentation. This is taken up in the next chapter.

Chapter 5

Characters and CharacterTables

In great mathematics there is a very high degree of unexpectedness, com-bined with inevitability and economy.

—Godfrey H. Hardy1

In the preceding chapter, we proved the Great Orthogonality Theorem,which is a statement about the orthogonality between the matrix ele-ments corresponding to different irreducible representations of a group.For many applications of group theory, however, the full matrix rep-resentations of a group are not required, but only the traces withinclasses of group elements—called “characters.” A typical applicationinvolves determining whether a given representation is reducible or irre-ducible and, if it is reducible, to identify the irreducible representationscontained within that representation.

In this chapter, we develop the mathematical machinery that is usedto assemble the characters of the irreducible representations of a groupin what are called “character tables.” The compilation of character ta-bles requires two types of input: the order of the group and the numberof classes it contains. These quantities provide stringent restrictions

1G.H. Hardy, A Mathematician’s Apology (Cambridge University Press, London,1941)

65

66 Characters and Character Tables

on the number of irreducible representations and their dimensionali-ties. Moreover, orthogonality relations derived from the Great Orthog-onality Theorem will be shown to provide constraints on characters ofdifferent irreducible representations, which considerably simplifies theconstruction of character tables.

5.1 Orthogonality Relations

The Great Orthogonality Theorem,∑α

(Akα)ij(Ak′α )∗i′j′ =

|G|dkδi,i′δj,j′δk,k′ (5.1)

is a relationship between the matrix elements of the irreducible repre-sentations of a group G. In this section, we show how this statementcan be manipulated into an expression solely in terms of the traces ofthe matrices in these representations. This will open the way to estab-lishing a sum rule between the number of irreducible representationsand the number of classes in a group.

We begin by setting j = i and j′ = i′ in (5.1),

∑α

(Akα)ii(Ak′α )∗i′i′ =

|G|dkδi,i′δk,k′ , (5.2)

where we have used the fact that δi,i′δi,i′ = δi,i′ . Summing over i and i′

on the left-hand side of this equation yields

∑i,i′

∑α

(Akα)ii(Ak′α )i′i′ =

∑α

[∑i

(Akα)ii

]︸︷︷︸

tr(Akα)

[∑i′

(Ak′α )∗i′i′

]︸︷︷︸

tr(Ak′α )∗

=∑α

tr(Akα)tr(Ak′α )∗ ,

and, by summing over i and i′ on the right-hand side of (5.2), we obtain

|G|dkδk,k′

∑i

∑i′δi,i′ =

|G|dkδk,k′

∑i

1︸︷︷︸dk

= |G|δk,k′ .

Characters and Character Tables 67

We have thereby reduced the Great Orthogonality Theorem to∑α

tr(Akα)tr(Ak′∗α ) = |G|δk,k′ . (5.3)

This expression can be written in a more useful form by observingthat matrices corresponding to elements in the same conjugacy classhave the same trace. To see this, recall the definition in Section 2.6of the conjugacy of two elements a and b group G. There must bean element g in G such that a = gbg−1. Any representation Aα,reducible or irreducible, must preserve this relation:

Aa = AgAbAg−1 .

This representation must also have the property that Ag−1 = A−1g . Thus

(Problem 2, Problem Set 4),

tr(Aa) = tr(AgAbA−1g ) = tr(A−1

g AgAb) = tr(Ab) .

We can now introduce the notation χkα for the trace corresponding toall of the elements of the αth class of the kth irreducible representation.This is called the character of the class. If there are nα elements in thisclass, then we can write the relation (5.3) in terms of characters as asum over conjugacy classes

C∑α=1

nαχkαχ

k′∗α = |G|δk,k′ , (5.4)

where C is the number of conjugacy classes. In arriving at this relation,we have proven the following theorem:

Theorem 5.1 (Orthogonality Theorem for Characters). The char-acters of the irreducible representations of a group obey the relation∑

α

nαχkαχ

k′∗α = |G|δk,k′ .

This orthogonality theorem can be used to deduce a relationshipbetween the number classes of a group and the number of irreduciblerepresentations. By rearranging (5.4) as∑

α

[(nα|G|

)1/2

χkα

][(nα|G|

)1/2

χk′∗α

]= δk,k′


and introducing the vectors

χk = |G|−1/2(√n1χ

k1,√n2χ

k2, . . . ,

√nCχ

kC) ,

we can write the orthogonality relation for characters as

χk ·χk′ = δk,k′ .

The χk reside in a space whose dimension is the number of classes C inthe group. Thus, the maximum number of a set of mutually orthogonalvectors in this space is C. But these vectors are labelled by an index kcorresponding to the irreducible representations of the group. Hence,the number of irreducible representations must be less than or equal tothe number of classes.

It is also possible2 to obtain an orthogonality relation with the rolesof the irreducible representations and classes reversed in comparison tothat in Theorem 5.1: ∑

k

χkαχk∗β =

|G|nα

δα,β . (5.5)

By following analogous reasoning as above, we can deduce that this or-thogonality relation implies that the number of irreducible representa-tions must be greater than or equal to the number of classes. Combinedwith the statement of Theorem 5.1, we have the following theorem:

Theorem 5.2. The number of irreducible representations of a group isequal to the number of conjugacy classes of that group.

Example 5.1. For Abelian subgroups each element is in a class byitself (Problem 6, Problem Set 3). Thus, the number of classes is equalto the order of the group, so, according to Theorem 5.2, the numberof irreducible representations must also equal the order of the group.When combined with the restriction imposed by Eqn. (4.19), which wecan now write as

|G|∑k=1

d2k = |G| ,

2M. Hamermesh, Group Theory and its Application to Physical Problems (Dover,1989, New York) pp. 106–110.


we have an alternative way (cf. Problem 4, Problem Set 5) of seeingthat all of the the irreducible representations of an Abelian group areone-dimensional, i.e., dk = 1, for k = 1, 2, . . . , |G|.

Example 5.2. For the group S3, there are three classes: e, a, b, c,and d, f (Example 2.9). Thus, there are three irreducible represen-tations which, as we have seen, consist of two one-dimensional repre-sentations and one two-dimensional representation.

5.2 The Decomposition Theorem

One of the main uses of characters is in the decomposition of a givenreducible representation into its constituent irreducible representations.The procedure by which this is accomplished is analogous to projectinga vector onto a set of complete orthogonal basis vectors. The theoremwhich provides the foundation for carrying this out with characters isthe following:

Theorem 5.3 (Decomposition Theorem). The character χα for theαth class of any representation can be written uniquely in terms of thecorresponding characters of the irreducible representations of the groupas

χα =∑k

akχkα ,

where

ak =1

|G|∑α

nαχk∗α χα .

Proof. For a reducible representation, the same similarity transfor-mation brings all of the matrices into the same block-diagonal form. Inthis form, the matrix Aα can be written as the direct sum of matricesAkj of irreducible representations:

Aα = Ak1α ⊕ Ak2

α ⊕ · · · ⊕ Aknα ,


where α = 1, 2, . . . , |G| and k1, k2, . . . kn label irreducible representa-tions. Given this, and the fact that similarity transformations leavethe trace invariant, we can write the character χi of this reducible rep-resentation corresponding to the ith class as

χα =∑k

akχkα , (5.6)

where the ak must be nonnegative integers. We now multiply bothsides of this equation by nαχ

k′∗α , sum over α, and use the orthogonality

relation (5.4):∑α

nαχk′∗α χα =

∑k

ak∑α

nαχkαχ

k′∗α︸︷︷︸

|G|δk,k′

= |G|ak′

Thus,

ak′ =1

|G|∑α

nαχk′∗α χα , (5.7)

so ak′ is the projection of the reducible representation onto the k′thirreducible representation. Note that, because the number of irreduciblerepresentations equals the number of classes, the orthogonal vectors ofcharacters span the space whose dimensionality is the number of classes,so this decomposition is unique.

The Decomposition Theorem reduces the task of determining the ir-reducible representations contained within a reducible representation toone of vector algebra. Unless a particular application requires the ma-trix forms of the representations, there is no need to block-diagonalizea representation to identify its irreducible components.

We can follow a procedure analogous to that used to prove the De-composition Theorem to derive a simple criterion to identify whethera representation is reducible or irreducible. We begin with the decom-position (5.6) and take its complex conjugate:

χ∗α =∑k′ak′χ

k′∗α , (5.8)


where we have used the fact that the ak are integers, so a∗k = ak. Wenow take the product of (5.6) and (5.8), multiply by nα, sum over α,and invoke (5.4):∑

α

nαχαχ∗α =

∑k,k′

akak′∑i

nαχkαχ

k′∗α︸︷︷︸

|G|δk,k′

= |G|∑k

a2k .

Thus, ∑α

nα|χ2α| = |G|

∑k

a2k . (5.9)

If the representation in question is irreducible, then all of the ak are zero,except for the one corresponding to that irreducible representation,which is equal to unity. If the representation is reducible, then therewill be at least two of the ak which are positive integers. We cansummarize these observations with a simple criterion for reducibility.If the representation is irreducible, then∑

α

nα|χα|2 = |G| , (5.10)

and if the representation is reducible,∑α

nα|χα|2 > |G| . (5.11)

Example 5.3. Consider the following representation of S3:

e =

(1 0

0 1

), a = 1

2

(1 −

√3

−√

3 −1

), b = 1

2

(1√

3√

3 −1

),

c =

(−1 0

0 1

), d = 1

2

( −1√

3

−√

3 −1

), f = 1

2

(−1 −√

3√

3 −1

).

There are three classes of this group, e, a, b, c, and d, f, so wehave n1 = 1, n2 = 3, and n3 = 2, respectively. The correspondingcharacters are

χ1 = 2, χ2 = 0, χ3 = −1 .


Forming the sum in (5.9), we obtain

3∑α=1

nα|χα|2 = (1× 4) + (3× 0) + (2× 1) = 6 ,

which is equal to the order of the group. Therefore, this representationis irreducible, as we have already demonstrated in Example 3.4 and inProblem 1, Problem Set 6.

Example 5.4. Another representation of S3 is

e = d = f =

(1 0

0 1

), a = b = c = 1

2

( −1 −√

3

−√

3 1

).

The characters corresponding to the three classes are now

χ1 = 2, χ2 = 0, χ3 = 2 .

Forming the sum in (5.9), we find

3∑i=1

n1|χi|2 = (1× 4) + (3× 0) + (2× 4) = 12 ,

which is greater than the order of the group, so this representation isreducible (cf. Problem 2, Problem Set 6). To determine the irreducibleconstituents of this representation, we use the decomposition theorem.There are three irreducible representations of S3: the one-dimensionalidentical representation, with characters

χ11 = 1, χ1

2 = 1, χ13 = 1 ,

the one-dimensional “parity” representation, with characters

χ21 = 1, χ2

2 = −1, χ23 = 1 ,

and the two-dimensional “coordinate” representation discussed abovein Example 5.3, with characters

χ31 = 2, χ3

2 = 0, χ33 = −1 .


We now calculate the ak using the expression in Equation (5.7). Thesedetermine the “projections” of the characters of the reducible represen-tation onto the characters of the irreducible representation. We obtain

a1 = 16

[(1× 1× 2) + (3× 1× 0) + (2× 1× 2)

]= 1 ,

a2 = 16

[(1× 1× 2) + (3×−1× 0) + (2× 1× 2)

]= 1 ,

a3 = 16

[(1× 2× 2) + (3× 0× 0) + (2×−1× 2)

]= 0 .

Thus, this reducible representation is composed of the identical repre-sentation and the “parity” representation, with no contribution fromthe “coordinate” representation. The block-diagonal form of this rep-resentation is, therefore,

e = d = f =

(1 0

0 1

), a = b = c = 1

2

(1 0

0 −1

),

which is the result obtained in Problem 5, Problem Set 5 by applyingmatrix methods.

5.3 The Regular Representation

Our construction of irreducible representations has thus far proceededin an essentially ad hoc fashion, relying in large part on physical argu-ments. We have not yet developed a systematic procedure for construct-ing all of the irreducible representations of a group. In this section, weintroduce a method, based on what is called the “regular” represen-tation, which enables us to accomplish this. However, our purposefor introducing such a methodology is not the determination of irre-ducible representations as such, since even for relatively simple groups,the approach we describe would present a computationally demandingprocess, but as a theoretical tool for proving a theorem. Moreover, wewill find that, for applications of group theory to quantum mechanics,the irreducible representations of the group of operations that leaveHamiltonian invariant will emerge naturally without having to rely onany auxiliary constructions.


The regular representation is a reducible representation that is ob-tained directly from the multiplication table of a group. As we will showbelow, this representation contains every irreducible representation ofa group at least once. The construction of the regular representationis based on arranging the multiplication table of a group so that theunit element appears along the main diagonal of the table. Within suchan arrangement the columns (or rows) of the table are labelled by thegroup elements, arranged in any order, and the corresponding order ofthe inverses labels the rows (or columns).

As an example, consider the multiplication table for S3 (Section 2.4)arranged in the way just described:

e a b c d fe = e−1 e a b c d fa = a−1 a e d f b cb = b−1 b f e d c ac = c−1 c d f e a bf = d−1 f b c a e dd = f−1 d c a b f e

The matrices of the regular representation are obtained by regardingthe multiplication table as an |G| × |G| array from which the matrixrepresentation for each group element is assembled by putting a ‘1’where that element appears in the multiplication table and zero else-where. For example, the matrices corresponding to the unit e and theelement a are

e→

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

, a→

0 1 0 0 0 0

1 0 0 0 0 0

0 0 0 0 0 1

0 0 0 0 1 0

0 0 0 1 0 0

0 0 1 0 0 0

with analogous matrices for the other group elements.


Our first task is to show that the regular ‘representation’ is indeeda representation of the group. First of all, it is clear that the mappingwe have described is one-to-one. For any two elements g1 and g2 ofthis group, we denote the matrices in the regular representation thatcorrespond to these elements as Areg(g1) and Areg(g2). Thus, to showthat these matrices form a representation of S3, we need to verify that

Areg(g1g2) = Areg(g1)Areg(g2) ,

i.e., that the multiplication table is preserved by this representation.We consider this relation expressed in terms of matrix elements:[

Areg(g1g2)]ij

=∑k

[Areg(g1)

]ik

[Areg(g2)

]kj. (5.12)

From the way the regular representation has been constructed, the ithrow index of these matrix elements can be labelled the inverse of theith group element g−1

i and the jth column can be labelled by the jthgroup element gj:

[Areg(g1g2)

]ij

=[Areg(g1g2)

]g−1i ,gj

=

1, if g−1i gj = g1g2;

0; otherwise

[Areg(g1)

]ik

=[Areg(g1)

]g−1i ,gk

=

1, if g−1i gk = g1;

0; otherwise

[Areg(g2)

]kj

=[Areg(g2)

]g−1k,gj

=

1, if g−1k gj = g2;

0; otherwise

Therefore, in the sum over k in (5.12), we have nonzero entries onlywhen

g1g2 = (g−1i gk)(g

−1k gj) = g−1

i gj ,

which gives precisely the nonzero matrix elements of Areg(g1g2). Hence,the matricesAreg(g1) preserve the group multiplication table and therebyform a faithful representation of the group.


Our main purpose in introducing the regular representation is toprove the following theorem:

Theorem 5.4. The dimensionalities dk of the irreducible representa-tions of a group are related to the order |G| of the group by∑

k

d2k = |G| .

This theorem shows that the inequality (4.19), which was deduced di-rectly from the Great Orthogonality Theorem is, in fact, an equality.

Proof. We first show, using Eqn. (5.9), that the regular represen-tation is reducible. To evaluate the sums on the left-hand side of thisequation, we note that, from the construction of the regular repre-sentation, the characters χreg,i vanish for every class except for thatcorresponding to the unit element. Denoting this character by χreg,e,we see that its value must be equal to the order of the group:

χreg,e = |G| .

Thus, ∑α

nα|χα|2 = χ2reg,e = |G|2 ,

which, for |G| > 1 is greater than |G|. Thus, for groups other than thesingle-element group e, the regular representation is reducible.

We will now use the Decomposition Theorem to identify the irre-ducible constituents of the regular representation. Thus, the charactersχreg,α for the αth class in the regular representation can be written as

χreg,α =∑k

akχkα .

According to the Decomposition Theorem, the ak are given by

ak =1

|G|∑α

nαχk∗α χreg,α .


We again use the fact that χreg,e = |G|, with all other characters vanish-ing. The corresponding value of χke is determined by taking the trace ofthe identity matrix whose dimensionality is that of the kth irreduciblerepresentation: χke = dk. Therefore, the Decomposition Theorem yields

ak =1

|G| × dk × |G| = dk ,

i.e., the kth irreducible representation appears dk times in the regularrepresentation: each one-dimensional irreducible representation appearsonce, each two-dimensional irreducible representation appears twice,and so on. Since the dimensionality of the regular representation is |G|,and since ak is the number of times the kth irreducible representationappears in the regular representation, we have the constraint∑

k

akdk = |G| ,

i.e., ∑k

d2k = |G| .

This sum rule, and that equating the number of classes to the num-ber of irreducible representations (Theorem 5.2), relate a property ofthe abstract group (its order and the number of classes) to a propertyof the irreducible representations (their number and dimensionality).The application of these rules and the orthogonality theorems for char-acters is the basis for constructing character tables. This is describedin the next section.

5.4 Character Tables

Character tables are central to many applications of group theory tophysical problems, especially those involving the decomposition of re-ducible representations into their irreducible components. Many text-books on group theory contain compilations of character tables for the


most common groups. In this section, we will describe the constructionof character tables for S3. We will utilize two types of information: sumrules for the number and dimensionalities of the irreducible represen-tations, and orthogonality relations for the characters. Additionally,the group multiplication table can be used to establish relationshipsfor one-dimensional representations. By convention, characters tablesare displayed with the columns labelled by the classes and the rows bythe irreducible representations.

The first step in the construction of this character table is to notethat, since |S3| = 6 and there are three classes (Example 2.9), there are3 irreducible representations whose dimensionalities must satisfy

d21 + d2

2 + d23 = 6 .

The unique solution of this equation (with only positive integers) isd1 = 1, d2 = 1, and d3 = 2, so there are two one-dimensional irreduciblerepresentations and one two-dimensional irreducible representation.

In the character table for any group, several entries can be made im-mediately. The identical representation, where all elements are equal tounity, is always a one-dimensional irreducible representation. Similarly,the characters corresponding to the unit element are equal to the di-mensionality of that representation, since they are calculated from thetrace of the identity matrix with that dimensionality. Thus, denotingby α, β, γ, and δ quantities that are to be determined, the charactertable for S3 is:

S3 e a, b, c d, fΓ1 1 1 1Γ2 1 α βΓ3 2 γ δ

where the Γi are a standard label for the irreducible representations.The remaining entries are determined from the orthogonality re-

lations for characters and, for one-dimensional irreducible representa-tions, from the multiplication table of the group. The orthogonalityrelation in Theorem 5.1, which is an orthogonality relation for the rowsof a character table, yield

1 + 3α + 2β = 0 , (5.13)


1 + 3α2 + 2β2 = 6 . (5.14)

The group multiplication table requires that

a2 = e, b2 = e, c2 = e, d2 = f .

Since the one-dimensional representations must obey the multiplicationtable, these products imply that

α2 = 1, β2 = β .

Substituting these relations into (5.14), yields 4 + 2β = 6, i.e.,

β = 1

Upon substitution of this value into (5.13), we obtain 3 + 3α = 0, i.e.,

α = −1

From the orthogonality relation (5.5), which is an orthogonality relationbetween the columns of a character table, we obtain

1 + α + 2γ = 0

1 + β + 2δ = 0

Substituting the values obtained for α and β into these equations yields

γ = 0, δ = −1

The complete character table for S3 is therefore given by

S3 e a, b, c d, fΓ1 1 1 1Γ2 1 −1 1Γ3 2 0 −1

When character tables are compiled for the most common groups,a notation is used which reflects the fact that the group elements cor-respond to transformations on physical objects. The notation for theclasses of S3 are as follows:


• e → E. The identity.

• a, b, c → 3σv. Reflection through vertical planes, where ‘verti-cal’ refers to the fact that these planes contain the axis of highestrotational symmetry, in this case, the z-axis. The ‘3’ refers tothere being three elements in this class.

• d, f → 2C3. Rotation by 23π radians, with the ‘2’ again referring

to the there being two elements in this class. The notation C23

is for rotations by 43π radians, so the ‘class’ notation is meant

only to indicate the type of operation. In general, Cn refers torotations through 2π/n radians.

Several notations are used for irreducible representations. One ofthe most common is to use A for one-dimensional representations, Efor two-dimensional representations, and T for three-dimensional rep-resentations, with subscripts used to distinguish multiple occurrencesof irreducible representations of the same dimensionality. The notationΓ is often used to indicate a generic (usually irreducible) representa-tion, with subscripts and superscripts employed to distinguish betweendifferent representations. With the first of these conventions, the char-acter table for S3, which is known as the group C3v when interpretedas the planar symmetry operations of an equilateral triangle, is

C3v E 3σv 2C3

A1 1 1 1A2 1 −1 1E 2 0 −1

5.5 Summary

This chapter has been devoted to characters and character tables. Theutility of characters in applications stems from the following:


1. The character is a property of the class of an element.

2. Characters are unaffected by similarity transformations, so equiv-alent representations—reducible or irreducible—have the samecharacters.

3. As shown in Equations (5.10) and (5.11), the characters of a repre-sentation indicate, through a straightforward calculation, whetherthat representation is reducible or irreducible.

4. Characters of irreducible representations obey orthogonality the-orems which, when interpreted in the context of character ta-bles, correspond to the orthogonality relations of their rows andcolumns.

5. According to the Decomposition Theorem, once the character ta-ble of a group is known, the characters of any representation canbe decomposed into its irreducible components.

Chapter 6

Groups and Representationsin Quantum Mechanics

The universe is an enormous direct product of representations of sym-metry groups.

—Steven Weinberg1

This chapter is devoted to applying the mathematical theory of groupsand representations which we have developed in the preceding chaptersto the quantum mechanical description of physical systems. The powerof applying group theory to quantum mechanics is that it provides aframework for making exact statements about a physical system witha knowledge only of the symmetry operations which leave its Hamil-tonian invariant, the so-called “group of the Hamiltonian.” Moreover,when we apply the machinery of groups to quantum mechanics, we findthat representations—and irreducible representations in particular—arise quite naturally, as do related concepts such as the importance ofunitarity of representations and the connection between the symmetryof a physical system and the degeneracy of its eigenstates. We willfollow the general sequence of the discussion in Sections 1.2 and 1.3,

1Steven Weinberg, Sheldon Glashow, and Abdus Salam were awarded the 1979Nobel Prize in Physics for their incorporation of the weak and electromagneticinteractions into a single theory.

83

84 Groups and Representations in Quantum Mechanics

beginning with the group of the Hamiltonian, using this to establish thesymmetry properties of the eigenfunctions, and concluding with a dis-cussion of selection rules, which demonstrates the power and economyof using character tables. As a demonstration of the usefulness of theseconstructions, we will prove Bloch’s theorem, the fundamental princi-ple behind the properties of wavefunctions in periodic systems such aselectrons and phonons (the quanta of lattice vibrations) in single crys-tals. The application of group theory to selection rules necessitates theintroduction of the “direct product” of matrices and groups, thoughhere, too, quantum mechanics provides a motivation for this concept.

6.1 The Group of the Hamiltonian

Recall the definition of a similarity transformation introduced in Section3.3. Two matrices, or operators, A and B are related by a similaritytransformation generated by a matrix (or operator) R if

B = RAR−1 .

The quantity B is therefore the expression of A under the transforma-tion R. Consider now a Hamiltonian H and its transformation by anoperation R

RHR−1 .

The Hamiltonian is said to be invariant under R if

H = RHR−1 , (6.1)

or, equivalently,

RH = HR . (6.2)

Thus the order in which H and the R are applied is immaterial, so Hand R commute: [H, R] = 0. In this case, R is said to be a symmetryoperation of the Hamiltonian.

Consider set of all symmetry operations of the Hamiltonian, whichwe will denote by Rα. We now show that these operations form a

Groups and Representations in Quantum Mechanics 85

group. To demonstrate closure, we observe that if Rα and Rβ are twooperations which satisfy (6.1), then

RαHR−1α = Rα(RβHR−1

β )R−1α = (RαRβ)H(RαRβ)−1 = H .

Thus, the product RαRβ = Rγ is also a symmetry operation of theHamiltonian. Associativity is clearly obeyed since these operationsrepresent transformations of coordinates and other variables of theHamiltonian.2 The unit element E corresponds to performing no op-eration at all and the inverse R−1

α of a symmetry operation Rα is theapplication of the reverse operation to “undo” the original transfor-mation. Thus, the set Rα forms a group, called the group of theHamiltonian .

6.2 Eigenfunctions and Representations

There are a number of consequences of the discussion in the preced-ing section for the representations of the group of the Hamiltonian.Consider an eigenfunction ϕ of a Hamiltonian H corresponding to theeigenvalue E:

Hϕ = Eϕ .

We now apply a symmetry operation Rα to both sides of this equation,

RαHϕ = ERαϕ ,

and use (6.2) to write

RαHϕ = HRαϕ .

Thus, we have

H(Rαϕ) = E(Rαϕ) .

If the eigenvalue is nondegenerate, then Rαϕ differs from ϕ by at mosta phase factor:

Rαϕ = eiφαϕ .

2The associativity of linear operations is discussed by Wigner in Group Theory(Academic, New York, 1959), p. 5.


The application of a second operation Rβ then produces

Rβ(Rαϕ) = eiφβeiφαϕ . (6.3)

The left-hand side of this equation can also be written as

(RβRα)ϕ = eiφβαϕ , (6.4)

Equating the right-hand sides of Eqs. (6.3) and (6.4), yields

eiφβα = eiφβeiφα ,

i.e., these phases preserve the multiplication table of the symmetry op-erations. Thus, the repeated application of all of the Rα to ϕ generatesa one-dimensional representation of the group of the Hamiltonian.

The other case to consider occurs if the application of all of thesymmetry operations to ϕ produces ` distinct eigenfunctions. Theseeigenfunctions are said to be `-fold degenerate. If these are the onlyeigenfunctions which have energy E, this is said to be a normal degen-eracy . If, however, there are other degenerate eigenfunctions which arenot captured by this procedure, this is said to be an accidental degen-eracy . The term “accidental” refers to the fact that the degeneracy isnot due to symmetry. But an “accidental” degeneracy can also occurbecause a symmetry is “hidden,” i.e., not immediately apparent, so thegroup of the Hamiltonian is not complete. One well-known example ofthis is the level degeneracy of the hydrogen atom.

For a normal degeneracy, there are orthonormal eigenfunctions ϕi,i = 1, 2, . . . , ` which, upon application of one of the symmetry op-erations Rα are transformed into linear combinations of one another.Thus, if we denote by ϕ the `-dimensional row vector

ϕ = (ϕ1, ϕ2, . . . , ϕ`) ,

we can write

Rαϕ = ϕΓ(Rα) ,

where Γ(Rα) is an `× ` matrix. In terms of components, this equationreads

Rαϕi =∑k=1

ϕk[Γ(Rα)]ki (6.5)


The successive application of operations Rα and Rβ then yields

RβRαφi = Rβ

∑k=1

ϕk[Γ(Rα)]ki =∑k=1

(Rβϕk)[Γ(Rα)]ki .

The operation Rβϕk can be written as in (6.5):

Rβϕk =∑j=1

ϕj[Γ(Rβ)]jk .

Thus,

RβRαφi =∑k=1

∑j=1

ϕj[Γ(Rβ)]jk[Γ(Rα)]ki

=∑j=1

ϕj

∑k=1

[Γ(Rβ)]jk[Γ(Rα)]ki

. (6.6)

Alternatively, we can write

RβRαϕi =∑j=1

ϕj[Γ(RβRα)]ji . (6.7)

By comparing (6.6) and (6.7) and using the orthonormality of the wave-functions, we conclude that

Γ(RβRα) = Γ(Rβ)Γ(Rα) ,

so the Γ(Ri) form an `-dimensional representation of the group of theHamiltonian. Since the eigenfunctions can be made orthonormal, thisrepresentation can always be taken to be unitary (Problem Set 8). Wewill now show that this representation is also irreducible. We firstconsider the effect of replacing the ϕi by a linear combination of thesefunctions, ψ = ϕU . Then the effect of operating with R on the ψ is

Rψ = RϕU = ϕΓU = ψU−1ΓU ,

i.e., the representation with the transformed wavefunctions is relatedby a similarity transformation to that with the original eigenfunctions,


i.e., the two representations are equivalent. Suppose that this represen-tation is reducible. Then there is a unitary transformation of the ϕjsuch that there are two or more subsets of the ψi that transform onlyamong one another under the symmetry operations of the Hamiltonian.This implies that the application of the Ri to any eigenfunction gen-erates eigenfunctions only in the same subset. The degeneracy of theeigenfunctions in the other subset is therefore accidental, in contradic-tion to our original assertion that the degeneracy is normal. Hence, therepresentation obtained for a normal degeneracy is irreducible and thecorresponding eigenfunctions are said to generate, or form a basis forthis representation.

We can summarize the results of this section as follows:

• To each eigenvalue of a Hamiltonian there corresponds a uniqueirreducible representation of the group of that Hamiltonian.

• The degeneracy of an eigenvalue is the dimensionality of this ir-reducible representation. Thus, the dimensionalities of the irre-ducible representations of a group are the possible degeneraciesof Hamiltonians with that symmetry group.

• Group theory provides “good quantum numbers,” i.e., labels cor-responding irreducible representations to which eigenfunctionsbelong.

• Although these statements have been shown for finite groups,they are also valid for continuous groups.

6.3 Group Theory in Quantum Mechanics

The fact that eigenfunctions corresponding to an `-fold degenerateeigenvalue form a basis for an `-dimensional irreducible representationof the group of the Hamiltonian is one of the fundamental principlesbehind the application of group theory to quantum mechanics. In thissection, we briefly describe the two main types of such applications,namely, where group theory is used to obtain exact results, and whereit is used in conjunction with perturbation theory to obtain approximateresults.


6.3.1 Exact Results

One of the most elegant applications of group theory to quantum me-chanics involves using the group of the Hamiltonian to determine the(normal) degeneracies of the eigenstates, which are just the dimen-sions of the irreducible representations. Because such a classification isderived from the symmetry properties of the Hamiltonian, it can be ac-complished without having to solve the Schrodinger equation. Amongthe most historically important of such applications is the classifica-tion of atomic spectral lines. The atomic Hamiltonian is comprisedof the sum of the kinetic energies of the electrons and their Coulombinteractions, so an exact solution is impractical, even for few-electronatoms such as He. Nevertheless, the spherical symmetry of the Hamil-tonian enables the identification of the irreducible representations ofatomic states from which are derived the angular momentum additionrules and multiplet structures. This will be explored further when wediscuss continuous groups. Another exact result is Bloch’s theorem,which is the basis for many aspects of condensed matter physics. Thistheorem uses the translational invariance of perfect periodic crystalsto determine the form of the eigenfunctions. As discussed in the nextsection, Bloch’s theorem can be reduced to a statement about the (one-dimensional) irreducible representations and basis functions of cyclicgroups.

The lowering of the symmetry of a Hamiltonian by a perturbationcan also be examined with group theory. In particular, the question ofwhether the allowed degeneracies are affected by such a perturbationcan be addressed by examining the irreducible representations of thegroups of the original and perturbed Hamiltonians. Group theory canaddress not only whether degeneracies can change (from the irreduciblerepresentations of the two groups), but how irreducible representationsof the original group are related to those of the perturbed group. Typi-cally, when the symmetry of a system is lowered, the dimensionalities ofthe irreducible representations can also be lowered, resulting in a “split-ting” of the original irreducible representations into lower-dimensionalirreducible representations of the group of the perturbed system.

Finally, on a somewhat more practical level, group theory can beused to construct symmetrized linear combinations of basis functions


to diagonalize a Hamiltonian. Examples where this arises is the low-ering of the symmetry of a system by a perturbation, where the basisfunctions are the eigenfunctions of the original Hamiltonian, the bond-ing within molecules, where the basis functions are localized aroundthe atomic sites within the molecule, and vibrations in molecules andsolids, where the basis functions describe the displacements of atoms.These applications are discussed by Tinkham.3

6.3.2 Approximate Results

The most common application of group theory in approximate calcula-tions involves the calculation of matrix elements in perturbation theory.A typical example is involved adding to a Hamiltonian H0 and pertur-bation H′ due to an electromagnetic field which causes transitions be-tween the eigenstates of the original Hamiltonian. The transition rateW is calculated from first-order time-dependent perturbation theory,with the result known as Fermi’s Golden Rule:4

W =2π

h%if |(i|H′|f)|2 ,

where %if is called the “joint density of states,” which is a measure of thenumber of initial and final states which are available for the excitation,and (i|H′|f) is a matrix element of H′ between the initial and finalstates. The application of group theory to this problem, which is thesubject of Section 6.6, involves determining when this matrix elementvanishes by reasons of symmetry.

6.4 Bloch’s Theorem∗

Bloch’s theorem is of central importance to many aspects of electrons,phonons, and other excitations in crystalline solids. One of the mainresults of this theorem, namely, the form of the eigenfunctions, can bederived solely from group theory. We will work in one spatial dimension,

3M. Tinkham, Group Theory and Quantum Mechanics (McGraw–Hill, New York,1964)

4L.I. Schiff, Quantum Mechanics 2nd edn (McGraw–Hill, New York, 1955)


but the discussion can be extended easily to higher dimensions. Weconsider a one-dimensional crystal where the distance between nearestneighbors is a and the number of repeat units is N (a large number fora macroscopic solid). Since this system is finite, it has no translationalsymmetry. However, by imposing a type of boundary condition knownas periodic, whereby the Nth unit is identified with the first unit—effectively forming a circle from this solid—we now have N discretesymmetries. The Schrodinger equation for a particle of mass m movingin the periodic potential of this system is[

− h2

2m

d2

dx2+ V (x)

]ϕ = Eϕ ,

where V (x+ a) = V (x).

6.4.1 The Group of the Hamiltonian

The translation of an eigenfunction by a will be denoted by Ra:

Raϕ(x) = ϕ(x+ a) .

The basic properties of translations originate with the observation thata translation through na,

Rnaϕ(x) = ϕ(x+ na) ,

can be written as the n-fold product of Ra:

Rnaϕ(x) = RaRa · · ·Ra︸︷︷︸

n factors

ϕ(x) = ϕ(x+ na) .

Moreover, because of the periodic boundary conditions, we identify theNth unit with the first, so

RNa = R0 ,

which means that no translation is carried out at all. Thus, the col-lection of all the translations can be written as the powers of a singleelement, Ra:

Ra, R2a, . . . , R

Na = E , (6.8)


where E is the identity. This shows that the group of the Hamiltonianis a cyclic group of order N . In particular, since cyclic groups areAbelian, there are N one-dimensional irreducible representations of thisgroup, i.e., each eigenvalue is nondegenerate and labelled by one of theseirreducible representations.

6.4.2 Character Table and Irreducible Represen-tations

Having identified the algebraic structure of the group of the Hamilto-nian, we now construct the character table. Since RN

a = E, and sinceall irreducible representations are one-dimensional, the character for Ra

in each of these representations, χ(n)(Ra) must obey this product:

[χ(n)(Ra)

]N= 1 .

The solutions to this equation are the Nth roots of unity (cf. Problem3, Problem Set 5):

χ(n)(Ra) = e2πin/N , n = 0, 1, 2, . . . N − 1 .

The character table is constructed by choosing one of these values foreach irreducible representation and then determining the remaining en-tries from the multiplication table of the group (since each irreduciblerepresentation is one-dimensional). The resulting character table is:

E Ra R2a · · · RN−1

a Γ1 1 1 1 · · · 1Γ2 1 ω ω2 · · · ωN−1

Γ3 1 ω2 ω4 · · · ω2(N−1)

......

......

......

ΓN 1 ωN−1 ω2N−2 · · · ω(N−1)2

where ω = e2πi/N . If we denote the eigenfunction corresponding to thenth irreducible representation by ϕn, then applying Ra yields

Raϕn(x) = ωn−1ϕn(x) = ϕn(x+ a) .


Since the characters of this group are pure phases, the moduli of theeigenfunctions are periodic functions:

|ϕn(x+ a)|2 = |ϕn(x)|2 .

Thus, the most general form of the ϕn is

ϕn(x) = eiφn(x)un(x) , (6.9)

where φn(x) is a phase function, which we will determine below, andthe un have the periodicity of the lattice: un(x + a) = un(x). By com-bining this form of the wavefunction with the transformation propertiesrequired by the character table, we can write

Rma ϕn(x) = ωm(n−1)ϕn(x) = ωm(n−1)eiφn(x)un(x) .

Alternatively, by applying the same translation operation directly to(6.9) yields

Rma ϕn(x) = ϕn(x+ma) = eiφn(x+ma)un(x) .

By equating these two ways of writing Rma ϕn(x), we find that their

phase changes must be equal. This, in turn, requires that the phasefunction satisfies

φn(x+ma) = φn(x) +2πm(n− 1)

N. (6.10)

Thus, φn is a linear function of m and, therefore, also of x+ma, sinceφn is a function of only a single variable:

φn(x) = Ax+B ,

where A and B are constants to be determined. Upon substitution ofthis expression into both sides of (6.10),

A(x+ma) +B = Ax+B +2πm(n− 1)

N,

and cancelling common factors, we obtain

φn(x) = knx+B ,


where

kn =2π(n− 1)

Na=

2π(n− 1)

L

and L = Na is the size system. The wavefunction in (6.9) therebyreduces to

ϕn(x) = eiknxun(x) ,

where we have absorbed the constant phase due to B into the definitionof un(x). This is called a Bloch function : a function un(x) with theperiodicity of the lattice modulated by a plane wave.5 This is one ofthe two main results of Bloch’s theorem, the other being the existenceof energy gaps, which is beyond the scope of the discussion here.

6.5 Direct Products

The direct product provides a way of enlarging the number of elementsin a group while retaining the group properties. Direct products oc-cur in several contexts. For example, if a Hamiltonian or Lagrangiancontains different types of coordinates, such as spatial coordinates fordifferent particles, or spatial and spin coordinates, then the symmetryoperations on the different coordinates commute with each other. Ifthere is a coupling between such degrees of freedom, such as particleinteractions or a spin-orbit interaction, then the direct product is re-quired to determine the appropriate irreducible representations of theresulting eigenstates. In this section, we develop the group theory as-sociated with direct products and their representations. We will thenapply these concepts to selection rules in the following section.

5A related issue which can be addressed by group theory is the nature of thequantity hkn. Although it has units of momentum, it does not represent a truemomentum, but is called the “crystal momentum.” The true momentum hk la-bels the irreducible representations of the translation group, which is a continuousgroup and will be discussed in the next chapter. The discrete translations of a pe-riodic potential form a subgroup of the full translation group, so the correspondingirreducible representations cannot be labelled by momentum.


6.5.1 Direct Product of Groups

Consider two groups

Ga = e, a2, . . . , a|Ga|, Gb = e, b2, . . . , b|Gb| ,such that all elements in Ga commute with all elements in Gb:

aibj = bjai ,

for i = 1, 2, . . . , |Ga| and j = 1, 2, . . . , |Gb|. We have defined a1 = e andb1 = e. The direct product of Ga and Gb, denoted by Ga⊗Gb, is the setcontaining all elements aibj:

Ga ⊗Gb = e, a2, . . . , a|Ga|, b2, . . . , b|Gb|, . . . , aibj, . . . . (6.11)

As shown in Problem 3 of Problem Set 8, the direct product is a groupof order |Ga||Gb|.

Example 6.1. Consider the symmetry operations on an equilateraltriangle that has a thickness, i.e., the triangle has become a “wedge.”Thus, in addition to the original symmetry operations of the planarequilateral triangle, there is now also a reflection plane σh. There arenow six vertices, which are labelled as in Example 2.1, except thatwe now distinguish between points which lie above, 1+, 2+, 3+, andbelow, 1−, 2−, 3−, the reflection plane. The original six operationsdo not transform points above and below the reflection plane into oneanother. The reflection plane, on the other hand, only transforms cor-responding points above and below the plane into one another. Hence,the 6 operations of a planar triangle commute with σh.

The symmetry group of the equilateral wedge consists of the original6 operations of a planar triangle, the horizontal reflection plane, andtheir products. Since the set with elements E, σh forms a group (andeach element commutes with the symmetry operations of an equilateraltriangle), the appropriate group for the wedge is thereby obtained bytaking the direct product

E, σv,1, σv,2, σv,3, C3, C23 ⊗ E, σh .

The 12 elements of this group are

E, σv,1, σv,2, σv,3, C3, C23 , σh, σhσv,1, σhσv,2, σhσv,3, σhC3, σhC

23 .


6.5.2 Direct Product of Matrices

The determination of the irreducible representations and the charactertable of a direct product group does not require a separate new calcu-lation of the type discussed in the preceding chapter. Instead, we canutilize the irreducible representations of the two groups used to formthe direct to obtain these quantities. To carry out these operationsnecessitates introducing the direct product of matrices.

The direct product C of two matrices A and B, written as A⊗B =C, is defined in terms of matrix elements by

aijbkl = cik;jl . (6.12)

Note that the row and column labels of the matrix elements of C arecomposite labels: the row label, ik, is obtained from the row labels ofthe matrix elements of A and B and the column label, jl, is obtainedfrom the corresponding column labels. The matrices need not have thesame dimension and, in fact, need not even be square. However, sincewe will apply direct products to construct group representations, wewill confine our discussion to square matrices. In this case, if A is ann× n matrix and B is an m×m matrix, C is an mn×mn matrix.

Example 6.2. For matrices A and B given by

A =

(a11 a12

a21 a22

), B =

b11 b12 b13

b21 b22 b23

b31 b32 b33

,

the direct product C = A⊗B is

A⊗B =

a11b11 a11b12 a11b13 a12b11 a12b12 a12b13

a11b21 a11b22 a11b23 a12b21 a12b22 a12b23

a11b31 a11b32 a11b33 a12b31 a12b32 a12b33

a21b11 a21b12 a21b13 a22b11 a22b12 a22b13

a21b21 a21b22 a21b23 a22b21 a22b22 a22b23

a21b31 a21b32 a21b33 a22b31 a22b32 a22b33

.


Another way of writing the direct product that more clearly displaysits structure is

A⊗B =

(a11B a12B

a21B a22B

).

The notion of a direct product arises quite naturally in quantummechanics if we consider the transformation properties of a product oftwo eigenfunctions. Suppose we have two eigenfunctions ϕi and ϕi′ ofa Hamiltonian H which is invariant under some group of operations.As in Section 6.2, the action of these operations on the eigenfunctionsof H is

Rϕi =∑j=1

ϕjΓji(R) ,

Rϕi′ =`′∑

j′=1

ϕj′Γj′i′(R) .

The question we now ask is: how does the product ϕiϕi′ transform underthe symmetry operations of the Hamiltonian? Given the transformationproperties of ϕi and ϕi′ noted above, we first observe that

R(ϕiϕi′) = R(ϕi)R(ϕi′) .

In other words, since R represents a coordinate transformation, its ac-tion on any function of the coordinates is to transform each occurrenceof the coordinates. Thus,

R(ϕiϕi′) =∑j=1

`′∑j′=1

ϕjϕj′Γji(R)Γj′i′(R)

=∑j=1

`′∑j′=1

ϕjϕj′Γjj′;ii′(R) ,

so ϕiϕi′ transforms as the direct product of the irreducible representa-tions associated with ϕi and ϕi′ .


6.5.3 Representations of Direct Product Groups

Determining the representations of direct products and the constructionof their character tables are based on the following theorem:

Theorem 6.1. The direct product of the representations of two groupsis a representation of the direct product of these groups.

Proof. A typical product of elements in the direct product group in(6.11) is

(apbq)(ap′bq′) = (apap′)(bqbq′) = arbr′ .

A representation of the direct product group must preserve the mul-tiplication table. We will use the notation that the matrix A(apbq)corresponds to the element apbq. Thus, we must require that

A(apbq)A(ap′bq′) = A(arbr′) .

By using the definition of the direct product of two matrices in Equation(6.12), we can write this equation in terms of matrix elements as[

A(apbq)A(ap′bq′)]ik;jl

=∑m,n

A(apbq)ik;mn︸︷︷︸A(ap)imA(bq)kn

A(ap′bq′)mn;jl︸︷︷︸(A(ap′ )mjA(bq′ )nl

=(∑

m

A(ap)imA(ap′)mj

)︸︷︷︸

A(apap′ )ij

(∑n

A(bq)]knA(bq′)nl

)︸︷︷︸

A(bpbp′ )kl

= A(ar)ijA(br′)kl

= A(arbr′)ik;jl .

Thus, the direct product of the representations preserves the multipli-cation table of the direct product group and, hence, is a representationof this group.

In fact, as shown in Problem 4 of Problem Set 8, the direct productof irreducible representations of two groups is an irreducible represen-tation of the direct product of those groups. An additional convenient


feature of direct product groups is that the characters of its represen-tations can be computed directly from the characters of the represen-tations of the two groups forming the direct product. This statementis based on the following theorem:

Theorem 6.2. If χ(ap) and χ(bq) are the characters of representationsof two groups Ga and Gb, the characters χ(apbq) of the representationformed from the matrix direct product of these representations is

χ(apbq) = χ(ap)χ(bq) .

Proof. From the definition of the direct product, a representationof the direct product group is

A(apbq)ij;kl = A(ap)ikA(bq)jl .

Taking the trace of both sides of this expression yields∑i,j

A(apbq)ij;ij︸︷︷︸χ(apbq)

=(∑

i

A(ap)ii

)︸︷︷︸

χ(ap)

(∑j

A(bq)jl

)︸︷︷︸

χ(bq)

.

Thus,

χ(apbq) = χ(ap)χ(bq) ,

which proves the theorem.

Since the characters are associated with a given class, the charactersfor the classes of the direct product are computed from the charactersof the classes of the original groups whose elements contribute to eachclass of the direct product. Moreover, the number of classes in the directproduct group is the product of the numbers of classes in the originalgroups. This can be seen immediately from the equivalence classes inthe direct product group. Using the fact that elements belonging tothe different groups commute,

(aibj)−1(akbl)(aibj)

−1 = (a−1i akai)(b

−1j blbj) .


Thus, equivalence classes in the direct product group must be formedfrom elements in equivalence classes in the original groups.

Example 6.3. Consider the direct product group of the equilateralwedge in Example 6.1. The classes of S3 are (Example 2.9), in thenotation of Example 5.5,

E, σv,1, σv,2, σv,3, C3, C23 ,

and the classes of the group E, σh are

E, σh .

There are, therefore, six classes in the direct product group, which areobtained by taking the products of elements in the original classes, asdiscussed above:

E, C3, C23, σv,1, σv,2, σv,3 ,

σh, σhC3, C23, σhσv,1, σv,2, σv,3 .

The structure of the character table of the direct product group cannow be determined quite easily. We denote the character for the αthclass of the jth irreducible representation of group Ga by χjα(ap). Simi-larly, we denote the the character for the βth class of the lth irreduciblerepresentation of group Ga by χlβ(bq). Since the direct products of irre-ducible representations of Ga and Gb are irreducible representations ofGa ⊗Gb (Problem 4, Problem Set 8), and since the classes of Ga ⊗Gb

are formed from products of classes of Ga and Gb, the character tableof the direct product group has the form

χjlαβ(apbq) = χjα(ap)χlβ(bq) .

In other words, with the character tables regarded as square matrices,the character table of the direct product group Ga ×Gb is constructedas a direct product of the character tables of Ga and Gb!


Example 6.4. For the direct product group in Example 6.3, the char-acter tables of the original groups are

E 3σv 2C3

A1 1 1 1A2 1 −1 1E 2 0 −1

and

E σhA1 1 1A2 1 −1

The character table of the direct product group is, therefore, thedirect product of these tables (cf. Example 6.1):

E 3σv 2C3 σh 3σhσv 2σhC3

A+1 1 1 1 1 1 1

A+2 1 −1 1 1 −1 1

E + 2 0 −1 2 0 −1A−1 1 1 1 −1 −1 −1A−2 1 −1 1 −1 1 −1E − 2 0 −1 −2 0 1

where the superscript on the irreducible representation refers to theparity under reflection through σh.

6.6 Selection Rules

One common application of direct products and their representationsis in the determination of selection rules. In this section, we will applythe techniques developed in this chapter to determine the conditionswhere symmetry requires that a matrix element vanishes.


6.6.1 Matrix Elements

As discussed in Section 6.3.2, the determination of selection rules isbased on using group theory to ascertain when the matrix element

Mif = (i|H′|f) =∫ϕi(x)∗H′ϕf (x) dx (6.13)

vanishes by reasons of symmetry. In this matrix element, the initialstate transforms according to an irreducible representation Γ(i) andthe final state transforms according to an irreducible representationΓ(f). It only remains to determine the transformation properties ofH′. We do this by applying each of the operations of the group ofthe original Hamiltonian H0 to the perturbation H′. If we retain onlythe distinct results of these operations we obtain, by construction, arepresentation of the group of the Hamiltonian, which we denote by Γ′.This representation may be either reducible or irreducible, dependingon H′ and on the symmetry of H0. If H′ has the same symmetry as H0,then this procedure generates the identical representation. At the otherextreme, if H′ has none of the symmetry properties of H0, then thisprocedure generates a reducible representation whose dimensionality isequal to the order of the group.

We now consider the symmetry properties under transformation ofthe product H(x)ϕf (x). From the discussion in Section 6.5.2, we con-clude that this quantity transforms as the direct product Γ′ ⊗ Γ(f).Since quantities that transform to different irreducible irreducible rep-resentations are orthogonal (Problem 6, Problem Set 8), the matrixelement (6.13) vanishes if this direct product is either not equal to Γ(i)

or, if it is reducible, does not include Γ(i) in its decomposition. We cansummarize this result in the following theorem:

Theorem 6.3. The matrix element

(i|H′|f) =∫ϕi(x)∗H′ϕf (x) dx

vanishes if the irreducible representation Γ(i) corresponding to ϕi is notincluded in the direct product Γ′ ⊗ Γ(f) of the representations Γ′ andΓ(f) corresponding to H′ and ϕf , respectively.


It is important to note that this selection rule only provides a con-dition that guarantees that the matrix element will vanish. It does notguarantee that the matrix element will not vanish even if the conditionsof the theorem are fulfilled.

6.6.2 Dipole Selection Rules

As a scenario which illustrates the power of group theoretical methods,suppose that H′ transforms as a vector, i.e., as (x, y, z). This situationarises when the transitions described by Fermi’s Golden Rule (Section6.3.2) are caused by an electromagnetic field. The form of H′ in thepresence of an electromagnetic potential A is obtained by making thereplacement6

p→ p− eA

for the momentum in the Hamiltonian. For weak fields, this leads to aperturbation of the form

H′ = e

mp ·A (6.14)

Since the electromagnetic is typically uniform, we can write the matrixelement Mif as

Mif ∼ (i|p|f) ·A

so the transformation properties of p = (px, py, pz), which are clearlythose of a vector, determine the selection rules for electromagnetic tran-sitions. These are called the dipole selection rules. The examination ofmany properties of materials rely on the evaluation of dipole matrixelements.

Example 6.6. Suppose the group of the Hamiltonian corresponds tothe symmetry operations of an equilateral triangle, i.e., C3v, the char-acter table for which is (Example 5.5)

6H. Goldstein, Classical Mechanics (Addison–Wesley, Reading, MA, 1950)


C3v E 3σv 2C3

A1 1 1 1A2 1 −1 1E 2 0 −1

To determine the dipole selection rules for this system, we must firstdetermine the transformation properties of a vector r = (x, y, z). Wetake the x- and y-axes in the plane of the equilateral triangle and thez-axis normal to this plane to form a right-handed coordinate system.Applying each symmetry operation to r produces a reducible repre-sentation because these operations are either rotations or reflectionsthrough vertical planes. Thus, the z coordinate is invariant under ev-ery symmetry operation of this group which, together with the fact thatan (x, y) basis generates the two-dimensional irreducible representationE, yields

Γ′ = A1 ⊕ E

We must now calculate the characters associated with the directproducts of between Γ′ and each irreducible representation to determinethe allowed final states given the transformation properties of the initialstates. The characters for these direct products are shown below

Γ′ = A1 ⊕ EC3v E 3σv 2C3

A1 1 1 1A2 1 −1 1E 2 0 −1

A1 ⊗ Γ′ 3 1 0A2 ⊗ Γ′ 3 −1 0E ⊗ Γ′ 6 0 0

Using the decomposition theorem, we find

A1 ⊗ Γ′ = A1 ⊕ E

A2 ⊗ Γ′ = A2 ⊕ E

E ⊗ Γ′ = A1 ⊕ A2 ⊕ 2E


Thus, if the initial state transforms as the identical representation A1,the matrix element vanishes if the final state transforms as A2. If theinitial state transforms as the “parity” representation A2, the matrixelement vanishes if the final state transforms as A1. Finally, there is nosymmetry restriction if the initial state transforms as the “coordinate”representation E.

6.7 Summary

This chapter has demonstrated how the mathematics of groups andtheir representations are used in quantum mechanics and, indeed, howmany of the structures introduced in the preceding chapters appearquite naturally in this context. Apart from exact results, such asBloch’s theorem, we have focussed on the derivation of selection rulesinduced by perturbations, and derived the principles behind dipole se-lection rules. A detailed discussion of other applications of discretegroups to quantum mechanical problems is described in the book byTinkham. Many of the proofs concerning the relation between quan-tum mechanics and representations of the group of the Hamiltonian arediscussed by Wigner.

Chapter 7

Continuous Groups, LieGroups, and Lie Algebras

Zeno was concerned with three problems . . . These are the problem ofthe infinitesimal, the infinite, and continuity . . .

—Bertrand Russell

The groups we have considered so far have been, in all but a few cases,discrete and finite. Most of the central theorems for these groups andtheir representations have relied on carrying out sums over the groupelements, often in conjunction with the Rearrangement Theorem (The-orem 2.1). These results provide the basis for the application of groupsand representations to physical problems through the construction andmanipulation of character tables and the associated computations thatrequire direct sums, direct products, orthogonality and decomposition.

But the notion of symmetry transformations that are based on con-tinuous quantities also occur naturally in physical applications. Forexample, the Hamiltonian of a system with spherical symmetry (e.g.,atoms and, in particular, the hydrogen atom) is invariant under allthree-dimensional rotations. To address the consequences of this in-variance within the framework of group theory necessitates confrontingseveral issues that arise from the continuum of rotation angles. Theseinclude defining what we mean by a “multiplication table,” determin-ing how summations over group elements are carried out, and deriving

107

108 Continuous Groups, Lie Groups, and Lie Algebras

the appropriate re-statement of the Rearrangement Theorem to enablethe Great Orthogonality Theorem and its consequences to be obtainedfor continuous groups. More generally, the existence of a continuumof group elements, when combined with the requirement of analyticity,introduces new structures associated with constructing differentials andintegrals of group elements. In effect, this represents an amalgamationof group theory and analysis, so such groups are the natural objectsfor describing the symmetry of analytic structures such as differentialequations and those that arise in differential geometry. In fact, theintroduction of analytic groups by Sophus Lie late in the 19th centurywas motivated by the search for symmetries of differential equations.

In this chapter we begin our discussion about the modifications toour development of groups and representations that are necessitated byhaving a continuum of elements. We begin in the first section with thedefinition of a continuous group and specialize to the most commontype of continuous group, the Lie group. We then introduce the ideaof an infinitesimal generator of a transformation, from which everyelement can be obtained by repeated application. These generatorsembody much of the structure of the group and, because there are afinite number of these entities, are simpler to work with than the fullgroup. This leads naturally to the Lie algebra associated with a Liegroup. All of these concepts are illustrated with the groups of properrotations in two and three dimensions. The representation of thesegroups, their character tables, and basis functions will be discussed inthe next chapter.

7.1 Continuous Groups

Consider a set of elementsR that depend on a number of real continuousparameters, R(a) ≡ R(a1, a2, . . . , ar). These elements are said to forma continuous group if they fulfill the requirements of a group (Section2.1) and if there is some notion of ‘proximity’ or ‘continuity’ imposedon the elements of the group in the sense that a small change in oneof the factors of a product produces a correspondingly small change intheir product. If the group elements depend on r parameters, this iscalled an r-parameter continuous group.

Continuous Groups, Lie Groups, and Lie Algebras 109

In general terms, the requirements that a continuous set of elementsform a group are the same as those for discrete elements, namely, clo-sure under multiplication, associativity, the existence of a unit, andan inverse for every element. Consider first the multiplication of twoelements R(a) and R(b) to yield the product R(c):

R(c) = R(a)R(b) .

Then c must be a continuous real function f of a and b:

c = f(a, b) .

This defines the structure of the group in the same way as the multi-plication table does for discrete groups. The associativity of the com-position law,

R(a)[R(b)R(c)

]︸︷︷︸R[f(b, c)]

= [R(a)R(b)]

︸︷︷︸R[f(b, c)]

R(c) ,

requires that

f [a, f(b, c)] = f [f(a, b), c] .

The existence of an identity element, which we denote by R(a0),

R(a0)R(a) = R(a)R(a0) = R(a) ,

is expressed in terms of f as

f(a0, a) = f(a, a0) = a .

The inverse of each element R(a), denoted by R(a′), produces

R(a′)R(a) = R(a)R(a′) = R(a0) .

Therefore,

f(a′, a) = f(a, a′) = a0 .

If f is an analytic function, i.e., a function with a convergent Tay-lor series expansion within the domain defined by the parameters, the


resulting group is called an r-parameter Lie group , named after SophusLie, a Norwegian mathematician who provided the foundations for suchgroups.

Our interest in physical applications centers around transformationson d-dimensional spaces. Examples include Euclidean spaces, where thevariables are spatial coordinates, Minkowski spaces, where the variablesare space-time coordinates, and spaces associated with internal degreesof freedom, such as spin or isospin. In all cases, these are mappings ofthe space onto itself and have the general form

x′i = fi(x1, x2, . . . , xd; a1, a2, . . . , ar), i = 1, 2, . . . , d .

If the fi are analytic, then this defines an r-parameter Lie group oftransformations.

Example 7.1 Consider the one-dimensional transformations

x′ = ax (7.1)

where a is an non-zero real number. This transformation correspondsto stretching the real line by a factor a. The product of two suchoperations, x′′ = ax′ and x′ = bx is

x′′ = ax′ = abx .

By writing x′′ = cx, we have that

c = ab , (7.2)

so the multiplication of two transformations is described by an analyticfunction that yields another transformation of the form in (7.1). Thisoperation is clearly associative, as well as Abelian, since the producttransformation corresponds to the multiplication of real numbers. Thisproduct can also be used to determine the inverse of these transfor-mations. By setting c = 1 in (7.2), so that x′′ = x, the inverse of(7.1) is seen to correspond to the transformation with a′ = a−1, whichexplains the requirement that a 6= 0. Finally, the identity is deter-mined from x′ = x, which clearly corresponds to the transformation


with a = 1. Hence, the transformations defined in (7.1) form a one-parameter Abelian Lie group.

Example 7.2 Now consider the one-dimensional transformations

x′ = a1x+ a2 , (7.3)

where again a1 is an non-zero real number. These transformations cor-responds to the stretching of the real line by a factor a1, as in thepreceding Example, and a translation by a2. The product of two oper-ations is

x′′ = a1x′ + a2 = a1(b1x+ b2) + a2 = a1b1x+ a1b2 + a2 .

By writing x′′ = c1x+ c2, we have that

c1 = a1b1, c2 = a1b2 + a2 ,

so the multiplication of two transformations is described by an analyticfunction and yields another transformation of the form in (7.1). How-ever, although this multiplication is associative, it is not Abelian, ascan be seen from the fact that the indices do not enter symmetricallyin c2. By setting, c1 = c2 = 1, the inverse of (7.3) is the transformation

x′ =x

a1

− a2

a1

.

The identity is again determined from x′ = x, which requires thata1 = 1 and a2 = 0. Hence, the transformations in (7.3) form a two-parameter (non-Abelian) Lie group.

7.2 Linear Transformation Groups

An important class of transformations is the group of linear transfor-mations in d dimensions. These can be represented by d× d matrices.For example, the most general such transformation in two dimensionsis x′ = Ax or, in matrix form,(

x′

y′

)=

(a11 a12

a21 a22

)(x

y

), (7.4)


where det(A) = a11a22 − a12a21 6= 0 (Example 2.4). With no furtherrestriction, and with the composition of two elements given by the usualrules of matrix multiplication, these matrices form a four-parameter Liegroup. This Lie group is called the general linear group in two dimensionsand is denoted by GL(2,R), where the ‘R’ signifies that the entriesare real; the corresponding group with complex entries is denoted byGL(2,C). In n dimensions, these transformation groups are denoted byGL(n,R), or, with complex entries, by GL(n,C).

7.2.1 Orthogonal Groups

Many transformations in physical applications are required to preservelength in the appropriate space. If that space is ordinary Euclideann-dimensional space, the restriction that lengths be preserved meansthat

x′21 + x′22 + · · ·+ x′2n = x21 + x2

2 + · · ·+ x2n . (7.5)

The corresponding groups, which are subgroups of the general lineargroup, are called orthogonal , and are denoted by O(n).

Consider the orthogonal group in two-dimensions, i.e., O(2), wherethe coordinates are x and y. By substituting the general transformation(7.4) into (7.5), we require that

x′2 + y′2 = (a11x+ a12y)2 + (a21x+ a22y)2

= (a211 + a2

21)x2 + 2(a11a12 + a21a22)xy + (a212 + a2

22)y2 .

For the right-hand side of this equation to be equal to x2 + y2 for all xand y, we must set

a211 + a2

21 = 1, a11a12 + a21a22 = 0, a212 + a2

22 = 1 .

Thus, we have three conditions imposed on four parameters, leaving onefree parameter. These conditions can be used to establish the followingrelation:

(a11a22 − a12a21)2 = 1 .


Recognizing the quantity in parentheses as the determinant of the trans-formation, this condition implies that

det(A) = ±1 .

If det(A) = 1, then the parity of the coordinate system is not changedby the transformation; this corresponds to a proper rotation. If det(A) =−1, then the parity of the coordinate system is changed by the transfor-mation; this corresponds to an improper rotation. As we have alreadyseen, both types of transformations are important in physical applica-tions, but we will first examine the proper rotations in two-dimensions.This group is called the special orthogonal group in two dimensions andis denoted by SO(2), where “special” signifies the restriction to properrotations. The parametrization of this group that we will use is

R(ϕ) =

(cosϕ − sinϕ

sinϕ cosϕ

), (7.6)

where ϕ, the single parameter in this Lie group, is the rotation angle ofthe transformation. As can easily be checked using the trigonometricidentities for the sum of two angles,

R(ϕ1 + ϕ2) = R(ϕ1)R(ϕ2) , (7.7)

so this group is clearly Abelian.

7.3 Infinitesimal Generators

A construction of immense utility in the study of Lie groups, which wasintroduced and extensively studied by Lie, is the infinitesimal generator .The idea behind this is that instead of having to consider the group asa whole, for many purposes it is sufficient to consider an infinitesimaltransformation around the identity. Any finite transformation can thenbe constructed by the repeated application, or “integration,” of thisinfinitesimal transformation.


7.3.1 Matrix Form of Generators

For SO(2), we first expand R(ϕ) in a Taylor series around the identity(ϕ = 0):

R(ϕ) = R(0) +dR

dϕ

∣∣∣∣ϕ=0

ϕ+1

2

d2R

dϕ2

∣∣∣∣ϕ=0

ϕ2 + · · · . (7.8)

The coefficients in this series can be determined directly from (7.6), buta more elegant solution may be found by first differentiating (7.7) withrespect to ϕ1,

d

dϕ1

R(ϕ1 + ϕ2) =dR(ϕ1)

dϕ1

R(ϕ2) , (7.9)

then setting ϕ1 = 0. Using the chain rule, the left-hand side of thisequation is [

dR(ϕ1 + ϕ2)

d(ϕ1 + ϕ2)

d(ϕ1 + ϕ2)

dϕ1

]∣∣∣∣ϕ1=0

=dR(ϕ2)

dϕ2

,

so Eq. (7.9) becomes

dR(ϕ)

dϕ= XR(ϕ) , (7.10)

where

dR(ϕ1)

dϕ1

∣∣∣∣ϕ1=0

=

(0 −1

1 0

)≡ X . (7.11)

Equations (7.10) and (7.11) allow us to determine all of the expan-sions coefficients in (7.9). By setting ϕ = 0 in (7.10) and observingthat R(0) = I, where I is the 2× 2 unit matrix,

I =

(1 0

0 1

),

we obtain

dR(ϕ)

dϕ

∣∣∣∣ϕ=0

= X . (7.12)


To determine the higher-order derivatives of R, we differentiate (7.10)n times, and set ϕ = 0:

dnR(ϕ)

dϕn

∣∣∣∣ϕ=0

= Xdn−1R(ϕ)

dϕn−1

∣∣∣∣ϕ=0

.

This yields, in conjunction with (7.12),

dnR(ϕ)

dϕn

∣∣∣∣ϕ=0

= Xn .

Substituting this expression into the Taylor series in (7.8) allows us towrite

R(ϕ) = I +Xϕ+ 12X2ϕ2 + · · ·

=∞∑n=0

1

n!(Xϕ)n

≡ eϕX ,

where X0 = I and the exponential of a matrix is defined by the Taylorseries expansion of the exponential. Thus, every rotation by a finiteangle can be obtain from the exponentiation of the matrix X, whichis called the infinitesimal generator of rotations. Since X2 = I, it is astraightforward matter to show directly from the Taylor series of theexponential (Problem 4, Problem Set 9) that

eϕX = I cosϕ+X sinϕ =

(cosϕ − sinϕ

sinϕ cosϕ

).

7.3.2 Operator Form of Generators

An alternative way of representing infinitesimal generators throughwhich connections with quantum mechanics can be directly made isin terms of differential operators. To derive the operator associatedwith infinitesimal rotations, we expand (7.6) to first order in dϕ toobtain the transformation

x′ = x cosϕ− y sinϕ = x− y dϕ ,

y′ = x sinϕ+ y cosϕ = x dϕ+ y .


An arbitrary differentiable function F (x, y) then transforms as

F (x′, y′) = F (x− y dϕ, x dϕ+ y) .

Retaining terms to first order in dϕ on the right-hand side of this equa-tion yields

F (x′, y′) = F (x, y) +(− y∂F

∂x+ x

∂F

∂y

)dϕ .

Since F is an arbitrary function, we can associate infinitesimal rotationswith the operator

X = x∂

∂y− y ∂

∂x.

As we will see in the next section, this operator is proportional to thez-component of the angular momentum operator.

The group SO(2) is simple enough that the full benefits of an in-finitesimal generator are not readily apparent. We will see in the nextsection, where we discuss SO(3), that the infinitesimal generators em-body much of the structure of the full group.

7.4 SO(3)

The orthogonal group in three dimensions is comprised of the trans-formations that leave the quantity x2 + y2 + z2 invariant. The groupGL(3,R) has 9 parameters, but the invariance of the length produces sixindependent conditions, leaving three free parameters, so O(3) forms athree-parameter Lie group. If we restrict ourselves to transformationswith unit determinant, we obtain the group of proper rotations in threedimensions, SO(3).

There are three common ways to parametrize these rotations:

• Successive rotations about three mutually orthogonal fixed axes.

• Successive about the z-axis, about the new y-axis, and then aboutthe new z-axis. These are called Euler angles .


• The axis-angle representation, defined in terms of an axis whosedirection is specified by a unit vector (two parameters) and arotation about that axis (one parameter).

In this section, we will use the first of these parametrizations todemonstrate some of the properties of SO(3). In the next chapter,where we will develop the orthogonality relations for this group, theaxis-angle representation will prove more convenient.

7.4.1 Rotation Matrices

Consider first rotations about the z-axis by an angle ϕ3:

R3(ϕ3) =

cosϕ3 − sinϕ3 0

sinϕ3 cosϕ3 0

0 0 1

.

The corresponding infinitesimal generator is calculated as in (7.11):

X3 =dR3

dϕ3

∣∣∣∣ϕ3=0

=

0 −1 0

1 0 0

0 0 0

.

These results are essentially identical to those for SO(2). However,for SO(3), we have rotations about two other axes to consider. Forrotations about the x-axes by an angle ϕ1, the rotation matrix is

R1(ϕ1) =

1 0 0

0 cosϕ1 − sinϕ1

0 sinϕ1 cosϕ1

and the corresponding generator is

X1 =dR1

dϕ1

∣∣∣∣ϕ1=0

=

0 0 0

0 0 −1

0 1 0


Finally, for rotations about the y-axis by an angle ϕ2, we have

R2(ϕ2) =

cosϕ2 0 sinϕ2

0 1 0

− sinϕ2 0 cosϕ2

and the generator is

X2 =dR2

dϕ2

∣∣∣∣ϕ2=0

=

0 0 1

0 0 0

−1 0 0

As can be easily verified, the matrices Ri(ϕi) do not commute, nor

do the Xi. However, the Xi have an additional useful property, namelyclosure under commutation. As an example, consider the productsX1X2 and X2X1:

X1X2 =

0 0 0

0 0 −1

0 1 0

0 0 1

0 0 0

−1 0 0

=

0 0 0

1 0 0

0 0 0

X2X1 =

0 0 1

0 0 0

−1 0 0

0 0 0

0 0 −1

0 1 0

=

0 1 0

0 0 0

0 0 0

Thus, the commutator of X1 and X2, denoted by [X1, X2] is given by

[X1, X2] ≡ X1X2 −X2X1 =

0 −1 0

1 0 0

0 0 0

= X3

Similarly, we have

[X2, X3] = X1, [X3, X1] = X2

The commutation relations among all of the Xi can be succinctly sum-marized by introducing the anti-symmetric symbol εijk, which takes the


value 1 for a symmetric permutation of distinct i, j, and k, the value−1 for an antisymmetric permutation, and is zero otherwise (i.e., if twoor more of i, j and k are equal). We can then write

[Xi, Xj] = εijkXk (7.13)

We will discuss the physical interpretation of these generators once weobtain their operator form in the next section.

7.4.2 Operators for Infinitesimal Rotations

As was the case in Section 7.3, an alternative to the matrix represen-tation of infinitesimal generators is in terms of differential operators.Proceeding as in that section, we first write the general rotation as anexpansion to first order in each of the ϕi about the identity. This yieldsthe transformation matrix

x′

y′

z′

=

1 −ϕ3 ϕ2

ϕ3 1 −ϕ1

−ϕ2 ϕ1 1

x

y

z

Substituting this coordinate transformation into a differentiable func-tion F (x, y, z),

F (x′, y′, z′) = F (x− ϕ3y + ϕ2z, y + ϕ3x− ϕ1z, z − ϕ2x+ ϕ1y)

and expanding the right-hand side to first order in the ϕi yields thefollowing expression:

F (x′, y′, z′) = F (x, y, z)

+(∂F

∂zy − ∂F

∂yz)ϕ1 +

(∂F

∂xz − ∂F

∂zx)ϕ2 +

(∂F

∂yx− ∂F

∂xy)ϕ3

Since F is an arbitrary differentiable function, we can identify the gen-erators Xi of rotations about the coordinate axes from the coefficientsof the ϕi, i.e., with the differential operators

X1 = y∂

∂z− z ∂

∂y


X2 = z∂

∂x− x ∂

∂z(7.14)

X3 = x∂

∂y− y ∂

∂x

Notice that X3 is the operator obtained for SO(2) in Section 7.3. Wecan now assign a physical interpretation to these operators by compar-ing them with the vectors components of the angular operators in thecoordinate representation, obtained from the definition

L = r × p = r × (−ih∇)

Carrying out the cross-product yields the standard expressions

L1 = −ih(y∂

∂z− z ∂

∂y

)

L2 = −ih(z∂

∂x− x ∂

∂z

)(7.15)

L3 = −ih(x∂

∂y− y ∂

∂x

)for the x, y, and z components of L, respectively. Thus, Li = −ihXi,for i = 1, 2, 3, and (7.13) becomes

[Li, Lj] = ihεijkLk

which are the usual angular momentum commutation relations. There-fore, we can associate the vector components of the angular momentumoperator with the generators of infinitesimal rotations about the cor-responding axes. An analogous association exists between the vectorcomponents of the coordinate representation of the linear momentumoperator and differential translation operations along the correspondingdirections.

7.4.3 The Algebra of Infinitesimal Generators

The commutation relations in (7.13) define a “product” of two gener-ators which yields the third generator. Thus, the set of generators is


closed under this operation. Triple products, which determine whetheror not this composition law is associative, can be written in a conciseform using only the definition of the commutator, i.e., in the form ofan identity, without any explicit reference to the quantities involved.Beginning with the triple product[

A, [B,C]]

= A[B,C]− [B,C]A

= ABC − ACB −BCA+ CBA

We now add and subtract the quantities BAC and CAB on the right-hand side of this equation and rearrange the resulting expression intocommutators to obtain[

A, [B,C]]

= ABC − ACB −BCA+ CBA

+BAC −BAC + CAB − CAB

= −C(AB −BA) + (AB −BA)C

+B(AC − CA)− (AC − CA)B

= −[[A,B], C

]+[[C,A], B

]A simple rearrangement yields the Jacobi identity :[

A, [B,C]]

+[B, [C,A]

]+[C, [A,B]

]= 0

Notice that this identity has been obtained using only the definition ofthe commutator.

For the infinitesimal generators of the rotation group, with the com-mutator in (7.13), each of the terms in the Jacobi identity vanishes.Thus, [

A, [B,C]]

=[[A,B], C]

]so the product of these generators is associative. In the more generalcase, however, products of quantities defined in terms of a commutatorare not associative. The Lie algebra associated with the Lie group from


which the generators are obtained consists of quantities A,B,C, . . .defined by

A =3∑

k=1

akXk, B =3∑

k=1

bkXk, C =3∑

k=1

ckXk, etc.

where the ak, bk, ck, . . . are real coefficients and from which linear com-binations αA + βB with real α and β can be formed. The product isgiven by

[A,B] = −[B,A]

and the Jacobi identity is, of course, satisfied.The formal definition of a Lie algebra, which is an abstraction of

the properties just discussed, is as follows.

Definition. A Lie algebra is a vector space L over some field F 1

(typically the real or complex numbers) together with a binary opera-tion [·, ·] : L × L → L, called the Lie bracket, which has the followingproperties:

1. Bilinearity .

[ax+ by, z] = a[x, z] + b[y, z]

[z, ax+ by] = a[z, x] + b[z, y]

for all a and b in F and x, y, and z in L.

2. Jacobi identity .[[x, y], z

]+[[z, x], y] +

[[y, z], x

]= 0

for all x, y, and z in L.

1A field is an algebraic system of elements in which the operations of addition,subtraction, multiplication, and division (except by zero) may be performed withoutleaving the system (closure) and the associative, commutative, and distributiverules, familiar from the arithmetic of ordinary numbers, hold. Examples of fieldsare the rational numbers, the real numbers, and the complex numbers. The smallestfield has only two elements: 0, 1. The concept of a field is useful for defining vectorsand matrices, whose components can be elements of any field.


3. Antisymmetry .

[x, y] = −[y, x]

for all x and y in L.

7.5 Summary

In this chapter, we have described the properties of Lie groups interms of specific examples, especially SO(2) and SO(3). With thisbackground, we can generalize our discussion to any Lie group. Anr-parameter Lie group of transformations on an n-dimensional space is

x′i = fi(x1, x2, . . . , xn; a1, a2, . . . , ar)

where i = 1, 2, . . . , n. If only one of the r parameters ai is changedfrom zero, while all the other parameters are held fixed, we obtain theinfinitesimal transformations Xi associated with this Lie group. Thesecan be expressed as differential operators by examining the effect ofthese infinitesimal coordinate transformations on an arbitrary differen-tiable function F :

dF =n∑j=1

∂F

∂xjdxj

=n∑j=1

∂F

∂xj

( r∑i=1

∂fj∂ai

∣∣∣∣a=0

dai

)

=r∑i=1

dai

( n∑j=1

∂fj∂ai

∣∣∣∣a=0

∂

∂xj

)F

We identify the differential operators Xi as the coefficient of dai in thisdifferential:

Xi =n∑j=1

∂fj∂ai

∣∣∣∣a=0

∂

∂xj

for r = 1, 2, . . . , r. These operators satisfy commutation relations ofthe form

[Xi, Xj] = ckijXk


where the ckij are called structure constants and are a property of thegroup. The commutator satisfies the Jacobi identity,[

Xi, [Xj, Xk]]

+[Xj, [Xk, Xi]

]+[Xk, [Xi, Xj]

]= 0

which places a constraint on the structure constants. The commutatorand the Jacobi identity, together with the ability to form real linearcombinations of the Xi endows these generators with the structure ofan algebra, called the Lie algebra associated with the Lie group.

Chapter 8

Irreducible Representationsof SO(2) and SO(3)

The shortest path between two truths in the real domain passes throughthe complex domain.

—Jacques Hadamard1

Some of the most useful aspects of group theory for applications tophysical problems stem from the orthogonality relations of charactersof irreducible representations. The widespread impact of these relationsstems from their role in constructing and resolving new representationsfrom direct products of irreducible representations. Direct products areespecially important in applications involving continuous groups, withthe construction of higher dimensional irreducible representations, thederivation of angular momentum coupling rules, and the characteriza-tion of families of elementary particles all relying on the formation anddecomposition of direct products.

Although the notion of an irreducible representation can be carriedover directly from our development of discrete groups through Schur’sfirst lemma, a transcription of Schur’s second lemma and the GreatOrthogonality Theorem to the language of continuous groups requiresa separate discussion. This is because proving the latter two theorems

1Quoted in The Mathematical Intelligencer 13(1), 1991.

125

126 Irreducible Representations of SO(2) and SO(3)

necessitates performing summations over group elements and invokingthe Rearrangement Theorem (Theorem 2.1). This theorem guaranteesthe following equality ∑

g

f(g) =∑g

f(g′g) , (8.1)

where the summation is over elements g in a group G, g′ is any other el-ement in G, and f is some function of the group elements. The crucialpoint is that the same quantities appear on both sides of the equa-tion; the only difference is the order of their appearance. To proceedwith the proofs of these theorems for continuous groups requires anequality analogous to (8.1):∫

f(R) dR =∫f(R′R) dR , (8.2)

where R and R′ are the elements of a continuous group and f is somefunction of these elements. To appreciate the issues involved, we writethe integral on the left-hand side of (8.2) as an integral over the pa-rameters ∫

f(R) dR =∫f(R)g(R) da , (8.3)

where g(R) is the density of group elements in parameter space in theneighborhood of R. The equality in (8.2) will hold provided that thedensity of group elements is arranged so that the density of the pointsR′R is the same as that of the points R. Our task is to find the form ofg(R) which ensures this. A related concept that will arise is the notionof the “order” of the continuous group as the volume of its elements inthe space defined by the parameters of the group.

This chapter is devoted to the characters and irreducible represen-tations of SO(2) and SO(3). For SO(2), we will show that the density ofgroup elements is uniform across parameter space, so the density func-tion reduces to a constant. But, for SO(3), we will need to carry outthe determination of the density function in (8.3) explicitly. This willillustrate the general procedure which is applicable to any group. Forboth SO(2) and SO(3), we will derive the basis functions for their irre-ducible representations which will be used to obtain the correspondingcharacters and to demonstrate their orthogonality

Irreducible Representations of SO(2) and SO(3) 127

8.1 Orthogonality of Characters for SO(2)

The structure of SO(2) is simple enough that many of the results ob-tained for discrete groups can be taken over directly with little or nomodification. The basis of this claim is that the Rearrangement Theo-rem for this group is, apart from the replacement of the sum by an in-tegral, a direct transcription of that for discrete groups which, togetherwith this group being Abelian, renders the calculation of characters astraightforward exercise.

8.1.1 The Rearrangement Theorem

We first show that the rearrangement theorem for this group is∫ 2π

0R(ϕ′)R(ϕ) dϕ =

∫ 2π

0R(ϕ) dϕ .

This implies that the weight function appearing in (8.3) is unity, i.e.,the density of group elements is uniform in the space of the parameterϕ. Using the fact that R(ϕ′)R(ϕ) = R(ϕ′ + ϕ), we have∫ 2π

0R(ϕ′)R(ϕ) dϕ =

∫ 2π

0R(ϕ′ + ϕ) dϕ .

We now introduce a new integration variable θ = ϕ′ + ϕ. Since ϕ′ isfixed, we have that dϕ = dθ. Then, making the appropriate changesin the upper and lower limits of integration, and using the fact thatR(ϕ+ 2π) = R(ϕ), yields∫ 2π

0R(ϕ′ + ϕ) dϕ =

∫ ϕ′+2π

ϕ′R(θ) dθ

=∫ 2π

ϕ′R(θ) dθ +

∫ ϕ′+2π

2πR(θ) dθ

=∫ 2π

ϕ′R(θ) dθ +

∫ ϕ′

0R(θ) dθ

=∫ 2π

0R(θ) dθ ,

which verifies our assertion.


8.1.2 Characters of Irreducible Representations

We can now use Schur’s first lemma for SO(2). Since SO(2) is anAbelian group, this first lemma requires all of the irreducible represen-tations to be one-dimensional (cf. Problem 4, Problem Set 5). Thus,every element is in a class by itself and the characters must satisfy thesame multiplication rules as the elements of the group:

χ(ϕ)χ(ϕ′) = χ(ϕ+ ϕ′) . (8.4)

The character corresponding to the unit element, χ(0), which mustmap onto the identity for ordinary multiplication, is clearly unity forall irreducible representations:

χ(0) = 1 . (8.5)

Finally, we require the irreducible representations to be single-valued,i.e., an increase in the rotation angle by 2π does not change the effectof the rotation. Thus,

χ(ϕ+ 2π) = χ(ϕ) . (8.6)

The three conditions in (8.4), (8.5), and (8.6) are sufficient to determinethe characters of all of the irreducible representations of SO(2).

We will proceed by writing Eq. (8.4) as a differential equation andusing (8.5) as an “initial condition” and (8.6) as a “boundary condi-tion.” In (8.4), we set ϕ′ = dϕ,

χ(ϕ)χ(dϕ) = χ(ϕ+ dϕ) ,

and expand both sides of this equation to first order in dϕ:

χ(ϕ)

[χ(0) +

dχ

dϕ

∣∣∣∣ϕ=0

dϕ

]= χ(ϕ) +

dχ

dϕdϕ .

Then, using (8.5) and cancelling common terms, this equation reducesto a first-order ordinary differential equation for χ(ϕ):

dχ

dϕ= χ′0χ(ϕ) ,


where χ′0 = χ′(0) is to be determined. The general solution to thisequation is

χ(ϕ) = A eχ′0ϕ ,

where A is a constant of integration which is also to be determined.In fact, by setting ϕ = 0 and invoking (8.5), we see that A = 1. Therequirement (8.6) of single-valuedness, when applied to this solution,yields the condition that

eχ′0(ϕ+2π) = eχ

′0ϕ ,

or,

e2πχ′0 = 1 .

The most general solution of this equation is χ′0 = im, where i2 = −1and m is any integer. This produces an infinite sequence of charactersof the irreducible representations of SO(2):

χ(m)(ϕ) = eimϕ, m = . . . ,−2,−1, 0, 1, 2, . . . (8.7)

The identical representation corresponds to m = 0. In contrast to thecase of finite groups, we see that SO(2) has an infinite set of irreduciblerepresentations, albeit one that is countably infinite.

8.1.3 Orthogonality Relations

Having determined the characters for SO(2), we can now examine thevalidity of the orthogonality theorems for characters which were dis-cussed for discrete groups in Theorem 5.1. We proceed heuristicallyand begin by observing that the exponential functions in (8.7) are or-thogonal over the interval 0 ≤ ϕ < 2π:∫ 2π

0ei(m

′−m)ϕ dϕ = 2πδm,m′ .

By writing this relation as∫ 2π

0χ(m)∗(ϕ)χ(m′)(ϕ) dϕ = 2πδm,m′ , (8.8)


we obtain an orthogonality relation of the form in Eq. (5.4), once weidentify the “order” of SO(2) as the quantity∫ 2π

0dϕ = 2π .

This is the “volume” of the group in the space of the parameter ϕ,which lies in the range 0 ≤ ϕ < 2π, given that the density function isunity, according to the discussion in thew preceding section. Note thatthe integration over ϕ is effectively a sum over classes.

Example 8.1. Consider the representation of SO(2) derived in Section7.2:

R(ϕ) =

(cosϕ − sinϕ

sinϕ cosϕ

). (8.9)

Since SO(2) is an Abelian group, this representation must be reducible.We can decompose this representation into its irreducible componentsby using either the analogue of the Decomposition Theorem (Section5.3) for continuous groups or, more directly, by using identities betweencomplex exponential and trigonometric functions:

χ(ϕ) ≡ tr[R(ϕ)]

= 2 cosϕ

= eiϕ + e−iϕ .

A comparison with (8.7) yields

χ(ϕ) = χ(1)(ϕ) + χ(−1)(ϕ) ,

so the representation in (8.9) is a direct sum of the irreducible repre-sentations corresponding to m = 1 and m = −1.2

2This example illustrates the importance of the field used in the entries of thematrices for SO(2). If we are restricted to real entries, then the representation in(8.9) is irreducible. But, if the entries are complex, then this example shows thatthis representation is reducible.


8.2 Basis Functions for Irreducible Rep-

resentations

We were able to determine the characters for all of the irreducible rep-resentations of SO(2) without any knowledge of the representationsthemselves. But this is not the typical case for continuous groups. Wewill see, for example, when determining the characters for SO(3) thatwe will be required to construct explicit representations of rotationscorresponding to different classes. The action of these rotations on thebasis functions will determine the representation of that class and thecharacter will be calculated directly from this representation. As anintroduction to that discussion, in this section we will determine thebasis functions of the irreducible representations of SO(2).

We begin by calculating the eigenvalues of the matrix in (8.9) fromdet(R− λI) = 0:

∣∣∣∣∣ cosϕ− λ − sinϕ

sinϕ cosϕ− λ

∣∣∣∣∣ = (cosϕ− λ)2 + sin2 ϕ

= λ2 − 2λ cosϕ+ 1 = 0 .

Solving for λ yields

λ = cosϕ± i sinϕ = e±iϕ . (8.10)

The corresponding eigenvectors are proportional to x ± iy. Thus, op-erating on these eigenvectors with R(ϕ) (see below) generates the irre-ducible representations corresponding to m = 1 and m = −1 in (8.7),i.e., the characters χ(1)(ϕ) and χ(−1)(ϕ).

Obtaining the basis functions for the other irreducible representa-tions of SO(2) is now a matter of taking appropriate direct products,since

χ(m)(ϕ)χ(m′)(ϕ) = χ(m+m′)(ϕ) .

In particular, the m-fold products (x± iy)m generate irreducible repre-sentations for the m-fold direct product, as discussed in Sec. 6.5. This


can be verified directly from the transformation (8.9) applied to x andy:

x′ = x cosϕ− y sinϕ ,

y′ = x sinϕ+ y cosϕ .

Then,

(x′ ± iy′)m =[x cosϕ− y sinϕ± i(x sinϕ+ y cosϕ)

]m=[x(cosϕ± i sinϕ)± iy(cosϕ± i sinϕ)

]m=[(x± iy) e±iϕ

]m= (x± iy)m e±imϕ .

Therefore, we can now complete the character table for SO(2), includingthe basis functions which generate the irreducible representations:

SO(2) E R(ϕ)

Γ±m : (x± iy)m 1 e±imϕ

We note for future reference that the basis functions (x± iy)m couldhave been derived in a completely different manner. Consider Laplace’sequation in two dimensions:

∂2u

∂x2+∂2u

∂y2= 0 .

This equation is invariant under all the elements of SO(2), as can beeasily verified. The general solution to this equation is

u(x, y) = F (x+ iy) +G(x− iy) ,

where F and G are arbitrary functions. Thus, if we are interestedin solutions which are homogeneous polynomials of degree m, we can


choose in turn solutions with F (s) = sm and G(s) = 0 and then withF (s) = 0 and G(s) = sm. We thereby obtain the expressions

u(x, y) = (x± iy)m (8.11)

as solutions of Laplace’s equations which are also the basis functions ofthe irreducible representations of SO(2). These functions are the ana-logues in two dimensions of spherical harmonics, which are the solutionsof Laplace’s equations in three dimensions. These will be discussed laterin this chapter.

8.3 Axis–Angle Representation of Proper

Rotations in Three Dimensions

The three most common parametrizations of proper rotations were dis-cussed in Section 7.4. For the purposes of obtaining the orthogonal-ity relations for the characters of SO(3), the representation in termsof a fixed axis about which a rotation is carried out—the axis–anglerepresentation—is the most convenient. We begin this section by show-ing how this representation emerges naturally from the basic propertiesof orthogonal matrices.

8.3.1 Eigenvalues of Orthogonal Matrices

Let A be any proper rotation matrix in three dimensions. Denoting theeigenvalues of A by λ1, λ2, and λ3, and the corresponding eigenvaluesby u1, u2, and u3, we then have

Aui = λiui

for i = 1, 2, 3. We can also form the adjoint of each equation:

u†iAt = λ∗u†i .

These eigenvalue equations imply

u†iui = u†iAtAui = |λ2

i |u†iui ,


which shows that |λ2i | = 1, i.e., that the modulus of every eigenvalue

of an orthogonal matrix is unity [cf. (8.10)]. The most general formof such a quantity is a complex number of the form eiϕ for some angleϕ. But these eigenvalues are also the roots of the characteristic equa-tion det(A − λI) = 0 so, according to the Fundamental Theorem ofAlgebra,3 if they are complex, they must occur in complex conjugatepairs (because the coefficients of this polynomial, which are obtainedfrom the entries of A, are real). Hence, the most general form of theeigenvalues of an orthogonal matrix in three dimensions is

λ1 = 1, λ2 = eiϕ, λ3 = e−iϕ . (8.12)

The eigenvector corresponding to λ1 = 1, which is unaffected by theaction of A, thereby defines the axis about which the rotation is taken.The quantity ϕ appearing in λ2 and λ3 defines the angle of rotationabout this axis.

8.3.2 The Axis and Angle of an Orthogonal Matrix

In this section, we show how the axis and angle of an orthogonal matrixcan be determined from its matrix elements. We take the axis of therotation to be a unit vector n, which is the eigenvector correspondingto the eigenvalue of unity:

An = n . (8.13)

This equation and the orthogonality of A (AAt = AtA = 1) enables usto write

Atn = AtAn = n . (8.14)

Subtracting (8.14) from (8.13) yields

(A− At)n = 0 .

3K. Hoffman and R. Kunze, Linear Algebra 2nd edn (Prentice–Hall, EnglewoodCliffs, NJ, 1971), p. 138.


In terms of the matrix elements aij of A and the components ni of n,we then have

(a12 − a21)n2 + (a13 − a31)n3 = 0 ,

(a21 − a12)n1 + (a23 − a32)n3 = 0 ,

(a31 − a13)n1 + (a32 − a23)n2 = 0 .

Notice that these equations involve only the off-diagonal elements of A.The solution of these equations yield the relations

n2

n1

=a31 − a13

a23 − a32

,n3

n1

=a12 − a21

a23 − a32

, (8.15)

which, when combined with the normalization condition

n·n = n21 + n2

2 + n23 = 1

determines n uniquely.The angle of the rotation can be determined from the invariance of

the trace of A under similarity transformations. Noting that the traceis the sum of the eigenvalues, and using (8.12), we have

a11 + a22 + a33 = 1 + eiϕ + e−iϕ = 1 + 2 cosϕ , (8.16)

so ϕ is determined only by the diagonal elements of A.

8.3.3 Normal Form of an Orthogonal Matrix

We conclude this section by deriving the form of a rotation matrix in anorthogonal coordinate system which naturally manifests the axis andangle. The diagonal form of a rotation matrix is clearly given by

Λ =

1 0 0

0 eiϕ 0

0 0 eiϕ

.

The eigenvector n corresponding to λ1 = 1 is the axis of the rotationand can always be chosen to be real. However, the eigenvectors of λ2 =


eiϕ and λ3 = e−iϕ are inherently complex quantities. An orthonormalset can be chosen as

n2 = 12

√2(0, 1, i), n2 = 1

2

√2(0, 1,−i) ,

respectively. Since we are interested in transformations of real coor-dinates, we must perform a unitary transformation from this complexbasis to a real orthogonal basis, in which case our rotation matrix Λ willno longer be diagonal. The required unitary matrix which accomplishesthis is

U =

1 0 0

0 12

√2 1

2i√

2

0 12

√2 −1

2i√

2

.

Thus,

R = U−1ΛU =

1 0 0

0 cosϕ − sinϕ

0 sinϕ cosϕ

. (8.17)

When expressed in this basis, the rotation matrix clearly displays theaxis of rotation through the entry R11 = 1, and the angle of rotationthrough a 2×2 rotational submatrix in a plane perpendicular to thisaxis.

8.3.4 Parameter Space for SO(3)

The axis-angle representation of three-dimensional rotations providesa convenient parametrization of all elements of SO(3). We have seenthat every element of SO(3) can be represented by a unit vector corre-sponding to the rotation axis and a scalar corresponding to the rotationangle. Thus, consider the space defined by the three quantities

(n1ϕ, n2ϕ, n3ϕ) , (8.18)

where n21 + n2

2 + n23 = 1. Every direction is represented by a point on

the unit sphere. Thus, defining an azimuthal angle φ and a polar angle


AB

C

D

E

O

Figure 8.1: Two-dimensional representation of the parameter space of SO(3)as the interior of a sphere of radius π. The point A represents a rotationwhose axis is along the direction OA and whose angle is the length of OA.The points at A, B and C correspond to rotations with the same angle butabout axis along different directions. This defines the classes of SO(3). Thediametrically opposite points at D and E correspond to the same operation.

θ according to the usual conventions in spherical polar coordinates, theparameter space of SO(3) can be represented as

(ϕ cosφ sin θ, ϕ sinφ sin θ, ϕ cos θ) , (8.19)

where

0 ≤ ϕ ≤ π, 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ π .

We can now see directly that this parameter space corresponds to theinterior of a sphere of radius π (Fig. 8.1). For every point within thesphere, there is a unique assignment to an element of SO(3): the direc-tion from the radius to the point corresponds to the direction of therotation axis and the distance from the point to the origin representsthe rotation angle. Two diametrically opposed points on the surface ofthe sphere (ϕ = π) correspond to the same rotation, since a rotationby π about n is the same as a rotation by −π about this axis which,in turn, is the same as a rotation by π about −n (whatever the senseof rotation).


Another useful feature of the axis-angle parametrization is the rep-resentation of classes of SO(3). Consider two elements of SO(3) whichhave the same angle of rotation ϕ but about different axes n and n′. Wedenote these operations by R(n, ϕ) and R(n′, ϕ). Let U(n,n′) denotethe rotation of n into n′. The inverse of this operation then rotates n′

into n. The relationship between R(n, ϕ), R(n′, ϕ), and U(n,n′) is,therefore,

R(n, ϕ) =[U(n,n′)

]−1R(n′, ϕ)U(n,n′) ,

i.e., R(n, ϕ) and R(n′, ϕ) are related by a similarity transformationand, therefore, belong to the same equivalence class. Referring toFig. 8.1, equivalence classes of SO(3) correspond to operations whichlie on the same radius. Thus, a summation over the classes of SO(3)is equivalent to a sum over spherical shells.

8.4 Orthogonality Relations for SO(3)

The axis-angle representation of rotations provides, in addition to aconceptual simplicity of elements of SO(3) in parameter space, a naturalframework within which to discuss the integration over the elementsof SO(3) and thereby to obtain the Rearrangement Theorem for thisgroup. In this section, we derive the density function g in (8.3) forthis group and then use this to identify the appropriate form of theorthogonality relations for characters

8.4.1 The Density Function

As discussed in the introduction, one of the basic quantities of interestfor continuous groups is the density of group elements as a functionof position in parameter space. To determine this function for SO(3),we first consider the elements in the neighborhood of the identity andthen examine the behavior of these points under an arbitrary elementof SO(3). Referring to the discussion in Section 7.4.2, these elementscorrespond to rotations by infinitesimal angles ϕ1, ϕ2, and ϕ3 abouteach of the three coordinate axes. The rotation matrix associated with


this transformation is

δR =

1 −ϕ3 ϕ2

ϕ3 1 −ϕ1

−ϕ2 ϕ1 1

.

The identity of SO(3) corresponds to the origin in the three-dimensionalparameter space, ϕ1 = ϕ2 = ϕ3 = 0, and is indicated by the pointO in Fig. 8.1. For infinitesimal rotation angles, the parameter spacespanned by δR is associated with an infinitesimal volume element inthe neighborhood of the origin.

We now follow the infinitesimal transformation δR by a finite trans-formation R(n, ϕ), i.e., we form the product RδR. This generates avolume element in the neighborhood of R and the product RδR can beviewed as transformation of the volume near the origin to that near R.The Jacobian of this transformation is the relative change of volumenear the origin to that near R or, equivalently, is the relative changeof the density of operations near the origin to that near R. Accordingto the discussion in the introduction, this is the information requiredfrom the density function for SO(3).

We have already seen that equivalence classes of SO(3) are com-prised of all rotations with the same rotation angle, regardless of thedirection of the rotation axis. Thus, the density function is expectedto depend only on ϕ. Referring to Fig. 8.1, this means that the densityof elements depends only on the “radial” distance from the origin, noton the direction, so we can choose R in accordance with this at ourconvenience. Therefore, in constructing the matrix RδR, we will usefor R a matrix of the form in (8.17). Thus,

RδR =

1 0 0

0 cosϕ − sinϕ

0 sinϕ cosϕ

1 −ϕ3 ϕ2

ϕ3 1 −ϕ1

−ϕ2 ϕ1 1

=

1 −ϕ3 ϕ2

ϕ3 cosϕ+ ϕ2 sinϕ cosϕ− ϕ1 sinϕ −ϕ1 cosϕ− sinϕ

ϕ3 sinϕ− ϕ2 cosϕ sinϕ− ϕ1 cosϕ −ϕ1 sinϕ+ cosϕ

.


We can now use (8.15) and (8.16) to determine the axis n′ and angleϕ′ of this product. The angle is determined from

1 + 2 cosϕ′ = 1 + 2 cosϕ− 2ϕ1 sinϕ ,

which, upon cancelling common factors, becomes

cosϕ′ = cosϕ− ϕ1 sinϕ .

Using the standard trigonometric formula for the cosine of a sum, wefind, to first order in ϕ1, that

ϕ′ = ϕ+ ϕ1 .

The unnormalized components of n′ are determined from (8.15) to be

n′1 = −2ϕ1 cosϕ− 2 sinϕ ,

n′2 = ϕ3 sinϕ− ϕ2(1 + cosϕ) ,

n′3 = −ϕ2 sinϕ− ϕ3(1 + cosϕ) .

To normalize the axis, we first determine the length based on thesecomponents. To first order in the ϕi, we find

|n′| = 2ϕ1 cosϕ+ 2 sinϕ .

Thus, the components of the normalized rotation axis of RδR are

n′1 = 1 ,

n′2 = −12ϕ3 + 1

2ϕ2

1 + cosϕ

sinϕ,

n′3 = 12ϕ2 + 1

2ϕ3

1 + cosϕ

sinϕ.

Expressed in terms of the parametrization in (8.18), RδR is given by

(n′1ϕ′, n′2ϕ

′, n′3ϕ′) =

ϕ+ ϕ1,12ϕ

(− ϕ3 + ϕ2

1 + cosϕ

sinϕ

), 1

2ϕ

(ϕ2 + ϕ3

1 + cosϕ

sinϕ

).


This defines the transformation from the neighborhood of the origin tothe neighborhood near RδR. The Jacobian J of this transformation,obtained from

J = det

∣∣∣∣∣∂(n′iϕ′)

∂ϕj

∣∣∣∣∣ , (8.20)

determines how the density of elements of SO(3) near the origin istransformed to the density of points near R. By taking the derivativesin (8.20) to obtain the entries (i, j) in the Jacobian matrix, we obtain

J =

∣∣∣∣∣∣∣∣∣∣∣∣∣

1 0 0

0 ϕ1 + cosϕ

2 sinϕ−1

2ϕ

0 12ϕ ϕ

1 + cosϕ

2 sinϕ

∣∣∣∣∣∣∣∣∣∣∣∣∣=

ϕ2

2(1− cosϕ).

Notice that

limϕ→0

J = 1 ,

so that the normalization of the volume in parameter space is suchthat the volume near the unit element is unity. Hence, the densityof elements in parameter space is the reciprocal of J , so the densityfunction g in (8.3) is

g(ϕ) =2

ϕ2(1− cosϕ) . (8.21)

8.4.2 Integrals in Parameter Space

The density function in (8.21) now permits us to carry out integral overthe group. Thus, for a function F (ϕ,Ω), where Ω denotes the angularvariables in the parametrization in (8.19), we have∫∫

g(ϕ)F (ϕ,Ω)ϕ2 dϕ dΩ ,

where we have used the usual volume element for spherical polar coor-dinates. Using the density function in (8.21), this integral becomes∫∫

2(1− cosϕ)F (ϕ,Ω)dϕ dΩ .


We can now establish the orthogonality relation for characters. Ifwe denote the characters for two irreducible representations of SO(3)by χµ(ϕ) and χν(ϕ), then we have∫∫

2(1− cosϕ)χµ(ϕ)χν(ϕ)dϕ dΩ = δµ,ν

∫∫2(1− cosϕ)dϕ dΩ .

The integral on the right-hand side of this equation, which has thevalue 8π2, corresponds to the volume of SO(3) in parameter space. Theintegral over the angular variables on the left-hand side yields 2× 4π,so cancelling common factors, we obtain∫ π

0(1− cosϕ)χµ(ϕ)χν(ϕ) dϕ = πδµ,ν . (8.22)

This is the orthogonality relation for characters of SO(3).

8.5 Irreducible Representations and Char-

acters for SO(3)

For SO(2), we were able to determine the characters of the irreduciblerepresentations directly, i.e., without having to determine the basisfunctions of these representations. The structure of SO(3), however,does not allow for such a simple procedure, so we must determine thebasis functions from the outset.

8.5.1 Spherical Harmonics

We proceed as in Section 8.2 by determining the homogeneous polyno-mial solutions of Laplace’s equation, now in three dimensions:

∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= 0 .

We seek solutions of the form

u(x, y, z) =∑a,b

cab(x+ iy)a(x− iy)bz`−a−b ,


which are homogeneous polynomials of degree `. In spherical polarcoordinates,

x = r cosφ sin θ, y = r sinφ sin θ, z = r cos θ ,

where 0 ≤ φ < 2π and 0 ≤ θ ≤ π, these polynomial solutions transformto

u(r, θ, φ) =∑a,b

cabr` sina+b θ cos`−a−b θ ei(a−b)φ . (8.23)

Alternatively, Laplace’s equation in spherical polar coordinates is

1

r2

∂

∂r

(r2∂u

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂u

∂θ

)+

1

r2 sin2 θ

∂2u

∂φ2= 0 .

When the method of separation of variables is used to find solutionsof this equation of the form u(r, θ, φ) = R(r)Θ(θ)Φ(φ), the stipulationthat the solution be single-valued with respect changes in φ by 2π,

u(r, θ, φ+ 2π) = u(r, θ, φ) ,

requires that

Φ(φ) ∝ eimφ ,

where m is an integer. Comparing this expression with the correspond-ing factor in (8.23), we see that a − b = m. Since the ranges of botha and b are between 0 and `, we see that there are 2` + 1 values ofm consistent with homogeneous polynomial solutions of degree `. Thecorresponding values of m are −` ≤ m ≤ `. The 2` + 1 independenthomogeneous polynomials of degree ` are called the spherical harmonicsand denoted by Y`m(θ, φ). Their functional form is

Y`m(θ, φ) ∝ P `m(θ) eimφ , (8.24)

where P `m(θ) is a Legendre function . In the following discussion, we will

utilize only the exponential factor in the spherical harmonics.


8.5.2 Characters of Irreducible Representations

The Y`m(θ, φ) form a (2` + 1)-dimensional representation of SO(3).Thus, for a general rotation R, we have

RY`m(θ, φ) =∑

m′=−`Y`m′(θ, φ) .Γ`m′m(R)

To determine the character of this representation, it is convenient toagain invoke the fact that the classes of SO(3) are determined only bythe rotation angle, not by the direction of the rotation axis. Thus, wecan choose a rotation axis at our convenience and we therefore focus onrotations through an angle ϕ about the z-axis. In this case, the formof (8.24) allows us to write

Rz(ϕ)Y`m(θ, φ) = Y`m(θ, φ− ϕ) = e−imϕY`m(θ, φ) .

Thus, the corresponding transformation matrix is given by

Γ`[Rz(ϕ)] =

e−i`ϕ 0 · · · 0

0 e−i(`−1)ϕ · · · 0...

.... . .

...

0 0 · · · ei`ϕ

. (8.25)

The character χ(`)(ϕ) of this class is obtained by taking the trace ofthis matrix:

χ(`)(ϕ) = e−i`ϕ + e−i(`−1)ϕ + · · ·+ ei`ϕ

= e−i`ϕ(

1 + eiϕ + e2iϕ + · · ·+ e2`iϕ)

= e−i`ϕ1− e−(2`+1)iϕ

1− eiϕ

=e(`+1/2)iϕ − e−(`+1/2)iϕ

eiϕ/2 − e−iϕ/2

=sin [(`+ 1

2)ϕ]

sin (12ϕ)

.


The orthogonality integral for these characters takes the form

∫ π

0(1− cosϕ)

sin [(`+ 12)ϕ] sin [(`′ + 1

2)ϕ]

sin2 (12ϕ)

dϕ .

Using the trigonometric identity

2 sin2 (12ϕ) = 1− cosϕ

enables us to write the orthogonality integral as∫ π

0sin [(`+ 1

2)ϕ] sin [(`′ + 1

2)ϕ] dϕ = 1

2πδ`,`′ ,

where the right-hand side of this equation follows either from (8.22) orfrom the orthogonality of the sine functions over (0, π).

It is possible to show directly, using Schur’s first lemma, that thespherical harmonics form a basis for (2` + 1)-dimensional irreduciblerepresentations of SO(3). However, this requires invoking properties ofthe Legendre functions in (8.24). If we confine ourselves to the matricesin (8.25) then we can show that a matrix that commutes with all suchrotation matrices must reduce to a diagonal matrix. If we then considerrotations about any other direction, which requires some knowledge ofthe Legendre functions, we can then show that this constant matrixmust, in fact, be a constant multiple of the unit matrix. Hence, accord-ing to Schur’s first lemma, these representations are irreducible. We cannow construct the character table for SO(3) with the basis functionswhich generate the irreducible representations:

SO(3) E R(ϕ)

Γ` : Y`m(θ, φ) 1sin [(`+ 1

2)ϕ]

sin (12ϕ)

8.6 Summary

In this chapter, we have shown how the orthogonality relations devel-oped for finite groups must be adapted for continuous groups, using


SO(2) and SO(3) as examples. For SO(2), which is a one-parameterAbelian group, this proved to be a straightforward matter. However,the corresponding calculations for SO(3) required us to determine ex-plicitly the density function to produce the appropriate form of theorthogonality relations. We found that the there are an infinite se-quence of irreducible representations of dimensionality 2` + 1, where` ≥ 0. Because of the connection between SO(3) and angular mo-mentum, the structure of these irreducible representations has severalphysical consequences:

• For systems that possess spherical symmetry, the energy eigen-states have degeneracies of 2`+1. The fact that there is a greaterdegeneracy for the hydrogen atom is due to a “hidden” SO(4)symmetry.4

• The formation and decomposition of direct products of the irre-ducible representations of SO(3) forms the basis of angular mo-mentum coupling rules (Problem 6, Problem Sets 10) and theclassification of atomic spectra.5

• When atoms are placed within crystals, the original sphericalsymmetry is lowered to the symmetry of the crystal. This causeslevels which were degenerate in the spherically-symmetric envi-ronment to split. Such “crystal-field” effects are important formany aspects for electrons in crystalline solids.6

4H.F. Jones, Groups, Representations and Physics (Institute of Physics, Bristol,1998), pp. 124–127.

5E.P. Wigner, Group Theory and its Application to the Quantum Mechanics ofAtomic Spectra (Academic, New York, 1959), pp. 177–194.

6M. Tinkham, Group Theory and Quantum Mechanics (McGraw–Hill, New York,1964), pp. 65–80.

Chapter 9

Unitary Groups and SU(N)∗

The irreducible representations of SO(3) are appropriate for describingthe degeneracies of states of quantum mechanical systems which haverotational symmetry in three dimensions. But there are many systemsfor which operations on classical coordinates must be supplementedby operations on “internal” degrees of freedom which have no classicalanalogue. For example, the Stern–Gerlach experiment showed thatelectrons are endowed with an internal degree of freedom called “spin”which has the properties of an angular momentum. The two spin statesare therefore inconsistent with the dimensionalities of the irreduciblerepresentations of SO(3), so another group—SU(2)—must be used todescribe these states. Since, as we will show in Section 9.2, SU(2) islocally isomorphic to SO(3), we can define a total spin S in an abstractthree-dimensional space, analogous to the total angular momentum inreal space. In particle physics, unitary symmetry was used to describethe approximate symmetry (called isospin) of neutrons and protonsand, more recently, to describe particle spectra within the frameworkof the quark model.

In this chapter, we introduce unitary groups and their irreduciblerepresentations in a similar manner to which we developed SO(3). Webegin by defining unitarity in terms of the invariance of an appropriatequantity and proceed to discuss the construction of irreducible repre-sentations of these groups in N dimensions. Higher-dimensional irre-ducible representations will be obtained with the aid of Young tableaux,which is a diagrammatic technique for determining the dimensionali-

147

148 Unitary Groups and SU(N)

ties and the basis functions of irreducible representations derived fromdirect products.

9.1 SU(2)

As with orthogonal matrices, the unitary groups can be defined in termsof quantities which are left invariant. Consider a general complex trans-formation in two dimensions, x′ = Ax which, in matrix form, reads:(

x′

y′

)=

(a b

c d

)(x

y

)

where a, b, c, and d are complex, so there are eight free parameters.The determinant of this matrix is nonzero to permit the constructionof inverses.

9.1.1 Unitary Transformations

Suppose we require the quantity |x|2 + |y|2 to be an invariant of such atransformation. Then,

|x′|2 + |y′|2 = |ax+ by|2 + |cx+ dy|2

= (ax+ by)(a∗x∗ + b∗y∗) + (cx+ dy)(c∗x∗ + d∗y∗)

= (|a|2 + |c|2)|x|2 + (ab∗ + cd∗)xy∗ + (a∗b+ c∗d)x∗y

+ (|b|2 + |d|2)|y|2

= |x|2 + |y|2

Since x and y are independent variables, this invariance necessitatessetting the following conditions on the matrix elements:

|a|2 + |c|2 = 1, |b|2 + |d|2 = 1, ab∗ + cd∗ = 0

These four conditions (the last equation provides two conditions be-cause it involves complex quantities) means that the original eight free

Unitary Groups and SU(N) 149

parameters are reduced to four. These conditions are the same as thoseobtained by requiring the A†A = 1, so the determinant of the result-ing matrix has modulus unity. These transformations are analogousto orthogonal transformations of real coordinates and, indeed, orthog-onal transformations are also unitary. The group comprised of unitarymatrices is denoted by U(2) and by U(N) for the N -dimensional case.

9.1.2 Special Unitary Transformations

If, in addition to the conditions above, we require that the determinantof the transformation is unity, the transformation matrix must have theform (

x′

y′

)=

(a b

−b∗ a∗

)(x

y

), |a|2 + |b|2 = 1 (9.1)

There are now three free parameters and the group of these matrices isdenoted by SU(2) where, as in our discussion of orthogonal groups, the‘S’ signifies ‘special’ because of the requirement of a unit determinant.

9.2 Relation between SU(2) and SO(3)

9.2.1 Pauli Matrices

If the matrix elements of the general unitary matrix in (9.1) are ex-pressed in terms of their real and imaginary parts, we can decomposethis matrix into the components of a “basis.” Thus, with a = ar + iaiand b = br + ibi, we have

U =

(ar + iai br + ibi

−br + ibi ar − iai

)

= ar

(1 0

0 1

)+ iai

(1 0

0 −1

)+ br

(0 1

−1 0

)︸︷︷︸ibr

(0 −ii 0

)+ibi

(0 1

1 0

)


Thus, any 2 × 2 unitary matrix can be represented as a linear combi-nation of the unit matrix and the matrices

σx =

(1 0

0 −1

), σy =

(0 −ii 0

), σz =

(0 1

1 0

)

These three (Hermitian) matrices are known as the Pauli matrices . Theysatisfy the following multiplication rules:

σ2i = I (i = x, y, z)

σiσj = −σjσi = iεijkσk (i, j, k = x, y, z)(9.2)

where I is the 2 × 2 unit matrix. These multiplication rules can beused to obtain a concise expression for the product of two matriceswritten as a ·σ and b ·σ, where a = (ax, ay, az), b = (bx, by, bz), andσ = (σx, σy, σz):

(a·σ)(b·σ) = (a·b)I + i(a× b)·σ (9.3)

9.2.2 Infinitesimal Generators

Moreover, if we define matrices Xi = −12iσi, for i = 1, 2, 3, then the sec-

ond of the multiplication rules in (9.2) yield the following commutationrelations:

[Xi, Xj] = εijkXk

These are identical to commutators of the infinitesimal generators ofSO(3) in (7.13). Thus, locally at least, there is an isomorphism betweenSO(3) and SU(2). Motivated by the discussion in Section 7.3, considerthe matrix

U = exp (− 12iϕn·σ)

where ϕn is the axis-angle representation of a rotation (Section (8.3).Since the exponential of a matrix is defined by its Taylor series expan-sion, we have

U =∞∑k=0

(−i)nn!

(12ϕ)n(n·σ)n


From Equation (9.3), (n·σ)2 = I, so

U = I∞∑k=0

(−1)n

(2n)!(1

2ϕ)2n − i(n·σ)

∞∑k=0

(−1)n

(2n+ 1)!(1

2ϕ)2n+1

= cos (12ϕ)I − i(n·σ) sin (1

2ϕ)

=

cos (12ϕ)− inz sin (1

2ϕ) −(ny + inx) sin (1

2ϕ)

(ny − inx) sin (12ϕ) cos (1

2ϕ) + inz sin (1

2ϕ)

(9.4)

This matrix is manifestly of the unitary form in (9.1) with unit deter-minant. The Pauli matrices are, therefore, the infinitesimal generatorsof SU(2) and form a representation of its Lie algebra.

9.2.3 Local and Global Mappings between SU(2)and SO(3)

The matrix in (9.4) is parametrized in the same way as rotations inSO(3), namely, in terms of a rotation angle ϕ and a rotation axis n.But, although the mapping between SU(2) and SO(3) is locally an iso-morphism, since their algebras are isomorphic, globally this relationshipis a homomorphism. The reason for this stems from the periodicity ofthe two groups: SO(3) has a periodicity of 2π, while SU(2) has a peri-odicity of 4π. In particular U(0,n) = I, but U(2π,n) = −I, so bothof these elements are associated with the identity of SO(3). Moreover,these elements form an invariant subgroup of SU(2) (Section 2.4) whichis isomorphic to the group Z2 = 1,−1 under ordinary multiplication.In general, using the trigonometric identities,

cos[

12(ϕ+ 2π)

]= − cos (1

2ϕ)

sin[

12(ϕ+ 2π)

]= − sin (1

2ϕ)

we find that

U(ϕ+ 2π,n) = −U(ϕ,n)


Thus, if we form the cosets of the subgroup U(0,n), U(2π,n), weobtain

U(0,n), U(2π,n)U(ϕ,n) =

U(ϕ,n), U(ϕ+ 2π,n)

Thus, the factor group SU(2)/Z2 is isomorphic to SO(3):

SU(2)/Z2 = SO(3)

In fact, this double-valuedness extends to characters as well. Takingthe trace of the matrix in 9.4) yields

2 cos (12ϕ)

If we compare this expression with that for χ(`)(ϕ) for SO(3) with ` = 12,

we find

χ(1/2)(ϕ) =sinϕ

sin (12ϕ)

= 2 cos (12ϕ)

so the two-dimensional (irreducible) representation of SU(2) generatedby the Pauli matrices corresponds to a representation of SO(3) with ahalf-integer index. The integer values of ` can be traced to the require-ment of single-valuedness of the spherical harmonics, so the double-valued correspondence between SU(2) and SO(3) results in this half-integer index.

9.3 Irreducible Representations of SU(2)

When we constructed the irreducible representations of SO(2) andSO(3), we used as basis functions obtained from the coordinates x, yand x, y, z, respectively, and to obtain higher-order irreducible rep-resentations from direct products. The basic procedure is much thesame for unitary groups, except that we can no longer rely on basisstates expressed in terms of coordinates. In this section, we carry outthe required calculations for SU(2) and then generalize the method forSU(N) in the next section.


9.3.1 Basis States

By associating the Pauli matrices with angular momentum operatorsthrough Ji = 1

2hσi, we choose as our basis states the vectors

u1 =

(1

0

), u2 =

(0

1

)

There are several physical interpretations of these states. For exam-ple, they can represent the two possible energy eigenstates of a spin-1

2

particle, such an electron or proton. Another possibility is that u1 andu2 represent the isospin eigenstates of an isospin-1

2particle, such as a

proton or a neutron. The fact that the proton and neutron are not ex-actly degenerate means that isospin symmetry is only an approximatesymmetry. A third interpretation of u1 and u2 is as “up” and “down”quarks which make up nucleons. We will discuss further refinements ofthe quark model in the context of SU(N) later in this chapter.

9.3.2 Multiparticle Systems and Direct Products

When using basis states of SU(2) to construct multiparticle statesthrough direct products, we must respect the indistinguishability ofthe particles. Thus, measurable properties of a quantum system can-not depend on the labelling of the particles, though wavefunctions, ofcourse, need not obey this invariance. Consider a two-particle system,with particle ‘1’ in state i and particle ‘2’ in state j. The correspondingwavefunction is ψi,j(1, 2). We require that

|ψi,j(1, 2)|2 = |ψi,j(2, 1)|2

which implies that

ψi,j(2, 1) = eiθψi,j(1, 2)

for some phase angle θ. Since a two-fold exchange restores the originallabelling,

ψi,j(1, 2) = eiθψi,j(2, 1) = e2iθψi,j(1, 2)


we must have that e2iθ = 1, or that θ = 0 or θ = π. In the first case,the wavefunction is symmetric under the interchange of particles,

ψi,j(2, 1) = ψi,j(1, 2)

while in the latter case, the wavefunction is antisymmetric under theinterchange of particles,

ψi,j(2, 1) = −ψi,j(1, 2)

Consider now a two-particle system each of which occupy one of thestates of SU(2). The basis of these two-particle states is comprised ofu1u1, u1u2, u2u1, u2u2, where we have adopted the convention that theorder of the states corresponds to the order of the particle coordinates,e.g., u1u1 ≡ u1(1)u1(2). But not all of these states are symmetric orantisymmetric under the interchange of particles. Hence, we constructthe new basis

u1u1, u1u2 + u2u1, u2u2︸︷︷︸symmetric

, u1u2 − u2u1︸︷︷︸antisymmetric

(9.5)

We can compare this result with that obtained from the two-fold directproduct representation of SU(2):

χ(1/2)(ϕ)χ(1/2)(ϕ) =[2 cos (1

2)]2

=(eiϕ/2 + e−iϕ/2

)2

=(eiϕ/2 + 1 + e−iϕ/2

)+ 1

= χ(1)(ϕ) + χ(0)(ϕ)

we see that the three symmetric wavefunctions for a basis for the ` = 1irreducible representation of SO(3) and the antisymmetric wavefunc-tions transforms as the identical representation (` = 0) of SO(3). If wethink of these as spin-1

2particles, the symmetric state corresponds to a

total spin S = 1, while the antisymmetric state corresponds to S = 0.We could proceed in this way to construct states for larger numbers ofparticles, but in the next section we introduce a technique which is farmore efficient and which can be applied to other SU(N) groups, wherethe direct method described in this section becomes cumbersome.


9.3.3 Young Tableaux

Determining the dimensionalities of the irreducible representations ofdirect products of basis states of SU(N) is a problem which is encoun-tered in several applications in physics and group theory. Young tableauxprovide a diagrammatic method for carrying this out in a straightfor-ward manner. In this section, we repeat the calculation in the precedingsection to illustrate the method, and in the next section, we describethe general procedure for applying Young tableaux to SU(N).

The basic unit of a Young tableau is a ‘box’, shown below

which denotes a basis state. If there is no entry in the box, then thistableau represents any state. An entry, signified by a number denotesone of the basis states in some reference order. Thus, for SU(2), wehave

u1 = 1 u2 = 2

The utility of Young tableaux centers around the construction ofdirect products. For the two-fold direct products of SU(2) in (9.5),there are two types of states, symmetric and antisymmetric. The Youngtableau for a generic two-particle symmetric state is

and the two-particle antisymmetric state is

The Young tableaux for three symmetric states in (9.5) are


1 1 1 2 2 2

and that for the antisymmetric state is

12

In the framework of Young tableaux, the two-fold direct product iswritten as

× = +

The three-fold direct product illustrates the conventions used in theconstruction of Young tableaux and their labelling. The generic tableauxare

× × = + +

The rules for constructing the “standard” arrangement of Young tableauxare as follows

• The rows are constructed from left to right

• The columns are constructed from top to bottom

• No row is longer than any row above it

• No column is longer than any column to the left of it

Thus, with these conventions, a typical tableau is shown below:


The states for the three-fold direct product are as follows. Thereare four symmetric states:

1 1 1 1 1 2 1 2 2 2 2 2

which correspond to a four-dimensional irreducible representation, andtwo “mixed” states:

1 12

1 22

which correspond to a two-dimensional irreducible representation. Thereare no totally antisymmetric three-particle states because we have onlytwo distinct basis states. Thus, the rules for entering states into Youngtableaux are:

• The numbers within rows are nondecreasing from left to right.

• The numbers within columns are increasing from top to bottom.

The two sets of rules for constructing Young tableaux of genericstates and identifying particular states enables the calculation of thedimensionalities in a straightforward manner, often by identifying ap-propriate combinatorial rules.

9.4 Young Tableaux for SU(N)

The groups SU(N) have acquired an importance in particle physicsbecause of the quark model. This necessitates calculating direct prod-ucts of basis states to determine the characteristics of particle spectra.


This, in turn, requires that we adapt the methodology of the Youngtableaux developed in the preceding section to SU(N), which turns outto be straightforward given the rules stated in the preceding section.There is no change to the construction of the generic tableaux; the onlychanges are in the labelling of the tableaux. Consider, for example thecase of a two-fold direct product of SU(3). There are six symmetricstates

1 1 1 2 1 3 2 2 2 3 3 3

and three antisymmetric states

12

13

23

As is evident from these constructions, the number of states asso-ciated with a tableau of a particular topology increases sharply withthe number of basis states. The rules in the preceding section allowthe number of such symmetric and antisymmetric states to be calcu-lated for SU(N). There 1

2N(N + 1) symmetric states and 1

2N(N − 1)

antisymmetric states.The only other modification to our discussion of SU(2) is that for

larger numbers of basis states, tableaux which make no contributionto SU(2), may make a contribution to SU(N). Consider, for example,the antisymmetric three-particle state. This state vanishes for SU(2)because there are only two basis states, but for SU(3), we have

123

In fact, this is a direct consequence of the rule for labelling Youngtableaux, and we see that, for SU(N), any column with more than Nboxes makes no contribution.


9.5 Summary

In this chapter, we have extended our discussion of orthogonal groupsto unitary groups. These groups play an especially important role inquantum mechanics because of their property of conserving probabilitydensity. We have constructed direct products of basis states, which arerequired in a number of applications of these groups. The use of Youngtableaux was shown to be an especially convenient way to determinethe dimensionalities of higher-dimensional irreducible representationsof unitary groups and their basis functions.

Group Theory

Problem Set 1 October 12, 2001

Note: Problems marked with an asterisk are for Rapid Feedback; problems marked with adouble asterisk are optional.

1. Show that the wave equation for the propagation of an impulse at the speed of light c,

1c2∂2u

∂t2=∂2u

∂x2+∂2u

∂y2+∂2u

∂z2,

is covariant under the Lorentz transformation

x′ = γ(x− vt), y′ = y, z′ = z, t′ = γ

(t− v

c2x

),

where γ = (1− v2/c2)−1/2.

2.∗ The Schrodinger equation for a free particle of mass m is

ih∂ϕ

∂t= − h2

2m∂2ϕ

∂x2.

Show that this equation is invariant to the global change of phase of the wavefunction:

ϕ→ ϕ′ = eiαϕ ,

where α is any real number. This is an example of an internal symmetry transformation,since it does not involve the space-time coordinates.

According to Noether’s theorem, this symmetry implies the existence of a conservationlaw. Show that the quantity

∫∞−∞ |ϕ(x, t)|2 dx is independent of time for solutions of the

free-particle Schrodinger equation.

3.∗ Consider the following sets of elements and composition laws. Determine whether theyare groups and, if not, identify which group property is violated.

(a) The rational numbers, excluding zero, under multiplication.

(b) The non-negative integers under addition.

(c) The even integers under addition.

(d) The nth roots of unity, i.e., e2πmi/n, for m = 0, 1, . . . , n− 1, under multiplication.

(e) The set of integers under ordinary subtraction.

4.∗∗ The general form of the Liouville equation is

ddx

[p(x)

dydx

]+[q(x) + λr(x)

]y = 0

where p, q and r are real-valued functions of x with p and r taking only positive values.The quantity λ is called the eigenvalue and the function y, called the eigenfunction, isassumed to be defined over an interval [a, b]. We take the boundary conditions to be

y(a) = y(b) = 0

but the result derived below is also valid for more general boundary conditions. No-tice that the Liouville equations contains the one-dimensional Schrodinger equation as aspecial case.

Let u(x;λ) and v(x;λ) be the fundamental solutions of the Liouville equation, i.e. uand v are two linearly-independent solutions in terms of which all other solutions maybe expressed (for a given value λ). Then there are constants A and B which allow anysolution y to be expressed as a linear combination of this fundamental set:

y(x;λ) = Au(x;λ) +Bv(x;λ)

These constants are determined by requiring y(x;λ) to satisfy the boundary conditions:

y(a;λ) = Au(a;λ) +Bv(a;λ) = 0

y(b;λ) = Au(b;λ) +Bv(b;λ) = 0

Use this to show that the solution y(x;λ) is unique, i.e., that there is one and only onesolution corresponding to an eigenvalue of the Liouville equation.

Group Theory


Note: Problems marked with an asterisk are for Rapid Feedback.

1.∗ Show that, by requiring the existence of an identity in a group G, it is sufficient to requireonly a left identity, ea = a, or only a right identity ae = a, for every element a in G,since these two quantities must be equal.

2.∗ Similarly, show that it is sufficient to require only a left inverse, a−1a = e, or only a rightinverse aa−1 = e, for every element a in G, since these two quantities must also be equal.

3. Show that for any group G, (ab)−1 = b−1a−1.

4.∗ For the elements g1, g2, . . . , gn of a group, determine the inverse of the n-fold productg1g2 · · · gn.

5.∗ Show that a group is Abelian if and only if (ab)−1 = a−1b−1. You need to show thatthis condition is both necessary and sufficient for the group to be Abelian.

6. By explicit construction of multiplication tables, show that there are two distinct struc-tures for groups of order 4. Are either of these groups Abelian?

7.∗ Consider the group of order 3 discussed in Section 2.4. Suppose we regard the rows ofthe multiplication table as individual permutations of the elements e, a, b of this group.We label the permutations πg by the group element corresponding to that row:

πe =

(e a b

e a b

), πa =

(e a b

a b e

), πb =

(e a b

b e a

)

(a) Show that, under the composition law for permutations discussed in Section 2.3,the multiplication table of the 3-element group is preserved by this association, e.g.,πaπb = πe.

(b) Show that for every element g in e, a, b,

πg =

(e a b

g ga gb

)Hence, show that the πg have the same multiplication table as the 3-element group.

(c) Determine the relationship between this group and S3. This is an example of Cayley’stheorem.

(d) To which of the operations on an equilateral triangle in Fig. 2.1 do these groupelements correspond?

Group Theory



1.∗ List all of the subgroups of any group whose order is a prime number.

2.∗ Show that a group whose order is a prime number is necessarily cyclic, i.e., all of theelements can be generated from the powers of any non-unit element.

3. Suppose that, for a group G, |G| = pq, where p and q are both prime. Show that everyproper subgroup of G is cyclic.

4.∗ Let g be an element of a finite group G. Show that g|G| = e.

5. In a quotient group G/H , which set always corresponds to the unit “element”?

6. Show that, for an Abelian group, every element is in a class by itself.

7. Show that every subgroup with index 2 is self-conjugate.

Hint: The conjugating element is either in the subgroup or not. Consider the two casesseparately.

8.∗ Consider the following cyclic group of order 4, G = a, a2, a3, a4 = e (cf. Problem 6,Problem Set 2). Show, by direct multiplication or otherwise, that the subgroup H =e, a2 is self-conjugate and identify the elements in the factor group G/H .

9.∗ Suppose that there is an isomorphism φ from a group G onto a group G′. Show that theidentity e of G is mapped onto the identity e′ of G′: e′ = φ(e).

Hint: Use the fact that e = ee must be preserved by φ and that φ(g) = e′φ(g) for all gin G.

Group Theory



1.∗ Given a set of matrices D(g) that form a representation a groupG, show that the matriceswhich are obtainable by a similarity transformation UD(g)U−1 are also a representationof G.

2.∗ Show that the trace of three matrices A, B, and C satisfies the following relation:

tr(ABC) = tr(CAB) = tr(BCA)

3. Generalize the result in Problem 4 to show that the trace of an n-fold product of matricesis invariant under cyclic permutations of the product.

4.∗ Show that the trace of an arbitrary matrixA is invariant under a similarity transformationUAU−1.

5. Consider the following representation of S3:

e =

(1 0

0 1

), a = 1

2

(1 −

√3

−√

3 −1

), b = 1

2

(1√

3√

3 −1

)

c =

(−1 0

0 1

), d = 1

2

(−1 −

√3

√3 −1

), f = 1

2

(−1

√3

−√

3 −1

)

How can these matrices be permuted to provide an equally faithful representation of S3?Relate your result to the class identified with each element.

6.∗ Consider the planar symmetry operations of an equilateral triangle. Using the matricesin Example 3.2 determined from transformations of the coordinates in Fig. 3.1, constructa three-dimensional representation of S3 in the (x, y, z) coordinate system, where the z-axis emanates from the geometric center of the triangle. Is this representation reducibleor irreducible? If it is reducible determine the irreducible representations which form thedirect sum of this representation.

7. Show that two matrices are simultaneously diagonalizable if and only if they commute.

Hint: Two matrices A and B are simultaneously diagonalizable if the same similaritytransformation brings both matrices into a form where they have only diagonal entries.Proving that simultaneous diagonalizability implies commutativity is straightforward. Toprove the converse, suppose that there is a similarity transformation which brings oneof the matrices into diagonal form. By writing out the matrix elements of the productsand using the fact that A and B commute, show that the same similarity transformationmust also diagonalize the other matrix.

8.∗ What does the result of Problem 7 imply about the dimensionalities of the irreduciblerepresentations of Abelian groups?

9.∗ Verify that the matrices

e =

(1 0

0 1

), a = 1

2

(−1

√3

√3 1

)

form a representation for the two-element group e, a. Is this representation reducibleor irreducible? If it is reducible determine the one-dimensional representations whichform the direct sum of this representation.

10.∗ Prove the relations in Eqns (3.9) and (3.11).

Group Theory

Problem Set 5 November 6, 2001


1. In proving Theorem 3.2, we established the relation BiB†i = I. Using the definitions in

that proof, show that this result implies that B†iBi = I as well.

Hint: Show that BiB†i = I implies that AiDA

†i = D.

2.∗ Consider the three-element group G = e, a, b (Sec. 2.4).

(a) Show that this group is Abelian and cyclic (cf. Problem 2, Problem Set 3).

(b) Consider a one-dimensional representation based on choosing a = z, where z is acomplex number. Show that for this to produce a representation of G, we mustrequire that z3 = 1.

(c) Use the result of (b) to obtain three representations of G. Given what you knowabout the irreducible representations of Abelian groups (Problem 8, Problem Set4), are there any other irreducible representations of G?

3.∗ Generalize the result of Problem 2 to any cylic group of order n.

4.∗ Use Schur’s First Lemma to prove that all the irreducible representations of an Abeliangroup are one-dimensional.

5.∗ Consider the following matrices:

e =

(1 0

0 1

), a = 1

2

(−1 −

√3

−√

3 1

), b = 1

2

(−1 −

√3

−√

3 1

)

c = 12

(−1 −

√3

−√

3 1

), d =

(1 0

0 1

), f =

(1 0

0 1

)

Verify that these matrices form a representation of S3. Use Schur’s first Lemma todetermine if this representation reducible or irreducible. If reducible, determine theirreducible representations that are obtained from the diagonal form of these matrices.

Group Theory



1.∗ Verify the Great Orthogonality Theorem for the following irreducible representation ofS3:

e =

(1 0

0 1

), a = 1

2

(1 −

√3

−√

3 −1

), b = 1

2

(1√

3√

3 −1

)

c =

(−1 0

0 1

), d = 1

2

(−1

√3

−√

3 −1

), f = 1

2

(−1 −

√3

√3 −1

)

2.∗ Does the following representation of the three-element group e, a, b:

e =

(1 0

0 1

), a = 1

2

(−1

√3

−√

3 −1

), b = 1

2

(−1 −

√3

√3 −1

)

satisfy the Great Orthogonality Theorem? Explain your answer.

3.∗ Specialize the Great Orthogonality Theorem to Abelian groups. When viewed as thecomponents of a vector in a |G|-dimensional space, what does the Great Orthogonal-ity Theorem state about the relationship between different irreducible representations?What bound does this place on the number of irreducible representations of an Abeliangroup?

4.∗ Consider the irreducible representations of the three-element calculated in Problem 2 ofProblem Set 5.

(a) Verify that the Great Orthogonality Theorem, in the reduced form obtained inProblem 3, is satisfied for these representations.

(b) In view of the discussion in Sec. 4.4, would you expect to find any other irreduciblerepresentations of this group?

(c) Would you expect your answer in (b) to apply to cyclic groups of any order?

5.∗ Consider any Abelian group. By using the notion of the order of an element (Sec. 2.4),determine the magnitude of every element in a representation. Is this consistent with theGreat Orthogonality Theorem?

Group Theory



1.∗ Identify the 12 symmetry operations of a regular hexagon.

2. Show that elements in the same class of a group must have the same order.

3.∗ Identify the 6 classes of this group.

Hint: You do not need to compute the conjugacy classes explicitly. Refer to the discus-sion for the group S3 in Example 2.9 and use the fact that elements in the same classhave the same order.

4.∗ How many irreducible representations are there and what are their dimensions?

5.∗ Construct the character table of this group by following the procedure outlined below:

(a) Enter the characters for the identical and “parity” representations. As in the caseof S3, the characters for the parity representation are either +1 or −1, dependingon whether or not the the operation preserves the “handedness” of the coordinatesystem.

(b) Enter the characters for the “coordinate” representation obtained from the actionon (x, y) for each group operation. Note that the character is the same for elementsin the same class.

(c) Use the products C3C23 = E and C3

3 = E to identify the characters for all one-dimensional irreducible representations for the appropriate classes. The meaningof the notation Cmn for rotations is discussed in Section 5.4.

(d) Use the result of (c) and the products C6C3 = C2 to deduce that the charactersfor the class of C6 and those for the class of C2 are the same. Then, use theorthogonality of the columns of the character table to compute these characters.

(e) Use the appropriate orthogonality relations for characters to compute the remainingentries of the character table.

Group Theory



1. Show that if two matrices A and B are orthogonal, then their direct product A⊗B isalso an orthogonal matrix.

2. Show that the trace of the direct product of two matrices A and B is the product ofthe traces of A and B:

tr(A⊗B) = tr(A) tr(B)

3.∗ Show that the direct product of groupsGa andGb with elementsGa = e, a2, . . . , a|Ga|and Gb = e, b2, . . . , b|Gb|, such that aibj = bjai for all i and j, is a group. What isthe order of this group?

4.∗ Use the Great Orthogonality Theorem to show that the direct product of irreduciblerepresentations of two groups is an irreducible representation of the direct product ofthose groups.

5.∗ For an n-fold degenerate set of eigenfunctions ϕi, i, 1, 2, . . . , n, we showed show thatthe matrices Γ(Rα) generated by the group of the Hamiltonian,

Rαϕi =n∑j=1

ϕjΓji(Rα)

form a representation of that group. Show that if the ϕj are chosen to be an orthonormalset of functions, then this representation is unitary.

6.∗ The set of distinct functions obtained from a given function ϕi by operations in thegroup of the Hamiltonian, ϕj = Rαϕi, are called partners . Use the Great OrthogonalityTheorem to show that two functions which belong to different irreducible representa-tions or are different partners in the same unitary representation are orthogonal.

7. Consider a particle of mass m confined to a square in two dimensions whose verticesare located at (1, 1), (1,−1), (−1,−1), and (−1, 1). The potential is taken to be zerowithin the square and infinite at the edges of the square. The eigenfunctions ϕ are ofthe form

ϕp,q(x, y) ∝

cos(kpx)

sin(kpx)

cos(kqy)

sin(kqy)

where kp = 12pπ, kq = 1

2qπ, and p and q are positive integers. The notation abovemeans that cos(kpx) is taken if p is odd, sin(kpx) is taken if p is even, and similarly forthe other factor. The corresponding eigenvalues are

Ep,q =h2π2

8m(p2 + q2)

(a) Determine the eight planar symmetry operations of a square. These operations formthe group of the Hamiltonian for this problem. Assemble the symmetry operationsinto equivalence classes.

(b) Determine the number of irreducible representations and their dimensions for thisgroup. Do these dimensions appear to be broadly consistent with the degeneraciesof the energy eigenvalues?

(c) Determine the action of each group operation on (x, y).

Hint: This can be done by inspection.

(d) Determine the characters corresponding to the identical, parity, and coordinaterepresentations. Using appropriate orthogonality relations, complete the charactertable for this group.

(e) For which irreducible representations do the eigenfunctions ϕ1,1(x, y) and ϕ2,2(x, y)form bases?

(f) For which irreducible transformation do the eigenfunctions ϕ1,2(x, y) and ϕ2,1(x, y)form a basis?

(g) What is the degeneracy corresponding to (p = 6, q = 7) and (p = 2, q = 9)? Is thisa normal or accidental degeneracy?

(h) Are there eigenfunctions which form a basis for each of the irreducible representa-tions of this group?

8.∗ Consider the regular hexagon in Problem Set 7. Suppose there is a vector perturbation,i.e., a perturbation that transforms as (x, y, z). Determine the selection rule associatedwith an initial state that transforms according to the “parity” representation.

Hint: The reasoning for determining the irreducible representations associated with(x, y, z) is analogous to that used in Section 6.6.2 for the equilateral triangle.

Group Theory

Problem Set 9 December 4, 2001


1.∗ Consider the group O(n), the elements of which preserve the Euclidean length in ndimensions:

x′21 + x′22 + · · ·+ x′2n = x21 + x2

2 + · · ·+ x2n .

Show that these transformations have 12n(n− 1) free parameters.

2. The condition that the Euclidean length is preserved in two dimensions, x′2 + y′2 =x2 + y2, was shown in lectures to require that

a211 + a2

21 = 1, a11a12 + a21a22 = 0, a212 + a2

22 = 1 .

Show that these requirements imply that

(a11a22 − a12a21)2 = 1 .

3. Rotations in two dimensions can be parametrized by

R(ϕ) =

(cosϕ − sinϕ

sinϕ cosϕ

).

Show thatR(ϕ1 + ϕ2) = R(ϕ1)R(ϕ2)

and, hence, deduce that this group is Abelian.

4.∗ We showed in lectures that a rotation R(ϕ) by an angle ϕ in two dimensions can bewritten as

R(ϕ) = eϕX ,

where

X =

(0 −1

1 0

).

Verify thateϕX = I cosϕ+X sinϕ ,

where I is the two-dimensional unit matrix. This shows that eϕX is the rotation matrixin two dimensions.

5.∗ Consider the two-parameter group

x′ = ax+ b

Determine the infinitesimal operators of this group.

6.∗ Consider the group C∞v which contains, in addition to all two-dimensional rotations,a reflection plane, denoted by σv in, say, the x-z plane. Is this group Abelian? Whatare the equivalence classes of this group?

Hint: Denoting reflection in the x–z plane by S, show that SR(ϕ)S−1 = R(−ϕ).

7. By extending the procedure used in lectures for SO(3), show that the infinitesimalgenerators of SO(4), the group of proper rotations in four dimensions which leave thequantity x2 + y2 + z2 + w2 invariant, are

A1 = z∂

∂y− y ∂

∂z, A2 = x

∂

∂z− z ∂

∂x, A3 = y

∂

∂x− x ∂

∂y

B1 = x∂

∂w− w ∂

∂x, B2 = y

∂

∂w− w ∂

∂y, B3 = z

∂

∂w− w ∂

∂z

8. Show that the commutators of the generators obtained in Problem 7 are

[Ai, Aj ] = εijkAk, [Bi, Bj ] = εijkAk, [Ai, Bj ] = εijkBk

9. Show that by making the linear transformation of the generators in Problem 7 to

Ji = 12 (Ai +Bi), Ki = 1

2 (Ai −Bi)

the commutators become

[Ji, Jj ] = εijkJk, [Ki,Kj ] = εijkKk, [Ji,Kj ] = 0

This shows that locally SO(4)=SO(3)⊗SO(3).

Group Theory

Problem Set 10 December 11, 2001


1.∗ Prove that a proper orthogonal transformation in an odd-dimensional space alwayspossesses an axis, i.e., a line whose point are left unchanged by the transformation.

2. Prove Euler’s theorem : The general displacement of a rigid body with one fixed point isa rotation about an axis.

3.∗ The functions (x ± iy)m, where m is an integer generate irreducible representationsof SO(2). Suppose we now consider the group O(2), where we now allow improperrotations. Use Schur’s lemma to show that these functions generate irreducible two-dimensional representations of O(2) for m 6= 0, but a reducible representation for m = 0.

Hint: The general improper rotation in two dimensions is(cosϕ sinϕ

sinϕ − cosϕ

)

4. Consider the rotation matrix obtained by rotating an initial set of axes counterclockwiseby φ about the z-axis, then rotated about the new x-axis counterclockwise by θ, andfinally rotated about the new z-axis counterclockwise by ψ. These are the Euler anglesand the corresponding rotation matrix is

cosψ cosφ− cos θ sinφ sinψ cosψ sinφ+ cos θ cosφ sinψ sinψ sin θ

− sinψ cosφ− cos θ sinφ cosψ − sinψ sinφ+ cos θ cosφ cosψ cosψ sin θ

sin θ sinφ − sin θ cosφ cos θ

Verify that the angle of rotation ϕ of this transformation is given by

cos(

12ϕ)

= cos[

12 (φ+ ψ)

]cos(

12θ)

5. Determine the axis of the transformation in Problem 4.

6.∗ Verify that the direct product of two irreducible representations of SO(3) has the fol-lowing decomposition

χ(`1)(ϕ)χ(`2)(ϕ) =`1+`2∑

`=|`1−`2|χ(`)(ϕ)

This is called the Clebsch–Gordan series and provides a group-theoretic statement ofthe addition of angular momenta.

7.∗ Determine the corresponding Clebsch-Gordan series for SO(2).

8.∗ Show that the requirement that xx∗+yy∗ is invariant under the complex transformation(x′

y′

)=

(a b

c d

)(x

y

)

together with the determinant of this transformation being unity means that the trans-formation must have the form(

x′

y′

)=

(a b

−b∗ a∗

)(x

y

)

where aa∗ + bb∗ = 1.

Group Theory

Solutions to Problem Set 1 October 6, 2000

1. To express the the wave equation for the propagation of an im-pulse at the speed of light c,

1

c2

∂2u

∂t2=∂2u

∂x2+∂2u

∂y2+∂2u

∂z2, (1)

under the Lorentz transformation

x′ = γ(x− vt), y′ = y, z′ = z, t′ = γ(t− v

c2x), (2)

where γ = (1− v2/c2)−1/2, we need to obtain expressions for thesecond derivatives in the primed variables. With u′(x′, y′, z′) =u(x, y, t), we have

∂u

∂x=∂u′

∂x′∂x′

∂x+∂u′

∂t′∂t′

∂x= γ

∂u′

∂x′− γ v

c2

∂u′

∂t′, (3)

∂2u

∂x2= γ2∂

2u′

∂x′2+ γ2v

2

c4

∂2u′

∂t′2, (4)

∂2u

∂y2=∂2u′

∂y′2, (5)

∂2u

∂z2=∂2u′

∂z′2, (6)

∂u

∂t=∂u′

∂t′∂t′

∂t+∂u′

∂x′∂x′

∂t= γ

∂u′

∂t′− γv∂u

′

∂x′, (7)

∂2u

∂t2= γ2∂

2u′

∂t′2+ γ2v2∂

2u′

∂x′2. (8)

Substituting these expressions into the wave equation yields

γ2

c2

∂2u′

∂t′2+γ2v2

c2

∂2u′

∂x′2= γ2∂

2u′

∂x′2+ γ2v

2

c4

∂2u′

∂t′2+∂2u′

∂y′2+∂2u′

∂z′2, (9)

2

which, upon rearrangement, becomes

γ2

c2

(1− v2

c2

)∂2u′

∂t′2= γ2

(1− v2

c2

)∂2u′

∂x′2+∂2u′

∂y′2+∂2u′

∂z′2. (10)

Invoking the definition of γ, we obtain

1

c2

∂2u′

∂t′2=∂2u′

∂x′2+∂2u′

∂y′2+∂2u′

∂z′2, (11)

which confirms the covariance of the wave equation under theLorentz transformation.

2. We begin with the Schrodinger equation for a free particle of massm:

ih∂ϕ

∂t= − h2

2m

∂2ϕ

∂x2. (12)

Performing the transformation

ϕ→ ϕ′ = eiαϕ , (13)

where α is any real number, we find

iheiα∂ϕ′

∂t= − h2

2meiα∂2ϕ′

∂x2, (14)

or, upon cancelling the common factor eiα,

ih∂ϕ′

∂t= − h2

2m

∂2ϕ′

∂x2, (15)

which establishes the covariance of the Schrodinger equation un-der this transformation.

We can derive the corresponding quantity multiplying Eq. (12)by the complex conjugate ϕ∗ and subtracting the product of ϕand the complex conjugate of Eq. (12):

ih(ϕ∗∂ϕ

∂t+ ϕ

∂ϕ∗

∂t

)= − h2

2m

(ϕ∗∂2ϕ

∂x2− ϕ∂

2ϕ∗

∂x2

). (16)

3

The left-hand side of this equation can be written as

ih(ϕ∗∂ϕ

∂t+ ϕ

∂ϕ∗

∂t

)= ih

∂

∂t(ϕϕ∗) = ih

∂

∂t|ϕ|2 , (17)

and the right-hand side can be written as

− h2

2m

(ϕ∗∂2ϕ

∂x2− ϕ∂

2ϕ∗

∂x2

)= − h2

2m

∂

∂x

(ϕ∗∂ϕ

∂x− ϕ∂ϕ

∗

∂x

). (18)

We now consider the solution to Eq. (12) corresponding to a wavepacket , whereby by ϕ and its derivatives vanish as x → ±∞.Then, integrating Eq. (16) over the real line, and using Eqns. (17)and (18), we obtain

ih∂

∂t

∫ ∞−∞|ϕ(x, t)|2 dx = − h2

2m

∫ ∞−∞

[∂

∂x

(ϕ∗∂ϕ

∂x− ϕ∂ϕ

∗

∂x

)]dx

= − h2

2m

(ϕ∗∂ϕ

∂x− ϕ∂ϕ

∗

∂x

)∣∣∣∣∞−∞

= 0 . (19)

Thus, the quantity∫∞−∞ |ϕ(x, t)|2 dx is independent of time for so-

lutions of the free-particle Schrodinger equation which correspondto wave packets.

3. (a) The multiplication of two rational numbers, m/n and p/q,where m,n, p and q are integers, yields another rational number,mp/nq, so closure is obeyed. The unit is 1, multiplication is anassociative operation, and the inverse of m/n is n/m, which is arational number. The set excludes zero, so the problem of findingthe inverse of zero does not arise. Hence, the rational numbers,excluding zero, under multiplication form a group.

(b) The sum of two non-negative integers is a non-negative in-teger, thus ensuring closure, addition is associative, the unit iszero, but the inverse under addition of a negative integer n is −n,

4

which is a negative integer and, therefore, excluded from the set.Hence, the non-negative integers under addition do not form agroup.

(c) The sum of two even integers 2m and 2n, where m and n areany two integers, is 2(m+n), which is an even integer, so closureis obeyed. Addition is associative, the unit is zero, which is aneven integer, and the inverse of 2n is −2n, which is also an eveninteger. Hence, the even integers under addition form a group.

(d) The multiplication of two elements amounts to the addition ofthe integers 0, 1, . . . , n−1, modulo n, i.e., the addition of any twoelements and, if the sum lies out side of this range, subtract n tobring it into the range. Thus, the multiplication of two nth rootsof unity is again an nth root of unity, multiplication is associative,and the unit is 1. The inverse of e2πmi/n is therefore

e−2πmi/n = e2πi(n−m)/n (20)

Thus, the nth roots of unity form a group for any value of n.

(e) For the set of integers under ordinary subtraction, the differ-ence between two integers n and m is another integer p, n−m = p,the identity is 0, since, n − 0 = 0, and every integer is its owninverse, since n− n = 0. However, subtraction is not associativebecause

(n−m)− p 6= n− (m− p) = n−m+ p

Hence, the integers under ordinary subtraction do not form agroup.

4. Let u(x;λ) and v(x;λ) be the fundamental solutions of the Li-ouville equation with the boundary conditions y(a) = y(b) = 0.Then there are constants A and B which allow any solution y tobe expressed as a linear combination of this fundamental set:

y(x;λ) = Au(x;λ) +Bv(x;λ) (21)

5

These constants are determined by requiring y(x;λ) to satisfy theboundary conditions given above. Applying these boundary con-ditions to the expression in (21) leads to the following equations:

y(a;λ) = Au(a;λ) +Bv(a;λ) = 0

y(b;λ) = Au(b;λ) +Bv(b;λ) = 0(22)

Equations (22) are two simultaneous equations for the unknownquantities A and B. To determine the conditions which guaranteethat this system of equations has a nontrivial solution, we writethese equations in matrix form:(

u(a;λ) v(a;λ)

u(b;λ) v(b;λ)

)(A

B

)=

(0

0

)(23)

Thus, we see that Equations (22) can be solved for nonzero valuesof A and B only if the determinant of the matrix of coefficients in(23) vanishes. Otherwise, the only solution is A = B = 0, whichyields the trivial solution y = 0. The condition for a nontrivialsolution of (22) is, therefore, given by∣∣∣∣∣u(a;λ) v(a;λ)

u(b;λ) v(b;λ)

∣∣∣∣∣ = u(a;λ)v(b;λ)− u(b;λ)v(a;λ) = 0 (24)

This guarantees that the solution for A and B is unique. Hence,the eigenvalues are non-degenerate.

Group Theory

Solutions to Problem Set 2 October 26, 2001

1. Suppose that e is the right identity of a group G,

ge = g (1)

for all g in G, and that e′ is the left identity,

e′g = g (2)

for all g in G. The choice g = e′ in the first of these equationsand g = e in the second, yields

e′e = e′ (3)

ande′e = e (4)

respectively. These equations imply that

e′ = e (5)

so the left and right identities are equal. Hence, we need specifyonly the left or right identity in a group in the knowledge thatthis is the identity of the group.

2. Suppose that a is the right inverse of any element g in a group G,

ga = e (6)

and a′ is the left inverse of g,

a′g = e (7)

Multiplying the first of these equations from the left by a′ andinvoking the second equation yields

a′ = a′(ga) = (a′g)a = a (8)

2

so the left and right inverses of an element are equal. The sameresult could have been obtained by multiplying the second equa-tion from the right by a and invoking the first equation.

3. To show that for any group G, (ab)−1 = b−1a−1, we begin withthe properties of the inverse. We must have that

(ab)(ab)−1 = e

Left-multiplying both sides of this equation first by a−1 and thenby b−1 yields

(ab)−1 = b−1a−1

4. For elements g1, g2, . . . , gn of a group G, we require the inverse ofthe n-fold product g1g2 · · · gn. We proceed as in Problem 2 usingthe definition of the inverse to write

(g1g2 · · · gn)(g1g2 · · · gn)−1 = e

We now follow the same procedure as in Problem 2 and left-multiply both sides of this equation in turn by g−1

1 , g−12 , . . . , g−1

n

to obtain

(g1g2 · · · gn)−1 = g−1n · · · g−1

2 g−11

5. We must prove two statements here: that for an Abelian groupG, (ab)−1 = a−1b−1, for all a and b in G, and that this equalityimplies that G is Abelian. If G is Abelian, then using the result

3

of Problem 3 and the commutativity of the composition law, wefind

(ab)−1 = b−1a−1 = a−1b−1

Now, suppose that there is a group G (which we must not assumeis Abelian, such that

(ab)−1 = a−1b−1

for all a and b in G. We now right multiply both sides of thisequation first by b and then by a to obtain

(ab)−1ba = e

Then, left-multiplying both sides of this equation by (ab) yields

ba = ab

so G is Abelian. Hence, we have shown that G is an Abeliangroup if and only if, for elements a and b in G, (ab)−1 = a−1b−1.

6. To construct the multiplication table of a four-element groupe, a, b, c we proceed as in Section 2.4 of the course notes. Theproperties of the unit of the group enable us to make the followingentries into the multiplication table:

e a b ce e a b ca ab bc c

We now consider the product aa. This cannot equal a, since thatwould imply that a = e, but it can equal any of the other el-ements, including the identity. However, this leads only to twodistinct choices for the product, since the apparent difference be-tween aa = b and aa = c disappears under the interchange of the

4

e a b ce e a b ca a eb bc c

e a b ce e a b ca a bb bc c

(1)

labelling of b and c. Thus, at this stage, we have two distinctstructures for the multiplication table:

We now determine the remaining entries for these two groups.For the table on the left, we consider the product ab. From theRearrangement Theorem, this cannot equal a or or e, nor can itequal b (since that would imply a = e). Therefore, ab = c, fromwhich it follows that ac = b. According to the RearrangementTheorem, the multiplication table becomes

e a b ce e a b ca a e c bb b cc c b

(2)

For the remaining entries of this table, we observe that b2 = aand b2 = e are equally valid assignments. These leads to twomultiplication tables:

e a b ce e a b ca a e c bb b c e ac c b a e

e a b ce e a b ca a e c bb b c a ec c b e a

(3)

Note that these tables are distinct in that there is no relabellingof the elements which transforms one into the other.

We now return to the other multiplication table on the right in(1). The Rearrangement Theorem requires that the second rowmust be completed as follows:

5

e a b ce e a b ca a b c eb b cc c e

(4)

Again invoking the Rearrangement Theorem, we must have thatthis multiplication table can be completed only as:

e a b ce e a b ca a b c eb b c e ac c e a b

(5)

Notice that all of the multiplication tables in (3) and (5) areAbelian and that the table in (5) is cyclic, i.e., all of the groupelements can be obtained by taking successive products of anynon-unit element.

We now appear to have three distinct multiplication tables forgroups of order 4: the two tables in (3) and the one in (5). How-ever, if we reorder the elements in the second table in (3) toe, b, a, c and reassemble the multiplication table (using the sameproducts), we obtain

e b a ce e b a cc b a c eb a c e ba c e b a

(6)

which, under the relabelling a 7→ b and b 7→ a, is identical to (5).Hence, there are only two distinct structures of groups with fourelements:

6

e a b ce e a b ca a e c bb b c e ac c b a e

e a b ce e a b ca a b c eb b c e ac c e a b

(7)

7. (a) All of the products involving the identity are self-evident. Theonly products that must be calculated explicitly are a2, ab, ba,and b2. These are given by

πaπa =

(e a b

a b e

)(e a b

a b e

)=

(e a b

b e a

)= πb

πaπb =

(e a b

a b e

)(e a b

b e a

)=

(e a b

e a b

)= πe

πbπa =

(e a b

b e a

)(e a b

a b e

)=

(e a b

e a b

)= πe

πbπb =

(e a b

b e a

)(e a b

b e a

)=

(e a b

a b e

)= πa

Thus, the association g → πg, for g = e, a, b preserves the prod-ucts of the three-element group.

(b) With the construction

πg =

(e a b

g ga gb

)(9)

for g = e, a, b, we can use the rows of the multiplication table onp. 19 to obtain

πe =

(e a b

ee ea eb

)=

(e a b

e a b

)

πa =

(e a b

ae aa ab

)=

(e a b

a b e

)

7

πb =

(e a b

be ba bb

)=

(e a b

b e a

)(10)

Thus, the association g → πg is one-to-one and preserves theproducts of the 3-element group. Hence, these groups are equiv-alent.

(c) The elements of the three-element group correspond to thecyclic permutations of S3. In other words, given a reference ordera, b, c, the cyclic permutations are a 7→ b, b 7→ c, and c 7→ a,yielding b, c, a, and then b 7→ c, c 7→ a, and a 7→ b, yieldingc, a, b.

(d) These elements correspond to the rotations of an equilateraltriangle, i.e., the elements e, d, f in Fig. 2.1.

Group TheorySolutions to Problem Set 3 November 2, 2001

1. According to Lagrange’s theorem, the order of a subgroup H ofa group G must be a divisor of |G|. Since the divisors of a primenumber are only the number itself and unity, the subgroups ofa group of prime order must be either the unit element alone,H = e, or the group G itself, H = G, both of which are im-proper subgroups. Therefore, a group of prime order has no propersubgroups.

2. From a group G of prime order, select any element g, which isnot the unit element, and form its period:

g, g2, g3, . . . , gn = e ,

where n is the order of g (Sec. 2.4). The period must includeevery element in G, because otherwise we would have constructeda subgroup whose order is neither unity nor |G|. This contradictsthe conclusion of Problem 1. Hence, a group of prime order isnecessarily cyclic (but a cyclic group need not necessarily be ofprime order).

3. For a group G with |G| = pq, where p and q are both prime, weknow from Lagrange’s theorem that the only proper subgroupshave order p and q. Since these subgroups are of prime order, theconclusion of Problem 2 requires these subgroups to be cyclic.

4. Since the period of an element g of a group G forms a subgroup ofG (this is straightforward to verify), Lagrange’s theorem requiresthat |g| must be a divisor of |G|, i.e., G = k|g| for some integerk. Hence,

g|G| = gk|g| = (g|g|)k = ek = e .

2

5. The identity e of a group G has the property that for every el-ement g in G, ag = ge = g. We also have that different cosetseither have no common elements or have only common elements.Thus, in the factor group G/H of G generated by a subgroup H,the set which contains the unit element corresponds to the unitelement of the factor group, since

e, h1, h2, . . .a, b, c, . . . = a, b, c, . . . .

6. The class of an element a in a group G is defined as the set ofelements gag−1 for all elements g in G. If G is Abelian, then wehave

gag−1 = gg−1a = a

for all g in G. Hence, in an Abelian group, every element is in aclass by itself.

7. Let H be a subgroup of a group of G of index 2, i.e., H hastwo left cosets and two right cosets. If H is self-conjugate, thengHg−1 = H for any g in G. Therefore, to show that H is self-conjugate, we must show that gH = Hg for any g in G, i.e., thatthe left and right cosets are the same. Since H has index 2, andH is itself a right coset, all of the elements in Hg must either bein H or in the other coset of H, which we will call A. There twopossibilities: either g is in H or g is not in H. If g is an elementof H, then, according to the Rearrangement Theorem,

Hg = gH = H .

If g is not in H, then it is in A, which is a right coset of H. Two(left or right) cosets of a subgroup have either all elements incommon or no elements in common. Thus, since the unit element

3

must be contained in H, the set Hg will contain g which, byhypothesis, is in A. We conclude that

Hg = gH = A .

Therefore,

Hg = gH

for all g in G and H is, therefore, a self-conjugate subgroup.

8. The subgroup H = e, a2 of the group G = e, a, a2, a3, a4 = ehas index 2. Therefore, according to Problem 7, H must be self-conjugate. Therefore, the elements of the factor group G/H arethe subgroup H, which corresponds to the unit element, so we callit E , and the set consisting of the elements A = a, a3: G/H =E ,A.

9. Let φ be an isomorphism between a group G and a group G′,i.e. φ is a one-to-one mapping between all the elements g of Gand g′ of G′. From the group properties we have that the identitye of G must obey the relation

e = ee .

Since φ preserves all products, this relation must in particular bepreserved by φ:

φ(e) = φ(e)φ(e) .

The group properties require that, for any element g of G,

φ(g) = e′φ(g) ,

where e′ is the identity of G′. Setting g = e and comparing withthe preceding equation yields the equality

φ(e)φ(e) = e′φ(e) ,

4

which, by cancellation, implies

e′ = φ(e) .

Thus, an isomorphism maps the identity in G onto the identityin G′.

Group Theory

Solutions to Problem Set 4 November 9, 2001

1. Let D′(g) = UD(g)U−1, where D(g) is a representation of a groupG with elements g. To show that D′(g) is also a representation ofG, it is sufficient to show that this representation preserves themultiplication table of G. Thus, let a and b be any two elementsof G with matrix representations D(a) and D(b). The product abis represented by

D(ab) = D(a)D(b) .

Therefore,

D′(ab) = UD(ab)U−1

= UD(a)D(b)U−1

= UD(a)U−1UD(b)U ′

= D′(a)D′(b) ,

so multiplication is preserved and D′(g) is therefore also a repre-sentation of G.

2. The trace of a matrix is the sum of its diagonal elements. There-fore, the trace of the product of three matrices A, B, and C isgiven by

tr(ABC) =∑ijk

AijBjkCki .

By using the fact that i, j, and k are dummy summation indiceswith the same range, this sum can be written in the equivalentforms ∑

ijk

AijBjkCki =∑ijk

CkiAijBjk =∑ijk

BjkCkiAij .

2

But the second and third of these are

∑ijk

CkiAijBjk = tr(CAB)

and

∑ijk

BjkCkiAij = tr(BCA) ,

respectively. Thus, we obtain the relation

tr(ABC) = tr(CAB) = tr(BCA) .

3. The trace of an n-fold product, A1A2 · · ·An is

tr(A1A2 · · ·An) =∑

i1,i2,...in

(A1)i1i2(A2)i2i3 · · · (An)ini1 .

Proceeding as in Problem 2, we observe that the ik (k = 1, . . . , n)are dummy summation indices all of which have the same range.Thus, any cyclic permutation of the matrices in the product leavesthe sum and, hence, the trace invariant.

4. From Problem 2, we have that

tr(UAU−1) = tr(U−1UA) = tr(A) ,

so a similarity transformation leaves the trace of a matrix invari-ant.

3

5. Given a faithful representation of a group, similarity transfor-mations of the matrices provide equally faithful representations.Since we wish to obtain permutations of a particular matrix repre-sentation, we base our similarity transformations on the non-unitelements in the group. Thus, consider the following similaritytransformations:

ae, a, b, c, d, fa−1 = e, a, c, b, f, d ,

be, a, b, c, d, fb−1 = e, c, b, a, f, d ,

ce, a, b, c, d, fc−1 = e, b, a, c, f, d , (1)

de, a, b, c, d, fd−1 = e, b, c, a, d, f ,

ee, a, b, c, d, fe−1 = e, c, a, b, d, f .

Thus, the following permutations of the elements e, a, b, c, d, fprovide equally faithful representations:

e, a, c, b, f, d , e, c, b, a, f, d ,

e, b, a, c, f, d , e, b, c, a, d, f ,

e, c, a, b, d, f .

Notice that only elements within the same class can be permuted.For S3, the classes are e, a, b, c, d, f.

6. In the basis (x, y, z) where x and y are given in Fig. 3.1 and thez-axis emanates from the origin of this coordinate system (thegeometric center of the triangle), all of the symmetry operations ofthe equilateral triangle leave the z-axis invariant. This is becausethe z-axis is either an axis of rotation (for operations d and f) orlies within the reflection plane (for operations a, b, and c). Hence,

4

the matrices of these operations are given by

e =

1 0 0

0 1 0

0 0 1

, a = 12

1 −

√3 0

−√

3 −1 0

0 0 2

,

b = 12

1√

3 0√

3 −1 0

0 0 2

, c =

−1 0 0

0 1 0

0 0 1

,

d = 12

−1 −

√3 0

√3 −1 0

0 0 2

, f = 12

−1

√3 0

−√

3 −1 0

0 0 2

.This representation is seen to be reducible and that it is the directsum of the representation in Example 3.2 (which, as discussed inExample 3.4, is irreducible) and the identical representation.

7. We must show (i) that two matrices which are simultaneously di-agonalizable commute and (ii) that two matrices which commuteare simultaneously diagonalizable. Showing (i) is straightforward.For two d× d matrices A and B which are simultaneously diago-nalizable, there is a matrix U such that

UAU−1 = DA and UBU−1 = DB ,

where DA and DB are diagonal forms of these matrices. Clearly,therefore, we have that

DADB = DBDA .

Hence, transforming back to the original basis,

(U−1DAU)︸︷︷︸A

(U−1DBU)︸︷︷︸B

= (U−1DBU)︸︷︷︸B

(U−1DAU)︸︷︷︸A

,

5

so A and B commute.

Now suppose that A and B commute and there is a transforma-tion that brings one of these matrices, say A, into the diagonalform DA:

UAU−1 = DA .

Then, with

UBU−1 = B′ ,

the commutation relation AB = BA transforms to

DAB′ = B′DA .

The (i, j)th matrix element of these products is

(DAB′)ij =

∑k

(DA)ik(B′)kj = (DA)ii(B

′)ij

= (B′DA)ij =∑k

(B′)ik(DA)kj = (B′)ij(DA)jj .

After a simple rearrangement, we have

(B′)ij[(DA)ii − (DA)jj] = 0 .

There are three cases to consider:

Case I. All of the diagonal entries of DA are distinct. Then,

(DA)ii − (DA)jj 6= 0 if i 6= j ,

so all of the off-diagonal matrix elements of B′ vanish, i.e., B′ is adiagonal matrix. Thus, the same similarity transformation whichdiagonalizes A also diagonalizes B.

Case II. All of the diagonal entries of DA are the same. In thiscase DA is proportional to the unit matrix, DA = cI, for somecomplex constant c. Hence, this matrix is always diagonal,

U(cI)U−1 = cI

6

and, in particular, it is diagonal when B is diagonal.

Case III. Some of the diagonal entries are the same and someare distinct. If we arrange the elements of DA such that the firstp elements are the same, (DA)11 = (DA)22 = · · · = (DA)pp, thenDA has the general form

DA =

(cIp 0

0 D′A

),

where Ip is the p × p unit matrix and c is a complex constant.From Cases I and II, we deduce that B must be of the form

B =

(Bp 0

0 D′B

),

where Bp is some p× p matrix and D′B is a diagonal matrix. LetVp be the matrix which diagonalizes B:

VpBpV−1p = D′′B .

Then the matrix

V =

(Vp 0

0 Id−p

)

diagonalizes B while leaving DA unchanged. Here, Id−p is the(d− p)× (d− p) unit matrix.

Hence, in all three cases, we have shown that the same transfor-mation which diagonalizes A also diagonalizes B.

8. The matrices of any representation A1, A2, . . . , An of an Abeliangroup G commute:

AiAj = AjAi

7

for all i and j. Hence, according to Problem 7, these matrices canall be simultaneously diagonalized. Since this is true of all repre-sentations of G, we conclude that all irreducible representations ofAbelian groups are one-dimensional, i.e., they are numbers withordinary multiplication as the composition law.

9. To verify that the matrices

e =

(1 0

0 1

), a = 1

2

(−1√

3√

3 1

)(2)

form a representation for the two-element group e, a, we needto demonstrate that the multiplication table for this group,

e ae e aa a e

is fulfilled by these matrices. The products e2 = e, ea = a, andae = a can be verified by inspection. The product a2 is

a2 = 14

(−1√

3√

3 1

)(−1√

3√

3 1

)=

(1 0

0 1

)= e ,

so the matrices in (2) form a representation of the two-elementgroup.

Since these matrices commute, they can be diagonalized simulta-neously (Problem 7). Since the matrix is the unit matrix, we candetermine the diagonal form of a, simply by finding its eigenval-ues. The characteristic equation of a is

det(a− λI) =

∣∣∣∣∣−12− λ 1

2

√3

12

√3 1

2− λ

∣∣∣∣∣= −(1

2− λ)(1

2+ λ)− 3

4= λ2 − 1 .

8

which yields λ = ±1. Therefore, diagonal form of a is

a =

(1 0

0 −1

), (3)

so this representation is the direct sum of the identical represen-tation 1, 1, and the “parity” representation 1,−1. Note that,according to Problem 8, every representation of the two-elementgroup with dimensionality greater than two must be reducible.

10. The relations in (3.9) and (3.11) can be proven simultaneously,since they differ only by complex conjugation, which preserves theorder of matrices. The (i, j)th matrix element of n-fold productof matrices A1, A2, . . . , An is

(A1A2 · · ·An)ij =∑

k1,k2,...,kn−1

(A1)ik1(A2)k1k2 · · · (An)kn−1j .

The corresponding matrix element of the transpose of this prod-uct is

[(A1A2 · · ·An)t]ij = (A1A2 · · ·An)ji .

Thus, since the ki are dummy indices,

[(A1A2 · · ·An)t]ij =∑

k1,k2,...,kn−1

(A1)jk1(A2)k1k2 · · · (An)kn−1i

=∑

k1,k2,...,kn−1

(At1)k1j(A

t2)k2k1 · · · (At

n)ikn−1

=∑

k1,k2,...,kn−1

(Atn)ikn−1(At

n−1)kn−1kn−2 · · · (At2)k2k1(At

1)k1j

We conclude that

(A1A2 · · ·An)t = AtnA

tn−1 · · ·At

1

9

and, similarly, that

(A1A2 · · ·An)† = A†nA†n−1 · · ·A†1

Group Theory


1. In proving Theorem 3.2, we established that BiB†i = I, where

Bi = D−1/2AiD1/2

To show that this result implies that B†iBi = I, we first use thedefinitions of Bi and B†i to write

B†iBi = D1/2A†iD−1AiD

1/2

We can find an expression for D−1 by first rearranging

BiB†i = D−1/2AiDA

†iD−1/2 = I

as

AiDA†i = D

Then, taking the inverse of both sides of this equation yields

D−1 = A†−1i D−1A−1

i

Therefore,

B†iBi = D1/2A†iD−1AiD

1/2

= D1/2A†i (A†−1i D−1A−1

i )AiD1/2

= I

2. (a) The multiplication table for the three-element group is shownbelow (Sec. 2.4):

2

e a be e a ba a b eb b e a

We can see immediately that a2 = b and that ab = ba = a3 = e,Thus, the three-element group can be written as a, a2, a3 = e,i.e., it is a cyclic group (and, therefore, Abelian).

(b) By choosing a one-dimensional representation a = z, for somecomplex number z, the multiplication table requires that a3 = e,which means that z3 = 1.

(c) There are three solutions to z3 = 1: z = 1, e2πi/3, ande4πi/3.The three irreducible representations are obtained by choosinga = 1, a = e2πi/3, and a = e4πi/3. Denoting these representationsby Γ1, Γ2, and Γ3, we obtain

e a bΓ1 1 1 1Γ2 1 e2πi/3 e4πi/3

Γ2 1 e4πi/3 e2πi/3

3. The preceding Problem can be generalized to any cyclic groupof order n. The elements of this group are g, g2, . . . , gn = e.By writing g = z, we require that zn = 1. The solutions to thisequation are the nth roots of unity:

z = e2mπi/n, m = 0, 1, 2, . . . , n− 1

Accordingly, there are n irreducible representations based on then choices z = e2mπi/n together with the requirements of the groupmultiplication table.

3

4. Suppose that we have a representation of an Abelian group ofdimensionality d is greater than one. Suppose furthermore thatthese matrices are not all unit matrices (for, if they were, the rep-resentation would already be reducible to the d-fold direct sum ofthe identical representation.) Then, since the group is Abelian,and the representation must reflect this fact, any non-unit ma-trix in the representation commutes with all the other matricesin the representation. According to Schur’s First Lemma, thisrenders the representation reducible. Hence, all the irreduciblerepresentations of an Abelian group are one-dimensional.

5. For the following matrices,

e = d = f =

(1 0

0 1

), a = b = c = 1

2

( −1 −√

3

−√

3 1

)

to be a representation of S3, their products must preserve themultiplication table of this group, which was discussed in Sec. 2.4and is displayed below:

e a b c d fe e a b c d fa a e d f b cb b f e d c ac c d f e a bd d c a b f ef f b c a e d

To determine the multiplication table of this representation, weuse the notation E for the unit matrix, corresponding to the el-ements e, d, and f , and A for the matrix corresponding to theelements a, b, and c. Then, by observing that A2 = E (as isrequired by the group multiplication table) the multiplication ta-ble of this representation is straightforward to calculate, and isshown below:

4

e a b c d fe E A A A E Ea A E E E A Ab A E E E A Ac A E E E A Ad E A A A E Ef E A A A E E

If we now take the multiplication table of S3 and perform themapping e, d, f 7→ E and a, b, c 7→ A, we get the same tableas that just obtained by calculating the matrix products directly.Hence, these matrices form a representation of S3.

From Schur’s First Lemma, we see that the matrix

12

( −1 −√

3

−√

3 1

)

commutes with all the matrices of the representation. Since thisis not a unit matrix, the representation must be reducible.

The diagonal form of these matrices must have entries of one-dimensional irreducible representations. Two one-dimensional ir-reducible representations of S3 (we will see later that these arethe only irreducible representations of S3) are (Example 3.2) theidentical representation,

Ae = 1, Aa = 1, Ab = 1,

Ac = 1, Ad = 1, Af = 1

and the ‘parity’ representation,

Ae = 1, Aa = −1, Ab = −1,

Ac = −1, Ad = 1, Af = 1

Since the diagonal forms of the matrices are obtained by perform-ing a similarity transformation on the original matrices, which

5

preserves the trace, they must take the form

e = d = f =

(1 0

0 1

), a = b = c =

(1 0

0 −1

)

i.e., this reducible representation contains both the identical andparity representations.

Group Theory


1. The Great Orthogonality Theorem states that, for the matrixelements of the same irreducible representation A1, A2, . . . , A|G|of a group G,

∑α

(Aα)ij(Aα)∗i′j′ =|G|dδi,i′δj,j′ .

Thus, we first form the vectors V ij whose components are the(i, j)th elements taken from each matrix in the representation insome fixed order. The Great Orthogonality Theorem can then beexpressed more concisely as

V ij · V ∗i′j′ =|G|dδi,i′δj,j′ .

For the given representation of S3,

e =

(1 0

0 1

), a = 1

2

(1 −

√3

−√

3 −1

), b = 1

2

(1√

3√

3 −1

),

c =

(−1 0

0 1

), d = 1

2

( −1√

3

−√

3 −1

), f = 1

2

(−1 −√

3√

3 −1

),

these vectors are:

V 11 =(1, 1

2, 1

2,−1,−1

2,−1

2

),

V 12 =(0,−1

2

√3, 1

2

√3, 0, 1

2

√3,−1

2

√3),

V 21 =(0,−1

2

√3, 1

2

√3, 0,−1

2

√3, 1

2

√3),

V 22 =(1,−1

2,−1

2, 1,−1

2,−1

2

).

2

Note that, since all of the entries are real, complex conjugation isnot required for substitution into the Great Orthogonality Theo-rem. For i = j and i′ = j′, with |G| = 6 and d = 2, we have

V 11 · V 11 = 1 + 14

+ 14

+ 1 + 14

+ 14

= 3 ,

V 12 · V 12 = 0 + 34

+ 34

+ 0 + 34

+ 34

= 3 ,

V 21 · V 21 = 0 + 34

+ 34

+ 0 + 34

+ 34

= 3 .

V 22 · V 22 = 1 + 14

+ 14

+ 1 + 14

+ 14

= 3 ,

all of which are in accord with the Great Orthogonality Theorem.For i 6= j and/or i′ 6= j′, we have

V 11 · V 12 = 0− 14

√3 + 1

4

√3 + 0− 1

4

√3 + 1

4

√3 = 0 ,

V 11 · V 21 = 0− 14

√3 + 1

4

√3 + 0 + 1

4

√3− 1

4

√3 = 0 ,

V 11 · V 22 = 1− 14− 1

4− 1 + 1

4+ 1

4= 0 ,

V 12 · V 21 = 0 + 34

+ 34

+ 0− 34− 3

4= 0 ,

V 12 · V 22 = 0 + 14

√3− 1

4

√3 + 0− 1

4

√3 + 1

4

√3 = 0 ,

V 21 · V 22 = 0 + 14

√3− 1

4

√3 + 0 + 1

4

√3− 1

4

√3 = 0 ,

which is also in accord with the Great Orthogonality Theorem.

2. For the following two-dimensional representation of the three-element group,

e =

(1 0

0 1

), a = 1

2

( −1√

3

−√

3 −1

), b = 1

2

(−1 −√

3√

3 −1

),

we again form the vectors V ij whose components are the (i, j)thelements of each matrix in the representation:

V 11 =(1,−1

2,−1

2

),

3

V 12 =(0, 1

2

√3,−1

2

√3),

V 21 =(0,−1

2

√3, 1

2

√3),

V 22 =(1,−1

2,−1

2

).

Calculating the summation in the Great Orthogonality Theorem,first with i = j and i′ = j′, we have

V 11 · V 11 = 1 + 14

+ 14

= 32,

V 12 · V 12 = 0 + 34

+ 34

= 32,

V 21 · V 12 = 0 + 34

+ 34

= 32,

V 11 · V 11 = 1 + 14

+ 14

= 32,

all of which are in accord with the Great Orthogonality Theoremwith |G| = 3 and d = 2. Performing the analogous summationswith i 6= j and/or i′ 6= j′, yields

V 11 · V 12 = 0− 14

√3 + 1

4

√3 = 0 ,

V 11 · V 21 = 0 + 14

√3− 1

4

√3 = 0 ,

V 11 · V 22 = 1 + 14

+ 14

= 32,

V 12 · V 21 = 0− 34− 3

4= −3

2,

V 12 · V 22 = 0− 14

√3 + 1

4

√3 = 0 ,

V 21 · V 22 = 0 + 14

√3− 1

4

√3 = 0 ,

which is not consistent with the Great Orthogonality Theorem,since all of these quantities must vanish. If there is even a singleviolation of the Great Orthogonality Theorem, as is the case here,the representation is necessarily reducible.

4

3. All of the irreducible representations of an Abelian group are one-dimensional (e.g., Problem 4, Problem Set 5). Hence, for Abeliangroups, the Great Orthogonality Theorem reduces to∑

α

AkαAk′∗α = |G|δk,k′ .

If we view the irreducible representations as |G|-dimensional vec-tors Ak with entries Akα,

Ak = (Ak1, Ak2, . . . A

k|G|) ,

then the Great Orthogonality Theorem can be written as a “dot”product:

Ak ·Ak′∗ = |G|δk,k′ .

This states that the irreducible representations of an Abeliangroup are orthogonal vectors in this |G|-dimensional space. Sincethere can be at most |G| such vectors, the number of irreduciblerepresentations of an Abelian group is less than or equal to theorder of the group.

4. (a) From Problem 2 of Problem Set 5, the irreducible representa-tions of the three element group are:

e a bΓ1 1 1 1Γ2 1 e2πi/3 e4πi/3

Γ2 1 e4πi/3 e2πi/3

In the notation of Problem 3, we have

A1 = (1, 1, 1), A2 = (1, e2πi/3, e4πi/3), A3 = (1, e4πi/3, e2πi/3) .

Note that some of these entries are complex. Thus, the distinctinner products between these vectors are

A1 ·A1∗ = 1 + 1 + 1 = 3 ,

5

A2 ·A2∗ = 1 + 1 + 1 = 3 ,

A3 ·A3∗ = 1 + 1 + 1 = 3 ,

A1 ·A2∗ = 1 + e−2πi/3 + e−4πi/3 = 0 ,

A1 ·A3∗ = 1 + e−4πi/3 + e−2πi/3 = 0 ,

A2 ·A3∗ = 1 + e−2πi/3 + e2πi/3 = 0 ,

all of which are consistent with the Great Orthogonality Theorem.

(b) In view of the fact that there are 3 mutually orthogonal vec-tors, there can be no additional irreducible representations of thisgroup.

(c) For cyclic groups of order |G|, we determined that the irre-ducible representations were based on the |G|th roots of unity(Problem 3, Problem Set 5). Since this produces |G| distinct irre-ducible representations, our procedure yields all of the irreduciblerepresentations of any cyclic group.

5. Every irreducible representation of an Abelian group is one-dimen-sional. Moreover, since every one of these representations is eithera homomorphism or isomorphism of the group, with the operationin the representation being ordinary multiplication, the identityalways corresponds to unity (Problem 9, Problem Set 3). Now,the order n of a group element g is the smallest integer for which

gn = e .

For every element in any group 1 ≤ n ≤ |G|. This relationshipmust be preserved by the irreducible representation. Thus, if Akgis the entry corresponding to the element g in the kth irreduciblerepresentation, then

(Akg)n = 1 ,

6

i.e., Akg is the nth root of unity:

Akg = e2mπi/n, m = 0, 1, . . . , n− 1 .

The modulus of each of these quantities is clearly unity, so themodulus of every entry in the irreducible representations of anAbelian group is unity.

This is consistent with the Great Orthogonality Theorem whenapplied to a given representation (cf. Problem 3):

∑α

AkαAk∗α =

∑α

∣∣∣Akα∣∣∣2 = |G| . (1)

Group Theory


1. A regular hexagon is shown below:

1

2

34

5

6

The following notation will be used for the symmetry operationsof this hexagon:

(1 2 3 4 5 6

a1 a2 a3 a4 a5 a6

),

where the first row corresponds to the reference order of the ver-tices shown in the diagram and the ai denote the number at theith vertex after the transformation of the hexagon.

The symmetry operations of this hexagon consist of the identity,rotations by angles of 1

3nπ radians, where n = 1, 2, 3, 4, 5, three

mirror planes which pass through opposite faces of the hexagon,and three mirror planes which pass through opposite vertices ofthe hexagon. For the identity and the rotations, the effect on thehexagon is

2

4

5

61

2

3 5

6

12

3

4 6

1

23

4

5

1

2

34

5

6 2

3

45

6

1 3

4

56

1

2

These operations correspond to

E =

(1 2 3 4 5 6

1 2 3 4 5 6

),

C6 =

(1 2 3 4 5 6

2 3 4 5 6 1

),

C26 = C3 =

(1 2 3 4 5 6

3 4 5 6 1 2

),

C36 = C2 =

(1 2 3 4 5 6

4 5 6 1 2 3

),

C46 = C2

3 =

(1 2 3 4 5 6

5 6 1 2 3 4

),

C56 =

(1 2 3 4 5 6

6 1 2 3 4 5

).

The three mirror planes which pass through opposite faces of thehexagon are

3

6

5

43

2

1 2

1

65

4

3 4

3

21

6

5

which correspond to

σv,1 =

(1 2 3 4 5 6

6 5 4 3 2 1

),

σv,2 =

(1 2 3 4 5 6

2 1 6 5 4 3

),

σv,3 =

(1 2 3 4 5 6

4 3 2 1 6 5

).

Finally, the three mirror planes which pass through opposite ver-tices of the hexagon are

1

6

54

3

2 3

2

16

5

4 5

4

32

1

6


σd,1 =

(1 2 3 4 5 6

1 6 5 4 3 2

),

σd,2 =

(1 2 3 4 5 6

3 2 1 6 5 4

),

4

σd,3 =

(1 2 3 4 5 6

5 4 3 2 1 6

).

That these 12 elements do, in fact, form a group is straightforwardto verify. The standard notation for this group is C6v.

2. The order n of an element a of a group is defined as the smallestinteger such that an = e. If group elements a and b are in thesame class, then there is an element g in the group such that

b = g−1ag .

The m-fold product of b is then given by

bm = (g−1ag)(g−1ag) · · · (g−1ag)︸︷︷︸m factors

= g−1amg .

If this is equal to the unit element e, we must have

g−1amg = e ,

or,

am = e .

The smallest value of m for which this equality can be satisfiedis, by definition, n, the order of a. Hence, two elements in thesame class have the same order.

3. For two elements a and b of a group to be in the same class,there must be another group element such that b = g−1ag. If thegroup elements are coordinate transformations, then elements inthe same class correspond to the same type of operation, but incoordinate systems related by symmetry operations. This fact,together with the result of Problem 2, allows us to determine theclasses of the group of the hexagon.

5

The identity, as always, is in a class by itself. Although all of therotations are the same type of operation, not all of these rotationshave the same orders: C6 and C5

6 have order 6, C3 and C23 have

order 3, and C2 has order 2. Thus, the 5 rotations belong to threedifferent classes.

The two types of mirror planes, σv,i and σd,i, must belong to differ-ent classes since there is no group operation which will transformany of the σv,i to any of the σd,i. To do so would require a rotationby an odd multiple of 1

6π, which is not a group element. All of

the σv,i are in the same class and all of the σd,i are in the sameclass, since each is the same type of operation, but in coordinatesystems related by symmetry operations (one of the rotations)and, of course, they all have order 2, since each reflection planeis its own inverse.

Hence, there are six classes in this group are

E ≡ E ,

2C6 ≡ C6, C56 ,

2C3 ≡ C3, C23 ,

C2 ≡ C2 ,

3σv ≡ σv,1, σv,2, σv,3 ,

3σd ≡ σd,1, σd,2, σd,3 .

4. As there are 6 classes, there are 6 irreducible representations, thedimensions of which must satisfy the sum rule

6∑k=1

d2k = 12 ,

6

since |C6v| = 12. The only positive integer solutions of this equa-tion are

d1 = 1, d2 = 1, d3 = 1, d4 = 1, d5 = 2, d6 = 2 ,

i.e., there are 4 one-dimensional irreducible representations and 2two-dimensional irreducible representations.

5. (a) For the identical representation, all of the characters are 1.For the parity representation, the character is 1 for operationswhich preserve the parity of the coordinate system (“proper” ro-tations) and −1 for operations which change the parity of thecoordinate system (“improper” rotations). Additionally, we canenter immediately the column of characters for the class of theunit element. These are equal to the dimensionality of each ir-reducible representation, since the unit element is the identitymatrix with that dimensionality. Thus, we have the followingentries for the character table of C6v:

C6v E 2C6 2C3 C2 3σv 3σdΓ1 1 1 1 1 1 1Γ′1 1 1 1 1 −1 −1Γ′′1 1Γ′′′1 1Γ2 2Γ′2 2

(b) The characters for one of the two-dimensional representationsof C6v can be obtained by constructing matrices for operationsin analogy with the procedure discussed in Section 3.2 for theequilateral triangle. One important difference here is that werequire such a construction only for one element in each class(since the all matrices in a given class have the same trace). Wewill determine the representations of operations in each class inan (x, y) coordinate system shown below:

7

x

y

Thus, a rotation by an angle φ, denoted by R(φ), is given by thetwo-dimensional rotational matrix:

R(φ) =

(cosφ sinφ

− sinφ cosφ

).

The corresponding character χ(φ) is, therefore, simply the sumof the diagonal elements of this matrix:

χ(φ) = 2 cosφ .

We can now calculate the characters for each of the classes com-posed of rotations:

χ(2C6) = χ(13π) = 1 ,

χ(2C3) = χ(23π) = −1 ,

χ(C2) = χ(π) = −2 .

For the two classes of mirror planes, we need only determine thecharacter of one element in each class, which may be chosen atour convenience. Thus, for example, since the representation of

8

σv,1 can be determined directly by inspection:(−1 0

0 1

),

we can obtain the character of the corresponding class as

χ(3σv) = 0 .

Similarly, the representation of σd,2 can also be determined di-rectly by inspection: (

1 0

0 −1

),

which yields the character

χ(3σd) = 0 .

We can now add the entries for this two-dimensional irreduciblerepresentations to the character table of C6v:

C6v E 2C6 2C3 C2 3σv 3σdΓ1 1 1 1 1 1 1Γ′1 1 1 1 1 −1 −1Γ′′1 1Γ′′′1 1Γ2 2 1 −1 −2 0 0Γ′2 2

(c) The one-dimensional irreducible representations must obeythe multiplication table, since they themselves are representationsof the group. In particular, given the products

C3C23 = E, C3

3 = E ,

if we denote by α the character of the class 2C3 = C3, C23, then

these products require that

α2 = 1, α3 = 1 ,

9

respectively. Thus, we deduce that α = 1 for all of the one-dimensional irreducible representations. With these additions tothe character table, we have

C6v E 2C6 2C3 C2 3σv 3σdΓ1 1 1 1 1 1 1Γ′1 1 1 1 1 −1 −1Γ′′1 1 1Γ′′′1 1 1Γ2 2 1 −1 −2 0 0Γ′2 2

(d) Since the character for all one-dimensional irreducible rep-resentations for the class 2C3 = C3, C

23 is unity, the product

C6C3 = C2 requires that the characters for the classes of C6 andC2 are the same in these representations. Since C2

2 = E, this char-acter must be 1 or −1. Suppose we choose χ(2C6) = χ(C2) = 1.Then, the orthogonality of the columns of the character tablerequires that the character for the classes E and 2C6 are orthog-onal. If we denote by β the character for the class 2C6 of therepresentation Γ′2, we require

(1× 1) + (1× 1) + (1× 1) + (1× 1) + (2× 1) + (2× β) = 0 ,

i.e., β = −3. But this value violates the requirement that∑α

nα|χα|2 = |G| . (1)

Thus, we must choose χ(2C6) = χ(C2) = −1, and our charactertable becomes

C6v E 2C6 2C3 C2 3σv 3σdΓ1 1 1 1 1 1 1Γ′1 1 1 1 1 −1 −1Γ′′1 1 −1 1 −1Γ′′′1 1 −1 1 −1Γ2 2 1 −1 −2 0 0Γ′2 2

10

(e) The characters for the classes 2C6, 2C3, and C2 of the Γ′2representation can now be determined by requiring the columnscorresponding to these classes to be orthogonal to the columncorresponding to the class of the identity. When this is done, wefind that the values obtained saturate the sum rule in (1), so thecharacters corresponding to both classes of mirror planes in thisrepresentation must vanish. This enables to complete the entriesfor the Γ′2 representation:

C6v E 2C6 2C3 C2 3σv 3σdΓ1 1 1 1 1 1 1Γ′1 1 1 1 1 −1 −1Γ′′1 1 −1 1 −1Γ′′′1 1 −1 1 −1Γ2 2 1 −1 −2 0 0Γ′2 2 −1 1 2 0 0

The remaining entries are straightforward to calculate. The factthat each mirror reflection has order 2 means that these entriesmust be either +1 or −1. The requirement of orthogonality ofcolumns leaves only one choice:

C6v E 2C6 2C3 C2 3σv 3σdΓ1 1 1 1 1 1 1Γ′1 1 1 1 1 −1 −1Γ′′1 1 −1 1 −1 1 −1Γ′′′1 1 −1 1 −1 −1 1Γ2 2 1 −1 −2 0 0Γ′2 2 −1 1 2 0 0

which completes the character table for C6v.

Group Theory

Solutions to Problem Set 8 December 7, 2001

1. A matrix A is said to be orthogonal if its matrix elements aijsatisfy the following relations:∑

i

aijaik = δj,k,∑j

aijakj = δij , (1)

i.e., the rows and columns are orthogonal vectors. This ensuresthat AtA = AAt = I.

The direct product C of two matrices A and B, denoted by C =A⊗B, is given in terms of matrix elements by

cik;jl = aijakl .

If A and B are orthogonal matrices, then we can show that C isalso an orthogonal matrix by verifying the relations in Eq. (1).The first of these relations is∑

ik

cik;jlcik;j′l′

=∑ik

aijaklaij′akl′ =(∑

i

aijaij′)(∑

k

aklakl′)

= δj,j′δl,l′ ,

where the last step follows from the first of Eqs. (1). The secondorthogonality relation is∑

jl

cik;jlci′k′;jl

=∑jl

aijaklai′jak′l =(∑

j

aijai′j

)(∑l

aklak′l

)= δi,i′δk,k′

where the last step follows from the second of Eqs. (1). Thus, wehave shown that the direct product of two orthogonal matrices isalso an orthogonal matrix.

2

2. The direct product of two matrices A and B with matrix elementsaij and bij is

cik,jl = aijbkl .

The trace of the direct product A⊗B is obtained by setting j = iand l = k and summing over i and k:

tr(A⊗B) =∑ik

cik,ik =∑ik

aiibkk =∑i

aii∑k

bkk = tr(A) tr(B) ,

which is the product of the traces of A and B.

3. We have two groups Ga and Gb with elements

Ga = ea, a2, a3, . . . , a|Ga|

and

Gb = eb, b2, b3, . . . , b|Gb| ,

such that aibj = bjai for all i and j. We are using a notationwhere it is understood that a1 = ea and b1 = eb. The directproduct Ga ⊗ Gb of these groups is the set obtained by formingthe product of every element of Ga with every element of Gb:

Ga ⊗Gb = e, a2, a3, . . . , ana , b2, b3, . . . , bnb , . . . , aibj, . . . .

To show that Ga⊗Gb is a group, we must demonstrate that theseelements fulfill each of the four requirements in Sec. 2.1.

Closure. The product of two elements aibj and ai′bj′ is given by

(aibj)(ai′bj′) = (aiai′)(bjbj′) = akbl ,

where the first step follows from the commutativity of elementsbetween the two groups and the second step from the group prop-erty of Ga and Gb.

3

Associativity. The associativity of the composition law followsfrom

(aibi′ajbj′)akbk′ =[(aiaj)ak

][(bi′bj′)bk′

]=[ai(ajak)

][bi′(bj′bk′)

]= aibi′(ajbj′akbk′) ,

since associativity holds for Ga and Gb separately.

Unit Element. The unit element e for the direct product group iseaeb = ebea, since

(aibj)(eaeb) = (aiea)(bjeb) = (eaai)(ebbj) = (eaeb)(aibj) .

Inverse. Finally, the inverse of each element aibj is a−1i b−1

j because

(aibj)(a−1i b−1

j ) = (aia−1i )(bjb

−1j ) = eaeb

and

(a−1i b−1

j )(aibj) = (a−1i ai)(b

−1j bj) = eaeb .

Thus, we have shown that the direct product of two groups isitself a group. Since the elements of this group are obtained bytaking all products of elements from Ga and Gb, the order of thisgroup is |Ga||Gb|.

4. Suppose we have an irreducible representation for each of twogroups Ga and Gb. We denote these representations, which maybe of different dimensions, by A(ai) and A(bj), and their matrixelements by A(ai)ij and A(bj)ij. Since these representations areirreducible, they satisfy the Great Orthogonality Theorem:

∑ai

A(ai)∗ijA(ai)i′j′ =

|Ga|da

δi,i′δj,j′ ,

∑bj

A(bj)∗ijA(bj)i′j′ =

|Gb|db

δi,i′δj,j′ ,

4

where da and db are the dimensions of the irreducible representa-tions of Ga and Gb, respectively. A representation of the directproduct of two groups, denoted by A(aibj), is obtained from thedirect product of representations of each group:

A(aibj)ik;jl = A(ai)ijA(bj)kl .

The sum in the Great Orthogonality Theorem for the direct prod-uct representation is∑

ai

∑bj

A(aibj)∗ik;jlA(aibj)i′k′;j′l′

=∑ai

∑bj

A(ai)∗ijA(bj)

∗klA(ai)i′j′A(bj)k′l′

=[∑ai

A(ai)∗ijA(ai)i′j′

]︸︷︷︸

|Ga|da

δi,i′δj,j′

[∑bj

A(bj)∗klA(bj)k′l′

]︸︷︷︸

|Gb|db

δk,k′δl,l′

=( |Ga||Gb|

dadb

)δi,i′δk,k′δj,j′δl,l′ .

This shows that this direct product representation is, in fact,irreducible. It has dimensionality dadb and the order of the directproduct is, of course, |Ga| × |Gb|.

5. If the ϕi are orthonormal, and if this property is required to bepreserved by the group of the Hamiltonian (as it must, to conserveprobability), then, in Dirac notation, we have

(i, j) ≡∫ϕi(x)∗ϕj(x) dx =

∫[Rϕi(x)]∗Rϕj(x) dx

= (i|R†R|j) = δi,j .

5

Therefore,

(i|R†R|j) =∑k,l

(k, l)Γ(R)∗kiΓ(R)lj

=∑k

Γ(R)∗kiΓ(R)kj

=∑k

[Γ(R)†

]ik

Γ(R)kj

=[Γ(R)†Γ(R)

]ij,

i.e., when written in matrix notation,

Γ(R)†Γ(R) = I .

Thus, the matrix representation is unitary.

6. We again use Dirac notation to signify basis functions ϕi andϕj corresponding to irreducible representations n and n′, respec-tively: |n, i) and |n′, j). Then, the operations R in the group ofthe Hamiltonian applied to these functions yield

R|n, i) =∑k

Γ(n)(R)ki|n, k) ,

R|n′, j) =∑l

Γ(n′)(R)lj|n′, l) .

Since the operators and their representations are unitary,

(n′, j|R† = (n′, j|R−1 =∑l

Γ(n′)(R)∗lj(n′, l| ,

we have

(n′, j|R−1R|n, i) = (n′, j|n, i)

=∑kl

Γ(n′)(R)∗kjΓ(n)(R)li(n

′, k|n, l) .

6

If we now sum both sides of this equation over the elements ofthe group of the Hamiltonian, and invoke the Great OrthogonalityTheorem, we obtain∑

R

(n′, j|n, i) = |G|(n′, j|n, i)

=∑kl

[∑R

Γ(n′)(R)∗kjΓ(n)(R)li

]︸︷︷︸

|G|dnδn,n′δk,lδi,j

(n′, k|n, l)

= |G|δn,n′δi,j(n′, k|n, k) ,

where |G| is the order of the group of the Hamiltonian and dn isthe dimension of the nth irreducible representation. Therefore,

(n′, j|n, i) = δn,n′δi,j ,

since (n, k|n, k) = 1.

7. (a) A square is shown below:

1

23

4

In analogy with the procedure described in Problem Set 7, wewill use the following notation for the symmetry operations of

7

this hexagon: (1 2 3 4

a1 a2 a3 a4

),

where the ai denote the number at the ith vertex after the trans-formation of the hexagon given in the indicated reference order.Thus, the identity operation, which identifies the reference orderof the vertices, corresponds to(

1 2 3 4

1 2 3 4

).

The symmetry operations on this square consist of the identity,rotations by angles of 1

2nπ radians, for n = 1, 2, 3, two mirror

planes which pass through opposite faces of the square, and twomirror planes which pass through opposite vertices of the square.For the rotations, the effect on the square is

4

12

3 3

41

2 2

34

1


C4 =

(1 2 3 4

4 1 2 3

),

C24 = C2 =

(1 2 3 4

3 4 1 2

),

C34 =

(1 2 3 4

2 3 4 1

).

8

The two mirror planes which pass through opposite faces of thesquare are

4

32

1 2

14

3

which correspond to

σv,1 =

(1 2 3 4

4 3 2 1

),

σv,2 =

(1 2 3 4

2 1 4 3

).

Finally, the two mirror planes which pass through opposite ver-tices of the square are

1

43

2 3

21

4


σ′v,1 =

(1 2 3 4

1 4 3 2

),

σ′v,2 =

(1 2 3 4

3 2 1 4

).

9

Elements in the same equivalence class must have the same orderand correspond to the same “type” of operation. Thus, there arefive equivalence classes of this group:

E, C2 = C24, 2C4, 2σv, 2σ′v .

Note that, as in the case of the regular hexagon (Problem Set7), all of the rotations need not belong to the same class, despitebeing the same “type” of operation because they must also havethe same order.

(b) The order of this group is 8 and there are 5 equivalence classes.Thus, there must be five irreducible representations whose dimen-sionalities must satisfy

d21 + d2

2 + d23 + d2

4 + d25 = 8 .

The only solution of this equation (with positive integer valuesfor the dk) is

d1 = 1, d2 = 1, d3 = 1, d4 = 1, d5 = 2 .

These dimensionalities imply that the energy levels for a Hamilto-nian with this symmetry are either nondegenerate or are two-folddegenerate. From the expression given for the energy eigenval-ues, we see immediately that the energy eigenvalues with p = qare non-degenerate, and those with p 6= q are two-fold degen-erate (but see below). Thus, the dimensions of the irreduciblerepresentations are consistent with these degeneracies.

(c) The simplest way to obtain a two-dimensional representationof this group is to consider the action of each group element onsome generic point (x, y). Then the action on this point of eachof the operations given above can be determined by inspection.We begin with the figure below:

10

From the diagrammatic representation of each symmetry opera-tion, we will be able to determine the corresponding representa-tion, simply by inspection. The action on this point by the threerotations acn be represented as

These rotations are thus seen to transform the point (x, y) into(−y, x), (−x,−y), and (y,−x), respectively. The two reflectionsthat pass through the center of faces are

so they transform the (x, y) into (−x, y) and (x,−y), respectively.Finally, for the two reflection planes which pass through vertices,

11

which transform the point (x, y) into (y, x) and (−y,−x), respec-tively. These transformations enable us to construct the char-acters corresponding to the “coordinate” representation. Then,together with the identical and parity representations, we havethe following entries of the character table for this group:

E C2 2C4 2σv 2σ′vΓ1 1 1 1 1 1Γ′1 1 1 1 −1 −1Γ′′1 1Γ′′′1 1Γ2 2 −2 0 0 0

The group multiplication table and the orthogonality of columnsallows us to immediately complete the entries for the classes C2and 2C4:

E C2 2C4 2σv 2σ′vΓ1 1 1 1 1 1Γ′1 1 1 1 −1 −1Γ′′1 1 1 −1Γ′′′1 1 1 −1Γ2 2 −2 0 0 0

The remaining four entries can be determined from the orthog-onality of either rows or columns and again invoking the groupmultiplication table:

12

E C2 2C4 2σv 2σ′vΓ1 1 1 1 1 1Γ′1 1 1 1 −1 −1Γ′′1 1 1 −1 1 −1Γ′′′1 1 1 −1 −1 1Γ2 2 −2 0 0 0

(e) The eigenfunctions ϕ1,1(x, y) and ϕ2,2(x, y) are given by

ϕ1,1(x, y) ∝ cos(12πx) cos(1

2πy)

and

ϕ2,2(x, y) ∝ sin(πx) sin(πy) .

Since ϕ1,1(x, y) is invariant under the interchange of x and y andunder changes in their signs, it transforms according to the iden-tical representation. However, although ϕ2,2(x, y) is invariant un-der the interchange of x and y, each sine factor changes sign iftheir argument changes sign. Thus, this eigenfunction transformsaccording to the parity representation.

(f) The (degenerate) eigenfunctions ϕ1,2(x, y) and ϕ2,1(x, y) are(ϕ1,2

ϕ2,1

)∝[

cos(12πx) sin(πy)

sin(πx) cos(12πy)

].

The transformation properties of these eigenfunctions can be de-termined from the results of part (c). This yields the followingmatrix representation of each symmetry operation:

E =

(1 0

0 1

), C4 =

(0 1

−1 0

), C2

4 =

(−1 0

0 −1

),

C34 =

(0 −1

1 0

), σv,1 =

(1 0

0 −1

), σv,2 =

(−1 0

0 1

),

σ′v,1 =

(0 1

1 0

), σ′v,2 =

(0 −1

−1 0

).

13

This produces the following characters:

E = 2, C2 = −2, 2C4 = 0, 2σv = 0, 2σ′v = 0 ,

which are the characters of the two-dimensional irreducible repre-sentation Γ2, which is the “coordinate” irreducible representation.

(g) We have that the energies E6,7 and E2,9 are given by

E6,7 = E7,6 =h2π2

8m(62 + 72) = 85

h2π2

8m

and

E2,9 = E9,2 =h2π2

8m(22 + 92) = 85

h2π2

8m,

so this energy is fourfold degenerate. However, since the groupoperations have the effect of interchanging x and y with possiblechanges of sign, the eigenfunctions ϕ6,7 and ϕ7,6 are transformedonly between one another, and the eigenfunctions ϕ2,9 and ϕ9,2

are transformed only between one another. In other words, thisfourfold degeneracy is accidental , resulting only from the numer-ical coincidence of the energies of two twofold-degenerate states.

(h) We have already determined that ϕp,p with p even transformsaccording to the identical representation, while if p is odd, ϕp,ptransforms according to the parity representation. Moreover, thepair of eigenfunctions ϕp,q where p is even and q is odd transformsaccording to the coordinate representation.

Consider now the case where the eigenfunctions are of the formϕp,q where both p and q are even. The matrices corresponding to

14

the symmetry operations are

E =

(1 0

0 1

), C4 =

(0 1

1 0

), C2

4 =

(1 0

0 1

),

C34 =

(0 1

1 0

), σv,1 =

(1 0

0 1

), σv,2 =

(1 0

0 1

),

σ′v,1 =

(0 1

1 0

), σ′v,2 =

(0 1

1 0

).

The corresponding characters are

E = 2, C2 = 2, 2C4 = 0, 2σv = 2, 2σ′v = 0 .

This representation must be reducible, since its characters do notcorrespond to those of any of the irreducible representations inthe table determined in Part (d). A straightforward applicationof the Decomposition Theorem (or simple inspection) shows thatthis representation is the direct sum of the Γ1 and Γ′′1 irreduciblerepresentations. This means that there is a linear combination ofthese eigenfunctions that diagonalizes the matrices correspondingto each symmetry operation of this group.

For the eigenfunctions of the form ϕp,q where both p and q areodd, the matrices corresponding to the symmetry operations are

E =

(1 0

0 1

), C4 =

(0 −1

−1 0

), C2

4 =

(1 0

0 1

),

C34 =

(0 −1

−1 0

), σv,1 =

(−1 0

0 −1

), σv,2 =

(−1 0

0 −1

),

σ′v,1 =

(0 1

1 0

), σ′v,2 =

(0 1

1 0

).

The characters are now

E = 2, C2 = 2, 2C4 = 0, 2σv = −2, 2σ′v = 0 .

15

which correspond to a reducible representation composed of thedirect sum of the Γ′1 and Γ′′′1 irreducible representations. Thus,all of the irreducible representations occur in the eigenfunctionsof the two-dimensional square well.

8. The character table of the regular hexagon is reproduced below:

C6v E 2C6 2C3 C2 3σv 3σdΓ1 1 1 1 1 1 1Γ′1 1 1 1 1 −1 −1Γ′′1 1 −1 1 −1 1 −1Γ′′′1 1 −1 1 −1 −1 1Γ2 2 1 −1 −2 0 0Γ′2 2 −1 1 2 0 0

A transformation properties of a vector perturbation can be de-duced in a manner analogous to that for the equilateral trian-gle (Section 6.6.2). Applying each symmetry operation to r =(x, y, z) produces a reducible representation because these oper-ations are either rotations or reflections through vertical planes.Thus, the z axis is invariant under every symmetry operation ofthis group which. Together with the fact that an (x, y) basis gen-erates the two-dimensional irreducible representation Γ2 [Problem5(b), Problem Set 7], yields

Γ′ = Γ1 ⊕ Γ2 .

The corresponding characters are

C6v E 2C6 2C3 C2 3σv 3σdΓ1 ⊕ Γ2 3 2 0 −1 1 1

to determine the selection rule for an initial state that transformsaccording to the parity representation (Γ′1, we must calculated

Γ′1 ⊗ Γ′ = Γ′1 ⊗ (Γ1 ⊕ Γ2) .

The characters associated with this operation are

16

C6v E 2C6 2C3 C2 3σv 3σdΓ′1 ⊗ (Γ1 ⊕ Γ2) 3 2 0 −1 −1 −1

Finally, either by inspection, or by applying the decompositiontheorem, we find that

Γ′1 ⊗ (Γ1 ⊕ Γ2) = Γ′1 ⊕ Γ2 ,

so transitions between states that transform according to the par-ity representation and any states other than those that transformas the parity or coordinate representations are forbidden by sym-metry.

Group TheorySolutions to Problem Set 9 December 14, 2001

1. The Lie group GL(n,R) has n2 parameters, because the transfor-mations can be represented as n× n matrices (with real entries).The requirement that the Euclidean length dimensions be pre-served by such a transformation leads to the requirement that,

x′21 + x′22 + · · ·+ x′2n = x21 + x2

2 + · · ·+ x2n .

Proceeding as in Sec. 7.2, we note that there are n conditionsfrom the requirement that the coefficients of xi, i = 1, 2, . . . , nbe equal to unity. Then, there are 1

2n(n− 1) conditions from the

requirement that the coefficients of the unique products xixj, i 6=j vanish. Thus, beginning with n free parameters for GL(n,R),there are

n2 − n− 12n(n− 1) = n2 − n− 1

2n2 + 1

2n = 1

2n(n− 1)

free parameters for O(n).

2. Beginning with the three conditions

a211 + a2

21 = 1, a11a12 + a21a22 = 0, a212 + a2

22 = 1 ,

we take the product of the first by the third equations and sub-tract the square of the second equation to obtain

(a211 + a2

21)(a212 + a2

22)− (a11a12 + a21a22)2

= a211a

212 + a2

11a222 + a2

21a212 + a2

21a222 − a2

11a212

−2a11a12a21a22 − a221a

222

= a211a

222 + a2

21a212 − 2a11a12a21a22

= (a11a22 − a12a21)2

= 1 .

2

Thus, the three constraints for orthogonal groups in two dimen-sions imply that the square of the determinant of such transfor-mation must be equal to unity.

3. Forming the product of the the matrices corresponding to R(ϕ1)and R(ϕ2) yields

(cosϕ1 − sinϕ1

sinϕ1 cosϕ1

)(cosϕ2 − sinϕ2

sinϕ2 cosϕ2

)

=

(cosϕ1 cosϕ2 − sinϕ1 sinϕ2 − cosϕ1 sinϕ2 − sinϕ1 cosϕ2

sinϕ1 cosϕ2 + cosϕ1 sinϕ2 − sinϕ1 sinϕ2 + cosϕ1 cosϕ2

).

By invoking the standard trigonometric identities for the sinesand cosines of the sum and difference of two angles,

cos(x± y) = cosx cos y ∓ sin x sin y ,

sin(x± y) = sin x cos y ± cosx sin y ,

we can write(cosϕ1 − sinϕ1

sinϕ1 cosϕ1

)(cosϕ2 − sinϕ2

sinϕ2 cosϕ2

)

=

[cos(ϕ1 + ϕ2) − sin(ϕ1 + ϕ2)

sin(ϕ1 + ϕ2) cos(ϕ1 + ϕ2)

].

Thus,

R(ϕ1 + ϕ2) = R(ϕ1)R(ϕ2) .

3

4. The expression

R(ϕ) = eϕX ,

where

X =

(0 −1

1 0

),

is defined by its Taylor series:

eϕX =∞∑n=0

1

n!(ϕX)n . (1)

Successive powers of X yield

X =

(0 −1

1 0

), X2 =

(−1 0

0 −1

),

X3 =

(0 1

−1 0

), X4 =

(1 0

0 1

),

whereupon this sequence is repeated. We can write this sequencein matrix form as X2 = −I,X3 = −X,X4 = I, . . ., where I isthe 2× 2 unit matrix. The powers of X are therefore given by

X2n =

I, n even

−I, n odd

for even powers and

X2n+1 =

X, n even

−X, n odd

for odd powers Thus, the Taylor series in (1) may be written as

eϕX =∞∑n=0

(−1)n

(2n)!ϕ2n

︸︷︷︸cosϕ

I +∞∑n=0

(−1)n

(2n+ 1)!ϕ2n+1

︸︷︷︸sinϕ

X

4

= I cosϕ+X sinϕ

=

(cosϕ − sinϕ

sinϕ cosϕ

),

which is the rotation matrix in two dimensions.

5. The two parameter group

x′ = ax+ b

was discussed in Example 7.1. The identity was found to cor-respond to the parameters a = 1 and b = 0. The infinitesimaltransformations are therefore given by

x′ = (1 + da)x+ db = x+ x da+ db .

If we substitute this into some function f(x) and expand to firstorder in the parameters a and b, we obtain

f(x′) = f(x+ x da+ db) = f(x) + x∂f

∂xda+

∂f

∂xdb ,

from which we identify the infinitesimal operators

X1 = x∂

∂x, X2 =

∂

∂x.

6. The group C∞v contains all two-dimensional rotations and a ver-tical reflection plane, denoted by σv, in the x-z plane. Since thisreflection changes the parity of the coordinate system, it changesthe sense of the rotation angle ϕ. Thus, a rotation by ϕ in theoriginal coordinate system corresponds to a rotation by −ϕ in the

5

transformed coordinate system. Denoting the reflection operatorby S, we then must have that

SR(ϕ)S−1 = R(−ϕ) . (2)

Since S = S−1, we can see this explicitly for the two-dimensionalrotation matrix R(ϕ):(

1 0

0 −1

)(cosϕ − sinϕ

sinϕ cosϕ

)(1 0

0 −1

)=

(cosϕ sinϕ

− sinϕ cosϕ

),

where the matrix on the right-hand side of this equation is R(−ϕ).Equation (2) shows that (i) the group is no longer Abelian, and(ii) the equivalence classes correspond to rotations by ϕ and −ϕ.

7. Proceeding as in Section 7.4, the infinitesimal rotations in fourdimensions which leave the quantity x2 + y2 + z2 + w2 invariantare

x′

y′

z′

w′

=

1 ϕ1 ϕ2 ϕ3

−ϕ1 1 ϕ4 ϕ5

−ϕ2 −ϕ4 1 ϕ6

−ϕ3 −ϕ5 −ϕ6 1

x

y

z

w

.

Substituting this coordinate transformation into a differentiablefunction F (x, y, z, w),

F (x′, y′, z′, w′)

= F (x+ ϕ1y + ϕ2z + ϕ3w, y − ϕ1x− ϕ4z + ϕ5w,

z − ϕ2x+ ϕ4y + ϕ6w,w − ϕ3x− ϕ5y − ϕ6z) .

and expanding to first order in the ϕi, yields

F (x′, y′, z′, w′) = F (x, y, z, w) + ϕ1

(y∂F

∂x− x∂F

∂y

)

6

= ϕ2

(z∂F

∂x− x∂F

∂z

)+ ϕ3

(w∂F

∂x− x∂F

∂w

)

= ϕ4

(y∂F

∂z− z∂F

∂y

)+ ϕ5

(w∂F

∂y− y∂F

∂w

)

= ϕ6

(w∂F

∂z− z∂F

∂w

).

From these equations and, if necessary, a change in sign of thecorresponding ϕi, we can identify the following differential oper-ators

A1 = z∂

∂y− y ∂

∂z, A2 = x

∂

∂z− z ∂

∂x, A3 = y

∂

∂x− x ∂

∂y,

B1 = x∂

∂t− t ∂

∂x, B2 = y

∂

∂t− t ∂

∂y, B3 = z

∂

∂t− t ∂

∂z.

8. With the infinitesimal generators calculated in Problem 7, we de-termine the commutators in the standard fashion. For the com-mutators between the Ai, we have

[A1, A2]f = A1(A2f)− A2(A1f)

=(z∂

∂y− y ∂

∂z

)(x∂f

∂z− z∂f

∂x

)−(x∂

∂z− z ∂

∂x

)(z∂f

∂y− y∂f

∂z

)

= xz∂2f

∂y∂z− z2 ∂

2f

∂y∂x− xy∂

2f

∂z2+ y

∂f

∂x+ yz

∂2f

∂z∂x

−x∂f∂y− xz ∂

2f

∂z∂y+ xy

∂2f

∂z2+ z2 ∂

2f

∂x∂y+ yz

∂2f

∂x∂z

= y∂f

∂x−−x∂f

∂y

= A3f ,

7

[A1, A3]f = A1(A3f)− A3(A1f)

=(z∂

∂y− y ∂

∂z

)(y∂f

∂x− x∂f

∂y

)−(y∂

∂x− x ∂

∂y

)(z∂f

∂y− y∂f

∂z

)

= z∂f

∂x+ yz

∂2f

∂y∂z− xz∂

2f

∂y2− y2 ∂

2f

∂z∂x+ xy

∂2f

∂z∂y

−yz ∂2f

∂x∂y+ y2 ∂

2f

∂x∂z+ xz

∂2f

∂y2− x∂f

∂z− xy ∂

2f

∂y∂z

= z∂f

∂x− x∂f

∂z

= −A2f ,

[A2, A3]f = A2(A3f)− A3(A2f)

=(x∂

∂z− z ∂

∂x

)(y∂f

∂x− x∂f

∂y

)−(y∂

∂x− x ∂

∂y

)(x∂f

∂z− z∂f

∂x

)

= xy∂2f

∂z∂x− x2 ∂

2f

∂z∂y− yz∂

2f

∂x2+ z

∂f

∂y+ xz

∂2f

∂x∂y

−y∂f∂z− xy ∂

2f

∂x∂z+ yz

∂2f

∂x2+ x2 ∂

2f

∂y∂z− xz ∂

2f

∂y∂x

= z∂f

∂y− y∂f

∂z

= A1f .

Thus, we can summarize these results as

[Ai, Aj] = εijkAk .

Similarly, for the Bi, we calculate the pertinent commutators as

[B1, B2]f = B1(B2f)−B2(B1f)

8

=(x∂

∂w− w ∂

∂x

)(y∂f

∂w− w∂f

∂y

)−(y∂

∂w− w ∂

∂y

)(x∂f

∂w− w∂f

∂x

)

= xy∂2f

∂w2− x∂f

∂y− xw ∂2f

∂w∂y− yw ∂2f

∂x∂w+ w2 ∂

2f

∂x∂y

−xy ∂2f

∂w2+ y

∂f

∂x+ yw

∂2f

∂w∂x+ xw

∂2f

∂y∂t− w2 ∂

2f

∂y∂x

= y∂f

∂x− x∂f

∂y

= A3f ,

[B1, B3]f = B1(B3f)−B3(B1f)

=(x∂

∂w− w ∂

∂x

)(z∂f

∂w− w∂f

∂z

)−(z∂

∂w− w ∂

∂z

)(x∂f

∂w− w∂f

∂x

)

= xz∂2f

∂w2− x∂f

∂z− xw ∂2f

∂w∂z− zw ∂2f

∂x∂w+ t2

∂2f

∂x∂z

−xz ∂2f

∂w2+ z

∂f

∂x+ zw

∂2f

∂w∂x+ xt

∂2f

∂z∂w− w2 ∂

2f

∂z∂x

= z∂f

∂x− x∂f

∂z

= −A2f ,

[B2, B3]f = B2(B3f)−B3(B2f)

=(y∂

∂w− w ∂

∂y

)(z∂f

∂w− w∂f

∂z

)−(z∂

∂w− w ∂

∂z

)(y∂f

∂w− w∂f

∂y

)

= yz∂2f

∂w2− y∂f

∂z− yw ∂2f

∂w∂z− zw ∂2f

∂y∂w+ w2 ∂

2f

∂y∂z

9

−yz ∂2f

∂w2+ z

∂f

∂y+ zw

∂2f

∂w∂y+ yt

∂2f

∂z∂w− w2 ∂

2f

∂z∂y

= z∂f

∂y− y∂f

∂z

= A1f .

These results can be summarized as

[Bi, Bj] = εijkAk .

Finally, for the commutators between the Ai and Bj, we note firstby inspection that

[Ai, Bi] = 0 ,

for i = 1, 2, 3, since Ai and Bi involve mutually exclusive pairs ofvariables. For the remaining commutator pairs, we have

[A1, B2] = A1(B2f)−B2(A1f)

=(z∂

∂y− y ∂

∂z

)(y∂f

∂w− w∂f

∂y

)−(y∂

∂w− w ∂

∂y

)(z∂f

∂y− y∂f

∂z

)

= z∂f

∂w+ yz

∂2f

∂y∂w− zw∂

2f

∂y2− y2 ∂2f

∂z∂w+ yw

∂2f

∂z∂y

−yz ∂2f

∂w∂y+ y2 ∂2f

∂w∂z+ zw

∂2f

∂y2− w∂f

∂z− yw ∂2f

∂y∂z

= z∂f

∂w− w∂f

∂z

= B3f ,

[A1, B3] = A1(B3f)−B3(A1f)

=(z∂

∂y− y ∂

∂z

)(z∂f

∂w− w∂f

∂z

)−(z∂

∂w− w ∂

∂z

)(z∂f

∂y− y∂f

∂z

)

10

= z2 ∂2f

∂y∂w− zw ∂2f

∂y∂z− y ∂f

∂w− yz ∂2f

∂z∂w+ yw

∂2

∂z2

−z2 ∂2f

∂w∂y+ yz

∂2f

∂w∂z+ w

∂f

∂y+ zw

∂2f

∂z∂y− yw∂

2f

∂z2

= w∂f

∂y− y ∂f

∂w

= −B2f ,

[A2, B3] = A2(B3f)−B3(A2f)

=(x∂

∂z− z ∂

∂x

)(z∂f

∂w− w∂f

∂z

)−(z∂

∂w− w ∂

∂z

)(x∂f

∂z− z∂f

∂x

)

= x∂f

∂w+ xz

∂2f

∂z∂w− xw∂

2f

∂z2− z2 ∂2f

∂x∂w+ zw

∂2f

∂x∂z

−xz ∂2f

∂w∂z+ z2 ∂2f

∂w∂x− w∂f

∂x+ xw

∂2f

∂z2− zw ∂2f

∂x∂z

= x∂f

∂w− w∂f

∂x

= B1f .

Thus,

[Ai, Bj] = εijkBk .

9. Consider the following linear combinations of the operators inProblem 7:

Ji = 12(Ai +Bi), Ki = 1

2(Ai −Bi) . (3)

11

We can now use the commutation relations derived in Problem 8to derive the commutation relations for the Ji and Kj. For theJi, we have

[Ji, Jj] = 14[Ai +Bi, Aj +Bj]

= 14

([Ai, Aj] + [Ai, Bj] + [Bi, Aj] + [Bi, Bj]

)= 1

4(εijkAk + εijkBk + εijkBk + εijkAk)

= εijk12(Ak +Bk)

= εijkJk ,

[Ki, Kj] = 14[Ai −Bi, Aj −Bj]

= 14

([Ai, Aj]− [Ai, Bj]− [Bi, Aj] + [Bi, Bj]

)= 1

4(εijkAk − εijkBk − εijkBk + εijkAk)

= εijk12(Ak −Bk)

= εijkKk ,

[Ji, Kj] = 14[Ai +Bi, Aj −Bj]

= 14

([Ai, Aj]− [Ai, Bj] + [Bi, Aj]− [Bi, Bj]

)= 1

4(εijkAk + εijkBk − εijkBk − εijkAk)

= 0 .

Group TheorySolutions to Problem Set 10 December 14, 2001

1. As shown in Section 8.3.1, the eigenvalues of an orthogonal ma-trix have modulus unity. These eigenvalues are also the rootsof the polynomial equation det(A − λI) = 0, so the Fundamen-tal Theorem of Algebra requires that, if these roots are com-plex, they must occur in complex conjugate pairs. Thus, only inan odd-dimensional space is there guaranteed to be a single realeigenvalue of unity. The corresponding eigenvector is the axis ofrotation.

2. If the fixed point is taken as the origin of the set of axes of thebody, then the displacement of the rigid body involves no trans-lation, but only a change of orientation, i.e., a rotation. Since, inthree dimensions, every rotation can be expressed in an axis-anglerepresentation, Euler’s theorem follows immediately.

3. The general improper transformation in two dimensions is(x′

y′

)=

(cosϕ sinϕ

sinϕ − cosϕ

)xy

.

Thus, for the functions (x± iy)m we have

(x′ ± iy′)m =[x cosϕ+ y sinϕ± i(x sinϕ− y cosϕ)

]m=[x( cosϕ± i sinϕ)∓ iy( cosϕ± i sinϕ)

]m= (x∓ iy)m e±imϕ ,

so they generate the representation[(x′ + iy′)m

(x′ − iy′)m

]=

(0 eimϕ

e−imϕ 0

)[(x+ iy)m

(x− iy)m

].

2

To determine whether or not this representation is reducible, weapply Schur’s first lemma. Suppose a matrix A commutes withall of the matrices of our two dimensional representation. Then,we have(a11 a12

a21 a22

)(0 eimϕ

e−imϕ 0

)︸︷︷︸ a12e−imϕ a11eimϕ

a22e−imϕ a21eimϕ

=

(0 eimϕ

e−imϕ 0

)(a11 a12

a21 a22

)︸︷︷︸ a21eimϕ a22eimϕ

a11e−imϕ a12e−imϕ

.

Thus, if m 6= 0, we must require that a12 = a21 = 0 and thata11 = a22, i.e., A is multiple of the 2×2 unit matrix and, accordingto Schur’s first lemma, this representation is irreducible. However,of m = 0, we need only require that a12 = a21 and a11 = a22, sothis is a reducible representation.

4. The rotation angle ϕ is calculated from the trace of the transfor-mation matrix:

1 + 2 cosϕ = cosψ cosφ− cos θ sinφ sinψ − sinψ sinφ

+ cos θ cosφ cosψ + cos θ

= (1 + cos θ)( cosφ cosψ − sinφ sinψ) + cos θ

= (1 + cos θ) cos(φ+ ψ) + cos θ .

Using the triginometric identity

1 + 2 cosϕ = 4 cos2 (12ϕ)− 1 ,

we obtain

4 cos2 (12ϕ) = (1 + cos θ)[1 + cos(φ+ ψ)]

= 4 cos2 (12θ) cos2 [1

2(φ+ ψ)] ,

3

or,

cos (12ϕ) = cos (1

2θ) cos [1

2(φ+ ψ)] .

5. The axis of the transformation in Problem 4 is determined fromthe equations derived in Section 8.3.2:

n2

n1

=a31 − a13

a23 − a32

,n3

n1

=a12 − a21

a23 − a32

.

The denominator of these expressions is

a23 − a32 = sin θ cosψ + sin θ cosφ = sin θ( cosψ + cosφ) .

We also have

a31 − a13 = sin θ sinφ− sin θ sinψ = sin θ(sinφ− sinψ)

a12 − a21 = cosψ sinφ+ cos θ cosφ sinψ

+ sinψ cosφ+ cos θ sinφ cosψ

= (1 + cos θ)(cosφ sinψ + sinφ cosψ)

= (1 + cos θ) sin(φ+ ψ) .

Thus, the (unnormalized) direction of the rotation axis is1,

sinφ− sinψ

cosψ + cosφ, 2

(1 + cos θ) sin(φ+ ψ)

sin θ(cosφ+ cosψ)

.

6. There are a number of ways of decomposing the direct productof irreducible representations of SO(3). The books by Tinkham

4

and Jones give two very different approaches. Below, we providea third method. We first calculate the direct product

χ(`)(ϕ)χ(1)(ϕ) =

( ∑m=−`

e−imϕ)(

1∑m1=−1

e−im1ϕ

).

By expanding the second summation and multiplying the firstsummation with each of the exponentials, we obtain( ∑

m=−`e−imϕ

)(1∑

m1=−1

e−im1ϕ

)=

∑m=−`

e−imϕ(eiϕ + 1 + e−iϕ

)

=∑m=−`

e−i(m−1)ϕ +∑m=−`

e−imϕ +∑m=−`

e−i(m+1)ϕ .

If, in the first summation on the right-hand side of this equation,we change the summation variable to m′ = m− 1 and in the lastsummation change the summation variable to m′ = m + 1, wehave ∑

m=−`e−i(m−1)ϕ +

∑m=−`

e−i(m+1)ϕ

=`−1∑

m′=−`−1

e−im′ϕ +

`+1∑m′=−`+1

e−im′ϕ

=`+1∑

m′=−(`+1)

e−im′ϕ +

`−1∑m′=−(`−1)

e−im′ϕ .

In fact, for any positive integer k, we have

∑m=−`

e−i(m−k)ϕ +∑m=−`

e−i(m+k)ϕ

=`−k∑

m′=−`−ke−im

′ϕ +`+k∑

m′=−`+ke−im

′ϕ

=`+k∑

m′=−(`+k)

e−im′ϕ +

`−k∑m′=−(`−k)

e−im′ϕ . (1)

5

Thus, we conclude that

χ(`)(ϕ)χ(1)(ϕ) = χ(`−1)(ϕ) + χ(`)(ϕ) + χ(`+1)(ϕ) .

Then, by using (1), we have, in the general case

χ(`)(ϕ)χ(`′)(ϕ) =∑m=−`

e−imϕ[ei`′ϕ + ei(`

′−1)ϕ + · · ·+ e−i`′ϕ]

= χ(`+`′)(ϕ) + χ(`+`′−1)(ϕ) + · · ·+ χ(`−`′) .

Therefore,

χ(`)(ϕ)χ`′(ϕ) =

`+`′∑m=`−`′

χ(m)(ϕ) ,

where, from our procedure, it is clear that `′ ≤ `.

7. The corresponding Clebsch–Gordan series for SO(2) is very simplebecause the group is Abelian. Since

χ(m)(ϕ) = eimϕ ,

then

χ(m1)(ϕ)χ(m2)(ϕ) = χ(m1+m2)(ϕ) .

8. Given the complex transformation

(x′

y′

)=

(a b

c d

)(x

y

), (2)

6

then the invariance of the quantity xx∗ + yy∗ yields

x′x′∗ + y′y′∗

= (ax+ by)(a∗x∗ + b∗y∗) + (cx+ dy)(c∗x∗ + d∗y∗)

= (aa∗ + cc∗)xx∗ + (ab∗ + cd∗)xy∗ + (a∗b+ c∗d)x∗y

+ (cc∗ + dd∗)yy∗ .

Maintaining equality for all independent variations of x and yrequires that

aa∗ + cc∗ = 1, ab∗ + cd∗ = 0, cc∗ + dd∗ = 1 . (3)

A fourth condition is that the determinant of the transformationin (2) is unity:

ad− bc = 1 (4)

If we take the second of equations (3), multiply by a∗, and thenuse the first of these equations and Equation (4), we obtain

a∗(ab∗ + cd∗) = aa∗b∗ + a∗cd∗

= (1− cc∗)b∗ + (1 + b∗c∗)c

= b∗ + c = 0

which yields

c = −b∗

The second of equations (4) then immediately yields

a = d∗ (5)

Thus, the transformation (2) must have the form(x′

y′

)=

(a b

−b∗ a∗

)(x

y

)

where aa∗ + bb∗ = 1.

(2001) d.d.vvedensky - group theory (lecture note)

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