Kusse B.R., Westwig E.A. Mathematical Physics: Applied Mathematics for Scientists and Engineers

616

INTRODUCTION

TO

GROUP

THEORY

set would

be

by trial and error.

A

more educational method is to derive it using the

orthogonality relations of the irreducible representations, a topic we explore later in

this

chapter.

Example

15.4

As

another example

of

representation reduction, consider the

03

group. We showed a

2

X

2

matrix representation for

this

group

in

Table 15.10.

An

attempt to diagonalize

this

representation

in

the same way we did for the previous

example fails. Therefore,

it

is

an

irreducible representation, which we will call

DY].

As

always,

Dyl

is the trivial one-dimensional irreducible representation.

A

3

X

3

matrix representation for

this

group can

be

constructed from the operator

representation developed

in

Table 15.12. For example, take the [231] operator.

It

generates the sequence

(ha)

from the sequence

(ah).

This

operation can also be

accomplished by a 3

X

3

matrix

(15.40)

This matrix is a valid representation for the

g3

element

of

D3.

The matrices for

all

the other group elements are constructed

in

a

similar manner. The result is shown in

Table 15.21 and we will refer

to

this

as

the

DY]

representation. Now we ask, is

this

representation reducible?

We can attempt to reduce

this

matrix by a similar process to the one used earlier.

We will diagonalize one of the group elements, and then check the other elements to

see if the same transformation puts them into a reduced form. We have no guarantee

that

this

will work

in

this

case, because there

is

the possibility that none of the

elements (besides the identity) are diagonal in the reduced representation. Later

in

the text, we will revisit

this

problem, using a more reliable technique based on the

orthogonality and completeness properties of the irreducible representations.

We have a small hint from looking back at the

2

X

2

Drl

representation. Its

gg

element is diagonalized. Since the

OF1

matrices are an irreducible representation, it

would make sense to attempt diagonalization of the same element

in

the 3

X

3 DY1

representation, and

see

what happens. Luckily,

this

process actually works, and the

transformed version of the set of

dtl

matrices is given in Table 15.22. Notice this is

just

the direct

sum

DF1

@

Df].

TABLE

15.21.

The

3

X

3

0,"'

Representation

I

g2

g3

g4

gs

g6

100

001

010

100

010

[x

t

:I

[:

:

xl

[;

x

tl

[:

:

tl

1:

x

:I

[8

x

bl

617

IRREDUCIBLE

MATRIX

REPRESENTATIONS

TABLE

15.22.

Block

Diagonalized

of

the

Dfl

Representation

g2

g3

I

g4

We already determined that

03

has

three

different classes. This means that there

should be three irreducible representations.

So

far, only two have been identified,

DY1

and

DY1.

In the next section, we determine the remaining irreducible representation

DY1,

using the orthogonality

of

the irreducible representations.

15.4.4

Orthogonality

of

Irreducible

Representations

There are several important rules that the irreducible matrix representations

obey:

1.

The number of irreducible matrix representations

for

a group is equal to the

2.

If is the dimension of the first irreducible matrix representation,

ni2]

the

number of different classes

of

the group.

dimension

of

the second, and

so

on, then

where the summation

is

over all the irreducible representations

and

h

is

the

order

of

the group.

3.

“Vectors” formed by the elements of the irreducible representation matrices are

orthogonal. In this context, a vector is formed by combining the elements

of

a representation in a very particular way. The components of one such vector,

for example, are formed by all the

1-2

matrix elements

of

the six

DY1

matrices.

618

INTRODUCTION TO GROUP THEORY

The components of another are formed by all the

2-2

elements of the

six

Dyl

matrices, etc.

4.

The magnitude squared of each of the “vectors” described in rule

#3

is equal to

h/n[‘l

where

dil

is

the dimension of the matrix representation

used

to construct

the vector.

When checking orthogonality or checking the magnitude

of

the vectors described

in rules

#3

and

#4,

be

aware that,

if

any of the components are complex, one set

of the vector components should

be

complex conjugated before performing any dot

products.

In both the examples of the previous section, we found there

was

one missing

irreducible representation. Instead of using guesswork to find the last representation,

it

is

instructive to use the above rules to determine it.

Example

15.5

For the

C4

cyclic group, three irreducible representations were iden-

tified:

C?],

CY],

and

Ci3].

There are four different classes in

this

group, and according

to rule

#

1, there must be another irreducible representation which we will call

Cfl.

First let’s determine the dimension of the

Ci4]

matrices. According to rule #2:

(#1)2

+

(n121)2

+

(n[31)2

+

(#J)2

=

4

1

+

1

+

1

+

(n[41)2

=

4

(15.42)

(15.43)

or

n~41

=

1.

(15.44)

So

the final irreducible representation for the

C4

group

is

composed of four 1

X

1

matrices. The identity element must be

1,

and according to the multiplication table

for

~4,

g3

is its

own

inverse,

so

grl

=

-t

1.

MSO,

g4

is the inverse of

g2,

so

if we let

grl

=

g,

then

gyl

=

l/g.

Notice we have given

a

the flexibility of being complex.

This information leads

us

to the partially specified Table 15.23.

Now

we just need

to

determine the value

of

g

and the sign of

gf].

A

table of

all

of the four irreducible representations for

C,

is shown in Table 15.24.

Notice all the representations are

1

X

1

matrices. Incidentally,

this

turns out

to

be a

general property of any Abelian group. Following the prescription described in rule

#3,

we form four, four-dimensional vectors, which we call

V

,

V

,

V

,

and

Vf4]:

-111

-121

-[31

TABLE

15.23.

The

Form

of

the

Missing

Irreducible

Representation

of

C,

I

‘I

J

IRREDUCIBLE MATRIX REPRESENTATIONS

619

TABLE 15.24.

Partially

Speci6ed

Table

of

the

Irreducible Representations

of

C4

I

-1

1

-

a

21

1

/c

1

-i

-1

i

TABLE

15.25.

The Irreducible Representations

for

C,

Ci*]

-1

q31

1

-i

-1

I

Cf'

1

-1

1

-1

-[41

.

According to the rule, these vectors must be orthogonal. If

V

is

orthogonal to

V1l1

with

grl

=

-

1,

we find that

g

=

+i.

Either choice for the sign of

g

does not lead

us

to a representation of

C,

that we have not seen before. But

if

instead we pick

grJ

=

+

1,

we obtain a new irreducible representation with

g

=

-

1.

Now

our set

of

irreducible representations is complete.

They

are summarized in Table

15.25.

According to rule

#4,

the magnitudes of

V

,

V

,

V

,

and

VL4]

should all be

equal to

4.

Taking the dot product of each vector with itself, remembering to take the

complex conjugate of one of the terms, shows

this

is indeed

true.

Example

15.6

The same procedure used above can

be

repeated for the

03

group.

This group is of order six, and has three classes. We have already identified two

inequivalent irreducible representations:

D:']

and

0k3].

That leaves a third, the

DYJ

representation, for

us

to find.

-111

-[Zl

331

Using rule

#2,

the dimension

of

DrJ

is given

by

(15.46)

(15.47)

620

INTRODUCTION

TO

GROUP

THEORY

Df]

g1

g2 g3

g4

g5

g6

1

?1

21

’1

or

(nq

=

1.

(15.48)

So

the irreducible representation we

seek

is a set of six

1

X

1

matrices. According to

the multiplication table for

D3,

the

g4, g5

and

g6

elements are

d

their own inverses,

and are

also

in the same class.

This

lets us write

gi2]

=

gyl

=

grl

=

5

1.

Also

according

to

the multiplication table,

gz

g4

=

g5

and

g3 g4

=

g6,

so

we must have

gfl

=

grl

=

+

1.

These properties give us the partially specified representation in

Table 15.26.

If

we take the positive sign for all these terms, we just get a repeat

of

the trivial representation

0:’’.

Therefore we take the negative signs instead, and the

complete

set

of irreducible representations for

03

are listed in Table 15.27.

It is educational to check that the other group rules hold for these representations.

Rule

#3

specifies that we can form “vectors”

from

the elements of the matrices, and

that these vectors should all

be

orthogonal to one another. There will be

six

of these

vectors. The

1

X

1

Dill

representation forms the first vector:

The second vector comes from the

Dk2’

representation:

TABLE 15.27.

The Irreducible Representations

for

D3

(15.49)

I

g2

g3

g4

g5

g6

DY1

1

I

Di2]

1

1 1

-1

IRREDUCIBLE

MATRIX

REPRESENTATIONS

--

1

-1

.

-1

-

-1

-

1

-1/2

1

/2

1

/2

-1

621

(15.50)

-

1

-

0

-I-

0

--

-&/2

&2

-

1/2

&/2

-&/2 -&2 -1/2

.

&/2 &/2

-&/2 -1/2 (15.51)

-1/2

1

0

-

--

The

2

X

2 DY1

representation generates the last four:

03

gl

g2

g3

g4

g5

g6

xI1l

x121

xr3,

1

2

1

-1

1

-1

1

-1

0

1 -1

0

1

-1

0

Taking the dot product of all fifteen possible pairs of these vectors shows they are

indeed orthogonal. According to rule

#4,

the vectors which come

from

OF1

and

DY1

should have magnitude squared values of

6

and those

from

DY1

should have

3.

This

is also obviously true,

15.4.5

Character Tables and Character

Orthogonality

Once

all

the irreducible representations

are

known for a particular group, it is useful

to

construct a

character table.

This table lists the character of each group element for

each of the irreducible representations. A character table for the

03

group is shown

in Table

15.28.

Similar to the orthogonality rules presented in the previous section, the character

tables have orthogonality rules of their own.

In

this

case,

the

“vectors” formed by

the columns of these tables are orthogonal to

one

another, while their magnitudes

squared are equal to the order

of

the

group.

It is

easy

to check that these rules hold

for the

03

character table.

622

INTRODUCTION

TO GROUP THEORY

4

TABLE

15.29.

A

More

Compact Character Table

for

D3

x[ll

xr21

x[31

I

g29

g3

g4r

gS9

g6

1

2

1

1 -1

1

-1

0

Since elements in the same class always have the same character, the character

table is often written more concisely by putting

all

the elements in the same class in a

single row. The compact character table for

D3

is given in Table

15.29.

When forming

dot products between the columns of these character tables, you need to weight each

term by the number of elements

in

each class.

15.4.6

Completeness

of

the

Irreducible

Representations

The real power of the irreducible representations stems from the fact that, using

the proper

similarity

transformation,

any

matrix representation can be written

as

a

direct

sum

of one or more of the irreducible representations.

A

particular irreducible

representation

may

appear once, more than once, or not at all.

In

other words, with

some

[TI,

any group representations

Gral,

can

be

written

This

is often called the

decomposition

of

a representation.

There

is

a remarkable result of group theory which says that the number

of

times a

particular irreducible representation appears in the direct sum of Equation

15.52

can

be

determined just by

looking

at the character table. Take the vector formed by the

characters of the representation you want to decompose. Then form the dot product

of

this

vector with one formed from the character values of a particular irreducible

representation. Divide the value of

this

dot product by the order of the group, and

the

result will

be

equal to the number of times the particular irreducible representation

appears

in

the decomposition. Notice how the irreducible representations are playing

a role

analogous

to that of the basis vectors of a coordinate system. Because all matrix

representations can be expanded

this

way, the irreducible representations are said to

be

complete.

Example

15.7

In

a previous example, we determined that the

3

X

3

Dyl

matrix

representation in Table

15.21

could be transformed and written as the direct

sum

Dyl

@

Oh3].

We arrived at the result using the diagonalization process and a little

guesswork. Let's see how the decomposition rule described above gives

us

this result

with much less work. The character vector for the Table

15.21

representation

for

Drl

IRREDUCIBLE MATRIX REPRESENTATIONS

623

is

(15.53)

Now take the dot product of this vector with each of the three vectors of the

03

character table listed in Table

15.28,

and divide by

6.

The

net

result is that there is one

DY1

representation, one

DY1

representation, and

no

DY1 representation in the block

diagonalization. This is exactly the result we found before.

15.4.7

Shur's

Lemma

Many practical applications of matrix representations center around an important

group theory result known as

Shur's

lemma.

Imagine we have a group

G,

and

GLX1

is

one of its matrix representations:

(15.54)

Now suppose we have a matrix

[

v]

which is invariant under similarity transformations

by

ull

these matrix elements of

GEXl.

That

is,

gj"I-l[~g,fXI

=

[VI

for

i

=

1,2,

*

-

.h.

(15.55)

Shur's lemma says if

[V]

satisfies Equation

15.55,

and

GLxl

is an irreducible repre-

sentation, then

[

V]

must be a multiple of the identity matrix. That is,

[Vl

=

AVI,

(15.56)

where

A

is

some constant. If [V] satisfies the conditions of Equation

15.55,

it

is said

to

commute

with the elements of

Grxl.

This terminology

is

used because an equivalent

way of writing Equation

15.55

is

[Vlgjx]

=

gj"][V].

(15.57)

If

Grxl

is reducible, we cannot use Shur's lemma directly. But remember from

our

previous discussion, any reducible representation can be decomposed into the direct

sum of two or more irreducible representations by using the appropriate coordinate

transformation. For example, imagine

the

decomposition of

GIX1

looks like

G[']

0 0

[T]-'C["'[T]

=

[

Gr

231]

,

(15.58)

624

INTRODUCTION

TO

GROUP

THEORY

where

GI1], G['],

and

Gf31

are irreducible representations

of

dimension

1,

2,

and 4

respectively, and

[TI

is the transformation matrix that puts the matrices into the block

diagonal form.

If

[VJ

commutes with

all

the elements of

Grx1,

the transformed matrix

[V']

=

[T]-'[VJ[T]

wiU

commute with

all

the elements of the transformed

Grxl

representation. Then, by Shur's

lemma,

the

[VJ'

matrix must be diagonal, with the

form

[V']

=

[T]-'[vJ[T]

=

AlOOOOOO

OA~OOOOO

OOhzOOOO

000h~000

0000h300

0

0 0

h3

0

000000h3

There are three different constants ;-mg the diagonal

of

[V'],

i

(15.59)

lifferent one

for

each block of the decomposition of

GLXl.

This

works because Shur's lemma applies

separately to each irreducible block.

Example

15.8

We now present

an

example that shows how many

of

the ideas

of

group theory can be applied to a problem from classical mechanics. Consider the

two-dimensional system of balls and springs shown in Figure 15.9. Each

of

the

three

balls

has

mass

m,

and the

three

springs have the

same

spring constant

k.

What

are

the

normal frequencies of vibration for

this

system? Because

of

the symmetry, we should

expect group

theory

to

be

useful

in

solving

this

problem. We

begin

with

a

review

of

the standard method of determining the nod modes of such a system and then

show how group theory simplifies

the

process.

Since there are

three

masses and we are working in

two

dimensions, we need

a

total

of

six

coordinates to describe

an

arbitrary state of the system. We will use

the coordinates

41

and

42

to describe the displacement of the mass in the lower-left

k

Figure

15.9

Coupled

Mass

Spring

System

IRREDUCIBLE MATRIX REPRESENTATIONS

-

5/4 &/4

1

0

-1/4 -fi/4-

&/4 3/4

0 0

-&/4 -3/4

0

-&/4

3/4 &/4 -3/4

-1/4 -4/4

-1/4

&I4

1/2

0

1

0

5/4 -&/4 -1/4 &/4

[VI

=

--&/4

-3/4

&/4

-3/4

0

3/2

A

625

'

Figure.lS.10

Coordinates

for

the

Mass

Spring

Problem

To determine the equations

of

motion, we can use Newton's second law to write

(15.63)

Kusse B.R., Westwig E.A. Mathematical Physics: Applied Mathematics for Scientists and Engineers

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