group theory - lunds universitethome.thep.lu.se/~malin/fytn13/grouptheoryferdi19.pdfchapter 1...

Group Theory

Ferdi AryasetiawanLund University, Mathematical Physics

20 January 2019

Contents

1 Abstract Group Theory 11.1 Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Abelian Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 Multiplication Table . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.7 Order of an Element . . . . . . . . . . . . . . . . . . . . . . . . . 41.8 Properties of Finite Groups . . . . . . . . . . . . . . . . . . . . . 41.9 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.10 Class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.11 Isomorphism and Homomorphism . . . . . . . . . . . . . . . . . . 61.12 Invariant Subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . 61.13 Direct Product Group . . . . . . . . . . . . . . . . . . . . . . . . 71.14 Summary of Questions . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Theory of Group Representations 92.1 De�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Equivalent Representation . . . . . . . . . . . . . . . . . . . . . . 92.3 Reducible and Irreducible Representation . . . . . . . . . . . . . 102.4 Equivalent Unitary Representation . . . . . . . . . . . . . . . . . 102.5 Schur�s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.6 Orthogonality Theorem . . . . . . . . . . . . . . . . . . . . . . . 142.7 The Characters of a Representation . . . . . . . . . . . . . . . . . 162.8 The Regular Representation . . . . . . . . . . . . . . . . . . . . . 172.9 The Character Table . . . . . . . . . . . . . . . . . . . . . . . . . 202.10 Properties of the Character Table . . . . . . . . . . . . . . . . . . 202.11 Class Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . 212.12 Direct Product Groups . . . . . . . . . . . . . . . . . . . . . . . . 222.13 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.14 Summary of Questions . . . . . . . . . . . . . . . . . . . . . . . . 26

iii

iv CONTENTS

3 Group Theory in Quantum Mechanics 293.1 Linear Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . 293.2 Symmetry Transformations . . . . . . . . . . . . . . . . . . . . . 313.3 Invariant Subspace and Its Generation . . . . . . . . . . . . . . . 323.4 Basis Functions for a Representation . . . . . . . . . . . . . . . . 323.5 Projection Operators . . . . . . . . . . . . . . . . . . . . . . . . . 343.6 Orthogonality of Basis Functions . . . . . . . . . . . . . . . . . . 363.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.8 Relationship to Quantum Mechanics . . . . . . . . . . . . . . . . 42

3.8.1 The Group of the Schrödinger Equation . . . . . . . . . . 423.8.2 Degeneracy and Invariant Subspace . . . . . . . . . . . . 433.8.3 Partial Diagonalisation of H . . . . . . . . . . . . . . . . 44

3.9 Irreducible Sets of Operators . . . . . . . . . . . . . . . . . . . . 443.10 Direct Product Representations of a Group . . . . . . . . . . . . 453.11 Clebsch-Gordan Coe¢ cients . . . . . . . . . . . . . . . . . . . . . 473.12 The Wigner-Eckart Theorem . . . . . . . . . . . . . . . . . . . . 483.13 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.14 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.15 Summary of Questions . . . . . . . . . . . . . . . . . . . . . . . . 55

4 Permutation Groups and Young Tableaux 594.1 Permutation Group Sn . . . . . . . . . . . . . . . . . . . . . . . . 594.2 Cayley�s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 614.3 Classes of Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.4 Young Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.4.1 De�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . 634.5 Young tabloids as basis for representations of Sn . . . . . . . . . 644.6 Specht Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4.6.1 The Specht modules are irreducible . . . . . . . . . . . . . 674.6.2 The Specht modules are distinct . . . . . . . . . . . . . . 714.6.3 Example: The irreducible representations of the permu-

tation group S4 . . . . . . . . . . . . . . . . . . . . . . . . 724.7 Summary of Questions . . . . . . . . . . . . . . . . . . . . . . . . 75

Chapter 1

Abstract Group Theory

1.1 Group

A group is a set of elements that have the following properties:

1. Closure: if a and b are members of the group, c = ab is also a member ofthe group.

2. Associativity: (ab)c = a(bc) for all a; b; c in the group.

3. Unit element: there is an element e such that ea = a for every element ain the group.

4. Inverse: to every element a there is a corresponding inverse element a�1

in the group such that a�1a = e.

The following properties follow from the above de�nition:

1. Left cancellation: If ax = ay then x = y for all a in the group.Proof: ax = ay ! a�1(ax) = a�1(ay) ! (a�1a)x = (a�1a)y ! ex =ey ! x = y.

2. Unit element on the right: ae = a = ea.Proof: a�1(ae) = (a�1a)e = ee = e = a�1a and using the left cancellationlaw we have ae = a.

3. Inverse element on the right: aa�1 = e = a�1a.Proof: a�1(aa�1) = (a�1a)a�1 = ea�1 = a�1 = a�1e. Using the leftcancellation law, aa�1 = e.

4. Right cancellation: If xa = ya then x = y for all a in the group.Proof: xa = ya ! (xa)a�1 = (ya)a�1 ! x(aa�1) = y(aa�1) ! xe =ye! x = y.

1

2 CHAPTER 1. ABSTRACT GROUP THEORY

We note the importance of associativity in the above proofs.The following identity is often useful:

(ab)�1 = b�1a�1

which follows from (ab)�1(ab) = 1! (ab)�1a = b�1 ! (ab)�1 = b�1a�1.

1.2 Abelian Group

If a group has a further property that ab = ba for all a; b in the group, the groupis called Abelian.

1.3 Subgroup

A subgroup is a set of elements within a group which forms a group by itself.Evidently, the unit element forms a subgroup by itself.

1.4 Examples

1. Integers under addition. The unit element e = 0 and the inverse of anelement a is a�1 = �a. This group is Abelian and in�nite.

2. A set of all n � n unitary matrices U under matrix multiplication. Theunit element is the unit matrix and the inverse of U is Uy by de�nition.We have to show that a product of two unitary matrices is unitary:

(U1U2)y = Uy2U

y1 = U�12 U�11 = (U1U2)

�1:

3. The set of permutation operations that take ABC into ABC, BCA, CAB,ACB, CBA, and BAC. The elements of the group are

e =

�A B CA B C

�� =

�A B CB C A

�� =

�A B CC A B

�

=

�A B CA C B

�� =

�A B CC B A

�� =

�A B CB A C

�The operation �� means: �rst do permutation � and then permutation� on the previous result. We show below that every permutation hasan inverse permutation and two successive permutations correspond to asingle permutation i.e. the permutations form a group.

We see from the above examples that "multiplication" can mean addition,matrix multiplication etc. or simply that one operation is performed on theresult of the preceeding operation like in the example 3.

1.5. MULTIPLICATION TABLE 3

1.5 Multiplication Table

The group multiplication table is a matrixMab = ab, where a and b are elementsof the group and the matrix element corresponding to row a and column b isgiven by the group multiplication ab. We illustrate this de�nition by construct-ing the multiplication table of the permutation group in the example 3 above.

e � � � �e e � � � �� e � � � � e � � � � � e � �� e �� e

Thus, for example, the inverse of � is � and vice versa. The group is clearlynot Abelian because e.g. � 6= �. The elements (e; ) form a subgroup, so do(e; �), (e; �) and (e; �; �). Obviously, the unit element e forms a subgroup byitself.The group multiplication table mathematically characterises the group. All

groups with the same multiplication table are mathematically identical withrespect to their group properties. The multiplication table has the followingproperties:

1. Rearrangement Theorem: In every row or column, each element mustappear once and once only and therefore each row (column) is di¤erentfrom any other row (column).Proof: Suppose element b appears twice in the row a. This means thatac = b and ad = b where c 6= d. But this implies that ac = ad ! c = dwhich is a contradiction. Thus element b cannot appear more than onceand since the size of the row or column is the same as the number ofelements in the group, it follows that each element must appear once andonce only.

2. The multiplication table is symmetric across the diagonal when the groupis Abelian because the elements commute with one another.

1.6 Cyclic Groups

A group of n elements is said to be cyclic if it can be generated from one element.The elements of the group must be

a; a2; a3; : : : ; an = e


n is called the order of the cyclic group. A cyclic group is evidently Abelianbut an Abelian group is not necessarily cyclic. It is shown below that everynon-cyclic group has at least a cyclic subgroup.Examples of cyclic groups are the subgroups of the permutation group in

the example 3. The subgroup (e; �; �) is the same as (�; �2 = �; �3 = e).

1.7 Order of an Element

Let a 6= e be an element of a group. Form the products a2; a3; : : :. a2 must beeither e or a di¤erent element from a because if a2 = a ! a = e. If a2 6= e wecontinue forming a3. By a similar argument, a3 must be either e or a di¤erentelement from a and a2. If a; a2; : : : ; an are distinct from each other and an = ethen n is called the order of element a. These elements form a cyclic group.Thus every group must have at least one cyclic subgroup. In the example 3above, � and � are of order 3 and , �, and � are of order 2.

1.8 Properties of Finite Groups

We summarise below the properties of �nite groups.

1. Every element a has a �nite order n such that an = e.

2. Rearrangement Theorem: Multiplying all elements in a group by an ar-bitrary element reproduces the group. This has been proven above in theproperties of the multiplication table.

3. Every non-cyclic group has at least a cyclic subgroup.

1.9 Cosets

If S is a subgroup of G and a is an element of G not in S, then the sets aSand Sa are called the left and right cosets of S respectively. a cannot be aunit element and therefore a coset can never be a group because it has no unitelement. It is also evident that aS or Sa has no element in common with S forotherwise, a should be included in S. Moreover, if b is an element of G which isneither in S nor in aS, then bS has no element in common with either S or aS.Proof: Let x and y be in S such that ax = by. We have axy�1 = b, but xy�1

is an element in S and therefore b is in aS which is a contradiction.Consequently, a �nite group can be factorised as

G = S + aS + bS + : : :

but the factorisation is not unique. It depends on the choice of S, a, b, : : :, etc.The number of factors must be �nite and the factorisation must exhaust thegroup. Thus we have the following theorem:

1.10. CLASS 5

Let h be the number of elements of a group G and g be the number of elementsof a subgroup S of G. Then

h

g= integer

As a consequence, if h is prime then G must be cyclic and it has no subgroupother than the trivial subgroup e.

1.10 Class

Two elements a and b of a group are conjugate to one another if there is anelement g in the group such that

a = gbg�1 (1.1)

(Since every element has an inverse, it also follows that b = g0ag0�1 where

g0 = g�1). The transformation gbg�1 is often called a similarity transformation.If a and b are conjugate to each other and b and c are conjugate to each other,then a and c are also conjugate to each other. This is because if a = g1bg

�11

and b = g2cg�12 then a = g1g2cg

�12 g�11 = g1g2c(g1g2)

�1.A class C is a set of elements which are conjugate to each other. The unit

element evidently forms a class by itself. If G is Abelian, each element alsoforms a class by itself. It is clear that a group may be broken up into classes

G = C1 + C2 + : : :

and an element cannot belong to more than one class.As an example, we work out one of the classes of the permutation group.

In working out the classes, it is obviously not necessary to consider similaritytransformations with either the unit element or the element itself. We use themultiplication table that we have constructed previously:

��1 = �� = �� = �

� �1= � = � = �

��1 = �� = �� = �

��1 = �� = � = �

Thus � and � belong to the same class. Similarly we can show that ( ; �; �)also form a class.Elements belonging to the same class usually have the same characteristic.

In the above example, � and � both correspond to cyclic permutations whereas , �, and � correspond to permutations with one member �xed.


1.11 Isomorphism and Homomorphism

Two groups are said to be isomorphic if they have the same multiplication table,by reordering the elements if necessary. If F is isomorphic to G then there is aone-to-one correspondence between the elements of F and G, fi ! gi, such thatif fifj = fk then gigj = gk. Two groups can only be isomorphic if they havethe same number of elements. As an example, the permutation group in theexample 3 is isomorphic to the symmetry operations that take an equilateraltriangle into itself. This isomorphism can be seen by labelling the corners of thetriangle with A;B;C.Homomorphism is similar to isomorphism except that the relationship is

many-to-one. Two groups G and G0 are said to be homomorphic if each elementof G can be associated with some elements of G0: a ! a0 = (a01; a

02; : : :), b !

b0 = (b01; b02; : : :), c! c0 = (c01; c

02; : : :), such that if ab = c then a0ib

0j = c0k i.e. for

every i; j, c0k lies in c0.

1.12 Invariant Subgroup

We prove the following theorem:

Theorem: If S is a subgroup of G and a is an element of G then S0 = aSa�1

is also a subgroup and it is isomorphic to S. We assume that the elements of Sare distinct.

Proof: We �rst show that the elements of S0 are distinct. Let asia�1 = asja�1

and i 6= j. Then by the cancellation law, si = sj , which is a contradiction.Thus the elements of S0 are distinct. We now show that if sisj = sk thens0is

0j = s0k. Proof: s

0is0j = asia

�1asja�1 = asisja

�1 = aska�1 = s0k. Thus S

and S0 are isomorphic with the same unit element e0 = e and the inverse of s0iis s0i

�1= (asia

�1)�1 = as�1i a�1 = (s�1i )0.If S0 is the same as S for every a in the group then the subgroup S is called

an invariant subgroup. This implies that the left coset of S is the same as itsright coset, i.e.

aS = Sa

If we break a group into its cosets,

G = S + aS + bS; : : :

with an invariant subgroup S, then these cosets form a group with the unitelement equal to S. Thus aSbS = abSS = abS = cS if ab = c. This group iscalled a factor group. It is an example of homomorphism where the invariantsubgroup S is associated with the elements of S and the coset aS is associatedwith the elements of aS.

1.13. DIRECT PRODUCT GROUP 7

1.13 Direct Product Group

Let F be a group with elements fi, i = 1; : : : ; hF , and G be a group withelements gi, i = 1; : : : ; hG, such that figj = gjfi for all i and j. The directproduct group F G is de�ned to be the set of all distinct elements figj . If Fand G have no common element, apart from the identity, then the order of thedirect product group will be hFhG.We show that F G is a group.

1. Closure: (figj)(fkgl) = (fifk)(gjgl) = frgs. Since fr is in F and gs is inG then by de�nition frgs is in F G.

2. Unit element is eF eG.

3. Inverse of figj is (figj)�1 = g�1j f�1i = f�1i g�1j .

4. Associativity: obvious.

We have the following theorem:

Theorem: The classes of the direct product group are given by the directproducts of the classes of the individual groups.

Proof: We label the elements of the product group aij = figj . Accordingto the de�nition of a class, we have

Cij = a�1rs aijars; for all r; s

= f�1r g�1s figjfrgs

= (f�1r fifr)(g�1s gjgs)

1.14 Summary of Questions

1. Rearrangement Theorem: Show that a row or column in a group multi-plication table is a rearrangement or permutation of the group elementsand each element appears once and once only. This also implies that eachrow or colum is unique.

2. What is a cyclic group? Is a cyclic group Abelian? Is an Abelian groupcyclic?

3. What is the order of a group element?

4. Cosets:

(a) If S is a subgroup of a group G show that all elements of aS or Sa,where a is a group element not in S, must be di¤erent from the elementsof S.

(b) Can aS or Sa be a subgroup?


(c) Show that bS, where b is not in S and aS, must contain elementsdi¤erent from those of S and aS. The same is true for the Sb.

(d) Hence show that if h is the order of G and g is the order of S thenh=g = integer.

5. Show that if the order of a group is a prime number then the group mustbe cyclic.

6. What is a class?

7. If a group of order h is Abelian, how many classes does it have?

8. Show that an element of a group cannot belong to more than one class.Hence show that a �nite group can always be decomposed wholly intoclasses.

9. What is isomorphism? Give some examples.

10. What is homomorphism? Give some examples.

11. What is an invariant subgroup?

12. What is a factor group?

13. What is a direct product group?

14. Show that the classes of a direct product group are given by the productsof the classes of the individual groups forming the product group.

Chapter 2

Theory of GroupRepresentations

2.1 De�nitions

A group fA;B;C; :::g may be represented by a set of square matrices

fT (A); T (B); T (C); : : :g:These matrices are said to form a representation for the group if they satisfythe same group multiplication rule:

AB = C ! T (A)T (B) = T (C) (2.1)

The identity element E is represented by a unit matrix and each matrix musthave an inverse. Some or all of the matrices may be the same. If the matricesare di¤erent, the representation is called faithful (isomorphic). The order of thematrices is called the dimension of the representation.

2.2 Equivalent Representation

If T is a representation thenT 0 = S�1TS (2.2)

with an arbitrary but non-singular S is also a representation because

T 0(A)T 0(B) = S�1T (A)SS�1T (B)S

= S�1T (A)T (B)S

= S�1T (AB)S

= T 0(AB)

T and T 0 are equivalent and there are an in�nite number of equivalent repre-sentations.

9

10 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS

2.3 Reducible and Irreducible Representation

A representation T is reducible if there exists a non-singular matrix S such thatthe equivalent representation T 0 = S�1TS has the form

T 0 =

0BBBB@T1

T2T3

::

1CCCCAfor all elements in the group. T1, T2, T3; : : : are square matrices and the rest ofthe elements are zero. The representation is irreducible if it cannot be reducedinto the above form.

2.4 Equivalent Unitary Representation

There are an in�nite number of equivalent representations but it is possible to�nd one which is unitary.Theorem 1:Given an arbitrary representation T , there is always an equivalent unitary rep-resentation (Maschke�s theorem).

Proof:Construct a Hermitian matrix

H =XG

T (G)T y(G)

According to matrix algebra, we can always diagonalise a Hermitian matrix bya unitary transformation

D = U�1HU

=XG

U�1T (G)T y(G)U

=XG

U�1T (G)UU�1T y(G)U

=XG

T 0(G)T 0y(G)

2.5. SCHUR�S LEMMA 11

U is made up of the eigenvectors of H and D is diagonal by de�nition and itsdiagonal elements are real and positive.

Djj =XG

Xk

T 0jk(G)T0ykj(G)

=XG

Xk

T 0jk(G)T0�jk(G)

=XG

Xk

��T 0jk(G)��2 � 0In fact, it can never be equal to zero because otherwise detT 0 = 0 and T 0 willhave no inverse. Since D is diagonal, we may de�ne a diagonal matrix D�1=2

with diagonal elements equal to D�1=2ii so that

1 = D�1=2XG

T 0(G)T 0y(G)D�1=2

Using the above result, the matrix T 00(G) = D�1=2T 0(G)D1=2 can be shown tobe unitary.

T 00(G)T 00y(G) = D�1=2T 0(G)D1=2[D�1=2

XG0

T 0(G0)T 0y(G0)D�1=2]

�D1=2T 0y(G)D�1=2

= D�1=2XG0

T 0(G)T 0(G0)T 0y(G0)T 0

y(G)D�1=2

= D�1=2XG0

T 0(G)T 0(G0)[T 0(G)T 0(G0)]yD�1=2

= D�1=2XG0

T 0(GG0)T 0y(GG0)D�1=2

= D�1=2XG00

T 0(G00)T 0y(G00)D�1=2

= 1

The second last step has been obtained by using the Rearrangement Theorem.Thus, given an arbitrary representation T , it is always possible to construct aunitary representation by forming

T 00 = D�1=2U�1TUD1=2 (2.3)

From now on we assume that we have a unitary representation.

2.5 Schur�s Lemma

Schur�s lemma is used in proving many of the theorems in group theory.Theorem 2 (Schur�s lemma):


Given that [M;T ] =MT � TM = 0 for all elements in the group, then(a) if T is irreducible, M = cI where c is a constant and I is a unit matrix.(b) if M 6= cI then T is reducible.

Proof: We �rst show that it is su¢ cient to prove the theorem for a HermitianM .

TM =MT

(TM)y = (MT )y

MyT y = T yMy

T (MyT y)T = T (T yMy)T

TMy =MyT

since T y = T�1 (unitary). Adding and substracting the �rst and the last equa-tions, we get

T (M +My) = (M +My)T

andT i(M �My) = i(M �My) T

Let H1 = M +My and H2 = i(M �My) so that [H1; T ] = [H2; T ] = 0. H1

and H2 are Hermitian and M = (H1 � iH2)=2. Thus, if the theorem is truefor a Hermitian matrix, it must be true for M . We can now assume M to beHermitian and perform a unitary transformation so that M becomes diagonal,D = U�1MU , and de�ne an equivalent representation T 0 = U�1TU . Then

T 0D = U�1TUU�1MU

= U�1TMU

= U�1MTU

= U�1MUU�1TU

= DT 0

We have to show that the diagonal elements of D are all identical if T , orequivalently T 0, is irreducible. Consider the ij; i 6= j; element,

T 0ijDjj = DiiT0ij

T 0ij(Djj �Dii) = 0 (2.4)

Let us order the diagonal elements of D such that Dii � Djj for i < j whichcan always be done by rearranging the columns of U . Suppose T 0 is an m�mmatrix and D11 = ::: = Dnn where n < m and the rest of the diagonal elementsare di¤erent from D11. Then it follows from Eq. (2.4) that T 0 must have theform

T 0 =

�X 00 Y

�

2.5. SCHUR�S LEMMA 13

where X is an n� n matrix, i.e. T 0 is reducible.Let us summarise the conclusions: if M = cI then obviously it commutes

with T whether it is reducible or irreducible. If M 6= cI then T must be re-ducible. Hence if T is irreducible then M must be equal to cI for otherwise Twould be reducible.

Theorem 3:Given T�M = MT � , where T� and T � are irreducible representations of di-mension l� and l� respectively and M is a rectangular l� � l� matrix, then(a) if l� 6= l� , M = 0 or(b) if l� = l� either M = 0 or jM j 6= 0. In the latter case, M will have aninverse so that T� and T � are equivalent.

Proof:

T�(G)M =MT �(G)

(T�(G)M)y = (MT �(G))y

MyT�y(G) = T �y(G)My

My[T�(G)]�1 = [T �(G)]�1My

MMy[T�(G)]�1 =M [T �(G)]�1My

MMyT�(G�1) =MT �(G�1)My

MMyT�(G�1) = T�(G�1)MMy

Since T� is irreducible, it follows from Schur�s lemma (Theorem 2) thatMMy =cI. Consider the case l� = l� .

jMMyj = jcIj; jM jjMyj = jjM jj2 = cl�

If c 6= 0, then jM j 6= 0 i.e. M�1 exists. It follows that T� and T � are equivalentsince

M�1(T�(G)M) =M�1(MT �(G)) = T �(G)

If c = 0, then MMy = 0. Let us look at the diagonal elements of MMy:Xk

MikMyki = 0X

k

MikM�ik = 0X

k

jMikj2 = 0

Since this is positive de�nite, each term must vanish i.e. M = 0. Consider nowthe case l� 6= l� ; l� > l� . We enlarge M into a l� � l� square matrix N with


the additional elements equal to zero. Then NNy = MMy = cI. But jN j = 0because one of its columns is zero.

NNy = cI ! jNNyj = cl� ! jN jjNyj = cl� ! c = 0

Hence NNy = 0 which implies N = 0. Since M is contained in N , then M = 0.

2.6 Orthogonality Theorem

We are now in a position to prove the Orthogonality Theorem which is centralin representation theory.

Theorem 4: Orthogonality Theorem

XG

T��ij (G)T�kl(G) = (h=l�)��ik�jl (2.5)

� and � label the irreducible representations, l� and l� are the dimensions ofthese irreducible representations, and h is the number of elements in the group.

Proof:De�ne a matrix

M =XG

T�(G)XT �(G�1)

where X is an arbitrary matrix and � 6= �. We want to show that this matrixsatis�es the postulates of Theorem 3. Consider

T�(A)M =XG

T�(A)T�(G)XT �(G�1)

=XG

T�(A)T�(G)XT �(G�1)T �(A�1)T �(A)

=XG

T�(AG)XT �(G�1A�1)T �(A)

=XG

T�(AG)XT �((AG)�1)T �(A)

=MT �(A)

We have again made use of the Rearrangement Theorem in the second last step.According to Theorem 3, M = 0 since we are considering the case of � 6= � sothat

Mij = 0 =XG

Xkl

T�ik(G)XklT�lj(G

�1)

2.6. ORTHOGONALITY THEOREM 15

Since X is arbitrary, we may set Xkl = �km�ln. ThenXG

T�im(G)T�nj(G

�1) =XG

T�im(G)T�nj

�1(G)

=XG

T�im(G)T�ynj (G)

=XG

T�im(G)T��jn (G)

= 0

When � = �, according to Schur�s lemma (Theorem 2)

Mij = c �ij =XG

Xkl

T�ik(G)XklT�lj (G

�1)

=XG

T�im(G)T�nj(G

�1)

Putting i = j and summing over i yields

c l� =XG

Xi

T�im(G)T�ni(G

�1)

=XG

T�nm(E)

= h�nm

Thus we have XG

T�im(G)T�nj(G

�1) = (h=l�)�nm�ij

Since T is unitary, XG

T�im(G)T��jn (G) = (h=l�)�nm�ij

Eqn. ( 2.5) has the form of a dot product with T�ij(G) as the G�th componentof a vector labelled by �, i and j, in the h dimensional space of the groupelements. This means that the number of distinct labels (�; i; j) cannot exceedh, i.e.

P� l

2� � h. It will be proven later that in factX

�

l2� = h

The result is very useful in working out the irreducible representations of agroup.


2.7 The Characters of a Representation

We have seen that a representation is not unique. We recall that if T is arepresentation, so is T 0 = S�1TS. Given T and T 0, how can we then tell thatthey are equivalent ? One way will be to �gure out if there is a matrix S suchthat T 0 = S�1TS, but this is a complicated procedure. What we are looking forare properties of a matrix which are invariant under a similarity transformation.The eigenvalues are one of them but they are too cumbersome to calculate. Asimpler quantity is the trace of a matrix or the sum of the diagonal elements:

��(G) = Tr T�(G) =Xi

T�ii (G) (2.6)

We show that the trace or character of a matrix is invariant under a similaritytransformation:

Tr T 0 =Xi

Xjk

S�1ij TjkSki

=Xjk

Xi

SkiS�1ij

!Tjk

=Xjk

�kjTjk

= Tr T

The characters of a representation � is a set of h numbers {��(G)}. It willbe shown later that if the representation is irreducible, its characters are unique.

Theorem 5:If elements A and B of a group belong to the same class, then the characters oftheir representations are the same.

Proof:Since A and B belong to the same class, there is an element C such thatA = C�1BC. Consequently,

Tr T (A) = Tr [T (C�1) T (B) T (C)]

= Tr [T (C) T (C�1) T (B)]

= Tr [T (E) T (B)]

= Tr T (B)

2.8. THE REGULAR REPRESENTATION 17

Theorem 6: First Orthogonality RelationXG

��(G)��(G) =Xk

��(Ck)��(Ck)Nk = h�� (2.7)

Ck labels a class with Nk elements.

Proof:Setting i = j and k = l in Eqn. ( 2.5) (Orthogonality Theorem) and summingover i and l we have X

G

��(G)��(G) = (h=l�)��Xil

�il�ilXk

��(Ck)��(Ck)Nk = h��

Eqn. ( 2.7) has the form of a weighted dot product with weight Nk in a spacewith dimension equal to the number of classes of the group. ��(Ck) is the k�thcomponent of vector �. It follows that the number of irreducible representationsmust be less than or equal to the number of classes.

2.8 The Regular Representation

We prove the following theorem:Theorem 7: X

�

l2� = h (2.8)

Proof:A proof of this theorem is provided by considering the so called regular repre-sentation. The regular representation is formed in the following way. We writethe multiplication table as follows:

e � � : : :e e��1 e��1 e �1 e

We form the regular representation of an element a by writing 1 wherever the


element a occurs in the table and zero otherwise. Let the group elements beai; i = 1; 2; : : : ; h, then the regular representation is given by

Tjk(ai) = 1 if a�1j ak = ai

= 0 otherwise

It is clear from the de�nition that the representation is faithful. To show thatit is really a representation, we have to prove thatX

k

Tjk(ai)Tkl(am) = Tjl(ap)

if and only if aiam = ap. By de�nition, Tjk(ai) 6= 0 only when k is such thatak = ajai and similarly Tkl(am) 6= 0 only when k is such that ak = ala

�1m .

ThereforeP

k Tjk(ai)Tkl(am) = 1 if and only if ak = ajai = ala�1m and zero

otherwise. However, ajai = ala�1m implies that a�1j al = aiam = ap.

The character is given by

�(ai) =

hXj=1

Tjj(ai) = h if ai = e

= 0 otherwise

This is clear by inspection of the multiplication table. We can now prove The-orem 7 by writing the character of the regular representation as a sum overcharacters of the irreducible representations:

�(ai) =X�

m��(ai)

From the orthogonality relation we have

m� =1

h

Xi

��(ai)�(ai)

Since ��(e) = l�, �(e) = h, and �(ai 6= e) = 0 it follows that m� = l�. Eachirreducible representation occurs in the regular representation a number of timesequal to the dimension of the irreducible representation. On the other hand,h =

P�m�l� so that

h =X�

l2�

This result together with the orthogonality theorem show that there are exactlyh orthogonal vectors T�ij .

Theorem 8: Second Orthogonality RelationX�

��(Ck)��(Cl) = (h=Nk)�kl (2.9)

2.8. THE REGULAR REPRESENTATION 19

Proof:We recall that the matrix representation T�ij can be regarded as a set of or-thogonal vectors in the h- dimensional space of the group elements. T�ij(G) isthe component of the vector T�ij along the "G-axis". T

�ij themselves can be re-

garded as a basis in the space of the group elements since the number of �; i; jis precisely h. Any vector in this space can therefore be expanded in T�ij :

� =X�ij

c�ijT�ij

We sum the components of � along G-axes which belong to a given class Ck,

�(Ck) =XG2Ck

�(G)

=XG2Ck

X�ij

c�ijT�ij(G)

=1

h

XG1

XG2Ck

X�ij

c�ijT�ij(G

�11 GG1)

=1

h

XG1

XG2Ck

X�ij

Xkl

c�ijT�ik(G

�11 )T�kl(G)T

�lj (G1)

=XG2Ck

X�ij

Xkl

c�ij�ij�klT�kl(G)=l�

=XG2Ck

X�i

c�ii��(G)=l�

=X�

a�Nk��(Ck)

where a� = (1=l�)P

i c�ii. The third step uses the fact that G�11 CkG1 = Ck for

every G1 in the group, the �fth step uses the orthogonality theorem, and the laststep uses the fact that the characters of elements belonging to the same classare identical. Thus the set of vectors f�(Ck)g are linearly independent becausethe group elements can be divided uniquely and completely into distinct classes.

Multiplying �(Ck) by ��(Ck) and summing over the classes k yieldsXk

�(Ck)��(Ck) =

X�

a�Xk

Nk��(Ck)�

��(Ck)

=X�

a�h��

= ha�


Consequently,

�(Cl) =X�

a�Nl��(Cl)

= (1=h)X�

Xk

�(Ck)��(Ck)Nl�

�(Cl)

0 =Xk

�(Ck)

24(Nl=h)X�

��(Ck)��(Cl)� �kl

35This is true for an arbitrary �(Ck) so that the quantity in the square bracketmust vanish which proves the theorem.Eqn. ( 2.9) has the form of a dot product in a space with dimension equal to

the number of irreducible representations. It follows that the number of classesmust be less than or equal to the number of irreducible representations. On theother hand by Theorem 6 the number of irreducible representations must be lessthan or equal to the number of classes. Thus we conclude the following corollary:

The number of irreducible representations is equal to the number of classes.Theorems 6 and 8 imply that the irreducible representations are uniquely

characterised by their characters and consequently two distinct irreducible rep-resentations cannot have the same set of characters.

2.9 The Character Table

The character table is a square matrix with dimension equal to the number ofclasses or irreducible representations:

Q�k = ��(Ck)

It has the form

C1 = CE N2C2 N3C3 : : :T 1 1 1 1T 2 l2 : : : : : :T 3 l3 : : : : : :

The row is labelled by the irreducible representations and the column by theclasses.

2.10 Properties of the Character Table

The properties listed below are useful for constructing the character table.

1. The number of irreducible representations is equal to the number of classes.

2.11. CLASS MULTIPLICATION 21

2. The �rst column contains the dimension of each representation and thesum of its square is the total number of elements in the group (Theorem7):

P� l

2� = h.

3. As a convention, the �rst row is the identity or unit representation con-sisting of 1.

4. Each row or column can be regarded as a vector. In this sense:

(a) The rows are orthogonal with weighting factor Nk (Theorem 6) andnormalised to h.

(b) The columns are orthogonal and normalised to h=Nk (Theorem 8).

5. If the character table of a factor group is known, then due to homomor-phism, the character of an element aS in the factor group may be assignedto the classes which are contained in aS. We recall that if a group has aninvariant subgroup S, then we can form a group S; aS; bS; : : : where a isnot in S, b is not in S and aS etc. This group is a called a factor groupwith S as the unit element and it is homomorphic with the group itself.An invariant subgroup means that x�1Sx = S for every element x in thegroup and by Theorem 9 discussed in the next section, S must consistwholly of classes. The homomorphism between the factor group and thegroup itself means that if T (aS) is a matrix representation of the elementaS of the factor group then the same matrix may be used to represent allelements in the coset aS. Thus, the characters of the classes in aS is justthe same as the character of T (aS).

The above �ve rules are often su¢ cient to construct the character tablebut the following rule, which is described in the next section, can be ofhelp.

6. The characters of the � representation are related by

Ni��(Ci)Nj�

�(Cj) = l�Xk

mijkNk��(Ck) (2.10)

wheremijk are the coe¢ cients of class multiplication, CiCj =P

kmijkCk.

2.11 Class Multiplication

Let C be a collection of classes. From the de�nition of a class we have

x�1Cx = C

for every element x in the group. This is evident from the fact that all elementsin a class are conjugate to one another and if two elements a and b are di¤erentthen x�1ax 6= x�1bx. Moreover, an element cannot belong to more than one


class. Thus the elements on the right hand side must be identical to the ele-ments in C.

Theorem 9:A set of elements C obeying x�1Cx = C for all x in the group must be com-posed wholly of classes.Proof:Suppose R is a set of elements in C which do not form a class. But x�1Rx mustbe equal to R itself since (C �R) form classes and by de�nition they will neverbe conjugate to elements in R. Thus R must be a class.

We now consider a product of two classes:

CiCj = x�1Cixx�1Cjx = x�1CiCjx

By Theorem 9, it follows that CiCj must consist wholly of classes. Therefore itmust be possible to write

CiCj =Xk

mijkCk (2.11)

where mijk are integers telling how often the class Ck appears in the productCiCj .Let us consider the class multiplication in terms of matrix representations. If

we let S(Ci) be the sum of matrices of all elements in the class Ci and T (x) be amatrix representation of the element x then we have T�1(x)S(Ci)T (x) = S(Ci)or S(Ci)T (x) = T (x)S(Ci). If the representation T is irreducible, then it followsfrom Schur�s Lemma that S(Ci) = ciI. Thus

cicj =Xk

mijkck

Taking the trace of S(Ci) and assuming we are in the irreducible representation� we get

Tr S(Ci) = Tr ciI = cil�

On the other hand,Tr S(Ci) = Ni�

�(Ci)

Thereforeci = Ni�

�(Ci)=l�

and Eq. ( 2.10) follows.

2.12 Direct Product Groups

If a group is a direct product of two groups, then the irreducible representationsand the character table of the product group can be worked out easily fromthose of the two individual groups with the help of the following theorem.

2.13. EXAMPLES 23

Theorem 10:If G = G1 G2 then T�� = T�1 T �2 is an irreducible representation of G.Moreover, for all T�1 and T

�2 these are all possible irreducible representations of

the product group. The direct product of an n � n matrix A and and m �mmatrix B is de�ned by

(AB)ii0;jj0 = AijBi0j0 (2.12)

(AB) is an (mn�mn) matrix and it is not the same as a matrix multiplicationof A and B.

Proof:We �rst show that T is a representation of G. Let a1; a2; : : : and b1; b2; : : : bethe elements of G1 and G2 respectively and let cij = aibj be the elementsof the direct product group G. Suppose cijckl = cmn which implies thataibjakbl = ambn or aiakbjbl = ambn since the elements of G1 and G2 com-mute. This means that aiak = am and bjbl = bn or T�(ai)T�(ak) = T�(am)and T �(bj)T �(bl) = T �(bn). Writing the last two matrix equations in compo-nent form, multiplying them and using the de�nition of direct product yield therequired result.Let M be a matrix which commutes with every T�� . M may be written in

the formM =M1M2 whereM1 andM2 have the same dimensions as T�1 andT �2 respectively. The commutivity of M with T�� implies the commutivity ofM1 with T�1 and M2 with T

�2 . By Schur�s lemma, M1 = k1I1 and M2 = k2I2

and hence M = k1k2I and no non-constant matrix M exists so that by Schur�slemma T�� is irreducible.To show the last part of the theorem, we recall that

P� l

21� = h1 andP

� l22� = h2. Let l�� = l1�l2� be the dimension of the irreducible representation

T�� . We have X��

l2�� =X�

l21�X�

l22� = h1h2

But h1h2 is the order of the direct product group and therefore there are nomore possible irreducible representations other than T�� for all T�1 and T

�2 .

2.13 Examples

1. We construct the character table of the permutation group of three ob-jects which we have considered previously. There are three classes C1 = e,C2 = (�; �), and C3 = ( ; �; �) and therefore by rule 1 there are threeirreducible representations. Rule 2 tells us that l21 + l22 + l23 = 6. There isalways an identity representation so we may take l1 = 1. Then l2 = 1 andl3 = 2 are the only possible solution and the character table must looklike as follows:


C1 2C2 3C3T 1 1 1 1T 2 1 a bT 3 2 c d

To determine a; b; c; and d we need to know one of them and we can �ndthe others by orthogonality (rule 4). The subgroup S = (e; �; �) = C1+C2is an invariant subgroup and the corresponding factor group has the mul-tiplication table

S SS S S S S S

where S = �S = �S = ( ; �; �) = C3. This is a group of two ele-ments and there are evidently two classes, the �rst class consists of theunit element S and the second class consists of the element S. The num-ber of irreducible representations is two and both must have dimensionone, by rule 2. The irreducible representations of the factor group mustbe (1; 1) and (1;�1). Then from rule 5 we may assign a = 1 and b = �1(a = 1 and b = 1 gives the identity representation). Note that we havetaken a and b and not c and d because the irreducible representations ofthe factor group are one dimensional and T 2 is also a one dimensionalrepresentation. The other two characters c and d can be easily obtainedfrom the orthogonality between columns.

1 � 1 + 1 � a+ 2 � c = 0! c = �11 � 1 + 1 � b+ 2 � d = 0! d = 0

Finally we have

C1 2C2 3C3T 1 1 1 1T 2 1 1 �1T 3 2 �1 0

2. Abelian groups: Every element in an Abelian group forms a class by it-self. Therefore the number of irreducible representations is simply equalto the number of elements in the group. Moreover, rule 2 implies that allirreducible representations have dimension one. The consequences are

2.13. EXAMPLES 25

(a) Unitarity of representations implies that the numbers representingan Abelian group have modulus one.

(b) If the number X represents an element x then X� represents x�1.

(c) The character table serves also as a representation table.

As an example we consider a non-cyclic group of fourth order with thefollowing multiplication table

e a b ce e a b ca a e c bb b c e ac c b a e

a2 = b2 = c2 = e so a, b, and c are represented either by �1. If a isrepresented by 1 then b = c = �1 because bc = a. If a = �1 then eitherb = 1 and c = �1 or b = �1 and c = 1. Thus we have

e a b cT 1 1 1 1 1T 2 1 1 �1 �1T 3 1 �1 1 �1T 4 1 �1 �1 1

3. Cyclic groups: These are Abelian groups which can be generated by asingle element: G = (a; a2; : : : ; ah = e). If a is represented by a number Akthen a2 is represented by A2k etc. and A

hk = 1 since e is always represented

by 1. Thus, Ak must be one of the h roots of unity, i.e. Ak = exp(i2�k=h)and the character table is given by

e a a2 : : : ah�1

T 1 1 1 1 : : : 1

T 2 1 A1 A21 : : : Ah�11

T 3 1 A2 A22 : : : Ah�12

. . . . : : : .

. . . . : : : .Th 1 Ah�1 A2h�1 : : : Ah�1h�1



1. If T (G) is a matrix representation of a group, can T (G1) = T (G2) forG1 6= G2?

2. What is meant by the statement that two representations are equivalent?

3. What are meant by reducible and irreducible representations?

4. What is Maschke�s theorem?

5. What is Schur�s lemma?

6. Given T�M =MT � , where T� and T � are irreducible representations ofdimension l� and l� respectively, and M is a rectangular l� � l� matrix,what is the implication on M

(a) if l� 6= l� ,

(b) if la = l� .

7. Write down the orthogonality theorem.

8. How are the dimensions of the irreducible representations related to theorder the group?

9. How can we interpret T�ij geometrically?

10. What are the characters of a representation f��g? Are they unique?

11. How can we interpret f��ggeometrically?

12. If two elements belong to the same class, what can be said about thecharacters of their representations?

13. Write down the orthogonality theorems between characters of two irre-ducible representations.

14. What is the regular representation and what is special about it?

15. Given the number of classes what is the number of irreducible representa-tions?

16. Describe the salient properties of the character table. What are the char-acter table for?

17. Given that a group has an invariant subgroup, describe how it can be usedto help construct the character table.

18. Given that g�1Cg = C, where g is any group element, show that C mustbe composed wholly of classes.

19. If two classes, Ci and Cj , are multiplied show that the resulting multipli-cation must be composed of classes, i.e., CiCj =

PkmijkCk.

2.14. SUMMARY OF QUESTIONS 27

20. Let the elements of a group G1 commute with the elements of anothergroup G2. Given that T�1 and T

�2 are irreducible representations of G1 and

G2, respectively, describe how to construct the irreducible representationof the direct product group G = G1 G2. Given the character tables ofG1 and G2, describe how to construct the character table of G.

21. If a group is Abelian, show that the irreducible representations can onlybe one dimensional.

22. If a group is cyclic, what are its irreducible representations?

Chapter 3

Group Theory in QuantumMechanics

3.1 Linear Vector Space

A set f~r1; ~r2; : : :g is said to form a linear vector space L if ~ri + ~rj is anothermember in the set for every i; j and c~ri is also another member in the set wherec is a complex constant. The quantities f~r1; ~r2; : : :g are called vectors and thereare an in�nite number of them.A set of vectors f~v1; ~v2; : : : ; ~vpg is said to be linearly independent if none

of them can be expressed as a linear combination of the others. If a set ofcoe¢ cients fckg can be found such that

pXk=1

ck~vk = 0

then the vectors are said to be linearly dependent. The largest number of vectorsin L which form a linearly independent set is called the dimension of L. This isthe same as the smallest number of vectors needed to describe every vector inL and these vectors are said to form a basis. We denote the dimension of L bys and the basis vectors by f~e1; ~e2; : : : ; ~esg so that any vector ~v may be writtenas

~v =sXi=1

vi~ei

In order to determine the coe¢ cients vi from a given vector ~v we introducethe concept of scalar or dot product between two vectors ~v1 and ~v2 which wedenote by (~v1; ~v2). The particular de�nition of the scalar product is arbitrarybut it must satisfy the following general conditions:

1. (~v1; ~v2) is a complex number

29

30 CHAPTER 3. GROUP THEORY IN QUANTUM MECHANICS

2. (~v1; ~v2) = (~v2; ~v1)�

3. (~v1; c~v2) = c(~v1; ~v2)! (c~v1; ~v2) = (~v2; c~v1)� = c�(~v1; ~v2)

4. (~v1 + ~v2; ~v3) = (~v1; ~v3) + (~v2; ~v3)

5. (~v1; ~v1) � 0. It is only zero when ~v1 = 0.

Examples:

1. Vectors in 3-dimensional space. The dot product is de�ned as (~v1; ~v2) =v1v2cos� where v1 and v2 are the lengths of the vectors and � is the anglebetween the two vectors.

2. Function spaces. The dot product is de�ned as ( i; j) =Rd3r w(~r) �i (~r) j(~r)

where w is a weight function which is usually real and positive de�nite sothat the last condition is satis�ed. We can also de�ne orthonormal ba-sis functions f�ig such that any function can be expanded in this basis: =

P1i=1 ci�i The linear space spanned by this basis is called the Hilbert

space.

We de�ne an operator T by

T~v = ~v0

where ~v is an arbitrary vector or function which is carried into another vectoror function ~v0. Both ~v and ~v0 are in L. The operator is called linear if

T (~v1 + ~v2) = T~v1 + T~v2

T c~v = cT~v

We will consider only linear operators.Since any vector in L can be expanded in the basis vectors, it is only neces-

sary to study the e¤ects of the operator on the basis vectors.

T~ej =Xi

~eiTij

The matrix Tij is said to form a representation for the operator T in the linearvector space L.It is usually convenient to choose an orthonormal basis, by this we mean

(�i; �j) = �ij . This can always be done by the Gramm-Schmidt orthogonalisa-tion procedure. We illustrate the method for three vectors. The generalisationto an arbitrary number of vectors is straightforward. Let �1; �2; �3 be threelinearly independent functions which are not necessarily orthogonal nor nor-malised. Let 1 = �1=

p(�1; �1) and construct ~ 2 = �2 � 1( 1; �2) which is

orthogonal to 1. We normalise ~ 2 to get 2 = ~ 2=q( ~ 2; ~ 2) and construct

~ 3 = �3 � 2( 2; �3) � 1( 1; �3) which is orthogonal to both 1 and 2 and

3.2. SYMMETRY TRANSFORMATIONS 31

which is then normalised to give 3 = ~ 3=q~ 3; ~ 3) In this way we have con-

structed an orthonormal set of functions and we can see that for a general casewe have

~ n = �n �n�1Xi=1

i( i; �n)

3.2 Symmetry Transformations

A symmetry transformation G with respect to a given function f(x; y; z) hav-ing the symmetry in question is a real linear coordinate transformation thatpreserves the length (real unitary transformation)

~r0 = G~r ! r0i =Xj

Gijrj ; ri = x; y; z

such that the function calculated with the new variables ri by the above substitu-tion of coordinates is the same as the function calculated with the old variables:f(x0; y0; z0) = f 0(x; y; z) = f(x; y; z)! f(G~r) = f(~r). In other words, the func-tion looks the same in the old and the new coordinate system. This also impliesthat

f(G~r) = f(~r) = f(G�1~r)

because f(~r0) = f(~r) = f(G�1~r0) and ~r0 is just a dummy variable.The action of a symmetry operator T (G) associated with a symmetry trans-

formation G on an arbitrary function is de�ned by

T (G)f(~r) = f(G�1~r)

It is important to note that the operator acts upon the coordinates ~r, and notupon the argument of f . Thus we mean that

T (G)f(A~r) = f(AG�1~r) 6= f(G�1A~r)

We also have G�1 instead of G in our de�nition. This is a natural choice becausetwo successive operations T (G1) and T (G2) on a function f(~r) correspond tothe same order of coordinate transformation:

T (G2)T (G1)f(~r) = T (G2)f(G�11 ~r) = f(G�11 G�12 ~r) = f([G2G1]

�1~r):

This can also be seen by letting g(~r) = T (G1)f(~r) = f(G�11 ~r) so that

T (G2)T (G1)f(~r) = T (G2)g(~r) = g(G�12 ~r) = f(G�11 G�12 ~r) = f([G2G1]�1~r):

Theorem 1:


If fGg is a set of all symmetry transformations of a function, then the associatedsymmetry operators fT (G)g form a group.

Proof:The set fGg evidently form a group because if G is a symmetry transformation,then so is G�1 as discussed above. A succession of two symmetry transfor-mations is itself a symmetry transformation and the identity transformation isobviously a symmetry transformation. From the de�nition, we have

T (G1)T (G2)f(~r) = T (G1)f(G�12 ~r)

= f(G�12 G�11 ~r)

= f([G1G2]�1~r)

= T (G1G2)f(~r)

Hence T (G1)T (G2) = T (G1G2) and the symmetry operators form a group.

3.3 Invariant Subspace and Its Generation

A set of functions or vectors f�ig is said to span an invariant subspace VS undera given set of symmetry operators fT (G)g if T (G)�i is also in VS for all G andi.An invariant subspace can be generated from an arbitrary function f and

a set of symmetry operators fT (G)g by applying each of the operators on f .The set of functions fT (G)fg form an invariant subspace under fT (G)g becauseT (G0)[T (G)f ] = T (G0G)f , which must lie in the subspace since T (G0G) is justanother member of fT (G)g. We note that the number of functions that spanthe subspace is not necessarily equal to the number of G. Some of the functionsmay be the same or can be expressed as linear combinations of the others. Thelargest number of linearly independent functions in VS is called the dimensionof the subspace.It may happen that we can construct a function f 0 in VS such that the

invariant subspace fT (G)f 0g has a smaller dimension than that of VS . In thiscase the subspace VS is said to be reducible. When such a function cannot befound than VS is said to be irreducible.

3.4 Basis Functions for a Representation

Let f�ig be a set of orthonormal functions that span an invariant subspaceVS . Then T (G)�i(~r) = �i(G

�1~r) is in VS by de�nition and it can therefore beexpanded as a linear combination of f�ig:

T (G)�i =Xj

�jTji(G) (3.1)

3.4. BASIS FUNCTIONS FOR A REPRESENTATION 33

If the basis functions are orthonormal, then T (G) is unitary because

(T (G)�i; T (G)�j) = �ij = (Xk

�kTki(G);Xl

�lTlj(G))

=Xkl

T �ki(G)Tlj(G)(�k; �l)

=Xk

T yik(G)Tkj(G)

Theorem 2:The matrices fT (G)g form a representation for fT (G)g

Proof:

T (G1)T (G2)�i =Xj

T (G1)�jTji(G2)

T (G1G2)�i =Xjk

�kTkj(G1)Tji(G2)Xk

�kTki(G1G2) =Xk

�kTkj(G1)Tji(G2)

Thus T (G1)T (G2) = T (G1G2). The functions f�ig are said to form a basis for arepresentation, which may be reducible. Evidently, if the subspace is irreducibleunder fT (G)g, then the representation is also irreducible. This simply followsfrom the fact that one always generates the entire subspace starting from anyfunction in the subspace. If the representation were reducible, then it wouldbe possible to start from a function that does not generate the entire subspace,in which case the subspace is reducible. We now see the relevance of matrixrepresentation theory studied in the previous chapter. Since the theorems weredeveloped for arbitrary matrices, they are also applicable here.

Theorem 3:A change of basis vectors j =

Pi �iSij corresponds to a similarity transfor-

mation of the original representation, which is an equivalent representation. Sis assumed to be unitary so that ( i; j) = �ij .

Proof:


T (G)�i =Xj

�jTji(G)

T (G)Xj

jS�1ji =

Xj

Xk

kS�1kj Tji(G)

Multiplying both sides by Sim and summing over i yields

T (G) m =Xk

kT0km(G)

where T 0km(G) =P

ji S�1kj Tji(G)Sim(G) or T

0 = S�1TS which is a similaritytransformation. It is clear that the reverse is also true, i.e. a similarity transfor-mation of the representation with a unitary matrix S corresponds to a unitarytransformation of the basis vectors.

3.5 Projection Operators

It is possible to construct a projection operator P � such that given an arbitraryfunction , P � is a component of in the irreducible subspace �, i.e. P � transforms according to the irreducible representation �. This is a very usefultool. In the following, we construct these projection operators.Suppose f �i g span an irreducible subspace transforming according to the

irreducible representation � of fT (G)g. That is

T (G) �i =Xj

�j T�ji(G)

Multiplying by T ��kl (G) and summing over G yieldsXG

T ��kl (G)T (G) �i =

XG

Xj

�j T�ji(G)T

��kl (G)

Using the Orthogonality Theorem on the right side, we getXG

T ��kl (G)T (G) �i =

Xj

�j (h=l�)�jk�il��

= (h=l�)�il�� k

We de�neP �kl � (l�=h)

XG

T ��kl (G)T (G)

so thatP �kl

�i = �il��

�k ! P�ki

�i = �k (3.2)

3.5. PROJECTION OPERATORS 35

Starting from a given function that belongs to the ith row of a given irreduciblerepresentation �, the projection operator can be used to generate the other"partner functions" corresponding to the other rows. These functions form abasis for the irreducible representation �.Moreover, we can also de�ne

P � �Xi

P �ii

� (l�=h)XG

��(G)T (G)

so that

P � �i = �� i (3.3)

The projection operators allow us to decompose the space, in which thesymmetry operators are de�ned, into irreducible subspaces. Any function cantherefore be written as follows

=X�

Xi

�i (3.4)

where �i belongs to the ith row of the irreducible representation �. We notealso that if ��i and

�i belong to the ith row of the irreducible representation

�, so does any linear combination a��i + b �i .

Suppose that a given representation T is reducible. Then there is a unitarymatrix S such that T 0 = S�1TS =

P��m�T

� has the block form and eachblock cannot be reduced any further. � labels the irreducible representationand m� tells us how many times the irreducible representation � appears in thereduction. We note that when m� > 1 the irreducible representations T� areassumed to be the same. This can always be done by means of similarity trans-formations in each subspace which transforms according to the same irreduciblerepresentation �. The new basis vectors according to Theorem 3 are given by j =

Pi �iSij and they may be labelled according to the irreducible subspaces

to which they belong. Any vector in VS can therefore be expanded as follows:

v =X�;t;i

�ti c�ti

=X�;i

v�i (3.5)

where t labels the di¤erent possible subspaces transforming according to thesame irreducible representation �, which is necessary when m� > 1, and i labelsthe basis functions in the subspace �; t. We have also de�ned v�i =

Pt

�ti c

�ti .

However, T (G)v�i is not necessarily equal toP

j v�j Tji(G) unless there is only

one t.


3.6 Orthogonality of Basis Functions

Theorem 4:Two vectors belonging to two inequivalent irreducible representations must beorthogonal. If they transform according to the same irreducible representation,they are still orthogonal if they belong to di¤erent rows.

Proof:Let ��i and

�j belong to the irreducible representations T

� and T � respectively.Then

T (G)��i =Xk

��kT�ki(G)

T (G) �j =Xl

�l T�lj(G)

Recalling that T is or may always be chosen to be unitary, we have

(��i ; �j ) = (T (G)�

�i ; T (G)

�j )

=Xkl

T��ki (G)T�lj(G)(�

�k ;

�l )

= (1=h)XG

Xkl


�k ; �

�l )

= (1=l�)��ijXk

(��k ; �k ) (3.6)

3.7 Examples

As an example, we consider the group D3 which is the group of permutationsof three objects. The elements are e = unit element, c1; c2 = anticlockwiserotations about the z�axis by 2�=3 and 4�=3 respectively, and d1; d2; d3 =rotations by � about axes lebelled d1; d2; d3 in Figure 3.1. The classes areCe = e, Cc = (c1; c2), and Cd = (d1; d2; d3). We choose a set of basis vectors orfunctions in order to form a representation forthe group elements. The choiceis arbitrary.

1. The �rst obvious choice is the three Cartesian unit vectors ~ex; ~ey; ~ez asshown in the �gure. Let us see how they transform under the group op-erations.

3.7. EXAMPLES 37

Figure 3.1: Rigid equilateral triangle representing the group D3.

~ex ~ey ~ezT (e) ~ex ~ey ~ez

T (c1) � 12~ex +

p32 ~ey �

p32 ~ex �

12~ey ~ez

T (c2) � 12~ex �

p32 ~ey

p32 ~ex �

12~ey ~ez

T (d1) �~ex ~ey �~ezT (d2)

12~ex +

p32 ~ey

p32 ~ex �

12~ey �~ez

T (d3)12~ex �

p32 ~ey �

p32 ~ex �

12~ey �~ez

We see that applying the group operations on ~ex or ~ey generate an in-variant subspace (~ex; ~ey) whereas ~ez forms an invariant subspace by itself.The set of Cartesian unit vectors (~ex; ~ey; ~ez) do form a basis for the rep-resentation of the group. The matrices can be calculated by using the


de�nition Tij(G) = (�i; T (G)�j):

T (e) =

0@ 1 0 00 1 00 0 1

1A T (c1) =

0B@ � 12 �

p32 0p

32 � 1

2 00 0 1

1CA

T (c2) =

0B@ � 12

p32 0

�p32 � 1

2 00 0 1

1CA T (d1) =

0@ �1 0 00 1 00 0 �1

1A

T (d2) =

0B@ 12

p32 0p

32 � 1

2 00 0 �1

1CA T (d3) =

0B@ 12 �

p32 0

�p32 � 1

2 00 0 �1

1CAThe characters are �(Ce) = 3, �(Cc) = 0, and �(Cd) = �1. As can beseen, the 3�3 representation is reducible into a 2�2 and a 1�1 represen-tations. The 1 � 1 representation is obviously irreducible, but it may bepossible to further reduce the 2�2 representation. We recall that the char-acter of a given representation can be written as � =

P�m��

� and byusing the �rst orthonormality relation between characters the coe¢ cientsm� can be calculated as follows:

m� =1

h

Xk

Nk��(Ck)�(Ck)

For convenience, we reproduce the character table of the group D3.

Ce 2Cc 3CdT 1 1 1 1T 2 1 1 �1T 3 2 �1 0

Thus

m1 =1

6[1 � 1 � 3 + 2 � 1 � 0 + 3 � 1 � (�1)] = 0

m2 =1

6[1 � 1 � 3 + 2 � 1 � 0 + 3 � (�1) � (�1) = 1

m3 =1

6[1 � 2 � 3 + 2 � (�1) � 0 + 3 � 0 � (�1) = 1

Hence the representation breaks up into T = T 2�T 3 and therefore (~ex; ~ey)do form an irreducible subspace for the representation T 3. We could have

3.7. EXAMPLES 39

arrived at this conclusion simply by calculating the characters of the 2�2representation which are �(Ce) = 2, �(Cc) = �1, and �(Cd) = 0 andwhich agree with the characters of the T 3 representation.

2. As a second example, we consider the same group but we choose a basiswhich consists of functions rather than the Cartesian vectors. We cantake an arbitrary function, say x2, and generate an invariant subspacefrom it. We construct an orthonormal basis that spans this subspaceand we form the representation for the group. We then use the projectionoperator method to �gure out which functions belong to a given irreduciblesubspace.

What we need to know is how x transforms under the group operations.We have de�ned T (G)f(~r) = f(G�1~r), i.e. T (G)x = T (G)(~ex � ~r) =(~ex � G�1~r) = (G~ex � ~r). Thus x; y and z tranform like the Cartesianvectors ~ex, ~ey, and ~ez and we have

T (e)x2 = x2

T (c1)x2 = (�1

2x+

p3

2y)2 =

1

4x2 +

3

2y2 �

p3

2xy

T (c2)x2 = (�1

2x�

p3

2y)2 =

1

4x2 +

3

2y2 +

p3

2xy

T (d1)x2 = (�x)2 = x2

T (d2)x2 = (

1

2x+

p3

2y)2 =

1

4x2 +

3

2y2 +

p3

2xy

T (d3)x2 = (

1

2x�

p3

2y)2 =

1

4x2 +

3

2y2 �

p3

2xy

These six functions form an invariant subspace for D3 but all of them canbe written as linear combinations of only three distinct functions �1 = x2,�2 = y2, and �3 = xy. Evidently none of these three functions can bewritten as a linear combination of the other two so that they are linearlyindependent and form a basis for the invariant subspace, although theyare neither orthogonal nor normalised.

We now use the Gramm-Schmidt orthogonalisation procedure to constructa set of orthonormal basis needed to construct a unitary representationfor the group. We can start with any of the three functions or any linearcombination of them. Let us start with �1 = x2 which we normalise insidea unit sphere so that

jAj2(�1; �1) = jAj2Z 1

0

dr r2Z �

0

d�sin �

Z 2�

0

d� ��1�1 = 1

which gives A =p35=4�. The normalised function is then

�1 = Ax2


The second function is constructed by using ~�2 = �2 � �1(�1; �2) whichyields after normalisation

�2 = A1

2p2(3y2 � x2)

The function xy is already orthogonal to both x2 and y2 due to antisym-metry so that the third orthonormalised function is

�3 = Ap3xy

To construct the representation matrices, we need to know how these basisfunctions transform into one another under the group operations.

�1 �2 �3T (e) �1 �2 �3T (c1)

12�1 +

1p2�2 � 1

2�31p2(�1 + �3)

12�1 �

1p2�2 � 1

2�3

T (c2)12�1 +

1p2�2 +

12�3

1p2(�1 � �3) � 1

2�1 +1p2�2 � 1

2�3

T (d1) �1 �2 ��3T (d2)

12�1 +

1p2�2 +

12�3

1p2(�1 � �3) 1

2�1 �1p2�2 +

12�3

T (d3)12�1 +

1p2�2 � 1

2�31p2(�1 + �3) � 1

2�1 +1p2�2 +

12�3

The matrix representations can be easily obtained and these are

T (e) =

0@ 1 0 00 1 00 0 1

1A T (c1) =

0B@12

1p2

12

1p2

0 � 1p2

� 12

1p2

� 12

1CA

T (c2) =

0B@12

1p2

� 12

1p2

0 1p2

12 � 1p

2� 12

1CA T (d1) =

0@ 1 0 00 1 00 0 �1

1A

T (d2) =

0B@12

1p2

12

1p2

0 � 1p2

12 � 1p

212

1CA T (d3) =

0B@12

1p2

� 12

1p2

0 1p2

� 12

1p2

12

1CAUnlike the previous example, these matrices are not in block form but we

know that they are reducible because there is no three dimensional represen-tation of D3. Let us �gure out which irreducible representations are contained

3.7. EXAMPLES 41

in these matrices by using the same formula as in the previous example. Thereducible characters are �(Ce) = 3, �(Cc) = 0, and �(Cd) = 1. We get

m1 =1

6[1 � 1 � 3 + 2 � 1 � 0 + 3 � 1 � 1] = 1

m2 =1

6[1 � 1 � 3 + 2 � 1 � 0 + 3 � (�1) � 1] = 0

m3 =1

6[1 � 2 � 3 + 2 � (�1) � 0 + 3 � 0 � 1] = 1

Hence the representation breaks up into T = T 1 � T 3.Since the representation contains an identity representation, it must be pos-

sible to construct a linear combination of �1, �2, and �3 such that it is invariantunder the group operations. We do this in a systematic way using the projectionoperator method. The projection operator is given by

P� =l�h

XG

��(G)T (G)

which when applied to an arbitrary function gives the component of the functionin the irreducible subspace �. Let us apply P 1 to �1, �2, and �3:

P 1�1 =1

6

��1 + (

1

2�1 +

1p2�2 �

1

2�3) + (

1

2�1 +

1p2�2 +

1

2�3)

+�1 + (1

2�1 +

1p2�2 +

1

2�3) + (

1

2�1 +

1p2�2 �

1

2�3)

�=

p2

3(p2�1 + �2)

P 1�2 =1

6

��2 + (

1p2(�1 + �3)) + (

1p2(�1 � �3))

+�2 + (1p2(�1 � �3)) + (

1p2(�1 + �3))

�=1

3(p2�1 + �2)

P 1�3 = 0

The linear combination (p2�1 + �2)=

p3 transforms according to the identity

representation. The component of �1 in the third irreducible subspace is givenby

P 3�1 =2

6

�2�1 � (

1

2�1 +

1p2�2 �

1

2�3)� (

1

2�1 +

1p2�2 +

1

2�3)

�=1

3(�1 �

p2�2)


Thus we may choose a new basis

1 =1p3(p2�1 + �2)

2 =1p3(�1 �

p2�2)

3 = �3

which brings the representation into block form with 1 transforming accordingto the identity representation T 1 and ( 2; 3) transforming according to theirreducible representation T 3. The new basis may be written as i =

Pj �jSji

where S is the unitary matrix that reduces the representation into block formwhich is given by

S =

0BB@q

23

q13 0q

13 �

q23 0

0 0 1

1CCA3.8 Relationship to Quantum Mechanics

A basic problem in quantum mechanics is �nding the eigenvectors and eigenval-ues of a Hamiltonian H. A general procedure is to calculate the Hamiltonianmatrix in an arbitrary basis and diagonalise it, but in most cases, this procedureis too di¢ cult to carry out. Instead one tries to �nd a basis in which the o¤diagonal elements of H are small or zero so that they can be neglected or treatedperturbatively. It is here that the study of the symmetry of the Hamiltonian isof great value because group theory can tell us which matrix elements are zeroand how the matrix elements are related to one another. The results are exactand depend only on the symmetry of the systems and not on the particular formof interactions or potentials.

3.8.1 The Group of the Schrödinger Equation

If fGg is a set of all symmetry transformations of a Hamiltonian, H, then ac-cording to Theorem 1 the associated symmetry operators fT (G)g form a groupwhich is called the group of the Schrödinger equation or in short the Schrödingergroup.

Theorem 5:Each element of the Schrödinger group commutes with the Hamiltonian. More-over, if (~r) is an eigenvector of H, then T (G) (~r) = (G�1~r) is also aneigenvector of H with the same eigenvalue.

Proof:By de�nition

T (G)H(~r) = H(G�1~r) = H(~r)

3.8. RELATIONSHIP TO QUANTUM MECHANICS 43

Let us investigate the consequences of this symmetry by considering the e¤ectof T (G) on H (~r) where is an arbitrary function .

T (G)H(~r) (~r) = H(G�1~r) (G�1~r)

= H(~r)T (G) (~r)hT (G)H(~r)� H(~r)T (G)

i (~r) = 0

Since (~r) is arbitrary, [T ; H] = 0. The second part of the theorem followsdirectly from this result:

H(~r)T (G) (~r) = T (G)H(~r) (~r) = �T (G) (~r)

3.8.2 Degeneracy and Invariant Subspace

The implication of Theorem 5 is that a degenerate set of eigenfunctions of Hform an invariant subspace under the symmetry operations of the Schrödingergroup and therefore form a basis for a representation of the group. Let f ig bea set of degenerate eigenfunctions with energy �. Since T (G) i(~r) = i(G

�1~r)is also an eigenfunction with the same energy �, it must be possible to expandT (G) i(~r) as a linear combination of f ig:

T (G) i =Xj

jTji(G)

The matrices fT (G)g form a representation for the Schrödinger group.The representation is normally irreducible but if it is reducible, then there

are two possibilities:

1. There is an accidental degeneracy so that the subspace can be brokenup into two or more irreducible subspaces. A vector in one irreduciblesubspace evidently cannot be brought into another irreducible subspaceby the symmetry operators.

2. The group is a subgroup of a larger group.

In practice, accidental degeneracies my be lifted up by varying parametersin the Hamiltonian without changing the symmetry of it. If the subspace isirreducible, the degeneracy is termed "normal".Assuming that there is no accidental degeneracy, each degenerate set of

eigenfunctions of H form a basis for an irreducible representation of the Schrödingergroup. We may label each eigenfunction by the row and the irreducible repre-sentation to which it belongs. In this sense, the row index and the irreduciblerepresentation provide "quantum numbers". The degeneracy is simply givenby the dimensionality of the irreducible representation. Thus by knowing thesymmetry of the Hamiltonian, we can tell the degrees of degeneracy possible


in any problem. It follows that these degeneracies can only be broken by aperturbation if it has a lower symmetry than the unperturbed Hamiltonian.Moreover, by knowing the symmetry of the perturbation, we can also tell howthe degeneracies are split.

3.8.3 Partial Diagonalisation of H

Theorem 6:The matrix element (��i ; H

�j ) is zero unless � = � and i = j.

Proof:

(��i ; H �j ) = (T (G)�

�i ; T (G)H

�j )

= (T (G)��i ; HT (G) �j )

=Xkl


�k ; H

�l )

=1

h

XG

Xkl


�k ; H

�l )

=1

l�

Xkl

�kl�ij��(��k ; H

�l )

=1

l��ij

Xk

(��k ; H �k )

This result is very useful because by choosing basis functions with the appropri-ate symmetry, the Hamiltonian is partially diagonalised with very little e¤ort.Also, (��i ; H

�i ) is independent of i. This is a special case of the so called

Wigner-Eckart theorem discussed later.

3.9 Irreducible Sets of Operators

We have studied the concept of irreducible subspace formed by a set of functionswhich transform among themselves under the group operations. A similar con-cept may be extended to operators and we de�ne an irreducible set of operatorsO�i by the closure property

O�0

i = T (G)O�i T�1(G) =

Xj

O�j T�ji(G) (3.7)

Thus the irreducible set of operators O�i transform among themselves underthe irreducible representation T�. The number of operators in such a set isequal to the dimension l� of the irreducible representation. An example of anirreducible set of operators is given by the position operators x, y, and z andthe momentum operators px, py, and pz under rotation. Consider taking matrixelement of T (G)xT�1(G) in the position eigenvector jri:

3.10. DIRECT PRODUCT REPRESENTATIONS OF A GROUP 45

DrjT (G)xT�1(G)jr

E=G�1rjxjG�1r

�= x �G�1r hrjri= Gx � r hrjri= (ax+ by + cz) � (xx+ yy + zz) hrjri= (ax+ by + cz) hrjri= hrjax+ by + czjri :

In operator equation this becomes

T (G)xT�1(G) = ax+ by + cz:

In general we have

T (G)riT�1(G) =

Xj

rjTji(G):

3.10 Direct Product Representations of a Group

The direct product of an n�n matrix A and and m�m matrix B is de�ned by

(AB)ii0;jj0 = AijBi0j0 (3.8)

(AB) is an (mn�mn) matrix and it is not the same as a matrix multiplicationof A and B.

Theorem 7:If T� and T � are two representations of a group, then T�� = T�T � is also arepresentation of the group. Note that we are not considering a direct productof two groups but rather the direct product representations of a group.

Proof:

[T��(G1)T��(G2)]ii

0; jj0 =Xkk0

T��ii0;kk0(G1)T��kk0;jj0(G2)

=Xkk0

T�ik(G1)T�i0k0(G1)T

�kj(G2)T

�k0j0(G2)

= T�ij(G1G2)T�i0j0(G1G2)

= T��ii0;jj0(G1G2)

Theorem 8:


The character of a direct product representation T�� is the product of thecharacters of T� and T � .

Proof:

�� =Xii0

T��ii0;ii0

=Xii0

T�iiT�i0i0

= ��

Theorem 9:If f �i g transform according to the representation T� and f �j g transform ac-

cording to the representation T � , then f �i �j g transform according to the direct

product representation T�� .

Proof:

T (G)( �i �j ) =

Xk

Xl

�k T�ki(G)

�l T

�lj(G)

=Xkl

( �k �l )T

��kl;ij

In general, the direct product representation is reducible even when thetwo constituent representations are irreducible. There is a unitary matrix U ,independent of the group elements, which transforms T�� into a block form:X

ij;i0j0

U mk;ijT��ij;i0j0U

yi0j0; 0m0k0 = � 0�mm0T kk0 (3.9)

The labelm distinguishes blocks with the same irreducible representation. Thus,we may write

T�� =X

�m T (3.10)

labels the irreducible representations and the coe¢ cientsm , which tells us thenumber of times the irreducible representation occurs in the decomposition,can be calculated using the �rst orthogonality relation of the characters

�� =X

m �

�� =X

m �

m = (1=h)Xk

Nk��(Ck)�

�(Ck)� �(Ck)

3.11. CLEBSCH-GORDAN COEFFICIENTS 47

3.11 Clebsch-Gordan Coe¢ cients

Since the set of product functions �i �j transform according to the product

representation T�� which is in general reducible, evidently the space formedby the product functions are also reducible and the product functions may thenbe projected into these irreducible subspaces:

�i �j =

X m

F ij(m)

where F ij(m) belongs to the irreducible subspace . The label m distinguishesfunctions with the same which is necessary only when m > 1 in the de-composition of the product representation. In other words, there may be morethan one subspace which transform according to the same irreducible represen-tation. F ij(m) itself may be expanded in an orthogonal basis which spans theirreducible subspace :

F ij(m) =Xk

f k (m)C k;ij(m)

and therefore �i

�j =

X m

Xk

f k (m)C k;ij(m) (3.11)

The coe¢ cients C are called the Clebsch-Gordan coe¢ cients and they dependon � and �.Apart from a possible phase factor, the Clebsch-Gordan coe¢ cients are ac-

tually identical to the unitary matrix U de�ned in the previous section. To seethis, we consider how the state k(m) =

Pij

�i

�j U

yij; mk transform under a

group symmetry operation:

T (G) k(m) =Xij;i0j0

�i0 �j0T

��i0j0;ijU

yij; mk

=Xi0j0k0

�i0 �j0U

yi0j0; mk0T

k0k

=Xk0

k0(m)T k0k

We have used Eq. ( 3.9) in the second step. Thus the states k(m) transformaccording to the irreducible representation . We may also write

�i �j =

Xk m

k(m)U mk;ij (3.12)

which shows that the Clebsch-Gordan coe¢ cients are identical to the unitarymatrix U , apart from a phase factor. By convention, they are normalised asfollows: X

ij

jC ijk(m)j2 = 1 (3.13)


Since C is unitary, it satis�es the following relationsXij

C k;ij(m)C� 0k0;ij(m

0) = � 0�mm0�kk0

X km

C k;ij(m)C� k;i0j0(m) = �ii0�jj0

3.12 The Wigner-Eckart Theorem

In physics, we are often faced with a problem of calculating matrix elements ofan operator O between two states and �: ( ; O�). In practice, the calculationof matrix elements can be very complicated and we would like to study if grouptheory can help us in simplifying this problem.First of all, from previous sections, we know that an arbitrary function be-

longing to a linear space L which transforms into itself under group operationscan be projected into the di¤erent irreducible components of the group. Conse-quently, we need only consider matrix elements of the type:

(f k ; O�i f

�j )

We study the transformation of the function ij = O�i f�j under a group sym-

metry operation:

T (G) ij = T (G)O�i f�j

= [T (G)O�i T�1(G)] T (G)f�j

=Xk

O�kT�ki(G)

Xm

f�mT�mj(G)

=Xkm

kmT��km;ij(G)

The new function ij may then be completely expanded into its irreduciblecomponents as in Eq. ( 3.12):

ij =X 0k0m

0

k0 (m)C 0

k0;ij(m)

The matrix element is then given by

(f k ; O�i f

�j ) =

X 0k0m

(f k ; 0

k0 (m))C 0

k0;ij(m)

=Xm

(f k ; k(m))C

k;ij(m)

The signi�cance of the above result is that the Clebsch-Gordan coe¢ cientsC k;ij(m) are obtained entirely from symmetry and they do not depend on the

3.13. APPLICATIONS 49

details of the potential of the system. Information about the details of the sys-tem is contained in the matrix elements (f k ;

k(m)) which do not depend on

i and j. In fact, it does not depend on k as shown in Eq. (3.6). These matrixelements are often referred to as reduced matrix elements and they are writtenas

(f k ; k(m)) = hf

jjO�jjf�imemphasising the fact that the reduced matrix elements are independent of i, j,and k. Thus we have

(f k ; O�i f

�j ) =

Xm

hf jjO�jjf�imC k;ij(m) (3.14)

This is known as the Wigner-Eckart theorem which tells us that the matrixelement (f k ; O

�i f

�j ) is zero unless the product representation T�� contains

the irreducible representation . This is a general form of selection rules. TheClebsch-Gordan coe¢ cients relate the matrix elements for a given �, �, and but with di¤erent i, j, and k to one another.We can see that theorem 7 follows immediately from the Wigner-Eckart

theorem because the Hamiltonian is invariant under the group symmetry oper-ations, i.e. T (G)HT�1(G) = H, so that H transforms according to the identityrepresentation.

3.13 Applications

We list a few of the common applications of group theory.

1. Selection rules: As an example we consider optical transitions in a mole-cule which has a D3 symmetry. This could correspond to a molecule withthree identical atoms at the corners of a triangle. Actually, the full sym-metry is higher than D3 but as an illustration it is su¢ cient to considerthe D3 symmetry. Optical transitions are governed by dipole operatorsx, y, and z which transform like the three Cartesian unit vectors. Thegroundstate of the molecule may be assumed to be non-degenerate andsymmetric so that it transforms according to the identity representation.From the previous example, the dipole operators transform according tothe representation T = T 2 � T 3 and therefore the �nal states can only bethose which transform according to (T 2 � T 3) T 1 = T 2 � T 3.

2. Symmetry breaking: In the above example, we know from symmetry alonethat the eigenfunctions can only be at most doubly degenerate. If we re-place one of the atoms by a di¤erent atom so that the symmetry becomesC2, then this degeneracy, if it exists, must split into two non-degeneratelevels because C2 is an Abelian group and the irreducible representationsare one dimensional. The actual reduction can be �gured out straight-forwardly using the orthogonality relation between the characters. Theordering of the energy levels, however, cannot be determined by grouptheory alone.


3. Variational principle: We are often interested in �nding the groundstateof a given Hamiltonian which usually cannot be solved analytically. Onemethod of obtaining an approximate groundstate is based on variationalprinciple. We start with some basis functions f�ig and approximate thetrue groundstate as a linear combination of these functions.

=Xi

ci�i

From the variational principle, the energy expectation value < E >=(; H) � E0, where E0 is the true groundstate energy. The coe¢ -cients fcig are determined by minimising < E > subject to the condition(;) = 1:

@

@c�i[< E > ��(;)] = @

@c�i

24Xjk

c�j (�j ; H�k)ck � �Xj

c�jcj

35=Xk

Hikck � �ci = 0

which is just an eigenvalue problem.

From Theorem 6, it is clear that we should sort out the basis functions,using the projection operator method for example, according to the rowsand irreducible representations f��mi g where m labels di¤erent possiblesubspaces transforming according to the same irreducible representation�. It is only necessary to express the trial function as follows: =P

m c�mi ��mi . When m = 1, the Hamiltonian is already diagonal and

when m > 1, we need only diagonalise an m�m matrix instead of an l� lmatrix where l =

P�mi 1, which is usually much larger than m.

4. Degenerate perturbation theory: In many cases, the Hamiltonian may bebroken up into two terms H = H0 + H1 such that H1 may be treated asa small perturbation to H0. H0 usually has a higher symmetry than H1.A typical example of this is an atom placed in a crystal where the crystal�eld symmetry is lower than the rotational symmetry of the atom. If theeigenfunctions of H0 are degenerate, we may use the variational principleto calculate the energy shifts. The eigenfunction of H is approximated asa linear combination of the degenerate eigenfunctions of H0. This leadsto an eigenvalue problem for H1 and the eigenvalues give the �rst ordershifts. As before, we should sort out the degenerate eigenfunctions of H0

according to the rows and irreducible representations of the symmetrygroup of H1 in order to minimise the size of the matrix to be diagonalised.

3.14 Examples

1. Bloch Theorem: A lattice is a parallelepiped formed by three linearlyindependent vectors ~a1;~a2;~a3. A set of vectors ~t = n1~a1 + n2~a2 + n3~a3

3.14. EXAMPLES 51

for all possible integers n1; n2; n3 generate lattice points which togetherwith the associated lattices form a crystal. The set t = f~tg clearly forma group under vector addition and the group is Abelian. In fact, it is adirect product of three groups

t = t1 t2 t3

where t1 = fn1~a1g, t2 = fn2~a2g, and t3 = fn3~a3g. It is customary to havea periodic boundary condition such that

T (N1~a1) (~r) = (~r +N1~a1) = (~r)! T (N1~a1) = T (0) = 1

Similarly for the other two groups. N1 is taken to be in�nite at the end.The groups t1, t2, and t3 become cyclic and the irreducible representationsare given by ei2�k1=N1 , k1 = 1; 2; : : : ; N1, and their powers. It is useful atthis stage to introduce the concept of reciprocal space de�ned by

~g1 = 2�~a2 � ~a3

~a1 � ~a2 � ~a3

~g2 = 2�~a3 � ~a1

~a1 � ~a2 � ~a3

~g3 = 2�~a1 � ~a2

~a1 � ~a2 � ~a3

so that~gi � ~aj = 2��ij

With this de�nition, the direct product representations can be written

T~k(~t) = ei

~k�~t

where ~k = (k1=N1)~g1 + (k2=N2)~g2 + (k3=N3)~g3. It follows that the eigen-functions of a Hamiltonian with translational lattice symmetry form basisfunctions for the irreducible representations of the group of translationsand they may be labelled by the irreducible representations ~k. This means

T (~t) ~k(~r) = ~k(~r +~t) = ~k(~r)e

i~k�~t

which is the Bloch theorem.

2. Normal modes (chapter 6 of Elliot and Dawber): When atoms in a mole-cule are displaced from their equilibrium positions, the potential energywill increase. If the atoms are let free, there will be an interchange betweenpotential and kinetic energies resulting in vibrations. These vibrationsmay be thought of as being composed of eigen modes or normal modes,each with a well de�ned frequency, in the same way that a wavefunctioncan be decomposed into a sum of eigenstates with well de�ned energies.


The eigen modes may be classi�ed according to the irreducible representa-tions of the symmetry group of the molecule, just like the the eigenstatesof the Hamiltonian.

The potential energy for arbitrary displacements of N atoms in a moleculeis given by

V = V0 +Xi

@V

@qiqi +

1

2

Xij

@2V

@qi@qjqiqj + :::

The sums run from 1 to 3N . At equilibrium, @V=@qi = 0 so that theHamiltonian for molecular vibrations in the harmonic approximation (quadraticin the displacements) is given by

H =Xi

1

2Mi _q

2i +

Xij

1

2Bijqiqj

where Bij = @2V=@qi@qj . It is conventional to absorb the masses into thedisplacement coordinates by de�ning

�i =pMiq1; Dij = Bij=

pMiMj ;

so that the Hamiltonian becomes

H =Xi

1

2_�2i +

Xij

1

2Dij�i�j

The matrix D is known as the dynamical matrix.

We can now introduce a 3N -dimensional abstract space with orthonormalbasis vectors f~eig. The 3N displacements of atoms in real space may bemapped into a vector in the 3N dimensional abstract space:

(q1; q2; : : : ; q3N )! ~q =Xi

~ei�i

The unit vector ~ei corresponds to a displacement qi = 1=pMi in real space

but we note that ~ei itself is not a displacement in real space but rathera displacement in the abstract space (c.f. the phase space in statisticalmechanics, in which (qi;pi) represent the canonical position and momen-tum of a particle but the phase space itself is abstract). We may de�ne anoperator D in the abstract space whose action on the base vector is givenby

D~ej =Xi

~eiDij ! Dij = (~ei; D~ej):

3.14. EXAMPLES 53

The potential energy can then be written as

V (~q) =1

2(~q; D~q)

=1

2

Xij

�i�j(~ei; D~ej)

=1

2

Xij

�i�jDij :

We would like to transform the Hamiltonian into the form H =P

i hi sothat the eigenvenctors of H can be written as a product of the eigenvectorsof h. This can be done by a unitary transformation

~uk =Xi

~eiaij

An arbitrary vector ~q may be written as

~q =Xi

~ei�i =Xk

~ukQk =Xki

~eiaikQk ! �i =Xk

aikQk

By choosing ~uk to be an eigenvector of D, i.e. D~uk = !k~uk orP

j Dijajk =!kaik, the Hamiltonian becomes

H =Xk

�1

2_Q2k +

1

2!kQ

2k

�which is the desired form. Quantum mechanically, we replace _Q2k with�@2=@Q2k and the solutions to hk = �@2=@Q2k + 1

2!kQ2k are

n(k) = AHn(k)(p!kQk)exp(�!kQk=2)

�n(k) = (n(k) +1

2)!k

We have set the Planck constant } = 1.So far we have not made any use of group theory and in principle we cancalculate the dynamical matrix D and diagonalise it to obtain the normalmodes and the corresponding eigen frequencies. The diagonalisation of Dis greatly simpli�ed by using group theory. The eigen frequencies, however,cannot be obtained from group theory alone because they depend on thestrength of the potential. Since [T (G); V ] = 0, we have

[T (G); D] = 0

for all G and it immediately follows from Theorem 5 that the degenerateeigenvectors of D (normal modes) form basis vectors for the irreduciblerepresentations of the group.


The 3N dimensional space forms a 3N � 3N representation which is ingeneral reducible:

�3N =X

m �

To calculate �3N , we �rst write the basis vector as ~ei(R), i = x; y; z,R = 1; 2; : : : ; N , which corresponds to a displacement of atom R along thei-axis. The e¤ect of the symmetry operation on the basis vector is givenby

T (G)~ei(R) =Xj

~ej(R0)TR

0Rji (G)

where the symmetry operation T (G) moves atom R to atom R0. Thecharacter is given by

�3N =XRi

TRRii = NR�V

where NR is the number of atoms unmoved by the symmetry operationT (G) and �V is the character of the three dimensional vector represen-tation. This result can be understood by observing that only when thesymmetry operation T (G) does not move atom R that it contributes tothe diagonal element of the matrix TR

0Rji (G). When atom R is not moved,

thenP

i TRRii = �V .

As an example, we consider a molecule with D3 symmetry (three atomsat the corners of an equilateral triangle). We have already worked outthe characters for the vector representation �V and they are given by�V (Ce) = 3, �V (Cc) = 0, and �V (Cd) = �1. Thus �3N (Ce) = 3 � 3 = 9,�3N (Cc) = 0, and �3N (Cd) = 1 � (�1) = �1. It is then straightforwardto decompose �3N into the di¤erent irreducible components using thecharacter table of D3. We get

m1 =1

6(1 � 1 � 9 + 2 � 1 � 0 + 3 � 1 � (�1)) = 1

m2 =1

6(1 � 1 � 9 + 2 � 1 � 0 + 3 � (�1) � (�1)) = 2

m3 =1

6(1 � 2 � 9 + 2 � (�1) � 0 + 3 � 0 � (�1)) = 3

There are three non-degenerate modes (the irreducible representations T 1

and T 2 are one-dimensional) and three doubly degenerate modes (the ir-reducible representation T 3 is two-dimensional). Actually, six of themcorrespond to rigid motions, i.e. there is no relative motion between theatoms. The signi�cance of classifying the vibrations into normal modes isthat when calculating absorption or emission spectra it is possible to �g-ure out which modes are allowed in the absorption process. The simplestabsorption processes are caused by electric dipole transitions from the


groundstate to some vibrational excited states. The groundstate usuallytranforms according to the identity representation and the dipole opera-tors transform according to the vector representation. Thus the modesallowed are given by those corresponding to the irreducible representa-tions occuring in the decomposition of the vector representation. In morecomplicated absorption processes, there are intermediate states involved.In this case the dominant operators are of the form xy, yz, zx, x2, y2, andz2 (Raman spectra). These are in fact products of x, y, and z, i.e. theytransform according to the product of the vector representations whichcan be decomposed into the irreducible components. Care, however, hasto be taken due to the fact that the product functions are not linearlyindependent. See chapter 6 of Elliot and Dawber for more examples anddetails.


1. Given a set of basis vectors feig such that

T (G)ei =Xj

ejTji(G);

show that the matrices T (G) form a representation of the group. If feigare orthonormal, what can be said about the representation?

2. Suppose fe0ig are not orthonormal and feig constructed from ei =P

j e0jSji

are orthonormal. Show that the representations generated by feig andfe0ig have nevertheless the same characters.

3. Given that T (G) jri = jGri, show that T (G)f(r) +DrjT (G)jf

E= f(G�1r).

4. Show that T (G1)T (G2)f(r) = f((G1G2)�1r):

5. Explain how to compute T (G)(xy), where G is a rotation or inversionabout a �xed point or a re�ection about a plane.

6. What is meant by an invariant subspace?

7. What is meant by an irreducible subspace?

8. Given an arbitrary function f(r), describe how one can generate an in-variant subspace. Is the so generated subspace irreducible?

9. Write down the projection operator P�ij .

10. Write down the projection operator P�.

11. Given a representation T , it can in general be decomposed into irreduciblerepresentations: T =

P m T

:Describe how to compute m .


12. Given an arbitrary function explain how to decompose it into its irre-ducible components.

13. Assuming unitary representations, what is��i ;

�j

�?

14. What is meant by the statement that a Hamiltonian is invariant under asymmetry group?

15. If a Hamiltonian is invariant under a symmetry groupnT (G)

owhat can

be said about its eigenvectors?

16. If an invariant subspace formed by a set of degenerate eigenvectors of aHamiltonian is reducible, what possibilities do we have?

17. If a Hamiltonian is invariant under a symmetry group, what is��i ;H

�j

�?

18. Suppose that a Hamiltonian is invariant under a symmetry group andthe group has a representation T which is reducible according to T =P

m T . Sketch the matrix T and the Hamiltonian matrix. What are

the sizes of the block Hamiltonians?

19. Give some examples of irreducible sets of operators:

O�0

i = T (G)O�i T�1(G) =

Xj

O�j T�ji(G) (3.15)

20. De�ne a direct product representation of a group (not to be confused witha direct product group). If T� and T � are two irreducible repesentations ofa group, show that T�� = T�T � is also a representation of the group.Is the representation T�� reducible in general? What is the character ofT��?

21. If f �i g transform according to the representation T� and f��j g transform

according to the representation T � , show that f �i ��j g transform accord-

ing to the direct product representation T�� .

22. Explain the meaning of the Clebsch-Gordan coe¢ cients.

23. Explain the meaning of the Wigner-Eckart theorem:

(f k ; O�i f

�j ) =

Xm

hf jjO�jjf�imC k;ij(m): (3.16)

24. Explain how the following result��i ;H

�j

�=1

l��ij

Xk

(��k ; H �k )

can be understood as a special case of the Wigner-Eckart theorem.


25. What is meant by symmetry breaking?

26. How do we understand Bloch�s theorem k(r+ t) = exp(ik � t) (r) interms of group theory?

27. In molecular vibrations, are the normal modes labelled by the irreduciblerepresentations of the symmetry group dependent on how the local Carte-sian coordinates on each atom are oriented?

Chapter 4

Permutation Groups andYoung Tableaux

Permutation groups are important for several reasons. Cayley�s theorem tellsus that any �nite group of order n is isomorphic to a subgroup of the permu-tation group Sn. This means the representations of a permutation group canalso be used as representations of any �nite group of the same order since theyshare the same multiplication table, irrespective of the type of multiplicationunder which the group is de�ned. In physics, permutation symmetry is associ-ated with identical fermions and bosons. Furthermore, permutation symmetryis intimately connected to the groups of SU(N) whose multiplets remarkablyprovide a mathematical framework for classifying elementary particles.In this chapter, we will construct the irreducible representations of the per-

mutation group Sn of any order n by making use of the Young tableaux.

4.1 Permutation Group SnA permutation of n objects is denoted by

p =

�1 2 3 ::: np1 p2 p3 ::: pn

�:

The meaning of this notation is element 1 is permuted or mapped to elementp1, 2 ! p2, etc. Note that the ordering of the columns is not important; whatis relevant is the information about the mapping of the elements. Thus,

p =

�1 2 3 ::: np1 p2 p3 ::: pn

�=

�3 2 1 ::: np3 p2 p1 ::: pn

�=

�2 3 1 ::: np2 p3 p1 ::: pn

�; etc:

59

60 CHAPTER 4. PERMUTATION GROUPS AND YOUNG TABLEAUX

The identity element is

e =

�1 2 3 ::: n1 2 3 ::: n

�and the inverse of p is

p�1 =

�p1 p2 p3 ::: pn1 2 3 ::: n

�:

Thus in p�1p, p permutes element i to pi and p�1 permutes pi back to i so thatp�1p = e. It is also to be noted that in general

p1p2 6= p2p1:

For example, for

p1 =

�1 2 32 3 1

�; p2 =

�1 2 33 2 1

�;

we �nd

p1p2 =

�1 2 32 3 1

��1 2 33 2 1

�=

�3 2 11 3 2

��1 2 33 2 1

�=

�1 2 31 3 2

�;

whereas

p2p1 =

�1 2 33 2 1

��1 2 32 3 1

�=

�2 3 12 1 3

��1 2 32 3 1

�=

�1 2 32 1 3

�:

An arbitrary permutation of n objects can be written as a product of cycles.As an illustration, consider the permutation

p =

�1 2 3 4 5 65 1 6 4 2 3

�:

In this permutation, 1 is sent to 5, 5 to 2, and 2 is sent back to 1. This formsa three-cycle which is written as (152). Similarly, 3 is sent to 6 and 6 is sentback to 3 and we write this permutation as a two-cycle (36). The element 4 isunpermuted and it forms a one-cyle (4). In the cycle notation

4.2. CAYLEY�S THEOREM 61

p = (152)(36)(4):

Note that within a given cycle only the relative ordering of the elements isrelevant. Thus, for example,

(152) = (521) = (215)

but

(152) 6= (125):

The identity element of n objects can be regarded as a product of n one-cycles.In the cycle notation, only the permuted numbers are shown. For example, forn = 5, the permutation (123)(4)(5) is written as (123) since 4 and 5 remainuntouched.

4.2 Cayley�s Theorem

Every �nite group G of order n is isomorphic to a subgroup of Sn.The proof can be understood by considering the multiplication table of a

group G = (g1; g2; :::gn), e.g., in example 2 of the previous chapter or in the�rst chapter in section Multiplication Table. We see that each row is a distinctpermutation of the elements in the �rst row (or any other row), which is aconsequence of the Rearrangement Theorem. No element in any given row orcolumn appears more than once. Let us represent a given element a 2 G as apermutation

a =

�g1 g2 g3 ::: gnag1 ag2 ag3 ::: agn

�:

From the Rearrangement Theorem the elements in the second row are simplya permutation of the elements in the �rst row. We wish to show that thisrepresentation ful�ls the multiplication table but before we do that we observethat for any element b 2 G


�=

�bg1 bg2 bg3 ::: bgn

a(bg1) a(bg2) a(bg3) ::: a(bgn)

�;

which follows from the fact that only the mapping information of the elements isrelevant so that the columns on the right hand side can be permuted to coincidewith the columns on the left hand side. Suppose that ab = c. Then


ab =


��g1 g2 g3 ::: gnbg1 bg2 bg3 ::: bgn

�=

�bg1 bg2 bg3 ::: bgn

a(bg1) a(bg2) a(bg3) ::: a(bgn)

��g1 g2 g3 ::: gnbg1 bg2 bg3 ::: bgn

�=

�g1 g2 g3 ::: gnabg1 abg2 abg3 ::: abgn

�=

�g1 g2 g3 ::: gncg1 cg2 cg3 ::: cgn

�= c:

Every �nite group of order n is therefore isomorphic to a subgroup of the per-mutation group Sn. It is a subgroup of Sn because in general there are otherpermutations of Sn which are not included in G since the total number of ele-ments in Sn is n! whereas the group G has n elements.

4.3 Classes of Sn

Elements of Sn with the same cycle structure belong to the same class.To show this, consider the conjugation bab�1:

bab�1 =

�1 2 ::: nb1 b2 ::: bn

��1 2 ::: na1 a2 ::: an

��b1 b2 ::: bn1 2 ::: n

�=

�a1 a2 ::: anba1 ba2 ::: ban

��b1 b2 ::: bna1 a2 ::: an

�=

�b1 b2 ::: bnba1 ba2 ::: ban

�We want to show that the last line has the same cycle structure as a. Considera two-cycle (ij) of a, which means ai = j and aj = i. This cycle is carried overinto bi ! bai = bj and bj ! baj = bi. This is evidently true for any number ofcycles and therefore a0 = bab�1 has the same cycle structure as a. Conversely,given two elements a and a0 having the same cycle structure, this result can alsobe used to construct b, such that a0 = bab�1. Therefore elements of Sn with thesame cycle structure belong to the same class.

4.4 Young Tableaux

Our aim is to work out the irreducible representations of the permutation groupSn for any n with the help of Young tableaux. We �rst introduce a number ofde�nitions.

4.4. YOUNG TABLEAUX 63

4.4.1 De�nitions

Partition: A partition � of n is de�ned according to:

� = (�1; �2; :::; �l); where �1 + �2 + :::+ �l = n and �1 � �2 � ::: � �l.

Young diagram: Given a partition � of n, a Young diagram of shape � is anarray of left-justi�ed boxes with �i boxes in row i. For example, for n = 5 thereare seven distinct Young diagrams:

From left to right, the shapes of the Young diagrams are

� = (5); (4; 1); (3; 2); (3; 1; 1); (2; 2; 1); (2; 1; 1; 1); (1; 1; 1; 1; 1):

We observe that a given Young diagram can be associated with a classof the permutation group. We recall that elements possessing the same cy-cle structure belong to the same class. A Young diagram with a given shape� = (�1; �2; :::; �l) can be associated with a cycle structure (�1)(�2):::(�l).The number of classes in a permutation group is therefore equal to the number

of distinct Young diagrams. Consequently, the number of ireducible representa-tions is given by the latter.Young tableau: A Young tableau of shape � is a Young diagram of the same

shape in which the boxes are �lled with numbers 1; 2;...; n in any order withno number repeated. The following is an example of Young tableaux of shape� = (3; 2):

1 4 23 5

3 1 45 2

:

Standard Young tableau: A young tableau �lled with numbers 1; 2; :::; n withincreasing order along any given row (from left to right) and any given column(from top to bottom). Examples:

1 2 43 5

1 3 42 5

:

Young tabloid : Two tableaux t1 and t2 of shape � are row-equivalent, de-noted by t1 � t2, if the corresponding rows of the two tableaux contain the sameelements. For example, the following tableaux are row-equivalent:

1 4 23 5

� 2 4 15 3

� 4 1 23 5

etc.


A tabloid of shape � is such an equivalence class denoted by ~t and drawn withoutthe bars to indicate that the elements in each row can be permuted withoutaltering the tabloid:

1 4 23 5

.

A tabloid can be thought of as a collection of tableaux generated from a tableauof a given shape by permuting the elements in each row in all possible ways.Therefore permuting the elements in a row of a tabloid does not alter the tabloid.Mathematically it is equivalent to symmetrising the elements in each row. Fora Young diagram of shape � = (3; 2) the distinct tabloids are

1 2 34 5

1 2 43 5

1 3 42 5

2 3 41 5

1 2 53 4

(4.1)

1 3 52 4

2 3 51 4

1 4 52 3

2 4 51 3

3 4 51 2

:

For a given partition � = (�1; �2; :::; �l) of n, the number of distinct tabloids isgiven by

n!

�1!�2!:::�l!: (4.2)

There are n! ways of assigning f1; 2; :::; ng to the boxes of a Young diagram ofshape �. A tabloid is not a¤ected by permutations of elements in the same rowand there are �i! ways of permuting the elements in row i. Therefore, there arein total �1!�2!:::�l! ways of permuting elements in the rows without a¤ectingthe tabloid and hence the number of distinct tabloids is n!=(�1!�2!:::�l!). Thenumber of distinct tabloids corresponding to � = (3; 2) is 5!=(3!2!) = 10, as inthe example above.We can de�ne the dot product between two tabloids as the inner product:�

~ti; ~tj�= �ij;

so the dot product is non-zero if the ~ti and ~tj are row equivalent.

4.5 Young tabloids as basis for representationsof Sn

We may regard Young tabloids as basis "vectors" for the representation of thepermutation group Sn. As an illustration, consider the action of a permutation(123) 2 S5 on a tabloid as follows:

(123)1 4 23 5

=2 4 31 5

:

4.5. YOUNG TABLOIDS AS BASIS FOR REPRESENTATIONS OF SN 65

Thus, the tabloid on the left-hand side is mapped to another tabloid, in thesame fashion as a vector is mapped to another vector under the action of agroup element, and can be written as:

g~ti =Xj

~tjM�ji;

where g is a group element and the matrix M� is a representation of g in thebasis of tabloids ~ti of shape �.Let � = (�1; �2; :::; �l) be a partition of n and M� be the representation of

Sn in the basis of Young tabloids. The size or dimension of M� is given by thenumber of distinct tabloids for a given shape �,

dimM� =n!

�1!�2!:::�l!:

Let g 2 Sn have a cycle structure (m1)(m2):::(mr), meaning that g is a productof an m1 cycle, an m2 cycle, ... and an mr cycle (Here all the cycles includingthe one cycles are included).The character or trace of M� representing g is given by the coe¢ cient of

x�11 x�22 :::x

�ll in the generating function

(x1; x2; :::; xl) =rYi=1

(xmi1 + xmi

2 + :::+ xmi

l ) :

To prove this formula, we �rst observe that the character of M� is given by thenumber of tabloids which map to themselves under the action of g:

g~ti =Xj

~tjM�ji = ~ti !M�

ji = �ji:

If g maps ~ti to a di¤erent tabloid then M�ii = 0. Thus, only those tabloids that

map to themselves contribute to the character of M�. In order to ful�ll g~ti = ~tithe elements of ~ti in each row must be such that they can be �tted into thecycles of g. In other words, a cycle of g cannot contain elements belonging totwo di¤erent rows of ~ti since in that case the action of g on ~ti would then mapit to a di¤erent tabloid. Let us write out :

= (xm11 + xm1

2 + :::+ xm1

l ) (xm21 + xm2

2 + :::+ xm2

l ) ::: (xmr1 + xmr

2 + :::+ xmr

l )

The meaning of the �rst bracket is an m1 cycle is placed in row 1, row 2, ... androw l. Similarly, the meaning of the second bracket is an m2 cycle is placed inrow 1, row 2, ... and row l and so on. After multiplying out the product, the termx�11 x

�22 :::x

�ll picks up a placement of cycles of g that �ts a tabloid. Therefore

the coe¢ cient of x�11 x�22 :::x

�ll is precisely the number of tabloids which �t the

cycles of g. These tabloids are not a¤ected by the action of g since for any ofthese tabloids each cycle of g lies within one of its rows (recall that a tabloid isinvariant under permutation of elements within a row).


Let us consider a couple of examples. For the tabloids of shape � = (3; 2) in(4.1) and g = (123) = (123)(4)(5) corresponding to m1 = 3 and m2 = m3 = 1,only the �rst tabloid is una¤ected by the action of g because the cycle (123) canonly �t the �rst row of the �rst tabloid. Hence, the character of M�(g) is equalto one. This agrees with the coe¢ cient of x31x

22 in the generating function

= (x31 + x32)(x1 + x2)(x1 + x2):

As a second example, let us consider g = (12)(34) = (12)(34)(5), correspondingto m1 = m2 = 2 and m3 = 1. Only the �fth tabloid on the �rst row and thelast tabloid in (4.1) are una¤ected by the action of g and hence the character ofM�(g) is equal to two. This agrees with the coe¢ cient of x31x

22 in the generating

function

= (x21 + x22)(x

21 + x

22)(x1 + x2):

The rest of the characters can be calculated readily yielding the following table:Class (1) (12) (123) (1234) (12345) (12)(34) (12)(345)�(3;2) 10 4 1 0 0 2 3

4.6 Specht Modules

Note that a representation is sometimes referred to as a module. We �rst de�nefor a given tableau t of shape � its associated polytabloid as follows:

et =X�2Ct

sign(�)�~t (4.3)

where Ct comprises the permutations of Sn which preserve the columns of t.The meaning of �~t is �rst operate � on t and then from the resulting tableaut0 = �t form a tabloid ~t0 by removing the bars in each row. For example, for

t =1 2 34 5

we have

et =1 2 34 5

� 4 2 31 5

� 1 5 34 2

+4 5 31 2

: (4.4)

The right-hand-side are tabloids in M�.The set of vectors et spans a module or forms a basis for a representation of

Sn known as Specht module or Specht representation S�. The Specht modulesare irreducible and form a complete list of distinct representations of Sn.To show that S� is a representation of Sn, we must verify that �et 2 S� for

every � 2 Sn. First, we observe that

C�t = �Ct��1:

Let us evaluate

4.6. SPECHT MODULES 67

�Ct��1 =

�1 2 ::: n�1 �2 ::: �n

��1 2 ::: nC1 C2 ::: Cn

��1 �2 ::: �n1 2 ::: n

�=

�1 2 ::: n�1 �2 ::: �n

��1 �2 ::: �nC1 C2 ::: Cn

�=

�C1 C2 ::: Cn�C1 �C2 ::: �Cn

��1 �2 ::: �nC1 C2 ::: Cn

�=

��1 �2 ::: �n�C1 �C2 ::: �Cn

�The last line is just what is meant by C�t. If we write Ct in cycle form andconsider a cycle (ijk:::), this cycle is replaced in C�t by (�i�j�k:::). Therefore

Ct =

�1 2 ::: nC1 C2 ::: Cn

�!�

�1 �2 ::: �n�C1 �C2 ::: �Cn

�= C�t:

Now consider

e�t =X�2C�t

sign(�)��~t

=X

�2�Ct��1sign(�)��~t

=X�02Ct

sign(��0��1)��0��1�~t

= �X�02Ct

sign(�0)�0~t

= �et: (4.5)

We have used the fact that sign(�) = sign(��1). The result shows that et ismapped by a permutation element � to another vector within S�, i.e., the setof et forms a basis for a representation of Sn.

4.6.1 The Specht modules are irreducible

We �rst consider a few examples of Specht modules as irreducible representationsof Sn. Consider a tableau t of shape � = (n):

t = 1 2 ::: n :

The associated polytabloid according to (4.3) is

et = 1 2 ::: n :


Evidently �et = et so that et forms the trivial one-dimensional irreducible rep-resentation.Consider now a tableau t of shape � = (1; 1; :::; 1):

t =

12:::n

Since there is only one element in each row, the associated polytabloid accordingto (4.3) is a combination of all possible permutations of the elements alongthe column with the appropriate permutation sign. Operating an arbitrarypermutation � on et gives

�et = e�t =X�2C�t

sign(�)��~t:

De�ning �� = ��, noting that C�t = Ct comprises all permutation elements,and utilising the Rearrangment Theorem yields

�et =X��2Ct

sign(��1)��~t

= sign(�)et:

Thus, et forms a basis for the one-dimensional sign representation, which isdi¤erent from the identity representation.Yet another example of an irreducible representation that can be readily

constructed is the standard representation. Consider a tableau

t =a i j :::b

:

From (4.3) the associated polytabloid is

et =a i j :::b

� b i j :::a

:

The tabloids are labelled by the number in the lower box and there are n inde-pendent of them. Due to the form of the polytabloids with a pair of tabloids ofopposite sign, the number of independent polytabloids, however, is n� 1. Thisprovides a basis for the (n�1)-dimensional irreducible standard representation.To prove that Specht modules are irreducible representations of Sn, we �rst

de�ne a projection operator for a given tableau t of shape �

�t =X�2Ct

sign(�)�;

which when acts on the tabloid ~t reproduces et. Let u =P

i ci~ti 2 U , where Uis a sub-module of M� and ~ti are tabloids of shape �. Operating �t on u gives


�tu =Xi

ciX�2Ct

sign(�)�~ti:

We will show that if �tu is not zero, it must be proportional to et, i.e., �tprojects out the components of u belonging to et. First, we note that

~eti =X�2Ct

sign(�)�~ti =X�2Ct

sign(�)��r~ti;

where �r is a permutation that shu es elements in the same rows of ti but doesnot permute elements belonging to di¤erent rows of ti. The above equation istrue because ~eti consists of tabloids. Assume that ti is such that each columnof �rti has the same elements as those in the corresponding column of t, butnot necessarily in the same order. Now let �c permute elements in the columnsof t such that �rti = �ct. We then have

~eti =X�2Ct

sign(�)��r~ti =X�2Ct

sign(�)��c~t:

Since � consists of column permutations, we �nd

~eti =X

�0��1c 2Ct

sign(�0��1c )�0~t

= sign(��1c )X�02Ct

sign(�0)�0~t = sign(�c)et:

The second step utilises the Rearrangment Theorem. If for a given ti there is no�r such that each column of �rti and the corresponding column of t share thesame elements then ~eti = 0. The reason is that for every � 2 Ct there is anotherelement belonging to Ct that produces the same tabloid but with opposite sign.This does not happen in the case of t because every � 2 Ct acting on t yieldsa new tabloid as can be seen in an example in (4.4). Some examples shouldclarify the point. Consider the following tableaux:

t =1 2 34 5

; t1 =2 3 41 5

; t2 =1 5 24 3

:

In this case Ct = e; (14); (25) and (14)(25). Applying �t on t1 yields

�tt1 =2 3 41 5

� 2 3 14 5

� 5 3 41 2

+5 3 14 2

= �et;

where et is given in (4.4). On the other hand

�tt2 =1 5 24 3

� 4 5 21 3

� 1 2 54 3

+4 2 51 3

= 0:


In the case of t1, �r = (234) will bring �rt1 to a tableau with the same elementsas t in each column but for t2 there is no such �r.To formally prove that �tti = 0 for any tableau ti that cannot be brought

into a tableau that is column-equivalent to t by a row-permutation �r, we notethat there must be numbers a and b in the same column of t and the same rowof ti. The tableau t2 above is such an example, with a = 2 and b = 5. We canthen write Ct in the form of cosets: Ct = �1; :::; �r; �1(ab); :::; �r(ab). However,�k and �k(ab) have opposite signs but otherwise yield the same tabloid whenacting on ti because a and b are in the same row. When applying �t on ti theterms �kti and �k(ab)ti cancel out and hence �tti = 0.We conclude that if �tu 6= 0 then it must be proportional to et, i.e., �t

projects out the et-component of u. Supposing that �tu is a non-zero multipleof et, then et 2 U and it follows that S� � U . On the other hand, suppose thatfor all u 2 U , �tu = 0. Then

(u; et) =

u;X�2Ct

sign(�)�~t

!=X�2Ct

sign(�)�u; �~t

�=

X�2Ct

sign(��1)��1u; ~t

�!=��tu; ~t

�= 0;

and it follows that U ��S��?. Thus, if U is a sub-representation of S�, then

either S� � U or U ��S��?

but since S� is non-zero then U = S�. In otherwords, S� is an irreducible representation of Sn.

Standard Young tableaux as basis for S�

Standard Young tableaux of a given shape � can be used to construct a basisfor the irreducible representation S�. The set fetg where t is a standard Youngtableau of shape � form a basis for S�. Unlike the tabloids, the number ofstandard Young tableaux corresponds exactly to the dimensionality of the irre-ducible representation, which can be calculated using the hook-length formuladescribed below. As an example, consider a tableau of shape � = (4; 1). Thereare 5!=(4!1!) = 5 corresponding tabloids labeled by the number in the secondrow:

1 2 3 45

;1 2 3 54

;1 2 4 53

;

1 3 4 52

;2 3 4 51

;

whereas there are 4 standard Young tableaux:

1 2 3 45

,1 2 3 54

;1 2 4 53

;1 3 4 52

:


The polytabloids generated from the fourth and �fth tabloids are equivalent,except for the sign, so that the number of independent polytabloids et is four,which is what we get if we start from the standard Young tableaux. In general,the standard Young tableaux remove double-counting present in the tabloidsarising from permutations in the columns by imposing the rule that numbersalong any given row (from left to right) or any given column (from top tobottom) is increasing.

The dimension of S�: the hook-length formula

The number of independent polytabloids et corresponding to a given shape � orthe dimension of the Specht module S�, which is the irreducible representationof Sn, is given by the hook-length formula. In each box draw an arrow to theright and an arrow to the bottom. Fill in the box with the number of boxescovered by the arrows. The dimension of the irreducible representation S� isgiven by

n!

m1m2:::mn

where mi is the number in box i. For example, for a given shape � = (3; 2)

! 4 3 12 1

the dimension is 5!=(4:3:1:2:1) = 5. The proof of the hook-length formula israther complicated and it is not included here.

4.6.2 The Specht modules are distinct

For two partitions � = (�1; :::; �r) and � = (�1; :::; �s), � is said to dominate �,� � �, if for each 1 � i � max(r; s) we have

�1 + :::+ �i � �1 + :::+ �i:

By convention, if r > s, we set �i = 0 for i > s, and vice versa.The following is an example for � � �:

� :

Dominance Lemma: Let t and s be tableaux of shape � and �. If for each1 � i � max(r; s) the elements of row i of s are all in di¤erent columns of t,then � � �.Since the elements of row i of s are in di¤erent columns of t, it is quite

evident that we can arrange the entries of each column of t so that the elementsin the �rst i rows of s are placed in the �rst i rows of t, which implies that


�1 + :::+ �i � �1 + :::+ �i

and therefore � � �.Let si be tabloids in M�. Consider

�t

Xi

cisi

!and suppose that �ts 6= 0 for some si appearing in the sum. If two numbers band c appear in the same row of s and the same column of t, then as before wecan write Ct in terms of cosets:

Ct = �1; :::; �r; �1(bc); :::; �r(bc):

Since b and c appear in the same row of s, (bc)s = s and since �i and �i(bc)have opposite signs, it follows that �ts = 0. This means no elements b and ccan be in the same row of s but in the same column of t and by the DominanceLemma � � �. Therefore the only irreducible representation in M� are S� for� � �.

4.6.3 Example: The irreducible representations of the per-mutation group S4

This group has 4! = 24 elements:

1(1); 6(12); 8(123); 6(1234); 3(12)(34).

The number in front of each cycle indicates the number of elements with thatcycle structure. The full tetrahedral group Td containing all the symmetryoperations of a regular tetrahedron, including re�ections, is isomorphic to S4,as can be understood by labeling the vertices of a regular tetrahedron by thenumbers 1; 2; 3; 4. An example is the molecule CH4 with the carbon atom C atthe centre of the regular tetrahedron and the hydrogen atoms H at the vertices.The Young diagrams are

:

As has been worked out before for the general case of Sn, the �rst and lastdiagrams correspond to the one-dimensional identity and sign representationsand the second to the standard representation of dimension 3. According to thehook-length formula, the third diagram generates a 4!=(3:2:2:1) = 2-dimensionalirreducible representation whereas the fourth generates a 4!=(4:1:2:1) = 3-dimensionalirreducible representation.


Let us construct the polytabloids corresponding to the third diagram withshape � = (2; 2). The corresponding standard Young tableaux are

t1 =1 23 4

t2 =1 32 4

:

From (4.3) we construct the polytabloids

e1 =1 23 4

� 3 21 4

� 1 43 2

+3 41 2

;

e2 =1 32 4

� 2 31 4

� 1 42 3

+2 41 3

:

which form basis vectors for a two-dimensional irreducible representation. Forexample, operating (12) on these vectors gives

(12) e1 =2 13 4

� 3 12 4

� 2 43 1

+3 42 1

= e1 � e2;

(12)e2 =2 31 4

� 1 32 4

� 2 41 3

+1 42 3

= �e2:

The matrix representation of (12) is then

(12) =

�1 0

�1 �1

�:

Operating (123) on e1 and e2 gives

(123) e1 =2 31 4

� 1 32 4

� 2 41 3

+1 42 3

= �e2;

(123) e2 =2 13 4

� 3 12 4

� 2 43 1

+3 42 1

= e1 � e2:


(123) =

�0 1

�1 �1

�; etc:

Let us now construct the irreducible representation corresponding to thefourth Young diagram. The standard Young tableaux are

1 234

1 324

1 423

:

The three polytabloids are


e1 =1 234

�1 243

�3 214

+3 241

�4 231

+4 213

,

e2 =1 324

�1 342

�2 314

+2 341

�4 321

+4 312

,

e3 =1 423

�1 432

�2 413

+2 431

�3 421

+3 412

.

Operating (12) on these vectors yields

(12) e1 =2 134

�2 143

�3 124

+3 142

�4 132

+4 123

= e1�e2+e3;

(12) e2 =2 314

�2 341

�1 324

+1 342

�4 312

+4 321

= �e2;

(12) e3 =2 413

�2 431

�1 423

+1 432

�3 412

+3 421

= �e3:


(12) =

0@ 1 0 0�1 �1 01 0 �1

1A ; etc.

The character table of S4 can now be constructed:

1(1) 6(12) 8(123) 6(1234) 3(12) (34)identity 1 1 1 1 1sign 1 -1 1 -1 1standard 3 1 0 -1 -1signstandard 3 -1 0 1 -1other 2 0 -1 0 2

References:Yufei Zhao "Young tableaux and the representations of the symmetric group"Jeremy Booher "Representations of the symmetric group via Young tableaux".



1. If p1 and p2 are two di¤erent permutations of n objects, is p1p2 = p2p1?

2. Write the permutation

p =

�1 2 3 4 5 65 1 6 4 2 3

�in the cycle notation.

3. Prove Cayley�s theorem.

4. Prove that permutations with the same cycle structure belong to the sameclass.

5. De�ne partition, Young diagram, Young tableau, and standard Youngtableau.

6. What is a tabloid? What does it mean mathematically?

7. What is a polytabloid?

8. Polytabloids generated from tabloids form a basis for an irreducible rep-resentation of the permutation group. Explain by means of an example,why some polytabloids generated from tabloids are redundant.

9. Starting from a standard Young tableau of shape �, describe how to con-struct an irreducible representation of the permutation group.

10. Using Young tableaux construct the irreducible representatios of the per-mutation group S3. Check that the character table agrees with that of D3

(rigid equilateral triangle) to which it is isomorphic.

Bibliography

[1] J. J. Sakurai and J. Napolitano, "Modern Quantum Mechanics" (2nd editionPearson Education).

77

group theory - lunds universitethome.thep.lu.se/~malin/fytn13/grouptheoryferdi19.pdfchapter 1...

Documents