GroupTheory-SERC2015

8/16/2019 GroupTheory-SERC2015

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Group Theory in Particle Physics

SERC THEP preparatory, IISER Bhopal, 2015

Anirban Kundu (University of Calcutta)

Joydeep Chakrabortty (IIT Kanpur)Tirtha Sankar Ray (IIT Kharagpur)

Contents

1 Why Should We Study Group Theory? 3

2 What is a Group? 3

3 Discrete and Finite Groups 53.1 Multiplication Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3.2 Isomorphism and Homomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.3 Conjugacy Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Representation of a Group 9

4.1 Reducible and Irreducible Representations . . . . . . . . . . . . . . . . . . . . . . . 11

4.2 Schur’s Lemma and the Great Orthogonality Theorem . . . . . . . . . . . . . . . . 12

5 Lie Groups and Lie Algebras 14

5.1 Algebra of Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

6 SU(2) 19

6.1 Product Representation of SU(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6.2 Weight and Root: A Pictorial Way of Composition . . . . . . . . . . . . . . . . . . 24

6.3 SU(2) and SO(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

6.4 Guiding Principle to Write the Lagrangian . . . . . . . . . . . . . . . . . . . . . . . 26

7 SU(3) 27

7.1 Example: Quark Model and the Eightfold Way . . . . . . . . . . . . . . . . . . . . . 28

7.2 The Method of Young Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

7.3 Three-dimensional Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 33


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7.4 SU(N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

7.5 More on Young Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

7.6 More on roots, weights, Cartan matrices, and Dynkin diagrams . . . . . . . . . . . . 36

8 Lorentz Group 37

8.1 More about Lorentz Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

8.2 The Inhomogeneous Lorentz Group . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

8.3 Euclidean Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

8.4 Symmetry Breaking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

This note is mostly based upon:Tinkham: Group Theory & Quantum MechanicsGeorgi: Lie Algebras in Particle Physics

This is an expanded version of the course on Group Thoery in Particle Physics, given in the SERCPreparatory School held in IISER Bhopal during June-July 2015. Subsections 3.3 and 4.2 were nottouched upon in the course and can be safely dropped at the rst reading. The problems form anintegrated part of the course; hints are provided for the comparatively tougher ones.

We do not discuss discrete groups in detail here. There are many beautiful textbooks on discretegroups, and their applications to physics. Consult them.

The various Grand Unied Theories (GUT) are based on higher groups that break down to the

Standard Model by some symmetry breaking mechanism. This is perhaps one of the most importantapplications of Group Theory in Particle Physics that we did not have time to touch. If you areinterested, look at any standard textbook for GUT groups, coset spaces, and symmetry breakingchains.

There can be typos. The responsibility is entirely ours. We will be thankful if you let us knowabout them. Send an e-mail to [email protected] .

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1 Why Should We Study Group Theory?

Group theory can be developed, and was developed, as an abstract mathematical topic. However,we are not mathematicians. We plan to use group theory only as much as is needed for physicspurpose. For this, we focus more on physics aspects than on mathematical rigour. All complicatedproofs have been carefully avoided, but you should consult the reference books if you are interested.

Almost every time, we have to use some symmetry property of the system under study toget more information (like the equations of motion, or the energy eigenfunctions) about it. Forexample, if the potential in the Schr¨ odinger equation is symmetric under the exchange x → −x(this is known as a parity transformation), even without solving, we can say that the wavefunctionsare bound to have a denite parity. Group theory is nothing but a mathematical way to study suchsymmetries. The symmetry can be discrete ( e.g., reection about some axis) or continuous ( e.g.,rotation). Thus, we need to study both discrete and continuous groups. The former is used morein solid state physics, particularly in crystallographic studies, while the latter is used exhaustivelyin quantum mechanics, quantum eld theory, and nuclear and particle physics.

2 What is a Group?

A group G is a set of discrete elements a,b, · · ·x alongwith a group operator 1, which we will denoteby , with the following properties:

• Closure : For any two elements a, b in G, a b must also be in G. We will try to avoid themathematical symbols as far as practicable, but let me tell you that in some texts this iswritten as ∀a, b∈G, a b∈G. The symbol ∀is a shorthand for “for all”. Another commonlyused symbol is ∃: ∃a∈G means “there exists an a in G such that”.

• Associativity : For any three elements a, b, c in G,a (b c) = ( a b) c, (1)

i.e. , the order of the operation is not important. Note that the positions of a, b, and c are thesame, a (b c) need not be the same as, say, b (a c).

• Identity : The set must contain an identity element e for which a e = e a = a.• Inverse : For every a in G, there must be an element b≡a−1 in G so that b a = a b = e (wedo not distinguish between left and right inverses). The inverse of any element a is unique;

prove it 2.1 In most of the texts this is called group multiplication or simply multiplication operator. Let me warn you that

the operator can very well be something completely different from ordinary or matrix multiplication. For example,it can be addition for the group of all integers.

2 Use reductio ad absurdum . Assume that there are two inverses of an element and show that this leads to acontradiction.

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These are essential properties of a group. Furthermore, if a b = b a for all a, b∈G, the groupG is said to be abelian . If not, the group is non-abelian . Note that no two elements of a group areidentical. If the number of group elements (this is also called the order of the group) is nite, it isa nite group; otherwise it is an innite group . Elements of a nite group are necessarily discrete(nite number of elements between any two elements). An innite group may have discrete orcontinuous elements.

Now some examples:

• The additive group of all integers. It is an abelian innite group. The group operation isaddition, the identity is 0, and the inverse of a is −a. Note that the set of all positive integersis not a group; there is no identity or inverse.• The multiplicative group of all real numbers excluding zero (why not including zero?).• Z n , the multiplicative group of n-th roots of unity (this is also called the cyclic group of ordern). For example, Z 2 = {1, −1}, Z 3 = {1,ω ,ω2} where ω = 11/ 3; Z 4 = {1, i, −1, −i}. Theoperation is multiplication and the identity is 1 (what is the inverse?). If the n-th root of

unity is denoted by x, the elements of Z n can be written as {x, x 2, x3, · · ·xn−1, xn (= 1) }. Suchgroups are called cyclic .• The n-object permutation group S n . Consider a 3-element permutation group S 3 = {a,b,c}.There are six elements: P 0, which is the identity and does not change the position of the

elements; P 12 , which interchanges positions 1 and 2, i.e. , P 12 → (b,a,c). Similarly, there areP 13 and P 23 . Finally, there are P 123 and P 132 , withP 123 (a,b,c) →(c,a,b) , P 132(a,b,c) →(b,c,a), (2)

i.e. , P 123 takes element of position 1 to position 2, that of position 2 to position 3 and that of position 3 to position 1. Obviously, P 123 = P 231 = P 312 , and similarly for P 132 . There are textsthat use some other denitions of the permutation operators, but that is just like renamingthe elements.

• C 4v , the symmetry group of a square. Consider a square in the x-y plane with corners at(a, a ), (a, −a), (−a, −a), and (−a, a ). The symmetry operations are rotations about the z axis by angles π/ 2, π and 3π/ 2 (generally, they are taken to be anticlockwise, but one cantake clockwise rotations too), reections about x and y axes, and about the diagonals.

• U (N ), the group of all N ×N unitary matrices. Thus, the elements of U (1) are pure phaseslike exp(iθ). It is the group of phase transformations. Apart from U (1), all U (N )s are non-abelian. When we say that an individual phase in the wave function does not have any physicalsignicance, we mean that the Lagrangian is so constructed that it is invariant under a U (1)transformation ψ →eiθ ψ.

• SU (N ), the group of all N ×N unitary matrices with determinant unity (also called unitaryunimodular matrices). This is the most important group in particle physics. All SU groupsare non-abelian, of which the simplest is SU (2). The simplest member of SU (2) is the two-

dimensional rotation matrix cosθ sin θ

−sin θ cosθ, characterised by different values of θ.

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• O(N ), the group of all N ×N orthogonal matrices. It is non-abelian for N ≥2.• SO(N ), the group of all N ×N orthogonal matrices with determinant unity. Thus, SO(2)and SO(3) are the familiar rotation groups in two and three dimensions respectively. The S

in SU (N ) and SO(N ) stands for special , viz., the unimodularity property.

The rst group is innite but discrete , i.e., there are only a nite number of group elementsbetween any two elements of the group. The second group is innite and continuous, since thereare innite number of reals between any two real numbers, however close they might be. Z n andS n are obviously discrete, since they are nite. The last four groups are innite and continuous ,which means that there are an innite number of group elements between two given elements of thegroup. We will be interested in only those continuous groups whose elements can be parametrisedby a nite number of parameters. Later, we will see that this number is equal to the number of generators 3 of the group.

Q. Show that the number of independent elements of an N ×N unitary matrix is N 2, and that of anN ×N unimodular matrix is N 2 −1. [Hint: U †U = 1, so det U det U † = 1, or |det U |2 = 1, so that thedeterminant must have modulus unity and a form like exp( iθ).]Q. How many independent elements are there in an N ×N orthogonal matrix, and an N ×N hermitianmatrix?Q. Show that the determinant of a 2 ×2 orthogonal matrix must be either +1 or −1. [Hint: Write thematrix as a bc d and construct the constraint equations. Show that this leads to ( ad −bc)2 = 1.]Q. Check whether the following are groups: (i) All integers except zero for multiplication; (ii) All 2 ×2orthogonal matrices with determinant −1; (iii) All purely imaginary numbers (excluding zero) for multi-plication, and the same (including zero) for addition.

Q. Check that the matrix eiα cos θ eiβ sin θ

−e−iβ sin θ e−iα cos θ is a member of SU (2). Note that α , β , and θ are all

real and independent. Later you will see that an SU (2) member has at most 3 independent elements, andan SU (N ) member has N 2 −1.

3 Discrete and Finite Groups

3.1 Multiplication Table

A multiplication table is nothing but a compact way to show the results of all possible compositionsamong the group elements. Obviously, this makes sense only for nite groups. For example, the

multiplication tables for Z 4 and S 3 are as shown in Tables 1 and 2 respectively.One should note a few things. First, Z 4 is obviously abelian, and hence the multiplication table

is symmetric, but S 3 is non-abelian (e.g., P 13P 23 = P 23P 13). For both of them, all elements occuronly once in each row or column. This is a general property of the multiplication table and is easyto prove. Suppose two elements a b and a c are same. Multiply by a−1, so b = c, contrary to

3 You do not yet know what a group generator is, so wait, or directly go to the Lie group section.

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1 i −1 −i1 1 i −1 −ii i −1 −i 1−1 −1 −i 1 i−i −i 1 i −1

Table 1: Multiplication table for Z 4.

P 0 P 12 P 13 P 23 P 123 P 132P 0 P 0 P 12 P 13 P 23 P 123 P 132P 12 P 12 P 0 P 132 P 123 P 23 P 13P 13 P 13 P 123 P 0 P 132 P 12 P 23P 23 P 23 P 132 P 123 P 0 P 13 P 12P 123 P 123 P 13 P 23 P 12 P 132 P 0P 132 P 132 P 23 P 12 P 13 P 0 P 123

Table 2: Multiplication table for S 3.

our assumption of all elements being distinct. However, number of elements in each row or columnis equal to the order of the group, so all elements must occur once and only once.

Second, (1, −1) or Z 2 is a subgroup (a subset of a group that itself behaves like a group underthe same operation) of Z 4, and (P 0, P 123 , P 132) is a subgroup of S 3. We of course exclude two trivialsubgroups that every group has, the identity element and the entire group itself.Q. There are three more subgroups of S 3 apart from ( P 0, P 123 , P 132 ). Find them.Q. Show that the identity of the bigger group must be the identity of the subgroup too.

Q. Given a member A of the cyclic permutation group,

A =

0 1 0 00 0 1 00 0 0 11 0 0 0

,

nd A2, A3, A4 and show that they form a cyclic group. Let I 4 be the 4 ×4 identity. Find A2, A3, A4 andshow that they form a cyclic group. [Hint: Construct A2, A3 and A4 explicitly. Check that A4 is the 4 ×4identity matrix. Then construct the multiplication table that looks likeA2 A3 A4 A

A3

A4

A A2

A4 A A2 A3

A A2 A3 A4

This Table is symmetric and the cyclic nature is reected in the table.]

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3.2 Isomorphism and Homomorphism

Consider the symmetry group C 3 of an equilateral triangle, with six elements (identity, three re-ections about the medians, and rotations by 2 π/ 3 and 4π/ 3). The multiplication table is identicalwith S 3. Only this (not just the number of elements) shows that the groups behave in an identicalway. This is known as isomorphism : we say that these two groups are isomorphic to one another.

Thus, isomorphism means a one-to-one correspondence between the elements of two groups so thatif a, b, c∈G and p, q, r∈H , and a, b, c are isomorphic to p, q,r respectively, then a b = c implies p q = r , for all a, b, c and p, q, r. The operations in G and H may be completely different. Try toconvince yourself that the identity of one group must be mapped on to the identity of the secondgroup 4.

Thus, S 3 is isomorphic to C 3. If we consider only rotational symmetries, then the three-membersubgroup of C 3 is isomorphic to Z 3 (and also to the ( P 0, P 123 , P 132) subgroup of S 3, which, becauseof isomorphism, we will call Z 3 from now on if there is no chance for any confusion). In fact, thisis a general property: the rotational symmetry group of any symmetric n-sided polygon, having2π/n rotation as a symmetry operation, is isomorphic to Z n . Check this for Z 4. This cannot be a

coincidence; what is the physical reason behind this?If the mapping is not one-to-one but many-to-one, the groups are said to be homomorphic

to one another. Obviously, in a many-to-one mapping, some information is lost. All groups arehomomorphic to the group containing the identity; but that is a very bad mapping, since noinformation about the group structure is retained. A better homomorphism occurs between Z 2 andZ 4, where (1, −1) of Z 4 is mapped to 1 of Z 2, and (i, −i) of Z 4 is mapped to −1 of Z 2. Isomorphismis only a special case of homomorphism, but in general, the two groups which are homomorphic toone another should be of different order, and the ratio of their orders n/m should be an integer k.In this case, set of k elements of G is mapped to one element of H . The set containing identity in Gmust be mapped to the identity of H : prove this. Also prove that this set itself must form a group.

Q. Which of the following sets form a group under conventional multiplication? Construct a group mul-tiplication table for those sets which form a group: (a) A = {1, −1}; (b) B = {1, −1, i, −i,j, − j, k, −k}where i2 = j 2 = k2 = −1, ij = − ji = k, jk = −kj = i, ki = −ik = j . [Hint: Multiplication table for Bis (ll up the empty places)

1 −1 i −i j − j k −k−1 1 −i i − j j −k k

−1 1 k −k − j j−1 −k k j − j

−1 1 i −i−1 −i i

−1 1−1This table is symmetric.

4 Show that if eG → p = eH , this leads to a contradiction.

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Q. Show that the eight matrices

± 1 00 1 , ±i

0 11 0 , ±i

0 −ii 0 , ±i 1 00 −1

.

form a group under matrix multiplication. Show that this group is isomorphic to group (b) of the previousproblem. [Hint: Let us dene the matrices as:

A± = ± 1 00 1 , B± = ±i

0 11 0 , C ± = ±i

0 −ii 0 , D± = ±i 1 00 −1

.

Now it is easy to see that A2± = I , B 2± = C

2± = D

2± = −I ; B±C ± = −D± = D∓, C ±D± = −B± =B

∓, D±C ± = B±.Thus we can conclude that {A±, B±, C ±, D±} form a cyclic multiplication table. and can be easilymapped to {1, −1, i, −i,j, − j, k, −k} set satisfying all the properties and this mapping is one-to-one.]

Q. Which of the following sets form a group under conventional multiplication? Construct a group multi-plication table for those sets which form a group.(a) F =

{1,

−1, i,

−i

} where i2 =

−1. [Hint: Multiplication table is following:

1 −1 i −i1 −i i−1 1

−1This table is symmetric.](b) G = {1, −1, i, −i,j, − j, k, −k,l, −l,m, −m,n, −n,o, −o} where i2 = j 2 = k2 = l2 = m2 = n2 = o2 =−1, products of different i , j ,k,l ,m,n,o anti-commute and the non-trivial products are ij = k, lk =m, mi = n, mj = l, io = l , jo = m, ko = n and their cyclic permutations. [Hint: Show that allmultiplications are not dened and hence it cannot be a group. It will be enough to nd just one suchundened operation.]

3.3 Conjugacy Classes

Consider a group G with elements {a,b,c,d, ···}. If aba−1 = c, (from now on we drop the symbolfor group operation), then b and c are said to be conjugate elements. If b is conjugate to both cand d, then c and d are conjugate to each other. The proof goes like this: Suppose aba−1 = c andhbh−1 = d, then b = a−1ca and hence ha−1cah−1 = d. But ah−1 is a member of the group, andhence its inverse, ha−1, too (why the inverse of ah−1 is not a−1h?). So c and d are conjugate toeach other.

It is trivial to show that the identity element in any group is conjugate only to itself, and for anabelian group all members are conjugate to themselves only.

Now, any discrete group can be separated into sets of elements (need not having the samenumber of elements) where all members of a set are conjugate to each other but no member of anyset is conjugate to another member of a different set. In that case, the sets are called conjugacy classes or simply classes .

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Let us take C 4v, the symmetry group of a square. The symmetry operations are 1 (identity),r π/ 2, r π , r 3π/ 2 (rotations, may be clockwise or anticlockwise), Rx , Ry (reections about x and y axes,passing through the centre of the square), and RNE and RSE , the reections about the NE andthe SE diagonal. [In some texts you will see different notations, but these are equally good, if notmore transparent.]

These eight elements form a group; that can be checked from the multiplication table. Thegroup is non-abelian. However, the rst four members form an abelian group isomorphic to Z 4.The eight elements can be divided into ve classes: (1), ( r π ), (r π/ 2, r 3π/ 2), (Rx , Ry), (RNE , RSE ). Itis left as an exercise to check the class structure.

What is the physical signicance of classes? In other words, can we guess which elements shouldbe in a particular class? The answer is yes: note that the conjugacy operation is nothing but asimilarity transformation performed with the group elements. This will be more obvious in the nextsection when we show how to represent the abstract group elements with matrices, in particularunitary matrices.

Identity must be a class by itself. rπ/ 2 and r3π/ 2 belong to the same class because there is an

element of the group which relates rotation by π/ 2 with rotation by 3 π/ 2: just the reection aboutthe x or y axis, which makes a clockwise rotation anticlockwise and vice versa. Similarly, Rx and Rybelong to the same class since there is an operation that relates them: rotation by π/ 2. However,Rx and RNE cannot belong to the same class, since there is no symmetry operation of r π/ 4.

4 Representation of a Group

The permutation group discussed above is an example of a transformation group on a physicalsystem. In quantum mechanics, a transformation of the system is associated with a unitary operatorin the Hilbert space (time reversal is the only example of antiunitary transformation that is relevantto us). Thus, a transformation group of a quantum mechanical system is associated with a mappingof the group to a set of unitary operators 5. Thus, for each a in G there is a unitary operator D(a),with identity 1 being the operator corresponding to the identity element of the group: D(e) = 1.This mapping must also preserve the group operation, i.e. ,

D(a)D(b) = D(a b) (3)

for all a and b in G. A mapping which satises eq. ( 3) is called a representation of the group G. Infact, a representation can involve nonunitary operators so long as they satisfy eq. ( 3).

For example, the mappingD(n) = exp( inθ ) (4)

is a representation of the additive group of integers, since

exp(imθ )exp( inθ ) = exp( i(n + m)θ). (5)

Also,D(e) = 1 , D(a) = e2πi/ 3 , D(b) = e4πi/ 3 (6)

5 If this mapping is one-to-one, the representation is called faithful . We will deal with faithful representations only.

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is a 1×1 representation of Z 3.Check that the following mapping is a representation of the 3-element permutation group S 3:

D(1) =1 0 00 1 00 0 1

; D(12) =0 1 01 0 00 0 1

; D(13) =0 0 10 1 01 0 0

;

D(23) =1 0 00 0 10 1 0

; D(123) =0 0 11 0 00 1 0

; D(321) =0 1 00 0 11 0 0

. (7)

Particularly, check the multiplication table.

In short, a representation is a specic realisation of the group operation law by nite or innitedimensional matrices. For abelian groups, their representative matrices commute.

Consider a n-dimensional Hilbert space. Thus, there are n orthonormal basis vectors. Let |ibe a normalised basis vector. We dene the ij -th element of any representation matrix D(a) as[D(a)]ij = i|D(a)| j , (8)

so thatD(a)| j =

i|i i|D(a)| j =

i

[D(a)]ij | j . (9)From now on, we will freely translate from one language (representations as abstract linear oper-ators) to the other (representations as matrices). Anyway, it is clear that the dimension of therepresentation matrices is the same as that of the Hilbert space.

Thus, the elements of a group can be represented by matrices. The dimension of the matriceshas nothing to do with the order of the group. But the matrices should be all different, and thereshould be a one-to-one mapping between the group elements and the matrices; that’s what wecall a faithful representation. Obviously, the representative matrices are square, because matrixmultiplication is dened both for D(a)D(b) and D(b)D(a). The representation of identity must bethe unit matrix, and if T (a) is the representation of a, then T −1(a) = T (a−1) is the representationof a−1. The matrices must be non-singular; the inverse exists.

However, the matrices need not be unitary. But in quantum mechanics we will be concerned withunitary operators, and so it is better to show now that any representation of a group is equivalent to a representation by unitary matrices. Two representations D1 and D2 are equivalent if they arerelated by a similarity transformation

D2(a) = SD1(a)S −1

(10)with a xed operator S for all a in the group G.

Suppose T (a) is the representation of a group G. T (a)s need not be unitary. Dene a hermitianmatrix

H =a∈

G

T (a)T †(a). (11)

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This matrix, being hermitian, can be diagonalised by a unitary transformation. Let U †HU = H d.H d is a real diagonal matrix whose elements are the eigenvalues of H . Using the form of H ,

H d = U †a∈G

T (a)T †(a)U =a∈G

U †T (a)U U †T †(a)U =a∈G

T (a)T †(a) (12)

where

T (a) is also a representation (remember U † = U −1). Take the k-th diagonal element of H

d:

[H d]kk ≡dk =a∈

G jT kj (a)T † jk (a) =

a∈

G j|T kj (a)|2. (13)

Thus dk ≥ 0. The case dk = 0 can be ruled out since in that case a particular row in all therepresentative matrices is zero and the determinants are zero, so all matrices are singular. So dk ispositive for all k, and we can dene a diagonal matrix H 1/ 2d whose k-th element is √ dk . Constructthe matrix V = U H 1/ 2d and the representation Γ( a) = V −1T (a)V . We now show that all Γ matricesare unitary, completing the proof.

We have Γ( a) = V −1T (a)V = H −1/ 2d U −1T (a)UH 1/ 2d = H −

1/ 2d T (a)H

1/ 2d , and

Γ(a)Γ†(a) = H −1/ 2d T (a)H 1/ 2d H

1/ 2d T †(a)H −

1/ 2d

= H −1/ 2d T (a)H dT †(a)H −1/ 2

d

= H −1/ 2d T (a)b∈

GT (b)T †(b)T †(a)H −

1/ 2d

= H −1/ 2db∈G

T (ab)T †(ab)H −1/ 2

d

= H −1/ 2d H dH −1/ 2

d = 1. (14)

Here we have used the rearrangement theorem: as long as we sum over all the elements of the group,how we denote them is immaterial.

This theorem depends on the convergence of a number of sums. For an innite group, this isnot so straightforward. However, for Lie groups (to be discussed in the next section) this theoremholds.

4.1 Reducible and Irreducible Representations

A representation D is reducible if it is equivalent to a representation D with block-diagonal form(or itself block-diagonal):

D (x) = SD(x)S −1 = D1(x) 00 D2(x). (15)

The vector space on which D acts breaks up into two orthogonal subspaces, each of which is mappedinto itself by all the operators D (x). The representation D is said to be the direct sum of D1 andD2:

D = D1⊕D2. (16)

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A representation is irreducible if it cannot be put into block-diagonal form by a similarity trans-formation. We will use the shorthand IR for irreducible representations.

One can add up several IRs to construct a bigger reducible representation. There is another wayto construct a bigger representation out of several IRs. Condider two IRs D1 and D2, where D1 ism-dimensional, acting on a vector space |i where i = 1, · · ·m and D2 is n-dimensional, acting onanother vector space

| p where p = 1,

· · ·n. We can make a mn dimensional space by taking basis

vectors labeled by |i and | p in an ordered pair |i, p , which can have mn possible values. Therepresentation D1⊗D2 is called a tensor product representation , which can be obtained by directmultiplication of two smaller representations:i, p|D1(g)⊗D2(g)| j, q ≡i|D1(g)| j p|D2(g)|q . (17)

Q. Show that the 3-dimensional representation of Z 3, given by D(1), D(123) and D(321) of (7), is com-pletely reducible with a similarity transformation by

S = 13

1 1 11 ω2 ω1 ω ω2

(18)

where ω = exp(2 πi/ 3).

4.2 Schur’s Lemma and the Great Orthogonality Theorem

Before closing the section on discrete groups, we will prove a very important theorem on theorthogonality of different representations of a group. This theorem tells you that if you havetwo different representations of a group which are both irreducible but inequivalent to each other,then the ‘dot product’ of these two representations is zero. What is a ‘dot product’? Each of theserepresentations form a g-dimensional vector space, where g is the order of the group. The ‘dot

product’ is something like taking a matrix from one space, taking its corresponding matrix from theother space, taking any two elements of these matrices, and then sum over all elements of the group.A more mathematical denition will soon follow, but before that we need to prove two lemmas bySchur.

Schur’s Lemma 1. If a matrix P commutes with all representative matrices of an irreduciblerepresentation, then P = c1, a multiple of the unit matrix.

Proof. Given, P T (a) = T (a)P for all a∈G, where we take T (a)s to be a unitary representationwithout loss of generality. Suppose the dimension of T (a) is n ×n; evidently, that should be thedimension of P . T (a)s possess a complete set of n eigenvectors. So does P , since [P, T (a)] = 0. Letψ j be some eigenvector of P with eigenvalue c j : P ψ j = c j ψ j . Then P T (a)ψ j = c j T (a)ψ j . So bothψ j = T (1)ψ j and T (a)ψ j , two independent eigenvectors, have same eigenvalue. How many suchdegenerate states exist? Obviously n, since if the number of such degenerate states be m < n , wehave an m-dimensional invariant subspace in the whole n-dimensional vector space, so the originalspace is not irreducible. Thus, all c j s are equal, and P = c1.

Schur’s Lemma 2. Suppose you have two irreducible representations T i(a) and T j (a) of dimensionli and l j respectively. If a matrix M satises T i(a)M = M T j (a) for all a∈G, then either (i) M = 0,a null matrix, or (ii) det M = 0, in which case T i and T j are equivalent representations.

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Proof. The dimension of M is li ×l j . If li = l j , M is a square matrix. If its determinant is notzero, T i and T j are obviously equivalent representations.First, we show, with the help of the rst lemma, that M †M is a multiple of the unit matrix.

Take the hermitian conjugate of both sides of the dening equation:

M †T i†(a) = T j †(a)M †

⇒ M †T i(a−1) = T j (a−1)M †,

M †T i(a−1)M = T j (a−1)M †M ⇒ M †MT j (a−1) = T j (a−1)M †M. (19)This is true for all the elements of G, so M †M = c1. If M is a square matrix and c = 0, then therepresentations are equivalent. If c = 0, then

(M †M )ii =k

|M ki |2 = 0 . (20)

This is true only if M ki = 0 for all k. But i is arbitrary and can run from 1 to n. So M = 0.

Now suppose li = l j . Let li < l j . Add l j −li rows of zero to M to get a square matrix M, whosedeterminant is obviously zero. But M†M= M †M , and since the rst one is zero, so is the secondone. Again take the ( i, i )-th element of M †M to get M = 0.The Great Orthogonality Theorem says that if T i and T j are two inequivalent irreduciblerepresentations of a group G, then

a∈G

T ikm (a)T jns (a−1) =

gli

δ ij δ ks δ mn , (21)

where li and l j are the dimensions of the representations T i and T j respectively, and g is the orderof the group.

Proof. Consider a matrix M constructed as

M =a∈

G

T i(a)XT j (a−1) (22)

where X is an arbitrary matrix, independent of the group elements. Note that if the representationsare unitary then T j (a−1) = T j†(a). Let T i and T j be two inequivalent irreducible representationsof dimensions li and l j respectively. Multiplying both sides of eq. ( 22) by T i(b), where b ∈ G, weget

T i(b)M =a

∈

G

T i(ba)XT j (a−1) =c

∈

G

T i(c)XT j (c−1b)

=c∈

G

T i(c)XT j (c−1)T j (b) = MT j (b), (23)

and so, by the second lemma, M = 0. But this is true for any X ; let X pq = δ pm δ qn , i.e. , only themn -th element is 1 and rest 0. Take the ks-th element of M :

M ks =a∈G p,q

T ikp (a)X pqT jqs (a−1) = 0 . (24)

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But this reduces to

a∈

G

T ikm (a)T jns (a−1) = 0 . (25)

Next, construct a matrix N = a∈G T i(a)XT i(a−1). By an argument similar to eq. ( 23), we getNT i(a) = T i(a)N for all a ∈ G, so N is a multiple of the unit matrix: N = c1. Taking the ks-thelement of N , and the same form of X , we nda∈G

= T ikm (a)T ins (a−1) = cδ ks . (26)

To get c, take the trace of N , which is a multiple of an li ×li dimensional unit matrix:T r(N ) = cli =

a∈

G k,p,q

T ikp (a)X pqT iqk (a−1)

= p,q

X pqa∈

G k

T iqk(a−1)T ikp (a)

= p,q

X pqa∈G

T iqp(1) = g p,q

X pqδ pq = g T r(X ), (27)

so that c = g T r(X )/l i . But T r(X ) = 0 if m = n, or T r(X ) = δ mn . Combining this with eq. ( 25),we get

a∈G

T ikm (a)T jns (a−1) =

gli

δ ij δ ks δ mn . (28)

5 Lie Groups and Lie Algebras

A Lie group is, for our purpose, a group whose elements are labeled by a set of continuous parameterswith a group operation law that depends smoothly on the parameters. This is also known as acontinuous connection to identity . We will be interested in compact Lie groups; in a certain sense,the volume of the parameter space of a compact group is nite. For example, translation is not acompact group, since you can go upto innity; but rotation is, since after a certain time you arebound to come back to the same point. There can be noncompact groups with a compact subgroup;as we will see later, the Lorentz group, consisting of noncompact boosts and compact rotations, hasthe 3-dimensional rotation group SO(3) as a subgroup.

Any representation of a compact Lie group is equivalent to a representation by unitary operators (we have proved an analogous statement for discrete groups earlier). Any group element whichcan be obtained from the identity by continuous changes of innitesimal real parameters a can be

written aslim n→∞(1 + i a X a )

n →exp(iα a X a ) (29)where a, which runs from 1 to N , has been summed over, and α = n is a nite parameter. Thelinearly independent hermitian operators X a , which are called group generator s, form a basis inthe space of all linear combinations αa X a 6. The dimension of this space is N ; this is the space

6 That these operators, as well as their representative matrices, must be hermitian can be seen form the fact thatthe group matrices are unitary and α parameters are real.

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of the group generators, and not to be confused with the Hilbert space on which the generatorsact, which may have a completely different dimension. Thus, a group may have different unitaryrepresentations, but the number of generators must be the same.

Consider a rotation in two dimensions. An innitesimal rotation by θ takes (x, y) to (x + yθ, −xθ + y), so that the only generator of this group is

X = 0 −ii 0 , (30)i.e. , the Pauli matrix σ2. Applying this innitesimal transformation a number of times, we cangenerate any rotation, i.e., any element of the 2-dimensional rotation group. We say that all suchelements are continuously connected to identity , and can be expressed as exp( iθσ2).

The generators, as we have already said, form a vector space; thus, any linear combination of them is a generator. They also satisfy simple commutation relations which determine almost thefull structure of the group, apart from some numerical normalisation. Consider the product

exp(iλX b)exp( iλX a ) exp(−iλX b)exp(−iλX a ) = 1 + λ2

[X a , X b] + · · · (31)Because of the group property, the product of a number of group elements is another group element;it can be written as exp( iβ cX c). As λ →0 we must have

λ2[X a , X b] = iβ cX c (32)

or, writing β c/λ 2 ≡f abc , [X a , X b] = if abcX c. (33)Note that a sum over c is implied; unless otherwise stated explicitly, we will always use the sum-mation convention. The real constants f abc are called the structure constants of the group. They

are completely determined by the group operation law.Q. Show that the 2-dimensional reection ( x, y) → (−x, −y) can be obtained by rotation too (so this iscontinuously connected to identity). Can you get a 3-dimensional reection ( x,y,z ) →(−x, −y, −z) fromrotations?Q. Show that ( 33) does not depend upon the specic choice of generators; an equivalent set of generators,obtained from the original set by some similarity transformation, will also satisfy ( 33).Q. If the generators are hermitian, show that f abc = −f bac .Q. You have earlier shown that the number of independent elements in an N ×N unitary matrix is N 2.How many of them are real angles and how many are phases? [Hint: Compare with an N ×N orthogonalmatrix.]

5.1 Algebra of Generators

The generators satisfy the Jacobi identity

[X a , [X b, X c]] + cyclic permutations = 0 . (34)

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It is obvious for a representation if we just write out the linear operators explicitly; it is also truefor the abstract group generators. In terms of the structure constants, eq. ( 34) becomes

f bcdf ade + f abd f cde + f cad f bde = 0. (35)

If we dene a set of matrices T a by

(T a )bc ≡ −if abc (36)then eq. ( 35) can be written as (note that the structure constants are antisymmetric in at least itsrst two indices; actually, as it will be shown later, it is completely antisymmetric in all its indices)

[T a , T b] = if abcT c. (37)

In other words, the structure constants themselves generate a representation of the group; this iscalled the adjoint representation . If the number of generators be N , the dimension of the adjointrepresentation matrices must be N ×N .

The generators and the commutation relations dene the Lie algebra associated with the Liegroup. Every representation of the group denes a representation of the algebra. The generatorsin the representation, when exponentiated, give the operators of the group representation. Thedenitions of equivalence, reducibility and irreducibility are the same for the algebra as for thegroup.

The structure constants depend on what basis we choose for the generators. To choose an ap-propriate basis, we use the adjoint representation dened by eq. ( 36). Consider the trace T r(T a T b).This is a real symmetric matrix, so we can diagonalise it by choosing appropriate real linear com-binations of the X a s and therefore of the T a s. Suppose we do that, and get

T r(T aT

b) = k

aδ

ab (no sum over a). (38)

We still have the freedom to rescale the generators, so we could choose all the nonzero ka s to havethe absolute value 1. However, we cannot change the sign of ka s.

In our future discussions, we will be concerned with algebras for which all ka s are positive. Thuswe write, with suitable normalisation to all generators,

T r(T a T b) = λδ ab (39)

for any convenient positive λ. We multiply both sides of eq. ( 37) by λ−1T c and take trace to nd

f abc =

−iλ −1Tr([T a , T b], T c) (40)

so that f abc is completely antisymmetric because of the cyclic property of the trace. Here T a s arehermitian; in fact, for compact Lie groups any representation is equivalent to a representation byhermitian operators and all irreducible representations are nite hermitian matrices. Thus, f abc isalways antisymmetric no matter what representation we choose.

An invariant subalgebra is some set of generators of a group G which goes either into itself orzero under commutation with any element of the whole algebra. Thus, if X is any generator of

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the invariant subalgebra and Y is any generator of the whole algebra, [ X, Y ] is a generator of theinvariant subalgebra (or it is zero). An algebra without any nontrivial invariant subalgebra (thewhole and the null set are trivial) is called simple . A simple algebra generates a simple group.

One must take note of the abelian invariant subalgebra s. The generators of an abelian groupcommute with everything; each of these generators is associated with what we call a U (1) factor of the group. At any rate, U (1) factors do not show up in the structure constants, because the structureconstants of an abelian group are all zero. Algebras without any abelian invariant subalgebras arecalled semisimple .

It is the semisimple Lie groups which play a crucial role in particle physics. It can be triviallyshown that any U (N ) (N = 1) group has one U (1) factor:

U (N ) →SU (N ) ×U (1). (41)SU (N ) groups are semisimple. The reason why we deal with SU (N )s rather than U (N )s is thatan arbitrary phase in a wavefunction does not play any role in quantum mechanics (unless we areinterested in interference phenomena, but that is a different story).

The generators, like the linear operators of the representations they generate, can be thought of as either linear operators or matrices:

X a |i = | j [X a ] ji . (42)In the Hilbert space, the group element exp( iα a X a ) transforms a ket |i to exp( iα a X a )|i ; thus, thecorresponding bra will transform like i| →i|exp(−iα a X a ). Any operator O transforms as

O →exp(iα a X a )Oexp(−iα a X a ). (43)From the commutation relation of the generators (eq. ( 33)), it is clear that if X a s (a = 1 to

N ) are the generators of a particular group, so also are −X ∗a s (take the complex conjugate of bothsides). If one can nd some similarity transformation for whichSX a S −1 = −X ∗a (44)

for all a, (i.e. , if they are equivalent) X a s are said to construct a real representation of the group;if not, they construct a complex representation . All irreducible representations of SU (2) are real(reducible representations are just direct sum of irreducible representations). This is not true forother SU (N ) groups; however, the adjoint representations are always real.

Given two representations D1 (dim m) and D2 (dim n) of a group G, we can form anotherrepresentation in two ways. One is to have a direct sum of dimension m+ n. This is a block-diagonal

matrix, which reduces to two irreducible dim- m and dim-n representations. The generators of therepresentation D1⊕D2 are

X D 1⊕D 2a =X D 1a 0

0 X D 2a. (45)

This operation is just the reverse of the process of reducing a representation.

As discussed before, we can also form a m ×n dimensional tensor product representation. If |i (i = 1 to m) is an orthonormal basis in the space on which D1 acts and | p ( p = 1 to n) is

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an orthonormal basis for D2, we can identify the orthonormal product basis |i | p in an m ×ndimensional space. On this space, the tensor product representation D1⊗D2 is(D1⊗D2)(a){|i | p }= D1(a)|i D2(a)| p . (46)

In the matrix language[(D1⊗D2)(a)]ip,jq = [D1(a)]ij [D2(a)] pq. (47)

The generators of the direct product representation are the sums[X D 1⊗D 2a ]ip,jq = [X

D 1a ]ij δ pq + δ ij [X

D 2a ] pq. (48)

Clearly we can form the direct product of an arbitrary number of representations.Q. Prove eq. ( 35).

Q. You know how the ket and the bra transform under the group element exp( iα a X a ). Differentiate withrespect to α and divide by i to show that the action of X a on the ket |i corresponds to the action of −X aon the bra i|, and the commutator [ X a , O] for the operator O.Q. Given the normalisation of the generators, how do you nd out the structure constants for a Lie algebra?[Hint: The generic Lie algebra can be written as [ X i , X j ] = if ijk X k . Now multiply this by any generator

X m , and take trace on both sides:T r ([X i , X j ]X m ) = if ijk T r(X kX m ) .

We can write T r(X kX m ) = λδ km using the orthogonality of the generators, where λ is the normalisationconstant, depending upon the dimensionality of the representation. Thus, f ijk = −iλ T r([X i , X j ]X k ) .]Q. Using the above result, prove that the structure constants are fully antisymmetric in all its indices.[Hint: Easy for the rst two indices. Else, f ikj = −iλ T r([X i , X k ]X j ) = iλ T r([X i , X j ]X k ) = −f ijk , usingthe cyclic property of trace.]Q. Show, for α = n̂θ , that

e−i2 σ· α = 1 cos θ

2 −iσ ·n̂ sin θ2

,

where n̂ is a unit vector in three dimensions and σis are the Pauli matrices. [Hint: Take rotation alongany direction, say z and use σ2i = I . Expand the exponential and collect real and imaginary parts.]

Q. Using Pauli spin matrices matrices, calculate 123 and 223 .

Q. If [A, B ] = B , calculate f (α ) ≡exp( iαA )B exp(−iαA ). This can be done in several ways.• Show that An B = B(1 + A)n . Use method of induction.• Show that ∂f (α )/∂α = f (α ) and solve using the boundary condition for α = 0.• Directly use the Baker-Campbell-Hausdorff identity

ea be−a = b + 1

1![a, b] +

1

2![a, [a, b]] + . . .

Q. Show that the structure constants indeed form a representation of the group. Use the fact that theyare antisymmetric in all their indices.

Q. Show that the adjoint representation of SU (N ) is real.

Q. Show that if a matrix M commutes with all the generators of an irreducible representation of a Liealgebra, M must be a multiple of the unit matrix. This is the Schur’s lemma for continuous groups. (Hint:assume M is hermitian and diagonalise it.)

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6 SU(2)

SU (2) is the simplest special unitary group; it is the group of all unitary unimodular 2 ×2 matrices.In fact, any N ×N unitary unimodular matrix U (i.e. , U U † = U †U = 1, detU = 1) can be writtenin terms of a hermitian matrix H :U = exp( iH ). (49)

From the identity det( eA) = exp( T r A) 7, it follows that H must be traceless. One can write onlyN 2 −1 linearly independent traceless hermitian matrices (can you prove that?), so SU (N ) will haveN 2 −1 generators. Among them, only N −1 can at most be diagonal. This number (of maximumpossible diagonal generators) is called the rank of the group; thus SU (2) is a rank-1 group, SU (3)is rank-2.

Any element of SU (N ) can be written as

U = exp iN 2 −1

a=1

α a X a (50)

where α a s are group parameters and X a s are the generators of the group.

For SU (2), there are only three αa s and three hermitian generators. They are convenientlytaken to be the three Pauli matrices divided by 2:

X 1 = 12

0 11 0 , X 2 =

12

0 −ii 0 , X 3 = 12

1 00 −1

. (51)

The factor of 2 is just to make the whole thing consistent with eq. ( 33) with the structure constantsidentied with the Levi-Civita symbols — note that it is completely antisymmetric. Another reasonis to make the whole thing work for spin-1/2 particles; the diagonal generator X 3 has 1/2 and −1/ 2as its diagonal elements.

Here one must carefully note a point. In the case discussed above, the generators themselvesare N ×N matrices, though they are not unimodular, and hence not members of SU (N ). However,this may not always be the case. For example, the generators in the adjoint representation are(N 2 −1) ×(N 2 −1) matrices. The representation where the generators are themselves of the samedimensionality as the members of the group they generate is called the fundamental representation .This is the lowest dimensional nontrivial representation of the group. We generally denote it by nfor an SU (N ) group. For fundamental representation, the generator normalisation is always set at12 , i.e., T r(X a X b) =

12 δ ab .

We have seen that if X a s generate a group, so also do −X ∗a s. the latter representation is denotedby n and is called the conjugate representation of n , but it is equally fundamental. For SU (2) only,2 and 2 are equivalent; thus it is a real representation. This can be veried from the relationshipsσ2σ1σ−12 = −σ1, σ2σ2σ−12 = σ2, σ2σ3σ−12 = −σ3. For SU (3) (and higher SU (N )s), the fundamentalrepresentation is complex, and is not equivalent to its conjugate.

7 This identity is true for hermitian matrices which can be diagonalized by unitary transformations. Suppose Ais hermitian, then UAU † can be diagonal if U is suitably chosen. Expanding eA , we can say that UeA U † = eA dwhere Ad is diagonal. So det A = det( UeA U †) = det[exp( UAU †)] = det[exp( Ad )] = det[ ea 1 , ea 2 , · · ·] = ea i =exp( a i ) = exp( T r A). We have used the fact that det( ABC ) = det( A)det( B ) det( C ) and det( U )det( U †) = 1.

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Let us now construct the representations of SU (2). This is so much analogous to the angularmomentum algebra that we present it in a rather sketchy way. Indeed, the angular momentumalgebra is nothing but the SU (2) algebra; the spin-1/2 representation is, as we will see shortly, thefundamental representation, and there are higher dimensional representations for higher angularmomentum systems. To make the symbols familiar to you, let us call our generators J a instead of X a .

We rst dene an operatorJ 2 = J iJ i . (52)

For the fundamental representation, J 2 = 34 1 = 12 (

12 + 1) 1. Such operators which commute with all

generators of a given group is called a Casimir operator . J iJ i , the sum of all generators squared,is the only quadratic ( i.e. , second power of the generators) Casimir for any SU (N ) group. ForSU (3), we will encounter cubic Casimirs. The number of independent Casimir operators ( J 4 is notan independent one) is equal to the rank of the SU (N ) group. Anyway, we have

[J 2, J i] = 0 (53)

for all i 8

. We dene the raising and lowering operatorsJ ± = J 1 ±iJ 2 (54)

so thatJ 2 = 12 (J + J − + J −J + ) + J

23 (55)

and[J + , J −] = 2J 3, [J ±, J 3] =∓J ±. (56)

Consider an eigenstate |λm of J 2 and J 3:

J 2

|λm = λ|λm , J 3|λm = m|λm . (57)It is trivial to show that the states J ±|λm are eigenstates of J 2 with eigenvalue λ and of J 3 witheigenvalues m ±1, so that we must haveJ ±|λm = C ±(λ, m )|λm ±1 (58)

where the C ±(λ, m ) are constants to be determined later. Since J 2 −J 23 ≥0, values of m for a givenλ are bounded:

λ −m2 ≥0. (59)Let j be the largest value of m, so that J + |λj = 0. Then

0 = J −J + |λj = ( J 2 −J 23 −J 3)|λj = ( λ − j 2 − j )|λm , (60)or λ = j ( j +1). Similarly, if j is the smallest value of m, λ = j ( j −1). This gives j ( j +1) = j ( j −1),whose only valid solution is j = − j (the other solution, j = j + 1, violates the assumption that

8 This is a general result for any SU (N ) group: [X 2 , X j ] = 0 for all j . To see this, write X 2 as X i X i and use therule for the commutator [ AB,C ]. Remember that the structure constants are completely antisymmetric. Completethe proof.

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|λj is the highest ket). Since J − lowers the value of m by one unit, j − j = 2 j must be an integer,or, in other words, j must be an integer or a half-integer.It is left to you as an exercise to show that

C + |λm = [( j −m)( j + m + 1)]1/ 2,C

−|λm = [( j + m)( j

−m + 1)]1/ 2. (61)

These states | jm with m = j, j −1, · · · − j + 1, − j form the basis of an SU (2) irreduciblerepresentation, characterised by j . The dimension of the representation is 2 j + 1 (thus, representa-tions which are not fundamental can very well be irreducible). The representation matrices can beworked out from the relations

J 3| jm = m| jm ,J ±| jm = [( j∓m)( j ±m + 1)]1/ 2| j m ±1 . (62)

For the fundamental representation, j = 1/ 2, m = ±1/ 2. Let us denote

|1/ 2, 1/ 2 = 10 , |1/ 2, −1/ 2 =01 (63)

so thatJ 3 = 12 σ3, J 1 =

12 σ1, J 2 =

12 σ2 (64)

which follows fromJ + =

0 10 0 , J − =

0 01 0 . (65)

This is the fundamental representation since the generators themselves are 2 ×2 matrices.For j = 1, m =

±1, 0. We denote

|1, 1 =100

, |1, 0 =010

, |1, −1 =001

(66)

so that

J 3 =1 0 00 0 00 0 −1

. (67)

From J + |1, 1 = 0, J + |1, 0 = √ 2|1, 1 , J + |1, −1 = √ 2|1, 0 , we have

J + = 0√

2 00 0 √ 20 0 0

. (68)

The hermitian conjugate is J −. Thus,

J 1 = 1√ 2

0 1 01 0 10 1 0

, J 2 = 1√ 2

0 −i 0i 0 −i0 i 0. (69)

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This is the adjoint representation of SU (2) — the dimension is 3 ×3 9.Q. Check that the generators in the adjoint representation satisfy the SU (2) algebra. Find the similaritytransformation matrix with which you could show that the adjoint representation is real.

Q. Consider an operator Oi (i = 1 , 2), which transforms according to the spin 1 / 2 representation of SU (2)as follows:[J p, Oi ] = i=1 ,2 O j

12(σ p) ji ,

where J p( p = 1 , 2, 3) are generators of SU (2) and σ p are the Pauli matrices. Given,

j = 3 / 2, m = −1/ 2, s|O1| j = 1 , m = −1, t = X,compute,

j = 3/ 2, m = −3/ 2, s|O2| j = 1 , m = −1, t .Here s, t are some quantum numbers other than angular momenta.

Q. Consider the Hamiltonian for a two dimensional simple HO,

H =

ω i=1 ,2 â†i â i + 12

where [â i , â j ] = [â†i , â† j ] = 0 & [â i , â† j ] = δ ij Now consider the operators

N =i

â†i â i and T b =ij

â†iσb2 ij

â j

Show that N and T b commute with H . Compute [ N, T b] and [T a , T b]. What algebra do these four operatorgenerate? Compute the commutators [ N, â†i ] and [T, â†i ]. Use this information to classify the energyeigenstate as irreducible representations of the algebra. Explain the degeneracy number of each energylevel.

6.1 Product Representation of SU(2)

In particle physics, we often face the situation where two or more particles combine to form a system(meson, baryon, atomic nuclei, two-electron systems, etc.). Under a particular symmetry group,one transforms as n and the other transforms as m . How does the total system transform?

Let us be more specic. Suppose we have two spin-1/2 particles; thus, both of them transformas 2 of SU (2) (we often speak loosely as two 2s of SU (2)). How does the combination transform?In other words, what may be the possible spin values of the combination? Here the answer is trivial:the total spin can be 0 or 1. Let us try to establish this from our knowledge of group theory.

Let r denote the rst particle and s the second; r 1 and r 2 are the spin projection +1 / 2 and thespin projection −1/ 2 components (in common parlance, spin-up and spin-down states respectively),and so for s. Under SU (2), they transform as

r i = U ( α)ij r j , sm = U ( α)mn sn , (70)9 This is not the same representation that you get from X a = −i abc , the rule for constructing the adjointrepresentation. But they are equivalent, so there’s no problem.

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where U ( α) = exp( iα a J a ), J a = σa / 2. Possible values of i,j ,m,n depend on the representationsthe states are in. The product representation will transform as

(r ism ) = U ( α)ij U ( α)mn (r j sn ) ≡D( α)im,jn (r j sn ). (71)Generally D( α) is reducible. To see what it decomposes into, take α 1 so that we can directlywork with the generators:

r i = (1 + iα a J (1)a )ij r j , sm = (1 + iα a J

(2)a )mn sn . (72)

Here J (1)a acts only on the rst particle and does not affect the second one; similarly for J (2)a . Thetotal angular momentum is

J = J (1) + J (2) . (73)

The combination with largest value of J 3 (J 3 = 1) is r 1s1. What is the total J of this combination?Apply J 2 on this combination:

J 2

|r 1s1 = [(J (1) )2 + ( J (2) )2 + 2 J (1) .J (2) ]

|r 1s1

= [(J (1) )2 + ( J (2) )2 + J (1)+ J (2)− + J (1)− J

(2)+ + 2J (1)3 J

(2)3 ]|r 1s1 (74)

to nd thatJ 2|r1s1 = 2 |r 1s1 , (75)

so that the total angular momentum of this state is 1. We thus identify it with |1, 1 . Applylowering operators sucessively to nd out

|1, 0 = 1√ 2|r 1s2 + r2s1 , |1, −1 = |r 2s2 . (76)

The remaining state |0, 0 must be orthogonal to |1, 0 :|0, 0 =

1√ 2|r 1s2 −r 2s1 . (77)

This is antisymmetric under 1 ↔ 2, while the earlier three states are symmetric. Thus we say, ingroup theoretic language,2⊗2 = 3⊕1 (78)

which means that two spin-1/2 particles, which are in 2 of SU (2), combine to give a triplet of SU (2) (spin-1) and a singlet of SU (2) (spin-0). The total number of states must remain the same.It also tells us that the representation matrices D of the combined representation is reducible into

two blocks, one of dimension 3×3 and the other of dimension 1 ×1.The fundamental representation of SU (2) is a doublet or 2, but it can have any n dimensionalrepresentation where n ≥ 2. To ascertain this, just think of a system with angular momentum(n −1)/ 2; the representation must be n-dimensional. This is, however, not true for higher SU (N )groups.

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6.2 Weight and Root: A Pictorial Way of Composition

Any state of SU (2) can be specied by only one quantum number, the entries in the diagonalgenerator X 3, provided we know the eigenvalue of X 2. The entries of X 3, i.e., the eigenvalues, arecalled the weight s. Thus, the fundamental representation has only two weights: +1 / 2 and −1/ 2. Itis customary to arrange the members of a multiplet by starting from the highest weight state andgoing downwards. For higher SU (N ) groups, there are more than one diagonal generators, and soeach state is to be specied by N −1 weights 10 . Again, arrangement of the members of the multipletfollows the same rule; if two states happen to have the same weight for the rst diagonal generator,then we take help of the second diagonal generator to remove the degeneracy. An example will beprovided when we discuss the SU (3) group.

Any representation of SU (2) can be shown pictorially by a line, the X 3 axis, with the states attheir respective weights. Thus, for the adjoint representation, there will be three states at +1 , 0and −1 respectively. The vectors that take one weight to the other are called root s 11. For SU (2),they are identical with the raising and the lowering operators. They will all have magnitude 1, butcan point either in the raising or in the lowering direction.

Take an isospin doublet N = pn . Under an innitesimal transformation in the isospin space,

they are transformed as pn −→

pn = [1 + iθσ2]

pn . (79)

What about the electric charges of p and n ? Well, that is an irrelevant question, since we aretalking about an isospin group whose members have different electric charge and hence one mustentertain the possibility of transformations that change the charge. If p and n are states of deniteelectric charge, p and n are not, but what matters is the expectation value of any operator withrespect to these states. However, the main point lies elsewhere. What happens if we take the charge

conjugated states? Apply the charge conjugation operator C on both sides of eq. (79) to get ¯ pn̄ −→

¯ pn̄ = [1 + iθσ2]

¯ pn̄ . (80)

But isospin projection is an additive quantum number, and hence it is −1/ 2 and +1 / 2 for ¯ p andn̄ respectively. Thus, the charge conjugated states are not arranged in descending order of theirweight. But we can interchange the two states; only the transformation matrix will be [1 −iθσ2].So, are 2 and 2̄ different? Again, this is only illusory, as we can put an extra minus sign in front of any member of the charge conjugated doublet:

−n̄

¯ p −→ −n̄

¯ p= [1 + iθσ2]

−n̄

¯ p. (81)

Thus, 2 and 2̄ transform in the same way (that is what we mean by a real representation), providedwe dene the antiparticle doublet as shown above. The treatment is identical for the ( u, d) quark

10 If there are at most p diagonal generators, every state is characterised by p good quantum numbers like(a 1 , a 2 , · · ·a p ). This is called a weight vector with p components.11 Given the roots, one can construct the representation and extract a lot of properties of the group. However, wewill not enter into that.

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doublet, and this gives us the composition of the π mesons, a triplet under SU (2) of isospin:

π+ ≡ −ud̄; π0 ≡ 1√ 2 uū −dd̄ ; π− ≡dū. (82)

There is a normalisation factor for π0, but the relative minus sign is important.

Getting a composite representation is easy. The steps are as follows:

• Suppose you want to combine two representations of dimensionality n1 and n2, i.e. , of multi-plicity 2n1 + 1 and 2 n2 + 1 respectively. Draw the weight diagram (the straight line indicatingthe position of the weights) of the rst representation and indicate the position of the weights.

• Find the centre of mass of the second weight diagram. Put it exactly on top of the weights of the rst representation. If the second representation has 2 n2 + 1 states and you put it onceon all 2n1 + 1 states of the rst representation, the total number of states thus generated is(2n1 + 1)(2 n2 + 1). What you have done is nothing but the generation of a direct productHilbert space. This is the number of states in the combined representation. The combinedrepresentation is reducible and our task is to get the irreducible representations.

• Start from the highest weight state of the combined representation. This must be unique.Its value will be n1 + n2; tick off all the weights of this representation, ending at the lowestweight −n1 −n2. Remember that the actual state of a given weight in a multiplet shouldbe a linear combination of all states of that particular weight that you got by direct product(unless this state is unique). Pictorially, of course, you do not perform the operation of linearcombination.

• Repeat this step till you are left with nothing. Remember that a solitary weight at X 3 = 0represents a singlet. If you like, check that the sum of the number of states in all the irreduciblerepresentations is (2 n1 + 1)(2 n2 + 1).

This seems to be like cracking a nut with a sledgehammer, but we will see soon what happensfor SU (3).

Q. Verify the following decompositions:

3⊗2 = 4⊕24⊗2 = 5⊕33⊗3 = 5⊕3⊕1

2

⊗

2

⊗

2 = 4

⊕

2

⊕

22⊗2⊗3 = 5⊕3⊕3⊕1. (83)

Can you nd out a shortcut way to check these decompositions from angular momentum additionrules?

Q. Suppose you want to combine two states with angular momentum j 1 and j 2. The dimensionalityof the reducible Hilbert space is obviously (2 j1 + 1)(2 j2 + 1). Show that the dimensionalities of allthe irreducible representations that you get add up to this.

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6.3 SU(2) and SO(3)

The group SU (2) is the group of angular momentum, i.e. , it has a close link with rotation. WhySU (2) and not our more familiar rotation group SO(3)? The answer is that SU (2) can take intoaccount half-integer spins and hence fermions, whereas SO(3) is good only for integer spin states.Let us clarify this statement.

Any element of SU (2) can be written as exp( iα a X a ). If we are in the fundamental representation,X a = 12 σa . Thus, e.g., the element with α 1 = α3 = 0 can be written as

exp iα22

σ2 = cos α22

+ iσ2 sin α22

= cos α 2

2 sin α 2

2

−sin α 22 cos α 22 (84)

(prove this!), which is nothing but the ordinary rotation matrix, but with half-angles. For α2 = 2π,the group element is −1 and not 1! Thus, a rotation of 2 π acting on the spin-1/2 wavefunctions willbring an extra minus sign: ψ → −ψ. This is something we do not expect for ordinary rotations:a rotation of 2 π should bring the system back to the original conguration. Here, for SU (2), youneed a rotation of 4 π to get back the same wavefunction.

The wavefunctions which pick up an extra minus sign under 2 π rotation are called spinors . Theyare multicomponent objects, since the multiplicity under spin has to be incorporated (this will beelaborated when we discuss Lorentz groups). It is also intuitively clear that for every two rotationsof SO(3), there is only one rotation of SU (2) (e.g., rotations by 2 π and 4π of SO(3) correspond toa rotation by 4 π of SU (2); in general, ( θ, 2π + θ)SO (3) →2θSU (2) .). This is why SU (2) is called thecovering group of SO(3): there is a homomorphism of 2-to-1 between these groups.

To prove homomorphism between Lie groups, we must show that they follow the same algebra(for the discrete groups, we checked the multiplication tables). We will take the generators of SO(3)to be those responsible for rotations about x, y, and z -axes. This is not the Eulerian choice, but solong as we take independent rotations, all are equally good.

For an innitesimal rotation by θ about the z -axis, we havex = x + yθ, y = −xθ + y, z = z, (85)

so that a vector r = ( x,y,z ) transforms to another vector r = ( x , y , z ) according to

r = (1 + iθJ 3)r , J 3 =0 −i 0i 0 00 0 0

. (86)

Similarly we can construct J 1 and J 2. It is left for you to show that [ J i , J j ] = i ijk J k , so that theysatisfy the same Lie algebra as SU (2). You should also show that in the fundamental representation,J 2 = J iJ i = 2, so this describes systems with angular momentum one (as j ( j + 1) = 2).

6.4 Guiding Principle to Write the Lagrangian

Any quantum mechanical system should contain wavefunctions that transform nontrivially undersome continuous group transformation. What about the total Lagrangian, or the Lagrangian den-sity, or the integrated action? Before proceeding further, let us just state the basic principle towrite any Lagrangian:

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The Lagrangian may contain elds which transform non-trivially under a given group,but the whole Lagrangian must be a singlet ( i.e, transform as 1) under the group.

This is true if the elds transform non-trivially under a number of different groups. Also notethat for groups with additive quantum numbers like electric charge or hypercharge, the singlet isgenerally denoted by 0 and not 1.

What does this mean? If you transform some eld in a non-trivial way, some other eld willtransform in such a way as to cancel that effect, and the whole Lagrangian is a scalar in the groupspace. This is analogous to why the Lagrangian, or the Hamiltonian for that matter, must bea scalar in ordinary 3-dimensional space, and why it must be a Lorentz scalar in 4-dimensionalspace-time. After all, a 3-dimensional rotation is the group SO(3); a 4-dimensional boost is a partof what is known as a Lorentz group.

Electromagnetism is a U (1) theory; U (1) is the group of Maxwellian gauge transformation,which is nothing but a phase transformation. Any eld which is a singlet under U (1) has a U (1)quantum number zero; this quantum number may be identied with the ordinary charge of U (1)em .Consider the Maxwell Lagrangian

L= eψ̄γ µψAµ (87)where ψ is the electron spinor. It has a charge −1, but ψ̄, which refers to a positron, has charge+1, and Aµ , the photon eld, has charge zero; so the total U (1) quantum number is zero andL is a singlet under U (1)em — in other words, electric charge is conserved under electromagneticinteraction.Q. Show that the Dirac Lagrangian L = ψ̄(iγ µ∂ µ −m)ψ is invariant under a global U (1) transformationψ →exp( iθ)ψ. Show that it is not invariant if the phase transformation is local: ψ →exp( iθ(x))ψ. Showthat the extra term can be cancelled by replacing ∂ µ with ∂ µ + iQeA µ(x), where Qe is some constant,and demanding a similar transformation for Aµ . Such local phase transformations are also called gaugetransformations and hence electromagnetism is called a U (1) gauge theory. There is only one generatorfor U (1), so there is only one force carrier, the photon, as the force carriers, also called gauge bosons, livein the adjoint representation of the group. The gauge groups of weak and strong interactions are SU (2)and SU (3) respectively, so there are 3 weak gauge bosons and 8 strong gauge bosons.

7 SU(3)

The next higher SU group is SU (3). As far as I know, the application of this group is limited onlyto particle physics, but there it has at least two very important applications. This is not a place todiscuss them; suffice it to say that this is the group for strong interaction, as well as the group for

the quark model of Gell-Mann that led to the famous eightfold way.SU (3) has eight generators, and two of them can be simultaneously diagonalised at most; this

is a rank-2 group. The generator matrices are conventionally written as X a = 12 λa , a running from1 to 8. The Gell-Mann matrices λa are as follows:

λ1 =0 1 01 0 00 0 0

, λ2 =0 −i 0i 0 00 0 0

, λ3 =1 0 00 −1 00 0 0

,

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λ4 =0 0 10 0 01 0 0

, λ5 =0 0 −i0 0 0i 0 0

, λ6 =0 0 00 0 10 1 0

,

λ7 =0 0 00 0 −i0 i 0

, λ8 = 1√ 3

1 0 00 1 00 0

−2

. (88)

Note that all matrices are traceless and hermitian, and just like the fundamental representation of SU (2), the generators are diagonalised according to T r(X iX j ) = 12 δ ij .

The fundamental representation 3 is complex, i.e. , inequivalent to 3̄. The states are labelled bythe weights, the eigenvalues corresponding to X 3 and X 8. Note that λ1, λ2, and λ3 constitute anSU (2) in the 1-2 block. The three Gell-Mann matrices λ2, λ5 and λ7 form the adjoint representationof SU (2).

SU (3) has a maximal subgroup SU (2)⊗U (1). Thus 3 of SU (3) can be decomposed into followingirreducible representations of SU(2): 3 = 2(1)⊕1(−2), where the U (1) charges are given in theparenthesis.

7.1 Example: Quark Model and the Eightfold Way

For concreteness, let us denote the three states of 3 by u, d, and s respectively, with orthonormaleigenvectors:

u =100

, d =010

, s =001

. (89)

These states are labeled by the third component of isospin, which is nothing but the entries of X 3,

and a quantum number called strangeness , which is a scaled variant of X 8. Strong hypercharge isdened as 2X 8/ √ 3, and subtracting baryon number (1/3 for all quarks) we get what is known asthe strangeness:

S = 2√ 3X 8 −

13

. (90)

Thus, for the three quarks, the I 3-S assignments are as follows:

u =12

, 0 , d = −12

, 0 , s = (0 , −1) . (91)

The weight diagram is shown in g. 1. For the antiquarks, which live in 3̄, the assignments are

just opposite. What happens when we combine 3 and 3̄? From a physics point of view, that is acombination of a quark and an antiquark, and should give some mesons. The question is: how dothese mesons transform under SU (3)?

We follow the same principle of composition: take the antiquark triangle and place its centreof mass on the three vertices of the quark triangle. The resulting regular hexagon is shown in g.2. There are six states at the corners of the hexagon, and three in the middle. The names of themesons are the standard ones used by particle physicists. There are a few points to note.

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!

" #$$

$

%&'()! %&'()

)( ! %

ud

s

!

" #

! %&'() %&'()

u d

s$ )(%

$$

_

_

_

Figure 1: 3 and 3̄ of SU (3) . Note that due to a scaling of X 8 to S , the centre of mass of the two triangles are not the same. This would have been the case if we drew the triangles in X 3-X 8 plane.

!!"!

!!

!!

"! #

"

"

!

#

#$$$$

!

"

#$%! $

$

! $

Figure 2: Composition of 3 and 3̄ of SU (3) .

• If we worked with only u and d quarks, that would have been an SU (2) doublet of isospin,and so among the composites, I should have a triplet. Experimentally, the middle memberof the triplet happens to be π0, see eq. (82) for the explicit wavefunctions. An SU (2) isospintransformation can be performed by the interchange of u and d, i.e., a shift along the horizontalline (that is obvious, S cannot change, since both quarks have zero strangeness). A look atg. 2 immediately tells you that there are two more SU (2) doublets: K 0 and K + , and theirantiparticles, K 0 and K −.

• The roots of SU (3) take one weight to the other. So they should act along the sides of thehexagon, and all members at the sides must belong to the same multiplet. By the isospinargument, π0 should also belong to the same multiplet. So what are the irreducible represen-

tations: 7⊕2, or 7⊕1⊕1, or what? The answer is that for any SU (n), n ⊗̄n = n2

−1⊕1.(How do we know this, at least for SU (3)? We will see soon how to have an algorithm forgetting the irreducible representations.) So the decomposition is 8⊕1. The singlet is onewhich does not transform under SU (3), and hence must be symmetric in the three quarksand antiquarks. This state is labeled as η = 1√ 3 uū + dd̄ + ss̄ . From the T 3 and S quantumnumbers, it is one of the states at the middle of the hexagon. The third state must be anorthogonal one to this and π0, and can be obtained by the standard Gram-Schmidt method

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12: η = 1√ 6 uū + dd̄ −2ss̄ .• These nine lowest mass meson states can be divided into an octet and a singlet. There arethree different sets of raising and lowering operators for the octet (they have no effect on the

singlet state). One of them is the set of isospin raising or lowering operators, obtained froma combination of X 1 and X 2. The other two sets are called U-spin and V-spin respectively.

V-spin operators are those obtained from X 4 and X 5, and can change an up quark to a strangequark and vice-versa. U-spin operators, obtained from X 6 and X 7, take a d quark to an squark and back. Note that only U-spin does not change the charge of the quarks and henceof the mesons.

7.2 The Method of Young Tableaux

This is the most elegant method to get the irreducible representations out of a composite. Themathematical background is somewhat complicated, and needs tensor algebra, so we will skip thathere. This forces us to state only the basic rules of the game. The algorithm goes like this:

• Each representation of SU (3) (well, there is nothing special about SU (3), it can be generalisedfor any SU group) is pictured as a set of boxes.Some simplest representations of SU (3) are like:

3 : , 3̄ : , 8 : , 1 : , 6 : , 10 :

This does not mean that a box corresponds to a wave function. Rather, the whole set, and itssymmetry and antisymmetry properties, are depicted by the box pattern. The boxes must be

– left-aligned;– arranged in such a way that the number of boxes does not increase as one goes down the

columns (it may decrease or remain constant);– arranged in not more than three rows (this number depends on what SU group you

choose; e.g., it will be ve for SU (5)).

• With such a pattern, let the difference between the number of boxes in the rst and the secondrows be p, and the difference between the second and the third rows be q . The dimensionalityof the representation, denoted also as ( p, q ), is given by

d = 12

[( p + 1)( q + 1)( p + q + 2)] . (92)

Thus, a single box corresponds to ( p, q ) = (1 , 0), and hence it has dimensionality 3; thisis the fundamental representation of SU (3). The arrangement of two boxes in a column is

12 I have cheated you a bit. The actual singlet and octet states are not the physical states η and η; rather, thelatter are a combination of the former states. However, the mixing angle is small, so for the group theory purpose,if not for particle physics purpose, the assignment that I use will do.

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( p, q ) = (0 , 1) and hence this is 3̄. The representations with p = q are the real representa-tions; setting aside p = q = 0, which is the singlet of dimensionality 1, the lowest possiblearrangement is ( p, q ) = (1 , 1)⇒8, the adjoint representation. (What is the dimensionality of

the representation ?)

• Three boxes in a column has p = q = 0. This is a singlet, and anything multiplied byunity is the same thing, so composition with a singlet is trivial. That is why this particularcombination is, so to say, irrelevant; to get the dimensionality of a representation, we canadd or subtract at our will as many number of three boxes in a column as we wish. In anycombination, if you happen to have any such arrangement, you can easily scratch that off (unless you require them to have a feeling of what q will be).

• Boxes in a row denote a symmetric combination of wavefunctions, and boxes in a columndenote antisymmetric combination. While trying a composition, one must remember thatsymmetrised wavefunctions cannot be antisymmetrised, and vice versa. Boxes which are notin a particular row or column can be placed anywhere. A singlet is a state where the wholewavefunction is antisymmetric, i.e. , it picks up a minus sign every time you interchange twostates:

|1 = 1√ 6 (uds −dus + dsu −sdu + sud −usd) . (93)

This is a linear combinaion of six wavefunctions; the rst means that u is at position 1, d at2, and s at 3. Interchange u and d and the whole wavefunction picks up a minus sign. Theprefactor is for normalisation.

• Double counting must be avoided.Fortunately, the most important applications of SU (3) Young tableaux are the most simple

ones. As an example, let us try to emulate Murray Gell-Mann in his search for the eightfold way.Fig. 3 shows the nal result, but we have to go through the intermediate steps.

• Unless this is a very simple composition (like joining two 3s), always label all the boxes of the second ( i.e. , the one on the right hand side) tableau 13. Label all boxes of a row in anidentical way, but different from boxes in another row. For example, label all boxes of therst row by a, and all in the second row by b (the third row never counts; a full column canalways be eliminated).

• Take one box from the rst row and join it in as many possible ways with the rst tableauas possible. Of course, you have to join it only on the right hand side or below. Rememberthat the number of boxes in each row of this combined representation (even if this is anintermediate step) should never increase as you go down the columns. The operation givesnothing but 3⊗3 = 6⊕3̄.

a ⊗ bc = a b ⊕ ab ⊗ c = a bc ⊕abc ⊕

a b c ⊕ a cb

Not allowed13 Tableau is the singular form, whose plural is tableaux.31


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• Repeat this step until you nish of all the boxes in the rst row. Then start the game with theboxes in the second row. Here you must be a little careful. Naively, you would have obtainedall the four combinations shown above, but two of them are not allowed: boxes b and c arein a column, i.e. , in an antisymmetric combination, and cannot be symmetrised, i.e. , put ina row. Also, in no tableau one should have c sitting above b (they can be in the same row,if initially they belong to different columns and hence without any antisymmetry property).This is nothing but avoiding double counting. Thus, two of the boxes are ruled out, and weget the required decomposition.

• However, a smarter way is⊗ = ⊕

as I can always take 3 as my second box. Then only two tableaux appear and there is noquestion of any double counting!

• As a further example, let us consider the combination of three quarks: 3⊗3⊗3.

a × b × c = a b + ab × c = a b c + a bc + a cb +abc

The process is a two-step one; rst we combine two 3s, and then combine the result with thelast 3. Mathematically, we write

3⊗3⊗3 = ( 6⊕3̄)⊗3 = 10⊕8⊕8⊕1. (94)

The 3 ×3 λ matrices satisfy the following relations:[λa , λ b] = 2if abcλc, {λa , λ b}= 43δ ab1 + 2 dabcλc, T r(λa λb) = 2 δ ab . (95)

The completeness relation is(λa )ij (λb)kl = −

23

δ ij δ kl + 2 δ il δ jk (96)

where i, j, k, l = 1, 2, 3. The quadratic and cubic Casimir operators are dened as

C 2(R)1 ≡8

a=1

(X a )2(R), C 3(R)1 ≡8

a,b,c =1

dabcX a (R)X b(R)X c(R), (97)

where R is the dimension of the representation (3 for fundamental) and X a = λa / 2. It can be shownthat

C 2( p, q ) = 1

3 p2 + pq + q 2 + 3 p + 3q ,

C 3( p, q ) = 118

( p−q )(2 p + q + 3)(2 q + p + 3) . (98)This gives, e.g., C 2(3) = 4 / 3, C 2(8) = 3.

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Now, some important relations. From eq. ( 96), we have

(λa )ij (λa ) jl = −23

+ 2 ×3 δ il = 4C 2(3)δ il . (99)

From the denition of the adjoint representation ( X a )bc = −if abc ,f acd f bcd = C 2(8)δ ab = 3δ ab . (100)

Q. Convince yourself that ( p, 0)⊗(1, 0) = ( p+ 1 , 0)⊕( p−1, 1). From the expression of the dimensionalityd, check that the dimensionalities match on both sides.Q. Perform the topnotch exercise: 8⊗8 = 27 ⊕10 ⊕10 ⊕8⊕8⊕1 . Be careful to avoid double counting.Also preserve the symmetry and the antisymmetry properties.

Q. Find ⊗ and ⊗ .Q. Show that f abcλbλc = iC 2(8)λa . (Hint : write λbλc = 12 [λb, λ c] since f abc is completely antisymmetric,

and then use eq. ( 100).)Q. Justify the steps:

λbλa λb = 1

2 (λb[λa , λ b]−[λa , λ b]λb + λbλbλa + λa λbλb)

= 4 C 2(3)λa + if abc [λb, λ c] = 4 C 2(3) − 12

C 2(8) λa . (101)

7.3 Three-dimensional Harmonic Oscillator

As another application of SU (3), consider the three-dimensional isotropic harmonic oscillator, whoseHamiltonian is given by

H = 12m

pi pi + 12

mω2xixi , (102)

where we have used the summation convention, with i = 1, 2, 3. The energy of the ( n1, n 2, n 3) stateis given by E n 1 n 2 n 3 = ω n1 + n2 + n3 + 32

14. The ground state is nondegenerate; the rst excitedstate is 3-fold degenerate ( E 100 = E 010 = E 001 ); the second excited state is 6-fold degenerate, andso on. To explore the underlying symmetry, let us write the Hamiltonian in terms of creation andannihilation operators:

ak = 1√ 2mω (mωxk + ipk) , a†k = 1√ 2mω (mωx k −ipk) , (103)as

H = ω a†kak + 32

. (104)

14 We will consistently use = 1.

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If U is some unitary matrix so that U ∗kl U km = δ lm , the Hamiltonian is invariant under a transfor-mation

ak →ak = U kl a l , a†k →a †k = U ∗kl a†l . (105)This, we may infer that the system has a U (3) symmetry. But note that we are interested inthe degeneracies of states belonging to a particular eigenvalue of H (or of a†kak). The unitarytransformation exp( iαa †

kak), where the sum over k is implied, just gives a common overall phase

which is irrelevant. This means that we have to factor out the U (1) part of U (3) and what is leftis SU (3), which is the underlying symmetry of the system.

The generators can be constructed out of a† and a: X ij = a†i a j . But only eight out of nine suchcombinations are independent, as a†kak = n1 + n2 + n3 is xed for a given level. The vacuum isunique, and any state with total energy E = ( N + 32 )ω can be constructed by N number of creationoperators acting on the vacuum

a†a†· · ·a†|0 . (106)All the a†s commute with each other, so this must be a symmetric representation of SU (3), with p = N and q = 0. Thus, the number of degeneracies for any given N is just the dimen

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