F R () N T I L R \ I ~ P ti Y S I (: S SECOND EDITION LIE ALGEBRAS 1N PARTICLE PHYSICS From Isospin to Unified Theories Ho\Yctrd (;corgi Lie Algebras in Particle Physics Second Edition Howard Georgi ',, __ __.____ / ~estJview \ /PIIIS.9 Advanced Book Prouram b"' "'--r--, A Member of the Perseus Books Group Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book and Perseus Books was aware of a trademark claim, the designations have been printed in initial capital letters. Library of Congress Catalog Card Number: 99-64878 ISBN: 0-7382-0233-9 Copyright© 1999 by Howard Georgi All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. Westview Press is a Member of the Perseus Books Group To Herman and Mrs. G Preface to the Revised Edition Lie Algebras in Particle Physics has been a very successful book. I have long resisted the temptation to produce a revised edition. I do so finally, because I find that there is so much new material that should be included, and so many things that I would like to say slightly differently. On the other hand, one of the good things about the first edition was that it did not do too much. The material could be dealt with in a one semester course by students with good preparation in quantum mechanics. In an attempt to preserve this advantage while including new material, I have flagged some sections that can be left out in a first reading. The titles of these sections begin with an asterisk, as do the problems that refer to them. I may be prejudiced, but I think that this material is wonderful fun to teach, and to learn. I use this as a text for what is formally a graduate class, but it is taken successfully by many advanced undergrads at Harvard. The important prerequisite is a good background in quantum mechanics and linear algebra. It has been over five years since I first began to revise this material and typeset it in IbTp(. Between then and now, many many students have used the evolving manuscript as a text. I am grateful to many of them for suggestions of many kinds, from typos to grammar to pedagogy. As always, I am enormously grateful to my family for putting up with me for all this time. I am also grateful for their help with my inspirational epilogue. Howard Georgi Cambridge, MA May, 1999 xi Contents Why Group Theory? 1 1 Finite Groups 2 1.1 Groups and representations 2 1.2 Example- Z3 . . . . " . . 3 1.3 The regular representation 4 l.4 Irreducible representations 5 1.5 Transformation groups .. 6 1.6 Application: parity in quantum mechanics 7 1.7 Example: S3 . . . . . . . . . · 8 1.8 Example: addition of integers . 9 1.9 Useful theorems . 10 1.10 Subgroups . . . . . . . . 11 1.11 Schur's lemma . . . . . 13 1.12 * Orthogonality relations 17 1.13 Characters . . . 20 1.14 Eigenstates ....... 25 1.15 Tensor products . . . . . 26 1.16 Example of tensor products . 27 1.17 * Finding the normal modes 29 1.18 * Symmetries of 2n+l-gons 33 1.19 Permutation group on n objects . 34 1.20 Conjugacy classes . . . . . . . 35 1.21 Young tableaux . . . . . . . . 37 l.22 Example - our old friend S3 . 38 1.23 Another example - S4 . . . . 38 1.24 * Young tableaux and representations of S11 38 xiii xiv 2 Lie Groups 2.1 Generators . 2.2 Lie algebras 2.3 The Jacobi identity 2.4 The adjoint representation 2.5 Simple algebras and groups . 2.6 States and operators .. 2.7 Fun with exponentials . 3 SU(2) 3.1 J3 eigenstates . . . . . . . . . 3.2 Raising and lowering operators 3.3 The standard notation 3.4 Tensor products 3.5 Ja values add 4 Tensor Operators 4.1 Orbital angular momentum 4.2 Using tensor operators .. 4.3 The Wigner-Eckart theorem 4.4 Example . . . . . . . . . . 4.5 * Making tensor operators 4.6 Products of operators 5 Isospin 5.1 Charge independence 5.2 Creation operators . 5.3 Number operators . . 5.4 Isospin generators . . 5.5 Symmetry of tensor products 5.6 The deuteron .... 5.7 Superselection rules . . . . . 5.8 Other particles . . . . . . . . 5.9 Approximate isospin symmetry . 5.10 Perturbation theory .... " ... 6 Roots and Weights 6. I Weights ... . . . . . . . . . . . 6.2 More on the adjoint representation . 6.3 Roots . . ~ . . . . 6.4 Raising and lowering " . . . . . . CONTENTS 43 43 45 47 48 51 52 53 56 56 57 60 63 64 68 68 69 70 72 75 77 79 79 80 82 82 83 84 85 86 88 88 90 90 91 92 93 CONTENTS xv 6.5 Lots of SU(2)s ......... . 93 6.6 Watch carefully - this is important! 95 1 SU(3) 98 7.1 The Gell-Mann matrices 98 7.2 Weights and roots of SU(3) 100 8 Simple Roots 103 8. I Positive weights . 103 8.2 Simple roots . . . 105 8.3 Constructing the algebra 108 8.4 Dynkin diagrams 111 8.5 Example: G2 ... 112 8.6 The roots of G2 .. 112 8.7 The Cartan matrix . 114 8.8 Finding all the roots . 115 8.9 The SU(2)s ..... 117 8.10 Constructing the G2 algebra 118 8.1 l Another example: the algebra 0 3 • 120 8.12 Fundamental weights .. 121 8.13 The trace of a generator . . . . . . 123 9 More SU(3) 125 9.1 Fundamental representations of SU(3) . 125 9.2 Constructing the states 127 9.3 The Wey! group . . . . . . . . .. 130 9.4 Complex conjugation ...... . 131 9.5 Examples of other representations 132 10 Tensor Methods 138 10.1 lower and upper indices . . . . . . . . , . 138 10.2 Tensor components and wave functions 139 10.3 Irreducible representations and symmetry 140 10.4 Invariant tensors . . . . . . . . . . 141 10.5 Clebsch-Gordan decomposition .. 141 10.6 Triality . . . . . . . . . . . . 143 I0.7 Matrix elements and operators 143 I0.8 Normalization . . . . . . 144 10.9 Tensor operators . . ... 145 lO.IOThe dimension of (n, m) 145 10.11 * The weights of (n, m) . 146 XVI l 0.12Generalization of Wigner-Eckart . . . . . . 10.13* Tensors for SU(2) . . . . . . . . . . . . 10.14 * Clebsch-Gordan coefficients from tensors 10.15* Spin s1 + s2 - 1 10.16 * Spin s 1 + s2 - k . . . . . . . . . . . . . 11 Hypercharge and Strangeness 11.1 The eight-fold way . . . . 11.2 The Gell-Mann Okubo fonnula . 11.3 Hadron resonances 11.4 Quarks . . . . . . . . . . . . . . 12 Young Tableaux 12.1 Raising the indices . . . . . . . . . 12.2 Clebsch-Gordan decomposition . 12.3 SU(3) -, SU(2) x U(l) . . . . . . 13 SU(N) 13.1 Generalized Gell-Mann matrices 13.2 SU(N) tensors . . . . . 13. 3 Dimensions . . . . . . . . . . . 13.4 Complex representations . . . . 13.5 SU(N) ® SU(M) E SU(N + M) 14 3-D Harmonic Oscillator 14.1 Raising and lowering operators . 14.2 Angular momentum . . . . . . . 14.3 A more complicated example . 15 SU(6) and the Quark Model 15.1 Including the spin . . . . 15.2 SU(N)@ SU(M) E SU(NM) 15.3 The baryon states . 15.4 Magnetic moments . . . . . . . 16 Color 16.1 Colored quarks . . . . . . . 16.2 Quantum Chromodynamics . 16. 3 Heavy quarks . . . . . . 16.4 Flavor SU(4) is useless! .. CONTENTS 152 154 156 157 160 166 166 169 173 174 178 178 180 183 187 187 190 193 194 195 198 198 200 200 205 205 206 208 210 214 214 218 219 219 CONTENTS xvii 17 Constituent Quarks 221 17.1 111e nonrelativistic limit . 221 18 Unified Theories and SU(5) 225 18. l Grand unification . . . . 225 18.2 Parity violation, helicity and handedness . . . . . . . . . . 226 18.3 Spontaneously broken symmetry . . . . . . . 228 18.4 Physics of spontaneous symmetry breaking . 229 18.5 Is the Higgs real? . . . . . . 230 18.6 Unification and SU(5) .. . 231 18.7 Breaking SU(5) . 234 18.8 Proton decay . . . . . . . . 235 19 The Classical Groups 237 19.1 The S0(2n) algebras . 237 19.2 The S0(2n + 1) algebras .. 238 19.3 The Sp(2n) algebras 239 19.4 Quaternions . . . . . . . 240 20 The Classification Theorem 244 20.l II-systems ...... . 244 20.2 Regular subalgebras .. 251 20.3 Other Subalgebras . . 253 21 S0(2n + 1) and Spinors 255 21.1 Fundamental weight of S0(2n + 1) 255 21.2 Real and pseudo-real . . . . 259 21.3 Real representations . . . . . 261 21.4 Pseudo-real representations . 262 21.5 R is an invariant tensor . 262 21.6 The explicit form for R . 262 22 S0(2n + 2) Spinors 265 22. l Fundamental weights of S0(2n + 2) 265 23 SU(n) c S0(2n) 270 23. l Clifford algebras . . . . . . . . 270 rm 23.2 and R as invariant tensors . 272 23.3 Products of rs . . . 274 23.4 Self-duality . . . . 277 23.5 Example: S0(10) . 279 xviii CONTENTS 23.6 The SU(n) subalgebra ........ . 279 24 SO(10) 282 24.1 SO(10) and SU(4) x SU(2) x SU(2) . . . . . . . . . . . 282 24.2 * Spontaneous breaking of SO(10) . . . . . . . . . . . . . . 285 24.3 * Breaking S0(10) --+ SU(5) . . . . . . . . . . 285 24.4 * Breaking SO(10) --+ SU(3) x SU(2) x U(l) . . . . . . 287 24.5 * Breaking SO(10) ~ SU(3) x U(l) . . . . . . . . . . . . 289 24.6 * Lepton number as a fourth color . . . . . . . . . . . . . . 289 25 Automorphisms 291 25.1 Outer automorphisms 291 25.2 Fun with S0(8) .. 293 26 Sp(2n) 297 26.1 Weights of SU (n) . 297 26.2 Tensors for Sp(2n) 299 27 Odds and Ends 302 27 .1 Exceptional algebras and octonians . . . . . . . . . . . . . 302 27.2 E6 unification . . . . . . . . . . . . 304 27.3 Breaking E5 . . . . . . . . . . . . . . . 308 27.4 SU(3) x SU(3) x SU(3) unification . 308 27.5 Anomalies . . . . . . . . . . . . . . . . 309 Epilogue 311 Index 312 Why Group Theory? Group theory is the study of symmetry. It is an incredible labor saving device. It allows us to say interesting, sometimes very detailed things about physical systems even when we don't understand exactly what the systems are! When I was a teenager, I read an essay by Sir Arthur Stanley Eddington on the Theory of Groups and a quotation from it has stuck with me for over 30 years: 1 We need a super-mathematics in which the operations are as unknown as the quantities they operate on, and a super-mathematician who does not know what he is doing when he perfonns these operations. Such a super-mathematics is the Theory of Groups. In this book, I will try to convince you that Eddington had things a little bit wrong, as least as far as physics is concerned. A lot of what physicists use to extract information from symmetry is not the groups themselves, but group representations. You will see exactly what this means in more detail as you read on. What I hope you will take away from this book is enough about the theory of groups and Lie algebras and their representations to use group representations as labor-saving tools, particularly in the study of quantum mechanics. The basic approach will be to alternate between mathematics and physics, and to approach each problem from several different angles. I hope that you will learn that by using several techniques at once, you can work problems more efficiently, and also understand each of the techniques more deeply. 1in The World of Mathematics, Ed. by James R. Newman, Simon& Schuster, New York, 1956. Chapter 1 Finite Groups We will begin with an introduction to finite group theory. This is not intended to be a self-contained treatment of this enormous and beautiful subject. We will concentrate on a few simple facts that are useful in understanding the compact Lie algebras. We will introduce a lot of definitions, sometimes proving things, but often relying on the reader to prove them. 1.1 Groups and representations A Group, G, is a set with a rule for assigning to every (ordered) pair of elements, a third element, satisfying: (1.A.1) If f,g E G then h = Jg E G. (l.A.2) For f, g, h E G, f (gh) = (Jg)h. = (l.A.3) There is an identity element, e, such that for all f E G, ef fe = f. (l.A.4) Every element f E G has an inverse, 1-1, such that f 1-1 = 1-1f = e. Thus a group is a multiplication table specifying g1g2 Vg1, g2 E G. If the group elements are discrete, we can write the multiplication table in the form \ I e I 91 I 92 I ··· e e 91 92 ... 91 91 9191 9192 ... (1. 1) 92 92 9291 9292 ... 2 1.2. EXAMPLE- Z3 3 A Representation of G is a mapping, D of the elements of G onto a set of linear operators with the following properties: l.B .1 D (e) = 1, where 1 is the identity operator in the space on which the linear operators act. l.B.2 D(91)D(92) = D(9192), in other words the group multiplica- tion law is mapped qnto the natural multiplication in the linear space on which the linear operators act. 1.2 Example - Z3 A group is finite if it has a finite number of elements. Otherwise it is infinite. The number of elements in a finite group G is called the order of G. Here is a finite group of order 3. \ I e Ia Ib I e eab a abe (1.2) b bea This is Z3, the cyclic group of order 3. Notice that every row and column of the multiplication table contains each element of the group exactly once. This must be the case because the inverse exists. An Abelian group in one in which the multiplication law is commutative = 9192 9291 · (1.3) Evidently, Z3 is Abelian. The following is a representation of Z3 D(e) = 1, D(a) = e21ri/3 , D(b) = e47ri/3 (1.4) The dimension of a representation is the dimension of the space on which it acts - the representation (1.4) is 1 dimensional. 4 CHAPTER 1. FINITE GROUPS 1.3 The regular representation Here's another representation of Z3 1 0 0) (0 D(e) = 0 1 0 (0 0 1 D(a) = 1 0 (0 1 0) D(b) = o O 1 (1.5) 1 0 0 This representation was constructed directly from the multiplication table by the following trick. Take the group elements themselves to form an orthonormal basis for a vector space, le), la), and lb). Now define (1.6) The reader should show that this is a representation. It is called the regular representation. Evidently, the dimension of the regular representation is the order of the group. The matrices of (1.5) are then constructed as follows. lei) =le), le2) = la), le3) = lb) (1. 7) [D(g)]ij = (eilD(g )lej) (1.8) The matrices are the matrix elements of the linear operators. (1.8) is a simple, but very general and very important way of going back and forth from operators to matrices. This works for any representation, not just the regular representation. We will use it constantly. The basic idea here is just the insertion of a complete set of intennediate states. The matrix corresponding to a product of operators is the matrix product of the matrices corresponding to the operators - [D(g1g2)]ij = [D(g1)D(g2)b = (eilD(gi)D(g2)lej) = :E(eilD(g1)lek) (eklD(g2)lej) (1.9) k = L[D(g1)]ik[D(g2)]kj k Note that the construction of the regular representation is completely general for any finite group. For any finite group, we can define a vector space in which the basis vectors are labeled by the group elements. Then (1.6) defines the regular representation. We will see the regular representation of various groups in this chapter. I .4. IRREDUCIBLE REPRESENTATIONS 5 1.4 Irreducible representations What makes the idea of group representations so powerful is the fact that they live in linear spaces. And the wonderful thing about linear spaces is we are free to choose to represent the states in a more convenient way by making a linear transformation. As long as the transformation is invertible, the new states are just as good as the old. Such a transformation on the states produces a similarity transformation on the linear operators, so that we can always make a new representation of the form D(g) --t D'(g) = s- 1D(g)S (1.10) Because of the form of the similarity transformation, the new set of operators has the same multiplication rules as the old one, so D' is a representation if D is. D' and D are said to be equivalent representations because they differ just by a trivial choice of basis. = Unitary operators (0 such that ot o- 1) are particularly important. A representation is unitary if all the D(g)s are unitary. Both the representations we have discussed so far are unitary. It will tum out that all representations of finite groups are equivalent to unitary representations (we'll prove this later it is easy and neat). A representation is reducible if it has an invariant subspace, which means that the action of any D(g) on any vector in the subspace is still in the subspace. In terms of a projection operator P onto the subspace this con- dition can be written as P D(g)P = D(g)P Vg E G (1.11) For example, the regular representation of Z3 ( 1.5) has an invariant subspace projected on by 1 D 1 1 ( 1.12) because D(g)P = P 't/g. The restriction of the representation to the invariant subspace is itself a representation. In this case, it is the trivial representa- tion for which D(g) = 1 (the trivial representation, D(g) = 1, is always a representation - every group has one). A representation is irreducible if it is not reducible. A representation is completely reducible if it is equivalent to a represen- 6 CHAPTER 1. FINITE GROUPS tation whose matrix elements have the following form: .. .") (1.13) where Dj(9) is irreducible Vj. This is called block diagonal form. A representation in block diagonal form is said to be the direct sum of the subrepresentations, Dj (9), (1.14) In transforming a representation to block diagonal form, we are decom- posing the original representation into a direct sum of its irreducible components. Thus another way of defining complete reducibility is to say that a completely reducible representation can be decomposed into a direct sum of irreducible representations. This is an important idea. We will use it often. We will show later that any representation of a finite group is completely reducible. For example, for (1.5), take IC 1 ;,) S= 3 ~ w2 (1.15) w where w = e21ri/3 ( 1.16) then G0 0) D'(e) = 1 0 G 1,)0 D'(a) = w 0 1 0 G ~)0 D'(b) = w2 0 ( 1.17) 1.5 Transformation groups There is a natural multiplication law for transformations of a physical system. If 91 and 92 are two transformations, 9192 means first do 92 and then do 91. I .6. APPLICATION: PARITY IN QUANTUM MECHANICS 7 Note that it is purely convention whether we define our composition law to be right to left, as we have done, or left to right. Either gives a perfectly consistent definition of a transformation group. If this transfonnation is a symmetry of a quantum mechanical system, then the transformation takes the Hilbert space into an equivalent one. Then for each group element g, there is a unitary operator D(g) that maps the Hilbert space into an equivalent one. These unitary operators form a representation of the transfonnation group because the transformed quantum states represent the transformed physical system. Thus for any set of symmetries, there is a representation of the symmetry group on the Hilbert space - we say that the Hilbert space transforms according to some representation of the group. Furthermore, because the transformed states have the same energy as the originals, D(g) commutes with the Hamiltonian, [D(g), H] = 0. As we will see in more detail later, this means that we can always choose the energy eigenstates to transform like irreducible representations of the group. It is useful to think about this in a simple example. 1.6 Application: parity in quantum mechanics Parity is the operation of reflection in a mirror. Reflecting twice gets you back to where you started. If p is a group element representing the parity reflection, this means that p 2 = e. Thus this is a transformation that together with the identity transformation (that is, doing nothing) fonns a very simple group, with the following multiplication law: ( 1.18) This group is called Z2. For this group there are only two irreducible representations, the trivial one in which D(p) = 1 and one in which D(e) = 1 and D (p) = -1. Any representation is completely reducible. In particular, that means that the Hilbert space of any parity invariant system can be de- composed into states that behave like irreducible representations, that is on which D (p) is either 1 or -1. Furthermore, because D (p) commutes with the Hamiltonian, D(p) and H can be simultaneously diagonalized. That is we can assign each energy eigenstate a definite value of D(p). The energy eigenstates on which D(p) = 1 are said to transform according to the trivial representation. Those on which D(p) = -1 transform according to the other representation. This should be familiar from nonrelativistic quantum me- chanics in one dimension. There you know that a particle in a potential that is 8 CHAPTER 1. FINITE GROUPS symmetric about x = 0 has energy eigenfunctions that are either symmetric under x --t -x (corresponding to the trivial representation), or antisymmetric (the representation with D(p) = -1). 1.7 Example: S3 The permutation group (or symmetric group) on 3 objects, called S3 where a1 = (1,2,3) a2 = (3,2,1) a3 = (1,2) a4 = (2,3) a5 = (3, 1) ( 1.19) The notation means that a 1 is a cyclic permutation of the things in positions 1, 2 and 3; a2 is the inverse, anticyclic permutation; a3 interchanges the objects in positions 1 and 2; and so on. The multiplication law is then determined by the transformation rule that 9192 means first do 92 and then do 91· It is e e a1 a2 a3 a4 a5 a1 a1 a2 e a5 a3 a4 a2 a2 e a1 a4 as a3 a3 a3 a4 as e a1 a2 a4 a4 as a3 a2 e a1 a5 a5 a3 a4 a1 a2 e (1.20) We could equally well define it to mean first do 91 and then do 92· These two rules define different multiplication tables, but they are related to one another by simple relabeling of the elements, so they give the same group. There is another possibility of confusion here between whether we are permuting the objects in positions 1, 2 and 3, or simply treating 1, 2 and 3 as names for the three objects. Again these two give different multiplication tables, but only up to trivial renamings. The first is a little more physical, so we will use that. The permutation group is an another example of a transformation group on a physical system. S3 is non-Abelian because the group multiplication law is not commuta- tive. We will see that it is the lack of commutativity that makes group theory so interesting. 1.8. EXAMPLE: ADDITION OF INTEGERS 9 Here is a unitary irreducible representation of S3 (-1 _r , (1 0) _v'3) Ji D(e) = O l , D(a1) = 1 -\ , ( _1 D(a2) = _ 1 -f\l) , ( 1 D (a4) = ( 1 -v2'.3) 2 -1 2 0 D(a3) = ( 0 1 ) , v'3) D (a5) = _1 ~ ½ (1.21) The interesting thing is that the irreducible unitary representation is more than 1 dimensional. It is necessary that at least some of the representations of a non-Abelian group must-be matrices rather than numbers. Only matrices can reproduce the non-Abelian multiplication law. Not all the operators in the representation can be diagonalized simultaneously. It is this that is responsible for a lot of the power of the theory of group representations. 1.8 Example: addition of integers The integers fonn an infinite group under addition. xy = x+y (1.22) This is rather unimaginatively called the additive group of the integers. Since this group is infinite, we can't write down the multiplication table, but the rule above specifies it completely. Here is a representation: D(x) = (~ ~) (1.23) This representation is reducible, but you can show that it is not completely reducible and it is not equivalent to a unitary representation. It is reducible because D(x)P = P (1.24) where • p = (~ ~) ( 1.25) However, D(x)(I - P) -::J (I - P) ( 1.26) so it is not completely reducible. 10 CHAPTER 1. FINITE GROUPS The additive group of the integers is infinite, because, obviously, there are an infinite number of integers. For a finite group, all reducible representations are completely reducible, because all representations are equivalent to unitary representations. 1.9 Useful theorems Theorem 1.1 Every representation ofa finite group is equivalent to a unitary representation. Proof: Suppose D(g) is a representation of a finite group G. Construct the operator L s = D(g)t D(g) (1.27) gEG 8 is hermitian and positive semidefinite. Thus it can be diagonalized and its eigenvalues are non-negative: ( 1.28) where dis diagonal (1.29) where dj 2: O Vj. Because of the group property, all of the dj s are actually positive. Proof - suppose one of the djs is zero. Then there is a vector .X. such that 8).. = 0. But then L = = )..fS).. 0 IID(g)>-.112 . gEG (1.30) Thus D(g)).. must vanish for all g, which is impossible, since D(e) = 1. Therefore, we can construct a square-root of 8 that is hermitian and invertible x = s112 =u-1 ( P~ i ~ O .. ') u ·.:.· ( 1.31) X is invertible, because none of the djs are zero. We can now define D'(g) = X D(g) x-1 ( 1.32) I. 10. SUBGROUPS 11 Now, somewhat amazingly, this representation is unitary! (1.33) but (L D(g)t8D(g) = D(g)t D(h)tD(h)) D(g) hEG L = D(hg)tD(hg) (1.34) hEG L = D(h)tD(h) = s = x 2 hEG where the last line follows because hg runs over all elements of G when h does. QED. We saw in the representation (1.23) of the additive group of the integers an example of a reducible but not completely reducible representation. The way it works is that there is a P that projects onto an invariant subspace, but (1 - P) does not. This is impossible for a unitary representation, and thus representations of finite groups are always completely reducible. Let's prove it. Theorem 1.2 Every representation ofa finite group is completely reducible. Proof: By the previous theorem, it is sufficient to consider unitary repre- sentations. If the representation is irreducible, we are finished because it is already in block diagonal form. If it is reducible, then 3 a projector P such that PD(g)P = D(g)P Vg E G. This is the condition that P be an invariant subspace. Taking the adjoint gives PD(g)tP = PD(g)t Vg E G. But because D(g) is unitary, D(g)t = D(g)- 1 = D(g-1 ) and thus since g- 1 runs over all G when g does, PD(g)P = PD(g) Vg E G. But this implies that (1 - P)D(g)(I - P) = D(g)(l - P) Vg E G and thus 1 - P projects onto an invariant subspace. Thus we can keep going by induction and eventually completely reduce the representation. 1.10 Subgroups A group H whose elements are all elements of a group G is called a subgroup of G. The identity, and the group G are trivial subgroups of G. But many groups have nontrivial subgroups (which just means some subgroup other than G or e) as well. For example, the permutation group, Eh, has a Z3 subgroup formed by the elements {e, a1, a2}. 12 CHAPTER 1. FINITE GROUPS We can use a subgroup to divide up the elements of the group into subsets called cosets. A right-coset of the subgroup H in the group G is a set of elements formed by the action of the elements of H on the left on a given element of G, that is all elements of the form Hg for some fixed g. You can define left-cosets as well. For example, {a3, a4, as} is a coset of Z3 in Eh in (1.20) above. The number of elements in each coset is the order of H. Every element of G must belong to one and only one coset. Thus for finite groups, the order of a subgroup H must be a factor of order of G. It is also sometimes useful to think about the coset-space, G / H defined by regarding each coset as a single element of the space. A subgroup H of G is called an invariant or normal subgroup if for every g E G gH=Hg ( 1.35) which is (we hope) an obvious short-hand for the following: for every g E G and hi E H there exists an h2 E H such that hig = gh2, or gh2g- 1 = h 1 . The trivial subgroups e and G are invariant for any group. It is less ob- vious but also true of the subgroup Z3 of Eh in (1.20) (you can see this by direct computation or notice that the elements of Z3 are those permuta- tions that involve an even number of interchanges). However, the set {e, a4 } is a subgroup of G which is not invariant. a5 {e, a4} = {a5, a1} while {e,a4}a5 = {a5,a2}. If H is invariant, then we can regard the coset space as a group. The multiplication law in G gives the natural multiplication law on the cosets, Hg: (1.36) But if H is invariant Hg 1Hg-;1 = H, so the product of elements in two cosets is in the coset represented by the product of the elements. In this case, the coset space, G/ H, is called the factor group of G by H. What is the factor group Eh/Z3 ? The answer is Z2. The center of a group G is the set of all elements of G that commute with all elements of G. The center is always an Abelian, invariant subgroup of G. However, it may be trivial, consisting only of the identity, or of the whole group. There is one other concept, related to the idea of an invariant subgroup, that will be useful. Notice that the condition for a subgroup to be invariant can be rewritten as gHg-1 = H\:/g E G ( 1.37) I.I I. SCHUR'S LEMMA 13 This suggests that we consider sets rather than subgroups satisfying same condition. ( 1.38) Such sets are called conjugacy classes. We will see later that there is a oneto-one correspondence between them and irreducible representations. A subgroup that is a union of conjugacy classes is invariant. Example- The conjugacy classes of Eh are {e}, {a1, a2} and {a3, a4, a5}. The mapping (1.39) for a fixed g is also interesting. It is called an inner automorphism. An isomorphism is a one-to-one mapping of one group onto another that preserves the multiplication law. An automorphism is a one-to-one mapping of a group onto itself that preserves the multiplication law. It is easy to see that (1.39) is an automorphism. Because g- 1g1g g- 1g2g = 9- 19192g, it preserves the multiplication law. Since g- 191g = g- 1929 ~ 91 = 92, it is one to one. An automorphism of the form ( 1.39) where g is a group element is called an inner automorphism). An outer automorphism is one that cannot be written as 9- 1Gg for any group element g. 1.11 Schur's lemma Theorem 1.3 If D1(g)A = AD2(g) \:lg E G where D1 and D2 are inequivalent, irreducible representations, then A = 0. Proof: This is part of Schur's lemma. First suppose that there is a vector Iµ) such that Alµ) = 0. Then there is a non-zero projector, P, onto the subspace that annihilates A on the right. But this subspace is invariant with respect to the representation D 2, because AD2(g)P = D 1(g)AP = 0 \:lg E G ( 1.40) But because D2 is irreducible, P must project onto the whole space, and A must vanish. If A annihilates one state, it must annihilate them all. A similar argument shows that A vanishes if there is a (vi which annihilates A. If no vector annihilates A on either side, then it must be an invertible square matrix. It must be square, because, for example, if the number of rows were larger than the number of columns, then the rows could not be a complete set of states, and there would be a vector that annihilates A on the 14 CHAPTER 1. FINITE GROUPS right. A square matrix is invertible unless its determinant vanishes. But if the determinant vanishes, then the set of homogeneous linear equations Alµ)= O ( 1.41) has a nontrivial solution, which again means that there is a vector that annihilates A. But if A is square and invertible, then (1.42) so D 1 and D2 are equivalent, contrary to assumption. QED. The more important half of Schur's lemma applies to the situation where D1 and D2 above are equivalent representations. In this case, we might as well take D1 = D2 = D, because we can do so by a simple change of basis. The other half of Schur's lemma is the following. Theorem 1.4 If D(g)A = AD(g) Vg E G where Dis a.finite dimensional irreducible representation, then A ex I. In words, if a matrix commutes with all the elements of a finite dimensional irreducible representation, it is proportional to the identity. Proof: Note that here the restriction to a finite dimensional representation is important. We use the fact that any finite dimensional matrix has at least one eigenvalue, because the characteristic equation det(A - >..I) = 0 has at least one root, and then we can solve the homogeneous linear equations for the components of the eigenvector Iµ). But then D(g)(A - >..I) = (A >..I)D(g) Vg E G and (A - >..I)Iµ) = 0. Thus the same argument we used in the proof of the previous theorem implies (A - >..I) = 0. QED. A consequence of Schur's lemma is that the form of the basis states of an irreducible representation are essentially unique. We can rewrite theorem 1.4 as the statement A-1D(g)A = D(g) Vg E G ~ A ex I (1.43) for any irreducible representation D. This means once the form of D is fixed, there is no further freedom to make nontrivial similarity transformations on the states. The only unitary transformation you can make is to multiply all the states by the same phase factor. In quantum mechanics, Schur's lemma has very strong consequences for the matrix elements of any operator, 0, corresponding to an observable that is invariant under the symmetry transformations. This is because the matrix xi elements (a, j, OI b, k, y) behave like the A operator in ( 1.40). To see this, I.I 1. SCHUR'S LEMMA 15 let's consider the complete reduction of the Hilbert space in more detail. The symmetry group gets mapped into a unitary representation g -t D(g) Vg E G ( 1.44) where D is the (in general very reducible) unitary representation of G that acts on the entire Hilbert space of the quantum mechanical system. But if the representation is completely reducible, we know that we can choose a basis in which D has block diagonal form with each block corresponding to some unitary irreducible representation of G. We can write the orthonormal basis states as la,j, x) ( I .45) satisfying (a, j, X I b, k, y) = 8ab 8jk 8xy (1.46) where a labels the irreducible representation, j = 1 to na labels the state within the representation, and x represents whatever other physical parameters there are. Implicit in this treatment is an important assumption that we will almost always make without talking about it. We assume that have chosen a basis in which all occurences of each irreducible representation a, is described by the same set of unitary representation matrices, Da(g). In other words, for each irreducible representation, we choose a canonical form, and use it exclusively In this special basis, the matrix elements of D (g) are (a, j, xi D(g) lb, k, y) = 8ab 8xy [Da(g)]jk ( 1.47) This is just a rewriting of (1.13) with explicit indices rather than as a matrix. We can now check that our treatment makes sense by writing the representa- tion D in this basis by inserting a complete set of intermediate states on both sides: L I= la,j,x)(a,j,xl a,j,x ( 1.48) Then we can write L L D(g) = la,j,x)(a,j,xlD(g) lb,k,y)(b,k,yl a,j,x b,k,y L = la, j, x) Oab Oxy [Da(g)]jk (b, k, YI a,j,x b,k,y L = la, j, x) [Da(g)b (a, k, xi a,j,k,x (1.49) 16 CHAPTER I. FINITE GROUPS This is another way of writing a representation that is in block diagonal form. Note that if a particular irreducible representation appears only once in D, then we don't actually need the x variable to label its states. But typically, in the full quantum mechanical Hilbert space, each irreducible representation will appear many times, and then the physical x variable distinguish states that have the same symmetry properties, but different physics. The important fact, however, is that the dependence on the physics in ( 1.47) is rather trivial - only that the states are orthonormal - all the group theory is independent of x and y. Under the symmetry transformation, since the states transform like Iµ) -t D(g) Iµ) (µI -t (µI D(g)t (1.50) operators transform like 0 -t D(g) 0 D(g)t ( 1.51) in order that all matrix element remain unchanged. Thus an invariant observ- able satisfies 0 -t D(g) 0 D(g)t = 0 (1.52) which implies that O commutes with D(g) [O, D(g)] = 0 Vg E G. (1.53) Then we can constrain the matrix element (a, j, xlOlb, k, y) (1.54) by arguing as follows: 0 = (a,j, xl[O, D(g)]lb, k, y) = :E(a,j, xlOlb, k', y)(b, k', y!D(g)lb, k, y) k' - L (a, j, xlD(g)la, j', x) (a, j', xlOlb, k, y) j' (1.55) Now we use (1.47), which exhibits the fact that the matrix elements of D(g) have only trivial dependence on the physics, to write 0 = (a,j,xl[O,D(g)]lb,k,y) = I:(a,j, xlOlb, k', y)[Db(g)]k'k k' - ~)Da(g)]jj'(a,j',xlOlb,k,y) j' ( 1.56) I. I2. * ORTHOGONALITY RELATIONS 17 Thus the matrix element (1.54) satisfies the hypotheses of Schur's lemma. It must vanish if a -/- b. It must be proportional to the identity (in indices, that is 81k) for a = b. However, the symmetry doesn't tell us anything about the dependence on the physical parameters, x and y. Thus we can write • (a,j, xlOlb, k, y) = fa(x, y) 8ab 8jk ( 1.57) The importance of this is that the physics is all contained in the function fa(x, y) - all the dependence on the group theory labels is completely fixed by the symmetry. As we will see, this can be very powerful. This is a simple example of the Wigner-Eckart theorem, which we will discuss in much more generality later. 1.12 * Orthogonality relations The same kind of summation over the group elements that we used in the proof of theorem 1.1, can be used together with Schur's lemma to show some more remarkable properties of the irreducible representations. Consider the following linear operator (written as a "dyadic") L AjJ = Da(g- 1)la,j)(b, llDb(g) gEG (1.58) where Da and Db are finite dimensional irreducible representations of G. Now look at L Da(gi)AjJ = Da(91)Da(g- 1)la,j)(b,l!IDb(g) gEG ( 1.59) L = Da(g1g-1)la,j)(b,l!IDb(g) gEG L = Da((gg11)- 1 )la,j)(b,l!IDb(g) gEG Now let g' = gg11 L = Da(g'-1)la,j)(b,l!IDb(g'g1) g'EG L Da(g'- 1 )la,j)(b, llDb(g')Db(g1) = AjZDb(gi) g'EG (1.60) (1.61) (1.62) (1.63) 18 CHAPTER 1. FINITE GROUPS = Now Schur's lemma (theorems 1.3 and 1.4) implies AJl 0 if Da and Db are different, and further that if they are the same (remember that we have chosen a canonical form for each representation so equivalent representations are written in exactly the same way) AJl ex I. Thus we can write L AJl = Da(g- 1 )la,j)(b,l!IDb(g) = <5ab>..JeI gEG ( 1.64) To compute >.Je, compute the trace of AJl (in the Hilbert space, not the in- dices) in two different ways. We can write (1.65) where na is the dimension of Da. But we can also use the cyclic property of the trace and the fact that AJ£ ex <5ab to write L Tr AJl = <5ab (a, llDa(g)Da(g-l )la,j) = N <5ab <5je gEG ( 1.66) where N is the order of the group. Thus >..Je = N '5je/ na and we have shown ( 1.67) Taking the matrix elements of these relations yields orthogonality relations for the matrix elements of irreducible representations. L ;; [Da(g- 1 )]kj[Db(g)]em = <5abc5jec5km gEG ( 1.68) For unitary irreducible representations, we can write L -;; [Da(g)]Jk[Db(g)]em = <5abc5jec5km gEG ( 1.69) so that with proper normalization, the matrix elements of the inequivalent unitary irreducible representations ( 1.70) are orthonormal functions of the group elements, g. Because the matrix elements are orthonormal, they must be linearly independent. We can also show I.12. * ORTHOGONALITY RELATIONS 19 that they are a complete set of functions of g, in the sense that an arbitrary function of g can be expanded in them. An arbitrary function of g can be writ- ten in terms of a bra vector in the space on which the regular representation acts: F(g) =(Fig)= (FIDR(g)le) (I.71) where L (Fl = F(g')(g'I g'EG (1.72) and DR is the regular representation. Thus an arbitrary F(g) can be written as a linear combination of the matrix elements of the regular representation. L L F(g) = F(g')(g'IDR(g)le) = F(g')[DR(g)] 91e g'EG g'EG ( I. 73) But since DR is completely reducible, this can be rewritten as a linear com- bination of the matrix elements of the irreducible representations. Note that while this shows that the matrix elements of the inequivalent irreducible representations are complete, it doesn't tell us how to actually find what they are. The orthogonality relations are the same. They are useful only once we actually know explicitly what the representation look like. Putting these results together, we have proved Theorem 1.5 The matrix elements ofthe unitary, irreducible representations of G are a complete orthonormal set for the vector space of the regular representation, or alternatively, for functions of g E G. An immediate corollary is a result that is rather amazing: (1.74) - the order of the group N is the sum of the squares of the dimensions of the irreducible representations ni just because this is the number of components of the matrix elements of the irreducible representations. You can check that this works for all the examples we have seen. = Example: Fourier series - cyclic group ZN with elements a1 for j = 0 to N - I (with a0 e) = a1ak a(j+k) mod N (1.75) The irreducible representations of ZN are = Dn(aj) e21rinj/N (1.76) 20 CHAPTER 1. FINITE GROUPS all I-dimensional. 1 Thus ( 1.69) gives _ L l N-1 e-21rin'j/N e21rinj/N = <>n'n N J.= o which is the fundamental relation for Fourier series. (1.77) 1.13 Characters The characters XD (g) of a group representation D are the traces of the linear -=- operators of the representation or their matrix elements: XD(g) = 'Ir D(g) = L[D(g)]ii (1.78) The advantage of the characters is that because of the cyclic property of the trace 'Ir(AB} = 'Ir(BA}, they are unchanged by similarity transformations, thus all equivalent representations have the same characters. The characters are also different for each inequivalent irreducible representation, Da - in fact, they are orthonormal up to an overall factor of N - to see this just sum e (1.69) over j = k and = m L ! L [Da(g)]jk[Db(g)]em = ]:_<>ab<>je<>km = <>ab 9ea j=k na j=k l=m l=m or N1~ " XDa(g) *XD&(g) = <>ab (1.79) gEG Since the characters of different irreducible representations are orthogonal, they are different. The characters are constant on conjugacy classes because (1.80) It is less obvious, but also true that the characters are a complete basis for functions that are constant on the conjugacy classes and we can see this by explicit calculation. Suppose that F(g1) is such a function. We already know 1We will prove below that Abelian finite groups have only I-dimensional irreducible representations. 1.13. CHARACTERS 21 that F(g1 ) can be expanded in terms of the matrix elements of the irreducible represent1:1tions - L F(g1) = cJk[Da(g1)]jk (1.81) a,j,k but since F is constant on conjugacy classes, we can write it as (l.82) and thus ! L F(g1) = cJk[Da(g- 1)]je[Da(g1)]em[Da(g)]mk a,j,k g,l,m ( 1.83) But now we can do the sum over g explicitly using the orthogonality relation, ( 1.68). ( 1.84) or ( 1.85) This was straightforward to get from the orthogonality relation, but it has an important consequence. The characters, Xa(g), of the independent irreducible representations form a complete, orthonormal basis set for the functions that are constant on conjugacy classes. Thus the number of irreducible representations is equal to the number of conjugacy classes. We will use this frequently. This also implies that there is an orthogonality condition for a sum over representations. To see this, label the conjugacy classes by an integer a, and let k0 be the number of elements in the conjugacy class. Then define the matrix V with matrix elements (1.86) where g0 is the conjugacy class a. Then the orthogonality relation ( 1.79) can vt be written as V = 1. But V is a square matrix, so it is unitary, and thus we also have vvt = 1, or (1.87) 22 CHAPTER 1. FINITE GROUPS Consequences: Let D be any representation (not necessarily irreducible). In its completely reduced form, it will contain each of the irreducible representations some integer number of times, ma, We can compute ma simply by using the orthogonality relation for the characters (1.79) N1 "~ XDa(g) *XD (g) = maD gEG ( 1.88) The point is that D is a direct sum m!; times a For example, consider the regular representation. It's characters are (1.89) XR(e) = N XR(g) = 0 for g-:/ e ( 1.90) Thus mf = Xa(e) = na (1.91) Each irreducible representation appears in the regular representation a num- ber of times equal to its dimension. Note that this is consistent with (1.74). Note also that ma is uniquely determined, independent of the basis. Example: Back to S3 once more. Let's determine the characters without thinking about the 2-dimensional representation explicitly, but knowing the conjugacy classes, {e}, {a1,a2} and {a3,a4,a5}. It is easiest to start with the one representation we know every group has - the trivial representa- tion, Do for which Do(g) = 1 for all g. This representation has characters x0 (g) = l. Note that this is properly normalized. It follows from the condition L n~ = N that the other two representations have dimensions 1 and 2. It is almost equally easy to write down the characters for the other I- dimensional representation. In general, when there is an invariant subgroup H of G, there are representations of G that are constant on H, forming a representation of the factor group, G / H. In this case, the factor group is Z2, with nontrivial representation 1 for H = {e, a1, a2} and -1 for {a3, a4, as}. We know that for the 2 dimensional representation, x3 (e) = n3 = 2, thus so far the character table looks like 011 1 1 1 1 -1 22? ? (l.92) 1.13. CHARACTERS 23 But then we can fill in the last two entries using orthogonality. We could actually have just used orthogonality without even knowing about the second representation, but using the Z2 makes the algebra trivial. 01 1 1 1 1 1 -1 2 2 -1 0 (1.93) We can use the characters not just to find out how many irreducible representations appear in a particular reducible one, but actually to explicitly decompose the reducible representation into its irreducible components. It is easy to see that if D is an arbitrary representation, the sum r;; L Pa= XDJg)* D(g) gEG (1.94) is a projection operator onto the subspace that transfonns under the represen- tation a. To see this, note that if we set j = k and sum in the orthogonality relation (1.69), we find r;; L XDJg)*[Db(g)]em = <5abOem gEG (1.95) Thus when D is written in block diagonal form, the sum in (1.95) gives 1 on the subspaces that transform like Da and O on all the rest - thus it is the projection operator as promised. The point, however, is that (1.94) gives us the projection operator in the original basis. We did not have to know how to transform to block diagonal form. An example may help to clarify this. Example - S3 again Here's a three dimensional representation of S3 0 G D, G D,(e) = 1 ~) 0 1 G G D3(a2) = 0 D n 0 0 D,(a1) = 0 1 1 D,(a,) = 0 0 (l.96) G D, G D 0 D3(a,) = 0 1 0 D,(a,) = 1 0 24 CHAPTER I. FINITE GROUPS More precisely, as usual when we write down a set of matrices to represent linear operators, these are matrices which have the same matrix elements that is (1.97) One could use a different symbol to represent the operators and the matrices, but its always easy to figure out which is which from the context. The important point is that the way this acts on the states, lj) is by matrix multiplication on the right, because we can insert a complete set of intermediate states D3(g)lj) = })k)(klD3(g)lj) = Llk)[D3(g)]kj ( 1.98) k k This particular representation is an important one because it is the defining representation for the group - it actually implements the permutations on the states. For example D3(a1)ll) = })k)[D3(a1)]k1 = 12) k D3(ai)l2) = Llk)(D3(ai)]k2 = /3) k D3(a1)13) = })k)[D3(a1)]k3 = II) (l.99) k thus this implements the cyclic transformation (1,2,3), or 1 --+ 2 --+ 3 --+ 1. Now if we construct the projection operators, we find ID (1.100) (1.101) (1.102) This makes good sense. Pa projects onto the invariant combination (11) + 12) + 13))/\,/3, which transforms trivially, while P2 projects onto the two dimensional subspace spanned by the differences of pairs of components, I1) - 12), etc, which transforms according to D3. This constructions shows that the representation D3 decomposes into a direct sum of the irreducible representations, (1.103) 1.14. EIGENSTATES 25 1.14 Eigenstates In quantum mechanics, we are often interested in the eigenstates of an invariant hermitian operator, in particular the Hamiltonian, H. We can always take these eigenstates to transform according to irreducible representations of the symmetry group. To prove this, note that we can divide up the Hilbert space into subspaces with different eigenvalues of H. Each subspace furnishes a representation of the symmetry group because D (g), the group represen- tation on the full Hilbert space, cannot change the H eigenvalue (because [D(g), H] = 0). But then we can completely reduce the representation in each subspace. A related fact is that if some irreducible representation appears only once in the Hilbert space, then the states in that representation must be eigenstates of H (and any other invariant operator). This is true because Hla, j, x) must be in the same irreducible representation, thus L H la,j,x) = Cy la,j, y) y (l.104) and if x and y take only one value, then la, j, x) is an eigenstate. This is sufficiently important to say again in the form of a theorem: Theorem 1.6 If a hermitian operator, H, commutes with all the elements, D(g), of a representation of the group G, then you can choose the eigenstates of H to transform according to irreducible representations of G. If an irreducible representation appears only once in the Hilbert space, every state in the irreducible representation is an eigenstate of H with the same eigenvalue. Notice that for Abelian groups, this procedure of choosing the H eigenstates to transform under irreducible representations is analogous to simulta- neously diagonalizing Hand D(g). For example, for the group Z2 associated with parity, it is the statement that we can always choose the H eigenstates to be either symmetric or antisymmetric. In the case of parity, the linear operator representing parity is hermitian, so we know that it can be diagonalized. But in general, while we have shown that operators representing finite group elements can be chosen to be unitary, they will not be hermitian. Nevertheless, we can show that for an Abelian group that commutes with the H, the group elements can simultaneously diagonalized along with H. The reason is the following theorem: Theorem 1.7 All of the irreducible representations of a finite Abelian group are I -dimensional. 26 CHAPTER I. FINITE GROUPS One proof of this follows from our discussion of conjugacy classes and from (1.74). For an Abelian group, conjugation does nothing, because g g' g- 1 = g' for all g and g'. Therefore, each element is in a conjugacy class all by itself. Because there is one irreducible representation for each conjugacy class, the number of irreducible representations is equal to the order of the group. Then the only way to satisfy (I.74) is to have all of the nis equal to one. This proves the theorem, and it means that decomposing a representation of an Abelian group into its irreducible representations amounts to just diagonalizing all the representation matrices for all the group elements. For a non-Abelian group, we cannot simultaneously diagonalize all of the D(g)s, but the procedure of completely reducing the representation on each subspace of constant H is the next best thing. A classical problem which is quite analogous to the problem of diago- nalizing the Hamiltonian in quantum mechanics is the problem of finding the normal modes of small oscillations of a mechanical system about a point of stable equilibrium. Here, the square of the angular frequency is the eigen- yalue of the M-1K matrix and the normal modes are the eigenvectors of M-1K. In the next three sections, we will work out an example. 1.15 Tensor products We have seen that we can take reducible representations apart into direct sums of smaller representations. We can also put representations together into larger representations. Suppose that D1 is an m dimensional representa- tion acting on a space with basis vectors lj) for j = 1 tom and D2 is an n dimensional representation acting on a space with basis vectors Ix) for x = I to n. We can make an m x n dimensional space called the tensor product space by taking basis vectors labeled by both j and x in an ordered pair lj, x). Then when j goes from 1 tom and x goes from I ton, the ordered pair (j, x) runs over m x n different combinations. On this large space, we can define a new representation called the tensor product representation D 1 0 D2 by multiplying the two smaller representations. More precisely, the matrix elements of D Di@D2 (g) are products of those of D1 (g) and D2(g): (1.105) It is easy to see that this defines a representation of G. In general, however, it will not be an irreducible representation. One of our favorite pastimes in what follows will be to decompose reducible tensor product representations into irreducible representations. 1.16. EXAMPLE OF TENSOR PRODUCTS 27 1.16 Example of tensor products Consider the following physics problem. Three blocks are connected by springs in a triangle as shown (1.106) Suppose that these are free to slide on a frictionless surface. What can we say about the normal modes of this system. The point is that there is an S3 symmetry of the system, and we can learn a lot about the system by using the symmetry and applying theorem 1.6. The system has 6 degrees of freedom, described by the x and y coordinates of the three blocks: ( XI YI X2 Y2 X3 Y3) (1.107) This has the structure of a tensor product - the 6 dimensional space is a product of a 3 dimensional space of the blocks, and the 2 dimensional space of the x and y coordinates. We can think of these coordinates as having two It indices. is three two dimensional vectors, 'G, each of the vector indices has two components. So we can write the components as Tjµ where j labels the mass and runs from 1 to 3 and µ labels the x or y component and runs from 1 to 2, with theconnection = ( X1 YI X2 Y2 X3 Y3) ( ru r12 r21 r22 r31 r32) (1.108) The 3 dimensional space transforms under S3 by the representation D 3. The n, 1 -=-n , 2 dimensional space transforms by the representation D 2 below: D,(e) = (~ D2(a1) = ( ~~ D2 (a2) = ( ~) , D,(a,) = ( ~l ~) , (1.109) 1 ~½ , -4 D2(a4) = ( 1 ,y] ) D2(a5) = ( 1 __v'3_2~3 ) 28 CHAPTER I. FINITE GROUPS This is the same as ( t.21 ). Then, using ( 1.105), the 6 dimensional representation of the coordinates is simply the product of these two representations: [D5(g)]jµkv = [D3(g)]jk[D2(g)]µv (1.110) Thus, for example, D5(a1) = 0 0 0 0 - 21 _v'3 2 0 0 0 0 v'3 2 - 21 -21 --vy'3 0 0 0 0 _y] 2 - 21 0 0 0 0 0 0 - 21 _ _y] 2 0 0 0 0 v'3 2 - 21 0 0 (1.111) This has the structure of 3 copies of D2 (a1) in place of the 1's in D3 (a1). The other generators are similar in structure. Because the system has the S3 symmetry, the normal modes of the system must transform under definite irreducible representations of the symmetry. Thus if we construct the projectors onto these representations, we will have gone some way towards finding the normal modes. In particular, if an irreducible representation appears only once, it must be a normal mode by theorem 1.6. If a representation appears more than once, then we need some additional information to determine the modes. We can easily determine how many times each irreducible representation appears in D5 by finding the characters of D6 and using the orthogonality relations. To find the characters of D5, we use an important general result. The character of the tensor product of two representations is the product of the characters of the factors. This follows immediately from the definition of the tensor product and the trace. (1.112) So in this case, X6(g) = I)D5(g)]jµjµ jµ = [D3(g)Jii[D2(g)]µµ = X3(g)x2(g) (1.113) 1.17. * FINDING THE NORMAL MODES 29 so that the product is as shown in the table below: D3 3 0 1 D2 2 -1 0 (1.114) D5 6 0 0 This is the same as the characters of the regular representation, thus this representation is equivalent to the regular representation, and contains Do and D1 once and D2 twice. Note that (1.113) is an example of a simple but important general relation, which we might as well dignify by calling it a theorem - Theorem 1.8 The characters ofa tensor product representation are the products of the characters of the factors. With these tools, we can use group theory to find the normal modes of the system. 1.17 * Finding the normal modes The projectors onto Do and D 1 will be 1 dimensional. Po is I: Po = -i xo(g) *D5(g) 6 gEG 1 4 _y] 12 - 41 _y] 12 0 _v'3 6 v'3 IT 1 12 _ _y] 12 1 12 0 _ _y] 6 - 41 _ _y] 12 1 4 _ _y] 12 0 _y] 6 v'3 IT 1 12 -1v2'3 1 12 0 _ _y] 6 0 0 0 0 0 0 - 6 v'3 -61 _y] 6 -61 0 1 3 (l.115) 1 ¥ ( ½ -½ 0 -~) (1.116) 30 corresponding to the motion CHAPTER 1. FINITE GROUPS (1.117) the so-called "breathing mode" in which the triangle grows and shrinks while retaining its shape. Pi is :E Pi=! x1(g)* D6(g) 6 gEG 1 _ _y] 1 12 12 12 _y] 12 -61 0 _ _y] 12 1 4 _ _y] 12 - 41 _y] 6 0 1 _ _y] 1 = 12 y3 12 12 - 41 12 _y] 12 _y] 12 -61 0 1 4 y3 -6 0 - 61 _y] 6 - 61 _.Y] 6 1 3 0 0 0 0 0 0 0 (1.118) = (-¥ ½ -f -½ ~ 0) (1.119) 1.17. * FINDING THENORMALMODES corresponding to the motion 31 (l.120) the mode in which the triangle rotates - this is a nonnal mode with zero frequency because there is no restoring force. Notice, again, that we found these two normal modes without putting in any physics at all except the symmetry! Finally, P2 is :E P2 = ~ x2(g)* D6(g) gEG 2 3 0 1 6 0 2 _y] 3 6 1 y3 2 =6 6 _y3 1 6 6 3 0 1 _y3 1 6 6 6 y3 1 _ _y] 6 6 6 _y] 1 y3 6 6 6 1 _ _y] 1 6 6 6 0 1 6 y3 -6 2 _y] 1 3 6 6 y3 2 6 3 0 1 6 0 2 3 (l.121) As expected, this is a rank 4 projection operator (Tr P2 = 4). We need some dynamical infonnation. Fortunately, two modes are easy to get - translations of the whole triangle. Translations in the x direction, for example, are projected by Tx = 1 3 0 0 0 1 ·o3 0 0 1 3 0 0 0 1 3 0 1 3 0 1 3 0 0 0 0 0 0 0 1 3 0 1 3 0 1 3 0 0 0 0 0 0 0 (1.122) 32 CHAPTER 1. FINITE GROUPS and those in the y direction by Ty= 0 0 0 0 0 0 0 1 3 0 1 3 0 1 3 0 0 0 0 0 0 0 1 3 0 1 3 0 1 3 0 0 0 0 0 0 0 1 3 0 1 3 0 1 3 So the nontrivial modes are projected by (1.123) P2 -Tx -Ty= 1 3 0 - 61 _y3 6 - 61 _y] 6 0 1 3 y3 6 - 61 _ _y] 6 - 61 - 61 _y] 6 1 3 0 - 61 y3 -6 y3 -6 - 61 0 1 3 _y] 6 - 61 - 61 y3 -6 - 61 _y] 6 1 3 0 y3 6 - 61 _ _y] 6 - 61 0 1 3 (1.124) To see what the corresponding modes look like, act with this on the vector ( 0 0 0 0 0 1 ) to get ("? -1{- -! 0 ½ ) (l.125) corresponding to the motion (1.126) Then rotating by 21r /3 gives a linearly independent mode. 1. 18. * SYMMETRIES OF 2N +1-GONS 33 1.18 * Symmetries of 2n+l-gons This is a nice simple example of a transfonnation group for which we can work out the characters (and actually the whole set of irreducible representa- tions) easily. Consider a regular polygon with 2n + 1 vertices, like the 7-gon shown below. (1.127) The grouf of symmetries of the 2n+1-gon consists of the identity, the 2n rotations by 2,;~{ for j = l to n, . rotations by ±21rj --- for . J = 1 to n 2n+ 1 (1.128) and the 2n+ 1 reflections about lines through the center and a vertex, as show below: reflections about lines through center and vertex (1.129) (1.130) Thus the order of the group of symmetries is N = 2 x (2n + 1). There are n + 2 conjugacy classes: 1 - the identity, e; 2 - the 2n+ 1 reflections; 3 to n+2 - the rotations by ~~{ for j = l to n - each value of j is a separate conjugacy class. The way this works is that the reflections are all in the same conjugacy class because by conjugating with rotations, you can get from any one reflection to any other. The rotations are unchanged by conjugation by rotations, but a conjugation by a reflection changes the sign of the rotation, so there is a ± pair in each conjugacy class. 34 CHAPTER I. FINITE GROUPS Furthermore, the n conjugacy classes of rotations are equivalent under cyclic permutations and relabeling of the vertices, as shown below: (1.131) (l.132) The characters look like 1 1 1 -1 2 0 1 1 2 COS 21rm 2n+l j=2 1···1 j=n 1 ... 1 1 ... 1 2 COS 41rm 2n+l ... 2cos 2nirm 2n+l (1.133) In the last line, the different values of m give the characters of the n different 2-dimensional representations. 1.19 Permutation group on n objects Any element of the permutation group on n objects, called Sn, can be written in term of cycles, where a cycle is a cyclic permutation of a subset. We will use a notation that makes use of this, where each cycle is written as a set of numbers in parentheses, indicating the set of things that are cyclicly permuted. For example: (l) means x1 --t x1 (I 372) means x1 --t X3 --t x7 --t x2 --t x1 Each element of Sn involves each integer from 1 to n in exactly one cycle. Examples: The identity element looks like e =(1)(2)· · ·(n) - n I-cycles - there is only one of these. 1.20. CONJUGACY CLASSES 35 An interchange of two elements looks like (12)(3)· · -(n) - a 2-cycle and n - 2 1-cycles - there are n(n - 1) /2 of these - (j132)(13) ···Un)- An arbitrary element has kj j-cycles, where n L, j kj = n j=l (1.134) For example, the permutation (123)(456)(78)(9) has two 3-cycles, 1 2-cycle and a 1-cycle, so k1 = k2 = 1 and k3 = 2. There is an simple (but reducible) n dimensional representation of Sn called the defining representation where the "objects" being permuted are just the basis vectors of an n dimensional vector space, 11) , 12) , .. · In) (l.135) If the permutation takes Xj to Xk, the corresponding representation operator D takes lj) to lk), so that D IJ) = lk) (1.136) and thus (1.137) Each matrix in the representation has a single 1 in each row and column. 1.20 Conjugacy classes The conjugacy classes are just the cycle structure, that is they can be labeled by the integers kj. For example, all interchanges are in the same conjugacy class - it is enough to check that the inner automorphism gg1g- 1 doesn't change the cycle structure of g1 when g is an interchange, because we can build up any permutation from interchanges. Let us see how this works in some examples. In particular, we will see that conjugating an arbitrary permutation by the interchange (12)(3)· ··just interchanges 1 and 2 without changing the cycle structure Examples - (12)(3)(4)·(1)(23)(4)·(12)(3)(4) (note that an interchange 36 CHAPTER I. FINITE GROUPS is its own inverse) 1234 ~ (12)(3)(4) 2134 ~ (1)(23)(4) 2314 ~ (12)(3)(4) 3214 1234 ~ (2)(13)(4) 3214 (12)(3)(4)-(1)(234)-(12)(3)(4) 1234 ~ (12)(3)(4) 2134 ~ (1)(234) 2341 ~ (12)(3)(4) 3241 1234 -1- (2)(134) 3241 (1.138) (1.139) If 1 and 2 are in different cycles, they just get interchanged by conjugation by (12), as promised. The same thing happens when 1 and 2 are in the same cycle. For example 1234 21+34 (12)(3)(4) ~ (123)(4) 1324 ~ (12)(3)(4) 3124 (1.140) 1234 ~ (213)(4) 3124 Again, in the same cycle this time, 1 and 2 just get interchanged. Another way of seeing this is to notice that the conjugation is analogous to a similarity transformation. In fact, in the defining, n dimensional representation of (1.135) the conjugation by the interchange (12) is just a change of basis that switches 11) t-t 12). Then it is clear that conjugation 1.21. YOUNG TABLEAUX 37 does not change the cycle structure, but simply interchanges what the permutation does to I and 2. Since we can put interchanges together to form an arbitrary permutation, and since by repeated conjugations by interchanges, we can get from any ordering of the integers in the given cycle structure to any other, the conjugacy classes must consist of all possible permutations with a particular cycle structure. Now let us count the number of group elements in each conjugacy class. Suppose a conjugacy class consists of permutations of the form of k1 Icycles, k2 2-cycles, etc, satisfying (1.134). The number of different permuta- tions in the conjugacy class is n! (l.141) because each permutation of number l to n gives a permutation in the class, but cyclic order doesn't matter within a cycle (123) is the same as (231) (1.142) and order doesn't matter at all between cycles of the same length (12)(34) is the same as (34)(12) (1.143) 1.21 Young tableaux It is useful to represent each j-cycle by a column of boxes of length j, topjustified and arranged in order of decreasing j as you go to the right. The total number of boxes is n. Here is an example: I II I I (1.144) is four I-cycles in S4 - that is the identity element - always a conjugacy class all by itself. Here's another: (1.145) is a 4-cycle, a 3-cycle and a I-cycle in Ss. These collections of boxes are called Young tableaux. Each different tableaux represents a different conjugacy class, and therefore the tableaux are in one-to-one correspondence with the irreducible representations. 38 CHAPTER 1. FINITE GROUPS 1.22 Example - our old friend S3 The conjugacy classes are ITIJ EP with numbers of elements 3! -3! = 1 3! -2 =3 8 3! - =2 3 (1.146) (l.147) 1.23 Another example - S4 If ~ EfTI EE I,--I,--I,--,I--,I with numbers of elements (l.148) 4! = 1 4! = 6 4! = 3 4! = 8 _4:1_ = 6 4! 4 8 3 4 (1.149) The characters of 84 look like this (with the conjugacy classes which label the columns in the same order as in (1.148)): conjugacy classes 11 1 1 1 3 1 -1 0 -1 2 0 2 -1 0 3 -1 -1 0 1 1 -1 1 1 -1 ( 1.150) The first row represents the trivial representation. 1.24 * Young tableaux and representations of Sn We have seen that a Young tableau with n boxes is associated with an irreducible representation of Sn. We can actually use the tableau to explicitly construct the irreducible representation by identifying an appropriate subspace of the regular representation of Sn. 1.24. * YOUNG TABLEAUX AND REPRESENTATIONS OF SN 39 To see what the irreducible representation is, we begin by putting the integers from l to n in the boxes of the tableau in all possible ways. There are n! ways to do this. We then identify each assignment of integers l to n to the boxes with a state in the regular representation of Sn by defining a standard ordering, say from left to right and then top down (like reading words on a page) to translate from integers in the boxes to a state associated with a particular pennutation. So for example ~ ~ ---+ 16532174) (1.151) where J6532174) is the state corresponding to the permutation 1234567 ---+ 6532174 (1.152) Now each of the n! assignment of boxes to the tableau describes one of the n! states of the regular representation. Next, for a particular tableau, symmetrize the corresponding state in the numbers in each row, and antisymmetrize in the numbers in each column. For example [ill] ---+ 112) + 121) (l.153) and w21 ---+ 1123) + 1213} _ /321} _ 1231) (l.154) Now the set of states constructed in this ways spans some subspace of the regular representation. We can construct the states explicitly, and we know how permutations act on these states. That the subspace constructed in this way is a representation of 8n, because a permutation just corresponds to starting with a different assignment of numbers to the tableau, so acting with the permutation on any state in the subspace gives another state in the subspace. In fact, this representation is irreducible, and is the irreducible representation we say is associated with the Young tableau. Consider the example of Eh. The tableau [I1J (1.155) gives completely symmetrized states, and so is associated with a one dimensional subspace that transforms under the trivial representation. The tableau § (1.156) 40 CHAPTER 1. FINITE GROUPS gives completely antisymmetrized states, and so, again is associated with a one dimensional subspace, this time transforming under the representation in which interchanges are represented by -1. Finally (1.157) gives the following states: [!]21 --t !123) + !213) - !321) - !231) (1.158) [1] 21--t !321) + 1231) - !123) - !213) (1.159) [1] 31--t !231) + !321) - !132) - !312) (1.160) 8] 31--t !132) + 1312) - !231) - !321) (1.161) [!j 11 --t /312) + 1132} - 1213} - 1123) (1.162) [j_] 11 --t !213) + !123) - !312) - !132) (l.163) Note that interchanging two numbers in the same column of a tableau just changes the sign of the state. This is generally true. Furthermore, you can see explicitly that the sum of three states related by cyclic permutations van- ishes. Thus the subspace is two dimensional and transforms under the two dimensional irreducible representation of S3. It turns out that the dimension of the representation constructed in this way is n! (1.164) H where the quantity H is the "hooks" factor for the Young tableau, computed as follows. A hook is a line passing vertically up through the bottom of some column of boxes, making a right hand tum in some box and passing out through the row of boxes. There is one hook for each box. Call the number of boxes the hook passes through h. Then H is the product of the hs for all hooks. We will come back to hooks when we discuss the application of Young tableaux to the representations of SU(N) in chapter XII I. This procedure for constructing the irreducible representations of Sn is entirely mechanical (if somewhat tedious) and can be used to construct all the representations of Sn from the Young tableaux with n boxes. 1.24. * YOUNG TABLEAUX AND REPRESENTATIONS OF SN 41 We could say much more about finite groups and their representations, but our primary subject is continuous groups, so we will leave finite groups for now. We will see, however, that the representations of the permutation groups play an important role in the representations of continuous groups. So we will come back to Sn now and again. Problems 1.A. Find the multiplication table for a group with three elements and prove that it is unique. 1.B. Find all essentially different possible multiplication tables for groups with four elements (which cannot be related by renaming elements). 1.C. Show that the representation (1.135) of the permutation group is reducible. 1.D. Suppose that D 1 and D 2 are equivalent, irreducible representations of a finite group G, such that D2(g) = S D1(g) s-1 Vg E G What can you say about an operator A that satisfies 1.E. Find the group of all the discrete rotations that leave a regular tetrahedron invariant by labeling the four vertices and considering the rotations as permutations on the four vertices. This defines a four dimensional representation of a group. Find the conjugacy classes and the characters of the irreducible representations of this group. *1.F. Analyze the normal modes of the system of four blocks sliding on a frictionless plane, connected by springs as shown below: 42 CHAPTER I. FINITE GROUPS just as we did for the triangle, but using the 8-element symmetry group of the square. Assume that the springs are rigidly attached to the masses (rather than pivoted, for example), so that the square has some rigidity. Chapter 2 Lie Groups Suppose our group elements g E G depend smoothly on a set of continuous parameters - g(a) (2.1) What we mean by smooth is that there is some notion of closeness on the group such that if two elements are "close together" in the space of the group elements, the parameters that describe them are also close together. 2.1 Generators Since the identity is an important element in the group, it is useful to param- eterize the elements (at least those close to the identity element) in such a way that a = 0 corresponds to the identity element. Thus we assume that in some neighborhood of the identity, the group elements can be described by a function of N real parameters, aa for a = l to N, such that g(a)la=O = e (2.2) Then if we find a representation of the group, the linear operators of the representation will be parameterized the same way, and D(a)la=O = 1 (2.3) Then in some neighborhood of the identity element, we can Taylor expand D (a), and if we are close enough, just keep the first term: D(da) = 1 +•idaaXa + ··· (2.4) 43 44 CHAPTER 2. LIE GROUPS where we have called the parameter da to remind you that it is infinitesimal. In (2.4), a sum over repeated indices is understood (the "Einstein summation convention") and Xa = -i aa D(a)I (2.5) aa a=O The Xa for a = l to N are called the generators of the group. If the parameterization is parsimonious (that is - all the parameters are actually needed to distinguish different group elements), the Xa will be independent. The i is included in the definition (2.5) so that if the representation is unitary, the Xa will be hermitian operators. Sophus Lie showed how the generators can actually be defined in the abstract group without mentioning representations at all. As a result of his work, groups of this kind are called Lie groups. I am not going to talk about them this way because I am more interested in representations than in groups, but it is a beautiful theoretical construction that you may want to look up if you haven't seen it. As we go away from the identity, there is enormous freedom to param- eterize the group elements in different ways, but we may as well choose our parameterization so that the group multiplication law and thus the multipli- cation law for the representation operators in the Hilbert space looks nice. In particular, we can go away from the identity in some fixed direction by simply raising an infinitesimal group element D(da) = 1 + idaaXa (2.6) to some large power. Because of the group property, this always gives another group element. This suggests defining the representation of the group elements for finite a as (2.7) In the limit, this must go to the representation of a group element because 1 + iaaXa/k becomes the representation of a group element in (2.4) as k becomes large. This defines a particular parameterization of the representations (sometimes called the exponential parameterization), and thus of the group multiplication law itself. In particular, this means that we can write the group elements (at least in some neighborhood of e) in terms of the generators. That's nice, because unlike the group elements, the generators form a vector space. They can be added together and multiplied by real numbers. In fact, we will often use the term generator to refer to any element in the real linear space spanned by the Xas. 2.2. LIE ALGEBRAS 45 2.2 Lie algebras Now in any particular direction, the group multiplication law is uncomplicated. There is a one parameter family of group elements of the form (2.8) and the group multiplication law is simply (2.9) However, if we multiply group elements generated by two different linear combinations of generators, things are not so easy. In general, (2.10) On the other hand, because the exponentials form a representation of the group (at least if we are close to the identity), it must be true that the product is some exponential of a generator, (2.11) for some o. And because everything is smooth, we can find Oa by expanding both sides and equating appropriate powers of a and /3. When we do this, something interesting happens. We find that it only works if the generators form an algebra under commutation (or a commutator algebra). To see this, let's actually do it to leading nontrivial order. We can write (2.12) I will now expand this, keeping terms up to second order in the parameters a and /3, using the Taylor expansion ofln(l + K) where = I( ei<', and r3 cos 0. Can you generalize this construction to arbitrary land explain what is going on? 4.C. Find where the X~ are given by (3.31) Hint: There is a trick that makes this one easy. Write aa Xal = aaAa xai where a= Jaaaa' You know that &aXJ has eigenvalues ±1 and 0, just like XJ (because all directions are equivalent). Thus (&aXJ) 2 is a projection operator and You should be able to use this to manipulate the expansion of the exponential and get an explicit expression for eiaaXJ. Chapter 5 Isospin The idea of isospin arose in nuclear physics in the early thirties. Heisenberg introduced a notation in which the proton and neutron were treated as two components of a nucleon doublet N = (~) (5.1) He did this originally because he was trying to think about the forces between nucleons in nuclei, and it was mathematically convenient to write things in this notation. In fact, his first ideas about this were totally wrong - he re- ally didn't have the right idea about the relation between the proton and the neutron. He was thinking of the neutron as a sort of tightly bound state of proton and electron, and imagined that forces between nucleons could arise by exchange of electrons. In this way you could get a force between proton and neutron by letting the electron shuttle back and forth - in analogy with an H;J ion, and a force between neutron and neutron - an analogy with a neutral H2 molecule. But no force between proton and proton. 5.1 Charge independence It was soon realized that the model was crazy, and the force had to be charge independent - the same between pp, pn and nn to account for the pattern of nuclei that were observed. But while his model was crazy, he had put the p and n together in a doublet, and he had used the Pauli matrices to describe their interactions. Various people soon realized that charge independence would be automatic if there were really a conserved "spin" that acted on the doublet of p and n just as ordinary spin acts on the two J3 components of a spin-1/2 representation. Some people called this "isobaric spin", 79 80 CHAPTER 5. ISOSPIN which made sense, because isobars are nuclei with the same total number of baryons, 1 protons plus neutrons, and thus the transformations could move from one isobar to another. Unfortunately, Wigner called it isotopic spin and that name stuck. This name makes no sense at all because the isotopes have the same number of protons and different numbers of neutrons, so eventually, the "topic" got dropped, and it is now called isospin. 5.2 Creation operators Isospin really gets interesting in particle physics, where particles are routinely created and destroyed. The natural language for describing this dynamics is based on creation and annihilation operators (and this language is very useful for nuclear physics, as we will see). For example, for the nucleon doublet in (5.1 ), we can write IP, a) = atN'21 ,o IO) = In, a) atN _1 IO) ' 2,0 (5.2) where the (5.3) are creation operators for proton(+½) and neutron(-½) respectively in the state a, and 10) is the vacuum state - the state with no particles in it. The N stands for nucleon, and it is important to give it a name because we will soon discuss creation operators for other particles as well. The creation operators are not hermitian. Their adjoints are annihilation operators, aN,±½,a (5.4) These operators annihilate a proton (or a neutron) if they can find one, and otherwise annihilate the state, so they satisfy (5.5) The whole notation assumes that the symmetry that rotates proton into neutron is at least approximately correct. If the proton and the neutron were not in some sense similar, it wouldn't make any sense to talk about them being in the same state. 'Baryons are particles like protons and neutrons. More generally, the baryon number is one third the number of quarks. Because, as we will discuss in more detail later, the proton and the neutron are each made of three quarks, each has baryon number I. 5.2. CREATION OPERATORS 81 Because the p and n are fermions, their creation and annihilation opera- tors satisfy anticommutation relations: tm' {aN,m,a, a ,/3} = '5mm' <5aj3 (5.6) {a~,m,a,a~,m',/3} = {aN,m,a,aN,m',/3} = 0 With creation and annihilation operators, we can make multiparticle states by simply applying more than one creation operator to the vacuum state. For example n proton creation operators at 1 N, 2 ,01 ···atN, 21,an 10) (5.7) ex /n protons; a1, ···,an) produces an n proton state, with the protons in states, a1 through an. The anticommutation relation implies that the state is completely antisymmetric in the labels of the particles. This guarantees that the state vanishes if any two of the as are the same. It means (among other things) that the Pauli exclusion principle is automatically satisfied. What is nice about the creation and annihilation operators is that we can construct states with both protons and neutrons in the same way. For example, n nucleon creation operators atN,m1,a1 ···atN,mn,°'n /0) ex In nucleon; m1, a1; · · ·; mn, an) (5.8) is an n nucleon state, with the nucleons in states described by the m variable (which tells you whether it is a proton or a neutron) and the a label, which tells you what state the nucleon is in. Now the anticommutation relation implies that the state is completely antisymmetric under exchange of the pairs of labels, m and a. In nucleon; m1, a1; m2, a2 · · ·; mn, an) = -/n nucleon; m2, a2; m1, a1; · · ·; mn, an) (5.9) If you haven't seen this before, it should bother you. It is one thing to assume that the proton creation operators anticommute, because two protons really cannot be in the same state. But why should proton and neutron creation operators anticommute? This principle is called the "generalized exclusion principleY Why should it be true? This is an important question, and we will come back to it below. For now, however, we will just see how the creation and annihilation operators behave in some examples. 82 CHAPTER 5. ISOSPIN 5.3 Number operators We can make operators that count the number of protons and neutrons by putting creation and annihilation operators together (the summation convention is assumed): at N,+½,a a N,+ 1 2 , a at a 1 N,-½,a N,-2•°' aNt ,m,a aN,m,a counts protons counts neutrons counts nucleons (5.10) Acting on any state with Np protons and Nn neutrons, these operators have eigenvalues Np, Nn and Np+ Nn respectively. This works because of (5.5) and the fact that for a generic pair of creation and annihilation operators (5.11) Notice that the number operators in (5.1 O) are summed over all the possible quantum states of the proton and neutron, labeled by a. If we did not sum over a, the operators would just count the number of protons or neutrons or both in the state a. We could get fancy and devise more restricted number operators where we sum over some a and not others, but we won't talk further about such things. The total number operators, summed over all a, will be particularly useful. 5.4 Isospin generators For the one-particle states, we know how the generators of isospin symmetry should act, in analogy with the spin generators: Or in terms of creation operators (5.12) (5.13) Furthermore, the state with no particles should transform like the trivial representation - (5.14) 5.5. SYMMETRY OF TENSOR PRODUCTS 83 Thus we will get the right transformation properties for the one particle states if the creation operators transform like a tensor operator in the spin 1/2 representation under isospin: [Ta, a~,m,a] = a~,m',a [JJl2]m'm = ~a~,m',a [aa]m'm It is easy to check that the following form for Ta does the trick: (5.15) Ta= a~m' a [JJl 2 ]m 1m aN,m,a + ··· = = -21 1 ' a~m' ' ' a ' -2atN,a aa [aa]m'm aN,m,a aN,a +··· + · · · (5.16) where · · · commutes with the nucleon creations and annihilation operators (and also annihilates )O} ). The last line is written in matrix form, where we think of the annihilation operators as column vectors and the creation operators as row vectors. Let us check that (5.16) has the right commutation relations with the creation operators so that (5.15) is satisfied. [Ta, atm,a] [JJ = [a~,m',,8 12]m1m" aN,m",,8, a~,m,a] = a~,m',,8 [JJl2]m1m" {aN,m",,8, a~,m,a} - {a~,m',,8• a~,m,a} [JJl 2 ]m1 m11 aN,m",,8 -_ aNt ,m',a [Jla/2] m'm (5.17) The advantage of thinking about the generators in this way is that we now immediately see how multiparticle states transform. Since the multiparticle states are built by applying more tensor (creation) operators to the vacuum state, the multiparticle states transform like tensor products - not a surprising result, but not entirely trivial either. 5.5 Symmetry of tensor products We pause here to discuss an important fact about the combination of spin states (either ordinary spin or isospin). We will use it in the next section to discuss the deuteron. The result is this: when the tensor product of two identical spin 1/2 representations is decomposed into irreducible representations, 84 CHAPTER 5. ISOSPIN the spin I representation appears symmetrically, while the spin Oappears antisymmetrically. To see what this means, suppose that the spin 1/2 states are ll/2, ±1/2, a) (5.18) where a indicates whatever other parameters are required to describe the state. Now consider the highest weight state in the tensor product. This is the spin 1 combination of two identical J3=1/2 states, and is thus symmetric in the exchange of the other labels: 11, 1) = ll/2, 1/2, a) ll/2, 1/2, ,8) = ll/2, 1/2, ,8) 11/2, 1/2, a) (5.19) The lowering operators that produce the other states in the spin l representation preserve this symmetry because they act in the same way on the two spin 1/2 states. 11,0) = ~(11/2,-1/2,a)ll/2,1/2,,8) ,8)) +11/2, 1/2, a) ll/2, -1/2, (5.20) 11, -1) = 11/2, -1/2, a) 11/2, -1/2, ,8) Then the orthogonal spin O state is antisymmetric in the exchange of a and ,8: 10,0) = ~(11/2,-1/2,a)ll/2,1/2,,8) -11/2, 1/2, a) 11/2, -1/2, ,8)) (5.21) 5.6 The deuteron The nucleons have spin 1/2 as well as isospin 1/2, so the a in the nucleon creation operator actually contains a J3 label, in addition to whatever other parameters are required to determine the state. As a simple example of the transformation of a multiparticle state, con- sider a state of two nucleons in an s-wave - a zero angular momentum state. Then the total angular momentum of the state is simply the spin angular momentum, the sum of the two nucleon spins. Furthermore, in an s-wave state, the wave function is symmetrical in the exchange of the position variables of the two nucleons. Then because the two-particle wave function is proportional to the product of two anticommuting creation operators acting on the vacuum state, it is antisymmetric under the simultaneous exchange of the isospin and spin labels of the two nucleons - if the spin representation is 5. 7. SUPERSELECTION RULES 85 symmetric, the isospin representation must be antisymmetric, and vice versa. When combined with the results of the previous section, this has physical consequences. The only allowed states are those with isospin l and spin O or with isospin O and spin 1. The deuteron is an isospin O conbination, and has spin l, as expected. 5.7 Superselection rules It appears, in this argument, that we have assigned some fundamental physical significance to the anticommutation of the creation operators for protons and neutrons. As I mentioned above, this seems suspect, because in fact, the proton and neutron are not identical particles. What we actually know directly al from the Pauli exclusion principle is that the creation operator, for any state of a particle obeying Fermi-Dirac statistics satisfies (5.22) If we have another creation operator for the same particle in another state, ab, al+ ab, we can form the combination which when acting on the vacuum creates the particle in the state a + f3 (with the wrong normalization). Thus the exclusion principle also implies (ai+a1)2 =0 (5.23) and thus at} = { at Ql /3 0 (5.24) This argument is formally correct, but it doesn't really make much physi- al a1 cal sense if and create states of different particles, because it doesn't really make sense to superpose the states - this superposition is forbidden by a superselection rule. A superselection rule is a funny concept. It is the statement that you never need to think about superposing states with different values of an exactly conserved quantum number because those states must be orthogonal. Anything you can derive by such a superposition must also be derivable in some other way that does not involve the "forbidden" superposition. Thus as you see, the superposition is not so much forbidden as it is irrelevant. In this case, it is possible to show that one can choose the creation operators to anticommute without running into inconsistencies, but there is a much stronger argument. The anticommutation is required by the fact that the creation operators transform like tensor operators. Let's see how this implies the stated result for the two nucleon system. 86 CHAPTER 5. ISOSPIN at± ( Call the creation operators for the baryons dropping the N for brevity) where the first sign is the sign of the third component of isospin and (a~+) the second is the sign of third component of spin. Since 2 = 0, there is no two nucleon state with T3 = 1 and h = 1. But this means that there is no state with isospin I and spin 1, since the highest weight state would have to have T3 = 1 and ]3 = 1. In terms of creation operators, for example (5.25) Similar arguments show that the operators must anticommute whenever they have one common index and the others are different. The argument for operators that have no index in common is a little more subtle. First compute (5.26) But the two terms in the sum must separately vanish because they are phys- ically distinguishable. There cannot be a relation like (5.26) unless the two operators {a~_, a~+} IO) (5.27) and (5.28) separately vanish, because these two operators, if they did not vanish, would do physically distinguishable things - the creation of a proton with spin up and a neutron with spin down is not the same as the creation of proton with spin down and a neutron with spin up. Thus the operators (5.27) and (5.28) must separately vanish. Thus, not only does the isospin 1, spin I state,(5.26) vanish but so also does the isospin 0, spin O state (5.29) 5.8 Other particles When isospin was introduced, the only known particles that carried it were the proton and neutron, and the nuclei built out of them. But as particle physicists explored further, at higher energies, new particles appeared that are not built out of nucleons. The first of these were the pions, three spinless bosons (that is obeying Bose-Einstein, rather than Fermi-Dirac statistics) with 5. 8. OTHER PARTICLES 87 charges Q = +1, 0 and - l, and T3 = Q, forming an isospin triplet.2 The creation and annihilation operators for the pions can be written as a1t T,m,o, a=,m,~ form= -1 to 1 11 l..l. They satisfy commutation, rather than anticommutation relations (5.30) [a1r,m,a, a~,m',/3] = Omm'Oa/3 [at,m,a,a~,m',/3] = [a1r,m,a,a1r,m',/3] = 0 (5.31) so that the particle states will be completely symmetric. They also commute with nucleon creation and annihilation operators. The isospin generators look like Ta= at,m,a [J~]mm' a1r,m1,a + ··· (5.32) where as in (5.16) the · · · refers to the contributions of other particles (like nucleons). Again, then the creation operators are tensor operators. There are many many other particles like the nucleons and the pions that participate in the strong interactions and carry isospin. The formalism of creation and annihilation operators gives us a nice way of writing the generators of isospin that acts on all these particles. The complete form of the isospin generators is particles .:t states o: T3 values m,m1 (5.33) where at,m,a and ax,m',a are creation and annihilation operators for x-type particles satisfying commutation or anticommutation relations depending on whether they are bosons or fermions, = [ax,m,a,a~,,m,,/3]± Omm'Oa,aOxx' [ai,m,a,at,,m',/3]± = [ax,m,a,ax',m',/3]± = 0 (5.34) The rule for the ± (+ for anticommutator, - for commutator) is that the anti- commutator is used when both x and x' are fermions, otherwise the commu- tator is used. The Jx in (5.33) is the isospin of the x particles. 2When these particles were discovered, it was not completely obvious that they were not built out of nucleons and their antiparticles. When very little was known about the strong interactions, it was possible to imagine, for example, that the 7r+ was a bound state of a proton and an antineutron. This has all the right quantum numbers - even the isospin is right. It just turns out that this model of the pion is wrong. Group theory can never tell you this kind of thing. You need real dynamical information about the strong interactions. 88 CHAPTER 5. ISOSPIN 5.9 Approximate isospin symmetry Isospin is an approximate symmetry. What this means in general is that the Hamiltonian can be written as H =Ho+ b:.H (5.35) where Ho commutes with the symmetry generators and b:.H does not, but in some sense b:.H is small compared to H0 . It is traditional to say in the case of isospin that the "strong" interactions are isospin symmetric while the weak and electromagnetic interactions are not, and so take Ho = Hs and b:.H = HEM + Hw where Hs, HEM and Hw are the contributions to the Hamiltonian describing the strong interactions (including the kinetic energy), the electromagnetic interactions, and the weak interactions, respectively. From our modem perspective, this division is a bit misleading for two reasons. Firstly, the division between electromagnetic and weak interactions is not so obvious because of the partial unification of the two forces. Secondly, part of the isospin violating interaction arises from the difference in mass between the u and d quarks which is actually part of the kinetic energy. It seems to be purely accidental that this effect is roughly the same size as the effect of the electromagnetic interactions. But this accident was important historically, because it made it easy to understand isospin as an approximate symmetry. There are so many such accidents in particle physics that it makes one wonder whether there is something more going on. At any rate, we will simply lump all isospin violation into b:.H. The group theory doesn't care about the dynamics anyway, as long as the symmetry structure is properly taken into account. 5.10 Perturbation theory The way (5.35) is used is in perturbation theory. The states are classified into eigenstates of the zeroth order, isospin symmetric part of the Hamilto- nian, H0 . Sometimes, just Ho is good enough to approximate the physics of interest. If not, one must treat the effects of b:.H as perturbations. In the scattering of strongly interacting particles, for example, the weak and electromagnetic interactions can often be ignored. Thus in pion-nucleon scattering, all the different possible charge states have either isospin 1/2 or 3/2 (because 1 0 1/2 = 3/2 EB 1/2), so this scattering process can be described approxi- mately by only two amplitudes. The mathematics here is exactly the same as that which appears in the decomposition of a spin-1/2 state with an orbital angular momentum 1 into 5.10. PERTURBATION THEORY 89 states with total angular momentum 3/2 and 1/2. The state with one pion and one nucleon can be described as a tensor product of an isospin 1/2 nucleon state with an isospin 1 pion state, just as the state with both spin and orbital angular momentum can be described as a tensor product, having both spin and angular momentum indices. Problems 5.A. Suppose that in some process, a pair of pions is produced in a state with zero relative orbital angular momentum. What total isospin values are possible for this state? 5.B. Show that the operators defined in (5.33) have the commutation relations of isospin generators. 5.C. ~ ++, ~ +, ~0 and ~ - are isospin 3/2 particles (T3 = 3/2, 1/2, -1/2 and -3/2 respectively) with baryon number 1. They are produced by strong interactions in 7r-nucleon collisions. Compare the probability of producing ~ ++ in 7r+ P --t ~ ++ with the probability of producing ~ 0 in 7l'- p-+ ~o.