Lectures on Symmetries in Physics

Apostolos PilaftsisDepartment of Physics and Astronomy, University of Manchester,

Manchester M13 9PL, United Kingdom

http://pilaftsi.home.cern.ch/pilaftsi/

0. Literature

Group Theory as the Calculus of Symmetries in Physics

1. Introduction to Group Theory

Definition of a Group G The Discrete Groups Sn, Zn and Cn Cosets and Coset Decomposition Normal Subgroup H and Quotient Group G/H Morphisms between Groups

2. Group Representations (Reps)

Definition of a Vector Space V Definition of a Group rep. Reducible and Irreducible reps (Irreps) Direct Products and ClebschGordan Series

3. Continuous Groups

SL(N,C); SO(N); SU(N); SO(N,M) Useful Matrix Relations in GL(N,C) Generators and Exponential rep of Groups

[ Examples: SO(2), U(1), SO(3), SU(2) ]

4. Lie Algebra and Lie Groups

Generators of a Group as Basis Vectors of a Lie Algebra The Adjoint Representation Normalization of Generators and Casimir Operators

5. Tensors in SU(N)

Preliminaries Young Tableaux Applications to Particle Physics

6. Lorentz and Poincare Groups

Lie Algebra and Generators of the Lorentz Group Lie Algebra and Generators of the Poincare Group Single Particle States

2

7. Lagrangians in Field Theory

Variational Principle and Equation of Motion Lagrangians for the Klein-Gordon and Maxwell equations Lagrangian for the Dirac equation

8. Gauge Groups

Global and Local Symmetries. Gauge Invariance of the QED Lagrangian Noethers Theorem YangMills Theories

9. The Geometry of Gauge Transformations (Trans)

Parallel Transport and Covariant Derivative Topology of the Vacuum: the BohmAharanov Effect

10. Supersymmetry (SUSY)

Graded Lie Algebra Generators of the Super-Poincare Group The WessZumino Model Feynman rules

3

Literature

In order of relevance and difficulty:

1. H.F. Jones: Groups, Representations and Physics (IOP,1998) Second Edition

2. L.H. Ryder, Quantum Field Theory (CUP, 1996) SecondEdition

3. T.-P. Cheng and L.-F. Li, Gauge Theory of ElementaryParticle Physics (OUP, 1984).

4. S. Pokorski, Gauge Field Theories (CUP, 2000) SecondEdition.

5. J. Wess and J. Bagger, Supersymmetry and Supergravity,(Princeton University Press, 1992) Second Edition

4

A list of related problems from H.F. Jones:

1. 2.5, 2.9, 2.12

2. 3.1, 3.3, 3.4, 3.6

3. 6.1, 6.2, 6.3

4. 9.1

5. 8.1, 8.3, 8.4, 8.5, 8.6, 8.7, 8.8, 8.9

6. 10.1, 10.2, 10.3

7. 10.4, 10.5, 10.6, 10.7, 10.8

8. 11.3, 11.5, 11.7, 11.8

Note that more problems as exercises are included in thesenotes.

5

1. Introduction to Group Theory

Definition of a Group G

A group (G, ) is a set of elements {a, b, c . . .} endowed witha composition law that has the following properties:

(i) Closure. a, b G, the element c = a b G.

(ii) Associativity. a, b, c G, it holds a (b c) = (a b) c

(iii) The identity element e. e G: e a = a, a G.

(iv) The inverse element a1 of a. a G, a1 G:a a1 = a1 a = e.

If a b = b a, a, b G, the group G is called Abelian.

The Discrete Groups Sn, Zn and Cn

Group G Multiplication Order Remarks

Sn: permutation Successive operation n! Non-Abelian

of n objects in general

Zn: integers Addition modn n Abelian

modulo n

Cn: cyclic group Unspecified product n Cn = Zn{e, a, . . . an1}with an = 1

6

Cosets and Coset Decomposition

Coset. Let H = {h1, h2, . . . , hr} be a proper (i.e. H 6= Gand H 6= I = {e}) subgroup of G.For a given g G, the sets

gH = {gh1, gh2, . . . , ghr} , Hg = {h1g, h2g, . . . , hrg}

are called the left and right cosets of H.

Lagranges Theorem. If g1H and g2H are two (left) cosetsof H, then either g1H = g2H or g1H g2H = .

Coset Decomposition. If H is a proper subgroup of G, thenG can be decomposed into a sum of (left) cosets of H:

G = H g1H g2H g1H ,

where g1,2,... G, g1 / H; g2 / H, g2 / g1H, etc.The number is called the index of H in G.

The set of all distinct cosets, {H, g1H, . . . , g1H}, is amanifold, the coset space, and is denoted by G/H.

7

Normal Subgroup H and Quotient Group G/H

Conjugate to H. If H is a subgroup of G, then the setH = gHg1 = {gh1g1, gh2g1, . . . , ghrg1}, for a giveng G, is called g-conjugate to H or simply conjugate to H.

Normal Subgroup H of G. If H is a subgroup of G andH = gHg1 g G, then H is called a normal subgroupof G.

Groups which contain no proper normal subgroups are termedsimple.

Groups which contain no proper normal Abelian subgroupsare called semi-simple.

Quotient Group G/H. Let G/H = {H, g1H, . . . , g1H}be the set of all distinct cosets of a normal subgroup H of G,with the multiplication law:

(giH) (gjH) = (gi gj)H ,

where giH, gjH G/H. Then, it can be shown that (G/H, )is a group and is termed quotient group.

Note that G/H is not a subgroup of G. (Why?)

8

Morphisms between Groups

Group Homorphism. If (A, ) and (B, ?) are two groups,then group homorphism is a functional mapping f from theset A into the set B, i.e. each element of a A is mappedinto a single element of b = f(a) B, such that the followingmultiplication law is preserved:

f(a1 a2) = f(a1) ? f(a2) .

In general, f(A) 6= B, i.e. f(A) B.

Group Isomorphism. Consider a 1 : 1 mapping f of (A, )onto (B, ?), such that each element of a A is mappedinto a single element of b = f(a) B, and conversely, eachelement of b B is the image resulting from a single elementof a A. If this bijectiv 1 : 1 mapping f satisfies thecomposition law:

f(a1 a2) = f(a1) ? f(a2) ,

it is said to define an isomorphism between the groups A andB, and is denoted by A = B.A group homorphism of A into itself is called endomorphism.

A group isomorphism of A into itself is called automorphism.

9

2. Group Representations (Reps)

Definition of a Vector Space V

A vector space V over the field of complex numbers C isa set of elements {vi}, endowed with two operations (+, ),satisfying the following properties:

(A0) Closure. u + v V u, v V .

(A1) Commutativity. u + v = v + u u, v V .

(A2) Associativity. u+(v+w) = (u+v)+w u, v, w V

(A3) The identity (null) vector. 0 V , such thatv + 0 = v , v V .

(A4) Existence of inverse. v V , (v) V , such thatv + (v) = 0.

(B0) u V C, u V .

(B1) (u + v) = u + v.

(B2) (1 + 2) u = 1 u + 2 v.

(B3) 1 (2 u) = (12) u.

(B4) 1 u = u.

10

Definition of a Group Rep.

Group Rep. A group representation T ,

T : g T (g) GL (N,C) g G ,

is a homomorphism of the elements g of a group (G, ) intothe group GL(N,C) of non-singular linear tranformations ofa vector space V of dimension N , i.e. the set of N N -dimensional invertible matrices in C.In addition, homomorphism implies that the groupmultiplication is preserved:

T (g1 g2) = T (g1)T (g2) .

Two reps. T1 and T2 are equivalent if there exists anisomorphism (1 : 1 correspondance) between T1 and T2.Such an equivalence is denoted as T1 = T2, or T1 T2.Two equivalent reps may be related by a similarity trans. S:T1(g) = ST2(g)S

1 g G and S independent of g.

Character of a rep T of a group G is defined as theset of all traces of the matrices T (g): = {(g)/(g) =

i[T (g)]ii g G}.Corrolary: Equivalent reps have the same character.Conversely, if two reps have the same character, they areequivalent.

11

Reducible and Irreducible Reps.

Reducible rep. A group rep. T (g) is said to be (completely)reducible, if there exists a non-singular matrixM GL(N,C)independent of the group elements, such that

M T (g)M1 =

T1(g) 0 0

0 T2(g)...

... . . . 00 0 Tr(g)

g G.T1(g), T2(g), . . . , Tr(g) divide T into reps. of lowerdimensions, i.e. dim (T ) =

ri=1 dim (Ti), and is denoted

by the direct sum:

T (g) = T1(g) T2(g) Tr(g) =T(i) .

Irreducible rep (Irrep). A group rep. T (g) which cannot bewritten as a direct sum of other reps. is called irreducible.

12

Direct Products and ClebschGordan Series

Direct Product of Groups. If A = {a1, a2, . . . , an} andB = {b1, b2, . . . , bm} are two groups with

aibj = bjai ai A, bj B ,

then a new direct-product group G = AB can be uniquelydefined with elements g = ab. The multiplication law in G is

(a1b1)(a2b2) = (a1a2)(b1b2) .

Remarks: (i) A and B are normal subgroups of G.

(ii) A = G/B = {a1B, a2B, . . . , anB}and B = G/A = {Ab1, Ab2, . . . , Abm}.

Direct Product of Irreps. If D(a) and D(b) are two irrepsof the group G, a direct product, denoted as D(ab)(g1g2) D(a)(g1)D(b)(g2), can be constructed as follows:

[D(ab)(g1g2)]ij;kl = [D(a)(g1)]ik [D(b)(g2)]jl .

Frequently, direct products of irreps are called tensor products.

It can be shown that D(ab) is an irrep of the (direct) productgroup GG.

13

ClebschGordan Series

If g1 = g2 = g, then the symmetry of the product groupGG is reduced to its diagonal G, i.e. GG G.In this case, D(a)(g)D(b)(g) may not be an irrep and canbe further decomposed into a direct sum of irreps of G:

D(a)(g)D(b)(g) =

acD(c)(g) .

Such a series decomposition is called a ClebschGordan series,and the coefficients ac are the so-called ClebschGordancoefficients.

Applications to reps of the continuous groups SO(2), SU(2)and SU(N) will be discussed in the next lectures.

14

3. Continuous Groups

SL(N,C); SO(N); SU(N); SO(N,M)

Group Properties No. of indep. Remarks

parameters

GL(N,C) detM 6= 0 2N2 General rep

SL(N,C) detM = 1 2(N2 1) SL(N,C) GL(N,C)

O(N,R)PN

i=1(xi)2 12N(N 1) OT = O1

=PN

i=1(xi)2

SO(N,R) as above + 12N(N 1) as abovedetO = 1

SU(N)PN

i=1 |xi|2 N2 1 U = U1=

PNi=1 |xi|2

detU = 1

SO(N,M)PN+M

i,j=1 xigijx

j ? Tg = g

=PN+M

i,j=1 xigijxj det = 1

gij = diag (1, . . . , 1| {z }Ntimes

, 1, . . . ,1)| {z }Mtimes

15

Useful Matrix Relations in GL(N,C)

Definitions:

(i) eM n=0

Mn

n!;

(ii) lnM n=1

(1)n+1 (M 1)n

n

=

10

du (M 1) [u(M 1) + 1]1 ,

where M GL(N,C), i.e. detM 6= 0.Basic properties: If [M1, M2] = 0 and M1,2 GL(N,C),then the following relations hold:

(i) eM1 eM2 = eM1+M2 , (ii) ln(M1M2) = lnM1+lnM2 .

Useful identity:

ln(detM) = Tr (lnM) .

This identity can be proved more easily if M can bediagonalized through a similarity trans: S1MS = M ,where M is a diagonal matrix, and noticing that lnM =S ln MS1. (Question: How?)

16

Generators and Exponential rep of Groups[ Examples: SO(2), U(1), SO(3), SU(2) ]

SO(2): Transf. of a point P (x, y) under a rotation through about z axis:(

x

y

)=

(cos sinsin cos

)

O()

(xy

).

Note that OT ()O() = 12 and hence x2 + y2 = x2 + y2,

i.e. O() is an orthogonal matrix, with detO=1.

SO(2) is an Abelian group, since O()O() = O(+ ) =O()O().

Taylor expansion of O() about 12 = O(0):

O() =

(1 00 1

)

: 12

i (

0 ii 0

)

: 2 = iO() |=0

+ O[()2] ,

with 22 = 12 and 2 = 2.

Exponential rep for finite :

O() = limN

[O(/N)]N = exp[i 2] .

The Pauli matrix 2 is the generator of the SO(2) group.

17

U(1): The 2-dim rep of SO(2) in (V,R) can be reduced in(V,C), by means of the trans:

M =

(12

12i

2i2

), M1 =

(12

i2

12

i2

),

i.e.

M1O()M =(ei 00 ei

)= D(1)() D(1)() .

Both reps, D(1)() = ei and D(1)() = ei, are faithfulirreps of U(1).

A gerenal irrep of U(1) is

D(m)() = eim ,

where m Z. (Question: What is the generator of U(1)?)Direct products of U(1)s:

D(m)()D(n)() = D(m+n)() .

18

Spatial rotation of a wave-function:

Unitary operator of rotation of a wave-function:

UR()(r, ) = (1 iX)(r, ) = (r, ) ,

where

X = i dd

=Jz~

is the z-component angular momentum operator.

19

SO(3): Group of proper rotations in 3-dim about a givenunit vector n = (nx, ny, nz) = (n1, n2, n3), with n

2 = 1.

Rotations about x, y, z-axes:

R1() =

0@ 1 0 00 cos sin

0 sin cos

1A , R2() =

0@ cos 0 sin0 1 0 sin 0 cos

1A ,

R3() =

0@ cos sin 0sin cos 0

0 0 1

1A .

The generators Xi = idRi()d

=0

of SO(3) are

X1 =

0@ 0 0 00 0 i

0 i 0

1A , X2 =

0@ 0 0 i0 0 0i 0 0

1A ,

X3 =

0@ 0 i 0i 0 0

0 0 0

1A .

Equivalently, they can be represented as

(Xk)ij = i ijk ; ijk =

8>>>:

1 for (i, j, k) = (1, 2, 3)

and even permutations,

1 for odd permutations,0 otherwise

where ijk is the Levi-Civita antisymmetric tensor.

General rep of the Group element of SO(3):

R(,n) = exp(in X) ,

with X = (X1,X2,X3).

20

Properties of the Generators of SO(3).

Commutation relations:

[Xi, Xj] XiXj XjXi = iijkXk .

(Need to use that (Xk)ij = iijk andijmklm = ikjl iljk.)Jacobi identity:

[X1, [X2, X3]] + [X3, [X1, X2]] + [X2, [X3, X1]] = 0 .

Irreps of SO(3). These are specified by an integer j (the so-called total angular momentum in QM) and are determined

by the (2j + 1) (2j + 1)-dim rep of the generators X (j)i :

[X(j)3 ]mm = jm|X3|jm = mmm ,

[X(j) ]mm = jm|X|jm =

(j m)(j m+ 1) m,m1 ,

with X(j) = X

(j)1 iX(j)2 and Xi = Li/~.

Exercise: Find the relation between X(1)i and Xi.

21

SU(2): Rotation of a complex 2-dim vector v = (v1, v2)(with v1,2 C) through angle about n:

v = U(,n)v ; v v = v v ,

with detU = 1 and

U(,n) = exp(in 12 ) = cos 12 i n sin 12 ,

where n2 = 1 and = (1, 2, 3) are the Pauli matrices.

Xi =12i are the generators of SU(2), with

1 =

(0 11 0

), 2 =

(0 ii 0

), 3 =

(1 00 1

).

Properties: (i) Tri = 0; (ii) ij = ij12 + i ijk k.

Commutation relation: [Xi, Xj] = i ijkXk i.e. the samealgebra as of SO(3).

Precise relation between SO(3) and SU(2):

Since R(0) and R(2pi) [with R(0) = R(2pi) = 13] map intodifferent elements U(0) = 12 and U(2pi) = 12, a faithful1 : 1 mapping is

SO(3) = SU(2)/Z2 ,

where Z2 = {12, 12} is a normal subgroup of SU(2).

22

4. Lie Algebra and Lie Groups

Generators of a Group as Basis Vectors of a Lie Algebra

A Lie algebra L is defined by a set of a number d(G) ofgenerators Ta closed under commutation:

[Ta, Tb] = Ta Tb Tb Ta = if cab Tc ,

where f cab are the so-called structure constants of L.In addition, the generators Tas satisfy the Jacobi identity:

[Ta, [Tb, Tc]] + [Tc, [Ta, Tb]] + [Tb, [Tc, Ta]] = 0 .

The set Ta of generators define a basis of a d(G)-dimensionalvector space (V,C).

In the foundamental rep, Ta are represented by d(F ) d(F )matrices, where d(F ) is the least number of dimensionsneeded to generate the continuous group.

Ex: (i) SO(3): Ta = Xa; (ii) SU(2): Ta =12a; (iii) U(1): ?

Exponentiation of Ta generates the group elements of thecorresponding continuous Lie group:

R(,n) = exp[in T] ,

with n2 = 1.

23

The Adjoint Representation

The Lie algebra commutator [Tc, ] (for fixed Tc) defines alinear homomorphic mapping from L to L over C:

[Tc, 1Ta + 2Tb] = 1[Tc, Ta] + 2[Tc, Tb] ,

Ta, Tb L.For every given Ta L, [Ta, ] may be represented in thevector space L by the structure constants themselves:

[DA(Ta)]cb = ifcab (= if cba) .

Such a rep of Ta is called the adjoint representation, denotedby A.The Killing product form is defined as

gab (Ta, Tb) Tr[DA(Ta)DA(Tb)] ( TrA(TaTb) ) .

gab = fdacf cbd is called the Cartan metric.The Cartan metric gab can be used to lower the index of f

cab:

fabc = fdab gdc .

Exercise: Show that fabc = iTrA([Ta, Tb]Tc) , and thatfabc is totally antisymmetric under the permutation of a, b, c:fabc = fbac = fbca etc.

24

General Remarks

If all f cabs are real for a Lie algebra L, then L is said tobe a real Lie algebra.

If the Cartan metric gab is positive definite for a real L,then L is an algebra for a compact group. In this case, gabcan be diagonalized and rescaled to unity, i.e. gab = 1ab.[Ex: the real algebras of SU(N) and SO(N)].

There is no adjoint representation for Abelian groups.(Why ?)

An ideal I is an invariant subalgebra of L, with[T Ia , Tb] I, T Ia I and Tb L,or symbolically [I, L] I.

Ideals I generate normal subgroups of the continuousgroup generated by L.

Lie algebras that do not contain any proper ideals arecalled simple (Ex: SO(2), SU(2), SU(3), SU(5), etc).

Lie algebras that do not contain any proper Abelian idealsare called semi-simple. (Question: What is the differencebetween a simple and a semi-simple Lie algebra?)

A semi-simple Lie algebra can be written as a direct sumof simple Lie algebras: L = I P .

25

Normalization of Generators and Casimir operators

The generators of a Lie group DR(Ta) of a given rep R arenormalized as

Tr [DR(Ta)DR(Tb)] = TR ab .

For example, in SU(N) [or SO(N)], TF =12 for the

fundamental rep and TA = N for the adjoint reps.

Casimir operators T2R of a Lie algebra of a rep R are matrixreps that commute with all generators of L in rep R.

A construction of a Casimir operator T2R in a given rep R ofSU(N) [or SO(N)] may be obtained by

(T2R)ij = TA

d(G)a,b=1

d(R)k=1

[DR(Ta)]ik gab [DR(Tb)]kj = ij CR ,

where gab is the inverse Cartan metric satisfying: gab gbc = ac .

Exercises:

Show that (i) [T2F , Ta] = 0;

(ii) TR d(G) = CR d(R);

(iii) CF =N212N and CA = N in SU(N).

26

5. Tensors in SU(N)

Preliminaries

Trans. of a complex vector i = (1, 2, . . . , n) in SU(N):

i i = Uijj (= U ji j) ,

where U U = UU = 1n and detU = 1.

Define the scalar product invariant under SU(N):

(, ) = i i (= i i) .

Hence, the trans. of the c.c. i is

i i i = Uijj (or i = U ijj) ,

with U ji = Uij, Uij = U

ij and U

ik U

kj = U

ikU

kj =

ij.

. . .

Higher-rank tensors are defined as those quantities that havethe same trans. law as the direct (diagonal) product ofvectors:

i1i2...ipj1j2...jq

= (U i1k1Ui2k2. . . U

ipkp

) (U l1j1 Ul2j2. . . U

lqjq

)k1k2...kpl1l2...lq

.

The rank of i1i2...ipj1j2...jq

is p + q, with p contravariant and qcovariant indices.

27

SU(N) trans. properties of the Kronecker delta ij and

Levi-Civita symbol i1i2...in:

Invariance of ij under an SU(N) trans:

ij = UikU

lj

kl = U

ikU

kj =

ij .

The Levi-Civita symbol i1i2...in:

i1i2...in =

1 if (i1, . . . in) is an evenpermutation of (1, . . . n)

1 if (i1, . . . in) is an oddpermutation of (1, . . . n)

0 otherwise

Note that i1i2...in is defined to be fully antisymmetric, suchthat ji2...in

ii2...in = (n 1)! ij.Invariance of i1i2...in (and

i1i2...in) under an SU(N) trans:

i1i2...in = Uj1i1

U j2i2 . . . Ujnin

j1j2...jn

= detU = 1

i1i2...in = i1i2...in .

28

Reduction of higher-rank tensors:

Lower-rank tensors can be formed by appropriate use of ijand i1i2...in:

i2...ipj2...jq

= j1i1 i1i2...ipj1j2...jq

,

i1 = i1i2...ini2...in ,

= i1i2...ini1i2...in ,

i1j1 = i1i2...in j1j2...jn i2...inj2...jn .

Since the Levi-Civita tensor can be used to lower or raiseindices, we only need to study tensors with upper or lowerindices.

Exercise: Show that is an SU(N)-invariant scalar.

29

Young Tableaux

Higher-rank SU(N) tensors do not generally define basesof irreps. To decompose them into irreps, we exploit thefollowing property which is at the heart of Young Tableaux.

An illustrative example. Consider the 2nd rank tensor ij,with the trans. property:

ij = Uki U

lj kl .

Permutation of i j (denoted by P12) does not change thetrans. law of ij:

P12ij =

ji = U

kj U

li kl = U

lj U

ki lk

= U lj Uki P12kl .

Hence, P12 can be used to construct the following irreps:

Sij =1

2(1 + P12)ij =

1

2(ij + ji ) ,

Aij =1

2(1 P12)ij = 1

2(ij ji) ,

with P12Sij = Sij and P12Aij = Aij, since there is nomixing between Sij and Aij under an SU(N) trans:

Sij = Uki U

lj Skl , A

ij = U

ki U

lj Akl .

30

Introduction to Young Tableaux

A complex (covariant) vector (or state) i in SU(N) isrepresented by a 2:

i i

The operation of symmetrization and antisymmetrization isrepresented as

(ij) i j [ij] ij

with (ij) =12 (1 + P12)ij = Sij = Sji and

[ij] =12 (1 P12)ij = Aij = Aji.

By analogy, for ijk we have

i j k

(ijk)

ijk

[ijk]

i jk

[(ij);k]

where (ijk) is fully symmetric in i, j, k,[ijk] is fully anti-symmetric in i, j, k and[(ij);k] = (1 P13) (1 + P12)ijk.

Exercise: Express (ijk) and [(ij);k] in terms of ijk.(Ans: [(ij);k] = ijk + jik kji jki.)

31

Rules for constructing a legal Young Tableau

A typical Young tableau for an (n-rank) tensor with nindices looks like:

Each row of a Young tableau must contain no more boxesthan the row above. This implies e.g. that

is not a valid diagram.

There should be no column with more than N boxes forSU(N). In this respect, a column with exactly N boxescan be crossed out. For example, in SU(3) we have:

= 1 =

(Why?)

32

How to find the dimension of a Young Tableau rep

Steps to be followed:

(a) Write down the ratio of two copies of the tableau:

(b) Numerator: Start with the number N for SU(N) in the top left box.

Each time you meet a box, increase the previous number by +1 when

moving to the right in a row and decrease it by 1 when going downin a column:

N N+1 N+2

N1 NN2 N1N3

(c) Denominator: In each box, write the number of boxes being to its

right + the number being below of it and add +1 for itself:

6 4 14 23 11

(d) The dimension d of the rep is the ratio of the products of the entries

in the numerator versus that in the denominator:

d = [N(N + 1)(N + 2)(N 1)N(N 2)(N 1)(N 3)]/ [6 4 4 2 3] .

33

Rules for Clebsch-Gordan series

The direct product of reps can be decomposed as a Clebsch-Gordan series (or direct sum) of irreps. This reduction canbe performed systematically by means of Young Tableaux,following the rules below:

(a) Write down the two tableaux T1 and T2 and label successiverows of T2 with indices a, b, c, . . .:

a a ab bc

(b) Attach boxes a, b, c . . . from T2 to T1 in all possibleways one at a time. The resulting diagram should bea legal Young tableau with no two as or bs beingin the same column (because of cancellation due toantisymmetrization).

(c) At any given box position, there should be no more bsthan as to the right and above of it. Likewise, thereshould be no more cs than bs etc. For example, the

tableau a b is not legal.

(d) Two generated tableaux with the same shape are differentif the labels are distributed differently.

34

An example in SU(3)

a ab

=(

a +a

+

a

) ab

=(

a a + aa

+ a

a

+a

a

) b= a a

b+ a a + a

a b+ a

a+ ab

+1

8 8 = 27 10 10 8 8 1

Exercise: Find the ClebschGordan decomposition of theproduct 8 10 in SU(3), represented by Young tableaux as

(Ans: 8 10 = 8 10 27 35)

35

Applications to Particle Physics

The SU(3) quark symmetry

Define the quark-basis states

qi =

uds

, qi = uds

.Then, qi 3 and qi qi 3.ClebschGordan series: 3 3 = 8 1:

qiqj = (qiq

j 13ji qkq

k) +1

3ji qkq

k .

In terms of YoungTableaux:

= +

The singlet state is 1 =13qiq

i = 13(uu+ dd+ ss).

The remaining 8 components represent the pseudoscalar octetP ji = (qiq

j 13 ji qkqk):

P ij =

12pi0 + 1

68 pi

+ K+

pi 12pi0 + 1

68 K

0

K K0 268

.

36

Baryons as three-quark states:

ClebschGordan series: 3 3 3 = 10 8 8 1Define qijk = qiqjqk, then

qijk = q(ijk) + q[(ij);k] + q[(ji);k] + q[ijk] .

For example, the baryon-octet may be represented by

B = q[(ij);k] =

0B@

120 + 1

68

+ p

120 + 1

6 n

0 26

1CA .

Exercise: Find the ClebschGordan decompositionfor 3 3 3, using YoungTableaux.What is the quark wave-function of p and n?

37

Particle assignment in an SU(5) unified theory

The particle content of the SM = SU(3)cSU(2)L U(1)Yconsists of three generations of quarks and leptons.

One generation of quarks and leptons in the SM contains 15dynamical degrees of freedom:(

ur,g,bLdr,g,bL

),

(LlL

), ur,g,bR , d

r,g,bR , lR .

In SU(5), the SM fermions are assigned as follows:

5 =

dr

dg

db

e

L

,

and

10 =

0 ub ug ur drub 0 ur ug dgug ur 0 ub dbur ug ub 0 edr dg db e 0

L

Exercise: Given that 5 is the complex conjugate rep i = i

of the SU(5) in the fundamental rep, find the tensor rep forthe 10-plet representing the remaining fermions of the SM.

38

6. Lorentz and Poincare Groups

Lorentz trans:

x2 = y

x1 = x

x3 = z

x0 = ct

O

= v/c

x2 = y

x1 = x

x3 = z

x0 = ct

O

x = ()x ,

where x = (ct, x, y, z), x = (ct, x, y, z) are thecontravariant position 4-vectors, and

=

0 0

0 00 0 1 00 0 0 1

, for ex .Given the metric g = diag (1,1,1,1), the covariant4-vector is defined as x = g x

= (ct,x,y,z).Under a Lorentz trans, we have xx = x

x or

x gx = xg

x

Tg = g ,

so SO(1,3), with det = 1.

39

Lie Algebra and Generators of the Lorentz Group

Generators and Lie Algebra of SO(1,3)

Generators of rotations J1,2,3:

J1 =

0BB@

0 0 0 00 0 0 00 0 0 i0 0 i 0

1CCA , J2 =

0BB@

0 0 0 00 0 0 i0 0 0 00 i 0 0

1CCA ,

J3 =

0BB@

0 0 0 00 0 i 00 i 0 00 0 0 0

1CCA .

Generators of boosts K1,2,3:

K1 =

0BB@

0 i 0 0i 0 0 00 0 0 00 0 0 0

1CCA , K2 =

0BB@

0 0 i 00 0 0 0i 0 0 00 0 0 0

1CCA ,

K3 =

0BB@

0 0 0 i0 0 0 00 0 0 0i 0 0 0

1CCA .

Commutation relations of the Lie algebra SO(1,3):

[Ji, Jj] = i ijk Jk ,

[Ji, Kj] = i ijkKk ,

[Ki, Kj] = i ijk Jk

40

SO(1,3)C = SU(2) SU(2) [or SO(1,3)R SL(2,C)]Define

X =1

2(J iK) ,

then

[X+i , X+j ] = i ijkX

+k ,

[Xi , Xj ] = i ijkX

k ,

[X+i , Xj ] = 0 .

Hence, SO(1,3) algebra splits into two SU(2) ones:

SO(1, 3)C= SU(2) SU(2) ,

where SO(1,3)C is the rep from a complexified SO(1,3)algebra. However, there is an 1:1 correspondence of thereps between SO(1,3)C and SO(1,3)R. In fact, we have thehomomorphism

SO(1, 3)R SL(2,C) ,

which is more difficult to use for classification of reps.

41

Classification of basis-states reps in SO(1,3)

We enumerate basis-state reps in SO(1,3) by (j1, j2), usingthe relation of SO(1,3) with SU(2)1 SU(2)2, where j1,2 arethe total spin numbers with respect to SU(2)1,2. The totaldegrees of freedom are (2j1 + 1)(2j2 + 1). In detail, we have

(0,0): This is a total spin zero rep, with dim one. (0, 0) representsa scalar field (x) satisfying the Klein-Gordon equation:(2 +m2)(x) = 0, where 2 = .

(12, 0): This a 2-dim rep, the so-called left-handed Weyl rep,e.g. neutrinos. It is denoted with a 2-dim complex vector, usually called the left-handed Weyl spinor. Under aLorentz trans, transforms as

= M ,

where M SL(2,C).

(0, 12): This is the corresponding 2-dim rep of the right-handedWeyl spinor and is denoted as , which transforms underLorentz trans as

= M ,

where M SL(2,C).

(12,12): This is the defining 4-dim rep, describing a spin 1 particle

with 4 components. One can use the matrix rep: A()or simply A, e.g. A = (/c, A) in electromagnetism.

42

Lie Algebra and Generators of the Poincare Group

The Poincare trans consist of Lorentz trans plus space-timetranslations:

x = x + a ,

where a is a constant 4-vector.

The generator of translations in a differential-operator rep is

P = i = i

x= i (

ct, ) ,

with P = i = i(ct , ), because

eiaPx = x + a . (Why?)

An analogous differential-operator rep of the 6-generators ofLorentz trans is given by the generalized angular momentumoperators:

L = xP xP ,with the identification

Ji =1

2ijkLjk , Ki = L0i .

Exercise: Show that Ji =12 ijkLjk and Ki = L0i satisfy

the SO(1,3) algebra.

43

The Lie Algebra of the Poincare Group:

The commutation relations defining the Poincare Lie algebraare

[P , P] = 0 ,

[P, L] = i (gP gP) ,[L, L] = i (gL gL + gL gL) .

In terms of J and K, the commutation relations read:

[P0 , Ji] = 0 ,

[Pi , Jj] = i ijk Pk ,

[P0 , Ki] = i Pi ,

[Pi , Kj] = i P0 ij .

Exercise: Prove all commutation relations that appear on thispage.

44

Single Particle States

The Poincare group has two Casimir operators: P 2 = PPand W 2 = WW, where

W = 12 L

P ,

with 0123 = 1, is the so-called PauliLubanski vector.

Classification of massive particle states

A single massive particle state |a can be characterized by itsmass and its total spin s, where s is defined in the rest ofmass system of the particle:

P 2 |a = m2 |a , W 2 |a = m2J2 |a = m2s(s+1)|a .

In addition, we use the 3-momentum P and the helicityH = J P operators to classify massive particle states:

P |a;m, s;p, = p |a;m, s;p, ,H |a;m, s;p, = |p| |a;m, s;p, .

Note that a massive particle state has (2s+ 1) polarizationsor helicities, also called degrees of freedom,i.e. = s,s+ 1, . . . , s 1, s.Examples: for an electron, it is = 12, and for a massivespin-1 boson (e.g. the Z-boson), we have = 1, 0, 1.

45

Classification of massless particle states

Massless particle states, for which P 2|a = 0 (m = 0),are characterized only by their 4-momentum p and helicity = P J.Alternatively, in addition to the operator P, one may use thePauliLubanski operator W:

W|a; p, = p |a; p, .

If the theory involves parity, then a massless state has onlytwo degrees of freedom (polarizations): .Examples of the above are the photon and the neutrinos ofthe Standard Model.

Exercises:

(i) Show that P 2 and W 2 are true Casimir operators,i.e. [P 2, P ] = [P

2, L ] = 0 , and likewise for W2;

(ii) In particles rest frame where p = (m, 0, 0, 0), show thatW0 = 0, Wi =

12mijk L

jk = mJi and W2 = m2J2;

(iii) Show that [J P,P] = 0, [P,W] = 0, and WP = 0;(iv) Calculate the commutation relation [W,W].

46

7. Lagrangians in Field Theory

Variational Principle and Equation of Motion

Classical Lagrangian Dynamics

The Lagrangian for an n-particle system is

L(qi, qi) = T V ,

where q1,2,...,n are the the generalized coordinates describingthe n particles, and q1,2,...,n are the respective timederivatives.

T and V denote the total kinetic and potential energies.

The action S of the n-particle system is given by

S[qi(t)] =

t2t1

dt L(qi, qi) .

Note that S is a functional of qi(t).

47

Hamiltons principle

Hamiltons principle states that the actual motion of thesystem is determined by the stationary behaviour of Sunder small variations qi(t) of the ith particles generalizedcoordinate qi(t), with qi(t1) = qi(t2) = 0, i.e.

S =

t2t1

dt

(qi

L

qi+ qi

L

qi

)=

t2t1

dt qi

(L

qi d

dt

L

qi

)= 0 .

The EulerLagrange equation of motion for the ith particle is

d

dt

L

qi L

qi= 0 .

Exercise: Show that the EulerLagrange equations of motionfor a particle system described by a Lagrangian of the formL(qi, qi, qi) are

d2

dt2L

qi d

dt

L

qi+

L

qi= 0 .

[Hint: Consider only variations with qi(t1,2) = qi(t1,2) = 0.]

48

Lagrangian Field Theory

In Quantum Field Theory (QFT), a (scalar) particle isdescribed by a field (x), whose Lagrangian has the functionalform:

L =

d3x L((x) , (x)) ,

where L is the so-called Lagrangian density, often termedLagrangian in QFT.

In QFT, the action S is given by

S[(x)] =

+

d4x L((x) , (x)) ,

with limx

(x) = 0.

By analogy, the EulerLagrange equations can be obtainedby determining the stationary points of S, under variations(x) (x) + (x):

L

() L

= 0 .

Exercise: Derive the above EulerLagrange equation for ascalar particle by extremizing S[(x)], i.e. S = 0 .

49

Lagrangians for the Klein-Gordon and Maxwell eqs

Lagrangian for the KleinGordon equation

LKG = 12

() () 1

2m2 2 ,

where (x) is a real scalar field describing one dynamicaldegree of freedom.

The EulerLagrange equation of motion is the KleinGordonequation

( + m2)(x) = 0 .

Lagrangian for the Maxwell equations

LME = 14F F

JA ,

where F = A A is the field strength tensor, andJ is the 4-vector current satisfying charge conservation:J

= 0.

A describes a spin-1 particle, e.g. a photon, with 2 physicaldegrees of freedom.

Exercise: Use the Euler-Lagrange equations for LME toshow that F

= J , as is expected in relativisticelectrodynamics (with 0 = 0 = c = 1).

50

Lagrangian for the Dirac equation

LD = (i m) ,

where

(x) =

((x)

(x)

), =

(0 ()

() 0

)and (x) ((x), (x) ), with = (12, ) and =(12, ).The and

are 2-dim complex vectors (also called Weylspinors) whose components anti-commute: 12 = 21,12 = 21, 12 = 21 etc.The EulerLagrange equation of LD with respect to is theDirac equation:

LD

= 0 (i m) = 0 .

The 4-component Dirac spinor (x) that satisfies the Diracequation describes 4 dynamical degrees of freedom.

Exercises:(i) Derive the EulerLagrange equation with respect to theDirac field (x);(ii) Show that up to a total derivative term, LD is Hermitian,i.e. LD = LD + j, with j = i.

51

Lorentz trans properties of the Weyl and Dirac spinors

The Dirac spinor is the direct sum of two Weyl spinors and with Lorentz trans properties:

= M ,

= M

,

= M1 , = M 1

.

with M SL(2,C).Duality relations among 2-spinors:

() = , () = , () = , () =

Lowering and raising spinor indices:

= , = , =

, = ,

with i2 =

0 11 0

= and i2 = .

Lorentz-invariant spinor contractions:

= = = = =

Likewise, () = = = = .Exercise: Given that MM

= and M1M1 =

, show that LD is invariant under Lorentz trans.

52

8. Gauge Groups

Global and Local Symmetries

Symmetries in Classical Physics and Quantum Mechanics:

Translational invariance in time Energy conservationt t+ a0 dEdt = 0Translational invariance in space Momentum conservation

r r + a dpdt = 0Rotational invariance Angular momentum conservationr R r dJdt = 0

Quantum Mechanics Degeneracy of energy states

[H,O] = 0 dOdt = i[H,O] = 0

Quantum Field Theory Noethers Theorem

(x) (x) + (x) ?

53

Global and Local Symmetries in QFT

Consider the Lagrangian (density) for a complex scalar:

L = () () m2 + ()2 .

L is invariant under a U(1) rotation of the field :

(x) (x) = ei (x) ,

where does not depend on x x.A transformation in which the fields are rotated about x-independent angles is called a global transformation. If theangles of rotation depend on x, the transformation is calleda local or a gauge transformation.

A general infinitesimal global or local trans of fields i underthe action of a Lie group reads:

i(x) i(x) = i(x) + i(x) ,

where i(x) = i a(x) (T a) ji j(x) , and T a are thegenerators of the Lie Group. Note that the angles or groupparameters a are x-independent for a global trans.

If a Lagrangian L is invariant under a global or local trans, itis said that L has a global or local (gauge) symmetry.Exercise: Show that the above Lagrangian for a complexscalar is not invariant under a U(1) gauge trans.

54

Gauge Invariance of the QED Lagrangian

Consider first the Lagrangian for a Dirac field :

LD = (i m) .

LD is invariant under the U(1) global trans:

(x) (x) = ei (x) ,

but it is not invariant under a U(1) gauge trans, when = (x). Instead, we find the residual term

LD = ((x)) .

To cancel this term, we introduce a vector field A in thetheory, the so-called photon, and add to LD the extra term:

L = LD eA .

We demand that A transforms under a local U(1) as

A A = A 1

e(x) .

L is invariant under a U(1) gauge trans of and A.

55

QED Lagrangian with an electron-photon interaction

The complete Lagrangian of Quantum Electrodynamics(QED) that includes the interaction of the photon with theelectron is

LQED = 14F F

+ (i 6 m e 6A) ,

where we used the convention: 6a a.

Exercises:

(i) Show that LQED is gauge invariant under a U(1) trans.(ii) Derive the equation of motions with respect to photonand electron fields.

(iii) How should the Lagrangian describing a complex scalarfield (x),

L = () () m2 ,

be extended so as to become gauge symmetric under a U(1)local trans?

56

Noethers Theorem

Noethers Theorem. If a Lagrangian L is symmetric under aglobal transformation of the fields, then there is a conservedcurrent J(x) and a conserved charge Q =

d3x J0(x),

associated with this symmetry, such that

J = 0 and

dQ

dt= 0 .

Proof:

Consider a Lagrangian L(i, i) to be invariant under theinfinitesimal global trans:

i = i a(T a) ji j ,

where T a are the generators of some group G.

Hence, the change of L is vanishing, i.e.

L = Li

i +L

(i)(i) = 0 .

This last equation can be rewritten as

L = [

L(i)

i

]+

[Li

L(i)

]i = 0 .

57

With the aid of the equations of motions for i, the lastequation implies that

[L

(i)i

]=

[

L(i)

Li

]i = 0 .

The conserved current (or currents) is

Ja, =L

(i)

ia

=L

(i)i (T a) ji j .

The corresponding conserved charges are

Qa(t) =

d3x Ja, 0(x) .

Indeed, it is easy to check that

dQa

dt=

d3x 0 J

a, 0(x) = d3x Ja(x)

= ds Ja 0 ,

because surface terms vanish at infinity.

Exercises: Find the conserved currents and charges for(i) QED;(ii) the gauge-invariant Langrangian with a complex scalar .

58

YangMills Theory

The Lagrangian of a YangMills (non-Abelian) SU(N) theoryis

LYM = 14F a F

a, ,

where

F a = Aa Aa + g fabcAbAc ,

and fabc are the structure constants of the SU(N) Lie algebra.

It can be shown that LYM is invariant under the infinitesimalSU(N) local trans:

Aa = 1

g

a + fabc bAc .

Examples of SU(N) theories are the SU(2)L group of the SMand Quantum Chromodynamics (QCD) based on the SU(3)cgroup.

The gauge (vector) fields of the SU(2)L are the W0 and W

bosons responsible for the weak force.

The gauge vector bosons of the SU(3)c group are the gluonsmediating the strong force between quarks.

Gauge bosons of YangMills theories self-interact!

Exercise: Show that LYM is invariant under SU(N) gaugetrans.

59

Interaction between quarks qi and gluons Aa in SU(3)c

If qi = (qred, qgreen, qblue) are the 3 colours of the quark,their interaction with the 8 gluons Aa is described by theLagrangian:

Lint = qi [ i 6 ji mji g 6Aa(T a) ji ] qj .

Exercise: Show that Lint is invariant under the SU(3) gaugetransformation:

Aa = 1

g

a + fabc bAc , qi = ia (T a) ji qj ,

where T a = 12 a are the generators of SU(3) and a are the

Gell-Mann matrices:

1,2,3

=

1,2,3 0

0 0

,

4=

0@ 0 0 10 0 0

1 0 0

1A ,

5

=

0@ 0 0 i0 0 0i 0 0

1A , 6 =

0@ 0 0 00 0 1

0 1 0

1A ,

7

=

0@ 0 0 00 0 i

0 i 0

1A , 8 =

0B@

13

0 0

0 13

0

0 0 23

1CA .

60

9. The Geometry of Gauge Transformations

Parallel Transport and Covariant Derivative

Simple Example:

e2

e1

v(t)

(t)

Time-dependent vector written in terms of t-dependent unitvectors:

v(t) = vi(t) ei(t) (with i = 1, 2).

The true time derivative of v(t) is

d

dtv(t) = lim

t0v(t+ t) v(t)

t.

To calculate this, we need to refer all unit vectors to t+ t:

ei(t) = ei(t+ t) t tei(t) .

61

Then, we have

d

dtv(t) =

1

t{ vi(t+ t) (vi(t) t vj(t) [ei tej])}

ei(t+ t)= [ tvi(t) + (ei tej) vj(t) ] ei(t) .

We can now define the covariant derivative to act only onthe components of v(t) as:

Dtvi(t) = tvi(t) + (ei tej)vj(t) ,= tvi(t) + 3ij vj(t) ,

with the obvious property ddtv(t) = ei(t)Dtvi(t). The secondterm is induced by the change of the coordinate axes, namelyafter performing a parallel transport of our coordinate systeme1,2(t) from t to t+ t.

Proper comparison of two vectors vi(t + t) and vi(t) canonly be made in the same coordinate system by means ofparallel transport. Differentiation is properly defined throughthe covariant derivative.

Exercise: Show that the covariant derivative satisfies therelation

Dtvi(t) = tvi(t) + ( v(t))i ,with = (t), which is known from Classical Mechanicsbetween rotating and fixed frames in 3 dimensions.

62

Differentiation in curved space

The notion of the covariant derivative generalizes to curvedspace as well. By analogy, the infinitesimal difference betweenthe 4-vectors V (x) and V (x) is given by

DV = dV + V ,

where dV is the difference of the 2 vectors in the samecoordinate system and V is due to parallel transport of thevector from x to x = x + x.

In the framework of General Relativity, we have

DV = (V + V

) dx ,

where is the so-called affine connection or the Christoffelsymbol.

63

Covariant derivative in the Gauge-Group Space

Consider the difference of a fermionic isovector field atx + x and x in an SU(N) gauge theory:

D = d + ,

where = ig T aAa dx

and the field Aa takes care of the change of the SU(N) axesfrom point to point in Minkowski space.

The covariant derivative of (x) is

D = ( + ig TaAa) ,

which is obtained from pure geometric considerations.

In analogy to General Relativity, the gauge field Aa Ta is

sometimes called the connection.

Exercise: Show that under a local SU(N) rotation of theisovector field: = U (with U SU(N)), itscovariant derivative transforms as

D D = U D ,

with

A = U AU +

i

g(U)U

.

64

A round trip in the SU(N) Gauge-Group Space

A B

CD

x

x

Keeping terms up to second order in x and x, we have

B = (1 + xD +

1

2xxDD)A,0 ,

C = (1 + xD +

1

2xxDD)B ,

D = (1 xD + 12

xxDD)C ,

A,1 = (1 xD + 12xxDD)D .

Hence,

A,1 = (1 + xx [D , D])A,0 ,

and A,1 6= A,0.Exercise: Show that

i

g[D , D] = F

a T

a

is the SU(N) Field-strength tensor.

65

Parallels between Gauge Theory and General Relativity

In General Relativity, a corresponding round trip of a vectorV in a curved space gives rise to

V =1

2R V

S ,

where S represents the area enclosed by the path andR is the RiemannChristoffel curvature tensor:

R = + .

Analogies:

Gauge Theory General Relativity

Gauge trans. Co-ordinate trans.Gauge field Aa T

a Affine connection, Field strength F Curvature tensor RBianchi identity: Bianchi identity:

,,cyclic

DF = 0

,,cyclic

DR = 0

66

Topology of the Vacuum: the BohmAharanov Effect

The BohmAharanov Effect:

B

dx: displacement of fringes

2

1

e source

Vector potential A and B field (with B = A) incylindrical polars:

Inside: Ar = Az = 0, A =Br

2,

Br = B = 0 , Bz = B ,

Outside: Ar = Az = 0, A =BR2

2r,

B = 0 ,

where R is the radius of the solenoid.

Although the electrons move in regions with E = B = 0, theB field of the solenoid induces a phase difference 12 of theelectrons on the screen causing a displacement of the fringes:

12 = 1 2 = e~

21

A dr = e~

B ds .

In regions with E = B = 0, it is A 6= 0, so the vacuum hasa topological structure! It is not simply connected due to thepresence of the solenoid.

67

Basic Concepts in Topology

Let a(s) and b(s) be two paths in a topological space Y bothstarting from the point P (a(0) = b(0) = P ) and ending at apossibly different point Q (a(1) = b(1) = Q). If there existsa continuous function L(t, s) such that L(0, s) = a(s) andL(1, s) = b(s), then the paths a and b are called homotopicwhich is denoted by a b.If P Q, the path is said to be closed.The inverse of a path a is written as a1 and is defined bya1(s) = a(1s). It corresponds to the same path traversedin the opposite direction.

The product path c = ab is defined by

c(s) = a(2s) , for 0 s 12 ,c(s) = b(2s 1) , for 12 s 1 .

If a b, then ab1 is homotopic to the null path: ab1 1.

Exercises:

(i) Consider the mappings S1 U(1): fn() = ei(n+a)(with a R and n Z), and show that they all are homotopicto those with a = 0.

(ii) Given that fn() 6 fm() for n 6= m, explain then whyL(t, ) = ei[n(1t)+mt] is not an allowed homotopy functionrelating fn to fm.

68

Homotopy Classes, Groups and the Winding Number

All paths related to maps X Y of two topological spacesX, Y can be divided into homotopy classes.

Homotopy Class. All paths that are homotopic to a givenpath a(s) define a set, called the homotopy class and denotedby [a]. For example, [fn] are distinct homotopy classes fordifferent n.

Winding Number. Each homotopy class may becharacterized by an integer, the winding number n (also calledthe Pontryargin index). For the case f() : S1 U(1), thewinding number is determined by

n =1

2pii

2pi0

d

(d ln f()

d

).

Homotopy Group. The set of all homotopy classes relatedto maps X Y forms a group, under the multiplication law

[a] [b] = [ab] ,

the so-called homotopy group piX(Y ).

Exercises:(i) Prove that the homotopy group satisfies the axioms of agroup.

(ii) Show that for S1 U(1), pi1[U(1)] = Z.

69

The BohmAharanov Effect Revisited

In regions with E = B = 0, A is a pure gauge: A = (Why?).

The configuration space X of the BohmAharanov effect isthe plane R2 with a hole in it, due to the solenoid. This istopologically equivalent ( homeomorphic) to R S1. Thespace X can be conveniently described by polar coords (r, ),with r 6= 0.It can be shown that (r, ) = const., which is a functionin the group space of U(1), i.e. Y = U(1).

Since functions mapping S1 onto R are all deformable to aconstant, the non-trivial part of is given by the map:

S1 U(1) .

Because pi1[U(1)] = Z, the electron paths cannot be deformedto a null path with a constant , implying A = 0 everywhereand the absence of the BohmAharanov effect.

Since pi1[SU(2)] = 1, there is no BohmAharanov effectfrom an SU(2) solenoid !

Exercises:(i) Show that (r, ) = 12BR

2 is a possible solution forE = B = 0, where B is the magnetic field and R the radiusof the solenoid.

(ii) Verify that 12 =e~[(2pi) (0) ].

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10. Supersymmetry (SUSY)

Graded Lie Algebra

Definition. A Z2-graded Lie algebra is defined on a vectorspace L which is the direct sum of two subspaces L0 andL1: L = L0 L1. The generators that span the space L areendowed with a multiplications law:

: L L L .

T (0) L0, T (1) L1, the generators satisfy the followingproperties:

(i) T(0)1 T (0)2 = (1)g

20 T

(0)2 T (0)1 = [T (0)1 , T (0)2 ] L0 ,

(ii) T (0)T (1) = (1)g0g1 T (1)T (0) = {T (0), T (1)} L1 ,(iii) T

(1)1 T (1)2 = (1)g

21 T

(1)2 T (1)1 = [T (1)1 , T (1)2 ] L0 ,

where g0 = g(L0) = 0 and g1 = g(L1) = 1 are the degreesof the graduation of the Z2-graded Lie algebra.

In addition, all generators of L satisfy the Z2-graded Jacobiidentity:

(1)gigk T (i) (T (j) T (k)) + (1)gkgj T (k) (T (i) T (j))+ (1)gjgi T (j) (T (k) T (i)) = 0 ,

where i, j, k = 0, 1.

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ZN-graded Lie algebra. The generalization of a Z2-gradedLie algebra L to ZN can be defined analogously. Let L bethe direct sum of N subalgebras Li:

L = N1i=0 Li .

Then, the multiplication law among the generators of Lcan be defined by

T (i) T (j) = (1)gigj T (j) T (i) L(i+j) modN .

The ZN-graded Jacobi identity is defined analogously withthat of Z2, where gi,j = 0, 1, . . . , N 1 is the degree ofgraduation of Li,j.

Exercise:??? Find the (anti)-commutation relations and thestructure constants of the Z2-graded Lie algebra of SU(2).

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Generators of the Super-Poincare Group

The generators super-Poincare algebra are P , L L0and the spinors Q , Q L1. They satisfy the followingrelations:

(i) [P , P] = 0 ,

(ii) [P, L] = i (gP gP) ,(iii) [L, L] = i (gL gL + gL gL) .(iv) {Q, Q} = {Q, Q} = 0 ,(v) {Q, Q} = 2()P ,(vi) [Q, P] = 0 ,

(vii) [L , Q] = i() Q ,(viii) [L , Q] = i() Q ,

where () =14 [ (

)() ()() ] and

()

= 14 [ ()() ()() ].

Exercise:? Prove the Z2-graded Jacobi identity:

[L , {Q , Q}] + {Q , [Q , L]}+ {Q , [Q , L]} = 0 .

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Consequences of the Super-Poincare Symmetry

Equal number of fermions and bosons.

Scalar supermultiplet ( , , F ), where is acomplex scalar (2), is a 2-component complex spinor (4),and F is an auxiliary complex scalar (2).

Vector supermultiplet V a (Aa , a, Da), where Aaare massless non-Abelian gauge fields (3), a are the 2-component gauginos (4), and Da are the auxiliary realfields (1).

The simplest model that realizes SUperSYmmetry (SUSY)is the WessZumino model. Counting on-shell degrees offreedom (dof), the Wess-Zumino model contains one complexscalar (2 dofs) and one Weyl spinor (2 dofs):

bosonic dofs = fermionic dofs

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The WessZumino Model

Non-interacting WZ model

Lkin = Lscalar + Lfermion= ()() + i() ; =

12

(1 + i2)

Consider + and + , with

= and = () = = ,

and infinitesimal anticommuting 2-spinor constant.

Lscalar Lscalar + Lscalar ,Lscalar = ()() + ()()

Try + and + , with

= i() and = i()

Lfermion Lfermion + Lfermion ,Lfermion = ()() + ()

= () + ()

[()]

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Exercise: Show that

{ + } = 2g , {+ } = 2g

Noticing that = and using the results of the aboveexercise, we get

Lfermion = () + ()= ()() ()()

+ [() + ()]

L = Lscalar + Lfermion = 0 !

But, we are not finished yet ! The difference of two successiveSUSY transfs. must be a symmetry of the Lagrangian as well,i.e. SUSY algebra should close.

(21 12) = i(12 21) iP (with = )

(21 12) = i(1)2 + i(2)1Fierz= i(12 21)

+ 12i 21i

Only for on-shell fermions, i = 0, the SUSY algebracloses.

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To close SUSY algebra off-shell, we need an auxiliary complexscalar F (without kinetic term) and add

LF = F F

to Lscalar + Lfermion, with

F = i() , F = i() = i()+ F , = i() + F

Exercise: Prove (i) that the Lagrangian

Lkin = ()() + i() + F F

is invariant under the off-shell SUSY transfs:

= , =

= i()+ F , = i() + F F = i() , F = i()

and (ii) that the SUSY algebra closes off-shell:

(21 12)X = i(12 21) X ,

with X = , , , , F, F .

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The interacting WZ model

LWZ = Lkin + Lint= ()() + i() + F F

12W + WF 1

2W + W

F

where

W () =m

2 +

h

6

is the so-called superpotential, and

W =W

= m +

h

22

W =2W

= m + h

Exercise: Show that up to total derivatives,

Lint = 12W + WF 1

2W + W

F

= 12

(m+ h) 12

(m+ h)

+ (m+h

22)F + (m +

h

22)F

remains invariant under off-shell SUSY transformations.

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Feynman rules

Equation of motions for the auxiliary fields F and F :

F = W , F = W ,

Substituting the above into LWZ, we get

LWZ = ()() + i() W W 1

2(W + W

)

and the real potential is

V = W W = m2 +

mh

2(2 + 2) +

h2

4()2

Exercise: If =

(

)is a Majorana 4-spinor, show that

the -dependent part of the WZ Lagrangian can be writtendown as

L = 12

i 12m

h2PL h

2PR ,

where PL,R = (14 5)/2 and 5 = diag (12 , 12).

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Summary

The complete WZ Lagrangian is

LWZ = ()() m2 + 12

i 12m

mh2

(2 + 2) h2

4()2

h2PL h

2PR ,

where the F -field has been integrated out.

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Feynman rules:

, p:

ip2 m2

, p: i6pm

: imh

: ih2

: ihPL

: ihPR

SUSY is such an elegant symmetry that it would be a pity ifnature made no use of it!

81