Lectures on Symmetries in Physics

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3. Continuous Groups Lectures on Symmetries in Physics –SL(N,C);SO(N); SU(N); SO(N,M) – Useful Matrix Relations in GL(N,C) – Generators and Exponential rep of Groups Apostolos Pilaftsis [ Examples: SO(2), U(1), SO(3), SU(2) ] Department of Physics and Astronomy, University of Manchester, Manchester M13 9PL, United Kingdom http://pilaftsi.home.cern.ch/pilaftsi/ 4. Lie Algebra and Lie Groups – Generators of a Group as Basis Vectors of a Lie Algebra 0. Literature – The Adjoint Representation – Group Theory as the Calculus of Symmetries in Physics – Normalization of Generators and Casimir Operators

1. Introduction to Group Theory 5. Tensors in SU(N) – Definition of a Group G –Preliminaries – Young Tableaux – The Discrete Groups Sn, Zn and Cn – Cosets and Coset Decomposition – Applications to Particle Physics – Normal Subgroup H and Quotient Group G/H – Morphisms between Groups 6. Lorentz and PoincaréGroups – Lie Algebra and Generators of the Lorentz Group 2. Group Representations (Reps) – Lie Algebra and Generators of the PoincaréGroup – Definition of a Vector Space V – Single Particle States – Definition of a Group rep. –ReducibleandIrreduciblereps(Irreps) – Direct Products and Clebsch–Gordan Series

2 7. Lagrangians in Field Theory Literature • – Variational Principle and Equation of Motion – Lagrangians for the Klein-Gordon and Maxwell equations In order of relevance and diﬃculty: – Lagrangian for the Dirac equation 1. H.F. Jones: Groups, Representations and Physics (IOP, 8. Gauge Groups 1998) Second Edition

– Global and Local Symmetries. 2. L.H. Ryder, Quantum Field Theory (CUP, 1996) Second – Gauge Invariance of the QED Lagrangian Edition –Noether’sTheorem – Yang–Mills Theories 3. T.-P. Cheng and L.-F. Li, Gauge Theory of Elementary Particle Physics (OUP, 1984). 9. The Geometry of Gauge Transformations (Trans) 4. S. Pokorski, Gauge Field Theories (CUP, 2000) Second – Parallel Transport and Covariant Derivative Edition. – Topology of the Vacuum: the Bohm–Aharanov Eﬀect 5. J. Wess and J. Bagger, Supersymmetry and Supergravity, 10. Supersymmetry (SUSY) (Princeton University Press, 1992) Second Edition – Graded Lie Algebra – Generators of the Super-Poincar´eGroup – The Wess–Zumino Model –Feynmanrules

3 4 A list of related problems from H.F. Jones: 1. Introduction to Group Theory

1. 2.5, 2.9, 2.12∗ – Deﬁnition of a Group G 2. 3.1, 3.3, 3.4, 3.6 A group (G, ) is a set of elements a, b, c . . . endowed with · { } 3. 6.1, 6.2, 6.3 a composition law · that has the following properties:

4. 9.1 (i) Closure. a, b G,theelementc = a b G. ∀ ∈ · ∈ 5. 8.1, 8.3, 8.4, 8.5, 8.6, 8.7, 8.8∗, 8.9∗ (ii) Associativity. a, b, c G, it holds a (b c)=(a b) c 6. 10.1, 10.2, 10.3 ∀ ∈ · · · · (iii) The identity element e. e G: e a = a, a G. 7. 10.4, 10.5, 10.6, 10.7, 10.8 ∃ ∈ · ∀ ∈

1 1 8. 11.3, 11.5, 11.7, 11.8 (iv) The inverse element a− of a. a G, a− G: 1 1 ∀ ∈ ∃ ∈ a a− = a− a = e. · · Note that more problems as exercises are included in these notes. 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 Successiveoperation n! Non-Abelian of n objects in general Zn:integers AdditionmodnnAbelian modulo n Cn:cyclicgroup Unspeciﬁed product nCn ∼= Zn n 1 · e, a, . . . a − { } with an =1

5 6 – Cosets and Coset Decomposition – Normal Subgroup H and Quotient Group G/H

Coset.LetH = h ,h ,...,h be a proper (i.e. H = G Conjugate to H.IfH is a subgroup of G,thentheset { 1 2 r} ̸ and H = I = e ) subgroup of G. H = gHg 1 = gh g 1,gh g 1,...,gh g 1 , for a given ̸ { } ′ − 1 − 2 − r − For a given g G,thesets g G,iscalledg{-conjugate to H or simply conjugate} to H. ∈ ∈

gH = gh1,gh2,...,ghr ,Hg= h1g, h2g, . . . , hrg Normal Subgroup H of G.IfH is a subgroup of G and { } { } 1 H = gHg− g G,thenH is called a normal subgroup ∀ ∈ are called the left and right cosets of H. of G. Groups which contain no proper normal subgroups are termed Lagrange’s Theorem.Ifg1H and g2H are two (left) cosets simple. of H,theneither g H = g H or g H g H = ∅. 1 2 1 ∩ 2 Groups which contain no proper normal Abelian subgroups Coset Decomposition.IfH is a proper subgroup of G,then are called semi-simple. G can be decomposed into a sum of (left) cosets of H: Quotient Group G/H.LetG/H = H, g1H, . . . , gν 1H { − } G = H g1H g2H gν 1H, be the set of all distinct cosets of a normal subgroup H of G, ∪ ∪ ··· ∪ − with the multiplication law: where g G, g / H; g / H, g / g H,etc. 1,2,... ∈ 1 ∈ 2 ∈ 2 ∈ 1 (giH) (gjH)=(gi gj) H, The number ν is called the index of H in G. · ·

where giH, gjH G/H.Then,itcanbeshownthat(G/H, ) The set of all distinct cosets, H, g1H, . . . , gν 1H ,isa ∈ · manifold, the coset space, and is{ denoted by G/H−. } is a group and is termed quotient group. Note that G/H is not a subgroup of G.(Why?)

7 8 – Morphisms between Groups 2. Group Representations (Reps) Group Homorphism.If(A, ) and (B,⋆) are two groups, then group homorphism is a functional· mapping f from the – Deﬁnition of a Vector Space V set A into the set B, i.e. each element of a A is mapped AvectorspaceV over the ﬁeld of complex numbers C is into a single element of b = f(a) B, such that∈ the following ∈ a set of elements vi , endowed with two operations (+, ), multiplication law is preserved: satisfying the following{ } properties: ·

f(a1 a2)=f(a1) ⋆ f(a2) . (A0) Closure. u + v V u, v V . · ∈ ∀ ∈ In general, f(A) = B,i.e.f(A) B. (A1) Commutativity. u + v = v + u u, v V . ̸ ⊂ ∀ ∈ (A2) Associativity. u+(v+w)=(u+v)+w u, v, w V Group Isomorphism. Consider a 1:1mapping f of (A, ) ∀ ∈ onto (B,⋆), such that each element of a A is mapped· ∈ (A3) The identity (null) vector. 0 V ,suchthat into a single element of b = f(a) B, and conversely, each ∃ ∈ ∈ v + 0 = v , v V . element of b B is the image resulting from a single element ∀ ∈ of a A.Ifthisbijectiv∈ 1:1mapping f satisﬁes the (A4) Existence of inverse. v V , ( v) V ,suchthat composition∈ law: v +( v)=0. ∀ ∈ ∃ − ∈ − f(a1 a2)=f(a1) ⋆ f(a2) , (B0) λ u V λ C, u V . · · ∈ ∀ ∈ ∀ ∈ it is said to deﬁne an isomorphism between the groups A and (B1) λ (u + v)=λ u + λ v. · · · B, and is denoted by A ∼= B. (B2) (λ1 + λ2) u = λ1 u + λ2 u. A group homorphism of A into itself is called endomorphism. · · · (B3) λ (λ u)=(λ λ ) u. A group isomorphism of A into itself is called automorphism. 1 · 2 · 1 2 · (B4) 1 u = u. ·

9 10 – Deﬁnition of a Group Rep. – Reducible and Irreducible Reps.

Group Rep.Agroup representation T , Reducible rep. A group rep. T (g) is said to be (completely) reducible, if there exists a non-singular matrix M GL (N,C) T : g T (g) GL (N,C) g G, independent of the group elements, such that ∈ → ∈ ∀ ∈ is a homomorphism of the elements g of a group (G, ) into T1(g)0 0 C · ··· . the group GL(N, ) of non-singular linear tranformations of 1 0 T2(g) . MT(g) M − = ⎛ . ⎞ g G. avectorspaceV of dimension N, i.e. the set of N N- . ... 0 ∀ ∈ × dimensional invertible matrices in C. ⎜ 0 0 T (g) ⎟ ⎜ ··· r ⎟ In addition, homomorphism implies that the group ⎝ ⎠ multiplication is preserved: T1(g), T2(g),...,Tr(g) divide T into reps. of lower r dimensions, i.e. dim (T )= i=1 dim (Ti), and is denoted T (g1 g2)=T (g1) T (g2) . by the direct sum: · !

··· T (g)=T1(g) T2(g) Tr(g)= T(i) . Two reps. T1 and T2 are equivalent if there exists an ⊕ ⊕ ··· ⊕ ⊕ isomorphism (1:1correspondance) between T1 and T2. ( Such an equivalence is denoted as T = T ,orT T . 1 ∼ 2 1 ∼ 2 Two equivalent reps may be related by a similarity trans. S: Irreducible rep (Irrep). A group rep. T (g) which cannot be 1 T (g)=ST (g)S− g G and S independent of g. written as a direct sum of other reps. is called irreducible. 1 2 ∀ ∈ Character χ of a rep T of··· a group G is deﬁned as the set of all traces of the matrices T (g): χ = χ(g)/χ(g)= [T (g)] g G . { i ii ∧ ∈ } !Corollary:Equivalentrepshavethesame character. Conversely, if two reps have the same character, they are equivalent.

11 12 – Direct Products and Clebsch–Gordan Series Clebsch–Gordan Series

Direct Product of Groups.If(A, )=( a ,a,..., a , ) If g = g = g, then the symmetry of the product group · { 1 2 n} · 1 2 and (B, ⋆)=(b1,b2,..., bm , ⋆) are two groups with G G is reduced to its diagonal G,i.e.G G G. composition laws{ and ⋆,respectively,} × × → · In this case, D(a)(g) D(b)(g) may not be an irrep and can then anewdirect-product group (G, )=(A B, ) can be ⊗ uniquely defined with elements g = a⊙ b. The× multiplication⊙ be further decomposed into a direct sum of irreps of G: law in G is defined as ⊗ ⊙ D(a)(g) D(b)(g)= a D(c)(g) . ⊗ c (a b ) (a b ) (a a ) (b ⋆ b ) . 1 ⊗ 1 ⊙ 2 ⊗ 2 ≡ 1 · 2 ⊗ 1 2 (⊕ Such a series decomposition is called a Clebsch–Gordan series, Remarks: (i) A and B are normal subgroups of G (Why?). and the coefficients ac are the so-called Clebsch–Gordan (ii) A = G/B = a B,a B, ...,a B ; ∼ { 1 ⊗ 2 ⊗ n ⊗ } coefficients. B = G/A = A b ,A b ,...,A b . ∼ { ⊗ 1 ⊗ 2 ⊗ m} Applications to reps of the continuous groups SO(2), SU(2) Direct Product of Irreps. If D(a) and D(b) are two irreps and SU(N) will be discussed in the next lectures. (a b) of the group G,adirect product, denoted as D × (g1g2) D(a)(g ) D(b)(g ), can be constructed as follows: ≡ 1 ⊗ 2

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

Frequently, direct products of irreps are called tensor products.

(a b) It can be shown that D × is an irrep of the (direct) product group G G. ×

13 14 C 3. Continuous Groups – Useful Matrix Relations in GL(N, ) Deﬁnitions: –SL(N,C);SO(N); SU(N); SO(N,M) ∞ M n (i) eM ; ≡ n! Group Properties No. of indep. Remarks n=0 parameters ( ∞ (M 1)n (ii) ln M ( 1)n+1 − 2 ≡ − n GL(N, C)detM =0 2N General rep n=1 ̸ (1 2 1 SL(N, C)detM =1 2(N 1) SL(N, C) = du (M 1)[u(M 1)+1]− , − GL(N, C) − − ⊂ )0

R N i 2 1 T 1 O(N, ) i=1(x ) 2N(N 1) O = O− where M GL(N,C),i.e.det M =0. N i 2 − ∈ ̸ =! i=1(x′ ) ! Basic properties: If [M ,M]=0and M GL(N,C), 1 2 1,2 ∈ SO(N, R)asabove+ 1N(N 1) as above then the following relations hold: 2 − detO =1 M1 M2 M1+M2 (i) e e = e , (ii) ln(M1M2)=lnM1+ln M2 . N i 2 2 1 SU(N) i=1 x N 1 U † = U − N| | i 2 − != x′ i=1 | | det!U =1 Useful identity:

N+M i j T SO(N, M) i,j=1 x gijx ? Λ gΛ = g ln(det M)=Tr(lnM) . N+M i j != i,j=1 x′ gijx′ det Λ =1 gij!=diag(1,...,1, 1,..., 1) This identity can be proved more easily if M can be − − N times M times 1 − − diagonalized through a similarity trans: S− MS = M, " #$ % " #$ % where M is a diagonal matrix, and noticing that ln M = 1 S ln MS− .(Question:How?) * * 15 * 16 – Generators and Exponential rep of Groups U(1): The 2-dim rep of SO(2) in (V,R) can be reduced in [ Examples: SO(2), U(1), SO(3), SU(2) ] (V,C), by means of the trans:

SO(2): Transf. of a point P (x, y) under a rotation through 1 1 1 i φ about z axis: √2 √2 1 √2 √2 M = i i ,M− = 1 i , − − 1 √2 √2 2 1 √2 √2 2 x′ cos φ sin φ x = − . y′ sin φ cos φ y i.e. + , + , + , O(φ) ≡ iφ 1 e 0 (1) ( 1) - ./ 0 M − O(φ) M = iφ = D (φ) D − (φ) . Note that OT (φ)O(φ)=1 and hence x2 + y2 = x 2 + y 2, 0 e− ⊕ 2 ′ ′ + , i.e. O(φ) is an orthogonal matrix, with detO=1. Both reps, D(1)(φ)=eiφ and D( 1)(φ)=e iφ,arefaithful SO(2) is an Abelian group, since O(φ)O(φ )=O(φ + φ )= − − ′ ′ irreps of U(1). O(φ′)O(φ). A general irrep of U(1) is Taylor expansion of O(φ) about 12 = O(0):

(m) im φ 10 0 i D (φ)=e , O(δφ)= i δφ − + [(δφ)2] , 01 − i 0 O + , + , Z : 1 ∂O(φ) where m .(Question: What is the generator of U(1)?) 2 : σ = i 2 ∂φ |φ=0 ∈ - ./ 0 - ./ 0 Direct products of U(1)’s: 2 with σ2 = 12 and σ2 = σ2†. D(m)(φ) D(n)(φ)=D(m+n)(φ) . Exponential rep for ﬁnite φ: ⊗

N O(φ)= lim[O(φ/N )] = exp[ iφσ2] . N →∞ −

The Pauli matrix σ2 is the generator of the SO(2) group.

17 18 Spatial rotation of a wave-function: SO(3): Group of proper rotations in 3-dim about a given 2 unit vector n =(nx,ny,nz)=(n1,n2,n3),withn =1. Unitary operator of rotation of a wave-function: Rotations about x, y, z-axes:

UR(δφ) ψ(r, θ)=(1 iδφXˆ) ψ(r, θ)=ψ(r, θ δφ) , 10 0 cos φ 0sinφ − − R (φ)=⎛ 0cosφ sin φ ⎞ ,R(φ)=⎛ 010⎞ , 1 − 2 0sinφ cos φ sin φ 0cosφ 3 ⎝ ⎠ ⎝ − ⎠ where cos φ sin φ 0 d Jˆ − Xˆ = i = z R3(φ)=⎛ sin φ cos φ 0 ⎞ . " 001 − dθ ⎝ ⎠ is the z-component angular momentum operator. dRi(φ) The generators Xi = i dφ φ=0 of SO(3) are

00 0 4 00i X = ⎛ 00 i ⎞ 4,X= ⎛ 000⎞ , 1 − 2 0 i 0 i 00 ⎝ ⎠ ⎝ − ⎠ 0 i 0 − X3 = ⎛ i 00⎞ . 000 ⎝ ⎠ Equivalently, they can be represented as

1 for (i, j, k)=(1, 2, 3) ⎧ and even permutations, (X ) = i ε ; ε = k ij − ijk ijk ⎪ 1 for odd permutations, ⎨ − 0otherwise ⎪ ⎩⎪ where εijk is the Levi-Civita antisymmetric tensor. General rep of the Group element of SO(3):

R(φ, n)=exp( iφ n X) , − ·

with X =(X1,X2,X3).

19 20 Properties of the Generators of SO(3). SU(2): Rotation of a complex 2-dim vector v =(v1,v2) (with v1,2 C) through angle θ about n: Commutation relations: ∈

v′ = U(θ, n) v ; v∗ v = v′∗ v′ , [X ,X] X X X X = iε X . · · i j ≡ i j − j i ijk k with det U =1and (Need to use that (Xk)ij = iεijk and − 1 1 1 εijmεklm = δikδjl δilδjk.) U(θ, n)=exp( iθn σ) = cos θ iσ n sin θ , − − · 2 2 − · 2 Jacobi identity: 2 where n =1and σ =(σ1, σ2, σ3) are the Pauli matrices. ∴ 1 [X1, [X2,X3]] + [X3, [X1,X2]] + [X2, [X3,X1]] = 0 . Xi = 2σi are the generators of SU(2), with

01 0 i 10 σ = , σ = , σ = . ··· 1 10 2 i −0 3 0 1 Irreps of SO(3). These are speciﬁed by an integer j (the so- + , + , + − , called total angular momentum in QM) and are determined Properties: (i) Tr σi =0;(ii)σiσj = δij12 + i εijk σk. (j) by the (2j + 1) (2j + 1)-dim rep of the generators Xi : × Commutation relation: [Xi,Xj]=i εijk Xk i.e. the same algebra as of SO(3). (j) ˆ [X3 ]m′m = jm′ X3 jm = m δmm′ , ⟨ | | ⟩ Precise relation between SO(3) and SU(2): (j) ˆ [X ]m m = jm′ X jm = (j m)(j m + 1) δm ,m 1 , Since R(0) and R(2π) [with R(0) = R(2π)=1 ]mapinto ± ′ ⟨ | ±| ⟩ ∓ ± ′ ± 3 5 diﬀerent elements U(0) = 12 and U(2π)= 12,afaithful (j) (j) (j) ˆ ˆ " − with X = X1 iX2 and Xi = Li/ . 1:1mapping is ± ± Exercise: Find the relation between X(1) and X . i i SO(3) ∼= SU(2)/Z2 ,

where Z = 1 , 1 is a normal subgroup of SU(2). 2 { 2 − 2} 21 22 – The Adjoint Representation 4. Lie Algebra and Lie Groups

The Lie algebra commutator [Tc, ] (for fixed Tc)definesa –Generators of a Group as Basis Vectors of a Lie Algebra linear homomorphic mapping from L to L over C: A Lie algebra L is defined by a set of a number d(G) of [T , λ T + λ T ]=λ [T ,T ]+λ [T ,T] , generators Ta closed under commutation: c 1 a 2 b 1 c a 2 c b

[T ,T]=T T T T = if c T , Ta,Tb L. a b a · b − b · a ab c ∀ ⊂ For every given Ta L, [Ta, ] may be represented in the c ∈ where fab are the so-called structure constants of L. vector space L by the structure constants themselves: In addition, the generators Ta’s satisfy the Jacobi identity: c c c [D (Ta)] b = ifab (= ifba) . A [Ta, [Tb,Tc]] + [Tc, [Ta,Tb]] + [Tb, [Tc,Ta]] = 0 . −

Such a rep of Ta is called the adjoint representation, denoted by . The set Ta of generators deﬁne a basis of a d(G)-dimensional A vector space (V,C). The Killing product form is deﬁned as

In the fundamental rep, Ta are represented by d(F ) d(F ) × gab (Ta,Tb) Tr[D (Ta)D (Tb)] ( Tr (TaTb)). matrices, where d(F ) is the least number of dimensions ≡ A ≡ A A ≡ A needed to generate the continuous group. d c gab = facfbd is called the Cartan metric. 1 − Ex: (i) SO(3): Ta = Xa; (ii) SU(2): Ta = 2σa; (iii) U(1): ? c The Cartan metric gab can be used to lower the index of fab: Exponentiation of Ta generates the group elements of the d corresponding continuous Lie group: fabc = fab gdc .

G(θ, n)=exp[ iθn T] , Exercise:Showthatfabc = i Tr ([Ta,Tb] Tc) , and that − · − A fabc is totally antisymmetric under the permutation of a, b, c: with n2 =1. fabc = fbac = fbca etc. − 23 24 General Remarks – Normalization of Generators and Casimir operators

c The generators of a Lie group DR(Ta) of a given rep R are If all fab’s are real for a Lie algebra L,thenL is said to • be a real Lie algebra. normalized as

If the Cartan metric g is positive deﬁnite for a real L, Tr [DR(Ta) DR(Tb)] = TR δab . • ab then L is an algebra for a compact group. In this case, gab can be diagonalized and rescaled to unity, i.e. g = 1 . 1 ab ab For example, in SU(N)[orSO(N)], TF = 2 for the [Ex: the real algebras of SU(N)andSO(N)]. fundamental rep and T = N for the adjoint reps. A 2 There is no adjoint representation for Abelian groups. Casimir operators TR of a Lie algebra of a rep R are matrix • (Why ?) reps that commute with all generators of L in rep R. T2 An ideal I is an invariant subalgebra of L,with A construction of a Casimir operator R in a given rep R of • I I SU(N)[orSO(N)] may be obtained by [Ta ,Tb] I, Ta I and Tb L, or symbolically⊂ ∀[I, L∈] I. ∀ ∈ ⊂ d(G) d(R) 2 ab Ideals I generate normal subgroups of the continuous (TR)ij = T [DR(Ta)]ik g [DR(Tb)]kj = δij CR , • A group generated by L. a,b(=1 k(=1

ab ab a Lie algebras that do not contain any proper ideals are where g is the inverse Cartan metric satisfying: g gbc = δc . • called simple (Ex: SO(2), SU(2), SU(3), SU(5), etc). Exercises: Lie algebras that do not contain any proper Abelian ideals Show that (i) [T2 ,T]=0; • are called semi-simple.(Question:Whatisthediﬀerence F a between a simple and a semi-simple Lie algebra?) (ii) TR d(G)=CR d(R); N2 1 (iii) CF = 2N− and C = N in SU(N). A semi-simple Lie algebra can be written as a direct sum A • of simple Lie algebras: L = I P . ⊕

25 26 i 5. Tensors in SU(N) SU(N) trans. properties of the Kronecker delta δj and Levi-Civita symbol εi1i2...in:

– Preliminaries i Invariance of δj under an SU(N)trans: Trans. of a complex vector ψi =(ψ1, ψ2,...,ψn) in SU(N): i i l k i k i δj′ = U kUj δl = U kUj = δj . j ψ ψ′ = U ψ (= U ψ ) , i → i ij j i j

i1i2...in where U †U = UU† = 1n and det U =1. The Levi-Civita symbol ε : Define the scalar product invariant under SU(N): 1 if (i1,...in) is an even i permutation of (1,...n) (ψ, φ)=ψ∗φi (= ψ φi) . ⎧ i εi1i2...in = ⎪ 1 if (i ,...i ) is an odd ⎪ − 1 n ⎪ permutation of (1,...n) Hence, the trans. of the c.c. ψi∗ is ⎨ 0 otherwise ⎪ i i i j ⎪ ψ∗ ψ ψ′∗ = U ∗ ψ∗ (or ψ′ = U ψ ) , ⎪ i ≡ → i ij j j ⎩⎪ Note that εi1i2...in is defined to be fully antisymmetric, such j ii2...in i i i k i k i that εji2...in ε =(n 1)! δj. with Ui = Uij, U j = Uij∗ and Uk U j = U kUj = δj. − ... i1i2...in Invariance of εi1i2...in (and ε )underanSU(N)trans: Higher-rank tensors are defined as those quantities that have j1 j2 jn the same trans. law as the direct (diagonal) product of ε′ = U U ... U ε i1i2...in i1 i2 in j1j2...jn vectors: =detU εi1i2...in = εi1i2...in . =1 i i ...ip i i ip l l lq k k ...kp ψ′ 1 2 =(U 1 U 2 ...U )(U 1U 2 ...U ) ψ 1 2 . j1j2...jq k k kp j1 j2 jq l l ...lq 1 2 1 2 - ./ 0 The rank of ψi1i2...ip is p + q, with p contravariant and q j1j2...jq covariant indices.

27 28 Reduction of higher-rank tensors: – Young Tableaux

i Lower-rank tensors can be formed by appropriate use of δj Higher-rank SU(N)tensorsdonot generally deﬁne bases and εi1i2...in: of irreps. To decompose them into irreps, we exploit the following property which is at the heart of Young Tableaux. ψi2...ip = δj1 ψi1i2...ip , j2...jq i1 j1j2...jq An illustrative example. Consider the 2nd rank tensor ψij, i1 i1i2...in with the trans. property: ψ = ε ψi2...in , i1i2...in ψ = ε ψi1i2...in , k l ψij′ = Ui Uj ψkl . i1j1 i1i2...in j1j2...jn ψ = ε ε ψi2...inj2...jn . Permutation of i j (denoted by P ) does not change the ↔ 12 Since the Levi-Civita tensor can be used to lower or raise trans. law of ψij: indices, we only need to study tensors with upper or lower k l l k indices. P12 ψij′ = ψji′ = Uj Ui ψkl = Uj Ui ψlk l k Exercise: Show that ψ is an SU(N)-invariant scalar. = Uj Ui P12 ψkl .

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

1 1 S = (1 + P ) ψ = ( ψ + ψ ) , ij 2 12 ij 2 ij ji 1 1 A = (1 P ) ψ = (ψ ψ ) , ij 2 − 12 ij 2 ij − ji

with P S = S and P A = A ,sincethereisno 12 ij ij 12 ij − ij mixing between Sij and Aij under an SU(N)trans:

k l k l Sij′ = Ui Uj Skl ,Aij′ = Ui Uj Akl .

29 30 Introduction to Young Tableaux Rules for constructing a legal Young Tableau

A complex (covariant) vector (or state) ψi in SU(N)is represented by a ✷: A typical Young tableau for an (n-rank) tensor with n • indices looks like:

ψi i ≡

The operation of symmetrization and antisymmetrization is represented as

i Each row of a Young tableau must contain no more boxes ψ(ij) i j ψ[ij] ≡ ≡ j • than the row above. This implies e.g. that

1 with ψ(ij) = 2 (1 + P12) ψij = Sij = Sji and ψ = 1 (1 P ) ψ = A = A . [ij] 2 − 12 ij ij − ji By analogy, for ψijk we have is not a valid diagram. i There should be no column with more than N boxes for i j k j i j • k k SU(N). In this respect, a column with exactly N boxes ψ(ijk) ψ[ijk] ψ[(ij);k] can be crossed out. For example, in SU(3) we have: where ψ(ijk) is fully symmetric in i, j, k, ψ[ijk] is fully anti-symmetric in i, j, k and = 1 = ψ =(1 P )(1+P ) ψ . [(ij);k] − 13 12 ijk (Why?) Exercise: Express ψ(ijk) and ψ[(ij);k] in terms of ψijk. (Ans: ψ = ψ + ψ ψ ψ .) [(ij);k] ijk jik − kji − jki 31 32 How to ﬁnd the dimension of a Young Tableau rep Rules for Clebsch-Gordan series

Steps to be followed: The direct product of reps can be decomposed as a Clebsch- Gordan series (or direct sum) of irreps. This reduction can (a) Write down the ratio of two copies of the tableau: be performed systematically by means of Young Tableaux, following the rules below:

(a) Write down the two tableaux T1 and T2 and label successive rows of T2 with indices a, b, c, . . .: (b) Numerator:StartwiththenumberN for SU(N)inthetopleftbox. Each time you meet a box, increase the previous number by +1 when aaa moving to the right in a row and decrease it by 1 when going down bb − in a column: × c N N+1 N+2 N 1 N − (b) Attach boxes a, b, c . . . from T2 to T1 in all possible N 2 N 1 − − ways one at a time. The resulting diagram should be N 3 − a legal Young tableau with no two a’s or b’s being in the same column (because of cancellation due to (c) Denominator:Ineachbox,writethenumberofboxesbeingtoits antisymmetrization). right + the number being below of it and add +1 for itself: 641 42 (c) At any given box position, there should be no more b’s 31 than a’s to the right and above of it. Likewise, there 1 should be no more c’s than b’s etc. For example, the tableau a b is not legal. (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) Two generated tableaux with the same shape are diﬀerent d =[N(N +1)(N +2)(N 1)N(N 2)(N 1)(N 3)] if the labels are distributed diﬀerently. − − − − / [6 4 4 2 3] . × × × ×

33 34 An example in SU(3) – Applications to Particle Physics The SU(3) quark symmetry aa = a + + a b a × b Deﬁne the quark-basis states × : a ; u u¯ i qi = d ,q= d¯ . = aa + a + a + b ⎛ s ⎞ ⎛ s¯ ⎞ a a × : a a ; ⎝ ⎠ ⎝ ⎠ Then, q 3 and qi q¯ 3¯. i ≡ ≡ i ≡ = aa + aa + a + a + a + 1 Clebsch–Gordan series: 3 3=8¯ 1: ⊗ ⊕ b a b a b 1 1 q qj =(q qj δj q qk)+ δj q qk . i i − 3 i k 3 i k 8 8 = 27 10 10 8 8 1 In terms of Young–Tableaux: × ⊕ ⊕ ⊕ ⊕ ⊕ = + × Exercise: Find the Clebsch–Gordan decomposition of the product 8 10 in SU(3), represented by Young tableaux as × The singlet state is η = 1 q qi = 1 (uu¯ + dd¯+ ss¯). 1 √3 i √3 The remaining 8 components represent the pseudoscalar octet × P j =(q qj 1 δj q qk): i i − 3 i k 1 0 1 + + (Ans: 8 10 = 8 10 27 35) π + η8 π K × ⊕ ⊕ ⊕ √2 √6 i π 1 π0 + 1 η K0 Pj = ⎛ − √2 √6 8 ⎞ . − 0 2 K− K¯ η8 ⎜ −√6 ⎟ ⎝ ⎠

35 36 Baryons as three-quark states: Particle assignment in an SU(5) uniﬁed theory

Clebsch–Gordan series: 3 3 3 = 10 8 8 1 The particle content of the SM = SU(3)c SU(2)L U(1)Y ⊗ ⊗ ⊕ ⊕ ⊕ consists of three generations of quarks and⊗ leptons. ⊗ Deﬁne qijk = qiqjqk,then One generation of quarks and leptons in the SM contains 15 dynamical degrees of freedom: qijk = q(ijk) + q[(ij);k] + q[(ji);k] + q[ijk] .

r,g,b uL νL r,g,b r,g,b r,g,b , ,uR ,dR ,lR . For example, the baryon-octet may be represented by d lL 1 L 2 + , 1 Σ0 + 1 Λ Σ+ p √2 √6 8 1 0 1 B = q = ⎛ Σ− Σ + Λ n ⎞ . [(ij);k] √2 √6 In SU(5), the SM fermions are assigned as follows: − 0 2 ⎜ Ξ− Ξ Λ ⎟ ⎝ −√6 ⎠ d¯r d¯g Exercise: Find the Clebsch–Gordan decomposition ⎛ ⎞ 5=¯ d¯b , for 3 3 3, using Young–Tableaux. ⎜ e ⎟ What⊗ is the⊗ quark wave-function of p and n? ⎜ ⎟ ⎜ ν ⎟ ⎜ − ⎟L ⎝ ⎠ and 0¯ub u¯g ur dr u¯b 0¯−ur ug dg ⎛ − ⎞ 10 = u¯g u¯r 0 ub db − ⎜ ur ug ub 0¯e ⎟ ⎜ − − − ⎟ ⎜ dr dg db e¯ 0 ⎟ ⎜ − − − − ⎟L ⎝ ⎠ ¯ i Exercise: Given that 5 is the complex conjugate rep ψi∗ = ψ of the SU(5) in the fundamental rep, ﬁnd the tensor rep for the 10-plet representing the remaining fermions of the SM.

37 38 – Lie Algebra and Generators of the Lorentz Group 6. Lorentz and Poincar´eGroups Generators and Lie Algebra of SO(1,3) Lorentz trans: 2 2 x = y x′ = y′ Generators of rotations J1,2,3: 0 000 0 0000 0 x′ = ct′ x = ct β = v/c ⎛ 000 0⎞ ⎛ 000i ⎞ J = ,J= , 1 000 i 2 0000 ⎜ − ⎟ ⎜ ⎟ O O′ ⎜ 00i 0 ⎟ ⎜ 0 i 00⎟ ⎝ ⎠ ⎝ − ⎠ 1 1 x = x x′ = x′ 00 0 0 3 3 ⎛ 00 i 0 ⎞ J = − . x = z x′ = z′ 3 0 i 00 ⎜ ⎟ ⎜ 00 0 0⎟ ⎝ ⎠

µ µ ν Generators of boosts K1,2,3: x′ = Λ ν(β) x , 0 i 00 00 i 0 µ µ − − where x =(ct, x, y, z), x′ =(ct′,x′,y′,z′) are the ⎛ i 000⎞ ⎛ 0000⎞ K = − ,K= , 1 0000 2 i 000 contravariant position 4-vectors, and ⎜ ⎟ ⎜ − ⎟ ⎜ 0000⎟ ⎜ 0000⎟ ⎝ ⎠ ⎝ ⎠ 000 i γ γβ 00 − ⎛ 0000⎞ − K = . µ γβ γ 00 3 0000 Λ ν = ⎛ − ⎞ , for β ex . ⎜ ⎟ 0010 ⎜ i 00 0⎟ ∥ ⎝ − ⎠ ⎜ 0001⎟ ⎜ ⎟ ⎝ ⎠ Commutation relations of the Lie algebra SO(1,3): Given the metric gµν = diag (1, 1, 1, 1), the covariant ν− − − 4-vector is defined as xµ = gµν x =(ct, x, y, z). − − − [J ,J]=i ε J , µ µ i j ijk k Under a Lorentz trans, we have x xµ = x′ xµ′ or [Ji,Kj]=i εijk Kk , µ ν β µ ν α T x gµνx = x Λ gµνΛ x Λ gΛ = g, [K ,K]= i ε J β α ⇒ i j − ijk k so Λµ SO(1,3), with det Λ =1. ν ∈ 39 40 SO(1,3)C = SU(2) SU(2) [or SO(1,3)R SL(2,C)] Classification of basis-states reps in SO(1,3) ∼ × ∼ Define We enumerate basis-state reps in SO(1,3) by (j1,j2),using 1 the relation of SO(1,3) with SU(2) SU(2) ,wherej are X± = (J i K) , 1 × 2 1,2 2 ± the total spin numbers with respect to SU(2)1,2. The total then degrees of freedom are (2j1 + 1)(2j2 + 1).Indetail,wehave

+ + + [Xi ,Xj ]=i εijk Xk , (0,0): This is a total spin zero rep, with dim one. (0, 0) represents ascalarfieldφ(x) satisfying the Klein-Gordon equation: [Xi−,Xj−]=i εijk Xk− , 2 µ (✷ + m )φ(x)=0,where✷ = ∂ ∂µ. + [Xi ,Xj−]=0. 1 (2, 0): This a 2-dim rep, the so-called left-handed Weyl rep, Hence, SO(1,3) algebra splits into two SU(2) ones: e.g. neutrinos. It is denoted with a 2-dim complex vector ξα,usuallycalledtheleft-handedWeylspinor.Undera Lorentz trans, ξ transforms as SO(1, 3) = SU(2) SU(2) , α C ∼ × β ξα′ = Mα ξβ , where SO(1,3)C is the rep from a complexified SO(1,3) algebra. However, there is an 1:1 correspondence of the β C where Mα SL(2, ). reps between SO(1,3)C and SO(1,3)R.Infact,wehavethe ∈ homomorphism 1 (0, 2): This is the corresponding 2-dim rep of the right-handed Weyl spinor and is denoted as η¯α˙ ,whichtransformsunder SO(1, 3)R SL(2, C) , Lorentz trans as ∼ β˙ η¯α′˙ = M † α˙ η¯β˙ , β˙ which is more difficult to use for classification of reps. where M † SL(2, C). α˙ ∈ 1 1 (2, 2): This is the defining 4-dim rep, describing a spin 1 particle µ with 4 components. One can use the matrix rep: A (σµ)αα˙ or simply Aµ, e.g. Aµ =(Φ/c, A) in electromagnetism.

41 42 – Lie Algebra and Generators of the Poincar´eGroup The Lie Algebra of the Poincar´eGroup:

The Poincarétrans consist of Lorentz trans plus space-time The commutation relations defining the PoincaréLie algebra translations: are µ µ ν µ x′ = Λ ν x + a , µ where a is a constant 4-vector. [Pµ ,Pν]=0, [P ,L ]=i (g P g P ) , The generator of translations in a differential-operator repis µ ρσ µρ σ − µσ ρ [Lµν,Lρσ]= i (gµρLνσ gµσLνρ + gνσLµρ gνρLµσ) . ∂ ∂ − − − P µ = i∂µ = i = i ( , ) , ∂xµ c∂t −∇ In terms of J and K, the commutation relations read: ∂ with Pµ = i∂µ = i(c∂t , ),because ∇ [P0 ,Ji]=0, ν ia Pν µ µ µ e− x = x + a . (Why?) [Pi ,Jj]=i εijk Pk ,

[P0 ,Ki]=iPi , An analogous diﬀerential-operator rep of the 6-generators of [P ,K]=iP δ . Lorentz trans is given by the generalized angular momentum i j 0 ij operators: L = x P x P , µν µ ν − ν µ Exercise: Prove all commutation relations that appear on this with the identiﬁcation page.

1 J = ε L ,K= L . i 2 ijk jk i 0i

1 Exercise: Show that Ji = 2 εijkLjk and Ki = L0i satisfy the SO(1,3) algebra.

43 44 – Single Particle States Classiﬁcation of massless particle states

2 µ 2 The Poincarégroup has two Casimir operators: P = P Pµ Massless particle states, for which P a =0(m =0), 2 µ | ⟩ and W = W Wµ,where are characterized only by their 4-momentum pµ and helicity λ = P J. · 1 νρ σ Wµ = εµνρσ L P , − 2 Alternatively, in addition to the operator Pµ, one may use the Pauli–Lubanski operator Wµ: with ε0123 =1, is the so-called Pauli–Lubanski vector. W a; p , λ = λ p a; p , λ . Classification of massive particle states µ| µ ⟩ µ | µ ⟩ A single massive particle state a can be characterized by its If the theory involves parity, then a massless state has only | ⟩ mass and its total spin s,wheres is defined in the rest of two degrees of freedom (polarizations): λ. mass system of the particle: ± Examples of the above are the photon and the neutrinos of P 2 a = m2 a ,W2 a = m2J2 a = m2s(s+1) a . the Standard Model. | ⟩ | ⟩ | ⟩ − | ⟩ − | ⟩ ··· P In addition, we use the 3-momentum and the helicity Exercises: H = J P operators to classify massive particle states: · (i) Show that P 2 and W 2 are true Casimir operators, P a; m, s; p, λ = p a; m, s; p, λ , i.e. [ P 2,P ]=[P 2,L ]=0, and likewise for W 2; µ | ⟩ µ | ⟩ µ ρσ H a; m, s; p, λ = λ p a; m, s; p, λ . | ⟩ | || ⟩ (ii) In particle’s rest frame where pµ =(m, 0, 0, 0),showthat 1 jk 2 2 2 W0 =0, Wi = mεijk L = mJi and W = m J ; Note that a massive particle state has (2s + 1) polarizations 2 − or helicities, also called degrees of freedom, (iii) Show that [J P, P]=0, [P ,W ]=0,andW P µ =0; · µ ν µ i.e. λ = s, s +1,...,s 1,s. − − − (iv) Calculate the commutation relation [Wµ,Wν]. 1 Examples: for an electron, it is λ = 2,andforamassive spin-1 boson (e.g. the Z-boson), we have± λ = 1, 0, 1. − 45 46 Hamilton’s principle 7. Lagrangians in Field Theory Hamilton’s principle states that the actual motion of the – Variational Principle and Equation of Motion system is determined by the stationary behaviour of S Classical Lagrangian Dynamics under small variations δqi(t) of the ith particle’s generalized coordinate qi(t),withδqi(t1)=δqi(t2)=0,i.e. The Lagrangian for an n-particle system is t2 ∂L ∂L δS = dt δq + δq˙ L(q , q˙ )=T V, i ∂q i ∂q˙ i i − )t1 + i i , t2 ∂L d ∂L where q1,2,...,n are the the generalized coordinates describing = dt δqi =0. t1 ∂qi − dt ∂q˙i the n particles, and q˙1,2,...,n are the respective time ) + , derivatives. The Euler–Lagrange equation of motion for the ith particle is T and V denote the total kinetic and potential energies. d ∂L ∂L ∴ =0. The action S of the n-particle system is given by dt ∂q˙i − ∂qi

t2 S[qi(t)] = dt L(qi, q˙i) . ··· )t1 Exercise: Show that the Euler–Lagrange equations of motion for a particle system described by a Lagrangian of the form Note that S is a functional of q (t). i L(qi, q˙i, q¨i) are

d2 ∂L d ∂L ∂L 2 + =0. dt ∂q¨i − dt ∂q˙i ∂qi

[Hint: Consider only variations with δqi(t1,2)=δq˙i(t1,2)=0.]

47 48 Lagrangian Field Theory – Lagrangians for the Klein-Gordon and Maxwell eqs

In Quantum Field Theory (QFT), a (scalar) particle is Lagrangian for the Klein–Gordon equation described by a ﬁeld φ(x), whose Lagrangian has the functional form: 1 µ 1 2 2 KG = (∂µφ)(∂ φ) m φ , L = d3x (φ(x) , ∂ φ(x)) , L 2 − 2 L µ ) where is the so-called Lagrangian density, often termed where φ(x) is a real scalar ﬁeld describing one dynamical L Lagrangian in QFT. degree of freedom.

In QFT, the action S is given by The Euler–Lagrange equation of motion is the Klein–Gordon equation + µ 2 ∞ (∂ ∂ + m ) φ(x)=0. S[φ(x)] = d4x (φ(x) , ∂ φ(x)) , µ L µ ··· )−∞ with lim φ(x)=0. Lagrangian for the Maxwell equations x →±∞ By analogy, the Euler–Lagrange equations can be obtained 1 µν µ ME = Fµν F JµA , by determining the stationary points of S, under variations L − 4 − φ(x) φ(x)+δφ(x): → where F = ∂ A ∂ A is the ﬁeld strength tensor, and µν µ ν − ν µ Jµ is the 4-vector current satisfying charge conservation: ∂ ∂ µ ∂µ L L =0. ∂µJ =0. ∂(∂µφ) − ∂φ

Aµ describes a spin-1 particle, e.g. a photon, with 2 physical ··· degrees of freedom. Exercise: Derive the above Euler–Lagrange equation for a Exercise: Use the Euler-Lagrange equations for ME to scalar particle by extremizing S[φ(x)],i.e.δS =0. µν ν L show that ∂µF = J , as is expected in relativistic electrodynamics (with µ0 = ε0 = c =1).

49 50 – Lagrangian for the Dirac equation Lorentz trans properties of the Weyl and Dirac spinors

µ The Dirac spinor ψ is the direct sum of two Weyl spinors ξ D = ψ¯ (i γ ∂µ m) ψ , L − and η¯ with Lorentz trans properties: where β β˙ ξα′ = Mα ξβ , η¯α′˙ = M † α˙ η¯β˙ , µ ξβ(x) 0(σ ) ˙ µ αβ α 1 α β α˙ 1˙α β˙ ψ(x)= β˙ , γ = µ αβ˙ ξ′ = M − ξ , η¯′ = M †− η¯ . η¯ (x) (¯σ ) 0 β β˙ + , + , and ψ¯(x) (ηα(x), ξ¯ (x)),withσµ =(1 , σ) and σ¯µ = with M SL(2, C). α˙ 2 ∈ (1 , σ).≡ 2 − Duality relations among 2-spinors: α˙ The ξα and η¯ are 2-dim complex vectors (also called Weyl α α˙ α α˙ spinors) whose components anti-commute: ξ1ξ2 = ξ2ξ1, (ξ )† = ξ¯ , (ξα)† = ξ¯α˙ , (¯ηα˙ )† = ηα , (η )† =¯η ˙ ˙ ˙ ˙ ˙ ˙ − η¯1η¯2 = η¯2η¯1, ξ η¯2 = η¯2ξ etc. − 1 − 1 Lowering and raising spinor indices: The Euler–Lagrange equation of with respect to ψ¯ is the LD Dirac equation: β α αβ β˙ α˙ α˙ β˙ ξα = εαβξ , ξ = ε ξβ , η¯α˙ = εα˙ β˙ η¯ , η¯ = ε η¯β˙ , ∂ LD =0 (i γµ∂ m) ψ =0. ¯ µ αβ 01 α˙ β˙ ∂ψ ⇒ − with ε iσ = = ε and ε iσ = ε ˙ . 2 0 101 αβ 2 α˙ β ≡ − − ≡ − The 4-component Dirac spinor ψ(x) that satisfies the Dirac Lorentz-invariant spinor contractions: equation describes 4 dynamical degrees of freedom. α α β β α β α β ξη ξ ηα = ξ εαβη = η εαβξ = η εβαξ = η ξβ = ηξ Exercises: ≡ − (i) Derive the Euler–Lagrange equation with respect to the ¯ α ¯ α˙ ¯α˙ ¯ Likewise, ξη¯ (ηξ)† = ξα† η † = ξα˙ η¯ =¯ηα˙ ξ =¯ηξ. Dirac field ψ(x); ≡ ν 1 1 (ii) Show that up to a total derivative term, D is Hermitian, Exercise: Given that MσµM † = Λ µ σν and M †− σ¯µM − = µ L ν i.e. = † + ∂ j ,withj = ψ¯ iγ ψ. Λ σ¯ν,showthat D is invariant under Lorentz trans. LD LD µ µ µ µ L 51 52 Global and Local Symmetries in QFT 8. Gauge Groups Consider the Lagrangian (density) for a complex scalar: – Global and Local Symmetries µ 2 2 =(∂ φ)∗ (∂ φ) m φ∗φ + λ(φ∗φ) . Symmetries in Classical Physics and Quantum Mechanics: L µ − is invariant under a U(1) rotation of the field φ: Translational invariance in time Energy conservation L dE t t + a0 dt =0 iθ → ⇒ φ(x) φ′(x)=e φ(x) , Translational invariance in space Momentum conservation → dp r r a µ + dt =0 where θ does not depend on x x . → ⇒ ≡ Rotational invariance Angular momentum conservation A transformation in which the fields are rotated about x- r R r dJ =0 → ⇒ dt independent angles is called a global transformation.Ifthe angles of rotation depend on x, the transformation is called Quantum Mechanics Degeneracy of energy states a local or a gauge transformation. d [H, ]=0 dtO = i[H, ]=0 O ⇒ O A general infinitesimal global or local trans of fields φi under the action of a Lie group reads: Quantum Field Theory Noether’s Theorem

φ(x) φ(x)+δφ(x) ? φ (x) φ′(x)=φ (x)+δφ (x) , → ⇒ i → i i i a a j a where δφi(x)= i θ (x)(T )i φj(x) , and T are the generators of the Lie− Group. Note that the angles or group parameters θa are x-independent for a global trans. If a Lagrangian is invariant under a global or local trans, it is said that hasL a global or local (gauge) symmetry. L Exercise: Show that the above Lagrangian for a complex scalar is not invariant under a U(1) gauge trans.

53 54 – Gauge Invariance of the QED Lagrangian QED Lagrangian with an electron-photon interaction

Consider ﬁrst the Lagrangian for a Dirac ﬁeld ψ: The complete Lagrangian of Quantum Electrodynamics (QED) that includes the interaction of the photon with the = ψ¯ (iγµ∂ m) ψ . electron is LD µ − 1 µν is invariant under the U(1) global trans: QED = Fµν F + ψ¯ (i ∂ m e A) ψ , LD L − 4 ̸ − − ̸

iθ ψ(x) ψ′(x)=e ψ(x) , where we used the convention: a γ aµ. → ̸ ≡ µ but it is not invariant under a U(1) gauge trans, when Exercises: θ = θ(x).Instead,wefindtheresidualterm (i) Show that is gauge invariant under a U(1) trans. LQED δ = (∂ θ(x)) ψγ¯ µψ . LD − µ (ii) Derive the equation of motions with respect to photon and electron fields. To cancel this term, we introduce a vector field Aµ in the theory, the so-called photon, and add to D the extra term: (iii) How should the Lagrangian describing a complex scalar L field φ(x), µ ψ = D eAµ ψγ¯ ψ . L L − µ 2 =(∂ φ)∗ (∂ φ) m φ∗φ , L µ −

We demand that Aµ transforms under a local U(1) as be extended so as to become gauge symmetric under a U(1) local trans? 1 A A′ = A ∂ θ(x) . µ → µ µ − e µ

is invariant under a U(1) gauge trans of ψ and Aµ. Lψ

55 56 – Noether’s Theorem With the aid of the equations of motions for φi, the last equation implies that Noether’s Theorem. If a Lagrangian is symmetric under a L global transformation of the ﬁelds, then there is a conserved ∂ ∂ ∂ µ 3 0 current J (x) and a conserved charge Q = d xJ (x), ∂µ L δφi = ∂µ L L δφi =0. ∂(∂ φ ) ∂(∂ φ ) − ∂φ associated with this symmetry, such that = µ i > = µ i i > < dQ ∂ J µ = 0 and =0. The conserved current (or currents) is µ dt

a, µ ∂ ∂δφi ∂ a j J = L a = L i (T )i φj . Proof: ∂(∂µφi) ∂θ ∂(∂µφi)

Consider a Lagrangian (φi, ∂µφi) to be invariant under the The corresponding conserved charges are inﬁnitesimal global trans:L

a 3 a, 0 a a j Q (t)= d xJ (x) . δφi = i θ (T )i φj , ) where T a are the generators of some group G. Indeed, it is easy to check that

Hence, the change of is vanishing, i.e. dQa L = d3x ∂ J a, 0(x)= d3x Ja(x) dt 0 − ∇ · ∂ ∂ ) ) δ = L δφi + L ∂µ(δφi)=0. a L ∂φi ∂(∂µφi) = ds J 0 , − · → ) This last equation can be rewritten as because surface terms vanish at inﬁnity. ∂ ∂ ∂ δ = ∂ L δφ + L ∂ L δφ =0. Exercises: Find the conserved currents and charges for L µ ∂(∂ φ ) i ∂φ − µ ∂(∂ φ ) i = µ i > = i µ i > (i) QED; (ii) the gauge-invariant Lagrangian with a complex scalar φ.

57 58 a – Yang–Mills Theory Interaction between quarks qi and gluons Aµ in SU(3)c

The Lagrangian of a Yang–Mills (non-Abelian) SU(N) theory If qi =(qred,qgreen,qblue) are the 3 colours of the quark, is their interaction with the 8 gluons Aa is described by the 1 µ = F a F a,µν , Lagrangian: LYM − 4 µν where =¯qi [ i ∂δj m δj g Aa(T a) j ] q . Lint ̸ i − i − ̸ i j F a = ∂ Aa ∂ Aa gfabc Ab Ac , µν µ ν − ν µ − µ ν and f abc are the structure constants of the SU(N) Lie algebra. Exercise: Show that int is invariant under the SU(3) gauge transformation: L It can be shown that YM is invariant under the inﬁnitesimal L 1 SU(N)localtrans: δAa = ∂ θa f abc θb Ac , δq = iθa (T a) j q , µ −g µ − µ i i j a 1 a abc b c δAµ = ∂µθ f θ Aµ . a 1 a a −g − where T = 2 λ are the generators of SU(3) and λ are the Gell-Mann matrices: Examples of SU(N)theoriesaretheSU(2)L group of the SM 1,2,3 001 and Quantum Chromodynamics (QCD) based on the SU(3) 1,2,3 σ 0 4 c λ = , λ = ⎛ 000⎞ , 0 001 group. 100 ⎝ ⎠ 0 The gauge (vector) ﬁelds of the SU(2)L are the W and W ± 00 i 000 5 − 6 bosons responsible for the weak force. λ = ⎛ 00 0⎞ , λ = ⎛ 001⎞ , i 00 010 ⎝ ⎠ ⎝ ⎠ The gauge vector bosons of the SU(3) group are the gluons c 1 00 mediating the strong force between quarks. 00 0 √3 7 8 ⎛ 0 1 0 ⎞ λ = ⎛ 00 i ⎞ , λ = √3 . − 2 0 i 0 ⎜ 00 ⎟ Gauge bosons of Yang–Mills theories self-interact! ⎝ ⎠ ⎝ −√3 ⎠

Exercise: Show that YM is invariant under SU(N) gauge trans. L

59 60 Then, we have 9. The Geometry of Gauge Transformations d 1 – Parallel Transport and Covariant Derivative v(t)= v (t + δt) (v (t) δtv (t)[e ∂ e ]) dt δt { i − i − j i · t j } Simple Example: e (t + δt) × i e e e e2 =[∂tvi(t)+( i ∂t j) vj(t)] i(t) . v(t) · We can now deﬁne the covariant derivative to act only on θ(t) the components of v(t) as: e1

D v (t)=∂ v (t)+(e ∂ e )v (t) , t i t i i · t j j = ∂tvi(t)+θε˙ 3ij vj(t) ,

d v e Time-dependent vector written in terms of t-dependent unit with the obvious property dt (t)= i(t)Dtvi(t). The second vectors: term is induced by the change of the coordinate axes, namely after performing a parallel transport of our coordinate system e1,2(t) from t to t + δt. v(t)=vi(t) ei(t)(with i =1, 2).

Proper comparison of two vectors vi(t + δt) and vi(t) can The true time derivative of v(t) is only be made in the same coordinate system by means of parallel transport.Differentiation is properly defined through d v(t + δt) v(t) v(t)= lim − . the covariant derivative. dt δt 0 δt → Exercise: Show that the covariant derivative satisfies the To calculate this, we need to refer all unit vectors to t + δt: relation D v (t)=∂ v (t)+(ω v(t)) , t i t i × i ei(t)=ei(t + δt) δt ∂tei(t) . with ω = θ˙(t), which is known from Classical Mechanics − between rotating and fixed frames in 3 dimensions.

61 62 Diﬀerentiation in curved space Covariant derivative in the Gauge-Group Space

The notion of the covariant derivative generalizes to curved Consider the difference of a fermionic isovector field ψ at space as well. By analogy, the infinitesimal difference between xµ + δxµ and xµ in an SU(N) gauge theory: µ µ µ the 4-vectors V (x′ ) and V (x ) is given by Dψ = dψ + δψ , DV µ = dV µ + δV µ , where a a µ where dV µ is the difference of the 2 vectors in the same δψ = ig T Aµ dx ψ µ a coordinate system and δV is due to parallel transport of the and the field Aµ takes care of the change of the SU(N) axes µ µ µ µ vector from x to x′ = x + δx . from point to point in Minkowski space. µ In the framework of General Relativity, we have The covariant derivative of ψ(x ) is

a a µ µ µ ν λ Dµψ =(∂µ + ig T Aµ) ψ , DV =(∂λV + ΓνλV ) dx ,

µ which is obtained from pure geometric considerations. where Γνλ is the so-called affine connection or the Christoffel symbol. a a In analogy to General Relativity, the gauge field Aµ T is sometimes called the connection. Exercise: Show that under a local SU(N) rotation of the isovector ψ field: ψ ψ′ = Uψ (with U SU(N)), its covariant derivative transforms→ as ∈

D ψ D′ ψ′ = UD ψ , µ → µ µ with i A′ = UA U † + (∂ U) U † . µ µ g µ

63 64 A round trip in the SU(N) Gauge-Group Space Parallels between Gauge Theory and General Relativity

D C In General Relativity, a corresponding round trip of a vector V µ in a curved space gives rise to δxµ 1 ABµ ∆V µ = Rµ V ρ∆Sσλ , ∆x 2 ρσλ Keeping terms up to second order in δx and ∆x,wehave where ∆Sσλ represents the area enclosed by the path and 1 µ ψ =(1+∆xµD + ∆xµ∆xν D D ) ψ , Rρσλ is the Riemann–Christoffel curvature tensor: B µ 2 µ ν A,0 µ 1 µ ν µ µ µ κ µ κ µ ψ =(1+δx D + δx δx D D ) ψ , Rρσλ = ∂λΓρσ ∂σΓρλ + ΓρσΓκλ ΓρλΓκσ . C µ 2 µ ν B − − µ 1 µ ν ψD =(1 ∆x Dµ + ∆x ∆x DµDν) ψC , − 2 Analogies: 1 ψ =(1 δxµD + δxµδxν D D ) ψ . A,1 − µ 2 µ ν D GaugeTheory GeneralRelativity Gauge trans. Co-ordinate trans. Hence, a a κ Gauge field Aµ T Affine connection, Γλν Field strength F µν Curvature tensor Rµ ψ =(1+δxµ∆xν [D ,D ]) ψ , ρσλ A,1 µ ν A,0 Bianchi identity: Bianchi identity: κ ρ,µ,ν DρFµν =0 ρ,µ,ν Dρ Rλµν =0 and ψ = ψ . cyclic cyclic A,1 ̸ A,0 ! ! Exercise: Show that

i [D ,D]=F a T a g µ ν µν is the SU(N) Field-strength tensor.

65 66 – Topology of the Vacuum: the Bohm–Aharanov Eﬀect Basic Concepts in Topology

The Bohm–Aharanov Effect: Let a(s) and b(s) be two paths in a topological space Y both starting from the point P (a(0) = b(0) = P )andendingata 1 possibly different point Q (a(1) = b(1) = Q). If there exists x:displacementoffringes e− source d a continuous function L(t, s) such that L(0,s)=a(s) and 2 L(1,s)=b(s),thenthepathsa and b are called homotopic B which is denoted by a b. ∼ Vector potential A and B field (with B = A)in If P Q,thepathissaidtobeclosed. cylindrical polars: ∇× ≡ 1 The inverse of a path a is written as a− and is defined by Br a 1(s)=a(1 s). It corresponds to the same path traversed Inside: A = A =0,A= , − r z φ 2 in the opposite− direction. Br = Bφ =0,Bz = B, The product path c = ab is defined by BR2 Outside: Ar = Az =0,Aφ = , c(s)=a(2s) , for 0 s 1 , 2r ≤ ≤ 2 B c(s)=b(2s 1) , for 1 s 1 . =0, − 2 ≤ ≤ 1 1 where R is the radius of the solenoid. If a b,thenab− is homotopic to the null path: ab− 1. ∼ ∼ Although the electrons move in regions with E = B = 0,the Exercises: B field of the solenoid induces a phase difference δφ12 of the 1 i(nθ+a) electrons on the screen causing a displacement of the fringes: (i) Consider the mappings S U(1): fn(θ)=e (with a R and n Z), and show→ that they all are homotopic e e A r B s ∈ ∈ δφ12 = φ1 φ2 = " d = " d . to those with a =0. − 2 1 · · ? − ) In regions with E = B = 0,itisA =0,sothevacuumhas (ii) Given that fn(θ) fm(θ) for n = m,explainthenwhy ̸ i[nθ(1 t)+m̸∼θt] ̸ a topological structure! It is not simply connected due to the L(t, θ)=e − is not an allowed homotopy function presence of the solenoid. relating fn to fm.

67 68 Homotopy Classes, Groups and the Winding Number The Bohm–Aharanov Eﬀect Revisited

All paths related to maps X Y of two topological spaces In regions with E = B = 0, Aµ is a pure gauge: Aµ = ∂µχ X, Y can be divided into homotopy→ classes. (Why?). Homotopy Class. All paths that are homotopic to a given The configuration space X of the Bohm–Aharanov effect is path a(s) define a set, called the homotopy class and denoted the plane R2 with a hole in it, due to the solenoid. This is by [a].Forexample,[f ] are distinct homotopy classes for topologically equivalent ( homeomorphic)toR S1.The n ≡ × different n. space X can be conveniently described by polar coords (r, φ), with r =0. Winding Number. Each homotopy class may be ̸ characterized by an integer, the winding number n (also called It can be shown that χ(r, φ) = const. φ, which is a function × the Pontryargin index). For the case f(θ): S1 U(1),the in the group space of U(1), i.e. Y = U(1). winding number is determined by → Since functions mapping S1 onto R are all deformable to a 2π constant, the non-trivial part of χ is given by the map: 1 d ln f(θ) n = dθ . 1 2πi dθ S U(1) . )0 + , →

Because π1[U(1)] = Z, the electron paths cannot be deformed Homotopy Group. The set of all homotopy classes related to a null path with a constant χ,implyingAµ =0everywhere to maps X Y forms a group, under the multiplication law → and the absence of the Bohm–Aharanov eﬀect.

[a][b]=[ab] , Since π1[SU(2)] = 1,thereisno Bohm–Aharanov effect from an SU(2) ‘solenoid’! the so-called homotopy group π (Y ). X Exercises: 1 2 Exercises: (i) Show that χ(r, φ)=2 BR φ is a possible solution for (i) Prove that the homotopy group satisfies the axioms of a E = B = 0,whereB is the magnetic field and R the radius group. of the solenoid.

1 e (ii) Show that for S U(1), π [U(1)] = Z. (ii) Verify that δφ12 = " [ χ(2π) χ(0) ]. → 1 ∼ − 69 70 Z -graded Lie algebra. The generalization of a Z -graded 10. Supersymmetry (SUSY) N 2 Lie algebra L to ZN can be deﬁned analogously. Let L be the direct sum of N subalgebras L : – Graded Lie Algebra i

N 1 Definition.AZ -graded Lie algebra is defined on a vector L = − L . 2 ⊕i=0 i space L which is the direct sum of two subspaces L0 and L : L = L L . The generators that span the space L are 1 0 ⊕ 1 Then, the multiplication law among the generators of L endowed with a multiplications law: can be defined by ◦

: L L L. T (i) T (j) = ( 1)gigj T (j) T (i) L . ◦ × → ◦ − − ◦ ∈ (i+j)modN T (0) L ,T(1) L , the generators satisfy the following Z ∀ ∈ 0 ∈ 1 The N -graded Jacobi identity is deﬁned analogously with properties: that of Z ,whereg =0, 1,...,N 1 is the degree of 2 i,j − (0) (0) 2 (0) (0) (0) (0) graduation of Li,j. g0 (i) T1 T2 = ( 1) T2 T1 =[T1 ,T2 ] L0 , (0)◦ (1) − − g g (1)◦ (0) (0) (1) ∈ ··· (ii) T T = ( 1) 0 1 T T = T ,T L1 , (1)◦ (1) − − 2 (1) ◦ (1) { (1) (1)} ∈ (iii) T T = ( 1)g1 T T =[T ,T ] L , 1 ◦ 2 − − 2 ◦ 1 1 2 ∈ 0 Exercise:⋆⋆⋆ Find the (anti)-commutation relations and the Z where g0 = g(L0)=0and g1 = g(L1)=1are the degrees structure constants of the 2-graded Lie algebra of SU(2). of the graduation of the Z2-graded Lie algebra.

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

( 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.

71 72 – Generators of the Super-Poincar´eGroup Consequences of the Super-Poincar´eSymmetry

The generators super-Poincaréalgebra are Pµ ,Lµν L0 ¯ ∈ Equal number of fermions and bosons. and the spinors Qα , Qα˙ L1. They satisfy the following • relations: ∈ Scalar supermultiplet Φ (φ , ξ ,F),whereφ is a (i) [Pµ ,Pν]=0, • complex scalar (2), ξ is a 2-component⊃ complex spinor (4), (ii) [Pµ,Lρσ]=i (gµρPσ gµσPρ) , and F is an auxiliary complex scalar (2). − 3 (iii) [L ,L ]= i (g L g L + g L g L ) . µν ρσ − µρ νσ − µσ νρ νσ µρ − νρ µσ Vector supermultiplet V a (Aa , λa,Da),whereAa (iv) Q ,Q = Q¯ , Q¯ ˙ =0, µ µ { α β} { α˙ β} • ⊃ a µ are massless non-Abelian gauge fields (3), λ are the 2- (v) Q , Q¯ ˙ =2(σ ) ˙ P , a { α β} αβ µ component gauginos (4),3 and D are the auxiliary real (vi) [Qα,Pµ]=0, fields (1). (vii) [L ,Q]= i(σ ) β Q , µν α − µν α β β˙ The simplest model that realizes SUperSYmmetry (SUSY) (viii) [Lµν , Q¯α˙ ]= i(¯σµν) Q¯ ˙ , − α˙ β is the Wess–Zumino model. Counting on-shell degrees of µν β 1 µ ν αβ˙ ν µ αβ˙ freedom (dof), the Wess-Zumino model contains one complex where (σ )α = 4 [(σ )αα˙ (¯σ ) (σ )αα˙ (¯σ ) ] and µν α˙ 1 µ αβ˙ ν ν −αβ˙ µ scalar φ (2 dofs) and one Weyl spinor ξ (2 dofs): (¯σ ) = [(¯σ ) (σ ) ˙ (¯σ ) (σ ) ˙ ]. β˙ 4 ββ − ββ ··· bosonic dofs = fermionic dofs ⋆ Exercise: Prove the Z2-graded Jacobi identity:

[L , Q , Q¯ ˙ ]+ Q , [Q ˙ ,L ] + Q¯ ˙ , [Q ,L ] =0. µν { α β} { α β µν } { β α µν }

73 74 – The Wess–Zumino Model Exercise:Showthat

Non-interacting WZ model σµσ¯ν + σνσ¯µ β =2gµνδ β , σ¯µσν +¯σνσµ α˙ =2gµνδα˙ { }α α { } β˙ β˙ = + Lkin Lscalar Lfermion µ µ 1 Noticing that ∂µ∂ν = ∂ν∂µ and using the results of the above =(∂ φ†)(∂µφ)+ξ¯iσ¯ (∂µξ); φ = (φ1 + iφ2) √2 exercise, we get

µ µ δ = θξ(∂ ∂ φ†)+ξ¯θ¯(∂ ∂ φ) Consider φ φ + δφ and φ† φ† + δφ†,with Lfermion µ µ → → µ µ = θ(∂ ξ)(∂ φ†) θ¯(∂ ξ¯)(∂ φ) − µ − µ δφ = θξ and δφ† =(θξ)† = ξ¯θ¯ = θ¯ξ¯, µ µ + ∂µ [θξ(∂ φ†)+ξ¯θ¯(∂ φ)] and θ inﬁnitesimal anticommuting 2-spinor constant. δ = δ + δ =0! ⇒ L Lscalar Lfermion

scalar scalar + δ scalar , But, we are not ﬁnished yet ! The diﬀerence of two successive ⇒ L → L L µ µ SUSY transfs. must be a symmetry of the Lagrangian as well, δ scalar = θ(∂ φ†)(∂µξ)+θ¯(∂ ξ¯)(∂µφ) L i.e. SUSY algebra should close.

Try ξα ξα + δξα and ξ¯α˙ ξ¯α˙ + δξ¯α˙ ,with → → (δ δ δ δ )φ = i(θ σµθ¯ θ σµθ¯ ) ∂ φ θ2 θ1 − θ1 θ2 − 1 2 − 2 1 µ µ µ µ µ µ δξ = i(σ θ¯) ∂ φ and δξ¯ = i(θσ ) ∂ φ† iϵ Pµφ (with ϵ ∗ = ϵ ) α − α µ α˙ α˙ µ ≡ µ ¯ µ ¯ (δθ2δθ1 δθ1δθ2)ξα = i(σ θ1)αθ2∂µξ + i(σ θ2)αθ1∂µξ fermion fermion + δ fermion , − − ⇒ L → L L Fierz µ µ ν µ ¯ µ ν ¯ = i(θ1σ θ¯2 θ2σ θ¯1) ∂µξα δ fermion = θσ σ¯ (∂µξ)(∂νφ†)+ξσ¯ σ θ(∂µ∂νφ) − − L − µ µ ν µ ¯ µ ν ¯ + θ1αθ¯2iσ¯ ∂µξ θ2αθ¯1iσ¯ ∂µξ = θσ σ¯ ξ(∂µ∂νφ†)+ξσ¯ σ θ(∂µ∂νφ) − ν µ ∂µ [θσ σ¯ ξ(∂νφ†)] µ − Only for on-shell fermions, iσ¯ ∂µξ =0, the SUSY algebra closes.

75 76 To close SUSY algebra oﬀ-shell, we need an auxiliary complex The interacting WZ model scalar F (without kinetic term) and add = + LWZ Lkin Lint F = F †F µ ¯ µ L =(∂ φ†)(∂µφ)+ξ iσ¯ (∂µξ)+F †F 1 1 to + ,with Wφφ ξξ + Wφ F W † ξ¯ξ¯ + W †F † Lscalar Lfermion − 2 − 2 φφ φ ¯ µ ¯ µ δF = iθσ¯ (∂µξ) , δF † = i(∂µξ)¯σ θ where − µ µ m h δξ = i(σ θ¯) ∂ φ + θ F, δξ¯ = i(θσ ) ∂ φ† + θ¯ F † W (φ)= φφ + φφφ α − α µ α α˙ α˙ µ α˙ 2 6 is the so-called superpotential, and Exercise: Prove (i) that the Lagrangian δW h 2 Wφ = = mφ + φ µ µ δφ 2 kin =(∂ φ†)(∂µφ)+ξ¯iσ¯ (∂µξ)+F †F L δ2W Wφφ = = m + hφ is invariant under the oﬀ-shell SUSY transfs: δφ δφ

δφ = θξ , δφ† = θ¯ξ¯ Exercise: Show that up to total derivatives, µ µ δξ = i(σ θ¯) ∂ φ + θ F, δξ¯ = i(θσ ) ∂ φ† + θ¯ F † α − α µ α α˙ α˙ µ α˙ µ µ δF = iθ¯σ¯ (∂ ξ) , δF † = i(∂ ξ¯)¯σ θ 1 1 µ µ = W ξξ + W F W † ξ¯ξ¯ + W †F † − Lint − 2 φφ φ − 2 φφ φ 1 1 and (ii) that the SUSY algebra closes oﬀ-shell: = (m + hφ)ξξ (m + hφ†)ξ¯ξ¯ − 2 − 2 µ ¯ µ ¯ h 2 h 2 (δθ2δθ1 δθ1δθ2)X = i(θ1σ θ2 θ2σ θ1) ∂µX, +(mφ + φ )F +(mφ† + φ† )F † − − − 2 2 ¯ with X = φ, φ†, ξ, ξ,F,F†. remains invariant under oﬀ-shell SUSY transformations.

77 78 – Feynman rules Summary

Equation of motions for the auxiliary ﬁelds F and F †: The complete WZ Lagrangian is

F = W † ,F† = W , − φ − φ µ 2 1 µ 1 =(∂ φ†)(∂ φ) m φ†φ + Ψ iγ ∂ Ψ m ΨΨ LWZ µ − 2 µ − 2 Substituting the above into WZ, we get L 2 mh 2 2 h 2 (φ†φ + φ† φ) (φ†φ) µ ¯ µ − 2 − 4 WZ =(∂ φ†)(∂µφ)+ξ iσ¯ (∂µξ) Wφ†Wφ L − h h 1 φ Ψ PL Ψ φ†Ψ PR Ψ , ( W ξξ + W † ξ¯ξ¯) − 2 − 2 − 2 φφ φφ where the F -ﬁeld has been integrated out. and the real potential is

2 2 mh 2 2 h 2 V = W †W = m φ†φ + (φ†φ + φ† φ)+ (φ†φ) φ φ 2 4

ξ Exercise:IfΨ = is a Majorana 4-spinor, show that ξ¯ the Ψ-dependent part+ of, the WZ Lagrangian can be written down as

1 1 = Ψ iγµ∂ Ψ m ΨΨ LΨ 2 µ − 2 h h φ Ψ P Ψ φ†Ψ P Ψ , − 2 L − 2 R where P =(1 γ )/2 and γ = diag (1 , 1 ). L,R 4 ± 5 5 2 − 2 79 80 Feynman rules:

φ,p i : p2 m2 − Ψ,p i : p m ̸ − : imh −

: ih2 −

: ihP − L

: ihP − R

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