Introduction to Supersymmetry and Supergravity

Home , Supergravity, Supersymmetry, Weak hypercharge

Simon Albino II. Institute for Theoretical Physics at DESY, University of Hamburg [email protected]

Every Wednesday and Thursday 8:30-10:00am, starting 3rd November, in room 2, building 2a, DESY

Lecture notes at www.desy.de/~simon/teaching/susy.html

Abstract

These lectures give a fairly formal developement of supersymmetry, beginning with some technical footing in symmetries (internal and external) in general in relativistic quantum mechanics, and a brief outline of the standard model and its GUT extensions. Following the Haag-Lopuszanski-Sohnius theorem, we allow for fermionic symmetry generators, and determine their properties and algebra using the restrictions of the Coleman-Mandula theorem. As an illustration we construct the supersymmetric field theory Lagrangian for the chiral supermultiplet, then discuss the more formal approach to constructing general supersymmetric field theories using superfields in superspace, including the development of supersymmetric gauge theories. Spontaneous supersymmetry breaking is discussed in some detail. Using the superfield formalism, the minimally supersymmetric standard model is developed. Next we develop supergravity, firstly in the weak field case and then to all orders. Finally, the more advanced topics of higher dimensions, extended supersymmetry and duality are discussed. The lectures will mostly follow Volume III of S. Weinberg’s “The Quantum Theory of Fields”. IMPORTANT

Lecture notes are at www.desy.de/~simon/teaching/susy.html. They will change! • If you get stuck, please email me at [email protected] • If you ﬁnd a mistake, please email me at [email protected] • Please let me know by email that you will attend this course so I can put you on the mailing list • (and thus notify you at the last minute of room changes, cancellations, help with problems etc.). Every second Thursday I will try to allocate time for you to ask more detailed questions, • for you to work through and present derivations outlined in the lectures, and (if time) for you to present solutions to interesting problems. There is alot of algebra to deal with in SUSY... • so please think about the ratio of algebraic explanations to material that you would like! My suggestion: Try to derive some results in detail at home, and present your derivations every second Thursday. • I am setting up an online forum where you can ask / answer SUSY questions. • Usually in the lecture notes, I give the result ﬁrst and then outline the derivation in smaller font below. • I use the Einstein summation convention, i.e. and e.g. 3 X Xµ X Xµ (and not just for spacetime indices). • i=0 µ → µ Any questions? P • Contents

1 Quantum mechanics of particles 2

1.1 Basic principles ...... 2

1.2 Fermionic and bosonic particles ...... 4

2 Symmetries in QM 6

2.1 Unitary operators ...... 6

2.2 (Matrix) Representations ...... 10

2.3 External symmetries ...... 16

2.3.1 Rotation group representations ...... 16

2.3.2 Poincar´eand Lorentz groups ...... 20

2.3.3 Relativistic quantum mechanical particles ...... 22

2.3.4 Quantum ﬁeld theory ...... 25

2.3.5 Causal ﬁeld theory ...... 27

2.3.6 Antiparticles ...... 28 2.3.7 Spin in relativistic quantum mechanics ...... 29

2.3.8 Irreducible representation for ﬁelds ...... 31

2.3.9 Massive particles ...... 32

2.3.10 Massless particles ...... 33

2.3.11 Spin-statistics connection ...... 35

2.4 External symmetries: fermions ...... 37

1 2.4.1 Spin 2 ﬁelds ...... 37

1 2.4.2 Spin 2 in general representations ...... 44

2.4.3 The Dirac ﬁeld ...... 49

2.4.4 The Dirac equation ...... 51

2.4.5 Dirac ﬁeld equal time anticommutation relations ...... 52

2.5 External symmetries: bosons ...... 53

2.6 The Lagrangian Formalism ...... 57

2.6.1 Generic quantum mechanics ...... 57

2.6.2 Relativistic quantum mechanics ...... 58 2.7 Path-Integral Methods ...... 61

2.8 Internal symmetries ...... 62

2.8.1 Abelian gauge invariance ...... 64

2.8.2 Non-Abelian gauge invariance ...... 66

2.9 The Standard Model ...... 68

2.9.1 Higgs mechanism ...... 70

2.9.2 Some remaining features ...... 74

2.9.3 Grand uniﬁcation ...... 76

2.9.4 Anomalies ...... 78

3 Supersymmetry: development 79

3.1 Why SUSY? ...... 79

3.2 Haag-Lopuszanski-Sohnius theorem and SUSY algebra ...... 83

3.3 Supermultiplets ...... 90

3.3.1 Field supermultiplets (the left-chiral supermultiplet) ...... 95

3.4 Grassman variables ...... 100 3.5 Superﬁelds and Superspace ...... 101

3.5.1 Chiral superﬁeld ...... 106

3.5.2 Supersymmetric Actions ...... 111

3.6 Superspace integration ...... 116

3.7 Supergraphs ...... 117

3.8 SUSY current ...... 124

3.9 Spontaneous supersymmetry breaking ...... 128

3.9.1 O’Raifeartaigh Models ...... 129

3.10 Supersymmetric gauge theories ...... 130

3.10.1 Gauge-invariant Lagrangians ...... 135

3.10.2 Spontaneous supersymmetry breaking in gauge theories ...... 139

4 The Minimally Supersymmetric Standard Model 143

4.1 Left-chiral superﬁelds ...... 143

4.2 Supersymmetry and strong-electroweak uniﬁcation ...... 147

4.3 Supersymmetry breaking in the MSSM ...... 149 4.4 Electroweak symmetry breaking in the MSSM ...... 157

4.5 Sparticle mass eigenstates ...... 162

5 Supergravity 164

5.1 Spinors in curved spacetime ...... 164

5.2 Weak ﬁeld supergravity ...... 168

5.3 Supergravity to all orders ...... 176

5.4 Anomaly-mediated SUSY breaking ...... 183

5.5 Gravity-mediated SUSY breaking ...... 184

6 Higher dimensions 187

6.1 Spinors in higher dimensions ...... 187

6.2 Algebra ...... 190

6.3 Massless multiplets ...... 194

6.4 p-branes ...... 197 7 Extended SUSY 198 [email protected] 2 1 Quantum mechanics of particles

1.1 Basic principles

Physical states represented by directions of vectors (rays) i in Hilbert space of universe. | i

Write conjugate transpose i † as i , scalar product j † i as j i . | i h | | i · | i h | i

Physical observable represented by Hermitian operator A = A† such that A = i A i . h ii h | | i Functions of observables represented by same functions of their operators, f(A).

Errors i A2 i A 2 etc. vanish when i = a , h | | i−h ii | i | i where A a = a a , i.e. a is A eigenstate, real eigenvalue a. a form complete basis. | i | i | i | i If 2 observables A, B do not commute, [A, B] = 0, eigenstates of A do not coincide with those of B. 6 If basis X(a, b) are A, B eigenstates, any i = C X obeys [A, B] i = 0 = [A, B]=0. | i | i X iX| i | i ⇒ P If A, B commute, their eigenstates coincide.

AB a = BA a = aB a , so B a a . | i | i | i | i∝| i [email protected] 3 Completeness relation: a a = 1 . a | ih | P Expand i = W a then act from left with a′ W ′ = a′ i , so i = a a i | i a ia| i h | −→ ia h | i | i a | ih | i P P 2 Probability to observe system in eigenstate a of A to be in eigenstate b of B: Pa b = b a . | i | i → |h | i| Iα Pa b are the only physically meaningful quantities, thus i and e i for any α represent same state. → | i | i

a Pb a = 1 for some state b as expected. b = a a b a . Act from left with b gives 1 = a a b b a . → | i | i h | i| i h | h | ih | i P P P Principle of reversibility: Pa b = Pb a. b a = a b ∗ → → h | i h | i

IHt Time dependence: Time evolution of states: i, t = e− i , H is Hamiltonian with energy eigenstates. | i | i 2 IHt Probability system in state i observed in state j time t later = Mi j , transition amplitude Mi j = j e− i . | i | i | → | → h | | i IHt IHt Average value of observable Q evolves in time as Q (t)= i e Qe− i . h ii h | | i Q is conserved [Q, H]=0 ( Q (t) independent of t). ⇐⇒ h i [email protected] 4 1.2 Fermionic and bosonic particles

Particle’s eigenvalues = σ. Particle states σ, σ′, ... completely span Hilbert space. Vacuum is 0 = . | i | i |i

σ, σ′, ... = σ′, σ, ... for bosons/fermions. Pauli exclusion principle: σ, σ, σ′... =0 if σ fermionic. | i ±| i | i

Iα 2Iα Iα σ, σ′,... and σ′,σ,... are same state, σ, σ′,... = e σ′,σ,... = e σ, σ′,... = e = 1. | i | i | i | i | i ⇒ ±

Creation/annihilation operators: aσ† σ′,σ′′, ... = σ, σ′,σ′′, ... , so σ, σ′, ... = aσ† a† . . . 0 . | i | i | i σ′ | i

[aσ† , a† ] =[aσ, aσ ] = 0 (bosons/fermions). σ′ ∓ ′ ∓

σ, σ′,... = a† a† ′ ... = σ′,σ,... = a† ′ a† ... | i σ σ | i ±| i ± σ σ| i a removes σ particle = a 0 =0. σ ⇒ σ| i

E.g. (a σ′, σ′′ )† σ′′′ = σ′, σ′′ (a† σ′′′ ) = 0 unless σ = σ′, σ′′′ = σ′′ or σ = σ′′, σ′′′ = σ′. i.e. a σ′, σ′′ = δ ′′ σ′ δ ′ σ′′ . σ| i ·| i h | σ| i σ| i σσ | i± σσ | i

[aσ, a† ] = δσσ . σ′ ∓ ′

e.g. 2 fermions a† ′′ a† ′′′ 0 : Operator a† a ′ replaces any σ′ with σ, must still vanish when σ′′ = σ′′′. σ σ | i σ σ

Check: (a† a ′ )a† ′′ a† ′′′ 0 = a† a† ′′ a ′ a† ′′′ 0 + δ ′ ′′ a† a† ′′′ 0 = δ ′ ′′′ a† a† ′′ 0 + δ ′ ′′ a† a† ′′′ 0 . σ σ σ σ | i − σ σ σ σ | i σ σ σ σ | i − σ σ σ σ | i σ σ σ σ | i [email protected] 5 ∞ ∞ † † Expansion of observables: Q = N=0 M=0 CNM;σ ...σ ;σ ...σ1 a ...a aσM ...aσ1. 1′ N′ M σ1′ σN′ P P Can always tune the CNM to give any values for 0 aσ′ ...aσ′ Qa† ...a† 0 . h | 1 L σ1 σK | i

Commutations with additive observables: [Q, aσ† ]= q(σ)aσ† (no sum) , where Q is an observable such that for σ, σ′,... , total Q = q(σ)+ q(σ′)+ . . . and Q 0 = 0 (e.g. energy). | i | i

Check for each particle state: Qa† 0 = q(σ)a† 0 + a† Q 0 = q(σ)a† 0 , σ| i σ| i σ | i σ| i

Qa† a† ′ 0 = a† Qa† ′ 0 + q(σ)a† a† ′ 0 = a† a† ′ Q 0 + q(σ′)a† a† ′ 0 + q(σ)a† a† ′ 0 = (q(σ) + q(σ′))a† a† ′ 0 etc. σ σ | i σ σ | i σ σ | i σ σ | i σ σ | i σ σ | i σ σ | i Note: Conjugate transpose is [Q, a ]= q(σ)a . σ − σ

Number operator for particles with eigenvalues σ is a† aσ (no sum). E.g. (a† aσ)a† a† a† ′ 0 = 2a† a† a† ′ 0 . σ σ σ σ σ | i σ σ σ | i

Additive observable: Q = σ q(σ)aσ† aσ . E.g. Q = H, (free) Hamiltonian, q(σ) = Eσ, energy eigenvalues. P [email protected] 6 2 Symmetries in QM

2.1 Unitary operators

Symmetry is powerful tool: e.g. relates diﬀerent processes.

Symmetry transformation is change in our point of view (e.g. spatial rotation / translation), does not change experimental results. i.e. all i U i does not change any j i 2. | i −→ | i |h | i| Continuous symmetry groups G require U unitary:

j i j U †U i = j i , so U †U = UU † = 1 , includes U = 1. h | i→h | | i h | i Wigner + Weinberg: General physical symmetry groups require U unitary,

or antiunitary: j i j U †U i = j i ∗ = i j , e.g. (discrete) time reversal. h | i→h | | i h | i h | i

A unaﬀected (and f(A) in general), so must have A UAU † . h i h i →

IHt IHt Transition amplitude Mi j unaﬀected by symmetry transformation, i.e. j U †e− U i = j e− i , → h | | i h | | i which requires [U, H] = 0 if time translation and symmetry transformation commute. [email protected] 7 Parameterize unitary operators as U = U(α), αi real, i = 1, ..., d(G). d(G) is dimension of G, minimum no. of paramenters required to distinguish elements.

Choose group identity at α = 0, i.e. U(0) = 1.

So for small α , can write U(α) 1 + It α . i ≈ i i ti are the linearly independent generators of G. Since U †U = 1, ti = ti†, i.e. ti are Hermitian .

[U, H]=0 = [t ,H] = 0, so conserved observables are generators. ⇒ i 1 Can replace all t t′ = M t if M invertible and real, because then α′ = α M − real. i → i ij j i j ji

In general, U(α)U(β)= U(γ(α, β)) (up to possible phase eIρ, removable by enlarging group). [email protected] 8 U(α) = exp [It α ] if Abelian limit is obeyed: whenever β α , U(α)U(β)= U(α + β) i i i ∝ i (usually true for physical symmetries, e.g. rotation about same line / translation in same direction).

Can write U(α)=[U(α/N)]N , then for N is [1 + It α /N + O(1/N 2)]N = exp[It α ] + O 1 . → ∞ i i i i N

Lie algebra: [ti, tj]= ICijktk , where Cijk are the structure constants

1 appearing in U(α)U(β)= U(α + β + 2ICαβ + cubic and higher)

2 2 (where (Cαβ)k = Cijkαiβj, and no O(α ) ensures U(α)U(0) = U(α), likewise no O(β )).

LHS: eIαteIβt 1 + Iαt 1(αt)2 1 + Iβt 1(βt)2 ≈ − 2 × − 2 1 + I(αt + βt) 1 (αt)2 + (βt)2 + 2(αt)(βt) ≈ − 2 h i I(α+β+ 1 ICαβ)t 1 1 1 2 RHS: e 2 1 + I(α + β + ICαβ)t (α + β + ICαβ)t ≈ 2 − 2 2 1 + I(αt + βt + 1ICαβt) 1 (αt)2 + (βt)2 + (αt)(βt) + (βt)(αt) , ≈ 2 − 2 h i 1 1 i.e. 2ICijkαiβjtk = 2 [αiti, βjtj].

Now take all α , β zero except e.g. α = β = ǫ IC t =[t ,t ] etc., gives Lie algebra. i j 1 2 → 12k k 1 2 [email protected] 9 In fact, Lie algebra completely speciﬁes group in non-small neighbourhood of identity.

This means that for U(α)U(β)= U(γ), γ = γ(α, β) can be found from Lie algebra.

We have shown this above in small neighbourhood of identity, i.e. to 2nd order in α, β, only.

Check to 3rd order: Write X = αiti, Y = βiti, can verify Baker-Hausdorﬀ formula 1 I exp[IX] exp[IY ] = exp[I(X + Y ) [X,Y ]+ ([X, [Y, X]]+[Y, [X,Y ]]) + quadratic and higher]. − 2 12 has the form Iγiti, γi real | {z }

Lie algebra implies: 1. C = C (antisymmetric in i, j). Cijk can be chosen antisymmetric in i,j,k (see later). ijk − jik

2. Cijk real.

Conjugate of Lie Algebra is t†,t† = IC∗ t† , which is negative of Lie Algebra because t† = t . j i − ijk k i i h i

Thus Cijk∗ tk = Cijktk, but tk linearly independent so Cijk∗ = Cijk.

2 Iα t 2 Iα t 2 3. t = tjtj is invariant (commutes with all ti, so transforming gives e i it e− k k = t ).

[t2,t ] = t [t ,t ]+[t ,t ]t = IC t ,t = 0 by (anti)symmetry in (j, i) j,k. i j j i j i j jik{ j k}

Examples: Rest mass, total angular momentum. tjtj doesn’t have to include all j, only subgroup. [email protected] 10 2.2 (Matrix) Representations

Physically, matrix representation of any symmetry group G of nature formed by particles:

( ) General transformation of aσ† : U(α)aσ† U †(α)= Dσσ (α)a† , with invariant vacuum: U(α) 0 = 0 . ′ σ′ | i | i

U(α)a† 0 is 1 particle state, so must be linear combination of a† ′ 0 , i.e. U(α)a† 0 = D ′ (α)a† ′ 0 . σ| i σ | i σ| i σσ σ | i

Thus U(α)a† a† ′ 0 = D ′′ (α)D ′ ′′′ (α)a† ′′ a† ′′′ 0 etc. σ σ | i σσ σ σ σ σ | i i D(α) matrices furnish a representation of G , matrix generators (t )σ′σ with same Lie algebra [ti, tj]σσ′ = ICijk(tk)σσ′.

Since U(α)U(β) = U(γ), must have Dσ′′σ′ (α)Dσ′σ(β) = Dσ′′σ(γ).

Iαiti If Abelian limit of page 8 is obeyed, Dσ σ(α)= e . ′ σ′σ

1 Similarity transformation (t ) (t′) = V (t ) (V − ) also a representation. i σσ′′′ → i σσ′′′ σσ′ i σ′σ′′ σ′′σ′′′ Particle states require V unitary.

a† 0 a′† 0 = V ′ a† ′ 0 , then othogonality 0 a′ ′ a′† 0 = δ ′ requires V †V = 1. σ| i → σ | i σσ σ | i h | σ σ | i σ σ But in general, representations don’t have to be unitary. [email protected] 11 1 In “reducible” cases, can similarity transform t V t V − such that t is block-diagonal matrix, i → i i each block furnishes a representation, e.g.:

(ti)jk 0 (ti)σ′σ = . 0 (ti)αβ σ′σ Particles of one block don’t mix with those of other — can be treated as 2 separate “species”.

Each block can have diﬀerent t2, corresponds to diﬀerent particle species.

Irreducible representation: Matrices (ti)σ′σ not block-diagonalizable by similarity transformation.

In this sense, these particles are elementary.

Size of matrix written as m(r) m(r), where r labels representation. × 2 2 2 Corresponds to single species, single value of t : t = C2(r)1 (consistent with [t , ti] = 0).

C2(r) is quadratic Casimir operator of representation r. [email protected] 12 Fundamental representation of G: Generators written (ti)αβ.

Matrices representing elements used to deﬁne group G (also called deﬁning representation).

Adjoint representation of G (r = A): Generators (tj)ik = ICijk , satisfy Lie algebra.

Use Jacobi identity [ti, [tj,tk]]+[tj, [tk,ti]]+[tk, [ti,tj]] = 0.

From Lie algebra, [t , [t ,t ]] = I[t ,C t ] = C C t , i j k i jkl l − jkl ilm m

so Jacobi identity is CjklCilm + CkilCjlm + CijlCklm = 0 (after removing contraction with linearly independent tm),

or, from C = C , IC IC + IC IC IC IC = 0, which from (t ) = IC reads [t ,t ] = IC (t ) . ijk − jik − kjl lim kil ljm − ijl klm j ik ijk i j km ijl l km

T Conjugate representation has generators t∗ = t (obey the same Lie algebra as t ). − i − i i Iα t Iα t Iα t If t∗ = Ut U † (U unitary), then e− i i∗ = e i i ∗ = Ue i iU †, − i i i.e. conjugate representation original representation, representation is real. ≡ −→

T For invariant matrix g (“metric”), i.e. eIαiti geIαiti = g, G transformation leaves φT gψ invariant. [email protected] 13 Semi-simple Lie algebra: no ti that commutes with all other generators (no U(1) subgroup).

Semi-simple group’s matrix generators must obey tr[ti]=0.

Make all tr[ti] = 0 except one, tr[tK], via ti′ = Mijti (this is just rotation of vector with components tr[ti]).

Determinant of U(α)U(β) = U(γ) is eIαitr[ti]eIβitr[ti] = eIγitr[ti], i.e. (α + β γ )tr[t ]=0. i i − i i But only tr[t ] = 0 (we assume), so α + β γ = 0. K 6 K K − K Must have C α β = 0 (recall γ α + β + C α β ), ijK i j K ≃ K K ijK i j

or CijK = CKij = 0 for all i, j, so from Lie group we have [tK,ti]=0 for all i — not possible for semi-simple group.

So assumption was wrong, must have tr[tK]=0. [email protected] 14 Normalization of generators chosen as tr[titj]= C(r)δij .

N =tr[t t ] becomes M N (M T ) after basis transformation t M t . ij i j ik kl lj i → ij j

T Nij components of real symmetric matrix N, diagonalizable via MNM when M real, orthogonal.

T Also N is positive deﬁnite matrix α Nα = α tr[t t ]α =tr[α t α t ] =tr[(α t )†α t ] 0 i i j j i i j j i i j j ≥

2 (because for any matrix A, tr[A†A] = A† A = A∗ A = A 0). βα αβ αβ αβ αβ | αβ| ≥ P After diagonalization, N = 0 for i = j and above implies N > 0 (no sum). ij 6 ii Then multiply each t by real number c , changes N c2N > 0. i i ii → i ii

Choose ci such that each Nii (no sum over i) all equal to positive C(r).

Representation dependence of quadratic Casimir operator: C2(r)m(r)= C(r)d(G).

2 Deﬁnition of quadratic Casimir operator gives tr[t ] = C2(r)m(r).

Normalization of generators tr[t t ] = C(r)δ = tr[t2] = C(r)d(G). i j ij ⇒

1 Example: 2 and 3 component repesentations of rotation group have diﬀerent spins (i.e. C2(r)) 2 and 1. [email protected] 15 Antisymmetric structure constants: C = I tr[[t , t ]t ]. ijk −C(r) i j k

From Lie algebra, tr[[ti,tj]tl] = ICijktr[tktl] = ICijlC(r).

Structure constants obey CjkiClki = C(A)δjl .

In adjoint representation, quadratic Casimir operator on page 11 is (t2) = C (A)δ = C C . jl 2 jl − jik kil

But C2(A) = C(A):

C2(A)m(A) = C(A)d(G) from representation dependence of quadratic Casimir operator on page 14,

and m(A) = d(G). [email protected] 16 2.3 External symmetries

2.3.1 Rotation group representations

(Spatial) rotation of vector v Rv preserves vT v, so R orthogonal (RT R = 1). → IJ θ Rotation θ about θ: U(θ)= e− · . | |

Lie algebra is [Ji, Jj]= IǫijkJk for generators J1, J2, J3 (see later).

Or use J3 and raising/lowering operators J = (J1 IJ2). ± ± [email protected] 17 Irreducible spin j representations:

(j) (j) (J3 )m m = mδm m and (J )m m = [(j m)(j m + 1)]δm ,m 1 , where m = j, j + 1, ..., j , ′ ′ ± ′ ∓ ± ′ ± − −

1 2 spin j = 0, 2, 1, ... , number of components n.o.c.= 2j + 1 and J = j(j +1).

Let m, j be J = m and J 2 = F (j) orthonormal eigenstates ([J , J 2] = 0). | i 3 3

J changes m by 1 because [J ,J3] = J , so J m, j = C (m, j) m 1,j . ± ± ± ∓ ± ±| i ± | ± i C (m, j) = F (j) m2 m (states absorb complex phase): ∓ − ± p 2 2 2 C (m, j) = m, j J J m, j (J † = J ) and J J = J J3 J3. | ∓ | h | ± ∓| i ± ∓ ± ∓ − ± Let j be largest m value for given F (j)

(m bounded because m2 = m, j J 2 m, j = F (j) m, j J 2 + J 2 m, j < F (j)). h | 3 | i −h | 1 2 | i

2 F (j) = j(j + 1) because J+ j,j = 0, so J J+ j,j = (F (j) j j) j,j = 0. | i − | i − − | i

Let j′ be smallest m: J j′,j = 0 = F (j) = j′(j′ + 1), so j′ = j (other possibility j′ = j + 1 >j). − −|− i ⇒ − So m = j, j + 1,...,j, i.e. n.o.c.= 2j + 1. Since n.o.c. is integer, j = 0, 1, 1,.... − − 2 [email protected] 18 Spin decomposition of tensors: e.g. 2nd rank tensor Cij (n.o.c.=9), representations are j = 0, 1, 2: scalar (n.o.c.=1) + antisymmetric rank 2 tensor (n.o.c.=3) + symmetric traceless rank 2 (n.o.c.=5) components.

C = 1δ C + 1(C C ) + 1(C + C 2δ C ). ij 3 ij kk 2 ij − ji 2 ij ji − 3 ij kk Component irreducible representations signiﬁed by 1 + 3 + 5.

Counting n.o.c. shows they are equivalent respectively to

the j = 0 (2j + 1 = 1), j = 1 (2j + 1 = 3) and j = 2 (2j + 1 = 5) representations:

1δ C 1δ C is like scalar spin 0, 3 ij kk → 3 ij kk ≡

1 1 1 2 ( (Cjk Ckj)) i=k,j = ǫijkCjk Ril ǫljkCjk (because ǫR = Rǫ) is like vector spin 1, 2 − | 6 2 → 2 ≡ 1(C + C 2δ C ) R R 1(C + C 2δ C ) is like rank 2 tensor spin 2. 2 ij ji − 3 ij ii → il jm 2 lm ml − 3 lm kk ≡ [email protected] 19 Direct product: 2 particles, spins j1,j2, also in a representation:

IJ θ IJ(j1,j2) θ (j1,j2) (j1) (j2) e · m1,j1; m2,j2 = (e · )m m m m m1′ ,j1; m2′ ,j2 where J = J δm m + δm m J . | i 1′ 2′ 1 2| i m1′ m2′ m1m2 m1′ m1 2′ 2 1′ 1 m2′ m2

IJ θ IJ (j1) θ IJ (j2) θ From e · m1,j1; m2,j2 = (e · ) ′ (e · ) ′ m′ ,j1; m′ ,j2 , i.e. total rotation is rotation of each particle in turn. | i m1m1 m2m2 | 1 2 i Thus J m ,j ; m ,j = (m + m ) m ,j ; m ,j , 3| 1 1 2 2i 1 2 | 1 1 2 2i (j ,j ) so m ,j ; m ,j is combination of J 1 2 , J (j1,j2)2 eigenstates 2; m + m ,j , | 1 1 2 2i 3 | 1 2 i with j = m1 + m2, m1 + m2 + 1, ..., j1 + j2. So jmax = j1 + j2.

Triangle inequality: representation for (j ,j ) contains j = j j , j j + 1,...,j + j . 1 2 | 1 − 2| | 1 − 2| 1 2 Number of orthogonal eigenstates m ,j ; m ,j = Number of orthogonal eigenstates 2; m, j , | 1 1 2 2i | i

j1+j2 b 1 i.e. (2j1 + 1)(2j2 + 1) = (2j + 1) (now use j = (b a + 1)(b + a)) so jmin = j1 j2 . j=jmin j=a 2 − | − | P P

Example: Representation for 2 spin 1 particles (j1,j2) = (1, 1):

From triangle inequality, this is sum of irreducible representations j = 0, 1, 2. ≡ Also from tensor representation on page 18: product of 2 vectors u v (2nd rank tensor) is 3 3 = 1 + 3 + 5. i j × [email protected] 20 2.3.2 Poincar´eand Lorentz groups

µ µ µ ν µ Poincar´e(inhomogeneous Lorentz) group formed by coordinate transformations x x′ = Λ x + a , → ν µ ν µ ν preserving spacetime separation: gµνdx dx = gµνdx′ dx′ . Implies:

µ ν 1 µ µ Transformation of metric tensor: gρσ = gµνΛ ρΛ σ or (Λ− ) ν = Λν .

µ µ µ Identity: Λ ν = δ ν, a =0.

Poincar´egroup deﬁned by U(Λ, a)U(Λ, a)= U(ΛΛ, Λa + a).

This is the double transformation x′′ = Λx′ + a = Λ(Λx + a) + a.

Poincar´egroup generators: J µν and P µ , appearing in U(1 + ω, ǫ) 1+ 1Iω J µν Iǫ P µ . ≃ 2 µν − µ

µ µ µ Obtained by going close to identity, Λ ν = δ ν + ω ν and aµ = ǫµ.

Choose J µν = J νµ . − Allowed because ω = ω : Transformation of metric tensor reads g = g (δµ + ωµ )(δν + ων ) g + ω + ω . ρσ − σρ ρσ µν ρ ρ σ σ ≃ σρ ρσ σρ [email protected] 21 µ µν µ µ ρ Transformation properties of P , J : U(Λ, a)P U †(Λ, a) = Λρ P

µν µ ν ρσ ρ σ σ ρ and U(Λ, a)J U †(Λ, a) = Λ Λ (J a P + a P ). ρ σ −

1 1 Apply Poincar´egroup to get U(Λ, a)U(1 + ω, ǫ) U †(Λ, a) = U(1 + ΛωΛ− , Λ(ǫ ωΛ− a)). − =U(Λ(1+ω),Λǫ+a) =U(Λ−1, Λ−1a) −

| {z1 } | {z } 1 1 1 Expand both sides in ω, ǫ: U(Λ, a)(1 + IωJ IǫP )U †(Λ, a) = 1 + IΛωΛ− J IΛ(ǫ ωΛ− a)P , equate coeﬃcients of ω, ǫ. 2 − 2 − −

µ So P transforms like 4-vector, Jij like angular momentum.

Poincar´ealgebra: I[J ρσ, J µν]= gσνJ ρµ gρµJ σν + gσµJ ρν + gρνJ σµ (homogeneous) Lorentz group, − − → I[P µ, J ρσ]= gµρP σ gµσP ρ , and [P µ, P ν]=0. − µ µ µ µ µν Obtained by taking Λ ν = δ ν + ω ν, aµ = ǫµ in transformation properties of P , J , to ﬁrst order in ω, ǫ gives

P µ 1Iω [P µ,J ρσ] + Iǫ [P µ,P ν] = P µ + 1ω gµσP ρ 1ω gµρP σ, and − 2 ρσ ν 2 ρσ − 2 ρσ J µν + 1Iω [J ρσ,J µν] Iǫ [P ρ,J µν] = J µν gρµǫ P ν + gρνǫ P µ + 1ω (gρνJ σµ gσνJ ρµ) + 1ω (gσµJ ρν gρµJ σν). 2 ρσ − ρ − ρ ρ 2 ρσ − 2 ρσ − [email protected] 22 2.3.3 Relativistic quantum mechanical particles

0 i 1 jk 23 31 12 Identify H = P , spatial momentum P , angular momentum Ji = 2ǫijkJ (i.e. (J1, J2, J3) = (J , J , J )).

[H, P ]=[H, J] = 0 P , J conserved. Rotation group [J , J ]= Iǫ J is subgroup of Poincar´egroup. → i j ijk k Also deﬁne boost generator K = (J 10, J 20, J 30), obeys [J , K ]= Iǫ K and [K , K ]= Iǫ J . i j ijk k i j − ijk k

K not conserved: [Ki,H]= IPi, because boost and time translation don’t commute.

µ IP aµ IK eˆβ IJ θ Explicit form of Poincar´eelements: U(Λ, a)= e− e− · e · × × translate aµ boost along eˆ by V =sinh β rotate θ about θ | | | {z } | {z } |{z} (V : magnitude of 4-velocity’s spatial part.)

Lorentz transformation of 4-vectors: (K )µ = I(δ δ + δ δ ) and (J )µ = Iǫ . i ν 0µ iν iµ 0ν i ν − 0iµν Then A A δω (δ A + δ A ) δθ ǫ A (1 δ ) as required. ν → ν − i iν 0 0ν i − i ijν j − 0ν

µ µ 2 µ IK eˆβ µ 2 IJ θˆθ e− · = 1 IKieî sinh β (Kieî) (cosh β 1) and e · = 1+ IJiθî sin θ Jiθî (1 cos θ) . ν − − − ν ν − − ν h i h i 3 3 Follows directly from (Kieî) = Kieî and (Jiθî) = Jiθî. [email protected] 23 µ 2 2 Particles: Use P , P = m eigenstates. Distinguish momentum p from list σ, i.e. a† a† (p). σ → σ ( ) Commutation relations for aσ† (p): [aσ† (p), a† (p′)] =[aσ(p), aσ (p′)] = 0 σ′ ∓ ′ ∓

0 (3) and [aσ(p), a† (p′)] = δσσ (2p )δ (p p′). σ′ ∓ ′ − As on page 4, but with diﬀerent normalization (Lorentz invariant).

( ) Application of general transformation of aσ† on page 10 to Lorentz transformation complicated by mixing between σ and p.

Simplify by ﬁnding 2 transformations, one that mixes σ and one that mixes p, separately. [email protected] 24 ( ) Lorentz transformation for aσ† (p): U(Λ)a† (p)U †(Λ) = Dσ σ(W (Λ, p))a† (Λp). σ ′ σ′

Same as general transformation on page 10, but because σ σ, p , mixing of σ with p must be allowed. → { } Construction of W : Choose reference momentum k and transformation L to mix k but not σ (deﬁnes σ):

µ ν µ L (p)k = p and U(L(p))a† (k) 0 = a† (p) 0 . ν σ | i σ | i

1 µ ν µ Then W (Λ, p)= L− (Λp)ΛL(p). W mixes σ but not k, i.e. belongs to little group of k: W νk = k .

General transformation is U(Λ)a† (p) 0 = U(Λ)U(L(p))a† (k) 0 , using deﬁnition of L. σ | i σ | i

1 Multiplying by 1 = U(L(Λp))U(L− (Λp)) gives

1 U(Λ)a† (p) 0 = U(L(Λp)) U(L− (Λp)) U(Λ)U(L(p))a† (k) 0 = U(L(Λp))U(W (Λ, p))a† (k) 0 . σ | i σ | i σ | i

But W doesn’t change k, so U(Λ)a† (p) 0 = U(L(Λp))D ′ (W (Λ, p))a† ′ (k) 0 . σ | i σ σ σ | i

Then L(Λp) changes k to Λp but doesn’t change σ′.

( ) IΛp a Poincar´etransformation for aσ† (p): U(Λ, a)a† (p)U †(Λ, a)= e− · Dσ σ(W (Λ, p))a† (Λp). σ ′ σ′

( ) IP µa In Lorentz transformation for aσ† (p), use U(Λ, a) = e− µ U(Λ) (from explit form of Poincar´eelements on page 22). [email protected] 25 2.3.4 Quantum ﬁeld theory

3 Lorentz invariant QM: H = d xH (x) , scalar ﬁeld H (x) (i.e. U(Λ, a)H (x)U †(Λ, a)= H (Λx + a) ), R obeys cluster decomposition principle (two processes with large spatial separation evolve independently).

Causality: [H (x), H (y)] = 0 when (x y)2 0 . Required for Lorentz invariance of S-matrix. − ≥ Intuitive reason: signal can’t propagate between 2 spacelike separated events.

( ) In QM (general): H built from aσ† . In QFT: H built from H (x) built from products of

c 3 c + 3 Fields ψl− (x)= D p vlσ(x; p)aσ†(p) and ψl (x)= D p ulσ(x; p)aσ(p).

R 3 d3p R4 2 2 0 3 3 Lorentz invariant momentum space volume D p = 2p0 = d pδ(p + m )θ(p ) obeys D Λp = D p.

These ﬁelds should obey

c To calculate u (x; p) and v (x; p), take single particle species in representation labelled j, allow a †(p) = a† (p). lσ lσ σ 6 σ [email protected] 26 Ip x Ip x x dependence of u, v: ul σ(x; p)= e · ul σ(p), vl σ(x; p)= e− · vl σ(p).

( ) Take Λ = 1 in Poincar´etransformation for ﬁelds and for aσ† (p) on page 25 and 24,

+ 3 3 Ip a e.g. U(1, a)ψl (x)U †(1, a) = D p ulσ(x; p)U(1, a)aσ(p)U †(1, a) = D p e · ulσ(x; p)aσ(p) R R + 3 Ip a = ψl (x + a) = D p ulσ(x + a; p)aσ(p), equate coeﬃcients of aσ(p) (underlined), gives ul σ(x; p)e · = ulσ(x + a; p). R (c) Klein-Gordon equation: (∂2 m2)ψ± (x)=0. − l

c 3 Ip x c 2 2 2 2 Act on e.g. ψ− (x) = D p e− · v (p)a †(p) with (∂ m ), use p = m . l l σ σ − − R (j) (j) 1 Transformation of u, v: Dll (Λ)ul σ(p)= D (W (Λ, p))ulσ (Λp), Dll (Λ)vl σ(p)= D (W − (Λ, p))vlσ (Λp). ′ ′ σ′σ ′ ′ ′ σσ′ ′

( ) E.g. consider v, use Poincar´etransformation for ﬁelds and for aσ† (p) on page 25 and 24.

c 3 c 3 IΛp a (j) c LHS: U(Λ, a)ψl− (x)U †(Λ, a) = D p vlσ(x; p)U(Λ, a)aσ†(p)U †(Λ, a) = D p e− · vlσ(x; p)Dσ′σ(W (Λ, p))aσ†′ (Λp), and R R 1 c 3 1 c 3 1 c RHS: D ′ (Λ− )ψ−′ (Λx + a) = D p D ′ (Λ− )v (Λx + a; p)a †(p) = D Λp D ′ (Λ− )v (Λx + a; Λp)a †(Λp)(p Λp). ll l ll lσ σ ll lσ σ → R R 3 3 c (j) 1 Use D Λp = D p, equate coeﬃcient of aσ†(Λp) (underlined), multiply by Dl′′l(Λ)Dσ′′σ(W − (Λ, p)). p p k p dependence of u, v: ulσ( )= Dll′(L(p))ul′σ( ) . ( dependence of v is the same.)

1 In transformation of u, v, take p = k so L(p) = 1, Λ = L(q) so Λp = L(q)k = q, then W (Λ, p) = L− (Λp)L(q)=1. [email protected] 27 2.3.5 Causal ﬁeld theory

+ c Since [ψ (x),ψ− (x′)] = 0, causality on page 25 only gauranteed by taking H (x) to be functional of l l′ ∓ 6 complete ﬁeld + c 1 + c ψl(x)= κψl (x)+ λψl− (x) (so representations Dll′(Λ− ) for ψl (x) and ψl− (x) the same), with

2 Causality: [ψl(x),ψl (y)] =[ψl(x),ψ† (y)] = 0 when (x y) > 0 by suitable choice of κ, λ. ′ ∓ l′ ∓ −

Now 0 H 0 = , i.e. consistency with gravity not gauranteed by QFT. h | | i ∞ In each term in H, last operator on right hand side is not always an annihilation operator.

3 Ip x Ip x c Complete ﬁeld: ψl(x)= D p κe · ulσ(p)aσ(p)+ λe− · vlσ(p)aσ†(p) . R [email protected] 28 2.3.6 Antiparticles

H (x) commutes with conserved additive Q: [Q, H (x)]=0.

Imposed in order to satisfy [Q, H]=0.

This is achieved as follows:

Commutation of ﬁelds with conserved additive Q: [Q, ψ (x)] = q ψ (x) , and l − l l

L1 L2 M M L1 L2 M M Field construction of H (x): H = ψ ψ ...ψ 1†ψ 2† . . . with q + q + . . . q 1 q 2 . . . = 0 l1 l2 m1 m2 l1 l2 − m1 − m2 − P (Mi, Lj label particle species).

Antiparticles: For every particle species there is another species with opposite conserved quantum numbers .

c c + + Commutation of ﬁelds with conserved additive Q implies [Q, ψ− (x)] = q ψ− (x) and [Q, ψ (x)] = q ψ (x), l − l l l − l l

c c c c c but since [Q, a †] = q a † and [Q, a ] = q a (no sum) from page 5, q = q and q = q , i.e. q = q . σ σ σ σ − σ σ l σ l − σ σ − σ [email protected] 29 2.3.7 Spin in relativistic quantum mechanics

Lorentz group algebra simpliﬁed by choosing generators A = 1(J IK ) and B = 1(J + IK ), i 2 i − i i 2 i i behaves like 2 independent rotations: [Ai,Aj]= IǫijkAk , [Bi,Bj]= IǫijkBk and [Ai,Bj]=0, i.e. relativistic particle of type (A, B) (i.e. in eigenstate of A2 = A(A + 1), B2 = B(B + 1))

2 particles at rest, ordinary spins A, B (in representation sense). ≡ In terms of degrees of freedom, (A, B) = (2A+1) (2B+1). ×

Triangle inequality: ordinary spin J = A + B , so j = A B , A B + 1, ..., A + B . | − | | − | Derived on page 19.

Can have eigenstates of A2 = A(A + 1), B2 = B(B + 1) and J 2 = j(j + 1) simultaneously,

2 2 2 2 because [J ,Ai]=[J ,Bi]=0. Use J = (A + B) and A, B commutation relations.

Simultaneous eigenstate also with any K = I(A B) or K2 not possible. − [email protected] 30 Example: (A, B) = (1, 1) is representation of 4-vector: (A )µ , (B )µ can have eigenvalues 1 only. 2 2 3 ν 3 ν ±2 From triangle inequality on page 29, j = 0, 1.

Also follows from tensor representation on page 18: 2 2 = 1 + 3. ×

N N 1 1 N 2 2 More generally, rank N tensor is (2, 2) = A=0 B=0(A, B), i.e. (1, 1)N = N , N +lower spins, where NP, N Ptraceless symmetric rank N tensor, j = 0, ..., N. 2 2 2 2 2 2 ≡ (N, 0) and (0,N) are purely spin j = N. [email protected] 31 2.3.8 Irreducible representation for ﬁelds

(c) 2 (c) (c) If particles created by aσ †(p) have spin j (i.e. [J , aσ †(p)] = j(j + 1)aσ †(p)), must take ﬁeld with same spin j but any (A, B) consistent with triangle inequality on page 29:

3 Ip x Ip x c ψab(x)= D p κ e · uab σ(p)aσ(p)+ λ e− · vab σ(p)aσ†(p) (l = ab, l′ = a′b′), R (A) (B) Lorentz transformation uses generators Aa b ab = J δb b , Ba b ab = δa aJ ′ ′ a′a ′ ′ ′ ′ b′b where a = A, A + 1, ..., A and b = B, B + 1, ..., B . − − − − [email protected] 32 2.3.9 Massive particles

Choose k = (0, 0, 0, m) (momentum of particle at rest) Then W is an element of the spatial rotation group.

L(p) in terms of K: L(p) = exp[ Ipˆ Kβ]. − · Then W (R, p)= R, i.e. rotation group apparatus of subsubsection 2.3.1 applies to relativistic particles too.

(j) (A) (B) (j) (A) (B) Conditions on u, v: J uab σ (0) = J ua b σ(0) + J uab σ(0) , J ∗vab σ (0) = J va b σ(0) + J vab σ(0) . σ′σ ′ aa′ ′ bb′ ′ − σ′σ ′ aa′ ′ bb′ ′ In transformation of u, v on page 26, take Λ = R and p = 0 (i.e. p = k)

1 1 (j) 1 (so p = k, Rp = p, L(p)=1, W (R, p) = L− (Rp)RL(p) = L− (p)R = R), so e.g. Dll′ (R)vl′σ(0) = Dσσ′ (R− )vlσ′ (0),

(j) 1 (j) (j) and use Dσσ′ (R− ) = Dσ′σ∗(R) because irreducible representations of R are unitary (see form of (Ji )σ′σ on page 17).

(j) (j) (j) Take l = ab, so D ′ ′ (R)v ′ ′ (0) = D ′ ∗(R)v ′ (0). Generators of D ∗(R), D(R) are respectively J ∗, A + B. aba b a b σ σ σ abσ −

j+σ u, v relation: vab σ(0) = ( 1) uab σ(0) up to normalization. − −

(j) σ σ′ (j) Conditions on u, v with (no sum over σ, σ′) Jσσ′∗ = ( 1) − J σ, σ′ from page 17 − − − −

σ gives vab σ(0) ( 1) uab σ(0), absorb proportionality constant into u, v. ∝ − −

j+σ The p dependence of u, v are the same, so vab σ(p) = ( 1) uab σ(p). − − [email protected] 33 2.3.10 Massless particles

Take reference vector k = (0, 0, 1, 1).

Little group transformation: W (θ,µ,ν) 1 + IθJ + IµM + IνN with M = J + K , N = J + K . ≃ 3 2 1 − 1 2

µ ν W has 3 degrees of freedom: For (ti) νk = 0, take ti = (J3,M,N), check with Lorentz transformation of 4-vectors on page 22.

Choice of states: (J ,M,N) k,σ = (σ, 0, 0) k,σ . 3 | i | i Since [M,N] = 0, try eigenstates for which M k,m,n = m k,m,n , N k,m,n = n k,m,n . | i | i | i | i Then m, n continuous degrees of freedom, unobserved: [J ,M] = IN, so M(1 IθJ ) k,m,n = (m nθ)(1 IθJ ) k,m,n , 3 − 3 | i − − 3 | i i.e. (1 IθJ ) k,m,n is eigenvector of M, eigenvalue m nθ. − 3 | i − Similarly, [J ,N] = IM, so (1 IθJ ) k,m,n is eigenvector of N, eigenvalue n + mθ. 3 − − 3 | i Avoid this problem by taking m = n = 0, so left with states J k, σ = σ k, σ . 3| i | i Since J = J kˆ, σ is helicity, component of spin in direction of motion. 3 · Iθσ Representation for massless particles: Dσ′σ(W )= e δσ′σ .

U(W ) k, σ = (1 + IθJ + IµM + IνN) k, σ = (1 + Iθσ) k, σ . For ﬁnite θ, U(W ) k, σ = eIθσ k, σ . | i 3 | i | i | i | i [email protected] 34 p k p As on page 26. p dependence of u: ul σ( )= Dll′(L(p))ul′ σ( ) . ( dependence of v is the same.)

k Iθ(W,k)σ k k Iθ(W,k)σ k Little group transformation of u, v: ul σ( )e = Dll′(W )ul′ σ( ), vl σ( )e− = Dll′(W )vl′ σ( ).

Iθ(Λ,p)σ Transformation of u, v on page 26 reads ul σ(Λp)e = Dll′ (Λ)ul′ σ(p). Take Λ = W , p = k.

Rotation of u, v: (J ) u (k)= σu (k), (J ) v (k)= σv (k). 3 ll′ l′ σ l σ 3 ll′ l′ σ − l σ Take W to be just rotation about 3-axis (W (θ, µ, ν) 1 + IθJ , i.e. µ = ν = 0) in little group transformation of u, v. ≃ 3 k k M, N transformation of u, v: Mll′ul′ σ( )= Nll′ul′ σ( ) = 0 . Same for v.

Take W = (1 + IµM + IνN) in little group transformation of u, v. u, v relation: vl σ(p)= ul∗ σ(p).

Implied (up to proportionality constant) by rotation (note (J3)ll′ is imaginary) and M, N transformation of u, v.

Allowed helicities for ﬁelds in given (A, B) representation: σ = (B A) for particle/antiparticle . ± −

3 J = A + B, and e.g. Aaba′b′ = aδaa′ δbb′ , so rotation of u is σuab σ(k) = (a + b)uab σ(k).

M, N transformation of u is Maba′b′ ua′b′ σ = Naba′b′ ua′b′ σ = 0. Using M = IA IB+ and N = A B+, − − − − −

where A = A1 IA2, B = B1 IB2 are usual raising/lowering operators, (A )aba′b′ ua′b′ σ = (B+)aba′b′ ua′b′ σ = 0, ± ± ± ± − i.e. must have a = A, b = B or u = 0. So σ = B A. Similar for v, gives σ = A B. − ab σ − − [email protected] 35 2.3.11 Spin-statistics connection

Determine which of for given j is possible for causality on page 27 to hold. Demand more general condition ∓ ˜ 3 Ip (x y) ˜ Ip (x y) 2 [ψab(x), ψ†˜(y)] = D p π ˜(p) κκ˜∗e · − λλ∗e− · − = 0 for (x y) > 0, a˜b ∓ ab,a˜b ∓ − R h i where ψ , ψ˜ for same particle species and π (p)= u (p)˜u (p)= v (p)˜v (p). ab a˜˜b ab,a˜˜b ab σ a∗˜˜b σ ab σ a˜∗˜b σ

( ) In ﬁeld on page 31, use commutation relations for aσ† (p) on page 23.

u (p)˜u (p) = v (p)˜v (p) holds for massive particles from u, v relation on page 32. ab σ a∗˜˜b σ ab σ a˜∗˜b σ

(u (p)˜u (p)=[v (p)˜v (p)] in massless case from u, v relation on page 34). ab σ ∗a˜˜b σ ab σ a˜∗˜b σ ∗

2A˜+2B Relation between κ, λ of diﬀerent massive ﬁelds: κκ˜∗ = ( 1) λλ˜∗ . ± −

2 2 Explicit calculation shows πab,a˜˜b(p) = Pab,a˜˜b(p) + 2 p + m Qab,a˜˜b(p), p where (P, Q)(p) are polynomial in p, obey (P, Q)( p) = ( 1)2A˜+2B(P, Q)(p). − − −

2 0 0 3 Ip x Take (x y) > 0, use frame x = y , write ∆ (x) = D pe · : − + R 2A˜+2B 2A˜+2B (3) [ψab, ψ˜ ˜] = κκ˜∗ ( 1) λλ˜∗ P ˜( I∇)∆+(x y, 0) + κκ˜∗ ( 1) λλ˜∗ Q ˜( I∇)δ (x y). a˜b ∓ ∓ − ab,a˜b − − ± − ab,a˜b − − h i h i Commutator must vanish for x = y, so coeﬃcient of P zero. 6 [email protected] 36 Relations between κ, λ of single ﬁeld: κ 2 = λ 2 and ( 1)2A+2B =1. | | | | ± −

For A = A˜, B = B˜, relation between κ, λ of diﬀerent ﬁelds reads κ 2 = ( 1)2A+2B λ 2. | | ± − | | Spin-statistics: Bosons (fermions) have even (odd) 2j and vice versa .

From triangle inequality on page 29,

2j j (A + B) is integer, so ( 1) = 1, i.e. in [ψab, ψ˜ ˜] , must have (+) for even (odd) 2j. − ± − a˜b ∓ − Relation between κ, λ of single massive ﬁeld: λ = ( 1)2AeIcκ , c is the same for all ﬁelds. −

Divide relation between κ, λ of diﬀerent massive ﬁelds on page 35 by κ˜ 2 = λ˜ 2: κ = ( 1)2A˜+2B λ = ( 1)2A+2A˜ λ. | | | | κ˜ ± − λ˜ − λ˜

Ic c Absorb κ into ﬁeld, e into aσ†(p) (does not aﬀect commutation relations on page 23).

2A c ( 1) can’t be absorbed into a †(p) since this is independent of A, − σ nor absorbed into v since this is already chosen. So

3 Ip x 2A Ip x c Massive irreducible ﬁeld: ψ (x)= D p e · u (p)a (p) + ( 1) e− · v (p)a †(p) , ab ab σ σ − ab σ σ R 3 (j) Ip x 2A+j+σ Ip x c or more fully as ψab(x)= D p D (L(p)) e · ua b σ(0)aσ(p) + ( 1) e− · ua b σ(0)aσ†(p) . aba′b′ ′ ′ − ′ ′ − R Use u, v relation on page 32. [email protected] 37 2.4 External symmetries: fermions

1 2.4.1 Spin 2 ﬁelds

2j is odd particles are fermions. j = 1 representations include (A, B)= 1, 0 and (A, B)= 0, 1 . −→ 2 2 2 In each case, group element acts on 2 component spinor X(A,B) (X(A,B) is e.g. a ﬁeld operator),

(A,B) with 2 components Xab :

1 (1,0) (1,0) (2,0) 2 2 X XL = X1 , X 1 (“left-handed”) and ≡ 2,0 2,0 −

1 (0,1) (0,1) (0,2) 2 2 X XR = X 1 , X 1 (“right-handed”). ≡ 0,2 0, 2 − “Handedness”/chirality refers to eigenstates of helicity for massless particles (see later). [email protected] 38 Lorentz transformation of spinors: U(Λ)XL/RU †(Λ) = hL/R(Λ)XL/R (D acts on the 2 a, b components).

1,0 / 0,1 1,0 / 0,1 IJ(2 ) ( 2) θ IK(2 ) ( 2) eˆβ From explicit form of Lorentz group elements on page 22, hL/R(Λ) = e · e− · , where

(1,0) (1,0) (0,1) (0,1) Lorentz group generators for spinors: J 2 = 1σ , K 2 = I 1σ , J 2 = 1σ , K 2 = I 1σ , i 2 i i 2 i i 2 i i − 2 i

0 1 0 I 1 0 where σi are the Pauli σ matrices σ1 = , σ2 = − and σ3 = , which obey 1 0 I 0 0 1 −

1 1 1 Rotation group algebra: 2σi, 2σj = Iǫijk 2σk . Follows from J = J (A) + J (B) and K = I(J (A) J (B)) on page 29, i i i i i − i

1 (0) ( 2 ) 1 and Ji = 0 and Ji = 2σi from irreducible representation for spin j on page 17.

1 1 I σiθi σieˆiβ Explicit form of hL/R: hL/R = e 2 e∓2 . Note σ matrices are Hermitian.

Product of σ matrices: σiσj = δij + Iǫijkσk . Follows by explicit calculation.

Direct calculation of h : h = (cos θ + Iσ θˆ sin θ )(cosh β σ eˆ sinh β ). L/R L/R 2 i i 2 2 ∓ i i 2

2 xT IxT From product of σ matrices, T = 1 where T = σieˆi (T = σiθˆi). Then e = cosh x + T sinh x (e = cos x + IT sin x). [email protected] 39 b Write hL = h and XL = X which has components Xa = (X1, X2) , which transforms as 1. Xa′ = ha Xb .

b˙ Can also deﬁne spinor transforming with h : Use dotted indices for h , so 2. X′† = h X† . ∗ ∗ a˙ a∗˙ b˙

1T Then hR = h∗− . From explicit form of hL/R on page 38.

1 a˙ 1˙ 2˙ Use upper indices for h− , and XR has components X† = (X† , X† ), Dotted indices because h∗ is used.

a˙ 1T a˙ b˙ T a a transformation is 4. X′† = h∗− b˙ X† , where we deﬁne (h ) b = hb . a 1T a b Conjugate of this turns dotted indices into undotted indices, so 3. X′ = h− b X .

a a˙ a˙ a Conjugate of spinors: (X )† = X† , (Xa)† = Xa†˙ . Deﬁnition of X† in terms of X , Xa†˙ in terms of Xa.

b b˙ Check that if X transforms as transformation 1., X† transforms as transformation 2.: X′† = (X ) = (h X ) = h X†. a a˙ a˙ a′ † a b † ∗a˙ b˙ [email protected] 40 0 1 Spinor metric: ǫab = , ǫ = ǫab. Note ǫ ǫbc = δ c. ǫ is unitary matrix. 1 0 ab − ab a −

Pseudo reality: ǫac(σ ) dǫ = (σT )a . i c db − i b

T Follows by explicit calculation. Since σi∗ = σi , shows rotation group representation by σ matrices is real (see page 12).

1 a˙ Xa†˙ is 0, 2 , i.e. right-handed, like X† .

˙ Pseudo reality above implies ǫa˙c˙h dǫ = (h 1T )a˙ , i.e. h , h same up to unitary similarity transformation. c∗˙ d˙b˙ ∗− b˙ ∗ R

1 1 For unitary representations, follows because A = B† from deﬁnition on page 29, so conjugation makes , 0 0, . 2 → 2

1 This means dotted indices are for right-handed ( 0, 2 ) ﬁelds.

1 (Similarly, undotted indices are for left-handed ( 2, 0 ) ﬁelds.)

Metric condition: h d˙ǫ (h T )b˙ = ǫ . e∗˙ d˙b˙ ∗ f˙ e˙f˙

Follows from ǫa˙c˙h d˙ǫ = (h 1T )a˙ result above. Thus ǫ is the group’s invariant matrix g (“metric”) on page 12. c∗˙ d˙b˙ ∗− b˙ [email protected] 41 a ab a Raising and lowering of spinor indices: X = ǫ Xb . This is deﬁnition of X in terms of Xa.

b Follows that Xa = ǫabX . Same deﬁnition / behaviour for dotted indices.

a a ab ab c Check that if Xa transforms as transformation 1. on page 39, X transforms as transformation 3.: X′ = ǫ Xb′ = ǫ hb Xc.

ab c 1T a ab c 1T a bc a 1T a bc 1T a b From pseudo reality, ǫ hb ǫcd = h− d, or ǫ hb = h− b ǫ , so X′ = h− b ǫ Xc = h− b X .

a˙ ab Right-handed from left-handed fields: X† = ǫ Xb † . So all fields can be expressed in terms of left-handed fields.

Scalar from 2 spinors: XY = XaY = Y Xa = Y aX = X Y a = Y X . a − a a − a

a 1T a c b 1 a b c a a ab b X ′Y ′ = h− X h Y = h− h X Y . X, Y anticommute (spinor operators). X Y = X Y because ǫ ǫ = δ . a c a b c a b a − a ac − c

a a a˙ Hermitian conjugate of scalar: (XY )† = (X Ya)† = (Ya)†(X )† = Ya˙†X† = Y †X† . [email protected] 42 1 0 4-vector σ matrices σµ : σi = (σ ) , σ0 = (σ ) and σ = . ab˙ ab˙ i ab˙ ab˙ 0 ab˙ 0 0 1

4-vector σ matrices with raised indices: σµ ab˙ = ǫbcσµ ǫa˙d˙ , so σi ab˙ = (σ )ab˙ , σ0ab ˙ = (σ )ab˙ . cd˙ − i 0 Second result follows from deﬁnition by explicit calculation.

Inner product of 4-vector σ matrices: g σω σρ ba˙ = 2δ aδ b˙ . ρω ef˙ − e f˙

Outer product of 4-vector σ matrices: σν σρ ba˙ = 2gνρ . ab˙ −

Xaσµ Y b˙ is a 4-vector . ab˙ †

a b˙ Need to show X aσµ Y b˙ = Λµ Xaσν Y b˙. Since X aσµ Y b˙ = h 1T Xcσµ h 1T Y d˙, ′ ab˙ ′† ν ab˙ † ′ ab˙ ′† − c ab˙ ∗− d˙ † c d˙ need to show h 1 σµ h 1T = Λµ σν . Contracting with σρ ba˙ and using outer product of 4-vector σ matrices − a cd˙ ∗− b˙ ν ab˙ ˙ µρ 1 ρ ba˙ 1 c µ 1T d 1 ρ 1 µ 1T gives Λ = σ h− σ h∗− = tr σ h− σ h∗− , which is equivalent because σ matrices linearly independent −2 a cd˙ b˙ −2 (or multiply this by g σω and use inner product of 4-vector σ matrices). Check group property: ρω ef˙

µ ν 1 µ 1 ν 1T 1 ν 1 ρ 1T 1 µ ν Λ νΛ′ ρ = 2tr σ h− (Λ)σ h∗− (Λ) 2tr σ h− (Λ′)σ h∗− (Λ′) = 2σijhjk(Λ)σµklhli† (Λ)σmnhnp(Λ′)σρpqhqm† (Λ′). From ˙ ν µ ν 1 µ ba˙ 1 1 c 1T 1T d inner product of 4-vector σ matrices, σ σ = 2δ δ , so Λ Λ′ = σ h− (Λ′)h− (Λ) σ ˙ h∗− (Λ)h∗− (Λ′) mn µkl − ml kn ν ρ −2 a ρcd b˙ [email protected] 43 Convenient to put left- (X) and right-handed (Z) ﬁelds together as 4 component spinor:

(1,0) (1,0) (0,1) (0,1) ψT = C X 2 ,C X 2 ,C Z 2 ,C Z 2 , where C , C scalar constants. 1 1,0 1 1,0 2 0,1 2 0, 1 1 2 2 −2 2 −2

1σ 0 I 1σ 0 Lorentz group generators: J = 2 i and K = 2 i . i 0 1σ i 0 I 1σ 2 i − 2 i Follows from Lorentz group generators for spinors on page 38.

This is the chiral (Weyl) representation,

1 1 others representations from similarity transformation: Ji′ = V JiV − , Ki′ = V KiV − , X′ = V X etc.

Preferable to express in terms of left-handed ﬁelds only:

Xa Xa ψ = bc = b˙ , where Y = Z† is left-handed like X. (ǫ Y )† Y c † [email protected] 44 1 2.4.2 Spin 2 in general representations

Any 4 4 matrix can be constructed from sums/products of gamma matrices (next page): × 0 1 0 σ 0 σµ Gamma matrices (chiral representation): γ0 = I and γi = I i , or γµ = I . − 1 0 − σ 0 − σµ 0 − i

µ µ 1 Gamma matrix in diﬀerent representations related by similarity transformation γ′ = Vγ V − .

Anticommutation relations for γµ: γµ,γν = 2gµν . This is representation independent. { } Check by explicit calculation in chiral representation, use rotation group algebra on page 38.

1 0 Deﬁne γ = = Iγ0γ1γ2γ3. Check last representation-independent equality explicitly in chiral representation. 5 0 1 − − 1 0 X X Then P = 1(1 + γ )= projects out left-handed spinor: P a = a , L 2 5 0 0 L (ǫbcY ) 0 c †

1 0 0 Xa 0 similarly PR = 2(1 γ5)= projects out right-handed spinor: PR bc = bc . − 0 1 (ǫ Y )† (ǫ Y )† c c 2 Note PL/R are projection operators: PL/R = PL/R, PL/RPR/L = 0. [email protected] 45 Any 4 4 matrix is linear combination of 1, γµ,[γµ,γν], γµγ , γ . × 5 5 Because these are 16 non-zero linearly independent 4 4 matrices. × Non-zero because their squares, calculated from anticommutation relations on page 44, are non-zero.

Linearly independent because they are orthogonal if we deﬁne scalar product of any two to be trace of their matrix product:

tr[1γµ] = 0, because tr[γµ] = 0 in chiral representation, therefore in any other representation.

tr[1[γµ,γν]] = 0 by (anti)symmetry. From anticommutation relations, tr[1γµγ ] = tr[γ γµ] = tr[γµγ ] = 0 5 − 5 − 5

0 and also tr[1γ5] =tr[γ5] = 0 by commuting γ from left to right.

Next, tr[γρ[γµ,γν]] = 0: From anticommutation relations, γ2 = γ0γ1γ2γ3γ0γ1γ2γ3 = γ0 2γ1 2γ2 2γ3 2 = 1. 5 − − So tr[γρ[γµ,γν]] =tr[γ2γρ[γµ,γν]] = tr[γ γρ[γµ,γν]γ ] =tr[γ γρ[γµ,γν]γ ]. 5 − 5 5 5 5 To show tr[γργµγ ] = 0, ﬁrst consider case ρ = µ. Then γργµ = gρρ, and result is tr[γ ]=0. 5 ∝ 5

ρ µ 0 3 µ If e.g. ρ = 1, µ = 2, tr[γ γ γ5] = Itr[γ γ ] = 0. tr[γ γ5] = 0 already shown.

µ ν ρ µ ν ρ µ µ µ tr[[γ ,γ ]γ γ5] = 0 from anticommutation relations. tr[[γ ,γ ]γ5] = 0 because tr[γ γ γ5] = 0. Finally tr[γ γ5γ5] =tr[γ ]=0.

So all spinorial observables expressible as representation independent sums/products of gamma matrices. [email protected] 46 Lorentz group generators from γµ: J µν = I [γµ,γν]. −4

1 10 20 30 Explicit calculation in chiral representation, Ji = 2ǫijkJjk, K = (J ,J ,J ),

Lorentz group generators on page 43, Gamma matrices on page 44.

Similarity transformation on page 43 equivalent to similarity transformation on page 44.

Inﬁnitesimal Lorentz transformation of γµ: I[J µν,γρ]= gνργµ gµργν , − Follows from anticommutation relations for γµ on page 44 and Lorentz group generators from γµ above.

µ µ 1 µ ν µ Lorentz transformation of γ : D(Λ)γ D− (Λ) = Λν γ , i.e. γ transforms like a vector.

Agrees with inﬁnitesimal Lorentz transformation of γµ, because for inﬁnitesimal case,

LHS is (1 + 1Iω J ρσ)γµ(1 1Iω J ωη) = γµ 1Iω [γµ,J ρσ], and RHS is (δ µ + ω gµσ)γν = γµ 1ω (gµργσ gµσγρ). 2 ρσ − 2 ωη − 2 ρσ ν νσ − 2 ρσ −

µ µ 1 Lorentz transformation of vector: γ (Λp)µ = D(Λ) γ pµ D− (Λ) .

From Lorentz transformation of γµ above.

µ γ pµ 0 1 0 µ Reference boost: = D(L(p))γ D− (L(p)) . Equivalent of p = L(p)k on page 24, γ m = γ k . m − − µ

µ µ 1 0 1 In Lorentz transformation of vector, take p = k, Λ = L(q). Then γ q = D(L(q)) γ k D− (L(q)) = D(L(q)) γ m D− (L(q)). µ µ − [email protected] 47 0 1 Parity transformation matrix: β = Iγ0 = . Note β2 = 1. 1 0

µν µν 1 Pseudo-unitarity of Lorentz transformation : J † = βJ β , D† = βD− β .

0 0 0 i i i µ µ µ βγ β = γ = γ † and βγ β = γ = γ †, or βγ β = γ †. Then use Lorentz group generators from γ on page 46. − − − −

1 µν 1 µν Inﬁnitesimal D† is 1 Iω J † = β(1 Iω J )β. − 2 muν − 2 muν

Adjoint spinor: X = X†β . Allows construction of scalars, vectors etc. from spinors:

Covariant products: XY is scalar , XγµY is vector .

First case: X′Y ′ = X′†βY ′ = X†D†βDY .

1 2 1 1 From pseudo-unitarity of Lorentz transformation, D†β = βD− β = βD− , so X′Y ′ = X†βD− DY = X†βY = XY .

µ µ 1 µ µ Second case: X′γ Y ′ = X†D†βγ DY = X†βD− γ DY . Lorentz transformation of γ on page 46 can be rewritten

1 µ 1 µ ν µ ν µ µ ν µ ν D− γ D = Λ− ν γ = Λ νγ , so X′γ Y ′ = Λ νX†βγ Y = Λ νXβγ Y . µ Vanishing products: ψR/LψL/R and ψL/Rγ ψL/R , where ψL/R = PL/Rψ (see page 44 for PL/R).

Using γ5† = γ5 in chiral representation and PL/Rβ = βPR/L gives ψRψL = ψ†PLβPLψ = ψ†βPRPLψ = 0 (because PL/RPR/L = 0).

µ µ µ µ µ Similarly, using γ γ = γ γ , one has ψ γ ψ = ψ†P βγ P ψ = ψ†βγ P P ψ = 0. 5 − 5 L L R L R L [email protected] 48 Problems 1

1: If the parameterization αi of group elements U(α), chosen such that U(0) = 1, can be chosen such that the Abelian limit U(α)U(β) = U(α + β) is obeyed for βi = cαi, show that U(α) = exp[Itiαi]. Hint: Use the Abelian limit to write k α N α αi 1 x x U(α) = U N , take N , expand U N = 1 + Iti N + O N 2 and use e = k∞=0 k! and the binomial theorem N N N! k →N ∞k (x + y) = k=0 k!(N k)!x y − . − P

2: Show that theP symmetry transformation U(α)a† a† ′ ... 0 = D ′′ (α)D ′ ′′′ (α)a† ′′ a† ′′′ ... 0 and U(α) 0 = 0 is uniquely σ σ | i σσ σ σ σ σ | i | i | i satisﬁed by the condition U(α)aσ† U †(α) = Dσσ′ (α)aσ† ′ . Show that the matrices D furnish a representation of the group deﬁned by U (namely γ(α, β) in U(γ(α, β)) = U(α)U(β)).

3: Show that a semi-simple group’s matrix generators ti all obey tr[ti]=0. 4: Use the Lorentz group algebra I[J ρσ,J µν] = gσνJ ρµ gρµJ σν + gσµJ ρν + gρνJ σµ to derive the rotation group algebra 23 31− 12 − [Ji,Jj] = IǫijkJk, where (J1,J2,J3) = (J ,J ,J ). µ 5: Using the explicit result for the generator of 4-vector boosts, namely (Ki) ν = I(δ0µδiν +δiµδ0ν), show that the explicit cal- µ IK eˆβ µ 2 culation of a general boost is e− · = 1 IKieî sinh β (Kieî) (cosh β 1) . Hint: Show by explicit calculation ν − − − ν 3 the result (Kieî) = Kieî and then use it. h i c 3 c c 6: Define a field ψl −(x) = D p vlσ(x; p)aσ†(p), where vlσ(x; p) is such that ψl −(x) has the transformation property c 1 c (j) 1 U(Λ, a)ψ (x)U (Λ, a) = D ′ (Λ )ψ ′ (Λx + a). Show that D ′ (Λ)v ′ (p) = D ′ (W (Λ, p))v ′ (Λp), where v (p) = l − † R ll − l − ll l σ σσ − lσ lσ vlσ(0; p). For a massive particle, for reference momentum k = (0, 0, 0, 1) and for L a pure boost, show that this implies (j) (A) (B) that J ′ ∗v ′ (0) = J ′ v ′ (0) + J ′ v ′ (0). Hint: Take Λ = R and p = k. − σ σ ab σ aa a b σ bb ab σ 1 1 ( 2 ,0) 1 ( 2 ,0) 1 7: Show that Ji = 2σi and Ki = I 2σi. Hint: Use J = A + B and K = I(A B), and the irreducible representation 1 − for A = 2 and B = 0. 8: A left-handed 2-component spinor operator X transforms according to U(Λ)XU †(Λ) = h(Λ)X, where the 2 2 matrix I 1 σ θ 1 σ eˆ β 1T I 1 σ θ 1 σ eˆ β × h = e 2 i i e− 2 i i . Show that h∗− = e 2 i i e 2 i i (the equivalent matrix for right-handed spinors). Hint: The Pauli matrices are Hermitian. µ µ 1 µν ρ νρ µ 9: Show that γ (Λp) = D(Λ) γ p D− (Λ). Hint: Consider the infinitessimal case, which follows if I[J ,γ ] = g γ µ µ − gµργν. To prove the latter, use J µν = I [γµ,γν] and γµ,γν = 2gµν. −4 { } µ 0 10: Show that Xγ Y is a 4-vector, where X = X†β with β = Iγ . [email protected] 49 2.4.3 The Dirac field

1 1 1 1 (A,B) ( ,0) ( ,0) (0, ) (0, ) Group 4 possibilities for u together as 4 component spinor: uT = u 2 , u 2 , u 2 , u 2 . ab σ σ 1,0,σ 1,0,σ 0,1,σ 0, 1,σ 2 −2 2 −2

1 1 1 1 T (2,0) (2,0) (0,2) (0,2) Likewise, vσ = v1 , v 1 , v 1 , v 1 . − 2,0,σ − 2,0,σ 0,2,σ 0, 2,σ − − First 2 v components multiplied by ( 1)2A = 1 to remove it from massive irreducible ﬁeld on page 36. − −

X 3 Dirac ﬁeld: ψ(x)= a = d p eIp xu (p)a (p)+ e Ip xv (p)ac (p) . (ǫbcY ) (2π)3 · σ σ − · σ σ† c † R (Note D3p d3p for convention.) →

1 3 (3) 0 Anticommutation relations for spin : [aσ(p), a† (p′)]+ = (2π) δσσ δ (p p′) (i.e. no 2p factor). 2 σ′ ′ −

1 m m p dependence of spin 2 u, v: uσ(p)= p0 D(L(p))uσ(0) and vσ(p)= p0 D(L(p))vσ(0) . q q This is just the p dependence of u, v on page 26. [email protected] 50 1 1 1 1 1 1 (0,2) 1 (0,2) 1 (0,2) 1 (0,2) Condition on spin u, v: σ∗ v (0) = σi bb v (0) , σi σ σu (0) = σi bb u (0) . 2 −2 i σ′σ 0b σ′ 2 ′ 0b′ σ 2 ′ 0b σ′ 2 ′ 0b′ σ From conditions on u, v on page 32 and Lorentz group generators on page 43. Recall A = 1(J IK ) and B = 1(J + IK ). i 2 i − i i 2 i i

1 1+σ 1+σ Spin u, v relation: v1 or 2 σ(0) = ( 1)2 u1 or 2 σ(0) and v3 or 4 σ(0) = ( 1)2 u3 or 4 σ(0) . 2 − − − − −

1 1 1 1 ( 2 ,0) 1 ( 2 ,0) (0, 2 ) 1 (0, 2 ) 2 +σ 2 +σ From u, v relation on page 32, va0 σ (0) = ( 1) ua0 σ(0) and v0b σ (0) = ( 1) u0b σ(0). − − − −

( 1 ,0) Then multiply v 2 (0) by ( 1)2A = 1 as discussed on page 49. a0 σ − − Form of u, v: uT (0) = 1 (1 0 1 0), uT (0) = 1 (0 1 0 1), vT (0) = 1 (0 1 0 1), vT (0) = 1 ( 1010). σ=1 √2 σ= 1 √2 σ=1 √2 σ= 1 √2 2 −2 2 − −2 −

1 1 1 1 1 (0, 2 ) (0, 2 ) (0, 2 ) (0, 2 ) Solution to condition on spin 2 v is v0, 1 , 1 = v0, 1 , 1 and v0, 1 , 1 = v0, 1 , 1 = 0. Components for u constrained similarly. − 2 2 − 2 − 2 2 2 − 2 − 2

1 1 1 Use spin 2 u, v relation, and adjust normalizations of 2, 0 and 0, 2 parts individually. Massless u, v: uT (0) = 1 (0 0 1 0), uT (0) = 1 (0 1 0 0), vT (0) = 1 (0 1 0 0), vT (0) = 1 (0010). σ=1 √2 σ= 1 √2 σ=1 √2 σ= 1 √2 2 −2 2 −2

1 From allowed helicities for ﬁelds in given (A, B) representation on page 34, e.g. for 2, 0 ﬁeld such as neutrino, only 1 1 3 ,0 ,0 1 1 d p Ip x ( 2 ) Ip x ( 2 ) c allowed σ = 0 2 for particle and σ = 2 0 for antiparticle, i.e. ψL(x) = (2π)3 e · u 1 (p)a 1 (p) + e− · v 1 (p)a 1†(p) . − − 2 − 2 2 2 − R c T b˙ Majorana particle = antiparticle: aσ†(p)= aσ† (p) , so ψM = Xa, X† (i.e. Ya = Xa on page 49). [email protected] 51 2.4.4 The Dirac equation

Representation independent deﬁnition of spin 1 u, v: (Iγµp + m)u (p) = 0 and ( Iγµp + m)v (p)=0. 2 µ σ − µ σ

µ γ pµ 1 m For u and v, reference boost on page 46 gives I u (p) = D(L(p))βD− (L(p))u (p) = 0 D(L(p))βu (0), − m σ σ p σ q 1 last step from p dependence of spin 2 u, v on page 49.

In the chiral representation, and therefore any other representation, βu (0) = u (0) and βv (0) = v (0), σ σ σ − σ

µ µ so I γ pµ u (p) = m D(L(p))u (0) = u (p), last step from p dependence of spin 1 u, v again, likewise I γ pµ v (p) = v (p). − m σ p0 σ σ 2 − m σ − σ q

µ (c) Dirac equation: (γ ∂µ + m) ψl± (x)=0.

µ 1 Act on Dirac ﬁeld on page 49 with (γ ∂µ + m), then use representation independent deﬁnition of spin 2 u, v.

2 2 (c) Consistent with Klein-Gordon equation ∂ m ψ± (x)=0. −

ν ν µ (c) From page 26. To check, act on Dirac equation from left with (γ ∂ m): 0 = (γ ∂ m)(γ ∂ + m) ψ± ν − ν − µ

ν µ 2 (c) 1 ν µ 2 (c) µν 2 (c) = γ γ ∂ ∂ m ψ± = γ ,γ ∂ ∂ m ψ± = g ∂ ∂ m ψ± ν µ − 2{ } ν µ − ν µ − using anticommutation relations for γµ on page 44. [email protected] 52 2.4.5 Dirac ﬁeld equal time anticommutation relations

1 µ 1 µ Projection operators from u, v: ul σ(p)ul σ(p)= 0 ( Iγ pµ + m) , vl σ(p)vl σ(p)= 0 ( Iγ pµ m) . ′ 2p − ll′ ′ 2p − − ll′

m 1 Deﬁne Nll′ (p) = ul σ(p) ul′ σ(p) = p0 [D(L(p))uσ(0)]l [uσ† (0)D†(L(p))β]l′ from p dependence of spin 2 u, v on page 49.

m 1 So from pseudo-unitarity of Lorentz transformation on page 47, N(p) = p0 D(L(p))N(0)D− (L(p)).

1 Explicit calculation from the form of u, v on page 50 gives N(0) = 2 (β + 1) which is true in any representation.

1 0 1 So N(p) = 2p0 D(L(p)) Iγ m + m D− (L(p)), then use reference boost on page 46.

(3) Equal time anticommutation relations: [ψl(x, t),ψ† (y, t)]+ = δll δ (x y). l′ ′ −

d3p Ip (x y) Deﬁne R ′ =[ψ (x,t),ψ†′ (y,t)] = 3 e · − [u (p)[u(p)β] ′ + v ( p)[v( p)β] ′ ], ll l l + (2π) l σ l σ l σ − − l σ R using Dirac ﬁeld and anticommutation relations for spin 1 on page 49. Then in“v” term, take p p. 2 →−

d3p Ip (x y) 0 0 0 0 (3) From projection operators from u, v, R = 3 0 e · − Iγ p Iγ p + m + Iγ p + Iγ p m β = 1δ (x y). (2π) 2p − · · − − R [email protected] 53 2.5 External symmetries: bosons

d3p Ip x Ip x c Scalar boson field: ψ(x)= 3 1 e · a(p)+ e− · a †(p) . (2π)2 (2p0)2 R From general form of irreducible field on page 36, where u(p) = u(0) and v(p) = u(p), absorb overall u(0) into field.

1 1 µ d3p Ip x µ Ip x µ c 0 0 Vector boson ﬁeld (2, 2), spin 1: ψ (x)= 3 1 e · uσ(p)aσ(p)+ e− · vσ(p)aσ†(p) , uσ(0) = vσ(0) = 0 . (2π)2 (2p0)2 R (j) µ µ ν (j) (j) µ µ ν Conditions on u, v on page 32 (using A + B = J) give e.g. (Jk )σ′σuσ′ = (Jk) νuσ, so (Jk Jk )σ′σuσ′ = (JkJk) νuσ.

(j) (j) i i 0 But (Jk Jk )σ′σ = j(j + 1)δσ′σ and (JkJk) j = 2δ j,(JkJk) µ = 0 from Lorentz transformation of 4-vectors on page 22,

i i 0 i 0 so j(j + 1)uσ(0) = 2uσ(0), j(j + 1)uσ(0) = 0, i.e. j = 0 and uσ(0) = 0 or j = 1 and uσ(0) = 0. [email protected] 54 1 1 µ ν µ ν µν pµpν Projection operator for vector boson (2, 2), spin 1: uσ(p)uσ∗(p)= vσ(p)vσ∗(p)= g + m2 .

j i (1) j (1) j 1 2 From conditions on u, v, page 32, (J3) i u0(0) = J3 uσ′ (0). But J3 = 0, (J3) i = Iǫ3ji, so u0(0) = u0(0) = 0, σ′0 σ′0 − 1/2 (1) (1) 1/2 i.e. u0(0) = (2m)− (0, 0, 1, 0) by choice of normalization. J1 IJ2 gives u 1(0) = (2m)− ( 1, I, 0, 0)/√2. ± ± ± −

µ ν Similarly, vσ(0) = uσ∗ (0). Explicitly show that uσ(0)uσ∗(0) projects onto space orthogonal to time direction,

1000

µ ν 0100 µ µ ν i.e. u (0)u ∗(0) =  . But u (p) = L (p)u (0) (p dependence of u, v, page 26), σ σ 0010 σ ν σ  0000      µ µ ν α ρ so Π ν(p) = uσ(p)(uν∗)σ(p) = L ρ(p)Lν (p)Π α(0) projects onto space orthogonal to p,

i.e. Π2(p) = Π(p) because Π2(0) = Π(0) and pΠ(p)q = 0 for any q:

µ ν ν ρ α ν ν 1 ν ρ pµΠ ν(p)q = pνL ρ(p)Π α(0)Lν (p)q But L ρ(p)pν = (L− )ρ (p)pν = kρ, and kρΠ α(0) = 0 because ki = 0.

µ ν Finally, write q = αp + βΠ ν(p)q and show β = 1:

q2 = α2p2 + β2Πµ (p)qνΠ ρ(p)q , i.e. Πµ (p)qνΠ ρ(p)q = 1 (q2 α2p2). ν µ ρ ν µ ρ β2 −

But Πµ (p)qνΠ ρ(p)q = qνΠ ρ(p)q = qν(q αp ) 1 = 1 (q2 α2p2), i.e. 1 = 1 . ν µ ρ ν ρ ν − ν β β − β2 β [email protected] 55 Projection operator shows there is problem for m 0. From allowed helicities for fields in given → 1 1 (A, B) representation on page 34, can’t construct (2, 2) 4-vector field, where helicity σ = 0, from massless helicity σ = 1 particle. But can construct 4-component field: ± µ 3 Ip x µ Ip x µ c Massless helicity 1 field: A (x)= D p e · u (p)a (p)+ e− · v (p)a †(p) , where σ = 1. ± σ σ σ σ ± R Iθσ 1 Lorentz transformation of massless helicity 1 polarization vector: e− u (p) = Λ− u (Λp)+ ω(W, k)p . ± σ σ Simplest approach: take p dependence and rotation of u on page 34 to be true, gives uµ(k) (1,Iσ, 0, 0). σ ∝ Then M, N transformation on page 34 cannot be true, in fact Mu (0, 0, 1, 1) k and likewise for N. σ ∝ ∝

Iθσ Then D(W )uσ(k) = (1 + IθJ3 + IµM + IνN)uσ(k) = uσ(k)(1 + Iσθ) + (µ + Iσν)(0, 0, 1, 1) = ... = e [uσ(k) + ω(W, k)k].

Iθσ 1 1 Iθσ 1 Since D(Λ) = Λ, multiplying this from the left by e− Λ− D(ΛL(p))D(W − ) = e− Λ− L(Λp) gives result.

µ µ ν µ Lorentz transformation of massless helicity 1 ﬁeld: U(Λ)A (x)U †(Λ) = Λ A (Λx)+ ∂ α(x). ± ν

µ 3 Ip x µ Iθ(Λ,p)σ Ip x µ Iθ(Λ,p)σ c Use Lorentz transformation of uµ σ above in U(Λ)A (x)U †(Λ) = D p e · uσ(p)e− aσ(Λp) + e− · vσ (p)e aσ†(Λp) . R As for Poincar´etransformation for ﬁelds on page 25, up to gauge transformation.

This implies F = ∂ A ∂ A is Lorentz covariant, as expected: µν µ ν − ν µ it is antisymmetric, i.e. is (1, 0) or (0, 1) if F µν = I ǫµνλρF , so from σ = (A B), can have σ = 1. ±2 λρ ± − ± [email protected] 56 Brehmstrahlung: Adding emission of massless helicity j boson with momentum q 0 ± ≃ µ µ1 µ2 j to process with particles n with momenta p modiﬁes amplitude by factor u (q) η gnpn pn ...pn , n µ1µ2...µj σ n n pn q ∝ · where g is coupling of these bosons to fermion n, and η = 1 for outgoing / incoming particles.P n n ±

µ µ1 µ2 j Lorentz invariance condition: q η gnpn pn ...pn =0. µ1 n n pn q · P Can show this for massless helicity 1 boson: Lorentz transformation of polarization vector on page 55 ± implies amplitude not Lorentz invariant unless this is true.

Einstein’s principle of equivalence: Helicity 2 bosons have identical coupling to all fermions . ±

For soft emssion of graviton from process involving multiple fermions of momentum pn,

µ2 µ2 Lorentz invariance condition reads n ηngnpn = 0. But momentum conservation is n ηnpn = 0, so gn same for all particles. P P

Constraint on particle spins: Massless particles must have helicity 2 and 2. ≤ ≥−

µ2 µj Lorentz invariance condition can be written n ηngnpn ...pn = 0. P For j > 2 this overconstrains 2 2 processes, since momentum conservation alone = it depends on scattering angle only. → ⇒ [email protected] 57 2.6 The Lagrangian Formalism

2.6.1 Generic quantum mechanics

Lagrangian formalism is natural framework for QM implementation of symmetry principles.

Can be applied to canonical ﬁelds (e.g. Standard Model):

Fields ψl(x, t) behave as canonical coordinates, i.e. with conjugate momenta pl(x, t) such that

3 [ψl(x, t), pl (y, t)] = Iδ (x y)δll and [ψl(x, t),ψl (y, t)] =[pl(x, t), pl (y, t)] = 0 as usual in QM. ′ ∓ − ′ ′ ∓ ′ ∓

In practice, ﬁnd suitable pl(x, t) by explicit calculation of [ψl(x),ψ† (y)] . l′ ∓

Lagrangian formalism: Deﬁne action A [ψ, ψ˙]= ∞ dtL[ψ(t), ψ˙(t)] , where L is Lagrangian. −∞ R d ∂L ∂L Laws of physics obeyed when action is stationary, i.e. coordinates obey ﬁeld equations ˙ = . dt dψl ∂ψl

δL 3 ˙ ˙ Alternatively, deﬁne pl = ˙ and Hamiltonian H[ψ(t), p(t)] = d x pl(x, t)ψl(x, t) L[ψ(t), ψ(t)] . δψl − R δH Then action stationary when coordinates obey ﬁeld equationsp ˙l = . −δψl [email protected] 58 2.6.2 Relativistic quantum mechanics

3 Take L[ψ(t), ψ˙(t)] = d xL (ψ(x),∂µψ(x)), R where Lagrangian density L (x) is scalar so A = d4xL (x) is Lorentz invariant. R In practice, determine L from classical ﬁeld theory, e.g. electrodynamics,

˙ ∂L (x) then L (ψ(x),∂µψ(x)) = pl(x)ψl(x) H (ψ(x), p(x)) and pl from L : pl(x)= ˙ . − ∂ψl(x)

∂L ∂L Stationary A requirement gives ﬁeld equations ∂µ = (Euler-Lagrange equations), ∂(∂µψl) ∂ψl

1 e.g. Klein-Gordon equation for free spin 0 ﬁeld, Dirac equation for free spin 2 ﬁeld etc.

(some deﬁnition of derivative with respect to operator ﬁeld ψl must be given here).

A must be real. Let A depend on N real ﬁelds. Stationary real and imaginary parts of A 2N ﬁeld equations. → [email protected] 59 Noether’s theorem: Symmetries imply conservation:

A invariant under ψ (x) ψ (x)+ IαF [ψ; x] conserved current J µ(x), ∂ J µ = 0, for stationary A . l → l l → µ 4 µ If α made dependent on x, A no longer invariant. But change must be δA = d xJ ∂µα R ∂A so that δA = 0 when α constant: Ignore l label, use discrete lattice in 1-D of N points so e.g. ψ(x) ψi and δA = Fiαi. → i ∂ψi P ∂A ∂A and write Fi = (Ji Ji+1) (no sum) for i = 1,...,N 1, and FN = (JN K). ∂ψi − − ∂ψN −

∂A But symmetry of action gives Fi = 0, so K = JN+1. (Indices are cyclical, i.e. XN+1 = X1, X0 = XN etc.) i ∂ψi P ∂A δA dα Thus δA = i Fiαi = i(Ji Ji+1)αi = i Ji(αi αi 1). Returning to continuous case, δA = dx F (x) = dxJ(x) . ∂ψi − − − δψ(x) dx P P P R R Thus δA = d4xα∂ J µ. Now take A stationary: δA = 0 even though α depends on x, so ∂ J µ = 0. − µ µ R µ ∂L Explicit result for Noether current when L is invariant: J = Fl . ∂∂µψl

4 4 ∂L ∂L ∂L ∂L A = d xL so, for x-dependent α, δA = I d x Flα + ∂µ(Flα) . But 0 = δL = Fl + ∂µFl Iα, ∂ψl ∂∂µψl ∂ψl ∂∂µψl R R h i 4 ∂L 4 µ so δA = d xI Fl∂µα, then compare with δA = d xJ ∂µα (from above). ∂∂µψl R R 3 0 dF Conserved charge: F = d xJ obeys dt = 0. R Use ∂ J µ = 0 and d3x J = 0 because J vanishes at inﬁnity. µ ∇ R [email protected] 60 Scalar boson (0, 0): L = 1∂ ψ∂µψ 1m2ψ2 . scalar −2 µ − 2

Scalar boson ﬁeld on page 53 implies ψ(x,t), ψ˙(y,t) = δ(3)(x y), i.e. p = ψ˙, consistent with p from L on page 58. − l h i− Field equations on page 58 give Klein-Gordon equation ∂2 m2 ψ = 0 as required. Note ψ is a single operator. −

Dirac fermion (1, 0) + (0, 1): L = ψ (γµ∂ + m) ψ . 2 2 Dirac − µ Recall ψ is a column of 4 operators, and covariant quantities on page 47.

(3) Recall equal time anticommutation relations on page 52, [ψ (x,t),ψ†′ (x,t)] = δ ′ δ (x y), so p = ψ†, l l + ll −

µ consistent with pl from L on page 58. Field equations give Dirac equation (γ ∂µ + m) ψ = 0 as required.

Vector boson (1, 1), spin 1: L = 1F F µν 1m2ψ ψµ, where F = ∂ ψ ∂ ψ . 2 2 spin 1 vector −4 µν − 2 µ µν µ ν − ν µ

1 1 From vector boson ﬁeld and projection operator for vector boson (2, 2), spin 1, on pages 53 and 54,

ψ (x,t), ψ˙ (y,t) + ∂ ψ0(y,t) = δ δ(3)(x y), i.e. conjugate momentum to ψ is p = ψ˙ + ∂ ψ0 = F i0, i j j ij − i i i i h i− consistent with p from on page 58. ψ is auxilliary ﬁeld because p = ∂L = F = 0. l L 0 0 ∂ψ˙ 0 00

Also ∂ ψµ = 0 and Klein-Gordon equation ∂2 m2 ψµ = 0, which is found from ﬁeld equations on page 58. µ − [email protected] 61 2.7 Path-Integral Methods

Follows from Lagrangian formalism. Assume H is quadratic in the pl.

Gives direct route from Lagrangian to calculations, all symmetries manifestly preserved along the way.

Can work in simpler classical limit then return to QM later.

Result is that bosons described by ordinary numbers, fermions by Grassmann variables.

LSZ reduction gives S-matrix from vacuum matrix elements of-time ordered product of functions of ﬁelds,

0, out T ψ (x ),ψ (x ),... 0, in dψ (x)ψ (x )ψ (x )...eIA [ψ] h | lA A lB B | i x,l l lA A lB B given by path integral as 0, out 0, in = IA [ψ] . R x,l dψl(x)e h | i Q R Q Contribution mostly from field configurations for which A is minimal, i.e. fluctuation around classical result.

Noether’s theorem again (see page 59): A invariant under ψ (x) ψ (x)= ψ (x)+ IαF [ψ; x]. l → l l l So if α dependent on x, dψ (x) exp[IA ] dψ (x) exp I A d4x∂ J µ(x)α(x) x,l l → x,l l − µ R R R assuming measure dψQ(x) invariant. This is justQ change of variables, so ∂ J µ(x) = 0. x,l l h µ i Q [email protected] 62 2.8 Internal symmetries

Consider unitary group representations.

Unitary U(N): elements can be represented by N N unitary matrices U (U †U = 1). × Dimension d(U(N))= N 2.

2 2 2N degrees of freedom in complex N N matrix, U †U = 1 is N conditions × or N 2 Hermitian N N matrices: N diagonal reals, N 2 N oﬀ-diagonal complexes but lower half conjugate to upper. × −

Special unitary SU(N): same as U(N) but U’s have unit determinant (det(U) = 1).

Thus tr[ti] = 0, i.e. group is semi-simple. d(SU(N))= N 2 1. − Fundamental representation denoted N.

1 1 Normalization of fundamental representation: tr[titj]= 2δij (i.e. C(N)= 2). [email protected] 63 U(1) (Abelian group): elements can be represented by phase eIqα.

One generator: the real number q (the charge).

1 SU(2): Fundamental representation denoted 2, spin 2 representation of rotation group.

σi SU(2) is actually the universal covering group of rotation group. 3 generators ti = 2 ,[ti, tj]= Iǫijktk.

Adjoint representation denoted 3. C(3)=2.

ǫ ǫ = (d(SU(2)) 1)δ = 2δ . jki lki − jl jl

σi∗ σi 2 representation is real, 2 = 2 (i.e. = U U †, pseudoreal), and g = ǫ and δ . − 2 2 αβ αβ αβ

λi SU(3): 8 generators 2 , structure constants fijk.

Fundamental representation 3: λ are 3 3 Gell-Mann matrices. Adjoint representation 8. C(8)=3. i × 3 representation is complex, 3 = 3. 6 Example: 3 3 = 8 + 1, i.e. quark and antiquark can be combined to behave like gluon or colour singlet. × [email protected] 64 2.8.1 Abelian gauge invariance

Global gauge invariance: Consider complex fermion / boson ﬁeld ψl(x), arbitrary spin.

µ Each operator in Lfree is product of (∂µ)ψl with (∂ )ψl†, invariant under U(1) transformation ψ eIqαψ (whence ∂ ψ eIqα∂ ψ ) l → l µ l → µ l if α independent of spacetime coords.

Local gauge invariance: Find L invariant when α = α(x). Leads to renormalizable interacting theory.

In L , ∂ ψ ∂ eIqαψ = eiqα ∂ ψ + Iq(∂ α)ψ = eIqα∂ ψ , ∴ replace ∂ by 4-vector “derivative” D , free µ l → µ l µ l µ l 6 µ l µ µ h i Iqα µ such that D gauge transformation D ψ e D ψ . Then D ψ (D ψ )† invariant. µ µ l → µ l µ l l Simplest choice: D ∂ is 4-component ﬁeld: µ − µ Covariant derivative: D = ∂ IqA (x). µ µ − µ Gauge transformation: A A + ∂ α whenever ψ eIqαψ . µ → µ µ l → l

iqα iqα Write transformation as A A′ (α). Require D ψ D′ e ψ = e D ψ , µ → µ µ l → µ l µ l

Iqα Iqα i.e. e ∂ ψ Iq(∂ α)ψ IqA′ ψ = e [∂ ψ IqA ψ ], so A′ = A + ∂ α. µ l − µ l − µ l µ l − µ l µ µ µ [email protected] 65 Use Dν to ﬁnd invariant (free) Lagrangian for Aµ, quadratic in (∂ν)Aµ:

From D gauge transformation, D D ...ψ eIqαD D ...ψ . Products D D . . . contain spurious ∂ s, but µ µ ν l → µ ν l µ ν ρ F from D : qF = I [D , D ], where electromagnetic ﬁeld strength F = ∂ A ∂ A . µν µ µν µ ν µν µ ν − ν µ [D ,D ] ψ = [∂ ,∂ ] +Iq([∂ ,A ] [∂ ,A ]) q2 [A ,A ] ψ . µ ν l µ ν µ ν − ν µ − µ ν l =0 =0 | {z } | {z } Fµν is gauge invariant.

Iqα Iqα F ψ F ′ e ψ = e F ψ , i.e. F ′ = F , or F F . Also check explicitly from F = ∂ A ∂ A . µν l → µν l µν l µν µν µν → µν µν µ ν − ν µ Conversely, choose Aµ to be massless helicity 1 ﬁeld, whose Lorentz transformation on page 55 ± implies Lorentz invariant free Lagrangian for Aµ must be gauge invariant.

F in representation of U(1): Since F F , F transforms in adjoint representation of U(1). µν µν → µν µν Example: QED Lagrangian for fermions: L = ψ (γµD + m) ψ 1F F µν. Dirac, QED − µ −4 µν µ LDirac, free+Iqψγ Aµψ | {z } Interactions due to IqψγµA ψ. Most general Lagrangian locally gauge invariant under U(1) (ψ eIqαψ, A A + ∂ α), µ → µ → µ µ assuming P,T invariance and no mass dimension > 4 terms (Wilson: no contribution). [email protected] 66 2.8.2 Non-Abelian gauge invariance

µ Global gauge invariance: Each term in Lfree proportional to (∂ )ψl γ(∂µ)ψ† , γ = 1,...,N. l′ γ

Then L invariant under ψ U ψ , where U = exp[Iα t ], α spacetime independent. free l γ → γδ l δ i i i

So ψγ is in fundamental representation of group G =SU(N) formed by matrices Uγδ, i = 1,...,d(G).

Local gauge invariance: Spacetime derivatives in Lagrangian appear as

Covariant derivative: D = ∂ IA (x) with Aµ = Aµt , µ µ − µ i i t contain couplings, Aµ for i = 1,...,d(G) are (for) massless helicity 1 gauge ﬁelds. i i ±

To achieve D ψ UD ψ, require µ → µ

Transformation of covariant derivative: D UD U † , which requires µ → µ

Transformation of gauge ﬁelds: A UA U † I (∂ U) U † . µ → µ − µ [email protected] 67 Non-Abelian ﬁeld strength: T F i = F = I[D , D ]= ∂ A ∂ A I[A ,A ]. i µν µν µ ν µ ν − ν µ − µ ν

F in adjoint representation: F UF U † . µν µν → µν In inﬁnitesimal case, F it F i [(1 + Iα t )t (1 Iα t )] = F i [t + Iα [t ,t ]] i αβ → k k i − k k αβ i k k i αβ

= F i [t + Iα (IC )t ] = F i t + Iα (tk )t = F it δ + Iα (tk ) = F it U = U F t , i.e. F i U F j. i k kij j αβ i αβ k ji j αβ j αβ ji k ji j αβ ji ij j i αβ → ij

Example: QCD Lagrangian for fermions:L = ψ (γµD + m) ψ 1F i F i µν. Dirac, QCD − α αβ µ β − 4 µν More general result is 1g F i F j µν, but can always diagonalize and rescale so g δ . −4 ij µν ij → ij [email protected] 68 2.9 The Standard Model

Symmetry of vacuum is G=SU(3) U(1) gauge group. colour× e.m. SM: At today’s collider energies, some “hidden” (broken) symmetries become apparent:

G=SU(3) SU(2) U(1) . colour× weak isospin× weak hypercharge Table 2.9.1: SM fermions and their SU(3) SU(2) U(1) representations, written as (SU(3) rep.,SU(2) rep.,U(1) hyper- C× L× Y C L Y charge = generator / [coupling Y ]). The SU(3) charges (3 for quarks, none for leptons) are not shown but, since SU(2) ≡ C L is broken, particles diﬀering only in T3 (component of weak isospin SU(2)L) are shown explicitly, namely uL / νe (T3 = 1/2) 1 + and d / e (T = 1/2). Recall ψ = (1 γ )ψ. Note e.g. u annihilates u− and creates u , and ν is left-handed. L L 3 − L/R 2 ± 5 L L R e

Names Label Representation under SU(3)C SU(2)L U(1)Y × × 1 Quarks QL =(uL,dL) (3, 2, 6 ) u† (3, 1, 2 ) R − 3 † 1 dR (3, 1, 3 ) 1 Leptons EL =(νe,eL) (1, 2, ) − 2 † eR (1, 1, 1)

λi SU(3): gµ = giµ gs 2

b σi Gauge ﬁelds are written as SU(2): Aµ = Aiµ g 2

b U(1): Bµ = Bµ g′Y .

b [email protected] 69 K We have only discussed the “1st generation” of fermions, in fact (EL, eR† , QL, uR† , dR† ) , K = 1, 2, 3.

1 2 3 1 2 3 1 2 3 1 2 3 where (e , e , e ) = (e, µ, τ), (νe , νe , νe ) = (νe, νµ, ντ ), (u , u , u ) = (u, c, t) and (d , d , d ) = (d, s, b).

Allow mixing between particles of diﬀerent generations with same transformation properties.

From Table 2.9.1, can construct the full Lagrangian by including all renormalizable invariant (1, 1, 0) terms.

K µ K These are all possible terms of form ψ γ Dµψ :

K K L = Q γµ[∂ I(g + A + B )]QK + uKγµ[∂ I(g + B )]uK + d γµ[∂ I(g + B )]dK. quark L µ − µ µ µ L R µ − µ µ R R µ − µ µ R b b b b b b b K L = E γµ[∂ I(A + B )]E K + eKγµ[∂ IB ]eK. lepton L µ − µ µ L R µ − µ R b b b K K More general ψ γµD RKM ψM for some constant matrix R not allowed, gives terms ψ γµ∂ ψM for K = M. µ µ 6

L = 1F i (g)F i µν(g) 1F i (A)F i µν(A) 1F (B)F µν(B). spin 1 −4 µν − 4 µν − 4 µν b b b b b b Recall only real part of L to be taken. [email protected] 70 2.9.1 Higgs mechanism

Mass terms mψ ψ are all (1, 2, 1) violate gauge symmetry and thus renormalizability. L/R R/L ±2 →

Instead introduce Yukawa coupling λφHψL/RψR/L which is (1, 1, 0), i.e. invariant (thus renormalizable), so φ is (1, 2 = 2, 1) scalar ﬁeld, then hide (“break”) symmetry so that λ 0 φ 0 = m. H 2 h | H| i

LHiggs = Lpure Higgs + LHiggs fermion . −

T + 0 T 0 + Writing D = ∂ I(A + B ) and SU(2) components φ = (φ , φ ) = (φ , φ ), (ǫφ† ) = (φ †, φ †), µ µ − µ µ L H H 1 H 2 H H H H − H

2 1 b bµ mH λ 2 L = (D φ )†D φ V (φ ), Higgs potential V (φ )= φ† φ + (φ† φ ) pure Higgs −2 µ H H − H H 2 H H 4 H H

KM K M KM K M KM K M LHiggs fermion = Ge E L aφH aeR Gu QL a(ǫφH)a†uR Gd QL aφH adR . − − − − All three terms are (1, 1, 0), i.e. invariant. Consider e.g. second term: From table 2.9.1, Y = 1 1 + 2 = 0. −6 − 2 3

1 I σiαi Write SU(2)L transformation of φH as φH′ a = UabφH b (U = e 2 from page 63),

1 T T T 1 so (ǫφ′ )† = (ǫφ )†U − ∗ from ǫ σ ǫ = σ∗ . (Same transformation as φ : φ′ = φ U = φ U − ∗.) H a H b ba ac cd db − ab H H a H b ba H b ba

Also QL′ a = UabQL b, so QL′† a = Uac∗ QL† c, so QL′† a(ǫφH′ )a† = QL† a(ǫφH )a† . Note uR is an SU(2)L singlet. [email protected] 71 2 mH > 0: Stationary L (vacuum) occurs when all ﬁelds vanish.

Spontaneous symmetry breaking (SSB): m2 < 0: tree level vacuum obeys ∂V (φH ) = 0 H ∂φ φH =φH0

m2 = φ2 = v2 = | H |. From Higgs potential on page 70. ⇒ | H0| λ

σi 0 Inﬁnite number of choices for φ = 0 φ 0 . Take general φ = eIvξi(x) 2 . 0H h | H| i H v + η(x)

Vacuum taken as ξi = η = 0, no longer invariant under symmetry transformations.

= m2 W W µ 1m2 Z Zµ , where m = v g2 + g 2 , m = v g = m cos θ , cos θ = g , Lgauge mass W µ† 2 Z µ Z 2 ′ W 2 Z W W 2 2 − − √g +g′ p W = 1 (A IA ), Z = cos θ A sin θ B . µ √2 1 µ − 2 µ µ W 3 µ − W µ

2 2 1 σi 1 0 1 g(A1 µ + IA2 µ) v From Lpure Higgs. Start with Lgauge mass = gAi µ + g′Bµ = ′ . −2 2 2 v −2 gA3 µ + g Bµ 2 −

L = 1 ∂µW ν ∂νW µ 2 1 ∂µZν ∂νZµ 2 1 ∂µAν ∂νAµ 2 , gauge dynamic −2| − | − 4| − | − 4| − | where Aµ = sin θW A3 µ + cos θW Bµ must be massless photon.

From L . Start with L = 1 ∂µAν ∂νAµ 2 1 ∂µBν ∂νBµ 2. spin 1 gauge dynamic −4| i − i | − 4| − | [email protected] 72 In Lquark and Llepton on page 69, between left-handed fermions:

e 1 + Y A + g cos θ g Y sin θ Z g W 2 µ 2 W ′ W µ √2 µ† (Aµ + Bµ)L = g − 1 g , W e + Y A + cos θ g′Y sin θ Z √2 µ −2 µ −2 W − W µ ! b b between right-handed: (A + B ) = eY A + ( g′ sin θ Y )Z , where e = g sin θ = g′ cos θ . µ µ R µ − W µ W W So charge/e is Q = T +bY , i.e.b charge of u , d , ν , e is 2, 1, 0, 1 and of u , d , e is 2, 1, 1. 3 L L e L 3 −3 − R R R 3 −3 − [email protected] 73 K L = eKmKM eM uKmKM uM d mKM dM , where mKM = GKM v . fermion mass − L e R − L u R − L d R ψ ψ

From LHiggs fermion on page 70. −

Can always transform uK = AKM uM , likewise for u , d , d , ν , e , e . R ′ uR R L L R e L R

K µ K A matrices must be unitary so that kinetic terms retain their previous forms, uR ′γ ∂µuR ′ etc.

Choose A matrices such that new mass matrices m = A m A etc. diagonal, entries mK : u′ uL u u† R u ′

K K K K K K K K K L = e ′m e ′ u ′m ′u ′ d ′m ′d ′. fermion mass K − L e R − L u R − L d R P K K µ K K µ K ′ µ KN N K µ K Then LW fermion dL γ WµuL + eL γ Wµνe = dL γ Wµ(V †) uL ′ + eL ′γ Wµνe ′′ − ∝ Ig 1 (proportionality constant is ), where CKM matrix V = Au A− . −√2 L dL

K 1 KN N Analogous leptonic matrix absorbed into νe ′′ = AeLAν−e νe ′.

K K (In contrast to uL , any combination of νe is mass eigenstate because mass matrix is zero.) [email protected] 74 2.9.2 Some remaining features

K Neutrino mass by adding to LHiggs fermion a term GνeE a (ǫφH)a†νeR νemνeνeR, where νeR is (1, 1, 0). − − →− Expect m v to be similar order of magnitude to quark and charged lepton masses. νe ∼ Also allowed SM invariant term 1νT M ν , can only come from higher scale symmetry breaking, −2 eR R eR so M v m , i.e. no right handed neutrinos at low energy. R ≫ ∼ νe 2 T mνe T 1 T 0 mνe νe 1 T 0 νe′ MR Gives Lneutrino mass = 2(νe νeR) 2(νe′ νeR′ ) − , − m M ν ≃− ν′ νe R eR 0 MR ! eR

2 mνe mνe i.e. seesaw mechanism: mass of (almost) left handed ν′ (i.e. mν suppressed by ). MR e e MR [email protected] 75 Invariance with respect to parity P , charge conjugation C and time reversal T transformations.

CPT conserved, but CP -violation due to phases in CKM matrix.

θ κλρσ i i CP and P violating terms 64π2 ǫ FκλFρσ allowed, but are total derivatives and therefore non-perturbative.

Current observation suggest θ consistent with zero (no CP violation in QCD).

Cancelled by (harmless) anomaly (subsubsection 2.9.4) of global symmetry ψ eIγ5αf ψ when α = 1θ, f → f f −2 Iθ P but this introduces unobserved CP violating phase e− on quark masses.

Peccei-Quinn mechanism: where (ǫφH)† in LHiggs fermion on page 70 is replaced with second Higgs, − which transforms differently to first Higgs and can soak up this phase at least at some GUT scale. [email protected] 76 2.9.3 Grand unification

SU(2) SU(3) U(1)Y L C Suppose SM uniﬁes to single group G at scale MX, then t (diagonal), ti , tj are generators of G.

Tracelessness requires sum of Y values (=elements of tU(1)Y ) to vanish, which is the case from Table 2.9.1.

SU(2) 2 SU(3) 2 U(1)Y 2 L c Normalization of generators as on page 14, so tr[t ] =tr[ti ] =tr[tj ], so

2 2 5 2 2 3 g (M )= g (M )= g′ (M ) (after dividing by 2 no. generations). Implies sin θ (M )= from page 71. s X X 3 X × W X 8 Within couplings’ exp. errors, unification occurs (provided N = 1 SUSY is included) at M = 2 1016 GeV. X × To find simplest unification with no new particles, note SM particles are chiral, require complex representations:

In general, deﬁne all particles fL to be left-handed, then antiparticles fR = fL† are right-handed.

Then if fL in representation R of some group G, fR is in representation R.

If fL, fR equivalent (have same transformation properties), then R = R, i.e. pseudoreal representation.

SM particles f = (E , e† , Q , u† , d† ) require complex representation because f = f † inequivalent, e.g. 3 = 3. L L R L R R R L 6

Pseudoreal representation possible if particle content enlarged to f F so that F , F = F † equivalent. L → L L R L e.g. in SO(10), can ﬁt 15 particles of each generation into real 16 representation, requires adding 1 νeR. [email protected] 77 SU(5) is simplest uniﬁcation.

Since SU(3) SU(2) U(1) SU(5), all internal symmetries accounted for by fermions ψ with α = 1, ..., 5. × × ⊂ α 0 0 00000 λi 0 0 00000  2    SU(3) SU(3)C SU(2) SU(2)L Choose ti = gs 0 0 = ti , i = 1, ..., 8, ti = g 00000 = ti , i = 1, ..., 3.      00000   0 0 0 σi       00000   0 0 0 2      U(1) generator must commute with generators above and be traceless. Tentatively take  1 3 00 0 0 0 1 0 0 0  3  U(1) 1 U(1)Y t = 2g′ 0 0 3 0 0 = 2t .  1   0 0 0 2 0   − 1   000 0   −2  Then fermions form 5 (fundamental) and 10representations of SU(5): 1 3 2 1 1 dR 0 uR uR uL dL 2 − 0 u1 u2 d2  dR   R L L  3 3 3 d , 0 uL dL .  R     e   0 eR   L   —“   ν   0   eL     −    (Note: all particles left-handed, 1,2,3 superscripts are colour indices, 10 matrix is antisymmetric).

In SO(10), 1 generation ﬁts into 16 = 1 + 5 + 10, and 1 is identiﬁed with right-handed neutrino. [email protected] 78 2.9.4 Anomalies

Gauge anomalies modify symmetry relations (Ward identities), spoils renormalizability and maybe unitarity.

4 Anomaly occurs because A = d xL respects symmetry, but not measure x,l dψl(x)= d[ψ]d[ψ]d[A]. R R Q Relevant example: Let A be invariant under ψα = Uk αβψβ, where Uk = exp[Iγ5αtk] is chiral symmetry (global) and each ψα is a Dirac ﬁeld.

Problem: although A is invariant, measure is not: Resulting change in path-integral d[ψ]d[ψ]d[A] exp[IA ]

µν ρσ R is “as if” L changes by αJ [A], where J = 1 ǫ F F tr[ t , t t ]. k k −16π2 µνρσ i j { i j} k Noether’s theorem on page 61: d[ψ]d[ψ]d[A] exp[IA ] d[ψ]d[ψ]d[A] exp[I A + d4xα(x)[J ∂ J µ(x)] ], → k − µ k R µ µν Rρσ R i.e. conservation violation: ∂ J (x) = 1 ǫ F F tr[ t , t t ] ( means no A integration). h µ k iA −16π2 µνρσ i j { i j} k hiA µ µν ρσ Anomalous (non-classical) triangle diagrams between Jk , Fi and Fj modify Ward identities.

Physical theories must be anomaly free (i.e. tr[ t , t t ] cancel), e.g. real representations: tr[ t , t t ]=0. { i j} k { i j} k

T T T tr[ t∗,t∗ t∗] =tr[ ( Ut U †), ( Ut U †) ( Ut U †)] = tr[ t ,t t ]. But tr[ t∗,t∗ t∗] =tr[ t ,t t ] =tr[ t ,t t ]. { i j } k { − i − j } − k − { i j} k { i j } k { i j } k { i j} k

SM is anomaly free. SM in 10 + 5 of SU(5), in real representation 16 of SO(10) (thus tr[ t ,t t ]5 = tr[ t ,t t ]10). { i j} k − { i j} k [email protected] 79 3 Supersymmetry: development

3.1 Why SUSY?

Attractive features of SUSY:

1. Eliminates ﬁne tuning in Higgs mass.

2. Gauge coupling uniﬁcation.

3. Radiative electroweak symmetry breaking: SUSY = Higgs potential on page 70 with m2 < 0. ⇒ H 4. Excess of matter over anti-matter (large CP violation, not in SM) possible from SUSY breaking terms.

5. Cold dark matter may be stable neutral lightest SUSY particle (LSP) = gravitino / lightest neutralino.

6. Gravity may be described by local SUSY = supergravity. [email protected] 80 SM is accurately veriﬁed but incomplete — e.g. does not + cannot contain gravity,

1/2 18 so must break down at / before energies around Planck scale M = (8πG)− = 2.4 10 GeV. P × In fact, SM cannot hold without modiﬁcation much above 1 TeV, otherwise we have

Gauge hierarchy problem: Since v = 0 φ 0 = 246GeV and λ = O(1), m = √λv = O(100) GeV . |h | H| i| | H| | |

If ΛUV > O(1)TeV, ﬁne tuning between

2 2 ∆mH from quantum loop corrections (Fig. 3.1) and tree level (bare) mH:

2 2 λφ 2 κt 2 ∆m = Λ 3 | | Λ + ... . H 8π2 UV − 8π2 UV colour“3′′ smaller terms φ |{z} ψ |{z}

φH φH

2 2 (a) Fermion ﬁeld ψ, Lagrangian term κψφH ψψ, giving 1-loop contribution to (b) Boson ﬁeld φ, Lagrangian term λφ φH φ , giving 1-loop contribution to 2 − λ | | | | |κψ | 2 φ 2 Higgs mass of 2 Λ . Higgs mass of 8π2 ΛUV. − 8π UV 2 Figure 3.1: Fermion and boson contributions to Higgs mass parameter mH .

λ (Largest from φ φ 4 and top quark (κ 1 because m v).) − 4 | H| t ≃ t ≃ 2 No similar problem for fermion and gauge boson masses, but these masses affected by mH. [email protected] 81 Avoid fine tuning by taking Λ 1 TeV, i.e. modify SM above this scale. UV ∼ One solution: Higgs is composite of new fermions bound by new strong force at Λ 1 TeV difficult. UV ≃ → Alternatively, forbid bare m2 φ 2 term by some new symmetry δφ = ǫ something. H| H| H × Various choices for “something” bosonic (leads to “little Higgs” models, extra dimensions).

IǫQa IǫQa For a standard symmetry, “something” would be I [Q , φ ], i.e. φ e φ e− . a H H → H 1 Try “fermionic” generator Qa, which must be a 2, 0 spinor (so ǫ a spinor of Grassmann variables). [email protected] 82 µ Relation with momentum: Q , Q† = 2σ P . { a b˙ } ab˙ µ

1 1 1 1 µ Q , Q† is , 0 0, = , from triangle inequality. Only candidate is P (see Coleman-Mandula theorem later). { a b˙ } 2 × 2 2 2 Lorentz invariance requires combination σµ P , factor 2 comes from suitable normalization of Q . ab˙ µ a

Note Q , Q† = 0 because for any state X , { a b˙ } 6 | i

2 2 X Q , (Q )† X = X Q (Q )† X + X (Q )† Q X = (Q )† X + Q X 0. If equality holds for all X , Q = 0. h |{ a a }| i h | a a | i h | a a| i a | i a| i ≥ | i a

At least one of Pµ non-zero on every state, so Qa aﬀects every state, not just Higgs, i.e. every particle has a superpartner with opposite statistics and spin diﬀerence of 1/2, together called a supermultiplet. This fermion-boson symmetry is supersymmetry.

For every fermion ﬁeld (component) ψ with κ φ ψ ψ , introduce boson ﬁeld φ with λ φ 2 φ 2. f − f H f f f − f | H| | f | λ κ 2 From Fig. 3.1, contribution of this supermultiplet (component) to ∆m2 is ∆m2 = f Λ2 | f | Λ2 . H H 8π2 UV − 8π2 UV boson fermion

Just requiring fermion-boson symmetry guarantees λ = κ 2, and ∆m2 = 0 (+ ﬁnite| {z terms)} | to{z all orders.} f | f | H [email protected] 83 3.2 Haag-Lopuszanski-Sohnius theorem and SUSY algebra

Reconsider symmetries: So far assumed generators are bosonic. Now generalize to include fermionic ones.

Generalize additive observable on page 5 to Q = Qσσ′aσ† aσ′.

(If Qσσ′ are components of Hermitian matrix, unitary transformation of particle states gives back original result.)

Since Q is bosonic, a† , a† both bosons or both fermions, i.e. Qσσ =0if a† bosonic, a† fermionic, or vice versa. σ σ′ ′ σ σ′

SUSY: Allow for Q’s containing fermionic parts to also be generators of symmetries that commute with S-matrix.

For convenience, distinguish between fermionic and bosonic parts of any Q.

Fermionic Q = Qσρaσ† aρ + Rσρaρ†aσ, where σ sums over bosonic particles, ρ over fermionic particles.

Such a generator converts bosons into fermions and vice versa.

E.g. action of Q on 1 fermion + 1 boson state using (anti) commutation relations on page 4 gives

Qa†′ a† ′ 0 = Q ′ a† a† ′ 0 + R ′ a†a†′ 0 . ρ σ | i σρ σ σ | i σ ρ ρ ρ | i But as a symmetry implies there are fermions and bosons with similar properties. [email protected] 84 Identify symmetry generators ti also with fermionic Q.

Graded parameters α , β obey α β = ( 1)ηiηj β α , where grading η = 0(1) for complex (Grassmann) α . i j i j − j i i i Graded generator t obeys α t = ( 1)ηiηj t α , where η = 0(1) for bosonic (fermionic) generator t . j i j − j i i i Iα t Iα t For transformations O e i iOe− i i to preserve grading of any operator O, α has same grading as t . → i i

Graded Lie algebra: [( 1)ηiηj t t t t ]= IC t . − i j − j i ijk k Repeating steps in derivation of Lie algebra on page 8 gives

1IC α β t = 1 [α t β t β t α t ] = 1 [( 1)ηiηj α β t t α β t t ]. 2 ijk i j k 2 i i j j − j j i i 2 − i j i j − i j j i

Fermionic generators Qi: U(Λ)QiU †(Λ) = Cij(Λ)Qj, so Qi furnishes representation of Lorentz group.

(A,B) (A,B) (A) (A,B) (A,B) (B) (A,B) Choose Qi = Q in (A, B) representation: [A, Q ]= J Q and [B, Q ]= J Q . ab ab − aa′ a′b ab − bb′ ab′

Anticommutators of fermionic generators can be used to build bosonic generators of various (A, B).

(A,B) Coleman-Mandula theorem puts limits on allowed bosonic generators, and hence allowed (A, B) for the Qab . [email protected] 85 Coleman-Mandula theorem: Only bosonic generators are of internal + Poincar´egroup symmetries.

Simple argument: Additional conserved additive rank 1 tensors constrain scattering amplitude too much. ≥ Only 1 4-vector, P µ: Consider 2 2 scattering, c.m. frame. → Conservation of momentum and angular momentum amplitude depends on scattering angle θ. → Second conserved additive 4-vector Rµ gives additional constraints unless Rµ P µ. ∝ µν Only 1 2nd rank tensor, J : Assume rank 2 conserved additive tensor Σµν.

2 2 Additive property means [Σµν, aσ† (p)] = Cµν σ(p)aσ† (p). Then Cµν(p,σ)= ασ(m )pµpν + βσ(m )gµν.

Lorentz transformation of RHS is U(Λ)[Σµν, aσ† (p)]U †(Λ) = [U(Λ)ΣµνU †(Λ),U(Λ)aσ† (p)U †(Λ)]

ρ δ ρ δ = [Λ µΛ νΣρδ,Dσ′σ(W (Λ, p))aσ† ′ (Λp)] = Dσ′σ(W (Λ, p))Λ µΛ νCρδ(Λp, σ)aσ† ′ (Λp),

ρ δ and of LHS is Cµν(p, σ)Dσ′σ(W (Λ, p))aσ† ′ (Λp), so Λµ Λν Cρδ(p, σ) = Cµν(Λp, σ). Only candidates are pµpν and gµν.

2 µ ν 2 µ ν 2 µ ν 2 µ ν In 2 2 scattering, conservation of Σ implies α (m )p p + α (m )p p = α (m )p′ p′ + α (m )p′ p′ . → µν σ1 1 1 1 σ2 2 2 2 σ1 1 1 1 σ2 2 2 2 µ µ µ µ µ µ µ µ µ 2 2 P conservation: p +p = p′ +p′ = p = p′ , i.e. no scattering (allowed p = p′ if α (m )= α (m )). 1 2 1 2 ⇒ 1,2 1,2 1,2 2,1 σ1 1 σ2 2 No higher rank tensors: Generalize last argument to higher rank tensors. [email protected] 86 (A,B) 1 1 1 1 Allowed representations for fermionic generators Qab : A + B = 2 , i.e. 2, 0 or 0, 2 , and j = 2.

(C,D) (C,D) (C+D,C+D) (C,D) QC, D, QC, D† = XC+D, C D : Firstly, Q † is of type (A, B) = (D,C) because B† = A. { − − } − −

(C,D) (D,C) (D,C) (C,D) Writing QC, D† = Qa,b , we ﬁnd b = C because [B , Qa,b ] = [A+, QC, D]† = 0 using B = A+† . Similarly a = D. − − − − − −

(C,D) (C,De) (C,D) (D,C) e Now QC, D, QC, D† = QC, D, QD, C , must have A3 = C + D and B3 = C D, i.e. A, B C + D. { − − } { − − } − − ≥ But since A must be (C + D) (frome triangle inequality), it must be = C + D. Similarly for B. ≤

(C+D,C+D) µ 1 1 Since X is bosonic, CM theorem means it must be P ( 2, 2 ), or internal symmetry generator ((0, 0)). Latter implies C = D = 0, not possible by spin-statistics connection. Final result is relation with momentum on page 82.

1 1 1 Take Q to be , 0 spinor, i.e. [A, Q ]= σ Q , [B, Q ]=0 (Q† will be 0, ). a 2 a −2 ab b a a 2 Qa not ruled out by reasoning of CM theorem, because no similar conservation law:

Take i , j to have deﬁnite particle number. j i =0 = even diﬀerence in fermion numbers, so j Q i = 0. | i | i h | i 6 ⇒ h | a| i Can have multiple generators Q , r = 1, ..., N simple SUSY is N = 1, extended SUSY is N 2. ar −→ ≥ Summary:

1 1 µ µν CM theorem: only bosonic generators are (0, 0) (internal symmetry), 2, 2 (P ), and (1, 0) and (0, 1) (J ).

1 1 HLS theorem: only fermionic generators are 2, 0 and 0, 2 (Qar). [email protected] 87 µ Relation with momentum for any N: Q , Q† = 2δ σ P . { ar bs˙ } rs ab˙ µ

µ From allowed representations for fermionic generators on page 86, Q , Q† = 2N σ P . N is Hermitian, { ar bs˙ } rs ab˙ µ rs

µ µ because Q , Q† † = Q , Q† = 2N ∗ σ P , but Q , Q† = 2N σ P . So N diagonalized by unitary matrix W . { ar bs˙ } { bs ar˙ } rs ba˙ µ { bs ar˙ } sr ba˙ µ

µ Writing Q′ ′ = W ′ Q gives Q′ , Q′† = 2n δ σ P (no sum over r on RHS), where n are eigenvalues of N . ar r r ar { ar bs˙ } r rs ab˙ µ r rs

Writing Qar = Qar′ /√nr gives result if nr > 0 (otherwise we have a factor -1):

Taking Q′† = (Q′ )† and operating from right and left with X(p) and X(p) where Q′ X(p) = 0 gives bs˙ ar | i h | ar| i 6

2 2 on LHS: X(p) Q′ , (Q′ )† X(p) = (Q′ )† X(p) + Q′ X(p) > 0, and h |{ ar ar }| i | ar }| i| | ar}| i| on RHS: 2n (p0 p3), where for a = 1, 2. If p0 p3, then n > 0 as required. r ± ± ≥∓ r

SUSY implies 0 H 0 = 0 for supersymmetric vacuum (Q 0 = Q† 0 = 0). h | | i ar| i ar˙ | i µ Commutation with momentum: [Qa, P ]=0.

µ 1 1 1 1 1 1 a˙ 1 [Q ,P ] is , 0 , = 1, + 0, . No 1, generator, but Q† is 0, . a 2 × 2 2 2 2 2 2 µ µ a˙ µ b µ a˙ µ µ ab˙ So [Q ,P ] = kσ Q† and therefore Q† ,P = k∗Q σ , or, using ǫ matrix, Q† ,P = k∗σ Q . b ba˙ a˙ − ba˙ b h i µ ν µ ν ν µ µ b˙ ν ν b˙ µ 2 µ ν b Jacobi identity: 0 = [[Q ,P ],P ]+[[P ,P ], Q ]+[[P , Q ],P ] = kσ [Q† ,P ] kσ [Q† ,P ] = k [σ , σ ] Q . a a a ab˙ − ab˙ | | a b

Since [σµ, σν] b = 0 for all µ, ν, must have k 2 = k = 0. a 6 | | [email protected] 88 Anticommuting generators: Q , Q = ǫ Z , with (0, 0) generators Z = Z . { ar bs} ab rs rs − sr Q , Q is (1, 0) + (0, 0). From CM theorem, only (1, 0) generator is J µν. But [ Q , Q ,P µ] = 0 (commutation with { ar bs} { ar bs} momentum, page 87), while (linear combinations of) J µν doesn’t commute with P µ (see Poincare algebra, page 21).

So only possibility is (0, 0) generators, which in general commute with P µ from CM theorem.

Lorentz invariance requires ǫ (= ǫ ), but whole expression symmetric under ar bs so Z = Z . ab − ba ↔ rs − sr Antisymmetry of Z vanish for N = 1. Z are central charges due to following commutation relations: rs → rs

Commutation with central charges: [Zrs, Qat]=[Zrs, Qat†˙ ]=0.

Jacobi identity 0 = [ Q , Q , Q† ]+[ Q , Q† , Q ]+[ Q† , Q , Q ]. { ar bs} ct˙ { bs ct˙ } ar { ct˙ ar} bs

2nd, 3rd terms vanish from commutation with momentum on page 87. Thus [Zrs, Qat†˙ ]=0.

Jacobi identity 0 = [Z , Q , Q† ] + Q† , [Z , Q ] Q , [Q† , Z ] . − rs { at bu˙ } { bu˙ rs at } − { at bu˙ rs }

µ 1st, 3rd terms vanish because Z commutes with P and Q† . rs bu˙

µ Commutator in 2nd term must be [Z , Q ] = M Q , so 0 = Q† , [Z , Q ] = 2M σ P , i.e. M =0so[Z , Q ]=0. rs at rstv av { bu˙ rs at } rstu ab˙ µ rstu rs at

Commuting central charges: [Z ,Z ]=[Z ,Z† ]=0. [Zrs, Ztu]=[ Q1r, Q2s , Ztu] = 0, etc. rs tu rs tu { }

R-symmetry: Z = 0 gives U(N) symmetry Q V Q (N = 1 case: Q eIφQ , always true). rs ar → rs as a → a [email protected] 89 µ µν CM theorem: [ti, P ]=[ti, J ] = 0 for generators ti of group G.

Since ti is (0, 0), must have [ti, Q1 ]= (ai)rsQ1 . Matrices ai represent G . 2r − 2s

Jacobi identity 0 = [[ti,tj], Q 1 r]+[[Q 1 r,ti],tj]+[[tj, Q 1 r],ti] = ICijk[tk, Q 1 r] + (ai)rt[Q 1 t,tj] (aj)rt[Q 1 t,ti] 2 2 2 2 2 − 2

= ICijk(ak)rsQ 1 s + (ai)rt(aj)tsQ 1 s (aj)rt(ai)tsQ 1 s, i.e. [ai, aj]rs = ICijk(ak)rs. − 2 2 − 2

Simple SUSY and internal symmetry generators commute . If N=1, numbers ai cannot represent G unless ai = 0.

Commutation of central charges and internal symmetry generators: [ti,Zrs]=0.

From anticommuting generators on page 88, [t , Z ]=[t , Q , Q ] = (a ) ′ Z (a ) ′ Z ′ . i rs i { 1r 2s} − i rr rs − i ss rs This has the form [t , Z ] = A Z , i.e. Z form invariant subalgebra of G if any A = 0 or are U(1) if all A = 0. i α iαβ β α iαβ 6 iαβ

But from CM theorem, generators of invariant subalgebra not allowed, only U(1) generators. So all Aiαβ = 0. [email protected] 90 3.3 Supermultiplets

Supermultiplets: Particles that mix under SUSY transformations, furnish representation of SUSY.

Irreducible representations of SUSY: doesn’t reduce to 2+ supermultiplets separately mixing under SUSY.

Action of Qar or Q†ar˙ converts one particle into another of the same irreducible supermulitplet (superpartners).

µ µ µ 2 Since [Qar, P ]=[Q†ar˙ , P ] = 0, all particles in supermultiplet have same P (and hence same mass P ).

Equal number of bosonic and fermionic degrees of freedom in supermultiplet: nB = nF .

Convention: over supermultiplet states, over complete basis. Choose given r for Q Q , Q† Q† below. X all X ar ≡ a ar˙ ≡ a˙

P µ P 2s Supermultiplet’s states X have same p , and ( 1) X = 1 X for spin s bosonic/fermionic X , | i − | i ± | i | i

2s so P ′ X ( 1) P X = p (n n ). We show P ′ = 0: From relation with momentum for any N on page 87, µ ≡ Xh | − µ| i µ B − F µ P µ 2s 2s 2s 2s 2σ P ′ = X ( 1) Q Q† X + X ( 1) Q†Q X = X ( 1) Q Q† X + X ( 1) Q† Y Y Q X . ab˙ µ Xh | − a b˙ | i Xh | − b˙ a| i Xh | − a b˙ | i X,all Y h | − b˙ | ih | a| i P P P P Since Q X , X in same supermultiplet, limit . Conversely, extend (so X X = 1). a| i | i all Y → Y X → all X all X | ih |

µ 2s P 2sP 2s P P P 2s 2σ P ′ = X ( 1) Q Q† X + Y Q X X ( 1) Q† Y = X ( 1) Q Q† X + Y Q ( 1) Q† Y . ab˙ µ h | − a b˙ | i all X,Y h | a| ih | − b˙ | i Xh | − a b˙ | i Y h | a − b˙ | i P P P 2s 2s µ 2s 2s But Q ( 1) Y = ( 1) Q Y , so 2σ P ′ = X ( 1) Q Q† X + Y ( 1) Q Q† Y = 0. a − | i − − a| i ab˙ µ Xh | − a b˙ | i − Y h | − a b˙ | i P P [email protected] 91 0 1 † 1 † Q r, Q1˙ = 4p δrs Q r, Q 1˙ = 0 1 2 3 0 { 2 s} { 2 s} Massless supermultiplets: In frame where p = p = 0, p = p , 2 −2 .  Q 1 , Q†˙ = 0 Q 1 , Q† ˙ = 0  2r 1s 2r 1s { − 2 } { − −2 }  

Q 1 , Q† ˙ annihilate supermultiplet states , so Zrs annihilate supermultiplet states . 2r 1r − −2

† 2 2 For any X in supermultiplet, 0 = X Q 1 , Q† X = Q 1 X + Q 1 X (no sum over r). 2 r 1˙ 2 r 2 r | i h |{ − 2 r}| i | − | i| | − | i| −

All supermultiplet states reached by acting on maximum helicity state λmax with the Q1 . | i 2r

Q† give no new states: Consider X not containing Q 1 . Then Q† X = 0 (Q† commutes across to act directly on λmax ). 1˙ 2 r 1˙ 1˙ 2 r | i 2 r| i 2 r | i

0 Then Q† Q 1 X = Q 1 , Q† X = 4p X (no sum over r), i.e. Q† just removes Q 1 . 1˙ 2 r 2 r 1˙ 1˙ 2 r 2 r | i { 2 r}| i | i 2 r

1 1 Recall [J3, Q† ] = Q† . Q† X (or Q1 X ) has helicity greater (or less) than X by . 1˙ r 2 1˙ r 1˙ 2r 2 2 2 2r| i | i | i

N! n N Range of helicities in supermultiplet: n!(N n)! helicity λmax 2 states , λmin = λmax 2 . − − −

Obtain supermultiplet states X by acting on maximum helicity state λmax with any n of Q 1 1, Q 1 2, ..., Q 1 N . | i | i 2 2 2

2 Order doesn’t matter since Q 1 a anticommute, each generator cannot appear more than once since Q 1 = 0. 2 2 r

Constraint on particle spins on page 56 implies λ 2 and λ 2, i.e. N 8. min ≥− max ≤ ≤ [email protected] 92 SM particles probably belong to massless supermultiplets. ∼ Superpartners (masses M) of SM particles ( m) not seen, i.e. M m, so SUSY is broken, at energy < m . ∼ ∼ ≫ SUSY SUSY restored at m ∴ m M m M (superpartner has same mass) ∴ supermultiplets massless. SUSY SUSY ≫ − ∼ ∼ 1 Most likely scenario is simple SUSY: quark, lepton (spin 2) superpartners are scalars (0): squarks, sleptons.

1 3 Higgs (0), gauge bosons (1), graviton (2) superpartners fermionic: Higgsino + gauginos (2), gravitino (2).

No alternatives: In simple SUSY, SM gauge bosons cannot be superpartners of SM fermions.

SM fermions, gauge bosons in diﬀerent representations (recall simple SUSY and SM SU(3) SU(2) commute from page 89). × Quarks, leptons cannot be in same supermultiplet as any beyond-SM vector (gauge) bosons.

1 Gauge bosons are in adjoint representation of a group. If e.g. helicity +2 fermion, +1 gauge boson in same supermultiplet,

1 fermion in adjoint = real representation. But SM is chiral, i.e. helicity +2 fermions belong to complex representations.

Superpartners of gauge bosons must be helicity 1, not 3, fermions. ±2 ±2 Helicity 3 particle couples only to Q (principle of equivalence on page 56: 2 only couples to P µ). ±2 a ± Extended SUSY probably cannot be realised in nature.

In extended SUSY, helicity 1 fermions either in same supermultiplet as gauge bosons (N 3), or each other (N = 2). ±2 ≥ [email protected] 93 Massive supermultiplets: For particles with masses M m , e.g. heavy gauge bosons in SU(5). ≫ SUSY

Q1 , Q† = 2Mδrs Q1 , Q† = 0 2r 1˙ 2r 1˙ 1 2 3 { 2s} { 2s} In frame where p = p = p = 0, p0 = M, − ,  Q 1 , Q†˙ = 0 Q 1 , Q† ˙ = 2Mδrs  2r 1s 2r 1s { − 2 } { − −2 }   1 so Q1 , Q† ˙ (or Q 1 , Q†˙ ) lower (or raise) spin 3 component by . 2r 1r 2r 1r 2 −2 − 2

1 a = Q are fermionic annihilation / creation operators: a , a† = δ , a , a = a† , a† = 0. A(a,r) √2M ar { A B} AB { A B} { A B} Deﬁne Cliﬀord “vacuum” Ω : a Ω = 0. Ω has given spin j, range of spin 3 is j σ j. | i A| i | i − ≤ ≤ N N Supermultiplet is all states a† ...a† Ω , spins ranging from Max(j , 0), . . ., j + . A1 An| i − 2 2 Simple SUSY: States with spin j 1, and 2 sets of states with spin j. ± 2 If j = 0, there are two bosonic spin 0 states and two fermionic states of spin 1 (i.e. spin 3 is 1). 2 ±2 [email protected] 94 Lower mass bound in extended SUSY: M 1 Tr√Z Z . ≥ 2N † Using anticommuting generators on page 88 and relation with momentum for any N on page 87,

0 0 Q ǫ U Q∗ , Q∗ ǫ U ∗ Q = 8NP 2Tr(ZU † + UZ†) for any unitary U. Act on supermultiplet at rest. ≤ { ar − ab rs bs ar − ac rt ct} − Polar decomposition theorem means any Z = HV , where H is positive Hermitian and V unitary.

1 2 Let U = V . Then use M TrH and Z†Z = H . ≥ 2N

When equality obeyed, supermultiplets are smaller (“short”) and similar to massless supermultiplets.

1 √ For states obeying M = 2N Tr Z†Z, underlined quantity above must be zero on these states.

Then there are just N independent helicity-lowering Q 1 r and N independent helicity-raising Q 1 r. 2 − 2 [email protected] 95 3.3.1 Field supermultiplets (the left-chiral supermultiplet)

1 Simplest case: N = 1, scalar φ(x) obeying [Q†, φ(x)] = 0 : Write [Q , φ(x)] = Iζ (x) , which is a , 0 ﬁeld. b˙ a − a 2 µ Q†, ζ (x) = 2σ ∂ φ(x) . Thus [Q, φ] ζ and Q†, ζ φ, i.e. φ and ζ are each others’ superpartners. { a˙ b } ba˙ µ ∼ { }∼

µ Q† , Iζ = Q† , [Q ,φ] =[ Q† , Q ,φ] = 2σ [P ,φ]. Poincar´etransformation for ﬁelds on page 25: [P ,φ] = I∂ φ. { a˙ − b} { a˙ b } { a˙ b} ba˙ µ µ µ Q , ζ (x) = 2Iǫ F (x). { a b } ab

Q , Iζ = Q , [Q ,φ] = Q , [Q ,φ] = Q ,Iζ ǫ . F is (0, 0) in 1, 0 1, 0 = (0, 0) + (1, 0). { a − b} { a b } −{ b a } −{ b a}∝ ab 2 × 2 [Qa, F (x)]=0. F has no superpartner, it is auxiliary ﬁeld.

2Iǫ [Q , F ]=[Q , Q ,ζ ] = [Q , Q ,ζ ] = 2Iǫ [Q , F ]. For a = c = b, 2Iǫ [Q , F ] = 2Iǫ [Q , F ], so [Q , F ]=0. ab c c { a b} − a { c b} − cb a 6 cb c − cb c c

a˙ µ ab˙ [Q† , F (x)] = σ ∂ ζ (x). − µ b

µ µ µ µ 2Iǫ [Q†, F ]=[Q†, Q ,ζ ]=[ Q†, Q ,ζ ] [Q , Q†,ζ ]=[2σ P ,ζ ] [Q , 2σ ∂ φ] = 2Iσ ∂ ζ 2Iσ ∂ ζ . ab c˙ c˙ { a b} { c˙ a} b − a { c˙ b} ac˙ µ b − a bc˙ µ ac˙ µ b − bc˙ µ a

da d da µ da µ da µ µ d d µ b Then ǫ ǫ [Q†, F ] = δ [Q†, F ] = ǫ σ ∂ ζ ǫ σ ∂ ζ = ǫ σ ∂ ζ σ ∂ ζ . Sum d = b: δ [Q†, F ]=2[Q†, F ] = 2σ ∂ ζ . ab c˙ b c˙ ac˙ µ b − bc˙ µ a ac˙ µ b − bc˙ µ d c˙ c˙ − bc˙ µ

a˙ a˙c˙ a˙c˙ µ bd µ ad˙ Then [Q† , F ] = ǫ [Q†, F ] = ǫ σ ǫ ζ = σ ζ , using 4-vector σ matrices with raised indices on page 42. c˙ − bc˙ d − d Conjugation of everything above gives right-chiral supermulitplet. [email protected] 96 X To simplify algebra, use 4-component Majorana spinors, ψ = a , with notation of page 43. X b˙ †

ab T ǫ 0 a From ψ = ψ β = (X , X†). Majorana conjugation: ψ = ψ γ5E where E = ab . † b˙ 0 ǫ ˙(= ǫ ) − a˙b Q (Anti)commutation relations: Q, Q = 2IγµP and [Q, P µ] = 0 , where Majorana Q = a . { } − µ Q b˙ = (ǫbcQ ) † c †

Directly from relations on page 87.

Inﬁnitesimal transformation of operator O: δO =[IαQ, O],

αa a b˙ where Grassmann spinor α = , so αQ = α Q + α†Q . Implies δO =[IαQ, O ]. α b˙ a b˙ † † † †

(δO)† =[IαQ, O]† =[IαQ, O†] = δ(O†), because, as for scalar from 2 spinors and Hermitian conjugate of scalar on page 41,

a b˙ a˙ b a˙ b a˙ b (αQ)† = (α Q + α†Q† )† = (Q† α† + Q α ) = ( α† Q† α Q ) = (α†Q† + α Q ) = αQ. a b˙ a˙ b − a˙ − b a˙ b Product rule for δ: δ(AB) = (δA)B + AδB , where A, B can be fermionic / bosonic.

[IαQ, AB] = I(αQAB ABαQ) = I(αQAB AαQB + AαQB ABαQ)=[IαQ, A]B + A[IαQ,B]. − − −

IαQ IαQ If Abelian limit on page 8 obeyed, ﬁnite transformation is O e Oe− . → [email protected] 97 (Anti)commutation relations on page 95 can be simpliﬁed with 4-component spinor notation

ζa and deﬁnitions φ = 1 (A + IB), ψ = 1 and F = 1 (F IG): √2 √2 ζ b˙ √2 − †

1. δA = αψ , 2. δB = Iαγ ψ . − 5

[Qa,φ] = Iζa [Qa,φ†] = [Q†a˙ ,φ]† = 0 [Q, φ] = b˙ b˙a˙ − and [Q, φ†] = b˙ b˙−a˙ , [Q† ,φ] = ǫ [Q†a˙ ,φ] = 0 [Q† ,φ†] = ǫ [Qa,φ]† = Iζ b˙ ! − − †

1 I so [Q, A]=[Q, (φ + φ†)] = Iψ, and [Q, B]=[Q, (φ† φ)] = γ ψ from deﬁnition of γ on page 44.. √2 − √2 − − 5 5 3. δF = αγµ∂ ψ , 4. δG = Iαγ γµ∂ ψ . µ − 5 µ

[Qa, F ] = 0 µ bc˙ µ cb˙ [Q, F ] = b˙ µ ba˙ and, using σ ∗ = σ , [Q† , F ] = σ ∂ ζ − µ a

b˙ µ bc˙ µ cb˙ µ cb˙ d˙ µ d˙ [Q , F ] = ǫ [Q , F ] = ǫ σ ∂ ζ† = ǫ σ ∂ ζ† = ǫ σ ǫ ˙∂ ζ = σ ∂ ζ [Q, ] = a † ab † † ab ∗ µ c˙ ab µ c˙ ab c˙d µ † ad˙ µ † , so F † − b˙ ba − [Q† , F †] = ǫ [Qa, F ]† = 0 ! −

Iσµ ∂ ζ b˙ [Q, F ]=[Q, 1 ( + )] = I ab˙ µ † = Iγµ∂ ψ and √2 F F † √2 − µ bd˙ µ − Iσ ∂µζd ! − −

Iσµ ∂ ζ b˙ [Q, G]=[Q, I ( )] = 1 ab˙ µ † = γ γµ∂ ψ. √2 F F † √2 µ bd˙ 5 µ − Iσ ∂µζd ! − − [email protected] 98 5. δψ = ∂ (A + Iγ B)γµα + (F Iγ G)α . µ 5 − 5

c d˙ c d˙c˙ α Q ,ζ + α† Q† ,ζ α Q ,ζ + α†ǫ Q†,ζ [IαQ,ψ] = I { c a} d˙{ a} = I { c a} d˙ { c˙ a} √2 c b˙ d˙ b˙ √2 c b˙d˙ d˙c˙ b˙a˙ α Q ,ζ† + α† Q† ,ζ† α Q†,ζ †ǫ + α†ǫ ǫ Q ,ζ † { c } d˙{ } ! { c˙ d} d˙ { c a} !

c d˙c˙ µ α (2Iǫ F ) + α†ǫ (2σ ∂ φ) = I ca d˙ ac˙ µ √2 ca µ b˙d˙ d˙c˙ b˙a˙ ǫ α (2σ ∂ φ†)ǫ α†ǫ ǫ (2Iǫ F †) a cd˙ µ − d˙ c˙a˙ !

α (IF ) α c˙(σµ ∂ φ) α (F IG) Iσµ α c˙∂ (A + IB) √ a † ac˙ µ a ac˙ † µ = I 2 − µ ba˙ − b˙ = b˙ − − µ ba˙ . α (σ ∂ φ†) α† (IF †) α† (F + IG) Iσ α ∂ (A IB) − a µ − − a µ −

For L = 1∂ A∂µA 1∂ B∂µB 1ψγµ∂ ψ + 1(F 2 + G2)+ m(FA + GB 1ψψ) −2 µ − 2 µ − 2 µ 2 − 2

+g[F (A2 + B2) + 2GAB ψ(A + Iγ B)ψ] , above transformations 1 — 5 leave action A = d4xL invariant . − 5 R For example, for m = g = 0, δL = ∂ δA∂µA ∂ δB∂µB (δψ)γµ∂ ψ + F δF + GδG. − µ − µ − µ We have replaced 1ψγµ∂ δψ 1(δψ)γµ∂ ψ, since diﬀerence with analogous term in hermitian conjugate (h.c.) of L −2 µ →−2 µ

(must be added to make L real) is total derivative ∂µf which doesn’t contribute to A :

µ µ µ µ µ µ µ (Xγ ∂ Z)† = (X†βγ ∂ Z)† = (∂ Z†)γ †β†X = (∂ Z†)βγ X = (∂ Z)γ X = Zγ ∂ X ∂ (Zγ X). µ µ µ − µ − µ µ − µ No derivatives of F , G (or F ) appear, so they are auxiliary ﬁelds,

∂L ∂L i.e. can be expressed in terms of the other supermultiplet fields by solving equations of motion ∂F = ∂G = 0. [email protected] 99 Problems 2 (3) 1 1 1: Derive the equal time anticommutation relations [ψl(x,t),ψl†′ (y,t)]+ = δll′ δ (x y) for a 2, 0 + 0, 2 field ψ(x) = d3p Ip x Ip x c − e u (p)a (p) + e v (p)a (p) where u and v obey the projection operator properties u (p)u ′ (p) = (2π)3 · σ σ − · σ σ† l σ l σ 1 µ 1 µ ( Iγ p + m) ′ and v (p)v ′ (p) = ( Iγ p m) ′ . R2p0 − µ ll l σ l σ 2p0 − µ − ll 1 1 2: Show that the canonically conjugate momenta to the 2, 0 + 0, 2 field ψ(x) for the Lagrangian density LDirac = µ (3) ψ (γ ∂ + m) ψ is consistent with the equal time anticommutation relations [ψ (x,t),ψ†′ (y,t)] = δ ′ δ (x y). − µ l l + ll − i 3: Under a local gauge transformation of a group G, the field strength F = F t transforms as F UF U †, where µν µν i µν → µν U and ti are respectively any element of G and the generators of G in any representation. By considering infinitessimal i j transformations, show that this implies that Fµν transforms in the adjoint representation of G, i.e. F UijF , where k → the generators of Uij are tji = ICkij.

4: How must the right-handed neutrino field νeR transform under the SM group in order for the Lagrangian term proportional K to E a (ǫφH )a† νeR to be invariant? 1 5: Suppose that there exists a two-component fermionic additive symmetry generator Qa which transforms in the 2, 0 µ representation. Using the Coleman-Mandula theorem, show that Q , Q† = 2σ P . What does this tell you about any { a b˙ } ab˙ µ particle? (Note that, as an additive symmetry generator, Q = Qσρaσ† aρ + Rσρaρ†aσ, where σ sums over bosonic particles, ρ over fermionic particles.) µ 6: For more than one spinor of such generators Q distinguished by the label r, show that Q , Q† = 2δ σ P . ar { ar bs˙ } rs ab˙ µ 7: How many different states are there in a massless supermultiplet for a given N? (Hint: working in a frame for which the particles of a massless supermultiplet have a momentum proportional to p = (0, 0, 1, 1), show that the Q 1 , r = 1,...,N, 2 r are the only generators that convert the maximum helicity state λ in this supermultiplet into any other state in this | maxi supermultiplet, because the other fermionic generators either annihilate λmax or just remove any Q 1 r generator.) | i 2 8: Starting with a field φ(x), the generators Q and the constraint [Q†,φ(x)] = 0, determine all the other fields in the same a b˙ supermulitplet as φ(x). 1 1 T 9: The adjoint of a 2, 0 + 0, 2 spinor ψ is defined as ψ = ψ†β. If ψ is a Majorana spinor, show that ψ = ψ γ5E. (Hint: T b˙ bc Write ψ = Xa, X† = (ǫ Xc)† .)

b˙ µ 10: Deﬁning Q = Q , Q† , show that the anticommutation relations above read Q, Q = 2Iγ P . a { } − µ [email protected] 100 3.4 Grassman variables

N independent Grassman variables θ are deﬁned to obey θ ,θ = 0. α { α β} 2 Thus θα = 0 (no sum), so for e.g. N = 3, any function f(θ) has the Grassman variable expansion f(θ)= A + Bθ1 + Cθ2 + Dθ3 + Eθ1θ2 + Fθ2θ3 + Gθ3θ1 + Hθ1θ2θ3.

∂ ∂ ∂ ∂ Deﬁne (no sums, β = α) θα = 1, θβ = 0, θαθβ = θβ and θβθα = θβ. 6 ∂θα ∂θα ∂θα ∂θα −

∂ ∂ ∂ Note (β can be equal to α now) that , = 0 and ,θβ = δαβ . ∂θα ∂θβ ∂θα By anticommuting Grassman variables θα, θβ to the left, all terms in the Grassman variable expansion of any function f

will be of the form g, θαg, θβg and θαθβg, where g is independent of θα, θβ but can depend on the other θγ.

g = 0 g = g ∂ ∂ + ∂ ∂ ∂θα ∂θβ ∂θβ ∂θα θαg = 0 ∂ ∂ θαg = θαg Then show   and (no sum, β = α) ∂θ θα + θα ∂θ  .  ∂ h ∂ i  θβg = 0 6 α α θβg = θβg ∂θ θβ + θβ ∂θ (β = α) α α 6  θ θ g = 0  h i  θ θ g = θ θ g   h i   α β   α β α β      [email protected] 101 3.5 Superﬁelds and Superspace

T b˙ Let Q generate translations in superspace (x, θ = θa,θ† ) (θa are Grassman) on superﬁeld S(x, θ) via µ µ µ [IαQ,S(x, θ)] = αQS(x, θ) , as P generates spacetime translations on ﬁeld φl(x) via [P , φl(x)] = I∂ φl(x).

Definition of superfield: δS = αQS . From infinitesimal transformation of operator O on page 96.

Condition on Q: Q , Q = 2γµ ∂ , where α, β run over the 4 spinor indices. { α β} αβ µ

From (anti)commutation relations on page 96, [ Q , Q ,S] = 2Iγµ [P ,S], which from above reads Q , Q S = 2γµ ∂ S. { α β} − αβ µ { α β} αβ µ

∂ µ ∂ µ ∂ µ Deﬁnition of Q: Q = + γ θ∂µ (explicitly, Qα = (γ5E)βα + γ θβ∂µ = (γ5E)αβ + γ θβ∂µ). −∂θ −∂θβ αβ ∂θβ αβ

For a (and b) Grassmann, ∂ is left derivative: anticommute a left then remove it, e.g. ∂ ba = ∂ ab = b. ∂a ∂a −∂a −

2 T ∂ ∂ This choice satisfies condition on Q: We use (γ5E) = 1,(γ5E) = γ5E (direct calculation), and = (γ5E)βα − − ∂θ α ∂θ β ∂ µ µ µT from Majorana conjugation on page 96. Now Q = Q† β = † (γ E) β + γ ∗ θ† β ∂ . But γ ∗ = βγ β, δ α αδ − ∂θ β 5 βα αδ αβ β αδ µ − 2 µ µ ∂ µ and β = 1 so that γ ∗ θ† β = θ γ . Thus, using γ E, β = 0, Q = (γ E) (γ E) θ (γ E) γ ∂ . αβ β αδ − α αδ { 5 } δ − ∂θ α 5 αβ 5 βδ − α 5 αβ βδ µ ∂ µ ∂ µ µ µ So Qδ = θα(γ5Eγ )αδ∂µ. Then from , θβ = δαβ, Qα, Qβ = δσρ(γ5E)ρα(γ5Eγ )σβ∂µ + γ ∂µ = 2γ ∂µ. ∂θ δ − ∂θα { } αβ αβ [email protected] 102 Superfields from superfields I: S = S1 + S2 , S = S1S2 , etc. are superfields when S1, S2 are superfields.

1st case clearly obeys SUSY transformation using Q above.

2nd case: From product rule for δ on page 96, δS1S2 = (δS1)S2 + S1δS2 = (αQS1)S2 + S1αQS2 = αQ(S1S2).

Superﬁelds from superﬁelds II: S′ = DβS , S′ = D βS , where superderivative D = ∂ γµθ∂ obeys D , Q = D , Q =0. −∂θ − µ { β γ} { β γ}

First case: δS′ = δD S =[IαQ, D S] = D [IαQ,S] since αQ is a commuting object. Then δS′ = D α Q S = α D Q S β β β β γ γ − γ β γ

since α , D anticommute. Then δS′ = α Q D S = αQS′. To show e.g. D , Q = 0, γ β γ γ β { β γ} first obtain D , Q = γµ θ , ∂ (γ E) α β ∂ f and use (γµγ E)T = γµγ E. { β γ} αγ γ ∂θ δ 5 δβ − ↔ µ 5 5 Summary: Function of superfields and their superderivatives is a superfield.

R quantum number assignments for superspace: θ has R quantum number R = 1. L/R ±

IR φ IR φ Recall definition of R-symmetry on page 88: Q e L Q and Q† e R Q† , where R = R = 1. a → a a → a L/R ± Assignments for superspace follow from definition of superfield (i.e. [Q, S(x, θ)] = IQS(x, θ)) and definition of Q on page 101. −

Note θ θ∗ . R ∼ L [email protected] 103 µ From now on, write e.g. Aµγ =A / .

General form of superfield: S(x, θ)= C(x) I[θγ ]ω(x) I θγ θ M(x) 1[θθ]N(x)+ I [θγ γ θ]V µ(x) − 5 − 2 5 − 2 2 5 µ I[(θγ θ)θ] λ(x)+ 1∂ω/ (x) 1[θγ θ]2 D(x)+ 1∂2C(x) , − 5 2 − 4 5 2 4 (Pseudo)scalar fields: C, M, N, D, 2 Spinor fields: ω, λ, 1 Vector field: V µ

(i.e. 8 bosonic and 8 fermionic degrees of freedom).

This is most general Taylor series in θT = (θ , θ , θ , θ ), or θ = θT γ E = ( θ , θ , θ , θ ), in manifestly Lorentz invariant form: 1 2 3 4 5 − 2 1 4 − 3 2nd term Iθγ ω(x) most general quantity linear in θ . − 5 α

Next 3 bilinears in θα is expansion for any bilinear B = B12θ1θ2 + B13θ1θ3 + B14θ1θ4 + B23θ2θ3 + B24θ2θ4 + B34θ3θ4:

θγ (1γ or 1 γ )θ = 2θ θ 2θ θ , θγ (1γ or 1 γ )θ = 2θ θ 2θ θ (gamma matrices on page 44), θ(1 or γ )θ = 2θ θ 2θ θ . 5 I 0 I 3 2 3 ± 4 1 5 I 1 I 2 2 4 ± 3 1 5 1 2 ± 4 3 6th term has all 4 possible cubics in θ :(θγ θ)θ = 2(θ θ θ , θ θ θ , θ θ θ , θ θ θ ). Last term [θγ θ]2 = 8θ θ θ θ . α 5 4 3 2 − 4 3 1 1 2 4 − 1 2 3 ∝ 5 1 2 3 4 Any higher products of θ vanish, since θ2 = 0 (no sum). E.g. θ θ θ θ θ = θ θ θ2θ = 0. α α 1 2 3 4 3 − 1 2 3 4 [email protected] 104 Superﬁeld transformation: δS = Iαγ ω + θ( ∂C/ + Iγ M + N Iγ V/ )α + I [θγ θ]α(λ +∂ω / ) 5 − 5 − 5 2 5

I [θθ]αγ (λ +∂ω / )+ I [θγ γµθ]αγ λ + I [θγ γνθ]α∂ ω −2 5 2 5 µ 2 5 ν

+1[(θγ θ)θ] I∂M/ γ ∂N/ I∂ Vγ/ µ + γ (D + 1∂2C) α I [θγ θ]2αγ (/∂λ + 1∂2ω). 2 5 − 5 − µ 5 2 − 4 5 5 2 From deﬁnition of superﬁeld and of Q, on page 101, i.e. δS = α ∂ + γµθ∂ S. −∂θ µ Note for M = 1, M = γ γ and M = γ , we have ∂ (θMθ) = 2Mθ. See also Weinberg III (hardback), page 63. 5 µ 5 ∂θ

Component ﬁeld transformations: δC = I(αγ ω), δω = ( Iγ ∂C/ M + Iγ N +V / )α , 5 − 5 − 5 δM = α(λ +∂ω / ), δN = Iαγ (λ +∂ω / ), δV = α(γ λ + ∂ ω), − 5 µ µ µ

1 µ δλ = (2[∂µV,γ/ ]+ Iγ5D)α and δD = Iαγ5∂λ/ .

Compare superfield transformation above with general form of superfield on page 103. [email protected] 105 2 D-component of general superfield (coefficient of [θγ5θ] ) is candidate for SUSY Lagrangian.

If L [S] D + 1∂2C, then δL δD + 1∂2δC = ∂ (Iαγ γµλ + 1∂µδC), i.e. a derivative, ∝ D ∝ 2 ∝ 2 µ 5 2 so action A = d4xL obeys δA = 0. R D-component of superﬁeld cannot be used to form kinematic Lagrangian.

Kinematic action must be quadratic in elementary ﬁelds and therefore in elementary superﬁelds, so can only try

µ 1 µ 1 µ µ [S∗S] = ∂ C∗∂ C (ωγ ∂ ω) + (∂ ω)γ ω + C∗D + D∗C ωλ λω + M ∗M + N ∗N V ∗V . D − µ − 2 µ 2 µ − − − µ Terms with D and λ constrain C and ω to vanish.

So introduce constraints on elementary ﬁelds. [email protected] 106 3.5.1 Chiral superﬁeld

Definition of chiral superfield X(x, θ): λ = D =0, Vµ = ∂µB , B is scalar field with δB = αω .

This is superﬁeld because these conditions are preserved by SUSY transformations: δD = Iαγ5∂λ/ = 0,

1 µ 1 µ µ ν µ and δλ = (2[∂µV,γ/ ] + Iγ5D)α = 2[∂µV,γ/ ]α = 0 because Vµ = ∂µB: [∂µV,γ/ ] = ∂µ∂νB[γ ,γ ]=0.

µ Finally, δV = α(γµλ + ∂µω) = ∂µαω, i.e. δB = αω, i.e. Vµ = ∂µB condition is preserved.

1 Chiral superﬁeld decomposition: X(x, θ)= [Φ+(x, θ) + Φ (x, θ)] with left / right-chiral superﬁelds √2 −

1 µ 1 1 2 2 Φ (x, θ)= φ (x) √2 θψL/R(x)+[θPL/Rθ]F (x) [θγ5γµθ]∂ φ (x) (θγ5θ)θ∂ψ/ L/R(x) [θγ5θ] ∂ φ (x), ± ± − ± ± 2 ± ∓ √2 − 8 ±

1 1 with ψL/R = PL/Rψ , φ = (A IB), F = (F IG). ± √2 ± ± √2 ∓

To make contact with notation on page 97, write C = A, ω = Iγ ψ, M = G, N = F , for later convenience − 5 − in general form of superﬁeld on page 103, so X = A θψ + 1θθF I θγ θG + I θγ γ θ∂µB + 1(θγ θ)θγ ∂ψ/ 1[θγ θ]2∂2A. − 2 − 2 5 2 5 µ 2 5 5 − 8 5

1 I I µ 1 1 2 2 Φ is obtained from Φ = A′ θψ′ + θθF ′ θγ5θG′ + θγ5γµθ∂ B′ + (θγ5θ)θγ5∂ψ/ ′ [θγ5θ] ∂ A′ with F ′ = G′, ± ± − 2 − 2 2 2 − 8 ±

B′ = IA′, ψ′ = √2ψ , i.e. by taking N = IM, B = IC, ω = left/right-handed spinor in general form of superﬁeld ∓ L/R ± ∓ ± on page 103, which is preserved under component ﬁeld transformations on page 104. [email protected] 107 µ Left / right transformation: δψL/R = √2(∂µφ γ PR/L + F PL/R)α , δF = √2 α∂ψ/ L/R , δφ = √2 αψL/R . ± ± ± ±

So left-chiral supermultiplet here = that on page 95 (ζa = ψL, φ = φ+, F = F+), right-chiral by conjugation.

T T Compact form: Φ (x, θ)= φ (x ) √2 θ E ψL/R(x ) θ EθL/R F (x ), ± ± ± ∓ L/R ± ± L/R ± ±

µ µ T µ where θL/R = PL/Rθ , x = x θREγ θL . ± ±

Recall ψT = 1 ζ ,ζ b˙ from page 97. This is left / right-chiral superﬁeld on page 106, by Taylor expansion in θT Eγµθ : √2 a † R L T 2 2 µ µ 1 µ Make use of θ = θγ5E (Majorana conjugation on page 96) and E = (γ5E) = 1. Also x = x 2θγ5γ θ − − ± ±

(see alternative form for x on page 108). Firstly, rewrite as Φ (x, θ) = φ (x ) √2 θPL/Rψ(x ) + θPL/RθF (x ), ± ± ± ± − ± ± ±

because θT Eψ = θγ E 1(1 γ )E 1(1 γ )ψ = θ 1( γ 1)1(1 γ )ψ = θP ψ. L/R L/R − 5 2 ± 5 2 ± 5 − 2 − 5 ∓ 2 ± 5 ± L/R

1 µ 1 1 µ 1 ν 1st term in underlined equation above: φ (x ) = φ (x) (θγ5γµθ)∂ φ (x) ( θγ5γ θ)( θγ5γ θ)∂µ∂νφ (x) ± ± ± ± 2 ± − 2 2 2 ±

1 µ 1 1 2 1 µ ν 1 µ ν µν = φ (x) (θγ5γµθ)∂ φ (x) ( θγ5θ) γ ,γ ∂µ∂νφ (x), then use γ ,γ = g . ± ± 2 ± − 2 2 2{ } ± 2{ }

1 µ 1 µ 2nd term: √2 θPL/Rψ(x ) = √2 θPL/Rψ(x) (θγ5γ θ)θ∂µPL/Rψ(x) = √2 θPL/Rψ(x) (θγ5θ)θγ ∂µPL/Rψ(x). − ± − ∓ √2 − ∓ √2

1 µ T µ T 3rd term: θPL/RθF (x ) = θPL/RθF (x) θγ5γ θθPL/Rθ∂µF (x) = θPL/RθF (x) + [(θREγ θL)(θ EθL/R)∂µF (x)], ± ± ± ± 2 ± ± L/R ±

term in square brackets vanishes due to 3 occurences of “2-component” θL/R. [email protected] 108 µ µ 1 µ Alternative form for x : x = x 2θγ5γ θ . ± ± ±

µ µ T µ 1 µ 1 T µ 1 T µ µ Recall x = x θREγ θL. Now 2θγ5γ θ = 2θ γ5Eγ5γ θ = 2θ γ5Eγ γ5θ using γ ,γ5 = 0. But γ5 = PL PR, ± ± − { } − so 1θγ γµθ = 1θT (P P )Eγµ(P P )θ = 1θT P EγµP θ + 1θT P EγµP θ 1θT P EγµP θ 1θT P EγµP θ. 2 5 2 L − R R − L 2 L R 2 R L − 2 L L − 2 R R

µ µ µ 3rd and 4th terms zero because PL/REγ PL/R = EPL/Rγ PL/R = Eγ PR/LPL/R = 0.

2nd term is 1θT P EγµP θ = (1θT P EγµP θ)T = 1θT P (Eγµ)T P θ ( sign because components of θ anticommute). 2 R L − 2 R L −2 L R − But (Eγµ)T = Eγµ, so 2nd term = 1st term, i.e. 1θγ γµθ = θT Eγµθ . − 2 5 R L

Relations between superspace directions: x+∗ = x . −

µ µ I.e. show θREγ θL = θγ5γ θ is “imaginary”.

µ µ T b˙ b T T x+∗ = x + θR† Eγ ∗θL∗ . But θR = (0, θ† ), so θR† = (0, θ ) = θLEβ, and θL = (θa, 0) so θL∗ = (θa†˙ , 0) = EβθR,

µ µT 2 2 T µ µ and γ ∗ = βγ β. These results together with β = E = 1, βE = Eβ and E = E give θ† Eγ ∗θ∗ = θ Eγ θ . − − − R L − R L

µ µ 1 µ µ µ Alternatively, use x = x 2θγ5γ θ and show (θγ5γ θ)† = θγ5γ θ: ± ± −

µ µ (θγ γ θ)† = (θ†βγ γ θ)† = θ†γ†γ βθ ( sign because components of θ anticommute). 5 5 − µ 5 −

2 µ µ µ Then inserting β = 1 in places, (θγ γ θ)† = (θ†β)(βγ† β)(βγ β)θ = θγ γ θ = θγ γ θ. 5 − µ 5 5 − 5 [email protected] 109 Compact form on page 107 is most general function of x and θL/R. ± So any function of left-chiral superﬁelds is left-chiral superﬁeld (likewise right).

Conjugate of right-chiral superfield is left-chiral superfield. Chiral superfield decomposition, page 106: Use (θγ )∗ = θγ . 5 − 5

Superspace “direction” of chiral superfield: D Φ = 0 , where superderivative D = PR/LD . ∓ ± ∓ Can be used to define left / right-chiral superfield.

µ ∂ µ ∂ Follows from D x = 0, using D α = Eαβ ∂θ (γ θL/R)α ∂xµ . D f(x, θ) = 0 implies f(x, θ) = f ′(x+, θL), i.e. f is left-chiral: ∓ ± ∓ ∓ R/L β − −

(a) (a) (a) Any function f(x, θ) = g(x+, θR, θL) = a g (x+)h (θR, θL), where each h is all possible products

P (a) (a) ∂ (a) of between 1 and 4 components of θ. Using ﬁrst D ,αg (x+) = 0 then D ,αh (θR, θL) = Eαβ h (θR, θL), − − − ∂θR β

(a) ∂ (a) (a) (a) we require EαβD ,βf = a g (x+) h (θR, θL) = 0. Therefore if h contains θR α, g (x+)=0. − ∂θR α P T Chiral superfield from superfield: For general superfield S, D αD βS (i.e. D ED S) is left / right-chiral. ± ± ± ±

D γ(D αD βS) = 0 because the D α anticommute and there are only 2 of them. ± ± ± ±

F -component of chiral superﬁeld (coeﬃcient of θPLθ) is candidate for SUSY Lagrangian.

4 If L [Φ ]F = F , then δL α∂ψ/ L/R, i.e. a derivative, so action A = d xL obeys δA = 0. ∝ ± ± ∝ R [email protected] 110 Any left/right-chiral superﬁeld Φ can be written Φ = D2 S respectively, ± ± R/L

2 where S is a general superﬁeld and D = EαβD ,αD ,β. R/L ∓ ∓

Expand S = f1(x+, θL) + f2(x+, θL)θR1 + f3(x+, θL)θR2 + f4(x+, θL)θR1θR2.

∂ µ ∂ But D ,αf(x+, θL) = 0 because D α = Eαβ (γ θL)α µ and D αx+ = 0. − − − ∂θR β − ∂x −

∂ 2 Thus D αS = Eαβ S, giving D S = 2f4(x+, θL), − − ∂θR β R/L −

where f4(x+, θL) is any function of (x+, θL), i.e. a general lef-chiral superﬁeld. [email protected] 111 3.5.2 Supersymmetric Actions

From now on, only work with left-chiral superﬁelds, write Φ+ = Φ, φ+ = φ, F+ = F .

4 4 4 1 Supersymmetric action from chiral superﬁelds: A = d x[f]F + d x[f]F∗ + d x2[K]D , R R R where f is left-chiral, K is real superﬁeld.

Superpotential f is polynomial in left-chiral superﬁelds.

Recall from page 109 that function of left-chiral superﬁelds (but not complex conjugates thereof) is left-chiral superﬁeld.

Superderivatives (and therefore derivatives) of left-chiral superﬁeld is not left-chiral. Let S be a general superﬁeld.

If (a term in) f contains a left-chiral superfield of form D αD βS (see chiral superfield from superfield on page 109), − −

2 can write f = DRh where h is product of S with left-chiral superﬁelds,

since D annihilates all left-chiral superﬁelds after diﬀerentiating with product rule. −

But D 2 h coeﬀ. of θT Eθ in h (neglecting spacetime derivatives which don’t contribute to A from now on). R ∝ R R Then [D 2 h] coeﬀ. of (θT Eθ )(θT Eθ ) = (θγ θ)2 in h, i.e. [h] , so d4x[D 2 h] d4x[h] , R F ∝ L L R R 5 D R F ∝ D

2 R R i.e. DRS terms not necessary in f, can be included in [K]D.

Kahler potential K depends on left / right-chiral superﬁelds but is more general. [email protected] 112 R quantum number assignments for potentials f and K: Rf =2, RK =0.

4 First case: Want d x[f]F to have RF = 0. From chiral superﬁeld decomposition on page 106, term in left-chiral superﬁeld Φ

R T containing F component is θPLθF = θ EθLF . Since RθT Eθ = 2 and want RF = 0, must have Rf = 2. ± L L L

4 Second case: Want d x[K]D to have RD = 0. From general form of superﬁeld on page 103,

R 2 1 2 term containing D component is [θγ5θ] (D + ∂ C). But R 2 = 0, so for RD = 0 must have RK = 0. ∝ 2 [θγ5θ]

For renormalizable theory of chiral superﬁelds:

Superpotential f is at most cubic in left-chiral superﬁelds Φn.

Operators O in L must have mass dimension d (O) 4 for renormalizability. From deﬁnition of D, d (θ) = 1. M ≤ M −2 So d ([S] ) = d (S)+1, d ([S] ) = d (S) + 2, so operators (represented by S here) in f (K) have d (S) 3(2). M F M M D M M ≤

Since dM (Φ ) = dM (φ )=1, f can only be cubic polynomial in Φn. ± ±

General form of Kahler potential: K(Φ, Φ∗)= mn gmnΦm∗ Φn , gmn Hermitian and positive-deﬁnite.

As noted above, operators in K have mass dimensionalityP 2. Φ Φ terms (or conjugates thereof) don’t appear, ≤ m n

since they are left-chiral so have no D term. gmn positive-deﬁnite to get correct commutation relations for ﬁelds. [email protected] 113 1 µ 1 µ 1 µ Kahler potential part of L : [K(Φ, Φ∗)] = ∂ φ∗∂ φ + F ∗F ψ γ ∂ ψ + (∂ ψ )γ ψ . 2 D − µ n n n n − 2 nL µ nL 2 µ nR nR

1 µ 1 µ 1 µ Direct calculation gives [K(Φ, Φ∗)] = g ∂ φ∗ ∂ φ + F ∗ F ψ γ ∂ ψ + (∂ ψ )γ ψ . 2 D mn − µ m n m n − 2 mL µ nL 2 µ mR nR Take Φm′ = NmnΦn, gmn′ = (N †gN)mn. Since gmn Hermitian, choose unitary N to diagonalize it.

µ Diagonal terms positive to get positive coeﬃcient for ∂ φ′∗∂ φ′ , absorb them into φ′ then drop primes. − µ n n n

1 ∂2f(φ) ∂f(φ) Superpotential part of L : [f(Φ)]F = ψ ψmL + Fn . Similar for f ∗. −2 ∂φn∂φm nR ∂φn

Use compact form on page 107 to write f(Φ) = f(φ (x ) √2 θT E ψ (x ) + θT Eθ F (x )). + + − L L + L L + + Taylor expand this in √2 θT E ψ (x ) + θT Eθ F (x ). Note e.g. θ θ = E θ θ for α, β 1, 2, but = 0 otherwise. − L L + L L + + Lα Lβ αβ L1 L2 ≤

T µ Replace x+ with x because θREγ θL gives zero when acting on expression with 2 θL factors.

µ 1 µ 1 µ L from chiral superﬁelds: L = ∂ φ∗∂ φ ψ γ ∂ ψ + (∂ ψ )γ ψ − µ n n − 2 nL µ nL 2 µ nR nR

1 ∂2f(φ) 1 ∂2f(φ) ∗ ψ ψmL (ψ ψmL)∗ V (φ), −2 ∂φn∂φm nR − 2 ∂φn∂φm nR − 2 ∂f(φ) ∗ ∂f(φ) where Fn = and scalar ﬁeld potential V (φ)= . − ∂φn n ∂φn

P ∂f(φ) ∗ Sum Kahler potential and superpotential parts (and conjugate) of L above, use ﬁeld equations to get Fn = . − ∂φn [email protected] 114 ∂f(φ) Zeroth order φn0 = 0 φn 0 at minimum of V (φ). If = 0 possible, V (φ0) = 0 and SUSY unbroken. h | | i ∂φn φ=φ0

Tree-level expansion of V : V (φ) = (M †M )mn∆φm† ∆φn ,

∂2f where ∆φn = φn φn0 , bosonic mass matrix Mmn = . − ∂φm∂φn φ=φ0

2 2 ∂f ∂2f V = n n ∆φm =∆φm† (M †M )mn∆φn. ∂φn ≃ ∂φm∂φn φ=φ0

P P

Matrix M is complex symmetric, so can be diagonalised by unitary matrix such that diagonal terms are real and positive.

From L from chiral superﬁelds on page 113, fermionic mass term is 1M ψ ψ , −2 mn nR mL so fermions and bosons have same mass, as required in SUSY.

µ 2 Free Lagrangian density: L = ∂ ∆φ∗∂ ∆φ m ∆φ∗∆φ 0 n − µ n n − n n n P 1 µ 1 µ 1 1 ψ γ ∂ ψ + (∂ ψ )γ ψ m ψ ψ m (ψ ψ )∗ . −2 nL µ nL 2 µ nR nR − 2 n nR nL − 2 n nR nL

µ 1 µ 1 µ 1 1 L = ∂ ∆φ∗ ∂ ∆φ (M †M ) ∆φ∗ ∆φ ψ γ ∂ ψ + (∂ ψ )γ ψ M ψ ψ M ∗ (ψ ψ )∗ 0 − µ n n − mn m n − 2 nL µ nL 2 µ nR nR − 2 mn nR mL − 2 mn nR mL from L from chiral superfields on page 113 and tree-level expansion of V above, then diagonalize M . [email protected] 115 Problems 3 T 2 T µ µT 2 1: By working in the chiral representation, show that E = E,(γ E) = 1,(γ E) = γ E, γ ∗ = βγ β, β = 1, − 5 − 5 − 5 − γ E, β =0, [E, β] = 0 and (Eγµ)T = Eγµ. { 5 } − T 2: Show that (ψ ) = ψγ E and γµψ = ψγµ, where ψ is a Majorana spinor. − 5 − ∂ µ ∂ µ µ 3: Using Qα = (γ5E)αβ + γ θβ∂µ and Qδ = θα(γ5Eγ )αδ∂µ, show that Qα, Qβ = 2γ ∂µ. ∂θβ αβ ∂θ δ − { } αβ ∂ µ 4: Show that Dβ, Qγ = 0, where Dα = (γ5E)αβ γ θβ∂µ. { } ∂θβ − αβ 5: Show that the result obtained by setting λ = D = 0 and Vµ = ∂µB, where B is scalar field, in a superfield is another superfield, called the chiral superfield. (Hint: show that these constraints are preserved under SUSY transformations.) 6: Show that the chiral superfield is the sum of two chiral superfields, called left- and right-handed chiral superfields respectively, for which N = IM, B = IC, ω = left/right-handed spinor. (Hint: show that these constraints are ± ∓ ± preserved under SUSY transformations. Be careful not to confuse the component fields N, M, ... of the three different superfields.) ∂ µ ∂ µ µ T µ 7: Show that D Φ = 0, where D α = Eαβ ∂θ (γ θL/R)α ∂xµ and x = x θREγ θL. ∓ ± ∓ ∓ R/L β − ± ± µ µ 8: Show that (θREγ θL)∗ = θREγ θL. T T 9: For a general superfield S, calculate D ED S and show that the result is the coefficient of θREθR in S up to spacetime − derivatives. −

10: Show that the D term of ΦΦ′ vanishes, where Φ and Φ′ are left-chiral superﬁelds. [email protected] 116 3.6 Superspace integration

Deﬁne integration over Grassman variables θ as (no sum over α = β = γ) dθ θ = 1, dθ = 0, α 6 6 α α α R R more generally e.g. dθβdθαθαθβθγ = θγ. R Because (θγ θ)2 = 8θ θ θ θ , then d4x[S] = 1 d4x d4θS(x, θ). 5 1 2 3 4 D −2 R R R

T 4 1 4 2 Because θLEθL = 2θ1θ2 in compact form on page 107, then d x[Φ]F = 2 d x d θLΦ(x, θ). R R R

(4) 4 (4) Deﬁne δ (θ′ θ) via d θ′δ (θ′ θ)f(θ′)= f(θ). − −

(4) R 1 T T Solution is δ (θ′ θ) = (θ′ θ )(θ′ θ )(θ′ θ )(θ′ θ )= (θ′ θ ) E (θ′ θ ) (θ′ θ ) E (θ′ θ ) . − 1− 1 2− 2 3− 3 4− 4 4 L − L L − L R − R R − R h i h i

∂ Note e.g. dθ1 f = 0, Try f = 1 amd f = θ1. ∂θ1 R ∂ ∂ and dθ1f g = dθ1 f g. Both sides zero for (f, g) = (1, 1) and (1, θ1), both sides 1 for (θ1, θ1). ∂θ1 ∂θ1 R R [email protected] 117 3.7 Supergraphs

(Super)graphs allow for preservation of Lorentz group symmetry (and SUSY) throughout calculations.

In e.g. theory of left-chiral superﬁelds Φn∗(x, θ) and their complex conjugates,

VEV of time-ordered product of component ﬁelds from

VEV of time-ordered product of superﬁelds:

0 T Φ (x ,θ ), Φ (x ,θ ) . . . 0 = dΦ (x, θ) eIA [Φ]Φ (x ,θ )Φ (x ,θ ) . . . , where h | n1 1 1 n2 2 2 | i n,x,θ n n1 1 1 n2 2 2 R hQ i

1 4 4 Action from spacetime integration: A [Φ] = 2 d x [Φn∗(x, θ)Φn(x, θ)]D + 2Re d x [f(Φ)]F . R R To get supergraph Feynman rules, ﬁrst write A [Φ] in form A = d4x d4θ something. × R R

2 Introduce superﬁelds Sn such that Φn = DRSn . No loss of generality - see page 110.

2 4 4 1 2 2 2 Action from superspace integration: A [D S]= d x d θ S∗D D S Re[f˜(D S,S)] , R −4 n L R n − R R R ˜ 2 2 ˜ 2 ˜ 2 where f deﬁned by f(DRS)= DRf(DRS,S). In f, all superﬁelds acted on by DR except one.

2 Recall DRSn is left-chiral because DRαDRβDRγ = 0.

2 2 2 2 2 2 So Φn∗ (x, θ)Φn(x, θ) = (DLSn∗)DRSn = Sn∗DLDRSn + EαβDLα((DLβSn∗)DRSn + Sn∗DLβDRSn).

2 2 Now for any superﬁeld T , [DL/RT ]D =0, so [Φn∗ (x, θ)Φn(x, θ)]D =[Sn∗DLDRSn]D.

2 n 2 2 n 1 2 2 2 2 ˜ 2 Because e.g. (DRT ) = DR(T (DRT ) − ) (follows from (DR) = 0) then f(DRS) = DRf(DRS,S).

4 T 4 4 2 ˜ 2 4 ˜ 2 From page 111, d x[D ED h]F d x[h]D, i.e. d x[DRf(DRS,S)]F = d x[f(DRS,S)]D. − − ∝ R R R R 4 1 2 2 ˜ 2 4 1 4 4 So A = d x Sn∗DLDRSn + 2Re f(DRS,S) . Then use d x[S]D = d x d θS(x, θ) from page 116. 2 D D −2 R h i R R R [email protected] 119 Also we can replace dΦ (x, θ) dS (x, θ) in path-integral. n,x,θ n → n,x,θ n Consider lattice ofQN points, ﬁelds φ Qψ = φ φ = C φ . Then dN ψ = det(C)dN φ = dN φ because det(C)=1. i −→ i i+1 − i ij j

Thus VEV of time-ordered product of superﬁelds on page 117 becomes

2 0 T Φ (x ,θ ), Φ (x ,θ ) . . . 0 = D 2 D 2 . . . dS (x, θ) eIA [DRS]S (x ,θ )S (x ,θ ) . . . h | n1 1 1 n2 2 2 | i 1R 2R n,x,θ n n1 1 1 n2 2 2 R hQ i

= D 2 D 2 . . . 0 T S (x ,θ ),S (x ,θ ) . . . 0 . 1R 2R h | n1 1 1 n2 2 2 | i Action invariant under S S + D X , where X is any superﬁeld. n → n R n n

In A [D 2 S], D 2 S D 2 S + D 2 D X but D 2 D = 0. R R → R R R n R R

Φ 1 2 2 (4) Propagator for left-chiral superﬁelds: ∆ = D D ′ ∆ (x y)δ (θ θ′)δ . nm 4 R L F − − nm

S 1 (4) Propagator for S is ∆ = ∆ (x y)δ (θ θ′)δ + D -terms (see page 120). nm 4 F − − nm R

S ∆nm is superpropagator for line created by superﬁeld Sm∗ (x′, θ′) and destroyed by Sn(x, θ),

2 2 Φ 2 2 S and Φm∗ (x′, θ′) = DL′ Sm∗ (x′, θ′) and Φn(x, θ) = DRSn(x, θ), so ∆nm = DRDL′ ∆nm. [email protected] 120 S 1 (4) Superpropagator: ∆ = ∆ (x y)δ (θ θ′)δ + D -terms . nm 4 F − − nm R Propagator usually comes from quadratic terms in action. If i labels spacetime position (on lattice) as well as spin etc.,

1 so A [φ] = D φ∗φ where D is Hermitian, then ∆ = D− . But D not invertible if A [φ] invariant under quad − ij i j quad φ φ + ξ , i.e. D ξ = 0. Let u be remaining eigenvectors of D, i.e. (no sum over ν unless explicitly indicated) i → i i ij j ν

Aquad[φ] Dijuνj = dνuνi, uνi∗ uν′i = δνν′ , uνi∗ ξi = 0. In path integral Pa...b... = i dφidφi∗e φa ...φb∗ ..., transform from φi R Q to φ(′ν) with Jacobian J , where φi = φ′ξi + ν φν′ uνi: With C = J dφ′dφ′∗ and, up to terms with 1+ factors of ξ,

′ 2 P R I ν dν φν Pa...b... = C ν dφν′ dφν′∗e− | | [ ν φν′ uνa] ... [ ν φν′ uνb]∗ ... + ξ terms = pairings[ I∆ab] + ξ terms, P − − −

R Q ∗ P P P uνauνb where propagator ∆ab = . Replacing Aquad with full gauge-invariant A , ξ-terms vanish ν dν P because φi contracted with objects Ji obeying Jiξi = 0. So, alternatively, solve Dac∆cb = Πab to get ∆ab,

2 where projection operator Πab = ν uνauνb∗ obeys Π = Π and Πξ = 0, because solution for ∆ab unique P up to these irrelevant ξ-terms. From action from superspace integration on page 118,

1 2 2 S (4) (4) 2 D D ∆ (x, θ; x′, θ′) = Pδ (x x′)δ (θ θ′)δ , where P obeys P = P and PD X = 0. 4 L R nm − − nm R n

1 1 2 2 2 2 2 2 2 Solution for P is P = − D D , because D D D = 16D and D D = 0, −16 L R R L R − R R R

S S 1 (4) so solution for ∆ is ∆ = ∆ (x y)δ (θ θ′)δ + D -terms. nm nm 4 F − − nm R [email protected] 121 4 IA [φ]+I d xφl(x)Jl(x) Generating functional: Z[J, A ]= x,l dφl(x)e . Contains all S-matrix elements. R R Q 0, out T φ (x ),φ (x ),... 0, in h | lA A lB B | i 1 δ δ From page 61, 0, out 0, in = Z[J,A ] I δJ (x ) I δJ (x ) ...Z[J, A ] . − lA A − lB A h | i J=0

Z[J, A ] = tree-level part of Z[J, Γ], where

4 δ( I ln Z[J,A ]) Quantum eﬀective action: Γ[φ]= d xφl(x)Jφl(x) I ln Z[Jφ, A ], Jφ is solution of φl = − . − − δJl(x) J=Jφ R

−1 4 g [IΓ[φ]+I d xφl(x)Jl(x)] Deﬁne Z[J, Γ, g] = x,l dφl(x)e , i.e. Z[J, Γ, 1] = Z[J, Γ]. Propagator for Z[J, Γ, g] is g, R ∝ R Q 1 1 because it is inverse of coeﬃcient of term in g− Γ[φ] which is quadratic in φ . Each vertex is g− . l ∝

L 1 (L) So for L = number of loops, Z[J, Γ, g] = ∞ g − Z [J, Γ]. Now take g 0, so integration in Z[J, Γ, g] is dominated L=0 →

P −1 4 g [IΓ[φJ ]+I d xφJl(x)Jl(x)] by φ-conﬁgurations that give stationary phase, i.e. Z[J, Γ, g] e where φJ such that ∝ R

δ 4 δ IΓ[φ] + I d xφl(x)Jl(x) = 0, or Γ[φ] = Jl(x). Thus Jφ = J because, from deﬁnition of Γ[φ] and Jφ, δφl(y) φ=φJ δφl(x) − J φ=φJ R

δ 4 δJφk(y) 4 δ( I ln Z[J]) δJφk(y) Γ[φ] = d yφk(y) Jφl(x) + d y − = Jφl(x). Proportionality constant in underlined equation δφl(x) − δφl(x) − δJk(y) δφl(x) J=Jφ R R

0 1 (0) 4 begins at O(g ) so, taking logs, coefficient of g− is I ln Z [J, Γ] = Γ[φ ] + d xφ (x)J (x) = I ln Z[J , A ] = I ln Z[J, A ]. − J Jl l − φJ − R Coefficients of products of fields in Γ[φ] from corresponding connected 1 particle irreducible (1PI) graphs. [email protected] 122 Γ[S,S∗] is local in fermionic coordinates.

Each incoming (outgoing) external line is attached to vertex labelled (x, θ), gives a factor Sn(x, θ)(Sn∗(x, θ)).

˜ 2 Each vertex gives factor equal to coeﬃcient of corresponding product of ﬁelds in f(DRS,S).

I (4) Each propagator from vertex (x, θ) to vertex (x′, θ′) gives δ (θ θ′)∆ (x x′). −4 − F −

˜ 2 Recall from page 118 that in f all superﬁelds acted on by DR except one.

2 2 So a DR (DL) acts on all lines but one that come into (go out of) a vertex.

But Γ[S,S∗] obtained by restricting to 1PI graphs and integrating over all xs and θs.

2 2 So use integration by parts to move all DR (DL) on propagators to external line factors Sn(x, θ)(Sn∗(x, θ)).

(4) Integrate out all δ (θ θ′) so that Γ[S,S∗] is integral over single 4-D θ. − [email protected] 123 Graphs in Γ[S,S∗] with loops are D-terms, the rest are F -terms.

2 2 For V (V ∗) vertices with n incoming (outgoing) lines, number of D (D ) is N = V (n 1) (N = V ∗(n 1)). n n R L R n n − L n n − P P Let there be E (E∗) incoming (outgoing) external line factors Sn(x, θ)(Sn∗(x, θ)), and I internal lines.

Because I + E(E∗) = nV (V ∗) and number of loops is L = I V V ∗ + 1, n n n − n n − n n P P P N = L + V ∗ + E 1 (N = L + V + E∗ 1). Thus a graph with any loops has N E and N E, R n n − L n n − R ≥ L ≥

P 2 2 P i.e. enough DR (DL) operators to act on all external line factors Sn(x, θ)(Sn∗(x, θ)) at least once

2 2 i.e. such graphs are integral over single 4-D θ of a functional of DRSn = Φn and DLSn∗ = Φn∗ , which is a D-term.

2 2 Graphs with N = E 1 D operators (N = E 1 D operators) have both L = 0 and V ∗ = 0 (V = 0), R − R L − L

2 2 so there can only be one vertex and no loops, and there is a single Sn(x, θ)(Sn∗(x, θ)) not acted on by DR (DL).

4 4 2 4 2 But recall from page 118 that d x d θg˜(DRS,S) = d x g(DRS) F , i.e. such graphs are F -terms. R R R

Superpotential f(S) not renormalized at any order in perturbation theory.

f(S) contributes in form of F -term, but F -terms in Γ[S,S∗] are free of loops (no inﬁnite renormalization).

Also there is just one vertex, which must be the integrated [f(S)]F (no ﬁnite renormalization). [email protected] 124 3.8 SUSY current

µ µ µ µ µ SUSY current: S = J + K (4-vector of Majorana spinors), where K appears in δL = α∂µK ,

µ ∂RL l J is the Noether current in explicit result for Noether current when L is invariant on page 59: αJ = l δχ . − ∂(∂µχ )

But L is bot invariant under SUSY, so add Kµ to deﬁnition of SUSY current. Then

µ Conservation of SUSY current: ∂µS =0.

µ µ ∂RL l ∂RL l (Following argument is general.) From deﬁnition of J above, α∂µJ = ∂µ l δχ l δ∂µχ . − ∂(∂µχ ) − ∂(∂µχ ) µ ∂RL l ∂RL l Using Euler-Lagrange equations on page 58, α∂µJ = l δχ l δ∂µχ = δL , − ∂χ − ∂(∂µχ ) −

µ cancels δL = α∂µK . [email protected] 125 Generator of SUSY transformations: d3xαS0, χl = Iδχl . R (Following argument is general.) Let L = L(q, q˙), δq = δq(q, q˙), canonical coordinates q are all ﬁelds χl at all spacetime points.

3 3 µ 3 dK0 ∂L n ∂L n dF 3 0 Recall L = d xL , so δL = d xα∂µK = d xα dt or ∂qn δq + ∂q˙n δq˙ = dt , where F = d xαK . R R R R But ∂L = d ∂L , so underlined equation is Q˙ = 0 where conserved charge Q = ∂L δqn + F . ∂qn dt ∂q˙n −∂q˙n

So Q = d3xαS0, because ∂L δqn = d3xαJ 0 from the deﬁnition of the Noether current. Must show [Q, χl] = Iδχl: −∂q˙n R R Now [Q, qm] = ∂L , qm δqn ∂L [δqn, qm]+[F, qm]. In 1st term, use QM result ∂L , qm = Iδm (assuming no constraints). − ∂q˙n − ∂q˙n ∂q˙n − n h i h i m n m For 2nd and 3rd terms consider general [f(q, q˙), q ]. We may write f = k∞=0 An1...nk (q)˙qn1 ... q˙nk . Using QM result [q , q ]=0,

m k P ∂f we ﬁnd [f(q, q˙), q ] = ∞ An ...n (q) q˙n ... q˙n [˙qn , qm]˙qn ... q˙n = [˙qn, qm] k=0 1 k r=1 1 r−1 r r+1 k ∂q˙n P P l (in last expression, insert [q ˙ , q ] in correct position amongq ˙ products). Then [Q, qm] = Iδqm + ∂F ∂L ∂δq [˙qn, qm]. n m ∂q˙n − ∂q˙l ∂q˙n ∂F ∂L ∂δql ˙ ˙ n ˙ dF But ∂q˙n = ∂q˙l ∂q˙n : In underlined equation above, replace X, where X = δq˙ on LHS and X = dt on RHS,

˙ l ∂X l ∂X ∂L n l ∂L ∂δqn l ∂L ∂δqn n ∂F n ∂F with chain rule X =q ˙ ∂ql +q ¨ ∂q˙l . Then ∂qn δq +q ˙ ∂q˙n ∂ql +q ¨ ∂q˙n ∂q˙l =q ˙ ∂qn +q ¨ ∂q˙n .

Match coeﬃcients ofq ¨n, which does not appear implicitly in any term.

Thus [Q, qm] = Iδqm. Taking the time derivative also gives [Q, q˙m] = Iδq˙m. Can be extended to cases with constraints. [email protected] 126

µ µ ∂f µ SUSY current for chiral supermultiplet: S = √2 (/∂φn)γ ψnR + γ ψnL + (φn φ∗,ψnR ψnL) . ∂φn → n ↔ h i µ 1 µ µ µ µ µ µ Noether current: J = [2(∂ φ∗ )ψ + 2(∂ φ )ψ + (/∂φ )γ ψ + (/∂φ∗ )γ ψ F γ ψ F ∗γ ψ ]: √2 n nL n nR n nR n nL − n nR − n nL Use explicit result for Noether current when L is invariant on page 59, where L from chiral superﬁelds on page 113 is

µ 1 µ 1 µ L = ∂ φ∗ ∂ φ ψ γ ∂ ψ + (∂ ψ )γ ψ + non-spacetime-derivative terms, − µ n n − 2 nL µ nL 2 µ nR nR change in ﬁelds from left / right transformation on page 107 (excluding δF ). ±

µ 1 µ ∂f ∂f ∗ Next, K = γ (/∂φn)ψnR (/∂φn∗ )ψnL + Fn∗ψnL + FnψnR + 2 ψnL + 2 ψnR : √2 − − ∂φn ∂φn h i µ Recall δL = α∂µK from page 125. In supersymmetric action from chiral superﬁelds on page 111,

use (from left / right transformation on page 107) δ[f]F = √2 α∂/[f]ψL ,

and (from component ﬁeld transformations on page 104) δ[K]D = Iαγ5∂/[K]λ. [email protected] 127 µ µ µ √2 µ ν Alternative SUSY current: S = K + J + 3 [γ ,γ ]∂ν (φnψnR + φn∗ψnL).

µν µ µν µν Can always add ∂νA to S (A is some antisymmetric tensor of spinors), because 1. it is conserved (∂µ∂νA = 0),

3 0ν 3 0i 2. generator of SUSY transformations on page 125 unchanged because d xα∂νA = d xα∂iA = 0. R R This gives

µ ∂2f ∂f Measure of scale invariance violation: γµS = 2√2 φm 2 ψnL + (L R, φ φ∗). − ∂φn∂φm − ∂φn → → ∂2f ∗ From L from chiral superﬁelds on page 113, Dirac equations are∂ψ / mL = ψnR − ∂φm∂φn µ µ and same with L R and φ φ∗. Use SUSY current for chiral supermultiplet on page 126 (= K + J ) ↔ → and alternative SUSY current above.

µ because γµS vanishes when f is 3rd order homogeneous polynomial (f = Cmnpφmφnφp), in which case coupling (Cmnp) is dimensionless. [email protected] 128 3.9 Spontaneous supersymmetry breaking

∂f(φ) Broken SUSY vacuum condition: 0 F 0 = 0 or = 0 or V (φ0) > 0. h | | i 6 ∂φn 6 φ=φ0

In left / right transformation on page 107, want to make one of 0 δΨ 0 = 0, where Ψ = ψ , φ or F . h | l| i 6 l nL n n Cannot make 0 ψ 0 = 0 since vacuum Lorentz invariant, nor 0 ∂ φ 0 = 0 since 0 φ 0 constant. h | nL| i 6 h | µ n| i 6 h | n| i Only possibility is 0 F 0 = 0 (whence 0 δψ 0 = √2 α 0 F 0 ), h | n| i 6 h | nL| i h | n| i which from L from chiral superﬁelds on page 113 is equivalent to last 2 statements.

Spontaneous SUSY breaking requires f such that ∂f(φ) = 0 has no solution. ∂φn φ=φ0

1 Spontaneous SUSY breaking gives rise to massless spin 2 goldstino.

2 ∂V ∂f ∂2f ∂f ∗ ∂f ∗ ∂f(φ) = 0. But V = , so 2 = 0, i.e. Mmn = 0. But one+ of = 0, ∂φn n ∂φn n ∂φm∂φn ∂φn n ∂φn ∂φn 6 φ=φ0 φ=φ0 φ=φ0 φ=φ0 P P P so M has at least one zero eigenvalue, so from free Lagrangian density on page 114, since M eigenvalues are the m n,

there is at least one linear combination of ψnL with zero mass.

In local SUSY, goldstino absorbed into longitudinal component of gravitino, gives it a mass. [email protected] 129 3.9.1 O’Raifeartaigh Models

Theories in which left-chiral ﬁelds Xn, Yi have R = 0, 2. Most general superpotential is f(X,Y )= i Yifi(X). P From R quantum number assignments for potentials f and K on page 112, i.e. Rf = 2 (X∗, Y ∗ not allowed since f left-chiral).

SUSY broken when no. ﬁelds X < no. ﬁelds Y .

∂f(x,y) Write scalar components of X, Y as x, y. Condition = 0 implies fi(x) = 0, i.e. more conditions than ﬁelds Xi, ∂yi

only possible to satisfy by careful choice of the fi(x).

2 2 ∂fi(x) Scalar ﬁeld potential: V = fi(x) + yi . i | | n i ∂xn

P P P

From general superpotential above and definition of scalar field pote ntial in L from chiral superfields on page 113.

Simplest (renormalizable) model: Fields X, Y1, Y2.

Then choice f (X)= X a, f (X)= X2 is general, for which V = x 4 + x a 2 + y + 2xy 2. 1 − 2 | | | − | | 1 2|

Renormalizability allows fi to be quadratic only. Then take linear combinations of Yi and shift and rescale X.

∂f ∂f ∂f = = = 0 only possible if we can make f1 = f2 = 0, only possible if a = 0. ∂X ∂Y1 ∂Y2 [email protected] 130 3.10 Supersymmetric gauge theories

Gauge transformation of left-chiral supermultiplet: (φ, ψ , F ) (x) exp[It ΛA(x)] (φ, ψ , F ) (x). L n → A nm L m Fields in same supermultiplet have same transformation properties under internal symmetry transformation U:

( ) ( ) ( ) Since [Q, a † ] a † , where a † are annihilation (creation) operators for bosonic and fermionic superpartners, B/F ∓ ∼ F/B B/F

( ) ( ) ( ) general transformation of aσ† on page 10 is same for aB† , aF† because [U, Q] = 0 from page 89.

So from complete ﬁeld on page 27, transformation same for all ﬁelds in supermultiplet.

Gauge transformation of left-chiral superﬁelds: Φ(x, θ) exp[It ΛA(x )]Φ(x, θ) (column vector Φ). → A + Implied by gauge transformation of left-chiral supermultiplet above and compact form on page 107.

A Gauge transformation of right-chiral superﬁelds: Φ†(x, θ) Φ†(x, θ) exp[ ItAΛ (x )] . → − −

A From conjugate of gauge transformation of left-chiral superﬁelds above. Note Λ (x) is real function of x and x+∗ = x . −

4 4 Global gauge invariance = local for d x[f(Φ)] but not for d x[K(Φ, Φ†)] . ⇒ F D

R A R No derivatives / conjugates of Φ in f, so argument y of Λ (y) irrelevant. But K also contains Φ†,

A A so e.g. Φn† Φn not invariant under gauge transformation of left and right-chiral superﬁeld above because Λ (x+) = Λ (x ). 6 − [email protected] 131 Deﬁne gauge connection Γ(x, θ), with transformation property

A A Gauge transformation of gauge connection: Γ(x, θ) exp[ItAΛ (x )]Γ(x, θ) exp[ ItAΛ (x+)] , → − − then “new” right-chiral superﬁeld Φ†(x, θ)Γ(x, θ) has transformation property

A Gauge transformation of new right-chiral superﬁeld: Φ†(x, θ)Γ(x, θ) Φ†(x, θ)Γ(x, θ) exp[ It Λ (x )] . → − A + 4 Global gauge invariance = local for d x[K(Φ, Φ†Γ)] . Also implies invariance under ⇒ D R Extended gauge transformations: Φ(x, θ) exp[It ΩA(x, θ)]Φ(x, θ), → A

A A A Γ(x, θ) exp[It Ω †(x, θ)]Γ(x, θ) exp[ It Ω (x, θ)] with any left-chiral superﬁelds Ω (x, θ). → A − A

We are extending ΛA(x ) to ΩA(x, θ) = ΛA(x ) √2 θT E ψ (x ) + θT Eθ F (x ) in compact form on page 107. + + − L ΛL + L L Λ+ +

A Φ(x, θ) is left-chiral (depends on x+, θL only), so cannot introduce x or θR in Ω (x, θ). −

1 1 Hermitian connection: Γ(x, θ)=Γ (x, θ). Take Γ (Γ + Γ†) or (Γ Γ†), note Γ, Γ† have same transformation. † → 2 2I −

Connection from gauge superﬁelds: Γ(x, θ) = exp[ 2t V A(x, θ)] , with real gauge superﬁelds V . − A A A Form preserved by gauge transformation of gauge connection above, new V independent of tA representation.

See Baker-Hausdorﬀ formula on page 9. [email protected] 132 A A I A A Extended gauge transformation of gauge superﬁelds: V (x, θ) V (x, θ)+ [Ω (x, θ) Ω †(x, θ)] + . . . , → 2 − where “. . .” = commutators of generators, which vanish for zero coupling and in U(1) (Abelian).

From extended gauge transformations and deﬁnition of gauge superﬁelds above.

Gauge superﬁelds in terms of components: V A(x, θ)= CA(x) Iθγ ωA(x) I θγ θM A(x) 1θθN A(x) − 5 − 2 5 − 2

+I θγ γµθV A(x) Iθγ θθ[λA(x)+ 1∂ω/ A(x)] 1(θγ θ)2(DA(x)+ 1∂2CA(x)) . 2 5 µ − 5 2 − 4 5 2 From general form of superﬁeld on page 103.

Transformation superﬁelds in terms of components: ΩA(x, θ)= W A(x) √2θP wA(x)+ W A(x)θP θ − L L +1θγ γ θ∂µW A(x) 1 θγ θθ∂P/ wA(x) 1(θγ θ)2∂2W A(x). 2 5 µ − √2 5 L − 8 5

From chiral superﬁeld decomposition on page 106, taking left-chiral part (Φ+ there).

Transformation of gauge supermultiplet ﬁelds: CA(x) CA(x) ImW A(x)+ . . . , → − ωA(x) ωA(x)+ 1 wA(x)+ . . . , V A(x) V A(x)+ ∂ ReW A(x)+ . . . , M A(x) M A(x) ReW A(x)+ . . . , → √2 µ → µ µ → −

N A(x) N A(x)+ImW A(x)+ . . . , λA(x) λA(x)+ . . . , DA(x) DA(x)+ . . . . → → → From all 3 above results. Again, “...” = commutators of generators, which vanish for zero coupling and in U(1) (Abelian). [email protected] 133 Wess-Zumino gauge: CA, ωA,M A,N A = 0 V A(x, θ)= I θγ γµθV A(x) Iθγ θθλA(x) 1(θγ θ)2DA(x). → 2 5 µ − 5 − 4 5

Transformation of supermultiplet ﬁelds on page 132: take ImW A(x) = CA(x), wA(x) = √2ωA(x), W A(x) = M A(x) IN A(x). − − Suﬃcient for Abelian case because “...” = 0. For non-Abelian case, add nth order (in gauge couplings) terms

to ImW A(x), wA(x), W A(x) to cancel nth order terms from commutators of m n 1th order terms. ≤ −

Wess-Zumino gauge is not supersymmetric.

A A A A A A Ensuring δC ,δω , δM ,δN = 0 in component ﬁeld transformations on page 104, requires also having Vµ , λ = 0,

A A A A A A which requires δVµ , δλ = 0. δVµ = 0 satisﬁed, but δλ = 0 requires D = 0, i.e. V = 0.

1 1 µ 1 µ 1 Gauge invariant L for chiral supermultiplet: [Φ†ΓΦ] = (D φ)†D φ ψ γ D ψ + F †F 2 D −2 µ − 2 L µ L 2

A 1 A A +I√2ψ t λ φ D φ†t φ + h.c. , where covariant derivative D = ∂ It V L A − 2 A µ µ − A µ (gauge transformation for gauge superﬁeld derived next on page 134).

In Wess-Zumino gauge, Γ(x, θ) = 1 Iθγ γ θt V A(x) 1θγ γµθθγ γνθt t V A(x)V B(x) + 2Iθγ θt θλA(x) + 1(θγ θ)2t DA(x). − 5 µ A µ − 2 5 5 A B µ ν 5 A 2 5 A Then use left-chiral superﬁeld in chiral superﬁeld decomposition on page 106.

Note D term is coefficient of 1(θγ θ)2 minus 1∂2 first “φ” term. −4 5 2 × [email protected] 134 A B C A Gauge transformation of gauge supermultiplet fields: δgaugeVµ (x)= CABCVµ (x)Λ (x)+ ∂µΛ (x)

(which is inﬁnitesimal version of transformation of gauge ﬁelds on page 66), and

A B C A B C A A δgaugeλ (x)= CABCλ (x)Λ (x), δgaugeD (x)= CABCD (x)Λ (x) (i.e. λ , D in adjoint representation).

Firstly ΛA(x ) = ΛA(x) + 1(θγ γ θ)∂µΛA(x) 1(θγ θ)2∂2ΛA(x). Then gauge transformation of gauge connection on page 131 + 2 5 µ − 8 5

A A A A (using connection from gauge superﬁelds) reads exp[ 2t V ′(x, θ)] = exp[It Λ †(x )] exp[ 2t V (x, θ)] exp[ It Λ (x )]. − A A + − A − A +

A a X b A I A A 1 2 A Write as exp[ 2t V ′(x, θ)] = e e e where X = 2t V (x, θ) = 2t [ θγ γ θV (x) Iθγ θθλ (x) (θγ θ) D (x)] , and − A − A − A 2 5 µ µ − 5 − 4 5

small quantities b + a = 2t ImΛA(x ) = Iθγ γ θt ∂µΛA(x) , b a = 2It ReΛA(x ) = 2It [ΛA(x) 1(θγ θ)2∂2ΛA(x)] . A + − 5 µ A − − A + − A − 8 5

A 1 To ﬁrst order in a, b, 2t V ′(x, θ) = X + [X,b a] + b + a + ... and “...” means higher order in a, b, − A 2 −

as well as terms of O(Xn)O(b a) with integer n 2 which vanish: − ≥ since (1,γ ) = P P , X θ θ ( further θ factors), while b + a θ θ , so O(X2)O(b a) θ3 θ3 ( ...)=0. 5 L ± R ∼ L R × ∼ L R − ∼ L R ×

A A B C I µ A So V ′(x, θ) = V (x, θ) + CABCV (x, θ)Λ (x) + 2 θγ5γµθ∂ Λ (x),

compare superﬁeld expansion in Wess-Zumino gauge on page 133.

Wess-Zumino gauge is gauge invariant under ordinary gauge transformations.

Last underlined equation above shows that δCA = δωA = δM A = δN A = 0. [email protected] 135 3.10.1 Gauge-invariant Lagrangians

Construct gauge-invariant Lagrangian from

A T A A Gauge-covariant spinor superﬁeld: 2tAWLα(x, θ)= D ED exp[2tAV (x, θ)]D+α exp[ 2tAV (x, θ)] − − − which is left-chiral because D WL = 0 (Dα Dβ Dγ = 0), i.e. − − − − Extended gauge transformation of W A: 2t W A (x, θ) exp[It ΩA(x, θ)]2t W A (x, θ) exp[ It ΩA(x, θ)] . L A Lα → A A Lα − A From extended gauge transformations and connection from gauge superﬁelds on page 131,

A T A A A A A A 2tAWLα D ED exp[ItAΩ ] exp[2tAV ]exp[ ItAΩ †] D+α exp[ItAΩ †]exp[ 2tAV ]exp[ ItAΩ ]. → − − − − −

A A A Use product rule for the D here. Since Ω is left-chiral, D Ω = D+Ω † = 0. ± −

A A T A A A A A Then 2tAWLα exp[ItAΩ ]D ED exp[2tAV ]exp[ ItAΩ †]exp[ItAΩ †]D+α exp[ 2tAV ]exp[ ItAΩ ] → − − − − −

A T A A A A = exp[ItAΩ ]D ED exp[2tAV ]D+α exp[ 2tAV ]exp[ ItAΩ ]. Not ﬁnished, because D+αexp[ ItAΩ ] = 0 − − − − − 6

T A T T T A (however, D ED exp[ ItAΩ ] does vanish). But D ED D+α = D+αD ED 4[PL∂/D ]α, so D ED D+α exp[ ItAΩ ]=0. − − − − − − − − − − − −

A A T A A A So 2tAWLα exp[ItAΩ ] D ED exp[2tAV ]D+α exp[ 2tAV ] exp[ ItAΩ ], → − − − −

where all derivatives in quantity between and evaluated before multiplying on left / right with exp[ It ΩA]. ± A [email protected] 136 Form of W A: W A(x, θ)= λA(x )+ 1γµγνθ F A (x )+ θT Eθ Dλ/ A(x ) Iθ DA(x ). L L L + 2 L µν + L L R + − L +

Note exp[ 2t V A(x, θ)] = 1 Iθγ γ θt V A(x) 1θγ γµθθγ γνθt t V A(x)V B(x) + 2Iθγ θt θλA(x) + 1(θγ θ)2t DA(x). − A − 5 µ A µ − 2 5 5 A B µ ν 5 A 2 5 A

A After performing all D in gauge-covariant spinor superﬁeld on page 135, choose gauge where Vµ (X) = 0 at given x = X, ±

gives W A(X, θ) = λA(X ) + 1γµγνθ (∂ V A(X ) ∂ V A(X )) + θT Eθ ∂λ/ A(X ) Iθ DA(X ). Then convert to L L + 2 L µ ν + − ν µ + L L R + − L + gauge covariant form consistent with this, i.e. ∂ V A ∂ V A F A (the non-Abelian ﬁeld strength of page 67) and∂ / D/ . µ ν − ν µ → µν →

A A A µνρσ Alternative form of WL : D+EWL = D EWR withD / ∂/, ǫ ∂ρfµν = 0. − →

A A A Note WR from WL via x+ x , R L, E E. Implies form of WL above, by direct calculation. → − ↔ →−

A Gauge supermultiplet Lagrangian: L = 1Re([W AT EW A] )= 1F A F Aµν 1λ (Dλ / )A + 1DADA . gauge −2 L L F −4 µν − 2 2

A From form of WL above.

2 2 A Additional non perturbative part: L = g θ Im [W AT EW A] = g θ Iλ Dγ/ λA 1ǫ F AµνF Aρσ , θ −16π2 L L F −16π2 5 − 4 µνρσ where g is coupling appearing in tA.

Fayet-Iliopolis term: LFI = ξD , where ξ is arbitrary constant, D is for Abelian supermultiplet..

Corresponding action is supersymmetric, because δD = Iαγ5∂λ/ is derivative. [email protected] 137 Explicit check that action from gauge supermultiplet Lagrangian Lgauge on page 136 is supersymmetric.

A A A A A In Vµ (X) = 0 gauge, component ﬁeld transformations on page 104 at x = X: δVµ = αγµλ , δD = Iαγ5∂λ/ , and

δλA = (1F A [γν,γµ] + Iγ DA)α where F A = ∂ V A ∂ V A + C V BV C is non-Abelian ﬁeld strength of page 67. 4 µν 5 µν µ ν − ν µ ABC µ ν Then using δ(F A F Aµν) = 2F A δF Aµν etc., δ(F A F Aµν) = 2F Aµνα(γ ∂ γ ∂ )λA, δ(DADA) = 2IDAαγ ∂λ/ A µν µν µν ν µ − µ ν 5

A B A B A B A B 1 A µ ν A A B C and δ(λ D/ ABλ ) = (δλ )D / ABλ + λ D/ ABδλ + λ (δD/ AB)λ = 2α[4Fµν[γ ,γ ] + Iγ5D]/∂λ + CABCλ δ(V / )λ

A (we will show C λ δ(V / B)λC = 0 later). For this last expression, use [γµ,γν]γρ = 2gµργν + 2gνργµ 2Iǫµνρσγ γ5 ABC − − σ (expand in 16 matrices on page 45, Lorentz and space inversion invariance limits this to these three terms.

1st coefficient by taking µνρ = 121: [γ1,γ2]γ1 = 2g11γ2, correct because from anticommutation relations for γµ on page 44, − γ1γ2 = γ2γ1 and γ1 2 = 2. Similarly for µνρ = 211 to get 2nd coefficient. 3rd coefficient from µνρ = 123: − [γ1,γ2]γ3 = 2Iǫ1230γ γ , LHS is 2γ1γ2γ3 = 2γ γ0γ1γ2γ3 = 2Iγ γ ). − 0 5 − 0 − 0 5

A So δ(λ ∂λ/ A) = F Aµνα(γ ∂ γ ∂ )λA IǫµνρσF A αγ γ ∂ λA + 2IDAαγ ∂λ/ A. 2nd term replaceable, after integration by parts, − ν µ − µ ν − µν σ 5 ρ 5

µνρσ A A µνρσ A A Aµν A A by Iǫ (∂ρFµν)αγσγ5λ , which vanishes since e.g. ǫ ∂ρ∂µVν = 0, 1st and 3rd terms cancel with δ(FµνF ) and δ(D D ).

A B C A C µ B Thus we are left with CABCλ δ(V / )λ = CABCλ γµλ αγ λ = 0,

where antisymmetry of CABC used in 1st step. Last step requires explicit calculation. [email protected] 138 Lagrangian for chiral and gauge supermultiplet ﬁelds:

µ 1 µ 1 µ 1 ∂2f T 1 ∂2f ∗ T L = (Dµφ)† (D φ)n ψ γ (DµψL)n + Dµψ γ ψnL ψ EψmL ψ EψmL ∗ − n − 2 nL 2 nL − 2 ∂φn∂φm nL − 2 ∂φn∂φm nL A A 1 A Aµν 1 A A g2θ Aµν Aρσ V (φ)+ I√2ψ (t ) λ φ I√2φ† λ (t ) ψ F F λ (Dλ / ) + ǫ F F , − nL A nm m − n A nm mL − 4 µν − 2 64π2 µνρσ

∂f(φ) ∂f(φ) ∗ 1 A A where potential V (φ)= + ξ + φ∗(t ) φ ξ + φ∗(t ) φ . ∂φn ∂φn 2 n A nm m k A kl l Sum of gauge invariant L for chiral supermultiplet on page 133, superpotential part of L on page 113 and 3 Lagrangians

1 A 1 AT A A A g2θ AT A on page 136 gives L = Φ† exp 2t V Φ + 2Re[f(Φ)] Re W EW ξ D 2 Im(W EW ) , 2 − A D F − 2 L L F − − 16π L L F T T T or explicitly (using Majorana conjugation on page 96, e.g. ψR = ψ†PRβ = ψPL = ψ Eγ5PL = ψ PLEγ5 = ψL Eγ5)

µ 1 µ ∂2f T ∂f(φ) A L = (Dµφ)n† (D φ)n ψnγ (Dµψ)n + Fn†Fn Re ψn Eψm + 2Re Fn 2√2Im(tA)nmψnLλ φm − − 2 − ∂φn∂φm ∂φn −

A A A A 1 A A 1 A Aµν 1 A A g2θ Aµν Aρσ +2√2Im(t ) ψ λ φ† φ† (t ) φ D ξ D + D D F F λ (Dλ / ) + 2 ǫ F F . A mn nR m − n A nm m − 2 − 4 µν − 2 64π µνρσ

∂f(φ) † A A Then use ﬁeld equations for auxiliary ﬁelds: Fn = and D = ξ + φn∗ (tA)nmφm. − ∂φn [email protected] 139 3.10.2 Spontaneous supersymmetry breaking in gauge theories

∂f(φ) A A Unbroken SUSY vacuum: F = =0, D = ξ + φ∗ (t ) φ = 0 V (φ)=0. n0 ∂φn 0 n0 A nm m0 − φ=φ0 ⇐⇒ h i A A A Can write V = Fn∗Fn + D D > 0, so V = 0 (if allowed) is a minimum. In this case, Fn0 = D0 = 0.

From left / right transformation on page 107 and component ﬁeld transformations on page 104,

0 δψ = δF = δφ = δλ 0 = 0. Argument holds in reverse. h | L | i

A Note: no overconstraining on M φ components: Fn0 = D0 =0 is M conditions,

not M + D, where D is dimensionality of group, i.e. A = 1,...,D,

because only M D conditions needed to satisfy F = ∂f(φ) = 0: f(Φ) invariant under extended gauge transformations − n0 ∂φ φ=φ0

It ΩB ∂f(Φ) ∂f(Φ) ∂([e B ]nmΦm) ∂f(Φ) ∂f(φ) on page 131, A = 0 = A C = (ItAΦ)n, i.e. (tAφ)n = 0, which is already D conditions. ∂Ω ∂Φn ∂Ω Ω =0 ∂Φn ∂φn

[email protected] 140 Existence of any supersymmetric field configuration = unbroken SUSY vacuum. ⇒ From above, SUSY field configuration has V = 0, which is absolute minimum so lower than V for non SUSY field configuration.

(Now let ξA = 0.) To check vacuum is unbroken SUSY, enough to check that ∂f(φ) = 0 can be satisﬁed. ∂φn

IΛAt A ∂f(φ) Λ IΛAt f(φ) has no φ†, so invariant under φ e A φ with Λ complex. If = 0 true for φ, true for φ = e A φ. → ∂φn

A Λ Λ e ∂ Λ Λ Λ e Λ Choose Λ such that φ †φ minimum (which exists because it is real and positive), i.e. A (φ †φ ) φ †(t ) φ = 0, ∂Λ ∝ n A nm m i.e. there exists a ﬁeld conﬁguration such that DA = 0, so unbroken SUSY vacuum condition obeyed. [email protected] 141 So break SUSY by

∂f 1. Making ∂φ = 0 impossible (already considered on page 128) or

2. Fayet-Iliopoulos term ξD. Simple example:

2 left-chiral Φ± with U(1) quantum numbers e, ± so spinor components are left-handed parts of electron / positron.

+ Then only possibility is f(Φ) = mΦ Φ−,

+ 2 + 2 2 2 2 + 2 2 2 2 so from V deﬁned on page 138, V (φ , φ−)= m φ + m φ− + (ξ + e φ e φ− ) , | | | | | | − | | which cannot vanish for ξ = 0, so SUSY broken. 6 m2 + Note U(1) symmetry intact for ξ < , since minimum at φ = φ− = 0. | | 2e2 [email protected] 142 Tree-level mass sum rule: C = 0 , where C = m2 m2. For unbroken/broken SUSY. fermions − bosons

Take new scalar ﬁelds ∆φ = φ φ . Then quadraticP part of V deﬁnedP on page 138 in unitarity gauge φ†(t φ )=0is n n − n0 A 0

∆φ † ∆φ M M + (t φ )(t φ ) + DAt ... V = 1 M 2 , where M 2 = ∗ A 0 A 0 † 0 A quad 2 ∆φ 0 ∆φ 0 ... MM + ((t φ )(t φ ) ) + DAtT ∗ ∗ ∗ A 0 A 0 † ∗ 0 A

(M deﬁned in tree-level expansion of V on page 114). From Lagrangian for chiral and gauge supermultiplet ﬁelds

T ψ ψ M I√2(t φ ) on page 138, terms bilinear in fermion ﬁelds are L = 1 L M L , where M = A 0 ∗ , 1/2 −2 λ λ I√2(t φ ) 0 A 0 ∗

M †M + 2(tAφ0)(tAφ0)† ... A 2 Bµ 2 so M †M = . For gauge ﬁelds, LV = Vµ (MV )ABV where (MV )AB = φ0† tB,tA φ0. ... 2φ†t t φ − { } 0 B A 0

2 2 A 2 Underlined results above give TrM0 = 2Tr(M ∗M )+TrMV + 2D0 TrtA and Tr(M †M) =Tr(M ∗M ) + 2TrMV .

2 2 A Eliminate Tr(M ∗M ) to get TrM 2Tr(M †M) + 3TrM = 2D Trt . 0 − V 0 A

2 2 2 A But trace = eigenvalues, so spin 0 mass 2 spin 1 mass + 3 spin 1 mass = 2D0 TrtA. − 2 P P P P 1 1 spin-3 component for scalar boson, 2 for spin-2 fermion, 3 for spin-1 gauge boson,

i.e. m2 m2 = 2DATrt fermions − bosons 0 A P P In non-Abelian theories, Trt = 0. Can have Trt = 0 for U(1) but gives graviton-graviton-U(1) anomaly. A A 6 C is coeﬃcient of quadratic divergence in vacuum energy, i.e. breaking SUSY does not aﬀect UV structure. 4 The Minimally Supersymmetric Standard Model

4.1 Left-chiral superﬁelds

Assign left-handed SM fermions to left-chiral superﬁelds according to table 4.1.1.

Table 4.1.1: MSSM equivalent of table 2.9.1 on page 68, for the left-chiral superﬁelds. Supermultiplets for e.g. U are written uL for the left-handed up quark andu ˜L for its scalar superpartner, the up squark. Likewise for U we write uR† andu ˜R† . Names Label Representation under SU(3)C SU(2)L U(1)Y × × 1 (S)quarks Q =(U, D) (3, 2, 6 ) U (3, 1, 2 ) − 3 1 D (3, 1, 3 ) (S)leptons L =(N,E) (1, 2, 1 ) − 2 E (1, 1, 1)

Baryon / lepton number violating terms:

N N N K M N [QK(ǫLM ) D ] = [(DKN M U KEM )D ] , [(EKN M N KEM )E ] , [D D U ] , a a F − F − F F SU(3) SU(2) U(1) invariant terms not in SM, × × allows proton decay p π0 + e+ in minutes, while experiment gives > 1032 years. → ∼ K M N Relevant processes are u d (˜s∗ or ˜b∗ ) e u and u u . (No u d d˜∗ : coupling is λ D D U R R → R R → L L L → L R R → R KMN

K M with λKMN antisymmetric in KM since D D antisymmetric in colour indices to get colour singlet coupling.) [email protected] 144 Rule out some / all baryon lepton number violating terms on page 143 by symmetry, e.g.:

1.) Rule out all by baryon number (B) and lepton number (L) assignments

B = B = 1, L = L = 0, B = B = 0, L = L = 1, B = B = L = 0. U,U D,D ±3 U,U D,D N,N E,E N,N E,E ± θL θR θL/R Or rule out some by requiring only conservation of linear combination, e.g. L, B, B L etc. − K M N 2.) Rule out [D D U ] by L = L = 0, L = L = L = L = 1, L = 2. F N E U D U D − E − “Conventional” L for quarks, leptons by taking L = 1. L for squarks, sleptons is then unconventional. θL/R ±

But continuous global symmetries are dubious in string theories.

3.) Rule out all by R parity, a discrete global symmetry, = 1 for SM particles and = 1 for their superpartners, − 2s 3(B L) i.e. Π = ( 1) ( 1) − . Lightest SUSY particle (LSP) (lightest particle with Π = 1) completely stable. R − − R − Cannot decay into other Π = 1 particles (heavier), or into Π = 1 particles only (violates R-parity conservation). R − R Colliders: sparticles produced in pairs.

Non-zero vacuum expectation values for scalar components of N K charged lepton, charge e quark masses → −3 2e via 1st, 2nd terms in baryon / lepton number violating terms on page 143, but charge 3 quarks remain massless. [email protected] 145 Spontaneous breakdown of SU(2) U(1) for massive quarks, leptons, W ±, Z by Higgs superﬁelds in table 4.1.2: × Table 4.1.2: Required Higgs superﬁelds in the MSSM.

Names Label Representation under SU(3)C SU(2)L U(1)Y × × 0 − 1 Higgs(ino) H1 =(H ,H ) (1, 2, ) 1 1 − 2 + 0 1 H2 =(H2 ,H2 ) (1, 2, 2 )

Higgs-chiral superﬁeld couplings in superpotential [f(Φ)]F : (K,M = generation)

D K 0 K M E K 0 K M U K + K 0 M h [(D H U H−)D ] , h [(E H N H−)E ] , h [(D H U H )U ] . KM 1 − 1 F KM 1 − 1 F KM 2 − 2 F

KM K M E.g. last term is just Qa(ǫH2)aU, leads to Gu QL a(ǫφH )a† uR term in LHiggs fermion on page 70. − −

1 1 Note second Higgs H2 needed for last term, because we need a (1, 2, 2) left-chiral superﬁeld. (H1† is (1, 2, 2), but is right-chiral.)

0 φ 0 0 = 0 gives mass to d-type quarks and charged leptons, 0 φ 0 0 = 0 gives mass to u-type quarks. h | H1 | i 6 h | H2 | i 6

1 ∂2f(φ) In superpotential part of L on page 113, fermions get mass from ﬁrst term ψnRψmL. −2 ∂φn∂φm

So from e.g. ﬁrst term of Higgs-chiral superﬁeld couplings in superpotential [f(Φ)]F above,

K ∂2f(φ) K K K K mass term for d from d d = φ 0 d d . ˜K ˜K∗ R L H1 R L ∂dL ∂dR

For more Higgs superﬁelds, number of H1 and H2 type superﬁelds must be equal.

2 1 2 1 1 2 1 1 2 Higgsinos produce SU(2)-SU(2)-U(1) anomalies: For H , anomaly t y = g g′ + g g′ = g g′, 1 ∝ 3 2 2 −2 2 2 P 2 1 2 1 1 2 1 1 2 for H , anomaly t y = g g′ + g g′ = g g′. No anomalies from gauginos, in adjoint representation. 2 ∝ 3 2 −2 −2 −2 −2 P [email protected] 146 Most general renormalizable Lagrangian for a gauge theory with R parity or B L conserved consists of −

1. sum of [Φ∗ exp( V )Φ] terms for quark, lepton and Higgs chiral superﬁelds, − D

2. sum of [ǫαβWαWβ]F for gauge superﬁelds,

3. sum of Higgs-chiral superﬁeld couplings in superpotential[f(Φ)]F on page 145, and

T + 0 0 4. µ term: L = µ H ǫH = µ H H− H H . µ 1 2 F 2 1 − 2 1 F µ has no radiative corrections due to perturbative non-renormalization theorem for F term on page 123.

Gauge hierarchy problem on page 80 explicitly solved as follows:

Unbroken SUSY: 1-loop correction to Higgs mass from any particle cancelled by that particle’s superpartner.

For broken SUSY, replace Λ2 with ∆mass2 of supermultiplet, UV ∼ 2 κt 2 2 | | so 1-loop correction to Higgs mass from top is 8π2 ∆ms, where ∆ms is mass splitting between top and stop.

No ﬁne-tuning if this is . 1 TeV, so since κ 2 1, stop mass is < √8π2 10 TeV. | t| ∼ ∼ Flavour changing processes suppressed to below experimental bounds if squark masses equal, ∼ so if gauge hierarchy problem solved by SUSY, all squark masses < 10 TeV. [email protected] 147 4.2 Supersymmetry and strong-electroweak uniﬁcation

Mentioned in grand uniﬁcation subsubsection, page 76. Assume SUSY unbroken in most of range < MX.

1-loop modiﬁcations to SM β (g (µ )) = µ d g (µ ) from new SUSY particles, with n Higgs chiral superﬁelds: i i r r dµr i r s

3 3 5ngg′ g′ 5ng ns 1 1 1 5ng ns MX β1 = 2 2 + = 2 = 2 + 2 + ln 36π → 4π 6 8 ⇒ g′ (µr) g′ (MX) 2π 6 8 µr 3 3 g 11 ng g 9 ng ns 1 1 1 3 ng ns MX β2 = 2 + 2 + + = 2 = 2 + 2 + + ln 4π − 6 3 → 4π −6 2 8 ⇒ g (µr) g (MX) 2π −2 2 8 µr 3 3 gs 11 ng gs 9 ng 1 1 1 9 ng MX β3 = 2 + 2 + = 2 = 2 + 2 + ln . 4π − 4 3 → 4π −4 2 ⇒ gs(µr) gs(MX) 2π −4 2 µr 5 2 Take µr = mZ, set 3g′(MX)= g(MX)= gs(MX), solution (where e deﬁned on page 72, sin θW on page 71) q 2 e (mZ) 18+3ns+ 2 (60 2ns) 1. sin2 θ (m )= gs(mZ) − W Z 108+6ns

2 8e (mZ) 2 1 2 MX 8π −3gs(mZ) 2. ln = 2 . mZ e (mZ) 18+ns 2 2 2 e (mZ) 1 gs(mZ) For measured values sin θW = 0.231, 4π = (128)− , 4π = 0.118, mZ = 91.2GeV, equation 1. above gives n = 2, then equation 2. above gives M = 2 1016GeV. s X × [email protected] 148

50 α −1 1

α−1 30 α −1 2 20

10 α −1 3 0 2 4 6 8 10 12 14 16 18

Log10(Q/1 GeV)

2 gi Figure 4.1: 2-loop RG evolution of inverse of αi = 4π in the SM (dashed) and MSSM (solid). α3(mZ ) is varied between 0.113 and 0.123, sparticle mass thresholds between 250 GeV and 1 TeV. [email protected] 149 4.3 Supersymmetry breaking in the MSSM

In eﬀective MSSM Lagrangian, SUSY must be broken.

SUSY breaking terms must not reintroduce hierarchy problem.

Terms with coupling’s mass dimension 0 κ 2 λ = O(ln Λ ) (see page 80), so δm2 Λ2 ln Λ . ≤ → | ψ| − φ UV H ∼ UV UV Superrenormalizable terms (coupling’s mass dimension > 0, power of some M) δm2 M 2 ln Λ , ∼ → H ∼ UV OK provided M m . 10 TeV. So SUSY breaking terms must be superrenormalizable, called soft terms. ∼ SUSY [email protected] 150 Soft SUSY breaking R parity / B L conserving SM invariant terms: − (Sum over SU(2), SU(3) indices and generations K,M)

S2 K M 1. L M φ †φ , where S = Q, U, D, L, E superﬁelds and φ their component scalars , SR ⊃ S − KM S S S P 2. L λ m λ , where X = gluino, wino, bino , SR ⊃ X X X X

P D D K T M D D K T M 3. Trilinear terms: L A h (φ ) ǫφ φ C h (φ ) φ∗ φ SR ⊃− KM KM Q H1 D − KM KM Q H2 D

E E K T M E E K T M A h (φ ) ǫφ φ C h (φ ) φ∗ φ − KM KM L H1 E − KM KM L H2 E

U U K T M U U K T M A h (φ ) ǫφ φ C h (φ ) φ∗ φ − KM KM Q H2 U − KM KM Q H1 U

D,E,U where hKM deﬁned in Higgs-chiral superﬁeld couplings in superpotential[f(Φ)]F on page 145,

1 T 1 2 1 2 4. LSR Bµφ ǫφH m φ† φH m φ† φH where µ deﬁned in µ term on page 146. ⊃−2 H2 1 − 2 H1 H1 1 − 2 H2 H2 2

T 2 2 Recall Hermitian conjugate is added to L . So LSR Re Bµφ ǫφH m φ† φH m φ† φH . ⊃− H2 1 − H1 H1 1 − H2 H2 2

T 2 2 Choose H1, H2 superﬁelds’ phases so Bµ real, positive: LSR BµRe φ ǫφH m φ† φH m φ† φH ⊃− H2 1 − H1 H1 1 − H2 H2 2 [email protected] 151 To respect approximate symmetries, choose AS , B 1: KM ∼ S AKM for chiral symmetry: Reﬂected by small Yukawa coupling of light quarks.

B for Peccei-Quinn symmetry: Reﬂected by small µ term on page 146.

S CKM terms involve scalar components of left- and right-chiral superﬁelds,

quadratic divergences = ﬁne-tuning and hierarchy problems. → ⇒ In fact these divergences from scalar tadpole graphs which disappear into vacuum, cannot occur since no SM invariant scalars.

Note SM superpartners acquire mass even if no electroweak symmetry breaking (i.e. if SM particles massless).

AS , B, CS are arbitrary and complex > 100 parameters even without CS terms. KM KM → KM But expect soft terms to arise from some underlying principle. [email protected] 152 SUSY breaking at tree-level for 3 generations ruled out.

Tree-level mass sum rule on page 142 holds for each set of colour and charge values.

Gives e.g. 2(m2 + m2 + m2) 2(5 GeV)2 = of all masses for bosonic degrees of freedom with charge e/3. d s b ≃ − P So each squark mass is < √2 5 GeV 7 GeV. ≃

+ Predicts squark mass(es) too small, would have eﬀect in accurately measured e e− hadrons. → This is good, otherwise get ﬁne-tuning: tree-level SUSY breaking would require mass parameter M in Lagrangian.

Would aﬀect all SM masses, so M must coincidentally be electroweak symmetry breaking scale v = 246 GeV. ∼ No SUSY breaking at tree-level implies no SUSY breaking at all orders. [email protected] 153 SUSY breaking at tree-level for 3 generations ruled out. ≥ General argument against tree-level SUSY breaking: Consider ﬁrst underlined equation on page 142.

A To conserve colour and charge, only allowed non-zero D0 terms are for y of U(1) (D1) and t3 of SU(2) (D2).

Squarks are colour triplets, so have zero vacuum expectation values. So for (˜uL, u˜R),

1 1 1 1 2 MU∗MU g′ 6D1 + g 2D2 ... ˜ ˜ 2 MD∗ MD g′ 6D1 g 2D2 ... M0U = − 2 . For (dL, dR), M0D = − − 1 . ... M M ∗ + g′D ... M M ∗ g′D U U 3 1 D D − 3 1

2e Second underlined equation on page 142 gives mass-squared matrix MU∗MU for charge 3 quarks

e 2 and M ∗ M for charge quarks. Let v , x = u, d be unit eigenvector for quark of lowest mass, i.e. M ∗ M v = m v . D D −3 x X X x x x

0 † 0 Then mass2 of lightest charge 2e squark < M 2 = m2 + 2g D , 3 v 0U v u 3 ′ 1 u∗ u∗

† 2 e 0 2 0 2 1 mass of lightest charge 3 squark < M0D = md 3g′D1 , − vd vd − ∗ ∗

so regardless of value / sign of D1 there is at least one squark lighter than u or d quark.

In fact, since D m2 , get squark with negative mass which breaks colour and charge conservation. 1 ∼ SUSY

2 2e e 2 2 2 1 Possible solution is new U(1) gauge superfield, so mass of lightest charge ( , ) squark < (m + g′D , m g′D ) +g ˜D˜. 3 −3 u 3 1 d − 3 1 [email protected] 2 154 G (µr) b Alternative breaking: Introduce new strong-force-like gauge field, asymptotically free coupling 2 , 8π ≃ ln µr/mSUSY so µ m gives G (µ ) 1 (SUSY unbroken), µ m gives G (µ ) 1 (SUSY broken). r ≫ SUSY r ≪ r ∼ SUSY r ∼ Allows for m M without introducing M into theory as new mass scale. SUSY ≪ X X SUSY broken either non-perturbatively or by scalar field potential with non-zero vacuum expectation value.

To get m 10TeV M , G (M ) does not have to be very small (no ﬁne-tuning). SUSY ∼ ≪ X X New force clearly does not interact with SM particles, so SUSY breaking occurs via 0 F 0 =0 in hidden sector of particles that interact via new force, h | hidden| i 6 communicated to observed particles by interactions felt by both hidden and observed particles, namely gravity (gravity-mediated SUSY breaking) or SM interactions (gauge-mediated SUSY breaking): [email protected] 155 0 Fhidden 0 11 S 1. Gravity-mediated SUSY breaking: mSUSY h | | i = 0 Fhidden 0 10 GeV. (No C terms.) ∼ MP ⇒ h | | i∼ KM p From dimensional analysis, subject to “no SUSY breaking conditions” m 0 as 0 F 0 0 or M . SUSY → h | hidden| i → P → ∞ 2. Gauge-mediated SUSY breaking: Messenger particle of mass M couples to 0 F 0 messenger h | hidden| i

αi 0 Fhidden 0 and to MSSM particles via SM interactions (loops). Then mSUSY h | | i . ∼ 4π Mmessenger

For 0 F 0 M , = 0 F 0 105 GeV. (Very small CS terms.) h | hidden| i∼ messenger ⇒ h | hidden| i∼ KM p p

Flavour blindness of above SUSY breaking mechanisms simpliﬁes soft terms in Lagrangian at µr = MX.

E.g. in minimal supergravity (mSUGRA), K on page 112, which appears in supergravity Lagrangian, is diagonal for hidden and observed sectors, i.e. K = Φ 2. i=observed, hidden | i| P Gives organising principle (for mSUGRA) (recall S = Q, D, U, L, E)

M S2 = m2δ , m2 = m2 = m2, m = m = m = m , AS = A δ , CS = 0. KM 0 KM H1 H2 0 gluino wino bino 1/2 KM 0 KM KM

This or similar principle expected: small experimental upper bounds on ﬂavour changing, CP violating processes. [email protected] 156 Flavour changing processes allowed because squark, slepton mass matrices not necessarily diagonalised in same basis as quark, lepton mass matrices.

0 Most stringent limits for quarks are on K0 K transitions involving d gluino + d˜K, d˜K gluino + s , − L → L L → L most stringent limits for leptons are on µ eγ decays. →

CP violating processes allowed due to many new phases in MSSM, can have large eﬀect at low SM energies.

Upper bound on electric dipole moments of neutron and atoms and molecules

2 requires CP violating phases . 10− or some superpartner masses & 1 TeV. [email protected] 157 4.4 Electroweak symmetry breaking in the MSSM

T T Consider Higgs H1 and H2 superﬁelds’ scalar components φH = (φ 0, φ ) and φH = (φ +, φ 0) . 1 H1 H1− 2 H2 H2

2 2 2 2 2 g g +g′ Scalar Higgs potential: V = φ† φH + φ† φH φ† φH 2 H1 2 8 H1 1 − H2 2

2 2 2 2 T +(m + µ )φ† φH + (m + µ )φ† φH + BµRe(φ ǫφH ). H1 | | H1 1 H2 | | H2 2 H1 2 From potential in Lagrangian for chiral and gauge supermultiplet ﬁelds page 138,

A T with ξ = 0 and f(H1, H2) = µH1 ǫH2 (the µ term on page 146),

together with part 4. of LSR in soft SUSY breaking terms on page 150.

V bounded from below: 2 µ 2 + m2 + m2 Bµ . | | H1 H2 ≥

Terms in V which are quartic in φH1 and φH2 are positive or zero. When positive, V has a minimum.

T T 2 2 2 2 2 However, quartic terms vanish when φH = (φ, 0) and φH = (0,φ) , and then V = (2 µ + m + m ) φ BµRe(φ ), 1 2 | | H1 H2 | | − which must not go to as φ . −∞ | | → ∞ [email protected] 158 Deﬁnition of vacuum: 0 φ + 0 = 0 φ 0 =0, 0 φ 0 0 = vi real , h | H2 | i h | H1−| i h | Hi | i

2 2 2 2 g +g′ 2 2 1 so (m + µ )v1 + (v1 v2)v1 Bµv2 = 0 and 1 2 . Around vacuum, φ 0 = vi + φi . H1 | | 4 − − 2 ↔ Hi

neutral Deﬁne V = V when charged φ = φ + = 0. H1− H2

∂V g2 Choose minimum φH+ = 0 by SU(2) rotation. = 0 = φH− = 0 (or Bµ = φ† 0 φ† 0 , but then vi = 0). 2 ∂φ + 1 2 H1 H2 H2 ⇒ −

neutral neutral g2+g′2 2 2 2 2 2 2 2 2 2 Then V = V , where V = φ 0 φ 0 + (m + µ ) φ 0 + (m + µ ) φ 0 BµRe(φ 0 φ 0 ). 8 | H1 | −| H2 | H1 | | | H1 | H2 | | | H2 | − H1 H2 neutral ∂V neutral Let stationary point of V be at φ 0 = vi, i.e. for φ 0 = vi + φi, must have = 0, Hi Hi ∂φi φi=0

2 2 g2+g′2 2 2 1 i.e. coeﬃcient of term φi vanishes. Gives (m + µ )v∗ + (v v )v∗ Bµv2 = 0 and 1 2. ∝ H1 | | 1 4 1 − 2 1 − 2 ↔

Last result implies v real if v real. Adjust relative phase between φ 0 and φ 0 so that v is real. 2 1 H1 H2 1

2 v2 2 2 2 2 Parameter relations with v1 and v2: Bµ = m sin 2β where tan β = , m = 2 µ + m + m , and A v1 A | | H1 H2

2 2 1 2 1 2 2 2 2 1 2 1 2 2 2 1 2 2 2 2 m + µ = m (m + m ) cos 2β , m + µ = m + (m + m ) cos 2β with m = (g + g′ )(v + v ). H1 | | 2 A − 2 A Z H2 | | 2 A 2 A Z Z 2 1 2

From deﬁnition of vacuum above. Compare mZ here with mZ on page 71.

Limit on β: 0 β π . Bµ = m2 sin2β above, but V bounded from below on page 157 is m2 Bµ, so 0 sin2β 1. ≤ ≤ 2 A A ≥ ≤ ≤ [email protected] 159 Quadratic (mass) terms in V neutral: V neutral = 1m2 cos 2β φ 2 φ 2 + m2 (Re(cos βφ sin βφ ))2 quad 2 Z | 1| − | 2| Z 1 − 2 +1m2 ( φ 2 + φ 2) 1(m2 + m2 ) cos 2β( φ 2 φ 2) m2 sin 2βRe(φ φ ). 2 A | 1| | 2| − 2 A Z | 1| − | 2| − A 1 2

Neutral Higgs particle masses: m2 , 0, m2 = 1 m2 + m2 + (m2 m2 )2 + 4m2 m2 sin2 2β , A H 2 A Z A − Z A Z q m2 = 1 m2 + m2 (m2 m2 )2 + 4m2 m2 sin2 2β (zero mass particle is Goldstone boson). h 2 A Z − A − Z A Z q neutral Real, imaginary parts of φi decouple in Vquad (no terms like e.g. Re(φ1)Im(φ2)).

neutral T 2 First two masses associated with Im(φ) = (Im(φ1), Im(φ2)): Write Im(φ) dependence of Vquad as Im(φ) MIm(φ)Im(φ),

1m2 (1 cos 2β) 1m2 sin2β then M 2 = 2 A − 2 A . Eigenvalues are m2 and 0, corresponding eigenstates C odd. Im(φ) 1m2 sin2β 1m2 (1 + cos 2β) A 2 A 2 A

2 ∂2V So m must be positive to ensure φi = 0 is local minimum, i.e. that eigenvalues of positive. Similarly, A ∂φi∂φj

1m2 (1 cos 2β) + 1m2 (1 + cos 2β) 1(m2 + m2 )sin2β last two masses are eigenvalues of M 2 = 2 A − 2 Z −2 A Z . Re(φ) 1(m2 + m2 )sin2β 1m2 (1 + cos 2β) + 1m2 (1 cos 2β) −2 A Z 2 A 2 Z −

2 2 2 2 2 2 Heaviest neutral scalar: mH > mA, mZ . Lightest neutral scalar: mh < mA, mZ .

Large top to bottom quark mass ratio suggests large tan β = v2 m2 Max(m2 , m2 ), m2 Min(m2 , m2 ). v1 → H ≃ A Z h ≃ A Z [email protected] 160 charged Deﬁne V = V when φ1 = φ2 = 0.

Quadratic (mass) terms in V charged:

charged 1 2 2 2 2 2 1 2 2 2 V = (m + m ) φ (1 cos 2β)+ φ + (1 + cos 2β)+2sin2βRe(φ φ +) , with m = g (v1 + v2). quad 2 W A | H1−| − | H2 | H1− H2 W 2 Compare mW here with mW on page 71.

2 2 2 Charged Higgs particle masses: 0, mC = mW + mA .

charged 2 T 2 1 2 2 1 cos 2β sin2β Vquad = (φH− ,φH+ )MC(φ∗ − ,φ∗ + ) , where MC = 2(mW + mA) − . 1 2 H1 H2 sin2β 1 + cos 2β

2 Above masses are eigenvalues of MC.

Results for neutral and charged scalar fields modified most significantly by radiative corrections from top quark, which has largest Yukawa couplings to Higgs. Most significant modification to above results are the increases m2 = 1 m2 + m2 +∆ + ((m2 m2 ) cos 2β +∆ )2 + (m2 + m2 )2 sin2 2β H 2 A Z t A − Z t A Z q

√ 4 2 2 1 2 2 2 2 2 2 2 2 2 3 2mt GF Mst mh = 2 mA + mZ +∆t ((mA mZ) cos 2β +∆t) + (mA + mZ) sin 2β , where ∆t = 2 2 ln 2 . − − 2π sin β mt q

By taking stop mass Mst > 300 GeV, tan β > 10, get mh > mZ. [email protected] 161 Condition which leads to electroweak symmetry breaking: 4(m2 + µ 2)(m2 + µ 2) (Bµ)2 . H1 | | H2 | | ≤

2 2 2 2 4 2 2 2 2 2 From parameter relations with v1 and v2 on page 158, 4(m + µ )(m + µ ) = m sin 2β m (m + 2m )cos 2β. H1 | | H2 | | A − Z Z A But Bµ = m2 sin2β, so 4(m2 + µ 2)(m2 + µ 2) = (Bµ)2 m2 (m2 + 2m2 )cos2 2β, so inequality follows. A H1 | | H2 | | − Z Z A

These inequalities mean that second derivative matrix of V has negative eigenvalue at SU(2) respecting point φH1 = φH2 = 0,

i.e. V unstable (not minimum) there: Quadratic part of scalar Higgs potential on page 157 can be written as four terms

m2 + µ 2 1Bµ ˜T 2 ˜ ˜ 2 H1 2 φ M φ, where φ is separately the imainary and real parts of (φ − ,φ + ) and (φH0 ,φH0 ), and M = 1 | | ± . φ H1 H2 1 2 φ Bµ m2 + µ 2 ±2 H2 | | So mass eigenvalues m2 obey (m2 + µ 2 m2)(m2 + µ 2 m2) 1(Bµ)2 = 0, H1 | | − H2 | | − − 4

solutions are 2m2 = 2 µ 2 + m2 + m2 (2 µ 2 + m2 + m2 )2 + (Bµ)2 4(m2 + µ 2)(m2 + µ 2). | | H1 H2 ± | | H1 H2 − H1 | | H2 | | q Inequality above implies one of these solutions is negative.

Example: For tan β = (β = π ), parameter relations with v and v on page 158 ∞ 2 1 2 imply m2 + µ 2 > 0 and m2 + µ 2 < 0, so electroweak symmetry broken. H1 | | H2 | | Alternatively, radiative corrections give d m2 = x hU + . . . (hU is top quark Yukawa coupling), x> 0. d ln µr H2 33 33

(Similarly for stop masses, smaller x.) So although m2 > 0 at µ = M , may have m2 < ( )0 at µ = v. H2 r X H2 ≪ r [email protected] 162 4.5 Sparticle mass eigenstates

Consider gauginos and higgsinos. Particles with diﬀerent SU(2) U(1) × but same U(1)e.m. transformation properties can mix after electroweak symmetry breaking.

0 Neutralinos: 4 neutral fermionic mass eigenstates χi , i = 1(lightest), ..., 4, mixtures of bino, neutral wino, neutral higgsinos. e

1 0 T 0 0 Bilinear terms in these ﬁelds appearing in Lagrangian can be written (λ ) Mχ0 λ , where λ = (λbino, λneutralwino, λ 0 , λ 0 ), −2 H1 H2 e m 0 c s m s s m bino − β W Z β W Z 0 mwino cβcW mZ sβsW mZ Mχ0 =  −  (sβ = sin β etc.). µ dependent part from µ term on page 146. c s m c c m 0 µ − β W Z β W Z −  s s m s c m µ 0  e  β W Z − β W Z −    m dependent part from 1. for Φ = H and V = SU(2) U(1) ﬁelds after electroweak symmetry breaking on page 146. Z i ×

mbino, mwino dependent terms from SUSY breaking terms for gauginos from 2. on page 150.

Mχ0 symmetric, diagonalized by unitary matrix.

0 χ1 can be LSP, and therefore candidate for cold dark matter. e [email protected] 163 Charginos: 4 charged fermionic mass eigenstates χi±, i = 1, 2, mixtures of charged winos and charged higgsinos.

1 + T Bilinear terms in these ﬁelds appearing in Lagrangian can be written (λ ) M c λ−, e −2 χ e + + mwino I√2mW sβ where λ = (λ , λ + ), λ− = (λ− , λ − ), Mχc = . charged wino H2 charged wino H1 I√2m c µ W β e

c + T 0 Mχ†c Deﬁne λ = (λ , λ−) , Mχc = , so contribution to Lagrangian including hermitian conjugate Mχc 0 e e e 1 c T c 2 Mχ†c Mχc 0 is 2(λ ) Mχc λ . Squared mass eigenvalues can be obtained from diagonalization of Mχc = , − 0 Mχc Mχ†c ! e e e e e e 2 2 1 2 2 2 2 2 2 4 2 2 2 2 gives m + = m − = 2 mwino + 2mW + µ (mwino µ ) + 4mW cos 2β + 4mW (mwino + µ 2mwinoRe(µ sin2β)) . χ1,2 χ1,2 | | ± −| | | | − e e p m2 mT 2 Slepton, squark mass matrix: m2 = Lsquark LRsquark , squark m2 m2 LRsquark Rsquark where m2 = M Q/U 2 + (m2 + m2 (T Q sin2 θ ) cos 2β)δ (no K sum), L/Rsquark,KM KM quark,K Z 3 − W KM

S2 MKM from 1. on page 150, mquark from unbroken SUSY condition msquark,KK = mquark,K (no K sum),

2 and term proportional to mZ from Higgs-chiral superﬁeld couplings in superpotential [f(Φ)]F on page 145.

2 2 U First term from 3. on page 150, second term from ∂f . mLRsquark,KM = (AKM δKM mquarkµ tan β)mquark . ∂φH0 − 1