QUANTUM FIELD THEORY? the Charged Particles, Are Ordinary Mechanical Point-Like Objects and Are Thus Not Associated to ﬁelds

Home , Bispinor, Generator (mathematics), Lorentz covariance, Riccardo Rattazzi, Weyl equation

Relativistic Quantum Fields

W +

¯b s¯

t¯ t¯

s s¯

Lecture notes

Based on the course given by Riccardo Rattazzi

First redaction, Johnny Espin, 2012 Second redaction, Riccardo Rattazzi, 2016 Contents

I Foundations and Matter5

1 Introduction 6

1.1 What is quantum ﬁeld theory?...... 6

1.2 Why quantum ﬁeld theory?...... 6

1.3 Units of measure...... 8

1.4 Overview of particle physics...... 9

2 Classical Field Theory 13

2.1 Lagrangian mechanics...... 13

2.2 From mechanics to ﬁeld theory...... 15

2.3 Least action principle in ﬁeld theory...... 16

2.4 Hamiltonian formalism...... 18

3 Symmetries 19

3.1 Group theory...... 19

3.1.1 Groups...... 19

3.1.2 Schur’s lemma...... 23

3.1.3 Lie algebras...... 25

3.2 Lorentz and Poincaré groups...... 27

3.2.1 Construction...... 27

3.2.2 Raising and lowering...... 30

3.2.3 Lie algebra...... 31

3.2.4 Representations...... 32

3.3 Noether’s theorem...... 35

3.3.1 Statement...... 35

3.3.2 The stress-energy tensor...... 39

1 4 Scalar ﬁelds 43

4.1 The Klein-Gordon ﬁeld...... 43

4.2 Quantisation of the Klein-Gordon ﬁeld...... 44

4.2.1 Continuum limit...... 45

4.2.2 Hilbert space...... 48

4.3 Properties of the Klein-Gordon ﬁeld...... 49

4.3.1 Relativistic normalisation of states and measure...... 49

4.3.2 States localised in space...... 51

4.3.3 Time evolution...... 51

4.4 The charged scalar ﬁeld...... 54

4.4.1 Construction...... 54

4.4.2 Internal current and charge...... 54

4.4.3 Quantisation...... 55

5 Spinor ﬁelds 57

5.1 Spinor representations of the Lorentz group...... 57

5.2 Miscellaneous about spinors...... 58

5.2.1 Parity...... 58

↑ 5.2.2 Relation between SL(2, C) and L+ ...... 61 5.3 Covariant wave equations...... 64

5.3.1 Weyl spinors...... 64

5.3.2 Majorana spinors...... 66

5.3.3 Dirac spinors...... 67

5.3.4 Lagrangians...... 68

5.4 Path towards quantisation...... 70

5.4.1 Classical solutions...... 70

5.4.2 Chirality and helicity...... 71

5.5 Quantisation of the Dirac ﬁeld...... 73

5.5.1 How one should not quantise the Dirac ﬁeld...... 73

5.5.2 The right quantisation method...... 75

5.5.3 Hilbert space...... 76

5.6 Properties of the Dirac ﬁeld...... 77

5.6.1 Time evolution...... 77

2 5.6.2 Internal charge...... 77

6 Causality in quantum ﬁeld theory 78

II Forces and Interactions 79

7 Vector Fields 80

7.1 Classical Electrodynamics...... 80

7.2 Exercise...... 82

7.3 A fresh look at Maxwell equations: counting degrees of freedom...... 83

7.4 A ﬁrst step towards quantization...... 85

7.5 Quantization à la Gupta-Bleuler...... 85

7.6 Quantization of massive vector ﬁelds...... 89

7.6.1 Massive Vector Fields...... 89

7.6.2 Canonical quantization...... 91

7.6.3 Heisenberg Picture Field...... 94

7.6.4 Polarization vectors...... 94

7.7 Polarisations of the vector ﬁeld...... 96

7.7.1 One-particle states...... 97

7.7.2 Operators and ﬁelds transformations...... 99

7.7.3 Polarisations...... 101

8 Discrete symmetries 103

8.1 Introduction...... 103

8.2 Parity...... 104

8.2.1 Foreword...... 104

8.2.2 Spin 0...... 105

8.2.3 Spin 1/2...... 107

8.2.4 Spin 1...... 109

8.3 Time reversal...... 110

8.3.1 Foreword...... 110

8.3.2 Spin 0...... 111

8.3.3 Spin 1/2...... 112

8.3.4 Spin 1...... 113

3 8.4 Charge conjugation...... 113

8.4.1 Foreword...... 113

8.4.2 Spin 0...... 114

8.4.3 Spin 1/2...... 114

8.4.4 Spin 1...... 115

9 Interacting Fields 116

9.1 Asymptotic States...... 116

9.2 The S-Matrix...... 119

9.2.1 Lippmann-Schwinger Equation...... 119

9.2.2 Symmetries of the S-Matrix...... 122

9.3 Phenomenology...... 124

9.3.1 Cross Sections...... 124

9.3.2 Decay Rates...... 124

9.4 Amplitudes in Perturbation Theory...... 126

9.4.1 Perturbative Expansion...... 126

9.4.2 Dyson Series Representation of the S-Matrix...... 127

9.4.3 Wick’s Theorem...... 131

9.4.4 Feynman Rules and Feynman Diagrams...... 134

A Proof of Wick’s Theorem 143

4 Part I

Foundations and Matter

5 Chapter 1

Introduction

Quantum ﬁeld theory (QFT) is the theoretical “apparatus” that is needed to describe how nature works at its most fundamental level, that is to say at the shortest distances we have explored. In such domain physics is described by elementary particles and by their interactions, and QFT beautifully accounts for that. It also turns out that the notion of Fundamentality is here paired with the notion of Simplicity. As we shall see towards the end of this course, after having digested the necessary set of mathematical and physical concepts (≡formalism), the laws of particle physics can be written in a few lines with absolute precision and greatest empirical adequacy. In a sense particle physics is orthogonal to the sciences that deal with complexity.

1.1 What is quantum ﬁeld theory?

Technically, quantum field theory is the application of quantum mechanics to dynamical systems of fields, very much like basic quantum mechanics concerns the quantisation of dynamical systems of particles. Therefore, while quantum mechanics deals with mechanical systems with a finite number of degrees of freedom, quantum field theory describes the quantisation of systems with infinitely many degrees of freedom. Specifically this course is devoted to relativistic QFT. Relativistic QFTs explain the existence of particles and describe their mutual interactions. The fact that nature at its most basic level consists of particles can thus be viewed merely as a consequence of relativistic QFT. The domains of application of the latter in modern physics are quite broad: from the study of collisions among elementary particles in high energy accelerators, to early Universe cosmology. For instance, the primordial density fluctuations that later gave rise to structures like galaxies, the origin of dark matter or black-hole radiation are all described by relativistic QFT. Nevertheless there are also applications of quantum field theory to non-relativistic systems, in particular in condensed-matter physics: superfluidity, superconductivity, quantum Hall effect, ...

1.2 Why quantum ﬁeld theory?

We just outlined what QFT describes, but why do we need a ﬁeld theory to describe particles? Is this a necessity? It basically is: relativistic QFT is the only way to reconcile ordinary quantum mechanics with special relativity. We can understand this necessity with several qualitative arguments:

• Special relativity postulates the existence of a limiting speed. This implies that instantaneous interactions between two particles separated in space are not possible. One needs a medium (a field) to “propagate” interactions. For example, two boats on the Lac Léman affect each other through the waves they generate at the surface of water. The water surface and its deformations represents in that case the field, which is however an effective macroscopic description of the complex microscopic dynamics of water molecules. The simplest example of a more fundamental and also relativistic field is instead given the electromagnetic field. So basically, interactions require field dynamics. However, in classical electrodynamics, the fields’ sources,

6 1.2. WHY QUANTUM FIELD THEORY? the charged particles, are ordinary mechanical point-like objects and are thus not associated to ﬁelds. With the addition of quantum mechanics, the only way things can be made compatible (consistent) is if matter as well is described at its basic level by ﬁelds.

The need to associate matter particles as well to ﬁelds, or in other words, the failure of the point-like picture in relativistic quantum mechanics is hinted by the following heuristic arguments:

• If one tries to generalise the Schrödinger equation to make it relativistically invariant in the most obvious way, one is forced with a Hamiltonian which is unbounded from below, i.e. negative energy states appear. The stability of the system is jeopardised!

2∇2 +i ∂ + ~ ψ(t, x) = 0 ⇒ − 2∂2 + c2 2∇2 + m2c4 ψ(t, x) = 0 (1.1) ~ t 2m ~ t ~ 2k2 p E = ~ ⇒ E = ±c m2c2 + 2k2 (1.2) 2m ~

• More indirectly, if one somehow disposes of the extra negative energy states, then one ﬁnds that with the single particle picture there is trouble implementing causality1.

2∇2 p +i ∂ + ~ ψ(t, x) = 0 ⇒ +i ∂ − c2 2∇2 + m2c4 ψ(t, x) = 0 (1.3) ~ t 2m ~ t ~ 2k2 p E = ~ ⇒ E = +c m2c2 + 2k2 > 0 (1.4) 2m ~ p corresponding to H = c m2c2 + p2 (p = ~k). But then:

ˆ hy|e−iHt/~|xi = A(x → y, t) (1.5)

is non-zero when the space-time event (t, y) sits outside the light cone of (0, x) (c2t2 −(x−y)2 < 0) indicating a violation of causality!

There is however a very basic reason why at the fundamental level a single particle description cannot work. Indeed the famous Einstein relation E = mc2 tells us that energy and mass are equivalent. Therefore, in a relativistic process, we have no right to assume that part of the energy cannot be used to create extra particles, as long other coservation laws such as the conservation of electric charge are respected. Quantum mechanics makes this possibility more urgent and relates to the fundamental properties of matter. Consider indeed the uncertainty principle with relativity for a particle of mass m:

p Relativity: ⇒ E = c p2 + m2c2 (1.6) Quantum mechanics: ⇒ ∆x∆p & ~ (1.7) so that if:

∆x < ~ = λ (1.8) mc Compton then:

∆E > mc2 (1.9)

1Causality: information cannot travel faster than light.

EPFL-ITP-LPTP Riccardo Rattazzi 7 1.3. UNITS OF MEASURE In other words, if we try to localise a particle to a distance shorter than its characteristic Compton wavelength, it becomes relativistic and the excess energy can lead to the production of extra particles! The production of these extra particles implies that the one-particle states of ordinary quantum mechanics stop making sense for localisations smaller than ∆x ' λCompton. Empirically, it is a fact that particles are indeed created and destroyed in fundamental processes.

The last argument can be made more concrete by considering the process of measurement of the position of an elementary particle, say an electron. In order to measure the position of the electron we need to "see" it, so we must have it interact with a light source. Light consists of the excitations of the electromagnetic field, a field whose quantum description we can assume to know. A result from the quantization of the electromagnetic field is that waves of wawelength λ consist of particles, the photons (normally indicated by γ), carrying momentum pγ = ~/λ and energy Eγ = cp = ~c/λ. Now, the precision in the measurement in the electron position ∆x is limited by the finite wavelength of the light we employ to detect it: ∆x >∼ λ. The energy of the photons will therefore grow as ∆x is made smaller: Eγ = ~c/λ >∼ ~c/∆x. As a consequence of that for small enough ∆x the photon will be sufficiently energetic to allow the production of additional particles. For instance one finds that for 2 Eγ > 4mec , where me the mass of the electron, the energy of the photon is high enough to lead to the production of an additional electron positron pair (e+ − e−). For such highly energetic photons the single electron picture stops making sense! If we try to make the measurement of the position of the electron more precise by using such short wavelength (and therefore energetic!) photons, we produce instead a different state with 3 particles (2 electrons and 1 positron). We thus conclude there exists a minimal length scale down to which the single electron picture makes sense: c c c 2 ~ > ~ > ~ 4mec > Eγ = ∼ ⇒ ∆x ∼ ≡ λC . (1.10) λ ∆x me −11 λC = 4 × 10 cm is known as the Compton wavelength of the electron. λC separates the domain of particle quantum mechanics from the domain of relativistic quantum field theory.

1.3 Units of measure

There are three dimensionful quantities in mechanics (CGS):

Length: [L] → cm (1.11) Time: [t] → s (1.12) Mass: [m] → g (1.13)

Energy and momentum are not fundamental quantities and are expressed in terms of the above three:

cm2 Energy: [E] → erg = g × (1.14) s cm Momentum: [p] → g × (1.15) s In relativistic QFT we have fundamental equations coming from both special relativity and quantum mechanics. The ﬁrst uniﬁes space and time through the speed of light:

cm c[t] = [L], c = 2.99 × 1010 (1.16) s and similarly uniﬁes mass, momentum and energy:

[p] [E] [m] = = . (1.17) c c2

EPFL-ITP-LPTP Riccardo Rattazzi 8 1.4. OVERVIEW OF PARTICLE PHYSICS The second associates a (wave-)length to a momentum through the Planck constant:

[L] = ~ , = 1.05 × 10−27 erg · s (1.18) [p] ~

Since these two theories form the basic pillar of QFT, it is natural to choose a system of units in which:

c = 1 = ~ (1.19) so that:

1 1 [t] = [L] = = (1.20) [p] [E]

In particle physics, it is customary to measure all quantities in units of an energy unit. We shall use as unit the GeV = 109 eV . Using 1 eV = 1.6 × 1012 erg, we can compute as an exercise the following relations:

Unit QFT system to CGS

Length GeV −1 = 1.97 × 10−14 cm Time GeV −1 = 6.58 × 10−25 s (1.21) Mass GeV = 1.78 × 10−24 g

Action GeV 0 (~ = 1) Velocity GeV 0 (c = 1)

We understand why it is equivalent to talk either about high-energy physics or about short-distance physics. High-energy accelerators are truly huge microscopes testing the short distance behaviour of matter and forces!

1.4 Overview of particle physics

The goal of the course is obviously to learn the basics of QFT, nonetheless before that it would be helpful to receive an overview of present understanding of the micro world.

There are four well known fundamental forces:

1. The electromagnetic force which is responsible for the stability of atoms, namely for binding nuclei and electrons into atoms. This is a long range force. 2. The weak force: a short-range force (L ≈ 10−16 cm) which is responsible for the beta-decay of nuclei, neutrons and for muon-decay:

− n → p + e +ν ¯e (1.22) − − µ → e +ν ¯e + νµ (1.23)

EPFL-ITP-LPTP Riccardo Rattazzi 9 1.4. OVERVIEW OF PARTICLE PHYSICS 3. The strong force is responsible for the existence of nuclei (for their stability). Also, for instance, more than 99% of the mass of nuclei is due to the strong force alone. The latter explains most of our mass. The strong force is also short-ranged (L ≈ 10−13 cm) 4. The gravitational force: it is very weak, however it is long ranged and universally attractive. For these reasons, gravity plays nonetheless an essential role in our life. In order to understand how weak gravity is compared to the other forces, we can calculate the ratio between the gravitational and electromagnetic forces’ amplitude between two protons:

2 FG GN mp −40 ' 2 ∼ 10 (1.24) FE e We see immediately that the diﬀerence is huge!

As we shall learn, in QFT forces are associated to ﬁelds and thus to particles. The electromagnetic force is mediated by the photon γ, the weak force by the W ± and Z0 and the strong force by eight gluons g. These particles are all helicity ±1 vector bosons. On the other hand, gravity is mediated by the graviton which is a helicity ±2 boson. This is what makes gravity so diﬀerent from the other forces. In addition, there is a mysterious “Higgs” force that remains to be studied. It might be mediated by one (or more) scalar particle (spin 0).

We saw which forces are present in Nature as far as we know. They all are mediated by bosonic particles. All other fundamental particles known so far are fermions with spin 1/2 and come into two classes:

1. The leptons: they only feel the electromagnetic and weak forces. They come into three distinct generations.

1st 2nd 3rd

electron e− muon µ− tau τ − Q = −1 (1.25) me = 0.511 MeV mµ = 105 MeV mτ = 1.777 GeV

Q = 0 νe νµ ντ

The masses of the neutrinos are non-zero but very small (mν . 1 eV ). In addition, every particle in this table comes in pairs with its antiparticle.

2. The quarks: they also feel the strong force. Because of this, quarks are strongly bound into hadrons and are always conﬁned into these ones. We have basically two types of hadrons: the baryons (like the proton) made up of three quarks and the mesons (like the pions) made up of a quark and an antiquark. They are also organised in three families.

1st 2nd 3rd

up u charm c top t Q = 2/3 mu ' 1.5 − 3 MeV mc ' 1.25 ± 0.1 GeV mt ' 171.4 ± 2 GeV (1.26) down d strange s bottom b Q = −1/3 md ' 3 − 7 MeV ms ' 95 ± 25 MeV mb ' 4.20 ± 0.07 GeV

EPFL-ITP-LPTP Riccardo Rattazzi 10 1.4. OVERVIEW OF PARTICLE PHYSICS We should understand how these observational facts fit into a coherent theoretical picture. More pragmatically, how well we describe the observed particles and their natural interactions. Particle physics, to a beginner, often gives the impression of being all about “finding particles and classifying them” much like botanics. Nothing could be more wrong than that! What is being sought for are indeed the fundamental laws of Nature. Underlying the above empirical fact, there is a well developed theoretical understanding; in certain aspects, our understanding is truly remarkable, while open problems and some mystery remain. What we understand well is the structure of the electromagnetic, weak and strong forces, and quarks and leptons are successfully described by one QFT called the Standard Model (SM). What do we mean by understanding something? Understanding means also being able to describe and predict: for example, some of the existing particles (the weak bosons the charm and the top quarks) have been predicted from logical (or mathematical) consistency of the theory before their experimental discovery. Also, the prediction of some observables is made and verified to a level of accuracy which is unmatched in any other field. One of the best results of our QFT formalism is the derivation of quantum effects in electrodynamics. One example is the study of the magnetic moment of a particle. The latter is proportional to the spin and is given by:

e ~µ = g ~s (1.27) 2m where g is the gyromagnetic ratio. Classically, we expect g = 1; for a relativistic electron in Dirac’s theory, we have g = 2. In QED, however, we can compute the digression from the value of 2: g − 2 α α 2 α 3 = + c + c + ... (1.28) 2 2π 2 2π 3 2π

The coeﬃcients ci are predicted by QED and are of order one. Alpha (α) is the ﬁne structure constant of the theory and is given by:

e2 1 α = ∼ (1.29) 4π0~c 137 The digression on g is measured experimentally with very high accuracy: g − 2 = 1159652180.85(76) × 10−12 (1.30) 2 exp Once this quantity is measure, we can use it to compute α with precision and then use this value to test the theory with other observable:

g − 2 → α−1 = 137.035999705(36) (1.31) 2 exp and for example, from atomic clocks (Rb and Cs):

α−1(Rb) = 137.03599878(91) (1.32)

These results are a great conﬁrmation that QFT is the correct description of the subatomic world. The success of QFT and of the SM make the open problems in the theory only more exciting! Here is a list if what we understand less well in decreasing order of understanding:

1. The SM necessarily predicts the existence of a ﬁfth force to explain the origin of the mass of the weak bosons as well as of the quarks and leptons. However, there are various possibilities in nature for this necessary force. The simplest is that it is mediated by a boson of spin zero, a scalar particle called the Higgs boson. However the presence of this boson has not been experimentally conﬁrmed2 and moreover many physicists question this simple picture proposing alternative theories. This is a very exciting time because before the end of this course, the Large Hadron Collider (LHC) at CERN should have had announced its discovery or it dismissal. 2As of when these notes were typed.

EPFL-ITP-LPTP Riccardo Rattazzi 11 1.4. OVERVIEW OF PARTICLE PHYSICS 2. Why are there three families of fermions? 3. What is the reason for the big ratios of masses?

4. What is the role of gravity? As far as we understand, gravity makes sense as a quantum ﬁeld theory only at suﬃciently low energy. A full quantum description is however missing. Possibly gravity will require to go beyond QFT (e.g. String theory or other alternative Quantum Gravity theories). The problem of gravity is not necessarily urgent. Indeed, the needed energy cannot be reached in any imaginable laboratory. It is however crucial to develop a theory of the very early Universe to answer question such as “How did the Universe start?” or “Was it inevitable?”. During our course, we shall neglect gravitational perturbations and work with a static Minkowski metric:

gµν → ηµν = diag(+1, −1, −1, −1) (1.33)

After this short introduction (for a very broad subject), we shall start learning its technicalities.

EPFL-ITP-LPTP Riccardo Rattazzi 12 Chapter 2

Classical Field Theory

2.1 Lagrangian mechanics

Non-relativistic quantum mechanics, in Schrödinger’s equation formulation, is based on the Hamiltonian. In relativistic quantum ﬁeld theory one instead normally uses the Lagrangian formulation. One reason for this, which will become clear below, is that in the Lagrangian approach relativistic invariance is manifest, while in the Hamiltonian it is not. In other words, the Hamiltonian depends on the reference frame while the Lagrangian does not.

According to the Lagrangian formulation of mechanics, the dynamics of a system of N degrees of freedom (generalised coordinates) q1, . . . , qN is characterised by a function:

L (q, q˙) (2.1) where by q and q˙ we indicate collectively all the qi and q˙i. Considering evolution from time ti to time tf one deﬁnes the action:

Z tf S [q] ≡ L (q, q˙) dt (2.2) ti

The action is a functional: a real valued function of the trajectory q(t) followed by the system between ti and tf . Throughout these notes we shall indicate by squared brackets (ex. S[q]) the funtional dependence and by round brackets the dependence on real (and complex) variables (ex. L(q, q˙)).

According to the Least Action Principle, the physical time evolution of the system correponds to the trajectory(- ries) q¯(t) that stationarize the action. More technically, that is to say that any solution q¯ of the dynamics satisﬁes

δS [¯q] = 0 (2.3) where δS is the linearized variation of S under the change q¯(t) → q¯(t)+δq(t) with δq satisfying δq(tf ) = δq(ti) = 0. This is concretely written as

0 = δS [q] ≡ (S [¯q + δq] − S [¯q]) (2.4) linearized Z tf ∂L ∂L = dt δqi + δq˙i (2.5) ti ∂qi ∂q˙i Z tf tf ∂L d ∂L ∂L = dt − δqi + δqi (2.6) ∂q dt ∂q˙ ∂q˙ ti i i i ti

where the second term vanishes when considering the most general variations with ﬁxed boundaries q(ti) and q(tf ). Stationarity of the action is then equivalent to:

13 2.1. LAGRANGIAN MECHANICS

∂L d ∂L − = 0 (2.7) ∂qi dt ∂q˙i

This set of N second order differential equations for the generalised coordinates is called the set of Lagrange equations. Equivalently, the dynamics of the system can be described in the Hamiltonian formulation where one ends up with 2N first order differential equations for the generalised coordinates and momenta. The Hamiltonian is defined as the Legendre transform of the Lagrangian with respect to the generalised momenta:

∂L pi ≡ (2.8) ∂q˙i This relation is then inverted and the velocities are expressed in terms of the coordinates and momenta:

q˙i → q˙i(qi, pi) (2.9)

The Hamiltonian is ﬁnally given by:

H ≡ H(qi, pi, t) = piq˙i − L(qi, q˙i) (2.10) Notice that we use throughout the whole course Einstein’s summation convention. The Lagrangian equations of motion read in the Hamiltonian formalism:

∂H q˙ = i ∂p i (2.11) ∂H p˙i = − ∂qi

Another important mathematical notion in Hamiltonian mechanics is the Poisson bracket of the functions of the coordinates and momenta:

∂A ∂B ∂A ∂B {A, B} ≡ − (2.12) ∂pi ∂qi ∂qi ∂pi

The Poisson brackets allow us to write Hamiltonian mechanics in algebraic form, e.g.:

q˙i = {H, qi} (2.13)

p˙i = {H, pi} (2.14)

while for any observable O ({qi} , {pi}), the time dependence is:

d ∂O ∂O O = p˙ + q˙ dt ∂p i ∂q i i i (2.15) ∂O ∂H ∂O ∂H = − + = {H,O} ∂pi ∂qi ∂qi ∂pi

EPFL-ITP-LPTP Riccardo Rattazzi 14 2.2. FROM MECHANICS TO FIELD THEORY Similarly, the Poisson brackets encapsulate the group structure of the symmetries. This algebraic property of Hamiltonian mechanics is unchanged when going to the quantum description of the system:

Classical Quantum

{·, ·} [·, ·] (2.16)

O ({qi} , {pi}) Oˆ ({qˆi} , {pˆi})

For example for the angular momentum operators’ algebraic structure:

{Li,Lj} = ijkLk → [Li,Lj] = i~ijkLk (2.17)

2.2 From mechanics to ﬁeld theory

The notion of a field arises when we consider a coordinate space X (e.g. Rn, Sn,. . . ) and we associate a number of dynamical variables φα(x, t) (α = 1,...,N) to each point x ∈ X . The time coordinate t encodes the dynamics of the system. From a mechanical perspective, the total number of degrees of freedom is given by N × dimX , where by dimX we indicate the number of points in X . Of course if X is a continuum, like Rn or Sn, then dimX is infinite. For example, consider the electric and magnetic fields:

E~ (~x,t) (2.18) B~ (~x,t) (2.19)

The coordinate space is here R3 3 ~x. This is infinite dimensional, both because the points form a continuum and because ~x extends to infinity. Thus each of these two fields has an infinite number of degrees of freedom! Ordinary mechanics concerns a finite number of dynamical variables qi(t) with i = 1,...,M. One can view field theory as a limit of mechanics where the discrete label i is replaced by (α, x), which is continuous, as long as x is a continuous coordinate. As we talk about quantum field theory, the reader should expect to be dealing with infinitely many degrees of freedom throughout this course. In particular, most of the time the coordinate space will be given by R3 unless specifically indicated. At this stage, the reader should have already dealt with fields other than the electromagnetic ones, in particular, in fluid mechanics:

ρ(~x,t) (2.20) P (~x,t) (2.21) ~v(~x,t) (2.22)

From now on we will drop the index α and consider R3 as our coordinate space. The generalisation of the action to the ﬁeld theory framework is trivial:

Z S [φ] = dtL φ, φ˙ (2.23)

We want the action of a field configuration to be defined as a global quantity that does not depend on the coordinate space points. Indeed, it should describe the whole “trajectory” regardless of its details. Much like

EPFL-ITP-LPTP Riccardo Rattazzi 15 2.3. LEAST ACTION PRINCIPLE IN FIELD THEORY the whole electric charge of a system of particles does not depend on where the particles are and how they move through it. Therefore, in field theory, we can define the Lagrangian such that, when evaluated on a specific field configuration, it is only a function of time:

Z L(t) ≡ d3x L φ(~x,t), φ˙(~x,t), ∇~ φ(~x,t) (2.24)

This involves a new quantity called the Lagrangian density L. Using the four-vector notation:

L ≡ L (φ(x), ∂µφ(x)) (2.25) where we denote the four-vector xµ by x for simplicity as it is clear from context that it is not a one-dimensional coordinate. Finally: Z 4 S [φ] = d x L (φ(x), ∂µφ(x)) (2.26) where d4x ≡ dtd3x. Notice that the Lagrangian density was taken to depend only on fields and their derivatives at each coordinate space point. This property is called locality. It is intuitively obvious that, if a field theory description is to help us to get rid of instantaneous interactions between spacelike separated objects, then its Lagrangian density must be local. Notice also that we stick to Lagrangian densities that depend only on one derivative with respect to the coordinates. This is analogous to the Lagrangian in ordinary mechanics. The systems we will be interested in have this property but it is straightforward to generalise the action principle to higher derivative Lagrangians. Finally, as it is clear from the context of field theory that we will always be dealing with local Lagrangian densities, we will simply call them “Lagrangians” from here onwards.

2.3 Least action principle in ﬁeld theory

Using the ﬁeld theory action, we can now generalise the least action principle.

In order to proceed, consider for deﬁniteness the action integrated over the solid cylinder Ω = [ti, tf ] × BR where BR is the spacial ball of radius R: |~x| ≤ R Z S[φ, Ω] ≡ d4x L(φ, ∂φ) (2.27) Ω and consider a small variation δφ vanishing on the boundary ∂Ω of Ω.

φ(x) → φ(x) + δφ(x), δφ(~x,ti) = δφ(~x,tf ) = 0 and δφ(~x,t)||~x|=R = 0 (2.28) the linearized variation of the action is Z 4 ∂L ∂L δS [φ, Ω] ≡ d x δφ(x) + δ(∂µφ)(x) Ω ∂φ ∂(∂µφ) Z Z tf Z tf Z 4 ∂L ∂L 3 ∂L 2 ∂L = d x − ∂µ δφ(x) + d x δφ(x) + dt d Σ ~n · δφ(x) (2.29) ∂φ ∂(∂µφ) ˙ ~ Ω BR ∂φ ti ti ∂BR ∂(∇φ) Z 4 ∂L ∂L = d x − ∂µ δφ(x) Ω ∂φ ∂(∂µφ) where in the last step we used that δφ|∂Ω = 0. The same result obviously holds true for any other choice of the bounded region Ω as long as δφ|∂Ω = 0. Similarly it holds true for Ω coinciding with the full Minkowsky space M, as long as δφ → 0 suﬃciently fast for xµ → ∞. The expression appearing in brackets in the last line deﬁnes the functional derivative of the functional S

δS ∂L ∂L ≡ − ∂µ (2.30) δφ(x) ∂φ ∂(∂µφ)

EPFL-ITP-LPTP Riccardo Rattazzi 16 2.3. LEAST ACTION PRINCIPLE IN FIELD THEORY so that we can write Z δS S[φ + δφ] − S[φ] = d4x δφ(x) + O(δφ(x))2 . (2.31) δφ(x) This result is in full analogy in analogy with ordinary derivatives of functions of many variables, where, given a function f(qi), we have

X ∂f f(q + δq) − f(q) = δq + O(δq )2 . (2.32) ∂q i i i i

The least action principle is now simply stated as follows: the solutions of the dynamics correspond to the ﬁeld ¯ ¯ conﬁgurations φ satisfying δS[φ, Ω] = 0 for any Ω and for any δφ satisfying δφ|∂Ω = 0. The dynamics is thus described by the Euler-Lagrange equations

δS ∂L ∂L ≡ − ∂µ = 0 (2.33) δφ(x) φ¯ ∂φ ∂(∂µφ) φ¯

The concept of functional derivatives is obviously straightforwardly generalised to arbitrary space dimensions and to densities that depend on coordinates as well. For example, deﬁne a functional F :

Z n F [φ] = d xf(φ, ∂iφ, x) (2.34)

Then:

δF ∂f ∂f ≡ − ∂i (2.35) δφ(x) ∂φ ∂(∂iφ) In particular, we can choose as density:

(n) f(φ, ∂iφ, x) = φ(x)δ (x − x0) (2.36)

So that F [φ] = φ(x0). This yields on one hand:

δF δφ(x ) = 0 (2.37) δφ(x) δφ(x) while on the other hand:

δF ∂f = = δ(n)(x − x ) (2.38) δφ(x) δφ(x) 0

Yielding ﬁnally an important generalisation of ∂qi = δi : ∂qj j

δφ(x ) 0 ≡ δ(n)(x − x ) (2.39) δφ(x) 0

Notice furthermore that, in the Lagrangian description of ﬁeld theory, time and space coordinates are treated on equal grounds. This is an advantage of the Lagrangian formalism when considering relativistic ﬁeld theory.

EPFL-ITP-LPTP Riccardo Rattazzi 17 2.4. HAMILTONIAN FORMALISM 2.4 Hamiltonian formalism

The Hamiltonian formalism can be developed along the same lines as in ordinary mechanics. This formalism is based on a speciﬁc choice of time slicing and on the corresponding spacially integrated lagrangian density L in eq. (2.24). The canonically conjugated momentum π(x) is deﬁned by δL ∂L π(x) ≡ = (2.40) δφ˙(x) ∂φ˙ which simply extends the ordinary derivative of eq. (2.8) to a functional derivative. The above local relation is then inverted at each point to express φ˙ in terms of π(x), φ(x) and ∇φ(x) φ˙(x) → φ˙(π(x), φ(x), ∇φ(x)) (2.41)

The Hamiltonian is deﬁned in analogy with mechanics by a Legendre transform:

Z H(π, φ) = d3x π(x)φ˙(x) − L(φ, π) (2.42) where:

H ≡ π(x)φ˙(x) − L(φ, π) (2.43) is the Hamiltonian density. The Hamiltonian is a local functional of the momenta and ﬁelds over three dimensional space! As expected, in the Hamiltonian formalism the time coordinate has been singled out so that in this formalism relativistic invariance is no longer manifest. Also, as we work with functionals, the formalism we developed in the previous section about functional derivatives becomes handy. “Equal time” Poisson brackets (we make the time dependence implicit from here on) can be generalised to functional by employing the latter. Given two functionals:

Z A[π, φ] = d3x a(π, φ) (2.44) Z B[π, φ] = d3x b(π, φ) (2.45) we have:

Z δA δB δA δB {A, B} ≡ d3x − (2.46) δπ δφ δφ δπ In particular, using Eq. 2.39, we have:

{π(~x,t), φ(~y, t)} = δ(3)(~x − ~y) (2.47) {π(~x,t), π(~y, t)} = 0 = {φ(~x,t), φ(~y, t)} (2.48)

Similarly to ordinary mechanics, the equations can be written as

δH ∂H ∂H π˙ = {H, π}= − = ∇~ · − δφ ~ ∂φ ∂(∇φ) (2.49) δH ∂H φ˙ = {H, φ} = = δπ ∂π

EPFL-ITP-LPTP Riccardo Rattazzi 18 Chapter 3

Symmetries

The study of symmetries is of capital importance in theoretical physics as they greatly simplify any problem we have to deal with. For example, in ordinary mechanics the study of ballistics, in the presence of gravity only, becomes a two dimensional problem since the system is symmetric under any translation along the direction perpendicular to both the gravitational ﬁeld and the direction of movement. There are many other possible symmetries in physics (e.g. rotational invariance) and their study falls into the name group theory. Very often (but not always) the symmetry operation involves a linear transformation and this naturally leads us to the idea of ﬁnding a set of matrices that realises it. This procedure falls into the topic of representation theory. Below, we will put these ideas into a mathematical language and study one of the most important groups in relativistic physics as well as the most beautiful consequence of the presence of symmetries in a system.

3.1 Group theory

3.1.1 Groups

Let us start with a short summary of basic group theory. It is highly suggested to any reader, familiar or not with the topic, to have a look at one among any reference book on group theory (e.g. Lie Algebras in Particle Physics by H. Georgi Chap. 1-3).

Deﬁnition. A group G is a set of elements {gi}, i = 1,..., |G|, where |G| is called the order of the group, endowed with a binary operation that assigns to each ordered pair of elements a third element of the group:

◦ : G × G 7→ G (3.1) (g1, g2) 7→ g1 ◦ g2 = g3 This binary operation is usually written with a multiplicative notation:

g1 ◦ g2 → g1g2 (3.2)

For G to be a group, the binary operation must satisfy the following axioms:

1. Associativity:

g1(g2g3) = (g1g2)g3, ∀ g1, g2, g3 ∈ G (3.3)

2. Existence of a (left-)identity:

∃ e ∈ G s.t. ∀ g ∈ G eg = g (3.4)

19 3.1. GROUP THEORY 3. Existence of a (left-)inverse:

∀ g ∈ G, ∃ g−1 ∈ G s.t. g−1g = e (3.5)

In many books, it is required both the identity and the inverse to act also from the right. It is actually enough to have a left-identity/inverse. Indeed, from these three axioms it is easy to prove the following corollaries: Corollary. The group axioms have the following consequences:

1. The left-inverse is also a right-inverse:

gg−1 = e (3.6)

2. The left-identity acts in the same way from the right:

ge = g (3.7)

3. Uniqueness of the identity:

∃! e ∈ G s.t. ∀ g ∈ G eg = g = ge (3.8)

4. Uniqueness of the inverse :

∀ g ∈ G, ∃! g−1 ∈ G s.t. gg−1 = e = g−1g (3.9)

The proofs are left as an exercise.

Two elements g1 and g2 are said to commute if g1g2 = g2g1. Moreover if g1g2 = g2g1 ∀ g1, g2 ∈ G, then G is called an abelian group.

If |G| is finite, then G is called a finite group. Examples of finite groups are the group of permutations of n objects 1 Sn which has order n! or the cyclic group Zn of order n defined as the group generated by one single element g such that gn = e. The cyclic group is an finite abelian group.

If |G| is infinite obviously the group is said to be infinite. One example the reader should be familiar with is the group of rotations in two dimensions SO(2). It is trivially infinite as we can label the group elements by gθ and for each θ ∈ [0, 2π) we have a different rotation. This is an example of a Lie group. We will define below what a Lie group is more specifically.

Let us now be more precise about why group theory is relevant in physics. Besides the rotations group in n dimensions, there are many other groups which have great importance. Take for example N coordinates {qi}, i = 1,...,N. The space of conﬁgurations {qi} is a manifold and there exists freedom in choosing coordinates (e.g. cartesian, spherical,. . . ). For example if {qi} ≡ θ ∈ [0, 2π), the manifold is a circle. A change of coordinates corresponds to a change in the “point of view” on the system. Mathematically:

0 qi 7→ qi = fi ({qi}) (3.10)

0 such that for any conﬁguration there is one and only one {qi}. Then f is a bijection and the set of all possible changes in coordinates is a group with respect to the binary operation which is the composition of functions! Obviously if f and g are two changes of coordinates, then f ◦ g is also a change of coordinates. Therefore, to prove that we deal with a group, we need only to check the three axioms that a group is required to obey:

1. Associativity. Let f, g, h be three changes of coordinates and let us work, without loss of generality, with one coordinate ({qi} ≡ q). Then: (f ◦ g) ◦ h(q) = f(g(h(q))) = f ◦ (g ◦ h(q)) (3.11) 1 The cyclic group Zn is isomorphic to the addition of integers modulo n and thus can also be deﬁned this way

EPFL-ITP-LPTP Riccardo Rattazzi 20 3.1. GROUP THEORY 2. The identity is simply:

q 7→ q (3.12)

3. Inverse. As we consider the set of bijective functions, by deﬁnition, for all changes of coordinates f there exists f −1, its inverse, such that:

f −1 ◦ f(q) = q = f ◦ f −1(q) (3.13)

Hence, the set of changes of coordinates endowed with the operation of functions’ composition is indeed a group. An important subgroup is the group of changes of coordinates that leave the laws of motion invariant. If we consider Newton’s second law:

d~p X = F~ (3.14) dt ext

The group that leaves the equations of motion invariant is called the Galileo group. Sets of transformations that do not change the form of the laws are called symmetries and in this case, the Galileo group is the group of symmetries which leave classical mechanics invariant! Galileo group consists of three dimensional rotations, uniform translations of spacetime as well as linear uniform boosts:

0 ~x 7→ ~x = R(α, β, γ)~x + ~v0t + ~x0 (3.15)

t 7→ t + t0 (3.16)

Transformations are generates by 10 parameters : α, β, γ,~v0, ~x0 and t0. Any of them has an infinite number of 3 possible values, for example, ~x0 ∈ R . It is therefore an infinite group. As we shall see shortly, Galileo group is a Lie group. In the language of Lie groups, it has dimension 10 as it is generated by 10 parameters. Be particularly aware of the fact that the dimension of the Lie group can be finite even if the group itself is infinite, “infinite” meaning here that it contains an infinite number of elements! From now on we shall always work with a group G that describes a symmetry of the system. Moreover, in relativistic QFT the most important groups are all Lie groups. Thus, unless specifically stated, we shall only work with infinite groups that are Lie groups.

Let us start deﬁning concretely what a Lie group is: Deﬁnition. A Lie group G is a group and at the same time a smooth manifold. It is a manifold because its group elements are labelled by continuous variables which, regarded as coordinates, span the latter. If G is spanned by N variables then it is an N-dimensional Lie group. It is a smooth manifold because its group operations are smooth in the sense that (without loss a generality we consider a one-dimensional Lie group):

g(α)g(β) = g(γ), γ = f(α, β) (3.17) g−1(α) = g(i(α)) (3.18) and both f and i are smooth functions of their parameters.

The parameters labelling a Lie group can be put into a vector form:

~α = (α1, . . . , αN ) (3.19) It is a convention to call the identity the group element which parameters correspond to the null vector:

g(~0) = e (3.20)

In general, we work with an explicit realisation of the group (realisation: how the group acts) and not with the abstract notion of a group (deﬁned through Eq. 3.17). A realisation is a concrete way of writing group elements

EPFL-ITP-LPTP Riccardo Rattazzi 21 3.1. GROUP THEORY in terms of transformations on the same space. For example, consider the one-dimensional colinear group on the real line. A realisation of the group is:

g(α1, α2)x = α1x + α2, x ∈ R, α1 > 0 (3.21)

It is a two-dimensional Lie group which manifold is the upper half-plane of R2.

An important realisation of Lie groups is the matrix representation (we drop the term “matrix” from here on). Deﬁnition. An n-dimensional representation of a group G is formally deﬁned to be a homomorphism from G to GL(V ), where V is n-dimensional vector space:

D : G → GL(V ) (3.22) g 7→ D(g) such that:

D(g1)D(g2) = D(g1g2), ∀ g1, g2 ∈ G and D(e) = 1 (3.23)

Then, the realisation of the Lie group on the vector space V is as follows:

∀ g ∈ G, ~v 7→ D(g)~v ∈ V, ∀ ~v ∈ V (3.24)

Let us give a list of important deﬁnitions attached to representations:

Deﬁnition. Consider a representation D of a group G on the vector space V .

1. D is reducible if there exists on non-trivial invariant subspace U 6= V, {0} of V . Formally:

∃ U ⊂ V, s.t. ∀ ~u ∈ U, D(g)~u ∈ U, ∀g ∈ G (3.25)

2. D is irreducible if it is not reducible. That is, the only invariant subspaces are V itself and {0}. 3. D is completely reducible if and only if we can decompose V into a non-trivial partition of invariant subspaces:

V = U1 ⊕ U2 ⊕ ... (3.26)

In other words, there exists a basis of V such that:

  D1(g) 0 0     ∀ g ∈ G, D(g) =  0 D2(g) 0  (3.27)  .  0 0 ..

where Di is a ni × ni matrix, n1 + n2 + ... = n and all Di are irreducible. D is said to be a direct sum of irreducible representations:

D = D1 ⊕ D2 ⊕ ... (3.28)

EPFL-ITP-LPTP Riccardo Rattazzi 22 3.1. GROUP THEORY

4. Two representations D1 and D2 are equivalent if:

−1 ∃ S ∈ GL(n, C), s.t. S D1(g)S = D2(g), ∀ g ∈ G (3.29)

S is said to be a change of basis. 5. D is unitary if it is equivalent to a representation in which :

D−1(g) = D†(g), ∀ g ∈ G (3.30)

The most important representations are those which are irreducible. It is however not a straightforward task to tell, just by looking at it, if a representation is irreducible or not. In order to do this, we shall develop now some tools allowing us to easily classify representations by their reducibility.

3.1.2 Schur’s lemma

The most useful result about the reducibility of representations comes from Schur’s lemma and two corollaries:

Schur’s lemma. Let D1 and D2 be two irreducible representations of G acting on V1 and V2 respectively and an intertwining operator Λ: V1 → V2 such that:

ΛD1 = D2Λ (3.31)

Then:

1. Either Λ ≡ 0 or Λ is invertible. In the second case, V1 and V2 have therefore the same dimension and −1 D1 = Λ D2Λ, that is to say D1 and D2 are equivalent representations.

Corollary: If, under the hypotheses of Schur’s Lemma, D1 and D2 are further assumed to be inequivalent, then it must be Λ = 0.

2. If V1 = V2 = V and D1 = D2 = D then Λ = λ1.

Now let D be a completely reducible representation and Λ an hermitian operator such that:

ΛD = DΛ (3.32)

3. Then Λ is block-diagonal in the same basis D is block-diagonal:

    D1 0 0 λ11 0 0         D =  0 D2 0  , Λ =  0 λ21 0  (3.33)  .   .  0 0 .. 0 0 ..

We give a short proof of the ﬁrst and second Schur’s lemmas:

EPFL-ITP-LPTP Riccardo Rattazzi 23 3.1. GROUP THEORY Proof. To prove the ﬁrst lemma, consider the two following objects:

[Ker(Λ) ≡ subspace of V1 annihilated by Λ] and [Im(Λ) ≡ ΛV1] (3.34)

Eq. 3.31 shows that Ker(Λ) ⊆ V1 and Im(Λ) ⊆ V2 are invariant subspaces for the representations D1 and D2 respectively:

∀ ~v1 ∈ Ker(Λ) : Λ(D1~v1) = D2(Λ~v1) = 0 ⇒ ∀ ~v1 ∈ Ker(Λ),D1~v1 ∈ Ker(Λ) (3.35) and

D2~v2 = D2Λ~v1 ∀ ~v2 ∈ Im(Λ), ∃ ~v1 ∈ V1, s.t. ~v2 = Λ~v1 : ⇒ ∀ ~v2 ∈ Im(Λ),D2~v2 ∈ Im(Λ) (3.36) D2~v2 = Λ(D1~v1)

This proves that the kernel and the image of the intertwining operator are invariant subspaces under D1 and D2 respectively. Now, since we assume these two representations to be irreducible (meaning that there is no non-trivial invariant subspace for any of them), either Λ ≡ 0 or these subspaces are trivial:

Ker(Λ) = {0}, Im(Λ) = V2 (3.37) In the latter case, Λ is injective and surjective, i.e. bijective and invertible.

The second lemma can be seen as a corollary since it uses the ﬁrst one to be proven. Eq. 3.31 becomes in this case:

ΛD = DΛ (3.38)

This equation remains obviously true if Λ is replaced by Λλ = Λ − λ1:

ΛλD = DΛλ (3.39)

Now, by Schur’s ﬁrst lemma we have that either Λλ ≡ 0 or Λλ is invertible. However, by the fundamental theorem of algebra,

det(Λλ) = det(Λ − λ1) (3.40) is a polynomial of order n and has at least one root. If we choose λ to be this root, the polynomial is zero and det(Λλ) = 0, meaning it is not invertible. Therefore, by Schur’s ﬁrst lemma:

Λλ ≡ 0 ⇒ Λ = λ1 (3.41)

The third lemma is proven in the same spirit using the ﬁrst two lemmas. The proof is left as an exercise.

The proof relies strongly on the fact that the representations we consider are irreducible. Therefore, Schur’s lemma can be seen as a tool to prove representations are reducible by ﬁnding a non-zero, non-invertible intertwiner satisfying Eq. 3.31.

Notice that, physically, Schur’s third lemma means that, if we ﬁnd an operator which commutes with all group symmetries, the eigenvalues of this operator can be used to label the diﬀerent representations of our theory! For

EPFL-ITP-LPTP Riccardo Rattazzi 24 3.1. GROUP THEORY example, in quantum mechanics, we know that we label the spin of the particles with their total spin operator 2 2 J = JiJi eigenvalue (0, 1/2,...). Formally, it is because J commutes with all generators of the algebra describing spin, the SU(2) algebra.

This closes this section devoted to the irreducibility of representations and we will now focus more speciﬁcally on the mathematics of Lie groups.

3.1.3 Lie algebras

We shall start this section with an important result from the theory of Lie groups: Ado’s theorem. Every ﬁnite dimensional Lie group can be represented faithfully2, at least locally, by ﬁnite dimensional matrices.

This implies that we do not lose any information by studying the matrix representation of the group. As we saw earlier, we choose to parametrise the elements in such a way that α = 0 corresponds to the identity element. Therefore if we ﬁnd a representation of the group, it will be parametrised in the same way:

D(g(0)) ≡ D(0) = 1 (3.42) where from here on we forget the abstract notation and will write simply D(α) for all elements. By Ado’s theorem, the tangent space at the identity fully describes the behaviour of the group, up to global eﬀects. This leads to the concept of Lie algebra: Deﬁnition. A Lie algebra A is an n-dimensional vector space endowed with a second mulitplication, the Lie bracket or Lie product:

[·, ·]: A × A → A (3.43) (X,Y ) 7→ [X,Y ] satisfying:

Antisymmetry

[X,Y ] = −[Y,X] (3.44)

Jacobi identity

[[X,Y ],Z] + [[Y,Z],X] + [[Z,X],Y ] = 0 (3.45)

Linearity

[aX + bY, Z] = a[X,Z] + b[Y,Z] (3.46)

Since it is a vector space, we can ﬁnd a basis Ti and, as for the standard product, it is enough to know the product rule for this basis vectors:

[Ti,Tj] = ifijkTk (3.47) where fijk are called the structure constants. It is a set of number deﬁning completely the Lie algebra, much like T the metric gij = ei · ej deﬁnes completely the geometry in standard Riemannian spaces. There is a set of three theorems due to Sophus Lie linking Lie algebras and Lie groups. We do not intend to be rigorous here and will only give a small overview of their content.

2For all element in the group, there is a diﬀerent matrix

EPFL-ITP-LPTP Riccardo Rattazzi 25 3.1. GROUP THEORY Lie theorems I,II,III. The ﬁrst of the theorems states that we can associate a Lie algebra to any Lie group.

Consider an n-dimensional Lie group and D a representation of the latter. The tangent space at α¯ can be obtained by Taylor expanding any matrix of the group:

∂ D(α) = D(¯α) + D(α) (αi − α¯i) + ... (3.48) ∂αi α=¯α In particular around the identity:

∂ D(α) = 1 + D(α) αi + ... ≡ 1 + iαiXi + ... (3.49) ∂αi α=0 where we put the complex unity i in evidence by (physicists’) convention. The quantities Ti, i = 1,...N (where N is the dimension of the group) are called the generators of the group and are given by:

∂ Xi ≡ −i D(α) (3.50) ∂αi α=0 The factor of i is removed from the generators because if the representation is unitary, the generators will be hermitean (something physicists appreciate!).

Lie theorem II states that the generators Xi form the basis of a Lie algebra:

[Xi,Xj] = ifijkXk (3.51) The structure constants follow from the group multiplication law and are independent of the representation. In particular if we consider the group structure:

D(α)D(β) = D(f(α, β)) (3.52) the function f can only depend on the structure constants.

Finally, Lie theorem III states that the tangent space Tg at any point g(¯α) is also a Lie algebra and:

−1 ∂ k iXi(¯α) ≡ D (¯α) D(α) = iXk(0)Vi (¯α) (3.53) ∂αi α=¯α where V is an invertible matrix. The last equality means that the generators at any point are simply a linear combination of the generators at the origin: any A(g) is isomorphic to A(e). This is not surprising since with the group action we can translate from the origin to any point!

Notice that Lie theorems imply that for any group G there is one Lie algebra. However, a Lie algebra A can give rise to diﬀerent groups (e.g. SU(2) and SO(3)). However, a theorem states that given any Lie algebra A there is one and only one connected and simply connected Lie group associated to it (e.g. SU(2) is simply connected and connected, SO(3) is only connected).

We will focus from this point on the Lie algebra around the origin. Consider the following quantities:

D(α)D(β) = 1 + i(αi + βi)Xi + ... (3.54)

D(λα) = 1 + i(λαi)Xi, ∀ λ ∈ C (3.55) We see that locally, the product of two group elements is linearised. This leads to the concept of exponential map. The exponential map is a particular parametrisation of the Lie group G. As we saw, close enough from the origin,

EPFL-ITP-LPTP Riccardo Rattazzi 26 3.2. LORENTZ AND POINCARÉ GROUPS the group product structure becomes linear and therefore we can reach any group element by proceeding step by step in the following way:

α D(α) = D()n, ≡ (3.56) n Taking smaller and smaller steps, we obtain:

α n n i 2 iαiXi D(α) = lim D() = lim 1 + i Xi + O(1/n ) ≡ e (3.57) n→∞ n→∞ n

The αi become a set of exponential coordinates, furthermore, this implies a particular function f describing the product:

D(α)D(β) = eiαiXi eiβj Xj = eifk(α,β)Xk = D(f(α, β)) (3.58) We can prove this equation using the Haussdorf-Campbell-Baker formula:

1 1 exp(A) exp(B) = exp(A + B + [A, B] + ([A, [A, B]] + [B, [B,A]]) + ...) (3.59) 2 12 when A = αiXi and B = βjXj each term will be linear in Xk with complicated factors coming from the structure constants. We recover the previous statement that the group product structure depends only on the fijk! Now, this parametrisation is valid close to the origin, but more precisely, how much of G can we recover with this exponential map? Theorem. If G is a compact and connected Lie group, then the exponential map covers the whole group.

Compact means that the coordinates αi take all values on a compact set, in other words, the Lie group’s manifold is compact. Connected refers as well to the manifold describing the Lie group. Intuitively, the theorem can be understood in the following way: compactedness implies that we do not risk to diverge from the expontial map as we try to reach distant points; connectedness implies that all points of the manifold can be reached without “jumps” from one side to another.

It is sometimes the case in physics that the groups we have to deal with are connected but not compact. In this case another theorem becomes useful. Theorem. If G is a connected but non-compact Lie group, then every element of G can be written as

Y iα(n)X e i i (3.60) n where, generally, the number of group elements needed to span the whole group is ﬁnite.

To summarise, when dealing with Lie groups, all we have to do is to ﬁnd a representation of the Lie algebra, and to exponentiate! We will now put all this machinery to work in one of the most important groups in high-energy physics.

3.2 Lorentz and Poincaré groups

3.2.1 Construction

Let us start with a bit of history. Before the 20th century, the physics of mechanical systems was governed by Newton’s laws of motion. These laws are left invariant by the action of a symmetry group: Galileo group. As we saw before, its realisation on space and time is as follows:

EPFL-ITP-LPTP Riccardo Rattazzi 27 3.2. LORENTZ AND POINCARÉ GROUPS

0 ~x 7→ ~x = R(α, β, γ)~x + ~v0t + ~x0 (3.61)

t 7→ t + t0 (3.62) which summarise space and time homogeneity (~x0, t0), space isotropy (R ∈ SO(3)) and invariance under translations of the center-of-mass (~v0). Galileo’s group is a 10 dimensional Lie group. This described pretty much all physics before Maxwell came up with his set of equations: these were not invariant under Galileo’ group! At that time, there were three possibilities:

i. Maxwell equations are wrong: this was crazy given the great success they had in describing electromagnetic phenomena. ii. Galilean relativity does not apply to electromagnetism: this implies that the notion of movement is not relativist for electromagnetic phenomena, in other words there is a preferential reference frame (ether). iii. Galileo’s group is not the group describing the symmetries of Nature: Maxwell is right but it is still true that relativity of observers holds. Therefore, the group must be diﬀerent (and as we shall see it is simply a deformation of Galileo’s group).

We know now that it is the third possibility that is correct and indeed, people started to work on this possiblity around the end of the 19th century. First Voigt (1887) and then Lorentz (1895) led the way but had only an incomplete solution in hand. It is Larmor (1897, 1900) and Lorentz (1899, 1904) who found the “Lorentz transformations”, but it is Poincaré (1905) who found the group structure of the latter who gave the transformations the name of Lorentz. During the same year, Einstein published what is now called special relativity and it his paper rederived the transformations using solely the two following principles of special relativity:

I. Spacetime is homogeneous, isotropic and the laws of Nature are the same for all inertial observers. II. The speed of light c is the same in all frames.

We will follow Einstein’s path to reconstruct the so-called Poincaré group and its subgroup the Lorentz group. The two principles of special relativity imply that the spacetime interval:

2 µ ν ds = ηµν dx dx , ηµν = diag(+1, −1, −1, −1) (3.63) is the same for all observers. Here, η is the metric of Minkowski spacetime. The symmetry group of spacetime will be the set of coordinate transformations that leave the line element invariant endowed with the composition of coordinate transformations as group product. Consider then two observers O and O0 whose frames are linked through a change of coordinates xµ 7→ x0µ(x). Then:

∂xµ ∂xν ds2 = η dxµdxν = η dx0αdx0β ≡ η dx0µdx0ν = ds02 (3.64) µν µν ∂x0α ∂x0β µν Hence the transformations must satisfy:

∂xµ ∂xν η ≡ η (3.65) µν ∂x0α ∂x0β µν This, in turn, implies that the transformations must be at most linear in x so that ∂x0/∂x does not depend on the latter. Finally, we ﬁnd the transformation rule that describes Poincaré transformations:

µ 0µ µ ν µ µ ν x 7→ x = Λ ν x + a , ηµν Λ αΛ β = ηαβ (3.66)

EPFL-ITP-LPTP Riccardo Rattazzi 28 3.2. LORENTZ AND POINCARÉ GROUPS Poincaré group P is a deformation of Galileo group such that Maxwell’s equations are invariant under its action. It is remarkable that before special relativity was developed, this set of equations already “incorporated” it indirectly! Let g1 = (Λ1, a1), g2 = (Λ2, a2) ∈ P. The group product structure is as follows:

g1g2 = (Λ1Λ2, Λ1a2 + a1) 6= g2g1 = (Λ1Λ2, Λ2a1 + a2) (3.67) and the inverse is given by:

−1 −1 −1 gi = (Λi , −Λi ai) (3.68) As it is a deformation of Galileo group, it must also have 10 parameters. Let us count them:

aµ → 4 (3.69) µ Λ ν → 16 (3.70)

It seems to have 20 parameters, however there are 10 constraints coming from

µ ν ηµν Λ αΛ β = ηαβ, (3.71) leaving us with exactly 10 parameters. It has an important subgroup called the Lorentz group which is simply the group consisting of homogeneous Poincaré transformations. As such, it is the isometry group of spacetime and is denoted by O(1, 3) much like O(3) is the isometry group of three dimensional space. More generally O(p, q) is the group of that leaves invariant the metric with signature:

gµν = diag(+1,..., +1, −1,..., −1) (3.72) | {z } | {z } p q

In other words, it is the group of orthogonal matrices (hence the “O”(p,q)) with respect to the metric 3.72.

We will focus now on the Lorentz group. We shall ﬁrst see what Eq. 3.71 implies on the latter. Taking the determinant of the equation, we obtain:

det(Λ) = ±1 (3.73)

Thus the group O(1, 3) is disconnected. The elements belonging to the class with det(Λ) = 1 form a subgroup called the proper Lorentz group denoted SO(1, 3). The so-called improper elements do not form a group as they do not contain the identity. Now, let us consider the “00” component of Eq. 3.71:

0 2 X i 2 0 2 1 = Λ 0 − Λ 0 ⇒ Λ 0 ≥ 1 (3.74) i

0 0 The value of Λ 0 tells us how time is measured in diﬀerent frames. In particular, for Λ 0 < 0, past and future are inverted. The constraint tells us that, again, there are two classes of elements in the Lorentz group. Those 0 ↑ for which Λ 0 ≥ +1 form the orthochronous Lorentz group O (1, 3). Similarly to the improper elements, the non- orthochronous elements do not form a group. To summarise, the Lorentz group has 4 disconnected components:

EPFL-ITP-LPTP Riccardo Rattazzi 29 3.2. LORENTZ AND POINCARÉ GROUPS

0 Λ 0 det(Λ) Group?

↑ ↑ L+ ≡ SO (1, 3) ≥ +1 +1 X ↓ (3.75) L+ ≤ −1 +1 ↑ L− ≥ 1 −1 ↓ L− ≤ −1 −1

↑ ↓ ↑ ↑ ↑ Notice that L+ ⊕ L+ = SO(1, 3) and L+ ⊕ L− = O (1, 3) are two subgroups. However, we ﬁll focus on the smallest subgroup of the Lorentz group, the proper-orthochronous Lorentz group. The three other components are connected to the latter by parity and time reversal transformations:

↑ P :(t, ~x) 7→ (t, −~x) ∈ L− (3.76) ↓ T :(t, ~x) 7→ (−t, ~x) ∈ L− (3.77) ↓ PT :(t, ~x) 7→ (−t, −~x) ∈ L+ (3.78) (3.79)

So that the full Lorentz group can be seen as the composition of the proper-orthochronous Lorentz subgroup and these spacetime transformations, schematically:

↑ O(1, 3) = L+ ◦ {1,P,T,PT } (3.80)

↑ ↑ From now on, when we talk about the Lorentz group we shall refer to L+ ≡ SO (1, 3). Also, some authors refer to SO(1, 3) or O(1, 3) when they actually mean SO↑(1, 3), we shall make the same misuse of notation and denote the (proper-orthochronous) Lorentz group by SO(1, 3) as it is now clear which subgroup we are working with.

3.2.2 Raising and lowering

We want here to briefly recall the notions of contravariant and covariant vectors and the action of lowering and raising indices at the aid of the Lorentz metric. Notice first of all that the mapping Λ → (ΛT )−1 furnishes an alternative representation of the Lorentz group. One can indeed easily check that the above mapping satisfies T −1 T −1 T −1 all the properties of representations, in particular Λ1Λ2 → [(Λ1Λ2) ] = (Λ1 ) (Λ2 ) . Moreover the defining property of Lorentz tranformations, written in matrix form as

ΛT ηΛ = η (3.81) gives (ΛT )−1 = ηΛη−1 , (3.82) implying that Λ and (ΛT )−1 are equivalent representations that are obtained one from the other by the simple action of the Minkowsky metric. Deﬁning η−1 as an objet with upper indices ≡ ηµν , the relation between the two µν equivalent representations conveniently reduces to raising and lowering the indices with the aid of ηµν and η . µν The fact that ηµν = η = diag(1, −1, −1, −1), is just an additional incidental fact. Putting indices back, eq. (3.82) thus reads

T −1 ν ρ σν ν [(Λ ) ]µ = ηµρΛ ση ≡ Λµ (3.83) which further implies µ ν ν ν τ τ Λ τ Λµ = δτ Λµ Λ ν = δµ . (3.84)

EPFL-ITP-LPTP Riccardo Rattazzi 30 3.2. LORENTZ AND POINCARÉ GROUPS µ 0µ µ ν A 4-vector V transforming according to V = Λ ν V is called contravariant, while a vector Wµ tranforming 0 ν ν according to W µ = Λµ Wν is called covariant. By eq. (3.83) we have that Vµ ≡ ηµν V is covariant, while µ µν µ µ W = η Wν is contravariant. It is easy to see that the 4-gradient ∂/∂x forms a covariant vector ∂µ ≡ ∂/∂x . 0µ µ ν Indeed, using x = Λ ν x one has ∂x0ρ ∂ = ∂0 = Λρ ∂0 . (3.85) ν ∂xν ρ ν ρ ν which upon multiplication by Λµ and use of eq. (3.84) becomes 0 ν ∂µ = Λµ ∂ν . (3.86)

3.2.3 Lie algebra

Let us now derive the Lie algebra of the Lorentz group, it is then easily extended to the full algebra of the Poincaré group. The Lorentz group is realised in a standard way on four vectors:

µ µ ν g(Λ)x = Λ ν x (3.87) Recall that to study the Lie algebra, we should consider group elements around the origin (represented by the identiy on spacetime):

µ µ µ 2 Λ ν = δ ν + ω ν + O(ω ), |ω| 1 (3.88)

We shall exploit the constraints imposed by the group to derive constraints on the Lie algebra. Eq. 3.71 reads at an inﬁnitesimal level:

α α 2 β β 2 2 ηµν = ηαβ(δ µ + ω µ + O(ω ))(δ ν + ω ν + O(ω )) = ηµν + ωµν + ωνµ + O(ω ) (3.89) Since this equation must be true order by order in ω, we obtain:

ωµν = −ωνµ (3.90)

Therefore the matrix of generators is a four-by-four antisymmetric matrix and thus has 6 components. Rewrite:

1 1 ω = (ω − ω ) = ω (δα δβ − δα δβ ) (3.91) µν 2 µν νµ 2 αβ µ ν ν µ

Now deﬁne:

αβ α β α β J µν ≡ i(δ µδ ν − δ ν δ µ) (3.92)

Then, at linear order:

i µ Λµ = δµ − ω J αβ (3.93) ν ν 2 αβ ν

This tells us that the six J s (labelled by the antisymmetric pair αβ) are the generators of the Lorentz group and that ωs are coordinates on the manifold. Using the exponential map, a general element of the Lorentz group will be written:

EPFL-ITP-LPTP Riccardo Rattazzi 31 3.2. LORENTZ AND POINCARÉ GROUPS

− i ω J αβ Λ = e 2 αβ (3.94)

If now, one writes down explicitly the generators, one notices that J ij generate rotations and J i0 are the boost generators. Take for example the following element:

  0 1 0 0  10 10   1 0 0 0  Λ = exp(−iω10J ) ≡ exp(−iηJ ) = exp −η   (3.95)   0 0 0 0  0 0 0 0 where η is called the rapidity. Using trivial hypergeometric relations, one ﬁnds:

 cosh η − sinh η 0 0   − sinh η cosh η 0 0  Λ =   (3.96)  0 0 1 0  0 0 0 1 which is nothing but a boost in the x direction. Now, with respect to the standard form of the boosts, we have the following relations:

cosh η = γ, tanh η = β (3.97) where, in units for which c = 1, β = v. Since |v| ≤ 1, we have from Eq. 3.97:

η ∈ (−∞, +∞) (3.98)

And thus, the Lorentz group is non-compact! The Lie algebra of the Lorentz group, denoted so(1, 3), can be derived from Eq. 3.92 (exercise):

[J µν , J ρσ] = i (ηµσJ νρ + ηνρJ µσ − ηµρJ νσ − ηνσJ µρ) (3.99)

3.2.4 Representations

Now that we have the Lie algebra in hand (Eq. 3.99), we must find its representations. We already know one: the defining representation Eq. 3.92. It is called the defining representation as the Lorentz group is defined through its action on four-dimensional spacetime (Lorentz transformations). In the same way, SO(3) is defined as the group of rotations in three dimensions and as such its defining representation will be the usual rotation matrices in three dimensions. We shall however classify all representations of the Lorentz group. In representation theory, one always tries to reduce the problem to another one that is already solved.

To make the Lie algebra more explicit, let us separate space and time. In other words, we separate the rotation generators (something we know about!) from the boost generators (less known) and we rewrite:

1 1 Rotations: J ≡ J jk, θ ≡ ω (3.100) i 2 ijk i 2 ijk jk i0 Boosts: Ki ≡ J , ηi ≡ ωi0 (3.101)

EPFL-ITP-LPTP Riccardo Rattazzi 32 3.2. LORENTZ AND POINCARÉ GROUPS The Lie algebra Eq. 3.99 becomes:

[Ji,Jj] = iijkJk (3.102)

[Ji,Kj] = iijkKk (3.103)

[Ki,Kj] = −iijkJk (3.104)

We recognise in Eq. 3.102 the SU(2) algebra of rotations for half-integer and integer spins3. Being embedded in a bigger algebra, it is here a subalgebra. Notice also that from Eq. 3.103, the boost generators transform as vectors under rotations.

We found that we have an SU(2) subgroup spanned by the rotation generators, however the full Lie algebra still mixes these generators with the boost generators. There is a crucial difference between these two sets of generators. µ Indeed, the spacetime realisation of the Lorentz group is given by matrices Λ ν defined in such a way that the rotations generators are hermitian and the boosts generators anti-hermitian4 (this reflects the non-compactedness of the boosts). It is therefore natural to look for an hermitian combination of generators that simplifies the Lie algebra we are studying. The result is:

1 J ± = (J ± iK ) (3.105) i 2 i i and the simpliﬁed Lie algebra becomes:

± ± ± [Ji ,Jj ] = iijkJk (3.106) ± ∓ [Ji ,Jj ] = 0 (3.107)

We managed to decompose the Lorentz algebra into two independent SU(2) subalgebras:

so(1, 3) ' su(2) ⊕C su(2) (3.108) where ⊕C denotes a complexiﬁed direct sum. This identity holds at the level of the algebra, much like the SO(3) Lie algebra is isomorphic to the SU(2) algebra. However it is not true that at the group level the Lorentz group decomposes into two SU(2) subgroups.

This decomposition of the algebra nonetheless implies that we can construct all the representations of the Lorentz group if we know how to build SU(2) representations. This is well known from spin quantum mechanics, and therefore we will label the representations of the Lorentz group with two half-integers:

(j1, j2) ≡ (j−, j+) (3.109)

± ± where j± are the eigenvalues of the casimir operators Ji Ji of the two SU(2)s. The dimension of the representation will be given by:

dim(j−, j+) = (2j+ + 1)(2j− + 1) (3.110)

Here is a short table describing the main representations of the Lorentz group:

3Note that the subgroup is SU(2) and not SO(3) (which have the same algebra) as we want to consider all spins and not only integer spins! 4 i0 µ i0 Be careful with spacetime indices: to see the non-hermicity, look at (J ) ν and not at (J )µν

EPFL-ITP-LPTP Riccardo Rattazzi 33 3.2. LORENTZ AND POINCARÉ GROUPS

(j−, j+) dim Type Example (0, 0) 1 Scalar π0, π±, Higgs? (1/2, 0) 2 Left-handed spinor Neutrinos (0, 1/2) 2 Right-handed spinor Anti-neutrinos (3.111) (1/2, 0) ⊕ (0, 1/2) 4 Dirac spinor e±, p, n (1/2, 1/2) 4 Vector γ, W ±,Z0, g (1, 1) 9 Traceless metric tensor Graviton?

From here, we could proceed as in spin quantum mechanics and define states labelled by the j± and their eigenvalues 3 for the operator J± given by m±. In this way we would construct algebraically the vector space on which the generators of the Lie algebra act. This would close the construction. However, we want to study quantum field theory. We shall thus study how the representations act on fields. We will see that each representation will act on a given type of particles as indicated in 3.111.

In order to proceed, we consider the simplest example consisting of a scalar field. The (0, 0) representation of the Lorentz group has dimension 1 and as such, being a matrix representation, acts trivially on the field (hence the name “scalar”). However, to gain a better understanding on how the transformation of spacetime changes the field, we shall consider the following argument: the principle of special relativity states that two observers whose reference frames are linked through a Lorentz transformation Λ should measure the very same things if the latter is taken into account in the calculations. In other words, if observer O measures a field value φ(x) at x, a second observer O0 should measure the very same value at that x transformed into his frame, i.e. at x0 = Λx. Therefore if we extrapolate the argument to all points in spacetime, the field distribution φ0 is linked to the original one by:

φ0(Λx) = φ0(x0) ≡ φ(x) (3.112)

From this rule we can extract how a Lorentz transformation acts on the ﬁeld:

0 −1 DΛ[φ](x) ≡ φ (x) = φ(Λ x) (3.113)

We can now check that the product rule of the Lorentz group is indeed satisﬁed :

−1 −1 −1 −1 DΛ2 [DΛ1 [φ]] (x) = DΛ1 [φ](Λ2 x) = φ(Λ1 Λ2 x) = φ((Λ2Λ1) x) = DΛ2Λ1 [φ](x) (3.114) and that the representation is linear:

DΛ[aφ + bψ](x) = aDΛ[φ](x) + bDΛ[ψ](x) (3.115)

This is generalised to any representation on ﬁelds:

0 b −1 Φa(x) 7→ DΛ[Φa](x) = Φa(x) ≡ Ma (Λ)Φb(Λ x) (3.116)

b where Ma are matrices belonging to some given representation of the Lorentz group and the field Φa is said to transform under that representation. We shall now shortly see how generators act on fields. In order to do so, we consider an infinitesimal Lorentz transformation:

EPFL-ITP-LPTP Riccardo Rattazzi 34 3.3. NOETHER’S THEOREM

Λ = 1 + ω, Λ−1 = 1 − ω (3.117)

Then:

i ∆ φ = D [φ](x) − φ(x) = φ(x − ωx) − φ(x) = −ω xµ∂ν φ(x) ≡ − ω J µν φ(x) (3.118) Λ Λ µν 2 µν so that:

J µν = i(xµ∂ν − xν ∂µ) (3.119)

Now, generators become linear operators acting on the ﬁelds! We can check that the commutation relations of the Lorentz Lie algebra are satisﬁed. Finally, we can add translations to the picture:

µ µ ∆aφ = Da[φ](x) − φ(x) = φ(x − a) − φ(x) = −a ∂µφ(x) ≡ ia Pµφ(x) (3.120)

We see that, just as in quantum mechanics, the generators of translations are the momenta Pµ. The remaining commutation relations of the Poincaré group (proper-orthochronous Lorentz group and translations) can be derived using the linear operator form of the generators and are given by:

[Pµ,Pν ] = 0 (3.121)

[Jαβ,Pµ] = i (ηµβPα − ηµαPβ) (3.122)

From this, it can be checked that P0 is a scalar under rotations and that Pi is a vector. We can with all this in hands write down how the representations of ﬁnite Poincaré transformations acts on ﬁelds:

−1 D(Λ,a)[φ](x) ≡ Da[DΛ[φ]](x) = DΛ[φ](x − a) = φ(Λ (x − a)) (3.123)

More details about the representations of the Lorentz group will be given when dealing with the ﬁrst non-trivial representation: the spinors.

Now that we have dealt with the symmetry group of spacetime, we can carry on and study the consequences of these symmetries.

3.3 Noether’s theorem

3.3.1 Statement

Noether’s theorem is a statement about the dynamical consequences of symmetry. But ﬁrst, let us deﬁne properly what a symmetry is. Let us consider a change of coordinates corresponding to a Lie group with parameters {αi}:

x0µ = f µ(x, α) 0 0 (3.124) φa(x ) = Fa(φ(x), α)

EPFL-ITP-LPTP Riccardo Rattazzi 35 3.3. NOETHER’S THEOREM In general the form of the Lagrangian density will change:

Z Z S = d4xL(φ(x), ∂φ(x)) = d4x0L0(φ0(x0), ∂0φ0(x0)) (3.125) Ω f(Ω,α) where in general L= 6 L0 as a function. The transformation 3.124 is said to be a symmetry if L = L0 and:

Z Z d4xL(φ(x), ∂φ(x)) = d4x0L(φ0(x0), ∂0φ0(x0)) (3.126) Ω f(Ω,α)

In other words, the equations of motion in the primed system have the same form as in the original system. Notice that this equality holds for all Ω, so that the requirement for Eq. 3.124 to be a symmetry reads:

L(φ(x), ∂φ(x))d4x = L(φ0(x0), ∂0φ0(x0))d4x0 (3.127)

0 0 Basically, this means that if φa(x) solves the equations of motion in the original system, so does φa(x ) in the primed system. More concretely, symmetries yield a degeneracy of solutions: if φa(x) is a solution, so will be 0 0 −1 0 µ φa(x ) = Fa(φ(f (x , α)), α) for all Fa, f satisfying Eq. 3.127. Examples. 1. Free scalar ﬁeld: 1 1 L = (∂φ)2 − m2φ2 (3.128) 2 2 Let us check that spacetime translations are indeed a symmetry:

x0µ = xµ − aµ (3.129) φ0(x0) = φ(x) The volume element is not aﬀected by translations:

d4x = d4x0 (3.130)

nor is the mass term of the Lagrangian:

1 1 m2φ2(x) = m2φ02(x0) (3.131) 2 2 It just remains to check the derivatve term:

∂xν ∂0 φ0(x0) = ∂0 φ(x) = ∂ φ(x) = δν ∂ φ(x) = ∂ φ(x) (3.132) µ µ ∂x0µ ν µ ν µ This implies that Eq. 3.127 is and spacetime translations are a symmetry. 2. Complex scalar ﬁeld: µ ∗ 2 ∗ L = ∂µφ∂ φ − m φφ (3.133) We will check that this Lagrangian respects a global U(1) symmetry: x0µ = xµ (3.134) φ0(x0) = eiαφ(x) The symmetry being global, the Lie parameter α is constant in spacetime and therefore is it enough to check:

φ0φ0∗ = φeiαe−iαφ∗ = φφ∗ (3.135)

EPFL-ITP-LPTP Riccardo Rattazzi 36 3.3. NOETHER’S THEOREM 3. Other examples of symmetries are Lorentz transformations, O(n) or U(n) symmetries when dealing with n real or complex ﬁelds, etc.

As we previously saw, in Lie groups all necessary information is encoded in the infinitesimal transformations. We shall do the same here and expand the finite transformations around the identity. It is a convention to defined the identiy as:

xµ = f µ(x, 0) (3.136) φa(x) = Fa(φ(x), 0) Then, at linear order in α we can write:

0µ µ µ i µ µ x = x − i (x)α ≡ x − (x) 0 0 i (3.137) φa(x ) = φa(x) + Eai(φ(x))α ≡ φa(x) + Ea(x) We will also need to consider the inﬁnitesimal change in the functional form of the ﬁeld. Consider:

0 0 0 µ 0 φa(x ) = φa(x) − (x)∂µφa(x) (3.138)

Then:

0 µ i i φa(x) = φa(x) + (Eai(φ(x)) + i (x)∂µφa(x)) α ≡ φa(x) + ∆ai(φ(x))α ≡ φa(x) + ∆a(x) (3.139) We are now ready to state Noether’s theorem.

Noether’s theorem. Let us consider a Lie group G with coordinates {αi}, i = 1,...,N which induces the transformation 3.124. If G is a symmetry group of the system, then there are N conserved quantities described by N currents:

µ ∂L µ µ Ji ≡ ∆ai − i L, ∂µJi = 0 (3.140) ∂(∂µφa)

Proof. The fact that G is a symmetry group implies:

Z Z S = L(φ, ∂φ)d4x = L(φ0, ∂0φ0)d4x0 (3.141) Ω Ω0=f(Ω,α)

Let us rewrite the second integral as an integral over x = f −1(x0, α). For an inﬁnitesimal transformation (we will work from here on at linear order in the parameters αi and thus we omit in all equation a O(α2)), the Jacobian will be given by:

0µ ∂x µ µ µ = det (δ − ∂ν ) = 1 − Tr (∂ν ) = 1 − ∂ (3.142) ∂xν ν

As the primed ﬁeld is a function of x0, so is the Lagrangian. Therefore we write:

0 0 0 0 0 0 0 µ 0 0 L(φ (x ), ∂ φ (x )) = L(φ (x), ∂φ (x)) − ∂µL(φ (x), ∂φ (x)) (3.143)

EPFL-ITP-LPTP Riccardo Rattazzi 37 3.3. NOETHER’S THEOREM This follows from5

0 0 µ 0 0 0 0 α 0 φa(x ) = (1 − ∂µ)φa(x) and ∂µφa(x ) = (1 − ∂α)∂µφa(x) (3.144) Thus:

Z Z 0 0 0 4 0 0 0 µ 0 0 4 L(φ , ∂ φ )d x = [L(φ , ∂φ ) − ∂µL(φ , ∂φ )] (1 − ∂) d x Ω0=f(Ω,α) Ω Z Z (3.145) 0 0 µ 0 0 4 4 = [L(φ , ∂φ ) − ∂µ ( L(φ , ∂φ ))] d x = L(φ, ∂φ)d x Ω Ω

Since this is true for all Ω we have:

0 0 0 0 µ ∂L δφb ∂L δ∂ν φb µ 0 = L(φ , ∂φ ) − L(φ, ∂φ) − ∂µ ( L) = 0 + 0 − ∂µ ( L) (3.146) ∂φb δφa ∂(∂ν φb) δ∂µφa

Recalling that:

0 δφb ≡ ∆aδab (no sum) (3.147) δφa and renaming the ﬁelds φ0 → φ for simplicty, we obtain:

∂L ∂L µ ∂L ∂L ∂L µ 0 = ∆a + ∂µ∆a − ∂µ ( L) = − ∂µ ∆a + ∂µ ∆a − L (3.148) ∂φa ∂(∂µφa) ∂φa ∂(∂µφa) ∂(∂µφa) So that when we consider ﬁeld conﬁgurations that satisfy the equations of motion:

∂L µ ∂L µ i µ i ∂µ ∆a − L = ∂µ ∆ai − i L α ≡ ∂µJi α = 0 (3.149) ∂(∂µφa) ∂(∂µφa) Which means that, when the Lie group G is a symmetry group of the system and the field configuration considered i µ satisfies the equations of motion, for each Lie symmetry α , there exists an associated current Ji which is conserved!

Let us make some remarks about this theorem. First, this result implies the conservation of the total charge µ associated to each current Ji :

Z 3 0 Qi ≡ d xJi (3.150)

Consider the spacetime region R3 × ∆T , where ∆T = [0, t]. Then, from Noether’s theorem:

Z t Z Z t Z Z Z t Z 3 µ 3 0 ~ ~ 3 0 0 ~ ~ 0 = dt d x∂µJi = dt d x(∂tJi + ∇ · Ji) = d x(Ji (t, x) − Ji (0, x)) + dt dΣ · Ji (3.151) 0 0 0

If the ﬁelds vanish fast enough at inﬁnity, then the space boundary integral vanishes, so that:

5Prove it!

EPFL-ITP-LPTP Riccardo Rattazzi 38 3.3. NOETHER’S THEOREM

Z Z 3 0 3 0 Qi(t) = d xJi (t, x) = d xJi (0, x) = Qi(0), ∀ t (3.152)

In other words, Noether’s theorem implies that for the total charge we have:

d Q (t) = 0 (3.153) dt i

Finally, Noether’s theorem implies a local conservations law as a result of a global symmetry. The locality of the conservation law suggests that a local derivation should exist. Indeed, consider a local transformation where the Lie coordinates αi ≡ αi(x) now depend on spacetime. Deﬁne:

0 i δφa(x) ≡ φa(x) − φa(x) = ∆aiα (x) (3.154) i i i δ∂µφa(x) ≡ ∂µ(∆aiα (x)) = (∂µ∆ai)α (x) + ∆ai∂µα (x) (3.155)

Now:

0 0 ∂L i ∂L i i δL = L(φ , ∂φ ) − L(φ, ∂φ) = ∆aiα + ((∂µ∆ai)α (x) + ∆ai∂µα (x)) (3.156) ∂φa ∂(∂µφa) Using the equations of motion and the invariance under the symmetry:

µ i ∂L i δL = ∂µ(i L)α + ∆ai∂µα (x) ∂(∂µφa) (3.157) µ i µ i = Ji ∂µα + ∂µ(i Lα )

So that for a local transformation we have:

Z µ i 4 δS = Ji ∂µα d x + boundary term (3.158) Ω

This shows how the current associated to a symmetry can be derived by promoting the global symmetry to a local one!

We will now give an important example and application of Noether’s theorem.

3.3.2 The stress-energy tensor

We will treat more speciﬁcally the invariance under spacetime translations and study its consequences. Firstly, we need:

L(φ(x), ∂φ(x)) = L(φ0(x0), ∂0φ0(x0)) (3.159)

If L depended explicitly on x this would not be true! The ﬁrst comment is therefore that to have spacetime translations invariance, the Lagragian must not depend explicitly on spacetime.

EPFL-ITP-LPTP Riccardo Rattazzi 39 3.3. NOETHER’S THEOREM Spacetime translations read:

0µ µ µ µ µ i x = x − a ≡ x − i (x)α 0 0 i (3.160) φa(x ) = φa(x) ≡ φa(x) + Eai(φ(x))α and

µ Eai(φ(x)) + i (x)∂µφa(x) ≡ ∆ai(φ(x)) (3.161)

We make the following indentiﬁcations:

µ µ ν µ µ ν i a = δν a ⇒ i ≡ δν , a ≡ α (3.162) Eai = 0 ⇒ ∆ai = ∆aµ ≡ ∂µφa (3.163)

µ We see that the Lie coordinates are simply the components of a spacetime four-vector. Therefore the currents Ji will actually form a rank 2 tensor called the stress-energy tensor:

µ ∂L µ Tν ≡ ∂ν φa − δν L (3.164) ∂(∂µφa)

The name comes from the fact that the conserved charges correspond to the total four-momentum of the system:

Z 0 3 Pµ ≡ Tµ d x (3.165)

Spacetime translations invariance of the action implies thus the conservation of energy and momentum:

d P = 0 (3.166) dt µ

There is yet another important symmetry of the action: Lorentz invariance. A ﬁnite Lorentz transformation reads:

0µ µ ν x = Λ ν x 0 0 b (3.167) φa(x ) = D(Λ)a φb(x) where

b b i µν D(Λ)a ≡ exp − ωµν Σ (3.168) 2 a and Σµν are a representation of the Lorentz algebra. At the inﬁnitesimal level:

0µ µ µ ν x = x + ω ν x 0 0 i µν b (3.169) φa(x ) = φa(x) − 2 (ωµν Σ )a φb(x)

EPFL-ITP-LPTP Riccardo Rattazzi 40 3.3. NOETHER’S THEOREM Which leads to:

µ µν = −ω xν (3.170) and

i ∆ = − (ω Σµν ) bφ (x) − ωµν x ∂ φ (x) a 2 µν a b ν µ a i = − ωµν (Σ + i(x ∂ − x ∂ )) b φ (x) (3.171) 2 µν µ ν ν µ a b i ≡ − ωµν (Σ + J ) b φ (x) 2 µν µν a b

This expression shows us how an inﬁnitesimal Lorentz transformation decomposes into the sum of a spin and orbital angular momentum contribution.

Now, for sake of simplicity, let us consider a single scalar ﬁeld φ(x). We have:

µ ∂L µν µν J = (−ω xν ∂µφ) + ω xν L (3.172) ∂(∂µφ)

Assuming Lorentz invariance of the Lagrangian and antisymmetrising the indices we obtain:

µ µ ∂L µ µ µ Ji ≡ Mαβ = xβ∂αφ − δαxβL − (α ↔ β) = xβTα − xαTβ (3.173) ∂(∂µφ)

Hence, we notice that we can express the currents arising from the Lorenrz invariance of the action as a function of the stress-energy tensor.

The conservation of the latter, as well as the conservation

µ ∂µMαβ = 0 (3.174) yields:

Tµν = Tνµ (3.175)

Therefore, in a theory which is translationally and Lorentz invariant, the stress-energy tensor is symmetric. Let us see what the conservation of the currents implies physically. The conserved charges are:

Z 3 0 Jµν ≡ d xMµν (3.176)

Take µν = ij, then:

Z Z 3 0 0 3 Jij = d x(xiTj − xjTi ) ≡ d x(xipj − xjpi) (3.177)

EPFL-ITP-LPTP Riccardo Rattazzi 41 3.3. NOETHER’S THEOREM where pi is the density of three-momentum. Deﬁne now:

1 Z J = J = d3x(~x ∧ ~p) (3.178) i 2 ijk jk i

Therefore, the spatial components of the conserved charges correrspond to the total angular momentum of the system! In other words, invariance under space rotations implies the conservation of angular momentum. Let us have a look at the space-time components (invariance under boosts):

Z 3 Ki ≡ Ji0 = d xxiρ − x0pi (3.179)

0 where ρ = T0 is the energy density. Using the deﬁnition of the center of mass:

R d3xx ρ XCM ≡ i (3.180) i R d3xρ

The conserved charges related to the invariance under boosts are:

CM Ki ≡ P0Xi − tPi (3.181)

Notice that the Ki depend explictly on time. So, what does its conservation imply?

d ˙ CM ˙ CM Pi 0 = Ki = P0Xi − Pi ⇒ Xi = = const. (3.182) dt P0

In other words, the center of mass moves at constant speed. This is the same result as in Galileo invariant systems.

EPFL-ITP-LPTP Riccardo Rattazzi 42 Chapter 4

Scalar ﬁelds

4.1 The Klein-Gordon ﬁeld

The goal of this chapter is to consider the simplest possible ﬁeld theory and study its quantisation. The simplest Lagrangian can be build of terms quadratic in the scalar ﬁeld φ(x). Working only up to quadratic order in the dynamical variable is the correct procedure to study small oscillations around the ground state. For example

00 2 m m 2 V (q ) m mω L = q˙2 − V (q) = δq˙ − 0 (δq)2 + O(δq)3 = (δq˙)2 − (δq)2 + O(δq)3 , (4.1) 2 |{z} 2 2 2 2 q=q0+δq where we have expanded around the minimum of V (q): q = q0 +δq. The most general Lorentz invariant Lagrangian for φ(x) involving at most two derivatives up to quadratic order in the ﬁeld reads

c c L = 1 ∂ φ∂µφ + 2 φ2 + c φ + c ∂2φ + c φ∂ ∂µφ , (4.2) 2 µ 2 3 4 5 µ

This Lagrangian can be simplified significantly. Firstly, we can rescale the field φ(x) to choose c1 = 1 which ensures a canonical normalisation. Secondly, the origin of the φ(x) space can be shifted such that c3 = 0. The term proportional to c4 corresponds to a total derivative and thus does not contribute, while the term proportional to c5 can be integrated by parts and becomes proportional to c1. The only remaining free parameter is thus the 2 mass of the scalar field c2 ≡ −m and we find the Klein-Gordon Lagrangian

1 m2 L = ∂ φ∂µφ − φ2 . (4.3) 2 µ 2 From dimensional analysis we know

Z d4xL = E0 , (4.4) and thus

d4x = E−4 , [L] = E4 , (4.5) [φ] = E, m2 = E2 .

1This chapter wasn’t typeset by the author. Any error shan’t be attributed to him. These are notes that were typeset on the ﬂy during lectures. For a more thorough and precise treatment, see Peskin.

43 4.2. QUANTISATION OF THE KLEIN-GORDON FIELD The above Lagrangian can be viewed as an expansion of

1 L = (∂ φ∂µφ) − V (φ) , (4.6) 2 µ 2 00 around the stationary minimum of the potential m ≡ V (φmin). The equations of motion are

∂L δL 2 ∂µ − = 0 → + m φ = 0 (4.7) ∂ (∂µφ) φ µ where ≡ ∂µ∂ . This is the Klein-Gordon equation. Deﬁning the conjugate momentum π ≡ ∂L = φ˙ we ﬁnd the Hamiltonian density ∂φ˙

1 1 1 1 H = πφ˙ − L = π2 + (∇φ)2 + V (φ) = π2 + (∇φ)2 + m2φ2 . (4.8) 2 2 |{z} 2 2 K.G Since the Hamiltonian H = R Hd4x is bounded from below the mass parameter is positive m2 > 0. The Klein- Gordon Lagrangian can thus be viewed as an expansion around the minimum of the potential.

4.2 Quantisation of the Klein-Gordon ﬁeld

We proceed by quantising the Klein-Gordon ﬁeld using the canonical formalism as done in ordinary quantum mechanics. This consists of the following substitutions This implies

· Replace π(x, t), φ(x, t) with hermitean operators acting on Hilbert space φ,ˆ πˆ · Replace Poisson bracket {-,-} with commutation relations [-,-]

Classical Quantum h i 3 ˆ 3 {π(y, t), φ(x, t)} = δ (x − y) φ(x, t), πˆ(y, t) = iδ (x − y) (4.9) n o {π, π} = {φ, φ} = 0 {π,ˆ πˆ} = φ,ˆ φˆ = 0

We now have to find the Hilbert space that realises this algebra. One suitable basis of this space is given by the eigenvectors of the Hamiltonian, just like for the harmonic oscillator in quantum mechanics. The Hamiltonian is given by Z Z 1 1 2 1 H = Hd3x = d3x π2 + ∇~ φ + m2φ2 . (4.10) 2 2 2 Note that we have dropped the hat on the fields. From now on we will always work with quantum fields. In this 2 1 ~ form, the Hamiltonian is not diagonal since 2 ∇ φ couples the field at neighbouring points. To diagonalize, we have to go to Fourier space. Let us first focus on the system in a square box of side lengths L (0 < x < L, 0 < y < L, 0 < z < L) with periodic boundary conditions φ(x + L, y, z) = φ(x, y, z). For finite values of L we will have to deal with discrete sum rather than integrals making the manipulations more familiar. Expanding in the complete set of orthonormalized periodic Fourier modes 1 i~kn·~x ψ~n = √ e (4.11) V where 1 Z ~ ~ < ψ , ψ >= ei(kn−km)·~xd3x = δ (4.12) ~n ~m V ~n,~m 3 ~ 2π 3 and V = L and kn = L ~n, ~n ∈ Z . The considered wave functions are X X φ(x, t) = ψ~n(x)φ~n(t) π(x, t) = ψ~n(x)π~n(t) (4.13) n n

EPFL-ITP-LPTP Riccardo Rattazzi 44 4.2. QUANTISATION OF THE KLEIN-GORDON FIELD and thus Z Z ∗ 3 ∗ 3 φn(t) = ψn(x)φ(t, x)d x πn = ψn(x)φ(t, x)d x (4.14)

The canonical commutation relations in Fourier space are

[φn, φm] = [πn, πm] = 0 (4.15)

And

Z Z 3 3 ∗ ∗ 3 ∗ ∗ [φn, πm] = d xd y ψn(x)ψm(y)[φ(t, x), π(t, y)] = i d xψn(x)ψm(x) i Z i (~n + ~m) 2π · ~x = d3x exp − = iδ ≡ iδ (4.16) V L ~n+ ~m,0 ~n,−~m

Notice that by the reality of ψ and π, we have

Z Z Z ∗ 3 ∗ ∗ 3 3 ∗ φ~n = d x (ψ~n) φ(x, t) = d x ψ~nφ(x, t) = d x ψ−~nφ(x, t) = φ−~n (4.17) and ∗ [φn, πm] = iδn,m (4.18)

Let us consider the Hamiltonian in terms of Fourier modes

Z R 2 X 3 ∗ ∗ V π (x) = d x ψn(x)ψm(x)πn(t)πm(t) n,m X = δn,−mπn(t)πm(t) n,m X ∗ = πn(t)πn(t) n X R φ2(x) = φ (t)φ∗ (t) V n n (4.19) n Z R 2 X 3 i i ∗ ∗ V (∇iφ(x)) = d x(iknikm) ψn(x)ψm(x)φn(t)φm(t) n,m X i i = (−knkm) δn,−mφn(t)φm(t) n,m X 2 ∗ = knφn(t)φn(t) n,m such that we get

Z 1 Z H = Hd3x = d3x π2 + (∇iφ)2 + m2φ2 2 1 X 2π (4.20) = π π∗ + (m2 + ~k2 )φ φ∗ ~k ≡ ~n 2 n n n n n n L n

p 2 2 This corresponds to an inﬁnite set of harmonic oscillators with a frequency ωn = kn + m .

4.2.1 Continuum limit

~ 2π In the continuum limit, L → ∞, the discrete eigenvalues kn = L ~n will form a continuum and we can thus replace the sums by integrals:

EPFL-ITP-LPTP Riccardo Rattazzi 45 4.2. QUANTISATION OF THE KLEIN-GORDON FIELD

3 X Z 1 Z (...) → d3n(...) = d3kV (...) (4.21) 2π n For instance we have

√ 1 X V Z φ(x) = √ expiknx ψ (t) → d3k expik·x φ(~k, t) (4.22) n (2π)3 V n √ where φ(k, t) = lim V φn(t) (with k = n/L = ﬁxed) and L→∞ Z φ(k, t) = d3x exp−ik·x φ(x, t) (4.23)

R 3 i(k1−k2)·x 3 3 where we used d x exp = (2π) δ (k1 − k2). We can hence also compute the commutation relations

Z [φ(k, t), φ(p, t)] = [π(k, t), π(p, t)] = 0 [φ(k, t), π(p, t)] = i d3x expi(k+p)x = i(2π)3δ3(p + k) (4.24)

So we have Discrete Continuum

1 X Z d3k V (2π)3 n Z (4.25) 1 ~ ~0 δ = ei(kn−km)·xd3x 1 R ei((k−k )·~x = 1 (2π)3δ3(k − k0) n,m V V V 1 1 X 1 Z d3k = δ (2π)3δ3(k − k0) V V n,m V (2π)3 n 3 3 Formally this means 1 = δn,n ∼ 1/V (2π) δ (0) and thus

V ∼ (2π)3δ3(0) (4.26) Again we can write

1 6 Z R d3xπ(x, t)2 = d3p d3k d3x expi(k+p)x π(k, t)π(p, t) 2π 1 3 Z = d3k π(k, t) π∗(k, t) (4.27) 2π 1 3 Z R d3x(∇iφ(x, t))2 = d3k k2π(k, t) π∗(k, t) 2π and thus the Hamiltonian is given by

1 Z d3k H = π(k)π∗(k) + (k2 + m2)φ(k)φ∗(k) (4.28) 2 (2π)3 Notice the agreement with the previous discrete sums through the substitution

X V Z → d3p (2π)2 √n (4.29) V πn → π(k, t) where k = 2π/Ln.

Since φ(−k) = φ∗(k) and π(−k) = π∗(k), the Hamiltonian contains the same term twice: π(k)π(−k) + π(−k)π(k). To ﬁnd the eigenvalues, wen can proceed like with a simple harmonic oscillator in quantum mechanics where the Hamiltonian is

EPFL-ITP-LPTP Riccardo Rattazzi 46 4.2. QUANTISATION OF THE KLEIN-GORDON FIELD

1 1 H = p2 + ω2q2 (4.30) SHO 2 2 By deﬁning

1 rω q = √ (a + a†) , p = − (a − a†) (4.31) 2ω 2 the Hamiltonian can be written in terms of the ladder operators 1 H = ω a†a + (4.32) SHO 2 with E = ω(n + 1 ). Note that the commutation relation [q, p] = i implies a, a† = 1. To find the spectrum of n 2 √ the Klein-Gordon equation we find similar ladder operators. Define ω ≡ ω(k) = k2 + m2 and thus

1 1 a(k) = √ (ωφ(k) + iπ(k)) , a†(k) = √ (ωφ(−k) − iπ(−k)) (4.33) 2ω 2ω and

1 rω φ(k) = √ a(k) + a†(−k) , π(k) = −i a(k) − a†(−k) (4.34) 2ω 2

The commutation relations become

s s 1 ω(k) 1 ω(p) [a(k), a(p)] = [φ(k), iπ(p)] + [iπ(k), φ(p)] 2 ω(p) 2 ω(k) s s ! (2π)3 ω(k) ω(p) = − δ(k + p) + δ(k + p) 2 ω(p) ω(k) (2π)3 = −δ3(k + p) + δ3(k + p) 2 = 0 (4.35) a†(k), a†(p) = 0

s s 1 ω(k) 1 ω(p) a(k), a†(p) = [φ(k), iπ(−p)] + [iπ(k), φ(−p)] 2 ω(p) 2 ω(k) (2π)3 = δ3(k − p) + δ3(k − p) 2 = (2π)3δ3(k − p)

Therefore, substituting the above expressions in the Hamiltonian, one gets

1 Z d3k hw w i H = a(k) − a†(−k) a†(k) − a(−k) + a(k) + a†(−k) a†(k) + a(−k) (4.36) 2 (2π)3 2 2 Simplifying the cross terms and noting that by an integration over all k one can drop the minus sign on (-k)

1 Z d3k H = ω(k) a(k)a†(k) + a†(−k)a(−k) 2 (2π)3 1 Z d3k = ω(k) a(k)a†(k) + a†(k)a(k) 2 (2π)3 (4.37) Z d3k ω(k) = 2a†(k)a(k) + a(k), a†(k) (2π)3 2 Z d3k 1 = ω(k) a†(k)a(k) + (2π)3δ(0) (2π)3 2

EPFL-ITP-LPTP Riccardo Rattazzi 47 4.2. QUANTISATION OF THE KLEIN-GORDON FIELD The last term in the ﬁnal expression is the divergent zero point energy. However, there is no need to worry, since energy is deﬁned up to an additive constant which is not observable. Notice that (2π)3δ3(0) ∼ V and thus

Z d3k ω(k) E = V (4.38) 0 (2π)3 2 The ﬁnite volume case can be easily obtained through the substitutions

√ a(k) = V an Z d3k X V → (4.39) (2π)3 n (2π)3δ3(0) → V which leads to X 1 H = ω(k ) a† a + (4.40) n n n 2 n By neglecting the zero point energy we can write

Z d3k H = ω(k)a†(k)a(k) (2π)3 (4.41) = continuous set of harmonic oscillators

4.2.2 Hilbert space

To ﬁnd the Hilbert space of states, notice that

h † i † H, ak = ω(k)ak , [H, ak] = −ω(k)ak (4.42)

† i.e. ak and ak respectively act as raising and lowering operators of the energy. If H|ψi = Eψ then

a(k)|0i = 0 ∀k (4.44) Since we have dropped the additive inﬁnite constant, also H|0i = 0. All other states can be built from the ground † state by acting with ak on |0i

† † |k1, . . . kni = a (k1) . . . a (kn)|0i (4.45) which satisﬁes H|k1, . . . kni = (ω(k1) + ··· + ω(kn)) |k1, . . . kni (4.46) By a generalisation of the harmonic oscillator, the Hilbert space is then

  Y †  Hilbert Space := a (kj)|0i, j = 0, 1, 2.... ≡ {|0i, |k1i, |k1, k2i, ...} (4.47)  j 

The states correspond to energies 0, ω(k1), ω(k1) + ω(k2),... , i.e. the ground state and excited states with E > 0. This is the Fock Space.

EPFL-ITP-LPTP Riccardo Rattazzi 48 4.3. PROPERTIES OF THE KLEIN-GORDON FIELD Having found the energy spectrum, what can we say about the 3-Momentum?

Z Z Z d3k P i = −P = − d3xφ∂˙ φ = − d3xπ∂ φ ≡ kia† a (4.48) i i i (2π)3 k k

This looks very similar the Hamiltonian. The operators a† and a are then also raising and lowering operators for the 3-momentum.

√ a†(k) creates energy = ω(k) = k2 + m2 = E k (4.49) momentum = ~k † ~ a (k) thus creates the 4-momentum kµ = (Ek, k). Note that

µ 2 k kµ = m (4.50) † † µ µ µ µ µ 2 Similarly, a (k1)a (k1) creates P = k1 + k1 where k1 (k1)µ = k2 (k2)µ = m It is quite natural to call these excitations (quanta) created by a†(k) particles since they are discrete entities with a ﬁxed energy momentum relation. Note that the a†(k) operators are also called creation operators while a(k) are called destruction operators.

Notice that the particle created by a†(k) is not localised in space. Its momentum is exactly determined and therefore position totally undetermined.

The Hilbert space is thus composed of states that are superpositions of

• vacuum - 0 particles: |0i • 1 particle: R d3kf(k)a†(k)|0i

R 3 3 † † • 2 particle: d k1d k2f(k1, k2)a (k1)a (k2)|0i • ...

R Qn 3 † † • n particle: i=1 d kif(k1, . . . kn)a (k1) . . . a (kn)|0i

† † † † where f(k) is the wave function. Notice that a (k1)a (k2)|0i = a (k2)a (k1)|0i by the commutation relations and thus

f(k1, k2) = f(k2, k1) (4.51) which implies that the wave function for 2-particle states is symmetric. The anti-symmetric part will always vanish when integrating. The particles are bosons and obey Bose-Einstein statistics. Similarly, also f(k1, . . . kn) is symmetric under the exchange of any two particle momenta. Note that a single momentum k can be occupied by many particles. We have thus proven that the particles associated to a relativistic quantum scalar ﬁeld obey Bose-Einstein statistics. As we shall see later the statistics followed by particles depends on their spin: integer spin implies Bose-Einstein statistics, while a half integer spin leads to Fermi-Dirac statistics.

4.3 Properties of the Klein-Gordon ﬁeld

4.3.1 Relativistic normalisation of states and measure

Z d3k P µ = kµa†(k)a(k) (4.52) (2π)3 P µ should be a Lorentz vector, however from this way of writing it, this property is less evident. In particular, the d3k measure of integration, (2π)3 , is rotation invariant but not Lorentz invariant (not invariant under Lorentz boosts). It is better to use a Lorentz invariant measure which is simply given by

EPFL-ITP-LPTP Riccardo Rattazzi 49 4.3. PROPERTIES OF THE KLEIN-GORDON FIELD

d3k 1 dΩk ≡ 3 (4.53) (2π) Ek √ 2 2 where Ek = k + m . To show this consider the measure

dk0d3k d4k δ(k2 − m2)θ(k0) = δ(k2 − m2)θ(k0) (4.54) (2π)3 (2π)3 which is manifestly invariant under the orthochronous Lorentz group: the ﬁrst two parts are obviously invariant, while the θ(k0) is not. The δ-function forces the 4-momentum to be inside either the forward or backward light cone. Under orthochronous Lorentz transformations, the forward light cone goes into itself, i.e. k0 does not change sign and θ(k0) is invariant. By integrating the measure in eq. 4.54 times a generic function f(k) we get

√ √ ! Z d4k Z d3k δ(k0 − k2 + m2) δ(k0 + k2 + m2) (2π)δ(k2 − m2)θ(k0)f(k) = dk0θ(k0) √ + √ f(k) (2π)4 (2π)3 2 k2 + m2 2 k2 + m2 Z d3k 1 = f(k) (2π)3 2E Z k = dΩkf(k) (4.55)

Using the new normalisation

Z 3 µ d k µ + p P = 3 k a¯ (k)¯a(k)a ¯(k) = 2Eka(k) (4.56) (2π) 2Ek and + 3 3 a¯(k), a¯ (p) = (2π) 2Ekδ (k − p) (4.57) which is also Lorentz invariant. Furthermore, the normalisation of the 1-particle state is

+ + 3 3 |ki ≡ a¯ (k)|0i hp|ki = h0|a¯(p)¯a (k)|0i = (2π) 2Epδ (p − k) (4.58) This normalisation diﬀers from the one we have used in non-relativistic quantum mechanics. The above normali- 0i i µ 0 0 µ sation is Lorentz invariant. Using p ≡ Λµp and E ≡ Λµp we can perform a Lorentz boost

0 0 3 0 3 0 0 3 3 hp |k i = (2π) 2Epδ (p − k ) ≡ (2π) 2Epδ (p − k) (4.59) This means that hp|ki = hp0|k0i. The advantage of this normalisation is that Lorentz transformations are realised without additional factors. All observers have the same normalisation convention.

How do we get |Λki ? Simply by acting with a representation of the Lorentz group UΛ. So we would have

3 3 + 2 0 0 2 3 3 0 0 (2π) 2Epδ (p − k) = hp|ki = hp|UΛ UΛ|ki = |cp,Λ| hp |k i = |cp,Λ| (2π) 2Ek0 δ (k − p ) (4.61) 2 This shows that |cp,Λ| = 1.

EPFL-ITP-LPTP Riccardo Rattazzi 50 4.3. PROPERTIES OF THE KLEIN-GORDON FIELD 4.3.2 States localised in space

Let us ﬁnally consider

Z 3 Z 3 d k + i~k·~x d k i~k·~x ~ φ(x)|0i = 3 a¯k +a ¯k e |0i = 3 e |ki (4.62) (2π) 2Ek (2π) 2Ek

Except for the factor 2Ek, this is the usual non-relativistic expression for the position eigenstate |xi. Indeed 2Ep both in dΩk and in the normalisation of det p is approximately constant for p m. We can thus propose the same interpretation and claim that φ(x) acting on the vacuum creates a particle at position ~x. Consistent with this interpretation we have

This is the usual result like in a non-relativistic theory. hx|pi = hp|xi∗ = eik·x is the position-space representation of the wave function of a single particle momentum eigenstate |pi.

4.3.3 Time evolution

We have so far considered the theory at a ﬁxed time t. We could now, given our Hamiltonian, consider the Schrdingier¨ picture evolution of our Hilbert space states

|ψ(t)i = exp−iHt |ψ(0)i (4.64) In ﬁeld theory it is often more convenient to use the Heisenberg picture of time evolution, for which the operators rather than the states evolve with time. We can go from the Schrdingier¨ to the Heisenberg picture (considering the operator Aˆ

A(t) = hψ(t)|Aˆ|ψ(t)i = hψ(0)|eiHtAeˆ −iHt|ψ(0)i (4.65) where now eiHtAeˆ −iHt ≡ Aˆ(t) is an operator in the Heisenberg picture. Let us ﬁnd

φ(t, ~x) ≡ φ(x) ≡ eiHtφ(0, ~x)e−iHt (4.66) π(t, ~x) ≡ π(x) ≡ eiHtπ(0, ~x)e−iHt Since φ and π are linear combination of a and a† we must ﬁnd

a(p, t) ≡ eiHta(p)e−iHt (4.67) This can be computed by solving an ordinary diﬀerential equation

d a(p, t) = eiHt(iHa(p) − ia(p)H)e−iHt dt iHt −iHt = ie [H, a(p)] e (4.68) iHt −iHt = ie (−Epa(p))e = −iEpa(p, t) The solution satisfying A(p, 0) = a(p) is just

a(p, t) = e−iEpta(p) (4.69) We could play the same game to compute a†(p, t). This is most trivially done by noting that

EPFL-ITP-LPTP Riccardo Rattazzi 51 4.3. PROPERTIES OF THE KLEIN-GORDON FIELD

† † a†(p, t) = eiHta†(p)e−iHt = eiHta(p)e−iHt = e−iEpta(p) = eiEpta†(p) (4.70) Our ﬁnal result is thus

Z i~p·~x −iEpt iEpt † φ(t, x) = dΩpe e a(p) + e a (−p) Z −i(Ept−~p·~x) i(Ept−~p·~x) † = dΩp e a(p) + e a (p) (4.71) Z −ip·x ip·x † = dΩp e a(p) + e a (p)

The ﬁrst term corresponds to the positive frequency modes (creation), while the second corresponds to the negative frequency modes (destruction). To conclude, we notice that φ(x) trivially solves the Klein-Gordon equation

Z 2 2 −ip·x 2 ip·x † ( + m )φ(x) = dΩp ( + m )e a(p) + ( + m )e a (p) Z 2 2 −ip·x ip·x † (4.72) = dΩp(−p + m ) e a(p) + e a (p) = 0 This is a generalisation of the well-known Ehrenfest theorem of quantum mechanics. The quantum operators in the Heisenberg picture solve the same equations of the classical theory.

iHt −iP~ ·~x µ µ Similarly to e , we can consider the action of e . Notice that [P , ak] = −k ak. Consider

j j −iP~ ·~b iP~ ·~b iP bj −iP bj e ake = e ake = ak(b) (4.73) and hence d j j iP bj j −iP bj j ak(b) = e i P , ak e = −ik ak(b) (4.74) dbj The solution to this diﬀerential equation is

j −ik bj i~k·~b ak(b) = e ak = e ak (4.75) We can thus write

φ(x) = ei(Ht−P~ ·~xφ(0)e−i(Ht−P~ ·~x (4.76) = eiP ·xφ(0)e−iP ·x µ µ where P · x = Pµx . From this result we also see directly that P acts as the generator of space-time translations. We can now consider the fate of Poincaré invariance in the quantum theory (with the ﬁeld φ(x) deﬁned over space-time). We have already found the expression for P µ. Now we need J µν . Using the expressions derived before, we have

Z J ij = d3x(xiT 0j − xjT 0i) Z † i ∂ j ∂ = −i dΩka (k) k − k a(k) ∂kj ∂ki Z ∂ (4.77) J j0 = i dΩ a†(k)ω a(k) − tP i k ∂k Z i † ∂ = dΩka (k) iωk − tki a(k) ∂ki Notice that in J i0, we should take a(k) ≡ a(k, t) = e−iωta(k) to see the time dependence J i(t) = eiHtJ i(0)e−iHt

Z j0 † iωt ∂ −iωt J = dΩka (k)e iω − tki e a(k) ∂ki Z (4.78) † ∂ = dΩka (k) iω a(k) ∂ki

EPFL-ITP-LPTP Riccardo Rattazzi 52 4.3. PROPERTIES OF THE KLEIN-GORDON FIELD To draw physical consequences we need the commutators

∂ ∂ J ij, a(p) = i pi − pj a(p) ∂pj ∂pi (4.79) ∂ J i0, a(p) = −iω(p) a(p) ∂pi With J ij we can compute the angular momentum of n-particle states. Note that J µν |0i = 0 since the vacuum has no angular momentum and is boost invariant.

R † k 1 kij ij Consider a linear combination of one-particle states |ψi = dΩkf(k)a (k)|0i. Using the deﬁnition J = 2 J we can compute

l ~ ~ l ~ ~ where J f = −i(k ∧ ∂k) f(k) = i(∂k ∧ k)f(k) like in ordinary quantum mechanics. The angular momentum is purely due to the orbital form of the wave function. Thus the quanta associated with φ carry no spin and have spin = 0.

Now let us study how J µν acts on the ﬁeld φ(x).

Recall: [P µ, φ(x)] = −i∂µφ(x) → P µ generates translations. Also [J µν , φ(x)] = −i (xµ∂ν − xν ∂µ) φ(x)

We thus find that the commutator of P µ and J µν with φ(x) act as the infinitesimal generators of the Poincaré group. One also verifies that the commutator of J µν and P ρ among themselves represents the Poincaré algebra. To find a finite Poincaré transformation we just need to exponentiate the infinitesimal transformation. Exactly as we did before we can compute

~ ~ ~ ~ µ e−iP ·ba(k)eiP ·b = eik bµ a(k) (4.81) Thus

Z −iP~ ·~b iP~ ·~b −i~k·(~x−~b) i~k·(~x−~b) † 0 e φ(x)e = dΩk e a(k) + e ak = φ(x − b) ≡ φ (x) (4.82) where φ0(x) is the ﬁeld seen by the observer using coordinates x0 = x + b. Note that U(b) = eiP~ ·~b is unitary. Note a0(k) = e−iP~ ·~ba(k)eiP~ ·~b = ei~k·~ba(k) and

Similarly for Lorentz transformations we have

i ωµν J − i ωµν J i µν 2 e 2 µν φ(x)e 2 µν = φ(x) + ω [J , φ(x)] + O(ω ) 2 µν 1 µ ν ν µ 2 = φ(x) + ωµν (x ∂ − x ∂ ) φ(x) + O(ω ) (4.84) 2 µ µ ρ 2 = φ x − ωρ x + O(ω ) = φ Λ−1x

EPFL-ITP-LPTP Riccardo Rattazzi 53 4.4. THE CHARGED SCALAR FIELD µ i ˆρσ µ − ωρσ J µ µ 2 ˆµν where we deﬁned the transformation Λν = e 2 = δν +ων +O(ω ) where J are 4x4 Lorentz generators ν in matrix representation. As before we can interpret the resulting ﬁeld as the one seen by the observer using coordinates x0 = Λx. And hence |ψi0 = U †(Λ)|ψi where U(Λ) = e−i/2ωµJµν represents Λ on the Hilbert space and is unitary since Jµν is hermitean. Now

φ0(x) ≡ U †(Λ)φ(x)U(Λ) = φ(Λ−1x) (4.85) † From this equation we can read the action on ak and ak.

Z † † −i~p·~x i~p·~x † U (Λ)φ(x)U(Λ) = U )Λ) dΩp e ap + e ap U(Λ) Z −i~p·~x 0 i~p·~x 0† ≡ dΩp e ap + e ap (4.86) Z −ip·Λ−1x ip·Λ−1x † = dΩp e ap + e ap

T Deﬁning pΛ ≡ k and use dΩp = dΩk we ﬁnd

Z † −ik·x ik·x † U (Λ)φ(x)U(Λ) = dΩk e aΛ−1k + e aΛ−1k (4.87) Hence

† 0 U (Λ)apU(Λ) = aΛ−1p = ap (4.88) The creation of a particle of momentum p in the farm of the primed observer is the creation of a particle of momentum Λ−1p in the original frame. The quantum ﬁelds of Lorentz observers are related by unitary transformations that realise a representation of the Lorentz (Poincaré) group. What we have just touched upon is one incarnation of Wigner’s theorem that states that a classical symmetry transformation g is realised at the quantum level by a unitary operator U(g) acting on the Hilbert space (and these unitary transformations on Hilbert space form a representation of the symmetry group). Fixing the sign conventions on Poincaré generators

4.4 The charged scalar ﬁeld

4.4.1 Construction

The charged scalar ﬁeld consists of two parts φ = φ1√+iφ2 where both φ , φ are real. We can write the Lagrangian 2 1 2

1 1 1 1 L = ∂ φ∗∂µφ − m2φ∗φ = ∂ φ ∂µφ − m2φ2 + ∂ φ ∂µφ − m2φ2 = L(φ ) + L(φ ) (4.89) µ 2 µ 1 1 2 1 2 µ 2 2 2 2 1 2 which is the sum of two Klein Gordon Lagrangians. This Lagrangian is up to total derivatives the most general quadratic Lagrangian invariant under Poincarré and under the symmetry φ(x) → φ0(x) = e−iαφ(x) (where α ∈ R).

More formally x0µ = xµ Symmetry (4.90) φ0(x0) = e−iαφ(x)

4.4.2 Internal current and charge

The conserved Noether currents is

 φ0(x) − φ(x) iαφ(x)  ∆φ = = − = −iφ(x) 0∗ α ∗ α ∗ (4.91) φ (x) − φ (x) iαφ (x) ∗  ∆φ∗ = = = iφ (x) α α

EPFL-ITP-LPTP Riccardo Rattazzi 54 4.4. THE CHARGED SCALAR FIELD Therefore,

µ ∗ ∗ ∗ ∗ ∗←→ J = −i∂µφ φ + i∂µφφ = i (φ ∂µφ − (∂µφ ) φ) ≡ iφ ∂µ φ = φ2∂µφ1 − φ1∂µφ2 (4.92)

The equations of motion are ∂L ∂L ∂µ ∗ − ∗ = 0 ∂∂µφ ∂φ (4.93) µ 2 ∂µ∂ φ − m φ = 0 µ which is just the Klein Gordon equation. It is easy to check that ∂µJ = 0 on the equations of motion. Later on we shall ﬁnd that the familiar conserved current J µ is precisely associated to a rephasing symmetry like the one just discussed.

4.4.3 Quantisation

∂L ∂L π = = φ˙† π∗ = = φ˙ (4.94) ∂φ˙ ∂φ˙∗ and hence the Hamiltonian is

Z H = d3x π†π + |∇φ|2 + m2φ†φ (4.95)

Thus, the commutation relations can be written in complex notation

[φ, φ] = φ†, φ† = φ, φ† = 0 [π, π] = π†, π† = π, π† = 0 (4.96) [φ(x), π(y)] = φ†(x), π†(y) = iδ3(x − y) Just another way of writing is

[φi, φj] = 0 [πi, πj] = 0 (4.97) 3 [φi, πj] = iδijδ (x − y) for i, j = 1, 2. Since L is the sum of the Lagrangian for φ1 and φ2, we can proceed with the canonical quantisation of φ1 and φ2 independently. Let us do so. We shall go back to complex notation at the end. Applying the results of the previous chapter we have in the Heisenberg picture

Z −i~p·~x i~p·~x † φ1 = dΩ e a1(p) + e a1(p) Z (4.98) −i~p·~x i~p·~x † φ2 = dΩ e a2(p) + e a2(p) with p0 = pm2 + φ2. Now

Z µ µ † † P = dΩp a1(p)a1(p) + a2(p)a2(p) (4.99)

† † The complex scalar ﬁeld consists of two types of particles, generated by a1 and a2, with the same mass m. In complex notation

  † Z  a (p) + ia (p) a+(p) − ia+(p)   −i~p·~x 1 2 i~p·~x 1 2  φ = dΩ e √ +e √   2 2  (4.100) | {z } | {z } a(p) b+(p) Z φ† = dΩ e−i~p·~xb(p) + ei~p·~xa†(p)

EPFL-ITP-LPTP Riccardo Rattazzi 55 4.4. THE CHARGED SCALAR FIELD The notation with a, b is more convenient to keep track of the charge of the states. The commutation relations are

[a, a] = [a, b] = [b, b] = 0 (4.101) a(k), a†(p) = b(k), b†(p) = (2π)32p0δ3(p − k)

So we can rewrite P in terms of these operators

Z P µ = dΩpµ a†(p)a(p) + b†(p)b(p) (4.102) and Z Z Q = J 0d3x = dΩ a†(p)a(p) + b†(p)b(p) (4.103) and + + + + Q, a (p) = a (p) Q, bp = −bp (4.104) The commutators are † † Q, ap = ap [Q, ap] = −ap † † (4.105) Q, bp = −bp [Q, bp] = bp † † Thus ap raises the charge by one unit (i.e. creates particles of charge +1). While bp decreases the charge by one unit (i.e. creates particle of charge −1). Notice that e−iαQ generates charge transformation

iαQ −iαQ −iα e ape = (1 + iαQ)ap(1 − iαQ) = ap − iα + .. = e ap iαQ −iαQ iα (4.106) e bpe = e bp iαQ −iαQ −iα † † From this√ follows e φe = e φ. The particles generated by a , b have exactly the same 4-momentum µ 2 2 i † p = k + m , k , mass, spin (here 0), but equal and opposite charge. It is conventional to say that ap † generates a particle, while bp generates an anti-particle. A deep symmetry of the theory requires the masses of particle and antiparticle to be the same, basically a and b are associated to the same complex ﬁeld φ.

Examples of particle - antiparticle pairs with spin-0 are π+, π− and K+,K−. Both pairs have opposite electromagnetic charge. In the case of K0 and K¯ 0 the particles have opposite strangeness. Notice that the original operators † † a1 and a2 create particles that are coherent quantum superpositions of particles and antiparticles. This state is not an eigenstate of the charge Q.

|ψi = (b† + a†)|0i (4.107) Q|ψi = (b† − a†)|0i= 6 |ψi

EPFL-ITP-LPTP Riccardo Rattazzi 56 Chapter 5

Spinor ﬁelds

5.1 Spinor representations of the Lorentz group

Recall that the irreducible representations of the Lorentz group are labelled by two (half)-integer numbers (j−, j+) and that the most general representation of Lorentz transformations on ﬁelds is:

b −1 Φa(x) 7→ Da (Λ) Φb(Λ x ) (5.1) | {z } | {z } spin part orbital part

b where Da (Λ) is a (j−, j+) representation on a finite dimensional vector space (they are matrices). As we saw, the i trivial representation (0, 0) corresponds to a scalar field with one index: φ(x). Indeed, j± = 0 ⇒ J± ≡ 0 so that b Da (Λ) ≡ 1. The first non trivial cases come from the (1/2, 0) and (0, 1/2) representations. Recall:

J i ± iKi J i ≡ , [J i ,J j ] = iijkJ k (5.2) ± 2 ± ± ±

This leads to the so-called left-handed spinors:

i i σi i σ J− = J = 2 (1/2, 0) ⇒ 2 ⇒ i (5.3) J i = 0 i σ + K = +i 2 and the right-handed spinors:

J i = 0 i σi − J = 2 (0, 1/2) ⇒ i ⇒ i (5.4) J i = σ i σ + 2 K = −i 2 where σi are the Pauli matrices deﬁned through their algebra:

σiσj = δij + iijkσk (5.5)

Finally, recalling:

57 5.2. MISCELLANEOUS ABOUT SPINORS

− i ω J µν −iθ~·J~+i~η·K~ D(Λ) = e 2 µν = e (5.6)

We obtain the left-handed and right-handed spinorial representations of the Lorentz transformations:

− 1 (~η+iθ)·σ ΛL ≡ D(1/2,0) = e 2 (5.7) − 1 (−~η+iθ)·σ ΛL ≡ D(0,1/2) = e 2 (5.8) which act on left-handed spinors and right-handed spinors:

ψL(x), ψR(x) (5.9)

The ﬁrst important fact is that these two representations are not unitary:

† −1 ΛL/R 6= ΛL/R (5.10)

This is due to the fact that finite dimensional representations of non-compact groups are not unitary. Among the finite dimensional representations, only the trivial one is unitary. One redily verifies:

−1 † ΛL = ΛR (5.11)

Another remark is that the two representations act on the two-dimensional complex space C2. Moreover, since:

det ΛL = det ΛR = 1 ⇒ ΛL, ΛR ∈ SL(2, C) (5.12)

Both the (1/2, 0) and the (0, 1/2) representations span the group SL(2, C). We will actually see later that:

SO(1, 3) ' SL(2, C)/Z2 (5.13)

In both representations, spatial rotations act in the same way:

−iθ~·~σ/2 −1 e ψL/R(Rθ x) (5.14)

So that the intrisic angular momentum (spin) is rotated by the Pauli matrices σi/2. This implies, we shall see it more explictly later on, that spinors represent particles that carry spin 1/2.

5.2 Miscellaneous about spinors

5.2.1 Parity

A ﬁrst question that arises when dealing with spinors is “why do we have two representations?”. There is a doubling because of a discrete symmetry, parity:

EPFL-ITP-LPTP Riccardo Rattazzi 58 5.2. MISCELLANEOUS ABOUT SPINORS

P :(t, ~x) 7→ (t, −~x) (5.15)

In particular, under parity the generators of Lorentz transformations transform as:

J i 7→ J i, axial vector P : (5.16) Ki 7→ −Ki, polar vector

The fact that the generators of spatial rotations (angular momentum) gather into an axial vector is well known and can be seen when we realise it on ﬁelds where:

J i = i(~x ∧ ∂~)i (5.17)

We can proceed similarly for the generator of boosts and show that it is a polar vector.

This means that under a parity transformation:

P :(j−, j+) 7→ (j+, j−) (5.18) Representations of the Lorentz group swap! This means that if we want to represent parity, we need to double the number of ﬁelds!1 Let us make this argument more explicit. Choose for example the left-handed representation of the Lorentz group:

~ ΛL(θ, ~η) (5.19)

Since a representation of parity changes the sign of the boost generators, we have:

~ ~ P ΛL(θ, ~η)P = ΛL(θ, −~η) (5.20) But the last quantity is nothing else than the right-handed representation, so that:

~ ~ P ΛL/R(θ, ~η)P = ΛR/L(θ, ~η) (5.21)

Therefore, when representing parity, we consider the so-called Dirac representation:

(1/2, 0) ⊕ (0, 1/2) : ΛD = ΛL ⊕ ΛR (5.22) and we deﬁne a Dirac spinor:

ΨD ≡ ψL ⊕ ψR (5.23) Since parity inverts representations, we deﬁne its action on a Dirac spinor:

P :ΨD(x) = ψL(x) ⊕ ψR(x) 7→ P ΨD(x) ≡ ψR(P x) ⊕ ψL(P x) (5.24) and it can be checked explictly (exercise) that with this deﬁnition, we have2:

1 Also, this implies that only representations with j− = j+ are irreducible representations of the full Lorentz group. 2 Hint: compare PD ~ [ΨD] to D ~ [P ΨD] ΛD (θ,~η) ΛD (θ,−~η)

EPFL-ITP-LPTP Riccardo Rattazzi 59 5.2. MISCELLANEOUS ABOUT SPINORS

~ ~ P ΛD(θ, ~η) = ΛD(θ, −~η)P (5.25)

∗ In what follows, we shall need the explicit form of ΛR/L and we will see that parity plays a role in charge conjugation for spinors. First, deﬁne the two-dimensional Levi-Civita tensor:

0 1 ≡ iσ = , 2 = −1 (5.26) 2 −1 0

From the Pauli algebra, we can derive an important relation:

∗ T σi = σi = σi (5.27) which leads to:

1 ~ T 1 ~ −1 † T 2 (~η−iθ)~σ − 2 (~η−iθ) ~σ −1 −1 −1 ΛR = e = e = ΛR = ΛL (5.28) so that:

−1 ∗ ΛR = ΛL (5.29)

Because of this we have:

∗ Lorentz ∗ ∗ ∗ −1 ∗ ∗ ψL 7→ ΛLψL = ΛL ψL = ΛR(ψL) (5.30) which means that:

∗ ψL/R ∼ ψR/L (5.31)

In other words, from Eq. 5.29, we see that the left-handed and the complex conjugate of the right-handed representations are related by a unitary transformation:

(1/2, 0) ∼ (0, 1/2)∗ (5.32) i.e., complex conjugation can be seen as a parity transformation. This leads us to deﬁne parity transformations on spinors as:

∗ P : ψL/R(x) 7→ ψL/R(P x) (5.33)

Parity transformations can be said to change the charge of the spinor and this is why Dirac spinors such as the electron embed both a left- and a right-handed component.

EPFL-ITP-LPTP Riccardo Rattazzi 60 5.2. MISCELLANEOUS ABOUT SPINORS ↑ 5.2.2 Relation between SL(2, C) and L+

↑ In order to better understand the relation between SL(2, C) and L+, we will consider the (1/2, 1/2) representation. But before doing so, we will make some comments about notation.

Aside: The spinor representations we have seen so far carry a spinor index that we have been omitting. Indeed, both the left- and right-handed representations are two-dimensional. As for Lorentz indices, it is sometimes convenient to omit them to simplify the notation, but sometimes it is clever to carry them along as they simplfy greatly calculations with many contracted terms. We denote the left- and right- handed spinors as3:

A0 (ψL)A , (ψR) (5.34) and the Lorentz transformation matrices:

B A0 (ΛL)A , (ΛR) B0 (5.35) so that, as a convention, we contract unprimed spinors from top-left to bottom-right & and primed spinors from bottom-left to top-right %. We can raise and lower spinor indices using the Levi-Civita tensor for primed and unprimed spinors:

AB A0B0 AB, and A0B0 , (5.36)

So that:

AB A B B0 B0A0 A0 (ψL)B = (ψL) , (ψL) BA = (ψL)A , and A0B0 (ψR) = (ψR)A0 , (ψR)B0 = (ψR) (5.37)

And ﬁnally, the operation of complex conjugation adds or removes primes, for example:

∗ ∗ A0∗ ∗ A (ψL)A = (ψL)A0 (ψR) = (ψR) (5.38) Eq. 5.29 now reads:

∗ B0 B ∗ (ΛR)A0C0 = A0B0 (ΛR) C0 = (ΛL)A BC = (ΛL)A0C0 (5.39)

???

Being familiar with spinor indices, we can write a ﬁeld on which the (1/2, 1/2) representation acts simply as:

ˆ A0 A0 VA ≡ (ψL)A ⊗ (ψR) (5.40) V is called a bispinor ﬁeld. It will transform under a Lorentz transformation as:

3Diﬀerent authors use diﬀerent conventions

EPFL-ITP-LPTP Riccardo Rattazzi 61 5.2. MISCELLANEOUS ABOUT SPINORS

B A0 ˆ B0 (ΛL)A (ΛR) B0 VB (5.41)

If we forget about the spinor indices, this is simply written as:

ˆ ˆ T V 7→ ΛLV ΛR (5.42)

We would like to deal with a bispinor that transforms under the left- or right-handed representations exclusively. Rewrite, for example:

ˆ B0 ˆ T VAA0 = A0B0 VA → V = V (5.43)

Then:

ˆ T T ˆ T † † V 7→ ΛLV ΛR = ΛLV ΛL = ΛLV ΛL (5.44)

We could also rewrite:

˜ AA0 AB ˆ A0 ˜ ˆ V = VB → V = V (5.45) which transforms as:

˜ ˆ T ∗ ˆ T ∗ T V 7→ ΛLV ΛR = ΛRV ΛR = ΛRV ΛR (5.46)

This is almost good, we deﬁne instead of V˜ :

0 0 V¯ A A = (V˜ AA )T → V¯ = Vˆ T = T V T (5.47)

Then:

¯ ¯ † V 7→ ΛRV ΛR (5.48)

We see how V and V¯ transform exclusively under the left- and right-handed representation respectively. Each of them contains 8 real degrees of freedom (or 4 complex). Moreover, it is easy to see that if either of them is hermitian, so will be its Lorentz transform. This means that Lorentz transformations on bispinors preserve (anti-)hermitity, leading to the fact that there are two invariant subspaces in this representation: it is reducible! Consider for sake of deﬁniteness the set of hermitian matrices:

V, V¯ ∈ MH(2, C) (5.49)

A basis for the set of hermitian matrices is given by:

EPFL-ITP-LPTP Riccardo Rattazzi 62 5.2. MISCELLANEOUS ABOUT SPINORS

σµ = (σ0, σi), σ0 = 1 (5.50)

We can then expand V on this basis:

v0 − v3 −v1 + iv2 V = η vµσν = (5.51) µν −v1 − iv2 v0 + v3

On the other hand, for the right-handed matrices, using Eqs. 5.27, 5.47:

µ ν V¯ = ηµν v σ¯ (5.52) where

σ¯ν = (σ0, −σi) (5.53)

The basis matrices have the following property:

σµσ¯ν + σν σ¯µ = 2ηµν =σ ¯µσν +σ ¯ν σµ (5.54)

What have we accomplished so far? We have two representations on bispinors, what do they correspond to? Notice:

ΛL, ΛR ∈ SL(2, C) (5.55)

And therefore, when we Lorentz transform our bispinors (either V or V¯ ), the determinant is preserved:

0 † det V = det ΛLV ΛL = det V (5.56) but:

µ ν 2 det V = ηµν v v ≡ v (5.57)

µ Which is nothing else but the Lorentz invariant norm! It can be checked that given ΛL(Λ), v transforms as a four-vector:

µ µ ν v 7→ Λ ν v (5.58)

To summarise, we have two diﬀerent representations of SL(2, C) (V and V¯ ), but they both lead to a unique four-vector representation of the Lorentz group, and this is indeed the (1/2, 1/2) representation we were looking for!

This result shows an important property of the Lorentz group:

EPFL-ITP-LPTP Riccardo Rattazzi 63 5.3. COVARIANT WAVE EQUATIONS

µ ∀ U ∈ SL(2, C) and X = x σµ (5.59)

The matrix X will transform as:

† 0 µ ν X 7→ UXU = X = (Λ ν x )σµ (5.60)

This deﬁnes a mapping:

↑ Λ: SL(2, C) → L+ (5.61) which is a homomorphism:

Λ(U1U2) = Λ(U1)Λ(U2) (5.62)

It is surjective and two-to-one:

Λ(U) = Λ(−U), ∀ U ∈ SL(2, C) (5.63)

In particular:

µ Λ(1) = Λ(1) = δν (5.64)

SL(2, C) is locally isomorphic to the Lorentz group, implying that they share the same Lie algebra, but the two ↑ groups are globally diﬀerent. SL(2, C) is connected and simply connected whereas L+ is connected but not simply connected. The above mapping describes the isomorphism:

↑ L+ ' SL(2, C)/Z2 (5.65)

SL(2, C) is the universal covering group of the proper orthochronous Lorentz group much like SU(2) is the universal covering group of SO(3). Spinors are those representations that transform under the bigger group SL(2, C). As we already are familiar from ordinary quantum mechanics, spinors ﬂip sign under a 2π rotation. This is why, in physical quantities, spinors enter twice: this phase is not observable. It is Nature (not us!) who chose to make full use of group theory to realise Lorentz.

5.3 Covariant wave equations

5.3.1 Weyl spinors

We have now all the ingredients to try and write down relativistic “wave equations” for spinors. Given ψL(x), we want to generalise the scalar field equation (Klein-Gordon) and find a differential equation that transforms covariantly under Lorentz transformations. To do so, it is convenient to work in Fourier space:

EPFL-ITP-LPTP Riccardo Rattazzi 64 5.3. COVARIANT WAVE EQUATIONS

Z 4 ipx ψL(p) = d xe ψL(x) (5.66)

Under Lorentz, it transforms as:

−1 ψL(p) 7→ ΛLψL(Λ p) (5.67)

Consider the two following quantities in Fourier space:

µ µ P ≡ pµσ = p · σ, P¯ ≡ pµσ¯ = p · σ¯ (5.68)

Their transformation properties under Lorentz are:

0 0 † −1 P = p · σ = (Λp) · σ = ΛL(p · σ)ΛL = ΛL(p · σ)ΛR (5.69) ¯0 0 † −1 P = p · σ¯ = (Λp) · σ¯ = ΛR(p · σ¯)ΛR = ΛR(p · σ¯)ΛL (5.70)

We have therefore a good candidate for the wave equation in Fourier space. Indeed, the quantity:

P¯ ψL(p) (5.71)

Transforms covariantly:

¯0 0 0 ¯ P ψL(p ) = ΛR P ψL(p) (5.72)

Which means that if:

P¯ ψL(p) = 0 (5.73) in one frame, then the equality will hold in any frame! Notice that the wave equation transforms as a right-handed spinor. Indeed, the left-handed ﬁeld belongs to (1/2, 0) and P¯ to (1/2, 1/2). We then contract the unprimed indeces to form a Lorentz scalar and it only remains a free primed index:

A0A A0 P¯ ψL(p) ≡ P¯ (ψL)A(p) = P¯ ψL (p) (5.74) The equation of motion for the spinors is therefore:

µ iσ¯ ∂µψL(x) = 0 (5.75)

This is the so-called Weyl equation to which we associate Weyl spinors. Notice further that:

¯ 2 P P ψL(p) ≡ p ψL(p) = 0 → −ψL(x) = 0 (5.76)

EPFL-ITP-LPTP Riccardo Rattazzi 65 5.3. COVARIANT WAVE EQUATIONS so that Weyl spinors are associated to massless particles. In Fourier space:

P¯ ψL(p) = 0 (5.77) has a non trivial solution iﬀ:

det(P¯) = 0 ⇔ p2 = 0 (5.78) which leads to the same conclusion. The right-handed Weyl spinors satisfy a similar wave equation:

µ iσ ∂µψR(x) = 0 (5.79) which, this time, transforms as a left-handed spinor.

5.3.2 Majorana spinors

We would like to describe something with mass and to do so we need to “add” something to the RHS of Eq. 5.75. Since the LHS transforms as a right-handed spinor, so must do the RHS. We saw earlier that we can get a right-handed spinor from a left-handed one by a parity transformation:

∗ ψL(x) ∈ (0, 1/2) (5.80) we therefore write:

µ ∗ Fourier ¯ ∗ iσ¯ ∂µψL(x) = cψL(x) → P ψL(p) = cψL(−p) (5.81)

From Eq. 5.27, we have:

−1 ∗ σ¯µ = σµ (5.82) This leads to the conjugate equation:

µ ∗ ∗ iσ pµψL(x) = c ψL(x) (5.83)

Finally, this implies:

2 2 P P¯ ψL(p) = p ψL(p) = |c| ψL(p) (5.84) Or in position space:

2 ( + |c| )ψL(x) = 0 (5.85)

This second wave equation corresponds to a particles of mass |c|2 = m2 and is the so-called Majorana equation describing Majorana spinors.

EPFL-ITP-LPTP Riccardo Rattazzi 66 5.3. COVARIANT WAVE EQUATIONS Neutrinos likely are, as far as we understand, Majorana particles. Notice that the Majorana spinor equation is not invariant under phase rotations:

iα ψL(x) 7→ e ψL(x) (5.86)

Thus, there is no conserved charge (electric or of any other type) associated to Majorana particles.

5.3.3 Dirac spinors

In order to describe charged particles, we must introduce a third type of spinor: the Dirac spinor. We want it to be massive as well, and therefore we need both a left-handed and right-handed spinor. Since, as we just saw, when we use the parity transformed ﬁeld as right-handed spinor we obtain the Majorana equation, we need to bring a second spinor, independent from the ﬁrst one to build the equation. It reads:

iσ¯ · ∂ψ (x) = cψ (x) L R (5.87) iσ · ∂ψR(x) =cψ ˜ L(x)

By rescaling the ﬁeld:

ψR → λψR, λ ∈ C (5.88) such that:

c˜ cλ = = m, m ∈ R (5.89) λ

The equations become:

iσ¯ · ∂ψ (x) = mψ (x) L R (5.90) iσ · ∂ψR(x) = mψL(x)

This set of equations has two symmetries. The ﬁrst one, is a global U(1) symmetry:

iα ψL(x) 7→ e ψL(x) iα (5.91) ψR(x) 7→ e ψR(x) Which tells us that there is a conserved charge associated to those particles. We will see later that when we couple this set of equations to electromagnetism, this leads to the conservation of the electric charge. The set is also invariant under parity transformations:

ψ (t, ~x) 7→ ψ (t, −~x) L R (5.92) ψR(t, ~x) 7→ ψL(t, −~x)

The systems described by this set of equations is suitable to describe representations of parity. We saw earlier that in terms of spinors, it is the Dirac spinor which carries a representation of that symmetry and indeed this set of equations can be rewritten in a more compact way. Recall:

EPFL-ITP-LPTP Riccardo Rattazzi 67 5.3. COVARIANT WAVE EQUATIONS

ΨD ≡ ψL ⊕ ψR (5.93)

Now deﬁne a set of matrices, called the Dirac matrices or gamma matrices:

0 σµ γµ ≡ (5.94) σ¯µ 0

They satisfy the Dirac algebra:

{γµ, γν } = 2ηµν (5.95)

The set of equations can then be rewritten in a compact form:

(i∂/ − m)ΨD(x) = 0 (5.96)

µ where we used the Feynman slash notation ∂/ ≡ γ ∂µ. This is the Dirac equation for four-components spinors. It describes charged particles with spin 1/2 (e.g. e−, µ−, p, n, . . .).

5.3.4 Lagrangians

We would like to have a Lagrangian formulation for our spinor dynamics. In order to do so, we must find spinor invariants, i.e. objects that are invariant under Lorentz transformations (also called Lorentz scalars). Let us go back to our Weyl fermions ψL and ψR. As it is the case for the charged scalar field, we would like to take the spinor field and a conjugate of it as independent variables so that we can build a second order Lagrangian which leads to the Dirac equation. However:

0 0 0† 0 † † † −1 ψL(x ) = ΛLψL(x), ψ L(x ) = ψL(x)ΛL = ψL(x)ΛR (5.97) and thus:

† ψLψL (5.98) is not an invariant. This is obvious if we recall the rules of complex conjugation for spinor indices. The left-handed spinor carries an unprimed index, while the complex conjugation turns it into a primed index, making therefore the construction of a Lorentz scalar impossible! However, if we consider a right-handed spinor and take its complex conjugate we can write:

∗ A † (ψR) (ψL)A ≡ ψRψL (5.99) This seems to be a Lorentz scalar. Let us check it explicitly:

0† 0 0 0 † † † ψ R(x )ψL(x ) = ψR(x)ΛRΛLψL(x) = ψR(x)ψL(x) (5.100)

EPFL-ITP-LPTP Riccardo Rattazzi 68 5.3. COVARIANT WAVE EQUATIONS Similarly:

0† 0 0 0 † ψ L(x )ψR(x ) = ψL(x)ψR(x) (5.101)

We also need, because of the four-derivative which transforms as a vector, to build Lorentz vectors out of spinors. ↑ Recall the mapping from SL(2, C) → L+:

† µ µ ν † µ µ ν ΛLσ¯ ΛL = Λ ν σ¯ , ΛRσ ΛR = Λ ν σ (5.102) We can make use of it here to build an SL(2, C) scalar but Lorentz vector:

† µ µ † ν ψL(x)¯σ ψL(x) 7→ Λ ν ψL(x)¯σ ψL(x) (5.103) † µ µ † ν ψR(x)σ ψR(x) 7→ Λ ν ψR(x)σ ψR(x) (5.104)

Again, this is obvious as:

µ µ AA0 µ µ σ ≡ (σ ) , σ¯ ≡ (¯σ )A0A (5.105)

This leads to two other scalars:

† † ψL(x)¯σ · ∂ψL(x), ψR(x)σ · ∂ψR(x) (5.106)

We have gathered all necessary ingredients to build the Lagrangians. Let us start with the Weyl Lagrangian. Simply:

Z Z 4 4 † SW = d xLW = d x iψL(x)¯σ · ∂ψL(x) (5.107)

Notice that the action is real up to boundary terms. The equations of motion lead as expected to the Weyl equation:

δSW 0 = † = iσ¯ · ∂ψL(x) (5.108) δψL

The next to simplest case is the Dirac Lagrangian. For the mass terms we simply add the ﬁrst scalars we built:

Z Z 4 4 † † † ∗ † SD = d xLD = d x iψL(x)¯σ · ∂ψL(x) + iψR(x)σ · ∂ψR(x) − mψL(x)ψR(x) − m ψR(x)ψL(x) (5.109)

Do we have a complex mass? No, we can simply redeﬁne:

−iθ iθ ψR 7→ e ψR, m = |m|e (5.110)

This is a so-called chiral rotation. To build the Majorana Lagrangian we use:

EPFL-ITP-LPTP Riccardo Rattazzi 69 5.4. PATH TOWARDS QUANTISATION

∗ ψR → ψL (5.111) in the Dirac Lagrangian. We then write:

Z Z m m∗ S = d4xL = d4x iψ† (x)¯σ · ∂ψ (x) − ψT ψ − ψ† ψ∗ (5.112) M M L L 2 L L 2 L L

Here comes the tricky part:

T AB BA BA T ψL ψL = (ψL)A (ψL)B = −(ψL)A (ψL)B = −(ψL)B (ψL)A = −ψL ψL (5.113)

Where we assumed in the last equality that spinors commute, just as classical variables. However, this implies T that ψL ψL = 0 so that the Majorana mass would not exist! The solution of this paradox relies on the fact that spinors do not have a classical limit! Spinors are Grassman variables that anti-commute:

{(ψL)A, (ψL)B} = 0 (5.114)

Notice that we did not need to use this fact in the construction of the Weyl and Dirac Lagrangians.

Finally, the Dirac Lagrangian can be rewritten in terms of a Dirac spinor. Deﬁne the Dirac conjugate of a Dirac ﬁeld ΨD:

¯ † 0 † † ΨD ≡ ΨDγ = ψR ⊕ ψL (5.115)

Then, the Dirac Lagrangian can be rewritten:

Z Z δS S = d4xL = d4x Ψ¯ (i∂/ − m)Ψ , 0 = D = (i∂/ − m)Ψ (5.116) D D D D ¯ D δΨD

Very much like we did with the charged scalar ﬁeld, one can prove that the Dirac Lagrangian is the most general Lagrangian that is Poincaré invariant, at most quadratic in ψL and ψR, with up to one-derivative terms and invariant under a global “charge” U(1) symmetry.

5.4 Path towards quantisation

5.4.1 Classical solutions

The dirac equation is (−i∂¡ − m)ψ = 0. We’re going to look for the most general plane wave solution ψp(x) = e−ipxu(p) with p = (p0, ~p)

−ipx µ µ (i∂¡ − m)ψp = e (γ pµ − m) u(p) = 0 from now on :p¡ = γ pµ (5.117) 4The rest of this chapter was not typeset by the author. Any error shan’t be attributed to him

EPFL-ITP-LPTP Riccardo Rattazzi 70 5.4. PATH TOWARDS QUANTISATION γµγν +γν γµ 2 In order to neglect trivial solutions, we require a null determinant. Note that p¡p¡ = 2 pµpν = p

2 2 2 ! 2 (p¡ + m)(p¡ − m) u(p) = p − m u(p) = 0 → p = m (5.118)

p 2 2 −ipx ipx So we get positive and negative energy solutions where p0 = m + p with ψp = e u(p) and ψpe v(p). What are v(p) and u(p) ?

The simplest case to study is when ~p = 0 which implies p0 = m which with little algebra implies

−1 1 ξ ξ ξ m γ0 − 1 u(p) = L = 0 → u = with ξ = 1 (5.119) 1 −1 ξR ξ ξ2

Where the 1’s in the matrix are two by two identities. Look at Peskin for more details. Let’s write ξ1 and ξ2 as 1 0 basis vectors (of which they were the indices in the above deﬁnition of ξ..). We deﬁne ξ = and ξ = 1 0 2 1 1 1 which will correspond to a σ3 = 2 and σ3 = − 2 .

√ σpξ √ √ q√ The most general solution is given by u(p) = √ where σp is deﬁned as σµp = 1 E + m − √ ~p~σ σpξ¯ µ 2 E+m

So let’s check it is at least a solution

√ −m pσ σpξ (p − m) u = √ = 0 (5.120) ¡ p pσ¯ −m σpξ¯

5.4.2 Chirality and helicity

We saw that the Dirac ﬁeld is putting together a left and right ﬁeld which are two irreducible representations

ψL ψD = (5.121) ψR 5 −1 0 −ψL . We deﬁne γ = so that γ5ψD = . We deﬁne the left and right projectors 0 1 ψR

1 − γ 1 0 1 + γ 0 0 P = 5 = P = 5 = (5.122) L 2 0 0 R 2 0 1

This will be useful in the concept of chirality, where eigenstates of the projector will be ’left’ or ’right’.

~p Now let’s deﬁne the concept of helicity. Helicity is the projection of the spin on the momentum ~s · |~p| .

~p · ~σ hˆ = (5.123) 2|~p|

hˆ 0 with the helicity operator h = 0 hˆ

q (1 − 2hˆ)ξ One can show that in the case m → 0, u(p) = E . What does the helicity do when acting on the 2 (1 + 2hˆ)ξ solution ?

EPFL-ITP-LPTP Riccardo Rattazzi 71 5.4. PATH TOWARDS QUANTISATION

r hˆ 0 E (hˆ − 1 )ξ − 1 0 u(p) = 2 = 2 u (5.124) ˆ ˆ 1 0 1 p 0 h 2 (h + 2 )ξ 2

So if we have a left handed spinor (upper component) it will be reversed under helicity, whereas the right handed spinor will not. Do solutions with deﬁnite chirality or helicity exist in the massless limit ? If you choose ξ such q ξ that ξ(1 + 2hˆ) = 0, then u(p) = E . 2 0

1 ˆ 2 0 0 1 For example, ~p = (0, 0, p), h = 1 , then we have, deﬁning ξ− = and ξ+ = 0 − 2 1 0

r r E ξ 1 E 0 1 u(p) = − Chirality = -1, helicity = - u(p) = Chirality = 1, helicity = (5.125) 2 0 2 2 ξ+ 2

In the massless limit, chirality and helicity are synonyms and are lorentz invariant. In the massive case, helicity is not lorentz invariant since if we overtake a particle in a boosted frame, it will look like it is going backwards with respect to us, but its spin will still be in the same direction, so helicity is not lorentz invariant.

The most general positive energy solution is

√ σpξ ψ(x) = e−ipx √ s s = 1,2 (5.126) σpξ¯ s

They have some orthogonality relations

√ s r s † r †√ †p 0 1 σpξs † u¯ (p)u (p) = u (p) γ0u (p) = (ξs σp, ξs σp¯ ) √ = 2mξsξr = 2mδrs (5.127) 1 0 σpξ¯ s

Moreover, s µ r µ u¯ (p)γ u (p) = 2p δrs (5.128)

Negative energy solutions are

√ ipx ipx σpξs ψ(x) = e vs(p) = e √ (5.129) − σpξ¯ s

Where s r v¯ (p)v (p) = −2mδrs (5.130)

s µ r µ v¯ (p)γ v (p) = 2p δrs (5.131)

Orthogonality

√ √ 0 1 pσξ u¯r(p)vs(p) = ξr† pσ, ξr†ppσ¯ √ s = 0 (5.132) 1 0 − pσξ¯ s

4 In other words, u1(p), u2(p), v1(p), v2(p) form a basis of C

EPFL-ITP-LPTP Riccardo Rattazzi 72 5.5. QUANTISATION OF THE DIRAC FIELD spin sums : Completeness relation √ αβ X α β X pσξs †√ †√ A = us (p)¯us (p) = √ ξs pσ¯ ξs pσ (5.133) pσξ¯ s s s

So m pσ A = = (p + m) (5.134) αβ pσ¯ m ¡ αβ αβ

So all in all, the most general ψ is

Z 3 X −ipx ipx ψgeneral(x) = d p as(p)us(p)e + bs(p)vs(p)e (5.135) s

Reminiscent of what we did in KG Fields, we’ll quantize the fourrier coeﬃcients to warp the theory to the quantum world.

5.5 Quantisation of the Dirac ﬁeld

5.5.1 How one should not quantise the Dirac ﬁeld

We’ll do exactly what we did with the klein gordon ﬁeld.

i ¯ ¯ µ ¯ L = ψ∂ψ¡ − ∂µψγ ψ − mψψ (5.136) 2

∂L ∂L T µ = ∂ ψ + ∂ ψ¯ − δµL (5.137) ν ν ¯ µ ν ∂∂µψ ∂∂µψ

Z Z Z 1 1 ← H = H = T 0 = ψ¯ −i~γ∇~ + m ψ + ψ¯ i∇γ + m ψd3x (5.138) 0 2 2 Z Z ¯ 3 † 3 = ψ −i~γ∇~ + m ψd x ≡ ψ hDψd x (5.139) |{z} ~ hD ≡γ0(−i~γ∇+m)

Z 1 Z Z P = T 0d3x = i2 ψγ¯ 0∂ ψ − ∂ ψγ¯ ψ = i ψ†∂ ψd3x (5.140) i i 2 i i 0 i

We recognize the 3-momentum from quantum mechanics, however with a sign ﬂip. Our ordinary 3-Momentum is i P not Pi Going back to the dirac hamiltonian

Z Z † 3 T 3 H = H = ψ hDψd x = ψ i∂0ψd x (5.141) |{z} using eq.o.m

So both expressions look quite similar and we’re happy. One has the 3 components (3-momentum) and the other one has the one that is left, the 0-th component. Now let’s quantize the whole thing. There’s a problem with the fact that the complex conjugated ﬁeld is troublesome for writing the canonical conjugate variable. Let’s take another Lagrangian which should be equivalent.

EPFL-ITP-LPTP Riccardo Rattazzi 73 5.5. QUANTISATION OF THE DIRAC FIELD

L = iψ∂ψ¯ − mψψ¯ (5.142)

The canonical variable conjugated to ψ

α ∂L ¯ 0α †α (πψ) = = i ψγ = iψ (5.143) ∂ψ˙ α

† So ψα, πα ≡ iψα are the canonically conjugated momenta. Let’s impose some commutation relations

3 h † i 3 [ψα(~x), ψβ(~y)] = 0 [ψα(~x), πβ(~x)] = iδαβδ (x − y) ψα(~x), ψβ(~y) = δαβδ (~x − ~y) (5.144)

Exercise: Show ψ˙(x) ≡ i [H, ψ(x)].

Moving on :

Z H = d3x ψ† −i~γ∇~ + m ψ (5.145)

Let’s now expand ψ into the basis of eigenfunctions to the hamiltonian so that it will trivially be diagonal. What is the eigenbasis of hD ? We had already found some stuﬀ :

−ipx ipx Positive energy : us(p)e Negative energy vs(p)e (5.146)

In particular for the positive energy solution

0 −ipx ipx ipx iγ ∂0 + i~γ∇~ − m us(p)e = 0 → − i~γ∇~ − m us(p)e = E(p)us(p)e t=0 (5.147)

For negative ones

0 ipx ipx ipx iγ ∂0 + i~γ∇~ − m vs(p)e = 0 → − i~γ∇~ − m vs(−p)e = −E(p)us(−p)e t=0 (5.148)

So we have 4 linearly independent eigenvectors (v1(−p), v2(−p), u1(p), u2(p)) when ﬁxing p (so the hamiltonian is a 4 by 4 matrix). By varying p, one spans the whole basis. The most general ψ(x) is

2 ! Z d3p X ψ(x) = eipx Qsu (p) + bs v (−p) (5.149) (2π)32E(p) p s −p s s=1

Remember for the scalar ﬁeld we had

Z d3p φ(x) = eipx a + a† (5.150) (2π)32E(k) p p | {z } φ(p)

Our choice of vs and us is very useful since as we will check, it diagonalizes the hamiltonian.

Z 1 X X u¯r(p)γ0 ψ(x)e−ipxd3x = u¯ (p)γ (a u (p) + b (−p)v (−p)) = δ a (p) = a (p) (5.151) 2E(p) r 0 s s s s rs s r s s Z r 0 −ipx 3 v¯ (p)γ ψ(x)e d x = br(−p) (5.152)

EPFL-ITP-LPTP Riccardo Rattazzi 74 5.5. QUANTISATION OF THE DIRAC FIELD So the ψ’s is a linear combination of a’s and b’s. So the commutation relations on ψ imply commutation relations on a and b.

† 3 3 † 3 3 ar(p), as(k) = (2π) 2E(p)δrsδ (k − p) br(p), bs(k) = (2π) 2E(p)δrsδ (k − p) (5.153)

Coming back to our sheep

Z Z 3 † X X † † i(k−p)x 3 H = d ψ hDψ = dΩkdΩp u¯s(p)as(p) +v ¯s(−p)bs(−p) (E(k)ur(k)ar(k) − E(k)vr(k)br(−k)) e d x s r Z X † † = dΩpEP a (p)as(p) − b (p)bs(p) (5.154) |{z} s s Trivial manipulation s

† † One can check H, as(p) = E(p)as(p) and [H, as(p)] = −E(p)as(p) while for the b’s it’s the other way around ! The daggered a is an energy raising operator but the daggered b is an energy lowering operator ! So the specturm of the Hamiltonian is not bound (if one assumes as(p)|0i = 0 and bs(p)|0i = 0). One can not fix this by redefining bdagger to b and conversely. This is because of their commutation relations. One should get bb† − b†b = 1 > 0. |{z} sort of If we would switch both definitions, one would get negative probability norms.

bb† − b†b = 1 (5.155)

Attempt to deﬁne b†|0i = 0 and b|0i= 6 0

h0|bb† − b†b|0i = h0|0i = −h0|b†b|0i = −||b|0i||2 (5.156)

So either the vacuum or the state where we created a particle has negative norm. We can’t do it. The question is : is the way to quantize the ﬁeld unique w.r.t to the commutation relations one imposes ? No ! There’s also the possibility of imposing anti-commutation relations !

5.5.2 The right quantisation method

So we impose the following commutation relations

3 n † o 3 {ψα(~x), ψβ(~y)} = 0 {ψα(~x), πβ(~x)} = iδαβδ (x − y) ψα(~x), ψβ(~y) = δαβδ (~x − ~y) (5.157)

But why can we do that ?

[A, BC] = B [A, C] + [A, B] C or = −B {A, C} + {A, B} C (5.158)

If the commutator that has the wrong sign is the one that vanishes it is ok !

We then have

† 3 3 † 3 3 ar(p), as(k) = (2π) 2E(p)δrsδ (k − p) br(p), bs(k) = (2π) 2E(p)δrsδ (k − p) (5.159)

The Hamiltonian is now the same as before, but the price one needs to pay to switch the roles of b and b† is free !

EPFL-ITP-LPTP Riccardo Rattazzi 75 5.5. QUANTISATION OF THE DIRAC FIELD 5.5.3 Hilbert space

Z Z X † † X † † H = dΩpEP as(p)as(p) − bs(p)bs(p) ≡ dΩpEP as(p)as(p) + Bs(p)Bs(p) (5.160) s s

Thus, we deﬁne |0i

Bs(p)|0i = as(p)|0i = 0 (5.161)

Consider the case (toy model) where one has only one set of operators. One could ask what is the equivalent of the harmonic oscillator with anti-commutation relations.

b, b† = 1 → bb† + b†b = 1 and {b, b} = b†, b† = 0 (5.162)

0 0 1 0 0 1 |0i = b† = b = |1i = (5.163) 1 0 0 1 0 0

But so far there’s no notion of what is the excited state and what is the ground state. However, one notices the exclusion principle is at work here.

† Coming back to the real thing, ar(p) creates a quantum of energy E(p) and 3-momentum ~p, polarization r. So

† † ~ ~ as(k)ar(p)|0i = |(~p,s); (k, r)i = −|(k, r); (~p,s)i = |ψi (5.164)

1. Notice that if you get identical quanta, then |ψi = −|ψi so that |ψi = 0. Therefrom follows exclusion principle. You cannot create two particles of the same quantum numbers.

2. Ex: a two particle state

Z X ~ ~ |Ψi = dΩkdΩp f((k, r); (~p,s))|(k, r); (~p,s)i (5.165) r,s

Since the ket is totally anti-symmetric, the only relevant piece in f is the anti-symmetric part, the symmetric part will drop out.

One should conclude that the Dirac ﬁeld quanta obeys Fermi-Dirac statistics.

The spin = 0 is quantized according to commutation relations so the resulting quanta satisﬁes Bose-Einstein statistics. 1 The spin = 2 is quantized according to anti-commutation relations so the resulting quanta satisﬁes Fermi-Dirac statistics.

These are the two examples of a theorem :

Spinor-statistics Theorem:

EPFL-ITP-LPTP Riccardo Rattazzi 76 5.6. PROPERTIES OF THE DIRAC FIELD Given a local-relativistic invariant unitary bounded Hamiltonian implies a connection between the spin of the quanta and their statistics : integer spin quanta follow bose-Einstein and half integer satisfy Fermi-Dirac.

5.6 Properties of the Dirac ﬁeld

5.6.1 Time evolution

5.6.2 Internal charge

EPFL-ITP-LPTP Riccardo Rattazzi 77 Chapter 6

Causality in quantum ﬁeld theory

78 Part II

Forces and Interactions

79 Chapter 7

Vector Fields

7.1 Classical Electrodynamics

Recall the classiﬁcation of the smallest representations of the Lorentz group:

(0,0) (1/2,0) (0,1/2) (1/2, 1/2) ... α β˙ αβ˙ φ χL χR A ...

We had previously seen that the field Aαβ˙ , a matrix with one left and one right spinorial index, can be packaged αβ˙ αβ˙ µ µ into a 4-vector according to A ≡ σµ A . We had also verified that A does transform as a 4-vector under the Lorentz group: αβ˙ αγ β˙δ˙ γδ˙ µ αβ˙ ν µ A −→ ΛL ΛR σµ A = σν Λ µA . (7.1) We would now like to construct the most general quadratic (≡ free) Poincaré invariant Lagrangian involving the field Aµ and at most two derivatives. Up to a total derivative it has the form:

µ ν µ 2 µ L = a1∂µAν ∂ A + a2(∂µA ) + a3AµA . (7.2) Notice that by assigning Aµ scaling dimension 1, the above terms have at most dimension 4. At this point it seems we have an embarassment of riches. On one hand the above Lagrangian appears to have more free parameters than one would expect to describe a particle of given spin and arbitrary mass: we could always rescale Aµ to make a1 = 1, but we would still be left with two free real parameters a2 and a3, which is one too many. On the other Aµ corresponds to 4 independent degrees of freedom (d.o.f.) describing in principle 4 independent polarizations. If we wanted to use Aµ to describe electromagnetic waves, which come in just two polarizations (left and right polarized light), we would need to somehow get rid of the extra polarizations.

The solution to both these problems comes from just one principle: symmetry. We can restrict the form of the Lagrangian (7.2) and at the same time lessen the number of d.o.f. demanding the action to be invariant under the following transformation:

0 0 Aµ −→ Aµ ≡ Aµ − ∂µα , S[A] = S[A − ∂α] = S[A ]. (7.3)

One can easily check that this request translates in a constraint on the coefficients of (7.2): a1 = −a2, a3 = 0. Finally we can determine the sign of a1 requiring the Hamiltonian of the theory to be positive and choose the overall normalization asking for a canonical kinetic term. This fixes a1 = −1/2 so that the final Lagrangian assumes the form 1 L = − F F µν ,F ≡ ∂ A − ∂ A . (7.4) 4 µν µν µ ν ν µ The equations of motion derived from the above Lagrangian reproduce the Maxwell equations (see Exercise Set 15). It is not immediately obvious to deduce from Maxwell equations that they imply the propagation of just two type of waves, the two transverse polarizations of light, but we already know from the study of classical electrodynamics that this is the case. In the following lectures we shall investigate this result in even more detail,

80 7.1. CLASSICAL ELECTRODYNAMICS including its quantum realization. It is remarkable how the Maxwell equations, and their physical implications regarding the property of light waves (and light quanta), can be derived by a symmetry principle.

Notice that the symmetry transformations (7.3) do not involve a finite set of parameters but an infinite one, namely a function α(x) over spacetime. This kind of symmetry is called a gauge symmetry. A gauge symmetry, more than a symmetry, is a redundancy in the description of the physical d.o.f of a system. Indeed if we declare that the only physical observables are gauge invariant quantities, then Fµν is and observable while Aµ, per se is not. Now Fµν is fully determined by Aµ, but only 3 out of the 4 components of Aµ contribute: the configuration µνρσ Aµ = ∂µα gives Fµν = 0 for any α. This can also be seen by making use of the Bianchi identity ∂µFνρ = 0. One finds that the longitudinal component of the magnetic field B~ L is identically vanishing while the transverse part B~ ⊥ is completely determined in terms of the curl of E~⊥ by a first order (in time) differential equation (see Exercise Set 15). Hence only the electric field is unconstrained: this corresponds to 3 unconstrained variables, as µν compared to the 4 carried by Aµ. Here we haven’t yet used the equations of motion ∂µF = 0: it is according to these other equations that 2 of these 3 variables turn out to be genuinely dynamical (corresponding to propagating waves), while the third variable, corresponding to a static Coulomb field, is not dynamical. (All this is clarified in the exercise on Coulomb gauge).

Gauge invariance ensures that the lagrangian and the physical quantities depend on less local variables than we have fields in our description. It thus allows to construct and describe systems that, is a certain sense, have a minimal number of d.o.f.. In order for this to make full sense, gauge symmetry should be extendable to interacting theories. Now, not only that is the case but, in addition, invariance under (7.3) turns out to fix in a very stringent way the form of the interaction between the vector field and the matter fields (bosons and fermions).

The matter Lagrangians are made gauge invariant by postulating the general transformation

0 iqaα(x) Ψa(x) −→ Ψa(x) ≡ e Ψa(x), (7.5) where Ψa is a general field and qa is real number called the charge of the field. It is clear that with the above definition the simple kinetic term that one is used to write is not invariant since the transformation is spacetime dependent and doesn’t commute with derivatives. On the other hand one can show that the combination

DµΨa ≡ (∂µ + iqaAµ)Ψa (7.6) transforms in the same way as the ﬁeld Ψa under gauge transformation:

iqaα(x) iqaα iqaα DµΨa −→ (∂µ + iqa(Aµ − ∂µα)) e Ψa = e (∂µ + iqaAµ − iqa∂µα + iqa∂µα)Ψa = e DµΨa. (7.7)

DµΨa is called covariant derivative of the ﬁeld Ψa and has the important property that transforms as the ﬁeld itself. At this point is straightforward to construct invariant Lagrangians. Let’s see two examples.

Fermions

µ µ L = i Ψa γ DµΨa −mΨaΨa = Ψa(i6 ∂ − m)Ψa + JµA , |{z} | {z } e−iqaα eiqaα µ µ µ J ≡ −qaΨaγ Ψa ≡ qaJNoether. (7.8)

Scalars

† µ 2 † † µ µ 2 † µ 2 † L = (DµΦa) (D Φa) − m ΦaΦa = ∂µΦa∂ Φa + JµA + qaΦaΦaAµA − m ΦaΦa, | {z } | {z } e−iqaα eiqaα † † µ Jµ ≡ −iqa Φa∂µΦa − Φa∂µΦa ≡ qaJNoether. (7.9) Notice that in the special case α = constant, the transformations (7.5) reduces to a global phase rotation. Therefore a necessary condition for gauge invariance is invariance under global phase rotations, which correspond to a U(1) symmetry. This implies, via Noether theorem, the existence of a conserved current Jµ. This current is exactly the

EPFL-ITP-LPTP Riccardo Rattazzi 81 7.2. EXERCISE

EM current that appears in Maxwell equations and is the one called Jµ in the previous examples. Indeed from a Lagrangian of the form 1 L = − F F µν + L , (7.10) 4 µν matter µν ν ν µν one can derive the equations of motion ∂µF = J , where J ≡ ∂Lmatter/∂Aν . The antisymmetric nature of F indeed also implies the conservation of the current:

µν ν ∂ν ∂µF = 0 = ∂ν J . (7.11) Current conservation, in this case, is both a consequence of Maxwell equations and of Noether theorem. In order to understand why the EM currents coincides with the Noether current associated to the global phase rotation symmetry, one could also recall the discussion in the Exercise Set 9 (Exercise 2). Given a matter Lagrangian a invariant under a global symmetry one can formally perform a transformation where the parameters qaα of the transformation are spacetime dependent. In this case the Lagrangian is, in general, no longer invariant and we have: µ δLmatter = Ja ∂µαa , (7.12) µ where Ja is exactly the Noether current that one would obtain with the standard procedure. In order to construct 0 µ a Lagrangian Lmatter invariant under gauge symmetry, one can start adding to Lmatter the term AµJ so that under gauge transformations we have

0 µ µ µ δLmatter ≡ δ(Lmatter + AµJ ) = δLmatter + J δAµ + AµδJ . (7.13)

µ 0 For simplicity let us assume that δJ = 0. Then using (7.12) and (7.3) one gets δLmatter = 0. The assumption δJ µ = 0 is true for the Noether current of fermions but not for the one for scalars. In the second case a further step ν 0 is needed but this will not aﬀect the discussion. We see than that the current J = ∂Lmatter/∂Aν appearing in 0 the Maxwell equation coincides with the Noether current extracted from the gauge invariant Lagrangian Lmatter.

7.2 Exercise

Consider the following two Lagrangians: 1 L = iψ¯6Dψ − F F µν , (7.14) QED 4 µν 1 L = (D φ)†Dµφ − F F µν . (7.15) sQED µ 4 µν The above Lagrangians are invariant under the transformations

ψ(x) −→ ψ0(x) = eiα(x)ψ(x) =⇒ δψ(x) = ieα(x)ψ(x), (7.16) φ(x) −→ φ0(x) = eiα(x)φ(x) =⇒ δφ(x) = ieα(x)φ(x), (7.17) 0 Aµ(x) −→ Aµ(x) = Aµ(x) − ∂µα(x). (7.18)

With these conventios the covariant derivative is deﬁned as Dµ = ∂µ + ieAµ. One can expand these Lagrangians: 1 L = iψ¯6 ∂ψ − eψγ¯ µψ A − F F µν , (7.19) QED µ 4 µν 1 L = ∂ φ†∂µφ + ie ∂µφ†φ − φ†∂µφ A + e2φ†φA Aµ − F F µν . (7.20) sQED µ µ µ 4 µν The ﬁrst term of both the Lagrangians coincides with the original "ungauged" Lagrangian, that is to say invariant under the global U(1) transformations (those with α = constant). The additional terms are indeed required to achieve the invariance under local transformations (those with α = α(x)). We recall the Noether currents associated to the global U(1) obtained from the old Lagrangians:

∂Lold old ¯ µ QED ¯ µ LQED = iψ6 ∂ψ =⇒ JNoether = ∆ψ = −ψγ ψ, (7.21) ∂(∂µψ) ∂Lold ∂Lold old † µ µ sQED sQED µ † † µ LsQED = ∂µφ ∂ φ =⇒ JNoether = ∆φ + † ∆φ† = i ∂ φ φ − φ ∂ φ . (7.22) ∂(∂µφ) ∂(∂µφ )

EPFL-ITP-LPTP Riccardo Rattazzi 82 7.3. A FRESH LOOK AT MAXWELL EQUATIONS: COUNTING DEGREES OF FREEDOM

If we consider the linear term in Aµ inside LQED, LsQED we notice that the combination of ﬁelds appearing are exactly the obtained currents. The motivation for this is related to how the gauge invariant Lagrangian has been constructed. Let us repeat brieﬂy the steps:

• We start with a Lagrangian invariant under a global symmetry: δL = 0. • We consider a transformation of the ﬁelds with parameters depending on the coordinates eα(x); in this case the variation of the Lagrangian induced by the transformation will not be zero. We have indeed shown that µ δL = eJNoether∂µα(x). µ • In order to make the Lagrangian invariant we add a term of the form eJNoetherAµ. Then the variation of the new Lagrangian:

µ µ µ µ δ(L + eJNoetherAµ) = eJNoether∂µα(x) − eJNoether∂µα(x) + e AµδJNoether (7.23)

• If the Noether current is gauge invariant then the procedure ends here. This is the case for fermions. The obtained Lagrangian is the covariantized Dirac Lagrangian. In the case of scalars the Noether current is not gauge invariant (since it contains derivatives) therefore one should iterate the procedure and add further terms to the Lagrangian. In the particular case of scalars indeed

µ µ † † µ † eAµδJNoether = ieA −iφ φ∂µα(x) − iφ φ∂µα(x) = 2eA φ φ ∂µα(x) (7.24)

2 µ † Hence adding to the Lagrangian a term like e AµA φ φ we can cancel this variation. Finally one observe that old µ 2 µ † no other source of non-invariance are introduced. Therefore the sum LsQED + eAµδJNoether + e AµA φ φ is gauge invariant.

This explains why the gauge vector couples to the Noether current of the global symmetry: we have engineered the construction in this way. Finally let us compute the Noether current associated to the global symmetry starting from the gauge invariant Lagrangian:

µ ∂LQED ¯ µ JQED = ∆ψ = −ψγ ψ, (7.25) ∂(∂µψ)

µ ∂LsQED ∂LsQED µ † † µ µ † µ † † µ JsQED = ∆φ + † ∆φ† = i ∂ φ φ − φ ∂ φ + 2eA φ φ = i (D φ) φ − φ D φ .(7.26) ∂(∂µφ) ∂(∂µφ ) The new currents are the equal to the old ones where we substitute normal derivatives with covariant derivatives and coincide exactly with the currents appearing in the Maxwell equations of motion:

µ ∂LQED µ ∂LsQED JQED = ,JsQED = . (7.27) ∂Aµ ∂Aµ

7.3 A fresh look at Maxwell equations: counting degrees of freedom

Before approaching the Hamiltonian description of the EM ﬁeld and its quantization, it is worth having a fresh look at the structure of its (classical) dynamics, and in particular at the counting of its dynamical degrees of freedom. For that purpose it is useful to recall that, given any vector ﬁeld V i(~x) on an euclidean space, we can i i i decompose it into longitudinal and transverse components V = VL + V⊥ according to ∇i∇j ∇i∇j V i ≡ ΠijV j ≡ V j ,V i ≡ ΠijV j ≡ δij − V j , =⇒ ∇iV i = ∇iV i , ∇iV i = 0. (7.28) L L ∇2 ⊥ ⊥ ∇2 L ⊥

In RD, with i = 1,...,D, the transverse and longitudinal ﬁelds have respectively D − 1 and 1 components. 3 Therefore in our R case, A⊥ is a vector determined by two independent functions.

µ 0 i i i In our case we can package the four components in A as (A ,AL,A⊥). Notice also that A⊥ (fully determining ij 2 0 i ˙ i 0i 0 i F ) and ∇ A + ∇ A (= ∂iF ) are gauge invariant, while any other linear combination of A and AL (for 0 i instance just A or AL) is gauge variant.

EPFL-ITP-LPTP Riccardo Rattazzi 83 7.3. A FRESH LOOK AT MAXWELL EQUATIONS: COUNTING DEGREES OF FREEDOM

Consider now the equations of motion for Aµ that are determined by the general lagrangian in eq. (7.10)

δS µν µ = 0 ⇒ ∂µF = J . (7.29) δAµ

The current J µ depends on the matter ﬁelds and their covariant derivatives (the latter only if there are scalars), 2 but not on derivatives of Aµ. The derivatives of Aµ only appear in the lagrangian via the Fµν term. Now, the structure of Fµν is such that the lagrangian does not depend on the time derivative of A0, indicating that this variable is not dynamical. Indeed its equation of motion reads 1

δS i0 2 0 i i 0 0 = =⇒ ∂iF = −∇ A − ∇ A˙ = J . (7.30) δA0 This equation, which is just the familiar Gauss Law (F i0 ≡ Ei), involves only terms with at most one time derivative acting on the ﬁelds. 2 It therefore represents a constraint, rather than a standard dynamical equation with second order time derivatives. In particular, initial conditions must satisfy this constraint! This equation can be solved to express, at any time, one variable in terms of the others. In particular it is convenient to solve for A0

A0 = −∇−2 J 0 + ∇iA˙ i . (7.31)

0 3 This shows that A is not an independent dynamical variable . One out of the four ﬁeld variables in Aµ is thus "eliminated" by Gauss law.

Consider next the equation for Ai, that reads

0i ij i 0 i 2 i i j j i ∂0F + ∂iF = ∇ A˙ + A¨ − ∇ A − ∇ ∇ A = J . (7.32)

Using current conservation and eq. (7.31) this equation can be written purely in terms of transverse ﬁelds

2 2 i i (∂t − ∇ )A⊥ = J⊥ . (7.33)

i i This is a consequence of the gauge invariance of eq. (7.32): A⊥ is gauge invariant while AL is not, so that once 0 i eq. (7.31) is used to eliminate A , the final equation can only depend on A⊥. Eq. (7.33) thus describes the dynamics of two d.o.f.: the two polarizations of light waves. At this stage we have not yet made any use of our i freedom to fix the gauge, and the single remaining field to play with is AL. Various options are given.

i i • Coulomb gauge: ∇iA = 0 ⇐⇒ AL = 0. For the purpose of quantization this is the best choice as it allows for the simplest Hamiltonian in terms of the 2 physical d.o.f. The problem with this choice is the lack of manifest Lorentz invariant. Of course the ﬁnal result of any computation respects Lorentz invariance, but that is not manifest in the intermediate stages.

0 0 i ˙ i • Temporal gauge: A = 0 ⇐⇒ J + ∇ AL = 0, slighly less simple from the Hamiltonian perspective and also not manifestly Lorentz invariant.

µ ˙ 0 i i • Lorentz gauge ∂µA = 0 ⇐⇒ A +∇ AL = 0; less straightforward Hamiltonian description but manifestly Lorentz covariant.

3 i • Axial gauge: A3 = 0 ⇐⇒ AL = −A⊥; not manifestly rotationally invariant.

0 i In the end the counting of degrees of freedom is already done: one linear combination of A and AL is made non dynamical by Gauss law, another can be eliminated by gauge ﬁxing, so that only two dynamical degrees of i freedom, which we can parameterize with A⊥, are left.

1 i i i i For ∇ we use the euclidean metric while for ∂i we use the Minkowski one: ∇ ≡ (∂/∂x ) ≡ ∂i = −∂ . 2 From a mechanical perspective we ould say it only depends on the qi (coordinates), the q˙i (velocities), but not on qï (accelerations). 3Notice that ∇2 only involves space derivative, so it inverse ∇−2 relates quantities at the same time, though not at the same space −1 position. The inverse of the D’alembertian would be a different story, as it relates quantities at different times (cfr. for instance the retarded or advanced Green’s functions in classical electrodynamics).

EPFL-ITP-LPTP Riccardo Rattazzi 84 7.4. A FIRST STEP TOWARDS QUANTIZATION 7.4 A ﬁrst step towards quantization

The simplicity (and beauty) of gauge theories implies however some technical difficulty when deriving the Hamil- tonian description of their dynamics. In turn, since canonical quantization is performed in the Hamiltonian formulation, this procedure is more involved than for scalar or fermionic fields. The origin of these complications is precisely the variable redundance discussed in the previous section: the vector field Aµ is determined by four local variables, but only two of them are dynamical. Let us see how this issues show up when trying to derive the Hamiltonian description for the general interacting case in eq. (7.10).

Consider now the conjugated momenta as deﬁned by eq. (7.10):

∂L Π0 ≡ = 0 , (7.34) ∂(∂0A0) ∂L Πi ≡ = −F 0i = Ei. (7.35) ∂(∂0Ai)

We have already discussed in the previous section the non-dynamical status of A0. The ﬁrst equation expresses 0 0 that fact in Hamiltonian languange: the conjugate momentum Π vanishes identically! This conﬁrms that (Π ,A0) does not form an ordinary pair of canonical Hamiltonian variables. But the peculiarity of A0 also determines the peculiarity of its equation of motion, Gauss law, that, as we have already seen, is a constraint rather than a dynamical equation. Indeed, using eq. (7.35), eq. (7.30) can be rewritten as

− ∇iΠi = J 0 , (7.36) which represents a constraint on the Hamiltonian variables. In particular it relates the space divergence of the conjugated momentum Πi to other Hamiltonian variables, associated with matter fields, and appearing in J 0. By i i i i this equation, the quantity ∇ Π ≡ ∇ ΠL is not an independent dynamical variable. This indicates that a second i i 0 0 Hamiltonian pair (AL, ΠL) should be eliminated in addition to (A , Π ), and thus reduces the number of genuine i local degrees of freeedom down to two. The elimination of AL is achieved via a choice of gauge, rather then by an equation of motion. Because of the above constraint, we cannot proceed to a straightforward quantization since a canonical commutation relation [Ai(~x,t), Πj(~y, t)] = iδijδ3(~x − ~y) (7.37) would not be in agreement with Gauss Law. In order to proceed we must eliminate the gauge redundancy of the Aµ field by "choosing a gauge". That corresponds to making a suitable transformation (7.3) in such a way that some linear combination of Aµ and its first derivatives vanishes. Some examples are:

i • Coulomb gauge: ∂iA = 0; it allows for the simplest Hamiltonian in terms of the 2 physical d.o.f but is not explicitly Lorentz invariant.

• Axial gauge: A3 = 0; not rotational invariant.

µ • Lorentz gauge ∂µA = 0; less straightforward Hamiltonian description but manifestly Lorentz covariant.

7.5 Quantization à la Gupta-Bleuler

In order to proceed to quantization in a way that is Lorentz invariant and allows the use of a simple Lagrangian, we will quantize the EM field using a different procedure: we will start with a Lagrangian that is inequivalent to the Maxwell Lagrangian but for which the the mentioned conditions are satisfied. The equivalence to the Maxwell theory will be obtained at the end selecting only the states |ψi for which the following gauge condition holds:

µ hψ|∂µA |ψi = 0. (7.38)

We will discuss extensively the meaning and the origin of the above constraint. The method we are going to discuss was ﬁrst proposed by Fermi and later formalized by Gupta and Bleuler.

We have seen in the past lecture that gauge invariance has the eﬀect of reducing the number of dynamical degrees of freedom in the theory and this introduces complication in the deﬁnition of the commutation relations. The idea

EPFL-ITP-LPTP Riccardo Rattazzi 85 7.5. QUANTIZATION À LA GUPTA-BLEULER is to make all the variables in Aµ formally dynamical by adding an explicit breaking of the gauge invariance to the theory so that our starting Lagrangian is 1 ξ 1 1 − ξ L = − F F µν − (∂ Aµ)2 = − ∂ A ∂µAν + (∂ Aµ)2, (7.39) GB 4 µν 2 µ 2 µ ν 2 µ where in the second equality we have integrated by part a term. The parameter ξ is a free parameter. The choice ξ = 1, that simpliﬁes a lot the form of the Lagrangian, is called Fermi’s or Feynman’s choice.

Now the action depends on all 4 d.o.f Aµ. In practice we have added a quadratic term for the component Aµ ∼ ∂µα which, being a pure gauge, in the Maxwell equation has an exactly vanishing action. For all the values ξ 6= 0 we can straightforwardly compute the conjugate momenta: ∂L 6= 0 for any µ. (7.40) ∂A˙ µ The equations of motion are analogously modiﬁed with respect to the Maxwell equations:

µν ν µ ν ∂µF + ξ∂ (∂µA ) = J . (7.41)

µ Let us define the scalar field χ = ∂µA . As we can observe from the above equation, the gradient of χ acts as a source for the EM fields, therefore the system is not in general equivalent to the Maxwell theory. On the other hand if we take the divergence of eq. (7.41) we get

ν µν ξχ = ∂ν J − ∂ν ∂µF = 0. (7.42) | {z } | {z } =0 by conservation =0 by antisymmetry

The above relation implies that the field χ is free, since nothing acts as source in its equation: the χ waves come and go without being affected by anything. This doesn’t mean that this field is completely decoupled because we have seen that the equation of motion for E~ and B~ are influenced by its presence. The important consequence of (7.42) is that it is consistent to choose χ ≡ 0 at all times since if it was vanishing at t = −∞ it will not be generated later on by the dynamics. When this is the case (7.41) goes back to Maxwell equations at all times. This is a intuitive justification of the condition (7.38).

Let us then proceed with the canonical quantization and let us work in the Feynman gauge ξ = 1. The Lagrangian 1 L = − ∂ A ∂µAν (7.43) GB 2 µ ν is equivalent to 4 scalars (A0,Ai), three of which have the proper kinetic term while one has a negative kinetic term. As consequence, the energy of the system will not be positive deﬁnite. We can introduce canonical commutation relations

ν ν 3 µ ˙ µ [Aµ(~x,t), Π (~y, t)] = iδµδ (~x − ~y), Π = −A 3 Aµ(~x,t), A˙ ν (~y, t) = −iηµν δ (~x − ~y), (7.44) while all the other commutators are zero. Notice the sign ﬂip between the commutator of the 0-component and the space components: it is a consequence of covariance. The Hamiltonian is thus:

3 1 1 X H = ΠµA˙ − L = − Π2 + (∂ A0)2 + ΠjΠj + (∂ Aj)(∂ Aj) = −H + H , (7.45) µ 2 0 i 2 i i 0 i i=1 The above Hamiltonian contains an oscillator with negative energy. Working in the Fourier space we can define, as we did for the Klein-Gordon fields, the expansion at a fixed time (t = 0): Z Z ~ 3 −i~k·~x ~ µ 3 µ −i~k·~x Aeµ(k) = d x Aµ(~x) e , Π(e k) = d x Π (~x) e , wk ≡ |k| , 1 a (~k) = w A (~k) − iΠ (~k) , µ 1/4 k eµ e µ (2wk) 1 a (~k)† = w A (−~k) + iΠ (−~k) . (7.46) µ 1/4 k eµ e µ (2wk)

EPFL-ITP-LPTP Riccardo Rattazzi 86 7.5. QUANTIZATION À LA GUPTA-BLEULER Notice the (−) sign in parenthesis in the second equation to reproduce the right combination as for the Klein- Gordon ﬁelds in the case of spatial components. Indeed the right conjugate momentum of Ai is −Πi. Following the same computation as in the Klein-Gordon case we can show that

h ~ i h † ~ † i aµ(k), aν (~p) = aµ(k), aν (~p) = 0 h ~ † i 3 ~ 3 aµ(k), aν (~p) = −ηµν δ (k − ~p)(2π) 2wk (7.47)

An important outcome is that the time photon (a0) has "wrong" sign. The inverse relation of (7.46) has the form Z ~ +i~k·~x † ~ −i~k·~x Aµ(~x, 0) = dΩ~k aµ(k) e + aµ(k) e . (7.48)

At this point one can plug the above expansion in the expression of the energy-momentum tensor and derive the associated conserved charge Pµ (see Exercise Set 17): Z Z µ µ † ~ ~ † ~ ~ µ † ~ ν ~ P = dΩ~k k ai (k)ai(k) − a0(k)a0(k) = dΩ~k k −aν (k)a (k) . (7.49)

µ Notice that we can use the expression of Aµ at t = 0 since the conserved charges like P are independent of time. At this point we can compute the vector potential in the Heisenberg picture

iHt −iHt Aµ(~x,t) = e Aµ(~x, 0) e . (7.50) For the spatial components the computation is the standard one. For the time component one should pay attention to the (−) signs in (7.45) and (7.47). These two factors (−) compensate and the result is simply Z ~ −ikx † ~ ikx Aµ(~x,t) = dΩ~k aµ(k) e + aµ(k) e . (7.51)

The result is consistent with Lorentz invariance: since Aµ is a four-vector, the integral measure and the exponential ~ in the integral are scalars, hence the operator aµ(k) must be a four-vector. This is also in accord with the commutation relations (7.47).

We are now ready to construct the Fock space of the theory. Insisting on Lorentz covariance we define the vacuum as the state annihilated by all destroying operators ~ ~ aµ(k)|0i = 0 , for any k . (7.52) The above requirement is surely consistent with the Lorentz invariance of the vacuum, U(Λ)|0i = |0i, but, as we will see in the following, it’s a stronger requirement. † ~ At this point the Fock space is constructed applying repeatedly the operators aµ(k). However we immediately † ~ encounter a serious issue: if we compute the norm of a single particle state aµ(k)|0i we get ~ † ~ 3 h0|aµ(k)aµ(k)|0i = −2k0(2π) ηµµ, no summation over µ. (7.53) † ~ Therefore for µ = i spatial the above norm is positive, while a0(k)|0i is a state with negative norm. This pathology, if not solved, would compromise the probabilistic interpretation of Quantum Mechanics. On the other hand we should recall that the system we are considering doesn’t coincide with Maxwell’s. As we saw at the very beginning, µ the equivalence can be achieved imposing the condition ∂µA = 0. This constraint has been derived at the classical level in order to reproduce Maxwell equations. At the quantum level the field Aµ is promoted to an operator and we cannot impose this condition naively at the operator level since it would be in conflict with the commutation µ relations (7.44). What we can do instead is to impose the vanishing of the expectation value of ∂µA . More precisely, given the Fock space F we define the subset F 0 of physical states as the states |Ψi satisfying

µ hΨ| ∂µA |Ψi = 0. (7.54)

We stress that the above condition is not a constraint on the ﬁeld Aµ, which is unconstrained, but a restriction of the states of F: only a subset of them is selected. Deﬁning

µ µ + µ − ∂ Aµ = ∂ Aµ + ∂ Aµ , Z Z µ + µ † ~ ikx µ − µ ~ −ikx ∂ Aµ = dΩ~k(ik )aµ(k)e , ∂ Aµ = dΩ~k(−ik )aµ(k)e , (7.55)

EPFL-ITP-LPTP Riccardo Rattazzi 87 7.5. QUANTIZATION À LA GUPTA-BLEULER µ − ρ ~ a sufficient condition for (7.54) is that ∂ Aµ |Ψi = 0. We can also define L(k) ≡ k aρ(k) as the Fourier transform µ − of the operator ∂ Aµ . Then we will postulate that a physical state is a state that satisfies the condition

L(k)|Ψi = 0 or equivalently hΨ|L†(k), for any ~k . (7.56)

µ † Let us see what this implies for a single particle state |Ψi ∼ ε aµ(~q)|0i. Imposing the above condition and using the commutation relations (7.47) we get

ρ µ † ~ 3 3 ~ µ L(q)|Ψi = q aρ(~q)ε aµ(k)|0i = −2k0(2π) δ (k − ~q)k εµ|0i = 0. (7.57)

Here εµ is the polarization of the photon and for a physical state it has to satisfy the transversality condition µ k εµ = 0. This means that the number of physical polarization is reduced by one. However we will now show that the vector εµ contains in fact only 2 physical polarizations. Indeed the general expression for the polarization µ vector satisfying k εµ = 0 is ⊥ ⊥ 1 2 εµ = εµ + aLkµ εµ = c1εµ + c2εµ, (7.58) ~ where 1 and 2 are the directions transverse w.r.t k. For example if kµ = (k, 0, 0, k) and εµ = (a, b, c, d) we have: a kµε = 0 ⇒ a = d, ε⊥ = (0, b, c, 0) , a = . (7.59) µ µ L k

The term proportional to kµ is called longitudinal polarization and its presence doesn’t affect any physical quantity. ⊥ ⊥ ~ Instead εµ is called transverse polarization and satisfies ~ε · k = 0. The first check of the irrelevance of the longitudinal polarization is given by the computation of the norm of a physical state: Z Z Z µ∗ ν † ∗ µ ⊥ 2 hΨ|Ψi = dΩ~pdΩ~q ε (~p)ε (~q)h0|aµ(~p)aν (~q)|0i = − dΩ~p εµ(~p)ε (~p) = dΩ~p|~ε (~p)| ≥ 0. (7.60)

µ ⊥ ⊥ In the last equality we have used kµk = 0, µ kµ = 0 and the fact that µ has zero time component. We conclude that the longitudinal polarization doesn’t aﬀect the norm of physical states: its only a redundancy of the theory. One can check that the same holds for any other physical quantity, for example energy and momentum: Z µ ρ∗ ν µ † ~ σ ~ † hΨ|P |Ψi = − dΩ~pdΩ~qdΩ~k ε (~p)ε (~q)k h0|aρ(~p)aσ(k)a (k)aν (~q)|0i Z Z ρ∗ ν ~ µ † ~ µ ⊥ = dΩ~pdΩ~kε (~p)ε (k)k h0|aρ(~p)aν (k)|0i = dΩ~p p |~ε (~p)|. (7.61)

Again the longitudinal polarizations doesn’t give any contribution. This property is a consequence of the cancel- ~ ~ ~ lation between the time photon a0(k) and the 3rd photon a3(k) (in general the photon aligned along k). This cancellation is enforced by the condition of being physical, as can be seen by considering the square of the vanishing state L(k)|Ψi; taking again kρ = (k, 0, 0, k) we have ~ ~ 0 = L(k)|Ψi =⇒ ka0(k)|Ψi + ka3(k)|Ψi (7.62) † ~ ~ † ~ ~ hΨ|a0(k)a0(k)|Ψi = hΨ|a3(k)a3(k)|Ψi. (7.63) The above relation expresses the fact that, for a physical state, the occupation number of the time photon is equal to the occupation number of 3rd component photons. Hence they are always equal in number and their contribution to observables quantities always cancel. Only transverse polarizations contribute to physical quantities.

Let us discuss more in details the form of the space of physical states. Given a physical state |Ψi and a second (even inﬁnite) set of other physical states {|ϕ1i, |ϕ2i, ...., |ϕni} we construct the state Z Z 0 † † † |Ψ i = |Ψi + dΩk1 f1(k1)L (k1)|ϕ1i + dΩk1 dΩk2 f2(k1, k2)L (k1)L (k2)|ϕ2i + ... . (7.64)

It’s easy to check that the above combination is a physical state: L(k)|Ψ0i = 0; indeed for instance

† µ ν ~ † ~ 3 0 3 ~ ~ µ [L(k1),L (k2)] = k1 k2 [aµ(k1), aν (k2)] = −(2π) 2k1δ (k1 − k2)k1 k2µ = 0, (7.65) Z Z † † dΩk1 f1(k1)L(q)L (k1)|ϕ1i = dΩk1 f1(k1)L (k1)L(q)|ϕ1i = 0 (7.66)

EPFL-ITP-LPTP Riccardo Rattazzi 88 7.6. QUANTIZATION OF MASSIVE VECTOR FIELDS 2 since q = 0. The same holds for the other terms. In addition we a have used the fact that the |ϕii are physical states. However, not only |Ψ0i is physical but it contributes to all the physical quantities in the same way as |Ψi. † † Indeed the states L(k1) ...L(ki) |ϕii are all states with null norm and are orthogonal to any other physical state. This means that the norm of |Ψi and |Ψ0i are identical and the scalar product of the two states with any other physical state coincide:

The above relations imply that the physical content of the tho states |Ψi and |Ψ0i is identical and therefore they describe the same state. This means that we can deﬁne an equivalence relation on the Hilbert space of physical states and consider equivalent two states if they coincide modulus a combination of states of the form † † L(k1) ...L(ki) |ϕii:

† † † † † |Ψi ∼ |Ψi + L (k1)|ϕ1i + L (k1)L (k2)|ϕ2i + ... + L (k1)...L (kn)|ϕni + ... . (7.68)

In the end any physical state is deﬁned by a tower of equivalent states; in particular also the physical vacuum is described by all the set of combinations

† † † † † |0i + a1L (k1)|ϕ1i + a2L (k1)L (k2)|ϕ2i + ... + anL (k1)...L (kn)|ϕni + ... . (7.69)

Let us now deﬁne in a more concrete way what is a physical observable. The basic requirement that must be satisﬁed is that the action of a physical observable on a physical state must produce a physical state. Then if O is an operator associated to a physical observable and |Ψi is a physical state we must have

L(k)|Ψi = 0,L(k)O|Ψi = 0. (7.70)

A suﬃcient condition to satisfy the above requirement is that the commutator [L(k),O] be proportional to some operator that has an L on the right:

[L(k),O] ∼ AL(k) , [O,L(k)†] ∼ L(k)†A†. (7.71)

It is clear that the action of O on a physical state produces again a physical state:

L(k)O|Ψi = [L(k),O]|Ψi = AL|Ψi = 0. (7.72)

In addition this requirement ensures that the expectation value of an observable O on two physical states is independent of the vector chosen to represent the states:

0 0 † hΦ |O|Ψ i ∼ (hΦ| + hϕ1|L(k1) + ...) O |Ψi + L(k2) |ϕ2i + ... = hΦ|O|Ψi. (7.73) where we have repeatedly made use of the commutation relations (7.71). Here we have defined observables all the operator satisfying equation (7.70). This definition is a stronger request than simply requiring equation (7.73), that instead is verified whenever [L(k),O] ∼ AL(k)+L†(k)B. The problem with this second definition is that is not preserved by the composition of observables. For instance the observable O2 is not independent of the vector that represents a physical state. In practice however we don’t have any simple example where B is different from 0.

7.6 Quantization of massive vector ﬁelds

7.6.1 Massive Vector Fields

Let us consider the Maxwell Lagrangian plus a real scalar ϕ: 1 1 L = − F F µν + ∂ ϕ ∂µϕ . (7.74) 4 µν 2 µ

The above Lagrangian is invariant under the shift symmetry ϕ → ϕ + c, as long as c is a constant. The ﬁeld Aµ doesn’t transform under this symmetry. The Noether current associated to this global symmetry is simply

Jµ = ∂µϕ , (7.75)

EPFL-ITP-LPTP Riccardo Rattazzi 89 7.6. QUANTIZATION OF MASSIVE VECTOR FIELDS µ and the conservation of the current is guaranteed by the equation of motion of the scalar field: ∂µJ = ϕ = 0. The above Lagrangian describes therefore 3 physical degrees of freedom: one is the real scalar field and the other two are described by the massless vector Aµ. We can now try to make the shift symmetry of ϕ local, exactly as we did for the phase symmetry in the case of spinor fields and complex scalar fields. In that case, imposing the invariance under the new local symmetry, we got a Lagrangian describing the interaction between electromagnetism and matter. What shall we obtain in this case? The local transformation must have the form:

ϕ → ϕ + Mα(x) . (7.76)

Notice that we were forced to add a parameter M with energy (mass) dimension [M] = E, because the function α(x), describing the transformation of Aµ in eq. (7.3), is by consruction dimensionless and ϕ has mass dimension 1. The Lagrangian (7.6.1), as it stands, is not invariant under the tranformation 7.76, since derivatives act non- trivially on the function α(x). Again we can deﬁne a new covariant derivative involving the vector ﬁeld Aµ to compensate for the additional terms. One can easily check that the covariant derivative

Dµϕ ≡ ∂µϕ + MAµ (7.77) is invariant under the combined set of transformations

ϕ → ϕ + Mα(x),Aµ → Aµ − ∂µα(x) . (7.78)

Indeed one has Dµϕ ≡ ∂µϕ + MAµ −→ ∂µ(ϕ + Mα) + M(Aµ − ∂µα) = Dµϕ . (7.79) so that the Lagrangian invariant under the local transformation 7.78 is 1 1 Linv = − F F µν + (∂ ϕ + MA )(∂µϕ + MAµ) . (7.80) 4 µν 2 µ µ In order to more directly extract the physical meaning of this Lagrangian we must perform a gauge fixing. For this system there exists an obvious choice: ϕ(x) α(x) = − , (7.81) M upon which ϕ simply vanishes! The resulting gauge fixed lagrangian 1 1 Lfixed = − F F µν + M 2A Aµ . (7.82) 4 µν 2 µ describes a free vector field of mass M, as we shall now discuss. In particle physics, this phenomenon, according to which a massless scalar shifting under gauge transformations implies that the photon obtains a mass, goes under the broad name of Higgs mechanism4. In condensed matter an analogous phenomenon leads to superconductivity. In that case the role of ϕ is played by the "phase" of the complex fermion bilinear field describing Cooper pairs. As already observed the original Lagrangian (7.6.1) describes 3 physical degrees of freedom, and it is a fair question to ask how many degrees of freedom the new Lagrangian (7.82) describes. One would argue that given one can go from to (7.82) by turning on a continuous parameter M, the number of degrees of freedom, an integer, should not change. It is instructive to check that is indeed the case. Let us then consider the equations of motion

µν 2 ν ∂µF + M A = 0 . (7.83)

Taking the diverge we obtain

µν 2 ν 2 ν 2 ν ∂ν ∂µF + M ∂ν A = M ∂ν A =⇒ M ∂ν A = 0 . (7.84)

The dynamics then forces the ﬁeld Aµ to be transverse. Expanding the ﬁeld strength Fµν and dropping the vanishing terms, the equations of motion can then be cast as the combined set

2 ν ν ( + M )A = 0, ∂ν A = 0 (7.85) 4The mechanism was fully understood in relativistic QFT in the 60’s. Notice that the famous Higgs boson does not correspond to the "unphysical" ϕ ﬁeld here introduced but to another degree of freedom that accompanies the ϕ in theories endowed with additional properties, in particular renormalizability. The association of a photon mass to a scalar shifting under a local symmetry goes back to the 30’s in the work of Stueckelberg.

EPFL-ITP-LPTP Riccardo Rattazzi 90 7.6. QUANTIZATION OF MASSIVE VECTOR FIELDS The first corresponds to four independent Klein-Gordon equations describing fields with mass M, while the second constrains one linear combination of these four Klein-Gordon fields to vanish. The plane wave solutions have the form ~ q µ µ ik0t−k·~x ~ 2 2 A (~x,t) = ε (k)e , k0 = |k| + M . (7.86) with polarization vector εν (k) constrained to be transverse by the second equation in (7.85) µ k εµ = 0 . (7.87)

This constraint implies that only 3 out of the 4 components of εµ are independent. For example in the special ~ case of k = 0, k0 = M, the polarization is of the form εµ = (0, ~ε). We thus conclude that the massive vector ﬁeld describes the propagation of 3 independent polarizations, precisely like the system we started with, for which the 3 polarizations where distributed among a massless vector (2 polarizations) and a massless scalar (1 polarization). The spin of the resulting particle is evident by considering the polarization 4-vector in the rest frame (~k = 0). There it reduces to a 3-vector transforming as the J = 1 representation of the group of rotations in the rest frame. Hence we conclude that the spin of the massive vector ﬁeld Aµ is 1. As usual the plane wave solution corresponds to an eigenstate of the 4-momentum, and the general solution is obtained by writing the most general superposition of plane waves.

7.6.2 Canonical quantization

We now proceed with the standard technique to quantize the theory: we first compute the conjugate momentum of the fields Aµ and the Hamiltonian, then we impose the canonical commutation relations. In the following we will work in the Schroedinger picture: the quantum states evolve in time while the operators do not. Hence all the fields, unless explicitly stated, will be evaluated at t = 0.

From the Lagrangian (7.82) we can extract the conjugate momentum: ∂L Πµ = = −F 0µ, ∂(∂0Aµ) 0 i i i i Π = 0 , Π = E = ∂ A0 − ∂0A . (7.88)

As in the massless case the ﬁeld A0 is not dynamical since its conjugate momentum vanishes. On the other hand, in the present case its equation of motion does not represent a constraint on the spatial components Ai but rather it allows to solve for A0 and express it in terms of the other degrees of freedom:

δS ν 2 = 0 =⇒ ∂ Fν0 + M A0 = 0, δA0(x) ∇~ · Π~ ∂i(F i0) − M 2A = 0 =⇒ A = − . (7.89) 0 0 M 2

The time component of the vector ﬁeld Aµ is completely determined in term of the others. Keeping the constraint (7.89) in mind we can start imposing the usual canonical commutation relations for the three pairs of unconstrained i variables (Ai, Π ): i j [Ai(0, ~x),Aj(0, ~y)] = [Π (0, ~x), Π (0, ~y)] = 0, j j 3 [Ak(0, ~x), Π (0, ~y)] = iδkδ (~x − ~y). (7.90)

We cannot adopt the standard prescription for the ﬁeld A0. At classical level this ﬁeld is indeed a function of the conjugate momenta, hence we cannot impose the vanishing of the commutator [A0,Ai]. Rather we would like to take into account the classical constraint at the quantum level. We impose therefore ∇~ · Π(0~ , ~x) i [A (0, ~x),A (0, ~y)] = [− ,A (0, ~y)] = ∂(x)δ3(~x − ~y). (7.91) 0 j M 2 j M 2 j Finally we can compute the Hamiltonian density associated to the Lagrangian density (7.82): ! ! ∇~ · Π~ 1 1 1 (∇~ · Π)~ 2 1 H = ΠiA˙ − L = Πi ∂ − − Π + Π Πi + F F ij − − M 2A Ai i i M 2 i 2 i 4 ij 2 M 2 2 i 1 1 (∇~ · Π)~ 2 1 1 = ΠiΠi + + F F + M 2A A + (total derivative). (7.92) 2 2 M 2 4 ij ij 2 i i

EPFL-ITP-LPTP Riccardo Rattazzi 91 7.6. QUANTIZATION OF MASSIVE VECTOR FIELDS When considering the Hamiltonian H = R Hd3x the last terms vanishes. Notice that we have substituted every- where the constraint (7.89) and we have expressed A˙ i in terms of the other variables using the relation (7.88). As consistency check we can compute the Hamilton equations of motion and show that they coincide with those derived with the Lagrangian formalism:

Z 1 H, Πj(0, ~x) ≡ −iΠ˙ j = d3y F mn F (0, ~y), Πj(0, ~x) − M 2Am A (0, ~y), Πj(0, ~x) 2 mn m Z 3 ij h (y) j i 2 m j = d y F ∂m An(0, ~y), Π (0, ~x) − M A Am(0, ~y), Π (0, ~x) Z 3 mj (y) 3 2 j 3 mj 2 j = i d y F ∂m δ (~x − ~y) − M A δ (~x − ~y) = −i∂m F (0, ~x) − iM A (0, ~x), (7.93) Z 1 [H,A (0, ~x)] ≡ −iA˙ = d3y −Π Πi(0, ~y),A (0, ~x) + (∂ Πm)[(∂ Πn)(0, ~y),A (0, ~x)] j j i j M 2 m n j Z 1 i = d3y iΠ (0, ~y)δ3(~x − ~y) − i (∂ Πm)∂ δ3(~x − ~y) = iΠ (0, ~x) + ∂ (∂ Πm)(0, ~x). j M 2 m j j M 2 j m (7.94)

Finally taking the time derivative of the second equation and using the ﬁrst one we get: 1 A¨j = −Π˙ j − ∂ ∂j(∂ Πm) = −Π˙ j + ∂ ∂jA = −∂ F mj − M 2Aj + ∂ ∂jA M 2 0 m 0 0 m 0 0 0 j j 0 mj 2 j =⇒ ∂0∂ A − ∂0∂ A + ∂m F + M A = 0. (7.95)

In order to construct the Fock space, we express all the ﬁelds and the Hamiltonian in Fourier space: Z Z ~ 3 −i~x·~k i ~ 3 i −i~x·~k Ai(k) = d x Ai(0, ~x)e , Π (k) = d x Π (0, ~x)e . (7.96)

The reality of the ﬁelds in the coordinate space translates in the condition

† ~ ~ i† ~ i ~ Ai (k) = Ai(−k) , Π (k) = Π (−k). (7.97) Thus we can straightforwardly express the Hamiltonian in terms of the new ﬁelds: ! Z d3k 1 (kiΠi(~k))(kjΠj(−~k)) h i H = Πi(~k)Πi(−~k) + + k A (~k) − k A (~k) k A (−~k) + M 2A (~k)A (−~k) (2π)3 2 M 2 i j j i i j i i ! Z d3k 1 |~k · Π(~ ~k)|2 = |Π(~ ~k)|2 + + |~k|2|A~(~k)|2 − |~k · A~(~k)|2 + M 2|A~(~k)|2 . (7.98) (2π)3 2 M 2

In order to make manifest the difference between the massive case and the massless case let us decompose the fields in their longitudinal and transverse part with respect to the direction of the momentum ~k. Introduce the projectors k k (P ) = i j , (P ) = δ − (P ) , k2 ≡ |~k|2 , L ij k2 ⊥ ij ij L ij 2 2 (PL) = PL , (P⊥) = P⊥ ,P⊥ + PL = 1 ,PLP⊥ = P⊥PL = 0, (7.99) and define the projected fields: ~ ~ ~ ~ ~ ~ Ai(k) = ALi(k) + A⊥i(k) , Πi(k) = ΠLi(k) + Π⊥i(k), (7.100) where: ~ ~ ~ ~ ALi(k) = (PL)ijAj(k) ,A⊥i(k) = (P⊥)ijAj(k), ~ ~ ~ ~ ΠLi(k) = (PL)ijΠj(k) , Π⊥i(k) = (P⊥)ijΠj(k), ~ ~ ~ ~ ~ k · A~⊥(k) = k · Π⊥(k) = 0. (7.101)

EPFL-ITP-LPTP Riccardo Rattazzi 92 7.6. QUANTIZATION OF MASSIVE VECTOR FIELDS Pugging the decomposition (7.100) in the Hamiltonian we obtain:

H = HL + H⊥ , Z d3k 1 w2 H = k |Π~ (~k)|2 + M 2|A~ (~k)|2 , L (2π)3 2 M 2 L L Z d3k 1 H = |Π~ (~k)|2 + w2|A~ (~k)|2 , ⊥ (2π)3 2 ⊥ k ⊥ 2 2 where wk = k + M . Notice that the transverse modes and the longitudinal ones decouple. Moreover the former have a standard Hamiltonian, schematically of the form H ∼ p2 + w2q2. The latter instead don’t seem to have a standard Hamiltonian, however it is sufficient a field redefinition to bring it in a more familiar form. Let us review the procedure first for the simple case of quantum mechanics. Suppose to start from a unidimensional system where w2 H = p2 + M 2q2 , [q, p] = i. (7.102) M 2 Define the canonical transformation w M p0 = p , q0 = q , [q0, p0] = [q, p] = i , H = p02 + w2q02. (7.103) M w Hence the starting Hamiltonian is equivalent to the above one and define the same spectrum. The same procedure can be followed for the quantum field theory of a massive vector. Hence we define the creation and annihilation operators for the transverse modes in the usual way: ~ ~ 1 ~ † ~ A⊥(k) = ~a⊥(k) + ~a⊥(−k) , 2wk i Π~ (~k) = ~a (~k) − ~a† (−~k) . (7.104) ⊥ 2 ⊥ ⊥ Instead, as suggested by the above discussion, we rescale the longitudinal mode: 1 A~ (~k) = ~a (~k) + ~a† (−~k) , L 2M L L ~ ~ iM ~ † ~ ΠL(k) = ~aL(k) − ~aL(−k) . (7.105) 2wk

In order to express the Hamiltonian in terms of the lowering and raising operator we need ﬁrst to translate the j † canonical commutation relation for Ai, Π to commutation relations between ai, aj. The ﬁrst straightforward step is to pass to Fourier space: Z ~ ~ 0 3 3 0 −i~x·~k−i~x0·~k0 0 3 3 ~ ~ 0 [Ai(k), Πj(k )] = d x d x e [Ai(~x), Πj(~x )] = iδij(2π) δ (k + k ),

~ ~ 0 ~ ~ 0 [Ai(k),Aj(k )] = [Πi(k), Πj(k )] = 0. (7.106) Then we can invert the decomposition ~ ~ 1 h ~ † ~ wk ~ † ~ i A(k) = ~a⊥(k) + ~a⊥(−k) + ~aL(k) + ~aL(−k) , 2wk M ~ ~ i ~ † ~ M ~ † ~ Π(k) = ~a⊥(k) − ~a⊥(−k) + ~aL(k) − ~aL(−k) , (7.107) 2 wk to obtain ~a(~k) in terms of A~(~k) and Π(~ ~k) w ~a(~k) = w A~ (~k) − iΠ~ (~k) + MA~ (~k) − i k Π~ (~k) , k ⊥ ⊥ L M L w ~a†(~k) = w A~ (−~k) + iΠ~ (−~k) + MA~ (−~k) + i k Π~ (−~k) , (7.108) k ⊥ ⊥ L M L and ﬁnally extract: ~ ~ 0 ~ ~ ~ 0 ~ ~ ~ 0 ~ ~ ~ 0 ~ ~ ~ 0 [~a(k), ~a(k )] = −iwk[A~⊥(k), Π⊥(k )] − iwk[Π⊥(k), A~⊥(k )] − iwk[A~L(k), ΠL(k )] − iwk[ΠL(k), A~L(k )] = 0, [~a†(~k), ~a†(~k0)] = 0, ~ † ~ 0 ~ ~ 0 ~ ~ 0 [ai(k), aj(k )] = iwk[A⊥i(k), Π⊥j(−k )] − iwk[Π⊥i(k),A⊥j(−k )] ~ ~ 0 ~ ~ 0 + iwk[ALi(k), ΠLj(−k )] − iwk[ΠLi(k),ALj(−k )] 3 3 ~ ~ 0 = 2wkδij(2π) δ (k − k ). (7.109)

EPFL-ITP-LPTP Riccardo Rattazzi 93 7.6. QUANTIZATION OF MASSIVE VECTOR FIELDS In the end the Hamiltonian expressed in terms of the new quantities reads: Z d3k 1 H = ~a (~k) · ~a† (~k) + ~a† (−~k) · ~a (−~k) , L (2π)3 4 L L L L Z d3k 1 H = ~a (~k) · ~a† (~k) + ~a† (−~k) · ~a (−~k) , ⊥ (2π)3 4 ⊥ ⊥ ⊥ ⊥ Z d3k 1 Z d3k X H = H + H = ~a(~k) · ~a†(~k) + ~a†(~k) · ~a(~k) = w ~a (~k)†~a (~k) L ⊥ (2π)3 4 2w (2π)3 k i i k i Z X ~ † ~ = dΩ~k wk ~ai(k) ~ai(k). (7.110) i

7.6.3 Heisenberg Picture Field

We can now consider the time dependent ﬁelds in the Heisenberg picture

∇~ · Π(0~ , ~x) A0(x) ≡ A0(t, ~x) ≡ eiHtA0(0, ~x)e−iHt = −eiHt e−iHt Ai(x) ≡ Ai(t, ~x) ≡ eiHtAi(0, ~x)e−iHt M 2 (7.111) The Hamiltonian straightfowardly dicatates the time dependende of creation and destruction operators: ai(t, k) = e−iωktai(k). By using eq. (7.107) we can then write ! Z ~k · ~a(k) A0(x) = dΩ e−ikx + h.c. k M Z h ω i A~(x) = dΩ ~a (k) + k ~a (k) e−ikx + h.c. (7.112) k ⊥ M L ~ ~ ~ where we used k · ~a⊥ = 0 and k · ~aL = k · ~a. We can check that the above quantum fields satisfy the equations µ ~ of motion 7.85. The first equation is obviously satisfied given k = (ωk, k) satisfies by construction the on-shell µ 2 µ condition k kµ = M . Consider then the diverge of ∂µA . We find ! Z ~k · ~a(k) A˙ 0(x) = −i dΩ ω e−ikx − h.c. (7.113) k k M Z h ω i ∇~ A~(x) = i dΩ ~k · ~a (k) + k ~a (k) e−ikx − h.c. (7.114) k ⊥ M L

~ 0 µ which, upon using k · ~a⊥ = 0, are easily seen to give A˙ + ∇~ A~ = ∂µA = 0.

7.6.4 Polarization vectors

Eq. (7.112) can be more compactly written Z µ µ i −ikx A (x) = dΩk i (k)a (k)e + h.c. (7.115)

µ µ with i (k) representing a basis of the 3-dimensional subspace of 4-vectors satisfying kµi (k) = 0. By comparison with eq. (7.86) we find that (like for all quantum fields we encountered so far) the quantum field Aµ is written as the most general linear superposition of plane wave solutions with quantized coefficients ai(k), ai†(k). Notice in µ α α particular that the i (k) are the analogue of the ur (k) and vr (k) of the Dirac field, where we made explicit the µ ~ 4-spinor index α. Like in the case of the Dirac field it is useful to write i explicitly. Consider first the case k = 0, µ ~ kCM = (M, 0) describing a particle in its rest frame. Comparing eq. (7.112) and eq. (7.115) we find the entries of µ i (kCM ) are given by 0 j j i (kCM ) = 0 i (kCM ) = δi . (7.116) i† i In the center of mass, the a (kCM ) and a (kCM ) are thus associated to quanta polarized in the direction i. These three polarization states transform as a J = 1 representation (a vector) under space rotations, corresponding to the particles having spin =1.

EPFL-ITP-LPTP Riccardo Rattazzi 94 7.6. QUANTIZATION OF MASSIVE VECTOR FIELDS The polarization vectors for arbitary kµ are obtained from those in the center of mass via a Lorentz transformation, precisely like in the case of the Dirac field the general ur(k) is obtained by boosting the result for the center of µ ~ ~ mass. This can be seen in a number of ways. Given k = (ωk, k) define first the boost in the direction −k

~k Λ(¯ k)µ ≡ (ei~η(k)·K~ )µ ~η(k) = tanh−1(|k|/ω ) (7.117) ν ν |k| k which, by construction, satisﬁes µ ¯ µ ν k = Λ(k) ν kCM . (7.118) For instance, in the case of ~k = (k, 0, 0) we have

 ωk k  M M 0 0 k ωk µ  0 0  Λ(¯ k) =  M M  . (7.119) ν  0 0 1 0  0 0 0 1

µ µ ν Considering that i (kCM ) satisfy the equation of motion ηµν kCM i (kCM ) = 0 in the CM, the general solutions can be obtained by a boost

µ ν ¯ µ0 ¯ ν0 µ ν µ0 ¯ ν0 0 = ηµν kCM i (kCM ) = ηµ0ν0 Λ(k) µΛ(k) ν kCM i (kCM ) = ηµ0ν0 k Λ(k) i . (7.120)

¯ ν0 where in the last step we used eq. (7.116). The Λ(k) i thus represent a basis of the solutions to the wave equation µ ¯ µ for arbitrary k. Comparing to eq. (7.112) one can indeed check that i (k) is precisely equal to Λ(k) i. The factors ~ ¯ µ ωk/M and k·/M in eq. (7.116) are precisely associated with the entries of the Lorentz boost matrix Λ(k) ν . Notice that, as consequence of this property, the polarization along the direction of motion grows with energy like ωk/M. This fact has an important impact on the high energy behaviour of the scattering amplitudes for massive vector particles: in a generic theory these amplitudes will grow large, signalling the onset of strong coupling. In special theories instead there exist additional compensating eﬀects that tame this growth with energy. The simplest way to implement this compensation is by the introduction of the so-called Higgs boson. The data produced by the Large Hadron Collider indicate that this is the way Nature chose to deal with the problem of growing amplitudes.

µ The polarization vectors i (k) satisfy a set of orthogonality and completeness relations, which are fully analogous α α to those satisﬁed by the Dirac polarization wave functions ur (k) and vr (k). These relation are readily derived µ ¯ µ using the explicit form i (k) = Λ(k) i:

• Orthogonality µ ¯ µ ¯ ν i (k)µj(k) = ηµν Λ(k) iΛ(k) j = ηij = −δij (7.121) • Completeness µ ν µ ν ¯ µ ¯ ν µ0ν0 µ0 ν0 µν k k (k) (k) = Λ(k) 0 Λ(k) 0 (−η + δ δ ) = −η + (7.122) i i µ ν 0 0 M 2 0 ¯ µ µ µ where we used Λ(k) µ0 δ0 = k /M.

One last issue to be discussed concerns the Lorentz transformation properties of ai(k), ai†(k) and of the corresponding Fock states. Notice that the Fock space furnishes an inﬁnite dimensional representation of the Lorentz group. Already the single particle states |~k, ii ≡ ai†(k)|0i form an inﬁnite dimensional representation, given the 3-momentum ~k constitutes a continuous parameter. Moreover, as the generators J µν , the conserved Noether charges, correspond to hermitian operators, the representation

µν U(Λ) = e−iωµν J /2 (7.123) is unitary. That is unlike the case of the defining representation in eq. (3.87) for which the infinitesimal boost generators J 0i in eq. (3.93) are not hermitean. Again this is just an instance of an important mathematical fact: for non-compact Lie groups (such as Lorentz), while finite dimensional representations are not unitary, infinite dimensional representations (such as the one on Fock space) can be unitary.

EPFL-ITP-LPTP Riccardo Rattazzi 95 7.7. POLARISATIONS OF THE VECTOR FIELD Consider then a Lorentz transformation of Aµ(x) Z † µ µ ν −1 h µ ν i −ik(Λ−1x) i U(Λ) A (x)U(Λ) = Λ ν A (Λ x) = dΩk Λ ν i (k)a (k)e + h.c. Z µ ν −1 i −1 −ikx = dΩk Λ ν i (Λ k)a (Λ k)e + h.c. (7.124)

i µ ν −1 µ −1 µ To deduce the transformation property of a (k) we must properly rewrite Λ ν i (Λ k). Using Λ ν = (Λ )ν µ (see eq. (3.84)) and the property pµi (p) = 0 for any p, we have

µ ν −1 −1 ν −1 kµΛ ν i (Λ k) = (Λ k)ν i (Λ k) = 0 . (7.125)

µ ν −1 µ Λ ν i (Λ k) is therefore a linear combination of i (k)

µ ν −1 µ s Λ ν i (Λ k) = s (k)R r(Λ, k) (7.126) where R is some 3 × 3 matrix. It turns out that R is orthogonal and thus describes a spacial rotation. In order to µ ¯ µ r prove that, notice now that, by using r (k) = Λ(k) r we can invert the above equation and write R s as a product r of Lorentz tranformations. More precisely R s corresponds to the space-space 3 × 3 submatrix of ρ ¯ −1 ρ µ ¯ −1 ν R σ(Λ, k) = (Λ (k)) µΛ ν Λ(Λ k) σ (7.127)

ρ σ ρ Now the crucial remark is that R σkCM = kCM . Indeed one has ¯ −1 ρ µ ¯ −1 ν σ ¯ −1 ρ µ −1 ν ¯ −1 ρ µ ρ (Λ (k)) µΛ ν Λ(Λ k) σkCM = (Λ (k)) µΛ ν (Λ k) = (Λ (k)) µk = kCM (7.128)

µ ¯ µ ν ρ where we repeatedly applied the deﬁning relation k = Λ(k) ν kCM . The matrix R σ thus belongs to the subgroup µ of O(3, 1) that leaves kCM ≡ (M, 0, 0, 0) invariant. This group is precisely given by matrices of the general form ρ 1 0 R σ = r (7.129) 0 R s

r where R s ∈ O(3), and simply corresponds to spacial rotations in the center of mass. We can at this point rewite eq. (7.124) as Z Z ν † r −ikx ν r s −1 −ikx dΩk r (k)U (Λ)a (k)U(Λ)e + h.c. = dΩk r (k)R s(Λ, k)a (Λ k)e + h.c. (7.130) which readily implies our long sought result

† r r s −1 U (Λ)a (k)U(Λ) = R s(Λ, k)a (Λ k) . (7.131) According to this equation, Lorentz transformations act, non-trivially, as O(3) rotations on the polarization 3- vector of the vector particle state † r −1 U (Λ)|r, ki = R s(Λ, k)|s, Λ ki (7.132) Notice this is non-trivial as R is a function of both the group element Λ and of the original momentum kµ of the particle. We should also stress that the orthogonality of R is mandated by the unitarity of U(Λ) and the normalization of our states, which in turn follows from the canonical commutation relations hs, p|r, ki = 3 0 s 3 ~ (2π) 2p δr δ (~p − k). The proof is left as an exercise. As a matter of fact we could have just invoked unitarity to conclude R must be orthogonal. Our longer derivation, however, also allowed us to write, at least formally, the expression for R. The matrix R deﬁnes what is called a Wigner rotation.

7.7 Polarisations of the vector ﬁeld

This section, which is partly work in progress, generalizes the previous results for the polarization of a spin-1 particle to the case of a particle of arbitary spin. A thorough discussion in given in volume 1 of Weinberg’s QFT monography.

EPFL-ITP-LPTP Riccardo Rattazzi 96 7.7. POLARISATIONS OF THE VECTOR FIELD 7.7.1 One-particle states

Let us start considering one-particle states whose particles are created and destroyed by two operators :

Annihilation : as(p) s.t. as(p)|0i ≡ 0 (7.133) † † Creation : as(p) s.t. as(p)|0i = |p, si (7.134) where s is a label denoting any other degree of freedom than momentum (e.g. spin). These one-particle states are eigenvectors of the momentum operator :

Pµ|p, si = pµ|p, si (7.135) 3 (3) ~ hp, s|k, ri = 2p0(2π) δ (~p − k)δsr the Lorentz invariant norm (7.136) where we denote by a capital letter an operator and by a lower-case letter its eigenvalue. We will study the transformation properties of the one-particle states under the inhomogenous Lorentz group. In particular we will see that their transformation will depend on their spin and mass (Wigner classiﬁcation). We will ﬁnally consider the case of a massive spin one particle.

Consider the state obtained after application of a quantum Lorentz transformation to a one-particle state : U[Λ]|p, si (7.137)

This new state must be an eigenvector of the momentum operator, its eigenvalue is given by :

−1 ν ν PµU[Λ]|p, si = U[Λ] U [Λ]PµU[Λ] |p, si = Λµ U[Λ]Pν |p, si = Λµ pν U[Λ]|p, si (7.138)

Its eigenvalue being given by Λp, the state must be a linear combination of state vectors with momentum Λp. As the only degree of freedom other than momemtum is s, we have :

X U[Λ]|p, si = Crs(Λ, p)|Λp, ri (7.139) r where the coefficients will depend naturally on the momentum as well as on the Lorentz transformation. We shall now conceptually simplify the problem. Let us assume from now on that the label s denotes exclusively the spin (e.g. for a spin one particle s can take the values −1, 0, +1). It is then clear that on a basis where particle states are labelled by spin, the coefficient functions Crs(Λ, p) cannot mix different spin species and therefore are block diagonal. Furthermore, inside a spin species they cannot mix different irreducible representations (irreps) 1 1 of the Lorentz group (e.g. there is no quantum Lorentz transformation which brings the ( 2 , 0) into a (0, 2 )). Schematically :

 C(0)  ( 1 )  2  Crs =  Cαβ  (7.140)  .  ..

1 (j) ( 2 ) where C is a matrix which mixes one-particle states of spin j. And, for example, Cαβ can be further expanded as :  1  C[( 2 , 0)] ( 1 ) 1 C 2 =  C[(0, 2 )]  (7.141)  .  ..

EPFL-ITP-LPTP Riccardo Rattazzi 97 7.7. POLARISATIONS OF THE VECTOR FIELD We must now identify these coeﬃcient functions for irreps of the inhomogeneous Lorentz group. We will use the method of induced representations to ﬁnd the transformations properties of the one-particle states under the inhomogenous Lorentz group. What we mean by induced representation should become clear shortly.

Lorentz transformations act on the momentum four-vector as :

ν pµ → Λµ pν (7.142)

Now, in the case of proper orthochronous Lorentz transformations there are two functions of pµ that are left 2 5 invariant : its norm squared p as well as the sign of p0 . This implies that we can classify four-momenta 2 2 according to the value p ≡ m and the sign p0 (sign of the energy, or in another interpretation, if it is a particle or an antiparticle). To each of these sets of momenta, we can ascribe a “standard” four-momentum kµ, that is a four-vector from which any four-momentum pµ in its class can be obtained by a Lorentz transformation :

ν pµ = λµ (p)kν (7.143) For example, for a particle of mass m2 ≥ 0 a typical standard four-vector would be its four-momentum in the rest frame. Let us assume without loss of generality that we work with particles only. There are then only three classes of momenta :

Standard momentum kµ 2 2 (a) p = m ≥ 0 (m, 0, 0, 0) (7.144) (b) p2 = 0 (k, 0, 0, k) (c) pµ = 0 (0, 0, 0, 0) By extension, we can deﬁne the states of momentum p to be obtained from the states of standard momentum k by a quantum Lorentz transformation :

|p, si = U[λ(p)]|k, si (7.145)

Consider now one more time the states (7.137):

U[Λ]|p, si = U[Λ]U[λ(p)]|k, si = U[Λ · λ(p)]|k, si = U[λ(Λp)]U[λ−1(Λp) · Λ · λ(p)]|k, si (7.146) where in the second equality we used the group property of quantum Lorentz transformations. The third manipulation seems unnecessary, nonetheless notice that, once the class of standard four-momentum is chosen, the −1 transformation λ (Λp) · Λ · λ(p) acting on kµ leaves it invariant. Indeed, it ﬁrst brings it into p with λ(p), then Lorentz rotates it into Λp to ﬁnally bring it back to k with λ−1(Λp). Why did we want to make this apparent? Be- cause this transformation belongs to the subgroup of the Lorentz group that leaves the corresponding kµ invariant. This subgroup is called the “little group” and has a great importance in many areas of theoretical physics. In our case it will help us to build the Lorentz group irreps’ transformations but it is also central in many applications of the Goldstone theorem which is the basic pilar of the Higgs mechanism. Let us denote a Lorentz transformation belonging to the little group as :

ν kµ → Lµ kν = kµ (7.147) This implies that for any quantum Lorentz transformation whose argument satisﬁes (7.147), we must have :

X U[L]|k, si = Drs[L]|k, ri (7.148) r

5 2 2 Note that the sign of p0 is only conserved if pµ is timelike, i.e. p = m ≥ 0. This means that we deal only with physical particles.

EPFL-ITP-LPTP Riccardo Rattazzi 98 7.7. POLARISATIONS OF THE VECTOR FIELD as the new quantum state can only be a linear combination of states with same momentum k but diﬀerent spin r. As can be easily checked, the coeﬃcients Drs[L] furnish a unitary representation of the little group (check that D[·] is a group homomorphism). Finally, if we return to (7.146) and denoting :

λ−1(Λp) · Λ · λ(p) = L(Λ, p) (7.149) we obtain :

X X U[Λ]|p, si = U[λ(Λp)]U[L(Λ, p)]|k, si = Drs[L(Λ, p)]U[λ(Λp)]|k, ri = Drs[L(Λ, p)]|Λp, ri (7.150) r r

We see now why we previously made a small detour through the little group construction. Indeed, the problem of ﬁnding irreps of the Lorentz group and by extension determining their transformation properties through the coeﬃcients Crs(Λ, p), can be now shrunk to considering the irreps of the little group which are in general much easier to construct and whose transformation properties are rather simple. This construction is the so-called “induced representation construction” as from the irreps of the little group we induce the transformation properties of a bigger group.

We now begin our example by considering a particle falling in class (a) in table (7.144). We must ask ourselves, what are the transformations that leave the four-vector kµ = (m, 0, 0, 0) invariant? That problem can be solved algebraically by writing down the most general four-by-four Lorentz transformation matrix, imposing on its parameters (7.147) and then finding to which group the matrix that we obtain corresponds. However there are in general much quicker and easier ways to find out. In our case, recall that Lorentz transformations are made out of boosts and three-dimensional rotations. On the one hand, boosts obviously do not leave kµ invariant. On the other hand, SO(3) rotations clearly do and the latter corresponds to our little group6. The unitary representations of the three-dimensional rotations group are well known and the block diagonal form of the coefficients Drs will be given by :

M (j) Drs = Drs (7.151) j where j is a positive semi-integer denoting the spin of the representation. For each block, the values s, r correspond- (j) ing to spin j run from −j, −j + 1, . . . j − 1, j. Matrices Drs are built from the SO(3) generators of inﬁnitesimal rotations Rik = δik + Θik, with Θik = −Θki :

(j) i (j) Drs [1 + Θ] = δrs + Θik Jik (7.152) 2 rs (j) (j) (j) (j) (j) Jik = iklJl [Jm ,Jn ] = imnlJl (7.153) We hence found the transformation properties of any massive particle of spin j :

X (j) U[Λ]|p, j, si = Drs [L(Λ, p)]|Λp, j, ri (7.154) r

7.7.2 Operators and ﬁelds transformations

With the transformation rule (7.154) in hand, we can ﬁnd how the creation and annihilation operators transform under quantum Lorentz transformations. Indeed, we have on one hand :

6This can be seen from a group theoretical point of view. Indeed any four-vector vµ decomposes under SO(3) representations as 0 ⊕ 1 where the scalar is v0 and the vector is vi. Since in our case the vector part is absent, the four-vector is clearly SO(3) invariant.

EPFL-ITP-LPTP Riccardo Rattazzi 99 7.7. POLARISATIONS OF THE VECTOR FIELD

X (j) X (j) † U[Λ]|p, j, si = Drs [L(Λ, p)]|Λp, j, ri = Drs [L(Λ, p)]ar(Λp)|0i (7.155) r r and on the other hand :

† † −1 † −1 U[Λ]|p, j, si = U[Λ]as(p)|0i = U[Λ]as(p)U [Λ]U[Λ]|0i = U[Λ]as(p)U [Λ]|0i (7.156) where in the last equality we used the fact that the vacuum is Lorentz invariant. Putting (7.155) and (7.156) together, we obtain :

† −1 X (j) † U[Λ]as(p)U [Λ] = Drs [L(Λ, p)]ar(Λp) (7.157) r This can be rewritten using the unitarity of the little group representations as :

−1 X (j) −1 U[Λ]as(p)U [Λ] = Dsr [L (Λ, p)]ar(Λp) r (7.158) † −1 X ∗(j) −1 † U[Λ]as(p)U [Λ] = Dsr [L (Λ, p)]ar(Λp) r

Finally, it remains to consider how the ﬁelds transform under the Lorentz group. Recall that any quantum ﬁeld can be written in terms of annihilation and creation operators as :

+ − ψa(x) = ψa (x)+ψa (x) (7.159) Z Z + X −ipx − X † +ipx ψa (x) = dΩpua(p, s)as(p)e ψa = dΩpva(p, s)as(p)e (7.160) s s

where the coeﬃcient functions ua(p, s) and va(p, s) are chosen and normalised so that under Lorentz transformations the ﬁeld is rotated by a Lorentz matrix dependent on its corresponding irrep :

−1 X −1 U[Λ]ψa(x)U [Λ] = Dab[Λ ]ψb(Λx) (7.161) b

The matrices Dab furnish a representation of the homogenous Lorentz group. Note that these do not have to be unitary. So far we have seen three of these representations, the scalar D[Λ] ≡ 1, the spinor D[Λ] ≡ Λ 1 and the 2 µ vector D[Λ] ≡ Λ ν . We will from now on only consider the real massive spin 1 vector case, so that :

ψa(x) →Aµ(x) (7.162) ∗ uµ(p, s) = vµ(p, s) = µ(p, s) (7.163)

We saw in equation (7.158) how creation and annihilation operators transform under quantum Lorentz transformations. If we add the transformation rule (7.161) of ﬁelds we obtain :

Z ! Z ! X −1 −ipx X ν −ip·(Λx) dΩp µ(p, s)Dsr[L (Λ, p)]ar(Λp)e + h.c. = dΩp Λµ ν (p, s)as(p)e + h.c. (7.164) sr s

EPFL-ITP-LPTP Riccardo Rattazzi 100 7.7. POLARISATIONS OF THE VECTOR FIELD where we dropped the index indicating that the matrices Drs belong to the spin 1 representation of SO(3). Since we want the extract the transformation properties of the coeﬃcient functions µ we must have the ladder operators as well as the Fourier exponentials evaluated at the same momenta. In order to do so, recall that dΩp is Lorentz invariant so that in the LHS we can simply replace dΩp by dΩΛp. Furthermore, in the RHS we can change the integration variable p → Λp recalling that p · (Λx) = (Λ−1p) · x so that :

Z ! Z ! X −1 −ipx X ν −ipx dΩΛp µ(p, s)Dsr[L (Λ, p)]ar(Λp)e + h.c. = dΩΛp Λµ ν (Λp, s)as(Λp)e + h.c. sr s (7.165)

We ﬁnally obtain7 :

X ν µ(Λp, r)Drs[L(Λ, p)] = Λµ ν (p, s) (7.166) r

This equation will allow us to compute the coeﬃcient functions µ.

7.7.3 Polarisations

We now compute the coeﬃcient functions µ. In order to do so, we ﬁrst explore the consequences of (7.166). If we take pµ = kµ and let Λ be a standard boost that brings the particle initially at rest into a state of momentum qµ, our little group transformation L will be given by :

L[Λ, p] = λ−1(Λp)Λλ(p) = λ−1(q)λ(q) = 1 (7.167) as λ(p = k) = 1. Now the matrix Drs[1] ≡ 1 so that :

ν µ(q, s) = λ(q)µ ν (k, s) (7.168)

This means that if we now the coeﬃcients µ at at zero spatial momentum, we can obtain their general form by the use of a standard boost. Now consider again pµ = kµ but let this time Λ be a pure rotation :

L[Λ, p] = λ−1(Λp)Λλ(p) = Λ ≡ R (7.169)

So that :

X ν µ(k, r)Drs[R] = Rµ ν (k, s) (7.170) r This equation can be written in inﬁnitesimal form using the generators of rotations :

X ~ ~ ν µ(k, r)Jrs = Jµ ν (k, s) (7.171) r

~ ν where Jµ are the angular momentum matrices in the vector representation. Their only non zero components are : 7The reader is strongly advised to understand all the steps necessary to go from equation (7.165) to equation (7.166)

EPFL-ITP-LPTP Riccardo Rattazzi 101 7.7. POLARISATIONS OF THE VECTOR FIELD

j (Jk)i = −iijk (7.172) 2 j j J i = 2δi (7.173) Using (7.171) twice we obtain :

X 2 0(k, r)Jrs = 0 (7.174) r X 2 i(k, r)Jrs = 2i(k, s) (7.175) r

2 which, recalling that Jrs = j(j + 1)δrs, gives us two possibilities : j = 0 or j = 1. As we previously said we consider here a spin 1 vector, so that we can discard the former case. It is nevertheless reassuring that we indeed recover the decomposition of a vector into 0 ⊕ 1 representations of SO(3). In the latter case, equations (7.174) and (7.175) imply that for the particle at rest only the spatial components i are non-zero. Now recall :

(J3)rs = sδrs (7.176) so that for s = 0 we have :

ν 0 = (J3)µ ν (k, 0) (7.177) which means that rotations around the z-axis leave the coeﬃcient function invariant, using a suitable normalisation of the ﬁelds8 :

µ(k, 0) = (0, 0, 0, −1) (7.178) It remains to ﬁnd the spin ±1 components. These are naturally obtained using the the raising and lowering operators (J1 ± iJ2) on µ :

1 µ(k, ±1) = √ (0, −1, ∓i, 0) (7.179) 2

We can now obtain these functions at any general momentum we wish by applying a standard boost to them : ν µ(p, s) = λ(p)µ ν (k, s) (7.180)

The three vectors µ(p, 0) and µ(p, ±1) are called the polarisations of the vector ﬁeld. They satisfy the transverse property : µ µ(p, s)p = 0 (7.181) which follows from the transversality of the vector ﬁeld. Another important property is :

X µ(p, s)µ∗(p, s) = −ηµν + pµpν /m2 ≡ Πµν (p) (7.182) s where m is the mass of the vector ﬁeld and Πµν (p) is the projector of the transverse space to p :

µν pµΠ (p)vν = 0 ∀vν (7.183)

8We follow here the conventions in Peskin & Schroeder

EPFL-ITP-LPTP Riccardo Rattazzi 102 Chapter 8

Discrete symmetries

8.1 Introduction

As we have seen, symmetries play an essential role in Quantum Field Theory (QFT). So far, we have mostly concentrated on continuous symmetries (i.e. Lie groups) for which the main dynamical consequences are Noether theorem and of course the presence of a degeneracy of solutions related by symmetry transformations. We have already brieﬂy encountered examples of discrete (as opposed continuous) symmetries when discussing the Lorentz group:

• Parity: P :(t, ~x) 7→ (t, −~x) (8.1) • Time Reversal: T :(t, ~x) 7→ (−t, ~x) (8.2)

These symmetries are familiar already from non-relativistic physics. In relativistic QFT there is a third important discrete symmetry which does not act on spacetime but which exchanges particles and antiparticles:

nParticleso Antiparticles xµ xµ • Charge Conjugation: C : 7→ ∗ (8.3) φa(x) φa(x)

The free field Lagrangians we have been considering so far are all invariant, not only under the proper orthochronous Poincaré group (proper orthochronous Lorentz group plus space-time translations), but also under the three discrete symmetries P , T and C. Basically, for the free lagrangians, the latter symmetries arise gratis from the invariance under the proper orthochronous Poincaré group. More specifically, the lagrangians for free scalar fields, free vector fields and free Dirac fields are all invariant under P , T and C, separately, while the lagrangian for the free massless Weyl spinor is invariant under T and under the combined action of parity and charge conjugation CP . The case of the Weyl spinor is special, in that C and P cannot even be independently defined. Indeed parity must act on it by exchanging handedness:

P ∗ ψL(x) ∈ (1/2, 0) −→ εψL(xP ) ∈ (0, 1/2) (8.4) where xP = (t, −~x). The original ﬁeld and the parity reﬂected one transform under U(1) (phase rotations) as:

iq ∗ −iq ∗ ψL → e ψL εψL → e εψL (8.5) We therefore see that the parity transformed ﬁeld has opposite charge (q → −q): P and C cannot act independently, but only together as a CP transformation. One can check that the free lagrangian for the Weyl ﬁeld is invariant under this CP transformation. The system is then only invariant under CP but not under P nor C separately!

103 8.2. PARITY The presence of the above discrete symmetries in free ﬁeld theories can however be considered an accidental fact. More generally, while preserving invariance under the proper orthochronous Poincaré group, all the three discrete symmetries can in principle be broken in an interacting theory. There is however a theorem stating that, in a local quantum ﬁeld theory invariant under the proper orthochronous Poincarè group, the combined transformation:

Θ = CPT (8.6) is always a symmetry. This is the CPT theorem (Schwinger ’51, Lüders and Pauli ’54). These are theoretical results obtained in the QFT framework, but how do they compare to experimental observations? The proper orthochronous Poincaré group as well as Θ are symmetries of Nature as far as we can see. Quantum Electrodynamics (QED) respects P , T and C at leading order as “accidental symmetries” very much like the free Lagrangians respect them. Quantum Chromodynamics (QCD, theory of strong interactions) respects both T and P . This is a surprising puzzle (Strong CP problem) as one could in principle add in QCD an interaction term that breaks T and P , but apparently Nature chose not to do so. However, neither T nor P are symmetries of the Standard Model (SM) of particle physics, which encompasses all the known interactions. Indeed, Electroweak (EW) interactions violate parity in a maximal fashion (as the building blocks of the SM are Weyl fermions for which we can define CP but not P ) and violate time reversal through small effects associated to quark flavor violation (for instance in Kaon oscillations).

8.2 Parity

8.2.1 Foreword

In classical mechanics a parity transformation reverses the direction of the canonical variables:

~q −~q P : 7→ (8.7) ~p −~p

Quantities transforming in the same way are called polar vectors or simply vectors. On the other hand, quantities like the angular momentum have a diﬀerent transformation rule:

P : L~ = ~q ∧ ~p 7→ L~ (8.8)

These are called axial vectors or pseudovectors. From here we can proceed to a canonical quantisation of the system passing from Poisson brackets to canonical commutation relations (CCRs):

{qi, pj} = δij → [ˆqi, pˆj] = iδij (8.9)

It is obvious that parity transformations will preserve the CCRs and therefore the operation P (which acts on spacetime) can be realised by an unitary operator UP on the Hilbert space (Wigner’s theorem):

† qi −qi UP UP = (8.10) pi −pi

Finally, parity can also be realised on quantum ﬁelds analogously to quantum Lorentz transformations:

† −1 P : φa(x) 7→ UP φa(x)UP = Pabφb(P x) (8.11) Now notice that P 2 = 1, indeed:

P P (t, ~x) −→ (t, −~x) −→ (t, ~x) (8.12)

EPFL-ITP-LPTP Riccardo Rattazzi 104 8.2. PARITY −1 so that P x = P x ≡ xP and the two elements {1,P } form the group Z2. A quantum realisation of the parity operation is a group homomorphism:

UP · UP = UP 2 = U1 = 1 (8.13) † UP = UP −1 = UP (8.14) and therefore the matrices Pab are a representation of the group :

† † PabPbcφc(x) UP UP φa(x)UP UP = (8.15) φa(x) so that:

PabPbc = δac (8.16)

We can diagonalise these matrices and use (8.16) to obtain:

(8.16) 2 Pab = ηaδab −→ ηa = 1 (8.17) where no sum is intended in the ﬁrst equation. Hence, we obtain the eigenvalues of the matrix:

ηa = ±1 (8.18)

These eigenvalues are called the intrinsic parities of the ﬁelds.

We should however add that when dealing with Majorana fermions parity cannot be represented via a “strict” 2 homomorphism. In that case UP satisfies UP = −1. The fact that acting twice with a parity transformation does not bring a Majorana fermion back to itself should not come as a complete shock: we already know fermions flip signs under a 2π rotation. Physical observables (like for instance the energy momentum tensor) are however composed of even powers of the fermion fields, so that parity acts in the standard way on them.

8.2.2 Spin 0

µ 0 µ 0 In the following we indicate the parity tranform of a 4-vector v ≡ (v ,~v) by vP ≡ (v , −~v). We start by considering the case of a real scalar:

† UP φ(x)UP = ηP φ(xP ) (8.19) where ηP = 1 corresponds to a scalar and ηP = −1 corresponds to a pseudoscalar. Notice that that the Klein- Gordon lagrangian is invariant under the above parity tranformation, regardless of the sign of ηP . Consequently the Klein-Gordon ﬁeld equation is also invariant, as one can easily check. More generally parity invariance of a lagrangian of the form (∂φ)2 + V (φ) requires

V (φ) = V (ηP φ) (8.20) which is always true for ηP = 1, while for ηP = −1 it requires V (φ) to be an even function of φ. Focussing on the Klein Gordon field, we now want to figure out how parity acts on the Fock space. In order to do so, consider the plane waves expansion of the fields:

EPFL-ITP-LPTP Riccardo Rattazzi 105 8.2. PARITY

Z † † −ikx UP φ(x)UP = dΩk UP akUP e + h.c. = ηP φ(xP ) (8.21)

Now it suﬃces to notice that the scalar product in the plane waves of the RHS can be rewritten simply:

k · xP = kP · x (8.22)

~ ~ so that by a change of variables in the integral of the LHS k ↔ −k (k ↔ kP ) we obtain :

† U akUP = ηP ak P P (8.23) U † a† U = η a† P k P P kP

The second equation can be obtained either by looking at the hermitian conjugate part of the plane waves expansion or simply by “taking the dagger” of the first equation and recalling that ηP is real. The case of many real scalar fields is a trivial generalisation of the latter but it is sensible to look at the charged scalar case. We define the parity transformation so that it does not flip the charge:

† UP φ(x)UP = ηP φ(xP ) (8.24) † ∗ ∗ UP φ (x)UP = ηP φ (xP ) (8.25) instead of:

† ∗ UP φ(x)UP = ηP φ (xP ) (8.26) Here particles and antiparticles have the same intrinsic parity. In free ﬁeld theory the vacuum is invariant under parity transformations:

UP |0i = |0i (8.27) and the quantum parity operator UP can be explicitly constructed out of the creation and annihilition operators. We can also check that the quantum operators of the Hilbert space transform as expected. For instance the four-momentum:

Z Z Z U † P µU = dΩ kµU † a† a U = dΩ η2 kµa† a = dΩ kµ a† a = P µ ≡ P (8.28) P P k P k k P k P kP kP k P k k P µ

Similarly, one can check that:

† i i Boosts: UP K UP = −K (8.29) † i i Rotations: UP J UP = J (8.30) So that we have:

† i i UP J±UP = J∓ (8.31)

EPFL-ITP-LPTP Riccardo Rattazzi 106 8.2. PARITY 8.2.3 Spin 1/2

Let us now focus on just one Dirac spinor. We want to implement a parity transformation on it:

† UP ψα(x)UP = ηP Pαβψβ(xP ) (8.32)

s where ηP is an overall intrisic parity factor. We need to ﬁnd the matrix Pαβ and the action of parity on bp and s dp. We already know from the discussion on the Dirac spinor how parity acts on the latter. Let us recall some facts. Since parity acts on the Lorentz group generators as:

i i P : J± 7→ J∓ (8.33) we have that:

P P : ψα ∈ (j1, j2) 7→ ψα ∈ (j2, j1) (8.34)

This implies that we cannot represent a parity transformed object from (j1, j2) into the same basis (unless j1 = j2). Therefore ψL and ψR separately do not form a basis for a representation of parity. This is why parity is broken in the Standard Model (SM) as these two representations appear diﬀerently in it. However there exists a spinor representation which is also a representation of parity, the Dirac spinor:

ψL ΛL 0 ΨD ≡ ψL ⊕ ψR = ΛD = (8.35) ψR 0 ΛR

Any parity transformation will act as:

ψL(x) ψR(xP ) P : 7→ ηP (8.36) ψR(x) ψL(xP )

So that:

† UP ΨD(x)UP ≡ ηP γ0ΨD(xP ) (8.37)

Thus we found our matrix Pαβ and it only remains to ﬁnd its action on the creation and annihilation operators. We proceed in the exact same way as before. We write (8.37) in terms of plane waves and do a change of variables in the LHS to obtain:

U † br U ur(k) = η br γ ur(k ) (8.38) P k P p kP 0 P U † d†rU vr(k) = η d†r γ vr(k ) (8.39) P k P p kP 0 P Now recall the form of the polarisation spinors we obtained:

√ k · σξs us(k) = √ (8.40) k · σξ¯ s √ k · σξs vs(k) = √ (8.41) − k · σξ¯ s

EPFL-ITP-LPTP Riccardo Rattazzi 107 8.2. PARITY and since kP · σ ≡ k · σ¯ it is straightforward to see that:

s s γ0u (kP ) = u (k) (8.42) s s γ0v (kP ) = −v (k) (8.43)

Hence we ﬁnally obtain:

† s s U b UP = ηpb P k kP (8.44) U † d†sU = −η d†s P k P p kP

These equations imply that particles and antiparticles have opposite intrinsic parities, fact that has big phe- nomenological consequences. Moreover the transformations we just derived imply that:

S~ S~ P (8.45) P~ −→ −P~

Let us now give an example of parity violation in Weak interactions. Consider the following decay of a pion into a muon and a muon-antineutrino:

π− → µ− +ν ¯µ (8.46)

The pion is a scalar particle and hence has spin 0. Both the muon and its antineutrino are fermions of spin 1/2. Since the pion has no orbital angular momentum in its center-of-mass frame we have:

X − µ 0 = J~init = J~final = L~ (i) + S~(i), i = µ , ν¯ (8.47) i

This equation can be projected on ~p(µ−) = −~p(¯νµ) ≡ ~p. Recalling that L~ (i) · ~p(i) = 0 we thus obtain

" # X 0 = ~p · L~ (i) + S~(i) = ~p(µ−) · S~(µ−) − ~p(¯νµ) · S~(¯νµ) (8.48) i and conclude that the muon and its antineutrino should have the same helicity

~p(µ−) · S~(µ−) ~p(¯νµ) · S~(¯νµ) h(µ−) ≡ = ≡ h(¯νµ) (8.49) |~p(µ−)| |~p(¯νµ)|

Now, as one is easily convinced, the composition of parity with a 180◦ rotation around any axis orthogonal to ~p, leaves the final state momenta unaffected while flipping the helicities. As the original state, a pion at rest, is invariant under such tranformation, if parity were a symmetry we would observe with the same probablity (50%) muons with the two opposite helicities: P (+1/2) = P (−1/2). Instead we only observe pions with helicity +1/2, indicating that parity is maximally violated in this process!

This experimental result has a simple theoretical explanation in the SM. There, neglecting the absolutely minis- cule eﬀects of the neutrino mass, the neutrino is described by a massless Weyl spinor, for which the particle (the neutrino) has helicity −1/2 and the antiparticle (the antineutrino) has helicity +1/2: in this case, as we already mentioned, only CP is deﬁned, and neither P nor C make sense independently. The practical manifestation of

EPFL-ITP-LPTP Riccardo Rattazzi 108 8.2. PARITY the impossibility to deﬁne parity is that antineutrinos come only in one helicity, so that by helicity conservation (eq. (8.49)) the accompanying muon must always carry the same helicity +1/2. We should add here that experimentally we mostly observe the muon as the neutrino interacts very weakly and is thus very hard to detect.

Four fermions bilinears are mostly important when dealing with the phenomenology of fermions in the Standard Model. Their transformation properties under parity are as follows:

Bilinear P : · 7→ ΨΨ¯ ΨΨ¯ scalar ¯ ¯ Ψγ5Ψ −Ψγ5Ψ pseudoscalar (8.50) ¯ ¯ µ ΨγµΨ Ψγ Ψ vector ¯ ¯ µ Ψγ5γµΨ −Ψγ5γ Ψ axial vector

8.2.4 Spin 1

We ﬁnally consider spin one particles. Parity acts on vectors as:

† µ UP VµUP = ηP V (8.51) where ηP = 1 corresponds to a polar vector and ηP = −1 corresponds to an axial vector. One more time, we use the plane waves expansion in terms of polarisation vectors to obtain:

U † ar U (k, r) = η ar (k , r) (8.52) P k P µ p kP µ P Recall that the polarisation vectors are given as a pure boost acting on the rest frame vectors (the standard momentum is denoted q here):

ν µ(k, s) = λ(k)µ ν (q, s) (8.53)

So that:

ν −1 ν µ(kP , s) = λ(kP )µ ν (q, s) = λ (k)µ ν (q, s) (8.54)

In order to to relate µ(k, s) to µ(kP , s), we need to use some identities of the Lorentz boosts:

−1 ν ν µ λ (k)µ = λ(k) µ ≡ λ(p) ν (8.55)

The second equality comes from the fact that for a boost the matrix is symmetric. Finally recalling that the rest frame polarisation vectors are purely spatial, we have:

ν ν (q, s) ≡ − (q, s) (8.56)

Therefore:

ν µ ν µ µ(kP , s) = λ(kP )µ ν (q, s) ≡ −λ(k) ν (q, s) ≡ − (k, s) (8.57)

EPFL-ITP-LPTP Riccardo Rattazzi 109 8.3. TIME REVERSAL Using this equation together with (8.51) and (8.52), we ﬁnally obtain:

U † ar U = −η ar (8.58) P k P p kP

We give now another derivation of the transformation properties of the ladder operators. Consider a gauge in which the vector is purely spatial. Then:

Z ω A~(x) = dΩ e−ikx ~a (k) + ~a (k) + h.c. (8.59) k ⊥ M L Parity transformations (8.51) become here:

† ~ ~ UP A(x)UP = −ηP A(xP ) (8.60)

Using the same change of variables as usual, it gives us:

† UP ~akUP = −ηP ~akP (8.61)

8.3 Time reversal

8.3.1 Foreword

In classical physics time reversal changes the direction of time while leaving space invariant as its name indicates. This implies that the transformation properties of the quantities of interest of the theory are:

~q ~q T : 7→ (8.62) ~p −~p The transformation property of the conjugate momenta can be easily understood as roughly they describe the change in time of their canonical variable. This implies that, for example:

T : L~ = ~q ∧ ~p 7→ −L~ (8.63)

So that the spin follows the same transformation law:

T : S~ 7→ −S~ (8.64)

Finally, the reader should have noticed that time reversal does not preserve the Poisson brackets between canonical variables:

T : {qi, pj} = δij 7→ {qi, −pj} = −δij (8.65) Hence, when we quantise the theory the operation T cannot be represented by an unitary operator. Nevertheless, by Wigner’s theorem it can be represented by an antiunitary operator AT . Let us recall some deﬁnitions:

EPFL-ITP-LPTP Riccardo Rattazzi 110 8.3. TIME REVERSAL Deﬁnition. An operator A : H 7→ H is antilinear if:

Let us list some properties antilinear operators sarisfy:

Properties. 1. If two operators A, B : H 7→ H are antilinear, then their product AB : H 7→ H is a linear operator.

2. If an operator A : H 7→ H is antilinear, then:

† hψ1|Aψ2i ≡ hψ2|A ψ1i (8.67)

The reader should prove these two properties as an exercise. Finally, we are interested in antiunitary operators:

Deﬁnition. An operator A : H 7→ H is antiunitary if it is antilinear and it satisﬁes:

A†A = AA† = 1 (8.68)

Now that we have these deﬁnitions in hand, we can study certain properties of time reversal transformations. Nonetheless, notice before that as for parity we have T 2 ≡ 1, so that T ≡ T −1 and:

AT · AT = AT 2 = A1 = 1 (8.69) † AT = AT −1 = AT (8.70)

Operators on the Hilbert space will transform under time reversal as:

† ∗ T : O 7→ AT OAT , ∀O ∈ H ⊗ H (8.71)

In particular, covariance of the time evolution of the system is in a one-to-one correspondance with the fact that its Hamiltonian is invariant under time reversal:

8.3.2 Spin 0

We start again considering the case of a real Klein-Gordon scalar:

† AT φ(x)AT = ηT φ(xT ) (8.73) where ηT = ±1 and xT = (−t, ~x) = −xP . We want to ﬁgure out how parity acts on the Fock space. In order to do so, consider the plane waves expansion of the ﬁelds:

EPFL-ITP-LPTP Riccardo Rattazzi 111 8.3. TIME REVERSAL

Z Z † † −ikx † +ikx AT φ(x)AT = dΩk AT ake AT + h.c. = dΩk AT akAT e + h.c. = ηP φ(xT ) (8.74)

Now it suﬃces to notice that the scalar product in the plane waves of the RHS can be rewritten simply:

k · xT = −k · xP = −kP · x (8.75) ~ ~ so that by a change of variables in the integral of the LHS k ↔ −k (k ↔ kP ) we obtain :

† A akAT = ηT ak T P (8.76) A† a† A = η a† T k T T kP

And states transform as:

A |ki = A a† A A |0i = η a† |0i = η |k i (8.77) T T k T T T kP T P

8.3.3 Spin 1/2

In the case of fermions, time reversal should reverse both spin and momentum. Let us take a guess working with a Weyl fermion:

χ1 χα = (8.78) χ2 where each component χα carries either spin up or down depending on the nature of the field (particle or antiparticle). We would like the transformed field to have opposite spin and hopefully we have a naturally well suited object at our disposal in order to invert the field components, the spinor metric αβ. As we previously saw, time reversal of x together with the antilinearity of AT implied the right transformation on p, thus an educated guess is:

† α αβ β α AT χ (x)AT = ηT χ (xT ) ≡ χT (8.79)

Let us see why this is a good guess:

• It exchanges the spinor components, which implies:

S~ → −S~ (8.80)

• It does not change the charge of the ﬁeld, under the QED U(1) gauge group:

iq iq χ 7→ e χ ⇔ χT 7→ e χT (8.81)

The reader will check as an exercise that this transformation indeed leaves the Weyl Lagragian invariant and will deduce the transformation of the Dirac spinor. In the end, the transformation law of the Dirac ﬁeld is:

† 0 2 AT ΨD(x)AT = ηT T ΨD(xT ) ≡ ηT iγ5γ γ ΨD(xT ) (8.82)

EPFL-ITP-LPTP Riccardo Rattazzi 112 8.4. CHARGE CONJUGATION and for the ladder operators:

† s sr r A b AT = ηT b T k kP (8.83) A† ds A = η srdr T k T T kP

8.3.4 Spin 1

The transformation rules of the spin one ﬁeld will be derived in a simple way below. For further details the reader should refer to the exercises. As in the case of parity transformations, when the ﬁeld is expressed only in terms of its spatial components, we have:

† ~ ~ AT A(x)AT = ηT A(xT ) (8.84)

Using the same change of variables as usual, it gives us:

† AT ~akAT = ηT ~akP (8.85)

8.4 Charge conjugation

8.4.1 Foreword

Charge is an internal quantity associated with any gauge symmetry of the theory. In particular, in the case of U(1) symmetry (e.g. QED), any general ﬁeld which is charged under the group will be (gauge-)rotated as:

U(1) iq φa(x) −→ e φa(x) (8.86)

Charge conjugation is deﬁned as:

∗ C : φa(x) 7→ ηC Cabφb (x) (8.87) q 7→ −q (8.88)

2 2 with C = 1, so that |ηC | = 1. More generally, charge conjugation maps all internal quantum numbers (e.g. electric charge, baryon number, lepton number, etc.) into their opposite. Notice that charge conjugation transformations commute with Lorentz transformations as the former do not aﬀect spacetime. It is represented by a unitary operator which as usual has the following properties:

UC · UC = UC2 = U1 = 1 (8.89) † UC = UC−1 = UC (8.90)

EPFL-ITP-LPTP Riccardo Rattazzi 113 8.4. CHARGE CONJUGATION 8.4.2 Spin 0

Firstly we consider the case of scalar fields. Real scalar fields are of no interest as they do not carry any charge. However their complex extension does and charge conjugation is defined as:

† ∗ UC φ(x)UC = ηC φ (x) (8.91)

As it is expected the transformation of the ladder operators is (exercise):

† UC akUC = ηC bk (8.92)

Notice that in general we could have ηC ∈ C, but without loss of generality we can have ηC ∈ R or even ηC = 1, indeed we can make a ﬁeld redeﬁnition:

iθ φ −→ e φ ≡ φθ (8.93) so that:

† iθ † iθ ∗ 2iθ ∗ UC φθUC = e UC φUC = e ηC φ = e ηC φθ (8.94)

2iθ 0 2iθ ∗ 0 Therefore we can deﬁne e ηC ≡ ηC ∈ R or e = ηC so that ηC = 1. Lastly, recall that the complex ﬁeld can be rewritten as:

φ ≡ φ1 + iφ2, φi ∈ R (8.95)

In terms of the two real ﬁelds, U(1) phase rotations become SO(2) rotations and charge conjugation acts as:

φ φ C : 1 7→ 1 (8.96) φ2 −φ2

Hence charge conjugation acts as a mirror symmetry on the ﬁeld space about their second component.

8.4.3 Spin 1/2

We start with two Weyl fermions since, as we saw in the introduction, charge conjugation cannot be represented on a single Weyl fermion. Charge keeps handedness but exchanges particles and antiparticles, therefore Weyl spinors will transform as:

∗ ψL ηLψR C : 7→ ∗ (8.97) ψR ηRψL

That the Weyl Lagrangian is invariant is left as an exercise. Its invariance, considering real ηs implies:

ηLηR = −1 (8.98)

EPFL-ITP-LPTP Riccardo Rattazzi 114 8.4. CHARGE CONJUGATION so that without loss of generality ηL = −ηR = 1. The extension of these transformations to Dirac spinors is straightforward:

2 ∗ 2 0 ¯ T ¯ T C :ΨD 7→ ηC iγ ΨD = ηC iγ γ ΨD ≡ ηC CΨD (8.99) where we define the matrix C ≡ iγ2γ0. The reader should be aware that the matrix C so defined does not correspond to a proper charge conjugation matrix (as it can be seen when we compute its hermitian conjugate or its square) but is a mathematical trick so that the charge conjugation of the field yields its “bar” conjugate which enters naturally the Dirac Lagrangian. A defining property of the matrix which allows us to compute it in any basis is:

CT γµT C = −γµ (8.100)

Equation (8.99) translates itself into the transformation rule for the ladder operators (with ηC = 1):

† s sr r UC bkUC = − dk (8.101) s s sr r As it is known, if bk destroys a particle of spin s, then dk destroys an antiparticle of spin −s, therefore − dk destroys an antiparticle of spin s as expected. Let us then deﬁne new creation and annihilation operators for the antiparticles:

sr r ˜s − dk → dk (8.102)

˜s so that dk eﬀectively destroys a antiparticle of spin s. The new transformation rules become:

† s ˜s UC bkUC = dk (8.103)

8.4.4 Spin 1

The transformation properties of the vector ﬁeld can be extracted very easily from the invariance of the QED Lagrangian under charge conjugation. The interacting term connects the four-vector potential to the current:

µ Lint ∝ J Aµ (8.104) Since by deﬁnition the current changes sign under charge conjugation:

C : J µ 7→ −J µ (8.105) we must have:

† µ µ UC A UC = −A (8.106) which straightforwardly leads to:

† r r UC akUC = −ak (8.107) This shows how the photon is its own antiparticle.

EPFL-ITP-LPTP Riccardo Rattazzi 115 Chapter 9

Interacting Fields

9.1 Asymptotic States

One basic result of our study of free ﬁeld theories is the structure of their Hilbert space H of states, the Fock space:

H = span (|0i, |ki, |k1, k2i,..., |k1, . . . kni,...) (9.1)

In reality, particles are associated, in position space, with suitable wave packets. The ith particle can be associated i with a blob around position x0. For example, for a one particle state we have:

Z ˆ |fi = dΩkf(k)|ki (9.2) where fˆ(k) represents the Fourier transform of its wave function. In position space, the distribution probability 2 2 2 will be given by hf|φ (x)|fi − h0|φ (x)|0i ∼ |f(x − x0)| as depicted in Fig. 9.1.

Figure 9.1: Blob associated to one particle in position space

One crucial feature of free field theory is that the presence of one particle does not affect the others, i.e. the superposition principle applies exactly. Indeed, the equations of motion of a free field theory (quadratic Lagrangian) are linear in the fields, so that if φ1 and φ2 are two solutions, so is φ = φ1 + φ2. This implies that even if two wave packets travel towards each other, Fig. 9.2, after overlapping their shape and direction is not affected and they will pursue their journey as if they never crossed, Fig. 9.3

116 9.1. ASYMPTOTIC STATES

x0 y0

Figure 9.2: Two wave packets travelling towards each other that will overlap in future.

y0 x0

Figure 9.3: After crossing each other, the wave packets follow their original path without being disturbed.

In physical reality, however, particles do interact and the wave packets of the previous example are affected by each other. On the other hand, if the range of the interaction is finite or decreases fast enough with distance, it is intuitive and reasonable to expect that the states are well described by individual particle wave packets of a free field theory, provided the mutual separation of these wave packets is sufficiently large. In other words, well separated wave packets are still described by a Fock space of states characterised by a set of particles types {A} with quantum numbers {mA, sA,QA,...} and associated to a Lagrangian and a Hamiltonian

X A X A ∃ L0 = L0 ⇒ H0 = H0 (9.3) A A

A A where by L0 (H0 ) we indicate the free Lagrangian (Hamiltonian) describing particle type A. Supported by empirical observation and by direct computation whenever applicable (basically in quantum ﬁeld theories that can be studied with the use of perturbation theory), what we stated just above will be our basic working assumption in the following.

The Hilbert space H0 = {|φαi} is built up such that:

µ µ P0 |φαi = pα|φαi (9.4) where α denotes all quantum numbers, continuous and discrete (we adopt here the notation of Weinberg, Chapter 3 of Volume I). Well separated wave packets are described by superpositions of these states:

Z |φi = dα g(α)|φαi (9.5)

These wave packets can meet at a certain time and thus interact. It is essential to think of wave packets since otherwise the states are fully delocalised, provoking interactions at all times. The typical situation which is realised in particle physics experiments is the following: at t → −∞ all separated wave packets (well described by the above state) move towards the collision point. At finite time, interactions (perhaps very complicated) take place. Finally, at t → +∞, the wave packets separate away from the collision point. In general the outgoing state is different with respect to the ingoing one, it is therefore a different superposition of |φαi.

EPFL-ITP-LPTP Riccardo Rattazzi 117 9.1. ASYMPTOTIC STATES

q out p out

p in p in

p out Interactions Interactions

k in

k in k out k out

Figure 9.4: Diﬀerent scattering processes.

We could for example consider the processes for which:

Z ˆ ˆ |φαiin = dΩpin dΩkin f(pin)f(kin)|pin, kini (9.6)

We can then have as a ﬁnal state:

Z ˆ ˆ |φαiout = dΩpout dΩkout f(pout)f(kout)|pout, kouti (9.7) as in the LHS of Fig. 9.4, but we could also have a final state with a different number of particles: whereas its final state is: Z ˆ ˆ ˆ |φαiout = dΩpout dΩkout dΩqout f(pout)f(kout)f(qout)|pout, kout, qouti (9.8)

We can formalise this physically interesting situation by deﬁning two complete sets of states of the full interacting + − theory, the in states |ψα i and the out states |ψα i in the following way:

+ • The in states |ψα i are such that for t → −∞, they become well described by |φαi. − • The out states |ψα i are such that for t → +∞, they become well described by |φαi.

+ − These are the so-called asymptotic states. In interacting theories, we have in general |ψα i= 6 |ψα i. For example, when we start before the interactions with two ﬁeld conﬁgurations φ1, φ2, we have in general φi 6→ φi after the interaction took place. The above statement can be made more formal:

Z Z −iHt ± t→∓∞ −iH0t e dα g(α)|ψα i −→ e dα g(α)|φαi (9.9)

where H and H0 are the fully interacting and free Hamiltonian respectively. Let us make a few comments. We have kept the wave packet smearing dα g(α) in order to stress that our states must be localised in time. Indeed, if they were localised in momentum (i.e. eigenstates of the Hamiltonian), they would not evolve in time (stationarity of the Hamiltonian eigenstates). However, the equality holds for any g(α) so that we could as well more sloppily t→∓∞ −iHt ± −iH0t write e |ψα i −→ e |φαi, so that:

± iHt −iH0t |ψα i = lim e e |φαi ≡ Ω(∓∞)|φαi (9.10) t→∓∞

EPFL-ITP-LPTP Riccardo Rattazzi 118 9.2. THE S-MATRIX where we deﬁned Ω(t) = eiHte−iH0t and Ω(±∞) are called the Möller wave operators. Let us derive an important property of the latter. If the limit in Eq. 9.10 exists, it should not be aﬀect by a shift in the origion of time (after all we chose t0 = 0 by convention, but t0 6= 0 is an equally good origin). Then:

Ω(∓∞) = lim eiH(t−t0)e−iH0(t−t0) (9.11) t→∓∞

Since the limit is independent of the origin of time, we have:

∂ 0 = Ω(∓∞) = −iHΩ(∓∞) + iΩ(∓∞)H0 (9.12) ∂t0 ⇒ HΩ(∓∞) = Ω(∓∞)H0 (9.13)

This has the following important implication:

± ± H|ψα i = HΩ(∓∞)|φαi = Ω(∓∞)H0|φαi = Eα|ψα i (9.14)

So that the asympotic states are eigenstates of the full interacting Hamiltonian when their corresponding free states are eigenstates of the free Hamiltonian, and both have the same eigenvalue. This fact is reassuring as it is natural to expect energy conservation when comparing both sets of states. The Möller wave operators transform a free state into its interacting analogue without changing the energy. Physically:

± ± H0|φαi = Eα|φαi H|ψα i = Eα|ψα i (9.15) | {z } | {z } Purely kinetic, no interaction Kinetic and potential

What physically happens is that the free states give up some kinetic energy for the interaction. One last property is that since the Möller operators are built up from the limit of an unitary operator, they are themselves unitary operators, so that the asymptotic states are normalised in the same way as the free states:

± ± † hψα |ψβ i = hφα|Ω (∓∞)Ω(∓∞)|φβi = hφα|φβi = δ(α − β) (9.16)

For sake of clarity we can redeﬁne1:

Ω± ≡ Ω(∓∞) (9.17)

9.2 The S-Matrix

9.2.1 Lippmann-Schwinger Equation

We will here derive an explicit solution for the Möller wave operators. Start deﬁning:

HI ≡ H − H0 (9.18)

From Eq. 9.14, we have:

1 Note that Ω± so deﬁned are the Möller wave operators one encounters in the litterature

EPFL-ITP-LPTP Riccardo Rattazzi 119 9.2. THE S-MATRIX

± ± ± (Eα − H) |ψα i = 0 ⇒ (Eα − H0) |ψα i = HI |ψα i (9.19)

± ± ± (Eα − H0) a |φαi + ∆|ψα i = HI |ψα i (9.20)

± HI ± ⇒ ∆|ψα i = |ψα i (9.21) Eα − H0 ± Added to this the limit where HI → 0 which implies a = 1, we obtain:

± HI ± |ψα i = |φαi + |ψα i (9.22) Eα − H0 ± i where the ±i factor is there to regulate the pole. This formal solution to the asymptotic states is the Lippmann- Schwinger equation. Recall that the kets entering the equation are actually wavepackets. Then the prescription ±i is important and chosen so that (Fig. 9.5):

Z Z −iHt ± t→∓∞ −iH0t e dα g(α)|ψα i −→ e dα g(α)|φαi (9.23)

The diﬀerent possible choices follow from Cauchy’s theorem. Explicitly:

Z Z Z H −iHt ± −iEαt ± −iEαt I ± e dα g(α)|ψα i = dα g(α)e |ψα i = dα g(α)e |φαi + |ψα i Eα − H0 ± i Z Z e−iEαt (9.24) −iEαt ± = dα g(α)e |φαi + dαdβ g(α) |φβi hφβ|HI |ψα i Eα − Eβ ± i | {z± } Tβα

± where Tβα is called the transition matrix and its meaning will become clear shortly. Now for the in states we must have:

Z Z −iHt + t→−∞ −iH0t e dα g(α)|ψα i −→ e dα g(α)|φαi (9.25) and therefore the second term in Eq. 9.24 must vanish. Consider a Cauchy integral in the complex plane. In our equation, we are integrating over real variables, but we can continue them analytically on the complex plane. If we want to determine the value of the integral on the real line, we choose a contour which contains the real line and, for example, a semi-circle which we choose to close either in the upper or the lower-half plane. This choice depends on where the poles of the intgral lay. Indeed, if they lay in the upper-half plane, we will choose to close our contour in the lower-half plane so that we can argue, using Cauchy’s theorem, that the value of the integral on the real line is equal to the value of the integral on the inﬁnite semi-circle (no poles inside the contour). In our case we have the freedom to choose where the poles lay (through the ±i prescription) and thefore we will ﬁrst focus on the fact that we want the integral to vanish.

+ - ΦΑ ΨΑ Interactions ΨΑ ΦΑ t®-¥ t®¥

Figure 9.5:H AsymptoticL states andH freeL states relation.

EPFL-ITP-LPTP Riccardo Rattazzi 120 9.2. THE S-MATRIX

As t → −∞, the exponential will behave as ∼ eiEα|t| so that we need a positive imaginary part to make the Cauchy integral converge and vanish. This implies that we will choose to close the contour in the upper-half plane. ± Then, to make the integral vanish, the poles must lay in the lower-half plane. Assuming smooth g(α) and Tβα the poles of the dα integral become relevant and are exponetially cancelled at ﬁnite Eα as t → −∞ if we choose the prescription +i. Indeed, this is achieved by selecting Eα = Eβ − i. Similarly, we choose the −i prescription for the out states.

On the other hand, as t → +∞ the +i contribution does not vanish. The Cauchy integral will be evaluated as the sum of the residues and we obtain:

+ Sβα = δ(α − β) − 2πiδ(Eα − Eβ)Tβα (9.28)

Indeed, what happens is that the in states, when evolved up until inﬁnite positive times will have passed through the interactions and thus we do not simply recover their corresponding free states but rather a superposition of them. The coeﬃcients mixing free states components correspond to the S-Matrix (matrix-)elements, Fig. 9.6.

+ ΨΑ Interactions dΒS ΒΑΦ Β t®¥

FigureH L 9.6: Asymptotic states and their relationÙ to the S-Matrix.

In practice, experiments have a deﬁnite constant states at t → −∞ and we measure what happens as t → +∞. Then, for α → β, we have:

− + Sβα = lim hψ |ψα i (9.29) t→+∞ β

So that the S-Matrix represents the array of complex amplitudes for any possible interaction. In particular, if there is no interaction, then Sβα = δ(α − β). The transition rate is thus deﬁned to be proportinal to:

2 2 + |Sβα − δ(α − β)| ∼ Tβα (9.30) We drop in general the limit in Eq. 9.29 as it is clear from the context that we consider late times. From that equation, we see that the S-Matrix connects two complete set of states and thefore it is unitary:

Z Z ∗ + − − + + + dβ Sβγ Sβα = dβ hψγ |ψβ ihψβ |ψα i = hψγ |ψα i = δ(γ − α) (9.31)

Similary S · S† = 1. Finally, as in non-relativistic quantum mechanics, we can derive the Born approximation. Indeed, for weak interactions we have:

EPFL-ITP-LPTP Riccardo Rattazzi 121 9.2. THE S-MATRIX

+ + Tβα = hφβ|HI |ψα i ' hφβ|HI |φαi ≡ Vβα (9.32) so that :

(1) Sβα = δ(α − β) − 2πiδ(Eα − Eβ)Vβα (9.33) where the upperscript (1) denotes the ﬁrst order Born appproximation. Finally, before studying the S-Matrix more into details, notice that instead of working with the latter we can deﬁne the scattering operator S such that:

Sβα ≡ hφβ|S|φαi (9.34) and thus:

S = Ω†(+∞)Ω(−∞) ≡ U(+∞, −∞) (9.35)

This expression will be useful in the next section to examine Lorentz invariance but also later to ﬁnd a formula in perturbation theory.

9.2.2 Symmetries of the S-Matrix

What is meant by the invariance of the S-Matrix under diﬀerent symmetries and what are the conditions on the Hamiltonian that will ensure such invariance properties?

Poincaré Invariance: Poincaré transformations act on spacetime as:

P : x → Λx + a (9.36)

We define an unitary operator U(Λ, a) for the interacting theory by how it acts on either in or out states. It should act in the same way as U0(Λ, a) (unitary operator of Poincaré transformations of the free theory) does on the free states. Our interacting theory will be called Poincaré invariant if the same U(Λ, a) acts on both sets of asymptotic states as U0(Λ, a) acts on the free states. In other words, the interactions do not affect the action of the Poincaré generators. We consider here for simplicity a theory of one interacting scalar field. We define U0(Λ, a) such that its action on the free states |φαi ≡ | {pi}i is given by:

µ P ν iaµΛ ν pi U0(Λ, a)| {pi}i = e i | {Λpi}i (9.37) and therefore, from the unitarity of U(Λ, a) we obtain the Poincaré covariance of the S-Matrix:

µ P ν P ν † iaµΛ ν pj − ki S ({ki} ; {pj}) = h{ki} , out| {pj} , ini = h{ki} , out|U · U| {pj} , ini = e j i S ({Λki} ; {Λpj}) (9.38)

+ where | {pj} , ini ≡ |ψα i and similarly for the out state and we required them both to transform in the same way as the free states. Since the LHS is independent of a, so must be the RHS. In other words:

EPFL-ITP-LPTP Riccardo Rattazzi 122 9.2. THE S-MATRIX

X ν X ν S 6= 0 ⇔ Pi ≡ pj = ki ≡ Pf (9.39) j i

This equation allows us to factorise a four-dimensional delta-function in the deﬁnition of the S-Matrix and rewrite:

4 (4) Sβα = δαβ − (2π) δ (Pf − Pi)iMβα (9.40)

We ﬁnally need to extract the condition on the Hamiltonian. In order to do so, recall:

− + hψβ |ψα i = Sβα = hφβ|S|φαi (9.41) Since the in and out states transform under “interacting” Poincaré transformations in the same way as the free states do under the “free” transformations, we have:

− † + † † † hψβ | U · U |ψα i = hφβ| U0 · U0 · S · U0 · U0 |φαi = hφβ|U0 · S · U0|φαi (9.42) | {z } | {z } 1 S where the last equality is imposed. This means that for the ﬁelds of the full theory to transform in the same way with U(Λ, a) as the free ﬁelds with U0(Λ, a), we must have:

[S, U0(Λ, a)] = 0 (9.43)

Therefore the set of free generators G0 = {H0, P~0, J~0, K~ 0} all commute with the scattering matrix. We can then deﬁne the exact generators G = {H, P,~ J,~ K~ } that generate the transformations of the asymptotic states. Their group structure will be the one of the Poincaré group. For interacting theories we have in general:

H0 → H0 + HI = H (9.44)

P~ = P~0 (9.45)

J~ = J~0 (9.46)

K~ 6= K~ 0 (9.47)

The ﬁrst two consequences are:

h i h i h i h i J,H~ = 0 = P,H~ → J~0,HI = 0 = P~0,HI (9.48)

h i so that the interactions should be rotational and translational invariant. Also from K,~ P~ ∼ H, since we already h i have P~ = P~0, we must have K~ = K~ 0 + ∆K~ 6= K~ 0 and ﬁnally, from K,H~ ∼ P~ = P~0, we have:

h i h i K,H~ I + ∆K,H~ 0 = 0 (9.49)

This equation for the exact boost generators allows us to build them. Finally, since we require the asymptotic ± ± states to transform in the same way as the free states we obtain, denoting U|ψα i = uα|ψα i and U0|φαi = uα|φαi:

± ± uαδ(α − β) = hφβ|U0|φαi hψβ |U|ψα i = ± † † ± † (9.50) hψβ |Ω± · Ω± · U · Ω± · Ω±|ψα i = hφβ|Ω± · U · Ω±|φαi

EPFL-ITP-LPTP Riccardo Rattazzi 123 9.3. PHENOMENOLOGY so that:

UΩ± = Ω±U0 (9.51)

This is the generalisation of Eq. 9.14 which implies that for any generator of the Poincaré group, we have:

GΩ± = Ω±G0 (9.52)

9.3 Phenomenology

9.3.1 Cross Sections

In order to have a derivation and a complete explanation of the physical meaning of the cross section, the reader is invited to follow the discussion in the book An Introduction to Quantum Field Theory (Peskin and Schroeder) Section 4.5.

The ﬁnal formula for the diﬀerential cross-section for the process a + b −→ ! + ... + n is:

|M |2 dσ = ab,n dΦ(n) (9.53) 4I where I is the ﬂux factor and is given by:

q 2 2 2 I = (pa · pb) − mamb (9.54)

9.3.2 Decay Rates

Consider a normalised one-particle wave-packet:

Z ˆ |φαi = dΩαfα(kα)|kαi, hφα|φαi = 1 (9.55)

Consider now the process:

α −→ 1 + ... + n (9.56)

In non-relativistic quantum mechanics the propability for the particle α to decay into n particles is given by:

X 2 P (α → 1 + ... + n) = |Mα,n| (9.57) all ﬁnal states In quantum ﬁeld theory, we have instead:

Y 2 dP (α → 1 + ... + n) = dΩf |h1 + ... + n|S|φαi| (9.58) f

EPFL-ITP-LPTP Riccardo Rattazzi 124 9.3. PHENOMENOLOGY where we leave the ﬁnal integration over all possible ﬁnal states undone. Recall:

4 (4) Sβα = δαβ − (2π) δ (Pf − Pi)iMβα (9.59)

In our case:

4 (4) Snα = −(2π) δ (Pf − kα)iMnα (9.60) as we consider the decay rate into a ﬁnal state diﬀerent from the inital one. This yields:

Z Y 4 (4) ˆ 4 (4) ∗ ˆ∗ dP (α → 1 + ... + n) = dΩf dΩαdΩβ(2π) δ (Pf − kα)Mnαfα(kα)(2π) δ (Pf − kβ)Mnβfβ (kβ) (9.61) f

~ ~ Using one three-dimensional delta-function which sets P~f = kα = kβ (⇒ Eα = Eβ) to integrate over dΩβ:

Z Y dΩα 3 (3) ~ 2 2 2 ˆ 2 dP (α → 1 + ... + n) = dΩf (2π) δ (P~f − kα)(2π) δ(Ef − Eα) |Mnα| |fα(kα)| (9.62) 2Eα f | {z } =δ(Ef −Eα)δ(0)

What does δ(0) mean? This factor appears because we are working with a particle that decays (an unstable particle) and unstable particles cannot be asymptotic states! Does this mean that our formalism is wrong? No! Indeed, we are simply considering a wrong observable (diﬀerential probability) for a decaying particle. This is the case because an unstable particle does not have inﬁnite time at its disposal and that the delta-function has to be replaced by a more suitable quantity. The delta-function Fourier representation is:

Z +∞ i(Ef −Eα)t 2πδ(Ef − Eα) = dt e (9.63) −∞ As we said, the decaying particle does not have inﬁnite time at its disposal. What are the other time scales of the problem? We can compare the lifetime of the particle to the time scale of interactions. The inﬁnity arising from the delta-function is due to the fact that in general:

τ tinteractions (9.64) where τ is the lifetime and it looks like its ﬁniteness can be neglected. Therefore instead of using an inﬁnite time we set a new time scale τ T tinteractions such that:

Z +∞ Z T/2 dt ei(Ef −Eα)t −→ 2πδ(0) ≡ dt = T (9.65) −∞ Ef =Eα −T/2

Finally, the right observable for a particle will be its decay rate:

dP dΓ ≡ (9.66) T

EPFL-ITP-LPTP Riccardo Rattazzi 125 9.4. AMPLITUDES IN PERTURBATION THEORY

Last step consists in considering a one-particle wave-packet at picked momentum ~pα:

ˆ 2 3 (3) ~ dΩα|fα(kα)| ∼ d kαδ (kα − ~pα) (9.67)

So that:

|M |2 dΓ = nα dΦ(n) (9.68) 2Eα and where dΦ(n) is the Lorentz invariant phase space factor (LIPS) and is given by:

(n) Y 4 (4) dΦ = dΩf (2π) δ (Pf − pα) (9.69) f

In order to determine the decay rate Γ we need to integrate over all possible ﬁnal states. Schematically the result will be:

1 Γ = × (Lorentz invariant term) (9.70) 2Eα The lifetime of the particle is then given by:

1 τ = (9.71) Γ

Notice that schematically it is given by:

τ = 2Eα × (Lorentz invariant term) = 2mαγ × (Lorentz invariant term) (9.72) where γ is the usual time dilation factor. This schematical formula shows us that the lifetime of a particle is frame dependent.

9.4 Amplitudes in Perturbation Theory

We saw in the last section that quantities of interest in quantum ﬁeld theory are all dependent on the amplitude squared extracted from the corresponding S-Matrix element. We derive now a perturbative expression for those amplitudes.

9.4.1 Perturbative Expansion

In order to understand why we use perturbation theory, let us start with an example. Consider a real scalar ﬁeld 2 Lagrangian which has Z2 symmetry :

2It is not necessary to add higher order terms as they arise in a renormalisation procedure.

EPFL-ITP-LPTP Riccardo Rattazzi 126 9.4. AMPLITUDES IN PERTURBATION THEORY

1 m2 λ L = (∂ φ)2 − 0 φ2 − 0 φ4 (9.73) 2 µ 2 4! | {z } | {z } L0 LI

The Hamiltonian of the system is:

1 1 m2 λ H = π2 + (∇~ φ)2 + 0 φ2 + 0 φ4 (9.74) 2 2 2 4! | {z } | {z } H0 HI

At leading order in λ0, this is a good choice for the interaction Hamiltonian. The system is weakly coupled as long as:

λ 0 1 (9.75) 8π2

In this regime, we make the hypothesis that we can work in perturbation theory in λ0, so that we can write for the S-Matrix operator:

(1) 2 (2) S = 1 + λ0S + λ0S + ... (9.76)

2 Some processes come at the leading order (LO) (at order λ0 in the Lagrangian or more precisely at order λ0 in the S-Matrix element as we square the amplitude of the process), and at that order we see that the splitting is good. However, at next to leading order (NLO) the splitting is not good! Indeed, interactions lead to corrections to the so-called bare parameters of the Lagrangian (m0, λ0) as well as to a renormalisation of the ﬁeld through a parameter called the ﬁeld renormalisation strength (φ → Z1/2φ). This implies for example that the physical mass will get a dependence on the coupling constant:

2 2 2 m = m0(1 + aλ0 + bλ0 + ...) (9.77) 2 2 2 And therefore to obtain a good splitting of our Lagrangian, we deﬁne Z = 1+δZ , δm = m0Z−m , δλ = λ0Z −λ where m and λ are the observable quantities. The Lagrangian becomes:

1 1 λ 1 1 δ L = (∂ φ)2 − m2φ2 − φ4 + δ ∂ φ∂µφ − δ φ2 − λ φ4 (9.78) 2 µ 2 4! 2 Z µ 2 m 4! | {z } | {z } L0 LI

Once this is done, we can proceed with our calculations in perturbation theory. Notice that for all practical purposes of these lectures we will work at LO so that δZ = δm = δλ = 0. From now on, let us assume that we have a good splitting of our theory so that:

H = H0 + HI (9.79)

9.4.2 Dyson Series Representation of the S-Matrix

As we saw, the S-Matrix operator is related to the limit of an operator U(t, t0) involving both the free and interaction Hamiltonian:

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0 0 U(t, t0) = eiH0(t−t0)e−iH(t−t )e−iH0(t −t0) (9.80)

This expression allows us to easily prove important properties of the evolution operator:

U(t1, t)U(t, t2) = U(t1, t2) (9.81) † −1 U (t1, t2) = U (t1, t2) = U(t2, t1) (9.82) We shall now find a perturbative solution to the latter. In order to do this, we can find a differential equation that U(t, t0) must satisfy and solve it with appropriate initial conditions. Being an evolution operator, we choose as initial conditions:

U(t0, t0) = 1 (9.83)

Now, a good guess would be to take its time derivative and see if we can recognise some solvable equation. Thus:

∂ 0 0 i U(t, t0) = eiH0(t−t0)(H − H )e−iH(t−t )e−iH0(t −t0) ∂t 0 0 0 iH0(t−t0) −iH0(t−t0) iH0(t−t0) −iH(t−t ) −iH0(t −t0) = e HI e e e e (9.84) | {z } HI (t) 0 = HI (t)U(t, t ) where HI (t) is the interaction Hamiltonian in the interaction picture. More simply:

∂ i U(t, t0) = H (t)U(t, t0) (9.85) ∂t I This is nothing but the Schrödinger equation for the interaction picture propagator! If the interaction picture Hamiltonian was time independent we would simply have the usual:

U(t, t0) = e−iHI ∆t (9.86)

However the situation is here more delicate. We will solve the problem pertubatively in HI (t) by writing:

X U(t, t0) = U (n)(t, t0) (9.87) n and to solve for U (n)(t, t0) we iterate n times the resolution of the diﬀerential equation starting from the zeroth order. For the latter:

∂ i U (0)(t, t0) = 0 (9.88) ∂t Integrating this equation from t0 to t and recalling the chosen initial condition, we have:

Z t ∂ dτ U (0)(τ, t0) = U (0)(t, t0) − U (0)(t0, t0) = 0 ⇒ U (0)(t, t0) = 1 (9.89) t0 ∂τ

EPFL-ITP-LPTP Riccardo Rattazzi 128 9.4. AMPLITUDES IN PERTURBATION THEORY At ﬁrst order, we have:

∂ i U (1)(t, t0) = H (t)U (0)(t, t0) = H (t) (9.90) ∂t I I Since we already used our initial condition, we set:

U (n)(t0, t0) = 0, ∀ n > 0 (9.91) and therefore: Z t (1) 0 U (t, t ) = −i dτHI (τ) (9.92) t0

The ﬁrst non-trivial step comes at second order:

Z t Z t1 Z t Z t1 (2) 0 2 U (t, t ) = −i dt1 HI (t1)(−i) dt2HI (t2) = (−i) dt1 dt2HI (t1)HI (t2) (9.93) t0 t0 t0 t0

We would like to ﬁnd an equivalent but simpler expression for this integral. In order to do so, notice that this 0 0 double integral, when represented on the t1 −t2 plane, covers only the lower-half of the square of size (t−t )×(t−t ) cut by its diagonal. In other words, the region in the square that satisﬁes t2 < t1. Mathematically, this can be represented by a θ-function:

Z t Z t1 Z t Z t1 2 2 (−i) dt1 dt2HI (t1)HI (t2) ≡ (−i) dt1 dt2 θ(t1 − t2)HI (t1)HI (t2) (9.94) t0 t0 t0 t0

Written with that θ function, the integral over t2 can be extended to any time beyond t1 as the value of the integrand is zero for times beyond the latter. In particular we can extend it up to t:

Z t Z t1 Z t Z t 2 2 (−i) dt1 dt2 θ(t1 − t2)HI (t1)HI (t2) ≡ (−i) dt1 dt2 θ(t1 − t2)HI (t1)HI (t2) (9.95) t0 t0 t0 t0

The last trick consists in noticing that written in this way we have that the integrals dt1dt2 are symmetric and hence only the symmetric part of the integrand will contribute:

Z t Z t 2 Z t 2 (−i) (−i) dt1 dt2 θ(t1 − t2)HI (t1)HI (t2) ≡ dt1dt2 (θ(t1 − t2)HI (t1)HI (t2) + θ(t2 − t1)HI (t2)HI (t1)) t0 t0 2 t0 (−i)2 Z t = dt1dt2 T {HI (t1)HI (t2)} 2 t0 (9.96)

We ﬁnally arrived at a suitable expression for the second order value of the evolution operator:

2 Z t (2) 0 (−i) U (t, t ) = dt1dt2 T {HI (t1)HI (t2)} (9.97) 2 t0

This result is now straightforwadly generalised for the nth order. Using the same trick, i.e. inserting n − 1 θ-functions to extend all the integrals up to time t and fully symmetrising the integrand, we obtain:

EPFL-ITP-LPTP Riccardo Rattazzi 129 9.4. AMPLITUDES IN PERTURBATION THEORY

n Z t ( n ) (n) 0 (−i) Y U (t, t ) = T dtiHI (ti) (9.98) n! 0 t i=1

In the end, the expression, in perturbation theory, for the evolution operator is:

∞ t ( n ) n Z R t 0 0 0 X (−i) Y −i dt HI (t ) U(t, t ) = 1 + T dtiHI (ti) ≡ T e t0 (9.99) n! 0 n=1 t i=1 where the last expression must be understood as the Taylor expansion of the exponential with the time-ordered product inside the multiple integrals and is just a notational short-hand. Finally, recalling the deﬁnition of the S-Matrix operator:

R ∞ † −i dtHI (t) S = Ω (+∞)Ω(−∞) = U(+∞, −∞) = T e −∞ (9.100)

This is the so-called Dyson series representation of the S-Matrix. Notice that the S-Matrix operator is independent of the time origin t0. Indeed, the latter can be changed by a simple shift in the integral inside the exponetial. We would like to study the Lorentz-invariance of the S-Matrix using the expression we just derived. Tha Hamil- tonian formalism is in this case not suitable as it requires a splitting of space and time. In order to do so, we should go back to a Lagrangian formulation. Recall the Legendre transform which brings us from the latter to the former:

˙ ∂L ˙ ∂L0 ˙ ∂LI H = φ − L = φ − L0 + φ − LI (9.101) ∂φ˙ ∂φ˙ ∂φ˙ | {z } | {z } H0 HI

We see that we simply have:

HI = −LI ⇔ No derivative couplings (9.102)

This is the case in many theories but it is not always true! One simple example is the case of scalar electrodynamics with a local U(1) symmetry:

1 L = − (F )2 + |D φ|2 − m2φφ∗,D φ = ∂ φ + ieqA φ (9.103) 4 µν µ µ µ µ

In most of the cases we are interested in, we have:

R 4 S = T ei d xLI (x) (9.104)

Now, all parts of this expression are manisfestly Lorentz invariant, except the time-ordering operator. However, the time-ordering of two spacetime points is Lorentz invariant unless their separation is spacelike. Indeed, if the points are spacelike separated, we can always go to a frame where their separation is of the form (0, ∆~x), so that the notion of ordering is not Lorentz invariant. Nevertheless, we are working with causal theories, i.e:

EPFL-ITP-LPTP Riccardo Rattazzi 130 9.4. AMPLITUDES IN PERTURBATION THEORY

[L(x), L(y)] = 0, |x − y|2 < 0 (9.105) and since the Lagrangians commute when the points are spacelike separated, the time ordering is unnecessary in that case. In short, our formalism is Lorentz invariant provided causality!3

A further comment should be made. In our formalism, we are working with the S-Matrix operator that connects free states. We never make use of the S-Matrix whose elements are the scalar products between asymptotic states. In the same spirit, the ﬁelds entering LI (x) are the free Heisenberg ﬁelds which evolve naturally with the free Hamiltonian and admit a plane-waves expansion.

9.4.3 Wick’s Theorem

We have now the basic ingredients: an explicit power series expansion in LI to compute the S-Matrix operator S as well as the interacting part of the Lagrangian itself LI (x) ≡ LI [φ(x)] with φ(x) the Heisenberg picture ﬁelds of the free theory and:

iH0t −iH0t iH0t −iH0t LI (x) = e LI (t = 0, ~x)e ≡ e LI [φ(t = 0, ~x)]e = LI [φ(x)] (9.106)

We just need an efficient way to compute the T-product of Lagrangians in the Dyson series and its matrix elements. We will develop this method step by step starting with some useful definitions. Let us start for simplicity with a single real scalar field. We define:

Z −ipx ipx † − + φ(x) = dΩp e ap + e ap ≡ φ (x) + φ (x) (9.107) with φ−(x), φ+(x) the destruction and annihiliation local ﬁelds respectively4. They satisfy the following commutation relations:

Z − + −ip(x−y) φ (x), φ (y) = dΩpe = D(x − y) · 1 (9.108)

φ±(x), φ±(y) = 0 (9.109) where 1 is the identity operator. Any functional of the ﬁeld φ can be written in terms of these two ﬁelds as:

Z n m + − X X Y 4 + Y 4 − O[φ] ≡ O[φ , φ ] = F (x1, . . . , xn; y1, . . . , ym) d xiφ (xi) d yjφ (yj) (9.110) n m i=1 j=1 i.e. with all creation fields to the left of the annihiliation fields. In particular the S-Matrix operator can also be written in this form. This leads us to the following definition:

5 Deﬁnition. The normal ordered product of bosonic operators O1,..., On is denoted as:

N {O1 ···On} ≡ : O1 ···On : (9.111) and is deﬁned as the product in which all creation operators are to the left of all destruction operators. 3For a more rigorous proof of the Lorentz invariance of the S-Matrix, read The Quantum Theory of Fields, Vol I (Weinberg) Section 3.5 4We follow here a diﬀerent convention than in the book by Peskin and Schroeder. 5Fermionic operators will be considered later on.

EPFL-ITP-LPTP Riccardo Rattazzi 131 9.4. AMPLITUDES IN PERTURBATION THEORY Examples. 1. For one scalar ﬁeld:

: φ :=: φ− + φ+ := φ− + φ+ = φ (9.112)

2. Two scalar ﬁelds:

+ + + − + − − − : φ1(x)φ2(y) := φ1 (x)φ2 (y) + φ1 (x)φ2 (y) + φ2 (y)φ1 (x) + φ1 (x)φ2 (y) (9.113)

Notice that for two scalar ﬁelds: − + − + φ1(x)φ2(y) = (φ (x) + φ (x))(φ (y) + φ (y)) 1 1 2 2 (9.114) + + + − + − − − − + = φ1 (x)φ2 (y) + φ1 (x)φ2 (y) + φ2 (y)φ1 (x) + φ1 (x)φ2 (y) + φ1 (x), φ2 (y) so that:

φ1(x)φ2(y) =: φ1(x)φ2(y) : +D12(x − y) · 1 (9.115) where Dij ∝ δij. Let us give some properties of the normal ordered product.

Properties. Let us consider a set of bosonic operators {On ≡ On[φn]} and of c-numbers an ∈ C

1. The normal ordered product is linear: X X : anOn := an : On : (9.116) n n

2. Since the creation ﬁelds commute among them and so do the annihilation ﬁelds, we have for any permutation π:

: φ1(x1) ··· φn(xn) :=: φπ(1)(xπ(1)) ··· φπ(n)(xπ(n)): (9.117) so that:

: O1 ···On :=: Oπ(1) ···Oπ(n) : (9.118)

3. The vacuum expectation value (VEV) of a normal ordered product vanishes:

h0| : O1 ···On : |0i = 0 (9.119)

Let us now deﬁne to operation on operators.

Deﬁnition. The contraction of two operators O1 and O2 is:

O1O2 ≡ h0|O1O2|0i (9.120)

Similarly, the chronological contraction is deﬁned as:

O1O2 ≡ h0|T O1O2|0i (9.121) Examples. In the case of free ﬁelds we have:

φ1(x)φ2(y) =: φ1(x)φ2(y) : +D(x − y) · 1 (9.122) where we drop from now on the indices on the propagator as it is clear from the context if it is zero or non-zero. This implies:

EPFL-ITP-LPTP Riccardo Rattazzi 132 9.4. AMPLITUDES IN PERTURBATION THEORY 1. Their contraction is given by:

φ1(x)φ2(y) ≡ D(x − y) (9.123)

2. Their chronological contraction is given by:

φ1(x)φ2(y) ≡ DF (x − y) (9.124)

where DF (x − y) is the Feynman propagator and is given by:

0 0 0 0 DF (x − y) = θ(x − y )D(x − y) + θ(y − x )D(y − x) (9.125)

Finally, let us give a property of the normal ordering of contracted operators:

Properties. Let us consider a set of bosonic operators {On ≡ On[φn]}, then:

c-number z }| { : O1 ···Oi−1OiOi+1 ···Oj−1OjOj+1 ···On : = : OiOj O1 ···Oi−1Oi+1 ···Oj−1Oj+1 ···On : (9.126)

≡ OiOj : O1 ···Oi−1Oi+1 ···Oj−1Oj+1 ···On :

All this machinery is necessary to prove a theorem that we state now:

Theorem. Let us consider a set of free ﬁelds {φn ≡ φn(xn)}. Then we have the following three results:

1. The time-ordering of n ﬁelds can be written as:

T φ1 ··· φn =: φ1 ··· φn : + : {all possible chronological contractions} : (9.127)

2. The product of n ﬁelds can be written as:

φ1 ··· φn =: φ1 ··· φn : + : {all possible contractions} : (9.128)

3. The product of two normal-ordered products of n = p + q ﬁelds can be written as:

: φ1 ··· φp :: φp+1 ··· φn :=: φ1 ··· φn : + : {all possible contractions} : (9.129)

where (multiple-)contractions arise only between pairs of fields with one field from the first p and and the other from the last q. Contractions between pairs of fields exclusively in the first p or last q do not arise.

This is the so called Wick’s theorem.

For a proof of the ﬁrst two results, see Peskin and Schroeder Section 4.3, for the proof of the third result, refer to appendixA.

We will now use this theorem to ﬁnd a convenient expression for the S-Matrix elements. They are of the form:

h0|a ··· a T {φ ··· φ } a† ··· a† |0i (9.130) p1 pi 1 n k1 kj

We consider the case for which pi 6= kj ∀i, j. Using 9.127, we obtain:

h0|a ··· a T {φ ··· φ } a† ··· a† |0i = h0|a ··· a : φ ··· φ : a† ··· a† |0i p1 pi 1 n k1 kj p1 pi 1 n k1 kj + h0|a ··· a : {all possible chronological contractions} : a† ··· a† |0i p1 pi k1 kj (9.131)

EPFL-ITP-LPTP Riccardo Rattazzi 133 9.4. AMPLITUDES IN PERTURBATION THEORY Finally, noticing that the ladder operators are normal ordered and thus using 9.129, we obtain our ﬁnal result:

† † X † † h0|ap ··· ap T {φ1 ··· φn} a ··· a |0i = h0| : ap ··· ap φ1 ··· φna ··· a : |0i 1 i k1 kj 1 i k1 kj (9.132) all (chrono-) contractions where only terms in which everything is contracted contribute since the VEV of normal ordered products vanishes. The following contractions appear in the sum:

ipx apφ(x) = h0|apφ(x)|0i = hp|φ(x)|0i = e (9.133) † † −ipx φ(x)ap = h0|φ(x)ap|0i = h0|φ(x)|pi = e (9.134)

φ(x)φ(y) = h0|T φ(x)φ(y)|0i = DF (x − y) (9.135)

To close this subsection, a word on fermions. All the results and properties we have seen so far apply to fermions as well. Indeed, the only diﬀerence is that fermions behave like grassman numbers and thus for each permutation we must count a factor (−1). For example:

2 π : ψ1ψ2ψ3 := (−1) : ψ2ψ1ψ3 := (−1) : ψ2ψ3ψ1 := (−1) : ψπ(1)ψπ(2)ψπ(3) : (9.136) where π denotes the parity of the permutation.

9.4.4 Feynman Rules and Feynman Diagrams

At this point we have deﬁned all ingredients to introduce the Feynman rules. Let us sketch the basic idea:

• The S-Matrix operator is a series of time-ordered products of free ﬁelds. • The in and out states correspond to applications of creation and annihiliation operators to the free vacuum. • By Wick’s theorem, we can write the time-ordered products in terms of normal ordered products which are already in a suitable form to take matrix elements with in and out states. • The set of Wick contractions within normal ordered matrix elements is eﬀectively described by a set of diagrammatic rules: Feynman diagrams.

Feynman diagrams are built out of three type of elements:

1. External lines from a† ··· a† |0i and h0|a ··· a . k1 ki p1 pj

2. Internal lines or propagators: φiφj. 3. Interaction vertices: interaction Lagrangian.

In order to illustrate the derivation of the Feynman rules as well as the computation of Feynman diagrams, let us start with a simple example, the λφ3 Lagrangian:

1 m2 λ L = (∂ φ)2 − φ2 − φ3 (9.137) 2 µ 2 3!

The S-matrix operator admits an perturbative expansion (Dyson series):

EPFL-ITP-LPTP Riccardo Rattazzi 134 9.4. AMPLITUDES IN PERTURBATION THEORY

X S = S(n) (9.138) n 1 Z λ n S(n) = d4x ··· d4x −i T φ(x )3 ··· φ(x )3 (9.139) n! 1 n 3! 1 n

The most general contribution to the matrix element:

iT ≡ hp1 ··· pj|S|k1 ··· kii (9.140) has the form:

Z n 1 4 4 λ 3 3 † † d x1 ··· d xn −i h0|ap ··· ap T φ(x1) ··· φ(xn) a ··· a |0i (9.141) n! 3! 1 j k1 ki and therefore we will have three types of contractions to which are added the interaction vertices:

† • Incoming lines: φ(x)ak

e-ikx k

• Outgoing lines: akφ(x)

eikx k

• Propagators: φ(x)φ(y)

DF x-y

H L

λ R 4 • Interaction vertices: −i 3! d x

Each possible Wick contraction corresponds to a diagram where the n vertices are connected among each other by internal lines and to in and out states (external lines). The amplitude is thus the sum over all possible topologies (each topology being weighted by its own combinatorical factor). Notice that physically, summing over all topologies is equivalent to summing over all possible histories of the incoming particles before being scattered into the outgoing particles. We see here the close link to the path integral formulation of quantum mechanics. The rules we just derived are the Feynman rules in position space. We will derive shortly similar rules in momentum space, but before doing so, let us warm ourselves up with a simple example.

EPFL-ITP-LPTP Riccardo Rattazzi 135 9.4. AMPLITUDES IN PERTURBATION THEORY Example. Let us compute the amplitude associated to the following Feynman diagram in λφ3 theory:

k1 p1

x y

k2 p2

Figure 9.7: A Feynman diagram contributing to the 2 to 2 scattering in λφ3.

From left to right, we have:

−ix(k1+k2) • Two external lines with momenta k1 and k2 arriving at the x vertex giving a factor e • The x vertex itself. Notice that the vertices have a factor 1/3!, but there are exactly 3! diﬀerent ways of contracting three lines and a vertex, hence exactly cancelling the factorial6. Therefore, a contracted vertex eﬀectively corresponds to the Feynman rule −iλ R d4x.

• The propagator (internal line) DF (x − y). • The second vertex and the last two external lines give a similar expression to previous one. • Finally, the diagram comes at second order in λ, there is therefore an extra factor 1/2. However, we see that it is symmetric in x and y. Indeed, we could have chosen to contract the incoming lines with the y vertex and the outgoing lines with the x vertex giving rise to the same diagram. It is a general feature that the 1/n! of the Dyson series exactly cancels the symmetry which exchanges vertices, we can therefore drop these two factors of the computation.

Putting everything together, we obtain:

Z 2 4 4 −ix(k1+k2)+iy(p1+p2) (−iλ) d xd yDF (x − y)e (9.142)

It is however often simpler to compute the amplitude and to express the Feynman rules in momentum space. This is done simply by performing the position integrals. For example a generic vertex with three incoming momenta corresponds to the integration:

Z 4 −ix(p1+p2+p3) 4 (4) d xe = (2π) δ (p1 + p2 + p3) (9.143)

Therefore, momentum ﬂow is conserved at each vertex. Electric circuits analogy give a useful intuitive picture of the set of constraints impsoed by the delta-functions (Kirchhof’s laws). The remaining rules are as follows:

• External lines: The exponential in momentum space becomes the identity operator. For a scalar ﬁeld:

External lines = 1 (9.144)

• Propagators: From the Lorentz invariant expression of DF we have:

Z d4p ie−ipx D (x) = (9.145) F (2π)4 p2 − m2

6In the Lagrangian, a φn interaction is generally normalised with a 1/n! combinatorical factor for this reason.

EPFL-ITP-LPTP Riccardo Rattazzi 136 9.4. AMPLITUDES IN PERTURBATION THEORY The exponential was previously included in the vertex integration leaving us with a factor:

Z d4p i (9.146) (2π)4 p2 − m2 We therefore need to integrate over all internal lines. Let us consider V vertice, I internal lines and I−V +1 = L loops. Thus, schematically:

Z 4 4 Z 4 4 d p1 d pI X X d p1 d pL ··· δ(4) ··· δ(4) = (2π)4δ(4)( k − p) ··· (9.147) (2π)4 (2π)4 | {z } (2π)4 (2π)4 V vertices in out So that we are left with an integration over the set of remaining free internal momenta that run in the loops. • Interaction vertices: After the integration is done, we are left with the coupling constant and the combinatorical factor. As we previously argued, the number of topologically equivalent diagrams exactly cancels the 1/3! so that for each vertex we simply associate a factor of (−iλ).

We can synthesise the result in a table of Feynman rules. We also add the position space rules for completeness. The algorithm is as follows:

1. Draw all possible diagrams contributing to the process you consider. 2. Choose whether to work in momentum or position space. 3. For each diagram, using the following Feynman rules, compute its amplitude:

Position space Momentum space

Ext. lines e−ix(±p) 1 (9.148) Int. lines DF (x − y) DF (p) Vertices −iλ R d4x −iλ

In addition, in momentum space, apply momentum conservation at each vertex, integrate over the momenta 4 (4) P P running in the loops and factor out an overall delta-function (2π) δ ( in k − out p). 4. Most of combinatorics factors have been treated with this algorithm, but there is an extra symmetry factor that can arise when we have closed loops inside the diagram. Compute it and divide the amplitude by this number7.

What remains, is the contribution to the reduced amplitude iM:

X X S ({k} → {p}) = 1 + (2π)4δ(4)( k − p)iM ({k} → {p}) (9.149) in out

Application: Consider the following process in λφ3:

k1, k2 −→ p1, p2 (9.150)

7This symmetry factor arises because, for example, in a closed loop, it does not matter how we connect internal lines inside it. For further details, see Peskin and Schroeder Section 4.4.

EPFL-ITP-LPTP Riccardo Rattazzi 137 9.4. AMPLITUDES IN PERTURBATION THEORY This is an elastic scattering process. Let us first compute explicitly all contractions and after that see how the algorithm we derived above leads us to the same result in a much faster way. We first bother doing it the brave way because it is important to understand very well all steps of the full computation. The algorithm can then be applied efficiently and, in the case of complicated diagrams, no symmetry factor will be left behind!

The in and out states are simply:

|ini = a† a† |0i (9.151) k1 k2

hout| = h0|ap1 ap2 (9.152)

The S-Matrix operator is given, up to second order, by:

λ Z 1 λ 2 Z S − 1 = iT = −i d4xT φ3(x) + −i d4xd4yT φ3(x)φ3(y) + O(λ3) (9.153) 3! 2 3!

By Wick’s theorem, only the fully contracted products contribute to the process. Therefore, at ﬁrst order, there is no contribution (we cannot fully contract an odd number of operators!)8. The ﬁrst contribution comes at second order. What are all possible contractions and what are their multiplicity? The matrix element we consider is:

2 Z 1 λ 4 4 X 3 3 † † −i d xd y h0| : ap ap φ (x)φ (y)a a : |0i (9.154) 2 3! 1 2 k1 k2 all (chrono-) contractions

As there are four ladder operators and six ﬁelds, there must be one chronological contraction. Let us ﬁrst consider the terms involving a chronological contraction of the form φ(x)φ(x) = DF (0) = φ(y)φ(y). Such a term would be for example:

Z 4 4 −ixk1−iy(k2−p1−p2) (4) (4) ∼ d xd yDF (x − x)e ∼ δ (k1)δ (k2 − p1 − p2) = 0 (9.155)

This vanishes as by assumption k1 6= 0. Thus, only terms with the chronological contraction φ(x)φ(y) = DF (x−y) contribute.

Let us start with the internal line. How many way of contracting one φ(x) with one φ(y) do we have? We ﬁrst must choose one φ(x): three possible choices. Then must contract it with one φ(y): three choices as well. The multiplicity for the contraction of the internal line is then 3 × 3 = 9. We are left with:

2 Z 1 λ 4 4 X † † 9 × −i d xd yDF (x − y) h0| : ap ap φ(x)φ(x)φ(y)φ(y)a a : |0i (9.156) 2 3! 1 2 k1 k2 all contractions

Let us now notice that since DF (x − y) = DF (y − x) (as it only depends on the causal distance) the full expression is x ↔ y symmetric. Therefore, in the sum, we can remove all contractions which diﬀer by an exchange x ↔ y together with the factor 1/2 of the Dyson series. We previously explained that the Taylor coeﬃcient 1/n! at nth order canceled exactly the combinatorical factor arising from the exchange of the n vertices. We see here an explicit example. Thus:

8It is a general feature that for a theory with a φn interaction, if n is even all orders of the Dyson series contribute to any elastic scattering and if n is odd only odd orders. What about a 2 → 3 inelastic scattering? 1 → 3 decay?

EPFL-ITP-LPTP Riccardo Rattazzi 138 9.4. AMPLITUDES IN PERTURBATION THEORY

2 Z λ 4 4 X † † 9 × −i d xd yDF (x − y) h0| : ap ap φ(x)φ(x)φ(y)φ(y)a a : |0i (9.157) 3! 1 2 k1 k2 all contractions x6↔y where, by x 6↔ y, we mean that we do not count contraction that are equivalent under the exchange of the vertices. What are the possible contractions left?

1. Contract a and a with φ(x) and a† and a† with φ(y). For the first contraction, we have two possibilities p1 p2 k1 k2 among φ(x)s, then the second annihilation operator is constrained to the remaining field. Similarly, for the first creation operator with start with two possibilities and the remaining one is constrained. This gives rise to a multiplicity 2 × 2 = 4 for these contractions. We choose one and we compute:

λ 2 Z 9 × 4 × −i d4xd4yD (x − y)eix(p1+p2)−iy(k1+k2) (9.158) 3! F Here, the contraction of the annihilition operators with the φ(y)s and of the creation operators with the φ(x)s falls in the class of equivalent diagrams under the exchange x ↔ y and thus have already been accounted for (notice how the integral is invariant under the echange of the two variables).

2. Contract a and a† with φ(x) and a and a† with φ(y). Again the multiplicity is 2 × 2 = 4. The result p1 k1 p2 k2 is:

λ 2 Z 9 × 4 × −i d4xd4yD (x − y)eix(p1−k1)−iy(k2−p2) (9.159) 3! F

3. The last possible contraction is a and a† with φ(x) and a and a† with φ(y). Again the multiplicity is p1 k2 p2 k1 2 × 2 = 4. The result is:

λ 2 Z 9 × 4 × −i d4xd4yD (x − y)eix(p1−k2)−iy(k1−p2) (9.160) 3! F

We see that in each of the three diﬀerent contractions a factor 9 × 4 = (3!)2 arises. This cancels exactly the normalisation of the interaction Lagrangian and corresponds to the fact that a vertex has 3! possible ways to be contracted with three lines.

The sum of all possible contractions gives:

Z 4 Z 2 d p i (−iλ) d4xd4ye−ip(x−y) eix(p1+p2)−iy(k1+k2) + eix(p1−k1)−iy(k2−p2) + eix(p1−k2)−iy(k1−p2) (2π)4 p2 − m2 (9.161) where we rewrote explicitly the propagator in terms of the four-dimensional momentum integral. Let us compute the ﬁrst term: Z d4p i Z Z d4p i Z Z d4xd4ye−ip(x−y)eix(p1+p2)−iy(k1+k2) = d4xeix(p1+p2−p) d4ye−iy(k1+k2−p) (2π)4 p2 − m2 (2π)4 p2 − m2 Z d4p i = (2π)4δ(4)(p + p − p)(2π)4δ(4)(k + k − p) (2π)4 p2 − m2 1 2 1 2

i 4 (4) = 2 2 (2π) δ (p1 + p2 − k1 − k2) (k1 + k2) − m (9.162)

EPFL-ITP-LPTP Riccardo Rattazzi 139 9.4. AMPLITUDES IN PERTURBATION THEORY and similarly for the other terms. Finally our result is: 4 (4) 2 i i i iT = (2π) δ (p1 + p2 − k1 − k2)(−iλ) 2 2 + 2 2 + 2 2 (9.163) (k1 + k2) − m (k1 − p1) − m (k1 − p2) − m so that:

2 i i i iM = (−iλ) 2 2 + 2 2 + 2 2 (9.164) (k1 + k2) − m (k1 − p1) − m (k1 − p2) − m

The computation was long and tedious but we arrived to the desired result: we have an expression for the reduced amplitude for the elastic scattering of two particles at second order in λφ3. Notice that we never talked nor used Feynman rules/diagrams as these are just tools to compute these expressions more eﬀectively but are not per se necessary. Nevertheless, let us now see how a powerful tool they are!

How do we proceed? We have a well deﬁned algorithm (set of Feynman rules) which we can follow (almost) blindly to compute any amplitude we wish.

1. Draw all Feynman diagrams that contribute to the considered process at the desired order. At ﬁrst order, we must join a 3-vertex to 4 external lines: topologically impossible. At second order, we have three possibilities:

k1 p1 k1 p2

k1 p1

k2 p2

k2 p2 k2 p1

Figure 9.8: The three Feynman diagrams contributing to the 2 to 2 scattering in λφ3.

2. Choose whether to work in momentum or position space. Notice that we have two vertices one internal line and no loops. Since there are no loops, the Feynman rules in momentum space are straightforward (no integration whatsoever). Hence, we choose the latter (even with loops the calculations are often simpler to carry out in momentum space). 3. Use the Feynman rules. Call p the momentum ﬂowing in the internal line, then the three diagrams will be given by the expression: i (−iλ)2 (9.165) p2 − m2 where for each diagram we need to apply a diﬀerent momentum conservation law at the vertices and we must sum the three diagrams. We must also add an overall four-dimensional delta-function and consider the symmetry factor. There is no particular symmetry factor for these diagrams (no internal loop or any other fancy topology) and therefore, we obtain:

2 i i i iM = (−iλ) 2 2 + 2 2 + 2 2 (9.166) (k1 + k2) − m (k1 − p1) − m (k1 − p2) − m

EPFL-ITP-LPTP Riccardo Rattazzi 140 9.4. AMPLITUDES IN PERTURBATION THEORY In a few straightforward steps we could derive the same result as we did before struggling with contractions and combinatorics. We this simple example we saw how the set of Feynman rules is a powerful algorithm to compute scattering processes. To conclude this subsection, we give a few deﬁnitions and properties of the reduced amplitude.

We studied here the two-to-two elastic scattering of the same particle. Imagine that we considered the four particles as incoming, then they would be completely undistinguishable one from another. Since the matrix element is a function of the four incoming momenta:

iM = f(k1, k2, k3, k4), k1 + k2 + k3 + k4 = 0 (9.167) it must be symmetric under the exchange of any two momenta. Moreover, it must be Lorentz invariant. Thus, we seek three independent Lorentz invariant quantities that can parametrise the amplitude, and under whose exchange the latter remains invariant. They are given by:

2 s = (k1 + k2) 2 t = (k1 + k3) (9.168) 2 u = (k1 + k4)

These are the Mandelstam variables. In a two-to-two scattering, they also exist and are given by the same formulas, 2 2 2 replacing k3 → −p1 and k4 → −p2. When pi = ki = m , we have:

s + t + u = 4m2 (9.169)

As we saw, in a two-to-two eleastic scattering we have three Feynman diagrams. As a function of the Mandelstam variables, we have:

i i i iM(s, t, u) = (−iλ)2 + + (9.170) s − m2 t − m2 u − m2

Each of the three diagrams corresponds to a diﬀerent channel and we see how the amplitude is (as we deal with four identical particles) the sum of the three channels. Knowing this, it would have been enough to compute, say, the s-channel and add the t and u contributions simply by changing channel. Indeed the t-channel diagram’s amplitude corresponds to the s-channel amplitude with s and t exchanged and u left invariant. Similarly we obtain the u-channel amplitude.

Let us compute the values of the three variables and see to what they correspond physically. Let us choose the center-of-mass (CM) frame of the incoming particles. Then, from the deﬁnition of the CM and the conservation of four-momentum:

~ k1 = (E, k) k = (E, −~k) 2 (9.171) p1 = (E, ~p) ~ p2 = (E, −~p), |k| = |~p|

Then:

EPFL-ITP-LPTP Riccardo Rattazzi 141 9.4. AMPLITUDES IN PERTURBATION THEORY

2 2 s = 4E = ECM 1 t = − (s − 4m2)(1 − cos θ) 2 (9.172) 1 u = − (s − 4m2)(1 + cos θ) 2 ~ where ECM is the energy of the CM and θ is the angle between k and ~p. We see that s really measures the energy of the CM and is a “ﬁxed” quantity, whereas t and u measure roughly the forward and backward transfer of momentum respectively and carry all the angular dependence of the process.

Now that we have all necessary ingredients, we can compute the cross-section for this process. Recalling Eq. 9.53, we have in terms of the Mandelstam variables:

2 |M(s, t, u)| (2) dσCM = √ dΦ (9.173) 2s s2 − 4m2 When all the particles have the same mass, the two-body LIPS simpliﬁes greatly and we obtain:

|M(s, t, u)|2 dσ = d(cos θ) (9.174) CM 32πs where we used the φ independence of the matrix element to integrate over the azimutal angle.

To summarise this subsection, we developped the tools necessary to compute scattering amplitudes and using a simple example we could carry out our ﬁrst complete QFT computation!

EPFL-ITP-LPTP Riccardo Rattazzi 142 Appendix A

Proof of Wick’s Theorem

We want to show that X : φa1 ··· φan :: φb1 ··· φbm : =: φa1 ··· φan φb1 ··· φbm : + : φa1 ··· φan φb1 ··· φbm : , (A.1) a,b where the sum runs over all the contractions (meaning also multiple contractions) between some φa’s and some φb’s (note that no contractions between φa’s and φa’s or φb’s and φb’s appear). Let’s show this by induction. We consider as the step 0 the one in which n = m = 1 (for either n = 0 or m = 0 the statement is trivially true):

+ + + − − + − − : φa :: φb : = φa φb + φa φb + φa φb + φa φb = : φaφb : + : φaφb : . (A.2)

We recall the following properties

+ φa φb = 0, (A.3) − φa φb = φaφb. (A.4)

To complete the proof, we now suppose the theorem to hold for n − 1 + m ﬁelds, and want to induce its validity for n + m. So we take as true X : φa2 ··· φan :: φb1 ··· φbm : = : φa2 ··· φan φb1 ··· φbm : + : φa2 ··· φan φb1 ··· φbm : . (A.5)

a>1,b Now we can write : φ ··· φ :: φ ··· φ : = φ+ : φ ··· φ :: φ ··· φ : + : φ ··· φ : φ− : φ ··· φ : a1 an b1 bm a1 a2 an b1 bm a2 an a1 b1 bm + − = φ : φa ··· φa :: φb ··· φb : + : φa ··· φa :: φb ··· φb : φ a1 2 n 1 m 2 n 1 m a1 (A.6) X + : φa2 ··· φan :: φb1 ··· φa1 φbi ··· φbm : .

a1,bi In the ﬁrst line of the previous equation we have used the relation

: φ ··· φ : = φ+ : φ ··· φ : + : φ ··· φ : φ− . (A.7) a1 an a1 a2 an a2 an a1 This can be deduced by simply rewriting

: φ ··· φ : = : φ+ ··· φ : + : φ− ··· φ :, (A.8) a1 an a1 an a1 an and considering that the the field φ+ is already on the far left, so the normal ordering has no effect on it; as for a1 the field φ− , the normal ordering will bring it always at the right of all the φ+ ’s, and since it commutes with all a1 ai the φ− ’s, this means that in every term of the expansion of : φ− ··· φ : in fields φ+ and φ− it can be commuted ai a1 an ai ai and appear on the far right. These two considerations, combined together, give the relation above. Another way of proving this relation is by means of Wick 3 applied to n fields (this is possible since we have supposed it to be

143 valid up to n − 1 + m > n ﬁelds).

Coming back to the decomposition for n + m ﬁelds, with the relation for n − 1 + m ﬁelds one rewrites it as X : φ ··· φ :: φ ··· φ : = φ+ : φ ··· φ φ ··· φ : + φ+ : φ ··· φ φ ··· φ : a1 an b1 bm a1 a2 an b1 bm a1 a2 an b1 bm a>1,b − X − + : φa ··· φa φb ··· φb : φ + : φa ··· φa φb ··· φb : φ 2 n 1 m a1 2 n 1 m a1 (A.9) a>1,b X + : φa2 ··· φan :: φb1 ··· φa1 φbi ··· φbm :

a1,bi The ﬁfth term can be split in two contributions: X X : φa2 ··· φan :: φb1 ··· φa1 φbi ··· φbm : = : φa2 ··· φan φb1 ··· φa1 φbi ··· φbm :

a1,bi a1,bi X X (A.10) + : φa2 ··· φan φb1 ··· ··· φa1 φbi ··· φbm : ,

a1,bi a>1,b so that the overall expression is X : φ ··· φ :: φ ··· φ : = φ+ : φ ··· φ φ ··· φ : + φ+ : φ ··· φ φ ··· φ : a1 an b1 bm a1 a2 an b1 bm a1 a2 an b1 bm a>1,b X + : φ ··· φ φ ··· φ : φ− + : φ ··· φ φ ··· φ : φ− a2 an b1 bm a1 a2 an b1 bm a1 a>1,b X (A.11) + : φa2 ··· φan φb1 ··· φa1 φbi ··· φbm :

a1,bi X X + : φa2 ··· φan φb1 ··· ··· φa1 φbi ··· φbm : .

a1,bi a>1,b

The sum of the ﬁrst and the third term is:

: φa1 ··· φbn : (A.12)

The sum of the second and fourth term is X : φa1 ··· φan φb1 ··· φbm : (A.13) a,b

where a1 is not involved. The sum of the last two terms is:

X : φa1 ··· φan φb1 ··· φbm : (A.14) a,b

where a1 is involved. Thus the sum of the second, fourth, ﬁfth and sixth term is:

X : φa1 ··· φan φb1 ··· φbm : (A.15) a,b as deﬁned in equation (A.1).

Since step 0 is true and step n − 1 + m implies step n + m, then the induction is complete, and the theorem (A.1) is proved.

EPFL-ITP-LPTP Riccardo Rattazzi 144