Quantum Electrodynamics with Application to Bhabha Scattering

Aristotle University of Thessaloniki Faculty of Sciences School of Physics

Gorantis Anastasios Quantum Electrodynamics with application to Bhabha Scattering

Supervisor: N.D. Vlachos

June 2014 Quantum Electrodynamics with application to Bhabha scattering Thesis (Final year project) Author: Gorantis Anastasios Supervisor: Nikolaos Vlachos Aristotle University of Thessaloniki Faculty of Sciences School of Physics Κβαντική Ηλεκτροδυναμική με εφαρμογή στη σκέδαση Bhabha Πτυχιακή Εργασία Συγγραφέας: Γοράντης Αναστάσιος Επιβλέπων Καθηγητής: Νικόλαος Βλάχος Αριστοτέλειο Πανεπιστήμιο Θεσσαλονίκης Σχολή Θετικών Επιστημών Τμήμα Φυσικής

1 2 Contents

Acknowledgments...... 5 Abstract...... 6 AbstractinGreek ...... 7 Conventionsandnotation ...... 9

1 Elements of Classical Field Theory 11 1.1 TheLagrangianFormalism ...... 11 1.2 TheHamiltonianFormalism...... 12 1.3 Noether’sTheorem...... 12 1.4 A few important conserved quantities ...... 13 1.5 Causality ...... 15

2 The Klein-Gordon Field 17 2.1 CanonicalQuantization ...... 17 2.2 TheKlein-GordonEquation ...... 18 2.3 TheQuantumHarmonicOscillator ...... 18 2.4 TheFreeScalarField ...... 19 2.5 ComplexScalarField...... 22 2.6 Causality and the Heisenberg picture ...... 23 2.7 TheFeynmanPropagator ...... 27

3 The Dirac Field 31 3.1 A short discussion on Lorentz invariance ...... 31 3.2 TheSpinorRepresentation ...... 32 3.3 TheDiracequation...... 34 3.4 TheDiracequationparaphernalia...... 36 3.5 TheDiracequationsolutions ...... 39 3.6 The Quantization of the Dirac ﬁeld ...... 42 3.7 TheFeynmanPropagator ...... 46

4 The Electromagnetic Field 49 4.1 Maxwell’sequations ...... 49 4.2 Gaugesymmetry ...... 50 4.3 Quantization of the electromagnetic ﬁeld ...... 51 4.4 Thephotonpropagator ...... 55

5 Interacting Fields 57 5.1 Theinteractionpicture...... 57 5.2 Dyson’sformula...... 58 5.3 Wick’stheorem...... 59

3 5.4 Scattering...... 60 5.5 Couplingvariousﬁelds ...... 63 5.6 Feynmandiagrams ...... 66 5.7 Tracetechnology ...... 67

6 Bhabha Scattering 69

Bibliography ...... 77

4 Acknowledgments

First of all, I would like to thank my supervisor, professor Nikolaos Vlachos for being so helpful and patient with me, especially at the time that I had doubts about which path I should follow in Physics. I would also like to express my gratitude towards professor Anastasios Petkou, for his valuable assistance. Furthermore, I want to thank all the professors in the Department of Physics that helped me understand Physics just a little deeper and ﬁlled me with inspiration. Last, but not least, I want to thank my parents, Anastasia Beka and Gorantis Konstantinos, for their unconditional love and support. This work is dedicated to them.

5 Abstract

In this thesis we provide a short introduction to Quantum Electrodynamics. First, we discuss the Klein-Gordon, the Dirac and the electromagnetic field as free fields. Then we add interaction terms in the Lagrangian of the fields and develop all the necessary tools to compute amplitudes with Feynman diagrams. Furthermore, we establish the connection of these amplitudes to differential scattering cross sections. Then we apply all these techniques we developed, to compute the differential cross section of an elementary scattering process in Quantum Electrodynamics, namely the Bhabha scattering, which includes the scattering of an electron and a positron to an electron and a positron: e+ + e− e+ + e− →

6 Abstract in Greek - Περίληψη

Η παρούσα πτυχιακή εργασία αποτελεί μια σύντομη εισαγωγή στην Κβαντική Ηλεκτροδυναμική, με εφαρμογή στη σκέδαση Bhabha. Στο πρώτο κεφάλαιο κάνουμε μια αναφορά σε διάφορα στοιχεία από την κλασική θεωρία πεδίου τα οποία θα χρησιμοποιήσουμε στη μελέτη της Κβαντικής Ηλεκτροδυναμικής. Αναφερόμαστε στις εξισώσεις Euler-Lagrange, στο φορμαλισμό του Hamilton, στο θεώρημα της Noether καθώς και σε κάποιες συνέπειες της αρχής της αιτιότητας. Το δεύτερο κεφάλαιο πραγματεύεται την κβάντωση του βαθμωτού πεδίου Klein-Gordon, α- κολουθώντας τη μέθοδο της κανονικής κβάντωσης. Εισάγουμε την εξίσωση Klein-Gordon για πραγματικό και μιγαδικό πεδίο, η λύση της οποίας συμπεριλαμβάνει μια επαλληλία κβαντικών αρ- μονικών ταλαντωτών που ταλαντώνονται ανεξάρτητα μεταξύ τους. Στη συνέχεια επιβάλουμε τις σχέσεις μετάθεσης, βρίσκουμε τις ιδιοτιμές του τελεστή Hamilton, και στρεφόμαστε στην εικόνα του Heisenberg για να διερευνήσουμε κατά πόσο τα αποτελέσματα της θεωρίας μας είναι αναλλοίωτα κάτω από μετασχηματισμούς Lorentz. Τέλος, υπολογίζουμε εκτενώς τον διαδότη Feynman για το βαθμωτό πεδίο Klein-Gordon. Το τρίτο κεφάλαιο αναφέρεται στην κβάντωση του πεδίου Dirac. Ξεκινάμε με μια γενική συ- ζήτηση για την ομάδα και την άλγεβρα που σχετίζεται με τους μετασχηματισμούς Lorentz και κατασκευάζουμε την αναπαράσταση των spinors. Στη συνέχεια εισάγουμε την εξίσωση Dirac και αναφερόμαστε σε μια πληθώρα αντικειμένων που σχετίζονται με αυτήν (κατασκευή διαφόρων διγραμ- μικών μεγεθών, εξίσωση Weyl και Weyl spinors, πίνακες γ5, τελεστής της ομοτιμίας, Majorana spinors, συζυγία φορτίου), προχωράμε στην εύρεση λύσεων της εξίσωσης Dirac με τη μορφή ε- πίπεδων κυμάτων και παραθέτουμε μια σειρά από ταυτότητες που σχετίζονται με εσωτερικά και εξωτερικά γινόμενα, οι οποίες αποδεικνύονται πολύ χρήσιμες παρακάτω. ΄Επειτα επιχειρούμε να κβαντώσουμε το πεδίο Dirac, αρχικά επιβάλλοντας σχέσεις μετάθεσης, γεγονός που οδηγεί σε α- συνέπειες, και αργότερα επιβάλλοντας σχέσεις αντιμετάθεσης, οι οποίες ταιριάζουν σε σωματίδια που ικανοποιούν τη στατιστική Fermi-Dirac. Τέλος, βρίσκουμε το διαδότη Feynman για το πεδίο Dirac. Στο τέταρτο κεφάλαιο αναπτύσσουμε τη θεωρία περί κβάντωσης του ηλεκτρομαγνητικού πεδίου. Αρχικά, κάνουμε μια σύντομη αναφορά στις εξισώσεις του Maxwell για το κενό και τις ξαναγρά- φουμε σε τανυστική μορφή, κατάλληλη για την Ειδική Θεωρία Σχετικότητας. ΄Επειτα, συζητάμε τη συμμετρία βαθμίδας η οποία παίζει καθοριστικό ρόλο στην κβάντωση. Προχωράμε στην κβάντωση του ηλεκτρομαγνητικού πεδίου, την οποία εκτελούμε μια φορά για τη βαθμίδα Coulomb (η οποία έχει το μειονέκτημα ότι δεν παραμένει αμετάβλητη κάτω από μετασχηματισμούς Lorentz) και μια φο- ρά για τη βαθμίδα Lorenz. Ακόμη, αναφερόμαστε στη συνθήκη Gupta-Bleuer η οποία εξασφαλίζει ότι οι καταστάσεις αρνητικού μέτρου αποκλείονται από το φυσικό χώρο Hilbert του συστήματος. Η συζήτηση για τα ελεύθερα πεδία κλείνει με την εξαγωγή του φωτονικού διαδότη Feynman. Στο πέμπτο κεφάλαιο διαπραγματευόμαστε τα αλληλεπιδρώντα πεδία. Από την εικόνα του Schr¨odinger και την εικόνα του Heisenberg που χρησιμοποιούσαμε μέχρι τώρα, μεταβαίνουμε στην εικόνα της αλληλεπίδρασης, η οποία επιτρέπει την ευκολότερη επίλυση συστημάτων με συναρτήσεις Lagrange οι οποίες περιέχουν διαταρακτικούς όρους αλληλεπίδρασης μεταξύ πεδίων. Στη συνέχεια αναφερόμαστε στη σχέση του Dyson και το θεώρημα του Wick. ΄Επειτα συζητάμε τη σκέδαση, εξάγουμε το Χρυσό Κανόνα του Fermi, την έκφραση της πιθανότητας διάσπασης, καθώς και την έκφραση της διαφορικής ενεργούς διατομής, την οποία μάλιστα θα χρησιμοποιήσουμε για τη μελέτη της σκέδασης Bhabha. Στη συνέχεια υποδεικνύουμε τον τρόπο σύζευξης του ηλεκτρομαγνητικού πεδίου με τα άλλα πεδία, και γράφουμε τη συνάρτηση Lagrange της Κβαντικής Ηλεκτροδυναμικής. Κατόπιν αναφερόμαστε στα διαγράμματα και τους κανόνες Feynman που πρέπει να χρησιμοποιη- θούν για την εύρεση των πλατών πιθανότητας. Τέλος, παραθέτουμε μια σειρά από σχέσεις που σχετίζονται με τα ίχνη διαφόρων συνδυασμών γινομένων των πινάκων γ, οι οποίες είναι απαραίτητες

7 για τη μετατροπή των πλατών πιθανότητας σε συνηθισμένες εκφράσεις για τη διαφορική ενεργό διατομή. Στο τελευταίο κεφάλαιο, εφαρμόζουμε τις τεχνικές που αναπτύξαμε στην εργασία στον υπολο- γισμό της διαφορικής ενεργούς διατομής για τη σκέδαση Bhabha, η οποία περιλαμβάνει τη σκέδαση ενός ηλεκτρονίου και ενός ποζιτρονίου σε ένα ηλεκτρόνιο και ένα ποζιτρόνιο:

e+ + e− e+ + e− →

8 Conventions and notation

Throughout this thesis we will use natural units:

~ = c = 1

Following the convention in the standard textbooks on Quantum Field Theory, we will use the Minkowski space metric: +1 0 0 0 0 1 0 0 ηµν =  −  0 0 1 0 −  0 0 0 1  −    Greek indices will run over 0, 1, 2, 3 while Latin indices will run over 1, 2, 3 (used solely for spatial components). We will abide to the Einstein convention, that is repeated indices shall be summed. The totally antisymmetric tensor (also referred to as the Levi-Civita tensor) ǫµνρσ will be deﬁned so that: ǫ0123 = +1 Note that this implies that ǫ = 1. Moreover, for four-vectors we have: 0123 − xµ = (x0 , ~x )

Consequently: x = g xν = (x0 , ~x ) µ µν − In this notation the derivative becomes: ∂ ∂ ∂ = = , µ ∂xµ ∂x0 ∇ Furthermore, we follow the usual conventions for Quantum mechanics: ∂ E = i , ~p = i ∂x0 − ∇ which in relativistic notation combine to

pµ = i∂µ

The Pauli sigma matrices are:

0 1 0 i 1 0 σ1 = σ2 = − σ3 = 1 0 i 0 0 1 − Finally, for electrodynamics, we use the Heaviside - Lorentz conventions. The ﬁne structure constant is: e2 e2 1 α = = 4π~c 4π ≃ 137 Lastly, the charge of the electron, denoted by e, will be a negative quantity.

9 10 Chapter 1

Elements of Classical Field Theory

Before delving into the quantum theory, we will begin by discussing a few concepts from Classical Field Theory that will be necessary later.

1.1 The Lagrangian Formalism

A field is a physical quantity that has a specific value for each point in space and time (~x, t). In Field Theory, the field φa = φa(~x, t) is the dynamical value, while both the subscript a and the position ~x are considered as mere labels. The field can be scalar (e.g. a temperature field), three-vector (e.g the electric field E~ (~x, t) or the magnetic field B~ (~x, t)), four-vector (e.g. the electromagnetic four-potential Aµ(~x, t) = (φ, A~), with µ = 0, 1, 2, 3), etc. While in classical mechanics we are dealing with systems of finite degrees of freedom, in Field Theory the degrees of freedom are infinite, as there is at least one for each point in space. The dynamics of the field can be described using the Lagrangian, which depends on the field, φ(~x, t), and its derivatives, ∂µφ(~x, t):

L(t)= d3x (φ , ∂ φ ) (1.1) L a µ a Z The quantity = (φ , ∂ φ ) is called Lagrangian density, or simply Lagrangian. Then, the L L a µ a action is: S = dtL(t)= d4x (φ , ∂ φ ) (1.2) L a µ a Z Z Then, the principle of least action dictates that when a system evolves from one state to another, the path it follows is the one on which the action is an extremum: δS = 0 (1.3) So: ∂ ∂ δS = d4x L δφ + L δ(∂ φ ) = ∂φ a ∂(∂ φ ) µ a Z a µ a ∂ ∂ ∂ = d4x L δφ ∂ L δφ + ∂ L δφ = 0 (1.4) ∂φ a − µ ∂(∂ φ ) a µ ∂(∂ φ ) a Z a µ a µ a The last term of the above equation is a surface integral over the boundary of the four- dimensional spacetime region of integration. As the initial and final configurations of the field are known (δφ = 0), this term vanishes. Thus, we get the Euler-Lagrange equation for the motion of the field: ∂ ∂ ∂ L L = 0 (1.5) µ ∂(∂ φ ) − ∂φ µ a a

11 1.2 The Hamiltonian Formalism

While the Lagrangian formalism is extremely useful in Quantum Field Theory, its main use lies in the path integral formulation. In this thesis, we will discus only the canonical quantization. Consequently, we will be using the Hamiltonian formalism quite frequently. So, a let’s deﬁne the momentum density π (x) conjugate to φa(x) as: ∂ πa(x)= L (1.6) ∂φ˙a Then, in complete analogy to the classical mechanics, the Hamiltonian density is given by:

= πa(x)φ˙ (x) (x) (1.7) H a −L where we replace the φ˙ (x) with πa(x) in . The Hamiltonian is then: a H H = d3x (1.8) H Z and Hamilton’s equations are: ∂H φ˙(~x, t)= (1.9) ∂π(~x, t) and ∂H π˙ (~x, t)= (1.10) −∂φ(~x, t) Notice that, while the Lagrangian formalism is manifestly Lorentz invariant and thus more suited to Field Theory, the Hamiltonian formalism doesn’t look Lorentz invariant Consequently, we should always be careful so that our ﬁnal results are Lorentz invariant, even if this isn’t clear in the intermediate steps.

1.3 Noether’s Theorem

Noether’s theorem allows us to use the symmetries that are hidden in the Lagrangian of our system, to create conservation laws.

Noether’s Theorem. Every continuous symmetry of the Lagrangian gives rise to a conserved current jµ(x), such that: µ ∂µj = 0 (1.11) or equivalently ∂j 0 + ~j = 0 (1.12) ∂t ∇ Proof. Let’s consider an inﬁnitesimal transformation transformation of the ﬁeld:

φ (x) φ′ (x)= φ (x)+ X (φ) (1.13) a → a a a We call this transformation a symmetry, if it leaves the equations of motion invariant. The mathematical condition that ensures this invariance of the equations of motion, is that the Lagrangian changes by a total derivative:

δ = ∂ µ (1.14) L µJ 12 for some µ = µ(φ). So, by making an arbitrary transformation of the ﬁeld δφ : J J a ∂ ∂ ∂ ∂ ∂ δ = L δφ + L δ(∂ φ )= L ∂ L δφ + ∂ L δφ (1.15) L ∂φ a ∂(∂ φ ) µ a ∂φ − µ ∂(∂ φ ) a µ ∂(∂ φ ) a a µ a a µ a µ a The term in the square brackets vanishes when the equation of motion are satisﬁed. Then:

∂ ∂ δ = ∂ L δφ = ∂ L X (φ) (1.16) L µ ∂(∂ φ ) a µ ∂(∂ φ ) a µ a µ a Combining equations (1.14) and (1.16) we get:

∂ ∂ L X (φ) µ(φ) = 0 (1.17) µ ∂(∂ φ ) a −J µ a So, comparing equations (1.11) and (1.17) we get:

µ ∂µj = 0 (1.18) with µ ∂ µ j = L Xa(φ) (φ) (1.19) ∂(∂µφa) −J

We should note here that the existence of the conserved current is much stronger than the existence of a conserved charge, since it implies the local conservation of charge. Consider a charge Q in a finite volume, V: Q = d3x j 0 (1.20) ZV Then: dQ ∂j 0 = d3x = d3x ~j = ~j dS~ (1.21) dt ∂t − ∇ − ZV ZV ZA where in the first equality we have made use of the equation (1.12) and in the second, the Gauss- Ostrogradsky theorem. This equation means that any charge leaving V must be accounted for by a flow of the current three-vector ~j out of the volume.

1.4 A few important conserved quantities

We will now apply Noether’s theorem to calculate some important conserved quantities that will be very useful later.

Energy-momentum tensor Let’s consider the inﬁnitesimal translation:

xν xν εν → − φ (x) φ (x)+ εν ∂ φ (x) (1.22) a → a ν a where the sign in the Taylor expansion has switched to plus, because we are dealing with an active transformation. So, when we apply this inﬁnitesimal translation, the Lagrangian transforms as (x) (x)+ εν∂ (x) (1.23) L →L ν L 13 We observe that the Lagrangian changes by a total derivative, so applying Noether’s theorem we get four conserved currents, one for each of the translations εν with ν = 0, 1, 2, 3:

µ ∂ µ (j )ν = L ∂νφa δν (1.24) ∂(∂µφa) − L

µ We deﬁne T ν as µ ∂ µ T ν L ∂νφa δν (1.25) ≡ ∂(∂µφa) − L µ T ν is called the energy-momentum tensor. It contains four conserved quantities (which can be µ obtained from T ν by raising the indices with the metric tensor):

E = d3xT 00 (1.26) Z and P i = d3xT 0i (1.27) Z where E is the total energy of the ﬁeld, and P i is the total momentum. Before moving on, we would like to point out the similarity to the respective concepts of particle mechanics, where translational invariance give rise to the conservation of momentum and invariance under time translations results in energy conservation.

Lorentz transformations and angular momentum In analogy to the the classical particle mechanics, we will investigate what is the effect of rotational invariance in Field Theory. Let us consider the infinitesimal form of the Lorentz transformations: µ µ µ Λ ν = δ ν + ω ν (1.28) µ with ω ν infinitesimal. As members of the SO(3,1) group, Lorentz transformations satisfy the equation: µ στ ν µν Λ ση Λ τ = η (1.29) and by substituting the equation (1.28) we get:

(δµ +ωµ )ηστ (δν + ων )= ηµν σ σ τ τ ⇒ ωµν + ωνµ = 0 (1.30)

Thus the inﬁnitesimal form ωµν of the Lorentz transformation must be anti-symmetric. Now, by applying the transformation (1.28) to a scalar ﬁeld we get:

φ (x) φ′ (x)= φ(Λ−1x)= a → a = φ (xµ ωµ xν)= a − ν = φ (xµ) ωµ xν∂ φ (x) (1.31) a − ν µ a from which we deduce that: δφ = ωµ xν∂ φ (x) (1.32) a − ν µ a Similarly, the Lagrangian transforms as

δ = ωµ xν∂ = ∂ (ωµ xν ) (1.33) L − ν µL − µ ν L 14 µ µ where the last equality follows from ω µ = 0, as ω ν is anti-symmetric. So, applying Noether’s theorem we get the conserved current:

µ ∂ ρ ν µ ν (j )ν = L ω νx ∂ρφa + ω νx = −∂(∂µφa) L ∂ = ωρ L xν ∂ φ δµ xν = ωρ T µ xν (1.34) − ν ∂(∂ φ ) ρ a − ρ L − ν ρ µ a where we have used the equation (1.25). Now we can create six diﬀerent currents, one for each µ choice of ω ν: ( µ)ρσ = xρT µσ xσT µρ (1.35) J − For ρ, σ = 1, 2, 3 the Lorentz transformation is a rotation and the three conserved charges give the total angular momentum of the ﬁeld:

Qij = d3x (xiT 0j xjT 0i) (1.36) − Z and for the boosts: Q0i = d3x (x0T 0i xiT 00) (1.37) − Z 1.5 Causality

We will discuss brieﬂy some features of the four-dimensional Minkowksi space and their consequences, and more speciﬁcally causality. The concept of causality will be central in the development of the theory. As we have mentioned before, for the study of the four-dimensional Minkowksi space we are using the metric: +1 0 0 0 0 1 0 0 ηµν =  −  (1.38) 0 0 1 0 −  0 0 0 1  −  Then, the line element is given by:  

ds2 = c2dt2 dx2 dy2 dz2 (1.39) − − − The line element is called interval. Note that in this section we will restore the speed of light, c, in our equations even thought we are using natural units, where c = 1. One of the most important consequences of the principle of Relativity, is that the interval between two events is the same for all inertial systems of reference:

ds2 = ds′ 2 (1.40)

This invariance allows to investigate on what conditions two events can be considered as happen- ing at the same place, or at the same time. Let x1,y1, z1,t1 and x2,y2, z2,t2 be the time-space ′ ′ ′ ′ ′ ′ ′ ′ coordinates of two events in one reference system S, and x1,y1, z1,t1 and x2,y2, z2,t2 be the respective coordinates of the two events in another reference system S′. We will use the following abbreviating notation: t = t t (1.41) 12 2 − 1 and l2 = (x x )2 + (y y )2 + (z z )2 (1.42) 12 2 − 1 2 − 1 2 − 1 15 So, the interval between the two events in the reference system S will be:

s2 = c2t2 l2 (1.43) 12 12 − 12 and the respective interval in the reference system S′ will be:

s′ 2 = c2t′ 2 l′ 2 (1.44) 12 12 − 12 So: c2t2 l2 = c2t′ 2 l′ 2 (1.45) 12 − 12 12 − 12 ′ ′ Let’s suppose now that the two events occur at the same point in S , so l 12 = 0. Then:

2 2 ′ 2 s12 = c t 12 > 0 (1.46)

Therefore, two events can occur at the same point in one system of reference, provided the interval between them is a real number. Real intervals are called timelike. On the other hand, ′ ′ 2 if we suppose that the two events occur simultaneously in S , so t 12 = 0, we get:

s2 = l2 < 0 (1.47) 12 − 12 Thus, two events can occur simultaneously in one system of reference, provided the interval between them is an imaginary number. Imaginary intervals are said to be spacelike. The division of the intervals in spacelike and timelike is an absolute concept, because of their invariance. Let’s consider now an event O that we will use as the origin of the timespace coordinates. For the sake of visualization, we shall restrict ourselves in a two dimensional timespace, with one dimension for space and one for time. The two lines between the axes that represent the uniform motion of light, whose speed, c, is the maximum speed possible, form the light cone. The line of motion for particles is restricted inside the light cone. It is quite straightforward to show that for the events inside the light cone c2t2 x2 > 0, so the interval between any event − in this region is timelike. The part of the light-cone that lies above the space axis, contains events that will happen after the event O in any reference system, and is called absolute future. The part of light-cone that lies beneath the time axis, involves events that happened before the event O in any reference system, and is called absolute past. On the other hand, the events outside the light cone are spacelike, and occur at diﬀerent points in space in every reference system; this area is called absolutely remote. As the area outside the light-cone is spacelike, the notions of “earlier”, “simultaneous” and “later” are relative. This fact has very important consequences. Two events can be related causally to one another, only if the interval between them is timelike (as no interaction can occur at speed higher than the speed of light). It is only for the events inside the light cone that the notions of “before” and “after” acquire absolute meaning. This is a necessary condition for the concept of cause and eﬀect to have meaning.

16 Chapter 2

The Klein-Gordon Field

2.1 Canonical Quantization

Paul A.M. Dirac developed in his doctoral thesis a very elegant method of quantization, the canonical quantization. According to this procedure, the transition from the Hamiltonian formalism of classical mechanics to quantum mechanics is simple: you take the classical degrees a of freedom, promote the generalized coordinates (qa) and their conjugate momenta (p ) to operators, and instead of the Poisson bracket relations of classical mechanics:

q ,q = 0 { i j} pi,pj = 0 (2.1) { } q ,pj = δj { i } i impose the following commutation relations:

[qi,qj] = 0 [pi,pj] = 0 (2.2) j j [qi,p ]= i δi where we have set ~ = 1. The transition from Classical Field Theory to Quantum Field Theory can be realized in b a similar way. We take the ﬁeld φa(~x) and its conjugate π (~x), and impose the following commutation relations

[φa(~x), φb(~y)] = 0 [πa(~x),πb(~y)] = 0 (2.3) [φ (~x),πb(~y)] = i δ(3)(~x ~y) δb a − a b The quantum fields, φa(~x) and π (~x), are now operator valued functions of space. The fact that the quantum fields depend only on space and that the commutation relations (2.3) do not depend on time has ruined Lorentz invariance. We are working in the Schrödinger picture, where all time dependence is on the states ψ , which obey the Schrödinger equation: | i d ψ i | i = H ψ (2.4) dt | i We will return to this point later, to mend the problem of Lorentz invariance.

17 2.2 The Klein-Gordon Equation

The dynamics of the real scalar free ﬁelds are governed by the Lagrangian: 1 1 1 1 1 = ηµν ∂ φ∂ φ m2φ2 = φ˙2 ( φ)2 m2φ2 (2.5) L 2 µ ν − 2 2 − 2 ∇ − 2 Then, using the Euler-Lagrange equation (1.5), we get:

µ 2 ∂µ∂ φ + m φ = 0 (2.6) which, written in a less intimidating form, is:

φ¨ 2φ + m2φ = 0 (2.7) −∇ This the Klein-Gordon equation. Let us now take the Fourier transform of the ﬁeld φ(~x, t):

d3p φ(~x, t)= ei~p~x φ(~p, t) (2.8) (2π)3 Z Then, equation (2.7) becomes:

∂2 + ~p 2 + m2 φ(~p, t) = 0 (2.9) ∂t2 This equation is just the diﬀerential equation of the quantum harmonic oscillator, with frequency: 2 2 ω~p = ~p + m (2.10) So, the most general solution of the Klein-Gordonp equation is a linear superposition of simple harmonic oscillators. In the following section, we will discuss brieﬂy Dirac’s approach to the quantum harmonic oscillator, so that we can apply a similar procedure to the Klein-Gordon equation.

2.3 The Quantum Harmonic Oscillator

Let us consider the simple harmonic oscillator, with Hamiltonian: 1 1 H = p2 + ω2q2 (2.11) 2 2 We deﬁne the creation and annihilation operators as

ω i ω i a = q + p a† = q p (2.12) 2 √ 2 − √ r 2ω r 2ω Then q and p can be expressed in terms of a and a†:

1 ω q = (a + a†) p = i (a a†) (2.13) √ − 2 − 2ω r Then, the commutation relation between q and p, [q,p]= i, becomes:

[a, a†] = 1 (2.14)

18 Now, substituting q and p in the Hamiltonian using equation (2.13) and taking into account the above commutation relation, we get: 1 1 H = (aa† + a†a)= ω(a†a + ) (2.15) 2 2 Using equation (2.15) we can verify the following commutation relations:

[H, a†]= ωa† [H, a]= ωa (2.16) − These equations tell us that we can jump between energy eigenstates using a and a†. So if E | i is an energy eigenstate with energy E:

H E = E E (2.17) | i | i then using the commutation relations (2.16):

Ha† E = (E + ω)a† E (2.18) | i | i and Ha E = (E ω)a E (2.19) | i − | i Thus, we have created a ladder that allows to jump between the energies:

...,E 2ω, E ω,E,E + ω, E + 2ω,... (2.20) − − If the energy has a lower bound, there must be a ground state 0 , which satisﬁes a 0 = 0. | i | i This has ground state energy: 1 H 0 = ω 0 (2.21) | i 2 | i which is usually called zero point energy. Then, we can create excited states by applying the creation operator, a†, to the ground state:

n = (a†)n 0 (2.22) | i | i The energy of this excited state is: 1 H n = H(a†)n 0 = (n + ) ω n (2.23) | i | i 2 | i Note that we have ignored the normalization factors.

2.4 The Free Scalar Field

In order to ﬁnd the energy spectrum of the Klein-Gordon equation, we will follow a procedure similar to the quantization of the harmonic oscillator. Here however, each mode of vibration will be treated as an independent oscillator with its own a and a†:

d3p 1 φ(~x)= (a ei~p~x + a† e−i~p~x) (2.24) (2π)3 2ω ~p ~p Z ~p d3p ω π(~x)= (pi) ~p (a ei~p~x a† e−i~p~x) (2.25) (2π)3 − 2 ~p − ~p Z r

19 It can be easily shown that the commutation relations (2.3) for φ(~x) and π(~x), now become commutation relations for the creation and annihilation operators, a† and a:

[a~p, a~q] = 0 † † [a~p, a~q] = 0 (2.26) [a , a†] = (2π)3δ(3)(~p ~q) ~p ~q − † Let us now express the Hamiltonian of the scalar ﬁeld in terms of a~p and a~p. First, lets compute the momentum density π conjugate to φ for the real Lagrangian (2.5):

∂ π(x)= L = φ˙ (2.27) ∂φ˙

Then, by replacing φ˙(x) with π(x) in the Lagrangian and applying equations (1.7) and (1.8), we get: 1 H = d3x π2 + ( φ)2 + m2φ2 (2.28) 2 ∇ Z Now, we must use the expressions (2.24) and (2.25) to ﬁnd the exact form of the Hamiltonian.

3 3 3 1 d x d p d q √ω~p ω~q H = (a ei~p~x a† e−i~p~x)(a ei~q~x a† e−i~q~x) 2 (2π)6 − 2 ~p − ~p ~q − ~q Z 1 i~p~x † −i~p~x i~q~x † −i~q~x + (i~pa~p e i~pa~p e )(i~qa~q e i~qa~q e ) 2√ω~p ω~q − − m2 + (a ei~p~x + a† e−i~p~x)(a ei~q~x + a† e−i~q~x) = 2 ω ω ~p ~p ~q ~q √ ~p ~q 1 d3p 1 = ( ω2 + ~p 2 + m2)(a a + a† a† ) + (ω2 + ~p 2 + m2)(a a† + a† a ) 4 (2π)3 ω − ~p ~p −~p ~p −~p ~p ~p ~p ~p ~p Z ~p Finally, we can use the equation (2.10) and the commutation relations (2.26), to get:

1 d3p d3p 1 H = ω a a† + a† a = ω a† a + (2π)3δ(3)(0) (2.29) 2 (2π)3 ~p ~p ~p ~p ~p (2π)3 ~p ~p ~p 2 Z Z This is a very troubling result. We have bumped into two different infinities. The most obvious infinity is the delta function which is evaluated at zero, where it has an infinite value. As if this wasn’t enough, the integral over ω~p diverges at large p. Let’s discuss each of those issues separately. First of all, in analogy to the harmonic oscillator, we define the vacuum 0 as: | i a 0 = 0 ~p (2.30) ~p | i ∀ Then: ω H 0 = E 0 = d3p ~p δ(3)(0) 0 (2.31) | i 0 | i 2 | i Z which is still infinite. Now, however, we get a little better understanding on the nature of the infinity. The delta function infinity arises because space is infinitely large. In Quantum Field Theory, infinities that are caused by long distances (or the infinitely small energies) are called infrared divergences. The delta function infinity that we met here however hardly qualifies to be considered as such. To remedy the problem, we can put the theory in a box, with sides of

20 length L, and impose periodic boundary conditions on the field. Then we can take the limit L . So: →∞ L/2 L/2 (2π)3δ(3)(0) = lim d3x ei~x~p = lim d3x = V (2.32) L→∞ ~p=0 L→∞ ZL/2 ZL/2 where V is the volume of the box. Thus, the delta function divergence arises because we compute the total energy instead of the energy density. To find the energy density we simply divide by the volume: E ω = 0 = d3p ~p (2.33) E0 V 2 Z Now we must confront the second infinity. We see that as ~p . This is a E0 → ∞ | |→∞ high frequency (or short distance) infinity, called ultraviolet divergence and corresponds to a much deeper problem. The root of this divergence is an assumption we made without actually realizing it: we have assumed that our theory is valid to arbitrarily high energies, or equivalently arbitrarily short distances. The integral should be cut off at high momentum, as our theory can be reliable only up to certain energies. Instead of cutting off the integral, we will follow a more practical solution. As in physics we are only interested in energy differences, and we are not able to measure the vacuum energy directly1, we can redefine the Hamiltonian as: d3p H = ω a† a (2.34) (2π)3 ~p ~p ~p Z Before we move on, it should be pointed out to the reader, that even though we can not measure the vacuum energy directly, we can measure forces that originate from vacuum energy fluctuations. This is the Casimir effect. According to our new definition, the vacuum energy is: H 0 = 0 (2.35) | i The reader might justifiably be wondering about the legitimacy of such an ad-hoc solution. This is indeed a slightly thorny issue in Quantum Field Theory, where infinity is very common sight. However, we can justify this re-definition of the Hamiltonian: if we write the classical Hamiltonian as: 1 1 1 H = p2 + ω2q2 = (ωq ip)(ωq + ip) (2.36) 2 2 2 − Then, after quantization we get: H = ωa†a (2.37) as in equation (2.34). The issue of the ordering ambiguity is a very common problem in field theories. The method we used above is called normal ordering. Normal ordering a string of operators is defined as placing all annihilation operators to the right. We will designate the normal ordering of the operators φ1(~x1), φ2(~x2),...,φn(~xn) as:

N (φ1(~x1), φ2(~x2),...,φn(~xn)) (2.38) Let’s turn our attention now to the particles that can emerge from the free scalar ﬁeld. Using equations (2.26) and (2.37) we ﬁnd that:

† † [H, a~p]= ω~p a~p (2.39) [H, a ]= ω a (2.40) ~p − ~p ~p 1This is not exactly accurate, as the sum of zero point energies should contribute the stress-energy tensor in Einstein’s equations. This is related to the Cosmological constant problem, which is still (2014) consider unsolved.

21 These equations help us form a ladder of energy eigenstates, just like we did in the quantum harmonic oscillator. Thus, we can create particle states by acting repeatedly on the vacuum with creation operators. Of course these ”particles” are by no means localized is space. For † example, by acting once with a~p on the vacuum we get:

~p = a† 0 (2.41) | i ~p | i This state has energy: H ~p = ω 0 = p2 + m2 0 = E 0 (2.42) | i ~p | i | i ~p | i where we have used the relativistic relation: p

2 2 2 E~p = ~p + m (2.43)

It can be shown that the particle that emerged from the quantization of the free scalar ﬁeld has spin zero. This remark brings us to the relevant issue of statistics. Let us consider a n-particle state of spin zero particles as the ones we described above:

† † ~p1,...,~pn = a . . . a 0 (2.44) | i ~p1 ~pn | i Turning to the equation (2.26), we see that the all the creation operators commute with each other. Consequently:

~p ,...,~p ,...,~p , . . . ~p = ~p ,...,~p ,...,~p , . . . ~p (2.45) | 1 i j n i | 1 j i n i for all i, j. This tells us that the particles that emerge from the quantization of the free scalar ﬁeld are bosons, that is they obey Bose-Einstein statistics.

Fock space The full Hilbert space of our theory is created by acting on the vacuum with all possible combinations of the creation operators. This space is called Fock space. The number of particles that belong to a given state in the Fock space is counted by the number operator N:

d3p N = a† a (2.46) (2π)3 ~p ~p Z When the number operator acts on a n-particle state, it gives us the number of the particles:

N ~p ,...,~p = n ~p ,...,~p (2.47) | 1 n i | 1 n i As a ﬁnal comment, we can say that even though the number operator commutes with the Hamiltonian (2.37), so the particle number is conserved, this won’t always be the case. In interacting theories particle number is not conserved. After all, that is one of the raisons d’ˆetre for Quantum Field Theory.

2.5 Complex Scalar Field

So far we have been considered only real scalar ﬁelds. Now, we will investigate what are the consequences when we introduce a complex scalar ﬁeld. Let’s study the Lagrangian:

= ∂ ψ∗∂µψ M 2ψ∗ψ (2.48) L µ − 22 where ψ(~x) is a complex scalar field. We could expand ψ by writing it as ψ = φ1 + iφ2, where ∗ φ1 and φ2 are two real scalar fields, but instead we will consider ψ and ψ as two different independent variables. Using the Euler-Lagrange equations we get:

µ 2 ∂µ∂ ψ + M ψ = 0 (2.49) µ ∗ 2 ∗ ∂µ∂ ψ + M ψ = 0 (2.50)

Now, we expand the complex ﬁeld operator like we did for the real scalar ﬁeld:

d3p 1 ψ = (b ei~p~x + c† e−i~p~x) (2.51) (2π)3 2ω ~p ~p Z ~p d3p 1 ψ† = p (b† e−i~p~x + c ei~p~x) (2.52) (2π)3 2ω ~p ~p Z ~p As the classical field ψ is complex, the correspondingp quantum field is not hermitian, so we get two different operators b~p and c~p. The commutation relations for the fields:

[ψ(~x),π(~y)] = iδ(3)(~x ~y) − [ψ(~x),π†(~y)] = 0 (2.53) become commutation relations for the annihilation and creation operators:

† [b~p, b~q] = [c~p, c~q] = [b~p, c~q] = [b~p, c~q] = 0 [b , b†] = (2π)3δ(3)(~p ~q) (2.54) ~p ~q − [c , c†] = (2π)3δ(3)(~p ~q) ~p ~q − So, quantizing a complex scalar field results into two different particle creation operators which correspond to two different particles, with mass M. Both particles have spin zero. Further- more, bearing in mind the commutation relations above, we conclude that these two particles are bosons. They are actually the particle and the antiparticle. This interpretation exists for the real field too, but there the antiparticle coincides with the particle. Similarly to the real scalar field, where we had the number operator, we can introduce the operator: d3p Q = (c† c b† b )= N N (2.55) (2π)3 ~p ~p − ~p ~p c − b Z which counts the number of antiparticles minus the number of particles. It can be shown that Q is a conserved quantity, and in fact it survives as such even when we add interaction terms in the Lagrangian.

2.6 Causality and the Heisenberg picture

As we have mentioned in the beginning of this chapter, the fact that we have separated time and space in the commutation relations (2.3) and (2.53) and the dependence of φ(~x) only on space, has cast doubts over the Lorentz invariance of our theory. In this section we will deal with this issue by switching to the Heisenberg picture. But before we do that, there are a couple of technical details we need to turn our attention to.

23 Relativistic Normalization Having deﬁned the vacuum 0 , we can demand: | i 0 0 = 1 (2.56) h | i as a normalization condition. However, the simplest normalization

~p ~q = (2π)3δ(3)(~p ~q ) (2.57) h | i − is not Lorentz invariant. We will now construct a Lorentz invariant measure. The quantity d4p is of course Lorentz invariant. Furthermore, another Lorentz invariant quantity comes from the relativistic dispersion relation: R p pµ = m2 µ ⇒ 2 2 2 2 p0 = E~p = ~p + m (2.58)

Combining these two: 3 4 2 2 2 d p d p δ(p0 ~p m ) = (2.59) − − p0>0 2E~p Z Z

So, we now have a Lorentz invariant measure. Using this, we can ﬁnd the correct normalization for every object we want. The relativistically normalized energy eigenstate is:

p = 2E a† 0 (2.60) | i ~p ~p | i Moreover, the Lorentz invariant normalizationp of the states is:

p q = (2π)3 2E δ(3)(~p ~q ) (2.61) h | i ~p − The Heisenberg Picture In the Heisenberg picture the time dependence is assigned to the operators and not the states, as it did in the Schrödinger picture. So, for an operator, , in the Heisenberg picture is: O = eiHt e−iHt (2.62) OH OS where the subscript H denotes the Heisenberg picture and the subscript S denotes the Schrödinger picture. Then: d OH = i[H, ] (2.63) dt OH In Quantum Field Theory, we will designate a field on the Schrödinger picture with φ(~x) and a field on the Heisenberg picture with φ(x). The two operators must agree at a fixed time, t0. Without loss of generality, lets suppose that t0 = 0. Then, in the Heisenberg picture the commutation relations in equation (2.3) become equal time commutation relations:

[φa(~x, t), φb(~y, t)] = 0 [πa(~x, t),πb(~y, t)] = 0 (2.64) [φ (~x, t),πb(~y, t)] = i δ(3)(~x ~y) δb a − a Now, we can use equation (2.63) to get:

φ˙(x)= i[H, φ(x)] = π(x) (2.65)

24 and π˙ (x)= i[H,π(x)] = 2φ(x) m2φ(x) (2.66) ∇ − Combining the two equations above we obtain:

µ 2 ∂µ∂ φ + m φ = 0 (2.67) which is of course the Klein-Gordon equation. As we have mentioned earlier when we discussed the real scalar ﬁeld (equations (2.39) and (2.40)):

† † [H, a~p]= E~p a~p (2.68) [H, a ]= E a (2.69) ~p − ~p ~p where we have replaced ω~p with E~p by taking into account the relativistic dispersion relation. Then:

iHt −iHt −iE~p t e a~p e = e a~p (2.70)

iHt † −iHt iE~p t † e a~p e = e a~p (2.71)

−iE~p t where we have used the Taylor expansion of the exponential e , and moved the operator a~p (or the operator a† in the second equation) to the left, picking up a E (or a E in the second ~p − ~p ~p equation) each time we do so. Thus, we get:

3 d p 1 −ipx † +ipx φ(~x, t)= 3 a~p e + a~p e (2.72) (2π) 2E~p Z where: p px = E t ~p ~x (2.73) ~p − · It will prove quite convenient later if we write φ(x)= φ(~x, t) as:

φ(x)= φ+(x)+ φ−(x) (2.74) where: 3 + d p 1 −ipx φ (x)= a~p e (2.75) (2π)3 2E Z ~p and p d3p 1 φ−(x)= a† e+ipx (2.76) (2π)3 2E ~p Z ~p A similar decomposition can also be performed forp complex ﬁelds: ψ = ψ+ + ψ− (2.77) ψ† = (ψ† )+ + (ψ† )− (2.78) where 3 3 + d p 1 −ipx − d p 1 † +ipx ψ = b~p e ψ = c e (2.79) (2π)3 2E (2π)3 2E ~p Z ~p Z ~p and p p 3 3 † + d p 1 −ipx † − d p 1 † +ipx (ψ ) = c~p e (ψ ) = b e (2.80) (2π)3 2E (2π)3 2E ~p Z ~p Z ~p p p 25 The issue of Causality and the Propagator Now it’s time to discuss the issue of causality. As we have seen in equation (2.64), φ and π satisfy equal time commutation relations. However, in order for our theory to be causal, absolutely separated events must not aﬀect each other. Hence, we must demand that all spacelike separated operators commute:

[ (x), (y)] = 0 for (x y)2 < 0 (2.81) O1 O2 − To see whether this is true for our theory, we must ﬁnd whether the commutators vanish outside the lightcone: 3 d p 1 −ip(x−y) ip(x−y) [φ(x), φ(y)] = 3 e e (2.82) (2π) 2E~p − Z If we take equal times, then (x y)2 = (~x ~y)2 < 0 and: − − − d3p 1 [φ(x), φ(y)] = ei~p (~x−~y ) e−i~p (~x−~y ) (2.83) (2π)3 2 ~p 2 + m2 − Z In the last term we can change the sign of p~p as it is an integration variable. Since [φ(x), φ(y)] is Lorentz invariant it can only depend on (x y)2 and must therefore vanish for all (x y)2 < 0. − − So our theory is indeed causal. Now, we introduce another big concept in Quantum Field Theory. Suppose we have a particle at spacetime point y and we want to know the probability to ﬁnd it at another point x. This probability is given by:

3 3 3 d p d q 1 † −ipx+iqy d p 1 −ip(x−y) 0 φ(x)φ(y) 0 = 0 a~p a 0 e = e (2.84) h | | i (2π)6 4E E h | ~q | i (2π)3 2E Z ~p ~q Z ~p We define: p d3p 1 D(x y) e−ip(x−y) (2.85) − ≡ (2π)3 2E Z ~p as the propagator. This propagator has a feature that is quite disturbing at first sight. For spacelike separated events ((x y)2 < 0) the propagator decays exponentially as e−m|~x−~y|. − So the propagator is non-zero outside the lightcone, even if it vanishes quite quickly. What happened to our precious causality? The concept of causality mandates not that particles can’t propagate over spacelike intervals, but that a measurement performed at one point can not affect a measurement made at another point which is spacelike separated from the first. As a result, we should compute the commutator [φ(x), φ(y)]:

d3p 1 [φ(x), φ(y)] = ei~p (~x−~y ) e−i~p (~x−~y ) = D(x y) D(y x) (2.86) (2π)3 2 ~p 2 + m2 − − − − Z But as we have seen, [φ(x),p φ(y)] equals to zero when (x y)2 < 0. The physical interpretation − of this results is that when an interval between two events is spacelike, there is no Lorentz invariant way to order the events. If the particles travels from x to y in one reference frame, then it can travel from y to x in another. These two amplitudes cancel each other, so such a propagation can not be measured. For the case of the complex ﬁeld, the same argument says that the amplitude for the particle to travel from x to y cancels the amplitude for antiparticle to travel from y to x. So our theory remains causal.

26 2.7 The Feynman Propagator

In this section we will introduce one of the most important concepts in Quantum Field Theory, the Feynman propagator. The Feynman propagator is deﬁned as:

D(x y), x0 >y0 ∆ (x y)= 0 T ψ(x)ψ†(y) 0 = − (2.87) F − h | { } | i D(y x), y0 >x0 − where T denotes the time ordering which is deﬁned as:

φ(x)φ(y), x0 >y0 T φ(x)φ(y) = (2.88) { } φ(y)φ(x), y0 >x0 The ﬁeld ψ(x) may belong to the real (which we used to denote with φ(x)) or the complex scalar Lagrangian. For x0 >y0, we have:

∆ (x y)= 0 ψ(x) ψ†(y) 0 = 0 ψ+(x)+ ψ−(x) (ψ†)+(y) + (ψ†)−(y) 0 = F − h | | i h | | i = 0 ψ+(x)(ψ†)−(y) 0 h | | i where we have just computed the product of the operators and used equation (2.61) to get rid of all the quantities that are equal to zero. Then:

∆ (x y)= 0 ψ+(x)(ψ†)−(y) 0 = 0 ψ+(x)(ψ†)−(y) 0 0 (ψ†)−(y)ψ+(x) 0 = F − h | | i h | | i − h | | i = 0 [ψ+(x), (ψ†)−(y)] 0 h | | i where we have added a term that equals to zero. For y0 > x0, following the same procedure, we get:

∆ (x y)= 0 ψ†(y) ψ(x) 0 = 0 (ψ†)+(y) + (ψ†)−(y) ψ+(x)+ ψ−(x) 0 = F − h | | i h | | i = 0 (ψ†)+(y) ψ−(x) 0 = 0 (ψ†)+(y) ψ−(x) 0 0 ψ−(x) (ψ†)+(y) 0 = h | | i h | |i − h | | i = 0 [(ψ†)+(y), ψ−(x)] 0 h | | i So, summarizing:

0 [ψ+(x), (ψ†)−(y)] 0 , x0 >y0 ∆ (x y)= h | | i (2.89) F − 0 [(ψ†)+(y), ψ−(x)] 0 , y0 >x0 h | | i We will use the symbol ∆+(x y) for the ﬁrst commutator: − ∆+(x y) = [ψ+(x), (ψ†)−(y)] (2.90) − Then, using the integral expression for each ﬁeld:

3 3 −ipx +iqy + + † − d p d q e e † ∆ (x y) = [ψ (x), (ψ ) (y)] = [b~p, b ]= − (2π)6 4E E ~q Z ~p ~q d3p e−ip(x−y) = p (2.91) (2π)3 2E Z ~p where we have used the commutation relations (2.64). On the other hand, for the second commutator ∆−(x y) we have: − ∆−(x y) = [(ψ†)+(y), ψ−(x)] (2.92) − 27 and using the integral expression, we get:

3 3 ipx −iqy − † + − d p d q e e † ∆ (x y) = [(ψ ) (y), ψ (x)] = [c~p, c ]= − (2π)6 4E E ~q Z ~p ~q d3p eip(x−y) = p (2.93) (2π)3 2E Z ~p

Summarizing: d3p e∓ip(x−y) ∆±(x y)= (2.94) − (2π)3 2E Z ~p Note that ∆±(x y) is no longer a q-number; it’s just a c-number, so it can move the left of − the equation (2.89), leaving 0 0 = 1. Now, we will express (2.94) as a contour integral using h | i Cauchy’s integral formula: 1 f(z′) f(z)= dz′ (2.95) 2πi z′ z IC − Then:

0 0 d3p e−ip(x−y) d3p e−iE~p (x −y ) i∆+(x y)= = ei~p (~x−~y) = − (2π)3 2E (2π)3 E + E Z ~p Z ~p ~p 3 −ip0(x0−y0) d p i~p (~x−~y) 1 e 0 = 3 e 0 0 dp = (2π) 2πi + (p E )(p + E ) Z ( IC − ~p ~p ) d4p e−ip(x−y) = i 4 2 (2.96) − + (2π) (p0)2 E ZC − ~p where the contour is taken counterclockwise. We also know that:

(p0)2 = p2 + ~p 2 (2.97) E2 = p2 + m2 (2.98)

Hence: 4 −ip(x−y) + d p e ∆ (x y)= i 4 2 2 (2.99) − − + (2π) p m ZC − Following the same procedure for the ∆−(x y), we get: −

0 0 3 ip(x−y) 3 iE~p(x −y ) − d p e d p −i~p (~x−~y) e ∆ (x y)= 3 = 3 e = − (2π) 2E~p (2π) E~p + E~p Z 0 0 Z d3p eiE~p (x −y ) = cos ~p (~x ~y) + i sin ~p (~x ~y) (2.100) (2π)3 E + E − − − − Z ~p ~p The cosine part is of course an even function, so we can change the ~p to ~p. The sine part on − the other hand is odd, so when it is integrated it will not contribute to the integral. Since the

28 sine part gives zero contribution, nothing changes if we set ~p instead of ~p. − 3 iE(x0−y0) − d p e ∆ (x y)= 3 cos ~p (~x ~y) + i sin ~p (~x ~y) = − (2π) E~p + E~p − − Z 0 0 d3p eiE(x −y ) = ei~p (~x−~y) = (2π)3 E + E Z ~p ~p 0 0 0 d3p 1 e−ip (x −y ) = ei~p (~x−~y) dp0 = (2π)3 2πi − ( p0 + E )(p0 + E ) Z ( IC − ~p ~p ) d4p e−ip(x−y) d4p e−ip(x−y) = i = i (2.101) − (2π)4 (p0)2 E2 − (2π)4 p2 m2 ZC − ~p ZC − where, again the contour is taken counterclockwise. Now we will merge the two parts of the propagator, ∆+(x y) and ∆−(x y), into one integral. To do this, let’s take a look at the − − ﬁgure below.

Figure 2.1: The contour for the Feynman propagator

The upper part is the integration contour that we follow for the ∆−(x y), which is coun- − terclockwise and it includes the pole E . The lower part is the integration contour for the − ~p ∆+(x y), which is clockwise and it includes the pole +E . The fact that the contour is − ~p clockwise compels us to add a minus sign. Thus, we can merge the two integrals into one:

d4p i ∆ (x y)= e−ip(x−y) (2.102) F − (2π)4 p2 m2 ZC − where C is the contour that is shown in the figure. Next, we take the contour to extend to infinity. Then, the only contribution to the integral that is not negligible, comes from the real axis, from to + . −∞ ∞ In order to simplify further the expression of the Feynman propagator, we will shift the poles off the real axis to an infinitesimal distance ǫ > 0, and take the contour along the real axis. In the limit ǫ 0 the integral has the same value. This is the ”iǫ prescription” or the → Feynman prescription.

Figure 2.2: The contour for the Feynman propagator in the ”iǫ prescription”.

29 Thus: d4p i ∆ (x y)= e−ip(x−y) (2.103) F − (2π)4 p2 (m iǫ)2 Z − − or d4p i e−ip(x−y) ∆ (x y)= (2.104) F − (2π)4 p2 m2 + iǫ Z − This is our ﬁnal result. The Feynman propagator for the scalar ﬁeld.

30 Chapter 3

The Dirac Field

In the previous chapter we have discussed the simplest relativistic ﬁeld equation. Now we will turn our attention to the Dirac equation, whose quantization will yield the Dirac Field, which describes particles of spin 1/2, that obey Fermi-Dirac statistics. Note that in order to obtain the Dirac equation we will not follow the path that its author did; we will instead use an approach based on symmetry.

3.1 A short discussion on Lorentz invariance

In the case of a scalar ﬁeld, a Lorentz transformation:

xµ x′ µ = Λµ xν (3.1) → ν results in: φ(x) φ′(x)= φ(Λ−1x) (3.2) → where we got the inverse of the transformation (Λ−1), because we are dealing with an active transformation. However, a ﬁeld can transform in a more general way:

Φa(x) Φ′ a(x)= M a (Λ) Φb(Λ−1x) (3.3) → b a where the matrix M b gives the Lorentz transformation. Then we impose two conditions: the ﬁrst is that the result of two successive Lorentz transformations must be another Lorentz transformation; this requirement translated into the mathematical language means that Lorentz a transformation must form a group. The second is that the matrices M b must preserve the structure of the Lorentz group, i.e.:

M(Λ1)M(Λ2)= M(Λ1Λ2) (3.4) where we have suppressed the indices as they serve no purpose. The equation (3.4) means that the matrices M(Λ) form a representation of the Lorentz group. So, now that we have formulated mathematically our problem, we can proceed to its study. Following a standard technique in the study of Lie Groups, we will work with the corresponding Lie algebra. As we have proven in the first chapter (equations (1.28)-(1.30)), if we take the infinitesimal form a Lorentz transformation: µ µ µ Λ ν = δ ν + ω ν (3.5) µ the matrices ω ν (which belong to the Lie algebra) are anti-symmetric. There are 6 linearly independent 4 4 anti-symmetric matrices. Now, we want to introduce a basis for these matrices. × 31 Such a basis would consist of 6 matrices (Sµ)ab, with µ = 1,..., 6. However, it will be very useful for later purposes if we take matrices with two anti-symmetric indices instead of one; that is, the basis consists of the matrices (J µν )ab with µ,ν = 0,..., 3, where the anti-symmetry of µ and ν ensures that we still have 6 basis elements. As we refer to a Lie algebra, the elements of the basis are called generators of the Lie algebra. To study this Lie algebra, it is necessary to compute the commutator of the generators. To do this, let’s remember that in non-relativistic quantum mechanics, for the rotation group, the generators of the algebra are the angular momentum operators Li, which satisfy: [Li,Lj]= i εijkLk (3.6) In order to produce this relation, one begins from the classical definition of angular momentum in three dimensional space and converts it to an operator relation:

L~ = ~r ~p = ~r ( i ) (3.7) × × − ∇ Of course, the use of the cross product is restricted to the three dimensional space, and in order to generalize it we should convert the operator into an anti-symmetric tensor:

Lij = i(xi j xj i) (3.8) − ∇ − ∇ The generalization to the four-dimensional Lorentz transformations is now straight-forward:

J µν = i(xµ∂ν xν∂µ) (3.9) − Now, we can just use equation (3.9) to ﬁnd the commutator of the generators of the Lie algebra. Doing so, we get:

[J µν , J ρσ]= i(ηνρJ µσ ηµρJ νσ ηνσJ µρ + ηµσJ νρ) (3.10) − − One representation that satisfies equation (3.10) is in terms of the 4 4 matrices: × ( µν ) = i(δµ δν δµ δν ) (3.11) J αβ α β − β α where µ,ν denote which matrix we refer to, and α, β denote the matrix elements. Now, a finite Lorentz transformation is given by: 1 Λ = exp ( i Ω µν ) (3.12) −2 µν J where Ωµν are just four number that specify which Lorentz transformation is performed. Now, we need to find matrices that satisfy the equation (3.10).

3.2 The Spinor Representation

To proceed, we will introduce the Dirac algebra (which is actually a Cliﬀord algebra):

γµ, γν γµγν + γνγµ = 2ηµν I (3.13) { }≡ with µ,ν = 0,... 4, where γµ are four matrices and I is the n n unit matrix. The simplest × matrices that comply with this requirement are 4 4 matrices. However there are many 4 4 × × matrices that obey equation (3.10). A very useful choice would be:

0 1 0 σi γ0 = γi = (3.14) 1 0 σi 0 −

32 where 1 is 2 2 unit matrix, and the σi are of course the Pauli matrices: × 0 1 0 i 1 0 σ1 = σ2 = − σ3 = (3.15) 1 0 i 0 0 1 − This representation is called the Weyl or chiral representation. Now consider the following matrices: i Sµν = [γµ, γν ] (3.16) 4 which can be equivalently written as:

0, µ = ν Sµν = (3.17) i γµγν, µ = ν 2 6 or even better: i i Sµν = γµγν ηµν (3.18) 2 − 2 We will prove that the matrices Sµν satisfy equation (3.10): i i i [Sµν ,Sρσ]= [Sµν , γργσ]= [Sµν , γρ]γσ + γρ[Sµν , γσ] 2 2 2 But i i i [Sµν , γρ]= [γµγν, γρ]= γµγν γρ γργµγν = 2 2 − 2 i i i i = γµ γν , γρ γµγργν γρ, γµ γν + γµγργν = 2 { }− 2 − 2{ } 2 = iγµηνρ iγν ηρµ (3.19) − Now, using (3.19): i i 1 1 1 1 [Sµν ,Sρσ]= [Sµν , γρ]γσ + γρ[Sµν , γσ]= γµγσηνρ + γνγσηρµ γργµηνσ + γργνησµ 2 2 −2 2 − 2 2 Rearranging equation (3.18), we get: 1 1 γµγν = iSµν + ηµν 2 − 2 Using this equation:

[Sµν ,Sρσ]= i(ηνρSµσ ηµρSνσ + ηνσSρµ ηµσSρν) (3.20) − − and taking into account the antisymmetry of Sµν , we ﬁnally get:

[Sµν ,Sρσ]= i(ηνρSµσ ηµρSνσ ηνσSµρ + ηµσSνρ) (3.21) − − So, the matrices Sµν indeed form a representation of the Lorentz algebra. Then, the representation of the Lorentz group is: 1 M(Λ) = exp ( i Ω Sµν ) (3.22) −2 µν In the Weyl representation the boost generators are

i i σi 0 S0i = [γ0, γi]= (3.23) 4 −2 0 σi − 33 and the rotation generators are

i 1 σk 0 Sij = [γi, γj]= εijk (3.24) 4 2 0 σk A ﬁeld ψa(x) with four components, that transforms under boosts and rotations according to the above equations is called a Dirac spinor. Note that the boost generators S0i are not Hermitian, thus the implementation of boosts is not unitary. As the Lorentz group is not compact, it has no faithful, ﬁnite-dimensional representations that are unitary.

3.3 The Dirac equation

In order to obtain the Dirac equation, we need to build a Lorentz invariant action. For this purpose, we need to construct some bilinears.

Constructing a few bilinears We will begin by trying to build a scalar object. The adjoint of a multi-component object is deﬁned as: ψ†(x) = (ψ∗)T (x) Now, if we try to build a scalar by taking the product ψ†ψ as we usually do in quantum mechanics, we see that under a Lorentz transformation this product transforms as:

ψ†(x)ψ(x) ψ†(Λ−1x)M †(Λ)M(Λ)ψ(Λ−1x) (3.25) → because each part of the product transforms as:

ψ(x) M(Λ)ψ(Λ−1x) ψ†(x) ψ†(Λ−1x)M †(Λ) → → But as we discussed a few lines above, M(Λ) is not unitary. So, ψ†ψ is not a scalar. If we pick a representation which satisﬁes (γ0)† = γ0 and (γi)† = γi, as the Weyl repre- − sentation does, then for µ = 0,..., 3 we get:

γ0γµγ0 = (γµ)† (3.26) and hence (Sµν )† = γ0Sµν γ0 (3.27) and 1 M †(Λ) = exp (+ i Ω (Sµν )†)= γ0M −1(Λ)γ0 (3.28) 2 µν Motivated by all this, we deﬁne the Dirac adjoint:

ψ¯(x)= ψ†(x)γ0 (3.29)

Let’s see now how the quantity ψψ¯ transforms under a Lorentz transformation:

ψ¯(x)ψ(x) ψ†(Λ−1x)M †(Λ)γ0M(Λ)ψ(Λ−1x)= ψ†(Λ−1x)γ0ψ(Λ−1x)= ψ¯(Λ−1x)ψ(Λ−1x) → So ψψ¯ is a scalar under Lorentz transformations. What about ψγ¯ µψ? By performing a Lorentz transformation we get: ψγ¯ µψ ψM¯ −1(Λ)γµM(Λ)ψ (3.30) → 34 where we have suppressed the x argument. Now, using equation (3.11) and (3.19), we get:

[γµ,Sρσ] = ( ρσ)µ γν (3.31) J ν or equivalently: i i i (1 + Ω Sρσ)γµ(1 Ω Sρσ) = (1 Ω ρσ)µ γν (3.32) 2 ρσ − 2 ρσ − 2 ρσJ ν Then, expanding equations (3.12) and (3.22) and keeping only ﬁrst order terms:

1 i Λ = exp ( i Ω µν ) 1 Ω µν (3.33) −2 µν J ≃ − 2 µν J 1 i M(Λ) = exp ( i Ω Sµν) 1 Ω Sµν (3.34) −2 µν ≃ − 2 µν Combining equations (3.32), (3.33) and (3.34) we get that:

−1 µ µ ν M (Λ)γ M(Λ) = Λ νγ (3.35)

Thus, ψγ¯ µψ transforms as a Lorentz vector. Following the same procedure, one can show that ψγ¯ µγν ψ transforms a Lorentz tensor.

The Dirac Equation Now, we are ready to form a Lorentz invariant action, the Dirac action. It should be noted though, that Dirac followed a different path to derive the equation of motion. The Dirac action is: S = d4x ψ¯(x)(iγµ∂ m)ψ(x) (3.36) µ − Z Notice that the Dirac Lagrangian is first order, in contrast to the Lagrangian we wrote for the scalar field, which was second order. This was actually one of Dirac’s goals when he wrote down the equation, as he wanted to avoid the negative energy solutions that the Klein-Gordon equation had. However, the negative energy solution are still there. We will return to this issue a little later. Now, using the Euler-Lagrange equation (varying the Dirac action with respect to ψ¯), we get: (iγµ∂ m)ψ = 0 (3.37) µ − This the celebrated Dirac equation. Using the Euler-Lagrange equation with respect to ψ, we get the conjugate equation: µ i∂µψγ¯ + mψ¯ = 0 (3.38) It is very common (and very useful), when we meet four-vectors which are contracted with the γµ matrices, to use the following notation

µ A/ = Aµγ (3.39) which is called the Feynman slash notation. Then Dirac’s equation can be written even more compactly as: (i∂/ m)ψ = 0 (3.40) − Using this notation, the Dirac Lagrangian can be written as:

= ψ¯(x)(i∂/ m)ψ (3.41) L − 35 One important feature of the solutions of the Dirac equation, ψ(x), is that each of its four components obeys the Klein-Gordon equation. This can be shown easily: (iγµ∂ + m)(iγν ∂ m)ψ = 0 µ ν − ⇒ (γµγν ∂ ∂ + m2)ψ = 0 − µ ν ⇒ 1 γµ, γν ∂ ∂ + m2 ψ = 0 − 2{ } µ ν ⇒ µ 2 (∂µ∂ + m )ψ = 0 which is of course the Klein-Gordon equation.

3.4 The Dirac equation paraphernalia

In this section we will include several topics around the Dirac equation that will be proven necessary or helpful later.

Weyl spinors and the Weyl Equation We saw earlier that the in the Dirac representation, the boost and rotation generators are block diagonal matrices (equations (3.23) and (3.24)). That means that the Dirac representation of the Lorentz group is reducible. Thus, we can decompose it into two irreducible representations, which will act on two-component spinors, which in the chiral representation are deﬁned as: ψ ψ = L (3.42) ψ R The two-component objects ψL and ψR in the above decomposition are called left-handed and right-handed Weyl spinors or chiral spinors. These two objects transform in the same way under rotations ( θ~ ), but oppositely under boosts ( β~ ). One can show that: ~σ ~σ ψ (1 iθ~ β~ )ψ (3.43) L → − 2 − 2 L ~σ ~σ ψ (1 iθ~ + β~ )ψ (3.44) R → − 2 2 R 1 Using group theory terminology, we can say that ψL is the ( 2 , 0) representation of the Lorentz group, ψ is the (0, 1 ) representation and the Dirac spinor, ψ, is the the ( 1 , 0) (0, 1 ) repre- R 2 2 ⊕ 2 sentation. Let’s see now what happens to the Dirac Lagrangian when we decompose the Dirac spinor: = ψ¯(i∂/ m)ψ = iψ† σµ∂ ψ + iψ† σ¯µ∂ ψ m(ψ† ψ + ψ† ψ ) (3.45) L − R µ R L µ L − L R R L where σµ = (1,σi) (3.46) and σ¯µ = (1, σi) (3.47) − where i = 1, 2, 3 and µ = 0, 1, 2, 3. A massive fermion requires both ψL and ψR, as they couple through the mass term. On the other hand, a massless fermion can be described by either ψL or ψR. Then the equation of motion are: µ iσ¯ ∂µψL = 0 (3.48) µ iσ ∂µψR = 0 (3.49) These equations are called the Weyl equations.

36 A short conversation on γ5 In the previous subsection we decomposed the Dirac spinor into two chiral spinors. However, we were able to do so, because the matrices of the boost and rotation generators were block diagonal. How can we deﬁne the chiral spinors if we choose an arbitrary representation of γµ of the Cliﬀord algebra? For such an arbitrary representation:

γµ UγµU −1 and ψ Uψ (3.50) → → We introduce another gamma matrix, the γ5, which is deﬁned as:

γ5 = iγ0γ1γ2γ3 (3.51) − Now, the set of matricesγ Ã = (γµ, iγ5), with A = 0, 1, 2, 3, 4 satisfies the five dimensional Clifford algebra γÃ, γ˜B = 2ηAB. It is easy to verify that: { } γµ, γ5 = 0 (3.52) { } which implies that 5 [Sµν , γ ] = 0 (3.53) So the γ5 matrix behaves as a scalar under boosts and rotations. Furthermore:

(γ5)2 = I (3.54)

In the Weyl representation, we can compute the γ5 matrix using its deﬁnition and the matrices γµ (equation (3.14)). 1 0 γ5 = (3.55) 0 1 − Now, we can build a Lorentz invariant projection operator using the γ5 matrix: 1 P = (I γ5) (3.56) ± 2 ±

2 2 which satisﬁes the equations: P+ = P+ and P− = P− and P+P− = 0. Now, we can use this projection operator to deﬁne the chiral spinors in a Lorentz invariant way:

ψ± = P±ψ (3.57)

In the discussion of the previous subsection, in the Dirac representation ψL corresponded to ψ+ and ψR corresponded to ψ−. The γ5 matrices are particularly useful in the study of weak interactions, but since in this thesis we restrict to Quantum Electrodynamics, we will not use them very much.

A brief look at parity So far we have considered only continuous transformations, which are also connected to the identity. There are another two discrete transformations which belong to the Lorentz group: the time reversal T : x0 x0 and xi xi (3.58) →− → (which we will not discuss in this thesis) and the parity

P : x0 x0 and xi xi (3.59) → →− 37 The spinors ψ± are related to each other via the parity transformation. More speciﬁcally, under the action of parity the left and right handed spinors are exchanged. In the Weyl representation we saw that: ~σ ~σ ψ (1 iθ~ β~ )ψ (3.60) L → − 2 − 2 L ~σ ~σ ψ (1 iθ~ + β~ )ψ (3.61) R → − 2 2 R and as under the action of partiy rotations do not change their sign, but boost do, we conclude that under parity left and right handed spinors are exchanged. More generally, in an arbitrary representation:

P : ψ (~x, t) ψ ( ~x, t) (3.62) ± → ∓ − Knowing how chiral spinors transform under parity, we see the Dirac spinor transforms as:

P : ψ(~x, t) γ0ψ( ~x, t) (3.63) → − Now, we can see that if ψ(~x, t) solves the Dirac equation, then γ0ψ( ~x, t) also solves it: − (iγ0∂ + iγi∂ m)γ0ψ( ~x, t)= γ0(iγ0∂ iγi∂ m)ψ( ~x, t) = 0 (3.64) t i − − t − i − − where the passing of γ0 to the left of γi, accounts for a minus sign which turns into plus, if we consider that the derivative acts on ~x instead of ~x. − Majorana spinors In the case of the scalar field that we discussed in the previous chapter, we saw that quantizing a complex scalar field gives rise to two kinds of particles, that we interpreted as particles and antiparticles. If we impose the condition that the scalar field must be real (φ∗ = φ), then the antiparticle becomes the same with the particle. Can we do the same now for the Dirac field? The spinors we have seen so far are complex objects. If we impose the condition ψ∗ = ψ, we will not accomplish a lot. The representation M(Λ) is also complex, so when we perform a Lorentz transformation, the spinors will not remain real. However, there is a way to render the spinors ψ real. We introduce another basis for the Clifford algebra, the Majorana basis:

0 σ2 iσ3 0 0 σ2 iσ1 0 γ0 = γ1 = γ2 = − γ3 = − (3.65) σ2 0 0 iσ3 σ2 0 0 iσ1 − These matrices satisfy the Cliﬀord algebra, and have the useful property that (γµ)∗ = γµ. µν i µ ν − This means that the generators of the Lorentz group S = 4 [γ , γ ] are pure imaginary, and consequently M(Λ) = exp ( 1 i Ω Sµν ) will be real. Now, if we impose the condition ψ∗ = ψ, it − 2 µν will remain true even after a Lorentz transformation. Such spinors are called Majorana spinors.

Charge conjugates Let’s generalize the previous conversation for a general basis of the Cliﬀord algebra. The only requirements for this general basis are the relations (γ0)† = γ0 and (γi)† = γi where − i = 1, 2, 3. We deﬁne the charge conjugate of a Dirac spinor ψ as:

ψ(c) = Cψ∗ (3.66)

38 where C is a 4 4 matrix that satisﬁes the equations: × C†C = I and C†γµC = (γµ)∗ (3.67) − Then, under a Lorentz transformation, ψ(c) transforms as:

ψ(c) CM ∗(Λ)ψ∗ = M(Λ)Cψ∗ = M(Λ)ψ(c) (3.68) → Furthermore, if ψ satisﬁes the Dirac equation, then ψ(c) satisﬁes it too:

(i∂/ m)ψ = 0 − ⇒ ( i∂/∗ m)ψ∗ = 0 − − ⇒ C( i∂/∗ m)ψ∗ = 0 − − ⇒ (i∂/∗ m)ψ(c) = 0 (3.69) − So, if we want to force ψ to be real in a Lorentz invariant way, we must impose the condition:

ψ(c) = ψ (3.70)

Doing so, we get a Majorana spinor, which upon quantization will yield a fermionic particle which coincides with its antiparticle.

3.5 The Dirac equation solutions

Let’s suppose that the solutions of the Dirac equation have the form:

ψ = u(~p ) e−ipx (3.71) where u(~p ) is a four-component spinor. Then, putting it in the Dirac equation, we get:

m p σµ − µ u(~p ) = 0 (3.72) p σ¯µ m µ − Let’s write the spinor u(~p ) as: u u(~p )= 1 (3.73) u 2 Then equation (3.72), becomes:

(p σ)u = mu and (p σ¯)u = mu (3.74) · 2 1 · 1 2 These two equations are equivalent, as follows by the identity:

(p σ)(p σ¯)= p2 p p σiσj = p2 p p δij = p pµ = m2 (3.75) · · 0 − i j 0 − i j µ Once again we will suppose that: u = (p σ) ξ′ (3.76) 1 · for some spinor ξ′. Then, using the second equation of (3.74) we get that:

′ u2 = mξ (3.77)

Thus, the spinor u(~p ) becomes: (p σ) ξ′ u(~p )= A · (3.78) mξ′ 39 where A is a constant. Now, we choose A = 1 and ξ′ = √p σξ¯ . Then the solution to the m · equation (3.72) becomes: √p σξ u(~p )= · (3.79) √p σξ¯ · where ξ is two-component spinor that we normalize as ξ†ξ = I. Now, if we suppose that another solution of the Dirac equation has the form:

ψ = v(~p ) e+ipx (3.80) where v(~p ) is a four-component spinor, following the same procedure, we get the matrix equation: m p σµ µ v(~p ) = 0 (3.81) p σ¯µ m µ whose solution is: √p σ η v(~p )= · (3.82) √p σ¯ η − · where η is a two-component spinor that we normalize as η†η = I. Solutions of the form of (3.71) are called positive frequency solutions (they oscillate as e−iEt) and solutions of the form of (3.80) are called negative frequency solutions ((they oscillate as e+iEt). However, we should stress that they both have positive energy. Finally, before we move on, we should mention that the two-component spinors ξ and η define the spin of the field. In a field with spin up (or down) along a specific direction, the spinors ξ and η are the eigenvectors of the corresponding Pauli matrices with eigenvalue +1 (or -1), just like in non-relativistic Quantum Mechanics. For instance the spinor:

1 ξ = (3.83) 0 corresponds to a ﬁeld with spin up along the z-axis.

Helicity The helicity operator is the projection of the angular momentum along the direction of momentum and is given by: 1 σi 0 h =p ˆ S~ = pˆ (3.84) · 2 i 0 σi where S~ is given by equation (3.24). A particle with h = +1/2 is called right-handed and a particle with h = 1/2 is called left-handed. For a massive particle, the helicity depends on − the frame of reference, as we can always move to a reference frame where momentum is in the opposite direction, while spin remains of course the same. However, for a massless particle which travels at the speed of light, we can not perform such a boost.

Inner and outer products In this section we will produce a few identities for the inner and outer product of the spinors u(~p ) and v(~p ) that will be very useful later. We choose a basis ξs and ηs with s = 1, 2 for the two-component spinors: (ξr)†ξs = δrs and (ηr)†ηs = δrs (3.85)

40 Now, for the positive frequency solution, we can write:

√p σξs us(~p )= · (3.86) √p σξ¯ s · The correct inner product for the spinors, that is the inner product which is Lorentz invariant, isu ¯ u. Here, we will also compute u† u, which we will use later. First, we compute u† u · · · √p σξs (ur(~p ))† us(~p )= (ξr)†√p σ , (ξr)†√p σ¯ · = · · · · √p σξ¯ s · = (ξr)†p σξs + (ξr)†p σξ¯ s = 2(ξr)†p ξs = 2p δrs (3.87) · · 0 0 and thenu ¯ u: · 0 1 √p σξs u¯ r(~p ) us(~p )= (ξr)†√p σ , (ξr)†√p σ¯ · = 2mδrs (3.88) · · · · 1 0 · √p σξ¯ s · For the negative frequency solutions:

√p σ ηs vs(~p )= · (3.89) √p σ¯ ηs − · following the same procedure, we get:

(vr(~p ))† vs(~p ) = 2p δrs andv ¯ r(~p ) vs(~p )= 2mδrs (3.90) · 0 · − Furthermore:

√p σ ηs (¯u r(~p ))† vs(~p )= (ξr)†√p σ , (ξr)†√p σ¯ γ0 · = · · · · · √p σ¯ ηs − · = (ξr)† (p σ¯)(p σ) ηs (ξr)† (p σ¯)(p σ) ηs = 0 (3.91) · · − · · Similarly: p p (¯v r(~p ))† us(~p ) = 0 (3.92) · Next, we will compute (ur(~p ))† vs( ~p ). Setting (p′)µ = (p0, ~p ) we get: · − − √p′ σ ηs (ur(~p ))† vs( ~p )= (ξr)†√p σ , (ξr)†√p σ¯ · = · − · · · √p′ σ¯ ηs · = (ξr)† (p σ)(p′ σ) ηs (ξr)† (p σ¯)(p′ σ¯) ηs (3.93) · · − · · The term under the second square rootp is: p

(p σ¯)(p′ σ¯) = (p + p σi)(p p σi)= p2 ~p 2 = m2 (3.94) · · 0 i 0 − i 0 − The same argument applies to (p′ σ¯)(p σ). Thus, we have: · · (ur(~p ))† vs( ~p ) = (vr(~p ))† us( ~p ) = 0 (3.95) · − · − Another quantity that we will need is the following:

2 2 √p σξs us(~p )u ¯ s(~p )= · (ξs)†√p σ¯ , (ξs)†√p σ (3.96) √p σξ¯ s · · · s=1 s=1 · X X 41 However: 2 s s † ξ (ξ ) = I2×2 (3.97) s=1 X where I is the unit 2 2 matrix. Thus: 2×2 × 2 m p σ us(~p )u ¯ s(~p )= · = p + m (3.98) p σ¯ m / s=1 X · Following the same procedure, we can also prove that:

2 vs(~p )v ¯ s(~p )= /p m (3.99) − s=1 X 3.6 The Quantization of the Dirac ﬁeld

In this section we will quantize the Dirac ﬁeld, by quantizing the Dirac Lagrangian (3.41), following the same procedure that we did for the scalar ﬁeld. However we will see that we must change a key element in our approach in order to achieve quantization.

A ﬁrst (ultimately failed) attempt at quantization Firstly, we compute the conjugate momentum for the Dirac Lagrangian: ∂ π = L = iψγ¯ 0 = iψ† (3.100) ∂ψ˙ Notice that the conjugate momentum does not depend on the time derivative of ψ as in the case of the scalar ﬁeld. Next, we impose the canonical commutation relations:

[ψα(~x ), ψβ(~y )] = 0 † † [ψα(~x ), ψβ(~y )] = 0 (3.101) [ψ (~x ), ψ† (~y )] = δ δ(3)(~x ~y ) α β αβ − thus promoting ψ and ψ† into operators. Then, we write these operators as sums of plane waves:

2 d3p 1 ψ(~x )= bs us(~p )e+i~p·~x + cs † vs(~p )e−i~p·~x (3.102) (2π)3 ~p ~p s=1 2E~p X Z h i 2 d3p p1 ψ†(~x )= bs † us †(~p )e−i~p·~x + cs vs †(~p )e+i~p·~x (3.103) (2π)3 ~p ~p s=1 2E~p X Z h i s † s p s where b~p and b~p are the creation and annihilation operators associated with the spinor u (~p ), s † s s and c~p and c~p are the creation and annihilation operators associated with the spinor v (~p ). As in the case of the scalar ﬁeld, these commutation relation for ψ and ψ† imply the following commutation relations for the creation and annihilation operators:

[br, bs †] = (2π)3δrsδ(3)(~p ~q ) ~p ~q − [cr, cs †]= (2π)3δrsδ(3)(~p ~q ) (3.104) ~p ~q − −

42 while all others commutators are equal to zero. Notice the strange minus sign in the second commutation relation. Now, we will construct the Hamiltonian of the Dirac ﬁeld and then turn it into an operator. Using equation (3.100) for the conjugate momentum, we get the Hamiltonian density: = πψ˙ = ψ¯( iγi∂ + m)ψ (3.105) H −L − i First, we will work with the term: 3 i d p 1 s i s +i~p·~x s † i s −i~p·~x ( iγ ∂i + m)ψ = [b ( γ pi + m)u (~p )e + c (γ pi + m)v (~p )e ] − (2π)3 2E ~p − ~p Z ~p where for the sake of brevityp we left the the sum over s = 1, 2 implicit. Now, we use the equations (3.72) and (3.81) to perform the following replacement: ( γip + m) us(~p )= γ0p us(~p ) (3.106) − i 0 and (γip + m) vs(~p )= γ0p vs(~p ) (3.107) i − 0 Thus: d3p E ( iγi∂ + m)ψ = ~p γ0 bs us(~p ) e+i~p·~x cs †vs(~p ) e−i~p·~x (3.108) − i (2π)3 2 ~p − ~p Z r h i Now, we can compute the Hamiltonian operator:

H = d3x ψ†γ0( iγi∂ + m)ψ = − i Z 3 3 3 d x d p d q E~p r † r † −i~q·~x r r † +i~q·~x s s +i~p·~x s † s † −i~p·~x = 6 b~q u (~q )e + c~q u (~q )e b~p v (~p )e c~p v (~p )e = (2π) s4E~q · − Z h i h i d3p 1 = br †bs [ur †(~p ) us(~p )] cr cs †[vr †(~p ) vs(~p )] br †cs † [ur †(~p ) vs( ~p )] (2π)3 2 ~p ~p · − ~p ~p · − ~p −~p · − Z + cr bs [vr †(~p ) us( ~p )] ~p −~p · − where in the last term we have relabeled ~p ~p. Using equations (3.87), (3.90) and (3.95) to →− replace the inner products in the above relation, we get: d3p d3p H = E (bs †bs cs cs †)= E bs †bs cs †cs + (2π)3δ(3)(0) (3.109) (2π)3 ~p ~p ~p − ~p ~p (2π)3 ~p ~p ~p − ~p ~p Z Z The delta function term was expected to come up, and can be dealt with by normal ordering. The term cs †cs though, is catastrophic to our theory. It means that the Hamiltonian is not − ~p ~p bounded below, as we could continually produce states of lower energy using c†. There must be something we missed. When we quantized the scalar field, the commutation relation: † † [a~p, a~q] = 0 (3.110) implied that: ~p, ~q = ~q, ~p (3.111) | i | i where ~p, ~q = a† a† 0 and ~q, ~p = a† a† 0 . This relation means that spin zero particles | i ~p ~q | i | i ~q ~p | i obey the Bose - Einstein statistics. Taking a look at equations (3.101) we see that we imposed the same commutation relations for the Dirac field. The Dirac field however produces spin 1/2 particles, which obey Fermi - Dirac statistics. Under the exchange of two particles, the quantum state picks up a minus sign. So the inconsistencies that we saw above, were the result of imposing the wrong commutation relations for the operators.

43 A second (ultimately correct) attempt at quantization As we mentioned above, when we exchange the two particles with half-integer spin, the quantum state picks up a minus sign. That gives us a hint on how to proceed with the quantization of fermions: we will use the anti-commutator instead of the commutator:

A, B = AB + BA (3.112) { } So, now we impose the following anti-commutation relations:

ψ (~x ), ψ (~y ) = 0 { α β } ψ† (~x ), ψ† (~y ) = 0 (3.113) { α β } ψ (~x ), ψ† (~y ) = δ δ(3)(~x ~y ) { α β } αβ − which in turn imply the following anti-commutation relations for the creation and annihilation operators:

br, bs † = (2π)3δrsδ(3)(~p ~q ) { ~p ~q } − cr, cs † = (2π)3δrsδ(3)(~p ~q ) (3.114) { ~p ~q } − while all the other possible commutators are equal to zero. The calculation of the Hamiltonian for the Dirac field with anti-commutation relations, proceeds in the same way as it did for the commutation relations until the penultimate line, where using the anti-commutation relations we get: d3p d3p H = E (bs †bs cs cs †)= E bs †bs + cs †cs (2π)3δ(3)(0) (3.115) (2π)3 ~p ~p ~p − ~p ~p (2π)3 ~p ~p ~p ~p ~p − Z Z The anti-commutation relations came to our rescue.1 Now, we define the vacuum 0 , so that is satisfies the relation: | i bs 0 = cs 0 = 0 (3.116) ~p | i ~p | i Furthermore, the creation and annihilation operators obey the following commutation relations with the Hamiltonian:

[H, br]= E br and [H, br †]= E br † ~p − ~p ~p ~p ~p ~p [H, cr ]= E cr and [H, cr †]= E cr † (3.117) ~p − ~p ~p ~p ~p ~p r † so we can once again build a ladder of energy eigenstates, by acting on the vacuum by b~p and r † c~p , to create particles and antiparticles. For instance, a two-particle state is:

~p , r ; ~p , r = br1 †br2 † 0 = ~p , r ; ~p , r (3.118) | 1 1 2 2i ~p1 ~p2 | i − | 2 2 1 1i as we expected from particles that obey Fermi - Dirac statistics. Furthermore:

~p, r; ~p, r = ~p, r; ~p, r | i − | i ⇒ ~p, r; ~p, r = 0 (3.119) | i which is the Pauli exclusion principle.

1Note also that the delta function term for the Dirac ﬁeld came out negative, in contrast to the scalar ﬁeld. This might provide us some hints in the issue of the Cosmological constant.

44 Propagators Now, we will move to the Heisenberg picture and deﬁne the spinors ψ(~x, t) at every point in spacetime such that they satisfy the equation: ∂ψ = i[H, ψ] (3.120) ∂t whose solution is:

2 3 d p 1 s s −ipx s † s +ipx ψ(x)= 3 b~p u (~p )e + c~p v (~p )e (3.121) (2π) 2E~p Xs=1 Z h i 2 3 † d p p1 s † s † +ipx s s † −ipx ψ (x)= 3 b~p u (~p )e + c~p v (~p )e (3.122) (2π) 2E~p Xs=1 Z h i Next, we deﬁne the fermionic propagatorp as: iS = ψ (x), ψ¯ (y) (3.123) αβ { α β } or we can omit the indices and write: iS(x y)= ψ(x), ψ¯(y) (3.124) − { } Using the solutions (3.121) and (3.122) the fermionic propagator becomes:

3 3 d p d q 1 s r † s r −i(px−qy) s r † s r +i(px−qy) iS(x y)= 6 b~p, b~q u (~p )¯u (~p )e + c~p, c~q v (~p )¯v (~p )e = − (2π) 4E~pE~q { } { } Z d3p 1 = p us(~p )¯us(~p )e−ip(x−y) + vs(~p )¯vs(~p )e+ip(x−y) = (2π)3 2E Z ~p 3 d p 1 −ip(x−y) +ip(x−y) = 3 (/p + m)e + (/p m)e (2π) 2E~p − Z where we have used equations (3.98) and (3.99). So we can write the fermionic propagator as: iS(x y) = (i∂/ + m) D(x y) D(y x) (3.125) − x − − − where D(x y) is propagator for the real scalar field, which is: − d3p 1 D(x y)= e−ip(x−y) (3.126) − (2π)3 2E Z ~p As we saw in the case of the scalar field, the propagator vanishes for spacelike seperated points, where (x y)2 < 0. This relation implied that for spin zero particles, [φ(x), φ(y)] = 0 for − (x y)2 < 0, which ensured that the theory is causal. However, in the case of spin 1/2 fields − we have anti-commutation relations: ψ (x), ψ (y) = 0 for (x y)2 < 0 (3.127) { α β } − What happened to causality? The best thing we can say is that, as the fermionic propagator is not an observable quantity, there is no reason why it should vanish outside the light cone. Before moving on to compute the Feynman propagator for the Dirac field, we will make a final comment. If we stay away from singularities, the propagator satisfies the Dirac equation: (i∂/ m)S(x y) = 0 (3.128) x − − which can be proven using (∂/2 + m2)D(x y) = 0 and the mass shell condition p2 = m2. x − 45 3.7 The Feynman Propagator

The calculation of the Feynman propagator for the Dirac field is quite similar to the respective calculation in the scalar field. First, we define the time ordering operator T for spinors as:

ψ(x)ψ¯(y), x0 >y0 T ψ(x)ψ¯(y) = (3.129) { } ψ¯(y)ψ(x), y0 >x0 − Notice the strange (at least on the first look) minus sign. It’s there to ensure Lorentz invariance. When (x y)2 < 0, there is no Lorentz invariant way to determine whether x0 >y0 or y0 >x0. − Thus, according to the definition we made, outside the light cone we get ψ(x), ψ¯(y) = 0. The { } minus sign occurs for any time ordered product; while inside a time ordered operator string, the bosonic operators commute, the fermionic operators anti-commute. Then, we define the spinor Feynman propagator as: S (x y)= 0 T ψ(x)ψ¯(y) 0 (3.130) F − h | { } | i Next, we will write the equations (3.121) and (3.122) as:

ψ(x)= ψ+ + ψ− (3.131) ψ¯(x)= ψ¯+ + ψ¯− (3.132) where

2 d3p 1 ψ+(x)= bs us(~p )e−ipx (3.133) (2π)3 ~p s=1 2E~p X Z 2 d3p p1 ψ−(x)= cs † vs(~p )e+ipx (3.134) (2π)3 ~p s=1 2E~p X Z 2 d3p p1 ψ¯+(x)= cs v¯s(~p )e−ipx (3.135) (2π)3 ~p s=1 2E~p X Z 2 3 p ¯− d p 1 s † s +ipx ψ (x)= 3 b~p u¯ (~p )e (3.136) (2π) 2E~p Xs=1 Z p So, for x0 >y0 we have:

S (x y)= 0 ψ(x)ψ¯(y) 0 = F − h | | i = 0 ψ+(x)ψ¯+(y) 0 + 0 ψ+(x)ψ¯−(y) 0 + 0 ψ−(x)ψ¯+(y) 0 + 0 ψ−(x)ψ¯−(y) 0 = h | | i h | | i h | | i h | | i = 0 ψ+(x)ψ¯−(y) 0 = 0 ψ+(x)ψ¯−(y) 0 + 0 ψ¯−(y)ψ+(x) 0 = h | | i h | | i h | | i = 0 ψ+(x), ψ¯−(y) 0 h | { } | i or making the matrix indices explicit:

S (x y)= 0 ψ+(x), ψ¯−(y) 0 (3.137) F αβ − h | { α β } | i In a similar way, for y0 >x0:

S (x y)= 0 ψ¯+(x), ψ−(y) 0 (3.138) F αβ − − h | { α β } | i 46 We will designate the part for x0 >y0 as S+(x y): − 2 d3p e−ip(x−y) S+(x y)= ur(~p )u ¯r(~p )= − (2π)3 2E ~p r=1 Z X d3p /p + m = e−ip(x−y) (3.139) (2π)3 2E Z ~p In a similar way, for y0 >x0, we designate the other part as S−(x y), and following the same − procedure we get:

d3p /p m S−(x y)= − e+ip(x−y) (3.140) − (2π)3 2E Z ~p Then, we can express the two parts in the following useful form:

∂ d3p 1 S+(x y)= iγµ + m e−ip(x−y) − ∂(xµ yµ) (2π)3 2E − Z ~p ∂ d3p 1 S−(x y)= iγµ + m e+ip(x−y) − ∂(xµ yµ) (2π)3 2E − Z ~p and taking into account equation (2.94) we get:

∂ S±(x y)= iγµ + m ∆±(x y) (3.141) − ∂(xµ yµ) − − Then, following the same procedure as we did for the scalar ﬁeld we get:

d4p i(/p + m) S (x y)= e−ip(x−y) (3.142) F − (2π)4 p2 m2 + iε Z − where ε< 1. As a ﬁnal remark, we note that the Feynman propagator satisﬁes the equation

(i∂/ m)S (x y)= iδ(4)(x y) (3.143) x − F − − Thus, it is a Green’s function for the Dirac operator.

47 48 Chapter 4

The Electromagnetic Field

In this section we will discuss the quantization of the electromagnetic ﬁeld.

4.1 Maxwell’s equations

Maxwell’s monumental equations in the absence of any sources are: E~ = 0 (4.1) ∇ · ∂E~ B~ = (4.2) ∇× ∂t B~ = 0 (4.3) ∇ · ∂B~ E~ = (4.4) ∇× − ∂t The process of solving these equations can be simplified with the introduction of the scalar potential φ(~x, t) (not to be confused with the real scalar field of the second chapter - this is not a quantum field, yet) and the vector potential A~(~x, t), such that:

∂A~ B~ = A~ and E~ = φ (4.5) ∇× −∇ − ∂t which ensure that equations (4.3) and (4.4) are true. In a more sophisticated approach, we can introduce the four-vector Aµ(~x, t) = (φ, A~), and derive Maxwell’s equations from the La- grangian: 1 1 = (∂ A )(∂µAν)+ (∂ Aµ)2 (4.6) L −2 µ ν 2 µ Then, using the Euler-Lagrange equation we get: ∂ ∂ L = ∂2Aν + ∂ν (∂ Aρ)= ∂ (∂µAν ∂νAµ)= ∂ F µν = 0 (4.7) µ ∂(∂ A ) − ρ µ − − µ µ ν where Fµν is the field strength which is defined as F ∂ A ∂ A (4.8) µν ≡ µ ν − ν µ Thus, the components of the field strength are:

0 Ex Ey Ez Ex 0 Bz By Fµν = − −  (4.9) E B 0 B − y z − x  E B B 0  − z − y x    49 Furthermore, we can rewrite the Lagrangian of the electromagnetic field using the field strength in an even more compact form: 1 = F F µν (4.10) L −4 µν The first two of the Maxwell’s equations are given by the Euler-Lagrange equation. The field strength also satisfies the Bianchi identity:

∂λFµν + ∂µFνλ + ∂νFλµ = 0 (4.11) which gives the last two of Maxwell’s equations.

4.2 Gauge symmetry

Notice that the Lagrangian for the electromagnetic field (equation (4.6) or (4.10)) does not depend on the term A˙ 0. This means that it is not a dynamical variable. Thus, if we know the initial conditions for Ai and A˙ i, where i = 1, 2, 3, then A0 can be determined by the equation E~ = 0, which, when expressed in terms of the potential, is: ∇ · ∂A~ 2A + = 0 (4.12) ∇ 0 ∇ · ∂t whose solution is: ( ∂A/∂t~ )(~x ′) A (~x )= d3x′ ∇ · (4.13) 0 4π ~x ~x ′ Z | − | We see that A0 is not independent. Furthermore, according to the definition of the field strength (4.8), if we make the following change: A (x) A (x)+ ∂ λ(x) (4.14) µ → µ µ where λ(x) can be any function that vanishes sufficiently quickly at spatial infinity, the field strength remains invariant:

F ∂ (A (x)+ ∂ λ(x)) ∂ (A (x)+ ∂ λ(x)) = F (4.15) µν → µ ν ν − ν µ µ µν This kind of symmetry is called gauge symmetry. This symmetry is very diﬀerent from the ones we have seen so far. Remember that all the symmetries we have exploited with Noether’s theorem were global symmetries: they acted the same at all points in spacetime. While the symmetries used in Noether’s theorem map one physical state to another which has the same properties, the gauge symmetry expresses the fact that two states which are related by gauge symmetry are identical. If we examine the equation (4.7):

[η (∂ρ∂ ) ∂ ∂ ]Aν = 0 (4.16) µν ρ − µ ν we see that the operator [η (∂ρ∂ ) ∂ ∂ ] is not invertible, as we have no way to distinguish µν ρ − µ ν between Aµ and Aµ + ∂µλ. Since the potential Aµ is not a physical object, we have no problem to identify these two quantities, but we should be aware that there is a redundancy in our description. Thus, the theory of electromagnetism is similar to an enlarged phase space which contains many gauge orbits; to proceed we need to make sure that the gauge we pick cuts all these orbits. Diﬀerent representative conﬁgurations of a physical state are called gauges. One of the two gauges we will use in this thesis is the Lorenz gauge:

µ ∂µA = 0 (4.17)

50 ′ ′ µ We can always choose such a gauge: if we have a gauge ﬁeld Aµ such that ∂µA = f(x) then we choose A = A′ + ∂ λ where ∂ ∂µλ = f(x); this equation always has a solution. The µ µ µ µ − Lorenz gauge is often called Lorenz gauge condition, as it does not completely ﬁx the gauge and µ we can make a further gauge transformation with ∂µ∂ λ = 0 which has non-trivial solutions. The second gauge we will use is the Coulomb gauge which is:

A~ = 0 (4.18) ∇ · In fact we can impose the Lorenz gauge condition and the Coulomb gauge. Using equation (4.13) we see that the Coulomb gauge implies that:

A0 = 0 (4.19)

However, Coulomb gauge has a major disadvantage: it isn’t Lorentz invariant.

4.3 Quantization of the electromagnetic ﬁeld

In this section we will proceed with the quantization of the electromagnetic field, first in Coulomb gauge and then in Lorenz gauge. µ The momentum π conjugate to the field Aµ is:

∂ π0 = L = 0 (4.20) ∂A˙0 ∂ πi = L = F 0i = Ei (4.21) ∂A˙i −

0 We see that π = 0, as A0 is not a dynamical variable.

Coulomb gauge

In Coulomb gauge we have the equation (remember that A0 = 0):

µ ∂µ∂ A~ = 0 (4.22) whose solution is: d3p A~ = ξ~(~p )eip·x (4.23) (2π)3 Z with p2 = ~p 2. Using the constraint A~ = 0, we get: 0 | | ∇ · ξ~ ~p = 0 (4.24) · so the vector ξ~ must be perpendicular to the momentum ~p, and can be expressed in terms of two orthonormal vectors ~ǫ1 and ~ǫ2 (which of course are perpendicular to the momentum), with ~ǫ (~p ) ~ǫ (~p )= δ . These two vectors are the two polarization states of the photon. We should r · s rs note that such a base can not be picked continuously for every ~p, due to Brouwer’s so called hairy ball theorem. The completeness relation for the polarization vectors is:

2 pipj ǫi (~p ) ǫj (~p )= δij (4.25) r r − ~p 2 r=1 X | | 51 Now, we proceed to quantization. Our initial attempt would be to impose the commutation relations:

i j [Ai(~x ), Aj(~y )] = [E (~x ), E (~y )] = 0 [A (~x ), Ej(~y )] = i δj δ(3)(~x ~y ) i i − However, these commutation relations are not consistent with the constraints:

A~ = E~ = 0 ∇ · ∇ · According to the commutation relations above, we get:

[ A~(~x ), E~ (~y )] = i 2δ(3)(~x ~y ) = 0 ∇ ∇ ∇ − 6 So, we must re-express the commutators so that they respect the constraints. The correct commutation relation are:

[Ai(~x ), Aj (~y )] = 0 [Ei~x ), Ej (~y )] = 0 (4.26) ∂ ∂ [A (~x ), E (~y )] = i δ i j δ(3)(~x ~y ) i j ij − 2 − ∇ Now, we can write A~ in the following expansion:

3 2 d p 1 r i~p·~x r † −i~p·~x A~(~x )= ~ǫr(~p ) a e + a e (4.27) (2π)3 2 ~p ~p ~p Z | | Xr=1 h i 3 p 2 ~ d p ~p r i~p·~x r † −i~p·~x E(~x )= 3 ( i) | | ~ǫr(~p ) a~p e a~p e (4.28) (2π) − r 2 − Z Xr=1 h i where ~ǫ (~p ) ~p = 0 and ~ǫ (~p ) ~ǫ (~p )= δ . Using the expansions (4.27) and (4.28) the commu- r · r · s rs tation relations (4.26) become:

r s [a~p , a~q ] = 0 r † s † [a~p , a~q ] = 0 (4.29) [ar , as †] = (2π)3 δrs δ(3)(~p ~q) ~p ~q − The Hamiltonian in Coulomb gauge is:

H = d3xπiA˙ (4.30) i −L Z which, after using the expansions (4.27) and (4.28) and applying normal ordering, becomes:

2 d3p H = ~p ar †ar (2π)3 | | ~p ~p Z Xr=1

52 Lorenz gauge If we impose the Lorenz gauge condition (4.17), the equations of motion become:

µ ν ∂µ∂ A = 0 (4.31)

In order to make this equation to arise naturally from the Lagrangian of the free electromagnetic field, we will change it to: 1 1 = F F µν (∂ Aµ)2 (4.32) L −4 µν − 2 µ from which we get the equation (4.31). Furthermore, we could try a more general Lagrangian of the form: 1 1 = F F µν (∂ Aµ)2 (4.33) L −4 µν − 2α µ with arbitrary α. The quantization of the electromagnetic field does not depend on the choice of α, which are referred to as different gauges. Here we will use α = 1, which is called the Feynman gauge. Another common choice is α = 0, which is called the Landau gauge. We proceed by calculating the momentum conjugate to Aµ:

0 ∂ µ π = L = ∂µA ∂A˙0 − ∂ πi = L = ∂iA0 A˙ i (4.34) ∂A˙i − No, we impose the commutation relations:

[Aµ(~x ), Aν (~y )] = 0 [πµ(~x ),πν (~y )] = 0 (4.35) [A (~x ),π (~y )] = iη δ(3)(~x ~y) µ ν µν − µ and then we expand Aµ(~x) and π (~x) to get:

3 3 d p 1 λ λ i~p·~x λ † −i~p·~x Aµ(~x )= ǫµ(~p ) a e + a e (4.36) (2π)3 2 ~p ~p ~p Z | | Xλ=0 h i 3 3 µ d p p ~p µ λ λ i~p·~x λ † −i~p·~x π (~x )= 3 | |(+i) (ǫ ) (~p ) a~p e a~p e (4.37) (2π) r 2 − Z λX=0 h i Here, we have four polarization vectors ǫλ(~p ) instead of three that we had in the Coulomb gauge. We pick ǫ0 to be timelike and ǫ1, ǫ2 and ǫ3 to be spacelike, with normalization:

′ ′ ǫλ ǫλ = ηλλ (4.38) · We choose ǫ1 and ǫ2 to be perpendicular to the momentum,and ǫ3 to be the longitudinal polarization. Then, the commutation relations above, using their expansion, become:

λ λ′ [a~p , a~q ] = 0 λ † λ′ † [a~p , a~q ] = 0 (4.39) ′ ′ [aλ , aλ †]= ηλλ (2π)3δ(3)(~p ~q ) ~p ~q − −

53 Notice, that for the timelike annihilation and creation operators we get a strange minus sign:

[a0 , a0 †]= (2π)3δ(3)(~p ~q ) (4.40) ~p ~q − − Now, we deﬁne the vacuum 0 as: | i aλ 0 = 0 (4.41) ~p | i Then, particles can be created by: ~p, λ = aλ † 0 (4.42) | i ~p | i For spacelike polarization states, everything is ﬁne. For timelike polarization though, the state ~p, 0 has negative norm(!): | i ~p, 0 ~q, 0 = 0 a0 a0 † 0 = (2π)3δ(3)(~p ~q ) (4.43) h | i h | ~p ~q | i − − A Hilbert space with negative norm implies negative probabilities. Something must be wrong. µ Remember that we have not used the the constraint equation ∂µA = 0. First, we decompose + − the potential as Aµ(x)= Aµ (x)+ Aµ (x) with:

3 3 + d p 1 λ λ −ip·x Aµ (x)= ǫµ(~p )a e (4.44) (2π)3 2 ~p ~p Z | | Xλ=0 3 3 − d p p1 λ λ † +ip·x Aµ (x)= ǫµ(~p )a e (4.45) (2π)3 2 ~p ~p Z | | Xλ=0 Note that the signs in the exponents havep switched comparing to the equations (4.36) and (4.37) as we have moved to four vectors and used the Minkowski metric. Now, we impose the Gupta-Bleuler condition: ∂µA+ Ψ = 0 (4.46) µ | i which ensures that: Ψ′ ∂µA Ψ = 0 (4.47) µ | i The physical states Ψ span a physical Hilbert space phys. H Suppose now that we have a basis for the Fock space of our theory. Each element of this basis can be written as Ψ = ψT φ , where ψT consists only of transverse photons (created 1 † 2 † | i | i | i | i 0 † by a~p and a~p ), and φ consist of timelike photons (created by a~p ) and longitudinal photons 3 † | i (created by a~p ). The Gupta-Bleuler condition implies that:

(a3 † a0 †) φ = 0 (4.48) ~p − ~p | i which means that the physical states consist of both timelike and longitudinal photons. So, φ is a linear combination of states φ which consists of n pairs of timelike and longitudinal | i | ni photons: ∞ φ = C φ (4.49) | i n | ni nX=0 where φ is the vacuum. | 0i All the states that involve timelike and longitudinal photons now have zero norm:

φ φ = δ δ (4.50) h m| ni n0 m0 which means that the inner product in the physical Hilbert space in positive semi-deﬁnite. The zero norm states that are left, are considered to be gauge equivalent to the vacuum. Thus, two

54 states whose only diﬀerence is the number of timelike and longitudinal photons are thought to be identical. Working in the same way as we did before, we can show that the Hamiltonian in the Lorenz gauge is: 3 3 d p i † i 0 † 0 H = 3 ~p a~p a~p a~p a~p (4.51) (2π) | | − ! Z Xi=1 The equation (4.48) implies that Ψ a3 †a3 Ψ = Ψ a0 †a0 Ψ , which means that the contribu- h | ~p ~p | i h | ~p ~p | i tion form the timelike and longitudinal photons cancel each other in the Hamiltonian. Thus, the Hamiltonian is positive deﬁnite as it should be.

4.4 The photon propagator

To derrive the propagator for the electromagnetic field, we work in the same way as we did for the scalar and the Dirac field. First, we define the time ordering operator T as:

Aµ(x)Aν (y), x0 >y0 T Aµ(x)Aν (y) = (4.52) { } Aµ(y)Aν (x), y0 >x0 which is the same as the time ordering operator for the scalar ﬁeld. Following the same procedure, we get the propagator:

d4p i ηµν D µν (x y)= − e−ip·(x−y) (4.53) F − (2π)4 p2 + iε Z and if had used an arbitrary α we would get:

d4p i p p Dµν (x y)= − ηµν + (α 1) µ ν e−ip·(x−y) (4.54) F − (2π)4 p2 + iε − p2 Z

55 56 Chapter 5

Interacting Fields

So far we have studied the three important free fields: the scalar field, the Dirac field and the electromagnetic field. Now it’s time to turn to interacting fields. The interactions will take the form of higher order terms in the Lagrangian, which will be small perturbations: λ = n φn (5.1) L L0 − n! nX≥3 where the coefficients λn are called coupling constants. In order to make the contribution of those terms small, so that we can use perturbative techniques, we could naively choose λ 1. n ≪ However, using dimensional analysis we see that the Lagrangian has dimensions: [ ] = 4 (5.2) L which implies that λn must have dimensions: [λ ] = 4 n (5.3) n − So λn is not a dimensionless quantity. In fact, the interaction terms can be put into three categories. The first contains terms with n = 3 and [λ3] = 1, so the dimensionless parameter is λ3/E, where E has dimensions of energy and is usually the energy scale of the process we study. Thus, the term λ ψ3/3! becomes a small perturbation when E λ , but large when E λ . Terms 3 ≫ 3 ≪ 3 that behave in such a way are called relevant, because they are of great importance for low energies. For such terms we require that λ m. 3 ≪ The second category consists of terms with n = 4 and [λ4] = 0. So these terms are dimensionless, and it suffices to require that λ 1. These terms are called marginal. 4 ≪ The last category contains all terms λ < 0 for n 5. The dimensionless parameter is n ≥ λ En−4, which is negligible for small energies and becomes significant for higher energies. n · These terms are called irrelevant. Typically, only the relevant and the marginal terms are significant for the low energy processes that we will study in this thesis. However, we should note that this categorization is not absolute, as quantum corrections can change the character of an operator.

5.1 The interaction picture

In the previous chapters we have used either the Schr¨odinger picture or the Heisenberg picture depending on what suited our needs. Now, we will use the interaction picture1 (which

1It should be noted that according to Haag’s theorem ([2]), the interaction picture does not exist in interacting Quantum Field Theory. Following the usual approach to Quantum Electrodynamics, we will ignore this result.

57 is sometimes also called the Dirac picture) which lies somewhere in between. The interaction picture is useful when we have small perturbations to a Hamiltonian whose energy spectrum is known. In other words, it is ideal for dealing with those interacting ﬁelds that we will consider in this thesis. In the Schr¨odinger picture the states ψ are time-dependent, but the the operators | iS OS are independent of time. On the other hand, in the Heisenberg picture the states ψ are | iH independent of time, but the operators do depend on time:

(t)= eiHt e−iHt (5.4) OH OS ψ = eiHt ψ (5.5) | iH | iS Let’s turn now to the interaction picture. Suppose we have a Hamiltonian of the form:

H = H0 + Hint (5.6) where H0 determines the time dependence of the operators and Hint determines the time dependence of the states. We suppose that we know the energy spectrum of the Hamiltonian H0. In the interaction picture, the states and the operators are given by:

ψ = eiH0t ψ(t) (5.7) | iI | iS (t)= eiH0t e−iH0t (5.8) OI OS

The last equation is also true for Hint, which is time dependent. In the interaction picture, the interaction Hamiltonian is:

iH0t −iH0t HI = (Hint)I = e (Hint)S e (5.9)

Then, in the interaction picture, the Schr¨odinger equation for states is:

5.2 Dyson’s formula

We wish to solve the equation (5.10). We write it’s solution in the following form:

ψ(t) = U(t,t ) ψ(t ) (5.11) | iI 0 | 0 iI where U(t,t0) is a time evolution parameter, which we demand to be unitary and to satisfy the equations:

U(t1,t2) U(t2,t3)= U(t1,t3) U(t,t) = 1

58 Putting (5.11) into the equation (5.10), we get: dU i = H (t) U (5.12) dt I t ′ ′ Differential equations of such form have solutions like exp( i HI (t ) dt ). However, this is an − t0 operator differential equation and we face ordering issues. The actual solution is: R t t t1 2 U(t,t0) = 1+( i) dt1 HI(t1) + ( i) dt1 dt2 HI(t1)HI (t2) − t0 − t0 t0 Z 2 Z Z t t1 t +( i)3 dt dt dt H (t )H (t )H (t )+ . . . (5.13) − 1 2 3 I 1 I 2 I 3 Zt0 Zt0 Zt0 which can be verified by differentiation. Note that the various quantities in each term are time ordered (operators evaluated at later times are put to the left). This fact allows us to write the above solution in a much more compact way. The third term for example can be written as: t t1 1 t t dt dt H (t )H (t )= dt dt T H (t )H (t ) 1 2 I 1 I 2 2 1 2 { I 1 I 2 } Zt0 Zt0 Zt0 Zt0 where the integral on the right hand side counts everything twice. The same identity is true for all the higher order terms: t t1 tn−1 1 t t dt dt dt H (t ) H (t )= dt dt T H (t ) H (t ) 1 2 · · · n I 1 · · · I n n! 1 · · · n { I 1 · · · I n } Zt0 Zt0 Zt0 Zt0 Zt0 although now it is not possible to visualize it. Thus, we can write equation (5.13) into the more compact form: t U(t,t )= T exp i H (t′) dt′ (5.14) 0 − I Zt0 This is Dyson’s formula.

5.3 Wick’s theorem

In Dyson’s formula the various components of each term are time ordered. However, as we have seen in the previous chapters, it is very convenient to arrange these term in normal order (placing the annihilation operators to the right). Wick’s theorem allows us to make the transition from time ordered products to normal ordered products. Remember that when we were dealing with the scalar ﬁeld, we decomposed it as: φ(x)= φ+(x)+ φ−(x) where: 3 + d p 1 −ipx φ (x)= a~p e (2π)3 2E Z ~p and p d3p 1 φ−(x)= a† e+ipx (2π)3 2E ~p Z ~p Then, for x0 >y0 we have: p T φ(x)φ(y)= φ(x)φ(y)= = φ+(x)+ φ−(x) φ+(y)+ φ−(y) = + + − + − + − − + − = φ (x)φ (y)+ φ(x)φ (y)+ φ (y)φ (x)+ φ (x)φ (y) + [φ (x), φ (y)]

59 Notice that the last line is normal ordered, but we have picked up the commutator [φ+(x), φ−(y)], which is the propagator D(x y). So: − T (φ(x)φ(y)) = N (φ(x)φ(y)) + D(x y) − Doing the same calculation for y0 >x0, we get:

T (φ(x)φ(y)) = N (φ(x)φ(y)) + D(y x) − and combining these two equations:

T (φ(x)φ(y)) = N (φ(x)φ(y))+∆ (y x) F − where ∆ (y x) is the Feynman propagator. F − Motivated by this, we deﬁne the contraction of a pair of ﬁelds in an operator product ...φ(x1) ...φ(x2) . . . , to mean the replacement of these two operators with the Feynman propagator, while leaving all the other operators untouched. We use the notation:

. . . φ(x1) ...φ(x2) . . .

For instance: φ(x)φ(y)=∆ (x y) F − Wick’s theorem. For any collection of ﬁelds φ1 = φ(x1), φ2 = φ(x2), ..., φn = φ(xn) we have:

T (φ1 ...φn)= N (φ1 ...φn + all possible contractions)

Proof. Wick’s theorem is proved by induction. We proved that it is true for n = 2 earlier. Suppose that it is true for φ2 ...φn. Next, we add φ1. Without loss of generality we will suppose that x0 x0 x0 (if it is not the case, we can just relabel the points). Then, we can 1 ≥ 2 ≥···≥ n place φ1 to the left of the time ordered product:

+ − T (φ1 ...φn) = (φ1 + φ1 )N (φ2 ...φn + all possible contractions)

− + The φ1 term is already where it should be. On the other hand, the φ1 term must move to − the right of the normal ordered string of operators. Each time it moves to the right of a φk operator, with k = 2,...,n, we get a factor of:

φ φ =∆ (x x ) 1 k F 1 − k Following this procedure we get all the possible contractions of the ﬁelds.

5.4 Scattering

All the machinery we have developed so far, can be used to calculate the quantum amplitudes for various scattering processes. In order to do so, we will make the assumption that the initial and final states of the scattering are eigenstates of the free theories, i.e. we consider the initial sate i at t and the final state f at t + to be eigenstates of the Hamiltonian of | i → −∞ | i → ∞ the free field H0. So, initially the particles are very far from each other and do interact, and as they approach each other they interact for a very brief amount of time. The amplitude from i | i to f is: | i lim f U(t+,t−) i f S i (5.15) t±→±∞ h | | i ≡ h | | i

60 where S is the S-matrix. However the assumption we made is a little weak, as in field theories particles are never alone, ever if they are very far from each other. Since every time we calculate a S-matrix element, we get a delta function factor which ensures the preservation of four momentum, we will define the amplitude by peeling of this Afi delta function: f S 1 i = i (2π)4δ(4)(p p ) (5.16) h | − | i Afi F − I where pI and pF are the initial and the final four momenta, and the i factor is just a convention.

Fermi’s Golden Rule As in non-relativistic quantum mechanics, the probabilities are the (modulus) square of the quantum amplitudes. As we mentioned above, all the S-matrix elements include a delta function, and consequently the probabilities include the square of a delta function, which is the result of working in an inﬁnite space. Consider two energy eigenstates m and n (with E = E ) for the leading order in the | i | i m 6 n interaction we get: t t m U(t) n = i m dt H (t) n = i m H n dt′ eiωt = h | | i − h | I | i − h | int | i Z0 Z0 eiωt 1 = m H n − (5.17) − h | int | i ω where ω = E E . Thus, the probability of transition from n to m in time t, is: m − n | i | i 1 cos ωt P (t)= m U(t) n 2 = 2 m H n 2 − (5.18) n→m | h | | i | | h | int | i | ω2 So most transitions occur in a region between energy eigenstates separated by ∆E = 2π/t, and as t , equation (5.18) becomes a delta function. The normalization is: →∞ +∞ 1 cos ωt dω − = πt ω2 ⇒ Z−∞ 1 cos ωt − πtδ(ω) as t ω2 → →∞ Next, we consider the probability of transition to a cluster of energy eigenstates of density ρ(E): 1 cos ωt P = dE ρ(E ) 2 m H n 2 − n→m m m | h | int | i | ω2 Z and in the limit t : →∞ P 2π m H n 2ρ(E )t as t (5.19) n→m → | h | int | i | m →∞ which gives us the probability of transition per unit time for states with approximately the same energy E E = E: m ∼ m P˙ = 2π m H n 2ρ(E ) (5.20) n→m | h | int | i | m which is of course Fermi’s Golden Rule. Notice that nowhere in this discussion did we come across the square of a delta function, as we mentioned above. However, if we weren’t that careful and tried to compute the amplitude from the state n at t to the state m at t + we would get: | i → −∞ | i → ∞ t=+∞ i m H (t) n = i m H n 2πδ(ω) (5.21) − h | I | i − h | int | i Zt=−∞ 61 Thus, taking the square of this expression to ﬁnd the probability we get the promised square of the delta function. Revisiting the whole argument, we realize that the extra delta function comes from the fact that the equation (5.21) refers to the transition amplitude in t . So, → ∞ if we write the delta function as:

(2π)2δ2(ω) = (2π)δ(ω)T (5.22) where T is shorthand for t . Then, diving by the time T to get the probability of transition →∞ per unit time: P˙ = 2π m H n 2δ(ω) (5.23) n→m | h | int | i | which after adding the density of energy eigenstates, is again Fermi’s Golden Rule. The method that used here can be used whenever we get the square of delta functions; we can re-interpret them as spacetime volume factors.

i i = (2π)3 2E δ(3)(0) = 2E V (5.25) h | i ~pI ~pI where we have replaced the delta function with volume of the three dimensional space. In a similar way: f f = 2E V (5.26) h | i ~pI ﬁnalY states If we boost ourselves to a inertial system where the initial particle is at rest, then ~pI = 0 and

E~pI = m, and the decay probability becomes:

2 fi 4 (4) 1 P = |A | (2π) δ (pI pF )V T (5.27) 2mV − 2E~pI V finalY states Next, we divide this expression by the time T to obtain the transition probability per unit time 3 3 and integrate over all possible momenta of the final state (V d pi/(2π) ), to get: 3R 4 (4) d pi 1 dΠ = (2π) δ (pF pI) 3 (5.28) − (2π) 2E~pi finalY states

Notice the presence of the Lorentz invariant term 1/2E~pi . Thus, we have a Lorentz invariant measure. Then, we sum over all final states with different types and numbers of particles, to finally get the the decay probability per unit time Γ = P˙ : 1 Γ= 2dΠ (5.29) 2m |Afi| finalX states Z Γ is called the width of the particle and relates to the half-life via: 1 τ = (5.30) Γ

62 Cross sections Now we will consider the collision of two particle beams. The cross section (σ) of the collision is the fraction of the times that they collide. The incoming ﬂux F is the number of incoming particles per unit area per unit time. The total number of scattering events then is:

N = Fσ (5.31)

Then, we deﬁne the diﬀerential cross section (dσ) as the probability for a given scattering process to happen in the solid angle (θ, φ), and using the relation for probability per unit time which we found above, we get:

1 1 2 dσ = fi dΠ (5.32) 4E1E2V F |A | where E1 and E2 are the energies of the incoming particles. If we boost ourselves to the center of mass frame of the collision, the ﬂux is given by:

~u ~u F = | 1 − 2| (5.33) V and putting this into equation (5.32), we get:

1 1 dσ = 2dΠ (5.34) 4E E ~u ~u |Afi| 1 2 | 1 − 2| In the special case where we have the scattering of two particles in the initial state that results into two particle in the final state, and all four particles have the same mass (including the case of the limit m 0), the differential cross section becomes: → dσ 2 = |Afi| (5.35) dΩ 64π2E2 which we can use to compute the differential cross section of any process we want. Now all that remains is to compute the the amplitude . Afi 5.5 Coupling various fields

It’s now time to start building Lagrangians which contain more than one free field. We want to construct a Lagrangian which couples the potential Aµ to scalar fields or spinor fields. We will do so, by introducing the Lagrangian:

1 = F F µν jµA (5.36) L −4 µν − µ where jµ is a function of the scalar or spinor ﬁeld. Then, the Euler Lagrange equations are:

µν µ ∂µF = j (5.37) and we impose the condition: µ ∂µj = 0 (5.38) which means that jµ must be a conserved current (and hence the choice of the symbol).

63 Coupling to fermions The Dirac Lagrangian: = ψ¯(i∂/ m)ψ (5.39) L − has the internal symmetry ψ e−iaψ and ψ¯ e+iaψ¯ (where a R). Thus, we can couple the → → ∈ electromagnetic ﬁeld to fermions via the Lagrangian: 1 = F F µν + ψ¯(i∂/ m)ψ eψγ¯ µA ψ (5.40) L −4 µν − − µ where we have introduced the coupling constant e. As we have discussed in a previous chapter, the Lagrangian for the electromagnetic ﬁeld has a very important symmetry, the gauge symmetry. Does the interacting theory still have it? We rewrite the Lagrangian as: 1 = F F µν + ψ¯(iD/ m)ψ (5.41) L −4 µν − where we have introduced the covariant derivative:

Dµψ = ∂µψ + ieAµψ (5.42)

Under a gauge transformation we have:

A A + ∂ λ µ → µ µ ψ e−ieλψ → where λ = λ(x) is an arbitrary function. Let’s see how Dµψ transforms:

D ψ ∂ (e−ieλψ)+ ie(A + ∂ λ)e−ieλψ = e−ieλD ψ (5.43) µ → µ µ µ µ So the covariant derivative under a gauge transformation changes only by a phase factor. Thus, the Lagrangian (5.41) is gauge invariant. Let’s return now to the coupling constant that we suspiciously called e. As the reader might have guessed, it has the interpretation of the electric charge of the fermion ψ. To see this, note that the Euler-Lagrange equations imply that:

µν ν ∂µF = j (5.44) and the j0component is the charge density. The total charge Q is:

Q = e d3x ψ¯(~x )γ0ψ(~x ) (5.45) Z which after quantization becomes:

2 d3p Q = e (bs †bs cs †cs) (5.46) (2π)3 ~p ~p − ~p ~p s=1 Z X which is the number of particles minus the number of antiparticles, multiplied by the charge. In Quantum Electrodynamics it is common to replace the constant of the electric charge e, to the ﬁne structure constant α, which is a dimensionless quantity given by:

e2 1 α = (5.47) 4π~c ≃ 137 64 So, in natural units ~ = c = 1, the electric charge is e = √4πα 0.3. ≃ Finally, before moving on, we will show that the charge conjugation matrix C that we introduced in the chapter of the free Dirac ﬁeld, performs charge conjugation. By taking the complex conjugate of the Dirac equation we get:

(iγµ∂ eγµA m)ψ = 0 µ − µ − ⇒ ( i(γµ)∗∂ e(γµ)∗A m)ψ∗ = 0 − µ − µ − and using the equation:

C†γµC = (γµ)∗ − and the deﬁnition

ψ(c) = Cψ∗ we get: (iγµ∂ + eγµA m)ψ(c) = 0 (5.48) µ µ − Thus, the charge conjugate spinor satisfies the Dirac equation with charge e instead of +e. − Coupling to scalars The real scalar field lacks a conserved current, and consequently it can not be coupled to a gauge field. The complex scalar field on the other does have such a conserved current. From here and on we will denote the complex scalar field by φ, so that we can distinguish it from the Dirac field. As we shall not use the real scalar field anymore, there won’t be any confusion. The complex scalar field has the symmetry φ e−iaφ. → In order to couple the electromagnetic field to the complex scalar field we use again the covariant derivative: φ = ∂ φ + ieA φ (5.49) Dµ µ µ which under a gauge transformation transforms as φ e−ieλ φ. Thus, a gauge invariant Dµ → Dµ Lagrangian for coupling the electromagnetic field to the complex scalar field is: 1 = F F µν + φ∗ µφ m2 φ 2 (5.50) L −4 µν Dµ D − | | The procedure we followed to make the previous two Lagrangians Lorentz invariant, is called minimal coupling. This approach is quite general: if we have a U(1) symmetry which we would like to couple to a gauge field, we can just replace all derivatives with covariant derivatives.

Quantum Electrodynamics The Lagrangian we use for the study of the interactions of electrons with photons (Quantum Electrodynamics) is: 1 = F F µν + ψ¯(iD/ m)ψ (5.51) L −4 µν − where Dµ = ∂µ + ieAµ (5.52) We will be working in the Lorenz gauge. It is this Lagrangian that we will use to study Bhabha scattering.

65 5.6 Feynman diagrams

There is very useful and undeniably elegant way to represent the expressions that result from Wick’s theorem using diagrams. We will present here some simple rules for turning these diagrams into analytic expressions (and vice versa). These diagrams are called Feynman diagrams. They represent the expansion of f S i (actually we want to compute f (S 1) i , h | | i h | − | i as we are not interested in processes that we have no scattering). Each of these diagrams is in one-to-one correspondence with the terms in the expansion for f (S 1) i . h | − | i To make a Feynman diagram:

We draw an external line for every particle in the initial state i and every particle • | i particle in the ﬁnal state f . We denote photons by wavy lines, scalars by dotted lines | i and fermions by solid lines. In each line we assign a directed momentum p and we use an arrow to denote its charge. We use incoming arrows for particles and outgoing arrows for antiparticles in the initial state and vise versa for the ﬁnal state.

We join the external lines together with trivalent vertices. • We draw all possible distinct diagrams with appropriate external legs and impose four- • momentum conservation at each vertex.

Feynman rules for scalars Then, in order to compute the amplitude i : Afi We write down a factor of ( ig) at each vertex. • − For each internal line we write down the propagator. • We integrate over momentum k ﬂowing through each loop d4k/(2π)4 • However in this thesis we will not consider diagrams with loops,R but we will restrict ourselves to tree level diagrams.

Feynman rules for fermions To each incoming fermion of momentum p and spin r, we associate a spinor ur(~p ), and • to each outgoing fermion we associateu ¯r(~p ).

To each incoming anti-fermion with momentum p and spin r, we associate a spinorv ¯r(~p ) • and to each outgoing anti-fermion we associate vr(~p ).

66 To each vertex we put a factor of iλ. • − To each internal line we put a factor of the relevant propagator: •

We impose momentum conservation at each vertex and integrate over undetermined loop • momenta.

We add extra minus signs to account for statistics. • Note that the arrows of the fermion lines must follow a consistent ﬂow throughout the diagram; this ensures the conservation of the fermion number.

Feynman rules for Quantum Electrodynamics Vertex: •

Photon propagator: •

Fermion propagator: •

For external lines in the diagrams we attach

Photons: We add a polarization vector ǫµ /ǫν for incoming/outgoing photons. • in out Fermions: We add a spinor ur(~p )/u¯r(~p ) for incoming/outgoing fermions. We add a spinor • v¯r(~p )/vr(~p ) for incoming/outgoing anti-fermions.

5.7 Trace technology

Now, we will turn to a diﬀerent subject. After the translation of the Feynman diagram into a quantum amplitude, we will have to calculate the traces of several combinations of products of the gamma matrices. Let’s suppose that we have a product of n such matrices. For our purposes, we will need to ﬁnd the trace of the product of two, three and four gamma matrices.

67 Of course in the case of n = 0, we have:

trI = 4 (5.53)

Then, for a single matrix, n = 1, using equation (3.14) we have:

tr γµ = 0 (5.54)

Next, we will compute the trace of a product of an odd number of matrices. Since (γ5)2 = I, we have: tr γµ = tr γ5γ5γµ = tr γ5γµγ5 − where we have used the fact that γµ, γ5 = 0. Then: { } tr γµ = tr γ5γµγ5 = tr γ5γ5γµ = tr γµ − − − where we have used the cyclic property of the trace. So, for n gamma matrices we would get n minus signs when we move the γ5 matrix to the right. Consequently:

tr[γµγν . . . γρ]=0 for n = 2κ + 1, κ Z ∈ n For n = 2, the trace is: | {z }

tr (γµγν)= tr(2ηµν I γνγµ) = 8ηµν tr (γµγν) · − − where we have once again used the cyclic property of the trace. Thus:

tr (γµγν ) = 4ηµν (5.55)

For n = 4 we have:

tr (γµγνγργσ)= tr (2ηµν γργσ γνγµγργσ) − = tr (2ηµν γργσ γν2ηµργσ + γνγρ2ηµσ γνγργσγµ) − − Then, using the cyclic property of the trace on the last term and moving it to the left hand side of the equation, we get:

tr (γµγνγργσ)= ηµν tr(γργσ) ηµρtr(γνγσ)+ ηµσtr(γνγρ) − = 4(ηµν ηρσ ηµρηνσ + ηµσηνρ) (5.56) − The same technique can be applied for the calculation of any even number of gamma traces, although for large n, this process can be tedious.

68 Chapter 6

Bhabha Scattering

In the last chapter of the thesis, we will use the techniques we developed in the previous chapters, to study an elementary process in Quantum Electrodynamics, the Bhabha scattering. The Bhabha scattering involves the scattering of an electron and a positron to an electron and a positron: e+ + e− e+ + e− → We will compute the unpolarized diﬀerential cross section of the Bhabha scattering in the ultra- relativistic limit, where E m , so we can ignore the electron mass. cm ≫ e Feynman diagrams There are two Feynman diagrams that contribute in the process:

e+ e+ e+ e+

k k′ k k′

q p p′ p p′

e− e− e− e− with amplitudes that we denote as (1) and (2) respectively. Using the Feynman rules we can A A translate the Feynman diagrams into amplitudes: iη i (1) =v ¯(k)( ieγµ)u(p) − µν u¯(p′)( ieγν )v(k′) (6.1) A − q2 − where q = p + k = p′ + k′, and iη i (2) =u ¯(p′)( ieγµ)u(p) − µν v¯(k)( ieγν )v(k′) (6.2) A − q2 − 69 where q = p p′ = k′ k. The total amplitude of the process is the sum of these two: − − i = i (1) + i (2) (6.3) A A A In order to simplify the expressions, we will introduce the Mandelstam variables:

2 ′ ′ 2 s = (p1 + p2) = (p1 + p2) t = (p p′ )2 = (p p′ )2 (6.4) 1 − 1 2 − 2 u = (p p′ )2 = (p p′ )2 1 − 2 2 − 1 ′ ′ where p1 and p2 are the momenta of the two initial particles and p1 and p2 are the momenta of the two ﬁnal particles. Then, the amplitudes can be written as:

ie2 i (1) = η v¯(k)γµu(p)u ¯(p′)γν v(k′) (6.5) A s µν ie2 i (2) = η u¯(p′)γµu(p)v ¯(k)γν v(k′) (6.6) A t µν Then, the square of the amplitude is:

2 = (1) 2 + (2) 2 + (1) (2) ∗ + (2) (1)∗ (6.7) |A| |A | |A | A A A A In most experiments, the beams of the initial particles are not polarized and the detector that we place to study the ﬁnal particles are blind to spin polarization. So, the measured cross section is an average over all possible spins of the ﬁnal and initial states. Thus, we actually have to calculate: 1 1 1 1 1 2 = (1) 2 + (2) 2 + (1) (2) ∗ + (2) (1)∗ (6.8) 4 |A| 4 |A | 4 |A | 4 A A 4 A A spinsX spinsX spinsX spinsX spinsX Now, we will work with each of these terms separately.

Computing (1) 2 |A | Let’s begin with the ﬁrst term:

1 1 ie2 ie2 (1) 2 = η v¯(k)γµu(p)u ¯(p′)γνv(k′) η v¯(k′)γρu(p′)¯u(p)γσv(k) 4 |A | 4 s µν − s ρσ spinsX spinsX e4 = η η v¯(k)γµu(p)u ¯(p′)γν v(k′)¯v(k′)γρu(p′)¯u(p)γσv(k) 4s2 µν ρσ spinsX e4 = η η v¯ (k)γµ u (p)u ¯ (p′)γν v (k′)¯v (k′)γρ u (p′)¯u (p)γσ v (k) 4s2 µν ρσ a ab b c cd d e ef f g gh h spinsX e4 = ( 1)14 η η γµ u (p)¯u (p)γσ v (k)¯v (k)γν v (k′)¯v (k′)γρ u (p′)¯u (p′) − 4s2 µν ρσ ab b g gh h a cd d e ef f c spinsX 4 e µ σ ν ′ ρ ′ = ηµν ηρσ γ (/p + m)bgγ (k/ m)haγ (k/ m)deγ (/p + m)fc 4s2 ab gh − cd − ef 4 e µ σ ν ′ ρ ′ = ηµν ηρσ tr γ (/p + m)γ (k/ m) tr γ (k/ m)γ (/p + m) 4s2 − − 4 e µ σ ′ ρ ′ ν = ηµν ηρσ tr (k/ m)γ (/p + m)γ tr (k/ m)γ (/p + m)γ (6.9) 4s2 − − 70 The factor ( 1)14 is the result of fourteen commutations of the fermionic ﬁelds. In the penul- − timate line we have used the equations (3.98) and (3.99). Next, we compute each of the traces above separately, using the trace technology we developed in the previous chapter: tr (k/ m)γµ(/p + m)γσ = tr(kγ/ µ/pγσ + kγ/ µmγσ mγµ/pγσ m2γµγσ) − − − = tr(kγ/ µ/pγσ)+ tr(kγ/ µmγσ) m tr(γµ/pγσ) m2 tr(γµγσ) − − = k p tr(γργµγνγσ)+ mk tr(γργµγσ) mp tr(γµγργσ) m2 tr(γµγσ) ρ ν ρ − ρ − = k p (4ηρµηνσ 4ηρν ηµσ + 4ηρσηµν ) 4m2ηµσ ρ ν − − = (4kµpσ 4(k p)ηµσ + 4kσpµ) 4m2ηµσ (6.10) − · −

The next trace term is:

′ tr (k/ m)γρ(/p′ + m)γν − We notice that this term is connected to the ﬁrst trace, if we perform the interchange: k k′ → and p p′ and µ ρ and σ ν. Thus: → → →

′ tr (k/ m)γρ(/p′ + m)γν = (4k′ ρp′ ν 4(k′ p′)ηρν + 4k′ νp′ ρ) 4m2ηρν (6.11) − − · − Replacing the traces that we have computed, to equation (6.9), we get:

1 e4 (1) 2 = η η (4kµpσ 4(k p)ηµσ + 4kσpµ) 4m2ηµσ 4 |A | 4s2 µν ρσ − · − spins X (4k′ ρp′ ν 4(k′ p′)ηρν + 4k′ νp′ ρ) 4m2ηρν − · − 4e4 = η η (kµpσ (k p)ηµσ + kσpµ) m2ηµσ (k′ νp′ ρ (k′ p′)ηνρ + k′ ρp′ ν) m2ηνρ s2 µν ρσ − · − − · − 4e4 = (p k′)(k p′) (k′ p′)(k p) + (p p′)(k k′) m2(k p) s2 · · − · · · · − · 4e4 (k p) (k′ p′) 4(k′ p′) + (k′ p′) 4m2 − s2 · · − · · − 4e4 + (p p′)(k k′) (k′ p′)(k p) + (k p′)(p k′) m2(k p) s2 · · − · · · · − · 4e4 m2 (k′ p′) 4(k′ p′) + (k′ p′) 4m2 (6.12) − s2 · − · · −

Collecting all the terms, we get:

1 8e4 (1) 2 = (p p′)(k k′) + (p′ k)(p k′)+ m2(k′ p′)+ m2(k p) + 2m4 (6.13) 4 |A | s2 · · · · · · spins X

71 Computing (2) 2 |A |

We proceed with the calculation for the next term:

1 1 i e2 i e2 (2) 2 = η u¯(p′)γµu(p)¯v(k)γν v(k′) η v¯(k′)γρv(k)¯u(p)γσu(p′) 4 |A | 4 t µν − t ρσ spinsX spinsX e4 = η η u¯(p′)γµu(p)¯v(k)γν v(k′)¯v(k′)γρv(k)¯u(p)γσu(p′) 4t2 µν ρσ spinsX e4 = η η u¯ (p′)γµ u (p)¯v (k)γν v (k′)¯v (k′)γρ v (k)¯u (p)γσ u (p′) 4t2 µν ρσ a ab b c cd d e ef f g gh h spinsX e4 = ( 1)14 η η γµ u (p)¯u (p)γσ u (p′)¯u (p′)γν v (k′)¯v (k′)γρ v (k)¯v (k) − 4t2 µν ρσ ab b g gh h a cd d e ef f c spinsX 4 e µ σ ′ ν ′ ρ = ηµν ηρσ γ (/p + m)bgγ (/p + m)haγ (k/ m)deγ (k/ m)fc 4t2 ab gh cd − ef − 4 e µ σ ′ ν ′ ρ = ηµν ηρσ tr γ (/p + m)γ (/p + m) tr γ (k/ m)γ (k/ m) (6.14) 4t2 − −

The factor ( 1)14 is the result of fourteen commutations of the fermionic ﬁelds. In the penulti- − mate line we have used the equations (3.98) and (3.99). Let’s compute now each trace separately, using the trace technology:

tr γµ(/p + m)γσ(/p′ + m) = tr γµpγ/ σ/p′ + mγµγσ/p′ + mγµ/pγσ + m2γµγσ µ σ ′ 2 µ σ = tr γ pγ/ /p + m tr (γ γ ) ′ µ λ σ τ 2 µσ = 4pλpτ tr γ γ γ γ + 4m η = 4p p′ ηµληστ ηµσηλτ + ηµτ ηλσ + 4m2ηµσ λ τ − = 4 pµp′σ (p p′)ηµσ + p′ µpσ + m2ηµσ (6.15) − ·

Following the same procedure for the computation of the other trace, we get:

tr γν (k/′ m)γρ(k/ m) = 4 k′ νkρ (k k′)ηνρ + kνk′ ρ + m2ηνρ (6.16) − − − · 72 Using these two equations, we get:

1 4e4 (2) 2 = η η pµp′ σ (p p′)ηµσ + p′ µpσ + m2ηµσ k′ νkρ (k k′)ηνρ + kνk′ ρ + m2ηνρ 4 |A | t2 µν ρσ − · − · spins X 4e4 = p p′ (k′ νkρ (k k′)ηνρ + kνk′ ρ + m2ηνρ) t2 ν ρ − · 4e4 (p p′)η k′ νkρ (k k′)ηνρ + kν k′ ρ + m2ηνρ − t2 · νρ − · 4e4 + p′ p k′ νkρ (k k′)ηνρ + kνk′ ρ + m2ηνρ t2 ν ρ − · 4m2e4 + η k′ νkρ (k k′)ηνρ + kν k′ ρ + m2ηνρ t2 νρ − · 4e4 = (p k′)(p′ k) (k k′)(p p′) + (p k)(p′ k′)+ m2(p p′) t2 · · − · · · · · 4e4 (p p′) (k k′) 4(k k′) + (k k′) + 4m2 − t2 · · − · · 4e4 + (p′ k′)(p k) (k k′)(p p′) + (p′ k)(p k′)+ m2(p p′) t2 · · − · · · · · 4m2e4 + (k k′) 4(k k′) + (k k′) + 4m2 t2 · − · · 8e4 = (p k′)(p′ k) (k k′)(p p′) + (p k)(p′ k′)+ m2(p p′)+ m2 (p p′) 2m2 (k k′) t2 · · − · · · · · − · − · 8e4 = (p k′)(p′ k) + (p k)(p′ k′) m2(p p′) m2(k k′) + 2m4 (6.17) t2 · · · · − · − · Computing (1) (2) ∗ A A The terms (1) (2) ∗ and (2) (1)∗ are slightly more complicated. We have: A A A A 1 1 ie2 ie2 (1) (2) = η v¯(k)γµu(p)u ¯(p′)γνv(k′) η v¯(k′)γσv(k)¯u(p)γρu(p′) 4 A A 4 s µν − t ρσ spinsX spinsX e4 = η η v¯ (k)γµ u (p)u ¯ (p′)γν v (k′)¯v (k′)γσ v (k)¯u (p)γρ u (p′) 4st µν ρσ a ab b c cd d e ef f g gh h spinsX e4 = ( 1)15 η η γµ u (p)¯u (p)γν v (k′)¯v (k′)γσ v (k)¯v (k)γρ u (p′)¯u (p′) − 4st µν ρσ ab b g cd d e ef f a gh h c spinsX 4 e µ ρ ′ ν ′ σ = ηµν ηρσk/ γ /p γ /p γ k/ γ −4st fa ab bg gh hc cd de ef 4 e µ ρ ′ ν ′ σ = ηµν ηρσ tr(kγ/ /pγ /p γ k/ γ ) (6.18) −4st Note that in the penultimate step we have dropped all the mass terms. To calculate this trace, we will use the fact that:

µ µ ν µν ν µ µ µ γ /p = γ γ pν = pν(2η γ γ ) = 2p /pγ (6.19) − − and

µ ν 1 µ ν 1 µν 2 p//p = pµpνγ γ = pµpν γ , γ = pµpν2η = p = 0 (6.20) 2 { } 2

73 Then, we use the conservation of momentum: k′ = p + k p′, to replace k′: − 4 1 (1) (2) e µ ρ ′ ν ′ σ = ηµν ηρσ tr(kγ/ /pγ /p γ k/ γ ) 4 A A −4st spinsX 4 e µ ρ ′ ν ′ σ = ηµν ηρσ tr kγ/ pγ/ /p γ (/p + k/ p/ )γ (6.21) −4st − Now we will compute each of the term in the above trace separately:

µ ρ ′ ν σ µ ρ ρ ′ ν ν σ ηµν ηρσ tr kγ/ /pγ /p γ /pγ =ηµν ηρσ tr kγ/ (2p γ p/)/p (2p pγ/ )γ − − ′ µ ′ ν ρ ′ σ =4 tr(k/p//p /p) 2ηµν tr(kγ/ /p pγ/ p/) 2ηρσtr(k//pγ /p/p γ ) − − µ ρ ′ ν σ + ηµν ηρσ tr(kγ/ γ /p/p /pγ γ ) ′ ′ µ ′ ν ν =4 tr(k/p/(2(p p ) /p/p )) 2ηµν tr(kγ/ /p (2p γ /p)/p) · − − − ρ ρ ′ σ µ ρ ′ ′ ν σ 2ηρσtr(k//p(2p /pγ )/p γ )+ ηµν ηρσ tr(kγ/ γ (2(p p ) p/ p/)/pγ γ ) − − · − Then, using p2 = m2 = 0:

µ ρ ′ ν σ ′ 2 ′ ′ µ ′ ν 2 ηµν ηρσ tr kγ/ /pγ /p γ /pγ =8(p p )tr(k/p/) 4tr(kp/ /p ) 4tr(k//p/p /p) + 2ηµν tr(kγ/ /p γ p ) · − − ′ 2 ρ ′ σ ′ µ ρ ν σ 4tr(k/p//p /p) + 2ηρσtr(kp/ γ /p γ )+ ηµν ηρσ2(p p )tr(kγ/ γ /pγ γ ) − · µ ρ ′ ν σ ηµν ηρσtr(kγ/ γ /p /p/pγ γ ) − ′ ′ ′ ′ µ ρ ν σ =32(p p )(k p) 4tr(k//p/p /p) 4tr(k//p/p /p) + 2(p p )ηµν ηρσtr(kγ/ γ /pγ γ ) · · − − · ′ ′ ′ ′ µ ρ ν σ =32(p p )(k p) 8tr(k/(2(p p ) /p /p)/p) + 2(p p )ηµν ηρσtr(kγ/ γ /pγ γ ) · · − · − · The last term becomes:

µ ρ ν σ µ µ ρ νσ σ ν ηµν ηρσ tr(kγ/ γ /pγ γ )=ηµνηρσ tr((2k γ k/)γ /p(2η γ γ )) − − ρ σ µ ρ ρ σ =4 tr(k//p) 2ηρσ tr(γ /pγ k/) 2ηµρ tr(γ kγ/ /p) + 4ηρσ tr(kγ/ pγ/ ) − − ρ σ =4 tr(k//p) (2 + 2 4)ηρσ tr(γ /pγ k/) − − =16 k p · Consequently:

µ ρ ′ ν σ ′ ′ ′ ′ ηµνηρσ tr kγ/ /pγ /p γ /pγ =32(p p )(k p) 16(p p )tr(k//p) + 8tr(k//p /pp/) + 32(p p )(k p) · · − · · · =32(p p′)(k p) 64(p p′)(p k) + 32(p p′)(k p) · · − · · · · =0 (6.22)

We proceed with the computation of the next trace term, which is relatively easier:

µ ρ ′ ν σ µ ρ ′ ν σ σ ηµν ηρσ tr kγ/ pγ/ /p γ kγ/ = ηµν ηρσ tr kγ/ /pγ /p γ (2k γ k/) − µ ′ ρ µ ρ ′ ν σ = 2ηµν tr(kγ/ /pk//p γ ) ηµν ηρσ tr kγ/ /pγ /p γ γ k/ − µ µ ′ ν = 2ηµν tr (2k γ k/)/pk/p/ γ − ′ µ ′ ν = 4 tr(/pk//p k/) 2ηµν tr(γ k//pk//p γ ) − = 4 tr(/pk//p′k/) 8 tr(k/p/k/p/′) − = 4 tr(/pk/(2(p′ k) k//p′)) − · − = 8(p′ k)tr(/pk/) + 4tr(/pk/k//p′) − · = 32(p′ k)(p k) (6.23) − · ·

74 We continue with the next term:

µ ρ ′ ν ′ σ ′ σ µ ρ ′ ν ηµν ηρσtr(kγ/ /pγ /p γ /p γ )= ηµν ηρσtr(/p γ kγ/ pγ/ /p γ ) (6.24)

We can calculate this trace easily if we notice that using the cyclic permutation: k p′ p k → → → and rename the dummy indices to get:

µ ρ ′ ν ′ σ ′ ′ ηµν ηρσtr(kγ/ pγ/ /p γ /p γ )= 32(p p )(k p ) (6.25) − · ·

Thus:

4 1 (1) (2) e µ ρ ′ ν ′ σ = ηµν ηρσ tr kγ/ /pγ p/ γ (/p + k/ /p )γ 4 A A −4st − spins X e4 = 0 32(p′ k)(p k) + 32(p p′)(k p′) −4st − · · · · 8e4 = (p p′)(k p′) (p′ k)(p k) (6.26) − st · · − · ·

Computing (2) (1)∗ A A The last term we need to compute is (2) (1)∗. We have: A A 1 1 ie2 ie2 (2) (1)∗ = η u¯(p′)γρu(p)¯v(k)γσv(k′) η v¯(k′)γν u(p′)¯u(p)γµv(k) 4 A A 4 t ρσ − s µν spinsX spinsX e4 = η η u¯ (p′)γρ u (p)¯v (k)γσ v (k′)¯v (k′)γν u (p′)¯u (p)γµ u (k) 4st µν ρσ a ab b c cd d e ef f g gh h spinsX e4 = ( 1)15 η η γρ u (p)¯u (p)γσ v (k′)¯v (k′)γν u (p′)¯u (p′)γµ u (k)¯v (k) − 4st µν ρσ ab b g cd d e ef f a gh h c spinsX e4 = η η γρ u (p)¯u (p)γµ u (k)¯v (k)γσ v (k′)¯v (k′)γν u (p′)¯u (p′) −4st µν ρσ ab b g gh h c cd d e ef f a spinsX 4 e ρ µ σ ′ ν ′ = ηµν ηρσ tr(γ /pγ kγ/ k/ γ /p ) −4st 4 e µ σ ′ ν ′ ρ = ηµν ηρσ tr(/pγ kγ/ k/ γ /p γ ) (6.27) −4st

When we computed the previous term, we calculated the quantity:

4 4 e µ ρ ′ ν ′ σ 8e ′ ′ ′ ηµν ηρσ tr(kγ/ /pγ /p γ k/ γ )= (p p )(k p ) (p k)(p k) 4st st · · − · · Thus, to calculate (6.27) we only need to switch p k and p′ k′: ↔ ↔ 1 8e4 (2) (1)∗ = (k k′)(p k′) (k′ p)(k p) (6.28) 4 A A − st · · − · · spinsX

75 Combining our results

Now, using equations (6.13), (6.17), (6.26) and (6.28), we get:

1 1 1 1 1 2 = (1) 2 + (2) 2 + (1) (2) ∗ + (2) (1)∗ 4 |A| 4 |A | 4 |A | 4 A A 4 A A spinsX spinsX spinsX spinsX spinsX 8e4 = (p p′)(k k′) + (p′ k)(p k′)+ m2(k′ p′)+ m2(k p) + 2m4 s2 · · · · · · 8e4 + (p k′)(p′ k) + (p k)(p′ k′) m2(p p′) m2(k k′) + 2m4 t2 · · · · − · − · 8e4 8e4 (p p′)(k p′) (p′ k)(p k) (k k′)(p k′) (k′ p)(k p) − st · · − · · − st · · − · · Then, we drop all the mass terms to get the ﬁnal expression for the amplitude 2: |A| 1 8e4 8e4 2 = (p p′)(k k′) + (p′ k)(p k′) + (p k′)(p′ k) + (p k)(p′ k′) 4 |A| s2 · · · · t2 · · · · spinsX 8e4 (p p′)(k p′) (p′ k)(p k) + (k k′)(p k′) (k′ p)(k p) (6.29) − st · · − · · · · − · ·

Diﬀerential cross section

Now, we move the center of mass frame of reference. We set the angle between the electrons of the initial and the ﬁnal state to be θ. The various four momenta of the particles of the process are:

p = (E, ~p ) k = (E, ~p ) p′ = (E, ~p ′) k′ = (E, ~p ′) (6.30) − −

Then, we set ~p 2 = E2 m2 = E2 and compute all the inner products and the quantities t and − s of the equation (6.29):

p k = E2 + ~p 2 = 2E2 • · p′ k = E2 + ~p ~p ′ = E2 + ~p 2 cos θ = E2(1 + cos θ) • · ·

k′ k = E2 ~p ~p ′ = E2(1 cos θ) • · − · − p p′ = E2(1 cos θ) • · − p k′ = E2(1 + cos θ) • · p′ k′ = E2 + ~p ′ ~p ′ = 2E2 • · ·

t = (p p′)2 = 2m2 2p p′ = 2E2(1 cos θ) • − − · − − s = (p + k)2 = 4E2 • 76 Now using equation (5.35) from the fifth chapter, the differential cross section of the Bhabha scattering is: dσ e4 2 4 + (1 + cos θ)2 e4 1 = (1 + cos2 θ)+ + 2(1 cos2 θ) 4(1 + cos θ) dΩ 64π24E2 (1 cos θ)2 64π24E2 1 cos θ − − − ! − e4 2 4 + (1 + cos θ)2 2 1 cos2 θ 4(1+cos θ) = (1 + cos2θ)+ + − 64π24E2 (1 cos θ)2 1 cos θ − 1 cos θ − − − ! e4 1 1 + cos4 θ θ 2 cos2 θ = (1 + cos2 θ)+ 2 + 2 cos2 2 64π22E2 2 sin4 θ 2 − sin2 θ 2 2 ! e4 1 1 + cos4 θ 2 cos4 θ = (1 + cos2 θ)+ 2 2 (6.31) 64π22E2 2 4 θ − 2 θ sin 2 sin 2 ! This is our final result: the ultra-relativistic limit of the Bhabha scattering cross section. The figure below shows the differential cross section of the Bhabha scattering. We have used E = 200 GeV, which is roughly the energy in which operated the Large Electron - Positron Collider (LEP) in CERN, before it was closed.

Bhabha Scattering Differential Cross Section (200 GeV)

(pb) 30 Ω /d σ d 20

0.13 0 π/2 π 3π/2 2π Angle θ

Figure 6.1: The diﬀerential cross section dσ/dΩ (in picobarns) of the Bhabha scattering at E = 200 GeV.

77 78 Bibliography

[1] Bjorken J. D. and Drell S. D. - Relativistic Quantum Mechanics, first edition, Mc Graw-Hill, 1964 [2] Haag R. - On Quantum Field Theories, Matematisk-fysiske Meddelelser, 29, 12, 1955 [3] Itzykson C. and Zuber J. B. - Quantum Field Theory, first edition, Dover Publications, 2006 [4] Klauber, R. D. - Student Friendly Quantum Field Theory, second edition, Sandtrove Press, 2013 [5] Landau L. D. and Lifshitz E. M. - The Classical Theory of Fields (fourth revised English edition, translation Morton Hamermesh), Butterworth-Heinemann, 1975 [6] Peskin M. E. and Schroeder D. V. - An Introduction To Quantum Field Theory, first edition, Westview Press, 1995 [7] Tong D. - Quantum Field Theory, Lecture Notes for the University of Cambridge Part III of the Mathematical Tripos, 2006-2007 [8] Zee A. - Quantum Field Theory in a Nutshell, second edition, Princeton University Press, 2010 [9] Wikipedia (Wikipedia, The Free Encyclopedia) articles used: Canonical quantization • URL: http://en.wikipedia.org/wiki/Canonical_quantization online, accessed on: 28 May 2014 Lorenz gauge condition • URL: http://en.wikipedia.org/wiki/Lorenz_gauge_condition online, accessed on: 28 May 2014 Interaction picture • URL: http://en.wikipedia.org/wiki/Interaction_picture online, accessed on: 28 May 2014 Haag’s theorem • URL: http://en.wikipedia.org/wiki/Haag’s_theorem online, accessed on: 28 May 2014 Bhabha scattering • URL: http://en.wikipedia.org/wiki/Bhabha_scattering online, accessed on: 28 May 2014 [10] James T. Wheeler, Bhabha Scattering (1936) URL: http://www.physics.usu.edu/Wheeler/QFT/PicsII/QFT10Mar05Bhabha.pdf online, accessed on: 16 May 2014

79 [11] James T. Wheeler, Electron-positron annihilation into muon-antimuon pairs URL: http://www.physics.usu.edu/Wheeler/QFT/PicsII/QFT10Feb23Muon.pdf online, accessed on: 16 May 2014

The Figures 2.1, 2.2 (page 29) and the Feynman diagrams of pages 66-67 were adopted from Dr David Tong’s Lecture Notes [7]. The cover page was inspired by a template mentioned in the webpage: http://tex.stackexchange.com/questions/85904/ showcase-of-beautiful-title-page-done-in-tex .