Quantum Field Theory.. A Diagrammatic Approach [2019 draft] [2019 Draft ed.] 9781108486217

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Quantum Field Theory A Diagrammatic Approach Ronald Kleiss

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Simon Partlic (T`´re` ´s´t,1590- ?, 1649) astronomer, mathematician and physician

”...The Ancients were wont to draw Diagrams & thus divine Predictions for future Happenings, by Arts magickal or conjectural... likewise the Savants of the Future will learn to employ Diagrams ; yet not by Arts magickal, rather by Arts arithmetickal, algebraickal & by Geometrie and the Quadrature will they study to foretell the Events of Nature...” (attributed to Simon Partli´c)

Contents 0 Introductory remarks 0.1 Preface . . . . . . . . . . . . . 0.2 Basic tools . . . . . . . . . . . 0.2.1 Units and fundamental 0.2.2 Planck units . . . . . 0.2.3 Charges . . . . . . . . 0.2.4 Conventions . . . . . . 0.3 The P 4 Hall of Fame . . . . . 0.4 Exercise . . . . . . . . . . . .

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1 QFT in zero dimensions 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Probabilistic considerations . . . . . . . . . . . . . . . . . . 1.2.1 Quantum field and action . . . . . . . . . . . . . . . 1.2.2 Green’s functions, sources and the path integral . . . 1.2.3 Connected Green’s functions . . . . . . . . . . . . . . 1.2.4 The free theory . . . . . . . . . . . . . . . . . . . . . 1.2.5 The ϕ4 model and perturbation theory . . . . . . . . 1.2.6 The Schwinger-Dyson equation . . . . . . . . . . . . 1.2.7 The Schwinger-Dyson equation for the field function . 1.3 Diagrammatics . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Feynman diagrams . . . . . . . . . . . . . . . . . . . 1.3.2 Feynman rules . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Symmetries and multiplicities . . . . . . . . . . . . . 1.3.4 Vacuum bubbles . . . . . . . . . . . . . . . . . . . . 1.3.5 An equation for connected graphs . . . . . . . . . . . 1.3.6 Semi-connected graphs and the SDe . . . . . . . . . . 1.3.7 The path integral as a set of diagrams . . . . . . . . 3

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2 On renormalization 2.1 Doing physics : mentality against reality . . . . . . . . . . . . 2.1.1 Physics vs. Mathematics . . . . . . . . . . . . . . . . . 2.1.2 The renormalisation program : an example . . . . . . . 2.2 A handle on loop divergences . . . . . . . . . . . . . . . . . . 2.2.1 A toy : the dot model . . . . . . . . . . . . . . . . . . 2.2.2 Nonrenormalizable theories . . . . . . . . . . . . . . . . 2.3 Scale dependence . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Scale-inpendent scale dependence . . . . . . . . . . . . 2.3.2 Low-order approximation to the renormalised coupling 2.3.3 Scheme dependence . . . . . . . . . . . . . . . . . . . . 2.3.4 Theories with more parameters . . . . . . . . . . . . . 2.3.5 Failure of the dot model . . . . . . . . . . . . . . . . . 2.4 Asymptotics of renormalisation in ϕ4 theory . . . . . . . . . . 2.5 The method of counterterms . . . . . . . . . . . . . . . . . . . 2.5.1 Counterterms in the action . . . . . . . . . . . . . . . . 2.5.2 Return to the dot model, and a preview . . . . . . . . 2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 More fields in zero dimensions 3.1 The action and the path integral . . . . . . . . 3.2 Connected Green’s functions and field functions 3.3 The Schwinger-Dyson equation . . . . . . . . . 3.4 The sum rules revisited . . . . . . . . . . . . . . 3.5 A zero-dimensional toy for QED . . . . . . . . 3.5.1 Fields and sources . . . . . . . . . . . .

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1.3.8 Dyson summation . . . . . . . . . . . . . . . Planck’s constant . . . . . . . . . . . . . . . . . . . 1.4.1 The loop expansion, and reverse engineering 1.4.2 The classical limit . . . . . . . . . . . . . . . 1.4.3 On second quantisation . . . . . . . . . . . . 1.4.4 Instanton contributions . . . . . . . . . . . . The effective action . . . . . . . . . . . . . . . . . . 1.5.1 The effective action as a Legendre transform 1.5.2 Diagrams for the effective action . . . . . . . 1.5.3 Computing the effective action . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . .

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3.5.2 Bald, furry and quenched toys . . . . . . . . . . . . . . 91 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

4 QFT in Euclidean spaces 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 4.2 One-dimensional discrete theory . . . . . . . . . . . . 4.2.1 An infinite number of fields . . . . . . . . . . 4.2.2 Introducing the propagator . . . . . . . . . . . 4.2.3 Computing the propagator . . . . . . . . . . . 4.2.4 Figments of the imagination : a sermon . . . . 4.3 One-dimensional continuum theory . . . . . . . . . . 4.3.1 The continuum limit for the propagator . . . . 4.3.2 The continuum limit for the action . . . . . . 4.3.3 The continuum limit of the classical equation 4.3.4 The continuum Feynman rules and SDe . . . . 4.3.5 Field configurations in one dimension . . . . . 4.4 The momentum representation . . . . . . . . . . . . . 4.4.1 Fourier transforming the SDe . . . . . . . . . 4.5 Doing it in momentum space . . . . . . . . . . . . . . 4.5.1 The Feynman rules . . . . . . . . . . . . . . . 4.5.2 Some example diagrams . . . . . . . . . . . . 4.6 More-dimensional theories . . . . . . . . . . . . . . . 4.6.1 The more-dimensional continuum . . . . . . . 4.6.2 The propagator, explicitly . . . . . . . . . . . 4.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 5 QFT in Minkowski space 5.1 Moving to Minkowski space : making time . . . . . 5.1.1 Distance in Minkowski space . . . . . . . . . 5.1.2 Farewell probability, hello SDe . . . . . . . . 5.1.3 A close look at almost nothing: i and − . . 5.1.4 The need for quantum transition amplitudes 5.1.5 Feynman rules for Minkowskian theories . . 5.1.6 The propagator, explicitly . . . . . . . . . . 5.2 Moving in Minkowski space : particles . . . . . . . 5.2.1 The Klein-Gordon equation . . . . . . . . . 5.2.2 Enter the particle ! . . . . . . . . . . . . . . 5.2.3 Unstable particles, i and the flow of time .

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CONTENTS 5.2.4 The Yukawa potential . . . . . . . . . . . . . . . . . 5.2.5 Kinematics and Newton’s First Law . . . . . . . . . . 5.2.6 Antimatter . . . . . . . . . . . . . . . . . . . . . . . 5.2.7 Counting states : the phase-space integration element Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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6 Scattering processes 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Incursion into the scattering process . . . . . . . . . . . . . . 6.2.1 Diagrammatic picture of scattering . . . . . . . . . . 6.2.2 The argument for connectedness . . . . . . . . . . . . 6.3 Building predictions . . . . . . . . . . . . . . . . . . . . . . 6.3.1 General formulæ for decay widhts and cross sections . 6.3.2 The truncation bootstrap . . . . . . . . . . . . . . . 6.3.3 A check on dimensionalities . . . . . . . . . . . . . . 6.3.4 Crossing symmetry . . . . . . . . . . . . . . . . . . . 6.4 Unitarity issues . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Unitarity of the S matrix . . . . . . . . . . . . . . . 6.4.2 The cutting rules . . . . . . . . . . . . . . . . . . . . 6.4.3 Infrared cancellations in QED . . . . . . . . . . . . . 6.5 Some example calculations . . . . . . . . . . . . . . . . . . . 6.5.1 The FEE model . . . . . . . . . . . . . . . . . . . . . 6.5.2 Two-body phase space . . . . . . . . . . . . . . . . . 6.5.3 A decay process . . . . . . . . . . . . . . . . . . . . . 6.5.4 A scattering process . . . . . . . . . . . . . . . . . . 6.6 The one-loop cookbook . . . . . . . . . . . . . . . . . . . . . 6.6.1 The one-loop calculation . . . . . . . . . . . . . . . . 6.6.2 Dispersion relations . . . . . . . . . . . . . . . . . . . 6.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7 Dirac particles 7.1 Pimp my propagator . . . . . 7.1.1 Down with dyads ! . . 7.1.2 The spin interpretation 7.2 The Dirac algebra . . . . . . . 7.2.1 The Dirac matrices . 7.2.2 The Clifford algebra . 7.2.3 Trace identities . . . .

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CONTENTS

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7.2.4 Pauli and Chisholm, their identities . . . . . . . 7.2.5 Hermite no, Dirac yes . . . . . . . . . . . . . . 7.2.6 Flipping in and flipping out . . . . . . . . . . . 7.2.7 A Fierz identity . . . . . . . . . . . . . . . . . . Dirac spinors . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Chirality spinors . . . . . . . . . . . . . . . . . 7.3.2 Chirality spinors are massless . . . . . . . . . . 7.3.3 Phase conventions, spinor products and Weyl . 7.3.4 General spinors and their dyads . . . . . . . . . 7.3.5 General spinors and Dirac spinors . . . . . . . . 7.3.6 Particular general spinors . . . . . . . . . . . . Dirac particles . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Casimir does tricks . . . . . . . . . . . . . . . . 7.4.2 The Dirac propagator, and a convention . . . . 7.4.3 Truncating Dirac particles : external Dirac lines 7.4.4 Lorentz transformations in Dirac space . . . . . 7.4.5 The spin of Dirac particles . . . . . . . . . . . . 7.4.6 Massless Dirac particles ; helicity states . . . . 7.4.7 The parity transform . . . . . . . . . . . . . . . The Feynman rules for Dirac particles . . . . . . . . . . 7.5.1 Dirac loops. . . . . . . . . . . . . . . . . . . . . . 7.5.2 . . . and Dirac loops only . . . . . . . . . . . . . 7.5.3 Interchange signs . . . . . . . . . . . . . . . . . 7.5.4 The Pauli principle . . . . . . . . . . . . . . . . The Dirac equation . . . . . . . . . . . . . . . . . . . . 7.6.1 The classical limit . . . . . . . . . . . . . . . . . 7.6.2 The free Dirac action . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

8 Helicity techniques for Dirac particles 8.1 The standard form for spinors . . . . . . . . . . . . . 8.1.1 Opting for helicities, opting for antisymmetry 8.1.2 The standard form for helicity spinors . . . . 8.1.3 Some useful identities . . . . . . . . . . . . . . 8.1.4 How to compute spinor products . . . . . . . 8.1.5 The standard form for massive particles . . . 8.1.6 The standard form for complex momenta . . . 8.2 Summary of tools for spinor techniques . . . . . . .

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CONTENTS 8.3

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Fermionic decays : the Fermi model . . . . . . . 8.3.1 The amplitude for muon decay . . . . . 8.3.2 Three-body phase space . . . . . . . . . 8.3.3 The muon decay width . . . . . . . . . . 8.3.4 Observable distributions in muon decay . 8.3.5 Charged pion decay: helicity suppression Exercises . . . . . . . . . . . . . . . . . . . . . .

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9 Vector particles 9.1 Massive vector particles . . . . . . . . . . . . . . . . . 9.1.1 The propagator . . . . . . . . . . . . . . . . . . 9.1.2 The Feynman rules for external vector particles 9.1.3 The spin of vector particles . . . . . . . . . . . 9.1.4 Polarisation vectors for helicity states . . . . . . 9.1.5 The Proca equation . . . . . . . . . . . . . . . . 9.2 The spin-statistics theorem . . . . . . . . . . . . . . . . 9.2.1 Spinorial form of vector polarisations . . . . . . 9.2.2 Proof of the spin-statistics theorem . . . . . . . 9.3 Massless vector particles . . . . . . . . . . . . . . . . . 9.3.1 Polarisations of massless vector particles . . . . 9.3.2 Current conservation from the polarisation . . . 9.3.3 Handlebar condition for massive vector particles 9.3.4 Helicity states for massless vectors . . . . . . . 9.3.5 The massless propagator : the axial gauge . . . 9.3.6 Gauge vector shift . . . . . . . . . . . . . . . . 9.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

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10 Quantum Electrodynamics 10.1 Introduction . . . . . . . . . . . . . . 10.2 Constructing QED . . . . . . . . . . 10.2.1 The QED vertex . . . . . . . 10.2.2 Handlebars : a first look . . . 10.2.3 Handlebar diagrammatics . . 10.2.4 The Ward-Takahashi identity 10.2.5 The charged Dirac equation . 10.2.6 The Gordon decomposition . 10.2.7 Furry’s theorem . . . . . . . . 10.3 Some QED processes . . . . . . . . .

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CONTENTS

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10.3.1 A classic calculation : muon pair production 10.3.2 Compton and Thomson scattering . . . . . 10.3.3 Electron-positron annihilation . . . . . . . . 10.3.4 Bhabha scattering . . . . . . . . . . . . . . 10.3.5 Bremsstrahlung in Mœller scattering . . . . Scalar electrodynamics . . . . . . . . . . . . . . . . 10.4.1 The vertices . . . . . . . . . . . . . . . . . . 10.4.2 Proof of current conservation in sQED . . . The Coulomb potential . . . . . . . . . . . . . . . . Electrons in external fields : g = 2 . . . . . . . . . . 10.6.1 The charged Klein-Gordon equation . . . . . 10.6.2 The relativistic Pauli equation . . . . . . . . 10.6.3 A constant magnetic field . . . . . . . . . . Selected topics in QED . . . . . . . . . . . . . . . . 10.7.1 Three-photon production . . . . . . . . . . . 10.7.2 The Thomson limit : scalar vs spinor . . . . 10.7.3 The Landau-Yang theorem . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . .

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11 Loop effects in QED 321 11.1 One-loop effects in QED . . . . . . . . . . . . . . . . . . . . . 321 11.2 The photon self-energy . . . . . . . . . . . . . . . . . . . . . . 321 11.2.1 Current conservation . . . . . . . . . . . . . . . . . . . 321 11.2.2 Using the optical theorem . . . . . . . . . . . . . . . . 322 11.2.3 Getting the divergence . . . . . . . . . . . . . . . . . . 323 11.2.4 The vacuum polarization . . . . . . . . . . . . . . . . . 324 11.2.5 Hadronic vacuum polarization . . . . . . . . . . . . . . 325 11.3 The fermion self-energy . . . . . . . . . . . . . . . . . . . . . . 327 11.3.1 A look at gauge invariance . . . . . . . . . . . . . . . . 327 11.3.2 Summing the self-energies . . . . . . . . . . . . . . . . 328 11.3.3 The loop calculation . . . . . . . . . . . . . . . . . . . 329 11.3.4 The curious incident of the divergences in the nighttime 331 11.4 Infrared singularities in Bremsstrahlung . . . . . . . . . . . . . 332 11.5 The vertex correction . . . . . . . . . . . . . . . . . . . . . . . 335 11.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

10 12 Quantum Chromodynamics 12.1 Introduction: coloured quarks and gluons . . 12.2 Quarks and gluons : first Feynman rules . . 12.2.1 The propagators . . . . . . . . . . . 12.2.2 The quark-gluon vertex . . . . . . . . 12.2.3 A closer look at the T matrices . . . 12.2.4 The Fierz identity for T matrices . . 12.3 The three-gluon interaction . . . . . . . . . 12.3.1 The need for three-gluon vertices . . 12.3.2 Furry’s failure . . . . . . . . . . . . . 12.3.3 The ggg vertex and its handlebar . . 12.3.4 On coupling quantisation . . . . . . . 12.4 The four-gluon interaction . . . . . . . . . . 12.4.1 Colourful manipulations . . . . . . . 12.4.2 A purely gluonic process . . . . . . . 12.5 Current conservation in QCD . . . . . . . . 12.5.1 More vertices ? . . . . . . . . . . . . 12.5.2 The Antkaz . . . . . . . . . . . . . . 12.5.3 Proof of current conservation . . . . 12.6 Selected topics in QCD . . . . . . . . . . . . 12.6.1 White and coloured states . . . . . . 12.6.2 The QCD Coulomb interaction . . . 12.6.3 The process q q¯ → gg . . . . . . . . . 12.6.4 The Landau-Yang theorem revisited . 12.7 Exercises . . . . . . . . . . . . . . . . . . . .

CONTENTS

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13 Electroweak theory 13.1 Muon decay . . . . . . . . . . . . . . . . . . . . . . 13.1.1 The Fermi coupling constant . . . . . . . . . 13.1.2 Failure of the Fermi model in µ− ν µ → e− ν e 13.2 The W particle . . . . . . . . . . . . . . . . . . . . 13.2.1 The IVB strategy . . . . . . . . . . . . . . . 13.2.2 The cross section for µ− ν µ → e− ν e revisited 13.2.3 The W W γ vertex . . . . . . . . . . . . . . . 13.3 The Z particle . . . . . . . . . . . . . . . . . . . . . 13.3.1 W pair production . . . . . . . . . . . . . . 13.3.2 The weak mixing angle for couplings . . . . 13.3.3 W, Z and γ four-point interactions . . . . .

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CONTENTS 13.4 The Higgs sector . . . . . . . . . . . . . . . . 13.4.1 The Higgs hypothesis . . . . . . . . . . 13.4.2 Predictions from the Higgs hypothesis 13.4.3 W, Z and H four-point interactions . . 13.4.4 Higgs-fermion couplings . . . . . . . . 13.4.5 Higgs self-interactions . . . . . . . . . 13.5 About anomalies . . . . . . . . . . . . . . . . 13.6 Conclusions and remarks . . . . . . . . . . . . 13.7 A look at non-minimal models . . . . . . . . . 13.7.1 Non-minimal Higgs sector . . . . . . . 13.8 Exercises . . . . . . . . . . . . . . . . . . . . .

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. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

394 394 400 401 403 405 409 411 412 412 418

14 Example computations 14.1 Neutrino production in e+ e− scattering . . . . . . . . 14.1.1 The cross section . . . . . . . . . . . . . . . . 14.1.2 Unitarity considerations . . . . . . . . . . . . 14.2 W pair production in e+ e− scattering . . . . . . . . . 14.2.1 Setting up the amplitude . . . . . . . . . . . . 14.2.2 Momenta and polarisations . . . . . . . . . . 14.2.3 Working out the amplitudes . . . . . . . . . . 14.2.4 W pair production at very high energy . . . . 14.3 Higgs coupling to massless vectors . . . . . . . . . . . 14.3.1 The γγH vertex . . . . . . . . . . . . . . . . . 14.3.2 The ggH amplitude and the Next Generation

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . .

421 421 421 424 425 425 426 427 430 431 431 435

. . . . . . . . . . . .

437 . 437 . 437 . 438 . 441 . 441 . 442 . 445 . 448 . 448 . 449 . 450 . 450

15 Appendices 15.1 Perturbative (non)convergence issues . . . . . . . . . . 15.1.1 Punishment at the singular point . . . . . . . . 15.1.2 Borel summation . . . . . . . . . . . . . . . . . 15.2 More on symmetry factors . . . . . . . . . . . . . . . . 15.2.1 The origin of symmetry factors . . . . . . . . . 15.2.2 Explicit computation of symmetry factors . . . 15.3 Derivation of the diagrammatic sum rules . . . . . . . . 15.4 Alternative solutions to the Schwinger-Dyson equation 15.4.1 Alternative contours for general theories . . . . 15.4.2 Alternative contours for ϕ3 theory . . . . . . . . 15.4.3 Alternative contours for ϕ4 theory . . . . . . . . 15.5 Diagram counting . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

12

CONTENTS 15.5.1 Tree graphs and asymptotics . . . . . . . . . . 15.5.2 Counting one-loop diagrams . . . . . . . . . . 15.6 Concavity of the effective action . . . . . . . . . . . . 15.7 Functional derivatives . . . . . . . . . . . . . . . . . . 15.8 Frustrated and unusual actions . . . . . . . . . . . . 15.8.1 Frustrating your neighbours . . . . . . . . . . 15.8.2 Increasing frustration . . . . . . . . . . . . . . 15.9 Newton’s First Law revisited . . . . . . . . . . . . . 15.9.1 Introduction : the matter of sources . . . . . . 15.9.2 Slow, fast and abrupt . . . . . . . . . . . . . . 15.9.3 Conclusion : general effect of the sources . . . 15.10 Unitarity bounds . . . . . . . . . . . . . . . . . . . . 15.10.1 Resonances . . . . . . . . . . . . . . . . . . . 15.10.2 Preliminaries : decay widths . . . . . . . . . 15.10.3 The rˆole of angular momentum conservation . 15.10.4 The unitarity bound . . . . . . . . . . . . . . 15.11 The fundamental theorem for Dirac matrices . . . . 15.11.1 Proof of the fundamental theorem . . . . . . 15.11.2 The charge conjugation matrix . . . . . . . . 15.12 States of higher spin . . . . . . . . . . . . . . . . . . 15.12.1 The spin algebra for integer spins . . . . . . . 15.12.2 Rank one for spin one . . . . . . . . . . . . . 15.12.3 Rank-2 tensors . . . . . . . . . . . . . . . . . 15.12.4 Rank-3 tensors . . . . . . . . . . . . . . . . . 15.12.5 Massless particles : surviving states . . . . . 15.12.6 The Bermuda triangle . . . . . . . . . . . . . 15.12.7 Massless propagators . . . . . . . . . . . . . . 15.12.8 Spin of the Kalb-Ramond state . . . . . . . . 15.12.9 Spin 1 from Dirac particles . . . . . . . . . . 15.12.10 Spin 3/2 particles . . . . . . . . . . . . . . . 15.12.11 The spin algebra for Dirac particles . . . . . 15.13 Generating three-particle kinematics . . . . . . . . . 15.14 The CPT theorem . . . . . . . . . . . . . . . . . . . 15.14.1 Transforming spinors . . . . . . . . . . . . . 15.14.2 CPT transformation on sandwiches . . . . . . 15.14.3 CPT transformation on diagrams . . . . . . . 15.14.4 How to kill CPT, and what it costs . . . . . . 15.15 Mathematical Miscellanies . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

450 455 461 463 464 464 467 469 469 471 473 473 473 474 475 476 477 477 480 480 480 482 483 485 489 490 491 493 494 496 498 499 500 501 502 503 504 506

CONTENTS 15.15.1 The Gaussian doubling trick . . . 15.15.2 Stirling’s approximation for n! . . 15.15.3 Manipulating asymptotic series . . 15.15.4 Gamma, Digamma and Bernoulli . 15.15.5 The Dirac delta distribution . . . 15.15.6 The principal-value distribution . 15.15.7 Generating the Bell numbers . . . 15.15.8 The exponential integral E1 and the 15.15.9 The Kramers-Kronig relation . . . 15.15.10 The dilogarithm function . . . . . 15.15.11 Some values of the ζ function . . 15.15.12 The Lagrange expansion . . . . . 15.15.13 Determinants from traces . . . .

13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bessel . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . K functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

506 507 508 511 513 514 514 515 518 519 521 522 524

14

CONTENTS

List of Figures 1.1

The developing of instanton effects . . . . . . . . . . . . . . . 53

2.1

Improvement factor for ϕ4 theory in zero dimensions . . . . . 81

4.1

Zooming out from a fractal path . . . . . . . . . . . . . . . . . 108

5.1 5.2 5.3 5.4 5.5 5.6

Closing the contour for positive times . . . . . . . . . Closing the contour for spacelike separation . . . . . Bessel functions K1 for real and imaginary argument Closing the contour for Newton’s first . . . . . . . . . Moving forward ; moving backward . . . . . . . . . . An annihilating spacetime path . . . . . . . . . . . .

6.1

The function R(m2 , s) of Eq.(6.89) . . . . . . . . . . . . . . . 178

7.1

A Mental Flip : one vase or two faces ? . . . . . . . . . . . . . 197

8.1

Elecrton-mass dependence of the muon decay width . . . . . . 239

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

127 129 130 137 140 142

10.1 Threshold behaviour for fermions and scalars . . . . . . . . . . 283 11.1 The fermion-induced vacuum polarization . . . . . . . . . . . 325 13.1 Monte Carlo investigation of cancellations in ZZ → ZZH . . 407 13.2 Monte Carlo investigation of cancellations in ZZ → HHH . . 409 14.1 The absolute value of F(M 2 /mH 2 ) defined in Eq.(14.58) . . . . 435 15.1 Example of Borel summation . . . . . . . . . . . . . . . . . . 440 15.2 Propagators for frustrated actions . . . . . . . . . . . . . . . . 469 15.3 Contour shift for a Gaussian source . . . . . . . . . . . . . . . 472 15

16

LIST OF FIGURES 15.4 The functions K0 , K1 , K2 and E1 . . . . . . . . . . . . . . . . 517 15.5 Integration contour for the ramers-Kronig relation . . . . . . . 518 15.6 The dilogarithm function . . . . . . . . . . . . . . . . . . . . . 520

List of Tables 13.1 Contributions to the triangle anomalies . . . . . . . . . . . . . 410 15.1 Exact and asymptotic number of tree diagrams in ϕ3/4 theory 454 15.2 Computing the average symmetry factors . . . . . . . . . . . . 460 15.3 Multiplication table for the Clifford algebra . . . . . . . . . . 478

17

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Chapter 0 Introductory remarks 0.1

Preface

In what follows, whatever is correct I owe to many other people ; that which is wrong I managed on my own. I am perpetually in need of, and grateful to, those pointing out typing or thinking errors in these notes1 . About the appendices : I have collected material that, when included in the main text, would interfere with the narrative flow. The appendices relevant to each chapter are indicated below the chapter heading. About the exercises : these are intended to encourage you towards competence in actually computing things. Theoretical knowledge without dirty hands is not very good2 . The exercises apposite to a topic are indicated by a box like Eℵ in the margin, where ‘ℵ’ is the number of the exercise, situated at the end of the chapter. Finally, a word of caution : I am a physicist, not E0 a mathematician : I will worry about mathematical rigour but not so much as to become paralysed3 .

1

I cordially invite all and sundry to do so. The P 4 Hall of Fame collects the names of friends who have helped me in learning about, formulating, contemplating, or execrating one or several issues. 2 Labori inutilis manus salubris (attr. to S. Partlic) 3 Rigor mathesis protractus mortis rigorem inducet (attr. to S. Partlic).

19

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0.2 0.2.1

Basic tools Units and fundamental units

The fundamental constants of relativistic quantum field theory are the speed of light in vacuo 4 : m , c = 299792458 sec and Planck’s (or rather Dirac’s) constant5 : ~ = 1.05457182 × 10−34 Joule sec . Compared to the scales of our everyday experiences, ~ is miniscule and c is huge ; in the world of elementary particles, they are just about right. We can see this as follows. It is useful to replace our human-scale meters6 , kilograms7 , and seconds8 , by what may be called fundamental HEP units of mass, length and time9 : Mf = 1.7827 10−27 kg , Lf = 1.9733 10−16 m , Tf = 6.5821 10−25 sec . In terms of these units, we have precisely ~=

Mf Lf 2 , Tf

c=

Lf , Tf

so that both ~ and c have the numerical value one ; and the unit of energy turns out to be Mf Lf 2 = 1.6022 10−10 Joule = 1 GeV . 2 Tf The mass and size of the proton are of the same order as Mf and Lf , respectively, and Tf is roughly the time scale of strong interactions. The use 4

This is exact since it establishes the definition of the metre as of October 21, 1983. This is exact since it establishes the definition of the kilogram as of May 20, 2019. 6 Typically, how far you can reach. 7 Typically, what you would not mind lifting. 8 Typically, how fast you can count things if you are not a teller. 9 I give these numbers with only a few digits, for simplicity. What matters is of course not the precise value, but rather that such units exist. 5

August 31, 2019

21

of fundamental units is attractive since you won’t have to write factors of c and ~, and can express both length and time in inverse GeV, and mass in GeV. This is customarily one of the first statements in advanced courses : “we shall adopt units such that c = ~ = 1”. This usage denies the various objects their dimensionality10 , and I have decided to try to retain the ~’s and c’s where they belong ; after all, it is much easier to erase them from formulæ than to put them back in11 .

0.2.2

Planck units

Along with ~ and c there exists a third fundamental constant of nature, namely Newton’s (or rather Cavendish’s) gravitational constant12 : G = 6.67408(31) 10−11

m3 . kg sec2

The truly, ultimately fundamental units of mass, length and time that can be recovered from c, ~ and GN are then the Planck units : r ~ = 2.1764 10−8 kg , MP = Gc r ~G LP = = 1.6162 10−35 m , 3 c r ~G = 5.3912 10−44 sec . (1) TP = 5 c These values are outrageously far removed from the typical scales of particle phenomenology. We may interpret this as an indication that in what follows the gravitational interaction will not play any part. In fact, in any case we do 10

This has the unfortunate consequence of making it impossible to check the (at least) grammatical correctness of an expression by dimensional analysis, and even leads to erroneous statements about ‘classical limits’ and the like : see our discussion of particle masses in chapter 6. 11 You would be surprised how difficult it is for even well-trained physicists to put the ~ and c back, when asked to do so in a hurry ! 12 Taken from P.J. Mohr, D.B. Newell, B.N. Taylor, CODATA recommended values of the fundamental physical constants: 2014, Rev. Mod. Phys. 88, July-September 2016. The uncertainty in the last two digits is bracketed.

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August 31, 2019

not (yet) have a satisfactory quantum theory of gravitation leading to specific and falsifiable predictions for particle phenomenology. To bring the Planck units close to the fundamental units we need to increase the value of G, thus the strength of gravity, by a factor of about 1038 . On a more philosophical note, consider that we have, basically, only the three measures, of length, of time, and of mass, to do physics with ; and there are precisely three constants of nature that we know of. Suppose a fourth one were discovered : it would be only natural that we would try very hard to relate it to the other three in some way ; this is how fundamental physics must work.

0.2.3

Charges

The electrostatic charge is adopted to the Gaussian system, so as to have no truck with the ‘permittivity of the vacuum’ and suchlike : that is, two charges e1 and e2 separated by a distance r feel a mutual Coulomb force F~ characterized by 1 | e1 e2 | . | F~ | = 4π r2 √ This implies that the charge has the dimensionality of ~c. It follows that, if we choose the proton charge as the unit charge e, the fine structure constant αe =

e2 4π ~ c

is a dimensionless number13 . Experimentally14 αe =

1 , 137.035999139(31)

which yields the result √ kg1/2 m3/2 e = 0.30282 ~c = 5.3844 10−14 . sec 13

Meaning that it has the same value in all possible systems of units ! An alien civilization in outer space will find the same value. 14 Taken from P.J. Mohr, D.B. Newell, B.N. Taylor, CODATA recommended values of the fundamental physical constants: 2014, Rev. Mod. Phys. 88, July-September 2016. The uncertainty in the last two digits is bracketed.

August 31, 2019

0.2.4

23

Conventions

Forgetting phases If two quantities A and B are equal up to a disregarded overall complex phase we shall write AlB . The step function The Whittaker (or step) function is a function of a real number :  1 if x ≥ 1 θ(x) = . 0 if x < 0 We extend this to a logical step function of a predicate P :  1 if P is true θ(P) = . 0 if P is false

(2)

(3)

The metric By convention, the Minkowski metric15 has the form g µν = gµν = diag(1, −1, −1, −1) . In many textbooks the metric tensor is introduced as a diagonal matrix. This is of course misleading since the covariant metric tensor has only lower indices, whereas a matrix has one upper and one lower index. Unfortunately, the ‘correct’ matrix form of the metric, which would be g µ ν , equals the unit matrix whatever the metric ! Kronecker’s symbol The Kronecker symbol is defined by  1 if α = µ α δ µ= 0 if α = 6 µ 15

.

In the usual Cartesian coordinate systems. Since we are not considering curved spacetime in these notes, we shall adhere to this simplest of coordinate system throughout, except when discussing phase space integration where polar coordinates often come in handy. But there we shall not refer to the metric.

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August 31, 2019

Note : in Minkowski space, Kronecker symbols tend to carry one upper, and one lower index. Kronecker symbols with two upper or two lower indices are slightly suspect and to be treated with care unless they refer to some other label such as color. The antisymmetrizer Using Kronecker symbols we can build the following antisymmetric objects :     µ1 µ2 µ1 µ1 = δ µ1 ν1 δ µ2 ν2 − δ µ1 ν2 δ µ2 ν1 , = δ ν1 , ν1 ν2 ν1   µ1 µ2 µ3 = δ µ1 ν1 δ µ2 ν2 δ µ3 ν3 + δ µ1 ν2 δ µ2 ν3 δ µ3 ν1 + δ µ1 ν3 δ µ2 ν1 δ µ3 ν2 ν1 ν2 ν3 − δ µ1 ν1 δ µ2 ν3 δ µ3 ν2 − δ µ1 ν2 δ µ2 ν1 δ µ3 ν3 − δ µ1 ν3 δ µ2 ν2 δ µ3 ν1 and so on. Here we encounter all signed permutations16 of the lower indices. These are compuationally handy as we see below. The Levi-Civita symbol The totally antisymmetric Levi-Civita symbol is defined by 0123 = + 1 hence 0123 = − 1 . This implies the following identities : ν1 ν2 ν3 ν4

µ1 µ2 µ3 µ4 

ν1 ν2 ν3 µ4

µ1 µ2 µ3 µ4 

ν1 ν2 µ3 µ4

µ1 µ2 µ3 µ4 

ν1 µ2 µ3 µ4

µ1 µ2 µ3 µ4 

 = −  = −

ν1 µ1

ν2 µ2

ν3 µ3

ν1 µ1

ν2 µ2

ν3 µ3 

 = −2  = −6

ν1 µ1 ν1 µ1

ν2 µ2 



ν4 µ4 

, ,

, = −6 δ ν1 µ1 ,

µ1 µ2 µ3 µ4 µ1 µ2 µ3 µ4 = −24 . 16

Even permutations occur with a +, and odd permutations with a – sign.

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25

In order not to lumber ourselves with too many explicit Lorentz indices, we shall use the following shorthand : µ (a, b, c) means µνρσ aν bρ cσ and so on. Finally, although it falls outside the scope of these notes, it is useful to note that the Kronecker and antisymmetrizer symbols are fully-fledged tensors in a much wider class of spaces than Minkowski space, but the Levi-Civita symbols themselves are not. For instance, if we move from Cartesian to polar coordinates, say, the Kronecker symbol remains unaffected but the Levi-Civita is changed. Minus in the momentum There is a subtlety : the contravariant partial derivative contains a possibly surprising minus sign :   1∂ ∂ µ ~ = , −∇ . (4) ∂ = ∂xµ c ∂t This explains why in nonrelativistic quantum mechanics the momentum op~ whereas in the relativistic theory we use pµ = i~ ∂ µ . erator is p~ = −i~ ∇

The polyparenthetophobia rule This deals with the notation for compound vector products. If pµ1,2,3,4 are four-vectors, then the expression (p1 + p2 · p3 + p4 ) must be understood to mean (p1 · p3 ) + (p2 · p3 ) + (p1 · p4 ) + (p2 · p4 ) The more rigorously correct form (p1 + p2 ) · (p3 + p4 )




is in my opinion less easily readable, unless by true parenthetophiliacs. In the same spirit, a slight tendency to periodophobia will lead us to write, e.g., (pq) as shorthand for (p · q) where no risk of confusion is likely. Also, k 2 occurs when it is clear that (k ·k) is intended rather than the second spacelike component of k µ .

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August 31, 2019

In favor of loose terminology Among particle physicists there exist a tendency to be sloppy with some terms. In particular this holds for the usage of the words ‘mass’ and ‘momentum’. Strictly speaking, in the Feynman rules to be discussed the ‘mass’ m and the ‘momentum’ k µ have dimensions of inverse length and therefore cannot be the same as the notions of the classical mass M , expressed in kg, and the classical momentum pµ , expressed in (kg m/sec) . As discussed in section 5.2.5, these various notions are related by m = M c/~ ,

k µ = pµ /~ .

Not wishing to succumb to pedantry, I shall use ‘mass’ and ‘momentum’ insouciantly. Experience shows that one easily gets used to it.

0.3

The P 4 Hall of Fame

Ernestos Argyres Wolfgang Hollik Dima Bardin Gerard ’t Hooft Wim Beenakker Staszek Jadach Frits Berends Frederick James Alain Blondel Tim Janssen Oscar Boher Luna Sijbrand de Jong Michael Borinsky Martijn Jongen Stefan Brinck de Heren Jos Kristof de Bruyn Marcel van Kessel Dirk van Buul Jochem Kip Sascha Caron Stefan Krieg Chris Dams Hans K¨ uhn Pertros Draggiotis Hans Kuijf Helmut Eberl Zoltan Kunszt Steve Ellis Achilleas Lazopoulos Raymond Gastmans Joep Leenaarts Edward Gibbon Yannis Malamos Walter Giele Mich. Mangano Jeroen de Groot John March-Russell Andr´e van Hameren Melvin Meijer Lisa Hartgring Mark Netjes Thank you for allowing me to think together

Harald Niederreiter Gijs van der Oord Kostas Papadopoulos Simon Partlic Giampiero Passarino Roberto Pittau Marcel Raas Frank Redig Robbert Rietkerk Chris Ripken Tom Rijken Bert Schellekens James Stirling John Swain Oleg Teryaev Theodor Todorov Tini Veltman Rob Verheyen Jos Vermaseren Tai Tsun Wu Sjoerd Ypma with you !

August 31, 2019

0.4

Exercise

Excercise 0 Would-be exercise This might have been an exercise.

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August 31, 2019

Chapter 1 QFT in zero dimensions 1.1

Introduction

For the description of elementary particles, a theory including both special relativity and quantum mechanics is necessary ; we shall introduce relativity further on, and concentrate in this chapter on the quantum-mechanical nature of nature. The fundamental object used for describing the particles is a quantum field. In many treatments quantum fields are considered to be operator-valued entities ; we shall rather adhere to Feynman’s approach and use what are called c-number fields. Such a field assigns one or more numbers to every point in spacetime, and is hence a pretty complicated subject the behaviour of which is not to be characterized trivially, especially when it also undergoes quantum fluctuations. It is therefore useful to first build up expertise in the various necessary techniques in a more controllable situation. To this end we shall simplify the whole four-dimensional spacetime arena of particle physics to a lower-dimensional system ; in fact, we shall reduce spacetime to a single point, hence a zero-dimensional arena. The quantum fields are then assignments of a single number ; the simplest quantum field is, in this case, a single stochastic, or random, number. Many of the techniques of quantum field theory do apply to this case : in particular the notion of path integrals, Green’s functions, the Schwinger-Dyson equation, and Feynman diagrams come up naturally. 29

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1.2 1.2.1

Probabilistic considerations Quantum field and action

Let us imagine a quantum field ϕ that takes its values on the whole real axis from −∞ to +∞. Since it is a random variable, the most we can specify about it is its probability density P (ϕ), which we write, for now, as
P (ϕ) = N exp − S(ϕ) . (1.1) The function S(ϕ) is called the action of the particular quantum field theory : in a sense, it defines the theory. For the probability density to be acceptable, S(ϕ) must go to infinity sufficiently fast as |ϕ| → ∞. The normalization factor N is defined by1 Z
−1 N = exp − S(ϕ) dϕ . (1.2)

1.2.2

Green’s functions, sources and the path integral

Since the quantum field is a random variable, the most that can be known about it2 is the collection of its moments, in the jargon called Green’s functions3 : Z
n Gn ≡ hϕ i ≡ N exp − S(ϕ) ϕn dϕ , n = 0, 1, 2, 3, . . . . (1.3) 1

If not explicitly indicated otherwise, integrals run from −∞ to +∞. You are here approaching a career decision ! You may decide simply to measure the value of ϕ : in that case you have decided to become an experimentalist rather than a theorist. 3 A clarifying remark is necessary here. In this text, the Green’s functions are simply defined to be expectation values. This may appear to contrast with the use of Green’s functions in the solution of inhomogeneous linear differential equations such as are encountered in classical electrodynamics where one uses them to compute the electromagnetic field configurations for given sources. The difference is only apparent since, as we shall recognize, the latter type of Green’s functions are in our treatment simply the two-point Green’s functions ; and for theories such as electrodynamics, where the electromagnetic fields do not undergo self-interaction, the two-point functions are in fact the only nonzero connected Green’s functions. Be not, therefore, misled into thinking that there are somehow two sorts of Green’s functions. The Green’s function formulation of electrodynamics will in fact appear as the classical limit of the Schwinger-Dyson equation discussed below. 2

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31

We shall assume that Gn exists for all n. By construction, we must always have

 G0 = ϕ0 = h1i = 1 . (1.4) The most fruitful way4 of discussing the set of all Green’s functions is in terms of a generating function : Z(J) =

X 1 J n Gn . n! n≥0

(1.5)

This is called the path integral, for reasons that will become clear later. It can be written as Z
Z(J) = N exp − S(ϕ) + Jϕ dϕ . (1.6) The number J, which here serves purely as a device to distinguish the various Green’s functions, is called a source, again for reasons that will become apparent later. Once Z(J) is known, an individual Green’s function is extracted by differentiation :   n ∂ Z(J) . (1.7) Gn = (∂J)n J=0

1.2.3

Connected Green’s functions

The path integral Z(J) contains all the information about the Green’s functions, and hence about the probability density P (ϕ). The same information is, therefore, also contained in its logarithm. We write W (J) = log Z(J) ≡

X 1 J n Cn . n! n≥1

(1.8)

The quantities Cn (with, obviously C0 = 0 since G0 = 1) are called the connected Green’s functions of the theory, and will play an important rˆole in what follows. The connected Green’s functions can be recognized to be the cumulants of the probability density : E1 4

Kids ! Do this at home. Whenever an infinite collection of objects with some kind of structure between them occurs, generating functions are always a good idea.

32

August 31, 2019 C0 C1 C2 C3 C4

=0 = hϕi = h(ϕ − hϕi)2 i = h(ϕ − hϕi)3 i = h(ϕ − hϕi)4 i − 3C2 2

the mean the variance the skewness the kurtosis

(1.9)

and so on. Since W (0) = C0 = 0, the same information about the probability density is also contained in the field function : X 1 ∂ W (J) = J n Cn+1 . (1.10) φ(J) ≡ ∂J n! n≥0 Since from its definition, we have −1  Z Z

exp − S(ϕ) + Jϕ dϕ , exp − S(ϕ) + Jϕ ϕ dϕ φ(J) = (1.11) we can say that φ(J) is the expectation value of the quantum field ϕ in the presence of sources: to denote this, we might write φ(J) = hϕiJ ,

(1.12)

which explains the similar typographies for the quantum field and the field function5 . We should not, however, forget the difference in status of these objects : ϕ is the physical entity, an unknowable, fluctuating random field ; but φ(J) is an perfectly well-defined function that contains all the information about the probability density of ϕ, and is6 computable once the action is given.

1.2.4

The free theory

The simplest probability density is probably7 the Gaussian one, given by the action 1 S(ϕ) = µ ϕ2 , (1.13) 2 5

Try this out : “phi of J equals phi of J”. In principle, if not in practice easily or completely. 7 A uniform density may be thought even simpler, but then it cannot run from ϕ = −∞ to ϕ = +∞. As a matter of fact, ask any mathematician or physicist to name you a nice proability density over the whole real line, and she will almost without fail quote the Gaussian. 6

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33

with µ a positive real number. For any action, we shall call the part quadratic in the fields (or bilinear in the case of several fields) the kinetic part. This action, called the free action, consists of only a kinetic part. The path integral is now simply computed by   Z 1 2 Z(J) = N exp − µϕ + Jϕ dϕ 2 !  2  2 Z 1 J J J2 = N exp − µ ϕ − dϕ = exp . (1.14) + 2 µ 2µ 2µ It is not even necessary8 to actually calculate the value of N . By Taylor expansion of the exponential, we immediately find that G2n =

(2n)! 1 , 2n n! µn

G2n+1 = 0 ,

n = 0, 1, 2, . . . , .

(1.15)

J , µ

(1.16)

The connected Green’s functions follow from W (J) = log Z(J) =

J2 , 2µ

φ(J) =

so that the only nonvanishing connected Green’s function is C2 = 1/µ. The fact that here only the two-point connected Green’s function is nonvanishing is the reason for calling this model the free theory. Again, things will become clearer later on, in a more realistic spacetime — after all, what does ‘freedom’ mean in a zero-dimensional prison ?

1.2.5

The ϕ4 model and perturbation theory

An action S(ϕ) may contain other terms than just the quadratic one. Such terms are called interaction terms : they may be linear, but more usually they are of higher power in the field ϕ. The simplest acceptable interacting theory in our probabilistic setting is therefore given by the action 1 1 S(ϕ) = µϕ2 + λ4 ϕ4 . 2 4!

(1.17)

The (nonnegative !) real number λ4 is called a coupling constant : this model is called the ϕ4 theory9 . E2 8

Because we must always have Z(0) = 1.

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August 31, 2019

Computing the path integral is now a much less trivial matter. A possible approach is to assume that, in some sense, the ϕ4 theory is close to a free theory, that is, in the same some sense, λ4 is a small number. We can then expand the probability density in powers of λ4 :   k  λ4 1 2 X 1 − ϕ4k . (1.18) exp(−S(ϕ)) = exp − µϕ 2 k! 24 k≥0 This procedure is called perturbation theory. Having thus reduced the problem to the previous case of the free theory, we cavalierly10 interchange the series expansion in λ4 with the integration over ϕ and arrive at the following expression for the Green’s functions : G2n = H2n /H0 ,  k (4k + 2n)! 1 X λ4 H2n = − . µn k≥0 22k+n (2k + n)!k! 24µ2

(1.19)

For example, we have 1 35 2 385 3 H0 = 1 − u + u − u + ··· , 8 384 3072 29 2 107 3 1 u + u + ··· , 1/H0 = 1 + u − 8 384 1024 E3

(1.20)

with u ≡ λ4 /µ2 . Note that, in this theory, also the normalization N has to be treated perturbatively, which explains the expression for 1/H0 . For the first few nonvanishing Green’s functions we find G0 = 1 ,   1 1 2 2 11 3 G2 = 1 − u + u − u + ··· , µ 2 3 8   33 2 68 3 1 3 − 4u + u − u + · · · , G4 = µ2 4 3   1 75 445 2 1585 3 G6 = 15 − u + u − u + ··· . µ3 2 4 4 9

(1.21)

An action in which ϕ3 is the highest power does not lead to a convergent integral over the real axis (see, however, Appendix 15.4). Of course, an action of the form S(ϕ) = µϕ2 /2 + λ3 ϕ3 /3! + λ4 ϕ4 /4! is perfectly acceptable, and we shall consider this ‘ϕ3/4 model’ later on. 10 And not with impunity ! See Appendix 15.1.

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35

The corresponding nonzero connected Green’s functions are given by   1 2 2 11 3 1 1 − u + u − u + ··· , C2 = µ 2 3 8   1 7 2 149 3 C4 = −u + u − u + ··· , µ2 2 12
1 2 3 10u − 80u + · · · . (1.22) C6 = µ3 Note that, whereas the Green’s functions all have a perturbation expansion starting with terms containing no λ4 , the connected Green’s functions of increasing order are also of increasingly high order in λ4 : the higher connected Green’s functions need more interactions than the lower ones.

1.2.6

The Schwinger-Dyson equation

Although the path integral is, generally, a very complicated function of J, it is nevertheless easy to find an equation describing it completely. This is the Schwinger-Dyson equation (SDe), which we construct as follows. Let the action be given by the general expression : X 1 λk ϕ k , (1.23) S(ϕ) = k! k≥1 where λ2 = µ11 . Now, from the observation that Z
∂p exp − S(ϕ) + Jϕ ϕp dϕ , p Z(J) = N (∂J)

p = 0, 1, 2, 3, . . . (1.24)

we immedately deduce that " # X λk+1 ∂ k Z(J) = −J + k! (∂J)k k≥0 " # Z X λk+1
= N exp − S(ϕ) + Jϕ −J + ϕk dϕ k! k≥0   Z
0 = N exp − S(ϕ) + Jϕ S (ϕ) − J dϕ = 0 , 11

(1.25)

The sum starts at 1 since a constant, ϕ-independent term in the action is always immediately swallowed up by the normalization factor N .

36

August 31, 2019

where in the last lemma we have used partial integration, and the fact that the integrand vanishes at the endpoints. Symbolically, we may write the SDe as     ∂ ∂ 0 S(ϕ) Z(J) = S Z(J) = JZ(J) . (1.26) ∂ϕ ∂J ϕ=∂/∂J E4

For our sample model, the ϕ4 theory, the SDe reads12 1 λ4 Z 000 (J) + µZ 0 (J) − JZ(J) = 0 . (1.27) 6 Using the series expansion of the path integral we can express this as a relation between different Green’s functions : λ4 Gn+3 + µGn+1 − nGn−1 = 0 , n ≥ 1 . (1.28) 6 This relation may usefully be rewritten as follows :   1 λ4 Gn = (n − 1)Gn−2 − Gn+2 , n≥2 . (1.29) µ 6 If we start by assigning to the Green’s functions the values Gn = δ0,n , then repeated applications of Eq.(1.29) will precisely reproduce the Green’s functions of Eq.(1.21)13 .

1.2.7

The Schwinger-Dyson equation for the field function

From the definition of φ(J) as the logarithmic derivative of the path integral, we can infer that  p ∂p ∂ 1 Z(J) = φ(J) + e(J) . (1.30) Z(J) (∂J)p ∂J E5

Here, e(J) is the unit function: e(J) ≡ 1. We immediately arrive at the form of the SDe for the field function: 12

The SD equation is, in general, of higher than the first order. It therefore has several independent solutions, only one of which corresponds to the usual perturbative expansion. The nature of the other solutions is discussed in Appendix 2. 13 The correct way to do this is to subsequently evaluate G2 , G4 , G6 , . . .. On the first iteration, the lowest-order expressions are obtained. Each subsequent iteration gives one higher order in perturbation theory. Note that if we want to obtain the k th order term in Gn , the (k + 1)th order term in Gn+2 is needed, and so on. It is therefore necessary to compute the lower-order terms for more Gn ’s.

August 31, 2019

37   ∂ e(J) = J . S φ(J) + ∂J 0

(1.31)

For the ϕ3/4 theory, it reads   J λ3 ∂ 2 φ(J) = − φ(J) + φ(J) µ 2µ ∂J   ∂ ∂2 λ4 3 φ(J) . φ(J) + 3φ(J) φ(J) + − 6µ ∂J (∂J)2

(1.32)

Although this leads to very nonlinear relations between the various connected Green’s functions this form of the SD equation is actually even simpler to apply : with φ(J) = 0 as a starting pont, iterating the assignment (1.32) then E6 results14 in the correct form of φ(J), giving the connected Green’s functions of Eq.(1.22).

1.3 1.3.1

Diagrammatics Feynman diagrams

An extremely useful tool for computing Green’s functions and connected Green’s functions is at hand in the form of Feynman diagrams. In this section we shall first introduce these diagrams and their concomitant Feynman rules. Only after that shall we prove that these diagrams do, indeed, correctly describe Green’s functions. Feynman diagrams are constructs of lines and vertices. A vertex is a meeting point for one or more lines. Diagrams are allowed in which one or more lines do not end in a vertex but, in a sense sie wandern ins Blaue hinein : such lines are called external lines. Lines that are not external lines, and end up at vertices at both ends, are called internal lines. Diagrams may be connected, in which case one can move between any two points in the diagram following lines of that diagram ; or they may be disconnected, in which case it consists of two or more disjoint pieces that are themselves connected. Any graph15 consists of a finite number of connected 14

For this approach to work in practice, it turns out to be useful to truncate φ(J) as a power series in J, the truncation order increasing by one with each iteration. If you don’t do this, each iteration triples the highest power in J, leading to very unwieldy expressions with only the first few terms being actually correct. 15 For us, the terms ‘diagram’ and ‘graph’ are interchangeable.

38

August 31, 2019

subgraphs. The ‘empty’ graph E, containing no lines or vertices whatsoever, also exists ; it does not count as connected16 . Diagrams containing one or more closed loops are perfectly allowed. Diagrams with no closed loops are called tree diagrams. Some examples of Feynman diagrams are

a connected graph

a disconnected graph

a connected tree graph

Note that the precise shape of the lines and the precise position of the vertices are irrelevant. The important thing is the way in which the lines are connected to the vertices17 .

1.3.2

Feynman rules

The noteworthy thing about Feynman diagrams is that they have an algebraic interpretation; that is, they correspond to numbers that may be added and multiplied. The assignment of a number to a Feynman diagram is governed by the Feynman rules, which postulate a numerical object for every ingredient of a Feynman graph. In the simple zero-dimensional theories that we consider here the Feynman rules are just numbers. We may use, for instance, the following rules : 16

Sophistry : it has no points between which one might wish to move. As you will discover, I have endeavoured in these notes to avoid drawing straight lines, or to draw blobs or closed loops as circles. Many texts do employ only straight lines and circles. This not only leads to awfully unæsthetic-looking pictures, but is also deeply misleading. There is a (natural) tendency to look at Feynman diagrams with the idea that the lines represent ‘particles moving freely through space’ so that the lines ‘ought’ to be straight according to Newton’s first law. That this is completely wrong becomes immediately clear if we realize that, in the zero-dimensional world we are dealing with for now, there cannot be any notion of movement yet, let alone any Newton to pronounce on it. In fact, Newton’s first law ought to be derived from our theory, and we shall do so in due course. 17

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39

↔ 1/µ ↔ −λ3 ↔ −λ4 ↔ +J Feynman rules, version 1.1

(1.33)

A vertex at which a single line ends (and which carries a Feynman rule factor +J) is called a source vertex. A disconnected diagram evaluates to the product of the values of its disjunct connected pieces. Because of this multiplicative rule, the value of the empty diagram E is taken to be unity. In addition, we assign to every Feynman diagram a symmetry factor. The symmetry factor is the single most nontrivial ingredient of the diagrammatic approach. We shall therefore devote a separate section to this issue.

1.3.3

Symmetries and multiplicities

Feynman diagrams have, in general, an ‘inner’ and an ‘outer’ part. The ‘inner’ part consists of the various vertices and internal lines : the ‘outer’ part is made up from the external lines (if any). The inner part concomitates with the symmetry factor of the diagram, and for the outer part we have what may be called the multiplicity, to be discussed below. Let us first turn to the symmetry factor. The rules are : • for every set of k lines that may be permuted without changing the diagram, there will be a factor 1/k! ; • for every set of m vertices that may be permuted without changing the diagram, there will be a factor 1/m! ; • for every set of p disjunct connected pieces that maybe interchanged without changing the diagram, there will be a factor 1/p!;

40

August 31, 2019 • a factor 1/k for every k-fold rotational symmetry18 ; • a factor 1/2 for every mirror symmetry.

External lines cannot be permuted without changing the diagram. Since external lines cannot be permuted, only vacuum diagrams, that is diagrams without any external lines, can have a rotational symmetry. It is important to note that the symmetry factor cannot be read off from the individual components of the diagram, but depends on the topology of the whole diagram19 . As our universe grows from zero to more dimensions, and as the particles considered acquire more properties, the Feynman rules will grow in complication ; but the symmetry factors remain the same20 . A few examples of diagram values are presented here. First, consider the diagram λ3 2 . (1.34) = µ5 In this case, the symmetry factor is 1, since for a tree diagram no internal lines or vertices can be interchanged with impunity. The similar-looking diagram 1 λ3 2 3 = J . (1.35) 2 µ5 has a symmetry factor 1/2! since the upper two one-point vertices are interchangeable. Then, there is the graph = −

1 λ4 2 µ3

Here, there is a symmetry factor 1/2 because the ‘leaf’ can be flipped over without changing the diagram. The diagram = 18

1 λ4 2 6 µ5

Note: 1/k, not 1/(k!). This is what makes the automated evaluation of diagrams a nontrivial task : component factors of diagrams can be easily assigned, but working out the symmetry factor of a diagram calls for for very complicated computer algorithms indeed. 20 This is only modified if we include lines of different types, or oriented lines. Then again, the more-dimensional diagrams have the same symmetry factors as their zero-dimensional siblings. 19

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41

carries a symmetry factor of 1/3! because the three internal lines are interchangeable. The graph = −

1 λ4 3 4 µ7

carries a symmetry factor (1/2!)(1/2!) since there are now only two interchangeable internal lines, and a single ‘leaf’. Finally, the diagram =

1 λ4 2 48 µ4

has a symmetry factor (1/4!)(1/2!) since there are 4 equivalent internal lines, E7 and moreover the diagram can be ‘flipped over’ without changing it. Next, we address the multiplicity. This is the number of different ways the external lines (that each have their own ‘individuality’) can be attached. To determine the multiplicity we must imagine that the whole diagram, or a part of it, can be ‘flipped over’ while retaining the same attachement of the external lines. To illustrate this, we temporarily denote the external lines with a letter, and then notice that the two diagrams a

c

d

and b

d

c

a b

are, in fact, identical ; the multiplicity of this graph is therefore 3, since there are 3 ways to group four letters into two groups of two without regard to ordering. We see that the diagram of Eq.(1.34) has, also, multiplicity 3, while that of Eq.(1.35) has multiplicity 1. We see that, if we include the multiplicity, the replacing of p external lines with p one-point source vertices induces a factor of 1/p!, which will become important later on. The determination of symmetry factors may appear somewhat fanciful21 , but of course it has a solid and unambiguous basis ; the symmetry factor 21

This is due to the fact that the line in the loop is not oriented: for oriented lines it will no longer hold. The discussion of symmetry factors of Feynman diagrams goes, in practice, with a lot of remarks like ‘... so you flip over this leaf, you wriggle this set of internal lines, you shove these vertices back and forth ... see ?’ Although the symmetry factor is totally unambiguous, the arguments for a symmetry factor often come with a lot of prestidigitatorial hand-waving and finger-wriggling in front of a blackboard.

42

August 31, 2019

(and the multiplicity) can always be computed. The procedure is somewhat involved, and is outlined in section 15.2.1.

1.3.4

Vacuum bubbles

Feynman diagrams exist that contain neither external lines nor source vertices. These are called vacuum bubbles!definition ; the empty graph E is obviously a vacuum bubble. We may consider the set of all vacuum bubbles, which we denote by H0 . Let us assume that only four-point vertices occur. Then, H0 , given by H0 = E +

+

+

+

+ ···

(1.36)

(where the ellipsis denotes diagrams with more four-vertices) evaluates to H0

 2 1 λ4 1 1 λ4 1 λ4 2 1 λ4 2 = 1− + + + ··· + 8 µ2 2 8 µ2 16 µ4 48 µ4 35 λ4 2 1 λ4 + + ··· , = 1− 8 µ2 384 µ4

(1.37)

which, indeed, looks suspiciously like H0 for the ϕ4 theory.

1.3.5

An equation for connected graphs

We shall now construct an equation for a special set of diagrams. We do this for the set of Feynman rules of section 1.3.2. First, let us denote by Cn the set of all connected graphs with no source vertices and precisely n external lines. Clearly this is a enumerably infinite set. Next, we define the object Ψ(J), denoted by the symbol Ψ(J) ≡

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

(1.38)

to be the set of all connected diagrams with precisely one external line, and any number of source vertices. The shading indicates that all the diagrams in the blob must be connected . Clearly, then, we have Ψ(J) =

X 1 J n Cn+1 , n! n≥0

(1.39)

August 31, 2019

43

where the extra factor 1/n! is the additional symmetry/multiplicity factor for n source vertices. Let us now consider what can happen if we enter the blob of Eq.(1.38) along the single external line. In the first place, we can simply encounter a source vertex, so that the diagram is =

J . µ

(1.40)

Alternatively, we may encounter a vertex. If this is a three-point vertex, the line splits into two. Taking one of these branches, we may be able to come back to the vertex via the other branch. In that case, the diagram has the form 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

On the other hand, it may happen that the two branches end up in disjunct connected pieces of the diagram, which then looks like 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

Note that these two alternative cases can be unambiguously distinguished because we have restricted ourselves to using only connected graphs. Another important insight is that, in the above diagram, the two final blobs (with their attached lines) are both exactly identical to the original Ψ(J) of Eq.(1.38), and therefore also to each other : a situation that is of course only possible because the blobs represent infinite sets of diagrams. In contrast, the closedloop blob of the first alternative is not equal to Ψ(J) since it has not one but two lines sticking out ; but then again these two lines are completely equivalent. If we encounter a four-point rather than a three-point vertex, the line splits into three, with three alternatives : no branches meeting again further on, all three meeting again, or only two out of the three. We find the diagrammatic equation 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

=

+

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

+

44

August 31, 2019 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111

+

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

+

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

.

(1.41)

Now, realize that 0000000 1111111 0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

and

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

=

X 1 ∂ J n Cn+2 = Ψ(J) n! ∂J n≥0

(1.42)

X 1 ∂2 J n Cn+3 = Ψ(J) , n! (∂J)2 n≥0

(1.43)

=

so that we can translate the diagrammatic equation (1.41) into an algebraic equation for Ψ(J) by carefully implementing the correct Feynman rules, including the symmetry factors for equivalent blobs and lines:   λ3 1 1 ∂ J 2 − Ψ(J) + Ψ(J) Ψ(J) = µ µ 2 2 ∂J   λ4 1 1 ∂ 1 ∂2 3 − Ψ(J) + Ψ(J) Ψ(J) + Ψ(J) . (1.44) µ 6 2 ∂J 6 (∂J)2

E8 E9 E10

Now Eq.(1.44), obtained from the Feynman diagrams via the Feynman rules, has exactly the same form as Eq.(1.32), valid for the field function φ(J) – note the importance of the symmetry factors ! Moreover, the iterative solution for φ(J) starts with φ(J) = J/µ, also identical to the diagrammatic starting . We therefore conclude that Ψ(J) = φ(J), in other words Cn = Cn point (n ≥ 1). This proves that connected Green’s functions can be obtained by the following recipe: to obtain Cn (n ≥ 1), write out all connected Feynman diagrams with no source vertices and precisely n external lines. Evaluate the diagrams using the Feynman rules, and sum them.

1.3.6

Semi-connected graphs and the SDe

A useful notion, which allows us to write SDe’s more compactly, is that of semi-connected graphs. We shall denote these with a lightly shaded blob, and they are defined as follows : a semi-connected graph with n ≥ 1 lines at the left is a general graph with n lines on the left (and any number of

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45

other external lines), with the constraint that each connected piece of the semi-connected graph is attached to at least one of the lines indicated on the left. This may sound more intimidating that is actually is : an example is 1

=

2

1

1 2

+

3

3

1

+

2 3

2

1

2

+

3

+

3

.

2

(1.45)

3

1

A single semi-connected graph with n indicated lines stands for B(n) diagrams with explicit connected graphs, where B(n) is the so-called Bell number : the number of ways to divide n distinct objects into non-empty groups22 . For ϕ3/4 theory, the SDe then becomes simply =

+

+

.

(1.46)

We shall use semi-connected diagrams to good effect in later chapters. Note that the sum of the symmetry factors of all connected diagrams arising from a ϕp vertex must be equal to B(p − 1)/(p − 1)!, which may serve as a check on your SDe’s.

1.3.7

The path integral as a set of diagrams

By affixing a source vertex to the single external line of Ψ(J), we immediately have the result that the generating function W (J) is the sum of all connected Feynman diagrams without external lines and at least one source vertex. If we explicitly indicate the source vertices, and recall that n source vertices in a diagram imply a factor 1/n!, we can write W (J) =

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

22

+

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

+

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

+ · · · , (1.47)

For small n we have B(0) = 1, B(1) = 1, B(2) = 2, B(3) = 5, B(4) = 15, and B(5) = 52 ; more general values can be obtained from the identity X xn B(n) = exp (ex − 1) . n! n≥0

which is derived in Appendix 15.15.7.

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August 31, 2019

where the ellipsis contains connected contributions with more source vertices. Vacuum bubbles do not contribute to W (J). By taking careful account of the symmetry factor assigned to identical connected parts of a disconnected diagram, we can see that 1 W (J)2 2!

=

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+ +

E11

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 111111 000000 000000 111111 000000 111111

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

(lots of other diagrams) .

+

00000 11111 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

(1.48)

Similar arguments hold for higher powers of W (J). In addition, W (J)0 = E = 1. From this it easy to see that the path integral Z(J) consists of all Feynman diagrams without external lines, and without vacuum bubbles, but including the empty diagram. We might wonder why the vacuum bubbles are so conspicously absent . Suppose that we would allow the inclusion of arbitrary numbers of vacuum bubbles in Z(J). Then the Green’s function G0 = 1 would be represented not by the single empty graph but by the whole set H0 discussed before: indeed, H0 is proportional to H0 . In fact, any Green’s function Gn would acquire exactly the same additional factor H0 . The normalization factor N , that must be chosen such as to make G0 equal to unity, therefore extracts exactly the factor H0 from any Green’s function. In the jargon, the vacuum bubbles ‘disappear into the normalization of the path integral’. This is not to say that vacuum diagrams are never important ; but in our approach to computing Green’s functions and connected Green’s functions they are indeed irrelevant. Another way of seeing this is very simple : if we take our diagrammatic prescription of Z(J) and then take J = 0, all diagrams disappear except the empty one, and we find Z(0) = E = 1, just as we must.

1.3.8

Dyson summation

Why is the Feynman rule for lines, stemming from the quadratic part of the action, so different from those for the vertices, that come from the nonquadratic terms ? To see that our treatment is actually a consistent one, let

August 31, 2019

47

us consider an action is given by 1 1 1 S(ϕ) = µϕ2 + λ2 ϕ2 + λ4 ϕ4 . (1.49) 2 2 4! If we wish, we may treat the λ2 term as an interaction, described by a vertex with two legs. the SDe is then seen to be 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

= +

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

+ 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111

+

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

+

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

.

(1.50)

corresponding to J λ2 λ4 φ(J) = − φ(J) − µ µ 6µ

  ∂ ∂2 3 φ(J) . (1.51) φ(J) + 3φ(J) φ(J) + ∂J (∂J)2

Multiplying the equation by µ and transposing the λ2 term to the left, we obtain   J ∂2 λ4 ∂ 3 φ(J) = φ(J) , − φ(J) + 3φ(J) φ(J) + µ + λ2 6(µ + λ2 ) ∂J (∂J)2 (1.52) precisely what we woud have obtained by taking the combination (µ + λ2 ) as the kinetic part from the start. This procedure, by which the effect of twopoint (effective) vertices is subsumed in a redefinition of the kinetic part, is called Dyson summation. In the present example, the summation is of course trivial ; but we shall see that two-point interactions can also arise from more complicated Feynman diagrams corresponding to higher orders in perturbation theory. The manner in which Dyson summation is usually treated is by explicitly writing out the propagator, ‘dressed’ with two-point vertices in all possible ways : + + + + ··· 1 1 1 1 1 1 1 1 1 1 = − λ2 + λ2 λ2 − λ2 λ2 λ2 + ··· µ µ µ µ µ µ µ µ µ µ  k 1 X λ2 = − µ k≥0 µ =

1 1 1 = , µ 1 + λ2 /µ µ + λ2

(1.53)

48

August 31, 2019

where it should come as no surprise that we cheerfully ignore all issues about convergence, in the spirit of perturbation theory. Every propagator line can (and must ! ) be dressed in this way once any two-point vertex (elementary of effective, that is, as the result of a collection of closed loops with two legs sticking out) occurs.

1.4 1.4.1

Planck’s constant The loop expansion, and reverse engineering

As we have seen, Green’s functions can be computed in a perturbative expansion in which the coupling constant λ4 is in some sense a small number. Now consider doing perturbation theory in the ϕ3/4 theory. We then have to decide on the relative order of magnitude of the two coupling constants λ3 and λ4 : are they of the same order, or should we take, say, λ4 to be of the same order as λ3 2 ? And what if even more coupling constants are involved ? We shall adopt the approach that the order of magnitude of the various diagrams should depend not on their coupling-constant content but, rather, on their complexity , in particular on the number of closed loops. That is, the more closed loops a diagram contains, the smaller it is considered to be ; and perturbation theory then prescribes the perturbation expansion to be truncated at a given number of closed loops. To quantify these ideas we shall assign to every closed loop a factor ~, where ~ is a (small) number23 . That is, we define the following ratios : 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

∼ ~ ,

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

∼ ~2 ,

etcetera. This implies, of course, a modification of the Schwinger-Dyson equation from the form (1.32) into   J λ3 ∂ 2 φ(J) = − φ(J) + ~φ(J) φ(J) µ 2µ ∂J   2 λ4 ∂ 3 2 ∂ − φ(J) + 3~φ(J) φ(J) + ~ φ(J) . (1.54) 6µ ∂J (∂J)2 23

As the notation suggests, it will develop into Planck’s (or Dirac’s) constant as our universe increases in complexity.

August 31, 2019

49

In turn, we shall have to modify everything else as well : we must re-define φ(J) = ~

∂ log Z(J) , ∂J

(1.55)

so that the SDe for the path integral must read   ∂ 0 Z(J) = JZ(J) . S ~ ∂J

(1.56)

The path integral must therefore be re-defined with inclusion of ~ :   Z
1 Z(J) = N exp − S(ϕ) − Jϕ dϕ , ~ and for the Green’s functions we have     n n n ∂ n ∂ Gn = ~ Z(J) , Cn = ~ log Z(J) . (∂J)n (∂J)n J=0 J=0

(1.57)

(1.58)

The Feynman rules must, therefore, take the form

↔

~ µ

↔ −

λ3 ~

↔ −

λ4 ~

↔ +

J ~

Feynman rules, version 1.2

(1.59)

The introduction of ~ as the perturbation expansion parameter allows us to determine the relative orders of magnitude of coupling constants. Since under our decree all tree diagrams are of the same order, the two graphs and

50

August 31, 2019

tell us that λ4 is of the same order as λ3 2 . Similarly, a k-point coupling constant λk is of the same order as λ3 k−2 . As a last point, you may note that the introduction of ~ does not influence the Dyson summation of sec.1.3.8, since every extra two-point vertex (with 1/~) also gives an extra propagator (with ~). Since in the Feynman rules ~ appears all over the place, it is advisable to check that the ~-behaviour of the Feynman graphs is indeed as desired. For an arbitrary given diagram let us define the characteristics E : the number of external lines, I : number of internal lines, Vq : the number of vertices of q-point type, L : the number of closed loops, and P : the number of disjunct connected pieces. There are precisely two ‘sum rules’ that involve linear combinations of these quantities. They are derived in section 15.3 and read X X qVq = 2I + E . (1.60) Vq = I + P − L , q

q

We are now able to read off the power of ~ associated with an arbitrary connected diagram (with P = 1). From the Feynman rules, we infer that every line contributes a factor ~ and every vertex a factor 1/~. The total power of ~ is, therefore X Vq = E + L − 1 . (1.61) E+I − q

Independently of its precise form, the power of ~ of any connected diagram depends only on the number of its external lines and the number of loops, and indeed each extra loop leads to an additional factor ~, as advertised.

1.4.2

The classical limit

Since in perturbation theory ~ is taken to be an infinitesimally small quantity, the limit ~ → 0 is of automatic interest. This limit has to be taken with some care since ~ = 0 strictly would imply that only Green’s functions with E + L = 1 would survive24 . Instead, the classical limit ~ → 0 is meant to be the result of leaving out diagrams containing closed loops. The diagrammatic 24

Later on, the discussion about truncation will clarify how this is not inconsistent.

August 31, 2019

51

SDe will, for the ϕ3/4 theory, then take the form 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

=

+

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 00001111 1111 00000000 1111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111

.

(1.62)

The corresponding solution will be denoted by φc (J) (with c for ‘classical’), and the classical SDe is written as λ3 λ4 J φc (J)2 − φc (J)3 . (1.63) φc (J) = − µ 2µ 6µ The classical field function is exclusively built up from tree diagrams : this is called the tree approximation. Note that it obeys an algebraic, rather than a differential, equation, that can be written as S 0 (φc (J)) = J .

(1.64)

This is called the classical field equation. This is not to be confused with equations from classical, nonquantum physics. In fact, the classical field equations will turn out to be the Klein-Gordon, Dirac, Proca and Maxwell equations. Of these, only the Maxwell equations can be considered classical, since they do not contain a particle mass.. Note that such equations have, in general, more than a single solution. Here, however, we are interested in that solution that vanishes as J → 0, which may be written out using Lagrange expansion25 :   n  J X 1 n−1 ∂ n−1 J µ S0 φc (J) = + . (1.65) n−1 µ n≥1 n! µ (∂J) Let us now look at the path-integral picture of the classical limit. When ~ becomes small, the fluctuations in the path integrand    1 S(ϕ) − Jϕ exp − ~ become extremely exaggerated. The main contribution to hϕi therefore comes from that value ϕc of ϕ for whichthe probability distribution attains its maximum, that is, hϕiJ ≈ ϕc , where S 0 (ϕc ) = J ,

S 00 (ϕc ) > 0 .

Also in the classical limit, we therefore have φc (J) = ϕc . 25

see Appendix 15.15.12.

(1.66)

52

1.4.3

August 31, 2019

On second quantisation

The ‘classical’ approximations of our quantum field theory are26 quantum equations. In fact, this is not so very surprising. In ordinary quantum mechanics, the classical variables such as position, momentum, etcetera are identified with the expectation values of their quantum-mechanical counterparts, and considered a useful approximation of reality as long as they are reasonably well-defined27 . So it is here as well : the field generating function φ(J) is considered as the expectation value of the quantum field ϕ, and it is identified with the quantum-mechanical wave function of whatever object it is we are studying. In this sense, to go from ϕ to a classical observable we have to ‘classicize’ two times. The transition from ordinary quantum mechanics to what we are doing here is therefore dubbed ‘second quantization’. Of course, from the point of view we have taken here, this is simply a matter of taking limits (expectation value upon expectation value), but if one comes in from the classical side it may look quite mysterious. This is another reminder that one should not try to build a more fundamental theory from a limiting case. Limiting cases are only hints.

1.4.4

Instanton contributions

As mentioned, for a non-free action S(ϕ), the equation (1.66) has, of couse, more than a single solution28 . Suppose that we have several such solutions, (0) (1) (2) denoted by ϕc , ϕc , ϕc ,. . ., and that the minimal value of S(ϕ) − Jϕ is (0) attained for ϕc . Then, the other classical solutions will give contributions that, relative to the dominant one, are suppressed by exponential factors of order    1 (k) (0) (k) (0) S(ϕc ) − S(ϕc ) − Jϕc + Jϕc , k = 1, 2, . . . . exp − ~ In figure 1.1 we plot the (normalized) form of exp(−S(φ)/~) for an action with two minima, for two values of ~ different by a factor 6. It is seen how the lowest minimum of S(ϕ) starts to dominate the integral as ~ becomes small ; the contribution from the subleading maximum decreases nonperturbatively 26

Will be found to be ; see the later chapters of these notes. With small uncertainty, that is, the variance of their statistical distribution around the expectation value. 28 Since the action is at least of order ϕ3 , the classical field equation is at least quadratic. 27

August 31, 2019

53

Figure 1.1: The developing of instanton effects

fast. Such subdominant solutions to the classical field equations are called instantons. Their contribution to Green’s functions do not have a series expansion around ~ = 0. Such effects are therefore not accessible using Feynman diagrams : they are called nonpertubative. This is not to say that they are irrelevant. Indeed, we usually have a finite value for ~ ; more (1) dramatically, if we let J vary as a parameter, ϕc , say, may for some value (0) of J take over from ϕc as the true maximum position of the probability (0) (1) density, causing a sudden shift in the value of φc (J) from ϕc to ϕc .

1.5 1.5.1

The effective action The effective action as a Legendre transform

Since perturbation theory presumes that higher orders in the loop expansion are small compared to lower orders, the following question suggests itself : is it possible to find, for a given action S(ϕ), another action, called the effective action, with the property that its tree approximation reproduces the full field function of the original action S ? If such an effective action, denoted by

54

August 31, 2019

Γ(φ), exists, we must have Γ0 (φ(J)) = J ,

(1.67)

where φ(J) is the full solution to the SDe belonging with S(ϕ). We can use partial integration to find Z Z Γ(φ) = J dφ = J φ − φ dJ = J φ − ~W , (1.68) where J is now to be interpreted as a function of φ. The transition from W (J) to Γ(φ) is called the Legrendre transform. In classical mechanics, we have the same situation : there, ~W would be the Lagrangian with J as the velocity and φ as the momentum, and then the effective action would turn out to be the Hamiltonian. An important fact to be noted about the effective action can be inferred as follows. Let us consider the derivative of φ(J). If we denote the probability density (including the sources) of the quantum field ϕ by PJ (ϕ), that is,   1 A(ϕ) , A(ϕ) = exp − (S(ϕ) − Jϕ) , (1.69) PJ (ϕ) = R ~ dϕA(ϕ) we can write this derivative as  R R PJ (ϕ) ϕ dϕ PJ (ϕ) ϕ2 dϕ 1 0 1 d R R φ (J) = = − ~ ~ dJ PJ (ϕ) dϕ PJ (ϕ) dϕ R PJ (ϕ1 )PJ (ϕ2 ) (ϕ1 2 − ϕ1 ϕ2 ) dϕ1 dϕ2 = .
2 R PJ (ϕ) dϕ


2 PJ (ϕ) ϕ dϕ
2 R PJ (ϕ) dϕ

R

(1.70)

By symmetry, we can replace the factor (ϕ1 2 − ϕ1 ϕ2 ) by (ϕ1 − ϕ2 )2 /2, so as to see that dφ(J)/dJ is positive. This implies that ∂2 dJ >0 . 2 Γ(φ) = dφ (∂φ)

(1.71)

In other words, the effective action is concave everywhere29 . Whereas one would assume that the effective action Γ would differ only slightly from the original action S, this can obviously no longer hold in situations where the action S is not concave. 29

This concavity persists in case there are more than just a single field involved. By extension, it also holds for Euclidean theories in more dimensions ; see also Appendix 15.6.

August 31, 2019

1.5.2

55

Diagrams for the effective action

A tree approximation consists of tree diagrams only. To see how the loop effects of the action S end up in Γ, we define a new concept, that of a oneparticle irreducible (1PI) diagram. A connected Feynman graph is 1PI if it contains no internal line such that cutting that line makes the diagram disconnected.

1PI diagrams

a non-1PI diagram

External lines, of course, do not enter in the 1PI criterion30 . Note that a diagram consisting in only external lines and a single vertex also counts as 1PI, since it does not have any internal lines to be cut. A typical one-loop 1PI diagram looks like this :

(1.72) Let us denote the set of all 1PI graphs with precisely n external lines by −γn /~, where the convention is that the Feynman factors for the external lines are not included. Consider, now, what happens if we enter the field function by way of its single external leg, as in the SDe. If we encounter a vertex, that vertex is part of a 1PI subdiagram (possibly consisting of only the vertex itself). Indicating the 1PI property with cross-hatches, we therefore obtain the diagrammatic equation 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

= +

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+ 111111 000000 000000 111111 000000 000000 111111 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 000000 111111 111111

+

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111

+

00000 11111 00000 11111 00000 11111 11111111 00000000 00000 11111 00000 11111 00000 11111 00000 11111 00000000 11111111 00000 11111 00000 11111 00000000 11111111 00000 11111 00000 11111 00000000 11111111 00000 11111 00000 11111 00000000 11111111 00000 11111 00000 11111 00000000 11111111 00000 11111 00000 11111 0000000011111 11111111 00000 11111 00000

· · · . (1.73)

00000 11111 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

Algebraically, it reads   J 1 1 1 2 3 φ(J) = − γ1 + γ2 φ(J) + γ3 φ(J) + γ4 φ(J) + · · · , µ µ 2! 3! 30

(1.74)

Including them would be silly, since any diagram falls apart if we chop through an external line.

56

August 31, 2019

in other words Γ0 (φ) = J ,

(1.75)

where 1 1 1 (γ2 + µ)ϕ2 + γ3 ϕ3 + γ4 ϕ4 + · · · . (1.76) 2! 3! 4! We conclude that the vertices of the effective action are determined by the 1PI diagrams. It must be noted that, in general, the effective action contains vertices with an arbitrarily large number of legs, even if the original action S goes up only to ϕ3 or ϕ4 , say. Γ(ϕ) = γ1 ϕ +

E12

1.5.3

Computing the effective action

We shall now describe a computation of the effective action Γ(φ) = Γ0 (φ) + ~Γ1 (φ) + ~2 Γ2 (φ) + · · · , from its Feynman diagrams, for a theory with arbitrary couplings : X λk 1 ϕk . S(ϕ) = µ ϕ2 + 2 k! k≥3

(1.77)

(1.78)

We start by considering a general one-loop 1PI diagram such as that of Eq.(1.72), and cutting through the loop at some arbitrary place. We then have a propagator ‘dressed’ with zero or more vertices where external lines are ‘radiated off’. If there are precisely n external lines we can denote this by n

Such an object has, of course, its own SDe. Taking careful account of all possibilities to attach external lines, we can write it as   n n−1 n = θ(n = 0) + 1     n−2 n−3 n n + + + · · ·(1.79) 2 3 We define the generating function for such dressed propagators as n X zn P (z) = = , n! n≥0

(1.80)

August 31, 2019

57

and see that the SDe reads P (z) =

~ λ3 z 2 λ4 z 3 λ5 − z P (z) − P (z) − P (z) − · · · , µ µ 2! µ 3! µ

(1.81)

in other words, P (z) =

~ S 00 (z)

.

(1.82)

We now close the loop again with an arbitrary vertex, at which vertex at least one other external line is included. By the same combinatorial arguments as above we can find the generating function L(z) for such loops :

L(z) = 1 = 2

+

+

+ ···

  λ4 z 2 λ5 S 000 (z) λ3 P (z) − · · · = − 00 (1.83) . − P (z) − z P (z) − ~ ~ 2! ~ 2S (z)

The symmetry factor 1/2 arises from the fact that the propagator is not oriented and thus we have to avoid double-counting. Considering that a propagator with n external legs leads to a closed loop with at least n + 1 external legs, we see that the one-loop effective action is given by Z ~ Γ1 (φ) = −~ dφ L(φ) = log(S 00 (φ)) . (1.84) 2 A few remarks are in order here. In the first place, we see that the effective action obtained in this way is only well-defined where the action itself is concave, in agreement with the discussion in 1.5. In the second place, the trick of closing the loop with an extra vertex, rather than just trying to ‘glue’ the endpoints of P (z) together, is technically useful since it avoids potential problems with the symmetry factors. In the last place, the above calculation is possible since all external lines are, so to speak, identical. In more dimensions, where external lines can carry momentum, this is no longer true. However, the effective potential, that is the effective action at zero momentum, does lend itself to such a calculation in higher dimensions31 . 31

For simple scalar theories. Of course external lines may carry more than just momentum information, that is, they can also carry spin/charge/colour· · · information. Then the calculation is again more difficult.

58

August 31, 2019

We can extend this treatment to higher loop orders as well. Let us denote a vertex where at least n + 1 lines come together by

n =

n+

n+1

+

n+2

+ ···

(1.85)

and assign to this dressed vertex the Feynman rule 1 ~

n = − S (n+1) (z) .

(1.86)

Now, we introduce the notion of a tadpole diagram : this is a connected diagram with precisely one external line and no source vertices. The effective action as given above then follows from writing out the 1PI tadpole diagrams, replacing propagators by dressed propagators and vertices by dressed vertices ; we can then simply read off the result. 111 000 S (3) (z) 000 111 000 111 = − 000 111 2S (2) (z) 000 111

→

⇒ Γ01 (φ) =

S (3) (φ) , 2S (2) (φ)

(1.87)

as before. In two loops, the 1PI tadpole is given by the diagrams +

+

+

Dressing these tadpole diagrams gives us 11111 00000 00000 11111 00000 11111 00 11 + 00 11 00 11 000000 111111 000000 111111

 = ~

111 000 000 111 000 111 000 111 000 000 111 111 000 111 000 111 000 111 000 111 000 000 111 111 000 111

+

1111 0000 0000 1111 000 111 000 111 000 111

111 000 000 111 000 111 000 111

1111 0000 0000 1111 0000 1111 +

111 000 000 111

S (3) (z) S (4) (z) S (3) (z)3 S (3) (z) S (4) (z) S (5) (z) − + − 6 S (2) (z)3 4 S (2) (z)4 4 S (2) (z)3 8 S (2) (z)2

 (1.88)

The two-loop contribution to the effective action is therefore d S (5) (φ) 5 S (3) (φ) S (4) (φ) S (3) (φ)3 Γ2 (φ) = − + . dφ 8 S (2) (φ)2 12 S (2) (φ)4 4 S (2) (φ)4 E13

(1.89)

The effective action itself, the integral over the above experession, does not

August 31, 2019

59

have so simple a form as in Eq.(1.84), but is of course calculable as soon as S(φ) is explicitly given ; moreover, we see that it, too, becomes undefined where S 00 (φ) vanishes. From our diagrammatic approach we see that this will persist in all loop orders32 .

1.6

Exercises

Excercise 1 Green’s functions and connected Green’s functions We have X X Jn Jn Z(J) = Gn , W (J) = Cn n! n! n≥0 n≥0 and W (J) = log(Z(J)) . Using that G0 = 1, and the expansion − log(1 − x) =

X xn n≥1

n

=x+

x2 x3 x4 x5 + + + − ··· , 2 3 4 5

express C0,...,5 in terms of the G’s. Excercise 2 The problem with ϕ4 theory By examining the case where λ is infinitesimally small, either positive and negative, argue that the limit λ → 0 establishes, in fact, an extremely singular theory. Excercise 3 Actually doing it for ϕ4 This exercise shows how to express the path integral for the ϕ4 theory in terms of ‘known functions’. We consider   Z µ 2 λ 4 H = dϕ exp − ϕ − ϕ 2~ 24~ 1. Show that the combination g=

~λ µ2

is dimensionless. 32

Because in all 1PI diagrams we have to dress the propagators, which implies lots of S (φ) in the denominators. 00

60

August 31, 2019 2. Perform the substitution s ϕ=


3~ 1/4 ψ − ψ −1/4 µg

to derive s H=

3~ 3/(4g) e 4µg

Z∞ dψ ψ

−3/4



3 exp − 8g

  1 ψ+ ψ

0

3. The so-called modified Bessel functions of the second kind , denoted by Kν (z), are discussed in appendix 15.15.8. Use this to express H in terms of these Bessel functions. 4. The function Kν (z) has two expansions: one, an unattractive-looking but convergent series in positive powers of z, and a nonconvergent, asymptotic series in terms of negative powers of z, given in Eq.(15.305). Show that H therefore has a regular series expansion in 1/g and an asymptotic one in g. Show that the leading term in the asymptotic expansion is independent of g (but not of µ), and compute the next two terms. Compare your result with Eq.(1.20). Excercise 4 The SDe for Z in another action Find the SDe for the path integral Z(J) for the action 6 µ 2 X λk k ϕ , S(ϕ) = ϕ + 2! k! k=3

Excercise 5 Writing the SDe for φ Prove Eq.(1.30). Do this using the fact that  Z  1 Z(J) = exp dJ φ(J) , ~ and then considering Z 0 , Z 00 and so on. Excercise 6 The SDe for φ in another action For the action of Exercise 4, derive the SDe for φ(J).

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61

Excercise 7 The symmetry factor of life, the universe, and everything Devise a diagram (or set of diagrams) that has a symmetry factor of 1/42. Excercise 8 Diagrammatic SDe for ϕ6 theory Give the diagrammatic SDE for ϕ6 theory, and write it out algebraically. Excercise 9 Actually doing it Consider the diagrammatic SDe of Eq.(1.41). Starting with the empty diagram, iterate this diagrammatic equation three or four times, and write down the resulting set of diagrams. Excercise 10 Some diagrams to consider Of the following 12 diagrams, determine the multiplicity factor, the symmetry factor, and the number of loops :

Excercise 11 Vacuum diagrams Consider a theory with couplings ϕn , n = 3, 4, 5, 6, . . .. Show that the single vacuum diagram at one loop is actually zero. At two loops there are 3 vacuum diagrams, and at 3 loops there are 15 vacuum graphs. Write these down, and determine the symmetry factors.

62

August 31, 2019

Excercise 12 SDe for the effective action The effective action is defined such that Γ0 (φ(J)) = J. Show that the SDe for pure ϕ4 theory can be written as   000 λ4 φ 0 3 2 Γ (φ) µφ = Γ (φ) − φ + 3~ 00 − ~ 00 3 6 Γ (φ) Γ (φ) Let the effective action have a perturbation expansion : Γ(φ) = Γ0 (φ) + ~Γ1 (φ) + ~2 Γ2 (φ) + · · · Use the SDe to find Γ0 , Γ1 and Γ2 . Compare the results with the corresponding 1PI diagrams with up to 6 external legs. Excercise 13 A wonderful logarithmic action Consider the action µ µ S(ϕ) = − 2 log(1 − aϕ) − ϕ , a a 1. The domain of ϕ is now no longer (−∞, ∞). Determine which domain is appropriate, and prove that the path integrand indeed vanishes at the endpoints of that domain. 2. Prove that this theory has ϕn interactions for all n = 3, 4, 5, . . . and that the vertex for a ϕn interaction reads −(n − 1)!µan−2 /~. 3. Prove that the SDe for this theory reads µ~Z 0 (J) = JZ(J) − a~Z(J) − a~JZ 0 (J) . 4. Prove that the field function is given by φ(J) =

J − a~ . µ + aJ

5. The field function has only one-loop corrections ! Verify that the twoloop correction to the propagator vanishes. 6. Show that the effective action is µ  µ Γ(φ) = − 2 + ~ log(1 − aφ) − φ a a It is also free from higher-order corrections beyond one loop !

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63

7. At J → −µ/a, the field function diverges. Show that this corresponds precisely to the situation where the path integrand no longer vanishes at one of the endpoints. Excercise 14 A wonderful exponential action Consider the action µ S(ϕ) = 2 (eaϕ − 1 − aϕ) a 1. Show that the SDe for the path integral takes the functional form  µ Z(J + a~) = J + Z(J) a 2. Show that the solution for the path integral has the form  Z(J) =

a2 ~ µ

aJ/~    µ + aJ µ  Γ /Γ a2 ~ a2 ~

3. Show that the field function reads   2    a~ µ + aJ 1 φ(J) = log +ψ a µ a2 ~ where the digamma function ψ is defined in appendix 15.15.4. 4. Show that the perturbation expansion of the field function is given by   X aJ a2k−1 Bk 1 a~ ~k − φ(J) = log 1 + − a µ 2(µ + aJ) k≥2 k(µ + aJ)k 5. Show that all odd loop orders beyond the first one vanish. 6. At J → −µ/a, the field function diverges. Show that this corresponds precisely to the situation where the path integrand no longer vanishes at one of the endpoints.

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August 31, 2019

Chapter 2 On renormalization 2.1 2.1.1

Doing physics : mentality against reality Physics vs. Mathematics

In this chapter, we digress a bit into a discussion of what it is that particle physicists claim to be doing : confronting theory with reality. This leads to some interesting subtleties. If we were mathematicians, the subject matter in the previous chapter might be formulated as the following task : given the parameters µ, λ3 and λ4 of the action, to compute the connected Green’s functions. This ‘mathematician’s scheme’ may be depicted as follows : µ , λ3 , λ 4

−→

C1 , C 2 , C 3 , C 4 , . . .

In this set-up, the parameters are supplied from outside the computational and experimental context. Since, however, as (I hope) we are physicists the situation is somewhat different : we first have to measure the values of the parameters from inside the experimental context, using some of the connected Green’s functions as measurement processes, and then predict some other connected Green’s functions, which we shall call prediction processes. This rather different situation may be depicted by the scheme Ek = Ck , k = 1 . . . 4

−→

µ , λ3 , λ 4

−→

C5 , C 6 , C 7 , . . .

Here, the quantities E1,2,3,... stand for the experimentally observed values of the connected Green’s functions : barring experimental errors or refinements, these numerical values do not change under any improvement of the theory. 65

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August 31, 2019

Now consider the fact that we are doing perturbation theory. That is, both the measurement and the prediction processes are known only as truncated series in ~. Let us suppose that by stolidity and perseverance a next higher order in perturbation theory for the prediction processes has become available. Is this any good ? Obviously not, unless a similar increased level of precision has been attained for the measurement processes. Only in that case a new ‘fit’ of the parameters of the action can be made, and improved values of the prediction processes can usefully be obtained. This order-by-order improvement is called renormalisation. Let us denote by a superscript the order to which the connected Green’s functions have been computed. The ‘physicist’s scheme’ can then be envisaged as follows : (0)

−→

µ(0) , λ3 , λ4

(1)

−→

µ(1) , λ3 , λ4

(2)

−→

µ(2) , λ3 , λ4

(3)

−→

µ(3) , λ3 , λ4 .. .

Ek = Ck , k = 1 . . . 4 Ek = Ck , k = 1 . . . 4 Ek = Ck , k = 1 . . . 4 Ek = Ck , k = 1 . . . 4 .. .

(0)

(0)

−→

C5 , C 6 , . . .

(0)

(0)

(1)

(1)

−→

C5 , C 6 , . . .

(1)

(1)

(2)

(2)

−→

C5 , C 6 , . . .

(2)

(2)

(3)

(3)

−→

C5 , C 6 , . . . .. .

(3)

(3)

Order by order, the parameters keep getting updated, but in the overall picture they are just bookkeeping devices that allow one to go from measurements to predictions of the more physically interesting connected Green’s functions. It should not come as a surprise that in the measurement-parameterprediction protocol, a higher-order correction in the parameters due to an improved measurement expression is cancelled again, to some extent, in the prediction. In fact, for certain classes of theories, which are called renormalisable, these cancellations may be quite extreme.

2.1.2

The renormalisation program : an example

As an example of the renormalisation program, we shall investigate ϕ4 theory, with coupling constant λ. In perturbation theory, the first few nonzero connected Green’s functions are given by   1 2 2 11 3 ~ 1 − u + u − u + ··· , C2 = µ 2 3 8

August 31, 2019 C4 = C6 = C8 = C10 =

67

  ~3 λ 149 2 197 3 7 − 4 1− u+ u − u + ··· , µ 2 12 4   ~5 λ2 1535 2 6405 3 10 − 80u + u − u + ··· , µ7 3 2   ~7 λ3 111755 2 3 − 10 280 − 3815u + u − 330925u + · · · , µ 3   ~9 λ4 12672800 2 3 15400 − 310940u + u − 49859600u + · · · , µ13 3 (2.1)

where u is the ubiquitous combination ~λ/µ2 . Now suppose that the two measurements give C2 = ~/m and C4 = −~3 g/m3 . In lowest order of perturbation theory, this would imply that the action parameters are µ = m and λ = g. However, in higher orders we would no longer have C2 = E2 and C4 = E4 . Instead, we must choose   7 2 1 3 1 µ = m 1 − y − y − y + ··· , 2 12 8   3 3 2 11 3 λ = g 1 + y + y + y + ··· , (2.2) 2 4 8 with y = ~g/m2 . The properly renormalised Green’s functions then become ~ , m ~3 g C4 = − 4 , m  5 2 ~g 375 3 3 C6 = y + ··· , 10 − 15y + 45y − m7 2   ~7 g 3 67725 3 2 C8 = − 10 280 − 1155y + 5775y − y + ··· , m 2
~9 g 4 2 3 C10 = 15400 − 118440y + 882000y − 6963075y + · · · . (2.3) m13 The difference between the ‘naive’ and the renormalised connected Green’s functions is quite evident. In particular C2 and C4 are completely free of higher-order corrections. For the other connected Green’s functions the coefficients in the perturbation expansion are smaller in absolute value than in the ‘naive’ expressions (see also below, section 2.4). C2 =

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August 31, 2019

This discussion is obviously only a drastically simplified example of a phenomenological situation that is in practice much more complicated. For instance, one does not, usually, renormalise connected Green’s functions but rather quantities extracted from scattering matrix elements, that are themselves not identical to, but extracted from connected Green’s functions. The experimental observables E therefore do not take the simple form given here. The higher-order corrections themselves are typically much more complicated, and not completely free from ambiguities, nor necessarily finite. Nevertheless, the operational scheme outlined above is essentially the same as those that are employed in real-life physics. In particular, I cannot stress often enough that the renormalisation procedure is necessary simply because we do perturbation theory, not because loop corrections may contain infinities1 .

2.2 2.2.1

A handle on loop divergences A toy : the dot model

Notwithstanding the above remarks on the per se necessity of renormalisation, the fact that, in nontrivial theories, loop diagrams often contain infinities makes the need to do something about them all the more urgent. Loop divergences arise from summation over internal degrees of freedom of Feynman diagrams. In zero dimensions there are no such internal degrees of freedom, and all diagrams are finite. We can, however, introduce the following toy model. Consider, as before, our working-horse ϕ3/4 theory. Let us assume that we introduce yet another Feynman rule : we shall apply a factor 1 + c1 to every closed loop that contains precisely one vertex, and a factor 1 + c2 to every closed loop that contains precisely two vertices. Loops with more vertices remain unaffected2 . The numbers c1 and c2 may depend on the parameters of the theory, or on other parameters. In the spirit of ‘loop divergences’ we shall envisage that c1,2 → ∞ at some stage. In terms of Feynman diagrams, this rule amounts to duplicating each one- or two-vertex 1

This insight is, even at present, not as endemic as it ought to be. This rule accords with ‘naive power counting’ for four-dimensional scalar theories without derivative couplings, the most direct four-dimensional extension of the zerodimensional theories we are discussing in this chapter. 2

August 31, 2019

69

loop with a ‘dotted’ loop : =

+

=

≡ c1 ×

,

+

,

≡ c2 ×

,

.

(2.4)

For example, under this rule the following two-loop diagrams are modified accordingly : →

+

+

= (1 + c1 )(1 + c2 ) →

+ ,

+

= (1 + c2 )

.

(2.5)

The Feynman diagrams are governed by the Schwinger-Dyson equation. Our new rule must therefore be implemented, somehow, into a modified SDe. Some reflection tells us that the necessary new ingredients are made up out of those Feynman diagrams that contain only dotted loops. Fortunately, E15 these form a manageable set, where we differentiate between 1PI diagrams with up to 4 legs3 : ≡

+

+

≡

+

+

+

+

+

,

≡ ≡ 3

(2.6)

+ ··· +

+ ··· (2.7)

+ +

+ ··· ,

+ +

+ ··· , + ··· .

(2.8) (2.9)

With 5 or more legs our rule does not allow for diagrams with only dotted loops.

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August 31, 2019

The only diagram that does not carry a ‘tower’ of loops on its back is the last diagram in the two-point dotted series. Using these artefacts, we can now rewrite the appropriate SDe for our ϕ3/4 theory with the added dotting rule : 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

=

+ 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111 0000 00001111 1111 0000 1111 0000 1111

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+ 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

+

000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

+

+

+ 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

+

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

+

00000 11111 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 00000 11111 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111

+

+

00000000 11111111 00000000 11111111 11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111

11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111

+

.

(2.10)

We can readily translate this SDe into algebraic form. If we take out the external propagators from the ‘black box’ graphs, we can write = B1 ,

= B2 ,

= B3 ,

= B4 .

(2.11)

We shall leave the actual evaluation of these sets of graphs for later (cf section 2.3.5): at this point, we shall simply treat them as effective vertices. The ‘dotted-loop’-modified SDe then reads, when we work out the graphs one after the other, in the order in which they are displayed above : φ =

J B1 B2 λ3 2 ~λ3 0 − − φ− φ − φ µ µ µ 2µ 2µ B3 B3 2 ~B3 0 ~B3 0 − φ2 − φ − φ − φ 2µ µ 2µ µ λ4 ~λ4 0 ~2 λ4 00 − φ3 − φφ − φ 6µ 2µ 6µ

August 31, 2019

71 −

B4 3 ~B4 0 ~B4 0 ~2 B4 00 φ − φφ − φφ − φ . 2µ 2µ µ 2µ

(2.12)

We can simply rewrite this SDe as (µ + B2 )φ = (J − B1 ) − (λ3 + 3B3 )(φ2 + ~φ0 ) −(λ4 + 3B4 )(φ3 + 3~φφ0 + ~2 φ00 ) .

(2.13)

But hold it ! This is just the SDe equation belonging to the action 1 1 1 S(ϕ) = B1 ϕ + (µ + B2 )ϕ2 + (λ3 + 3B3 )ϕ3 + (λ4 + 3B4 )ϕ4 . (2.14) 2 6 24 Therefore, the spirit of renormalisation tells us that in every application the bare parameters µ, λ3 and λ4 will never occur on their own, but always only R in the renormalised combinations µR = µ + B2 , λR 3 = λ3 + 3B3 , and λ4 = λ4 + 3B4 ; and that therefore, whatever the values of B2,3,4 , the combination E17 will automatically be finite if the experimental quantities in which they enter are finite. We can therefore choose the action’s parameters such that all Green’s functions come out finite ; and the remaining B1 can always be completely compensated by an extra ϕ1 in the action4 . Indeed, this is the way in which the notorious ‘loop divergences’ are absorbed into the bare action : infinite loop corrections are compensated for by infinite bare parameters5 .

2.2.2

Nonrenormalizable theories

The significant point in the discussion above is the fact that all dottedloop contributions can be absorbed into a finite number of terms of the bare action. We may formulate the requirement of a renormalizable theory as that which states that a finite number of measured quantities6 suffice to make all other predictions of the theory well-defined. If an infinite number of measured quantities would be necessary, the theory would be called non-renormalizable : but, worse, from the operational point of view it would be worthless7 . As an example of a non-renormalizable situation, let us 4

Here we introduce the notion of counterterms. More about this in section 2.5. Or, in other words, at any loop order there are only finitely many counterterms to be determined. 6 Think of E2,3,4, . 7 This train of thought might be relaxed. If the necessary additional experimental values are only relevant at some very high energy scale, the theory would be effectively renormalizable. It is a matter of taste whether you feel comfortable with this, or not. 5

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August 31, 2019

consider a Feynman rule in which a loop with three vertices acquires a dotted counterpart : that is, we would have a (potentially infinite) contribution of the form

This can, of course, be repaired by introducing into the bare action a ϕ6 interaction ; but in that case there would arise dotted loops with eight external legs :

which would necessitate a ϕ8 interaction — and so on. A theory would arise in which an infinite number of measured quantities would be needed before any consistent8 prediction could be made : a non-renormalizeable situation ! The same problem occurs in a theory with a bare ϕ6 interaction. It is seen that the requirement of renormalizability puts constraints on the bare action9 .

2.3 2.3.1

Scale dependence Scale-inpendent scale dependence

As mentioned above, the parameters of the action have to be determined by comparison to experimentally measured quantities. Such measurement experiments do not take place in some abstract realm10 , but rather in a concrete physical situation. This experimental context partially determines the measurement result. A very concrete example is the measurement of the coupling constant using a scattering process : in that case, one of the determining factors is the energy at which the scattering takes place. Also choices made in the theoretical computation of the measured quantities play their rˆole : for example, in dimensional regularization11 an energy scale must be introduced, and this scale can to a large extent be chosen arbitrarily. We 8

i.e. finite in high orders of perturbation theory. It must come as no surprise that the Higgs potential of the Standard Model has no interaction terms for the Higgs field (which is scalar) more complicated than the four-point coupling. 10 Pace Platone. 11 To be discussed later on. 9

August 31, 2019

73

shall lump all these effects together into a quantity s, which we shall call the scale. It must be stressed that the scale also contains the (regularized) loop divergences, and may be expected to become infinite at some stage. Let us consider a theory with only one parameter : an example of such a theory is massless QCD, that is the theory of massless quarks and gluons and their interactions. The single parameter is then the coupling constant. Let the bare parameter, as it occurs in the action, be denoted by v. The renormalised parameter, extracted from experiment, will be denoted by w. The renormalised coupling is then given by the bare coupling and the experimental context, embodied by the scale s : w = F (v, s) .

(2.15)

This relation ought to be invertible12 so that we can find v given w : v = G(w, s) .

(2.16)

Obviously we have w = F (G(w, s), s) ,

v = G(F (v, s), s) .

(2.17)

By differentiation we find the following relations between the derivatives of F and G : ∂v F (v, s) ∂w G(w, s) = 1 ,

∂s F (v, s) + ∂v F (v, s) ∂s G(w, s) = 0 ,

(2.18)

where ∂x stands for the partial derivative ∂/∂x. Let us now consider an infinitesimal change in the scale. Since v is independent13 of s, the value of w has to adapt itself in a manner prescribed by F : d w = ∂s F (v, s) . (2.19) ds This expression contains the (divergent) scale s and the (divergent) value of the bare coupling v. We can, of course, express everything in terms of w : d w = ∂s F (G(w, s), s) . ds 12

(2.20)

Usually, F is given as a (formal) series in v, for instance w = v + α2 v 2 + α3 v 3 + · · ·, and then the inverse always exists , v = w − α2 w2 + (2α22 − α3 )w3 + · · · 13 After all, a parameter in the action couldn’t possibly anticipate which experiment is going to measure it !

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August 31, 2019

This expression now contains the finite number w and the divergent scale s : it can only make sense if s actually drops out. Note that this is not a proof, it is a requirement that we make of the theory ! We therefore demand that F (v, s) be such that
∂s F (G(w, s), s) = 0 . (2.21) In other words, ∂s2 F (v, s) + ∂s ∂v F (v, s) ∂s G(w, s) = 0 . Using Eq.(2.18) and dividing by ∂v F we can write this as   ∂s F ∂s ∂v F ∂s F ∂s2 F − = ∂v =0 : ∂v F (∂v F )2 ∂v F

(2.22)

(2.23)

this means that there is a function h(v) of v only, such that ∂s F (v, s) = h(v) ∂v F (v, s) .

(2.24)

By separation of variables, Eq.(2.24) is easily solved, and we find w = F (v, s) = f s + ρ(v)




,

1 d ρ(v) = , dv h(v)

(2.25)

for some function f . There must be some value s0 for s, such that w and v precisely coincide. This value can, of course, depend on v, so we write it as s0 (v). We therefore have
f s0 (v) + ρ(v) = v , s0 (v) + ρ(v) ≡ j(v) , (2.26) so that f and j are each other’s inverse : f (j(v)) = v. Applying j to Eq.(2.25) we see that s + ρ(v) = j(w) = s0 (w) + ρ(w) . (2.27) In this equation, both terms on the left-hand side are divergent, but on the right-hand side they are finite. We can now determine the scale dependence of w. This dependence, called the beta function of the theory, is given by
d ∂ w = f s + ρ(v) ds ∂s
1 h(w) = f 0 j(w) = 0 = . j (w) 1 + h(w)s00 (w)

β(w) ≡

(2.28)

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75

All reference to the bare coupling has been removed : we see that the renormalised coupling has a definite, predictable dependence on the energy scale of the measuring experiment14 . Note, also, that whereas we introduced h as a function of the bare parameter, it enters in the beta function as a function of the renormalised parameter. Finally, as we shall see in the next section, the h functions usually starts at second order (O (v 2 )) in perturbation theory, so that the two first terms in the beta function are independent of the form of s0 (v) as long as this is non-singular for v = 0.

2.3.2

Low-order approximation to the renormalised coupling

Let us examine the possible shape of the function F (v, s) in some more detail. In the spirit of perturbation theory, it will be given by a series expansion like F (v, s) = v + v 2 α1 (s) + v 3 α2 (s) + v 4 α3 (s) + · · · .

(2.29)

Let us assume that the functions αj (s) vanish at s = 0, so that s0 (v) = 0 and h(v) = β(v). The h function is given by v 2 α10 (s) + v 3 α20 (s) + v 4 α30 (s) + · · · F2 (v, s) , h(v) = = F1 (v, s) 1 + 2vα1 (s) + 3v 2 α2 (s) + · · ·

(2.30)

so that we see that the beta function must start with v 2 : β(v) = β0 v 2 + β1 v 3 + β2 v 4 + · · ·

(2.31)

The requirement that the beta function depend not on s governs the form of the functions αj (s) : to low order in v we have from Eq.(2.30)
(2.32) β(v) = v 2 α10 (s) + v 3 α20 (s) − 2α1 (s)α10 (s) + · · · , 14 A remark is in order here. What, in these notes, is called the scale is usually understood to be the logarithm of the actual energy scale : indeed, whereas the energy scale has the dimension of energy (obviously), the number s is, strictly speaking, dimensionless. If we denote the scale by the conventional symbol µ, the derivative dw/ds should then be rewritten as d d w → µ w ds dµ .

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August 31, 2019

so that we can derive α1 (s) = β0 s ,

α2 (s) = (β0 s)2 + β1 s , . . .

(2.33)

It is easily derived that the leading term in αn (s) is (β0 s)n . Let us assume that the beta function is dominated by its lowest-order term, that is, β(v) = β0 v 2 . Then ρ(v) = −1/(β0 v) and we find 1 1 = − β0 s . w(s) v

(2.34)

We can exchange the bare parameter v for the measured value of w at some fixed scale s0 , and then the running is given by 1 1 = − β0 (s − s0 ) , w(s) w(s0 )

(2.35)

w(s0 ) . 1 − β0 w(s0 )(s − s0 )

(2.36)

or w(s) =

At this point we may start to distinguish between different theories. The renormalised, physical parameter w is a priori unknown, and has to be determined by experiment ; but the number β0 is perfectly computable from inside the theory15 . The running of the coupling is therefore determined as soon as the action has been sufficiently specified. Now, it may happen that β0 is positive : in that case, the effective coupling w(s) increases with increasing s, and will eventually become infinite at some high scale. On the other hand, when β0 is negative, the effective coupling decreases with increasing energy scale. This is called asymptotic freedom. It is the phenomenon that has saved the theory of strong interactions : in the 1960’s when the typical energy scales of experiments where low, the effective coupling was so high (of order 10) as to cast doubts on the usefulness of perturbation theory, whereas at the high energies current from around 197516 the effective coupling has become small enough (of the order of 0.1) to warrant the use of perturbation techniques. 15

The number β0 is a combinatorial factor with the addition of some powers of π, and simple numbers depending on the ingredients and quantum numbers of the particles in the theory. 16 I take the commissioning of the PETRA (Hamburg, BRD) and PEP (Stanford, USA) colliders as the definitive starting point of the relevance of perturbative QCD.

August 31, 2019

2.3.3

77

Scheme dependence

We must recognize that not only the scale of a given measurement process is important, but of course also the nature of the measurement process. That is, we may define the measured coupling constant w in two different ways, on the basis of two different measurement processes17 : let us denote the two results by w and w. ˜ We say that such different values have been obtained using different renormalisation schemes. In all cases I have encountered, two such schemes agree at the tree level18 , and the results are therefore perturbatively related : w˜ = w + t1 w2 + t2 w3 + t3 w4 + · · · ,

(2.37)

with t1,2,3,... computable numbers ; and conversely

w = w˜ − t1 w˜ 2 + 2t1 2 − t2 w˜ 3 − 5t1 3 − 5t1 t2 + t3 w˜ 4 + · · ·

(2.38)

Having computed the beta function for w, we can now simply obtain it for w˜ : β(w) ˜ =

dw˜ dw dw˜ = ds dw ds 

  2 3 4 = 1 + 2t1 w + 3t2 w + 4t3 w + · · · β0 w + β1 w + β2 w + · · ·
= β0 w2 + (β1 − 2t1 β0 ) w3 + β2 − 2t1 β1 + 6t1 2 β0 − 3t2 β0 w4 + · · ·
= β0 w˜ 2 + β1 w˜ 3 + β0 t1 2 − β0 t2 + t1 β1 + β2 w˜ 4 + · · · (2.39) 2

3

The two beta functions can be transformed from one scheme to another ; for any scheme dependence for which Eq.(2.37) holds, the first two coefficients, β0 and β1 , are seen to be independent of the actual scheme, as was to be expected (cf Eq.(2.28)) : the two schemes correspond to different functions s0 (v) as defined in section 2.3. 17

In practice, this difference can be quite small, as between the so-called MS and MS schemes. With ‘different measurement processes’, we here mean two different, complete operational schemes that both lead to a well-defined value for coupling constants. 18 This rules out possible but, for a practicing physicist useless and/or irrelevant, differences such as for instance obtained by defining w ˜ = 2w. Get a life !

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2.3.4

Theories with more parameters

We can extend the investigation of section 2.3.1 to theories with n ≥ 2 parameters, labelled by a superscript. We then have wi = F i (v, s) and v i = Gi (w, s) ,

i = 1, 2, . . . , n ,

(2.40)

with v = (v 1 , v 2 , . . . , v n ) and w = (w1 , w2 , . . . , wn ). The generalisations of Eq.(2.18) then read : ∂` F i (v, s) ∂j G` (w, s) = δji ,

∂s F i (v, s) + ∂j F i (v, s) ∂s Gj (w, s) = 0 ,

∂` Gi (w, s) ∂j F ` (v, s) = δji ,

∂s Gi (w, s) + ∂j Gi (w, s) ∂s F j (v, s) = 0 , (2.41)

where ∂j stands for ∂/∂v j , and we adhere to the Einstein summation convention. Note that the pullback function Pji (v, s) ≡ ∂j Gi (F 1 (v, s), . . . , F n (v, s), s)

(2.42)

is the inverse of the derivative of F : Pji (v, s) ∂i F ` (v, s) = δj` = Pj` (v, s) ∂` F i (v, s) .

(2.43)

The scale dependence of wk is given by d k w = ∂s F k (w, s) = ∂s F k (G(w, s), s) , ds

(2.44)

and as before we require the last form to be actually independent of s :
0 = ∂s ∂s F k (G(w, s), s) = ∂s2 F k (v, s) + ∂s ∂` F k (v, s) ∂s G` (w, s) = ∂s2 F k (v, s) − ∂s ∂` F k (v, s) Pj` (v, s) ∂s F j (v, s) .

(2.45)

Multiplying this by Pkj and using Eq.(2.43) twice we can derive
0 = Pkj (v, s) ∂s2 F k (v, s) − Pkj (v, s) ∂s ∂` F k (v, s) Pj` (v, s) ∂s F j (v, s)

= Pkj (v, s) ∂s2 F k (v, s) + ∂s Pkj (v, s) ∂` F k (v, s) Pj` (v, s) ∂s F j (v, s)
= Pkj (v, s) ∂s2 F k (v, s) + ∂s Pkj (v, s) ∂s F k (v, s) , (2.46) as the generalisation of Eq.(2.23). Thus there must be functions hi (v) (i = 1, . . . , n) such that Pkj (v, s) ∂s F k (v, s) = hj (v) . (2.47)

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79

Upon multiplication by ∂j F m (v, s) we then find the analogue of Eq.(2.24) : ∂ ∂ m F (v, s) = hj (v) j F m (v, s) , m = 1, 2, . . . , n . (2.48) ∂s ∂v Notice the occurrence of only a single differential operator hj (v)∂j !

2.3.5

Failure of the dot model

Let us apply the above to the ϕ4 theory in the dot model of section 2.2.1. We can write Eq.(2.48) as follows: ∂s K = p(u) λ

∂ ∂ K + q(u) µ K ∂λ ∂µ

(K = λR , µR ) ,

(2.49)

and u = ~λ/(2µ2 ). In the dot model, the sums of dotted diagrams in Eqns.(2.7,2.9) yields closed forms for the renormalised parameters :     uc1 (s) 2 2 3 R 2 R , µ =µ 1− + u c2 (s) . λ = λ −2 + 1 + uc2 (s) 1 + uc2 (s) 3 (2.50) Applying Eq.(2.49) to λR then tells us that
−3u c02 (s) = p(u) 1 − 4u c2 (s) + 2u2 c2 (s)2 + 6q(u) u c2 (s) . (2.51) It is easy to see that p(u) = αu + O (u2 ), with some constant α, and q(u) = O (u). The terms proportional to u1 in Eq.(2.51) demand that c2 (s) = −α(s − s0 )/3 (with arbitrary s0 ). But in that case the right-hand side of Eq.(2.51) cannot be made s-independent unless p(u) = q(u) = 0, which implies that c1,2 do not depend on s at all. The dot model exhibits a curious half-renormalisability, where the dots can be absorbed into renormalised parameters, but the running of these parameters cannot be made scale-independent. For a viable, fully renormalisable theory the divergence structure of the Feynman diagrams must be much more complicated and subtle than simple ‘dotting’ !

2.4

Asymptotics of renormalisation in ϕ4 theory

As mentioned in section 2.1.2, the perturbation expansion of the connected Green’s functions appears to have numerically smaller coefficients than that

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of the unrenormalised ones. We can investigate this for the ϕ4 theory systematically as follows. The bare connected Green’s functions are given by  n X (k) ~λ ~ (−u)n−1 t2n (u) , u = 2 , t2n (u) = C2n = t2n uk . (2.52) µ µ k≥0 We employ a renormalisation scheme in which V2 and C4 are free of corrections : ˆ ~2 ~λ ~ , C4 = − 2 uˆ , uˆ = 2 , (2.53) C2 = µ ˆ µ ˆ µ ˆ where the renormalised quantities are indicated with a hat. Using uˆ = −

u t4 (u) C4 2 = t2 (u)2 C2

(2.54)

we express uˆ = u − 5u2 /2 + · · · as a series in u, which can be inverted to yield u = uˆ s(ˆ u) ,

(2.55)

and the series s(ˆ u) starts with unity. The bare and renormalised µ’s are then related by
µ=µ ˆ τ2 (ˆ u) , τ2 (ˆ u) = t2 uˆ s(ˆ u) , (2.56) and C2,4 are indeed free of corrections. For the higher connected Green’s functions C2n , n ≥ 3 we then have  n ~ (−ˆ u)n−1 τ2n (ˆ u) , C2n = µ ˆ
 n−1 X t2n uˆ s(ˆ u) s(ˆ u) (k) τ2n uˆk . (2.57) τ2n (ˆ u) = = τ2 (ˆ u) τ2 (ˆ u) k≥0 All these manipulations are well suited to computer algebra. We can then (k) (k) compute the ‘improvement factor’ t2n /τ2n . The results for C6,8,10,12 are depicted in figure 2.1 as a function of k, the loop order, up to 155 loops. The improvement is less for higher connected Green’s functions but at high order the factors approach an asympotic value. We see that the asymptotic character of perturbation theory is completely unchanged by renormalisation, since neither the superexponential (n!) nor the purely exponential (cn ) or even the polynomial (na ) behaviour of the coefficients is affected. The asymptotic value of all improvement factors is exp(15/4) = 42.521 · · ·, which is explained in appendix 15.15.3.

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81

Figure 2.1: Improvement factor for ϕ4 theory in zero dimensions

2.5 2.5.1

The method of counterterms Counterterms in the action

We can view the renormalization program in a slightly different way, by insisting that the original parameters and the renormalized ones coincide, and shifting the burden onto the action itself. That is, we include interaction terms that are of higher order in ~, the so-called counterterms. For the ϕ4 theory this leads to µ + δµ 2 λ + δλ 4 ϕ + ϕ , 4! X2 X (k) δµ = ~k δµ(k) , δλ = ~ k δλ .

S(ϕ) →

k≥1

(2.58)

k≥1

Inspecting the dimensions we see that we can write δµ(n) = δn

λn µ2n−1

,

(n)

δλ = ηn

λn+1 , µ2n

(2.59)

where the δn and ηn are pure numbers. In the rest of this section we can therefore use λ = µ = 1 without causing confusion. The counterterms (including the δµ ) are treated as extra interactions, and we can choose them so that they give us our favourite renormalization scheme, for instance we can

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again decide to have C2 and C4 free of loop corrections. Obviously it will be E16 sufficient to restrict ourselves to the 1PI diagrams. Denoting the counterterm interactions (and their formal loop order) by dots, we start by computing the one-loop correction to C2 : 1 = − − δ1 2

1

+

⇒

δ1 = −

1 . 2

(2.60)

Similarly, for the one-loop correction to C4 we have +

1

=

3 − η1 2

⇒

3 . 2

η1 =

(2.61)

And so we continue : at two loops we have + =

+

1 1 δ1 η1 + + − − δ2 4 6 2 2

1

1

+ ⇒

δ1 = − 1

+ + + 3 3 = − − − 3 − 3δ1 − 3η1 − η2 4 2 E18 E19

2

+

+ ⇒

7 , 12 1

η2 =

+ 3 . 4

2

(2.62)

The resulting values for δ1,2 and η1,2 are of course precisely those we already encountered in Eq.(2.2), only now we can compute them in a systematic, diagrammatic way.

E20

2.5.2

Return to the dot model, and a preview

If we compare our dotted-loop model with the counterterm approach, it becomes clear why the dot model fails : in essence, it only considers one-loop couterterms, or in other words it assumes that all loops can only be divergent in a single way. If (as we shall do from chapter 4 onwards) we recognize that the loop divergences arise from momenta in the various loop reaching all the way to infinity, then more-loop diagrams contain several loop momenta and we will have to worry in which way they go to infinity, all together or one after the other. In chapter 4 we shall come back to this issue.

August 31, 2019

2.6

83

Exercises

Excercise 15 Some reflection Verify the claim made below Eq.(2.5) by showing that every diagram containing dotted loops can be assigned its place in the modified SDe of Eq.(2.10). Excercise 16 Argue it yourself Give the argument for restricting ourselves to the 1PI diagrams only in the computation of the counterterms as discussed in section 2.5. Excercise 17 Dotting in ϕ3 Show that for pure ϕ3 theory (never mind its ill-defined character) there are only two dotted 1PI diagrams possible. There are therefore no ‘infinite towers’ of dotted diagrams. Write the appropriate SDe in this dotted model. Such theories, where only a finite number of diagrams diverge, are called super-renormalizable. Show that, in this case, we can afford to also dot loops with 3 propagators, and end up with a renormalizable theory19 . Excercise 18 Counterterms at three loops The three-loop 1PI contribution to C2 has 5 distinct diagrams without counterterms, 6 with one 1-loop counterterm, 2 with one 2-loop counterterm, and 2 with two 1-loop counterterms : and, of course, 1 diagram with the 3-loop counterterm. Write these out, and compute their symmetry factors and multiplicities, and compute from them that δ3 = −1/8. 19

This would correspond with a ‘realistic’ ϕ3 theory in 6 dimensions, and this is why a 6-dimensional ϕ3 theory makes a nice playground to study renormalization.

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Excercise 19 Counterterm at three loops (cont’d) The three-loop 1PI contribution to C4 has 12 distinct diagrams without counterterms, 14 with a single one-loop counterterm, 2 with one 2-loop counterterm, and 4 with two 1-loop counterterms : and, of course, 1 diagram with the 3-loop counterterm. Write these out, and compute their symmetry factors and multiplicities, and compute from them that η3 = 11/8. Excercise 20 Employing the counterterms Write the one-loop diagrams for C6 , including the counterterm diagrams. From that, compute the renormalized value of the one-loop correction to C6 , and compare to Eq.(2.3).

Chapter 3 More fields in zero dimensions 3.1

The action and the path integral

We now have established an overview of the quantum-field theoretic behaviour of a single field ϕ in zero dimensions. Generalising to a theory with more than one field is fairly straightforward, and we shall do so in this chapter. We shall assume that there are K distinct fields, labelled ϕj , j = 1, 2, . . . , K. The number K can be taken as large as we please, and even infinite provided that the fields form a countable set1 . These fields have a combined probability density given by   1 (3.1) P (ϕ1 , ϕ2 , . . . , ϕK ) = N exp − S(ϕ1 , ϕ2 , . . . , ϕK ) , ~ where we have immediately introduced ~ since we are now familiar with it. In the special case where the action is separable, that is, S(ϕ1 , ϕ2 , . . . , ϕK ) = S1 (ϕ1 ) + S2 (ϕ2 ) + · · · + SK (ϕK ) , the fields are actually independent random variables ; the theory is just so many copies of the single-field one, and in the following we shall disregard that uninteresting situation. In concert with our convention of having a coupling constant λn accompanied by a factor 1/n!, we shall let the coupling of several fields be accompanied by combinatorial factors for each field separately, so the action may contain a term λ1,3,7 ϕ1 2 ϕ3 2 ϕ7 4 , (2!)(2!)(4!) 1

This notion will be relaxed later on, as we move to higher dimensions.

85

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August 31, 2019

which gives the Feynman rule −λ1,3,7 /~ for this vertex. The Green’s functions are defined by Gn1 ,n2 ,...,nK = hϕ1 n1 ϕ2 n2 · · · ϕK nK i

(3.2)

In order to be able to keep the various fields apart, we have to assign to each of them its own source Jj , j = 1, 2, . . . , K. The path integral therefore reads2 X J n1 · · · J nK 1 K Gn1 ,...,nK Z(J1 , . . . , JK ) = n ! · · · n ! 1 K n1,...,K ≥0 !! Z K X 1 S(ϕ1 , . . . , ϕK ) − Jj ϕj dϕ1 · · · dϕK . (3.3) = N exp − ~ j=1 The extraction of the Green’s functions is then performed as   ∂ nK ∂ n1 n1 +···nK Z(J1 , . . . , JK ) . Gn1 ,...,nK = ~ ··· (∂J1 )n1 (∂JK )nK J1 =···=JK =0 (3.4)

3.2

Connected Green’s functions and field functions

The relation between the Green’s functions and their connected counterparts is again given by straightforward generalization:
W (J1 , . . . , JK ) = log Z(J1 , . . . , JK ) X J n1 · · · J nK 1 K = Cn1 ,...,nK (3.5) n ! · · · n 1 K! n ≥0 1,...,K

The precise expression of the G’s in terms of the C’s is of course now somewhat more involved : for instance, for K = 3 we have G1,0,0 = C1,0,0 , G1,1,0 = C1,0,0 C0,1,0 + C1,1,0 , G1,1,1 = C1,0,0 C0,1,0 C0,0,1 + C1,1,0 C0,0,1 + C1,0,1 C0,1,0 + C0,1,1 C1,0,0 + C1,1,1 . 2

(3.6)

Where possible, we denote multiple integral by a single integration sign. This usually does not lead to confusion.

August 31, 2019

87

We now have K field functions, one for each field ; they are given by φj (J1 , . . . , JK ) = ~

∂ W (J1 , . . . , JK ) . ∂Jj

(3.7)

An important thing to note is that, since the field functions are derivatives, ∂ ∂ φj (J1 , . . . , JK ) = φk (J1 , . . . , JK ) . ∂Jk ∂Jj

3.3

(3.8)

The Schwinger-Dyson equation

The SDe for the path integral can be summarized as follows :   ∂ Z(J1 , . . . , JK ) = Jk Z(J1 , . . . , JK ) , S(ϕ1 , . . . , ϕK ) ∂ϕk ϕj =~∂/∂Jj

(3.9)

as can easily be verified. For the field functions, the SDe is best illustrated with an example. Suppose that we have the following action for K = 2 : 1 1 λ S(ϕ1 , ϕ2 ) = µ1 ϕ1 2 + µ2 ϕ2 2 + ϕ1 2 ϕ2 2 . 2 2 4

(3.10)

This time, the coupling constant λ carries a factor 1/(2!)/(2!) since there are not four identical fields ‘meeting’ at the vertex, but rather two pairs of identical fields, as mentioned above. We indicate the field type with either ‘1’ or ‘2’. The Feynman rules for this case are ~ , µ1 J1 ↔ , ~ ↔

1

1

~ , µ2

↔

2

2

1

2

1

2

−λ , ~

↔

J2 . ~

↔

(3.11)

There are two coupled Schwinger-Dyson equations, one for each field : 1

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 00000 11111 0000 1111 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

1

=

1

+

1

2

2 000000 111111 000000 111111

+

000000 1 111111 000000 111111 000000 111111

1

000000 111111 111111 000000 000000 111111

2

2

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

+

2

1

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

1

2

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

000000 111111 1 111111 000000 000000 111111

+

1

2

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

2

,

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August 31, 2019

2

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

0000 1111 1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

2

=

2

+

2 1

+

2

2

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

1

1

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 00000 11111 1111 0000 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

1

+

000000 111111 000000 111111

000000 111111 1 111111 000000 000000 111111

2

000000 111111 111111 000000 000000 111111

2

1

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

+

2

000000 1 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 000000 111111 1 111111 000000 111111 000000 111111 000000 000000 111111 2111111 000000 111111

,

(3.12) E21

with the following analytical representation for the field functions φj = φj (J1 , J2 ) (j = 1, 2) :   2 λ J1 ∂ ∂ 2 2 ∂ − φ1 φ2 + ~φ1 φ2 + 2~φ2 φ1 + ~ φ1 , φ1 = µ1 2µ1 ∂J2 ∂J2 (∂J2 )2   2 λ J2 ∂ ∂ 2 2 ∂ − φ2 φ1 + ~φ2 φ1 + 2~φ1 φ2 + ~ φ2 = φ2 . µ2 2µ2 ∂J1 ∂J1 (∂J1 )2 (3.13) The effective action must of course be a two-variable function Γ(φ1 , φ2 ) such that ∂ Γ(φ1 , φ2 ) = Jj , j = 1, 2 . (3.14) ∂φj

E22 E23

This effective action is also concave. The two-field case can, obviously, be extended to the case of arbitrarily many fields, provided the couplings are unambiguously defined.

3.4

The sum rules revisited

Chapter 1 discusses the diagrammatic sum rules for a theory with only one field (derived in appendix 15.3). Of course, we have to extend this also. Let us therefore assume various fields, labelled by an index j. Then, E (j) and I (j) are the number of external and internal lines of type j, respectively, and (j) by Vq we denote the number of vertices that have precisely q legs of type j sticking out. The generalization of Eq.(15.21) then reads X qVq(j) = 2I (j) + E (j) ∀ j , (3.15) q

August 31, 2019

89

and that of Eq.(15.22) takes the form X X Vq(j) = I (j) + P − L . j,q

(3.16)

j

As before, these sum rules are valid in any nonzero dimension as well.

3.5 3.5.1

A zero-dimensional toy for QED Fields and sources

We consider the following action for three fields, including sources : 1 ¯ − ϕJ S(ϕ, ϕ, ¯ B) = µB 2 + mϕϕ¯ + eϕBϕ ¯ − Jϕ ¯ − HB . 2

(3.17)

The field ϕ¯ is not automatically 3 the ‘conjugate’ of ϕ in any sense : it is just different from ϕ. Note the absence of symmetry factors since all the fields in the three-point vertex are distinct. Also the two-point interaction term mϕϕ ¯ carries no factor of 1/2. Such an action can stand for an extremely primitive model for QED, the theory of electrons and photons. The action has three partial derivatives : ∂ S(ϕ, ϕ, ¯ B) = mϕ¯ + eϕB ¯ − J¯ , ∂ϕ ∂ S(ϕ, ϕ, ¯ B) = mϕ + eBϕ − J , ∂ ϕ¯ ∂ S(ϕ, ϕ, ¯ B) = µB + eϕϕ ¯ −H . ∂B The SDe’s for the path integral are therefore   2 ∂ 2 ∂ ¯ ¯ J, H) = 0 , ~m + e~ − J Z(J, ∂J ∂J∂H   2 ∂ 2 ∂ ¯ J, H) = 0 , ~m ¯ + e~ ¯ − J Z(J, ∂J ∂ J∂H   2 ∂ 2 ∂ ¯ J, H) = 0 . ~µ + e~ ¯ − H Z(J, ∂H ∂ J∂J 3

See, however, section 3.5.2.

(3.18)

(3.19)

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August 31, 2019

The field-generating functions (the ‘field functions’) are, of course, each a function of J, J¯ and H, and are given by ∂ ψ = ~ ¯ log Z , ∂J

∂ ψ¯ = ~ log Z , ∂J

A=~

∂ log Z , ∂H

(3.20)

so that

∂ ∂ ∂ Z = AZ , ~ ¯Z = ψ Z , ~ Z = ψ¯ Z , ~ ∂J ∂H ∂J and Eq.(3.18) can be written as   1 e ∂ ψ = J− ψ , Aψ + ~ m m ∂H   ∂ ¯ 1 ¯ e ¯ ¯ J− ψA+~ ψ , ψ = m m ∂H   1 e ¯ ∂ A = H− ψψ+~ ψ . µ µ ∂J

(3.21)

(3.22)

We may rewrite these SDe’s since ∂ ∂ ψ = ¯A , ∂H ∂J

∂ ¯ ∂ ψ= H , ∂H ∂J

∂ ∂ ψ = ¯ψ¯ . ∂J ∂J

(3.23)

The Feynman rules are, for this action, as follows : ψ

E24

ψ

~ , m ~ , ↔ µ e ↔ − . ~ ↔

(3.24)

A few things are of interest here. In the first place, we have here the first instance of an important concept : that of oriented lines. The propagator is oriented, it runs from ϕ to ϕ. ¯ In the second place, in the action we find the ¯ two terms Jϕ and ϕJ, ¯ which would suggest that J is the source in the SDe ¯ and J¯ is the source in the SDe for ψ ; but it is actually the other way of ψ, around ! What is the source for a given field function is seen by taking the derivative of the action, and inspecting which field then occurs as a linear term, and which source term is left by itself after the differentiation.

August 31, 2019

3.5.2

91

Bald, furry and quenched toys

We can try to recast the above ‘QED’ in the form of a path integral. We then immediately run into problems ! The integral over B is doable, but the integral of exp(−mϕϕ/~) ¯ diverges if we simply take ϕ and ϕ¯ to be unrelated real fields. The way out is to say that, after all, ϕ¯ is the complex conjugate of ϕ so that ϕϕ ¯ is nonnegative. The action can then be written as 2 2 µ 2 J − |J| S(ϕ, ϕ, ¯ B) = B + (m + eB) ϕ − − HB 2 e + mB m + eB |J|2 µ 2 2 B + (m + eB)r − − HB , (3.25) = 2 m + eB where we have shified ϕ to ϕ0 = ϕ − J/(m + eB) as integration variable, then wrote dϕ dϕ¯0 ∝ r dr with r = |ϕ|. We can now integrate over the ϕ, ϕ¯ fields :     Z Z 1 1 2 Z ∼ dB d(r ) exp − S(ϕ ϕ, ¯ B) ∼ dB exp − S(B) , ~ ~   2 µ |J| eB S(B) = B 2 − HB − + ~ log 1 + . (3.26) 2 m + eB m Here we have dropped overal constant factors at will, and reabsorbed the result of the r2 integral into the exponent. For reasons that shall appear below, we can call this Bald QED. How can we interpret the term ~ log(1 + eB/m) ? Let us consider the set of all diagrams with a single closed ϕ loop and n B legs attached, with the E25 B propagators removed : +

+

+ ···

(3.27)

The diagram with n B legs evaluates to (n − 1)!(−e/m)n . Suppose that we wish to avoid these ϕ loops altogether. This can be done by introducing counterterms as described in chapter 2 : these counterterms are n-point B self-interactions vertices, that can be summed :   X  e n B n eB ~ − = −~ log 1 + . (3.28) m n m n≥1 We see that the logarithmic term in S(B) simply embodies the closed ϕ loops ! As we shall see in section 10.2.7, in ‘grown-up’ QED we have Furry’s

92

August 31, 2019

theorem that states that ϕ loops with an odd number of B legs must vanish when summed correctly. We can implement this by using counterterms for E26 only these odd-n loops :  2k+1      X ~ eB ~ eB eB − = − log 1 + − log 1 − . (3.29) 2k + 1 m 2 m m k≥0 Thus we arrive at three formulations for ‘zero-dimensional QED’ : Bald, with no modifications ; Furry, with Furry’s theorem implemented ; and Quenched, where we suppress all closed ϕ loops4 . Their respective S(B) actions are :   µ 2 eB |J|2 Bald : S(B) = B − HB − + ~ log 1 + , 2 m + eB m   |J|2 ~ e2 B 2 µ 2 + log 1 − Furry : S(B) = B − HB − , 2 m + eB 2 m2 |J|2 µ . (3.30) Quenched : S(B) = B 2 − HB − 2 m + eB

3.6

Exercises

Excercise 21 Two-field action Consider the two-field action, now with the sources included : S(ϕ1 , ϕ2 ) =


λ µ ϕ1 2 + ϕ2 2 + ϕ1 2 ϕ2 2 − J1 ϕ1 − J2 ϕ2 2 4

1. Determine the 2 SDe’s for the path integral. 2. From these, determine the 2 SDe’d for the field functions φ1 and φ2 . 3. Verify that this is agreement with the diagrammatically obtained result. Excercise 22 A O(N ) symmetric theory Consider a theory containing N fields ϕ1,2,...,N , with the following action : S(~ ϕ) =

µ 2 λ |~ ϕ| + |~ ϕ|4 , 2 24

and with sources J1,2,...,N . Since the action is invariant under any real rotation in the space of vectors (ϕ1 , ϕ2 , . . . , ϕN ) it is called O(N ) symmetric. 4

Leaving plenty of other loops !

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93

1. Show by using diagrams that the SDe can be written as   ∂ ~2 λ 2 2~ 2 ~ ~ φk |~ ϕ| + ~φk (∇ · φ) + ~ |φ| + ~ ∇ φk µφk = Jk − 6 ∂Jk Notice that to obtain this form one may have to use the fact that ∂φk /∂Jm = ∂φm /∂Jk . 2. Show that the same result can be obtained algebraically, by working out de SDe in the following form :   ~ + ~∇ ~ e = Jk Sk φ where Sk stands for the partial derivative of the action S(~ ϕ) to ϕk , and ∇k stands for ∂/∂Jk . ~ alone. We can therefore write 3. By symmetry, W must depend on |J|   ~ 2 /2 φk = Jk F |J| for some function F . Show that this function obeys 
λ 2xF (x)3 + ~ (N + 2)F (x)2 + 6xF (x)F 0 (x) µF (x) = 1 − 6 
2 0 00 + ~ (N + 2)F (x) + 2xF (x) Excercise 23 The 123 theory We consider a theory containing not 1 but 3 fields, labeld ϕ1,2,3 . The action, including the sources, is given by
µ ϕ1 2 + ϕ2 2 + ϕ3 2 + gϕ1 ϕ2 ϕ3 − J1 ϕ1 − J2 ϕ2 − J3 ϕ3 S(ϕ1 , ϕ2 , ϕ3 ) = 2 In the following we shall concentrate on Feynman diagrams and will not worry about the convergence of the path integral. 1. Prove that there are now 3 different SDe’s, of the form µ~

∂ ∂ ∂ Z + g~2 Z = Ji Z ∂Ji ∂Jj ∂Jk

where i, j, k is a permutation of 1,2,3.

94

August 31, 2019 2. There are now of course also 3 field functions φ1,2,3 . Prove that ∂ ∂ φj = φi , ∂Ji ∂Jj

i, j ∈ (1, 2, 3)

3. Give the SDe for the field functions using diagrams. 4. Prove that for any diagram with nj external lines of type j (j = 1, 2, 3) the following must hold: the nj are either all even, or all odd. 5. By Cij we denote the connected Green’s function with two external legs, one of type i and one of type j. Prove that Cij = 0 , i 6= j , and furthermore that there are no tadpole diagrams. Excercise 24 Symmetry factors in QED For the model of section 3.5, write the three SDe’s in diagrammatic form. Show that all symmetry factors are equal to unity for diagrams with at least one external leg. Show that this implies the same for actual Quantum Electrodynamics, that is, the same model but extended to Minkowski space and with more complicated propagators and vertices. Excercise 25 Throwing in the loops in QED Compute the diagrams in Eq.(3.27), and prove the result in the text. Keep track of the symmetry and multiplicity factors ! Use this to show the correctness of the counterterms in Eq.(3.28). Excercise 26 Furry’s theorem in higher loops Prove the following : if Furry’s theorem holds at one-loop order, it holds to all loop orders.

Chapter 4 QFT in Euclidean spaces 4.1

Introduction

The main characteristic of a space(-time) of more than zero dimensions is the fact that the quantum field is defined at more than one point ; in fact, at an infinity of points. The possibility of sending signals from one point to another one requires the existence of correlations between the field values at different points. The nature of this correlation, and its reflection in the appropriate Feynman rules, is our subject now.

4.2 4.2.1

One-dimensional discrete theory An infinite number of fields

We shall consider a theory of a countably infinite set of fields in zero dimensions. We denote by {ϕ} the set of all these fields : {ϕ} = . . . , ϕ−3 , ϕ−2 , ϕ−1 , ϕ0 , ϕ1 , ϕ2 , ϕ3 , . . . where the field labels run from −∞ to +∞. The collection of all their corresponding sources is denoted by {J}. We shall, as a working example, consider a theory where the interaction consists of four fields with the same label meeting at one point. Moreover, we shall assume the kinetic terms to 95

96

August 31, 2019

be uniform in the field labels. Thus, the action will be1 :  X 1 λ4 4 2 S({ϕ}, {J}) = µϕn − γϕn ϕn+1 + ϕn − Jn ϕn , 2 4! n

(4.1)

where we include the sources in the action2 . If γ were zero, the action would be separable and the theory would be an uninteresting series of replicas of the zero-dimensional action for a single field. We shall consider positive values of γ ; in that case, the action tends to minimize if ϕn and ϕn+1 carry the same sign : a positive correlation between ‘neighbour’ fields is the result. Note, moreover, that the action is, by choice, invariant under the relabelling of n by n + K with any fixed K : this is called translation invariance, in this case translation by a fixed increment in labelling3 . The model is also invariant under the relabelling of n by −n : this is called parity invariance. The Feynman rules are easily derived from the action of Eq.(4.1) :

n

n

m

↔ +

n n

n

~ µ

↔


γ δm,n+1 + δm,n−1 ~

n n

↔ −

↔ +

λ4 ~

Jn ~

Feynman rules, version 4.1 1

(4.2)

If not indicated explicitly otherwise, sums will run from −∞ to +∞. Both µ and γ are independent of n simply because I choose them so. I might choose differently, only I don’t want to. 3 This will lead to momentum conservation later on. Note however that, as indicated above, momentum conservation is a consequence of our choice, or in practice of our belief in the translation invariance of our physical laws. Other models are possible and not a priori wrong : they are simply much more complicated. 2

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The identity of the field is indicated by its label. Alternatively, the fourvertex and the source vertex may be labelled. The SDe now takes the following form, for any n : n

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

n

=

+ 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

n

+

n

0000 1111 1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

n+1

n

+

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111

n

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

n−1

+

n

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

,

(4.3)

or, in terms of the field functions φn ({J}), that depend on all sources : φn =

4.2.2

Jn γ + (φn−1 + φn+1 ) µ µ   λ4 ∂ ∂2 2 3 − φn . φn + ~ φn + 3~φn 6µ ∂Jn (∂Jn )2

(4.4)

Introducing the propagator

The Schwinger-Dyson equation (4.3) can be cast in another, more useful form. Consider the fact that, upon entering the field function via its external leg, one must encounter either zero or more two-point functions before encountering a source vertex or a four-vertex. Let us denote by Πm,n ≡

m

n

(4.5)

the total set of diagrams that contain only two-point vertices (or no vertices), and have fields n and m at its external legs4 . The SDe can then be rewritten as follows :  000000 111111 000000 111111 000000 111111 X 000000 111111 n 000000 111111 k k n 000000 111111 000000 111111 = + 000000 111111  000000 111111 000000 111111 000000 111111 000000 111111

k

n

 +

4

n

k

+

n

k

   , 

(4.6)

To go from ϕn to ϕm one needs, of course, at least |n − m| vertices, but more vertices are also possible.

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where we must sum over the label of the field exiting the Π. Therefore, we have X φn = Πn,k ×   k λ ∂ ∂2 Jk 3 2 φk . − φk + 3~φk φk + ~ µ 6µ ∂Jk (∂Jk )2

(4.7)

The object Πn,k describes to what extent the fields ϕn and ϕk influence one another : it will be called the propagator from now on.

4.2.3

Computing the propagator

From the translation and parity invariance of the model we have discussed, we can infer that Πn,k can actually only depend on |n − k|, so that we can restrict ourselves to Π0,n ; we denote this by Π(n). For Π(n), we have a very simple Schwinger-Dyson equation : n

0

=

0

n

+

n 0

or ~ γ Π(n) = δ0,n + µ µ

1

+

n 0

−1

  Π(n + 1) + Π(n − 1) .

,

(4.8)

(4.9)

The easiest way to solve this set of equations is by Fourier transform. We define5 X Π(n) e−inz , (4.10) R(z) = n 6

from which the propagator may be recovered using 1 Π(n) = 2π

Z+π

e+inz R(z) dz .

(4.11)

−π 5

We are forced to use exp(−inz), with absolute value unity ; any other absolute value would make the generating function divergent either as n → ∞ or n → −∞. 6 We choose e−inz rather than e+inz in Eq.(4.10) by convention. Although this may not be completely, glaringly obvious at this point, this convention is ultimately related to the fact that, in nonrelativistic quantum mechanics, the Schr¨odinger equation has been ˆ ˆ chosen to read i~∂|ψi/∂t = H|ψi rather than −i~∂|ψi/∂t = H|ψi.

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Multiplying both sides of Eq.(4.9) by exp(−inz) and summing over n leads to
~ γ + R(z) eiz + e−iz µ µ ~ ~u = = iz −iz µ − γ (e + e ) µu − γ(u2 + 1)

R(z) =

(4.12)

where we have introduced u = eiz . This allows us to write the integral (4.11) as I un ~ du , (4.13) Πn = − 2iπγ (u − u+ )(u − u− ) |u|=1

where u± are the two roots of the quadratic form µu − γ(u2 + 1) : u± =

 p 1  µ ± µ2 − 4γ 2 . 2γ

(4.14)

Provided that µ exceeds 2γ, the two poles of the integrand are real, and 0 < u− < 1 < u+ . We can then contract the contour around the point u = u− , upon which we find Π(n) = ~

u− n , γ(u+ − u− )

n≥0 .

(4.15)

The general solution for the propagator is therefore7 Π(n) = p

~ µ2

−

4γ 2

u− |n| .

(4.16)

Unsurprisingly, the propagator falls off exponentially with |n|. Note that if E27 γ were negative, then u− would also be negative, and the propagator would oscillate between positive and negative correlations. Moreover if µ were 2γ or smaller the poles of the integrand would lie on the unit circle |u| = 1, making the integral ill-defined. With this explicit form of the propagator, we can now switch to a new set of Feynman rules : 7

This derivation is valid for n ≥ 0. For negative n, Cauchy’s theorem on which it is based does not hold immediately : but in that case we can perform the variable transform from u to 1/u and obtain the result.

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August 31, 2019

m

n

↔ Π(m − n) n

n n

n

n

↔ −

↔ +

λ4 ~

Jn ~

Feynman rules, version 4.2

(4.17)

The difference with the previous set of rules is that now the line denotes a propagator running between n and m. The SDe is now very similar to that of the zero-dimensional ϕ4 theory :  n

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

X  =   m

n

m

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 000 111 0000 1111 000 111 000 111 000 111 000 111 000 111 0000 1111 000 111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

m

n

+

 +

4.2.4

n

m

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111

+

n

m

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

   . 

(4.18)

Figments of the imagination : a sermon

The concept of an infinite number of fields all huddling together at a single point simply cries out for a better visualization. The most useful picture is that of each field occupying its own point. Indicating by a line those fields that have a direct coupling, we arrive at a picture like

ϕ ϕ ϕ ϕ −1 0 1 2

ϕ3

where we now need to introduce a new notion, that of distance. In our sensorial experience, distances are, in their essence, measured by the sending and receiving of signals, and the weaker the signal from one point to another,

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the further those points are deemed to be apart ; in the language of these notes, the smaller Π(m − n), the larger the ‘distance’ between n and m. We can therefore dress up our picture by introducing a fundamental distance ∆, subsequent field locations being separated by this distance,

∆

∆

∆

∆

∆

} } } } } }

∆

ϕ ϕ ϕ ϕ ϕ3 −1 0 1 2 where the distances between the successive points are all equal since the couplings γ are all equal. We have, as it were, constructed a one-dimensional universe. It may come as a surprise that the concept of space is here presented as a visualization device. If we reflect, however, on how someone who (like a new-born infant) has no a priori concept of spacelike separations would have to envisage the workings of the physical world, we shall conclude that that person had better invent space in order not to go insane pretty quickly. In its essence, space, like so much else in the world around us, is simply a mental construction that allows us to come to grips with, and control, our environment8 . After all this has been said, we must acknowledge the empirical fact that to our knowledge space seems not to be made up from single points9 . Therefore we have to assume that ∆ must be much smaller than the smallest distances that can, at present, be resolved10 . We therefore introduce the continuum limit : we assume that the theories we consider are such that the limit ∆ → 0 can be taken in a sensible manner, yielding sensible results. This sidesteps the interesting question of whether ∆ is really zero or not. Indeed, we do not know. Any theoretical result that depends sensitively on whether ∆ = 0 or ∆ 6= 0 would be extremely important since experimental information about it would allow us a look at the fundamental structure of space ; but for us it is safer to construct theories the predictions of which do not hinge on this unknown. As we shall see, this can be made to work ; 8

See also Peter L. Berger and Thomas Luckmann, The Social Construction of Reality : A Treatise in the Sociology of Knowledge (Garden City, New York: Anchor Books, 1966). Additionally, the thought of all of us sharing a single point with every particle of our bodies clashes with my sense of personal space to say the least. 9 Nor does it appear to be one-dimensional – but that is easily repaired, as we shall see. 10 About 10−19 meter.

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as an added bonus, we can feel free from misgivings about the mathematical rigour of taking the continuum limit — we may not be at the limit anyway.

4.3

One-dimensional continuum theory

4.3.1

The continuum limit for the propagator

Having identified the positions occupied by the various fields with points in space (or time), we define the distance between points m and n by x = (n − m)∆ .

(4.19)

The dimension of x is that of ∆, that is, a length L. The continuum limit is, then, that where ∆ → 0 and |n − m| → ∞ while x remains fixed. The propagator is now a function of x, so we redefine it as Π(x) ← Π(x/∆) . This means that ~ Π(x) = 2π

Z+π exp(ixz/∆) dz , µ − 2γ cos(z)

(4.20)

−π

where µ > 2γ as before. A corresponding change in the integration variable z is now in order : we write z = k∆ , (4.21) The dimension of k is therefore L−1 . The propagator becomes ~∆ Π(x) = 2π

+π/∆ Z

dk

exp(ixk) µ − 2γ cos(k∆)

−π/∆

~∆ ≈ 2π

Z dk

exp(ixk) . (µ − 2γ) + γ∆2 k 2

(4.22)

In the last line, we have taken ∆ to be very small indeed. Note that the approximation cos(z) ≈ 1 − k 2 ∆2 /2 is, of course only justified as long as k is finite ; but for very large k the integrand is extremely oscillatory and contributes essentially nothing11 . Now, in order to avoid a propagator that 11

This handwaving argument is justified by the fact that we get the right propagator in the continuum limit.

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103

either blows up or vanishes, we must define the ∆-dependence of µ and γ such that γ ∼ 1/∆ and µ − 2γ ∼ ∆. We shall take γ→

m2 ∆ 1 − , ∆ 4

µ→

2 m2 ∆ + , ∆ 2

(4.23)

with m2 a positive number (remember that we need µ > 2γ). We shall also take m itself to be positive. We then find the exact results p 1 − m∆/2 µ − 2γ = m2 ∆ , µ2 − 4γ 2 = 2m , u− = . (4.24) 1 + m∆/2 The propagator takes the form12 Z eixk ~ ~ dk 2 = exp(−m|x|) . Π(x) = 2π k + m2 2m

(4.25)

To check that this result is indeed the correct one, we can consider the continuum limit directly for the propagator result (4.16) :  |x/∆| 1 − m∆/2 ~ ~ exp(−m|x|) , (4.26) → Π(n) → 2m 1 + m∆/2 2m as desired.

4.3.2

The continuum limit for the action

In the action (4.1), we shall want to replace the sum over n by an integral over x : Z X ∆ → dx . n

It is therefore necessary that every term in the action acquires a factor ∆. Now, the action depends on the quantum fields ϕn . As we let the distance between the points shrink to zero, the collection of values {ϕ} turns into a function ϕ(x). The precise correspondence between {ϕ} and ϕ(x) is something that, in the end, we have to decide for ourselves. Out of the several possibilities we shall adopt the following :     1 1 0 ϕ(x) = ϕn+1 + ϕn , ϕ (x) = ϕn+1 − ϕn . (4.27) 2 ∆ 12

To obtain this result we can close the integral in the complex-k plane, circling around either the pole at k = im or that at k = −im.

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This assignment is called the Weyl ordering. Its converse reads, of course, ϕn = ϕ(x) −

∆ 0 ϕ (x) , 2

ϕn+1 = ϕ(x) +

∆ 0 ϕ (x) . 2

(4.28)

In a sense, the field value ϕ(x) is sitting ‘in between’ the points ϕn and ϕn+1 . Other assignments can be proposed, for instance ϕn = ϕ(x). However, these are less attractive13 . Upon careful application of Weyl ordering and the assumed continuum limits for µ and γ, the kinetic part of the action (4.1) has the following continuum limit :  i X 1 X hµ ∆2 2 2 0 2 ϕn − γϕn ϕn+1 = (µ − 2γ)ϕ(x) + (µ + 2γ)ϕ (x) = 2 2 8 n n   Z  X 1 1 0 2 1 0 2 1 2 2 2 2 m ϕ(x) + ϕ (x) ∆ = m ϕ(x) + ϕ (x) dx . (4.29) 2 2 2 2 n The interaction and source terms in the path integral do not have a factor ∆ coming out naturally, but we may simply define the continuum limits by redefining the objects in the action : λ4 → ∆λ4 ,

Jn → ∆J(x) ,

(4.30)

so that the continuum limit of the full action, including this time also the sources, becomes14  Z  1 0 2 λ4 1 2 2 4 m ϕ(x) + ϕ (x) + ϕ(x) − J(x)ϕ(x) dx . (4.31) S[ϕ, J] = 2 2 4! Note the notation with square brackets: the action is now no longer a number depending on (a countably infinite set of) numbers, but rather on the functions ϕ(x) and J(x) ; this is called a functional. 13

For example, consider a function ϕ(x) that vanishes for x → ±∞. The integral 2ϕ(x)ϕ0 (x) dx then vanishes upon partial integration. Weyl ordering tells us that 2ϕ(x)ϕ0 (x) = (ϕn+1 2 − ϕn 2 )/∆, leading to the correspondence Z X 2ϕ(x)ϕ0 (x) dx ↔ ∆ (ϕn+1 2 − ϕn 2 ) , R

n

where the sum also vanishes explicitly after relabelling. For the alternative assignment ϕn = ϕ(x) the vanishing cannot be proven. 14 Strictly speaking, the Weyl ordering requires the replacement of Jn not by ∆J(x) but by ∆J(x) + ∆2 J 0 (x)/2. The additional term, however, vanishes in the continuum limit as ∆ → 0, as do the higher powers of ∆ involved in the ϕn 4 term.

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4.3.3

105

The continuum limit of the classical equation

For the discrete action, there is an obvious classical equation : ∂ S({ϕ}) = 0 ∂ϕn

for every n ,

(4.32)

where, again, the source terms have been subsumed into the action. For the ϕ4 model of Eq.(4.1), the classical equation is therefore ∆λ4 3 ϕ = ∆Jn (4.33) 3! n for all n, and the extra factor ∆ in the coupling constant and the sources have been taken into account. The Weyl prescription leads us to write µϕn − γ(ϕn+1 + ϕn−1 ) +

µϕn − γ(ϕn+1 + ϕn−1 ) ≈ m2 ∆ϕ(x) − ∆ϕ00 (x) ,

(4.34)

so that the continuum limit of the classical field equation takes the form λ4 ϕ(x)3 = J(x) . (4.35) 3! This is precisely the Euler-Lagrange equation, that may also be obtained immediately from the continuum form of the action by taking functional derivatives ; we shall not really have a use for them in these notes, but for good measure they are discussed in appendix 15.7. m2 ϕ(x) − ϕ00 (x) +

4.3.4

The continuum Feynman rules and SDe

Let us have a look again at the SDe for the discrete model, for simplicity taking the ϕ4 model again : X Π(n − m) φn = m



λ × Jm − 6



φ3m

∂ ∂2 + 3~φm φm φm + ~2 ∂Jm (∂Jm )2

 . (4.36)

Going over to the continuum limit entails, as we have seen, the following substitutions : φn , φm → φ(x), φ(y) , Jm → ∆J(y) , Π(n − m) → Π(x − y) , Z X 1 ∂ δ λ4 → ∆ λ4 , → dy , → . (4.37) ∆ ∂Jm δJ(y) m

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With this, the SDe becomes   Z λ4 φ(x) = dy Π(x − y) J(y) − φ(y)3 6  δ δ2 2 +3~φ(y) φ(y) + ~ φ(y) . δJ(y) (δJ(y))2

(4.38)

This version of the SDe is not meant to be solved, rather it tells us the Feynman rules in the continuum limit :

x

y

↔

Π(x − y)

↔

−

↔

x

Z dx

λ4 ~ J(x) + ~ for a vertex at position x

Feynman rules, version 4.3

(4.39)

This comes with the understanding that the positions of all vertices are to be integrated over, and that the field function φ is now a functional of the source J. For a free theory there are no interactions, and we find Z φ(x) = dy Π(x − y) J(y) . (4.40) We see that the free field is the sum of its responses to the source, weighted by the correlation between the position where the field is measured and that of the strength of the source at all points. It is this property that establishes the propagator as the ‘differential-equation’ Green’s function; but this correspondence is only valid for non-interacting theories (see also footnote 3 in Chapter 1).

4.3.5

Field configurations in one dimension

Before entering spaces of more dimensions, we may have a look at the field variables. The zero-dimensional variable ϕ, with its integration element, is

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107

in the discrete one-dimensional formulation replaced by the whole set ϕ, for which the path integration element reads, of course, Dϕ =

Y

dϕn

n

The continuum limit of this object is defined to be the continuum-formulation path integration element, however badly defined this may be. The assigning a functional value S[ϕ] to a given field ϕ(x) is not problematic ; rather it is the prescription of how all field configurations are to be summed over that makes it so hard to define path integrals rigorously15 . It is instructive to consider the nature of the dominant contributions. Consider the part of the path integrand that governs the point-to-point variation of the paths: it is   1 2 (ϕn+1 − ϕn ) . K∆ (ϕn , ϕ+1 ) ≡ exp − 2~∆ It is clear that the majority of values (ϕn+1 − ϕn )2 will be of order O (~∆), as usual for Gaussian distributions. This means that ϕn+1 and ϕn must approach each other as ∆ → 0, so the contributing fields are continuous. On the other hand, the approach is not too fast, since by ϕn+1 − ϕn ≈ ∆ϕ0 (x) we see that the derivative ϕ0 (x) diverges as ∆−1/2 , hence the contributing functions are nowhere differentiable. This is not to say that differentiable fields are not allowed : rather, the nondifferentiable ones are the overwhelming majority. Two conclusions follow. In the first place, the use of continuum-formulation objects like ϕ0 (x) or ϕ00 (x) in the action are to be treated as highly symbolic, almost purely mnemonic, concepts. In the second place, the classical solution, which is typically almost everywhere differentiable, is itself not the dominant contribution to the path integral ; rather, it is the bundle of fields close to the classical one that constitutes the lowest-order approximation to the behaviour of the theory. To gain some insight in the structure of a typical path (field configuration), let us consider the interrelation of three consecutive fields : it is given by K∆ (ϕ0 , ϕ1 )K∆ (ϕ1 , ϕ2 ). For simplicity, we neglect the rest of the action. The positions of these three fields are separated by ∆. The ‘typical’ jumps 15

In fact, the mathematical definition of continuum path integrals relies on the discrete formulation !

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Figure 4.1: Zooming out from a fractal path

√ in field values are of order ∆, as mentioned above. Now imagine ‘zooming out’, that is, disregarding the value of ϕ1 , and inspecting only ϕ0 and ϕ2 , which are now separated by 2∆. This is obtained by integrating over R ϕ1 : dϕ1 K∆ (ϕ0 , ϕ1 )K∆ (ϕ1 , ϕ2 ) = K2∆ (ϕ0 , ϕ2 ) , where the proportionality constant is absorbed in the normalization of the path integral. The typical √ jump from ϕ0 to ϕ2 is now of order 2∆. We conclude that, if we resolve the continuum path down to√a scale ∆, the typical fluctuations over this scale will always be of order ∆. The typical path has a fractal structure. Such behaviour, with zigs and zags at every length scale, is encountered in Brownian motion – and in the behaviour of the stock market16 . In figure 4.1 we plot a typical fractal path running over 10,000 points separated by a distance of 0.01, with ∆ = 1. The first plot shows all points ; in the second, only every 10th point is used, and in the third plot only every 100th point is used. The qualitative form of the three paths remains the same, as expected for a fractal path. The average absolute value of the point-to-point jumps are 0.80, 2.49, and 6.97, √ respectively : the ratios between these numbers are indeed roughly equal to 10.

16

Note that this qualitative picture holds only for one-dimensional theories (and, luckily, the price of stocks, bonds, futures etc is expressed in one-dimensional currency). In more dimensions, the paths’ behaviour is even more wild.

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4.4 4.4.1

109

The momentum representation Fourier transforming the SDe

We are now ready to make an important technical change. So far, we have considered the fields and their expectation values as functions of position. It will turn out to more practical to consider them as functions of momentum or, in the one-dimensional case, of wave number 17 . There are a number of good reasons for doing so. In the first place, in the free theory the various momentum modes are independent of one another, in contrast to the fields at different space points18 : propagators are simpler in momentum language than in position language. In the second place, there is a law of momentum conservation operative in the universe, and not a law of conservation of position. In the third place, momenta are more directly the physical characteristics that are controlled and measured in actual particle physics E28 experiments. The transition from position to momentum is nothing but applying Fourier transforms : we already had, from Eq.(4.25), Z Π(x − y) =

dk ~ exp(ik(x − y)) , 2 2π k + m2

and we now introduce19 Z Z dk dk φ(x) = φ(k) exp(ikx) , J(x) = J(k) exp(ikx) . 2π 2π

(4.41)

(4.42)

We now have to figure out what the correct Feynman rules are in this new language. To do so we use (what else ? ) the SDe. It suffices to restrict ourselves to ϕ3 theory at the tree level, where it reads   Z λ3 1 2 J(y) − φ(y) . (4.43) φ(x) = dy Π(x − y) ~ 2~ 17

Recall the discussion on loose terminology in Chapter 0. Indeed, the more-dimensional theories have been constructed expressly to make fields at different points correlate to one another! 19 We use the same notation for the position-dependent quantities and their momentumdependent Fourier transforms. This will not lead to confusion since we shall soon drop the position-dependent ones anyway. 18

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Inserting the Fourier representation leads to Z Z Z dk dk ik(x−y) ~ ikx φ(k) e = dy e × 2 2π 2π k + m2  Z Z dk1 dk2 dk3 J(k3 ) ik3 y λ3 i(k1 +k2 )y . e − φ(k1 ) φ(k2 ) e 2π ~ 2~ 2π 2π We find the following form of the SDe in momentum language :  ~ J(k) φ(k) = 2 − 2 k +m ~  Z λ3 dk1 dk2 φ(k1 ) φ(k2 ) (2π) δ(k − k1 − k2 ) . 2~ 2π 2π

4.5 4.5.1

(4.44)

(4.45)

Doing it in momentum space The Feynman rules

On the basis of the above we can now formulate the Feynman rules for our theory in momentum space (for the example of ϕ4 theory) :

k k1

k2

k4

↔

~ k 2 + m2

↔

X λ − (2π) δ kj ~ j

↔

1 + J(q) (2π) δ (q + k) ~

k3 k

q Z∞

dk 2π

!

for every momentum k

−∞

Feynman rules, version 4.4

(4.46)

Where before we had to integrate over the position of every vertex, we now have to integrate over every momentum. It is of course possible (and this is in fact the most common situation) that the source contains only a

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111

single momentum mode. In that case the external legs in a diagram carry a single momentum ; but all momenta of the internal lines have to be integrated over. Also note that the vertices now carry Dirac deltas imposing momentum conservation. This is a direct consequence of our choosing the vertices of the theory to be position-independent20 . In addition, it has becomes necessary to indicate how the momenta involved in the vertices are to be counted. It is usual to count all the momenta either incoming or outgoing. The precise convention is unimportant, but it is important that you use it consistently.

4.5.2

Some example diagrams

Here we present some diagrams, evaluated according to the rules we have formulated so far. The first one,

q

k (4.47)

with a source on only one endpoint, evaluates to Z dk 1 ~ 1 J(q) (2π) δ(q − k) 2 = J(q) 2 . 2 2π ~ k +m q + m2

(4.48)

When we add another source,

q1

q2 (4.49)

this gives us 1 1 J(q1 ) 2 J(q2 ) (2π) δ(q1 + q2 ) . (4.50) ~ q1 + m2 we see here an important fact : every connected diagram contains one Dirac delta informing us that overall momentum must be conserved. The diagram q1 q2 20

q3

(4.51)

This means that the homogeneity of space(-time) can be investigated by very carefully checking momentum(-energy) conservation in interactions. Of course, if vertices take on different values very far away in space or time these effects may be undetectable.

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August 31, 2019

bears this out : it reads Z dk1 dk2 dk3 J(q1 ) J(q2 ) J(q3 ) 2π 2π 2π ~ ~ ~ ~ ~ ~ × 2 2 2 2 2 k1 + m k2 + m k3 + m2 ×(2π)δ(q1 − k1 )(2π)δ(q2 − k2 )(2π)δ(q3 − k3 ) −λ3 (2π)δ(k1 + k2 + k3 ) × ~ 1 1 1 = J(q1 )J(q2 )J(q3 ) 2 q1 + m2 q2 2 + m2 q3 2 + m2 −λ3 (2π)δ(q1 + q2 + q3 ) . × ~

(4.52)

Next, we consider the one-loop diagram k2 q1

q2

k1 k4 k3

(4.53)

for which we have to write down (including the symmetry factor ! ) Z dk1 dk2 dk3 dk4 J(q1 ) J(q2 ) 2π 2π 2π 2π ~ ~ ~ ~ ~ ~ 1 × 2 × 2 2 2 k1 + m2 k2 + m2 k3 + m2 k4 + m2 2 ×(2π) δ(q1 − k1 ) (2π) δ(q2 − k4 )     −λ3 −λ3 × (2π) δ(k1 − k2 − k3 ) (2π) δ(k2 + k3 − k4 ) ~ ~ 1 1 λ3 2 = J(q1 )J(q2 ) 2 (2π) δ(q1 − q2 ) 2 2 2 q1 + m q2 + m2 Z dk2 1 1 (4.54) × 2 2 2π k2 + m (q1 − k2 )2 + m2

E29

In addition to the overall momentum conservation delta, we see here the other significant fact : every closed loop involves a momentum that is not fixed by conservation, and that demands an integral over that momentum.

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4.6 4.6.1

113

More-dimensional theories The more-dimensional continuum

Our choosing a labelling of fields with a single integer index is, of course, arbitrary. We can consider an alternative in which the fields are labelled by D integer indices : ϕn → ϕ~n , ~n = (n1 , n2 , . . . , nD ) . In the interest of brevity, we introduce the notation ~n ± k = (n1 , . . . , nk−1 , nk ± 1, nk+1 , . . . , nD ) . An appropriate action for this choice would be # " D X X 1 λ 4 ϕ~n ϕ~n+k + ϕ~n 4 − J~n ϕ~n . µϕ~n 2 − γ S({f }) = 2 4! k=1

(4.55)

(4.56)

~ n

The obvious visualization for this choice is that of a space rather than a line, covered with a regular square grid of fields, each connected to 2D nearest neighbors: the corresponding continuum picture, therefore, is that of a theory in D equivalent dimensions. The propagator of this theory obeys, of course, the SDe  D D  ~Y γX Π(~n) = δn ,0 + Π(~n + k) + Π(~n − k) , µ k=1 k µ k=1

(4.57)

with the solution ~ Π(~n) = (2π)D

Z+π −π

dD z

exp(i(n1 z1 + · · · + nD zD )) , µ − 2γ cos(z1 ) · · · − 2γ cos(zD )

(4.58)

so that, now, µ must exceed 2Dγ. The continuum limit takes a different form than in the one-dimensional case. We define ~x = (x1 , x2 , . . . , xD ) , xj = nj ∆ , ~k = (k 1 , k 2 , . . . , k D ) , k j = zj /∆ ,

(4.59)

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The simplest nontrivial choice is then to approach the continuum as follows : γ → ∆D−2 , µ → 2Dγ + m2 ∆D , λ4 → ∆D λ4 , ϕn1 ,n2 ,...,nD → ϕ(~x) , Jn1 ,n2 ,...,nD → ∆D J(~x) .

(4.60)

The propagator21 takes the continuum form

Π(~x) =

~ (2π)D

Z

dD k

  exp i~x · ~k ~k · ~k + m2

.

(4.61)

The continuum form of the action is  Z  1 2 1 ~ λ4 2 2 4 S[ϕ, J] = m ϕ(~x) + (∇ϕ(~x)) + ϕ(~x) − J(~x)ϕ(~x) dD x , 2 2 4! (4.62) The Feynman rules are seen to be

x

↔

Π(~x − ~y )

↔

−

y

↔

x

λ4 ~ J(~x) + ~

Feynman rules, version 4.5

(4.63)

and also the SDe is a straightforward generalization of the one-dimensional case :  Z D φ(~x) = d y Π(~x − ~y ) × J(~y )   λ4 δ δ2 3 2 − φ(~y ) + 3~φ(~y ) φ(~y ) + ~ φ(~y ) . (4.64) 6 δJ(~y ) (δJ(~y ))2 21

The propagator is still a smooth function ; but, as you can easily check by comparing to the discussion in section 4.3.5, for D = 2 the field function ϕ is no longer continuous. For D ≥ 3, the discontinuities are not even guaranteed to be finite, and only the interaction terms can keep the field from jumping infinitely wildly !

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The classical field equation, ~ 2 ϕ(~x) + λ4 ϕ(~x)3 = J(~x) , m2 ϕ(~x) − ∇ 3!

(4.65)

can be obtained directly from the continuum action by the functional EulerLagrange equation

δ ~ · S[ϕ, J] − ∇ δϕ(~x)

!

δ ~ x) δ ∇ϕ(~

S[ϕ, J]

=0 .

(4.66)

The propagator only depends on |~x| and is therefore rotationally invariant : this is a larger symmetry22 than that of the original lattice, that only allows rotations over multiples of π/2. The mutual influence of things happening at a point ~x and at a nearby point ~x + d~x is governed by ds(d~x), the real distance between points, on which the physics depends. In this case,

2

2

ds(d~x) = |d~x| =

D X

(dxj )2 ,

(4.67)

j=1

the Euclidean distance between the points ; this type of quantum field theory is therefore said to be Euclidean. As before, it will turn out to be more useful to go over to a momentum formulation of the theory. This is performed by a completely straightforward generalization of what we did in the one-dimensional case, and we can give the Feynman rules (in D dimensions) without more ado :

22

The increase in symmetry depends on an interplay between the lattice action and the form of the continuum limit ; it is possible to construct actions in which the continuum symmetry is not larger than that of the lattice theory. But I prefer to choose actions that (will come to) look like the physics of the real world.s

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k k1

k2

k4

↔

~ 2 ~ |k| + m2

↔

λ − (2π)D δ ~

k3 k

q

~kj

j

  1 D ~ + J(q) (2π) δ ~q + k ~

↔ Z∞

! X

dD k (2π)D

for every momentum k

−∞

Feynman rules, version 4.6

4.6.2

(4.68)

The propagator, explicitly

It is possible to express the Euclidean propagator Π(~x) in terms of known functions, using a Gaussian representation : ~ Π(~x) = (2π)D

Z∞

Z dt

  dD k exp i~x · ~k − t~k · ~k − tm2

0

~ = (2π)D ~ = (2π)D

Z∞

−m2 t

dt e 0 Z∞

dk j exp −z(k j )2 + ik j xj




j=1 −m2 t

dt e 0

D Z Y

  D   Y (xj )2 π 1/2 exp − t 4t j=1

  Z∞ ~ |~x|2 −D/2 2 = dt t exp −m t − (4π)D/2 4t 0  1−D/2 ~ 2π|~x| = K1−D/2 (m|~x|) . 2π m

(4.69)

The function K is the so-called modified Bessel function of the second kind, discussed in appendix 15.15.8 ; here we have used its integral representation

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117

given by Eq.(15.296). For large m|~x|, the propagator decreases exponentially, while for small m|~x|, we have
~ log m|~x| , D = 2 , 2π ~Γ (−1 + D/2) 2−D Π(~x) ≈ |~x| , D≥3 . 4 π D/2

Π(~x) ≈ −

In every dimension, the propagator is normalized in the same way :   Z Z Z ~k · ~x exp i ~ Π(~x) dD x = dD x dD k D (2π) |~k|2 + m2 Z D D ~ ~ ~ D (2π) δ (k) = = . d k D 2 2 (2π) m2 |~k| + m

4.7

(4.70)

(4.71)

Exercises

Excercise 27 Guessing the correlator For the ‘one-dimensional discrete model’ we found for the correlator the recursion relation   γ ~ Πn+1 + Πn−1 Πn = δ0,n + µ µ Make the following Ansatz: Πn = A B |n| and compute A and B. Compare to the correct answers. Excercise 28 Fourier transformation of the action and the rules Consider the one-dimensional continuum action   Z 1 0 2 m2 λ 2 4 S = dx ϕ (x) + ϕ(x) + ϕ(x) − ϕ(x)J(x) 2 2 4! Here, ϕ(x) and J(x) are real fields. 1. We define the Fourier transformations by Eq.(4.42). how that ϕ(k) = ϕ(−k), and J(k) = J(−k).

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2. Compute the transformed form for the action. 3. Determine the Feynman rules in the transformed formulation. Excercise 29 Being clever in one dimension In one dimension, loop diagrams can be conveniently computed using Cauchy integration. We shall do this in the momentum representation for amputated diagrams, that is, the Feynman factors for external lineas are not included. 1. Consider the tadpole diagram in ϕ3/4 theory :

k where the loop momentum is indicated. (a) Show that the external line carries no momentum. (b) Write down the expression for the tadpole diagram. By direct integration over k from −∞ to +∞, show that the result is −λ3 π/(2m). (c) Redo the above calculation in another way : the integrand has poles at k = ±im. Close the integration contour in the complex k plane and contract it around a pole. Show that it does not matter which pole you choose. 2. Consider the following diagram : p−k p

k

where the external momentum and the loop momenta are indicated. Show, using the same contour technique as above, that it evaluates to 1/(2m)/(p2 + 4m2 ). Show that again the choice of the upper or the lower half-plane is irrelevant. 3. Apply the same technique twice to show that =

1 (8m2 )(p2

+ 9m2 )

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119

4. Show that p2 p1

p3

=

p1 2 + p2 2 + p3 2 + 24m2 (p1 2 + 4m2 )(p2 2 + 4m2 )(p3 2 + 4m2 )

where all external momenta are counted ingoing, so that p1 +p2 +p3 = 0.

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Chapter 5 QFT in Minkowski space 5.1

Moving to Minkowski space : making time

Since the known space in which particle physics takes place is not of a Euclidean but rather of a Minkowskian nature1 , it behooves us to make the transition to this new type of space. Essentially, this involves singling out one of the coordinate for a special rˆole.

5.1.1

Distance in Minkowski space

Whereas the real distance, that is that notion distance measure that actually governs the relative influence of fields at different points, is in Eulidean space given by the Euclidean square distance of Eq.(4.67), we know that in the spacetime in which we actually live and do physics, the real distance is quite different. In particular, one of the coordinate directions represents time. That is, events in spacetime taking place at position ~x = (x1 , x2 , x3 ) and time t relative to some freely chosen origin are denoted by four coordinates: xµ = (x0 , x1 , x2 , x3 ) ,

x0 = ct ,

(5.1)

where c is the universal constant providing the exchange rate between units of distance and units of time2 ; it is the necessary velocity of massless particles3 , 1

We shall not involve ourselves in the complications that arise in curved space ; a consistent theory of quantum gravity is not — at present — relevant to particle phenomenology. 2 See section 0.2.1. 3 As soon as we have defined what we mean by a particle. It is customary to specify ‘the speed in vacuo’ but since particles inside a medium with which they interact are no

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August 31, 2019

and the real distance between two events with coordinates xµ and xµ + dxµ is given by 2

0 2

s(dx) = (dx ) −

3 X

(dxj )2 ≡ gµν dxµ dxν ≡ dxµ dxµ ≡ dx · dx ,

(5.2)

j=1

(summation over repeated indices implied), where gµν is the covariant metric tensor4   1 if µ = µ = 0 -1 if µ = ν ∈ {1, 2, 3} gµν = diag(1, −1, −1, −1) ≡ (5.3)  0 otherwise We also have the contravariant metric tensor g µν , defined by g µα gαν = δ µ ν ,

(5.4)

so that g µν is numerically equal5 to gµν . The metric tensors allow for the raising or lowering of indices : for instance, xµ = gµν xν

: x0 = x0 , xj = −xj (j = 1, 2, 3) .

(5.5)

The special rˆole of time in physics is evidenced by the relative minus sign in the metric tensor.

5.1.2

Farewell probability, hello SDe

Up to now, the action of our theory was real-valued, and the path integrand a real probability density. In de derivation of the SDe, however, and the consequent use of our Feynman diagrams, we have not used that fact anywhere ; the only requirement for the validity of the SDe is that the path integrand go to zero sufficiently fast at the endpoints. As long as this is guaranteed we may generalize the parameters of the action (including the sources) to complex values : indeed we shall take them to be purely imaginary, with one exception. longer massless, this may not be necessary. 4 See section 0.2.4. 5 By coincidence. Even in the flat Minkowski space, another set of coordinates (spherical ones, for instance) would lead to a g µν quite different from gµν . However, we shall always use the sensible (pseudo)Cartesian coordinates in these lectures.

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123

We can repeat the treatment of chapter 4 in this different setting. In what follows we shall use the labelling ~n = (n0 , n1 , . . . , nD−1 ), and use the notation of Eq.(4.55). For the action we choose " # D−1 X 1 X λ 4 S({ϕ}) = µϕ~n 2 − γ0 ϕ~n ϕ~n+0 − γ ϕ~n ϕ~n+k + ϕ~n 4 − J~n ϕ~n , 2 4! k=1 ~ n (5.6) with a disctinction between the ‘time’ component and the ‘space’ conponents. We ensure the validity of the Schwinger-Dyson equations by letting µ have a real part. This can be arbitrarily small, as long as it is positive. The SDe for the propagator now reads   D−1 γ0 ~ Y δn ,0 + Π(~n + 0) + Π(~n − 0) Π(~n) = µ k=0 k µ  D−1  γX + Π(~n + k) + Π(~n − k) . (5.7) µ =1 For the generating function we choose X
R(~z) = Π(~n) exp in0 z 0 − i(n1 z 1 + · · · nD−1 z D−1 ) .

(5.8)

~ n

Note the special treatment of the zeroeth component, which we again put in by hand 6 We can now solve Eq.(5.7) Z
1 D 0 0 1 1 D−1 D−1 d z exp −in z + i(n z + · · · + n z Π(~n) = R(z) , (2π)D !−1 D−1 X R(z) = i~ µ − 2γ0 cos(z 0 ) − 2γ cos(z k ) . (5.9) k=1

We now proceed to the continuum limit ; we choose, as ∆ → 0, ~n = 6

1 µ x , xµ = (x0 , x1 , . . . , xD−1 ) , ~z = ∆ k µ , k µ = (k 0 , k 1 , . . . , k D−1 ) , ∆ (5.10)

This sidesteps the question of whether we can safely move from Euclidean to Minkowski space by some kind of ‘analytic continuation’. At any rate such an analytic continuation must be carefully tailored to give the Minkowskian theory that we want ; we may as well put that effort into setting up a Minkowskian theory from the start, as we do here.

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and, in addition to ϕ~n → ϕ(x), γ = −γ0 = i∆D−2 , µ = 2γ0 + 2(D − 1)γ + im2 ∆D +  , λ4 → i∆D λ4 , J~n → i∆D J(x) . (5.11) The continuum propagator now has the form Z i~ e−ik·x D Π(x) = d k (2π)D k 2 − m2 + i

(5.12)

Here k · x = k 0 x0 − ~k · ~x and k 2 = (k 0 )2 − |~k|2 . By a slight redefinition7 , the path integral can then be written as   Z i Z[J] = N Dϕ exp S[ϕ] , (5.13) ~ E31

with the final, full Minkowski action for ϕ4 theory :   Z 1 µ m2 − i λ4 D 2 4 S[ϕ] = d x ∂ ϕ(x)∂µ ϕ(x) − ϕ(x) − ϕ(x) + J(x)ϕ(x) . 2 2 4! (5.14)

5.1.3

E32

A close look at almost nothing: i and −

In the above, the introduction of the quantity  deserves attention. We have seen that it is necessary in order to maintain the validity of the SDe’s by making the path integrand vanish at the endpoints. Another issue is the correctness of Eq.(5.9) ; in order to stay well away from singularities, we would rather have |µ| > 2(D − 1)|γ| + 2|γ0 | rather than |µ| > 2|(D − 1)γ + γ0 |. On the other hand, the very presence of  ensures that, in the integration over ~z, the singularities are bypassed and the integral is well-defined. We may also notice that (as can be seen from our treatment of the one-dimensional case in chapter 4) the essential contributions to the integral come from z ∼ u− ∼ 1. The choice of Eq.(5.11) is therefore justified. But, it may be argued, a new parameter  appearing in the theory, even if it becomes infinitesimal, must surely have a physical meaning ? Indeed it has, and we shall come back to this point later on. In the meantime, the  is understood and almost never written explicitly in the action ; indeed ‘real’ physics assumes that we take 7

We take a factor −i out of S.

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125

 → 0 in the end of any discussion anyway. As we have seen, the Minkowskian nature of the theory has been imposed, rather than derived, at two separate moments : first, the distinguishing of γ0 and γ in Eq.(5.6), and second, the definition (5.8). Of course, there is no a priori reason why the universe would have Minkowskian rather than Euclidean symmetry8 . So the difference between γ and γ0 can be understood from a phenomenological point of view. It is somewhat surprising, then, that a second ‘by hand’ intervention, having z 0 act differently from z 1,...,D−1 , is necessary to have position x and momentum k have the right Minkowskian product.

5.1.4

The need for quantum transition amplitudes

We now find ourselves in a new interpretational situation. Since the exponent in the path integrand is now no longer real but complex-valued a straightforward probabilistic picture of the path integral is no longer possible. Indeed, every path gives a contribution which is a complex phase factor, with the same absolute value, namely precisely one9 . In fact, all possible dynamics must now arise from interference effects. The leading contribution still comes from the bundle of paths around the classical solution (that is still given by the Euler-Lagrange equation), because there the phases are to first order approximation constant. Further away from the classical solution the phases of nearby path fluctuate wildly as ~ → 0 and these paths contribute very little10 . As mentioned before, the complex-valued character of the action does not prevent us from keeping the machinery of Green’s functions, connected Green’s functions and the Feynman diagrams to compute them, since we have ensured that the SDe’s still hold. But we shall have to reinterpret them. In accordance with standard quantum mechanical practice, we postulate that the (connected) Green’s functions are related to the quantum-mechanical transition amplitudes. The squared modulus of 8

Although you may wonder what the world would look like if time was indistinguishable from space. But, even in approaches to spacetime that attempt to develop a picture of quantum gravity, such as causal dynamical triangulations, a foliation that preconceives something like a special rˆ ole for time has to be put in, in fact, by hand. 9 In the limit  → 0. 10 The remarks about instantons remain valid also in Minkowski space.

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such an amplitude is the transition probability, to be used in the computation of cross sections and decay rates. The precise nature of the Green’s function-amplitude relation will be elucidated later.

5.1.5

Feynman rules for Minkowskian theories

Having deduced the propagator in four-dimensional Minkowski space, we can now formulate the provisional Feynman rules for Green’s functions with fixed external wave vectors :

k

↔

k1

k2

k4

k3 k1 k2

Z∞

dk 2π

↔ ↔

i~ k · k − m2 + i i − λ4 (2π)4 δ 4 (k1 + k2 + k3 + k4 ) ~ i + J(k2 )(2π)4 δ 4 (k1 + k2 ) ~ for every momentum k

(5.15)

−∞

At the vertices the momenta must be counted either all incoming or al outgoing. Feynman rules, version 5.1 The vertices also pick up an additional factor i, and all vectors from now on are assumed to be Minkowskian four-vectors.

5.1.6

The propagator, explicitly

We can find position-space expressions for the Feynman propagator by actually doing the Fourier transform : Z i~ e−ik·x 4 Π(x) = d k (2π)4 k 2 − m2 + i Z 0 0 ~ i~ e−ik x eik·~x 0 3 = dk d k 0 2 , (5.16) (2π)4 (k ) − ω(k)2

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127

Im k0 Re k0

Figure 5.1: Closing the contour for positive times

where q q 2 2 ~ ω(k) = |k| + m − i ≈ |~k|2 + m2 − i

(5.17)

in accordance with our idea of having  infinitesimal but positive. Note that Π(x) is Lorentz invariant : it will not change if we perform a Lorentz transformation on xµ the Minkowski vector separating the two endpoints of our propagator. This simplifies our work considerably, and we need to consider only a few cases. Timelike separation : x2 > 0 In this case we can adopt a Lorentz frame in which ~x = 0 and we denote s = x0 , while we assume s > 0 for the moment. Our integral hence becomes i~ Π(x) = (2π)4

Z

dk 0 d3 k

exp(−ik 0 s) . (k 0 − ω(k))(k 0 + ω(k))

(5.18)

The integrand has two simple poles, one at ω(k) and the other at −ω(k). For s > 0 the contour must be closed in the lower half plane in order to make the exponent exp(−ik 0 s) vanish at infinity (cf figure 5.1). We can now perform the k 0 integral : ~ Π(x) = (2π)3

Z

d3 k

1 exp(−isω(k)) . 2ω(k)

(5.19)

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The integration over the solid angle of ~k is trivial, and we are left with a single integral : ~ Π(x) = (2π)2

Z∞

k2 exp(−isω) , dk ω

ω = ω(k) .

(5.20)

0

Using the fact that k dk = ω dω we can switch to a new variable τ = ω/m : ~m2 Π(x) = (2π)2

Z∞ dτ

√

τ 2 − 1 exp(−imsτ ) = −i

~m K1 (ims) , (2π)2 s

(5.21)

1

where we have used Eq.(15.296) and partial integration. For negative s the same result is found, but with −s instead of s. So we can formulate the final form as ~m K1 (im|s|) . (5.22) Π(x) = −i (2π)2 |s| Spacelike separation : x2 < 0 In this case we can adopt a Lorentz frame in which x0 = 0, and we may take |~x| = s. Now, we can choose either contour for the k 0 integral to obtain Z exp(i~k · ~x) i~ 0 3 dk d k Π(x) = (2π)4 (k 0 )2 − ω(k)2 Z∞ Z1 Z2π k2 ~ dk = dc dϕ exp(iskc) (2π)3 2ω(k) −1

0

−i~ = (2π)2 s

Z∞

0

k ~ dk sin(ks) = − 2 ω(k) 8π s

Z∞ dk

k exp(iks) . (5.23) ω(k)

−∞

0

The integral of k over the real axis can be deformed as indicated in figure 5.2. The integrand has two cuts along the imaginary axis in conformity with the usuals choice where the cut of z 1/2 lies along the negative real axis. Since s > 0 we have to move around the cut in the upper half plane. The result is Π(x) =

~m K1 (ms) . (2π)2 s

(5.24)

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129

im

−im

Figure 5.2: Closing the contour for spacelike separation

It is gratifying that, in spite of the very different way in which they are arrived at, the Minkowskian propagator for spacelike separations and the Euclidean propagator of Eq.(4.69) (where all separations are spacelike anyway) exactly coincide ! In conclusion, we see that the propagators connecting points with a spacelike or with a timelike separation are related to one another by an analytic continuation11 . It may be instructive to have a look at limiting cases. For E33 small s we find the same result in both cases (using Eq.(15.297)) : Π(x) ∼ −

(2π)2

~ , (x · x) → 0 . (x · x)

(5.25)

For large values of ms, we can use the asymptotic expansion of Eq.(15.305) to arrive at Π(x) ∼ 11

m1/2 exp(−im|s|) for x · x = s2 > 0 , |s|3/2

Incidentally, this can easily be obscured since K1 (iz) is sometimes written as (2) (2) (π/2)H1 (z), where H1 is a so-called Hankel function. The world of Bessel functions can be bewildering...

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Figure 5.3: Bessel functions K1 for real and imaginary argument

Π(x) ∼

m1/2 exp(−m|s|) for x · x = −s2 < 0 . |s|3/2

(5.26)

In figure 5.3 we give the two functions |K1 (ix)| and K1 (x) for real x > 0. For imaginary argument, (i.e. in the timelike region) it shows a power-damped oscillatory behaviour while for real argument (i.e. the spacelike region) it falls off exponentially ; similar to the behaviour of, say, exp(x). This practically prohibits any faster-than-light signalling by particles obeying this propagator over distances much larger than 1/m.

(Almost) lightlike separation: x2 ≈ 0 We have seen that for small x · x of either sign, the propagators coincide and are independent of m. We can understand this as follows. If x · x ≈ 0, then |x0 | ≈ |~x| can be made very small indeed using an appropriate Lorentz transformation12 . In that case the k integral will be dominated by very large values, for which ω(k) ≈ k and hence the value of m becomes unimportant. 12

If x · x = 0 then x0 can be brought arbitrarily close to zero.

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131

Let us therefore put m = 0, upon which the propagator reads ~ Π(x) = (2π)3

Z∞

k2 dk 2(k − i)

Z dΩ exp(−ik · x)

0

≈

~ (2π)2

Z∞ dk

e−ikx+ − e−ikx− 2i|~x|

,

x± = |x0 | ± |~x| .

(5.27)

0

In the sprit of the i prescription we regulate this by a limiting procedure : Z∞

e−ikx+ − e−ikx− −ηk ~ 1 e = lim η↓0 (2π)2 (η + ix+ )(η + ix− ) 2i|~x| 0   −~ x · x + iη ~ 1 = lim =− + iπδ(x · x) . (5.28) P η↓0 (2π)2 (x · x)2 + η 2 (2π)2 x·x

~ Π(x) = lim η↓0 (2π)2

dk

The P stands for the principal-value distribution (see appendix 15.15.6). We recover the result (5.25), and now in addition that for x2 = 0. Note, however, that the propagator is defined in terms of distributions rather than of functions : no problem since in the SDe the propagator occurs inside integrals anyway !

5.2 5.2.1

Moving in Minkowski space : particles The Klein-Gordon equation

For a free theory the SDe is again quite simple. In position, rather than momentum, representation we have Z i d4 y Π(x − y) J(y) φ(x) = ~ Z exp(−ik · (x − y)) 1 = − d4 y d4 k J(y) . (5.29) 4 (2π) k · k − m2 + i Forgetting for the moment about the i, we can apply a differential operator13 on φ(x) :
∂ µ ∂µ + m2 φ(x) = 13

Note that in Eq.(5.30), ∂ stands for a derivative to x, not to y !

132

August 31, 2019 Z µ 2 −ik·(x−y) 1 4 4 (∂ ∂µ + m ) e = − d y d k J(y) (2π)4 k 2 − m2 Z 2 2 −ik·(x−y) 1 4 4 (−k + m ) e = − d y d k J(y) (2π)4 k 2 − m2 Z Z 1 4 4 −ik·(x−y) dydke J(y) = d4 y δ 4 (x − y) J(y) ,(5.30) = (2π)4

since the integral over k has become trivial14 . The free-field function φ(x) is therefore seen to obey the equation
∂ µ ∂µ + m2 φ(x) = J(x) , (5.31) and this is known as the Klein-Gordon equation. In more conventional treatments, this equation is the starting point for a relativistic quantum field theory, being introduced as a direct relativistic adaptation of the nonrelativistic Schr¨odinger equation ; for us, it is a fairly unimportant15 result following from the Feynman rules. You might fret about just leaving out the i in this derivation. Indeed we can keep it, and the the Klein-Gordon equation would contain m2 − i instead of just m2 . This more puristic approach is certainly more correct, but in writing down the Klein-Gordon equation no-one really can be bothered.

5.2.2

Enter the particle !

Equation (5.29) leads to an equation for φ(J), which I find not very interesting. What is important, however, is the light it sheds on the source J : the natural interpretation is, indeed, for J to be a physical source, generating the field φ via Huygens’ principle. The propagator takes the rˆole of the Green’s function as used in the solution of inhomogeneous differential equations. This is the place where the notion of particle comes in ! We picture the source influencing its environment by sending out (or receiving) signals that carry information such as energy, (angular)momentum, charge, colour, etcetera. Under the right circumstances these signals obey Newton’s first law, and that is what justifies our particle picture. The field function φ is now interpreted as the quantummechanical wave function of the particle. 14

This is of course solely due to our taking just the right differential operator. Unimportant in the sense that we shall not derive any consequences from it. The same will be seen to hold for the Dirac, Proca and Maxwell equations. 15

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5.2.3

133

Unstable particles, i and the flow of time

We are now in a position to investigate the physical meaning of the i prescription. In order to do so, let us assume that  is not infinitesimal, but rather of fixed value γ. That is, we shall use a propagator Z i~ exp(−ik · (x − y)) , γ>0 . (5.32) Πγ (x − y) = d4 k 4 (2π) k 2 − m2 + iγ Moreover, let us choose a source that emits particles simultaneously16 at time t = 0, all over space : we take17 J(x) ∼ δ(x0 ) .

(5.33)

The response of the field can be written as Z i φ(x) = d4 y Πγ (x − y) J(y 0 ) ~   0 0 Z ~k · ~y δ(y 0 ) exp −ik · x + ik y − i 1 d4 y d4 k = − 4 (2π) (k 0 )2 − |~k|2 − m2 + iγ Z −1 exp(−ix0 k 0 ) = dk 0 0 2 . (5.34) 2π (k ) − m2 + iγ The integrand has poles in the complex k 0 plane at  p γ  0 2 k = ± m − iγ ≈ ± m − i , 2m where we have assumed that γ is small compared to m2 . For times later than t = 0, the integration contour must be closed along the lower half complex plane (just as in 5.1.6), and we find  γ 0 φ(x) ∝ exp −imx0 − x . (5.35) 2m In accordance with the quantum-mechanical interpretation of our theory, |φ(x)|2 must be (related to) the probability of finding particles. In the present case, we have    γ  t 1 γc 2 0 |φ(x)| ∝ exp − x = exp − , ≡ . (5.36) m τ τ m 16

Simultaneity is an ambiguous concept in Minkowski space : here, we mean simultaneous in our frame : a ‘Little Bang’. 17 We do not worry about normalization issues here.

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That is, the probability of finding particles anywhere decreases exponentially as time goes on. This is what one expects for unstable particles with a mean lifetime equal to τ . We shall write γ = mΓ, where Γ is called the total decay width of the particle, so that 1/τ = c Γ18 . We see that a Feynman rule is now available for unstable particles :

1 k · k − m2 + imΓ The propagator for an unstable particle with mean lifetime 1/(Γc).

k

↔ i~

Feynman rules, version 5.1 (addendum)

(5.37)

The i prescription is seen to just mean that we should treat stable particles as the infinitely-long-lifetime limit of unstable particles. Equation (5.35) describes particles at rest since there is no space dependence in the wave function. The lifetime/width is therefore that of particles at rest, which is indeed the usual definition. For particles in motion, time dilatation indicates that the lifetime be increased by a factor p0 /m ; we shall encounter this situation in the next chapter. An issue that appears resolved is the direction of time flow. Whereas Minkowski space itself, being essentially static, does not assign any preferred direction associated with the time coordinate, the direction of time flow is now defined to be that direction in which unstable particles disappear , rather than appear19 . Another point to be noted is the following. The unstable propagator by itself is seen to lead to a decreasing overall probability : that contradicts the normal unitary evolution of quantum mechanics. But it is not the whole 18

Remember that Γ has the dimension of inverse length, not kilograms ! In practice, it is customary to give total decay widths, like particle masses, in GeV, and then τ = ~/Γ. 19 Attractive as the above argument appears, a drawback comes from the case x0 < 0. In that case, the contour integral must be closed along the upper half plane, so that the pole k 0 = −m + iγ/(2m) becomes the significant one. We find φ(x) ∝ exp(−|t|/τ ), which is to be interpreted as a particle density that starts out as zero at t = −∞, and grows to a crescendo at t = 0 ; this lacks an obvious interpretation. We ascribe this to the use of the simple form (5.33). A better source, needed for a more rigorous treatment, can be simply constructed. Notice that this really means that the direction of time is governed by the sources !

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story : for a particle to be unstable it must be able to go over into other particles, that is, there must be interactions. These have been left out of our discussion. In a more complete treatment, we shall of course see that, as the unstable particles disappear, the density of other particles will increase, and total probability will be preserved. In other words, the decay width must be consistently computable from the interactions present in the theory. The assumption that γ is considerably smaller than m2 implies that Γ is small compared to m. Indeed, if we assume that Γ becomes nonzero due to interactions, the very spirit of perturbation theory argues that Γ is relatively small. Rigorous upper limits on the width of any given particle cannot easily be given ; but let us imagine a particle of mass M (in kilograms, not inverse meters !). Its natural ‘size’ is given by its Compton wavelength λc = ~/(M c). If Γ (a quantity with the dimension of inverse length) were larger than 1/λc , this would mean that such a particle would, upon production, decay even before a lightlike signal could have crossed its diameter : it is as if the particle would vanish before it was even aware that it existed. In general, the situation Γ > m is held to signal a breakdown of the concept of a particle as a more or less identifiable entity.

5.2.4

The Yukawa potential

For another application, we can consider a static point-like source : J(x) ∼ δ 3 (~x) .

(5.38)

The response of the field is then Z Z e−ik·x i 3 i~ 4 4 d k δ (~y ) φ(x) = dy 4 2 (2π) k · k − m + i ~ Z ~ 1 eik·~x 3~ = dk . (2π)3 |~k|2 + m2

(5.39)

The i term in the denominator can safely be neglected here. Writing |~x| ≡ r ≥ 0 and k ≡ |~k|, and going over to polar coordinates for ~k, we have 1 φ(x) = (2π)3

Z∞ dk k 0

2

Z2π

Z1 dϕ

0

d cos θ −1

eikr cos θ k 2 + m2

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August 31, 2019 1 = 4iπ 2 r 1 = 4iπ 2 r

Z∞

eikr e−ikr dk k − k 2 + m2 k 2 + m2 

Z0 dk

k2



k eikr . + m2

(5.40)

For r > 0 we can close the integration contour in the upper half of the complex-k plane, and we find φ(x) =

E34

1 exp(−mr) . 4π r

(5.41)

This is the so-called Yukawa potential, introduced in the 1930’s as a model for the strong nucleon-nucleon force, with m the mass of the pion. The Compton wavelength of the pion is, indeed, roughly the range of the nuclear forces. If we take m → 0 we find the Coulomb potential of a static electric source ; but the real propagator of the photon field, responsible for the Coulomb interaction, is more complicated, so that the above derivation is more or less just handwaving in the case of electromagnetism. A fuller treatment is given in section 10.5.

5.2.5

Kinematics and Newton’s First Law

Let us see to what extent the picture of the source as an object that, in a sense, emits particles can be reconciled with standard ideas in classical relativistic mechanics. That is, we want to measure positions and times, as well as energies, velocities and momenta, as well as possible. To this end, we shall choose the source to be    i |x0 | |~x|2 0 0 − 2− p x − ~x · p~ . (5.42) J(x) ∼ exp − σ0 4σ ~ That is, the source is active for a period σ0 /c around t = 0, and in a region of volume σ 3 around the spatial origin. Its Fourier transform, "   #−1  2 ! 0 2 1 p p ~ + k0 − , (5.43) J(k) ∼ exp −σ 2 ~k − σ0 2 ~ ~ shows that it emits particles with all kinds of wave vectors k µ = (k 0 , ~k), centered around values pµ /~, with pµ = (p0 , p~). For a bridge to non-quantum

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Im k0

Re k

0

Figure 5.4: Closing the contour for Newton’s first

physics to be built, both the position and wave representation of the source should be adequately localized ; σ0 and σ should be neither too large nor too small. For now, we do not assume any particular relation between p0 and p~. We want to study the response of the field to this source for positive times. We have   0 0 Z ~ exp −ik x + ik · ~x J(k) . (5.44) φ(x) ∼ d4 k (k 0 )2 − |~k|2 − m2 + i In figure 5.4 we exhibit the k 0 integral contour in the complex plane. For x0 > 0, the contour is to be closed in the lower half complex-k 0 plane. The integrand has simple poles at the loci i p0 0 ~ k = ω(k) − i , k = − , ~ σ0 0

p0 i k 0 = −ω(~k) + i , k 0 = + , ~ σ0 the latter two lying outside the contour. The k 0 integral therefore leads to the following expression for φ(x) :  2 ! Z p ~ φ(x) ∼ d3~k exp i~x · ~k − σ 2 ~k − A(k 0 , ~k) , ~

138

August 31, 2019   0 ~k) exp −ix ω( 1 A(k 0 , ~k) =  2 2ω(~k) p0 /~ − ω(~k) + 1/σ0 2 +

iσ0 exp(−ix0 p0 /~ − x0 /σ0 ) . 2 (p0 /~ − i/σ0 )2 − ω(~k)2 + i

(5.45)

The second term decays exponentially at the same rate as the source. Since we are interested in the behaviour of the field when it is free, i.e. unaffected by any interactions, we can only study that behaviour once the source has died out, and then so has this term20 . The first term describes Fourier modes of the field that obey the dispersion relation k 0 = ω(~k), together with the resonance condition that tells us that the field can only be appreciable if both p0 /~ ≈ ω(~k) and p~/~ ≈ ~k. We therefore expect any fruitful resonance in the field, which can allow for the transmission of signals over macroscopic distances, only if   p~ p0 ≈ω . (5.46) ~ ~ If we relate the zero component p0 (with dimension kg m/s) to an energy E by writing p0 = E/c , (5.47) we find that the only particle modes emitted by the source that have a chance of propagating over distances much further than σ must satisfy E≈

q |~p|2 c2 + M 2 c4 ,

m=

Mc . ~

(5.48)

This is the mass shell condition, which prescribes the relation between the energy E (in Joule), momentum p~ (in kg m/s), and mechanical mass M (in kg) of a particle moving freely through spacetime. We recognize the quantity m that we have been using so far as the inverse Compton wavelength of the particle. A particle is called on-shell if its momentum pµ satisfies Eq.(5.48) ; if not, it is called off-shell. Off-shell particles are not exotic or improbable ; they are just not visible as the result of an experiment since they cannot 20

This is comparable with what you would do classically: studying the trajectory of a thrown ball to see whether Newton’s laws are obeyed only makes sense once the ball has definitively left your hand.

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propagate well21 . Note the occurrence of ~ in the denominator : interpreted strictly, this implies that a truly classical limit leads to m → ∞. The EulerLagrange equations are therefore not classical at all except for the Maxwell equations holding for massless photons. Given that the particle is emitted on its mass shell, the integral φ(x) is not yet automatically appreciable. The complex phase in Eq.(5.44) will lead to extremely rapid oscillatory behaviour of the integrand, and an essentially vanishing result, except for those regions where the phase of the integrand is stationary. This happens if     ~k ∂ ∂ 0 0 0 x k − ~x · ~k = x ω(~k) − ~x · ~k = x0 − ~x = 0 . (5.49) ~ ~ ~ ∂k ∂k ω(k) That is, φ(x) is appreciable on a line in spacetime given by ~x = t

c p~ p~ = ~v t , ~v = c 0 : 0 p p

(5.50)

the particle moves along a straight line, with constant velocity ~v . This is Newton’s First Law. One might envisage other time-dependences of the source. Two additional cases are discussed in Appendix 15.9 ; the conclusions (although based on slightly different mathematics) are the same. From this simple investigation we may conclude that (a) free particles moving over macroscopic distances must be on-shell ; and that (b) the motion of such particles follows Newton’s first law.

5.2.6

Antimatter

We again consider the free SDe : Z Z d3~k exp(−ik · x) dk 0 0 J(k 0 , ~k) , φ(x , ~x) ∼ 3 0 2 2 ~ 2π (2π) (k ) − ω(k) + i q ω(~k) = |~k|2 + m2 . 21

(5.51)

In popular literature, off-shell particles are often dicussed with a lot of mumbling about ‘uncertainty relations’, ‘borrowing energy from the vacuum’, and so on. Do not allow yourself to be misled ! When a theorist starts invoking the uncertainty principle as a reason for something, keep your hand on your wallet. The ‘uncertainty principle’ is not a principle, not a reason, but a result.

140

August 31, 2019 x

x

B

B E0 A

A t

t

Figure 5.5: Moving forward ; moving backward

If x0 > 0, the integration contour can be closed through the lower half of the complex k 0 plane : Z

0

φ(x , ~x) ∼

d3~k (2π)3 2ω(~k)

  exp −i(x0 ω(~k) − ~x · ~k) J(ω(~k), ~k) .

(5.52)

If, on the other hand, x0 < 0, the closure must be over the upper half of the plane, and then 0

φ(x , ~x) ∼

Z

d3~k (2π)3 2ω(~k)

  0 ~ ~ exp −i(−x ω(k) − ~x · k) J(−ω(~k), ~k) . (5.53)

We see that the propagator essentially describes plane waves, with the following characteristic: positive energies travel towards the future, and negative energies travel towards the past. While the concept of particles with positive kinetic energy, moving from past to future, conforms to our everyday experience, the idea of negative (kinetic) energies and movement backwards in time is not only æsthetically repellent but may lead to splitting headaches in the verbal description of physical processes. When, however, we consider more closely how such a situation will appear, it becomes clear that negative energies moving backwards in time are indistinguishable from positive energies moving forward. Some simple bookkeeping with the help of the two diagrams in figure 5.5 comes in handy here. Consider two loci in space, denoted by A and B. In the

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first diagram a particle moves forward in time, with positive energy, from A to B. As a result the energy at A decreases, and that at B increases. In the second diagram, a particle with negative kinetic energy starts at B, and moves backwards in time to A. The net effect on the energies at A and B is exactly the same ! The two situations are indistinguishable from the point of view of the energy balance22 . There may still be a difference, of course ; if the particles have additional properties such as electric charge, the backwards-moving particles will appear with the opposite charge. For instance, a negatively charged electron moving backwards will appear as a positively charged positron moving forward, as can be seen from the two diagrams above. Such re-interpreted time-reversed particles are called antiparticles 23 . Every particulate object whose propagator contains the denominator of Eq.(5.51) is seen to contain both the regular particles and their antiparticles. Moreover, we find the fundamental result that particles and their antiparticles must have exactly the same mass and lifetime. The antiparticle interpretation is just the way we surrender to a prejudice about motion in time24 . Particles and their antiparticles may be identical, the photon being an example. Such particles must, of course, be electrically neutral. On the other hand, not all neutral particles are their own antiparticles ; neutrons and antineutrons are distinct from one another25 . We have thus found the following result for free particles : if we (a) replace all particles by their antiparticles and vice versa, the so-called charge conjugation operation C, (b) inverse all space directions26 , the so-called parity transformation P, and (c) invert the direction of time, the so-called time reversal operation T, then the world will look exactly the same ! This is (a restricted form of) the CPT theorem, valid for the propagation of free particles. The more interesting, 22

At this point you may wonder about lots of the particles that constitute you and the world actually moving backwards in time. Try to formulate, say, your own digestive processes in this light — it doesn’t bear thinking about. 23 A commonly made error is to say that antimatter moves backwards in time. This is wrong. Antimatter is invented precisely to make everything move forward in time ! 24 Physicists from some alien civilization might have less problems with the other interpretation. 25 Once the neutron is seen to be a collection of charged quarks, the distinction becomes obvious. On the other hand, neutrinos, while electrically neutral, are not equal to antineutrinos, and are yet believed to be elementary. 26 Since, as can be seen from our diagrams, inverting the direction of the motion through time will simultaneously change motion towards the left (say) into motion towards the right, and so on.

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x C B A t0

t

Figure 5.6: An annihilating spacetime path

real CPT theorem, valid also for interacting particles, needs more tools than we have at our disposal right now : its proof is referred to Appendix 15.14. Let us now consider the (classically depicted) path a particle tracks out in spacetime, as given by the space-time diagram given in figure 5.6. In one description, the particle starts at A and moves to B, where at time t0 it reverses its time direction, and moves backwards in time to C. In the alternative description, a particle starts at A and its antiparticle starts at C, and the pair collides at B at time t0 . For times later than t0 , the particle and/or its antiparticle have disappeared ; but because of momentum conservation their combined energy has to be transferred onto one or more other particles (not depicted). The two descriptions are completely equivalent, but the second one conforms much better to the way we tend to view the world. At the ‘collision/reversal-point’ B the particle coming from A must dump its energy, and even an additional amount since its energy must become negative for it to start moving backwards to C. Therefore, particle-antiparticle annihilation releases energy, often in the form of photons27 . For instance, 27

It is sometimes stated that particles can only annihilate with their own antiparticle. This is a somewhat restricted point of view, since for instance electrons can annihilate with anti-neutrinos into W particles, as we shall see. It may be more appropriate to say that it needs particles with their own antiparticles to annihilate into something that has quantum numbers (electric charge, fermion number, etcetera) equal to those of the vacuum. Neutrinos and their antineutrinos cannot easily annihilate into photons, being electrically neutral : but they can annihilate into one or more Z bosons.

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when positrons meet electrons, the usual result28 is e− e+ → γ γ. We also see that nothing forbids the opposite process, in which available energy turns into particle-antiparticle pairs : γ γ → e− e+ .

5.2.7

Counting states : the phase-space integration element

The treatment of the previous section is also useful in that it provides a hint on how to count the wave-vector states. For on-shell particles of mass m we use the integration element q d3~k 1 ~ , ω(k) = ~k 2 + m2 . (2π)3 2ω(~k) This object has dimension L−2 . It is not explicitly Lorentz-covariant, but we can write it also in a more attractive form : E36
1 1 d3~k 4 2 2 = d k δ k − m θ(k 0 ) . (2π)3 2ω(~k) (2π)3

(5.54)

Note that if k 0 is positive for an on-shell particle in any given inertial frame, it is positive in all intertial frames that can be reached by Lorentz transformations from the first one. This ensures that the step function θ(k 0 ) always has the same value, irrespective of any Lorentz boosts we may care to make. Lorentz covariance of the phase space integration element is thus guaranteed. We shall use the density of states (5.54) for all on-shell particles in the calculation of cross sections and lifetimes. If, for a given scattering process, the final state contains N particles with masses mj , j = 1, 2, . . . , N , and momenta pµ1 , pµ2 , . . . , pµN , the combined phase-space integration element is dV (P ; p1 , p2 , . . . , pN ) ≡ ! ! N N X Y 1 d4 pj δ(pj 2 − mj 2 ) (2π)4 δ 4 P − pj , 3 (2π) j=1 j=1

(5.55)

Note that the simpler-seeming process e− e+ → γ is kinematically impossible if the resulting photon is to be on its mass shell. On the other hand, a single off-shell photon can be produced, but such a photon must immediately decay again, in for instance a particle-antiparticle pair of some kind. 28

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where P µ is the total momentum of the scattering system. The four-dimensional Dirac delta forces the overall conservation of wavevector29 . The condition θ(p0 > 0) imposing positive energy for the outgoing particles is, here and in the following, always understood.

5.3

Exercises

Excercise 30 Dangerous  Explain why, in Eq.(5.11), it is important that  > 0. Excercise 31 Zero Minkowski dimensions Once we have discarded the purely probabilistic interpretation of the path integral, we may as well return to zero dimensions, with an action S(ϕ) =

m2 − i 2 λ 4 ϕ − ϕ − Jϕ 2 4!

1. Using the i prescription, show that Z∞

  r m2 2 2π~ −iπ/4 dϕ exp −i ϕ = e 2~ m2

−∞

2. Find the Feynman rules for the propagator and the vertices. 3. Write down the SDe for this model. Excercise 32 Useful  Show that the integral Z I0 =

dx exp(ixy)

where x runs over the whole real axis and y is an arbitrary real number, is ill-defined. Show that the integral Z    I = dx exp ixy − x2 2 29

Conservation of total energy and momentum.

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where  is an infinitesimally small but positive real number, is well-defined. Compute I , and show that lim ↓0 I = 2π δ(y) . This handling of the integral is precisely what we also do by including an  in the action. Excercise 33 careful, careful... Perform the steps leading to Eq.(5.22) and to Eq.(5.24). Especially in the latter case you have to be quite careful in the complex plane ! Excercise 34 Starting out at Yukawa Consider the Yukawa potential in three space dimensions: V (~x) =

1 −mr e , 4π r

r = |~x|

Compute its (three-dimensional) Fourier transform explicitly, to show that Z 1 ~ d3 x V (~x) ei~x·k = 2 |~k| + m2 Excercise 35 Short-range weak forces From our discussion of the Yukawa potential, estimate the effective range of the static weak-interaction potential, mediated by particles of masses 80 to 90 GeV/c2 . Excercise 36 Counting states Prove Eq.(5.54). Excercise 37 Auxiliary fields In Eq.(5.14) the action for ϕ4 theory was given. Let us consider a slightly different one, involving two fields : Z S = d4 x L(ϕ, ψ) , 1 µ m2 − i µ2 − i ∂ ϕ(x)∂µ ϕ(x) − ϕ(x)2 − ψ(x)2 2 2 2 g 2 − ψ(x)ϕ(x) + J(x)ϕ(x) . 2 The field ψ does not occur with a derivative term : such fields are called auxiliary fields. This exercise examines some of their effects. L(ϕ, ψ) =

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1. Show that the path integral over ψ(x) can be performed independently at every point in Minkowski space. 2. Compute the result of the ψ integration for the path integral. This is called integrating out the field ψ. 3. Show that after having integrated out ψ we are left with the ‘original’ ϕ4 theory, provided we choose λ to be a combination of g and µ. Find this combination. 4. Write out the coupled SDe’s for the ϕ−ψ theory in configuration space, using the formalism of functional derivatives. Insert the SDe for ψ into that for ϕ and show that the correct ϕ4 -SDe for ϕ is again obtained. 5. Show that, if we insist that ψ obeys its classical (Euler-Largange) equation, the same result is obtained. 6. Show that the method of auxiliary fields does not work in Euclidean space since there the path integral is not defined (it is easiest to do this in zero dimensions).

Chapter 6 Scattering processes 6.1

Introduction

In this chapter we turn our attention to the bread-and-butter subject of particle phenomenology : the description of scattering processes. We shall discuss the way in which Feynman diagrams and their evaluation are postulated to predict the probability for finding specified final states given specified initial states. We also investigate the consequences of the claim that our approach describes quantum physics and is therefore of a probabilistic nature : that is, we can only compute probabilities, which are necessarily bounded1 . This leads to the notion of unitarity and the use (and usefulness) of cutting rules.

6.2 6.2.1

Incursion into the scattering process Diagrammatic picture of scattering

To a large extent, particle phenomenology can be viewed as the study of scattering processes, in which some initial state is prepared and allowed to time-evolve, and finally an observation is made in which the system is seen to have resulted in some final state. A useful example is provided by the current practice in high-energy colliders : here the initial state is prepared by machine physicists operating the collider, and it consists of two (beams of) particles with more or less well-defined momenta coming out of the beam pipes. The interesting part of the time-evolution of the system is that during 1

After all, the probability of a certain scattering process occurring cannot exceed 100%.

147

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August 31, 2019

which the initial-state particles approach one another and meet, hopefully2 , in the interaction point, where the dynamics takes place. The final state is observed by the detector operated by the particle physicists. Since not only the scattering itself but also the initial-state preparation and the final-state observation are quantum processes, all these parts of the process must, according to our assumptions, be described by Feynman diagrams in a manner still to be established. The diagrammatic form of the complete process will then look as follows :

and here (and in the following) we adopt the convention that the initial state appears on the left-hand side of the diagrams, and the final state on the righthand side3 . This does not imply any spatial or timelike relation between any of the vertices in the diagram: indeed, they are supposed to be integrated over all of spacetime4 . Another observation on the above diagram is also relevant : the initial-state preparation and the final-state observation should contain physics that is better understood than the scattering part, and there should be a clear notion of precisely which particles constitute the initial and final states. This is indicated by the identifiable propagators connecting the various ingredients of the process. We therefore adopt the idealization that the only relevant part of the scattering should reside in the central, or 2

In the idealised setting in which particles with perfectly well-defined momenta form plane waves of infinite spatial extent, they can hardly avoid meeting. In practice, the momenta and spatial extensions of the particles’ wave packets are of course more limited. 3 This convention is based on not much more than the fact that in Latin writing you move from left to right. Indeed a not uncommon alternative is to move upwards, so that the incoming state is at the bottom. I do not know of any script that moves like that. It could be worse : fortunately no-one employs boustrophedonic diagrams ! 4 Of course, if there is any justice the contribution from paths in which a vertex is very far out ought to be small.

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scattering part, in this case

We now have to confront the two following questions. In the first place, which Feynman diagrams should occur in the scattering part ? And secondly, in actual experiments the initial- and final-state particles travel over many meters between preparation, scattering, and detection. These particles should therefore be on their mass shell, but isn’t this precisely the case in which their propagators blow up ? The situation obviously calls for some reinterpretation and additional Feynman rules, to which we shall come. Before finishing this section, let us remark that also initial states consisting of only a single particle occur :

In this case, we simply study the decay properties of the particle, such as its total or partial decay width.

6.2.2

The argument for connectedness

Let us consider the set of all Feynman diagrams describing a decay process. As discussed in section 1.3.7, we omit all vacuum bubbles. The set can then be split up into its connected pieces, for instance =

0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

+

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

+ · · · (6.1)

where as before the shading indicates connected diagrams. Now, recall that every vertex in any diagram contributes a Dirac delta imposing energymomentum conservation. Therefore, every connected diagram has an overall

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Dirac delta imposing overall energy conservation. That, however, implies that a satellite diagram such as 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

(6.2)

asks for particles carrying positive energy to originate (by some interactions) from the vacuum. Such contributions therefore vanish by energy conservation, and the only contributing diagrams are contained in the totally connected blob. Next, consider two-particle scattering. Since we forbid satellite diagrams the only possible contributions are given by =

E38

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

+

0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111

+ · · · (6.3)

Now, the second term here is in principle possible but only if a) the two incoming particles are inherently unstable5 and b) the outgoing particles arrange themselves in precisely two groups according to the indicated decay patterns. Leaving aside such special cases, we conclude that the scattering amplitude is given by the connected Feynman diagrams. Note that the restriction to connected diagrams only arises here from simple energy considerations, and not from any deep inherent superiority of connected diagrams over disconnected ones : in essentially all cases of interest, the result of the disconnected diagrams vanish anyway6 . We may imagine situations where particles can be created from the vacuum. This is the cases in ‘field theories at high temperature’ where processes take place in a heat bath which can deliver energy to create particles, so energy conservation is relaxed. In such a picture the heat bath is the ‘vacuum’ of the theory, and diagrams such as that of Eq.(6.2) are not automatically zero. Another delicate situation is that of more incoming particles : for instance, we might consider four particles scattering into four, in which we 5

This makes the notion of particles ‘coming in from infinity’ conceptually dubious in this scattering. 6 There are plans on the table for a muon collider. Muons are unstable particles, but their lifetime (especially at high energies) is extremely long as particle lifetimes go, so we can treat them as essentially stable in most processes. However, it is possible that final states are produced where one of the muons actually decays ; then we have to take more care (see also exercise 38).

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might recognize two groups of two particles scattering into two :

=

0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

+

00000000 11111111 11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

+ · · · (6.4)

In this case, the only argument to disregard the disconnected diagrams is an appeal to the special kinematics.

6.3 6.3.1

Building predictions General formulæ for decay widhts and cross sections

Consider a ‘slightly unstable’ particle of mass m at rest, with momentum P µ . We shall adopt the following prescription for its differential decay width into n particles with momenta pµ1 , pµ2 , . . . , pµn : dΓ = ΦΓ h|M|2 i dV (P ; p1 , p2 , . . . , pn ) Fsymm .

(6.5)

Here, M stands for the transition amplitude, which we still have to establish. The symbol h|M|2 i indicates that in accordance with quantum-mechanical practice we have to square the absolute value of M in order to arrive at a probability, and the brackets indicate summation and/or averaging over degrees of freedom other than the momenta : at present such degrees of freedom are not in our theory yet, but they will arrive ! The symbol ΦΓ denotes the collection of factors that must be included to account for the density of states for the incoming particle, etcetera. That such a factor must be present is clear from the fact that dΓ must have dimension 1/L. dV denotes the phase-space integration element discussed in section 5.2.7. The total decay width in this particular decay channel is obtained by integrating over the available phase space7 . The momentum P µ is that of the incoming particle at rest. The symmetry factor Fsymm is included to handle identical particles in the final state. In quantum mechanics, the statement that two particles are identical means that an interchange of these particles leads to the physically identical final state, so that an unconstrained summation over 7

This can become very challenging !

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their momenta (and other quantum numbers) would lead to over-counting. We therefore prescribe that Fsymm contains a factor 1/k! for every group of precisely k indentical particles in the final state8 . For example, a final state containing precisely 2 photons, 3 electrons and 1 positron leads to Fsymm = 1/(2!)(3!)(1!) = 1/12. Note that the decay width is inversely proportional to the particle’s lifetime. This means that for a moving particle the decay width must decrease by a factor m/P 0 to account for time dilatation. In the case of two stable incoming particles with momenta pµa and pµb , we rather talk about the transition rate per unit flux, that is, the cross section for their scattering. It has dimension L2 , and must be given by a formula of the form dσ = Φσ h|M|2 i dV (pa + pb ; p1 , p2 , . . . , pn ) Fsymm ,

(6.6)

with its own flux factor, also still to be determined. We see that, in order to get the formulae (6.5) and (6.6) to actually work, we have to establish • the flux factors ΦΓ and Φσ ; • the algorithm to derive from the connected Green’s function the amplitude. In particular this calls for a special treatment of the external lines. We shall solve these issues in the next section.

6.3.2

The truncation bootstrap

We have come to one of the centrally important steps in our treatment of scattering. We shall solve the questions of the previous section by assuming that we have solved them and drawing consequences : a bootstrap mechanism Consider the process in which two particles with momenta pa and pb scatter and yield j + n stable particles in the final state, whose momenta we label by k1 , k2 , . . . , kj and q1 , q2 , . . . , qn . The distinction between these groups lies in the fact that, whereas the k’s emerge ‘directly’ from the scattering, the q’s are in fact the decay products of an unstable particle that was ‘directly’ produced √ Some authors choose to include a factor 1/ k! in the transition amplitude M. I am opposed to this since such a prescription introduces a distinction between particles in the initial and those in the final state, which may destroy the crossing symmetry of the amplitude. 8

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together with the k’s. Nevertheless, the complete final state consists of both the k’s and the q’s. The relevant diagrams thus look like

pa

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

}k

1,2,...

A

M=

pb

p

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

B

}q

1,2,...

(6.7)

Note that the connected blobs may themselves contain many different individual diagrams. By separating the blobs A and B we indicate that the unstable particles is actually quite long-lived so that the place where it is produced and that where it decays tend to be clearly separated. Now, we shall assume that we have, somehow solved the problem of how to go from connected Green’s function to amplitude, and that we have applied this procedure to the above process, in particular to the external lines. We then have for the amplitude the form M=A

p2

i~ B , − m2 + imΓ

(6.8)

where p = q1 + · · · + qn is the momentum of the (internal !) line corresponding to the unstable particle, and p2 = p · p. The unstable particle’s mass is m, and its total decay width is Γ. The symbols A and B stand for the ‘processed’ connected Green’s functions for the ‘production’ process A and the ‘decay’ process B, but with the Feynman factors for the unstable intermediate particle removed. Assuming, for simplicity, that Fsymm = 1, we then have for the differential cross section the form dσ = Φσ |A|2 |B|2

~2 dV (P ; k1 , . . . , kj , q1 , . . . , qn ) , (p2 − m2 )2 + m2 Γ2

(6.9)

where P = pa + pb . In order to emphasize that p is the sum of the q’s, we may write this also as dσ = Φσ |A|2 |B|2 dV (P ; k1 , . . . , kj , q1 , . . . , qn ) ~2 d4 p (2π)4 δ 4 (p − Σq) , (p2 − m2 )2 + m2 Γ2 (2π)4 with obvious notation for the sum over the wavevectors q.

(6.10)

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August 31, 2019

Now, we let the unstable particle approach stability, so that the location where it decays becomes widely separated from that where it is produced. That is, we examine the case that Γ becomes very, very small, and we may approximate 1 (p2

−

m2 )2

+

m2 Γ2

→

π δ(p2 − m2 ) . mΓ

(6.11)

We can then use this to rewrite dV (P ; k1 , . . . , kj , q1 , . . . , qn ) d4 p (2π)4 δ 4 (p − Σq) 2 4 2 2 2 2 (2π) (p − m ) + m Γ

(6.12)

as d4 p δ(p2 − m2 ) 1 dV (P ; k1 , . . . , kj , q1 , . . . , qn ) (2π)4 δ 4 (p − Σq) 2mΓ (2π)3 1 = dV (P ; k1 , . . . , kj , p) dV (p; q1 , . . . , qn ) . (6.13) 2mΓ Inserting this in Eq.(6.10) we see that the cross section now takes the form
dσ = ~ |A|2 dV (P ; k1 , . . . , kj , p)
1 1 ~ |B|2 dV (p; q1 , . . . , qn ) . (6.14) Γ 2m Now, step back and consider what it is we are actually computing here : it is the cross section for producing an almost-stable particle p, together with the k’s in a specified configuration, followed by the decay of the particle p into a specified configuration of q’s. Under the usual ideas of conditional probability, this is the same as first computing the cross section for the production of p and the k’s, followed by the conditional probability that, given p, we see it decay into the q’s. This conditional probability, called the (differential) branching ratio, is the partial decay width for p to go into the q’s (computed in the p rest frame), divided by the total decay width, in this case Γ (also defined in the rest frame). We conclude that • ~ |A|2 is h|M|2 i for the process pa + pb → k1 + · · · + kj + p ; • ~ |B|2 is h|M|2 i for the process p → q1 + · · · + qn ; • ΦΓ is given by 1/(2m).

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In a sense, we have managed to cut through the p line, and interpret the process rather as it would be given by the diagrams

pa

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

A

M →

pb

}k

1,2,...

p

p

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

B

}q

1,2,...

(6.15)

A point to be noted here has been somewhat hidden so far. The connected Green’s functions contain factors (2π)4 δ 4 (· · ·) for overall momentum conservation. This conservation has been imposed already, however, in our choice of the phase space integration elements dV . We therefore have to remove these factors as well in the transition from connected Green’s function to M.

What about the treatment of the external lines ? In the above discussion we started with p as an internally ocurring unstable particle, carrying its own propagator. As we let it become stable, the propagator has disappeared into the phase space counting, leaving only a residue of a factor ~2 . At the end of the story the particle p has become a stable particle occuring as an external line in the blob A. This, therefore, must be the prescription for the external lines ! This is called truncation or amputation of external lines. An external line√must apparently carry, instead of its undefined propagator, simply a factor ~. We arrive at the following, expanded set of rules for the calculation of scattering amplitudes M (as opposed to Green’s functions) :

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k 111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111

↔

k

k1

k2

k4

k3 k1 k2

Z∞

↔ ↔ ↔

i~ internal k · k − m2 + i √ ~ external i − λ4 (2π)4 δ 4 (k1 + k2 + k3 + k4 ) ~ i + J(k2 )(2π)4 δ 4 (k1 + k2 ) ~

dk 2π

for every momentum k

(6.16)

−∞

At the vertices the momenta must be counted either all incoming or al outgoing. Drop the 2π 4 δ(· · ·) for overall momentum conservation. Feynman rules, version 6.1 Let us now turn to the two-particle initial state. The flux factor ΦΓ for particle decay has been found to be 1/(2m). It is related to how we count the density of states of the incoming particle9 . We can directly translate this to the case of two-particle scattering. Let us work in the Lorentz frame in which particle b is at rest while particle a impinges upon it10 . Keeping in mind the effect of Lorentz transformations on the density of states we see that whereas mb remains, ma has to be replaced by p0a , in accordance with the discussion in section 5.2.3. The density-of-states factor for the two-body initial state is therefore 1/4p0a mb . Since, however, we are asking for a cross section rather than a transition rate, we have to divide this by the velocity of particle a in b’s rest frame, that is, by a factor |~pa |/p0a . The flux factor therefore becomes Φσ = (4mb |~pa |)−1 . This expression, being given in a specific Lorentz frame, is not very attractive. R If we fix the particle to be at rest, we must count the states as d3 (~k) d3~k/2ω(~k) = 1/2m. 10 Not the most practical frame, to be sure : at the LHC it is a frame in which the whole ATLAS detector, say, moves at 99.99999882% of the speed of light. 9

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We can, however, write it in an explicitly Lorentz-invariant form :
−1/2 1 , Φσ = λ (pa + pb )2 , p2a , p2b 2

(6.17)

where we have introduced the K¨all´en function λ(x, y, z) ≡ x2 + y 2 + z 2 − 2xy − 2xz − 2yz = (x − y − z)2 − 4yz . (6.18) It often happens that the colliding particles have masses that are negligible compared to their combined invariant mass, the square of which is commonly denoted by the Mandelstam variable s. In that case, we may write Φσ ≈

1 , 2s

s ≡ (pa + pb )2 .

(6.19)

This finishes our bootstrap treatment of the relation between connected Green’s functions and scattering amplitudes, or matrix elements.

6.3.3

A check on dimensionalities

It is instructive to check that the widths and cross section expressions that we have derived do, indeed, have the correct dimensionality. By dim[ ] we shall denote the dimensionality of objects, and L will denote R a length. In the first place, from the fact that the action S, that contains d4 x (∂ϕ)2 , must have the same dimension as ~, we can immediately derive the dimensionality of the fields11 :   dim[ϕ] = dim[φ] = dim ~1/2 /L . (6.20) A connected Green’s function with n external lines is the expectation value of ϕn and must therefore have dimension12   dim[Cn ] = dim ~n/2 /Ln . (6.21) The Dirac delta function imposing wavevector conservation has dimensionality       dim δ 4 (k) = dim k −4 = dim L4 . (6.22) 11

In dimensions! In d dimensions it would read dim[ϕ] =   four spacetime dim ~1/2 L1−d/2 . 12 Higher-order contributions to Green’s functions contain, of course, additional powers of ~: but these must occur only in dimensionless combinations with the coupling constants of the theory.

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To go from the connected Green’s function Cn to the n-point matrix element Mn , we have to extract the external propagators as well as the overall wavevector conservation delta function, and assign a factor ~1/2 to each external line: therefore,     Cn n/2 = dim Ln−4 . (6.23) ~ dim[Mn ] = dim n 4 (C2 ) δ (k) The n-particle phase-space integration element dVn has dimensionality L4−2n as we have seen. Taking into account that the flux factor ΦΓ = 1/2m must have the dimensionality of 1/m, that is, L, the dimensionality of the decay width of a single particle into n particles is given by     dim[Γ(1 → n)] = dim (1/m)|Mn+1 |2 dVn = dim L−1 , (6.24) as required. Similarly, for the cross section of two particles going into n particles we have     dim[σ(2 → n)] = dim (1/m2 )|Mn+2 |2 dVn = dim L2 , (6.25) again as required. Note that the above analysis is kept simple because we have restricted ourselves to the use of wavevectors rather than mechanical momenta, which would introduce additional factors of ~ in the calculation. The other natural constant, c, need not enter here.

6.3.4

Crossing symmetry

In our treatment of antimatter in the previous chapter we have seen that the production (absorption) of a particle is, in a sense, æquivalent to the absorption (production) of its antiparticle. We can make this even more specific as a relation between various scattering amplitudes : this goes by the name of crossing symmetry. Consider a generic 2 → 2 scattering process : a(p1 ) + b(p2 ) → c(q1 ) + d(q2 ) where we have indicated the momenta of the particles. Let us write the corresponding amplitude as M(p1 , p2 , q1 , q2 ). By moving particles from the initial to the final state13 , or vice versa, we can then find the amplitudes for 13

You can visualize this by taking an outgoing particle, say, and dragging its external leg from the final to the initial state.

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the crossing-related processes, for example : a+b→c+d a + c¯ → ¯b + d a + d¯ → ¯b + c c¯ + d¯ → a ¯ + ¯b

: : : :

M(p1 , p2 , q1 , q2 ) , M(p1 , −p2 , −q1 , q2 ) , M(p1 , −p2 , q1 , −q2 ) , M(−p1 , −p2 , −q1 , −q2 ) .

(6.26)

Since the momenta of all (anti)particles have positive energy, the minus signs yield momenta with negative energy. Depending on the type of the particle14 , this may involve an analytic continuation of the amplitude function M.

6.4 6.4.1

Unitarity issues Unitarity of the S matrix

If M is to be a correct form of the scattering amplitude for a given initial state to be observed, after time evolution, as a given final state, it must obey the constraints of unitarity which we shall now discuss. In a more traditional quantum-mechanical parlance, the initial state is given to us at some time in the far past, where the incoming particles are supposed to be so widely separated that they are essentially free : the state of the system is then |in, t = −∞i We now let nature take its course : the incoming particles approach one another, the interaction is ‘switched on’, and the system evolves into some, possibly very complicated, superposition of free-particle states : |in, t = −∞i

→

|in, t = +∞i

Finally, the final state is observed to be a particular free-particle state (assuming the final-state particles have been able to move very far away from one another), that is, |out, t = +∞i . The probability amplitude for this to happen is of course M = hout, t = +∞|ins, t = +∞i ≡ hout, t = +∞| S |in, t = −∞i , (6.27) 14

Especially for Dirac particles.

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where S is the matrix describing the time evolution of the incoming state from t = −∞ to t = +∞. An important consideration here is the conservation of probability. That is, any initial state |ii must go to some final state |f i with 100% probability15 ; of course |f i = |ii may also be one of the possibilities. Writing this using the S matrix we have X X    1= | hf |S| ii |2 = i S † f f S i = i S † S i . (6.28) f

f

Conversely, any final state |f i must have come from some initial state |ii with 100% probability, so that X X    1= | hf |S| ii |2 = f S i i S † f = f SS † f . (6.29) i

E40

i

Since this must hold for all states |i, f i we have S † S = SS † = 1 ,

(6.30)

and the S matrix is unitary. Note that we have had to assume that the set of initial states |ii and final states |f i are complete16 . The free-particle states are natural choices for complete orthonormal bases, and we see that M is simply a matrix element of the S matrix. We shall investigate this in some more detail. For simplicity, let us assume that we can label the initial states with a discrete label i, and the final states by a similar discrete label f . We can then write the S matrix element as S f i = δ f i + Mf i ,

(6.31)

where the Kronecker delta embodies what would happen if there were no interactions : the only possible observed final state would in that case be identical to the initial state (two particles, say, continuing on their way without having interacted). The remainder Mf i is the object described by Eq.(6.27) ; it is the result of the interactions of the theory, and is described by the Feynman diagrams. Note that Mii 6= 0 is quite possible ; it corresponds to the case where the final state happens to reproduce the initial state, so to speak 15

Try to imagine a world in which this does not hold ! Conservation of probability is a dogma — but a reasonable one. 16 Again, try to imagine what things would look like if that were not the case. . .

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161

in spite of the interactions. This is called the forward scattering amplitude. Now, the unitarity of the S matrix is expressed17 as SS † = S † S = 1, or X ∗ Skf Ski = δf i , (6.32) k

or, in terms of M : Mf i + M∗if +

X

M∗kf Mki = 0 .

(6.33)

k

As a special case, we can consider f = i : we then have the optical theorem, X 2 Re (Mii ) + |Mki |2 = 0 , (6.34) k

which immediately shows that the forward scattering amplitude must have negative real part18 . In Eq.(6.34), the first term is called the dispersion term, and the second one the absorptive term, terms which originate in the study of waves moving through a medium. Also, in our ‘discrete’ formulation, |Sf i | ≤ 1 for all i, f ,

(6.35)

which implies that Mf i can not be arbitrarily large ; we shall employ this idea extensively later on. In appendix 15.10 we derive a more precise bound, called partial-wave unitarity, but that calls for Feynman rules we still have to develop. A an elementary illustration of the optical theorem we consider the following physical process. We start with an empty initial state i (that is, a state containing no particles). At some moment a source kicks in, producing an unstable particle with wavevector p, mass m and total width Γ. Sometime later, the same source absorbs the particle, and at the end the final state f is empty again. The simple Feynman diagram describing this is

J 17

p

J

Since S may be an infinite matrix, both conditions are necessary, whereas for a finite matrix one would suffice. 18 A word of caution : in much of the literature, the statement reads that the amplitude must have positive imaginary part. This is simply due to the fact that in those texts, the S matrix element is written not δ + M but δ + iM. I do not see any particular virtue in this.

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and since the initial and final state coincide (f = i) this is a forward scattering amplitude ; it must obey the optical theorem. We shall now verify this. The matrix element is given by19     J i~ J Mii = i i , (6.36) 2 2 ~ p − m + imΓ ~ so that

mΓ J2 , (6.37) ~ (p2 − m2 )2 + m2 Γ2 which is indeed negative. Now, we consider the matrix elements Mki as used in Eq.(6.34). These describe the initial state i going over in any final state k, that is, they describe the decay of the particle after it has been produced by the source : Re (Mii ) = −

J p

111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111

and we shall denote them by Mki = −i

J ~ D , 2 ~ p − m2 + imΓ

(6.38)

where iD is the contribution of the ‘decay blob’. We then have X k

|Mki |2 =

X J2 |D|2 . 2 2 2 2 2 (p − m ) + m Γ k

The optical theorem (6.34) will therefore be satisfied if 1 X ~ |D|2 . Γ= 2m k

(6.39)

(6.40)

But this is, of course, precisely the prescription for the decay width of the particle, if we realize that the final state k indicates not only all possible final states, but also that the summation over k should include the phasespace integration. This short excercise illustrates both the optical theorem and the computational prescriptions arrived at before. Note that the factor ~ corresponds √ precisely with the Feynman rule that an external line should carry a factor ~. You might opt for the ‘absorbing’ source to be J ∗ rather than J. The conclusions are the same. 19

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6.4.2

163

The cutting rules

We shall now consider how the unitarity relation (6.33) can be made useful in the language of Feynman diagrams. To start, we realize that this equation contains, in addition to the ‘standard’ matrix element Mf i for initial state i and final state f , also M∗if which describes the (complex conjugate) matrix element for initial state f going over into final state i, that is, the timereversed process. We shall embody this in a useful manner by introducing a cutting line. A cutting line cuts across diagrams separating them into a ‘left’ and ‘right’ piece. Any diagram to the left of a cutting line is interpreted in the usual manner ; any diagram to the right of a cutting line is interpreted to be the complex conjugate of the time-reversed version of the diagram. That is,  ∗ 0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

=

0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

,

=

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111



If the diagram contains oriented lines, the time-reversal also inverts the orientation of those lines (if the orientation is indicated by an arrow, we reverse the arrow). We can write Eq.(6.33) diagrammatically as

i

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

f

+

i

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

f +

X k

i

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

k k

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

f

=0 .

(6.41)

It is possible to sharpen this equation to make it more useful. In the first place, Eq.(6.41) holds for whole matrix elements, evaluated to all orders in perturbation theory. This implies that it must also hold for each order separately20 . However, even at some fixed order, Mf i can contain very many diagrams. Consider a somewhat-complicated Feynman diagram in ϕ3 theory : (6.42) 20

If Eq.(6.41) were not to hold order-by-order, this would imply subtle relations between coupling constants, ~, and the like. We would then be in a position to actually compute coupling constants from first principles, which would be good — too good to be true, in fact, if we believe that we are allowed some freedom in choosing the values of the coupling constants while retaining a consistent theory.

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The corresponding Lagrangian reads

E41

1 1 1 (6.43) L = (∂ µ ϕ)(∂µ ϕ) − m2 ϕ2 − λϕ3 . 2 2 6 The unitarity structure of the above Feynman diagram is not immediately obvious since there are, at this order of perturbation theory, no less than 58 diagrams that contribute to this amplitude. We can, however, employ the following trick. Let us assign a different label to each line in the diagram, in an arbitrary manner, for instance 2

5 6

1

4

7

(6.44)

3 8

9

and let us pretend that each line corresponds to a different field. This diagram can then be interpreted as coming from a theory with 9 distinct fields (with identical mass) and Lagrangian  9  X 1 2 2 1 µ (∂ ϕn )(∂µ ϕn ) − m ϕn − V , L = 2 2 n=1 V

= λ123 ϕ1 ϕ2 ϕ3 + λ245 ϕ2 ϕ4 ϕ5 + λ349 ϕ3 ϕ4 ϕ9 +λ567 ϕ5 ϕ6 ϕ7 + λ789 ϕ7 ϕ8 ϕ9 .

(6.45)

Nothing forbids us to assign to the various ϕϕϕ couplings precisely the numerical value λ later on. Now, it is easily seen that, in order λ123 λ245 λ349 λ567 λ789 , the diagram (6.44) is the only diagram that can contribute in this theory21 ! We can do even more : by inspection of all possibilities, we can simply realize that the only final states k in the unitarity condition (6.41) must be precisely k = {2, 3}, {5, 9}, {2, 4, 9} or {3, 4, 5}, if we want to end up with the right order in perturbation theory22 . In other words, + 21

+

The secret resides in the fact that in V the external fields 1,6 and 8 occur precisely once, and the other fields precisely twice. 22 Note that, for instance, the choice k = {5, 7, 8} would result in the right-hand half of the diagram being disconnected ; the choice k = {2, 4, 7} is inconsistent since both 6 and 8 are in the final state.

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+

+

+

= 0 , (6.46)

where we have omitted the line labellings : indeed, the same identity must hold for the original diagram (6.42). This establishes the so-called cutting rules (also called the Cutkosky rules), which can be most simply expressed in words : take a diagram and move the cutting line through it from right to left in all possible manners, making sure that the two halves in which the diagram is cut remain connected and that neither the inital state or the final state is dissected. The particles described by internal lines through which the cut runs must be assumed to be on their mass shell23 . The sum of all the possible contributions then vanishes24 . It must be stressed, however, that we have not proven the Cutkosky rules in this section : rather we have established that a theory must satisfy them if it is to be a good, probability-conserving theory.

6.4.3

Infrared cancellations in QED

As an illustration of how the cutting rules may be applied we shall make a slight jump ahead and consider quantum electrodynamics, that is the theory of photons and electrons. Their Feynman rules will be dicussed later ; for now it is sufficient to know that the only interaction vertex in the theory is the three-point vertex

where the oriented lines stand for electrons and positrons, and the wavy line denotes the photon. Let us consider the 1PI two-loop corrections to the photon propagator. These are given by 111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111

23

=

+

+

This may mean that the situation thus described fails to meet the restrictions of momentum/energy conservation ; then, that contribution vanishes. 24 You might object that in a theory with many different fields the symmetry factors of the diagrams will, in general, be different from those of a theory with only a single field, and this is true : however, in the summation over the ‘intermediate states’ k we must of course also include the ‘indentical-particle’ symmetry factor Fsymm , which precisely repairs the correspondence — another illustration of the crucial rˆole of the symmetry factors !

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August 31, 2019

By applying the cutting rules we can investigate the real part of this two-loop contribution: ∗  + ↔ 000000000 111111111 111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111

↔

000000000 111111111 111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111

+

+

+

+

+

+

+

+

.

This set of 10 cut diagrams is, as we can see, equal to25   ∗ + + + (c.c.) 2 , + +

+

(6.47)

(6.48)

where integration over the final state is implied. As we shall see, the presence of a photon in the final state leads to a so-called infrared (IR) divergence arising from the fact that the probability of emitting an on-shell photon goes to infinity as the photon energy goes to zero. The process described by the last two diagrams has therefore an infrared divergence. This divergence is neatly cancelled by a compensating divergence in the diagrams with a virtual photon in the first line. This is a well-known fact26 ; but it is instructive to see that the statement about the cancellation of the infrared divergences can be replaced by the simpler statement that the photon propagator is free from infrared divergences27 . This is one example of a useful rule of thumb : when 25

Recall the remark about oriented lines just above Eq.(6.41). And a fortunate one. 27 Two remarks are in order here. In the first place, the virtual-photon diagrams do contain divergences related to the loop momentum going to infinity : these are ultraviolet (UV) divergences. ß The photon propagator is therefore still ultraviolet divergent, and this is cured in the usual manner by renormalization. In the second place, the cancellation of IR divergences takes place even when we restrict the phase space for the outgoing particles, provided that zero-energy photons are admitted. This last point is, of course, not proven by our argument ; but if it were not true you would see really infinite physical cross sections by making suitable cuts on the final state ! Nature, fortunately, is more friendly than that. 26

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you encounter loop diagrams, try to envisage the physics that is described by cutting them. In fact, the cancellation can be pinpointed further ; the single statement that the single diagram

is IR-finite means that the IR divergences in 2 and



!

!∗ + (c.c.)

must cancel between them.

6.5 6.5.1

Some example calculations The FEE model

As an example of an application of what we have learned so far, we shall investigate at theory that contains two particle types, one of mass m, denoted by E, and another denoted by F , of mass M . The Lagrangian density of this theory is given by L =

m2 2 1 µ (∂ ϕE ) (∂µ ϕE ) − ϕE 2 2 1 M 2 2 mλ + (∂ µ ϕF ) (∂µ ϕF ) − ϕF − ϕF ϕE 2 . 2 2 2

(6.49)

There exists a single coupling between two E’s and one F . Note that the Feynman rule for the vertex is given28 by −imλ/~ ; we have introduced a factor m in order to ensure that   dim[λ] = dim ~−1/2 with no length scale. 28

It is customary to leave out the (2π)4 δ 4 () of momentum conservation, since it is present in all vertex Feynman rules for translation-invariant interactions.

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6.5.2

Two-body phase space

Since we shall consider processes ending in a two-body final state, it is expedient first to work out the corresponding two-body phase space. For the sake of generality we shall do this for a final state containing two momenta q1,2 µ with general masses m1,2 . Furthermore we shall write P µ = q1 µ + q2 µ ,

s = P µ Pµ .

(6.50)

The phase space (and with it widths and cross sections) is often most easily evaluated in the rest frame of P µ , in which ~q1 = −~q2 . The phase space integration element is given by29 1 d4 q1 δ(q1 2 − m1 2 ) d4 q2 δ(q2 2 − m2 2 ) δ 4 (P − q1 − q2 ) . (2π)2 (6.51) As a first step, we cancel d4 q2 against the four-dimensional Dirac delta, and write the q1 integration in its not-explicitly-covariant form : dV (P ; q1 , q2 ) =


1 d3 ~q1 2 2 δ (P − q ) − m dV (P ; q1 , q2 ) = . 1 2 (2π)2 2q1 0

(6.52)

Now, the q1 integration element can be expressed in polar coordinates as |~q1 |2 d|~q1 | dΩ 1 d3 ~q1 = = |~q1 | dq1 0 dΩ , 0 0 2q1 2q1 2

(6.53)

where we denote the ~q1 solid angle by dΩ = d cos θ dφ

(6.54)

with a polar angle θ running from 0 to π and an azimuthal angle φ running from 0 to 2π, and use the fact that |~q1 | d|~q1 | = q1 0 dq1 0 .

(6.55)

The Dirac delta imposing the mass shell condition on q2 can be written as
√
δ (P − q1 )2 − m2 2 = δ s + m1 2 − m2 2 − 2q1 0 s   1 s + m1 2 − m2 2 0 √ = √ δ − q1 , (6.56) 2 s 2 s 29

It is usual not to include the step functions that require the energies to be positive.

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169

where the rest frame of P has been used. We immediately find that q1 0 = and

s + m1 2 − m2 2 √ , 2 s

q2 0 =

s + m2 2 − m1 2 √ , 2 s

1 |~q1 | = |~q2 | = √ λ(s, m1 2 , m2 2 )1/2 , 2 s

(6.57)

(6.58)

where the K¨all´en function crops up again. In the P µ rest frame, the phase space integration element is therefore given by  1/2 m1 2 m2 2 1 λ 1, , dΩ . dV (P ; q1 , q2 ) = 32π 2 s s

6.5.3

(6.59)

A decay process

As a first application, we shall assume that M > 2m so that the F particle can decay into a pair of E’s: F (P ) → E(q1 ) E(q2 ) . In lowest order, its single Feynman graph is given by q1 P

q2

The corresponding matrix element is quite simple : M = −i

√ mλ √ 3 ~ = −imλ ~ , ~

(6.60)

so that it has dimensionality dim[1/L] as it should. The decay width is therefore 1 1 |M|2 dV (P ; q1 , q2 ) 2M 2! r 2 2 2 mλ~ 4m = 1− dΩ . 2 128π M M2

dΓ(F → EE) =

(6.61)

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August 31, 2019

Note the occurrence of the symmetry factor 1/2 arising from the fact that the two final-state E particles are indistinguishable. The angular integration is of course trivial in this simple case, and we immediately find the total width r m2 λ2 ~ 4m2 Γ(F → EE) = 1− , (6.62) 32πM M2 with the correct dimensionality dim[Γ] = dim[1/L].

6.5.4

A scattering process

As a second application, we F to be massless, M = 0. We now have an extremely primitive picture of the electron-photon system, where E is the electron and F the photon. We consider the process of ‘Compton scattering’ : E(p1 ) F (p2 ) → E(q1 ) F (q2 ) which, to lowest order, is given by two Feynman diagrams:

p1 M=

p2

q2 p1 q1 + p2

q2 q1

(6.63)

We start by defining the various momenta. It is usually best to work p in the centre-of-mass frame. If we denote the total scattering energy by (s), we can write the momenta as follows, using Eqs.(6.57,6.58) : pµ1 = (p, q~ep ) , pµ2 = (q, −q~ep ) , q1µ = (p, q~eq ) , q2µ = (q, −q~eq ) . (6.64) Here ~ep,q are the √ give the directions of p~1 and ~q1 , and √ unit vectors2 that 2 p = (s + m )/2 s, q = (s − m )/2 s. The tree-level matrix element is then   1 1 2 2 M = −i~λ m + (p1 + p2 )2 − m2 (p1 − q2 )2 − m2 ~λ2 m2 (1 − cos(θ)) = =i , (6.65) s + m2 + (s − m2 ) cos(θ) where a = 2s/(s + m2 ), b = (s − m2 )/(s + m2 ), and θ is the angle between ~ep and ~eq . At large energies, b is close to one and the angular distribution is strongly peaked in the backwards direction, θ ∼ π ; at lower energies if

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flattens out. We can also see that in the strictly forward direction, θ = 0, the amplitude vanishes. The flux factor and the phase space integration element read Φσ =

1 1 s − m2 , dV (p + p ; q , q ) = d cos(θ) dϕ , (6.66) 1 2 1 2 2(s − m2 ) 32π 2 s

with ϕ the azimuthal angle, on which M does not depend. The differential cross section thus becomes ~2 λ4 m4 dσ = 64π 2 s



1 − cos(θ) (s + m2 ) + (s − m2 ) cos(θ)

2 d cos(θ) dϕ .

(6.67)

The angular integral readily performed, and we find the total cross section ~2 λ4 m4 σ(s) = 16πs(s − m2 )2

 1−

  s  s 2s log + 2 . s − m2 m2 m

(6.68)

This expression would appear to diverge at the threshold, s = m2 , but by E42 carefully expanding in powers of s − m2 we find σ(m2 ) =

~2 λ4 m4 , 48πm2

(6.69)

which can be considered the cross section for static scattering. Let us consider this expression more closely. In the first place, the factor λ4 and consequently the factor ~2 could have been foreseen from the start. The fact that the cross section must have dim[σ] = dim[L2 ] implies that at the threshold, where m is the only length scale in the problem, there must also be an overall factor 1/m2 . Moreover, n body phase space contains a power π 4−3n from its definition ; and also it contains n − 1 solid angles to be integrated over, each giving rise to30 a factor π. This means that the total cross section for an E43 n-body final state will contain a factor π 3−2n . In this way, almost the whole E44 cross section formula is determined, and all the calculational effort is only used to find the numerical factor 1/48. E45 30

This is to say that the angular integral does not necessarily evaluate to π, but rather that a factor π invariably arises in the result of a solid-angle integral.

172

6.6 6.6.1

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The one-loop cookbook The one-loop calculation

In the computation of loop corrections, many more technicalities and tricks are required ; here we present a first introduction to them. As a working example we consider the diagram in ϕ3 theory : 2

L(p ) =

k

p

p

(6.70) p+k

The momentum flowing through the external propagators is pµ . The diagram, which excludes the external propagators, is given by Z 1 1 2 L(p ) = . (6.71) d4 k 2 4 2 2(2π) (k − m + i)((k + p)2 − m2 + i) We have dropped the coupling constants, but included the symmetry factor 1/2. We start by combining the two denominators by using the so-called Feynman trick, which we shall now discuss. The Feynman trick Consider n positive real numbers aj and real numbers νj (j = 1..n). We can write !   Z∞ Y n n Y 1 dzj zj νj −1 n = exp − Σj=1 zj aj . (6.72) a νj Γ(νj ) j=1 j=1 j 0

In this integral, we may define σ as the sum of the z’s, and define xj as zj /σ, as follows: !   Z∞ Y n n Y dzj zj νj −1 1 = exp − Σj zj aj νj a Γ(ν ) j j j=1 j=1 0 !   Y  zj  dσ δ σ − Σj zj dxj δ xj − . (6.73) σ j

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We can now eliminate the z’s in favor of the x’s, and integrate out σ :    Z∞ Y n n  Y 1 dxj xj νj −1 δ 1 − Σj xj = a νj Γ(νj ) j=1 j j=1 0

Z∞ ×

dσ σ

P −1+ j νj

  exp − σ Σj xj aj

0

P

Γ = Q

j νj



j Γ(νj )

Q

Z1

P

j

 0

(dxj xj νj −1 ) δ(1 −

j

x 1 a1 + x 2 a2 + · · · + x n an

xj ) Pj νj . (6.74)

In our example we have n = 2 and ν1 = ν2 = 1 : 1 = a1 a2

Z1 0

dx xa1 + (1 − x)a2


2 .

(6.75)

Incidentally we now also have a proof of the so-called Euler’s integral : Q Z1 Y n

j Γ(νj ) νj −1 , (6.76) dxj xj δ − 1 + Σ j xj = P Γ( ν ) j j j=1 0

by simply putting a1 = a2 = · · · = an = 1. Loop momentum shift After the Feynman trick, the loop integral (6.71) is given by 1 L(p2 ) = 2(2π)4

Z1

Z dx

d4 k k 2 + 2x(pk) + xp2 − m2 + iη


−2

,

(6.77)

0

where we have switched the notation for the infinitesimal width from  to η. Replacing the loop integral momentum k by k − xp we obtain 1 L(s) = 2(2π)4

Z1

Z dx

d4 k k 2 + x(1 − x)s − m2 + iη


−2

,

s = p2 ,

0

(6.78) and we see explicitly that the loop integral in this case depends only on s (and m).

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Dimensional regularization The k integral, containing as many positive powers of k as negative ones, is logarithmically divergent, and we have to regularize it somehow. The most popular method of doing so is nowadays that of dimensional regularization31 . We change the number of dimensions of spacetime from 4 to D = 4 − 2 where we keep  finite up to the end of the calculation, and only then (after renormalization, say) take the limit  → 0. The k integral is then no longer d4 k but rather dD k. However, since this changes the engineering dimension of the diagram this would lead to loop-corrected amplitudes with inconsistent units ; therefore we introduce a quantity µ with the dimension of k 32 , and write µ2 L(s) = 2(2π)4−2

Z1

Z dx

dD k k 2 + x(1 − x)s − m2 + iη


−2

(6.79)

0

It is important to realise the flexibility that we allow ourselves. If we interpret  as a real number then choosing  > 0 will make the above integral finite. In other situations (cf section 11.1)  > 0 mihgt be more ‘natural’. The power of dimensional regularization is that we may assign these properties to  in any way we like ; dimensional regularization treats the integral’s dimensionality as a complex number, and the integral amenable to analytic continuation. A particularly striking instance of dimensional doublethink is found in exercise 46.

The Wick rotation Now we let the iη come into its own. The k integral can be written as dD k = dk 0 dD−1~k, and in terms of k 0 the integrand skirts along a singularity if (k 0 )2 ∼ ~k 2 − sx(1 − x) + m2 . we see that we can change the integration contour from s(−∞, +∞) to (−i∞, +i∞) by a counterclockwise rotation. That is, we may write dk 0 = idk 4 and let k 4 run over the real axis. This (a) gives an overall factor of +i and (b) changes every occurrence of k 2 into 31

This is mainly due to the fact that the dimensionality of a theory usually has only a small influence on properties such as causality or current conservation. 32 That is, inverse length. Loosely speaking, an energy scale.

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−kE2 with kE2 = (k 4 )2 + |~k|2 . Our integral therefore becomes iµ2 L(s) = 2(2π)4−2

Z1

Z dx

dD k kE2 + m2 − x(1 − x)s − iη


−2

.

(6.80)

0

The t-shell formula The integrand in Eq.(6.80) depends on k µ only through kE2 . It is therefore sensible to first compute Z R(D; t) = dD k δ(kE2 − t) . (6.81) Since we want to do this for possibly non-integer D this is on necessity somewhat symbolic. We write R(D; t) = 2D

Z∞


dk 1 dk 2 · · · dk D δ (k 1 )2 + (k 2 )2 + · · · + (k D )2 − t

0

= t

D/2−1

Z∞

dy1 · · · dyD y1 −1/2 · · · yD −1/2 δ(y1 + · · · + yD − 1)

0

= tD/2−1

Γ(1/2)D . Γ(D/2)

(6.82)

√ To get this result we have written k j = yj t and used Euler’s formula, Eq.(6.76). Our loop integral thus becomes, with D = 4 − 2 : iµ2 L(s) = 2(4π)2− Γ(2 − )

Z∞

Z1 dx 0

dt 0

t1− . (t + m2 − x(1 − x)s − iη)2

(6.83)

The t integral By writing t = a(1 − y)/y and using again Eq.(6.76) we can compute the t integral : Z∞ 0

tα dt = a1+α−β (t + a)β

Z1

dy y β−α−2 (1−y)α = a1+α−β

Γ(β − α − 1)Γ(α + 1) . Γ(β)

0

(6.84)

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August 31, 2019

Thus we get, in our example, i µ2 Γ() L(s) = 2(4π)2−

Z1

dx m2 − x(1 − x)s − iη


−

.

(6.85)

0

Incidentally, when a vanishes so does the t integral : we say that Z∞

dt t1− tn = 0 ,

(6.86)

0

E46

simply because the integral has a dimensionality but there is no quantity available to carry the appropriate power. The  expansion Keeping in mind that at some point we want to take  → 0 it is sensible to expand in relevant powers of . First, we use Eq.(15.274),
1 Γ() = (1 − γE  + O 2 ) , 

(6.87)

where γE ≈ 0.5772 is Euler’s constant33 , to see that our integral diverges as −1 whern  vanishes. This implies that in all other factors we must keep the first-order terms but may discard higher orders34 . For the other factors we may use
A = exp( log(A)) = 1 +  log(A) + O 2 , (6.88) and putting everything together we arrive at the final, intermediate result :
i 2 R − R(m , s) + O () ,  2(4π)2 1 R = − γE + log(4π) + log(µ2 ) ,  Z1 R(m2 , s) = dx log(m2 − x(1 − x)s − iη) . L(s) =

(6.89)

0 33

Lots and lots of mathematical things are called after Euler. To spread the credit somewhat, it is also called the Euler-Mascheroni constant. See also appendix 15.15.4 34 Higher-loop integrals typically diverge with higher powers of 1/, and figuring out what to keep becomes accordingly more complicated.

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So far the computation has been quite straightforward, in that many of its steps are the same for all one-loop calculations. In particular we can see that 1/ always occurs with −γE + log(4π) + log(µ2 ) in the combination R . The x integral : the hard work In the evaluation of the x integral, it of course becomes important to keep careful track of the logarithm’s singularity structure. We can distinguish three cases, depending on the value of s = p2 . 1. s < 0. In this case m2 − sx(1 p − x) is always positive. Writing x = (1 + y)/2 we have, with β = 1 − 4m2 /s (so that β > 1) : R(m2 , s) =

Z1

dx log(m2 + |s|x − |s|x2 )

0

1 = 2

Z1

dy log(m2 + |s|/4 − |s|y 2 /4)

−1

Z1

    |s| 2 2 = dy log + log(β − y ) 4 0   β+1 2 = log(m ) − 2 + β log . β−1

(6.90)

Limiting cases are R(m2 , 0) = log(m2 ) and R(0, s) = −2 + log |s|. 2. 0 < p s < 4m2 . Proceeding in a similar way as above, but now with η = 4m2 /s − 1, we can write Z1

2

R(m , s) = log(s/4) +

dy log(y 2 + η 2 )

0

= log(m2 ) − 2 + 2η arctan(1/η) .

(6.91)

3. s > 4m2 . This is the more tricky case since the argument crosses zero twice as xpmoves between 0 and 1. The two roots are given as x0,1 , with β = 1 − 4m2 /s (but now 0 < β < 1) : x1 = (1 + β)/2 + iη ,

x0 = 1 − x1 = (1 − β)/2 − iη .

(6.92)

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August 31, 2019

Figure 6.1: The function R(m2 , s) of Eq.(6.89)

The iη becomes important when taking logarithms : log(−x1 ) = log(x1 ) − iπ ,

log(−x0 ) = log(x0 ) + iπ .

(6.93)

We evaluate the integral as follows : R(m2 , s) = log(s) +

Z1


dx log(x − x0 ) + log(x − x1 )

0



1

= log(s) + (x − x0 ) log(x − x0 ) + (x − x1 ) log(x − x1 ) − 2x

0 = log(s) + x1 log(x1 ) + log(−x1 ) + x0 log(x0 ) + log(−x0 ) − 2 = log(s) + 2x1 log(x1 ) + 2x0 log(x0 ) + iπ(x0 − x1 )     1+β 2 = log(m ) − 2 + β log − iπ . (6.94) 1−β The form of R(m2 , s) in the three different regions can be seen as analytic continuations of each other, but of course precisely how to analytically continue is the tricky point here ! In figure 6.1 we plot the function R(m2 , s) for

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179

m = 3.14. It is continuous both at s = 0 and s = 4m2 . The real part of R is positive (since log(3.142 ) > 2), its imaginary part is negative. At s = 4m2 the imaginary part ‘switches on’ suddenly. This indicates that R(s) has a cut along the real axis starting at that value ; and of course the ‘kink’ in the real part tells us the same. We also see that the real part of L(p2 ) only develops above the ‘threshold’ value p2 = 4m2 , and is positive there, as required by unitarity.

6.6.2

Dispersion relations

Kramers and Kronig rule! We can place the above treatment into context by considering it as part of an amplitude. Since the momentum p cannot possibly be on-shell for all values of p2 , we consider the diagram of the FEE model Mf =

(6.95)

This diagram is therefore given by35 Mf =

~2 λ2 Σ(s) , (s − M 2 )2

Σ(s) =


iλ2 R − R(m2 , s) , 2 16π

(6.96)

where we recognize M as the F mass, and m that of the E particle. We take the final state to be identical to the initial state36 . In that case, R(m2 , s) has negative imaginary part, and the real part of M is negative as required by the optical theorem of Eq.(6.34). Next, we concentrate on the imaginary part of R(m2 , s), for which we had found p (6.97) =R(m2 , s) = −π β θ(s > 4m2 ) , β = 1 − 4m2 /s . By the Kramers-Kronig relations of appendix 15.15.9 we can relate this to its real part by an integral : p Z∞ Z∞ 1 − 4m2 /z 1 =R(m2 , z) 2 0, p · p = m2 , p · s = 0 and s · s = −1. Excercise 56 Building a Dirac spinor Let ξ be an arbitrary spinorial object (i.e. a four-component column). Show that (/p + m)(1 + γ 5 s/)ξ is proportional to u(p, s). Excercise 57 Spin basis for given momentum Consider the two spinors u(p, s) and u(p, −s). Show that any other spinor with momentum p can be written as u(p, s0 ) = a+ u(p, s) + a− u(p, −s) for some coefficients a± , and write a± as spinor products. Excercise 58 Rotating over 60 degrees and so on p µ = xµ ,

q µ = cos(θ) xµ + sin(θ) y µ

Compute Σ(p → q) for θ = π/3, and from there, by taking powers, for θ = 2π/3 and finally for θ = 2π. Excercise 59 Dirac FEE model We reconsider the FEE model,but this time the E are Dirac particles. This

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227

means that we have not only E particles but also E¯ particles. There is one vertex :

E E

i

F ↔ ~λ

where λ is a dimensionless coupling constant. As before, F has mass M and E has mass m. 1. There are now 3 diagrammatic SDe’s for this model. Write them out. 2. Assume M > 2m. Compute the decay width Γ(F → EE) at the tree level. 3. Assume 0 < M < 2m. ¯ 2 ) → E(q1 ) E(q ¯ 2) (a) Write the tree diagrams for the process E(p1 ) E(p by F exchange. We have indicated the momenta of the E particles. (b) Compute h|M|2 i for this process, where we sum over the finalstate spins and average over the initial-state spins of the E’s. ¯ as a function of (c) Compute the total cross section σ(E E¯ → E E), the total invariant mass-squared s ≡ (p1 + p2 )2 . This is best done in the centre-of-mass frame. 4. In the above process, assume now M = m = 0 (and still Γ = 0). Determine h|M|2 i, using momentum conservation to simplify the result as much as possible. 5. Assume M = 0 and m > 0. (a) Write down the 2 diagrams for E(p1 ) F (k1 ) → E(p2 ) F (k2 ) at the tree level. (b) Work out h|M|2 i for this process. (c) Compute the total cross section as a function of s = (p1 + k1 )2 . This is best done in the centre-of-mass frame. (d) Compute the total cross section in the limit s → m2 . 6. Make no assumption on m or M .

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August 31, 2019 (a) Write down the 24 diagrams for EF → EF at one loop. (b) Assume that there are also F F F and F F F F vertices. i. Write the 3 diagrams for EF → EF at the tree level. ii. Write the 13 diagrams for F → EE at the one-loop level. iii. Determine the number of Feynman diagrams at the tree level for the following processes: ¯ A. E E¯ → E EF ¯ F B. E E¯ → E EF C. E E¯ → EE E¯ E¯ D. E E¯ → EEE E¯ E¯ E¯

Excercise 60 Dirac FED model We consider a variation on the FEE model: there are now 2 types of Dirac particles, E and D, with masses mE and mD , respectively. The F particle is massless. There are now 2 vertices : E

D E

F

D

F

where both vertices have Feynman rule iλ/~ as before. We shall assume mE > mD . 1. Compute the decay width Γ(E → F D) at the tree level. ¯ at the tree level, as a function 2. Compute the cross section σ(F F → E E) of the total invariant mass-squared s.

Chapter 8 Helicity techniques for Dirac particles 8.1

The standard form for spinors

In order to perform calculations involving Dirac particles we need to develop computational techniques that are better1 than the Casimir trick. This chapter discusses how we can compute things like amplitudes without invoking explicit representations of Dirac matrices. This goes by the general appellation of helicity methods, although chirality methods might be more appropriate2 .

8.1.1

Opting for helicities, opting for antisymmetry

The case of massless Dirac particles is of special interest, both since calculations typically become easier if the masses vanish and because in high-energy interactions the masses can often be neglected as a good approximation3 . Therefore, helicity states are the theorist’s favoured choice in such calculations. As we have seen before, the complex phase of spinors is not well defined, and we can make use of this freedom to endow spinor products with special symmetry properties. Let us start by realizing that, for massless p and q, we can write the helicity spinors (possibly with a suitable normalization 1

Or at least faster ! I think that this really is because ‘chirality’ pronounces less fluenty than ‘helicity’. 3 At the LHC, electrons are almost perfectly massless. 2

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factor) u± (p) ∼ p/ ξ∓ ,

ξ∓ = ω∓ ξ∓ ,

(8.1)

where the helicity spinors ξ∓ can be chosen almost4 arbitrarily. Note that this is a phase convention ! Then, the spinor product can be written in the form s+ (p, q) = u+ (p) u− (q) ∼ ξ¯− p/q/ ξ+ . (8.2) You might try to find a spinor product that is symmetric in p ↔ q, but that is useless since it would give 1 1 s+ (p, q) = (s+ (p, q) + s+ (q, p)) ∼ ξ¯− (/pq/ + q/p/) ξ+ = (p · q) ξ¯− ξ+ , (8.3) 2 2 and such a spinor product is simply the vector product. The alternative is to choose the spinor product antisymmetric, so that s+ (p, q) + s+ (q, p) = 0, and we see that this requires ξ¯− ξ+ = 0 : the helicity spinors ξ± must be orthogonal. Then we shall have s+ (p, q) = −s+ (q, p) = −s− (p, q)∗ = s− (q, p)∗ .

8.1.2

(8.4)

The standard form for helicity spinors

Let us flesh out the discussion of the previous section by introducing what we shall call the standard form for massless spinors. Let pµ be the momentum of the spinor, so that p2 = 0. We now choose two basis vectors k0µ and k1µ , which satisfy k0 · k0 = k0 · k1 = 0 , k1 · k1 = −1 . (8.5) Furthermore we require that k0 ·p is nonzero5 for any massless momentum pµ encountered in the problem at hand ; this is usually not difficult to arrange. Since k 0 is massless, it may serve to define the basis spinor u0 ≡ u− (k0 )

⇒

u0 u0 = ω− k/0 .

(8.6)

We can then also define6 u+ (k0 ) = k/1 u0 , 4

(8.7)

With extremely bad luck you might find, say, a ξ+ that will give a vanishing product. If so, simply pick another one ! 5 And preferably not too small, for your favourite definition of ‘too small’. 6 Be careful here ! / k1 u− (k0 ) = u+ (k0 ) but /k1 u+ (k0 ) = −u− (k0 ).

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and easily verify that u− (k0 )u+ (k0 ) = 0. Reversal ‘flips’ the helicity : (u− (k0 )u− (k0 ))R = (ω− k/0 )R = k/0 ω− = ω+ k/0 = u+ (k0 )u+ (k0 ) .

(8.8)

Using the basis spinor, we now define all other massless spinors by u± (p) = √

1 p/ u∓ (k0 ) . 2p · k0

(8.9)

This is by construction a valid form since u± (p)u± (p) =

1 1 p/ ω∓ k/0 p/ = ω± p/k/0 p/ = ω± p/ . 2p · k0 2p · k0

(8.10)

This choice is at the basis of the so-called spinor techniques, or helicity methods. we shall use Eq.(8.9) extensively to good effect.

8.1.3

Some useful identities

At this point we prove a few results that often
turn out to be useful. In the first place, from the property Tr (Γ) = Tr ΓR , we can see that u+ (p1 ) γ µ u+ (p2 ) = K u0 p/1 γ µ p/2 u0 = K Tr (u0 u0 p/1 γ µ p/2 )
= K Tr p/2 γ µ p/1 (u0 u0 )R = K Tr (/p2 γ µ p/1 k/1 u0 u0 k/1 ) = K u0 k/1 p/2 γ µ p/1 k/1 u0 , (8.11) with K = (4p1 · k0 p2 · k0 )−1/2 , which leads to the useful spinor reversal : u+ (p1 )γ µ u+ (p2 ) = u− (p2 )γ µ u− (p1 ) .

(8.12)

In fact this can easily be generalized to uλ1 Γ uλ2 (q) = λ1 λ2 u−λ2 (q) ΓR u−λ1 (p) ,

λ1,2 = ± .

(8.13)

The Chisholm identity of Eq.(7.41) can be applied to simple spinor sandwiches so as to yield     µ u± (p1 )γ u± (p2 ) γµ = 2 u± (p2 )u± (p1 ) + u∓ (p1 )u∓ (p2 ) . (8.14) Finally, we have the so-called Schouten identity. It is based on the truism u+ (p1 )/p2 p/3 u− (p4 ) + u+ (p1 )/p3 p/2 u− (p4 ) = 2(p2 · p3 ) u+ (p1 )u− (p4 ) .

(8.15)

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Writing this out in terms of spinor products, we have s+ (p1 , p2 )s− (p2 , p3 )s+ (p3 , p4 ) + s+ (p1 , p3 )s− (p3 , p2 )s+ (p2 , p4 ) = s+ (p2 , p3 )s− (p3 , p2 )s+ (p1 , p4 ) .

(8.16)

Using the antisymmetry property of s, and dividing out the factor s− (p2 , p3 ), we obtain the so-called Schouten identity : s+ (p1 , p2 )s+ (p3 , p4 ) + s+ (p1 , p3 )s+ (p4 , p2 ) + s+ (p1 , p4 )s+ (p2 , p3 ) = 0 . (8.17) E61

Note the cyclicity in p2,3,4 . Obviously, the identity holds for s− as well. At the end of this chapter we give a summary table of all these spinor techniques.

8.1.4

How to compute spinor products

To evaluate amplitudes numerically we need a recipe to compute spinor products numerically, just like the recipe to compute vector products. For spinors in the standard form we can use the Casimir trick : s+ (p, q) = (4(p · k0 )(q · k0 ))−1/2 u0 p/ q/ k/1 u0 = (4(p · k0 )(q · k0 ))−1/2 Tr (ω− k/0 p/q/k/1 )  1 = p (p · k0 )(q · k1 ) − (p · k1 )(q · k0 ) (p · k0 )(q · k0 )  µ ν α β − iµναβ k0 k1 p q . (8.18) You can easily check the spinor products’ properties by explicit inspection. From |sλ (p, q)|2 = 2(pq) we can see that spinor products are square roots of vector products. Actually, in practical applications it pays to compute the spinor products of massless momenta in your problem anyway. As an example, if two massless momenta pµ and q µ make a small angle θ, their vector product reads (pq) = pp q 0 (1 − cos θ) ∼ θ2 . For their spinor product, on the other hand, we have |s+ (p, q)| ∼ θ so the numerical cancellations are much less severe in te spinor product than in the vector one. The vector product can then be obtained as a simple square. µ We may consider an explicit choice for the vectors k0,1 : k0 µ = (1, 1, 0, 0) ,

k1 µ = (0, 0, 1, 0) :

(8.19)

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this gives the explicit form for the spinor product s s 0 − q1

q p0 − p1 2 3 s+ (p, q) = p2 + ip3 − q + iq , p0 − p1 q0 − q1

(8.20)

which is very useful for actual numerical applications. Note that this choice E62 presupposes that none of the light-like vectors in the problem is oriented E63 exactly along the x-axis. Since the ‘special’ direction in many problems is traditionally chosen to be the z-axis, this is usually safe. E64 E65

8.1.5

The standard form for massive particles

The standard form for Dirac spinors given in Eq.(8.9) can be generalised to the case of massive particles. Let pµ be the momentum of such a particle, and let m be its mass. We then simply define 1 (/p + m)u∓ (k0 ) , 2p · k0 1 (/p − m)u∓ (k0 ) . v± (p) = √ 2p · k0

u± (p) = √

(8.21)

We can find out the spin vector for these two cases : writing u± (p) = u(p, ±s0 ) we obtain 1 u+ (p) γ 5 γ µ u+ (p) 2m
1 Tr ω− k/0 (/p + m)γ 5 γ µ (/p + m) = − 4m p · k0 1 µ m = p − k0 µ , m (pk0 )

s0 µ = −

(8.22)

which is indeed the only vector built from p and k0 that can have the right properties s0 2 = −1 and (ps0 ) = 0. Note that for small(ish) m and generally positioned k0 , ~s0 points in the general direction of p~. Therefore we call u+ (p) a right-handed spinor, and u− (p) a left-handed spinor. In addition, from the fact that, for the antispinor v± (p), 1 v + (p) γ 5 γ µ v+ (p) = −s0 µ , 2m

(8.23)

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we see that v+ (p) is a left-handed antispinor and v− (p) is a right-handed antispinor7 .

8.1.6

The standard form for complex momenta

We have seen that for a helicity spinor u+ there is a momentum q such that u+ u+ = ω+ q/, and that this momentum is real and massless, with positive energy. This is, of course, just what the actual momentum of a massless particle is supposed to look like. But in the course of actually computing amplitudes it frequently happens that we would like to have spinors for momenta p that have negative energy, or are complex-valued8 . The question is then : can we find two spinors u and w such that uw = ω+ p/ ? Obviously, now w 6= u but that doesn’t matter as long as we don’t need to suddenly conjugate a spinor inside an amplitude9 . Note that uw† = ω+ p/γ 0 has precisely the form of the dyad Θ of sect.7.3.2, and therefore the restriction p2 = 0 holds. The solution to the task we set ourselves above is actually almost trivial : we use the same basis spinor u0 as before, and the same (real-valued !) vectors k0 and k1 . We then simply define u+ (p) = N p/u0 , w+ (p) = N u0 p/ ,

(8.24)

p with N = 1/ 2(p · k0 ). Note that N is now no longer purely real ; but as long as we take it equal in the definition of u and w it is directly seen that they do the trick. Of course, for the other helicity we can use u− (p) = N p/k/1 u0 , w− (p) = N u0 k/1 p/ .

(8.25)

Remember the ‘not too small’ in
sect.8.1.2 ? In a frame where ~k0 and p~ are parallel the dot product (k0 p) is of order O m2 and the ‘subleading’ term with k0 in Eq.(8.22) is no longer subleading. Whereas normally the spinor u+ (p) looks a lot like a positive-chirality spinor u+ for small m, it then looks more like u− : a continuous source of worry especially in large computer programs that deal with many fermions numerically. 8 In cases like this, you will usually hear your lecturer going ‘mumble mumble analytic continuation mumble mumble’ but that is unhelpful unless you know how to analytically continue. This section, in fact, explains how. 9 I have never encountered this need. Of course, the complex conjugate amplitude occurs in the cross section, but that can be accounted for by replacing every u by w and vice versa, if you really insist on computing the complex conjugate amplitude separately (why would you ?). 7

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We have now found the dyadic forms we wanted. In what way do they differ from the Dirac case ? The spinor products s+ (p, q) = w+ (p)u− (q) ,

s− (p, q) = w− (p)u+ (q)

(8.26)

are still antisymmetric, but we no longer have s− = −(s+ )∗ . On the other hand, it is still true that s+ (p, q)s− (q, p) = 2(p · q) ;

(8.27)

and, in fact, it is easily checked that all the other relations such as the reversal formula, the Chisholm identity, and the Schouten identity still hold.

8.2

Summary of tools for spinor techniques

Here we present the definitions and identities useful for performing calculations with massless spinors. • Dirac equations : p/ uλ (p) = 0 ,

ωλ uλ (p) = uλ (p) ,

ω−λ uλ (p) = 0

(8.28)

• projecting operator : uλ (p) uλ (p) = ωλ p/

(8.29)

• Spinor products : uλ (p) uλ (q) = 0 sλ (p, q) = uλ (p) u−λ (q) = −sλ (q, p) = −s−λ (p, q)∗ |sλ (p, q)|2 = sλ (p, q) s−λ (q, p) = 2(p · q)

(8.30)

• Elimination of repeated indices10 : γα uλ (p) uλ (q) γ α = −2 u−λ (q) u−λ (p) γα uλ (p) u−λ (q) γ α = 2 s−λ (q, p) ω−λ 10

This is also valid for p = q.

(8.31)

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• Chisholm identity :     uλ (p) γα uλ (q) γ α = 2 uλ (q) uλ (p) + u−λ (p) u−λ (q)

(8.32)

• Reversal : uλ1 (p) Γ uλ2 (q) = λ1 λ2 u−λ2 (q) ΓR u−λ1 (p)

(8.33)

• Schouten identity : sλ (q, p1 ) sλ (p2 , p3 ) + sλ (q, p2 ) sλ (p3 , p1 ) + sλ (q, p3 ) sλ (p1 , p2 ) = 0 (8.34)

8.3

Fermionic decays : the Fermi model

After all the mathematics it is high time to do some actual phenomenology. Up to now all interacting theories have been toys to acquaint ourselves with Feynman rules and the like, but our goal ought of course to be to be able to say something about real, observable processes. Fortunately, there are a few processes that can be formulated (almost) purely in terms of Dirac particles.

8.3.1

The amplitude for muon decay

An example of an actually occurring process involving only Dirac particles is provided by muon decay in the Fermi model. The process is11 µ− (p)

→

e− (q) νµ (k1 ) ν e (k2 )

and is pictured by the single Feynman diagram k1

p

M=

q

(8.35)

k2

Here, a muon at rest undergoes a three-particle decay into an electron, a muon neutrino and an electron antineutrino. We shall assume the neutrinos 11

In this section, the vector k1 is a momentum, and has nothing to do with the auxiliary vector of section 8.1.

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to be massless. The Fermi amplitude introduced to describe the phenomenology of this process contains only a single pointlike vertex where four fermions √ meet with a coupling constant called GF / 2, and is given by GF ~ M=i √ 2

u(q) (1 + γ 5 )γα v(k2 )

u(k1 ) (1 + γ 5 )γ α u(p) .

(8.36)

The decision to ‘hook up’ the muon and the muon neutrino is in principle arbitrary12 , but as we have seen in section 7.2.7 we may easily interchange the muon neutrino and the electron, and end up with the matrix element in the charge retention form : GF ~ M = −i √ 2

u(q) (1 + γ 5 )γα u(p)

u(k1 ) (1 + γ 5 )γ α v(k2 ) .

(8.37)

The amplitude (8.36) implies that the neutrinos must have negative helicity13 : we can write 4GF ~ M=i √ 2

u(q) γα v− (k2 ) u− (k1 ) γ α u(p) .

(8.38)

By eliminating the repeated indices (cf sect.8.2) we can arrive at the very compact form 8GF ~ M = −i √ u(q) u+ (k1 ) v + (k2 ) u(p) . 2

(8.39)

The transition rate can now easily be computed with a few simple traces :

 1 X |M|2 = 16 GF 2 ~2 Tr ((/q + me )ω+ k/1 ) Tr ((/p + mµ )ω+ k/2 ) |M|2 = 2 = 64 G2F ~2 (q · k1 ) (p · k2 ) . (8.40) Here, the sum is over the electron and muon spins. It is practical to evaluate this in the muon rest frame. We shall write E1,2 for k1,2 0 in this frame. Then (p · k2 ) is equal to mµ E2 , and by momentum conservation we find (q · k1 ) = 12


1
1 (q + k1 )2 − me 2 = (P − k2 )2 − me 2 = mµ (K − E2 ) , 2 2 (8.41)

Unless lepton flavour number is invoked. In the standard form of spinors, the helicity for antispinors is reversed. The antineutrino therefore actually has positive chirality. 13

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where

mµ 2 − me 2 . 2mµ The transition rate then takes the form

 |M|2 = 64 GF 2 ~2 mµ 2 E2 (K − E2 ) , K=

(8.42)

(8.43)

and for the partial decay width we find dΓ(µ → eνµ ν e ) = 32 GF 2 ~2 mµ E2 (K − E2 ) dV (p; q, k1 , k2 ) .

8.3.2

(8.44)

Three-body phase space

The phase space for the muon decay process reads 1 dV (p; q, k1 , k2 ) = d4 q d4 k1 d4 k2 δ 4 (p − q − k1 − k2 ) (2π)5 δ(q 2 − me 2 ) δ(k1 2 ) δ(k2 2 ) .

(8.45)

Since the rate depends only on E2 , we shall implicitly integrate over all other phase space variables. By cancelling the q integration against the Dirac delta for momentum conservation, we arrive at 1 E1 E2 dE1 dE2 dΩ1 dΩ2 δ((p − k1 − k2 )2 − me 2 ) . 5 (2π) 4 (8.46) The Dirac delta function can be written as
δ mµ 2 − me 2 − 2mµ E1 − 2mµ E2 + 2E1 E2 − 2E1 E2 cos θ , dV (p; q, k1 , k2 ) =

where θ is the angle between the neutrino momenta. Hence we can integrate trivially over the other polar, and the two azimuthal, angles (leading to a factor 8π 2 ), and the integral over θ is resolved by the delta function. The result is π2 dV (p; q, k1 , k2 ) = dE1 dE2 . (8.47) (2π)5 In terms of these variables, the phase space is perfectly flat14 . Since |cos θ| cannot exceed unity, we also have the restrictions mµ 2 − me 2 − 2mµ E1 − 2mµ E2 ≤ 0 , 14

This flatness does not depend on the masslessness of the neutrinos. For massive neutrinos the same phase space density is found, only the boundaries of the phase space become (horribly) complicated.

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Figure 8.1: Elecrton-mass dependence of the muon decay width

mµ 2 − me 2 − 2mµ E1 − 2mµ E2 + 4E1 E2 ≥ 0 ,

(8.48)

which we can work into bounds on E1 : 2 2 ˆ 2 ) ≡ mµ − me − 2mµ E2 , K − E2 ≤ E1 ≤ K(E 2(mµ − 2E2 )

(8.49)

while E2 is seen to run from 0 to K.

8.3.3

The muon decay width

After the simple integration over E1 , we have the muon partial decay width d ˆ 2 ) + E2 − K) . (8.50) Γ(µ → eνµ ν e ) = π 2 GF 2 ~2 mµ E2 (K − E2 )(K(E dE2 The remaining integral over E2 can now be performed, and the final result is GF 2 ~2 mµ 5 F (me 2 /mµ 2 ) , 192 π 3 F (x) = 1 − 8x + 8x3 − x4 − 12x2 log(x) .

Γ(µ → eνµ ν e ) =

(8.51)

The function F (x), shown in figure 8.1, is strictly decreasing since with increasing me /mµ the available phase space shrinks. For the realistic values

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of me and mµ F (x) is smaller than 1 by about 2 × 10−4 . The effects of nonzero electron mass are therefore completely negligible, certainly if we realize that we have not included any loop diagrams, the contribution of which is very considerably larger than this. Before finishing, it is instructive to inspect the muon width formula

E67

GF 2 ~2 mµ 5 Γ(µ → eνµ ν e ) = 192 π 3 from the point of view of dimensional analysis. In the first place, the matrix element M, being of 2 → 2 type, must be strictly dimensionless (see Eq.(6.23)). Since every spinor carries half a power of momentum15 , the Fermi coupling constant GF must carry dimension momentum−2 . Since decay widths carry the dimension of momentum, as do masses like mµ , and the only mass scale in the problem is mµ if we neglect the electron mass, the width is necessarily proportional to GF 2 mµ 5 . The discussion at the end of section 6.5.4 shows that the factor 1/π 3 was also to be expected. It is a somewhat sobering thought that all the work of this section amounts to no more than computing the number 1/192 !

8.3.4

Observable distributions in muon decay

We can look more closely into the behavior of the electron in muon decay, in particular its angular and energy distribution : in fact, these are the only quantities of interest since the neutrinos can essentially never be detected. If the sample of decaying muons is unpolarised, no particular direction is favored in its rest frame, and the electrons may be expected to emerge isotropically. Let us therefore assume that the muons are fully polarised, with polarisation vector sµ = (0, ~s) in their rest frame. Since we are interested in distributions rather than the overall decay rate, we can afford to be sloppy with pre-factors in this computation. Additionally we shall assume me = 0. The matrix element is now given by M ∼ u− (q) γµ v− (k2 ) u− (k1 ) γ µ u(p, s) ∼ u− (q)u+ (k1 ) u+ (k2 )u(p, s) . Squaring this, we obtain
|M|2 ∼ Tr (ω− q/k/1 ) Tr ω+ k/2 (1 + γ 5 s/)(/p + mµ ) 15

Since the spin sum of uu contains /p.

(8.52)

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∼ (q · k1 ) (k2 · p) − mµ (k2 · s) ∼ k20 (mµ − 2k20 ) (1 + cos(θ) cos(θ2 ) + sin(θ) sin(θ2 ) cos(φ2 )) , (8.53) where θ is the angle between ~q and ~s, and θ2 , φ2 are the polar and azimuthal angles of ~k2 with respect to ~q. As we have seen above, the phase space integration element is, up to overall factors, dq 0 dk20 d cos(θ) dφ dφ2 , where φ is ~q ’s azimuthal angle around ~s, while cos(θ2 ) = 1 −

mµ (q 0 + k20 − mµ /2) . q 0 k20

(8.54)

Performing the integral over the (unobservable) angle φ2 , and disregarding the trivial φ dependence, we therefore have    mµ (q 0 + k20 − mµ /2) 0 0 dΓ ∝ k2 (mµ −2k2 ) 1 + cos(θ) 1 − dq 0 dk20 d cos(θ) . q 0 k20 (8.55) The integral over k20 from mµ /2 − q 0 to mµ /2 is straightforward, leading to the following, properly normalised distribution :
1 d2 Γ = y 2 3 − 2y + cos(θ)(1 − 2y) , Γ dy d cos(θ)

y=

2q 0 . mµ

(8.56)

The overall angular distribution is
1 dΓ 1 = 3 − cos(θ) , Γ d cos(θ) 6

(8.57)

and the overall distribution of the electron energy reads 1 dΓ = 2y 2 (3 − 2y) . Γ dy

(8.58)

Let us now consider the parity properties of the decay. Under the parity transform, Eq.(7.149) tells us that ~q → −~q and ~s → ~s, so that effectively cos(θ) → − cos(θ). We see that the muon decay amplitude, which is not

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symmetric in cos(θ), displays parity violation. This is most prominent when the electron has maximal energy :  1 d2 Γ = 1 − cos(θ) . (8.59) Γ dy d cos(θ) y=1 Qualitatively, we can understand this as follows. At y = 1, the two neutrinos recoil in parallel against the electron. Since νµ is left-handed and ν¯e is righthanded (because of the factors 1 + γ 5 in the coupling), their spins cancel, and the spin of the electron must therefore be in the direction of the muon spin by conservation of angular momentum. If the electron were to be emitted in the direction of the muon spin, it would therefore be right-handed, which is again forbidden by the coupling16 . Hence the amplitude must vanish at θ = 0.

8.3.5

Charged pion decay: helicity suppression

Another example of a process that we can now compute is that of pion decay. The charged pion π − is a bound state of a d quark and a u antiquark, which are fermions. These annihilate into a charged lepton ` (e− or µ− ) and the corresponding anti-neutrino, assumed to be massless as before. The process therefore reads π − (p) → `− (q1 ) ν ` (q2 ) where we have indicated the observable momenta. We assume that the interaction vertex is again of Fermi type, so that the lepton and antineutrino couple with the familiar17 (1 + γ 5 )γ µ coupling. But for the quarks it is different : they are certainly not free particles, since they form a bound state, and the best we can do is to assume that they conspire to form some object with a Lorentz index. The only possibility (since the pion is spinless) is that somehow they form something proportional to the pion momentum pµ . The phenomenological amplitude therefore reads √ GF M = i ~ √ fπ pµ u(q1 )(1 + γ 5 )γ µ v(q2 ) . 2

(8.60)

Here the totality of our lack of knowledge about the quarks’ goings-on is embodied in the pion decay constant fπ . We can use the spinors’ properties 16 17

Remember that we took me = 0. By now, I hope.

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243

to simplify u(q1 ) (1 + γ 5 ) p/ v(q2 ) = u(q1 ) (1 + γ 5 ) (/q1 + q/2 ) v(q2 ) = m` u(q1 ) (1 − γ 5 ) v(q2 ) . (8.61) We therefore find


|M|2 = 2~ G2F fπ2 m2` m2π − m2` , (8.62) and for the 2-body phase space we have dV (p; q1 , q2 ) =

m2π − m2` 1 λ1/2 (m2π , m2` , 0) dΩ = dΩ . (2π)2 m2π 8 32π 2 m2π

(8.63)

Consequently, 2

~ G2F fπ2 m2` (m2π − m2` ) Γ(π → ` ν ` ) = 8π m3π −

−

.

(8.64)

This tells us that a muon is much more likely to decay into a muon and its antneutrino than into an electron and its antineutrino :  2 m2µ m2µ Γ(π − → µ− ν µ ) ≈ 2 1− 2 ∼ 7.8 × 103 (8.65) Γ(π − → e− ν e ) me mπ The smallness of the electronic decay mode is thus assigned to the mechanism of helicity suppression. Since the pion has no spin, the spins of the charged fermion and the antineutrino must be opposite. The antineutrino is righthanded and therefore the charged lepton has right-handed helicity. On the other hand, the weak interactions couple only to the left-handed chirality part of the charged lepton. For massless leptons helicity and chirality are the same, and the decay would be impossible. The more massive the lepton E68 is, the larger is the overlap between the right-handed helicity and the lefthanded chirality components.

8.4

Exercises

Excercise 61 Helicity operator for massless spinors Show that, for a general timelike or lightlike momentum pµ , the operator Σp = cos(θ/2) + i sin(θ/2)

1 5 γ [/p, γ 0 ] 2|~p|

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corresponds to a rotation around the p~ axis over an angle θ. For massless pµ , prove that it acts as a helicity probe : Σp uλ (p) = exp(iλθ/2) uλ (p) Excercise 62 Spinor products and vector products Using the form of Eq.(8.20) for s+ (p, q) to prove, by direct computation, that |s+ (p, q)|2 = 2(p · q) Excercise 63 Long Dirac strings Let pµj , (j = 1, 2, 3, . . .) be massless momenta. Write the expression A = u+ (p1 )/p2 p/3 p/4 p/5 u− (p6 ) in terms of spinor products, and compute |A|2 . Excercise 64 Why is it zero ? for massless momenta (as in the previous exercise), let A1 = u(p1 )/p2 p/3 u(p7 ) ,

A2 = u(p1 )/p4 p/5 p/6 u(p7 )

where you can chooce the helicities of p1,7 yourself. Show that in all cases A1 A2 ∗ = 0 Excercise 65 Levi-Civita with spinor currents Let pµ1,2,3,4 be massless vectors. Prove that µναβ u+ (p1 )γα u+ (p2 ) u+ (p3 )γβ u+ (p4 )
= i u+ (p1 )γ µ u+ (p4 ) u+ (p3 )γ ν u+ (p2 ) − (µ ↔ ν) Excercise 66 Towards a real theory prediction We consider the process e+ (p1 ) e− (p2 ) → e+ (q1 ) e− (q2 ) where the electrons/positrons are massless spin-1/2 particles. In Quantum Electrodynamics (QED), the amplitude for this process is given by M(λ1 , λ2 , ρ1 , ρ2 ) =  1 2 i~Qe uλ1 (p1 )γ µ uλ2 (p2 ) uρ2 (q2 )γµ uρ1 (q1 ) 2 (p1 + p2 )  1 µ − uλ (p1 )γ uρ1 (q1 ) uρ2 (q2 )γµ uλ2 (p2 ) (p1 − q1 )2 1

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where (as usual for massless fermions) we have disregarded the distinction between spinors and antispinors. The quantity Qe is a fixed coupling constant. The helicities λ1,2 and ρ1,2 are explicitly indicated. There are, in total, 16 helicity combinations. 1. Using spinor techniques, compute M for the sixteen helicity cases. 2. Assume that we are in the centre-of-mass frame, where p~1 + p~2 = 0. Let θ be the angle between p~1 and ~q1 , and E = p1 0 . Give the simplest form you can find for h|M|2 i. 3. Show that h|M|2 i is really divergent for θ = 0, and that this has nothing to do with neglecting the electron mass. Excercise 67 Muon decay revisited We reconsider muon decay, again with vanishing neutrino and electron mass. We shall investigate the result of leaving out the factors (1 + γ 5 ) in the amplitude, so that we take it to have the form M = i~ G u(k1 ) γ α u(p) u(q) γα v(k2 ) and all fermion can be in two spin states. 1. Compute h|M|2 i, summed over the outgoing fermions’ helicities and averaged over the muon spin. 2. Use the results for three-body kinematics of section 8.3.2 to write the above expressions in terms of q 0 , k2 0 , and mµ . 3. After the angular integrations, as explained in section 8.3.2 there remains the energy integral m Zµ /2

dq 0

0

m Zµ /2

dk2 0

mµ /2−q 0

The electron energy q 0 is observable, but the antineutrino energy k2 0 is not. Integrate over k2 0 to get the normalised electron energy spectrum (dΓ/dq 0 )/Γ.. 4. Compare this spectrum to that of Eq.(8.58) and show how one can experimentally infer the presence of the 1 + γ 5 .

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Excercise 68 It’s different for us who are vectors Let us reexamine pion decay of section 8.3.5. If the decaying particle is not a scalar but a vector particle (as we shall see, that corresponds to unit spin instead of zero spin), the part of the decay amplitude involving the leptons has a vectorial character, for instance u(q1 )(1 + γ 5 )γ µ v(q2 ) Show that in that case there is no helicity suppression.

Chapter 9 Vector particles 9.1 9.1.1

Massive vector particles The propagator

In the last chapter we have studied the consequences of embellishing the scalar propagator by endowing it with a numerator linear in the momentum. The next obvious generalization is to let T (p) depend on two powers of the momentum. That is, we assume it to be of the form T (p) → T (p)µν = Ag µν + Bpµ pν , B 6= 0 , for some A and B that may depend on p2 . The numerator now carries two Lorentz indices, one of each to be contracted with a corresponding index in the vertices between which the propagator runs. The discussion of the last chapter leads us to require that for momenta on the mass shell T (p) must be a projecting operator : T (p)µα T (p)α ν = kT (p)µν if p2 = m2 ,

(9.1)

for some k 6= 0, in other words A2 = kA ,

B 2 m2 + 2AB = kB .

(9.2)

We might choose the solution A = 0, but then the resulting form T (p)µν ∼ pµ pν would be factorizable. It follows that A must equal −m2 B, and therefore we shall use1 1 (9.3) T (p)µν = −g µν + 2 pµ pν . m 1

This choice of T is called the unitary gauge.

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The first Feynman rule for these particles, that we call vector particles since they carry a Lorentz index, is therefore established :

µ

p

↔

ν

i~

−g µν + pµ pν /m2 p2 − m2 + i

internal lines

Feynman rules, version 9.1 Note that this propagator is even in p and therefore has no orientation2 . Another thing, to which we shall refer to often enough, is that the propagator of a vector particle, in contrast to that of a Dirac particle, can to some extent be chosen out of several possibilities. These are denoted gauges. The above choice is called the unitary gauge.

9.1.2

The Feynman rules for external vector particles

From the form of T (p) we must be able to derive the form of the external-line factors. Indeed, let us assume pµ to be in its rest frame, pµ = (m, ~0). There, we have T (p)µν = −g µν + g 0µ g 0ν = diag(0, 1, 1, 1) , (9.4) that is, the unit tensor in the spatial sector of Minkowski space. We see that we can write   T (p)µν = − xµ xν + y µ y ν + z µ z ν

,

(9.5)

which means that, for the objects U, W three mutually orthogonal choices can be made, for instance U (1) = x, U (2) = y, and U (3) = z. Of course, complex linear combinations of these are also possible : in general, we can say that there can be found three polarisation vectors µλ , with λ = −1, 0, 1, such that µ

(λ ) ( )µ = −δ λ0

λ,λ0

, T (p)

µν

=

1 X

ν

(λ )µ (λ ) .

(9.6)

λ=−1

E69

If pµ is not in the rest frame, we simply boost both p and the various  to 2

That is, its spacetime part is unoriented. There may of course be other properties such as charge that do impose a distinction between production and decay of the particle.

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the actual Lorentz frame. We can now go once more through the truncation argument of chapter 6, with the obvious result that the polarisation vectors are to be assigned to the external lines, and we immediately arrive at the full set of Feynman rules for massive vector particles :

µ

p

ν

↔

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

i~

↔

↔

−g µν + pµ pν /m2 p2 − m2 + i

internal lines

√ ~ λ µ

incoming lines

√

outgoing lines

~ λ µ

Feynman rules, version 9.2

(9.7)

Owing to the lack of orientation, the rules for the external lines are quite simple, and fortunately no Dirac indices appear, nor do any curious and cumbersome minus signs. The fact that T (p)µν corresponds precisely to sum of the polarization vectors’ dyads, with no terms left out or left over, justifies the name unitary gauge.

9.1.3

The spin of vector particles

To ascertain the spin of vector particles3 , we need to establish the form of the Lorentz transformation in the space of the polarisation vectors, i.e. Minkowski space. We can do this conveniently using the transform in Clifford space, as follows. Let us denote by Λ(p; q)µ ν the representation of the minimal Lorentz transformation between pµ and q µ in Minkowski space : that is, if an arbitrary vector aµ is transformed into bµ , we have Λ(p; q)µ ν aν = bµ .

(9.8)

Since a / and b/ encode exactly the same information as do aµ and bµ , consistency requires that b/ = Λ(p; q)µ ν aν γµ = Σ/aΣ = Σ aν γν Σ , 3

(9.9)

The fact that there are three polarisation vectors of course suggests that the spin is unity, but again suspicion does not equal proof.

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with Σ as defined in section 7.4.5 ; since this must hold for arbitrary a, we have the relation (9.10) Λ(p; q)µ ν γµ = Σ γν Σ , E70

By multiplying with γ α on both sides and taking the trace, we immediately find the form of Λ(p; q) in Minkowski space :
1
1 Tr Λ(p; q)µ ν γµ γ α = Tr Σ γν Σ γ α 4  4    2 p p/q/ q/p/ = Tr 1 + 2 γν 1 + 2 γ α 4(p + q)2 p p 2 2 = δαν − (p + q)α (p + q)ν + 2 q α pν . 2 (p + q) p

Λ(p; q)α ν =

E71

(9.11)

Let us now specialize to the case of rotations from pµ = xµ , to q µ = cos(θ)xµ + sin(θ)y µ , as in section 7.4.5 : the rotation matrix now reads Λµ ν = (tµ tν − z µ zν ) − cos(θ) (xµ xν + y µ yν ) + sin(θ) (xµ yν − y µ xν ) . (9.12)

E72

In the case of Dirac particles, a rotation over 2π resulted in a sign change : here, such a full rotation is seen to be simply the identity. For infinitesimal θ this reads
Λ(p; q)µ ν = δ µ ν + θ (xµ yν − y µ xν ) + O θ2 , (9.13) so that the generators of the rotation group must in this case have the form (Tz )µ ν = β(xµ yν − y µ xν ) and cyclic permutations ,

(9.14)

with the constant β again to be determined from the commutation algebra : [Tx , Ty ]µ ν = β 2 (xµ yν − y µ xν ) = β(Tz )µ ν .

(9.15)

We conclude that β = i~ in Minkowski space. We find (Tx 2 )µ ν = −~2 (y µ yν + z µ zν ) ,

(9.16)

etcetera, so that the total-spin operator takes the form   1 2 µ 2 µ µ µ 2 µ µ ~ ) = −2~ (x xν + y yν + z zν ) = 2~ (L −δ ν + 2 p pν . (9.17) ν m

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we conclude that the spin is indeed unity. The total spin operator contains, as it must, the projection of all vectors on the spatial subspace. In words: to be a good polarisation vector, µ must be properly normalised and satisfy the Lorenz condition4 :  · ∗ = −1 ,

·p=0 .

(9.18)

Any part of a polarisation vector that is parallel to pµ does, of course, not transform under rotations in the space orthogonal to pµ (in our case, the spatial part of Minkowski space since pµ is at rest). That part, therefore, corresponds to a scalar degree of freedom. Returning to T (p) we may interpret the form 1 (9.19) T (p)µν = −g µν + 2 pµ pν m as a propagator in which a priori four degrees of freedom propagate (the g µν part), and where the scalar part (the pµ pν term) is carefully excised. The pµ pν term is sometimes loosely called the longitudinal part of the propagator, but this is incorrect ; we should do better by calling it the scalar part.

9.1.4

Polarisation vectors for helicity states

As usual, the helicity of a state refers to its spin as measured along the direction of its motion. For definitiveness, let us assume that our massive vector particle moves along the z direction. If we boost carefully5 back to the rest frame, p~ of course vanishes, but we shall remember that to go back to the original situation we must boost along the z direction. The operator for the helicity is therefore Tz in this case. Good polarisation vectors for helicity E73 1,0 and -1 are then 1 + µ = √ (xµ + iy µ ) , 2

0 µ = z µ ,

1 − µ = − √ (xµ − iy µ ) , 2

(9.20)

The vectors ± are said to describe transverse polarisation, and the vector 0 is called longitudinal. If we now perform the boost back to the original system 4

Note the spelling ! This does not refer to the famous Dutchman Hendrik Antoon Lorentz (1853-1928) of transformation fame, but to the Dane Ludvig Valentin Lorenz (1829-1891), quite another person. A relation between the polarizability and the refractive index of a medium goes by the funky name of the Lorentz-Lorenz equation. 5 And minimally !

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in which pµ is moving along the z direction, the transverse polarisations remain unaffected, while the longitudinal one takes the form6     m |~p| µ µ p + zµ . (9.21) 0 → mp0 p0 Very fast-moving particles, for which m  p0 ≈ |~p|, have longitudinal polarisation vector   1 µ m µ 0 → p + O . (9.22) m p0 This is, of course, precisely what we also found for Dirac particles, since the algebraic conditions (9.18) are the same as those of Eq.(7.80).

9.1.5

The Proca equation

Massive vector particles have their own ‘classical’ equation, which we shall now uncover. The coupling of a massive vector particle to a source is given by the following Feynman rule for position space :

µ

i − J µ (x) ~

↔

(9.23)

The SDe for a free vector particle’s field function V µ is then again very simple : 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

x

=

x

,

(9.24)

or, more explicitly, 1 V (x) = (2π)4 µ

Z

e−ik·(x−y) dydk 2 k − m2 4

4

 −g

µν

1 + 2 kµkν m

 Jν (y) .

(9.25)

We can then form the following derivative operator acting on V µ : ∂ α ∂α V µ (x) − ∂ µ ∂α V α (x) + m2 V µ (x) = Z −ik·(x−y) 1 4 4 e = d y d k W µν Jν (y) , (2π)4 k 2 − m2 6

(9.26)

In a somewhat simpler notation, if pµ = (p0 , p~), with p = |~ p| and ~e = p~/p, then the longitudinal polarisation vector reads 0 µ = (p, p0~e)/m.

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where W µν can be evaluated as     1 µ ν 1 α ν µν 2 2 µν µ αν W = (−k + m ) −g + 2 k k + k kα −g + 2 k k m m 2 2 µν = (k − m )g . (9.27) The remaining integrals over y and k now lead immediately to the so-called Proca equation for V µ : ∂ · ∂ V µ − ∂ µ ∂ · V + m2 V µ = J .

(9.28)

This is the ‘Maxwell equation’ for massive vector fields. It is instructive to examine this equation in empty space, that is, for J = 0. Multipying it by ∂µ , we find that the first two terms cancel, and we recover the Lorenz condition ∂ · V = 0 : all physical polarisations must be orthogonal to the momentum, as we had already found. Reinserting this condition in Eq.(9.28), we are left with the Klein-Gordon equation (∂ ·∂ +m2 )V µ = 0, which essentially requires the particles to be on the mass shell. Note that this nicely compact way of enforcing the Lorenz condition only works for m 6= 0 : for massless vector particles, it must be put in by hand. We can also write down the Lagrangian corresponding to the Proca equation, that is, that Lagrangian that has the Proca equation as its EulerLagrange equation. It reads 1 1 1 (∂µ Vν )(∂ µ V ν ) − (∂µ Vν )(∂ ν V µ ) + m2 V µ Vµ 2 2 2 1 µν 1 2 µ = F Fµν + m V Vµ , 4 2

L =

(9.29)

where the field strength tensor is defined as F µν = ∂ µ V ν − ∂ ν V µ .

9.2 9.2.1

(9.30)

The spin-statistics theorem Spinorial form of vector polarisations

Although there may not (now !) appear to be a special need for it, we can define the polarisation vectors for a massive vector particle using Dirac

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spinors. Let the momentum of the vector particle be pµ and its mass m. We can choose two massless momenta k1µ and k2µ that sum to pµ : pµ = k1µ + k2µ ,

k1,2 2 = 0 , 2 (k1 · k2 ) = m2 .

(9.31)

Three orthonormal states can now be constructed by standard-form spinors as follows : 1 √ u+ (k1 )γ µ u+ (k2 ) , m 2
1 = u+ (k1 )γ µ u+ (k1 ) − u+ (k2 )γ µ u+ (k2 ) , 2m 1 √ u− (k1 )γ µ u− (k2 ) . = m 2

+ µ = 0 µ − µ

(9.32)

In fact, the longitudinal polarisation 0 µ can (by the Casimir trick, as usual) be seen to be nothing else than 0 µ =

1 (k1 − k2 )µ . m

(9.33)

This polarisation, then, is properly normalized and orthogonal to ± µ . Furthermore, from Eq.(8.31) it follows that + · − =

1 u+ (k1 )γ µ u+ (k2 ) u− (k2 )γµ k− (p1 ) = 0 . 2m2

(9.34)

The polarisation vectors are therefore all orthogonal to each other. To check the normalization of + , we write 1 1 u+ (k1 )γ µ u+ (k2 ) u+ (k2 )γµ u(k1 ) = u(k1 )γ µ k/2 γµ u+ (k1 ) 2 2m 2m2 2(k1 · k2 ) 1 = −1 . (9.35) = − 2 u+ (k1 )/k2 u+ (k1 ) = − m m2

+ · + =

We can now have a look at the helicities of these polarisation states. For ± we can compute
1 1  0 0 ~ ~  0 0 |± 0 |2 = Tr ω k / γ k / γ = k k + k1 · k2 , (9.36) ± 1 2 2m2 m2 1 2 and this only vanishes in a Lorentz frame in which the ~k1,2 are antiparallel, so these states can only have pure helicity in frames where ~k1,2 are aligned

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with ~q. Similarly, ~0 is only parallel to p~ if ~k1,2 are antiparallel, therefore in general 0 is not a purely longitudinal state. Nevertheless, the representations (9.32) form a perfectly acceptable orthonormal set of polarisations, and in µ particular (as we shall see) the freedom in choosing k1,2 can often be used to good effect. In a frame in which ~k1,2 are antiparallel we can find the helicities as follows. Using the result of exercise 72, we can write the rotation over an angle θ around the common direction as uλ (k1 ) γ µ uλ (k1 ) →  
1 − cos θ µ ν sin θ µν µ ν µν  (k1 , k2 ) uλ (k1 ) γν uλ (k2 ) cos θg + k1 k2 + k2 k1 − k1 · k2 k1 · k2 (9.37) and since we can write
i µν (k1 , k2 )uλ (k1 ) γν uλ (k2 ) = − Tr γ 5 γ µ γ ν k/1 k/2 uλ (k1 ) γν uλ (k2 ) 4
i 5 µ = − Tr γ γ u−λ (k1 )u−λ (k2 ) k/1 k/2 = −i(k1 k2 ) u−λ (k2 )γ 5 γ µ u−λ (k1 ) 2 = −iλ (k1 k2 ) uλ (k1 )γ µ uλ (k2 ) , (9.38) this rotation results in just a phase factor exp(iλθ). E74 Before finishing this section we point out that also the (trivial) externalE75 line Feynman factor for scalar particles can be written in terms of spinors. µ For a massive scalar with momentum q µ , the same choice of k1,2 is of course possible. We simply note that we can always find a complex phase eiϕ such √ that the external-line factor h can be cast in a form containing two spinors : √ √ eiϕ √ s+ (k1 , k2 ) . (9.39) ~ → ~ 2 p1 · p2 It should not come as a surprise that an external integer-spin particle can conventiently be represented by a spinor-antispinor pair. After all, this is precisely the way in which particles like the W and Z (and, to a lesser extent, the Higgs as well) are most often seen in experiment : namely, through their decay into a fermion-antifermion pair.

9.2.2

Proof of the spin-statistics theorem

The treatment of the previous section may appear somewhat academic, but it has an interesting consequence. Integer-spin particles (scalars and vectors)

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can be represented in their external lines with an even number of spinors, that is an even number of Dirac particles. Particles with half-integer spin are represented by an odd number of Dirac particles. This persists : spin-3/2 particles can be formulated using 3 spinors, spin-2 particles by 4 spinors, and so on. This implies that the interchange of two external half-integerspin particles involves the interchange of an odd number of Dirac particles, and will therefore lead to a minus sign. The interchange of two external integer-spin particles involves the interchange of an even number of Dirac particles, and hence no minus sign. These particles, therefore, obey opposite statistics : integer-spin particles are bosons, half-integer spin particles are fermions7 .

9.3 9.3.1

Massless vector particles Polarisations of massless vector particles

Let us reconsider the helicity states of Eq.(9.20). These are defined in the rest frame of the particle, with the understanding that we have to boost back to the frame in which the particle moves, in our case along the z axis. Under this boost the longitudinal polarisation takes the form of Eq.(9.21). Let us now imagine that the particle approaches masslessness, that is, we let m/p0 decrease towards zero. The boost necessary to reach the original frame then becomes enormous, and the longitudinal polarisation will go to infinity when the particle becomes massless. The only way to avoid matrix elements becoming arbitrarily large, and hence violating unitarity sooner or later, is to arrange the interactions of the theory in such a way that the effect of longitudinal polarisation are suppressed by a factor of order O (m/p0 ) : we shall use this extensively later on. In the strictly massless case, the longitudinal polarisation vector must decouple completely, and we arrive at the result that for massless particles, only the two states of maximal helicity are physical8 . 7

Traditionally, the spin-statistics theorem, like the CPT theorem, is considered to be very deep and difficult. Make up your mind. 8 This can also be proven for particles of higher spin. There is a single exception, the so-called Kalb-Ramond state. See Appendix 15.12.

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9.3.2

257

Current conservation from the polarisation

A photon is a vector particle ; as far as we know it is massless9 . Its polarisation vectors must therefore be transverse. For a photon moving in the z direction, any possible √ polarisation vector must be a superposition of √ µ µ (x + iy) / 2 and (x − iy) / 2. If k µ is the photon momentum, and µ its polarisation, we must therefore have not only k ·  but also 0 = 0 , ~k · ~ = 0 .

(9.40)

However, a problem immediately arises: for the above equations are not invariant under Lorentz boosts. If we boost k µ and µ to a generically other frame, they no longer hold. Let us assume that we are in such a frame ; there we have the Lorentz-invariant conditions (k 0 )2 = |~k|2 , (0 )2 − |~|2 = −1 ,

k 0 0 = ~k · ~ .

(9.41)

We can decompose ~ into a parallel and a perpendicular part : ~ = ~k + ~⊥ , ~k // ~k , ~⊥ · ~k = 0 .

(9.42)

Inserting this into the last equation of Eq.(9.41), we find immediately that 0 = |~k |, and the second equation then gives |~⊥ | = 1. We see that, whatever the value of µ , we can always write  0  µ µ kµ , (9.43)  = ⊥ + k0 where ⊥ µ does satisfy Eq.(9.40). We can therefore have a consistent and unitary theory of massless vector particles, provided that the k µ term decouples from the physics. Now, any matrix element involving an external massless vector particle with momentum k µ and polarisation vector µ will be of the form M = J (k)µ µ , (9.44) where J µ (k) stands for the rest of the amplitude. Note that J µ does not carry any information about µ , but it does know what k µ is, by momentum conservation. Our requirement then is that the interactions of the theory be such that J µ (k) kµ = 0 . (9.45) 9

The current limit is about 10−36 GeV/c2 .

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That is, if we replace the polarisation vector by the momentum, the amplitude must vanish. Diagrammatically, we may indicate the replacing of polarisation by momentum by attaching a handlebar 10 to the external line, so that we may write =M , = Mc→k . (9.46) 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

We shall use the convention that the momentum under the handlebar is counted outgoing. The requirement for strictly massless external vector particles then becomes =0 . (9.47) 0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

What, finally, is the physical content of the requirement ? This is simply answered if we let our massless vector particle be a photon. The object J µ is then seen as a source of photons, that is, an electromagnetic current11 . If we now briefly return from a momentum-language formulation to a positionlanguage one, we see that the Fourier transform of the requirement (9.45) is written as ∂µ J (x)µ = 0 . (9.48) We see that our requirement is nothing but current conservation in the case of electromagnetism ! The fact that electric charge is conserved ensures that longitudinally polarised photons are safely absent from our experience12 . A similar message can be gotten from the propagator. After all, the massive-vector propagator i~

−g µν + k µ k ν /m2 k 2 − m2

clearly becomes horribly singular at m = 0. The solution, as before, is to require that in our theory the k µ k ν term should drop out. There is a catch, 10

The term ‘handlebar’ is my own, biker’s privilege. You might use the term ‘taking the divergence’ but that obscures the very mechanical way by which we investigate the suitability of amplitudes. 11 One may for instance have the source J represent a charge whose momentum changes, thereby emitting radiation. 12 Whether they exist is another question ; at any rate we cannot produce them, nor observe them.

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however: whereas external vector particles must be on the mass shell, the momentum of internal lines is off the mass shell. We therefore arrive at the sharper requirement that Eq.(9.47) must hold even if the particle is off-shell.

9.3.3

Handlebar condition for massive vector particles

Let us examine the situation where a vector particle does have a mass, but the mass m is very small compared to the vector particle’s energy E or its momentum. Clearly, it would be unacceptable13 if the limit m → 0 would be singular while the case m = 0 is not14 . We shall therefore require that, for massive vector particles partaking in a process at high energy, the handlebar condition (9.47) holds in a milder form : 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

= O (m) .

(9.49)

The meaning of this condition is the following. The longitudinal polarisation vector of a massive vector boson has energy behaviour different from its two transverse ones : it grows at high energy E  m with an extra power of E. If for transverse polarisation the amplitude is well-behaved at high energy it may not be so for longitudinal polarisation. The requirement implied by the handlebar condition is that the extra power E inserted into the expression because of longitudinal polarisation is softened, by cancellations over at least one order of magnitude in terms of E/m. We shall presently see that this condition is sufficiently strict to determine, to a large extent, the possible couplings of a theory containing such particles.

9.3.4

Helicity states for massless vectors

The spinor-based helicity states for massive vector particles of section 9.2.1 are apparently not well suited to the massless case. Note, however, that we may generalize the method of Eq.(9.31) as follows : q µ = p1 + α p 2 , 13

p1,2 2 = 0 ,

m2 = 2α (p1 · p2 ) .

(9.50)

Or at least embarassing — after all, we do not know for certain if the mass of the photon is strictly zero or just a paltry 10−137 kilograms. 14 Note that we do not even insist that m → 0 gives the same result as m = 0, only that the limit is nonsingular.

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Using the fact that the spinors of massless particles are homogeneous of degree 1/2 in the argument : u± (αp2 ) =

√

α u± (p2 ) ,

(9.51)

we see that (for instance) the polarisation vector + can be written, in analogy to Eq.(9.32), as 1 u+ (p1 ) γ µ u+ (p2 ) . (9.52) + µ = √ 2 p1 · p2 Since α does not occur in the polarisation vector, we may consider the limit α → 0. In that case, q = p1 is massless, and the only condition on the massless vector p2 is that (p1 · p2 ) must not vanish. By a judicious choice of overall complex phase, this leads us to propose, for a massless vector particle with momentum k µ , states of definite helicity as follows, where the spinors are again in the standard form : uλ (k)γ µ uλ (r) λ µ = √ , λ 2 s−λ (k, r)

λ=± .

(9.53)

Here, the vector rµ is an arbitrarily chosen massless vector not parallel to k µ ; it is called the gauge vector. We can ascertain that + · − =

1 u+ (k)γ µ u+ (r) u− (r)γµ u− (k) = 0 , 4k · r

(9.54)

Also, 1 −1 u+ (k)γ µ u+ (r) u+ (r)γµ u+ (k) = u+ (k) r/ u+ (k) = −1 . 4k · r 2k · r (9.55) These, then, are acceptable helicity states. A few useful properties of these polarisation vectors are √ √ λ 2 λ 2 ωλ /λ = uλ (r)uλ (k) , ω−λ /λ = u−λ (r)u−λ (k) (9.56) s−λ (k, r) s−λ (k, r) + · + =

and k/ /λ = λ

√

2 u−λ (k) uλ (k) ,

and this object is explicitly gauge-invariant.

(9.57)

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9.3.5

261

The massless propagator : the axial gauge

We can perform the sum over the physical polarisation states of a massless vector from the helicity states : X

1 uλ (k)γ µ uλ (r) uλ (r)γ ν uλ (k) 4k · r λ=± X 1 1 = Tr (/k γ µ r/ γ ν ) uλ (k)γ µ r/γ ν uλ (k) = 4k · r 4k · r λ=±

λ µ λ ν =

λ=±

X

= −g µν +


1 kµ rν + rµ kν . k·r

(9.58)

The form of the massless vector propagator in which only physical degrees of freedom propagate is therefore given by the following Feynman rule :

µ

k

µν µ ν µ ν ν ↔ i~ −g + (k r + r k )/(k · r) k 2 + i

Feynman rules, version 9.3

massless internal lines

(9.59)

Note the appearance of the arbitrary vector r. This way of writing the propagator is called the axial gauge. The propagator is explicitly orthogonal to rµ whatever the value of k : The vector r acts as an ‘axis’ with respect to which the field is always orthogonal, hence the name. The fact that the vector r is arbitrary is of course bothersome, in the same way that the arbitrariness of the representation chosen for the Dirac matrices in the case of Dirac particles is bothersome. We solve it in the same way, by insisting that we ought to be able to remove r from the final expressions for matrix elements. This can of course not be by virtue of any property of r itself, but must come from the handlebar condition, since every term containing r also contains k. Two things are worthy of remark here. In the first place, the propagator is homogeneous of degree zero in r, so any result cannot depend on the length of r anyway. In the second place, in contrast to the propagator proposed before, with pµ pν /m2 , the propagator in the axial gauge does not diverge.

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9.3.6

Gauge vector shift

Let us consider helicity states for massless vector particles as defined in sect.9.3.4. We shall denote these by λ µ (k, r). If we change the gauge vector r from one value into another, another perfectly acceptable helicity state is obtained. What is the relation between these states ? To answer this we simply compute the difference between the states with different gauge vector : λ µ (k, r1 ) − λ µ (k, r2 ) =   λ uλ (k)γ µ uλ (r1 ) uλ (k)γ µ uλ (r2 ) √ − s−λ (k, r1 ) s−λ (k, r2 ) 2
µ λ u−λ (r1 )γ u−λ (k)u−λ (k)uλ (r2 ) + u−λ (r1 )uλ (k)uλ (k)γ µ uλ (r2 ) = −√ s−λ (r1 , k)s−λ (k, r2 ) 2 µ µ √ λ u−λ (r1 )(γ k/ + k/ γ )uλ (r2 ) s−λ (r1 , r2 ) = √ =λ 2 k µ . (9.60) s−λ (k, r1 ) s−λ (k, r2 ) 2 s−λ (k, r1 ) s−λ (k, r2 ) we see that the two states differ only by the vector particle’s momentum. In any current-conserving set of diagrams we may therefore choose the gauge vector at will ; there is no risk of picking up a phase difference if two different gauge vectors are used for two different current-conserving sets of diagrams. As an illustration of how the gauge vector can disappear from a currentconserving object, let us consider   q p − , λ · 2k · p 2k · q with p and q two massless momenta. The form of section 9.3.4 turns this into   λ sλ (k, p)s−λ (p, r) sλ (k, q)s−λ (q, r) √ − 2k · p 2k · q 2 s−λ (k, r)   λ s−λ (p, r) s−λ (q, r) = √ − 2 s−λ (k, r) s−λ (p, k) s−λ (q, k) λ s−λ (p, r)s−λ (q, k) − s−λ (q, r)s−λ (p, k) = √ (9.61) s−λ (k, r)s−λ (p, k)s−λ (q, k) 2 Now, the Schouten identity tells us that s−λ (p, r)s−λ (q, k) + s−λ (p, k)s−λ (r, q) = −s−λ (p, q)s−λ (k, r)

(9.62)

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263

so that the gauge vector indeed drops out, and   p q λ s−λ (p, q) λ · − = −√ . 2k · p 2k · q 2 s−λ (k, p)s−λ (k, q)

(9.63)

One can easily check that the same form is obtained without using the Schouten identity if we choose either r = p or r = q. E76

9.4

Exercises

Excercise 69 Polarisation averages If, for vector particles, we take not a spin sum but a spin average, we find   3 1X µ ν 1 pµ p ν µν j ¯j = −g + 2 3 j=1 3 m for an on-shell vector particle of momentum p and mass m. We can also consider it classically, in the rest frame where p~ = 0. In that case there is a polarisation vector µ = (0,~) (we only consider real polarisations). Show that, if we also average the polarisation vector over all possible orientations, we find   Z 1 pµ pν 1 µ ν µν dΩ   = −g + 2 4π 3 m as well. Excercise 70 A character-building calculation Let bµ = Λ(p; q)µ ν aν for some arbitrary vector a. Use the explicit form (9.11) of the minimal Lorentz transform to prove by explicit calculation that b · b = a · a. Excercise 71 Rewriting the unity Proce that using the unit vectors t, x, y, and z we can write δ µ ν = tµ tν − xµ xν − y µ yν − z µ zν Excercise 72 Another minimal Lorentz transform ? µ Let k1,2 be two massless vectors and θ an angle, and define the tensor Λµν = cos θ g µν +


1 − cos θ µ ν sin θ µν k1 k2 + k2µ k1ν −  (k1 , k2 ) k1 · k2 k1 · k2

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Show that Λµ ν is the minimal Lorentz transformation that is a rotation along their common direction in any frame in which ~k1,2 are parallel or antiparallel. Excercise 73 Checking helicities For the polarisaton vectors +,0,− defined in section 9.1.4, verify that (Tz )µ ν + ν = ~+ µ , (Tz )µ ν 0 ν = 0 , (Tz )µ ν − ν = −~− µ Excercise 74 Another helicity check We can check the helicity of the polarisation representation of Eq.(9.32) another way. 1. Show that the Clifford-space operator Σ = cos(α/2) + i

sin(α/2) 5 0 γ [γ , p/] 2|~p|

satisfies ΣΣ = 1, and that it is a minimal Lorentz transformation describing a rotation around the axis p~ over an angle α (it may help to consult Eq.(7.132)). 2. Show that this holds in any Lorentz frame. µ 3. Let k1,2 be two massless vectors with k1µ + k2µ = pµ and let ~k1 and −~k2 point in the direction of p~. Show that

Σu+ (k1 ) = exp(−iα/2) u+ (k1 ) and Σu+ (k2 ) = exp(+iα/2) u+ (k2 ) 4. Show that, therefore, the object u+ (k1 )γ µ u+ (k2 ) acquires a phase exp(iα) under this rotation. Excercise 75 An orthogonality proof For a massive vector particle with momentum pµ and mass m, choose two µ with k1 + k2 = p, and define, as in Eq.(9.32), massless vectors k1,2 µ± =

1 1 √ u± (k1 ) γ µ u± (k2 ) , µ0 = (k1µ − k2µ ) m m 2

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1. Use the Chisholm identity to prove that p/ /+ /− − /− /+ p/ = 2 γ 5 (/k1 − k/2 ) 2. Use this to show that the vectors pµ and ±,0 form an orthogonal set Excercise 76 Current-conserving radiation factor Check the statement below Eq.(9.63).

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Chapter 10 Quantum Electrodynamics 10.1

Introduction

In this chapter we shall start to work our way to realistic theories about the actual elementary particles encountered in nature1 . Almost all elementary particles seen so far have nonzero spin2 . We shall defer the discussion of charged spin-1 particles to a later chapter ; at this point we shall only discuss how to set up a consistent theory of spin-1/2 particles (charged leptons and/or quarks) and photons. This is the theory of quantum electro-dynamics, or QED.

10.2

Constructing QED

10.2.1

The QED vertex

Since the propagators of spin-1/2 particles and of the massless spin-1 photon have already been fixed, the only ingredient which we still have to determine is the coupling between them ; and on this coupling rests the burden of ensuring the current-conservation requirement as embodied in Eq.(9.47). The vertex that connects two Dirac particles must have one upper, and one lower 1

It is of course possible that the elementary particles discussed in this text are not truly elementary and that a yet deeper level of substructure is at some moment discovered. In that case, please insert in whatever follows the addendum (as of August 31, 2019). 2 The Higgs particle is the only known exception.

267

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Dirac index3 : and since the photon is involved, it must also carry a Lorentz index. The simplest, and – as we shall see – indeed the correct form of the vertex is that of a Dirac matrix. We therefore propose the following Feynman rule :

Q ~

µ ↔ i γµ Feynman rules, version 10.1

QED vertex (10.1)

Here Q is the strength of the fermion-photon coupling : the charge of the fermion4 . By dimensional analysis, we see that is has dimension     dim Q = dim ~−1/2 . (10.2) The Dirac delta function imposing momentum conservation is implied. As is conventional, we shall employ wavy lines to indicate photons. This choice of vertex can only been argued to be reasonable if the photon current is conserved, so this is now our task.

10.2.2

Handlebars : a first look

Let us now start to investigate the requirements of current conservation for our theory. One of the simplest possible processes is the decay of a photon into a fermion-antifermion pair, shown below : p1 q

p2

Of course the photon has to be off-shell here, but that is no problem since also off-shell photons must obey current conservation. The part of the amplitude 3

Since we cannot form constant (i.e. momentum-independent) objects with one Dirac index, or with, say, two upper Dirac indices, we are essentially forced to have conservation of the orientation of Dirac lines in each vertex. This is modified to some extent if Majorana particles are present in the theory. 4 Or, rather, it is related to the charge. The precise form of this relation must, of course, be established by investigating the coupling in a well-defined physical situation.

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269

depicted is given by5 M = i Q u(p1 )γ µ v(p2 ) ,

(10.3)

where the index µ of the photon is coupled to a corresponding index somewhere else in the larger Feynman diagram. Let us now attach the handlebar, so that we get p1

q

p2

With the convention, to which we shall try to adhere, that the momentum assigned in the handlebar must be counted outgoing from the vertex, so in this case should read −q, the handlebarred M becomes Mc = −iQ u(p1 ) q/ v(p2 ) .

(10.4)

Note that we indicate the handlebar algebraically by the symbol c. Now we apply momentum conservation that tells us that q = p1 + p2 :
(10.5) Mc = −iQ u(p1 ) p/1 + p/2 v(p2 ) . To the expression in the middle we add zero in a clever way :
Mc = −iQ u(p1 ) p/1 − m + p/2 + m v(p2 ) ,

(10.6)

where m is the mass of the fermion. Now, we know that the spinors u(p1 ) and v(p2 ) satisfy the Dirac equations (/p1 − m)u(p1 ) = 0 and (/p2 + m)v(p2 ) = 0

(10.7)

for on-shell momenta, so that half of the expression 10.6 ‘cancels to the left’ and the other half ‘cancels to the right’, and we end up with Mc = 0 .

(10.8)

We shall see that this is the general mechanism by which unitarity and current conservation are ensured. The above is of course only the simplest example of current conservation in QED, and in the following we shall in fact study all conceivable QED 5

√ The 1/~ from the vertex cancels against the ( ~)2 from the external Dirac lines.

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process at once, but already we can learn a few useful things. In the first place, a possible alternative coupling, with γ 5 γ µ instead of γ µ , is ruled out since we cannot obtain two Dirac equations : γ 5 q/ = −/p1 γ 5 + γ 5 p/2 = −(/p1 ± m)γ 5 + γ 5 (/p2 ± m) ,

E78

so that either the cancellation to the left would be spoiled, or that to the right. In the second place, it is necessary that both fermions have precisely the same mass. Since all known different fermion types have different masses, this means that the QED interaction must conserve fermion type, or ‘flavour’. Electromagnetic muon decay, µ → eγ, is therefore forbidden, not by conservation of the electric charge (which is indeed the same for muons and electrons) but by conservation of the whole electromagnetic current6 .

10.2.3

E77

(10.9)

Handlebar diagrammatics

The argument for current conservation in the previous section went through because both fermions were on their mass shell. Since fermions in internal lines in Feynman diagrams are not on the mass shell, we have to extend our approach to off-shell fermions. Consider an arbitrary diagram in which a fermion of mass m propagates and couples to a photon, k q

where the fermion momenta p and q are indicated and for the photon momentum k we have k µ = (p − q)µ . The momenta p and q may be on-shell (in which case the corresponding blob isn’t actually there), but any of them may be off-shell, and hence leads into a further piece of Feynman diagram. In that case the blobs stand for the other vertices, where the fermion is created and absorbed7 . The part of the diagram between the blobs is      q/ + m Qγ µ p/ + m i~ 2 i i~ 2 , q − m2 ~ p − m2 6

Fortunately, the decay µ → eγ has never been observed, and the branching ratio is smaller than about 10−11 . 7 Actually, the p and q lines are attached to a semi-connected graph rather than two separate connected ones, but here the distinction is irrelevant.

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271

where µ is the index belonging to the photon line ; in an actual process, µ may be coupled to the photon’s polarisation vector if the photon is external, or to the photon’s propagator if the photon happens to be an internal line. In case p, say, is on-shell we have to write     √  q/ + m Qγ µ  ~ . i~ 2 i u(p) q − m2 ~ Let us now put the handlebar on the photon leg, k p

q

and see what happens. Algebraically, we must multiply the above expression by kµ , and then     q/ + m µ p/ + m iQ 2 γ 2 (i~) kµ = ~ q 2 − m2 p − m2   p/ + m iQ q/ + m (/p − q/) 2 = (i~)2 2 2 ~ q −m p − m2     iQ p/ + m /+m 2 q = (i~) 2 (/ p − m) − (/ q − m) ~ q − m2 p2 − m 2     iQ p/ + m q/ + m = (i~)2 − 2 . (10.10) 2 2 ~ q −m p − m2 We see that under the handlebar the double propagator splits up into two single ones. Note that, for this to be possible, it is again essential that the mass of the fermion does not change at the vertex. We may write this operation diagrammatically as k p

q

−

=

,

(10.11)

where we have introduced two new diagrammatic ingredients: a slashed fermion line, with a trivial Feynman rule : ↔

i~ ,

(10.12)

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and a new vertex, also carrying a trivial rule : Q . (10.13) ~ The handlebarred photon line is replaced by a dotted line which evaluates trivially to unity, but we do not want to leave it out of the diagram since the dashed propagator still carries momentum, so that without it momentum conservation would not hold at the new vertex. Like the handlebar this rule is not intended to represent some physical interaction, but serves only as a computational device. For external Dirac lines we find even simpler rules, since the external spinors satisfy the Dirac equation : ↔

i

= 0 for on-shell external lines ,

(10.14)

where the external line may belong to the initial or final state, and the arrow orientation may be also reversed. An important result follows immediately from the triviality of our new Feynman-rule tools :

=

10.2.4

.

(10.15)

The Ward-Takahashi identity

We shall now prove that the Feynman rule (10.2.1) is a good one, in the sense that a handlebar on any photon gives a zero result, both for on-shell (external) and off-shell (internal) photon lines. This is quite a tall order, since we have to consider a literal infinity of possible processes. We shall base the proof on - what else ? - the SDe’s of the theory. Throughout this section we shall use semi-connected diagrams only. Let us consider a general Green’s function that contains r fermion lines flowing in, and s fermion lines flowing out, together with any number (≥ 1) of photon lines, one of which we single out : p1 pr

q1

≡

qs

(10.16) k

The fermions’ momenta are reckoned along their respective arrows, and the photon momentum k is counted outgoing. Note that, since we are considering

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273

here a Green’s function and not an amplitude, any of the external lines may be off-shell. It is this object that we shall submit to a handlebar operation. The SDe’s of QED are, in our notation : =

+

,

=

+

,

=

+

.

(10.17)

We therefore have =

,

(10.18)

and the handlebar operation gives us −

=

.

(10.19)

Each term on the right-hand side can be subjected to its own SDe, to give =

X

pj

−

j

+

qj

X j

r;s

−

r;s

.

(10.20)

By virtue of Eq.(10.15) the last two terms cancel precisely, and we are left with the Ward-Takahashi identity : =

X j

pj

−

X j

qj

.

(10.21)

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August 31, 2019

We can conveniently express this in a more analytic form. Let us denote our Green’s function by Gµ (p1 , . . . , pr ; q1 , . . . , qs ; k) ≡

,

(10.22)

where the explicit Lorentz index is that of the photon line, and the same Green’s function, only with the special photon removed, by G0 (p1 , . . . , pr ; q1 , . . . , qs ) ≡

.

(10.23)

Taking into account the flow of the momenta and the fact that the two trivial Feynman rules (10.12) and (10.13) together yield a factor of (iQ/~)(i~) = −Q, we can write the diagrammatic Ward-Takahashi identity (10.21) as follows8 : Gµ (p1 , . . . , pr ; q1 , . . . , qs ; k) kµ = s X Q G0 (p1 , . . . , pr ; q1 , . . . , qj + k, . . . , qs ) j=1 r X

−Q

G0 (p1 , . . . , pj − k, . . . , pr ; q1 , . . . , qs ) .

(10.24)

j=1

It is this result that proves that, indeed, the choice of the vertex (10.2.1) leads to an acceptable theory. So far we have considered the general case, with no constraint on the external momenta. If we now specialize to amplitudes, in which all external momenta except perhaps for k µ , must be on their mass shell, the rule (10.14) applies, and we find the even more attractive Ward identity : = 0 . 8

(10.25)

In many, or even most, cases of interest fermion number is conserved, which means that in every (fundamental or effective) vertex the number of incoming and outgoing fermions is the same ; in that case we have r = s. But since we have nowhere used this, the WardTakahashi identity may be expected to hold also for processes in which fermion number is not conserved, for example in supersymmetry where Majorana fermions occur. Note, however, that Majorana fermions are necessarily neutral and themselves do not couple to photons.

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10.2.5

275

The charged Dirac equation

We still have to determine the precise relation between the coupling constant Q in the Feynman rule, and the classical electric charge q of the particle. We shall do this by establishing a relation with classical electrodynamics. The classical (i.e. non-loop) SDe for ψ in the presence of a photon field A is given by 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

in other words Z ψ(x) = d4 y

1 (2π)4

Z

4

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

=

dke

−ik·(x−y)

,

k/ + m i~ 2 k − m2 + i

(10.26)



Q i ~



γµ ψ(y)Aµ (y) , (10.27)

whence
i/∂ − m + QA(x) / ψ(x) = 0 ,

(10.28)

which is the Dirac equation in the presence of an electromagnetic field. Let us work this expression towards classical physics. In the first place, the derivative is, by the standard assignment rules for quantum mechanics, related to the momentum operator : pµ = i~ ∂ µ , (10.29) and the mass m to the mechanical mass M by (as we have seen) m=

Mc . ~

The Dirac equation can therefore be written as
(pµ + ~QA(x)µ )γµ − M c ψ(x) = 0 ,

(10.30)

(10.31)

which is to be compared with the standard expression for the electromagnetic momentum if a charged particle in classical electrodynamics: q pem µ = pµ − Aµ . c

(10.32)

where q is the classical charge of the particle and Ac the classical electromagnetic field. In the Gaussian system of units, the charges have dimensionality dim[q 2 ] = kg m3 /sec2 and the Coulomb field strength E therefore obeys

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August 31, 2019

dim[E] = dim[q] /m2 . Since this is the gradient of the classical e.m. vector potential Ac we have dim[Ac ] = dim[q] /m, and because the photon field A has dimensionality dim[A2 ] = kg/sec, it follows that the correct relation between the photon field and the classical e.m. field must read Ac 2 = c A2 .

(10.33)

From this it follows that the coupling Q and the charge q are related by √ Q = −q/(~ c) , (10.34) √ which implies the correct dimensionality dim[Q] = dim[1/ ~] ; moreover, we find immediately that, for particles with unit electric charge, Q2 =

4π α , ~

(10.35)

where α stands for the electromagnetic fine structure constant : α ≈ 1 / 137.036 .

(10.36)

Since in QED every next loop order contains two extra powers of Q and one (effective) power of ~, the loop expansion is in QED equivalent to an expansion in powers of α. What we have√done here is a pure √ dimensional analysis, it does not decide between Q = q/~ c and Q = 2q/~ c. Later on, the discussion of Thomson scattering will assure us that we have made the right choice. Of course, any extra numerical factor in Q can be compensated for by a rescaling of A.

10.2.6

The Gordon decomposition

Consider a charged Dirac particle that scatters by emitting (or absorbing) a single photon. The corresponding current reads Jµ =

ie u(q) γ µ u(p) , ~

(10.37)

where p is the incoming, and q the outgoing momentum. By the properties of the Dirac spinors we can write this as Jµ =


ie u(q) q/γ µ + γ µ p/ u(p) . 2m~

(10.38)

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277

Since q/γ µ = q µ − iqα σ αµ ,

γ µ p/ = pµ + ipα σ αµ ,

(10.39)

the current takes the form Jµ =


ie u(q) (p + q)µ + i(p − q)α σ αµ u(p) . 2m~

(10.40)

This is called the Gordon decomposition : the vertex is split up into a piece that we shall later recognize as the vertex by which a scalar particle interacts with photons, which is called the convection term, and a tensorial part, called the spin term. Both terms vanish individually under the handlebar operation. The Gordon decomposition is especially useful in nonrelativistic situations, where q µ and pµ are both dominated by their time components. In that case u(q)u(p) ≈ 2m, and the vertex becomes simply (p + q)µ . Note that for antispinors we find exactly the same result, namely v(p)γ µ v(q) ≈ (p + q)µ in the nonrelativistic situation9 .

10.2.7

Furry’s theorem

An interesting observation concerns closed fermion loops in QED. Let us consider a fermion loop that is attached by three QED vertices to the rest of a Feynman diagram: k

D− =

µ

p2 p1

ν

p3

(10.41)

λ

Here, we have indicated the Lorentz indices on the photon lines, and the momenta across the photon lines are considered incoming into the loop. In addition to this diagram, there is also a similar diagram in which the orientation of the loop is reversed : k

D+ = 9

µ

p2 p1

ν

p3 λ

Since the Dirac equation reads / q v(q) = −mv(q), say, which gives us a minus sign : on the other hand, v(p)v(q) now evaluates to approximately −2m, thereby cancelling the minus sign.

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Note that these graphs cannot be twisted into one another. For loops with only one or two vertices they can be so twisted, and then do not count as separate diagrams ; for three or more vertices, there are two distinct ones. Without pretending to evaluate the whole loop, let us concentrate on the Dirac structure of their numerators. The first diagram contains the trace10 D− → Tr (/k + m) γ µ (/k − p/1 + m) γ λ (/k + p/2 + m) γ ν




≡ T− , (10.42)

whereas the corresponding trace for the other diagram reads D+ → Tr (−/k + m) γ ν (−/k − p/2 + m) γ λ (−/k + p/1 + m) γ µ




≡ T+ . (10.43) Note that the rest of the loops, and in particular the propagator denominators, are identical for both graphs. By using the reversibility inside traces of Clifford algebra elements, we can write T+ = − Tr (/k − m) γ ν (/k + p/2 − m) γ λ (/k − p/1 − m) γ µ




= − Tr (/k − m) γ µ (/k − p/1 − m) γ λ (/k + p/2 − m) γ ν = − T− ,


(10.44)

since no terms with an odd power of m survives the trace. We see that the two loops cancel each other precisely ! This can obviously be extended to loops with more vertices, and we find Furry’s theorem : fermion loops with an odd number of vector vertices11 and opposite orientation cancel each other ; with an even number of vector vertices, they are identical12 . Furry’s theorem does not hold if one or more of the vertices are of axial-vector type, and so it is not generally valid for the weak interactions. For QCD, in which the quark-gluon couplings have the Diracmatrix form as in QED, Furry’s theorem holds in a more restricted form : the spacetime part of the two quark loops with even(odd) number of vertices are equal(opposite), but the additional colour structures of the diagrams are different. This implies, for instance, that the two quark loops with three gluon vertices do not cancel completely. We shall come back to that case later on. 10

By the rules of Dirac particles, closed loops automatically evaluate to traces. That is, vertices consisting of a single Dirac matrix, such as in QED. 12 Furry’s theorem is usually proved by invoking the charge-conjugation matrix, discussed in section 15.11.2. However, as we see this is not necessary. 11

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Now consider a diagram (or set of diagrams) N with only one single external line which is a photon :

p N ≡

,

(10.45)

where we have indicated the momentum of the photon. Such objects go under the name of tadpoles 13 . By Lorentz covariance we see that, whatever goes on inside the blob, N must always be of the form N µ = pµ f (p2 ) .

(10.46)

But . . . by momentum conservation, pµ = 0 since no momentum is coming out on the other side of the blob. Therefore N = 0 ; photon tadpoles vanish identically, even if Furry’s theorem does not apply in this case. It is easy to see that the same must hold for the tadpoles of any other vector particle. For scalar particles it does not apply, since N = f (0) has no particular reason to vanish. Also, for particles that are described not by vectors but by symmetric tensors, the tadpole N µν = g µν f (0) does not necessarily vanish14 .

10.3

Some QED processes

10.3.1

A classic calculation : muon pair production

We are now in a position to compute, for the first time, a realistic cross section. We shall follow the classic steps that lead to our final result. Description of the process with momentum assignments The simplest calculation is that of the cross section for muon pair production in e+ e− collisions: e− (p1 ) e+ (p2 )

→

µ− (q1 ) µ+ (q2 ) .

since it is described, at tree level, by only a single diagram15 . 13

Of course, ‘spermatozoon’ would be more appropriate. . . This is not the graviton, since that is described by a symmetric traceless tensor ; it is in fact another representation of a scalar particle. 15 It might be thought that processes involving only electrons would be simpler, but as we shall see these contain always at least two diagrams. 14

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Drawing and writing out the diagram(s) The single lowest-order Feynman diagram is given by p

q1

1

M =

(10.47) q2

p

2

Both the electron and muon are Dirac particles. We shall denote the electron charge by Qe , and the muon charge by Qµ , and their masses by me and mµ , respectively. The total invariant mass squared is conventionally denoted by s, and of course momentum is conserved : p1 α + p 2 α = q 1 α + q2 α ,

s = (p1 + p2 )2 = (q1 + q2 )2 .

(10.48)

The amplitude corresponding to the Feynman diagram is M=i

~Qe Qµ v(p2 ) γ α u(p1 ) u(q1 ) γα v(q2 ) , s

(10.49)

and is strictly dimensionless: dim[M] = dim[1], as it ought to be for a 2 → 2 process at tree order. Squaring and averaging, eliminating Dirac stuff In a typical muon pair production process, we shall accept muons with any polarisation in the final state ; also, usually the beams of incoming electrons and positrons are unpolarised. The amplitude, squared and averaged over the incoming electron and positron spins can be evaluated using the Casimir trick :

 1 X |M|2 |M|2 = 4 spins

=

~2 Qe 2 Qµ 2 4s2

X

v(p2 )γ α u(p1 ) u(p1 )γ β v(p2 )

spins

×

X spins

u(q1 )γα v(q2 ) v(q2 )γβ u(q1 )

August 31, 2019 =

=

281
~2 Qe 2 Qµ 2 Tr (/p2 − me )γ α (/p1 + me )γ β 2 4s
Tr (/q1 + mµ )γα (/q2 − mµ )γβ 4~2 Qe 2 Qµ 2 s2

4~2 Qe 2 Qµ 2 = s2


p2 α p1 β + p1 α p2 β − (p1 · p2 )g αβ − me 2 g αβ
q1 α q2 β + q2 α q2 β − (q1 · q2 )gαβ − mµ 2 gαβ  2(p1 · q1 )(p2 · q2 ) + 2(p1 · q2 )(p2 · q1 )  2 2 + s(me + mµ ) (10.50)

Choosing a Lorentz frame, working out dot products We shall work in the centre-of-mass frame of the colliding electron-positron pairs. In that frame, we shall write     E E  0     , pµ2 =  0  , pµ1 =  (10.51)  0   0  p −p and 

  E E    q sin θ cos ϕ  µ  −q sin θ cos ϕ q1µ =   q sin θ sin ϕ  , q2 =  −q sin θ sin ϕ q cos θ −q cos θ

   , 

(10.52)

where s = 4E 2 ,

p2 = E 2 − me 2 ,

q 2 = E 2 − mµ 2 .

(10.53)

The various vector products are therefore given by (p1 · p2 ) = 2E 2 − me 2 , (q1 · q2 ) = 2E 2 − mµ 2 , (p1 · q1 ) = (p2 · q2 ) = E 2 − pq cos(θ) (p1 · q2 ) = (p2 · q1 ) = E 2 + pq cos(θ) ,

(10.54)

where θ is the polar scattering angle, that is, the angle between p~1 and ~q1 . We also use the fact that Qµ and Qe are the negative of the unit charge, so that Qµ Qe = 4πα/~.

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The final transition rate We now arrive at 

16π 2 α2 |M|2 = s2

 s2 (1 + cos(θ)2 ) + 4s(me 2 + mµ 2 ) sin(θ)2  2 2 2 (10.55) + 16me mµ cos(θ)

The differential cross section Using what we have already learned about the flux factor and the two-body phase space, we can write the differential cross section as  1/2

 1 s − 4mµ 2 2 dσ = |M| dΩ 64π 2 s s − 4me 2  1/2  α2 s − 4mµ 2 me 2 + mµ 2 2 sin(θ)2 = (1 + cos(θ) ) + 4 2 4s s − 4me s  2 2 me mµ cos(θ)2 dΩ . + 16 (10.56) s2 This cross section therefore only depends on s and the polar scattering angle: there is, for unpolarised incoming beams, no azimuthal direction singled out and there is therefore no azimuthal angle dependence16 . The total cross section The total cross section is obtained by simple     4π α2 me 2 mµ 2 σ= 1+2 1+2 3s s s

angular integration, and reads  1/2 s − 4mµ 2 . (10.57) s − 4me 2

The cross section is only nonzero above the muon pair-production threshold, s > 4mµ 2 . Since the muon mass mµ is much larger than the electron mass me , we may accurately approximate by putting me ≈ 0 :    1/2 4π α2 mµ 2 mµ 2 σ≈ 1+2 1−4 . (10.58) 3s s s 16

This could be different, e.g. in the case of transversely polarised beams.

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Figure 10.1: Threshold behaviour for fermions and scalars

For large s, furthermore, we have 4π α2 σ≈ 3s

  mµ 4 1 − 6 2 + ··· . s

(10.59)

By accidental cancellation of the leading mµ 2 /s terms, the large-s limit is reached quite rapidly. Note that the cross section does not depend on ~. For massless incoming electrons and final-state particles p of mass m,2 the cross section at beam energy E = xm goes as (1 + 1/(2x)) 1 − 1/x2 /x for fermions, but as (1 − 1/x2 )3/2 /x2 for scalars in the final state. Studying the threshold behaviour of the cross section can therefore inform us about the spin of newly pair-produced particles : the peak cross sections differ by a factor of about 2.8. as shown in figure 10.1

10.3.2

Compton and Thomson scattering

We next consider the Compton scattering process, an elastic collision between a photon and an electron : e− (p) γ(k1 )

→

e− (q) γ(k2 )

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Now, there are two Feynman diagrams,

p

p

q

M =

k2

+

k1

k2

(10.60)

k1

q

The amplitude is given by M = M1 + M2 , A1 M1 = −i~Qe 2 , A1 = u(q) /2 (/p + k/1 + m) /1 u(p) , 2(p · k1 ) A2 M2 = +i~Qe 2 , A2 = u(q) /1 (/q − k/1 + m) /2 u(p) , 2(q · k1 ) (10.61) where 1,2 are the polarisation vectors of the respective photons. Taking into account the averaging factor 1/4, we find17 (with m for me )

|A1 |2



=  |A2 |2 =



=

= hA1 A∗2 i = = =


1 Tr (/q + m) γ α (/p + k/1 + m) γ β (/p + m) γβ (/p + k/1 + m) γα 4 16m4 − 8(pq)m2 + 8(pk1 )(qk1 ) + 16(pk1 )m2 − 8(qk1 )m2 ,
1 Tr (/q + m) γ β (/q − k/1 + m) γ α (/p + m) γα (/q − k/1 + m) γβ 4 16m4 − 8(pq)m2 + 8(pk1 )(qk1 ) + 8(pk1 )m2 − 16(qk1 )m2 , hA2 A∗1 i
1 Tr (/q + m) γ α (/p + k/1 + m) γ β (/p + m) γα (/q − k/1 + m) γβ 4 8(pq)(pk1 ) − 8(pq)(qk1 ) + 16(pq)m2 − 8(pq)2 −4(pk1 )m2 + 4(qk1 )m2 . (10.62)

We can most easily evaluate this in the photon-electron centre-of-mass frame18 . In this frame, we have p0 = q 0 = 17

s + m2 K √ , |~p| = |~q| = |~k1 | = |~k2 | = √ , 2 s 2 s

(10.63)

Both the incoming electron and the incoming photon have 2 degrees of freedom, hence (1/2)(1/2)=1/4. 18 In the actual experiment, the photon will of course be impingeing on the stationary electron ; but since the cross section is invariant we may choose any frame we want.

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where K = 2(pk1 ) = s − m2 : and the angle between ~q and ~k1 is denoted by θ. Putting everyhting together, we find  2

 8m 8m4 2m4 4m2 2 2 2 |M| = 16π α + 2 + − K K (qk1 )2 (qk1 )  K 4(qk1 ) 8m4 + + . (10.64) − (qk1 )K (qk1 ) K The phase space integration element is given by dV (p + k1 ; q, k2 ) =

1 1 K dΩ , (2π)2 8 s

(10.65)

where Ω is the solid angle of the emitted electron. The flux factor is 1 1 = . 2 1/2 2λ(s, m , 0) 2K The only nontrivial quantity in the computation is     K 0 0 2 (qk1 ) = k1 q − |~q| cos θ = (s + m ) − K cos θ , 4s and we can find the angular averages Z 1 dΩ (qk1 ) = 4π Z 1 1 dΩ = 4π (qk1 ) Z 1 1 dΩ = 4π (qk1 )2

K(s + m2 ) , 4s   2s K log 1 + 2 , K2 m 4s . m2 K 2

(10.66)

(10.67)

(10.68)

We therefore have for the transition rate, now also averaged over the scattering angle : 

 m2 m2 s 2 2 2 |M| = 16π α 1 + + 16 2 s K     2 4 ms ms s K + −8 2 − 16 3 + 2 log 1 + 2 (10.69) K K K m

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August 31, 2019

The total cross section σ=

 1

|M|2 . 16πs

(10.70)

It is interesting19 to note that the ‘static’ limit K → 0 is well-defined : lim σ =

K→0

8π α2 . 3 m2

(10.71)

This is called the Thomson cross section. It may serve as the ‘measurement’ prediction by which the electric charge of the electron is defined. Note that, just as in the case of muon pair production, the cross section does not depend on ~. This means that this cross section had better coincide with the prediction from classical electromagnetism — as, indeed, it does. Returning to the arguments that led us to make the identifications q Q= √ , ~ c

Acl 2 = c A2 ,

(10.72)

we see that we have made the correct choice.

10.3.3

Electron-positron annihilation

The process e+ (p1 ) e− (p2 )

→

γ(k1 ) γ(k2 )

(10.73)

is related by crossing to Compton scattering, and is described at the tree level by the two Feynman diagrams

p2

p2

k1

M =

k2

+

p1

k2

.

p1

(10.74)

k1

We shall study it in the context of the way it is actually observed at highenergy e+ e− colliders, that is, in the centre-of-mass frame with the photons emerging an nonnegligible angles with respect to the electron and positron beams. In that case, no invariant vector products are small, and we may 19

And comforting.

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287

neglect the electron mass. We then have an example of a process in which helicity-spinor techniques can be usefully employed. We shall do that, slowly and explicitly. The amplitude is given by M(ν, λ1 , λ2 ) = i~Qe 2 (A1 (ν, λ1 , λ2 ) + A2 (ν, λ1 , λ2 )) , (/p2 − k/1 ) /λ1 (k1 ) uν (p2 ) , A1 (ν, λ1 , λ2 ) = uν (p1 ) /λ2 (k2 ) 2(p2 k1 ) (/p2 − k/2 ) A2 (ν, λ1 , λ2 ) = uν (p1 ) /λ1 (k1 ) /λ2 (k2 ) uν (p2 ) . (10.75) 2(p2 k2 ) Since me = 0 we may as well employ the symbol u for both the positron and the electron. Also, the helicity of the electron fixes that of the positron, and both are indicated by ν. The helicities of the two photons are denoted by λ1,2 . We start by taking ν = +. Then Eq.(9.56) tells us that we then have to use √ √ 2 − 2 u− (kj )u− (r+ ) , ω− /− (kj ) = u− (r− )u− (kj ) , /+ (kj )ω+ = s− (kj , r+ ) s+ (kj , r− ) (10.76) µ where we may choose the gauge vectors r± as we please, for each photon and for each polarisation case individually. The best choice is then r+ = p2 , r− = p1

(10.77)

since it makes lots of terms vanish. Indeed, if λ1 = λ2 = + both A’s contain u− (p2 )u+ (p2 ) as a factor, and for λ1 = λ2 = − we likewise find u+ (p1 )u− (p1 ) as a common factor, and therefore M(ν, λ, λ) = 0

(10.78)

which puts paid to 4 out of 8 helicity configurations20 . Of course the fact that E79 these amplitudes vanish cannot depend on our choice of r± , but the proof that they vanish is now very simple ! The next case is that of λ1 = −λ2 = +. We see that A1 still vanishes, and we can compute A2 as follows (using (p2 k2 ) = (p1 k1 )) : √ √ 1 2 − 2 A2 = 2(p1 k1 ) s− (k1 , p2 ) s+ (k2 , p1 ) 20

For ν = 0 it suffices to interchange s+ and s− , and include a minus sign for each polarization vector.

288

August 31, 2019 × u+ (p1 )u− (k1 )u− (p2 ) (/p2 − k/2 ) u− (p1 )u− (k2 )u+ (p2 ) u+ (p1 )u− (k1 )u− (p2 ) k/2 u− (p1 )u− (k2 )u+ (p2 ) = (p1 k1 ) s− (k1 , p2 )s− (k2 , p1 ) s+ (p1 , k1 )s− (p2 , k2 )s+ (k2 , p1 )s− (k2 , p2 ) = (p1 k1 ) s− (k1 , p2 )s− (k2 , p1 ) s− (k2 , p2 )2 (10.79) = −2 s− (k1 , p1 )s− (k1 , p2 )

The case λ1 = −λ2 = − amounts to simply interchanging k1 and k2 . The four nonvanishing amplitudes are therefore −2i~Qe 2 s−ν (k2 , p2 )2 −2i~Qe 2 s−ν (k1 , p2 )2 , M(ν, −ν, ν) = , s−ν (k1 , p1 )s−ν (k1 , p2 ) s−ν (k2 , p1 )s−ν (k2 , p2 ) (10.80) and the spin-averaged transition rate is  

 k1 · p1 k2 · p1 2 2 2 |M| = 32π α + . (10.81) k2 · p1 k1 · p1 M(ν, ν, −ν) =

The computation of the cross section is left as an excercise. We have discussed this process, rather, to show how spinor techniques may be usefully employed to compute amplitudes for massless-particle processes in a fast and efficient manner ; moreover, we can gain results (such as the vanishing of the amplitude when the photons helicities are equal) that are not so easily obtained by more traditional approaches21 .

10.3.4

Bhabha scattering

Our final 2 → 2 QED process is that of Bhabha scattering, the process e+ (p1 ) e− (p2 ) 21

→

e+ (q1 ) e− (q2 , ) ,

(10.82)

A word of caution is in order here. The Minkowski products (pi kj ) can become small if the photons are emitted collinearly. In that case these products are of order m2 rather than of order s. It is therefore not adviseable to blindly put m = 0 in any process in which photons are emitted, since then we might miss terms looking like m2 /(pi kj )2 . As can be seen from the matrix element for Compton scattering, in this case the double-pole term is actually suppressed by m4 rather than by m2 , and therefore at high energies we do not have to worry about double poles for this process. For other Bremsstrahlung processes such as e+ e− → µ+ µ− γ, the double poles are important : see section 10.3.5.

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described by the two following Feynman graphs :

p1

q1

M =

p1

q1

+

p2

q2

(10.83) p2

q2

We shall use, in addition to s, the following conventional invariants : t = (p1 − q1 )2 = (p2 − q2 )2 ,

u = (p1 − q2 )2 = (p2 − q1 )2 .

(10.84)

For me = 0 we have s + t + u = 0 by momentum conservation. As before, E80 we shall work in the high-energy limit so that me is neglected. The helicitydependent amplitude is M(λ1 , λ2 , ρ1 , ρ2 ) = i~Qe 2 A(λ1 , λ2 , ρ1 , ρ2 ) , 1 A(λ1 , λ2 , ρ1 , ρ2 ) = uλ (p1 ) γ µ uλ2 (p2 ) uρ2 (q2 ) γµ uρ1 (q1 ) s 1 1 − uλ (p1 ) γ µ uρ1 (q1 ) uρ2 (q2 ) γµ uλ2 (p2 ) .(10.85) t 1 Note the relative minus sign between the two diagrams ! By the Chisholm identity, we can now evaluate the various helicity configurations : 2 2 s+ (p1 , q2 )s− (q1 , p2 ) − s+ (p1 , q2 )s− (p2 , q1 ) s  t  1 1 u2 ∼ 2u + , ∼ 2 s t st 2 t A(+, +, −, −) = s+ (p1 , q1 )s− (q2 , p2 ) ∼ 2 , s s 2 s A(+, −, +, −) = − s+ (p1 , p2 )s− (q2 , q1 ) ∼ 2 , (10.86) t t A(+, +, +, +) =

where the symbol ∼ denotes our throwing away unimportant complex phases. The other helicity configurations with λ1 = + give zero, and those with λ1 = − follow again trivially by conjugation. We find  2 4 4 4

 3 + cos2 θ 4 s +t +u 2 2 2 2 |M| = 2~ Qe = 16π α , (10.87) s2 t2 1 − cos θ where θ is the angle between p~1 and ~q1 in the centre-of-mass frame in which most e+ e− scattering experiments are performed. Note that, in this case,

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August 31, 2019

the singularity is not due to our neglecting the electron mass ; indeed, for nonzero mass we have t = (p1 − q1 )2 = 2m2 − 2(p1 0 )2 + 2|~p1 |2 cos θ = −2|~p1 |2 (1 − cos θ) . E81 E82

(10.88)

To this order in perturbation theory, the total cross section for Bhabha scattering is therefore indeed divergent.

10.3.5

Bremsstrahlung in Mœller scattering

The nonradiative process Mœller scattering is the mutual scattering of two electrons : e− (p1 ) e− (p2 ) → e− (q1 ) e− (q2 ) and is just a crossed version of Bhabha scattering. The relevant expression is therefore, for negligible electron mass,

 s4 + t4 + u4 (10.89) |M|2 = 2~2 Qe 4 s2 u 2 The radiative process We shall now consider the so-called Bremsstrahlung22 process : e− (p1 ) e− (p2 ) → e− (q1 ) e− (q2 ) γ(k) At the tree level, it is described by the eight Feynman diagrams k

q1

M =

p1 p2

k

+

q2

p1

k q2

+

+

p1 p2

+

p

q2

k

+

p1 p2

q2

k

q2 q1

q1

p2 1

q1

q2 k

q1 p2

p1 p2

q1

q1

p2 p1

p2

q1 k q2

+

p1 k

q2

,

(10.90) 22

The term originated in studies of the motion of charged particles through a medium ; they may lose energy by emitting photons, and slow down, or ‘brake’, or – in the language of early-twentieth-century physics, which was German rather than American English – ‘Bremsen’.

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291

which we may conveniently put in four groups of two diagrams each : M =

4 X

Mi ,

i=1

M1 = −i(Qe

√

  /1 − k/ + me q/1 + k/ + me α αp ~) u(q1 ) / γ −γ / u(p1 ) 2q1 · k 2p1 · k 3

1 u(q2 )γα u(p2 ) , (p2 − q2 )2 = M1 cp1 ↔p2 , q1 ↔q2 , = − M1 cp1 ↔p2 , M4 = − M2 cp1 ↔p2 . ×

M2 M3

(10.91)

Note the Fermi minus sign between M1,2 and M3,4 . The four pairs of diagrams are separately current-conserving, i.e. Mi c→k = 0 , i = 1, 2, 3, 4 .

(10.92)

The Soft-photon approximation Since the emitted photon is a massless particle, its energy can be arbitrarily low. A useful result can be obtained if we take this limit, that is, the photon energy is taken to be negligible with respect to the other particle energies. Consider an arbitrary process in which a fermion with momentum q and mass m is produced during a scattering : M0 =

q

A

(10.93)

this amplitude can be written as M0 ≡

√ ~ u(q) A(q) ,

(10.94)

where A denotes the rest of the diagram(s). The corresponding radiative process will (amongst others) contain diagrams in which the photon is emitted by this particular fermion : Ms =

A

q k

(10.95)

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which evaluates to √ q/ + k/ + m A(q + k) . Ms ≡ −(Qe ~) u(q) / 2q · k

(10.96)

Notice that the denominator q · k goes to zero as the photon energy vanishes, and hence the diagram diverges in the limit of soft Bremsstrahlung . In the soft-photon approximation ( and assuming that the object A does not depend on q in too drastic a manner23 ) we have √ q/ + m Ms ≈ −(Qe ~) u(q) / A(q) . 2q · k

(10.97)

Anticommuting / and q/, and using the property of the Dirac spinor, which tells us that u(q)/q = mu(q), we then find √ q· u(q)A(q) , (10.98) Ms ≈ −(Qe ~) q·k that is, the diagram factorizes into the nonradiative result and an ‘infrared factor’24 . We can repeat this procedure for those diagrams in which the photon is emitted by the other external particles. There are, of course, also (possibly) diagrams in which the photon is emitted from internal lines ; but, as can easily be checked, such diagrams do not diverge as k 0 → 0. In the soft-photon approximation, they do therefore not contribute. For radiative Mœller scattering, we therefore have the nicely factorized form √ M = −(Qe ~) (VIR · ) M0 , q1µ q2µ pµ1 pµ1 µ VIR = + − − , (10.99) q1 · k q2 · k p1 · k p1 · k where M0 is the amplitude for the nonradiative process ; and, using the polarisation sum rule Σµ ν = −g µν , we find

 s4 + t4 + u4 |M|2 = −2Qe 6 ~3 (VIR · VIR ) , t2 u2 m2e m2e m2e m2e −VIR · VIR = − − − − (p1 k)2 (p2 k)2 (q1 k)2 (q2 k)2 23

This assumption fails, for instance, close to a resonance. However, since every resonance has a finite width, the soft-photon approximation is formally correct for infinitesimal photon energies. 24 Since infrared light has low energy compared to visible light.

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293 2(p1 p2 ) 2(q1 q2 ) 2(p1 q1 ) + − (p1 k)(p2 k) (q1 k)(q2 k) (p1 k)(q1 k) 2(p1 q2 ) 2(p2 q1 ) 2(p2 q2 ) − − − . (10.100) (p1 k)(q2 k) (p2 k)(q1 k) (p2 k)(q2 k)

+

As has already been intimated, the double poles are indeed suppressed by a factor me 2 . Hard Bremsstrahlung: massless case Next, we consider ‘hard Bremsstrahlung’ (i.e. any photon emission which is not soft) in the limit of vanishing electron mass. It is then most useful to assign definite helicities to the electrons, so that the scattering process is e− (p1 , µ1 ) e− (p2 , µ2 ) → e− (q1 , ν1 ) e− (q2 , ν2 ) γ(k, λ) with µ1,2 , ν1,2 , λ = ±. The amplitude is then a function of the helicities, and we write M(µ1 , µ2 ; ν1 , ν2 ; λ). We first consider M1 (+, +; +, +; +). Using Eq.(9.56) this can be written as √ √ (Qe ~)3 2 M1 (+, +; +, +; +) = i 2(p2 · q2 )s− (k, r)   q/1 + k/ α /1 − k/ αp γ −γ u− (k)u− (r) u+ (p1 ) × u+ (q1 ) u− (k)u− (r) 2k · q1 2k · p1 (10.101) × u+ (q2 )γα u+ (p2 ) , and since M1 is current-conserving by itself we may choose r at will ; in this case r = p1 appears to be optimal since it kills the second term. Applying standard (hopefully, by now) spinor techniques we arrive at M1 (+, +; +, +; +) = √ √ s+ (q1 , k)u− (p1 )(/q1 + k/ )u− (q2 )s− (p2 , p1 ) . (10.102) i(Qe ~)3 8 (2p2 · q2 )(2k · q1 )s− (k, p1 ) We may employ momentum conservation and masslessness for a further manipulation : u− (p1 )(/q1 + k/ )u− (q2 ) = u− (p1 )(/q1 + k/ + q/2 )u− (q2 ) = u− (p1 )(/p1 + p/2 )u− (q2 ) = s− (p1 , p2 )s+ (p2 , q2 ) ,

(10.103)

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so that √ √ M(+, +; +, +; +) = i(Qe ~)3 8

s− (p1 , p2 )2 . s− (p2 , q2 )s− (k, p1 )s− (k, q1 )

(10.104)

Finally, we can make use of the identity of Eq.(9.63) to arrive at the form √

s− (p1 , p2 )2 ~) s− (p1 , q1 )s− (p2 , q2 )

 + · p1 + · q1 M1 (+, +; +, +; +) = −2i(Qe − . k · p1 k · q1 (10.105) The infrared factor also appears in this case ! Performing the appropriate subtitutions we can write the complete amplitude as 3



√ M(+, +; +, +; +) = −2i(Qe ~)3 s− (p1 , p2 )2 (VIR · + )   1 1 − × . s− (p1 , q1 )s− (p2 , q2 ) s− (p1 , q2 )s− (p2 , q1 )

(10.106)

The minus sign in the last term is the Fermi sign ; it helps us to simplify our expression even further using the Schouten identity, and the final form for the amplitude is √ M(+, +; +, +; +) = 2i(Qe ~)3

s− (p1 , p2 )3 s− (q1 , q2 ) (VIR · + ) . s− (p1 , q1 )s− (p2 , q2 )s− (p1 , q2 )s− (p2 , q1 ) (10.107) For the other helicity configurations, the above treatment can be repeated straightforwardly. The resulting nonvanishing amplitudes turn out to all have quite similar forms : M(µ1 , µ2 ; ν1 , ν2 ; λ) = (VIR · λ ) σ s−λ (a, b)3 s−λ (c, d) 2i~3/2 Q3 , (10.108) s−λ (p1 , q1 )s−λ (p2 , q2 )s−λ (p1 , q2 )s−λ (p2 , q1 ) where the sign σ and the identities of the four vectors a, b, c and d are given in the following table :

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295 ν1 + + + + + + – – – – – –

µ1 + + + + – – – – – – + +

ν2 + + – – – – – – + + + +

µ2 + + – – + + – – + + – –

λ + – + – + – + – + – + –

σ a + p1 + q1 + q2 – q1 – q2 + q1 – q1 – p1 + q1 – q2 – q1 + q2

b p2 q2 p1 p2 p2 p1 q2 p2 p2 p1 p1 p2

c q1 p1 q1 q2 q1 q2 p1 q1 q2 q1 q2 q1

d q2 p2 p2 p1 p1 p2 p2 q2 p1 p2 p2 p1

A few things may be noted, apart from the surprising simplicity of the amplitudes. In each case, only s+ or s− occur. Moreover, when the electrons have equal spins the amplitudes are explicitly antisymmetric in p1 ↔ p2 or q1 ↔ q2 , as required by Fermi statistics25 . The spin-averaged matrix element squared therefore has the following form in the strictly massless case :

 |M|2 me =0 = −2Qe 6 ~3 (VIR · VIR ) ss0 (s2 + s02 ) + uu0 (u2 + u02 ) + tt0 (t2 + t02 ) (10.109) , uu0 tt0 with s = (p1 +p2 )2 , s0 = (q1 +q2 )2 , t = (p1 −q1 )2 , t0 = (p2 −q2 )2 , u = (p1 −q2 )2 , and u0 = (p2 − q1 )2 . The final result is surprisingly simple. It consists of the ‘soft-photon’ factor VIR 2 (evaluated for non-soft photon momenta), multiplying a ‘symmetrized’ form of the nonradiative cross section. ×

Double-pole terms at high energy We have already mentioned that putting me = 0 strictly may be too strict since there are invariant products of momenta that may become equally small. To see how this works, let us again inspect the radiation emitted from a produced fermion, as given in figure 10.95, that can be written as √ q/ + k/ + m A(q + k) (10.110) Mc ≡ −(Qe ~) u(q)/ 2q · k 25

If the incoming electrons have unequal spin they are distinguishable, and no antisymmetry is required.

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where, as before, A stands for the rest of the diagram(s). We shall not assume the soft-photon limit, however. Let us assume that the photon is emitted as small angle θ with respect to the fermion momentum. We then find, assuming the fermion energy to be large compared to its mass m :
(k · q) = k 0 q 0 − |~q| cos θ  2 !
me 1 ≈ k 0 (q 0 − |~q|) + |~q|θ2 /2 ≈ k 0 q 0 θ2 + (10.111) , 2 q0 where we have used the fact that q 0 − |~q| = me 2 /(q 0 + |~q|) ≈ me 2 /(2q 0 ). we conclude that as soon as θ is of order me /q 0 or smaller26 , the product (k · q) becomes of order m2e ; and this means that in that case the ‘single pole’ (k · q)−1 and the ‘double pole’ me 2 (k · q)−2 are of the same order27 . The squared matrix element (summed over fermion and photon spins) contains of course

 Qe 2 ~ |Mc |2 = − 4(k · q)2 × A(q + k)(/q + k/ + m)γ α (/q + m)γα (/q + k/ + m)A(q + k)(10.112) Using standard Dirac algebra we can write (/q + k/ + m)γ α (/q + m)γα (/q + k/ + m) = 4m2 (/q + k/ + m) + 4(k · q)(m − k/ ) .

(10.113)

The second term in this expression enters into the ‘massless’ result since it will give rise only to single-pole terms, whereas the first term tells us that the double-pole term coming from this Mc must read

2

|Mc |



m2 A(q + k)(/q + k/ )A(q + k) , = −Qe ~ (k · q)2 2

(10.114)

where we have again discarded terms of order m. The nonradiative transition rate was given by (A)/qA(q), and in this expression we have now substituted q+k for q. We can, by momentum conservation, always express the invariants s, t and u in Eq.(10.89) into a form that does not contain q, and this then 26 27

This is sometimes called the ‘ultra-collinear case’. For this reason we use the subscript c which stands for ‘collinear’.

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297

gives us the double-pole terms : keeping all four collinear situations in sight, we can write the transition rate including the double-pole terms as

2

|M|



6

2



= Qe ~

ss0 (s2 + s02 ) + tt0 (t2 + t02 ) + uu0 (u2 + u02 ) tt0 uu0  2(q1 · q2 ) 2(p1 · q1 ) 2(p1 · p2 ) − + × − (k · p1 )(k · p2 ) (k · q1 )(k · q2 ) (k · p1 )(k · q1 )  2(p2 · q2 ) 2(p1 · q2 ) 2(p2 · q1 ) + − − (k · p2 )(k · q2 ) (k · p1 )(k · q2 ) (k · p2 )(k · q1 ) me 2 2(s04 + t04 + u04 ) me 2 2(s04 + t4 + u4 ) − − (k · p1 )2 t02 u02 (k · p2 )2 t2 u2  me 2 2(s4 + t4 + u02 ) me 2 2(s4 + t04 + u2 ) − − (10.115) (k · q1 )2 t02 u2 (k · q2 )2 t2 u02 This is our final expression for unpolarised Mœller scattering ; it is accurate in the limit of small me even for collinear28 photon emission. E83 Radiation cones and Yennie’s crater For every charged fermion with momentum p in a scattering process with emission of a photon with momentum k over an angle θ between p~ and ~k there are terms in the cross section that go as 1 2 1
, = 0 0 ∼ p·k k (p − |~p| cos θ) k 0 p0 (m/p0 )2 + θ2

(10.116)

where we have assumed that m/p0 and θ are small. This has been the reason for statements such as ‘the radiation is essentially confined to a cone of angle m/p0 around the fermion direction’, the so-called radiation cone. This is E84 quite wrong since the polar angular integration variable is d cos θ ∼ d(θ2 )/2 and not dθ. Suppose that the photon is strictly parallel to the fermion, ~k//~p. Since the photon polarisation is strictly orthogonal, ~ · ~k = 0, we must also have ~ · p~ = 0, and the soft-photon approximation 10.99 then tells us that the diagram with denominator (p · k) vanishes in the soft-photon limit : the

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angular distribution shows a sharp dip known as Yennie’s crater. For photons E85 that are not infinitesimally soft, the crater is less deep but still there.

10.4

Scalar electrodynamics

10.4.1

The vertices

We can also consider the possibility of interactions between photons and charged scalar particles29 . The simplest vertex is then given by

q p

µ

where the charge flow is indicated by the arrow. The photon index is µ. The momenta p and q are counted along the arrow. Note that the propagator of scalar particles may be unoriented, but the vertices do not have to be so if there is a quantum number, such as charge, that distinguishes between particle and antiparticle. Since no Dirac matrices occur the only quantities in this vertex that carry a Lorentz index are the momenta p and q (and of course the photon’s own momentum, but that is fixed by p and q). We therefore propose a Feynman rule of the form

q p

µ

↔

i

Q (c1 pµ + c2 q µ ) , ~

with constants c1,2 to be determined. This is simple, since we can study the annihilation of the charged scalar-antiscalar pair into an off-shell photon : under the handlebar operation, the amplitude becomes

p1 p2

28

√ = iQ ~ (c2 p2 µ − c1 p1 µ ) kµ

k

√
= iQ ~ (c2 p2 µ − c1 p1 µ ) p1 µ + p2 µ √ (c2 − c1 ) = iQ ~(p1 + p2 )2 . 2

(10.117)

Or ultra-collinear. Elementary charged scalar particles have to date not been observed, although they are predicted in extensions of the standard model. We include them here since they will provide indications on how to treat charged vector particles. 29

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299

We see that c1 = c2 is required, and therefore the first Feynman rule for scalar electrodynamics (sQED) reads

µ ↔ i Q (p + q)µ ~

q p

sQED vertex

sQED Feynman rules, version 10.1

(10.118)

Let us now consider the more complicated process of annihilation into two on-shell photons. With the above vertex two diagrams are involved :

p1 M =

k1 p2

k2 +

p1

k2 p2

k1

(10.119)

The amplitude is then given, with m indicating the scalar’s mass, by (p1 + (p1 − k1 )) · 1 ((p1 − k1 ) + (−p2 )) · 2 + (k1 ↔ k2 ) (p1 − k1 )2 − m2   (p1 · 1 )(p2 · 2 ) (p1 · 2 )(p2 · 1 ) 2 + = −2i~Q (10.120) (p1 · k1 ) (p2 · k1 )

M = −i~Q2

The test of current conservation now fails, since Mc1 →k1 = −2i~Q2 ((p2 · 2 ) + (p1 · 2 )) = −2i~Q2 (k1 · 2 ) .

(10.121)

The solution is to introduce a four-point vertex into the Feynman rules. For reasons lost in the mists of time, such a vertex is called a sea-gull vertex30 . The improved Feynman rules are now : 30

To me it does not look very gully nor even particularly birdy.

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µ ↔ i Q (p + q)µ ~

q p

µ ↔ 2i

ν

sQED 3-vertex

Q2 µν g ~

sQED 4-vertex

sQED Feynman rules, version 10.2

(10.122)

Now we find immediately the desired current conservation : +

+

= 0 .

(10.123)

You might worry that annihiliation into three photons will necessitate a fivepoint vertex, and so on. Fortunately, the above two vertices are sufficient to guarantee current conservation in all sQED processes, as we shall now show using some more handlebar diagrammatics.

10.4.2

Proof of current conservation in sQED

Consider a charged scalar propagator somewhere in a Feynman diagram, and assume a photon attached to it : µ k p

i~ = 2 p − m2

q

  i~ Q µ . i (p + q) 2 ~ q − m2

As in our proof for regular QED, none of these lines is necessarily on-shell. . Momentum conservation again fixes the photon momentum to be k = p − q. In analogy to regular QED we can now invent some handlebar diagrammatics as follows : (p − q) · (p + q) (p2 − m2 )(q 2 − m2 )     i~ Q Q i~ = 2 i (i~) + (i~) i 2 2 q −m ~ ~ p − m2 = −iQ~

=

−

,

(10.124)

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301

with the trivial auxiliary rules = i~ ,

=i

Q . ~

(10.125)

These rules are very similar to those we adopted in regular QED : however, in general we have 6=

(10.126)

since the scalar-scalar-photon vertex still depends on the various momenta. We now turn to the second vertex, with two photon lines. Not denoting the two scalar propagators, we have µ k p

q

Q2 (p − q − k)sµ ~         Q Q Q Q µ µ = i (i~) i (2q + k) − i (2p − k) (i~) i , ~ ~ ~ ~ (10.127)

= 2i

in other words, −

=

.

(10.128)

The proof of current conservation again relies on the SDe’s for this model : =

+

+

,

=

+

+

,

=

+

+

, (10.129)

where again we have used semi-connected graphs. The handlebar operation is now seen to lead to =

+

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August 31, 2019 −

=

−

+

(10.130) If we now iterate the SDe cleverly for the first two of these four diagrams, we obtain =

+ −

− −

+ = 0 ,

(10.131)

since we do have =

,

(10.132)

owing to the simple, momentum-independent structure of the seagull vertex. Comparing the lines of the proof for sQED with that of regular QED, the general proof strategy becomes clear : if in a diagram a slashed propagator occurs as one of the indicated lines of a (semi-)connected graph, we must iterate de SDe for that line, and then we can collect the various canceling contributions.

10.5

The Coulomb potential

The Feynman rules that we have constructed are designed to describe dynamical processes. To say something about the static character of QED we return to our treatment of the Yukawa potential of section 5.2.4. There, we used the result Z ~ x) 1 1 exp(−mr) 3~ exp(ik · ~ d k = , r = |~x| (10.133) 3 (2π) 4π r |~k|2 + m2

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303

to derive (up to an overall factor) the Yukawa potential. But we used a single source and looked at the field’s response to it : now, we have to consider two particles, say, scattering ever so slightly. Let us start with two scalars31 with masses M1,2 and charge Qe . Using the sQED Feynman rules we find for the tree amplitude : p

q

1

M=

p2

k

1

q2

=

i~Q2 (p1 + q1 · p2 + q2 ) . k 2 + i

(10.134)

Since p01 = q10 and p02 = q20 in the centre-of-mass frame we have k 0 = 0 and we can write i~Q2e (4M1 M2 ) , (10.135) M = M(~k) = − |~k|2 where we have assumed the static limit (SL) in the numerator ; and this we have to Fourier-transform to find the interaction energy V (r). The correct convention (as we shall see) is to use Z i 1 d3~k M(~k) exp(i~k · ~x) , (10.136) V (r) = 4M1 M2 (2π)3 so that we find, for this scattering, V (r) = α/r, a repulsive interaction32 . If we now change the direction of the charge flow in the (p2 , q2 ) line in Eq.(10.134), the amplitude changes sign by the sQED Feynman rules, and the interaction energy becomes attractive. Therefore, like charges repel, and opposite charges attract33 . We may also consider the ‘original’ Yukawa interaction, mediated by a scalar particle. Its propagator has numerator i~ rather than the −i~g µν for the photon, with no sign change in the interaction vertex if we replace particle by antiparticle. We conclude that in the exchange of a scalar particle rather than a vector one, the interaction is always attractive. Let us now turn to the case of regular QED, and consider e− µ− scattering : M(~k) = 31

p

q

1

p2

k

1

q2

=

−i~Q2e u(q1 )γ µ u(p1 ) u(q2 )γµ u(p2 ) . (10.137) 2 ~ |k|

We shall assume that they are not identical so as to avoid trouble with having to deal with two diagrams rather than one. 32 What about the factor 4M1 M2 ? It is just the wave function normalization factor for two nonrelativistic particles. 33 If we had chosen the other convention in Eq.(10.136), with −i rather than i, we would have found that like charges attract, contrary to experience. The point here is that the interaction can change sign between like-charge and opposite-charge.

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By the Gordon decomposition (see section 10.2.6) this evaluates to exactly the result (10.135), hence like-sign fermions also repel each other. But let us now consider fermion-antifermion scattering, e− µ+ : we would write M(~k) =

p

q

1

p2

k

1

q2

=

−i~Q2e u(q1 )γ µ u(p1 ) v(p2 )γµ v(q2 ) , 2 ~ |k|

(10.138)

and find, from the Gordon decomposition, that opposite-sign fermions also repel ! But we have forgotten the Fermi minus sign lurking backstage. The correct Feynman rules are those of Eq.(7.123) rather than the ‘prettified’ ones of Eq.(7.151), and we obtain an extra sign that tells us that oppositelycharged fermions attract each other just like oppositely-charged scalars34 .

10.6

Electrons in external fields : g = 2

10.6.1

The charged Klein-Gordon equation

A point of particular interest is the way in which electrons react to external fields. Here ‘external’ is used for fields that are not of the fluctuating quantum type, but rather applied ‘from outside’, under our experimental control ; this is the sense in which it was used in section 10.2.5. In the absence of an explicit source, the Dirac equation with external field A reads (cf. Eq.(10.28)) :
γµ (i ∂ µ − eAµ (x)) − m ψ(x) = 0 (10.139) in the position representation, where e denotes the coupling. Let us compare this to the behaviour of a scalar electron. For a charged scalar in an external field we can write down a classical (tree-level) SDe 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

=

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

+

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

(10.140)

or, by explicit use of the Fourier transforms of the fields : Z Z i~ 1 4 φ(x) = dy d4 k e−ik·(x−y) 2 4 (2π) k − m2 + i 34

There is a tendency among theorists to take the prettified rules as the ‘real’ ones, with the Fermi sign occurring only in loops. As we see that is dangerous ! Note also that it does not matter whether we assign the Fermi sign to the v or to the v, as indeed it should not.

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305



Z e 1 −i d4 pd4 qe−ip·y−iq·y (2p + q)µ Aµ (q) φ(p) ~ (2π)8  2  1 e µ + 2i Aµ (y)A (y)φ(y) . (10.141) 2 ~

Note the occurence of the symmetry factor 1/2 in the last line. Applying the differential operator −∂ 2 − m2 , and using the expressions for derivatives in Fourier representation we arrive at
− ∂ 2 − m2 φ(x) = 2ieAµ (x)∂µ φ(x) +
ie ∂µ Aµ (x) φ(x) − e2 Aµ (x)Aµ (x)φ(x) (10.142) or

 i∂ − eA(x)


2

2

−m

 φ(x) = 0 .

(10.143)

This is the Klein-Gordon equation for charged scalar fields. We see that the same ‘minimal substitution rule’ pµ → pµ − eAµ as in the Dirac case is employed to account for the presence of the e.m. field ; and we see that the charge coupling constant e is defined in the same way for both scalar and Dirac particles.

10.6.2

The relativistic Pauli equation

We can cast the Dirac equation for spin-1/2 electrons in a form closely resembling that of the Klein-Gordon equation, by multiplying Eq.(10.139) :   

µ µ ν ν 0 = γµ i∂ − eA (x) + m γν i∂ − eA (x) − m ψ(x)   2 2 2 2 = − ∂ + e A(x) − m ψ(x) −ieγ µ γ ν ∂µ Aν (x)ψ(x) − ieγ ν γ µ Aν (x)∂µ ψ(x) . The second line can be rewritten as  
µν µν −ie g − iσ ∂µ Aν (x) ψ(x) − 2ieAµ (x) ∂µ ψ(x)

(10.144)

(10.145)

We find, for spin-1/2 electrons, the relativistic Pauli equation, the KleinGordon equation with an extra spin term, the so-called Stern-Gerlach term added on : E86  
2
i∂ − eA(x) − m2 − eσ µν ∂µ Aν (x) φ(x) = 0 . (10.146)

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10.6.3

A constant magnetic field

The spin of particles manifests itself most clearly in a magnetic field. Since the notions of electric and magnetic fields are of dubious Lorentz covarience, we adopt the following. Let tµ be a constant vector with t2 = 1, and let B µ be a constant vector with (B · t) = 0. We now choose the external A field as follows : 1 Aµ (x) = εµ (t, B, x) . (10.147) 2 ~=B ~ × ~x/2, which In the ‘rest frame’ where ~t = 0 this reduces to A0 = 0 , A ~ corresponds to a constant magnetic field of strength B. In the Klein-Gordon equation, the term linear in A is given by35 −2ieA(x) · ∂ψ(x) = −ie ε(∂, t, B, x) ψ(x) = e ε(t, B, x, i∂) ψ(x) = −e Bµ εµ (t, x, i∂) ψ(x) . (10.148) We can introduce the angular momentum operator Lµ = εµ (t, x, i∂)

(10.149)

~ = i~x × ∂~ = ~x × p~. In the rest frame we since, for ~t = 0, this reduces to L therefore have the interaction term ~ ·L ~ ψ(x) −2ieA(x) · ∂ψ(x) = e B

(10.150)

which represents the coupling between the magnetic field and the angular momentum of the moving charge. For the Stern-Gerlach term, we have e µν σ εµναβ tα B β 2 = ieγ 5 σαβ tα B β = −eB / γ 5 t/ .

−e σ µν ∂µ Aν (x) =

(10.151)

We now make the approximation that the particle is moving nonrelativistically slow in the rest frame ~t = 0. In that case we may write, in momentum language, 1 t/ u(p, s) → p/ u(p, s) = u(p, s) , (10.152) m so that the Stern-Gerlach interaction term in this limit reads −e σ µν ∂µ Aν (x) ψ(x) = eBµ γ 5 γ µ ψ(x) . 35

The term proportional to (∂ · A) vanishes because of the Lorenz condition.

(10.153)

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307

Now, as we have seen in Eq.(7.111), the operator −γ 5 γ µ describes the spin vector sµ , which is normalized to s2 = −1. However, as we have seen the actual spin S is one-half this amount since the electron has total spin one half. We therefore write ~ ·S ~ ψ(x) , −e σ µν ∂µ Aν (x) ψ(x) = 2e B

(10.154)

~ = −γ 5~γ /2 is the operator for the spin of a nonrelativistic electron. where S We see that the coupling of a (nonrelativistic) electron to an external magnetic field is proportional to ~ + ge S) ~ ·B ~ , (L

ge = 2 ,

where the factor ge comes somewhat as a surprise : it is called the gyromagnetic ratio of the electron36 . The prediction ge = 2 was one of the first and most welcome results from Dirac’s description of the electron. There is a way to pinpoint the gyromagnetic behaviour of an electron in a more precise and useful manner, using the Gordon decomposition of section 10.2.6, which reads   1 µ µ µ ν u(q) γ u(p) = u(q) (p + q) + (p − q)ν [γ , γ ] u(p) . (10.155) 2m Since the convection term (p + q)µ describes the coupling of a scalar to the photon, it is the spin term which must be responsible for the result ge = 2 for the electron. The above implies that, by calculating loop corrections to the electron-photon vertex, we can isolate the σ part of the loop-corrected vertex and infer the loop corrections to ge .

10.7

Selected topics in QED

10.7.1

Three-photon production

A fine example of a quite nontrivial computation using standard-spinor (‘helicity’) techniques is provided by the process of three-photon annihiliation in e+ e− collisions : e+ (p1 ) e− (p2 ) → γ(k1 ) γ(k2 ) γ(k3 ) . 36

(10.156)

Composite spin-1/2 particles can have different values for g : the proton and neutron have about 5.586 and -3.826, respectively.

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We shall again alculate this for massless electrons. At the tree level there are 6 Feynman diagrams, so that the amplitude reads M(λ; λ1 , λ2 , λ3 ) = i~3/2 e3 (/k3 − p/1 ) (/p2 − k/1 ) × uλ (p1 )/λ3 (k3 ) /λ2 (k2 ) /λ (k1 )uλ (p2 ) 2(k3 p1 ) 2(k1 p2 ) 1 + (perm) . (10.157) Here we have explicitly indicated the various helicities, and ‘perm’ stands for the other 5 permutations of the photons. For λ = + we take the photon polarisations as in section 10.3.3, and that immediately tells us that the amplitude vanishes if λ1 = λ2 = λ3 (as was already anticipated in exercise 79). We see that the only helicity amplitude that we have to work hard on is, say, M(+; − + +). Neglecting, for now, the overall factor i(2~e2 )3/2 we can write M(+; − + +) = u− (k3 )u− (p2 ) (/k3 − p/1 ) u− (k2 )u− (p2 ) (/p2 − k/1 ) u− (p2 )u− (k1 ) u+ (p1 ) u+ (p2 ) s− (k3 , p2 ) 2(k3 p1 ) s− (k2 , p2 ) 2(k1 p2 ) s+ (k1 , p2 ) + (perm) . (10.158) Actually, only two diagrams contribute here, namely the one written down and the one where k2 and k3 are interchanged. With

and

u+ (p1 )u− (k3 ) 1 = 2(k3 p1 ) s− (k3 , p1 )

(10.159)

u− (p2 )(/p2 − k/1 )u− (p2 ) = −1 2(k1 p2 )

(10.160)

we can rewrite u− (p2 )(/k3 − p/1 )u− (k2 ) s− (k1 , p2 ) + (k2 ↔ k3 ) . s− (k3 , p1 ) s− (k3 , p2 ) s− (k2 , p2 ) s+ (k1 , p2 ) (10.161) We can simplify further : M(+; − + +) = −

u− (p2 )(/k3 − p/1 )u− (k2 ) = u− (p2 )(/k2 + k/3 − p/1 − p/2 )u− (k2 ) = −u− (p2 )/k1 u− (k2 ) , (10.162)

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Making the denominator symmetric in k2 and k3 gives us M(+; − + +) u− (p2 )/k1 u− (k2 ) s− (k2 , p1 ) s− (k1 , p2 ) + (k2 ↔ k3 ) . (10.163) = s− (k3 , p1 ) s− (k3 , p2 ) s− (k2 , p1 ) s− (k2 , p2 ) s+ (k1 , p2 ) After yet another manipulation : u− (p2 )/k1 u− (k2 ) s− (k2 , p1 ) + (k2 ↔ k3 ) = u− (p2 )/k1 k/2 u+ (p1 ) + (k2 ↔ k3 ) = u− (p2 )/k1 (/k2 + k/3 )u+ (p1 ) = u− (p2 )/k1 p/2 u+ (p1 ) = s− (p2 , k1 )s+ (k1 , p2 )s− (p2 , p1 ) , (10.164) we arrive at M(+; − + +) =

s− (k1 , p2 )2 s− (p1 , p2 ) ; s− (k3 , p1 )s− (k3 , p2 )s− (k2 , p1 )s− (k2 , p2 )

(10.165)

putting it more symmetrically, and reinserting the overall factor, 2 3/2

M(+; − + +) = i(2~e )

s− (k1 , p2 )3 s− (k1 , p1 ) . s− (p1 , p2 ) 3 Q s− (kj , p1 )s− (kj , p2 )

(10.166)

j=1

We can easily infer the other helicity amplitudes. The final answer is 3 P

 |M|2 = 2~2 e4 (p1 p2 )

(kj p1 )(kj p2 ) [(kj p1 )2 + (kj p2 )2 ]

j=1 3 Q

.

(10.167)

(kj p1 )(kj p2 )

j=1

Looking at this result, we notice two interesting things. In the first place, it is very simple, something you would not have guessed right off, and certainly would have had to work on very hard using the classical Casimir-trick approach. In the second place, Eq.(10.166) contains only s− and no s+ , as in the two-photon annihiliation case of section 10.3.3. This is a general feature : conceding that M(+; +++) = 0 is the simplest possible amplitude, the ‘next-simplest’, in our case M(+; − + +), is both simple and holomorphic in the spinor products37 . Such amplitudes are called maximal helicity violating (MHV) and are an object of research in their own right, occurring in many theories with massless particles such as QCD with massless quarks and gluons, and even in gravity. 37

In the sense that s+ is the complex conjugate of s− up to a sign.

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The Thomson limit : scalar vs spinor

It is interesting to note that the Thomson cross section of Eq.(10.71) is actually a classical object, derivable within classical, nonrelativistic electrodynamics38 . It should therefore not depend on the fact that the electron is a Dirac particle : the amplitudes must be the same for spin-0 and spin-1/2 electrons in the low-energy limit. We shall now investigate this in some detail, at the tree level. In the static limit the kinetic energies are negligible with respect to the rest energies of massive particles. We shall, for our process e(p) γ(k1 ) → e(q) γ(k2 ) ,

(10.168)

define the momenta as follows, in the centre-of-mass frame: pµ = (E, 0, 0, k) , q µ = (E, 0, k sin θ, k cos θ) , k1 µ = (k, 0, 0, −k) , k2 µ = (k, 0, −k sin θ, −k cos θ) , (10.169) where E 2 = k 2 + m2 for electron mass m, the polar scattering angle of the photons is θ, and we have arbitrarily fixed the irrelevant azimuthal scattering angle. The static limit (SL) is defined by k → 0 and E → m. We start with the scalar case. The scattering is described by the three diagrams

p

p

q

M =

+

k1

k2

k1

k2 q

+

p

q

k1

k2

in which the charge flow is indicated. The amplitude reads

2 p · 1 q · 2 2 M(λ1 , λ2 ) = −ie ~ − (p · k1 ) !


2 q · 1 p · 2 − 2 1 · 2 . (q · k1 ) 38

(10.170)

(10.171)

After all, Thomson didn’t know either about quantum physics at the time, nor about relativity.

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Here e is the electron charge. The photon helicities are denoted by λ1,2 , and we shall use the representation 1 (λ)α =

uλ (k1 )γ α uλ (k2 ) uλ (k2 )γ α uλ (k1 ) p , 2 (λ)α = p , 4(k1 · k2 ) 4(k1 · k2 )

(10.172)

so that (j · p) = (j · q) (j = 1, 2) and 2 (λ) = 1 (λ)∗ . The first amplitude we consider is     1 1 2 2 − +2 . (10.173) M(+, +) = −ie ~ 2|p · 1 (+)| (pk1 ) (qk1 ) Using |p · 1 (+)|2 = and

Tr (ω+ k/1 p/k/2 p/) (pk1 )(qk1 ) m2 = − 4(k1 k2 ) (k1 k2 ) 2

(10.174)

1 1 (k1 k2 ) − =− (pk1 ) (qk1 ) (pk1 )(qk1 )

(10.175)

and we find the exact result M(+, +) = −ie2 ~

m2 (k1 k2 ) (E − k)(1 − cos θ) = −ie2 ~ . (pk1 )(qk1 ) E + k cos θ

(10.176)

In the SL this becomes M(+, +)SL = −ie2 ~(1 − cos θ) .

(10.177)

Since the amplitude must be dimensionless in energy units and m is the only scale in the SL, it is not surprising that m drops out altogether. The second amplitude is given by  
2 1 1 2 M(+, −) = −2ie ~ − p · 1 (+) (pk1 ) (qk1 ) (k1 k2 ) = 2ie2 ~ eiφ |p · 1 (+)|2 (pk1 )(qk1 ) 2(pk1 )(qk1 ) − m2 (k1 k2 ) = ie2 ~ eiφ (pk1 )(qk1 ) (E + k)(1 + cos θ) = ie2 ~ eiφ . (10.178) E + k cos θ

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Here we have introduced the phase factor u+ (k1 ) p/ u+ (k2 ) p · 1 (+) . = ∗ p · 1 (+) u− (k1 ) p/ u− (k2 )

(10.179)

M(+, −)SL = −ie2 ~ eiφ (1 + cos θ) .

(10.180)

eiφ = In the SL we arrive at

We now turn to the spin-1/2 case. In section 10.3.2 we already computed the cross section using the Casimir trick, but now we want the individual spin amplitudes. The amplitude, now given by two diagrams, reads  /2 (λ2 ) (/p + k/1 + m) /1 (λ1 ) 2 M = −ie ~ u(q) 2(pk1 )  /1 (λ2 ) (/q − k/1 + m) /1 (λ2 ) − u(p) . (10.181) 2(qk1 ) We do not bother to indicate the spins of the electrons explicitly, although we shall have to discuss them later. As before, we start with  ++  A1 A++ −ie2 ~ 2 − , M(+, +) = (k1 k2 ) 2(pk1 ) 2(qk1 )
A++ = u(q) u+ (k1 )u+ (k2 ) + u− (k2 )u− (k1 ) (/p + k/1 + m) × 1
u+ (k2 )u+ (k1 ) + u− (k1 )u− (k2 ) u(p) ,
A++ = u(q) u+ (k2 )u+ (k1 ) + u− (k1 )u− (k2 ) (/q − k/1 + m) × 2
(10.182) u+ (k1 )u+ (k2 ) + u− (k2 )u− (k1 ) u(p) . Using the Dirac equation and some algebra, we can rewrite
= u(q) 2(qk2 )ω+ k/1 + 2(pk1 )ω− k/2 − mω+ k/1 k/2 − mω− k/2 k/1 u(p) A++ 1
= u(q) 2(pk1 )/k2 − 2m(k1 k2 )ω+ u(p) ,
u(q) 2(qk )ω k / + 2(pk )ω k / − mω k / k / − mω k / k / A++ = u(p) 1 + 2 2 − 1 − 1 2 + 2 1 2
= u(q) 2(qk1 )/k1 − 2m(k1 k2 )ω+ u(p) . (10.183) Putting everything back together :   k/2 − k/1 m m 2 M(+, +) = −ie u(q) − ω+ + ω+ u(p)(10.184) . (k1 k2 ) (pk1 ) (qk1 )

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Since

u(q) k/2 − k/1 u(p) = u(q) p/ − q/ u(p) = 0 ,

(10.185)

we arrive at the exact result M(+, +) = −ie2 ~


m(k1 k2 ) u(q, s0 ) 1 + γ 5 u(p, s) , 2(pk1 )(qk1 )

(10.186)

where we have indicated the spins. In the SL, we may in this expression approximate q by p. The γ 5 term then drops out, and the amplitude vanishes takes again the form (10.177), and that if s0 = −s. We conclude that M++ SL the electron spin is not influenced in the SL. The second amplitude is given by  +−  −ie2 ~ A+− A1 2 M(+, −) = − , (k1 k2 ) 2(pk1 ) 2(qk1 )
A+− = u(q) u (k )u (k ) + u (k )u (k ) (/p + k/1 + m) × − 1 − 2 + 2 + 1 1
u+ (k2 )u+ (k1 ) + u− (k1 )u− (k2 ) u(p) ,
A+− = u(q) u+ (k2 )u+ (k1 ) + u− (k1 )u− (k2 ) (/q − k/1 + m) × 2
(10.187) u− (k1 )u− (k2 ) + u+ (k2 )u+ (k1 ) u(p) . Again using the various standard-form techniques we can establish that A+− = u(q)u− (k1 )u− (k2 )/pu− (k1 )u− (k2 )u(p) 1 + u(q)u+ (k2 )u+ (k1 )/pu+ (k2 )u+ (k1 )u(p)
= eiφ u(q) ω− k/1 p/k/2 + ω+ k/2 p/k/1 u(p) ,

(10.188)

where eiφ is the phase factor of Eq.(10.179). For A+− we find the exact same 2 result, and thus
ie2 ~ eiφ u(q) ω− k/1 p/k/2 + ω+ k/2 p/k/1 u(p) . M(+, −) = 2(pk1 )(qk1 )

(10.189)

In this expression, the denominator is of order O (k 2 ) which is already compensated by the occurrence of k/1 and k/2 in the numerator. In the SL we can therefore again replace q by p since q = p + O (k), and use ω− k/1 p/k/2 + ω+ k/2 p/k/1 = (pk1 )/k2 + (pk2 )/k1 − (k1 k2 )/p − iγ α α (k1 , k2 , p) (10.190) which yields 4(pk1 )(pk2 ) − 2m2 (k1 k2 ) when sandwiched between u(p, s) and u(p, s), but zero when sandwiched between u(p, −s) and u(p, s). We see that

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also M(+, −)SL is the same as in the scalar case, while the spin again remains unaffected by the Thomson scattering. We have thus established that Thomson scattering will not distinguish between scalar or Dirac electrons. It is interesting to note, however, that the amplitudes depend on θ even in the SL, while of course in that precise limit (zero photon energy) the value of θ becomes undetermined ! We see that Thomson scattering (and consequently the determination of the electron charge by this process) is only meaningful if there is some momentum transfer, no matter how small39 .

10.7.3

The Landau-Yang theorem

The photon polarisation revisited As stated above, any good amplitude for processes in which a photon is absorbed or produced must vanish under the handlebar operation. That means that, provided the amplitude is acceptable, we may add to any photon polarisation a piece of photon momentum. Let us consider a process with several photons present, with momenta qi µ and polarisation vectors i µ . We have, obviously, (qi · qi ) = (qi · i ) = 0 and (i · i ) = −1. From the above, we see that, if we wish, we may employ instead of i the more complicated object (p · i ) µ (10.191) qi , ηi µ = i µ − (p · qi ) E87

where p is any vector not proportional to qi . This has the properties (ηi · qi ) = (ηi · p) = 0 ,

ηi 2 = −1 .

(10.192)

In numerous applications, η is actually more profitable to use than . But we should note that, in any amplitude described by more than one Feynman diagram, the shift from  to η simply means that parts of some Feynman diagrams are ‘transferred’ to other diagrams : the total result must, of course, be the same. The most important difference between  and η is in the handlebar, since η then vanishes : ηi ci →qi = 0 ; (10.193) 39

This situation is, of course, the same in classical physics : to determine the elasticity modulus of a spring you will have to stretch or compress it, no matter by how small an amount.

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therefore, any expression written in terms of η’s vanishes automatically under the handlebar. On the one hand this is, of course, nice ; on the other hand, it deprives us of a powerful check on the correctness of our diagrams, since almost any mistake made in writing them out will show up as a failure under the handlebar. The Landau-Yang result Although this may seem to fall somewhat outside the province of QED, we can consider the decay of a spin-1 particle into photons. But even within QED this can be envisaged, since we may have a bound state of electron and positron (positronium ) that may, of course, have some angular momentum. Such a positronium state can, unless we look really closely, be considered a single particle40 . In its ground state, positronium comes in two varieties : parapositronium in which the electron and positron’s spin are antiparallel and hence form total spin zero, and orthopositronium in which the spins are parallel, leading to a total spin of one. Without knowing anything much about the bound-state structure of ortho-positronium, let us consider the amplitude for its decay into a pair of photons. Let us denote by P µ the ortho-positronium momentum (in its rest frame), and by q1,2 and 1,2 the photon momenta and polarisations. We shall define q µ = (q1µ − q2µ )/2. In addition, the positronium being a spin-1 particle, we need its polarisation vector 0 . Any amplitude for the decay must necessarily be linear in 0 , 1 and 2 ; and current conservation is imposed by replacing 1,2 by ηi = i − (P · i )/(P · qi )qi . Since also (P · 0 ) = 0, the three polarisations (as well as the vector q) have no timelike component. Noting that, in this case, (q · η1,2 ) = 0 as well, we see that to build an amplitude M we actually have but a very few structures that we can use41 : M = A1 (q ·0 )(η1 ·η2 )+A2 ε(P, 0 , η1 , η2 )+A3 ε(P, q, η1 , η2 )(q ·0 ) . (10.194) The coefficients A are of course undetermined, but they can only depend on P 2 , q 2 and (P · q). This last product is zero, and P 2 = −4q 2 = M 2 where M is the positronium mass, so the A’s are effectively just constants. We now come to the main observation : under interchange of the two photons we 40

Looking really closely would in this case mean splitting it up, and then the positronium is gone. 41 You might be tempted to write down a term like ε(q, 0 , η1 , η2 ) but since all these vectors have vanishing zeroth component, this Levi-Civita product is simply zero.

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have η1 ↔ η2 and q → −q. It is immediately seen that all possible terms in M are antisymmetric under this operation, and hence cannot occur if we are to have Bose statistics. It is obvious that this results holds to all orders of perturbation theory, neither is it restricted to the case of ortho-positronium. We conclude that a spin-1 particle cannot decay into two photons, which is the Landau-Yang theorem. And so it is : para-positronium has a lifetime of 1.25 × 10−10 seconds, while ortho-positronium, having to perform the much more cumbersome decay into three photons, lives for as long as 1.39 × 10−7 seconds. In the literature and most textbooks, the Landau-Yang theorem, especially when applied to positronium, appears to be based on fairly complicated reasonings having to do with the charge-conjugation properties of the various states. In our more simple-minded approach, we see that it is simply a consequence of the relative paucity of building blocks available when you start to imagine what a decay amplitude could look like. Indeed, as soon as you envisage three-photon decay, a host of terms can be written down that respect Bose symmetry, so that it is easily understood why three-photon decay is not forbidden42 .

10.8

Exercises

Excercise 77 Compton Current Conservation Consider the process e− (p1 ) γ(k1 , 1 ) →

e− (p2 ) γ(k2 , 2 )

where the momenta and polarisations are indicated, and write the two diagrams that describe it at tree level. Then, substitute 1 → k1 and show that the amplitude, so treated, vanishes. Excercise 78 Rare or impossible ? The process µ → e γ The process µ− (p) → e− (q) γ(k) is not allowed in standard QED since it violates current conservation. Nevertheless, it could be possible with a different interaction vertex coming from some ‘new physics’. The amplitude would then read √
M = i ~g u(q) cos θ + sin θγ 5 k/ / u(p) 42

In the words of Feynman, ‘everything that is not explicitly forbidden is allowed’.

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where g is the coupling in the new physics, and the angle θ simply parametrizes the relative weight of the two alternative couplings. 1. Show that this amplitude is current-conserving by applying the handlebar. 2. Show that the coupling g must have the following dimensionality : L dim[g] = √ ~ We shall therefore write

e Λ with e the QED coupling constant and Λ the scale of the new physics. g=

3. Compute h|M|2 i using the Casimir trick and the trace identities. 4. Compute the total decay width Γ(µ → eγ). 5. Current limits on this process are expressed as Γ(µ → eγ) ≤B , Γ(µ → all)

B ≈ 10−11 .

We shall assume that the decay µ → eνµ ν¯e is by far the dominant one. Show that we can relate this to B as follows : K Λ≥ √ . B Compute K, and find the current lower limit on Λ. Excercise 79 Multi-photon production Consider the process e+ e− → n γ, the n-photon analogue of the process 10.73. Show the following : 1. There are, at the tree level, n! diagrams. 2. The amplitude vanishes if all photons have the same helicity. 3. If all photon helicities except one are equal, only (n − 1)! diagrams contribute if we choose the gauge vectors cleverly.

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Excercise 80 Relating Mandelstams In a 2 → 2 scattering process, the momentum notation may read p 1 + p2 → q1 + q2 We take the incoming particles to have mass m and the outgoing ones to have mass M . From the conservation of energy and momentum, pµ1 + pµ2 = q1µ + q2µ and the definitions s = (p1 + p2 )2 , t = (p1 − q1 )2 , u = (p1 − q2 )2 , derive the relation s + t + u = 2m2 + 2M 2 Excercise 81 Bhabha scattering got wrong In section 10.3.4 it is mentioned in a footnote how important the correct application of the Fermi minus sign can be. Investigate this by re-computing the cross section for Bhabha scattering using the wrong sign, and finding the ratio between the two expressions ; then find the maximum ratio and the scattering angle at which this is reached. Excercise 82 Bhabha scattering the hard way Compute h|M|2 i for Bhabha scattering at tree level, this time keeping me nonzero. To do this, use the Casimir (trace) method. Excercise 83 It’s a chore but someone’s got to do it Compute the transition rates, that is, h|M|2 i, for the processes e+ e− → µ+ µ− γ and e+ e− → e+ e− γ, taking the fermions to be massless, but keeping the double pole terms in the radiative amplitude. This involves using spinor techniques extensively! Excercise 84 Measuring the radiation cone At the LEP accelerator in its latest stages, the energy E of the incoming electron beam was around 100 GeV, while the electron mass is me ∼ 5 × 10−4 GeV. 1. Show that the ‘canonical’ radiation cone angle derived from Eq.(10.116) is about 5 × 10−6 radians.

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2. The angular distribution of the photon around the electron is roughly (1+δ−cos θ)−1 , where δ = m2e /(2E 2 ). Show that a cone that contains a fraction f of the radiation is bounded by an angle α such that cos(α) = 1 − δ(2/δ)f . 3. Show that α ∼ 3 × 10−3 radians for f = 0.5, i.e. half the radiation is emitted at larger angles than this. Excercise 85 Exploring Yennie’s crater Consider a neutral particle (the Z 0 boson or the Higgs) decaying into a fermion-antifermion pair (with equal masses m but opposite charges) with momenta p1,2 and a soft photon with momentum k. 1. Show that the partial decay width contains the soft-photon factor S=

m2 m2 2(p1 p2 ) − − (p1 k)(p2 k) (p1 k)2 (p2 k)2

2. Let the angle between p~1 and ~k in the centre-of-mass frame have cosine equal to c. Show that S=

2(E 2 + p2 ) 2m2 (E 2 + p2 c2 ) − E 2 − p 2 c2 (E 2 − p2 c2 )2

where p01,2 = E and |~p1,2 | = p. 3. Show that S = 0 if c = ±1. Excercise 86 Deriving the Pauli equation Verify that Eq.(10.146) is correctly derived. Excercise 87 Optimising the polarisation for Landau-Yang Show, by explicit computation, that for massless p the transformation of Eq.(10.191) is in fact precisely the adoption of p as the gauge vector. Note, however, that in section 10.7.3 we may actually take p to be almost any vector, so that η is actually more general. Excercise 88 The Landau-Yang theorem in action

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1. Consider the proces e+ (p1 , s1 ) e− (p2 , s2 ) → γ(k1 , 1 ) γ(k2 , 2 ) where the polarisations and spins are indicated along with the momenta. Write out the amplitude at the tree level, keeping the electron mass m nonzero. 2. Choose the gauge vectors such that not only (k1 1 ) = (k2 2 ) = 0 but also (k1 2 ) = (k2 1 ) = 0. Show that in that case 1,2 0 = 0 in the centre-of-mass frame. 3. Take the static limit, in which p1 and p2 become equal. Show that in this limit the amplitude is proportional to µ (1 , k1 − k2 , 2 ) v(p1 , s1 ) γ 5 γµ u(p1 , s2 ) 4. Show that the amplitude vanishes if s1 = s2 , but does not vanish if s1 = −s2 . Hint: show that µ (1 , k1 − k2 , 2 ) ∝ p1 µ .

Chapter 11 Loop effects in QED 11.1

One-loop effects in QED

We shall discuss a number of one-loop calculations for QED. On the one hand, these are of course phenomenologically relevant, and on the other hand they form a nice demonstration of real-life quantum field theory at work, including lots of gory detail.

11.2

The photon self-energy

We start with the photon self-energy due to a fermion loop : k p

F µν (m2 , p) =

ν .

µ

(11.1)

p+k

The fermion mass is m, and its charge is denoted by Q.

11.2.1

Current conservation

First, we prove that the electromagnetic current is conserved in this diagram : using what we have learned in section 10.2.3, we can write = =

− = 321

, =

.

(11.2)

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Therefore, pµ F µν (m2 , p) = 0 so that
F µν (m2 , p) = pµ pν − s g µν K(m2 , s) ,

s = p2 .

(11.3)

The quantity K, being a scalar, can only depend on p via p2 .

11.2.2

Using the optical theorem

We shall obtain information about K(m2 , s) by embedding it in a physical amplitude. For this, we choose the annihilation of a massless charged scalarantiscalar pair (with charge Q0 ) into a fermion-antifermion pair1 , s(p1 ) s¯(p2 ) → f (q1 ) f¯(p2 ), which is the absorptive part of the optical theorem (6.34) : p2

q2

Ma = p 1

q1

=

i~QQ0 u(q1 ) (/p1 − p/2 ) v(q2 ) . s

Summing over the final-state spins we have X
|Ma |2 = 2~2 Q2 Q02 1 − β 2 (cos θ)2 ,

β=

p 1 − 4m2 /s ,

(11.4)

(11.5)

spins

where θ is the scattering angle between p~1 and ~q1 in the centre-of-mass frame. The full sum over the final state of course includes the phase-space integration : Z X ~2 Q2 Q02 β(3 − β 2 )θ(s > 4m2 ) . (11.6) |Ma |2 dV (p1 + p2 ; q1 , q2 ) = 12π spins The forward scattering amplitude is that for s(p1 ) s¯(p2 ) → s(p1 ) s¯(p2 ) with the fermion loop inserted : p2 p1

p2 p1

~2 Q02 (p1 − p2 )µ F µν (m2 , p1 + p2 )(p1 − p2 )ν s2 = ~2 Q02 K(m2 , s) . (11.7)

=

Note that the pµ pν term in Eq.(11.3) drops out here because of current conservation. The optical theorem 6.34 then tells us that < K(m2 , s) = −

α β(3 − β 2 ) θ(s > 4m2 ) , 6~

α=

Q2 ~ . 4π

(11.8)

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323

The imaginary part of K can now be computed using the Kramers-Kronig relations of appendix 15.15.9, just as we did in section 6.6.2 : Z∞ 1 < K(m2 , z) = K(m2 , s) = − P dz . (11.9) π z−s 4m2

The z integral diverges since < K(m2 , z) goes to a constant for z → ∞. Therefore we cannot make much sense of K, but we can figure out how it depends on s by subtracting its value at s = p 0, say. This subtracted dispersion relation is best computed by using v = 1 − 4m2 /z as an integration variable, and writing β 2 = 1 − 4m2 /s : Z∞ v(3 − v 2 ) α s P dz = K(m2 , s) − = K(m2 , 0) = 6π~ z(z − s) 4m2

Z1

 2  α β (3 − β 2 ) 2 2 = P dv +3−β −v 3π~ v2 − β 2  0 β + 1 β(3 − β 2 ) α 8 2 −β − log = (β 2 > 0) , 3π~ 3 2 β − 1    1 α 8 2 2 − β − |β|(3 − β ) arctan (β 2 < 0) . (11.10) 3π~ 3 |β|

11.2.3

Getting the divergence

We still have to compute K(m2 , 0). Eq.(11.1) can be written as2 Z −Q2 T (p, k) µν 2 F (m , p) = d4 k 2 , 4 2 (2π) (k − m + iη)((p + k)2 − m2 + iη)
T (p, k) = Tr (/k + m)γ µ (/k + p/ + m)γ ν . (11.11) We can employ the Feynman trick and the momentum shift from section 6.6.1 to arrive at 1 Z 2 Z −Q T (p, k − xp) µν 2 F (m , p) = dx d4 k 2 . (11.12) 4 (2π) (k + sx(1 − x) − m2 + iη)2 0 1

Why introduce charged scalars here ? Because we can. It works equally well with fermions in the initial state, see exercise 89. 2 The Fermi minus sign is really important here !

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Now, T (p, k − xp) = −8x(1 − x)pµ pν + (other terms) ,

(11.13)

and we see that3 Q2 K(m , s) = (2π)4 2

Z1

Z dx

d4 k

(k 2

8x(1 − x) + sx(1 − x) − m2 + iη)2

(11.14)

0

And we must compute this only for s = 0. In dimensional regularization as described, again, in section 6.6.1 we have iQ2 µ2 K(m2 , 0) = (4π)2− Γ(2 − )

Z∞

Z1 dx 8x(1 − x) 0

dt

t1− (t + m2 )2

0


iα = R − log m2 + O () , 3π~

(11.15)

where the divergent constant R was defined in Eq.(??)

11.2.4

The vacuum polarization

The photon self-energy can of course be Dyson-summed. Denoting the dressed propagator Πµν by a thick line, we can write the SDe as =

+

,


−i~ g µν −i~ g µα + −s gαβ K(s) Παν (s) s s −i~ g µν −i~ g µν
= . = s (1 − Π(s)) s 1 − i~K(s)

Πµν (s) =

(11.16)

We have dropped the pµ pν terms because of current conservation, and introduced the vacuum polarization function :     α β(3 − β 2 ) β+1 8 2 2 −R + log m + log +β − , (11.17) Π(s) = 3π 2 β−1 3 3

This saves us a lot of unnecessary work : the ‘other terms’ contain k µ k ν , g µν k 2 and suchlike, with integrals more cumbersome by far.

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Figure 11.1: The fermion-induced vacuum polarization

with β = (1 − 4(m2 − iη)/s)1/2 . For large s values, we have     |s| 5 α 2 log − − iπθ(s > 4m ) . Π(s) − Π(0) ≈ 3π m2 3

(11.18)

Note that we have defined β with the mass m2 augmented by its ‘natural’ extension into m2 −iη (η ↓ 0); the iη takes care of the correct imaginary part, as it should. Plot 11.1 shows the behaviour of Π(s) − Π(0) as a function of s/m2 . It is continuous both at s = 0 and at s = 4m2 , and we see the opening of a branch cut for s ≥ 4m2 . I used the value η = 10−8 and so obviated the need to distinguish between the three different cases s < 0, 0 < s < 4m2 and s > 4m2 .

11.2.5

Hadronic vacuum polarization

The vacuum polarization Π(s) discussed above can reliably be computed for charged leptons since their masses are quite accurately known, and higherloop QED contributions can be expected to be small. How different for quarks ! Not only do we not know the quark masses to anything like the same precision as the leptonic ones, but also we expect enormous higher-

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order effects from strong interactions4 . But all is not lost ! Let us write the hadronic self-energy of the photon as5

µ

ν

= (pµ pν − s g µν ) H(s) ,

s = p2 .

(11.19)

p Let us now consider the forward amplitude for e+ e− → e+ e− mediated by a hadronically corrected photon. We take the electrons as massless and denote their helicity by λ, to write Mf =

p1

p1

p2

p2

Q2 ~2 uλ (p1 )γµ uλ (p2 ) (pµ pν − s g µν ) H(s)uλ (p2 )γµ uλ (p1 ) 2 s = 2Q2 ~2 H(s) , (11.20)

=

which is easily computed using the helicity techniques of chapter 8. On the other hand, we can write Z p1 p 2

2 dV = 4s σhad (s)

(11.21)

+ − where σhad (s) stands for the total → √ cross section for the process e e hadrons at total collision energy s, and dV includes a sum over all hadronic final states6 . The optical theorem (6.34) then tells us that

1 in the text.

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Consider the following diagram :

b

c

a j

The colour part of this diagram reads X (T j )a b (T j )b c j,b

and colour conservation/democracy hence demands that X  2 a Tj = k δab

(12.5)

b

j

for some constant k. Similarly, the diagram

b

j

k

a contains the colour factor X

(T j )a b (T k )b a = Tr T j T k




,

a,b

and using the normalization freedom we may take
1 Tr T j T k = δ jk . 2

(12.6)

Since colour must be conserved, a gluon cannot lose its colour charge and therefore gluons and photons cannot mix : in all diagrams of the form a

j

photon b

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we must have b = a since the colour is conserved and the photon is colourless ; therefore the T matrices must be traceless :
(12.7) Tr T j = 0 for all gluon colours j . Finally, we consider the following two-loop self-energy diagram of the photon : j

Here, the fermions are quarks and the internal line labelled j is a gluon of colour type j : of course, we have to sum over all j values. If we compare this diagram to the corresponding QED one, we see that apart from the overall charges (g 2 instead of Q2 ) the only difference is the colour factor, in this case X
Tr T j T j j

Now, if our theory is to be unitary, it must obey the Cutkosky rules, and therefore we demand that

+

j

+

+

j

+

j

j

j

=0 .

(12.8)

For the QED diagram, this indeeds holds. In the coloured case, however, the colour structures of the diagram cut in the various ways are no longer the same : the three lines in Eq.(12.8) are proportional to, respectively,  X X  X  † †
j j j j† , and Tcc = Tr T j T j . T = Tr T T , Tc = Tr T T j

j

j

Unitarity can therefore only be safe if these three different traces are, in fact, equal to one another. We may therefore write   X
† 2Tc − T − Tcc = Tr Aj Aj = 0 , Aj = i T j − T j . (12.9) j

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The matrices Aj are Hermitean, so that Eq.(12.9) can also be written as X



j

Tr A A

j

j†



N XX j a 2 (A ) = 0 , = b j

(12.10)

a,b=1

hence all Aj are actually identically zero, and the matrices T j must be Hermitean. The number of different gluon colours type is therefore N 2 − 1, and the constant k of Eq.(12.5) is equal to (N 2 − 1)/2N .

12.2.4

The Fierz identity for T matrices

We have now zoomed in quite efficiently on the matrices T j . On the other hand, just as in the case of Dirac particles we would prefer if predictions for cross sections and the like dit not depend on the particular choice of the matrices5 . We can, in fact, derive a relation between the T ’s that holds independently of any representation : it goes under the name of the Fierz identity6 . Any N × N matrix M can be written7 as X M = a0 1 + aj T j . (12.11) j

By taking traces we can determine the coefficients :
Tr (M ) = a0 N , Tr M T k = ak /2 .

(12.12)

Therefore we have M =2

X j


1 Tr T j M T j + Tr (M ) , N

or, in terms of the matrix components, X 1 M dc δcb δad = 2 M d c (T j )c d (T j )a b + M d c δ c d δ a b , N j 5

(12.13)

(12.14)

In the Dirac case this was indispensable since any dependence would destroy Lorentz invariance. In the present case one might argue that the T j could, in principle, just be measured. Nevertheless having a representation-independent theory just feels so much more comfortable. 6 Same Fierz. 7 You might be tempted to think that this holds only for Hermitean matrices. But since iT j is antiHermitean we can accomodate any M provided the a’s can be complex.

August 31, 2019 whence the following, representation-independent identity :   1 a c 1 a c j a j c (T ) b (T ) d = δ dδ b− δ bδ d . 2 N

347

(12.15)

Since8 the colour of quarks and gluons cannot be observed, any cross section will involve a summation over all colours, and therefore every cross section is expressed as (a product of) traces of strings of T matrices, in which every matrix T k occurs exactly twice, and the index k is summed over. The Fierz identity comes in useful here, since we can write (with summation implied)  

1 1 j j Tr (AB) − Tr (A) Tr (B) , Tr T A Tr T B = 2 N  
1 1 j j Tr (A) Tr (B) − Tr (AB) . (12.16) Tr T A T B = 2 N With these trace identities we can simplify and compute any set of colour traces without recourse to any explicit representation, especially if we recall that Tr (1) = N, Tr (T j ) = 0 and T j T j = (N 2 − 1)/2N times unity. As an illustration, we work out the following cases in detail :

Tr T j T k T l Tr T j T k T l  


1 1 k l k l k l k l = Tr T T T T − Tr T T Tr T T 2 N  




1 1 2 l l l l l l = Tr T Tr T − Tr T T + 2 Tr T Tr T 4 N N 2
N −1 1 Tr T l T l = − , (12.17) = − 2N 4N and

Tr T j T k T l Tr T j T l T k  


1 1 k l l k k l l k = Tr T T T T − Tr T T Tr T T 2 N  



1 2 1 k k k k k k Tr T T Tr (1) − Tr T T + 2 Tr T Tr T = 4 N N 2 2 2
N −2 (N − 1)(N − 2) = Tr T k T k = . (12.18) 4N 8N 8

Empirically.

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12.3

The three-gluon interaction

12.3.1

The need for three-gluon vertices

It is now time to investigate our theory using handlebars. In the first place, in the process g → q q¯ the current is conserved in the same way as in QED, since there is only a single Feynman diagram and the colour structure is therefore irrelevant to any cancellation. The situation becomes more delicate in the case of more complicated interactions, so let us consider the process q¯(p1 , a) q(p2 , b) → g(q1 , 1 , j) g(q2 , 2 , k) where we have explicitly indicated the momenta, polarisations, and colours. We have at least the following two diagrams9 :

p1

a

j

p

q1

M=

a jq 1

1

q2 + p2

p2

k

b

kq

,

(12.19)

2

b

They read q/1 − p/1 + m /2 u(p2 ) (T j T k )a b , −2(q1 · p1 ) p/2 − q/1 + m /1 u(p2 ) (T k T j )a b . = −i~g 2 v(p1 )/2 −2(q1 · p2 )

M1 = −i~g 2 v(p1 )/1 M2

(12.20)

Let us now put the handlebar on gluon 1, so that we replace µ1 by q1µ . By the same reasoning as in chapter 10, we arrive at the handlebar rule

−

=

,

(12.21)

g i (T j )a b . ~

(12.22)

with the auxiliary Feynman rules j

b 9

a

↔

i~ δ

a

b

,

b

a

↔

In the second diagram the two gluon lines form an overpass, without a vertex.

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As before, slashed propagators vanish on external lines. Applying this to the two Feynman diagrams of Eq.(12.19) yields10

j

j

M1+2 c1 →q1 =

+ k k j

j

−

=

+

−

j k

k

j k

k

j

−

=

j k

k

= −i~g 2 v(p1 )/2 u(p2 ) [T j , T k ]a b ,

(12.23)

where the square brackets denote, of course, the commutator of the matrices T j and T k . Because of the colour structure we have a non-vanishing result, and current conservation is in trouble ! The remedy must be to introduce a third diagram, with a nontrivial ggg vertex,

p1

a

j q1

n p2

q2 b

k

and it is now our job to determine the form of the new three-gluon vertex. We shall do this by investigating loop diagrams.

10

In calculations such as this one it is often sufficient to simply label the gluons by their colour, thus simplifying the typography.

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12.3.2

Furry’s failure

Consider the following Feynman diagram : p −q 2

l

ν k

K ∼

ρ q3

q2

(12.24) p + q1

p j

q1 µ

Here three gluons are effectively coupled by a quark loop. We have explicitly indicated the momentum flows. Note especially that the gluon momenta are all counted flowing out of the vertex, so that we have q1 + q2 + q3 = 0 .

(12.25)

Apart from overall coupling constants and the like, the loop diagram is given by Z
Tr ((/p + m)γ µ (/p + q/1 + m)γ ρ (/p − q/2 + m)γ ν ) j l k Tr T T T . K ∼ d4 p (p2 − m2 )((p + q1 )2 − m2 )((p − q2 )2 − m2 ) (12.26) There is also a loop diagram in which the quark runs counterclockwise instead of clockwise. In our discussion of Furry’s theorem in sect. 10.2.7, we have seen that the space-time part of the second diagram is exactly opposite to the one of the first, so that in QED these two diagrams cancel. In QCD they do not : the second diagram contains the
colour matrices in the opposite order,
j k l j l k that is to say it contains Tr T T T instead of Tr T T T . The sum of the two diagrams, if we take into account the Lorentz-covariant nature of the loop integral, and the fact that out of q1 , q2 and q3 only two momenta are independent, must be of the form
K ∼ Y (q1 , µ; q2 , ν; q3 , ρ) Tr T j [T k , T l ] , (12.27) with Y (q1 , µ; q2 , ν; q3 , ρ) = (a1 q1 + a2 q2 )ρ g µν + (a3 q2 + a4 q3 )µ g νρ + (a5 q3 + a6 q1 )ν g ρµ , (12.28)

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351

for some numbers a1 , . . . , a6 . For large p, each of the three propagators goes as 1/p, and the loop integral is therefore divergent. We see that indeed a threegluon counterterm must be allowed, and therefore a three-gluon coupling must appear in the action, otherwise the theory would not be renormalizable ; and the form of the three-gluon vertex must be that of Eq.(12.28). Without evaluating the loop integral completely, we can glean all the information we need. Consider the following transformation on K : q1 ↔ −q2 ,

q3 → −q3 , µ ↔ ν .

(12.29)

This transformation leaves the momentum conservation law (12.25) intact, and also preserves the value of T (by the reversal property (7.29) of Dirac traces). The same holds, of course, for the transformations q1 ↔ −q3 , q2 ↔ −q3 ,

q2 → −q2 , µ ↔ ρ , q1 → −q1 , ν ↔ ρ .

(12.30)

The function Y must therefore satisfy Y (q1 , µ; q2 , ν; q3 , ρ) = Y (−q2 , ν; −q1 , µ; −q3 , ρ) = = Y (−q3 , ρ; −q2 , ν; −q1 , µ) = Y (−q1 , µ; −q3 , ρ; −q2 , ν) ; (12.31) and by inspection we then find that c1 = c3 = c5 = −c2 = −c4 = −c6 . We shall therefore from now on use the Yang-Mills three-boson vertex Y (q1 , µ; q2 , ν; q3 , ρ) ≡ (q1 − q2 )ρ g µν + (q2 − q3 )µ g νρ + (q3 − q1 )ν g ρµ .

(12.32)

Note that this form is antisymmetric in the interchange of any two gluons, E100 and therefore invariant under a cyclic permutation11 .

12.3.3

The ggg vertex and its handlebar

On the basis of the previous section, we see that the only reasonable form of the three-gluon vertex Feynman rule is

ρ

q3

j

l

µ q1

q2

↔

i g3 Y (q1 , µ; q2 , ν; q3 , ρ) hjkl ~

(12.33)

ν k

11 I have adopted the notation ‘Y ’ for this vertex since it reminds us both of the name Yang(-Mills), and of the fact that in such a vertex three bosons meet.

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Note that the gluon momenta are counted outgoing from the vertex. The value of g3 must be determined, as well as the colour factor hjkl . The Y function’s total antisymmetry strongly suggests that we take the h symbols antisymmetric as well, and later on we shall show that this is indeed the case. The following object will turn out to be useful : ∆(q)αβ ≡ q α q β − q 2 g αβ ,

(12.34)

∆(q)αβ = ∆(q)βα , ∆(q)αβ qβ = 0 .

(12.35)

∆(q)αβ β = q α (q · ) − q 2 α = 0

(12.36)

for which Also, if  is the polarisation vector of an on-shell gluon with momentum q. We now come to an important result. Let us consider the vertex of Eq.(12.33), and let us put a handlebar on gluon q3 . We find, using momentum conservation in the form q3 = −q1 − q2 , Y (q1 , µ; q2 , ν; q3 , q3 ) ≡ = = =

Y (q1 , µ; q2 , ν; q3 , ρ) q3 ρ (q1 − q2 · q3 )g µν + (q2 − q3 )µ q3 ν + (q3 − q1 )ν q3 µ (q2 − q1 · q2 + q1 )g µν − q2 µ (q1 + q2 )ν + q1 ν (q1 + q2 )µ ∆(q1 )µν − ∆(q2 )µν . (12.37)

Some algebra tells us, moreover, that in the axial gauge Πµα (q1 ) ∆αβ (q1 ) Πβν (q2 ) = (i~g µα ) (gαβ ) Πβν (q2 ) ,

(12.38)

so that we find the handlebar rule j

j

=

m

j

+

m k

k

,

(12.39)

m k

with the auxiliary Feynman rules j µ

k

↔ ν

i~ g µν δ jk

(12.40)

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353

and m

↔

β k

i

α

j

g3 αβ jkm g h . ~

(12.41)

It is now time to return to the q q¯ → gg process. The new available Feynman diagram, given by p

a

q1

1

n

p2

j

q2 k

b

reads M3 = −gg3 v(p1 )γµ u(p2 )Πµν (p1 + p2 )Y (q1 , 1 ; q2 , 2 ; −q1 − q2 , ν) hjkn (T n )a b . (12.42) with summation over the colour n implied. Putting the handlebar on gluon 1 as before, we get n

j k

=

n

j

+

j

n

k

(12.43)

k

so that M3 c = −i~gg3 v(p1 )/2 u(p2 ) hnkj (T n )a b .

(12.44)

The total handlebarred amplitude now reads
a M1+2+3 c = i~g v(p1 )/2 u(p2 ) g3 hjkn T n − g[T j , T k ] b

(12.45)

The colour current will therefore be conserved if we choose g3 = g

(12.46)

[T j , T k ] = hjkn T n .

(12.47)

and

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Note that since the matrices T are Hermitean, the constants h must be purely imaginary12 . Moreover, we can compute them, using Eq.(12.6), as
(12.48) hjkl = 2 Tr T j T k T l − T l T k T j , from which we see that the h symbols must be totally antisymmetric. Since they are related to commutators, we can use the Jacobi identity to find relations between them13 : 0 = [[T j , T k ], T l ] + [[T k , T l ], T j ] + [[T l , T j ], T k ] = hjk n [T n , T l ] + hkl n [T n , T j ] + hlj n [T n , T k ] = hjk n hnl m T m + hkl n hnj m T m + hlj n hnk m T m ,

(12.49)

which after a few interchanges of indices leads to hjk n hlm n + hjl n hmk n + hjm n hkl n = 0 .

(12.50)

More information comes from colour conservation/democracy in the diagram m

j

k n

from which we find the requirement that X hmnj hmnk = C δ j k ,

(12.51)

m,n

with some constant C. Eq.(12.51) is the gluonic equivalent of the property (12.5) of the T matrices. It does not follow from the Jacobi identity. But since we have already defined the h symbols by Eq.(12.48), it is not an extra condition bur rather has to be proven : since



hmnj hmnk = 8 Tr T m T n T j Tr T m T n T k − Tr T m T k T n , (12.52) 12

It is customary to write [T j , T k ] = i f jk n T n . The f ’s are then called the structure constants, and the set of T matrices are then the generators of the Lie algebra of the group SU (N ). The i is then combined with the overall i of the vertex to give a Feynman rule without any i. This is of course a matter of taste. 13 Here and in the following, raising or lowering colour labels has no physical meaning ; I do it only for typographical reasons.

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and the reduction formulæ (12.16) give

1 jk δ , Tr T m T n T j Tr T m T n T k = −  4N 

N 1 m k n m n j = Tr T T T Tr T T T − δ jk , 8 4N

(12.53)

we arrive at hmnj hmnk = −N δ j k .

12.3.4

(12.54)

On coupling quantisation

In the previous chapter we have discussed QED, characterized by the fact that only fermion-fermion-photon couplings occur. The coupling constant Qf for a given fermion is not constrained in any way, and there is no a priori reason why different fermion species ought to have the same charge14 . In QCD the situation is different. In deriving the ggg vertex we have started with the process q q¯ → gg, without specifying the quark type, only saying that its coupling to the gluon has coupling constant (‘colour charge’) g. We then find that the three-gluon vertex has the same coupling constant g. But that implies that all quark types must have the same coupling constant to gluons, since two different quark types must lead to the same value of the Yang-Mills coupling. We see that the presence of bosonic self-interactions enforces a uniformity on the quark colour charges that was not there before. In the next chapter we shall see how the presence of a W W γ vertex forces the electron and muon charges to be, indeed, identical.

12.4

The four-gluon interaction

12.4.1

Colourful manipulations

Before proceeding it is useful to prepare some groundwork. Let us define, first, the ‘four-colour’ object [jklm] = hjk a hlm a = hjk a ha lm . 14

(12.55)

Of course, leptons and quarks have different charge ; but I rather refer to the ratio of √ muon and electron charge, which could be π for all that QED cares.

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This has the symmetries15 [jklm] = −[kjlm] = −[jkml] = [lmjk] ;

(12.56)

and the Jacobi identity reads [jklm] + [jlmk] + [jmkl] = 0 .

(12.57)

In addition, we have the ‘five-colour’ object [jklmn] = [jkla] hamn = hjk a ha l b hb mn .

(12.58)

We can easily verify the symmetries [jklmn] = −[kjlmn] = −[jklnm] = −[mnljk]

(12.59)

and the two Jacobi identities [jklmn] + [jkmnl] + [jknlm] = [jklmn] + [kljmn] + [ljkmn] = 0 . (12.60) With such an arsenal of identities quite a few results can be derived: for instance, by using each of the three symmetries and the two Jacobi identities once, we can prove that [jklmn] − [jklmn] + [lmjkn] − [lmjkn] = 0 , [jlkmn] + [kmjln] − [jlmkn] − [kmljn] = 0 .

12.4.2

(12.61)

A purely gluonic process

We have seen that gluons, in contrast to photons, exhibit self-interactions. We can therefore consider the process g(q1 , 1 , j) g(q2 , 2 , k) → g(q3 , 3 , l) g(q4 , 4 , m) which so far is given (at the tree level) by three Feynman diagrams16 : M =

j k

15

j

l n

+ m

k

l n m

j

+

l n

k

(12.62)

m

These are precisely those of the Riemann-Christoffel tensor, familiar from the theory of general relativity. 16 Since in this section we only consider gluons anyway, I shall here denote them by smooth rather than wriggly lines ; this makes them somewhat easier to read, and certainly easier to draw.

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As before, we put a handlebar on gluon 1 : j

Mc1 →q1 =

j

l

+

n

m

k

k

l

+

n m

j

l n

(12.63)

m

k

and this combination does not obviously vanish. The three-gluon vertex is somewhat cumbersome, but we can streamline our calculations a bit by introducing a ‘partial’ three-gluon vertex : λ

n

k β q2

g i (q1 + q2 )λ g αβ hjkn . ~

↔

j q1 α

(12.64)

Here, the momenta are counted in the direction of the arrows. Note that reversing the arrows leaves the vertex unchanged owing to the antisymmetry of the h symbol. There is therefore no ambiguity. We may write =

+

+

.

(12.65)

This makes it easier to single out particular terms in Eq.(12.63). For instance, the terms proportional to (2 · 4 ) are given by the diagrams j k

j

l n

+ m

k

j

l

+

n m

l n

k

m

that, apart from an overall factor −i~g 2 (2 · 4 ), evaluate to (3 · q2 + q4 ) hmkn hnlj + (3 · q1 + q2 + q4 ) hmnl hnkj + (3 · 2q2 − q3 ) hnkl hnmj = (3 · q2 + q4 ) [mklj] − (3 · q1 + q2 + q4 ) [mlkj] + (3 · 2q2 − q3 ) [klmj]
= (3 · q2 + q4 ) −[mljk] − [mjkl] +(3 · q1 + q2 + q4 ) [mljk] − (3 · 2q2 − q3 ) [mjlk] = [mljk] (3 · q1 + q2 + q4 − q2 − q4 ) + [mjlk] (3 · q2 + q4 − 2q2 + q3 )
= (q1 · 3 ) [mjlk] + [mljk] . (12.66)

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I have displayed this computation in detail in order to emphasize that we only use momentum conservation, and not for instance the property (3 · q3 ) = 0. In addition, in the third line the Jacobi identity comes into play. The other terms are of course treated in the same way, so that we find 
2 Mc1 →q1 = −i~g (q1 · 2 )(3 · 4 ) [jlmk] + [jmlk]
+ (q1 · 3 )(2 · 4 ) [jmkl] + [jkml] 
+ (q1 · 4 )(2 · 3 ) [jklm] + [jlkm] . (12.67) The handlebar requirement can now be satisfied by introducing a fourgluon interaction vertex as follows: α

µ m

j β k

ν

↔

n

i

g2 X(α, j; β, k; µ, m; ν, n) ~

(12.68)

with X(α, j; β, k; µ, m; ν, n) = g αβ g µν [jmnk] + [jnmk]




+ g αµ g βν [jknm] + [jnkm]




+ g αν g µβ [jkmn] + [jmkn]




.

(12.69)

In addition, we have found the handlebar rule

= −

−

−

12.5

Current conservation in QCD

12.5.1

More vertices ?

.

(12.70)

After having introduced the three-gluon vertex, we have seen that we also need a four-gluon vertex in order to save current conservation in qq → qq. What if we now considergg → ggg ? Wil we also need a five-gluon vertex ? And then what about gg → gggg ? Fortunately, as we shall see, no such

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bad luck : with the three- and four-gluon vertices all amplitudes will vanish under the handlebar. We shall prove this in the same manner as for QED, only the proof will be obviously somewhat more involved. Before we turn to our reliable working horse, the SDe, we first need one more result.

12.5.2

The Antkaz

Consider the four-gluon vertex in the following form : α

j k

β

l µ m ν

and let us now attach a fifth gluon in all possible ways, and slash the intermediate propagators. This leads to the following expression : j

V

=

j

n k l

k

+ m

l m

j

j

n

k

+

k

n l m

l

+

n

. (12.71)

m

In V , no gluon momenta enter. Let us now concentrate on the term with g αβ g µν , say. The colour part of this term reads Vαβ;µν = [plmk] hpjn + [pmlk] hpjn + [jlmp] hpkn + [jmlp] hpkn + [jpmk] hpln + [jmpk] hpln + [jlpk] hpmn + [jplk] hpmn = −[mkljn] − [lkmjn] + [jlmkn] + [jmlkn] +[mkjln] − [jmkln] − [jlkmn] + [lkjmn] (12.72) We now call upon the various symmetry properties and Jacobi identities : Vαβ;µν = +[mklnj] + [lkmnj] + [jlmkn] + [jmlkn] +[mkjln] + [lkjmn] + [jlknm] + [jmknl] = −[mknjl] − [jlnmk] − [lknjm] − [jmnlk] = 0 . (12.73) So we find that V vanishes completely, and we shall employ that fact in E96 what follows next.

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12.5.3

August 31, 2019

Proof of current conservation

Let us consider an amplitude with a gluon line sticking out, and see what the SDe has to tell us about it =

+

+

(12.74)

Let us now apply the handlebar rules we have developed in this chapter : −

=

−

+

.

(12.75)

It is important to realize that the internal lines in the SDe take on all possible identities, and therefore the third diagram stands for both graphs in Eq.(12.39), and the fourth stands for all three diagrams in Eq.(12.70)17 . We now iterate the SDe for the slashed propagators in the first three diagrams, and this gives us −

=

+ −

+

=

+

0 ,

(12.76)

since the first diagram on the second line cancels against those of the first line, and the last diagram on the second line vanishes all by itself, as we have seen in the previous section. This establishes the proof of current conservation in QCD. It is important to realise that the above proof relies on our use of the axial gauge for the gluon propagator. Indeed, that choice is what makes the identity (12.39) possible. 17

Are you worried about possible double-counting here ? Don’t worry, be happy : the symmetry factors are there for precisely that reason.

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12.6

Selected topics in QCD

12.6.1

White and coloured states

So far we have simply assumed quarks to have some colour label running from 1 to N for quarks/antiquarks, and gluons to have another label running from 1 to N 2 − 1 ; in the rest of this section we shall use the observed value N = 3. The whole idea of quark (and gluon) colours implies that we assume our QCD physics to be invariant if we ‘rotate’ in colour space, or rather, if we apply to the quarks and antiquarks a unitary operation : q a → Rba q b , q¯a − → q¯b (R† )ba ,

(12.77)

with the summation convention implied. Here the R are unitary 3×3 matrices with unit determinant : they constitute the group SU(3). The quarks are said to be triplets (3) under colour, and the antiquarks to be anti-triplets (¯3). It pays to examine what kind of colour combined states of quarks and antiquarks can have18 . In the first place, the combination q¯a q a transforms as q¯a q a → q¯b (R† )ba Rca q c = q¯b δcb q c = q¯c q c :

(12.78)

this combination doesn’t change colour at all, it is the ‘white’ singlet. The other eight possible (orthogonal) q q¯ colour combinations are the colour octet (8). In group-theory speak we say that 3 ⊗ ¯3 = 1 ⊕ 8. A typical octet state is, for instance, q¯1 q 2 . The fact that the matrices T j are traceless combinations of a colour and and anticolour index tells us that gluons are colour octets as well. We can also investigate how the special quark-quark combination abc q b q c transforms, where abc is the three-dimensional Levi-Civita symbol : b abc q b q c → abc Rm Rnc q m q n .

(12.79)

b We start by observing that the object abc R`a Rm Rnc is totally antisymmetric in its three indices `, m, n, and must therefore itself be proportional to the Levi-Civita symbol ; furthermore, we have

1 `mn b Rnc = det (R) = 1 ,  abc R`a Rm 3! 18

(12.80)

We shall not delve deeply into group theory here, but only consider some of the representations of SU(3).

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so that it actually is the Levi-Civita symbol. Judiciously multiplying with R† we then find that abc q b q c → `mn q m q n (R† )`a :

E97

(12.81)

these quarks are also in the antitriplet, ¯3 state ! The remaining orthogonal quark-quark colour combinations are the sextet (6), so that we can say that 3 ⊗ 3 = ¯3 ⊕ 6. A typical sextet state is, for instance, q 3 q 3 . Finally we can observe that the colour combination abc q a q b q c is also a colour singlet, a ‘white’ state. The empirical fact that we see only mesons (q q¯), baryons (qqq), and antibaryons (¯ q q¯q¯) occurring as more-or-less stable, more-or-less free hadrons can therefore be nicely summed up by the postulate that only ‘white’ (singlet) colour states can occur freely.

12.6.2

The QCD Coulomb interaction

Just like we did for QED (cf section 10.5) we can also investigate what QCD tells us about almost-static scattering of quarks. Let us start with q q¯ scattering, using the same notation as in 10.5 and indicate the various colours as well19 : p

a

q

i~g 2 ac u(q1 )γ µ u(p1 ) v(p2 )γµ v(q2 ) Cbd , p2 q2 |~k|2 c d   1 a c 1 a c j a j c = (T )b (T )d = δ δ − δ δ . (12.82) 2 d b N b d

M(~k) = ac Cbd

b

1

k

j

1

=

The attraction/repulsion discussion now centers on the colour structure of the q q¯ state. If the quark-antiquark pair is in the singlet state, we must compute20  2   1 1 a b c d N2 − 1 1 d b ac a d c b √ δa δc Cbd = (δd δa )(δb δc ) − δb δc δd δa = , (12.83) 2N N 2N N and the interaction energy is negative. On the other hand, for the alternative octet states we may choose a = b 6= c = d and C then evaluates to −1/2N . 19

And remember the discussion of the Fermi sign ! Note that, even with conservation of colour, the colour assignment 1¯1 → 2¯2 is allowed. We must therefore sum √ over the initial- as well as over the final-state colour singlet combination. The factors 1/ N are the normalization factors for the singlet state. 20

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We conclude that a q q¯ pair in a colour singlet state attracts, but in the octet state repels. And indeed, colour octet states are never observed21 . We can also consider qq scattering : p

b

a

1

M(~k) = p 2

k d

j c

q

1

q2

=

−i~g 2 ac u(q1 )γ µ u(p1 ) u(q2 )γµ u(p2 ) Cbd . (12.84) 2 ~ |k|

Two quarks cannot form a singlet state, but they can form an antitriplet state, an antisymmetric combination of two colours, for which we may take 1 and 3, say. The colour factor now evaluates to22 2 
1 N +1 13 13 31 31 √ C13 (12.85) − C31 − C13 + C31 =− 2N 2 and the interaction is attractive. The remaining six colour combinations are the symmetric sextet ones, for which we may simply choose a = b = c = d = 33 = (N − 1)/2N . A quark pair in such a state feels a 3, say, and then find C33 23 repulsive interaction .

12.6.3

The process q q¯ → gg

The amplitude For an example of a QCD calculation we may look to the process q¯(p1 )q(p2 )

→

g(q1 , λ1 , j) g(q2 , λ2 , k) ,

in the high-energy approximation where the quark mass is negligible ; we have indicated the momenta and helicities of the quarks and gluons, as well as the gluon colour labels. We have already studied this process in order 21

This is of course horribly handwaving since we rely on tree-level perturbation theory here, which is known to be a very poor approximation at very small scattering momentum. Nevertheless, as a suggestion is comes out right ! 22 This expression is a little bit too pretentious : the antitriplet state only makes sense if N = 3 (otherwise the Levi-Civita symbol in Eq.(??) would not make sense. The only √ really sensible value for the colour factor in Eq.(12.85) is therefore -2/3, and the 1/ 2 factors are again the state’s normalization. 23 Quark pairs in a triplet state have not been observed. However, inside a proton, say, all three possible quark pairs are in an antisymmetric colour combination since the whole state is totally antisymmetric : again only a suggestion, but again in the right direction.

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to establish the three-gluon vertex, but what interests us here is not how the unwanted terms vanish, but rather what are the acceptable terms that are left. As we have seen, at the tree level the amplitude is described by 3 Feynman diagrams : p1

q1

p1

q2

p1

q1

p2

q2

p2

q1

p

q2

M=

(12.86) 2

and we can write it as M(λ1 , λ2 ) = A1 (λ1 , λ2 ) = A2 (λ1 , λ2 ) = A3 (λ1 , λ2 ) =

  1 1 1 j k k j j k ig ~ − A1 T T − A2 T T + A3 [T , T ] , t u s u+ (p1 )/1 (/q1 − p/1 ) /2 u+ (p2 ) , u+ (p1 )/2 (/q2 − p/1 ) /1 u+ (p2 ) , u+ (p1 )γα u+ (p2 ) Y (q1 , 1 ; q2 , 2 ; −q1 − q2 , α) , (12.87) 2

where as usual s = (p1 + p2 )2 , t = (p1 − q1 )2 , u = (p1 − q2 )2 , and s + t + u = 0. Note that we consider only the + helicity case for the quarks. This is quite allright, since the QCD couplings do not contain any γ 5 and overall interchange of all positive and negative helicities can only result in an overall phase change of the amplitude. It is at this point that in calculations like these a handlebar check like 1 → q1 is very useful to verify e.g. the correctness of the signs. Since we have already used the handlebar in section 12.3.3, we shall not do it now. We have to calculate three A’s, with four helicity cases for each. The polarisations The first (and perhaps most crucial) step24 is to make a clever choice for the polarisation vectors. It turns out to be smart indeed to take u+ (qi )γ µ u+ (p2 ) , i (+)µ = √ 2 s− (qi , p2 )

u− (qi )γ µ u− (p1 ) i (−)µ = √ , 2 s+ (qi , p1 )

just like we did in QED, since then √ √ 2 u− (qi )u− (p2 ) 2 u− (p1 )u− (qi ) ω− /i (+) = , ω− /i (−) = ; s− (qi , p2 ) s+ (qi , p1 ) 24

(12.88)

(12.89)

Again, you must realise that the choice of polarisation vector cannot change the final result for the amplitude : but making the intermediate steps easy or even trivial can save a lot of sweat and tears (if not blood).

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which, if we insert the polarisation vectors into the diagrams, leads to /i (+)u+ (p2 ) = u+ (p1 )/i (−) = 0 .

(12.90)

In addition we have 1 (±) · 2 (±) = 0 , s− (q2 , p2 )s+ (q1 , p1 ) 1 (+) · 2 (−) = , s− (q1 , p2 )s+ (q2 , p1 ) s− (q1 , p2 )s+ (q2 , p1 ) . 1 (−) · 2 (+) = s− (q2 , p2 )s+ (q1 , p1 )

(12.91)

Spinorial workout It is easy to see that out of the twelve cases, eight vanish straight away25 : An (±, ±) = A1 (−, +) = 0 = A2 (+, −) = 0 (n = 1, 2, 3) ,

(12.92)

The four remaining cases are readily evaluated in terms of spinor products26 : s− (p2 , q2 ) , s+ (p1 , q2 ) A3 (+, −) = −A1 (+, −) , s− (p2 , q1 ) , A2 (−, +) = 2s+ (p1 , q2 )2 s+ (p1 , q1 ) A3 (−, +) = A2 (−, +) . A1 (+, −) = 2s+ (p1 , q1 )2

(12.93)

Perhaps surprisingly, the A3 ends up looking just like the A1,2 ! If we neglect E98 overall complex phases, we have p p A1 (+, −) ∼ 2t t/u , A2 (−, +) ∼ (12.94) = 2u u/t , and so we find   1 j k 1 j k − T T − [T , T ] t s 2 p
2g ~ = t/u u T j T k + t T k T j . s

p M(+, −) ∼ 2g ~ t t/u 2

25

(12.95)

Note that, since in this calculation, u(p1 )/i u(p2 ) is always zero, we here have Y (q1 , 1 ; q2 , 2 , −q1 − q2 , α) ∼ (q1 − q2 )α (1 2 ). 26 In the A3 results, it may be useful to note that there are alternative forms : u+ (p1 )(/q1 − q2 )u+ (p2 ) equals 2s+ (p1 , q1 )s− (q1 , p2 ) as well as −2s+ (p1 , q2 )s− (q2 , p2 ). /

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Here we have used −1/t − 1/s = u/ts. In exactly the same way we find M(−, +) ∼


2g 2 ~ p u/t u T j T k + t T k T j . s

(12.96)

In these expressions, replacing T j,k by unity gives us the QED result for e+ e− → γγ if we remember that u + t = −s. Summing and averaging, and symmetry In order to compute h|M|2 i we must count a factor 1/4 for the spin average, a factor 1/N 2 for the colour average, and a factor 2 for the so-far neglected case of negative quark helicity. We find  

 u 2g 4 ~2 t 2 |M| + = × N 2 s2 u t


(t2 + u2 )Tr T j T k T k T j + 2utTr T j T k T j T k . (12.97) E99

The colour sums are most easily handled using the Fierz identities (12.16) : (N 2 − 1)2 , 4N
(N 2 − 1) Tr T j T k T j T k = − . 4N

Tr T j T k T k T j




=

The final result is  2 

 g 4 ~2 (N 2 − 1) 2 (N − 1) 2N 2 2 2 |M| = (t + u ) − 2 . 2N 3 ut s The differential cross section is

 1 dσ 2 = |M| : dΩ 128π 2 s

(12.98)

(12.99)

(12.100)

note the symmetry factor Fsymm = 1/2 for the 2-gluon final state !

12.6.4

The Landau-Yang theorem revisited

A loophole in Landau-Yang ? We may return to the Landau-Yang theorem discussed in the context of QED in section 10.7.3. It states that no spin-1 particle can decay into two photons. The reason was seen to be that each possible contribution to the decay

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amplitude (10.194) was antisymmetric in the interchange of the photons, in violation of their Bose statistics. Now suppose we replace the photons by gluons, that have an additional colour property. Can we build an amplitude that is also antisymmetric in the gluon colour labels ? In that case the total amplitude would obey Bose statistics and would hence not be forbidden. To investigate this, we shall look into quarkonium decay. Threshold amplitudes for q q¯ annihilation We shall compute (at tree level) the amplitude for the annihilation process of a (now massive ! ) quark-antiquark pair into two gluons : q(p) q¯(p) → g(q1 ) q(q2 ) , where the quark and the antiquark, of mass m, have the same momentum, i.e. they annihilate at rest. As before, we shall employ the modified polarisation vectors (q2 1 ) µ q (12.101) η1µ = µ1 − (q1 q2 ) 1 (and vice versa), and q µ = (q1µ − q2µ )/2. With the colour labels of gluon 1 and 2 denoted by j and k, respectively, the amplitude is again given by the diagrams (12.86) but now it reads ig 2 ~ v(p)/η1 (/q1 − p/) η/2 u(p) T j T k M = − 2 2 (p − q1 ) − m ig 2 ~ v(p)/η2 (/q2 − p/) η/1 u(p) T k T j − 2 2 (p − q2 ) − m ig 2 ~ + v(p)γα u(p) Y (q1 , η1 ; q2 , η2 ; −2p, α) [T j , T k ] (2p)2
ig 2 ~ = + 2 v(p)/η1 q/η/2 u(p) {T j , T k } + [T j , T k ] 4m
ig 2 ~ − 2 v(p)/η2 q/η/1 u(p) {T j , T k } − [T j , T k ] 4m ig 2 ~ + 2 v(p)/qu(p) (η1 η2 ) [T j , T k ] . (12.102) 2m The first two lines corresponds to the ‘QED-like’ diagrams with quark exchange, and the third line derives from the s-channel diagram containing the

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three-gluon vertex. Because (q · η1,2 ) = 0 we can write, e.g., η/1 q/η/2 = −/q(η1 η2 ) + iγ 5 γµ µ (q, η1 , η2 ) ,

(12.103)

and, since q and the η1,2 are orthogonal to p, γ 5 γµ µ (q, η1 , η2 ) =

1 (p, q, η1 , η2 ) γ 5 p/ 2 m

(12.104)

and we see two things happening : the terms with the commutator [T j , T k ] cancel amongst themselves, and for the terms with the (symmetric) anticommutator {T j , T k } we have M=−

g2~ (p, q, η1 , η2 ) {T j , Y k } v(p)γ 5 u(p) . m3

(12.105)

We now consider the total spin of the q q¯ system. If it has spin-1, the spin vectors of the quark and antiquark must be the same, and then we find
1 v(p, s)γ 5 1 + γ 5 s/ u(p, s) 2
1 v(p, s) 1 − γ 5 s/ γ 5 u(p, s) = 0 . = 2

v(p, s)γ 5 u(p, s) =

(12.106)

For opposite spin vectors, the amplitude is nonzero : |v(p, s)γ 5 u(p, −s)|2 = 4m2 . Thus, a quark-antiquark pair at relative rest in a spin-1 state cannot decay into two gluons, and the Landau-Yang theorem appears to hold here as well - at the tree level. We might turn QCD into QED by replacing the T matrices by unity, and so find the proof for orthopositronium. Landau and Yang come down from the trees We note that the vanishing of the colour-antisymmetric term (with [T j , T k ]) in the above relied on a cancellation between three contributions. If we now consider loop effects, we see that the s-channel graph is modified by quark loops in the gluon propagator and in the three-gluon vertex27 ,

27

This last diagram is discussed in section 12.3.2.

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and that the t- and u-channel diagrams do not pick up similar quark loops. Since the number (and certainly the masses) of the other quarks are not fixed by any principle there is, therefore, no possibility of a similar cancellation holding at the one-loop level and beyond ; and we conclude that the LandauYang theorem does not hold for gluons28 .

12.7

Exercises

Excercise 96 Prove the Antkaz ! Prove the result (12.73) by using the properties of the [jklmn] symbols. Excercise 97 A four-colour problem ? Suppose that the number of QCD colours were 4 rather than 3. Use what you know about fermions versus bosons, the Pauli exclusion principle, and general relativity to argue that in that case almost all hadronic matter in the universe would disappear into a few huge black holes, and we wouldn’t be here to do this exercise. . . Excercise 98 QCD spinorial workout Using standard-spinor techniques, prove Eq.(12.93). Excercise 99 Colour traces for q q¯ → gg Prove the two colour traces of Eq.(12.98). Excercise 100 Fun with Furry’s Failure Consider the diagram K of section 12.3.2. If one of the couplings were not of vector type (with γ µ ) but of axial-vector type (with γ 5 γ µ ), then the integral would change sign under the above transformations. Show that in that case the function Y would read −q3ρ g µν − q1µ g νρ − q2ν g ρµ and show that this is completely transverse to any external polarisation vector. 28

Note, however, that this violating would show up only at second loop order, since the interference of the tree-level result with the one-loop result would of course still vanish. Experimentally, therefore, the violation is very modest.

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Excercise 101 Gluonic MHV amplitudes We consider the purely gluonic process g(q1 , 1 ) g(q2 , 2 ) → g(q3 , 3 ) g(q4 , 4 ) . . . , g(qn , n ) at the tree level. 1. Show, using Eq.(6.23), that the amplitude, if correct, must be strictly prtoportional to E 4−n , where E is the total invariant energy of the system. 2. Show that in every contributing diagram the maximum number of propagators equals n − 3. 3. Show that the denominator of every diagram goes at most as E 2n−6 . 4. Show that the numerator of each diagram consists of three types of factors : (p·p), (p·), and (·), where the p stand for some combinations of the momenta q1 , q2 , . . . , qn , and  for some of the polarization vectors of the external gluouns. 5. Show that the maximum number of p’s in the numerator is n − 2, and that therefore every diagram contains at least one ( · ). 6. Choose the gluon polarizations according to Eq.(9.53), with gauge vector rj for gluon j. Show that (a) if gluon j and gluon k have the same helicity λ, then (j · k ) is proportional to sλ (qj , qk )s−λ (rj , rk ), (b) if gluon j has helicity λ and gluon k has helicity −λ then (j · k ) is proportional to sλ (qj , rk )s−λ (rj , qk ). 7. By choosing the gauge vectors cleverly, show that the amplitude vanishes if all n gluons have the same helicity. 8. By choosing the gauge vectors cleverly, show that the amplitude vanishes if n−1 gluons have the same helicity and one the opposite helicity. 9. Show that, when two gluons have helicity opposite from the other n − 2 gluons we cannot choose all gauge vectors such that the amplitude vanishes. These amplitudes, with 2 gluons having polarization opposite from the other ones, are called MHV amplitudes, see also section 10.7.1.

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Excercise 102 Other MHV amplitudes In the process q q¯ → gg · · · g the tree-level diagrams have the generic form

We conider the case where all gluons have the same helicity. 1. Let us first concentrate on a single ‘blob’ where an off-shell gluon coming from the q − q¯ line decays into n gluons: µ

We indicate the Lorentz index. We shall denote the momenta of the outgoing gluons generically by p, and their polarization vectors by . Show that every term in this off-shell amplitude is of one of two possible forms: (a) : pµ (p · )α ( · )β (p · p)−γ , (b) : µ (p · )α ( · )β (p · p)−γ 2. Show that the energy dependence of this amplitude is p1−n . 3. Show that the maximum number of gluon propagators in each term is equal to n − 1, so that γ ≤ n − 1 4. For case (a), show that α + 1 − 2γ = 1 − n and α + 2β = n, and that this implies β ≥ 1. 5. Show that all terms of the form (a) vanish by a suitable gauge vector choice. 6. For case (b), show that α − 2γ = 1 − n and α + 1 + 2β = n, and that that implies β ≥ 0. 7. Show that all nonvanishing terms of the form (b) are therefore of the generic form µ ( · p)n−1 (p · p)1−n . 8. Show that this implies the MHV result that the amplitude for q q¯ → gg · · · g vanishes at the tree level if all gluons have the same helicity.

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Chapter 13 Electroweak theory 13.1

Muon decay

13.1.1

The Fermi coupling constant

Let us return to the Fermi model of muon decay as discussed in chapter 7. There, the (phenomenological) amplitude for this decay was proposed to be of the form of Eq.(8.36). The resulting width was Γµ ≡ Γ(µ− → e− ν e νµ ) =

GF 2 ~2 mµ 5 . 192π 3

(13.1)

The measured values of the mechanical mass Mµ and the lifetime τµ of the muon are Mµ ≈ 1.884 10−28 kg ,

τµ ≈ 2.197 10−6 sec ;

(13.2)

the muon mass may be more familiar under its appellation of Mµ c2 0.106 GeV. From these we can construct the more useful quantities mµ =

≈

1 Mµ c ≈ 5.354 1014 m−1 , Γµ = ≈ 1.518 10−3 m−1 . (13.3) ~ cτµ

From Eq.(13.1) we then find

or

GF ~ ≈ 4.532 10−37 m2 ,

(13.4)

GF ≈ 1.164 10−5 GeV−2 . 2 ~c

(13.5)

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We can therefore derive the ‘energy scale’ of the interaction responsible for muon decay1 : s ~c2 ≈ 292.5 GeV . (13.6) ΛW = GF

13.1.2

Failure of the Fermi model in µ− ν µ → e− ν e

If the phenomenologically motivated Fermi interaction were to have any claim to global validity, it should also govern the process µ− (p1 ) ν µ (p2 )

→ e− (q1 ) ν e (q2 ) ,

(13.7)

which amounts to the previous process, only with the outgoing muon neutrino moved to an incoming anti-muon neutrino. No matter that we cannot, at present, build µν µ colliders ; the very, very, very early universe did provide such processes, and their description must be correct. By the rules of the Fermi model, the amplitude is given by GF ~ v(p2 )(1 + γ 5 )γ µ u(p1 ) u(q1 )(1 + γ 5 )γµ v(q2 ) M = i √ 2 4 = i √ GF ~ v − (p2 )γ µ u− (p1 ) u− (q1 )γµ v− (q2 ) 2 8 = i √ GF ~ s− (p2 , q1 ) s+ (q2 , p1 ) 2

(13.8)

Here, we have neglected both the muon and the electron mass since the scattering takes place at high energy, and we have applied the Chisholm identity in order to remove the contracted Lorentz index. Disregarding phases and using momentum conservation, we then find Mli

16 GF ~ √ (p1 · q2 ) . 2

(13.9)

Neutrinos2 have only one helicity state, and therefore, in the centre-of-mass system,

 |M|2 = 64 GF 2 ~2 (p1 · q2 )2 = 4 GF 2 ~2 s2 (1 + cos θ)2 , (13.10) 1

What precisely √ constitutes the scale is of course to some extent a matter of taste. If √ we include a factor 2 in GF the scale is reduced by a factor ( 2)1/2 to 246 GeV, which is the more commonly used number. 2 We shall assume, in this section, that neutrinos are strictly massless.

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where θ is the angle between the muon and electron momenta. By taking also the angular average we obtain



|M|2



=

16 GF 2 ~2 s2 . 3

(13.11)

The total cross section is therefore given by GF 2 ~2 s σ(µ ν µ → e ν e ) = 3π −

−

(13.12)

As we have seen before, only the factor 1/3 is not immediately obvious in this expression but has to be computed from the Feynman diagrams. The scattering cross section rises linearly with s, and will therefore violate the unitarity bound at sufficiently high energy. Since the the muon and its antineutrino couple with a single γ µ , they must be in a J = 1 state. The unitarity bound on this cross section (cf appendix 15.10) is therefore σ(µ− ν µ → e− ν e ) ≤

1 16π 24π (2J + 1) = , 2 s s

(13.13)

which leads to a fundamental failure √ of the Fermi model (at least, at the tree level) at a scattering energy of s ≈ 1.5 TeV.

13.2

The W particle

13.2.1

The IVB strategy

We have to modify the Fermi model in such a way that its success in the low-energy description of muon decay is preserved, while at high energies unitarity remains inviolate. One possible way out might be to simply make GF depend on the energy scale of the process so that it decreases at high energies, making the µ− ν µ → e− ν e cross section well-behaved. For this, we would need to replace GF → GF (0) f (s/Λ2 )

(13.14)

with f (x) decreasing at least as fast as 1/x for large arguments, and Λ some constant energy. Such energy-dependent couplings, called form factors,

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are employed in for instance ‘low-energy’ hadronic physics ; in such cases, however, this approach is generally viewed as an admission of ignorance of, and an attempt to cope with, some underlying and simpler physics at a smaller distance scale 1/Λ ; that is, one would basically be invoking some ‘new physics’ that would make everything work out right 3 . The more elegant, and (as it turns out) the correct way to go is to make the Fermi model look more ‘QED-like’: instead of using a contact interaction between four fermions, we postulate the existence of a new particle, the socalled W boson. This couples to fermion-antifermion pairs in a way reminiscent of the photon. The four-fermion interaction then resolves into two f f¯W interactions, with the W boson mediating between the two vertices ; the corresponding Feynman diagram for the process µ− (p) to e− (q) νµ (k1 ) ν e (k2 ) is therefore given by

p M =

k1 Q

q k2

(13.15)

At the time this model was first seriously discussed, it went under the name of Intermediate Vector-Boson (IVB) hypothesis. We take the W to couple to the fermion pairs eνe and µνµ , so that (as we shall check!) the W must be electrically charged, and assume that the coupling is in both cases of equal strength4 (for now). We therefore postulate the following Feynman rules :

3

In the present case, GF , being dimensionful, sets such a scale by itself. At this point, these are of course just assumptions. Since 1983, when the W boson was first freely produced, they have been tested with great accuracy. The alternative scenario of the charge retention form, with an electrically neutral W , is completely ruled out by the fact that the decay W → e+ µ− is never seen. The equality of the couplings is verified by the fact that the branching ratios for W → eν e and W → µν µ are the same up to computable mass effects. 4

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µ

k

377

ν ↔ i~

µ

↔

−g µν + k µ k ν /mW 2 k 2 − mW 2 + i


i gW 1 + γ 5 γ µ ~

internal W lines

f f 0 W vertices

EW Feynman rules, part 13.1

(13.16)

The W propagator is the standard one for a vector particle. Note that the occurrence of the (1 + γ 5 ) in the vertex is suggested by the form of the Fermi interaction ; and that the two fermions meeting in the vertex must be of different type. The values of mW and gW are to be determined. Another attractive property of this model is that here the coupling constant, gW , has the same dimensionality as the QED one, and does not formally contain a length scale. With the above Feynman rules, the muon decay amplitude can now be written as  i~gW 2 u(k1 )(1 + γ 5 )γ α u(p) u(q)(1 + γ 5 )γα v(k2 ) M = Q2 − mW 2  1 5 5 − u(k1 )(1 + γ )/Qu(p) u(q)(1 + γ )/Qv(k2 ) , (13.17) mW 2 where the momentum of the internal W is given by Qµ = (p − k1 )µ = (q + k2 )µ .

(13.18)

The last term in Eq.(13.17) appears to deviate significantly from the spinorial structure of the first term, which coincides with the Fermi model. Hoewever, notice that u(k1 )(1 + γ 5 )/Qu(p) = u(k1 )(1 + γ 5 )(/p − k/1 )u(p)
= u(k1 ) − k/1 (1 − γ 5 ) + (1 + γ 5 )/p u(p) = mµ u(k1 )(1 + γ 5 )u(p) (13.19) upon application of the Dirac equation to the external spinors ; and since, in the same way, u(q)(1 + γ 5 )/Qv(k2 ) = me u(q)(1 − γ 5 )v(k2 ) ,

(13.20)

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the second term in Eq.(13.17) is actually suppressed by a factor (me mµ )/mW 2 , which is small if mW is sufficiently large5 . Neglecting this term, we see that the Fermi-model amplitude √ is recovered with the single replacement of the coupling constant GF / 2 by gW 2 /(Q2 − mW 2 ). Now, the maximum value that Q2 can take in this process is mµ 2 , which is attained in the improbable case that the muon neutrino emerges with zero momentum from the decay. If, therefore, we assume that mW is large compared to mµ , we see that the successes of the Fermi model in describing muon decay will be completely reproduced provided GF gW 2 √ , l (13.21) mW 2 2 which we may also write in purely dimensionless terms as     gW 1 mW c2 √ . (13.22) = 1/4 2 ΛW c ~

13.2.2

The cross section for µ− ν µ → e− ν e revisited

We can now study the modification that the IVB hypothesis makes in the cross section for the process µ− ν µ → e− ν e , where the Fermi model fails. In this case the total invariant mass is (assumed to be) much larger than the W mass, so that the modified prediction can immediately be seen to be  2 2~2 gW 4 s ~2 GF 2 s mW 2 − − σ(µ ν µ → e ν e ) = = , 3π (s − mW 2 )2 3π s − mW 2 (13.23) and this cross section does decrease as 1/s for large s. Of course, the unitarity limit (13.13) still has to be observed, which puts an upper limit6 on the useful values of mW : mW c2 ≤ (72π 2 )1/4 ΛW ≈ 1.5 TeV .

(13.24)

However, from Eq.(13.22) we see that for such large values the dimensionless coupling constant is so large that the tree-level approximation for the cross section is questionable. −7 In fact, for the actual values of the √ masses the suppression factor is about 10 . This value is close to the value of s at which unitairy breaks down in the unmodified Fermi model, see Eq.(13.13). This is not a coincidence. Whatever we do to the electroweak interactions, 1.5 TeV appears to be the energy r´egime where things get tricky. 5

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379

One may wonder what happens at s = mW 2 . There, the cross section would seem to diverge ! We must realise, however, that at that energy we are, in fact, producing an on-shell W that decays into a fermion-antifermion pair : that is to say, the W is an unstable particle, and has a decay width. We ought, therefore, to include the decay width into the propagator, so that in the neighbourhood of the resonance at s ≈ mW 2 the cross section reads σ(µ− ν µ → e− ν e ) =

2~2 gW 4 s . 2 2 3π (s − mW ) + mW 2 ΓW 2

(13.25)

This is well below the unitarity limit. The IVB hypothesis therefore indeed E103 cures the unitarity problem in this process. E104 Because of these successes, we shall adopt the notion of an existing W particle of spin 1 (and hence obeying the lines laid out in chapter 9), coupling to pairs of fermions separated by one unit of charge7 .

13.2.3

The W W γ vertex

Minimal coupling Since the W particle couples to fermion pairs of unequal charge, it must itself also be charged8 , which means that it must couple to the photon in (at least !) a W W γ vertex. It is our aim now to find the form of such a vertex. Both W ’s and photons are characterised by the fact that, in addition to their momentum, they carry also a polarisation vector, i.e. a Lorentz index: the W W γ vertex must therefore carry no fewer than 3 Lorentz indices. As a first attempt, we can simply view the W particles as a kind of funny scalars, and adopt the sQED vertex dressed up with a metric tensor to take care of the W indices. That is, the Feynman rule for the vertex is taken to be

ν W (p1 )

γ(p3 )

+

ρ

_

W (p2 )

↔

i QW (p1 − p2 )ρ g µν ~

(13.26)

µ

where the coupling constant (the W charge) is to be determined, and the particles are considered to be outgoing from the vertex. To this end, let us 7 8

Note that this automatically rules out couplings between a W , a lepton, and a quark. At pain of charge nonconservation, i.e. at pain of pain.

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examine the process D(q1 ) U (q2 ) → γ(k1 , ) W + (k2 , + ) .  and W denote the polarisation vectors of the photon and the W , respectively, and we have indicated the particle momenta. Here, and in the following, we shall denote by U and D two fermions of which the U has an electric charge one unit higher than the D: for instance, U = νe and D = e, or U = u and D = d. Their respective charges are QU and QD . At the tree level, we then have three Feynman diagrams : q1 q1 q1 k1 k k2

M =

+

q2

k2

2

+ q2

k1

q2

k1

(13.27)

The three diagrams correspond to the three partial matrix elements k/1 − q/1 + mD (1 + γ 5 ) /W u(q2 ) , M1 = −i~gW QD v(q1 )/ (k1 − q1 )2 − mD 2 q/2 − k/1 + mU M2 = −i~gW QU v(q1 ) (1 + γ 5 ) /W / u(q2 ) , (q2 − k1 )2 − mU 2 M3 = +i~gW QW v(q1 ) (1 + γ 5 ) γα u(q2 ) g αβ − P α P β /mW 2 W β (2k2 + k1 · ) , (13.28) s − mW 2 where s = P 2 , P = q1 + q2 = k1 + k2 . Since this process involves a produced photon, the handlebar identity must hold : if we replace µ by k1 µ the amplitude must vanish. We shall investigate this in some detail, by explicit computation rather than by the diagrammatic manipulations of the previous chapters. In the first place, we perform some simple Dirac algebra to note that  v(q1 )/(/k1 − q/1 + mD ) →k1 = v(q1 )/k1 (/k1 − q/1 + mD )
= v(q1 ) k1 2 − 2(q1 · k1 ) + (/q1 + mD )/k1
= (k1 − q1 )2 − mD 2 v(q1 ) , (13.29) where in the second line we have used anticommutation between k/1 and q/1 , and in the third line the Dirac equation for v(q1 ). We see that M1 c→k1 = −i~gW QD v(q1 ) (1 + γ 5 ) /W u(q2 ) ,

(13.30)

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and similarly M2 c→k1 = +i~gW QU v(q1 ) (1 + γ 5 ) /W u(q2 ) .

(13.31)

For the third diagram we find M3 c→k1 = +i~gW QW v(q1 ) (1 + γ 5 ) /W u(q2 ) −i~gW QW (k1 · W ) v(q1 ) mU (1 + γ 5 ) − mD 1 − γ 5





u(q2 ) . (13.32)

If we were allowed to consider only the first of the two terms of the result (13.32), we could obtain the desired cancellation : % 3 X Mj = 0 ⇒ QW = QD − QU : (13.33) j=1

→k1

but the second term in Eq.(13.32) spoils this idea by having a quite different algebraic structure ; no tuning of coupling constants is going to ensure that a W W γ vertex of the form (13.26) can do the job. The W -photon coupling Treating the W W γ vertex as a prettified sQED vertex does not work. It means that the photon-W interactions cannot be obtained by the minimalsubstitution rule. This should not come as a surprise since the vertex (13.26) is only designed for graceful behaviour towards longitudinal photons, not towards longitudinal W ’s. We therefore propose to replace Eq.(13.26) by a vertex of the form i


QW (a1 p1 + a2 p2 )ρ g µν + (a3 p2 + a4 p3 )µ g νρ + (a5 p3 + a6 p1 )ν g ρµ . (13.34) ~

Of course, we have encountered a similar vertex when considering the threegluon coupling in QCD (cf section 12.3). However, we are now dealing with a much less symmetric situation since two out of the three bosons are massive, and it behooves us to be careful. Note that because of momentum conservation each of the three terms need contain only two of the momenta; the constants a1,...,6 are to be determined. This we shall do by considering several situations.

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First, we consider the process of decay of a photon in a W + W − pair : γ ∗ (q) → W + (k+ , + ) W − (k− , − ) . Kinematically this is only possible if the photon is quite off-shell, and therefore we do not give it a polarisation vector but leave its Lorentz index µ free. The matrix element is given by M = i~1/2 QW Aµ , Aµ = (a1 k+ + a2 k− )µ (+ · − ) +((a3 k− − a4 q) · + )− µ + ((−a5 q + a6 k+ ) · − )+ µ = (a1 k+ + a2 k− )µ (+ · − ) +(a3 − a4 )(q · + )− µ + +(a6 − a5 )(q · − )+ µ , (13.35) where in the last line we have used q = k+ + k− and (k± · ± ) = 0. Since even for off-shell photons the current must be strictly conserved we require that 1 Aµ qµ = q 2 (a1 + a2 )(+ · − ) + (a3 − a4 − a5 + a6 )(q · + )(q · − ) = 0, (13.36) 2 which leads to the following relations between the six constants : a1 + a2 = 0 ,

a3 − a 4 = a 5 − a 6 .

(13.37)

In the second place, we return to the process DU → γW + discussed in the previous section. The third Feynman diagram now reads differently :
1 Zα , M3 = +i~gW QW v(q1 ) (1 + γ 5 ) γα u(q2 ) 2(k1 · k2 )
Z α = δ α β − P α Pβ /mW 2  ((a1 k2 − a2 P ) · )+ β + ((−a3 P + a4 k1 ) · + )β +(a5 k1 + a6 k2 )β (+ · )
= δ α β − P α Pβ /mW 2  (a1 − a2 )(k2 · )+ β + (a4 − a3 )(k1 · + )β +(a5 k1 + a6 k2 )β (+ · ) . (13.38) The replacement  → k1 now leads, after some simple algebra (and use of momentum conservation !) to the form Z α c→k1 = (a1 − a2 )(k1 · k2 )+ α + T α , T α = (−a3 + a4 + a5 − a6 )k1 α (k1 · k2 ) (a1 − a2 − a3 + a4 + a5 + a6 )P α . (13.39) − mW 2

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Now a complete cancellation of all diagrams in this case is only possible if only the first term in Z α c survives. Using the assignment9 QW = QD − QU , we then come to the following additional relations between the a’s : a1 − a2 = 2 ,

a1 − a2 = a3 − a4 − a5 − a6 = 0 .

(13.40)

A third result is obtained by considering the process U D → γW − . Because of the symmetry between this amplitude and the previous one, we can establish that also a1 − a2 = a5 − a6 + a3 + a4 . (13.41) For the last necessary piece of information we must turn to the handlebar operation for the produced W rather than the photon. We can rewrite the three Feynman diagrams as
~gW QD 5 v(q )/  (/ q − k / + m ) (1 + γ ) /+ u(q2 ) , M1 = −i 1 2 2 D (q2 − k2 )2 − mD 2
~gW QU 5 M2 = −i v(q ) (1 + γ ) /+ (/k2 − q/1 + mU ) /u(q2 ) , 1 (k2 − q1 )2 − mU 2
~gW QW 5 M3 = +i v(q ) (1 + γ ) γα u(q2 ) Z α , (13.42) 1 2 s − mW with Z α as in Eq.(13.38). The handlebar operation on + now gives the slightly more complicated result
 v(q1 )/ (/q2 − k/2 + mD ) (1 + γ 5 ) /+ u(q2 ) + →k2 =

= − (q2 − k2 )2 − mD 2 v(q1 ) (1 + γ 5 ) /u(q2 )


− v(q1 ) mU (1 + γ 5 ) − mD 1 − γ 5 /k/2 u(q2 )

(13.43) + mU 2 − mD 2 v(q1 ) (1 + γ 5 ) /u(q2 ) . Of these three lines, the second is suppressed with respect to the first one by a factor (mass/energy), and the third line even by (mass/energy)2 . In the high-energy limit, therefore, the second and third line will not contribute to any unwanted high-energy behaviour of the amplitude : we shall call such terms safe terms10 . We can therefore write
(13.44) M1 c+ →k2 = +i~gW QD v(q1 ) (1 + γ 5 ) /u(q2 ) + · · · , 9

Any common factor in the a’s is always absorbed in the value of QW so this is no loss of generality. 10 Which is not to say that they are negligible ! The point here is that they do not contribute to any condition on the coupling constants. At the end of a cross section calculation it is the safe terms that we want !

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where the ellipsis denotes safe terms. For the second diagram, we find in a similar way :
M2 c+ →k2 = −i~gW QU v(q1 ) (1 + γ 5 ) /u(q2 ) + · · · , (13.45) For the third graph we find, after some algebra, Z α c+ →k2 = (a4 − a3 )(k1 · k2 )α − a3 mW 2 α (k2 · )(k1 · k2 ) (a1 − a2 − a3 + a4 + a5 + a6 )P α mW 2 + (k2 · )(−a1 + a2 + a5 − a6 )k1 α . (13.46)

−

Requiring M3 to cancel against M1 + M2 up to safe terms therefore leads to yet more relations between the a’s : a3 − a4 = 2 ,

a1 − a 2 = a 5 − a 6 .

(13.47)

Combining the requirements (13.37), (13.40), (13.41) and (13.47) we find the unique solution a1 = a3 = a5 = 1 ,

a2 = a4 = a6 = −1 .

(13.48)

We find that the spacetime part of the vertex has, after all, exactly the same form as that of the three-gluon vertex, Eq.(12.33) ! We have thus established the W W γ vertex to be

µ W (p1 ) +

_

W (p2 ) ν

γ (p3 ) ↔ ρ

i QW Y (p1 , µ; p2 , ν; p3 , ρ) ~

All particles and momenta counted outgoing EW Feynman rules, part 13.2

13.3

The Z particle

13.3.1

W pair production

(13.49)

Unitarization from extra fermions In the previous section we have investigated how the possible coupling between W ’s and photons are restricted by the requirements of the handlebar.

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We shall now pursue the same strategy for different processes. Since we shall be interested in the high-energy behaviour of amplitudes we shall allow ourselves to neglect particle masses wherever possible. Let us consider the process U (p1 ) U (p2 ) →

W + (q+ , + ) W − (q− , − )

With the vertices available so far, we have the following two Feynman diagrams _ _ _ U + U W W D γ _ W+ U W U which contribute to the amplitude as follows :
~gW 2 5 v(p ) (1 + γ ) /− (/p2 − q/+ ) /+ u(p2 ) , 1 (p2 − q+ )2 ~QU QW = i v(p1 )γµ u(p2 ) Y (q+ , + ; q− , − , −q+ − q− , µ)(13.50) . (q+ + q− )2

M1 = −2i M2

Here we have neglected the masses as announced. The high-energy behaviour can be investigated by putting a handlebar on the W + , say ; we then obtain M1 c+ →q+ = 2i~gW 2 v(p1 )(1 + γ 5 )/− u(p2 ) , M2 c+ →q+ = i~QU QW v(p1 )/− u(p2 ) ,

(13.51)

and we see that these two diagrams cannot possibly cancel one another. We must therefore introduce an additional ingredient in the model. A possible approach is the following. In the analogous process U U → γγ the handlebar requirement is satisfied because there are two diagrams, with the photons interchanged. We might do the same for the W by postulating the existence of another fermion type U 0 , with charge one unit higher than QU , and the existence, in addition to the U DW vertex, of a U 0 U W vertex with vector and axial-vector couplings. We then have a third diagram at hand : _ U W+ U’ _ W U

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with its own contribution ~ v(p1 )ω/+ (/q+ − p/1 ) ω/− u(p2 ) , (p1 − q+ )2 ω = g1 + g2 γ 5 .

M3 = −i

(13.52)

The mass of the U 0 is also neglected, and g1,2 are to be determined. We have M3 c+ →q+ = −i~ v(p1 )ω 2 /− u(p2 ) ,

(13.53)

so that 3 X j=1

% Mj

=0

⇒

g1 + g2 γ 5


2

= 2gW 2 (1 + γ 5 ) + QU QW . (13.54)

+ →q+

We see that it is in principle possible to attain good high-energy behaviour in the process U U → W + W − , at the cost of introducing new fermion types ; and the same is possible for DD → W + W − . But a very serious conundrum immediately arises. Having postulated the existence of the U 0 , we of course 0 also have to consider high-energy behaviour in the process U 0 U → W + W − . It is easy to see that that can only be cured by postulating also a fermion U ”, of again one unit of charge higher . . . An infinite tower of fermions with higher and higher charge becomes unavoidable. Not only is this extremely unattractive11 , but as the charges grow without bound perturbation theory is bound to break down since it is based on the assumption that the couplings governing the interactions are not large. The Z boson to the rescue Since introducing additional Dirac particles does not seem a viable way to ensure good high-energy behaviour in U U → W + W − , we shall investigate the alternative of an additional boson. That is, we shall postulate the existence of a neutral spin-1 particle, coupling to W + W − pairs and to fermionantifermion pairs. This particle, denoted by Z (or Z 0 ) is supposed to cure the high-energy behaviour in both U U → W + W − and DD → W + W − simultaneously12 . For the W W Z vertex it stands to reason to employ the useful 11

Even leaving aside the fact that no higher-charge fermions have been found to date. This is the simplest scenario. Other possibilities could be explored, in which there is more than one type of Z, perhaps one type for the U fermions and one type for the D fermions. Experiment, however, has taught us that the simplest option appears, as usual, to be the one chosen by nature. 12

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387

Yang-Mills form Y (· · ·), with a coupling constant to be determined. Since the diagram with the Z must cancel against a combination of the purely vectorial photon diagram and the D-exchange diagram with its (1 + γ 5 ) structure, the Z must couple to the fermions with a nontrivial mixture of vector and axial-vector terms. We therefore arrive at the following putative Feynman rules :

µ + W (p1 ) _ W (p2 ) ν

Z(p3 ) i ρ → ~ gWWZ Y (p1 , µ; p2 , ν; p3 , ρ) ,

U Z

i ~


vU + aU γ 5 γ µ ,

→

i ~


vD + aD γ 5 γ µ ,

µ

U D Z D

→

µ

where as before in the Yang-Mills vertex every participant is counted in the outgoing manner. With these vertices a new Feynman diagram is available in the process U U → W + W − :

_ U

W+ Z

U

_ W ,

which evaluates to M3 = i


~ gWWZ 5 v(p ) v + a γ γµ u(p2 ) 1 U U (q+ + q− )2 − mZ 2 Y (q+ , + ; q− , − ; −q+ − q− , µ) .

(13.55)

Note that nothing has been neglected in this expression ; the second term in the massive-boson propagator drops out when we multiply it into the YangMills vertex. Since this diagram is so similar to M2 it is easy to perform the handlebar operation :
M3 c+ →q+ ≈ i~ gWWZ v(p1 ) vU + aU γ 5 /− u(p2 ) , (13.56)

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where we have assumed that s = (q+ + q− )2 is also much larger than mZ 2 , and neglected safe terms. We now see that the high-energy behaviour is acceptable provided that the non-safe terms cancel under the relations 0 = vU gWWZ + 2gW 2 + QU QW , 0 = aU gWWZ + 2gW 2 .

(13.57)

We can perform precisely the same procedure for the process DD → W + W − and obtain 0 = vD gWWZ − 2gW 2 + QD QW , 0 = aD gWWZ − 2gW 2 .

(13.58)

A final piece of information is obtained if we realise that, the Z being a massive spin-1 particle, it must obey its own handlebar relations ; we can therefore investigate the process U D → W + Z, which gives a single extra condition 0 = vD + aD − vU − aU − gWWZ . (13.59)

13.3.2

The weak mixing angle for couplings

We can handle (if not completely solve) the system of constraints as follows. Let us subtract Eqs.(13.57) from Eqs.(13.58). We then obtain (vD + aD − vU − aU )gWWZ + (QD − QU )QW = 8gW 2 .

(13.60)

Using Eq.(13.59) and the definition of QW , we find a relation between three couplings : gWWZ 2 + QW 2 = 8gW 2 . (13.61) There must, therefore, exist an angle θW such that √ √ QW = 8 gW sin θW , gWWZ = 8 gW cos θW .

(13.62)

In the following we shall use the notation sW = sin θW and cW = cos θW . This angle is called the weak mixing angle13 , and it parametrizes essentially all of the minimal model of electroweak interactions we are constructing here. In 13

The subscript W was originally used to refer to S. Weinberg, one of the early proponents of the electroweak model, but nowadays it appears more fair to take it to mean just ‘weak’.

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the first place, we know that the charge of the W must be equal to the charge of the electron (since neutrinos are neutral) and therefore we might prefer to write QW (13.63) gW = √ 8 sW which leads to a parametrization of the W mass itself14 : (~ c mW )2 = √

πα 1 GeV2 , 2 −5 2 1.16 10 sW

(13.64)

or

37.3 GeV . (13.65) sW As we see, the assumption of the existence of a single, neutral Z boson immediately implies that the W has a mass of at least 37.3 GeV. Notice that no prediction for the mass of the Z is obtained, however. The other unknowns in our treatment can now be expressed in terms of θW . Adopting the usual convention of denoting by e the positive unit charge, we find by straightforward algebra ~ c mW =

QW = −e ,

gWWZ = −e

cW , sW

e , aU = −aD = 4sW cW   QU vU = aU 1 − 4sW 2 , e   2 QD . vD = aD 1 + 4sW e

(13.66)

We note here that θW is defined at this stage as a relation between coupling E105 constants ; later on we shall encounter it in another guise !

13.3.3

W, Z and γ four-point interactions

The 2 → 2 processes involving either four fermions or two fermions and two bosons have led us to postulate W and Z particles and their interactions with fermions, as well as their mutual three-point vertices. Since we have pretty 14

To arrive at this expression we have used the definition (13.5) for GF , and the result (10.35) of α.

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much quarried all possible information15 about this sector, we now turn to the 2 → 2 processes involving four bosons. First we consider the process W + (p1 , 1 ) γ(p2 , 2 ) → W + (p3 , 3 ) γ(p4 , 4 ) With the available vertices we have two Feynman diagrams for this process : W+

W+

γ

M =

+ γ

W+

W

+

γ

(13.67)

γ

with the respective contributions ~QW 2 Y (p3 , 3 ; p2 − p3 , ν; −p2 , 2 ) (p2 − p3 )2 − mW 2
× g µν + (p1 − p4 )µ (p2 − p3 )ν /mW 2 × Y (p1 − p4 , µ; −p1 , 1 ; p4 , 4 ) ~QW 2 − mW 2 (2 · 3 )(1 · 4 ) = −i 2(p2 · p3 )
+Y (p3 , 3 ; p2 − p3 , µ; −p2 , 2 )Y (p1 − p4 , µ; −p1 , 1 ; p4 , 4 ) ,

M1 = i

~QW 2 Y (p3 , 3 ; −p3 − p4 , ν; p4 , 4 ) (p3 + p4 )2 − mW 2
× g µν + (p1 + p2 )µ (−p3 − p4 )ν /mW 2 × Y (p1 + p2 , µ; −p1 , 1 ; −p2 , 2 ) ~QW 2 = i − mW 2 (3 · 4 )(1 · 2 ) 2(p3 · p4 )
+Y (p3 , 3 ; −p3 − p4 , µ; p4 , 4 )Y (p1 + p2 , µ; −p1 , 1 ; −p2 , 2 ) , (13.68)

M2 = i

where we have already used Eq.(??) in the internal W lines, as well as the fact that (pj · j ) = 0, j = 1, 2, 3, 4. Let us now proceed to check current conservation for the outgoing photon. The following algebra applies to M1 : Y (p3 , 3 ; p2 − p3 , µ; −p2 , 2 )Y (p1 − p4 , µ; −p1 , 1 ; p4 , 4 )c4 →p4 =
= Y (p3 , 3 ; p2 − p3 , µ; −p2 , 2 ) (p4 · 1 )(p2 − p3 )µ + 2(p2 · p3 )1 µ = 2(p2 · p3 )Y (p3 , 3 ; p2 − p3 , 1 ; −p2 , 2 ) + mW 2 (p4 · 1 )(2 · 3 ) , (13.69) 15

As long as the fermion masses are neglected, see later.

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so that M1 c4 →p4 = −i~QW 2 Y (p3 , 3 ; p2 − p3 , 1 ; −p2 , 2 )

(13.70)

In the same manner we arrive at M2 c4 →p4 = i~QW 2 Y (p1 + p2 , 3 ; −p1 , 1 ; −p2 , 2 )

(13.71)

Adding these last two results we obtain % 2 X Mj = j=1

4 →p4 2

= i~QW (2(1 · 3 )(2 · p4 ) − (1 · 2 )(3 · p4 ) − (2 · 3 )(1 · p4 )) . (13.72) We might also have chosen choose to put the handlebar on 2 instead ; the result would then have been % 2 X Mj = j=1

2 →p2 2

= i~QW (2(1 · 3 )(p2 · 4 ) − (1 · p2 )(3 · 4 ) − (p2 · 3 )(1 · 4 )) . (13.73) Going to the limit of large energies, we can also envisage putting a handlebar on 1 or 3 . Neglecting safe terms leads to % 2 X Mj = j=1

1 →p1 2

= i~QW (2(p1 · 3 )(2 · 4 ) − (p1 · 2 )(3 · 4 ) − (2 · 3 )(p1 · 4 )) , (13.74) and 2 X j=1

% Mj

= 3 →p3 2

= i~QW (2(1 · p3 )(2 · 4 ) − (1 · 2 )(p3 · 4 ) − (2 · p3 )(1 · 4 )) . (13.75)

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We can repair the high-energy behaviour of the amplitude, for all these cases at once, by introducing a four-boson vertex : W+

µ

ν

_

i − QW 2 X µναβ ~

(13.76)

X µναβ = 2 g µν g αβ − g µα g νβ − g µβ g να .

(13.77)

W γ α

↔

γ β

where The occurrence of such a four-point vertex should not surprise us, with our experience of a similar vertex in sQED and QCD. Its precise algebraic structure can, of course, not be inferred from that example16 . From the similarity between the W W γ and W W Z vertices we can also immediately conclude that the analogous processes W Z → W γ and W Z → W Z will necessitate the existence of the following four-point vertices : W+

µ

_ W

Z W+

ν

α µ Z α

γ β ν _ W

↔

i cW µναβ − QW 2 X ~ sW

↔

i cW 2 − QW 2 2 X µναβ . ~ sW

Z β

(13.78)

fFinally, we consider the process W + (p1 , 1 ); W − (p2 , 2 ) → W + (p3 , 3 ) W − (p4 , 4 ) , for which we have, so far, the four diagrams

W+

W+ Ζ,γ

M =

W 16

_

W+

Z,γ

W+

+

W

_

.

W

_

W

(13.79)

_

Except, perhaps, the idea that it contains only the metric tensor, and not any of the momenta : but momenta would only worsen the high-energy behaviour

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It will turn out to be useful to take the γ and Z exchanges together so that we have two contributions : M1 = i~QW 2 Y (p3 , 3 , −p1 , 1 , p1 − p3 , µ)   g µν cW 2 g µν − (p1 − p3 )µ (p1 − p3 )ν /mZ 2 + (p1 − p3 )2 sW 2 (p1 − p3 )2 − mZ 2 Y (−p2 , 2 , p4 , 4 , p2 − p4 , ν) , M2 = i~QW 2 Y (−p2 , 2 , −p1 , 1 , p1 + p2 , µ)   g µν cW 2 g µν − (p1 + p2 )µ (p1 + p2 )ν /mZ 2 + (p1 − p3 )2 sW 2 (p1 + p2 )2 − mZ 2 Y (p3 , 3 , p4 , 4 , p2 − p4 , ν) . (13.80) Because the masses of the external particles are all equal, the second term in the Z propagator can be seen to drop out exactly. We can therefore afford to take the limit s  mZ 2 without more ado, and combine the γ and Z propagators to arrive at the following high-energy form of the contributions : M1

M2

1 ~QW 2 Y (p3 , 3 , −p1 , 1 , p1 − p3 , µ) = i 2 sW (p1 − p3 )2 Y (−p2 , 2 , p4 , 4 , p2 − p4 , µ) , ~QW 2 1 = i Y (−p2 , 2 , −p1 , 1 , p1 + p2 , µ) 2 sW (p1 + p2 )2 Y (p3 , 3 , p4 , 4 , p2 − p4 , µ) .

(13.81)

Let us now take the outgoing W − longitudinal, i.e apply the handlebar on 4 , and drop safe terms : Y (p3 , 3 , −p1 , 1 , p1 − p3 , µ)Y (−p2 , 2 , p4 , 4 , p2 − p4 , µ)c4 →p4 = Y (p3 , 3 , −p1 , 1 , p1 − p3 , µ)
× (p1 − p3 )µ ((p1 − p3 ) · 2 ) − ((p1 − p3 )2 − mW 2 )2 µ ≈ −(p1 − p3 )2 Y (p3 , 3 , −p1 , 1 , p1 − p3 , 2 ) (13.82) so that M1 c4 →p4 = −i

~QW 2 Y (p3 , 3 , −p1 , 1 , p1 − p3 , 2 ) ; sW 2

(13.83)

and the exactly analogous treatment gives M2 c4 →p4 = −i

~QW 2 Y (−p2 , 2 , −p1 , 1 , p1 + p2, 3 ) . sW 2

(13.84)

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The total result of the handlebar operation is given by M1 + M2 c4 →p4 = −i

~QW 2 (2(p4 · 1 )(2 · 3 ) − (p4 · 2 )(1 · 3 ) − (p4 · 3 )(1 · 2 )) : sW 2 (13.85)

we arrive at precisely the same algebraical structure as before, and we can immediately conclude that, in addition to the W W γγ, W W Zγ and W W ZZ couplings there must also be a W W W W coupling : W+ W

µ _ α

ν

W+

W β

_

↔

i QW 2 µναβ X ~ sW 2

(13.86)

Note, however, a slight difference of this vertex as compared to the previous ones. There, the term that couples the two W Lorentz indices carries the factor 2 ; here, it is the term that couples the two W + ’s that is ‘special’.

13.4

The Higgs sector

13.4.1

The Higgs hypothesis

Fully longitudinal scattering Having pursued the consequences of unitarity in processes where a single external spin-1 particle is longitudinally polarised, we must of course also face the more taxing case in which, perhaps, all external spin-1 particles are longitudinally polarised : surely this is the most dangerous case from the point of view of unitarity. In doing so, we must however take into account the fact that the notion of longitudinal polarisation is not strictly a Lorentz-invariant one since a generic Lorentz boost will mix longitudinal and transverse degrees of freedom. We must therefore specify in which particular Lorentz frame the particles are assumed to be longitudinally polarised. To this end we introduce a timelike vector cµ with c · c = 1 ; the frame in which ~c = 0 is defines the appropriate Lorentz frame. In these notes we shall take cµ to be proportional to the total momentum involved in the scattering process, that

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is, the external vector particles are assumed to be purely longitudinal in the centre-of-mass frame of the scattering17 . The longitudinal polarisation of an on-shell vector particle with momentum pµ and mass m is then given by   m2 m2 µ NL µ µ , NL −2 = 1 − p − c L = , (13.87) m c·p (c · p)2 which expression is well-defined as long as p~ 6= 0. We see that, as before, L = p/m + O (m/p0 ). In the cases studied so far, the subleading terms in L have only led to safe terms so that they could be neglected18 ; now, this is no longer automatically the case. Before continuing we may note that another possible choice for longitudinal polarisation is to take a lightlike vector tµ and define m µ 1 t . (13.88) µL = pµ − m p·t In many applications this is useful. However, the boson is purely longitudinal only in a frame where ~t k p~. In scattering processes, not all bosons can then be strictly longitudinal at the same time. In this section we therefore opt for cµ rather than tµ W W → ZZ The first Gedanken process19 is W + (p1 , 1 ) W − (p2 , 2 ) → Z 0 (p3 , 3 )Z 0 (p4 , 4 ) So far, we have the following three Feynman graphs available at the tree level : 1 W Z 3 1 W Z 4 1 W Z 4 M =

W 2

17

W

+

Z 4

W 2

W

+

Z 3

W 2

,

(13.89)

Z 3

That this is not a trivial point becomes clear when we realise that in ‘W W scattering’ at the LHC, say, the centre-of-mass frame of the scattering does not coincide with the laboratory frame, in which the detector is at rest, and in which the polarisation analysis of the produced bosons is presumably performed. 18 From the point of view of restoring unitarity, not that of actually getting the cross section right! 19 As usual, with improving technology and the commissioning of machines like the LHC, Gedanken processes are gradually being turned into actual ones.

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and the following contributions : Nj , j = 1, 2 , ∆j Y (p1 − p3 , µ; −p1 , 1 ; p3 , 3 )   1 µν µ ν × −g + (p1 − p3 ) (p1 − p3 ) mW 2 × Y (−p2 , 2 ; p2 − p4 , ν; p4 , 4 ) , (p1 − p3 )2 − mW 2 = mZ 2 − 2(p1 · p3 ) , Y (p1 − p4 , µ; −p1 , 1 ; p4 , 4 )   1 µν µ ν × −g + (p1 − p4 ) (p1 − p4 ) mW 2 × Y (−p2 , 2 ; p2 − p3 , ν; p3 , 3 ) , (p1 − p4 )2 − mW 2 = mZ 2 − 2(p1 · p4 ) , −i~gWWZ 2 N3 , X(1 , 2 , 3 , 4 ) .

Mj = −i~gWWZ 2 N1 =

∆1 = N2 =

∆2 = M3 = N3 =

(13.90)

Owing to the work we have done so far, we may already anticipate some cancellations between the diagrams when we make all bosons longitudinal and the safe terms are therefore not the subleading ones, but rather the subsubleading ones. We have to proceed carefully20 . Denoting by the subscript L the ‘fully longitudinal’ case, it appears best to write the result as % 3 X N123 = −i~gWWZ 2 , Mj ∆ 12 j=1 L

N123 = N1 ∆2 + N2 ∆1 + ∆12 N3 = −4E 6 ∆12 = ∆1 ∆2 = 4E 4 (sin θ)2 + · · · ,

mZ 2 (sin θ)2 + · · · , mW 4 (13.91)

where E = p1 0 = p2 0 = p3 0 = p4 0 and θ = ∠(~p1 , p~3 ), all evaluated in the centre-of-mass frame. As before, the ellipses denote contributions that can only give rise to safe terms, and that therefore do not interest us here. Note that we have disregarded also the normalization factors NL ; since the polarisation vectors are overall factors in the scattering amplitude, the NL 20

This is most safely done using computer algebra, using e.g. FORM.

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can never play a rˆole in any dynamical cancellation, and their subleading terms are therefore always safe. The non-safe contribution from our three Feynman graphs is therefore % 3 X mZ 2 + ··· , (13.92) Mj = i~ gWWZ 2 E 2 4 m W j=1 L

and it violates unitarity at sufficiently large E. Note that each individual Mj will go as E 4 at high energy so, as already anticipated, some cancellation has already taken place, but not enough ; and since the vertices have already been fixed before, we have to introduce a new ingredient into the theory. The Minimal Higgs approach We shall assume that, in addition to the three graphs used so far, there is a fourth one available, mediated by a new particle type. We assume this to be a neutral, scalar particle, denoted by H, that couples to W + W − and ZZ as follows : µ

W H

↔

i gWWH g µν ~

H

↔

i gZZH g µν ~

ν W µ Ζ ν

(13.93)

Ζ

A fourth Feynman diagram is now possible :

3

1 M4 =

2

H

=

−i~ gWWH gZZH (1 · 2 ) (3 · 4 )

4

4E 2

1 . − mH 2 (13.94)

Its contribution to the fully longitudinal scattering reads M4 cL = −i~ E 2 gWWH gZZH

1 2

mW mZ 2

(13.95)

and good high-energy behaviour will be restored in the process W W → ZZ provided that mZ 4 gWWH gZZH = gWWZ 2 . (13.96) mW 2

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Before we proceed to the next Gedanken process, a few remarks are in order. In the first place, the choice for a scalar Higgs particle is almost unavoidable. It certainly cannot be a fermion ; if it were a vector particle, its propagator would contain unwanted higher powers of the energy E, the W W H and ZZH would presumably be of Yang-Mills type hence also E-dependent. The vertices given above are essentially the only ones possible for the interactions between two vectors and a scalar if we want them to be energy-independent. Note that gWWH and gZZH may √ both be expected to contain a mass, that is, they are of dimension L−1 / ~. The assumption that there is just one type of neutral scalar involved is, of course, based on nothing but a prejudice in favour of simplicity. Finally, at high energy all contributions from mH end up in safe terms, and we do not expect to glean any information on the Higgs mass from our considerations. W W → W W scattering Another four-boson scattering process of interest is W + (p1 , 1 )W + (p2 , 2 ) → W + (p3 , 3 )W + (p4 , 4 ) for which we have five purely vector-boson diagrams : 1

1 3

M =

γ,Ζ 4

2

1

4

+

+

γ,Ζ 2

3

2

3

(13.97)

4

whose contributions can be conviently written as M1 = −i~ Y (p3 , 3 ; −p1 , 1 ; p1 − p3 , µ)   µν −gµν + (p1 − p3 )µ (p1 − p3 )ν /mW 2 2 2 −g × QW + gWWZ (p1 − p3 )2 (p1 − p3 )2 − mZ 2 × Y (p4 , 4 ; −p2 , 2 ; p2 − p4 , ν) , M2 = M1 cp3 ,3 ↔ p4 ,4 , M3 = i~

QW 2 X(3 , 4 , 1 , 2 ) . sW 2

By the same methods as used in the previous section we arrive at % 3 X
~ E 2 QW 2 Mj =i −4mW 2 + 3mZ 2 cW 2 + · · · 4 2 mW sW j=1 L

(13.98)

(13.99)

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The Higgs hypothesis now provides for two additional diagrams : 1

1 3

M4,5 =

H

4

,

4

2

(13.100)

H 2

3

with the contributions M4 = −i~ gWWH 2 M5 = −i~ gWWH 2

(1 · 3 )(2 · 4 ) , (p1 − p3 )2 − mH 2 (1 · 4 )(2 · 3 ) , (p1 − p4 )2 − mH 2

(13.101)

so that 5 X j=4

% = i~ E 2

Mj

gWWH 2 + ··· mW 4

(13.102)

L

In this process, then, good high-energy behaviour is obtained under the condition
QW 2 gWWH 2 = 4mW 2 − 3mZ 2 cW 2 . (13.103) 2 sW Again, no restrictions on mH occur. HZ → W W scattering We have now run out of four-vector Gedanken processes. ZZ → ZZ scattering has no Yang-Mills contributions21 , and any four-vector process involving photons will have vanishing amplitudes under a handlebar on any photon. However, in the same spirit by which we boldly proposed the process U D → W Z as soon as the Z was hypothesised, we can consider the process H(p1 )Z 0 (p2 , 2 ) → W + (p3 , 3 )W − (p4 , 4 ) Since only three out of four particles can become longitudinal here, the unitarity violations are not so bad, and the safe terms are of sub- rather than 21

Under the Higgs hypothesis ZZ → ZZ scattering is described by three diagrams each containing Higgs exchange : no Higgs means no scattering at all ! The amplitude is safe by itself and hence does not lead to any constraints.

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of sub-sub-leading type. We have three diagrams, 1

3

3

M =

+ 2

3 1

4

2 1

+ 4

(13.104) 2

4

that contribute as M1 = −i~ gWWZ gWWH Y (p3 , 3 ; p2 − p3 , µ; −p2 , 2 ) −g µν + (p2 − p3 )µ (p2 − p3 )ν × (4 )ν , (p2 − p3 )2 − mW 2 M2 = −i~ gWWZ gWWH Y (p2 − p4 , µ; p4 , 4 ; −p2 , 2 ) −g µν + (p2 − p4 )µ (p2 − p4 )ν (3 )ν , × (p2 − p4 )2 − mW 2 M3 = −i~ gWWZ gZZH Y (p3 , 3 ; p4 , 4 ; −p3 − p4 , µ) −g µν + (p1 + p2 )µ (p1 + p2 )ν × (2 )ν . (p1 + p2 )2 − mZ 2

(13.105)

The kinematics of this process is a little different from that of the two previous ones, since mH and mZ cannot be assumed to be equal. Still, at high energy we may apply massless kinematics since we only have to cancel the leading non-safe terms. Neglecting, therefore, mW , mZ and mH in the kinematics22 we find %   3 X 1 mZ 2 − gZZH Mj +· · · (13.106) = i~ E cos θ gWWZ gWWH mW 4 mZ mW 2 j=1 L

and find the final requirement gWWH

mZ 1 = gZZH 4 mW mZ mW 2

(13.107)

if good high-energy behaviour is to emerge.

13.4.2

Predictions from the Higgs hypothesis

The Higgs hypothesis has given us the three conditions of Eqs.(13.96), (13.103) and (13.107). If we consider gWWH and gZZH as the two unknowns, this system is overconstrained, and we obtain additional information. The system 22

But not, of course, in the longitudinal polarisations !

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of conditions can easily be solved and we find the two couplings gWWH =

Q W mW , sW

gZZH =

QW mZ , s W cW

(13.108)

and, in addition, the interesting relation mW = mZ cW .

(13.109)

It is apposite to dwell on this last result. The weak mixing angle θW was E106 introduced to parametrize the system of coupling constants, as discussed in section 13.3.2 : we now see it come back here as a relation between masses instead ! From the treatment of the Electroweak Standard Model presented in these notes, it also becomes clear that the mixing angle as a description of coupling constants is, in a logical sense, prior to that as a description of masses. The assumption of a single Z 0 particle determines the couplings as described in section 13.3.2 : but it takes the supposition of a single, neutral Higgs particle to obtain Eq.(13.109). If the Higgs sector of the Standard Model turns out to be different, with more Higgs-like particles, say, the W and Z mass become uncorrelated ; but the couplings of W and Z with the fermions and each other remain unaffected. In the usual textbook derivation of the model this distinction tends to be obscured by the simultaneous obtention of all couplings at once after symmetry breaking. As a final comment we remark that, if unitarity is restored by any Higgslike phenomenon whatsoever, the weak mixing angle must always obey the bound 4 mW 2 (13.110) cW 2 < 3 mZ 2 as can be seen from Eq.(13.103)23 .

13.4.3

W, Z and H four-point interactions

The class of bosonic four-particle scattering amplitudes is not yet completely exhausted. We can consider the process Z 0 (p1 , 1 )Z 0 (p2 , 2 ) → H(p3 )H(p4 ) 23

For the actually observed values of W and Z mass this bound is itself somewhat larger than unity, and therefore not so significant; but it is nice to have it even so.

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given by two diagrams so far, 1

1

3

M =

4

+ 4

2

(13.111) 2

3

and the following amplitude : M1+2 = −i~ gZZH 2 (1 )µ (2 )ν  µν −g + (p1 − p3 )µ (p1 − p3 )ν /mZ 2 (p1 − p3 )2 − mZ 2  −g µν + (p1 − p4 )µ (p1 − p4 )ν /mZ 2 + . (p1 − p4 )2 − mZ 2

(13.112)

In the fully longitudinal case the non-safe terms are M1+2 cL = −i~ E 2

gZZH 2 + ··· mZ 4

(13.113)

and the remedy ought to be straightforward by now. We introduce yet another vertex, involving two Z’s and two H’s :

Z

µ

H ↔

H

ν Z

i gZZHH g µν ~

(13.114)

upon which we have a third diagram, whose nonsafe part is trivial : gZZHH + ··· (13.115) M3 cL = 2i~ E 2 mZ 2 We see that the four-point coupling constant must be given by gZZHH =

gZZH 2 QW 2 = . 2mZ 2 2sW 2 cW 2

(13.116)

As in the case of sQED and YM, this four-point coupling does not contain a length scale, in contrast to the ZZH coupling. For the case of W W → HH scattering, exactly the same treatment holds. It suffices to replace mZ by mW and gZZH by gWWH . We find that also a W W HH vertex is required :

µ W

H ↔

H ν W

i gWWHH g µν , ~

(13.117)

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with gWWHH =

13.4.4

gWWH 2 QW 2 = . 2mW 2 2sW 2

(13.118)

Higgs-fermion couplings

Let us return to the process U (p1 )U (p2 ) → W + (q+ , + )W − (q− , − ) which was used in section 13.3.1 to argue the existence of the Z boson. This time, however, we shall not neglect the fermion masses ; and we shall take both W ’s longitudinal. It can be seen that each individual diagram will go as E 2 when the energy E of the W ’s in their centre-of-mass frame becomes large. This means that, in the longitudinal polarisation of Eq.(13.87), the second term will only contribute to the safe terms, and we may simply write (± )L = q± /mW , so that  s (q+ − q− )µ + · · · (13.119) Y (q+ , + ; q− , − ; −q+ − q− , µ) ≈ − 2 2mW L where once more the ellipsis denotes safe terms. In fact, the restriction to nonsafe terms in our treatment means that we may neglect the boson masses in the kinematics : every occurrence of boson masses from the kinematics is quadratic and hence gives safe terms. For the fermions this is not the case as we shall see. Let us revisit the diagrams of our process. The first one now reads q/− − p/1 + mD (1 + γ 5 )/+ u(p2 ) . (13.120) M1 = −i~gW 2 v(p1 )(1 + γ 5 )/− 2 2 (q− − p1 ) − mD Note that the mD in the numerator drops out by virtue of the (1 + γ 5 )’s. We can now perform some Diracology, using the Dirac equation and dropping safe contributions wherever opportune : M1 c L = A = → = → →

−i~gW 2 v(p1 ) A u(p2 ) , mW 2 ((q− − p1 )2 − mD 2 ) (1 + γ 5 ) q/− (/q− − p/1 ) (1 + γ 5 ) q/+ 2(1 + γ 5 ) q/− (/q− − p/1 ) (/q+ − p/2 + mU ) 2(1 + γ 5 ) q/− (/q− − p/1 ) (/p1 − q/− + mU )
2(1 + γ 5 ) −(q− − p1 )2 q/− + (/q− − p/1 − mU )(/q− − p/1 )mU 2(1 + γ 5 ) (mU − q/− ) (q− − p1 )2 ; (13.121)

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so that the fully longitudinal case gives for this diagram M1 cL = 2i~gW 2 v(p1 )(1 + γ 5 )(/q− − mU )u(p2 ) + · · ·

(13.122)

For the third diagram we can perform a similar analysis : ~ gWWZ −s v(p1 ) B u(p2 ) , M3 c L = i s − mZ 2 2mW 2 B = (vU + aU γ 5 )(/q+ − q/− ) → (vU + aU γ 5 )(/q+ − q/− − p/2 + mU − p/1 ) − mU (vU − aU γ 5 ) = −2(vU + aU γ 5 )/q− + 2mU aU γ 5 ; (13.123) and up to safe terms, we therefore have
~gWWZ 5 5 M3 c L = i v(p ) (v + a γ )/ q − m a γ u(p2 ) + · · · (13.124) 1 U U − U U mW 2 To obtain the contribution from the second diagram, we simply put gWWZ → QW , vU → QU , and aU → 0 in the third diagram : ~QW QU M2 c L = i v(p1 ) q/− u(p2 ) + · · · (13.125) mW 2 If we add the three diagrams, the contributions with v/q− u cancel precisely, as they should since that was what we imposed in section 13.3.1. We are left with terms proportional to mU :
~mU 2 5 5 v(p ) −2g (1 + γ ) − g a γ M1+2+3 cL = i u(p2 ) + · · · 1 W WWZ U mW 2 ~ QW 2 mU = −i v(p1 )u(p2 ) + · · · (13.126) mW 2 4sW 2 so that an energy behaviour of E 1 at high energy is still uncompensated. The Higgs boson is usefully applied here as well. We may assume the U U H vertex U i H ↔ gUUH 1 , (13.127) ~ U where we must realise that the Dirac unit matrix is involved24 . For the process U U → W W we then have a fourth available diagram : M4 = 24

(13.128)

In fact, the observation that the nonsafe part in this process is proportional to vu is the strongest argument in favour of a scalar Higgs.

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which contributes to the amplitude the amount M4 = −i~ gUUH gWWH v(p1 )u(p2 )

1 (+ .− ) . s − mH 2

E107 (13.129)

In the fully longitudinal case we therefore have M4 c = − i

~ gUUH gWWH v(p1 )u(p2 ) + · · · , 2mW 2

(13.130)

and the following requirement on gUUH is obtained : QW 2 mU ~ gUUH gWWH + =0 , 4sW 2 2mW 2

(13.131)

or

QW mU . (13.132) 2sW mW This discussion can of course be applied to any fermion type25 , and we find E108 the general Feynman rule gUUH = −

f

H

f

13.4.5

↔

i e mf 1 ~ 2sW mW

(13.133)

Higgs self-interactions

The triple H coupling There remains the issue of possible self-interactions of the Higgs particle. To this end we examine not a 2 → 2 but a 2 → 3 process, namely Z(p1 , 1 ) Z(p2 , 2 ) → Z(p3 , 3 ) Z(p4 , 4 ) H(p5 ) . At the tree level, this process is described by 21 Feynman diagrams provided we allow for three-point couplings between H’s. These belong to one of the three following types : M1,2,3 =

,

,

(13.134)

Note that for D-type fermions, aD has opposite sign ; but also the W + and W − are interchanged in the first diagram. 25

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where as usual the dotted lines denotes Z’s and the solid lines stand for H particles, and we have to take into account the appropriate permutations of the external Z particles. The amplitude is given by the three corresponding contributions : M1 = + + + M2 = + M3 = A1 (i1 , i2 , i3 , i4 , i5 ) = A2 (i1 , i2 , i3 , i4 , i5 ) = A3 (i1 , i2 , i3 , i4 , i5 ) = Πµν (q) = ∆Z (q) =

A1 (1, 2, 3, 4, 5) + A1 (2, 1, 3, 4, 5) + A1 (3, 4, 1, 2, 5) A1 (4, 3, 1, 2, 5) + A1 (1, 3, 2, 4, 5) + A1 (3, 1, 2, 4, 5) A1 (2, 4, 1, 3, 5) + A1 (4, 2, 1, 3, 5) + A1 (1, 4, 3, 2, 5) A1 (4, 1, 3, 2, 5) + A1 (3, 2, 1, 4, 5) + A1 (2, 3, 1, 4, 5) , A2 (1, 2, 3, 4, 5) + A2 (3, 4, 1, 2, 5) + A2 (1, 3, 2, 4, 5) A2 (2, 4, 1, 3, 5) + A2 (1, 4, 3, 2, 5) + A2 (3, 2, 1, 4, 5) . A3 (1, 2, 3, 4, 5) + A3 (1, 3, 2, 4, 5) + A3 (1, 4, 2, 3, 5) , i~3/2 gZZH 3 i1 µ Πµν (pi1 + pi5 ) i2 ν (i3 · i4 ) × ∆Z (pi1 + pi3 ) ∆H (pi3 + pi4 ) , −i~3/2 gZZH gZZHH (i1 · i2 )(i3 · i4 ) × ∆H (pi3 + pi4 ) , i~3/2 gZZH 2 gHHH (i1 · i2 )(i3 · i4 ) × ∆H (pi1 + pi2 ) ∆H (pi3 + pi4 ) , 1 qµ qν , −gµν + mZ 2
−1
−1 q 2 − mZ 2 , ∆H (q) = q 2 − mH 2 . (13.135)

Here we have, for once, taken all momenta outgoing, which means that the momenta of the incoming Z’s have negative zeroth component. For this 5particle process the phase space is of course more complicated, and here we demonstrate a numerical method to investigate cancellations. This can be quite sensitive if done right26 . Although na¨ıvely each diagram A1 and A2 grow quadratically with the energy in the fully longitudinal case, both M1 and M2 actualy become energy-independent at sufficiently high energy E. But this is not safe : a 2 → 3 amplitude must go at most as E −1 , and therefore cancellations between (M1 + M2 ) and M3 are still necessary. We find that the required HHH coupling is given by ↔ 26

i

gHHH , ~

gHHH = −

3 QW mH 2 2 mW sW

(13.136)

An algorithm for sampling the relevant phase space is given in appendix 15.13

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Figure 13.1: Monte Carlo investigation of cancellations in ZZ → ZZH

if the necessary cancellations are to arise. In figure 13.1 we have, somewhat arbitrarily, chosen mW c2 = 80.4 GeV, mZ c2 = 91.2 GeV, mH c2 = 125 GeV. The plot shows − M3 cL / M1+2 cL for varying total energy w. The sampling is performed as described in appendix 15.13. The two contributions to the amplitude are seen to balance one another precisely, and the combined amplitude goes as w−1 , provided the right choice of gHHH is made. The amplitudes themselves are heavily dependent on the various scattering angles: but their ratio is not. A word of caution is in order on the interpretation of this picture. The high-energy limit is, strictly speaking, only obtained if all products of momenta grow large with respect to all masses involved. In a sampling over phase space it can always happen that some momentum products are comparable to squared masses ; these cases are responsible for the outlying dots in the plot at large values of the energy scale.

The quartic H coupling The last Gedanken process needed is Z(p1 , 1 ) Z(p2 , 2 ) → H(p3 ) H(p4 ) H(p5 )

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which is described by 25 Feynman diagrams in six types: M1,...,6 =

,

,

,

,

,

.

(13.137) where we have already anticipated a quartic Higgs coupling in the last diagram. The contributions to the amplitude are M1 = + M2 = + M3 = M4 = + M5 = M6 = B1 (i1 , i2 , i3 , i4 , i5 ) = B2 (i1 , i2 , i3 , i4 , i5 ) = B3 (i1 , i2 , i3 , i4 , i5 ) = B4 (i1 , i2 , i3 , i4 , i5 ) = B5 (i1 , i2 , i3 , i4 , i5 ) =

B1 (1, 2, 3, 4, 5) + B1 (1, 2, 4, 5, 3) + B1 (1, 2, 5, 3, 4) B1 (1, 2, 5, 4, 3) + B1 (1, 2, 3, 5, 4) + B1 (1, 2, 4, 3, 5) , B2 (1, 2, 3, 4, 5) + B2 (1, 2, 4, 5, 3) + B2 (1, 2, 5, 3, 4) B2 (2, 1, 3, 4, 5) + B2 (2, 1, 4, 5, 3) + B2 (2, 1, 5, 3, 4) , B3 (1, 2, 3, 4, 5) + B3 (1, 2, 4, 5, 3) + B3 (1, 2, 5, 3, 4) , B4 (1, 2, 3, 4, 5) + B4 (1, 2, 4, 5, 3) + B4 (1, 2, 5, 3, 4) B4 (2, 1, 3, 4, 5) + B4 (2, 1, 4, 5, 3) + B4 (2, 1, 5, 3, 4) , B5 (1, 2, 3, 4, 5) + B5 (1, 2, 4, 5, 3) + B5 (1, 2, 5, 3, 4) , −i~3/2 gZZH gHHHH (1 · 2 ) ∆H (p1 + p2 ) , i~3/2 gZZH 3 1 µ Πµ λ (pi1 + pi3 ) Πλν (pi2 + pi5 ) 2 ν × ∆Z (pi1 + pi3 ) ∆Z (pi2 + pi5 ) , i~3/2 gZZH 2 gHHH i1 µ Πµν (pi2 + pi5 ) i2 ν × ∆Z (pi2 + pi5 ) ∆H (pi3 + pi4 ) , i~3/2 gZZH gHHH 2 (i1 · i2 ) ∆H (p1 + p2 ) ∆H (pi4 + pi5 ) , −i~3/2 gZZHH gZZH i1 µ Πµν (pi2 + pi5 ) i2 ν × ∆Z (pi2 + pi5 ) , −i~3/2 gZZHH gHHH (i1 · i2 ) ∆H (pi3 + pi4 ) , . (13.138)

A treatment analogous to that of the previous paragraph leads to the following, final Feynman rule : ↔

i 3 QW 2 mH 2 gHHHH , gHHHH = − ~ 4 mW 2 sW 2

(13.139)

In figure 13.2 we plot the ratio − M6 cL / M1+···+5 cL obtained in the same manner as in the previous paragraph. Again, the choice of the factor 3/4 in gHHHH is justified by the fact that the ratio goes to 1 with great accuracy as the total energy w increases. At high energy there are no outliers : M6 depends only on w.

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Figure 13.2: Monte Carlo investigation of cancellations in ZZ → HHH

13.5

About anomalies

In sections 10.2.7 and 12.3.2, we have discussed triangle diagrams in which three vector particles couple to fermions running around a loop, V1

V3

V2

which is always accompanied by the same diagram with orientation of the fermion lines reversed. In three or more spacetime dimensions, such diagrams are problematic. A fermion propagator at large momentum pµ goes as 1/p, while the integration element in D dimensions goes as pD , leading to a divergence if d ≥ 3. In four dimensions, the diagrams are therefore linearly divergent, but this leads to trouble : there is no unambiguous integration prescription. If we cavalierly shift integration momenta and use dimensional regularization, then the triangle diagram will have a logarithmic UV divergence of the usual type, and might be balanced with a counterterm : but E109 this only works if the counterterm has a counterpart in the bare action. This can mean that we are forced to change the original action, with possibly disastrous consequences. For instance, if the three vector particles V1,2,3 are

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August 31, 2019 V1 , V2 , V3 γ, γ, Z γ, Z, Z

terms P 2 Qf af P Q f v f af

γ, W, W Z, Z, Z

P 2 P Qf2gW vf af

Z, Z, Z Z, W, W

P 3 P af 2 vf gW

Z, W, W Z, g, g

P 2 P af gW2 af gs

total contribution
Nc QU 2 − QD 2 + Qν 2 − QL 2
Nc QU (1 − σQU ) + QD (1 + σQD ) + Qν (1 − σQν ) + QL (1 + σQL ) Nc (QU + QD ) + Qν + QL
Nc (1 − σQU )2 − (1 + σQD )2 + (1 − σQν )2 − (1 + σQL )2 trivially zero
Nc (1 − σQU ) − (1 + σQD ) + (1 − σQν ) − (1 + σQL ) trivially zero trivially zero

Table 13.1: Contributions to the triangle anomalies

γ, Z, and Z a γZZ coupling in the bare action implies that the electrically neutral Z’s are, actually, not neutral at all. In that case, what does electric charge mean ? If all three fermion-fermion-vector couplings are of vector type, Furry’s theorem (cf section 10.2.7) shows how the two diagrams have opposite sign, all other things being equal. Therefore, in QED, where the three vector particles are photons, (V1 , V2 , V3 ) = (γ, γ, γ), such diagram pairs will completely cancel. In QCD, with three gluons, (V1 , V2 , V3 ) = (g, g, g), the two diagrams form a colour-antisymmetric contribution to the one-loop correction of the three-gluon vertex. Two mixed cases are (V1 , V2 , V3 ) = (γ, γ, g), where the diagrams vanish individually because Tr (T j ) = 0 for gluonic colour matrices T j , and (V1 , V2
, V3 ) = (γ, g, g)
where they again cancel against one another since Tr T j T k = Tr T k T j . We shall now investigate how things stand if we include the electroweak sector. In particular we have to worry about diagrams containing one or three γ 5 ’s, since in such cases the two diagrams with oppositely-oriented fermion loops have the same sign. Since we shall be interested only in the possible ultraviolet behaviour of the theory, we may assume that all fermions have the same (negligible) mass. Also, we consider the contribution of a single fermion family, that is a charged lepton (with charge QL ), a neutrino (with charge Qν , which we alllow to be nonzero for the moment), and an up-type quark (with charge QU ) and a down-type quark

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(with charge QD ). The quarks are supposed to occur in Nc colour types. We consider the collective effect of all these fermions together in table ??, where we have a look at all possibly problematic axial-vector contributions. We use the shorthand σ for 4sW 2 /e, sum over all fermions in a family, and drop all common overall factors. Inspection tells us that in fact two conditions suffice to let the sum over fermions vanish :
Nc QU 2 − QD 2 = QL 2 − Qν 2 ,
Nc QU + QD = − QL − Qν . (13.140) The ratio of these two result in the requirement QU − QD = Qν − QL (= −QW ) ,

(13.141)

which is nothing else than the charge conservation we already encountered in Eq.(13.33). We see that the single condition, necessary to ensure all the cancellations, is that the quark charges conform to QU =

Qν − QL Qν + QL − , 2 2Nc

QD =

QL − Qν Qν + QL − , 2 2Nc

(13.142)

In the Standard Model, with Nc = 3, we have indeed Qν = 0 ,

2 QU = − QL , 3

1 QD = QL . 3

(13.143)

This is one of those curious instances where the universe appears to arrange itself to make the theory as well-behaved as possible. Whether there is a deeper reason for this I do not pretend to know, but I am grateful nonetheless.

13.6

Conclusions and remarks

We have now derived all vertices of the electroweak Standard Model. That is to say, the more usual textbook derivations arrive at precisely the set of Feynman rules that we have also obtained. There are, however, a number of differences between the treatment given here and the usual one. • We have not invoked any symmetry principle, but rather the (underlying) SU (2) × U (1) symmetry has spontaneously emerged from our choices for the ‘minimal’ solution, for instance by insisting on only a single Z particle while we could have opted for more.

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• Since we have not invoked any symmetry, there is also no need to explain its ‘breaking’ in order to arrive at massive W ’s and Z’s. Instead, we have simply faced the observed fact of their massiveness and come to grips with it with the help of a (minimal) Higgs sector. • There is, as we have already discussed, a logical distinction between the two uses of the weak mixing angle, in which the ratio of coupling constants is logically ‘prior’ to the ratio mW /mZ . • We have not needed to introduce any Higgs doublet, but rather only a single, physically observable H particle. This approach elegantly sidesteps the question whether , and if so how the Higgs field configuration is ‘spontaneously broken’. This would indicate that the Higgs particle is also, in a sense, logically prior to a complete Higgs doublet.

13.7

A look at non-minimal models

13.7.1

Non-minimal Higgs sector

The Minimal Standard Model we have derived here is, of course, by no means the only possibility. An alternative is presented here. Let us inspect the process U (p1 ) D(p2 ) → W + (p3 , + ) Z(p4 , ) for which, at the tree level, we have the following Feynman graphs: Z W

W

W

Z Z

These diagrams are given by, respectively, p/4 − p/2 + mD M1 = −i~gW v(p2 )(vD + aD γ 5 )/ (1 + γ 5 )/+ u(p1 ) , (p4 − p2 )2 − mD 2 p/1 − p/4 + mU M2 = −i~gW v(p2 )(1 + γ 5 )/+ (vU + aU γ 5 )/u(p1 ) , 2 2 (p1 − p4 ) − mU M3 = −i~gW gWWZ v(p2 )(1 + γ 5 )γα u(p1 ) −g αβ − (P )α (−P )β /mW 2 Y (p3 , + ; −P, β; p4 , ) , (13.144) × s − mW 2

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413

with P = p1 + p2 = p3 + p4 . Using
v(p2 )(1 + γ 5 )/P u(p1 ) = v(p2 ) p/2 (1 − γ 5 ) + (1 + γ 5 )/p1 u(p1 ) (13.145) = mU v(p2 )(1 + γ 5 )u(p1 ) − mD v(p2 )(1 − γ 5 )u(p1 ) and Y (p3 , + , −P, −P ; p4 , ) = (mW 2 − mZ 2 )(+ · )

(13.146)

we can rewrite M3 as  ~ gW gWWZ M3 = i v(p2 )(1 + γ 5 )γα u(p1 ) Y (p3 , + ; −P, α; p4 , ) s − mW 2   
mZ 2 5 5 . (+ · ) mU v(p2 )(1 + γ )u(p1 ) − mD v(p2 )(1 − γ )u(p1 ) + 1− mW 2 (13.147) In the fully longitudinal case all three diagrams actually grow as E 2 at most, at high energy E. From this point on, therefore, we take the bosons to be massless since their masses can only lead to safe terms. The fully longitudinal expressions, in which we systematically ignore safe terms, can then be evaluated as follows. ~gW v(p2 ) Σ u(p1 ) , 2(p2 · p4 )mW mZ (vD + aD γ 5 )/p4 (−/p2 + mD )(1 + γ 5 )/q3 −2(p2 · p4 )(vD + aD γ 5 )(1 + γ 5 )/p3 +(vD + aD γ 5 )(/p2 + mD )/p4 p/3 (1 − γ 5 ) −2(p2 · p4 )(vD + aD )(1 + γ 5 )/p3 +2mD aD aD γ 5 p/4 p/3 (1 − γ 5 ) −2(p2 · p4 )(vD + aD )(1 + γ 5 )/p3 −4(p2 · p4 )mD aD aD (1 − γ 5 ) ,

M1 c L = i Σ = = = =

(13.148)

where we have used the fact that, up to safe terms, we can write p/4 p/3 = p/4 (/p3 + p/4 − p/1 ) = p/4 p/2 ; so that for the first diagram we have  ~gW M1 c L = i − (vD + aD ) v(p2 )(1 + γ 5 )/p3 u(p1 ) mW mZ  5 − 2aD mD v(p2 )(1 − γ )u(p1 ) . (13.149)

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In the same way we arrive at M2 c L

~gW = i mW mZ



(vU + aU ) v(p2 )(1 + γ 5 )/p3 u(p1 )  5 − 2mU aU v(p1 )(1 + γ )u(p1 ) .

(13.150)

Finally, we can derive27 that  ~gW gWWZ M3 c L = i v(p2 )(1 + γ 5 )/q3 u(p1 ) mW mZ 
mZ 2 5 5 mU v(p2 )(1 + γ )u(p1 ) − mD v(p2 )(1 − γ )u(p1 ) . − 2mW 2 (13.151) These diagrams each contain a term that goes as E 2 , and these contributions cancel one another under the condition vU + aU − vD − aD + gWWZ = 0

(13.152)

which we have already encounteered in Eq.(13.59). We are left with an amplitude that grow as E 1 ~gW v(p2 ) A u(p1 ) , mW mZ   mZ 2 A = −mU gWWZ + 4aU ω+ mW 2   mZ 2 − 4aD ω− . +mD gWWZ mW 2

M123 cL = i

Using the results obtained in section 13.3.2, we have   mZ 2 mZ 2 e mZ 2 cW 2 gWWZ + 4aU = gWWZ − 4aD = 1− , mW 2 mW 2 cW s W mW 2

(13.153)

(13.154)

so that unitarity is respected provided that mW = cW mZ . What if the observed W and Z masses have a different ratio? In the first place, we must then 27

In the computation of this diagram, it is important to notice that the unsafe terms going as E 1 come from terms in the brackets, and not just from the second term.

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415

conclude that there moust be more than just one single type of neutral Higgs boson, for it was this assumption that gave us the relation 13.109, and it would now be seen to be incorrect. In the second place, there must be a Higgs-like object that compensates the unsafe behaviour in U D → W + Z, and it must be charged: a charged Higgs particle! We would need to introduce couplings of this object to fermions and vector bosons as follows: D

H+

↔

U

i gUDH (mU ω+ − mD ω− ) ~

+ ν W _ H µ Z

↔

i gWZH g µν ~

so that a fourth diagram becomes available:

with the contribution M4 = −i~ gUDH gWZH

(+ · ) v(p2 ) (mU ω+ − mD ω− ) u(p1 ) . s − M2

(13.155)

The fully longitudinal form up to safe terms is given by M4 cL = −i


~gUDH gWZH v(p2 ) mU (1 + γ 5 ) − mD (1 − γ 5 ) u(p1 ) , 4mZ mW

and we see that unitarity restoration is possible provided that   mZ 2 gUDH gWZH + 4aU + =0 . gW gWWZ 2 mW 4

(13.156)

(13.157)

The algebraic form of this coupling is, in fact, the same as that found in supersymmetric versions of the Standard Model, that contain charged Higgses in addition to at least 3 neutral ones. We see that the form of the coupling is not dependent on which supersymmetric extension of the Standard Model is preferred, but is dictated by the simple existence of a charged Higgs particle.

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At this point we shall assume that there is just a single type of charged Higgs H ± in the model, but we shall allow for more than one type of neutral Higgs, denoted therefore by Hj0 , where j can run over several values. Let us now find some more couplings. In the first place, since the H ± is a charged scalar, we know immediately what the structure of its couplings to photons must be:

q

µ ↔ p

i Qc (p − q)µ , ~

where the momenta are counted along the arrows, and of course

µ ↔

ν

i (2Qc 2 ) g µν ~

As discussed in section 10.4. However, we must also allow for the possibility of charged bosons coupling to H 0 H ± pairs. Simple analogy then motivates the choices of two additional vertices:

q

µ

↔

p µ

↔ ν

i gWcj (p − q)µ , ~ i gWcjγ g µν . ~

Here the momenta are counted along the arrow as before. The coupling constants can be restricted by considering the process H + (p1 )Hj0 (p2 ) → W + (p3 , 3 ) γ(p4 , 4 ) , for which we now have three Feynman diagrams at the tree level:

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The three contributions are given, with P = p1 + p2 = p3 + p4 , by   ~gWcj QW 1 αβ α β M1 = −i Y (p3 , 3 ; −P, α; p4 , 4 ) −g + (p1 − p2 )β , P P 2(p3 · p4 ) mW 2 (p1 · 4 )(p2 · 3 ) , M2 = −2i~gWcj Qc (p1 · p4 ) M3 = i~gWcjγ (3 · 4 ) . (13.158) The handlebar operation on the photon, which tests for current consaervation, gives    M123 c4 →p4 = i~ (p1 · 3 ) QW gWcj + gWcjγ   + (p2 · ) − QW gWcj + gWcjγ − 2gWcj Qc ,(13.159) from which we derive the relations, valid for any j: gWcjγ = −gWcj QW ,

Qc = −QW .

(13.160)

Note that we needed a process involving W in order to establish a relation between Qc and QW ; and, that the four-point vertex does not contain a factor 2 as it does in sQED. Another process of interest is H + (p1 )H − (p2 ) → W + (p3 , 3 )W − (p4 , 4 ) in which we encounter two new vertices:

W µ ν W

↔

Z

q

µ p

↔

i gWWcc g µν ~ i gZcc (p − q)µ ~

so that again three diagrams are available:

γ,Ζ

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The contributions are X (p1 · 3 )(p2 · 4 ) M1 = 4i~gWcj 2 , 2−m 2 (p − p ) 1 3 j j M2 = i~gWWcc (3 · 4 ) ,   QW Qc gWWZ gZcc M3 = i~ + Y (p3 , 3 ; p4 , 4 ; −P, p1 − p2 ) (13.161) . s s − mZ 2 Putting a handlebar on 3 results in    M123 c3 →p3 = i~ (p1 · 4 ) gWWcc + QW Qc + gWWZ gZcc " #) X +(p2 · 4 ) − 2 gWcj 2 + gWWcc − QW Qc − gWWZ gZcc (.13.162) j

13.8

Exercises

Excercise 103 The width of the W Assume that the weak coupling constant gW is universal, i.e. independent of the fermion’s flavour. 1. Compute (at the tree level) the decay width for the decay W − → e− ν¯e 2. Assuming that all quarks and leptons are essentially massless compared to the W mass mW , with of course the exception of the top quark. Determine the total W decay with ΓW . Hint : quarks have colour ! 3. Insert your result into Eq.(13.25), and verify that unitarity is not violated in this process. For the unitarity limit, see Appendix 15.10. Excercise 104 The W width with Cabibbo mixing In the simplest form of the standard model the W couples with universal coupling to ud and cs quark-antiquark pairs (and, of course to tb but we disregard that here). In fact, its couplings are more complicated, and we have the following pattern of couplings : W ud : gW cos(θc ) , W us : gW sin(θc ) , W cd : −gW sin(θc ) , W cs : gW cos(θc ) .

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Here, θc ≈ 13o is the so-called Cabibbo angle. Show that this refinement does not change the total W decay width. Excercise 105 The width of the Z Compute the total Z decay width at the tree level, with the same approximation for the masses as in exercise 103. Excercise 106 Landau-Yang strikes again, sneakily! From the fact that there is a ZW W vertex in the electroweak model, and Eq.(13.109), you might hope to circumvent a ‘weak’ variation of the LandauYang theorem by letting cW go to zero, so that the W mass vanishes and the Z could actually decay into two massless spin-1 particles. Show that this is impossible. Hint : this is simple. Excercise 107 The width of the H Before the Higgs was found in 2012, its mass was unknown. Let us assume that mH is sufficiently large for decays into heavy bosons to be possible. 1. Compute the decay width of the Higgs into a fermion-antifermion pair. Show that these widths are proportional to mH . 2. Compute the widths Γ(H → W + W − ) and Γ(H → ZZ). Show that these widths are proportional to mH 3 . 3. Esimate the value of mH for which ΓH ≈ mH . This was considered to be the largest realistic value of mH . Excercise 108 Unitarity-violating fermions Assume that there are some extremely heavy fermions around, called F1 and F2 , both with mass M , much larger than mH . 1. Compute the total cross section for F1 F¯1 → F2 F¯2 at energies large compared to M . 2. Compare your result with the unitarity limit and find a restriction on M. Excercise 109 Linear or logarithmic ? Write out a Feynman diagram where three spin-1 particles V1,2,3 are attached to a fermion loop (possibly with three different fermion types in the three

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propagators). Perform the Feynman trick to combine the three propagator denominators into one, and shift the integration momentum k µ appropriately so that this denominator contains no term linear in k. Show that the leading divergence, with three powers of k in the numerator, always vanishes upon integration, and that therefore the actual divergence is logarithmic.

Chapter 14 Example computations Relevant appedix : 15.15 In this chapter we shall go through several actual computations of a number of tree-level processes and some loop diagrams. The theory lives in four Minkowski dimensions ; but in the loop calculations we shall use dimensional regularization throughout.

14.1

Neutrino production in e+e− scattering

14.1.1

The cross section

In this section we consider the process e− (p1 , λ) e+ (p2 , λ)

→

νj (q1 ) ν¯j (q2 )

(j = e, µ) ,

where both the e± and the neutrinos are taken to be massless, and we have indicated the momenta and the electronic helicities. At the tree level there are either one or two diagrams : M =

q1

p1 p2

Z

q

2

+

p1 p2

W

q1 q2

(14.1)

For j = e (electron neutrinos) the process is described by the two diagrams. For j = µ (muon or tau neutrinos) only the first diagram contributes. We shall neglect the width of the Z boson1 . Using the looser notation for spinors 1

Since the W is exchanged with negative invariant mass and hence cannot decay at all, its width is naturally zero in this case anyway.

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allowed in the massless case we can write the two diagrams as follows, where we indicate their dependence on the helicity: i~ aν uλ (p2 )(ve + ae γ 5 )γ α uλ (p1 ) u− (q1 )(1 + γ 5 )γα u− (q2 ) , s − mZ 2 i~ gW 2 M2 (λ) = u− (q1 )(1 + γ 5 )γ α uλ (p1 ) uλ (p2 )(1 + γ 5 )γα u− (q2 ) . t − mW 2 (14.2)

M1 (λ) =

We have introduced the Mandelstam variables s = (p1 + p2 )2 = (q1 + q2 )2 , t = (p1 − q1 )2 = (p2 − q2 )2 , u = (p1 − q2 )2 = (p2 − q1 )2 = − s − t . (14.3) The first step is to get rid of the explicit γ 5 ’s : 2i~ aν (ve − λae ) uλ (p2 )γ α uλ (p1 ) u− (q1 )γα u− (q2 ) , s − mZ 2 4i~ gW 2 M2 (λ) = u− (q1 )γ α uλ (p1 ) uλ (p2 )γα u− (q2 ) . (14.4) t − mW 2 M1 (λ) =

Using the Chisholm identity we can compute the explicit helicity forms : 4i~ aν (ve − ae ) s+ (p2 , q2 ) s− (p1 , q1 ) , s − mZ 2 4i~ aν (ve + ae ) M1 (−) = s− (p2 , q1 ) s+ (q2 , p1 ) , s − mZ 2 M2 (+) = 0 , 8i~ gW 2 M2 (−) = s− (q1 , p2 ) s+ (q2 , p1 ) . t − mW 2 M1 (+) =

(14.5)

Keeping in mind the antisymmetry of the spinor products, and the Fermi minus sign, we find that up to an irrelevant overall complex phase the amplitudes are given by aν (ve − ae ) t , s − mZ 2   aν (ve + ae ) 2gW 2 M(−) = 4~ + u . s − mZ 2 t − mW 2 M(+) = 4~

(14.6)

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We arrive at  2 

aν (ve − ae )2 2 aν 2 (ve + ae )2 2 2 2 = 4~ |M| t + u (s − mZ 2 )2 (s − mZ 2 )2  4gW 4 4gW 2 aν (ve + ae ) 2 2 u + . (14.7) + u (s − mZ 2 )(t − mW 2 ) (t − mW 2 )2 The centre-of-mass frame is the obvious choice to work in ; in this frame s t = − (1 − cos θ) , (14.8) 2 where θ is the angle between p~1 and ~q1 , and the cross section has no azimuthalangle dependence. We may therefore write the phase space as follows : d cos θ 1 d cos θ dφ → = dt , (14.9) dV (p1 + p2 ; q1 , q2 ) = 2 32π 16π 8πs with the integration interval being t ∈ [−s, 0]. The various integrals are easily worked out ; we have      Z0 s mW 2 s u2 =s +2−2 1+ log 1 + , (t − mW 2 )2 mW 2 s mW 2 −s

Z0

u2 = s2 2 t − mW

 2  ! 3 mW 2 mW 2 s + − 1+ log 1 + , 2 s s mW 2

−s

Z0

2

Z0

t dt = −s

1 u2 dt = s3 . 3

(14.10)

−s

Putting everything together2 we obtain for the total cross section the expression σ(e+ e− → νe ν¯e ) = ( 2
2 2 s ~2 2 2 aν v e + ae 4πs 3 s − mZ 2    ! 2 2 2 s 3 m m s W W + 4gW 2 aν (ve + ae ) + − 1+ log 1 + s − mZ 2 2 s s mW 2      s mW 2 s 4 + 4gW +2−2 1+ log 1 + . (14.11) mW 2 s mW 2 2

And not forgetting the flux factor !

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For muon (or tau) neutrinos only the first line remains : 2 s ~2 2 2 + − 2 . σ(e e → νµ ν¯µ ) = aν (ve + ae ) 6πs s − mZ 2

14.1.2

(14.12)

Unitarity considerations

The above neutrino cross section lends itself to a few interesting observations. At the Z peak As it stands, the cross section diverges for s = mZ 2 . This is of course due to our neglecting ΓZ . To remedy this, we may replace mZ 2 by mZ 2 − imZ ΓZ , and neglect the second (W -exchange) diagram3 . Close to the Z pole, the cross section for each neutrino type is then given by σ(e+ e− → ν ν¯) =

mZ 2 ~2 aν 2 (ve 2 + ae 2 ) 6π (s − mZ 2 )2 + mZ 2 ΓZ 2

(s ≈ mZ 2 ) , (14.13)

while at the very peak we have4 σ(e+ e− → ν ν¯) =

~2 aν 2 (ve 2 + ae 2 ) 6πΓZ 2

(s = mZ 2 ) ,

(14.14)

This can be cast in an instructing form, using the fact that Γ(Z → e+ e− ) =

~ (ve 2 + ae 2 ) mZ ~ aν 2 mZ , Γ(Z → ν ν¯) = . 12π 6π

The cross section at the peak can therefore be written as    Γ(Z → ν ν¯) 12π Γ(Z → e+ e− ) + − σ(e e → ν ν¯) = s ΓZ ΓZ

(14.15)

(s = mZ 2 ) ,

(14.16) which is exactly the form demanded by unitarity for an intermediate state (the Z) with unit spin (see section 15.10). 3

This is really justified ! In the sense in which ΓZ 6= 0 comes about by interactions, ΓZ is formally of higher order in perturbation theory, and then a factor ΓZ −1 actually lowers the order of such diagrams. Thus, around the Z pole, the W -exchange diagram is formally of higher order in perturbation theory. 4 Remember, this is all strictly tree level...

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Very high and very, very high energy We may also consider what happens at very high energies, s  mZ , mW . In that case we may approximate the cross section by σ(e+ e− → νe ν¯e ) =    ~2 gW 4 s 2gW 4 + gW 2 aν (ve + ae ) log . − π mW 2 s mW 2

(14.17)

At extremely high energies the cross section becomes very simple : ~2 gW 4 (s → ∞) . (14.18) π mW 2 This is indeed a nice, simple expression — but it is a constant. Does this not conflict with unitarity, that demands a 1/s behaviour ? The solution appears if we realize that the unitarity behaviour is required in each individual angular momentum channel, or alternatively in a situation where s becomes very large together with all other momentum transfers : in this case, a fixed t/s ratio. As we can see, at high energies the process is completely dominated by t values close to zero, so the ratio t/s approaches zero5 . Another way of putting this is to say that in the infinite-energy limit, all intermediate angular momentum channels contribute. A quick inspection of Eq.(14.6) shows that for s → ∞ at fixed t/s the amplitudes are indeed just simple, energy-independent (but angle-dependent) quantities6 . σ(e+ e− → νe ν¯e ) =

14.2

W pair production in e+e− scattering

14.2.1

Setting up the amplitude

In sec. 13.3.1 we have saved unitarity in the process e+ e− → W + W − by introducing the Z particle, upon which the dangerous terms that lead to faulty high-energy behaviour are cancelled. But of course, the well-behaved remainder is also of interest, and its computation is a nice example of getyour-hands-dirty theoretical work. In detail, the process is described as e+ (p1 , λ) e− (p2 , λ) → W + (q1 , ρ11 ) W − (q2 , ρ12 ) Since the (t−mW 2 )−1 propagator peaks for vanishing t. The distribution proportional to (t − mW 2 )−2 , with −s < t < 0, gives expectation value hti ≈ mW 2 − mW 2 log(s/mW 2 ) so that for very large s the ratio is typically t/s ∼ − log(s)/s. 6 Since u = −s − t, when t/s is fixed then so are u/s and u/t. 5

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where we explicitly indicate the momenta p1,2 (taken to be massless) and q1,2 , and the polarisation vectors 1,2 . The electron helicity is λ = ± and the W polarisations are denoted by ρ1,2 that can take the values ± or 0, the latter standing for the longitudinal polarisation. There are thus 18 different amplitudes to be considered. We write the amplitude as M(λ, ρ1 , ρ2 ) = i~

2 X

Aj (λ) Bj (λ; ρ1 , ρ2 ) ,

(14.19)

j=1

where the index 1 refers to the νe -exchange diagram, 2 to the combined γ/Z diagrams. The A’s collect the coupling constants and propagators : 1 δλ,− , ∆ e2 mZ 2 4gW 2 e2 gWWZ (ve − λae ) = − + δλ,− (14.20) . A2 (λ) = + s s − mZ 2 s(s − mZ 2 ) s − mZ 2 A1 (λ) = −4gW 2

where P = p1 + p2 = q1 + q2 and s = P 2 , ∆ = (p1 − q1 )2 . We see that A2 (+) goes asymptotically as 1/s2 at high energies, whereas A1,2 (−) decreases as 1/s. The real work of calculation is in the ‘spacetime’ objects B1 (−; ρ1 , ρ2 ) = u− (p1 ) /ρ11 (/q1 − p/1 ) /ρ22 u− (p2 ) , B2 (λ; ρ1 , ρ2 ) = uλ (p1 ) γα uλ (p2 ) Y (q1 , ρ11 ; q2 , ρ22 ; −q1 − q2 , α) = (ρ11 · ρ22 ) uλ (p1 ) (/q1 − q/2 ) uλ (p2 ) + 2 (q2 · ρ11 ) uλ (p1 ) /ρ22 uλ (p2 ) − 2 (q1 · ρ22 ) uλ (p1 ) /ρ11 uλ (p2 ) , (14.21) where in the last expression we have used (qj · j ) = 0.

14.2.2

Momenta and polarisations

We shall, of course, work in the centre-of-mass frame, where the momenta are chosen as follows : pµ1 = (E, E~ep ) , pµ2 = (E, −E~ep ) , q1µ = (E, q~e) , q2µ = (E, −q~e) . (14.22) The beam energy is E so that s = 4E 2 , and the W velocities are β = q/E, where q 2 = E 2 − mW 2 . The ~ep and ~e are unit vectors, with ~ep · ~e = c, the

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cosine of the scattering angle between e+ and W + . In addition we define two massless vectors7 aµ = (1, ~e ) , bµ = (1, −~e ) (14.23) For the transverse W polarisations we shall use ± 1,2 = ± and 1 (± )µ = √ u± (a) γ µ u± (b) . 8 The two longitudinal polarisations are of different : we write     1 2mW 2 2mW 2 1 0 0 1 = P P q1 − , 2 = q2 − . mW β s mW β s

14.2.3

(14.24)

(14.25)

Working out the amplitudes

It is enlightening to see explicitly how the various elements of M are computed. First, we can use /ρ /ρ = 0 to write B1 (−; +, +) = − u− (p1 ) /+ p/1 /+ u− (p2 ) = −2(p1 · + ) u− (p1 ) /+ u− (p2 ) 1 = − u+ (a) p/1 u+ (b) u− (p1 ) u+ (b)u+ (a) u− (p2 ) 2 1 = − s− (p1 , b)2 s+ (a, p1 ) s+ (a, p2 ) , 2 1 B1 (−; −, −) = − s− (p1 , a)2 s+ (b, p1 ) s+ (b, p2 ) . (14.26) 2 Similarly, 1 u− (p1 ) u+ (b)u+ (a) (/q1 − p/1 ) u+ (a)u+ (b) u− (p2 ) 2 = (a · q1 − p1 ) u− (p1 ) b/ u− (p2 ) ∆ + m2 u− (p1 ) q/2 u− (p2 ) = 2q q 2 2(∆ + m ) = u− (p1 )/q2 u− (p2 ) = B1 (−; −, +) .(14.27) β 2s

B1 (−; +, −) =

7

Note that the definition of a and b can be made Lorentz invariant by requiring them to be massless linear combinations of q1 and q2 .

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The next one is a bit more involved : B1 (−; +, 0) = =

=

= =

  1 2mW 2 u− (p1 ) /+ (/q1 − p/1 ) q/2 − p/1 u− (p2 ) mW β s  1 2mW 2 u− (p2 ) −∆/+ − (2p1 · q1 )/+ mW β s  2mW 2 + (2p1 · + )/q1 u− (p2 ) s   2mW 2 s− (p1 , b)s+ (a, p2 ) −∆ − (2p1 · q1 ) s  2mW 2 + s+ (a, p1 )s− (p1 , b) q s− (p1 , a)s+ (a, p2 ) 2  
s− (p1 , b)s+ (a, p2 ) 2mW 2 √ (2p1 .q1 ) − 2q(a · p1 ) −∆ − s mW β 2
1 − u− (p1 ) /+ u− (p2 ) ∆ + mW 2 (1 − β) . (14.28) mW β

This is the most non-trivial of them all : notice the juggling with the spinor products. Nevertheless the result factorizes nicely, and that does not depend on the use of the standard form. The other such cases are
1 u− (p1 ) /+ u− (p2 ) ∆ + mW 2 (1 + β) , B1 (−; −, 0) = − mW β B1 (−; 0, ±) = −B1 (−; ±, 0) . (14.29) The last case for B1 is B1 (−; 0, 0) =     1 2mW 2 2mW 2 u− (p1 ) q/1 − p/2 (/q1 − p/1 ) q/2 − p/1 u− (p2 ) mW 2 β 2 s s   1 2mW 2 2mW 2 4mW 4 u− (p1 ) ∆/q2 − ∆ p/1 + ∆ p/2 + p/2 q/1 p/1 u− (p2 ) = mW 2 β 2 s s s2   1 4mW 4 = . (14.30) u− (p1 ) q/2 u− (p2 ) ∆ + mW 2 β 2 s Note that we have used momentum conservation, u(p1 )(/q1 + q/2 )u(p2 ) = 0. Computationally, for B2 things are simpler. First of all, we have trivially B2 (λ; +, +) = B2 (λ, −, −) = 0 ,

(14.31)

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and, almost equally trivially, B2 (λ; +, −) = B2 (λ; −, +) = 2 uλ (p1 ) q/2 uλ (p2 ) ,

(14.32)

owing to our choice of the polarisations. Next, by
q1,2 · 02,1 =


1 sβ (q1 q2 ) − mW 2 = mW β 2mW

(14.33)

we have B2 (λ, ±, 0) = −

sβ uλ (p1 ) /± uλ (p2 ) , = −B2 (λ, 0, ±) . mW

(14.34)

And, finally, s + 2mW 2 uλ (p1 ) q/2 uλ (p2 ) . (14.35) mW 2 We see that the nonzero B2 always carry the same spinor sandwich as do the B1 . Indeed they had better if they want to show (partial) cancellation between them. As expected, the B’s behave, for large energy, as E 2 if both W ’s are transverse, while they go as E 3 and E 4 if one or both become longitudinally polarised, respectively. Of the various spinorial objects we can compute the absolute values : B2 (λ; 0, 0) =

|s− (p1 , b)2 s+ (a, p1 )s+ (a, p2 )|2 = s2 (1 + c)3 (1 − c) , |s− (p1 , a)2 s+ (b, p1 )s+ (b, p2 )|2 = s2 (1 − c)3 (1 + c) , 1 |uλ (p1 )/q2 uλ (p2 )|2 = β 2 s2 (1 − c2 ) , 4 1 2 |uλ (p1 )/−λ uλ (p1 )| = s(1 + c)2 , 2 1 2 |uλ (p1 )/λ uλ (p1 )| = s(1 − c)2 . (14.36) 2 Using these results, we can easily evaluate the absolute values of the amplitudes. Some observations are in order here. First, there are two vanishing amplitudes : M(+; +, +) = M(+; −, −) = 0. This is, of course, due to our choice of ± for both transverse polarisations. Secondly, the nonzero amplitudes do not look nice, what with the occurence of the two couplings gW and e, two masses mZ and mW , the denominator ∆ here and there, and β cropping up in various places. There is a message here : real phenomenology is messy. It is only in the more Platonic realm of very high energies that we may hope to see our results simplify, and this we shall now investigate.

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14.2.4

W pair production at very high energy

As has been amply discussed in these notes, the high-energy behaviour of amplitudes often requires diagrams to partially cancel against one another. Indeed, that is what led us to introduce the Z boson in W W production in the first place. However, here we are interested in what is actually left after the cancellations have taken place. We shall consider the limit where E is very much larger than mZ , mW , at fixed c. In that limit, we may approximate ∆ ≈ −2E 2 (1 − c). We shall neglect overall complex phases in what follows. The amplitude for this 2 → 2 process should have no energy dimension (cf. sec. 6.3.3), so any contribution going as E −1 or lower can be neglected. Of course all contributions going as E +1 or higher had better cancel8 ! This actually leaves only a few surviving amplitudes. In the first place, ~e2 (1 + c) (1 − c2 )1/2 , 2sW 2 (1 − c) ~e2 (1 − c2 )1/2 . M(−; −, −) ≈ 2 2sW M(−; +, +) ≈

(14.37)

The apparent singularity at c = 1 is of course due to our approximating ∆. Note also that e2 ~ is actually the properly dimensionless quantity 4πα, see sec. 10.2.5. Next we have M(+; 0, 0) ≈

~e2 mZ 2 (1 − c2 )1/2 . 2 mW 2

(14.38)

The most difficult one is M(−; 0, 0) = ~ u− (p1 )/q2 u− (p2 ) K ,   4gW 2 4mW 2 K = − 2 ∆+ β mW 2 ∆ s   2 2 s + 2mW e mZ 2 4gW 2 + − + . (14.39) mW 2 s(s − mZ 2 ) s − mZ 2 Performing the cancellations inside the factor K with care we find    ~e2 1 mZ 2 1 2 1/2 M(−; 0, 0) ≈ (1 − c ) − −1 . 2 sW 2 mW 2 2sW 2 8

This is what we did all the hard work for, after all.

(14.40)

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We have now proven that the amplitudes surviving the high-energy limit are dimensionless and well behaved ; but they still depend on the ratio mZ /mW . Since we have taken the fermions to be massless, the Higgs boson is absent at the tree level here. Nevertheless (but this is, as we see, an additional input !) we may take mW /mZ = cW from our discussion of the minimal Higgs sector in sec. 13.4.2, and then we find the even more pleasing-looking results M(−; +, +) ≈ 4πα (1 − c2 )1/2 M(−; −, −) ≈ 4πα (1 − c2 )1/2 M(+; 0, 0) ≈ 4πα (1 − c2 )1/2 M(−; 0, 0) ≈ 4πα (1 − c2 )1/2

1 1+c , 2sW 2 1 − c 1 , 2sW 2 1 , 2cW 2 1 . 4sW 2 cW 2

14.3

Higgs coupling to massless vectors

14.3.1

The γγH vertex

(14.41)

Although the photon is massless and therefore has no direct coupling to the Higgs, such a coupling is effectively realised by fermion loops9 . At one-loop order, we then have two contributing diagrams :

MγγH =

+

,

(14.42)

which differ in the orientation of the fermion line. If our theory is to be consistent, this amplitude must be ultraviolet-finite (since otherwise we would have to introduce a counterterm which would mean a direct photon-photonHiggs coupling after all) and it must obey current conservation. We shall verify this point first, by putting a handlebar on one of the photons : + 9

And it provided the first mechanism for observing Higgs particles at the LHC by measurement of the invariant mass of photon pairs.

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−

=

−

+

(14.43) By using the identities =

,

=

(14.44)

we see that the third and fourth diagram are actually equal to the first and second one (flipped over), so that the sum vanishes and current conservation (gauge invariance) is assured. The process we investigate is, more explicitly, given as γ(q1 , 1 ) + γ(q2 , 2 )

→

H

where we have explicitly given the momenta and polarisations of the (onshell) photons. We denote the Higgs mass by mH and the fermion mass by M . The fermion charge is denoted by Q, and its coupling to the Higgs by gffH . Although we do not expect (or hope to see) divergences, we shall still work in a general dimension 4 − 2 for reasons that will be come clear later on : but we shall take  → 0 wherever we can. One of our diagrams10 is given by Z N (−1)i6 Q2 ~3/2 gffH d4 k , M1 = 4 (2π) D


D = k 2 − M 2 + iη (k − q2 )2 − M 2 + iη (k + q1 )2 − M 2 + iη ,   N = Tr (/k + M )/2 (/k − q/2 + M )(/k + q/1 + M )/1 . (14.45) The other diagram, M2 , is obtained by interchange of the two photons. Using the shorthand δxy as in section 11.5, we can write Z 1 2 (14.46) = δxy
3 ; D k 2 + 2x(k · q1 ) − 2y(k · q2 ) − M 2 + iη accordingly we shall shift k µ = kˆµ − xq1 µ + yq2 µ . 10

In fact, the second one.

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We now turn to N . Since the whole amplitude is current-conserving, we are allowed to replace the polarisations by the explicitly current-conserving combinations we already encountered in section 10.7.3 : (q1 · 2 ) µ q2 , (q1 · q2 ) (14.47) with the nice properties that ηj · qk = 0. After shifting the momenta and ˆ we have dropping terms linear in k, 1 µ → η1 µ = 1 µ −

(q2 · 1 ) µ q1 , (q1 · q2 )

2 µ → η2 µ = 2 µ −

N → 16M (kˆ · η1 )(kˆ · η2 )   + 4M (η1 · η2 ) − 4kˆ2 + 4M 2 + mH 2 (4xy − 2) .

(14.48)

Considering the first term, we notice that in our integral we can again, just like in section 11.5, replace   1  ˆ2 αβ 1 ˆ2 αβ α ˆβ ˆ k g ∼ + k g . (14.49) k k → 4 − 2 4 8 Thus effective form of N is   2 2 2 ˆ N → M (η1 · η2 ) 4k + 4M + mH (4xy − 2) ,

(14.50)

and the diagram is therefore M1 = Q2 gffH M (η1 · η2 ) A , Z Z 2 2 ˆ2 2µ2 4−2 ˆ 4k + 4M + mH (4xy − 2) A = δ d k 3 (14.51)  xy (2π)4−2 kˆ2 − M 2 + xy mH 2 + iη We see that in fact M2 = M1 , owing to our use of the η’s rather than the ’s11 . Proceeding with the one-loop cookbook of section 6.612 we can write i A= 2 8π

Z∞

Z δxy

0 11

dt t1−

4t − 4M 2 − mH 2 (4xy − 2)
3 t + M 2 − xy mH 2 − iη

(14.52)

You might start to worry at this point. Weren’t the two diagrams necessary for the cancellations proving current conservation ? How, then can they be equal ? The answer is that by going from  to η we have exchanged contributions between the diagrams, and indeed in the η formulation they are separately current-conserving. 12 And taking  → 0 wherever that is justified by the event.

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Since Z∞

t2− 1 dt ≈ + O (1) , 3 (t + a) 

0

Z∞ dt

t1− 1 + O () , ≈ 3 (t + a) 2a

(14.53)

0

we find

Z i 1 − 4xy (14.54) A = 2 δxy 2 8π (M /mH 2 ) − xy − iη We can also prettify the product η1 · η2 by introducing the field strength tensor 13 Fiµν ≡ µi qiν − qiµ νi (i = 1, 2) (14.55) so that

1 F1 µν F2µν . mH 2 The whole amplitude then takes the form η1 · η2 =

√ 1/2 µν iQ2 ~3/2 MγγH = 2) F1 F2µν F (G F 12π 2 Z 3τ (1 − 4xy) F(τ ) = δxy . τ − xy − iη

(14.56)



M2 mH 2

 , (14.57)

where we have used Eq.(13.133)) to rewrite gffH /M . The limit of large fermion mass is interesting. For very large τ , F(τ ) approaches unity. The amplitude thus becomes independent of M for very large fermion mass ! That this happens is due to the fact that in Eq.(14.54) the M 2 terms in the numerator cancel. But for that to happen, the first term, proportional to , has to be there : were we to set d = 4 from the outset, it would be absent. Of course, the fact that the numerator would contain a kˆ2 term and the loop integral would be divergent if d would differ ever so slightly from 4 should give us pause. This is the reason why we have to stick to variable dimension in this calculation : if we don’t, then the amplitude will be proportional to M 2 ! For finite values of τ we can straightforwardly find

F(τ ) = 3τ 2 + (1 − 4τ ) Li2 (1/x+ ) + Li2 (1/x− ) ,   p 1 1 ± 1 − 4τ + iη , (14.58) x± = 2 and we plot it in figure 14.1. 13

Perhaps more familiar in its Fourier-transformed form ∼ ∂ µ Aν − ∂ ν Aµ .

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Figure 14.1: The absolute value of F(M 2 /mH 2 ) defined in Eq.(14.58)

14.3.2

The ggH amplitude and the Next Generation

We can easily extend the computation of the previous section to the case of gluon fusion, that is the process g(q1 , 1 , j) g(q2 , 2 , `) → H where we have indicated the momenta, polarizations, and colours of the gluons. It is in fact this process that enabled the famous 2012 Higgs discovery at the LHC14 . It suffices to replace Q by the strong coupling constant g, and to include the colour factor
1 (14.59) Tr T j T ` = δj` . 2 The gluon-gluon-Higgs amplitude is  2 √ 1/2 µν iαs ~1/2 M MggH = (GF 2) F1 F2µν δj` F , (14.60) 6π mH 2 where we used g 2 = 4παs /~. 14

Since it involves both a loop-induced production of the Higgs by gluon fusion and a loop-induced decay of the Higgs into photon pairs, this process may also be considered as a ‘proof’ of the actual reality of loop effects ; and as a true ‘quantum’ process since at the tree level it is impossible.

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An interesting observation can be made here. The Standard Model15 contains three families of quarks and leptons. Since these are so similar (except for their masses), it is natural to ask whether a fourth, fifth,. . . generation may exist, with particles that are simply too heavy to have been produced so far16 . The Higgs is very helpful here, since we can see that each extra doublet of super-heavy U, D quarks would enhance Higgs production by gluon fusion by an amount that is about twice that delivered by the top quark, independently of how heavy these newcomers are ! From the observed rate of Higgs production we can be quite sure that the three generations we know are all there are17 .

15

At least, its version of August 31, 2019. Indeed, if you present the Standard Model to a lay audience, you should always try to elicit this question. 17 You might construct theories in which the extra generations are somehow masked by some cancellation mechanism - but then they are no longer copies of the known ones. 16

Chapter 15 Appendices 15.1

Perturbative (non)convergence issues

15.1.1

Punishment at the singular point

Let us reinspect the Green’s functions G2n for zero-dimensional ϕ4 theory, in perturbation theory : G2n = H2n /H0 , with s  n X  λ4 ~ k (n) 2π~ ~ H2n = H2n , H2n = − ak , 2 µ µ 24µ k≥0 √ (4k + 2n)! (n) k n−1 2 2 ≈ k! 16 (4k) , (15.1) ak = 2k+n 2 (2k + n)! k! π where in the last line we have used the Stirling approximation (cf appendix 15.15.2). Although we have treated the expressions for the H’s as if they were well-defined objects, in fact these series do not converge ! For large k and (n) fixed n the coefficient ak increases superexponentially with k, so that the series has a radius of convergence equal to zero. This should not come as a surprise. For, in the discussion of the perturbation expansion we have assumed the coupling constant λ4 to be small, but positive. If, on the other hand, it was small but negative, perturbation theory would look very different : in fact it would look like nothing at all since for negative λ4 the path integral is completely undefined. Therefore, the perturbative expansion is not regular around λ4 = 0, and in the set of all ϕ4 theories the point λ4 = 0 constitutes an essential singularity. And this is what we are expanding around ! Small wonder that things go haywire. 437

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15.1.2

Borel summation

All may not be lost, however. The method of Borel summation sometimes1 enables us to assign a value to a sum with vanishing radius of convergence. Suppose that a function of a positive variable x is given by the sum X f (x) = ck x k , (15.2) k≥0

where the coefficients ck grow superexponentially. Clearly it is difficult to make sense of such a sum ; but it may be possible to make sense of a related sum : X ck xk , (15.3) g(x) = k! k≥0 simply because the coefficients do not grow as rapidly. Let us suppose that this is indeed the case. We then may employ the formula Z∞

dy exp(−y) (xy)n = n! xn ,

n = 0, 1, 2, . . .

(15.4)

0

to arrive at the rule

Z∞ f (x) =

dy e−y g(xy) .

(15.5)

0

Notice that here, we have again interchanged summation and integration, thus in a sense repairing the damage done when we arrived at the perturbation expansion in the first place. This approach is called Borel summation. We can illustrate this in a simple example, using ck = 1 : 1 f (x) = → g(x) = ex → 1−x

Z∞

dy e−y exy =

1 . 1−x

(15.6)

0

However, an important observation is to be made here. The sum for f (x) converges (conditionally) for the region |x| ≤ 1, whereas the sum for g(x) converges everywhere, and the Borel integral converges in this case as long 1

In zero dimensions this will work. In four-dimensional Minkowski space things are not nearly as simple. . .

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as 0 . (15.7) k≥0

for which Borel-summed form is therefore Z∞ f (x) =

−y

dy e 0

X k≥0

k

Z∞

(−xy) =

1 exp(−y) = e1/x E1 dy 1 + xy x

  1 x

(15.8)

0

where the function E1 , the exponential integral, is discussed in appendix 15.15.8. F (x) is a function that starts (obviously) at F (0) = 1 and then gently decreases. Borel summation works ! But how do we actually compute the series F (x) ? The theory of asymptotic functions provides an answer. Let us consider not the infinite sum F (x) as given in Eq.(15.7) but its truncated version K−1 X k! (−x)k (15.9) FK (x) = k=0

It can be shown that the difference between f (x) and fK (x) isK of
the or2 der of the first neglected term : F (x) − FK (x) = O K! (−x) . Taking ‘order’ to mean ‘roughly equal in magnitude, barring accidents’3 we might therefore conclude that the optimal value of K is that for which the error term is minimal, that is, we truncate around K ≈ 1/x. In that case k!xK behaves roughly as exp(−1/x), so that the numerical error can be very small indeed for small x. small indeed for small x. As an illustration, figure 15.1 shows the actual and asymptotically-inspired-truncated result for the function (15.7).The exact and truncated results for the function F (x) of (15.7). The smooth curve is the exact, the zigzagging one the truncated result. The approximate value oscillates around the true one, but for small x the difference is negligible. This shows that, even if a sum is divergent, it may still 2

Also this statement needs interpretation. In the theory of asymptotic series it means that the difference will go to zero at least as fast as the first neglected term goes to zero, not that these two numbers must be necessarily comparable in magnitude. As an example, the object 1012 /x2 is formally of the order of 1/x as x → ∞, but x has to be really large for them to be of equal size. Fortunately, it often happens that the difference and the neglected term are of similar magnitude. 3 Only to be justified by its success.

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Figure 15.1: Example of Borel summation

be possible to make sense out of it by Borel summation. Note that in our example we have required x to be positive, so that (−x)n oscillates in sign. That this is essential becomes clear when we try to Borel-sum X F (x) = n! (x)n , x > 0 : (15.10) n≥0

the Borel integral reads Z∞ F (x) =

dy e−y

1 , 1 − xy

(15.11)

0

and this integral runs into problems around y = 1/x. One may of course extend the integral to complex y values, and then skirt around the singularity. But it is not clear whether we should pass the point y = 1/x above, or below, the real axis. The ambiguity, that is, the difference between the results from the alternative contours, is of course given by the number I e−y e−1/x , dy = 2πi 1 − xy x y∼1/x

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and, since during the integration we might decide to circle around the singularity any number of times, arbitrary multiples of the ambiguity may be added. We see that the Borel integral becomes ambiguous : it may be some consolation that the ambiguity is nonperturbative in nature, i.e. it has no series expansion for infinitesimal but real and positive x. We conclude that the function F (x) is given by Z∞ F (x) = P

dy

e−1/x e−y − (2n + 1)iπ , 1 − xy x

0

(15.12) where n is an undetermined integer. The real part of F is a principal-value integral (cf appendix 15.15.6) that evaluates to the finite result Z∞ P

−y

dy

−1/x

e e = 1 − xy x

0

 Z1/x z −z −E1 (1/x) + dz e − e  . z 

(15.13)

0

15.2

More on symmetry factors

15.2.1

The origin of symmetry factors

In this section we shall return to the ‘simple’ world of zero dimensions, since symmetry factors do not depend on the dimensionality of the theory. Let us consider again the ϕ3/4 theory, with a path integrand  
µ 2 λ3 3 λ4 4 exp −S(ϕ) + Jϕ = exp − ϕ − ϕ − ϕ + Jϕ 2! 3! 4!    n 3 n3 X
n1 1 −λ3 ϕ 1 −λ4 ϕ4 4 1 −µϕ2 /2! = e Jϕ n3 ! 3! n4 ! 4! n1 ! n 1,3,4≥0

(15.14) We see that a diagram with n3 three-vertices, n4 four-vertices and n1 source vertices carries an a priori factor of 1 . n1 ! n3 ! n4 ! (3!)n3 (4!)n4

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We then have to consider the number of ways in which a particular diagram can be formed by connecting external lines and the vertices in appropriate ways. The best way to learn this is, of course, to see how it is done.

15.2.2

Explicit computation of symmetry factors

Let us start with the very simple diagram

It is built up from 2 external lines (distinguishable !) and one four-vertex : the a-priori factor is therefor 1/24. We can lay out the ingredients as a tool kit4 : We can denote this as follows : 1 4! To build the diagram, we first connect one of the external lines to the vertex : there are 4 possible ways to do so since all legs of the vertex are indistinguishable. We then have 4 4! To attach the other external leg, there are now 3 choices : 4×3 4! For the remaining operation of linking the two other legs of the vertex there is of course only one possibility. The resulting symmetry factor is 4×3/24 = 1/2 as advertised. We next turn to a slightly more complicated diagram :

4

If you have ever played with K’NEX this may look familiar

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The tool kit is now 1 2! × 4! × 4! The first external leg can now be attached in 8 ways, since the vertices and all their legs are indistinguishable : 8 2! × 4! × 4! For the other external line there are now 4 possibilities : 8×4 2! × 4! × 4! Now we arbitrarily consider the upper vertex leg on the left. It has to be connected to one on the right, which can be done in 3 ways : 8×4×3 2! × 4! × 4! The next leg on the left now has 2 ways to go : 8×4×3×2 2! × 4! × 4! and then the diagram is closed uniquely, leading to a symmetry factor of 8.4.3.2/2!4!4! = 1/6. The symmetry factors of vacuum diagrams are no exception, as we can see from the construction of

The tool kit is now 1 4! × (3!)4 We arbitrarily pick one of the vertices and attach to an arbitrarily chosen leg one of the 9 remaining vertex legs : 9 4! × (3!)4

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The other two vertices are attached in 6 and 3 ways, respectively : 9×6×3 4! × (3!)4 Now take one of the outside legs. It must be attached to another outside leg that comes from another vertex, hence 4 possibilities : 9×6×3×4 4! × (3!)4 In the same way, for the next outside leg there are possibilities : 9×6×3×4×2 4! × (3!)4 and then the diagrams is closed uniquely. The symmetry factor is therefore 1/24, corresponding to the symmetry of permuting the vertices. We finish with a truly nontrivial one :

Note that the two crossing lines do not touch but form an ‘overpass’ : this diagram is nonplanar in the sense that an overpass remains no matter how you try to draw it. The toolkit is now even more impressive : 1 5! × (3!)5 We can attach the external lines to different 3-vertices in 15, 12, and 9 ways, respectively : 15 × 12 × 9 5! × (3!)5

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Now the leftmost object can be attached to the two remaining 3-vertices in 6 and then in 3 ways : 15 × 12 × 9 × 6 × 3 5! × (3!)5 The topmost leg of the object on the left has to be connected to both the upper and the lower objects on the right, for which there are thus 4 and 2 options, respectively : 15 × 12 × 9 × 6 × 3 × 4 × 2 5! × (3!)5 By now, it ought to be obvious that there are 2 ways to finish off the diagram, leading to a symmetry factor of (15.12.9.6.3.4.2.2)/(5!(3!)5 ), in simpler terms 1/2. The actual symmetry is, once again, the permutation of the two innermost vertices.

15.3

Derivation of the diagrammatic sum rules

In this section we shall derive the diagrammatic sum rules For an arbitrary given diagram let us define the characteristics E I Vq L P

= = = = =

number number number number number

of of of of of

external lines, internal lines, vertices of q-point type, closed loops, disjunct connected pieces.

An example is E = 2 , I = 6 , V1 = 1 , V3 = 3 , V4 = 1 , P = 1 , L = 2 . We now look for linear combinations T of these numbers that are the same for all diagrams. That is, whatever we do to a diagram, the value of T must remain unchanged. It is easy to see that any diagram can be transformed into

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any other diagram by application of the following four basic transformations, or their inverse : (i) coalescing a q-vertex and a 3-vertex : ↔ (ii)

adding an external line onto any other line : 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

(iii)

cutting through a line such that the graph falls apart : 0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

(iv)

↔

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

↔

0000000 1111111 1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

cutting through a line which is part of a loop : 0000000000000000 1111111111111111 1111111111111111 0000000000000000 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111

↔

0000000000000000 1111111111111111 1111111111111111 0000000000000000 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111

These four operations modify the characteristics as follows : (i) (ii) (iii) (iv)

: : : :

V3 → V3 − 1 , Vq → Vq − 1 E → E+1 , I → I +1 , I → I −1 , E → E+2 , I → I −1 , E → E+2 ,

, V3 P L

Vq+1 → Vq+1 + 1 , → V3 + 1 ; → P +1 ; → L−1 .

I → I −1 ;

If the combination T = αE E + αI I +

X

αq Vq + αL L + αP P

(15.15)

q

is to be invariant under the four basic transformations, then the coefficients α must obey (i) (ii) (iii) (iv)

: : : :

−αq + αq+1 − α3 − αI = 0 , αE + αI + α3 = 0 , αI − 2αE − αP = 0 , αI − 2αE + αL = 0 .

(15.16)

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Adding (i) and (ii) we find −αq + αq+1 + αE = 0 ,

(15.17)

αq = β − qαE .

(15.18)

αI = 2αE − β

(15.19)

with the general solution (ii) then gives us and (iii) and (iv) yield αP = −αL = −β. The invariant T can therefore be written as ! ! X X Vq + I + P − L , (15.20) qVq + E + 2I − β − T = αE − q

q

where αE and β are undetermined. We see that we have precisely two diagrammatic sum rules. By inspection of an arbitrary5 diagram we see that T = 0, so that the sum rules are X qVq = 2I + E , (15.21) q

X

Vq = I + P − L .

(15.22)

q

These two sum rules each have their own ‘physical’ interpretation. In the first place, the sum rule 15.21 simply tells us that external lines have one endpoint to be attached to a vertex, and internal lines have two endpoints that must be accomodated. The second rule, 15.22, comes into its own when we assign momenta to flow along the lines. The momenta in the external lines are assumed to be given In every vertex the total momentum is conserved, and each connected piece must contain an overall momentum conservation. We see that in that picture L is simply the number of internal lines whose momentum is not fixed by conservation. This derivation of the sum rules given above is to convince you that there is not a third type of sum rule, based on yet again different arguments, lurking about. 5

Arbitrary, except that it must contain at least one vertex. There are two connected diagrams without vertices: the first one, , conforms to the sum rules by choosing I = −1, and the second one, , fits in if we choose I = 0. But these choices are obviously somewhat forced.

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15.4

Alternative solutions to the SchwingerDyson equation

15.4.1

Alternative contours for general theories

In section 1.2.5, it was mentioned that ϕ3 theory is not well-defined for real fields since the action will go to infinity whenever ϕ → +∞ or ϕ → −∞. It is instructive to lift the requirement that ϕ be real. In that case, we see that different integration contours become available for which the path integral is well-defined (albeit not necessarily real). Let us consider a zero-diensional theory with general action m X λp p ϕ . S(ϕ) = p! p=1

(15.23)

The requirement for the path integral to be defined is that at both endpoints (still assumed to be at infinity in some complex direction) the real part of the action goes to positive infinity. That is, 0

=

X n≥0

1 n! (m!)n



∂ ∂J

n−1

J mn

(mn)! J mn−n+1 . n!(m!)n (mn − n + 1)!

(15.35)

The nonvanishing N ’s are therefore Nn(m−1)+1 =

(mn)! , n = 0, 1, 2 . . . n!(m!)n

(15.36)

As expected, for m > 2 some connected Green’s functions vanish identically at the tree level since no diagrams contribute. Asymptotic methods For asymptotically large n, we can estimate the form of Nn by realizing that these must be given by the behaviour of Φ(J) near that of its singularities that lies closest to the origin in the complex-J plane. Now, if Φ(J) is singular, then Φ0 (J) is divergent10 , so that dJ/dΦ must vanish. We therefore solve the equation ∂ J = 1 − F 00 (Φ) = 0 (15.37) ∂Φ for Φ. If the highest power of interaction in the theory is ϕm , this equation has m − 2 complex roots Φ1 , Φ2 , . . . , Φm−2 , and Jp = Φp − F 0 (Φp ) , 10

p = 1, 2, . . . , m − 2 .

(15.38)

The divergence might also show up in higher derivatives only, but in every actual case that I have studied the divergence shows up in Φ0 .

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Now, single out that Jp that has the smallest absolute value11 , which we shall call J0 , and its corresponding Φp will be writtten Φ0 . For J and Φ very close to the values J0 and Φ0 , respectively, we may use Taylor expansion to write 1 J ≈ J0 − F 000 (Φ0 )(Φ0 − Φ)2 , 2 since the linear term vanishes by definition. Hence  1/2 s J 2J0 Φ ≈ Φ0 − 1 − 000 J0 F (Φ0 )

(15.39)

(15.40)

close to the singularity. From the standard Taylor expansion12 1−

X √ 1−x= n≥0

(2n)! xn+1 (n + 1)!n!22n+1

we then recover the asymptotic form for Nn : s (2n − 2)! 1 8J0 Nn ≈ . n 000 (n − 1)! (4J0 ) F (Φ0 )

(15.41)

(15.42)

This estimate grows roughly as n!, as ought to have been immediately obvious from the fact that Φ(J) has a finite radius of convergence ; the above, more careful, treatment gives an estimate that is quite good even for non-huge n. As an application, we may consider purely gluonic QCD. In this theory, the only interactions are between 3 or 4 gluons, and the theory is equivalent, as far as counting is concerned, to the ϕ3/4 theory, with F (ϕ) =

1 3 1 4 ϕ + ϕ . 3! 4!

(15.43)

The solutions of Eq.(15.37) and the corresponding J values are Φ1 = −1 + 11

√ 4 √ 3 , J1 = − + 3 ; 3

Φ2 = −1 −

√

4 √ 3 , J2 = − − 3 , 3 (15.44)

The case that there are several such values is discussed in the next paragraph. This can be proven by applying the Lagrange expansion to the object u = y + u2 /2 = √ 1 − 1 − 2y, and putting y = x/2. 12

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Nn (exact) 1 1 4 25 220 2485 34300 559405 10525900

Nn (asymptotic) 0.85 1.07 4.01 25.17 220.94 2493.60 34397.35 560754.85 10547973.57

Table 15.1: Exact and asymptotic number of tree diagrams in ϕ3/4 theory √ √ √ so that J0 = 3 − 4/3, Φ0 = 3 − 1, and F 000 (Φ0 ) = 3. In table 15.5.1 we give the exact number Nn , and its asymptotic estimate. The approximation is better than one per cent for n ≥ 3. The non-polynomial (that is, n!) growth of the number of diagrams with n can be seen as an immediate indication of the failure of perturbation theory as a convergent series, as discussed in Appendix 1. Coarse-graining effects In the above we have assumed that there is only a single J0 . This is indeed usually the case ; for pure ϕp theories, however, Eq.(15.37) reads 1 q ϕ =1 , q!

q =p−2 ,

(15.45)

and this has solutions 1/q

φn = (q!)

  n exp 2iπ , q

n = 1, 2, . . . , q ;

(15.46)

the corresponding values for J are Jn = 1 −

1 q φn q+1 = φn , (q + 1)! q+1

n = 1, 2, . . . , q ,

(15.47)

and these have all the same absolute value. The thing to do is therefore to take the asymptotic contributions from all these q singular points into

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account, and sum them. We then obtain Nk ≈

q X (2k − 2)!

s (4Jn )−k

8(q − 1)!Jn φn q−1

(k − 1)!  k r X q 8 (2k − 2)! q + 1 φn −(k−1) . = (k − 1)! 4q q n=1 n=1

(15.48)

The sum over the n values of φ will vanish completely, except when k − 1 is a multiple of q, and then it evaluates to q/(q!)k−1 ; this is exactly the behaviour we found using Lagrange expansion. We might have proceeded otherwise, by simply taking the single real solution φq = (q!)1/q as the only singular point. The number of diagrams Nk will then be nonvanishing for every k value, while in the asymptotic expression (15.48) the sum over n φ’s is replaced by φq −(k−1) , that is precisely q times smaller than the nonvanishing sums of Eq.(15.48). We see that the taking into account of only the single, real solution causes the asymptotic values of Nk to be ‘smeared out’ ; Nk is then never zero anymore, but its average value13 is still correct.

15.5.2

Counting one-loop diagrams

The SDe approach to counting diagrams has a number of interesting or useful applications, one of which we discuss here. We can extend the treatment of the previous section as follows. For the case of purely gluonic QCD the number of one-loop diagrams including their symmetry factors can be counted by iterating the appropriate Schwinger-Dyson equation : 1 ~ 1 Φ(J) = J + Φ2 + Φ3 + (1 + Φ) Φ0 2 6 2

(15.49)

and taking care to discard terms of order ~2 or higher. As an example, the gluonic 20-point function is given by
N (19) = N0 (19) + ~ N1 (19) + O ~2 , N0 (19) = 11081983532721088487500 , N1 (19) = 2900013601350201168582750 . (15.50) 13

For the correct definition of ‘average’.

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The number N0 (19) is the actual number of diagrams since tree diagrams always have unit symmetry factor ; but the number N1 (19) underestimates the actual number of diagrams since the symmetry factors are not trivial. We can see, however, that the only possible nonntrivial symmetry factor at the one-loop level is 1/2, as evidenced by the factor ~/2 in Eq.(15.49). Inspection tells us that in this theory the only elementary Feynman diagrams that have symmetry factor 1/2 are E1 =

,

E4 =

E2 = ,

E5 =

,

E3 = .

, (15.51)

All diagrams that contain one of these elementaries as a subgraph will have a symmetry factor 1/2, and it will suffice to determine their number and multiply it by two14 . Alternatively, we may get rid of all such diagrams, and work with the difference. This is the more useful approach ; and it illustrates how we may go about using counterterms to impose constraints on the structure of Feynman diagrams. The procedure is best explained by going through it step by step. In the first place, it will become necessary to again distinguish betwee three- and four-point vertices. We therefore modify Eq.(15.49) be reinserting labels for these couplings: Φ(J) = J +

g3 2 g4 3 ~ Φ + Φ + (g3 + g4 Φ) Φ0 2 6 2

(15.52)

Iterating this gives for the first N : ~ g3 , 2   1 2 N (1) = 1 + ~ g4 + g3 ,   7 2 N (2) = g3 + ~ 4g3 + g4 g3 , 2   7 2 59 2 4 2 g4 + 24g3 + g4 g3 . N (3) = g4 + 3g3 + ~ 2 2 N (0) =

14

(15.53)

This relies, of course, on the fact that there can be no diagrams containing two (or more) of the elementaries, since that would be a two-loop diagram (or even higher).

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We can now start to remove graphs. We shall get rid of all diagrams with a tadpole by introducing a tadpole counterterm ~T in the SDe: Φ(J) = J +

g3 2 g4 3 ~ Φ + Φ + (g3 + g4 Φ) Φ0 − ~T 2 6 2

(15.54)

We see that this amounts to replacing J by J − ~T , and the N ’s become   1 g3 − T , N (0) = ~ 2   1 2 N (1) = 1 + ~ g4 + g3 − g3 T ,   7 2 2 , N (2) = g3 + ~ 4g3 + g4 g3 − g4 T − 3g3 T 2   59 7 2 4 3 2 2 . g4 + 24g3 + g4 g3 − 10g4 g3 T − 15g3 T N (3) = g4 + 3g3 + ~ 2 2 (15.55) The tadpole N (0) is removed by choosing T = g3 /2 ; and by the recursive structure of the SDe all diagrams containing the elementariy E1 are removed as well. The remaining low-order N s are now   1 1 2 N (1) = 1 + ~ , g4 + g3 2 2   5 3 N (2) = g3 + ~ 3g4 g3 + g3 , 2   33 4 7 2 49 2 2 N (3) = g4 + 3g3 + ~ g4 + g4 g3 + g3 . (15.56) 2 2 2 Next, we want to get rid of the two self-energy bubbles E2 and E3 . To this end, we again modify the SDe: Φ(J) = J +

g3 2 g4 3 ~ ~B Φ + Φ + (g3 + g4 Φ) Φ0 − ~T + Φ , 2 6 2 1 + ~B

(15.57)

where the strange-looking form of the counterterm is justified by the fact that we can rewrite Eq.(15.57) into    g3 2 g4 3 ~ 0 Φ(J) = J + Φ + Φ + (g3 + g4 Φ) Φ − ~T 1 + ~B , (15.58) 2 6 2

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which lends itself better to the purpose of iteration. We then obtain   1 1 2 N (1) = 1 + ~ g4 + g3 + B , 2 2   5 3 N (2) = g3 + ~ 3g4 g3 + g3 + 3Bg3 , 2   7 2 49 33 4 2 2 2 . N (3) = g4 + 3g3 + ~ g4 + g4 g3 + g3 + 4Bg4 + 15Bg3 2 2 2 (15.59) Requiring N (1) = 1 leads to B = −(g4 + g3 2 )/2, and we are left with   3 3 N (2) = g3 + ~ g4 g3 + g3 , 2   3 2 2 2 4 N (3) = g4 + 3g3 + ~ g + 15g4 g3 + 9g3 . (15.60) 2 4 Now, the one-loop contribution to the three-point function N (2) must not be completely cancelled, since it contains the diagram

which has symmetry factor 1 and must be retained. We therefore add a counterterm to the three-point coupling in the SDe:    (g3 − ~δ3 ) 2 g4 3 ~ 0 Φ(J) = J + Φ + Φ + (g3 + g4 Φ) Φ − ~T 1 + ~B , 2 6 2 (15.61) where the counterterm is needed only at one place since we are working to one-loop accuracy. The result of the iteration is   3 3 N (2) = g3 + ~ g4 g3 + g3 − δ3 , 2   3 2 2 2 4 g + 15g4 g3 + 9g3 − 6δ3 g3 . (15.62) N (3) = g4 + 3g3 + ~ 2 4 The condition now is that N (2) = g3 + ~g3 3 ,

(15.63)

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which requires δ3 = 3g4 g3 /2 to remove all elementaries E4 , and leads to   3 2 2 2 4 . (15.64) N (3) = g4 + 3g3 + ~ g4 + 6g4 g3 + 9g3 2 The same trick can be applied to the four-point coupling: the SDe is then    (g3 − ~δ3 ) 2 (g4 − ~δ4 ) 3 ~ 0 Φ(J) = J + Φ + Φ + (g3 + g4 Φ) Φ − ~T 1 + ~B , 2 6 2 (15.65) which gives   3 2 2 4 2 g4 + 6g4 g3 + 9g3 − δ4 . (15.66) N (3) = g4 + 3g3 + ~ 2 For the four point coupling, we only want to retain the diagrams ,

, and

,

which occur respectively 3,6, and 3 times. Therefore, δ4 = 3g4 2 /2 removes all occurrences of E5 . With these choices, the SDe Eq.(15.65) can be iterated (and truncated to one-loop order!) to give all diagrams that do not contain any of the elementaries E1,...,5 as subdiagrams15 . For the 20-point gluonic amplitude we find that the number of diagrams with symmetry factor unity is given by ˆ (19) = N0 (19) + ~M1 (19) , N

M1 (19) = 2013070318716871853439000 . (15.67) The total number of one-loop diagrams is therefore given by
ˆ1 (19) = M1 (19) + 2 N1 (19) − M1 (19) = 3786956883983530483726500 . N (15.68) In table 15.5.2 we give the results for the amplitudes from two to twenty external lines. It is seen that the ratio of one-loop to tree diagrams increases with the number of external legs ; while the average symmetry factor per one-loop diagram seems to slowly approaches unity. It can indeed be proven that asymptotically it does do so. 15

The actual implementation of the approach described here in computer algebra mayhave to be somewhat modified in the interest of speed : simply iterating Eq.(15.65) as it stands may lead to unwieldily large expressions.

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n+1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

N0 (n) 1. 1. 4. 25. 220. 2485. 34300. 5.594 105 1.053 107 2.244 108 5.349 109 1.409 1010 4.064 1012 1.274 1014 4.315 1015 1.569 1017 6.101 1018 2.525 1020 1.108 1022

ˆ1 (n)/N0 (n) N 3. 14. 24.75 37.88 52.09 67.47 83.86 101.2 119.4 138.5 158.3 178.9 200.2 222.2 244.9 268.2 292.1 316.6 341.7

avg.symm. 0.5000 0.5357 0.5758 0.6066 0.6309 0.6506 0.6672 0.6813 0.6936 0.7044 0.7140 0.7226 0.7305 0.7376 0.7441 0.7502 0.7558 0.7609 0.7658

Table 15.2: Computing the average symmetry factors

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The above strategy can of course be applied to other problems as well. For instance, we may remove all one-loop three- and four point elementaries instead of just those with symmetry factor one-half : in that case we are essentially renormalising the theory. It should also be clear that in that case, in which we just want to remove subdiagrams rather than count them, it is easy to go to more loops in an order-by-order approach.

15.6

Concavity of the effective action

In the zero-dimensional case of a single field variable, the effective action is concave. Let us now investigate whether this persists in case of more fields. Let the collection of all fields be denoted by {ϕ} as before, and the collection of all sources, one for each field, by {J}. We shall denote the combined probability density of all fields, including the effects of the sources, by P ({ϕ}, {J}). The effective action is now that function of the collection of all field functions {φ} that has the correct classical equation : ∂ Γ({φ}) = Jn . ∂φn

(15.69)

Concavity of the effective action in the many-field case means that the matrix Γnm ≡

∂ ∂ ∂ Γ({φ}) = Jn ∂φn ∂φm ∂φm

(15.70)

has only positive eigenvalues. If this is the case, then also its inverse, the matrix ∂ Hmn = φn (15.71) ∂Jm must have only positive eigenvalues16 . That is, for any eigenvector a of H the eigenvalue λ must be positive : X Hmn an = λ am , λ > 0 . (15.72) n

In turn, this is guaranteed if X

Hmn am an > 0

(15.73)

m,n 16

Since Γmn is symmetric, so is Hmn although this is not obvious from the form it is written here.

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August 31, 2019

for any vector a. Now, we have R Q ( n dϕn ) P ({ϕ}, {J}) ϕm φm = R Q , ( n dϕn ) P ({ϕ}, {J})

(15.74)

and therefore R Q ( n dϕn ) P ({ϕ}, {J}) ϕm ϕn 1 R Q Hmn = ~ ( n dϕn ) P ({ϕ}, {J}) R Q R Q ( n dϕn ) P ({ϕ}, {J}) ϕm ( n dϕn ) P ({ϕ}, {J}) ϕn R Q R Q (.15.75) − ( n dϕn ) P ({ϕ}, {J}) ( n dϕn ) P ({ϕ}, {J}) We now employ the following trick : duplicate the set of fields {ϕ} by the addition of another set of fields, {ϕ}, ˆ with the combined probability density P ({ϕ}, {ϕ}, ˆ {J}) = P ({ϕ}, {J})P ({ϕ}, ˆ {J}) .

(15.76)

By this construction, the random variables ϕ and ϕˆ are statistically independent. We can then write the matrix H as 1 Hmn = hϕm ϕn − ϕm ϕˆn i , (15.77) ~ with the average taken with respect to the new probability density. Using the fact that this density is symmetric in ϕ ↔ ϕ, ˆ we can write this as 1 1 Hmn = hϕm ϕn − ϕm ϕˆn − ϕˆm ϕn + ϕˆm ϕˆn i ~ 2 1 = h(ϕm − ϕˆm )(ϕn − ϕˆn )i , (15.78) 2 and we arrive at * !2 + X X ~ Hmn am an = (ϕn − ϕˆn )an , (15.79) 2 m,n n which is necessarily positive. The matrix H has, therefore, only positive eigenvalues, and the effective action is always concave. It is of course possible (and even likely in the case of continuum theories that have a noncountable infinity of field values) that the eigenvalue is actually infinite. In that case the effective action contains flat directions. So perhaps the more careful statement is that the effective action cannot be convex anywhere. A final point to note is that our proof relies only on the fact that the ϕ values are randomly distributed over some nonvanishing region, no matter how small. Of course, by restricting the values that the ϕ are allowed to take we will change the effective action ; but it will never be convex.

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15.7

463

Functional derivatives

To understand how functional derivatives work, let us assume that the action of a theory, as in section 4.3.3, can be written as   Z 0 S[ϕ] = F ϕ(x); ϕ (x) dx . (15.80) Upon ‘discretisation’ using the Weyl ordering, this becomes   X 1 1 S = (ϕk+1 + ϕk ); (ϕk+1 − ϕk ) ∆F 2 ∆ k   1 1 (ϕn+1 + ϕn ); (ϕn+1 − ϕn ) + = ∆F 2 ∆   1 1 (ϕn + ϕn−1 ); (ϕn − ϕn−1 ) + ∆F 2 ∆ lots of terms not containing ϕn . The classical equation then reads ∂S/∂ϕn = 0, with   1 1 1 ∂ 1 S = F1 (ϕn+1 + ϕn ); (ϕn+1 − ϕn ) ∆ ∂ϕn 2 2 ∆   1 1 1 + F1 (ϕn + ϕn−1 ); (ϕn − ϕn−1 ) 2 2 ∆   1 1 1 − F2 (ϕn+1 + ϕn ); (ϕn+1 − ϕn ) ∆ 2 ∆   1 1 1 + F2 (ϕn + ϕn−1 ); (ϕn − ϕn−1 ) , ∆ 2 ∆

(15.81)

(15.82)

where Fj denotes the partial derivative of F with respect to its j-th argument. Re-inserting the Weyl ordering, we can write this equation as     1 1 0 0 F1 ϕ(x); ϕ (x) + F1 ϕ(x − ∆); ϕ (x − ∆) 0 = 2 2     1 1 0 0 − F2 ϕ(x); ϕ (x) + F2 ϕ(x − ∆); ϕ (x − ∆) (15.83) . ∆ ∆ By Taylor expansion we get, for arbitrary f :     0 0 0 00 f ϕ(x − ∆); ϕ (x − ∆) ≈ f ϕ(x) − ∆ϕ (x); ϕ (x) − ∆ϕ (x)

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       0 0 0 00 0 ≈ f ϕ(x); ϕ (x) − ∆ ϕ (x)f1 ϕ(x); ϕ (x) + ϕ (x)f2 ϕ(x); ϕ (x)     d 0 0 = f ϕ(x); ϕ (x) − ∆ f ϕ(x); ϕ (x) . (15.84) dx The classical equation thus takes the form     d 0 0 F1 ϕ(x); ϕ (x) − F2 ϕ(x); ϕ (x) = 0 . dx

(15.85)

we can cast this in the formal language of functional derivatives : we define, using the Dirac delta function, δϕ0 (y) δϕ(y) = δ(x − y) , =0 , δϕ(x) δϕ(x) δϕ(y) δϕ0 (y) = δ(x − y) , =0 , 0 δϕ (x) δϕ0 (x)

(15.86)

where, as we see, ϕ(x) and ϕ0 (x) are treated as independent variables. Applying these rules to the continuum form of the action, we find that the formal form of the classical field equation is therefore that of the Euler-Lagrange equation. The language of functional derivatives is, in these notes, treated as an effective method, valid in the continuum limit, of writing the more fundamental discrete classical field equation. In the functional formalism, the Euler-Lagrange equation reads   d δ δ S[ϕ, J] − S[ϕ, J] = 0 . (15.87) δϕ(x) dx δϕ0 (x) For ϕ4 theory, the Euler-Lagrange equation takes precisely the form (4.35).

15.8

Frustrated and unusual actions

15.8.1

Frustrating your neighbours

The one-dimensional action we have studied was based on ‘nearest-neighbour’ interactions. We can, of course, extend this treatment to include ‘next-tonearest-neighbour’ interactions as well. Let us take  X 1 2 S({ϕ}) = ∆ µϕn − γ1 ϕn ϕn+1 − γ2 ϕn ϕn+2 2 n

August 31, 2019 =

465 X n

 ∆

1 1 (µ − 2γ1 − 2γ2 )ϕn 2 − (γ1 + 4γ2 )(ϕn+1 − ϕn )2 2 2  1 2 − γ2 (ϕn+2 − 2ϕn+1 + ϕn ) , (15.88) 2

with the continuum behaviour of µ, γ1 and γ2 to be determined. We disregard any other interactions since we shall only be interested in the propagator. Setting up the SDe for the discrete propagator is trivial: we have   ~ Π(n) = δn,0 + γ1 Π(n + 1) + Π(n − 1) µ   +γ2 Π(n + 2) + Π(n − 2) , (15.89) so that Fourier transformation gives us I un−1 ~ Π(n) = du , 2iπ f (u) |u|=1     1 1 2 f (u) = µ − γ1 u + − γ2 u + 2 . u u

(15.90)

In the continuum limit, the only relevant poles of the integrand are those at values of u such that |u| = 1 − O (∆). Let uj (j = 1, 2, . . .) be these poles: then X uj |x|/∆ . (15.91) Π(x) = ~ f 0 (uj ) j Writing u = 1 − v∆, we can approximate f (u) = (µ − 2γ1 − 2γ2 ) − (γ1 + 4γ2 )(v 2 ∆2 + v 3 ∆3 )
−(γ1 + 5γ2 )v 4 ∆4 + O ∆2 .

(15.92)

There are now two possible continuum limits. In the first case, we can assume that γ1 + 4γ2 does not vanish. In that case, we can take γ1 + 4γ2 ∼ 1/∆, and the resulting continuum limit is indistinguishable from the nearest-neighbour case. For later reference we shall denote this propagator by P1 (x) =

~ exp(−m|x|) . 2m

(15.93)

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The more curious solution is provided by the special choice γ1 = −4γ2 . The only sensible continuum limit in that case is to take γ1 + 5γ2 ∼

1 ∆3

→

γ1 ∼

4 1 , γ2 ∼ − 3 , 3 ∆ ∆

(15.94)

and µ = m4 ∆ + 2γ1 + 2γ2 ∼ m2 ∆ +

6 . ∆3

(15.95)

The poles of the integrand are therefore approximately given by

f (u) = ∆ m4 + v 4 + O ∆2 = 0 ,

(15.96)

so that the solutions are  uk ≈ 1 − ∆m

1+i √ 2

2k−3 , k = 1, 2, 3, 4 .

(15.97)

Only u1 and u2 are inside the unit circle, and we obtain the propagator       ~ −m|x| m|x| m|x| √ exp √ Π(x) = cos √ + sin √ (15.98) , m3 8 2 2 2 which we shall denote by P2 (x): it has the interesting property that Π2 (x) is negative for mx between 3π/4 and 7π/4, modulo 2π. An discrete action such as the one belonging to this continuum limit, in which nearest-neighbour and next-to-nearest-neighbour couplings have opposite sign, are called frustrated17 . The continuum limit of the propagator can also be written as ~ P2 (x) = 2π

Z

exp(ikx) dk , k 4 + m4

(15.99)

and that of the action reads Z  S[ϕ] = 17

1 4 1 m ϕ(x)2 + ϕ00 (x)2 2 2

 .

Frustrated in the sense that ‘not all couplings can have it their own way’.

(15.100)

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15.8.2

467

Increasing frustration

It is quite possible to construct even more frustrated actions, as follows. Let us suppose that the action is given by # " p X X 1 γj ϕn ϕn+j . (15.101) µϕn 2 − S({ϕ}) = 2 n j=1 The propagator is given by Eq.(15.91), where now   p X 1 j γj u + j f (u) = µ − . u j=1

(15.102)

We shall now arrange for the only the highest possible power of 1 − u to survive in this expression. We first put u = exp(ik∆), so that the function f (u) becomes f (u) = µ −

p X

2γj cos(jk∆) = µ −

j=1

Br ≡

p X 2(−)r j=1

(2r)!

X (k∆)2r Br , r≥0

j 2r γj .

(15.103)

We now seek to find the γ’s such that B1 = B2 = · · · = Bp−1 = 0 ,

Bp = −

1 ∆2p−1

.

(15.104)

In that case, we can take arbitrary constants cr , with cp = 1, and always have p p X X cr Br = γj Q(j) = Bp , (15.105) r=1

with Q(j) =

j=1 p X 2(−)r r=1

(2r)!

cr j 2r .

(15.106)

The polynomial Q(j) is even and of degree 2p in j, and Q(0) = 0. We can now, for any preassigned q with 1 ≤ q ≤ p, choose the numbers cr such that Q(0) = · · · = Q(q − 1) = Q(q + 1) = · · · = Q(p) = 0 ,

Q(q) 6= 0 , (15.107)

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August 31, 2019

upon which γq = Bp /Q(q) .

(15.108)

Obviously, the polynomial Q(j) is given by 2(−)p Q(j) = (2p)!



Y

2

j −n

2

 ,

(15.109)

0≤n≤p n 6= q

from which we derive γq =

(−)q−1 (2p)! , 1≤q≤p . ∆2p−1 (p − q)!(p + q)!

The continuum limit of the propagator is, then Z ~ exp(ikx) Πp (x) = dk 2p 2π k + m2p The poles of the integrand are located at k = mωj , where   2j + 1 ωj = exp iπ , j = 0, 1, 2, . . . , 2p , 2p

(15.110)

(15.111)

(15.112)

so that Cauchy integration gives p −i~ X ωj exp(iωj m|x|) . Πp (x) = 2pm2p−1 j=0

(15.113)

We may even investigate the limit p → ∞: in that case we may approximate  −2p 1 m if −m < k < m ≈ (15.114) 2p 2p 0 elsewhere k +m so that the propagator takes the form ~ Πp (x) ≈ 2πm2p

Zm dk exp(ikx) =

1 m2p−1 π

sin(mx) . mx

(15.115)

−m

Figure 15.2 shows the propagators Πp (x) for ~ = m = 1, as a function of

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469

Figure 15.2: Propagators for frustrated actions

x. The values of p are 1,3,10, and also the asymptotic form of Eq.(15.115) is plotted. For large p the asymptotic form is approximated smoothly. The higher the value of p, the more frustrated the lattice is, and the more difficult it becomes for momentum modes with high wave number to propagate through the lattice, as is evident from Eq.(15.111). For the totally frustrated lattice, all wave numbers smaller than m propagate equally, and all wave nubers larger than m do not propagate at all.

15.9

Newton’s First Law revisited

15.9.1

Introduction : the matter of sources

In our discussion of Newton’s first law in section 5.2.5 we have used a particular expression for the shape of the time-dependent part of the source, motivated by mathematical convenience. Here we shall redo the analysis of 5.2.5 but with several different time dependences of the source. The response of the field function to the source is Z 0 0 ~ e−ik x +ik·~x 1 0 3 0 dk d k 0 2 Jt (k 0 ) Js (~k) , (15.116) φ(x , ~x) = 4 2 (2π) k − ω + i

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p with ω = ~k 2 + m2 as usual. The space part of the source will be Gaussian : in position language it reads   ~x2 i 2 −3/4 (15.117) Js (~x) = (2πσ ) exp − 2 + p~ · ~x 4σ ~ which corresponds to unit strength, Z d3 x |Js (~x)|2 = 1 .

(15.118)

In momentum language, we have 2 !  Z 1 ~ 3 −i k·~ x 2 3/4 2 . (15.119) Js (~k) = d x Js (~x) e = (8πσ ) exp −σ ~k − p~ ~ For the time dependence of the source we examine three alternatives, which may be called slow, fast, and abrupt, respectively : in position language,   1 |x0 | i 0 0 (1) 0 Jt (x ) = exp − − px , (a1 )1/2 a1 ~ ! 2 x0 i 0 0 1 (2) 0 exp − 2 − p x , Jt (x ) = (2πa22 )1/4 4a2 ~  
i 0 0 1 (3) 0 0 θ −a3 < x < a3 exp − p x Jt (x ) = . (15.120) (2a3 )1/2 ~ These three sources are all normalised to unit strength : Z (j) dx0 |Jt (x0 )|2 = 1 , j = 1, 2, 3 .

(15.121)

To compare the spread of the sources in the time domain we can use D E Z a1 2 a3 2 2 (j) = a2 2 = . (15.122) x0 = dx0 (x0 )2 |Jt (x0 )|2 = 2 3 In momentum language we have (1)

Jt (k 0 ) = (2)

2a1 1/2 , ∆2 + 1/a1 2 2

2

Jt (k 0 ) = (8πa22 )1/4 e−a2 ∆ ,
−i (3) ia3 ∆ −ia3 ∆ Jt (k 0 ) = e − e , (2a3 )1/2 ∆

(15.123)

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with ∆ = k 0 − p0 /~. The response of the field to the timelike part of the source is Z 0 0 e−ik x (j) 0 ψj ≡ dk 0 2 Jt (k 0 ) , (15.124) k − ω 2 + i and this is what we now investigate for positive times : x0 > 0.

15.9.2

Slow, fast and abrupt

Slow source: exponential behaviour The case j = 1 is what we have already investigated. Recalling the discussion of how to close the contour in the k 0 plane, we have  √ exp(−iωx0 ) ψ1 = −2iπ a1 ω (ω − p0 /~)2 + 1/a1 2  ia1 ω exp(−x0 /a1 − ip0 x0 /~) + (15.125) (p0 /~ − i/a1 )2 − ω 2 As we have remarked before, the second term dies out with the source, so that for large enough times x0  a1 we can disregard it. Fast source : Gaussian behaviour The second, ‘moderate’ time dependence has to be treated more carefully. 2 This is due to the fact that the exponential exp(−k 0 ) diverges when k 0 → ∞ if the argument of k 0 lies in (π/4, 3π/4) or (5π/4, 7π/4), so we cannot close the contour simply as in the previous case. Instead, we write ! 02 x ψ2 = a2 (8πa2 )1/4 exp − 2 − ix0 p0 /~ A2 , 4a2 Z∞ exp (−(y + iτ )2 ) A2 = dy , (y − b+ + i)(y − b− − i) −∞

b± = a2 (±ω − p0 /~) ,

τ = x0 /(2a2 ) ,

(15.126)

and we have chosen y = a2 (k 0 −p0 /~). The y integral runs over the real axis ; we may shift it downwards by an amount τ provided we include a contour integral around the point b+ − i, as indicated in the figure 15.9.2. The poles at b± ∓ i are indicated.

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August 31, 2019

→

Figure 15.3: Contour shift for a Gaussian source

The result of this operation on A2 is I exp (−(y + iτ )2 ) dy A2 = − (y − b+ + i)(y − b− − i) y∼b+ −i ∞ Z exp(−z 2 ) + dz (z − b− iτ )(z − b− − iτ ) −∞


2π exp −(b+ + iτ )2 b+ − b−   Z∞ 1 exp(−z 2 ) 1 − + dz . (15.127) b+ − b− z − b− − iτ z − b+ − iτ

= −i

−∞

For large times (τ → ∞) the two integral terms in the second both behave as ∼ 1/τ so, as in the previous case, they give rise to a contribution that dies out with the source. We therefore have
lim ψ2 ∼ a2 (8πa2 )1/4 exp −a22 (ω − p0 /~)2 − iωx0 . (15.128) τ →∞

As before, the on-shell condition implied by ω ≈ p0 /~ is enforced. Abrupt source : Heavyside behaviour (3)

This case is the easiest one to analyse once we realise that Jt is perfectly regular at ∆ = 0 notwithstanding the denominator. So for x0 > 0 we may

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simply close the contour in the lower half complex k 0 plane to find 0

ψ3 =

−π eia3 (ω−p /~) − e−ia3 (ω−p √ exp(−iωx0 ) √ a3 (ω − p0 /~) ω 2

0 /~)

.

(15.129)

The numerator in the last factor remains bounded in absolute value. There√ fore, as long as ω 6= p0 /~, ψ3 goes to zero as 1/ a3 for large a3 , while at √ ω = p0 /~ it approaches infinity as a3 : yet another situation in which the on-shell condition is enforced.

15.9.3

Conclusion : general effect of the sources

We have seen that all three type of sources exhibit a large-time behaviour 0

ψj ∼ e−iωx dj (aj ; ω − p0 /~) ,

(15.130)

characterised by a plane wave exp(−iωx0 ), with the contributing values of ω governed by a distribution dj such that for aj becoming large (hence the source becoming broader in time) the energies ω are closely clustered around the central value p0 /~. The reasoning leading to the mass-shell condition E 2 ≈ |~p|2 c2 + M 2 c4 , as well as the motion along straight trajectories, ~x ≈ p~ t/p0 , follow in each case as discussed in section 5.2.5.

15.10

Unitarity bounds

15.10.1

Resonances

In this appendix we shall establish bounds on total cross sections as implied by the unitarity of the theory. We are interested in upper bounds on cross sections, that is we want to investigate the most efficient way to get rid of the initial state in favour of some final state. Now, as is known from the elementary theory of coupled oscillators, the most efficient way to pump energy (i.e. the energy content of the initial-state particles) into another state is by resonance. In our language, this means that we shall consider two initial-state particles colliding and coupling to another particle with just the right energy to put that particle on its mass shell. Unavoidably, if the new particle can be made in such a way it can also decay, and it therefore must have a nonzero decay width which protects its propagator from exploding. We shall investigate this process in some detail.

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15.10.2

Preliminaries : decay widths

We shall investigate the unitarity bound on the cross section for a given initial two-particle state 1 to evolve into a given n-particle state 2 by way of a resonant particle X of rest mass M and total decay width Γ. This means that particle X must couple both to 1 and to 2. There is therefore a possible decay X→1, described by MX→1 =

X

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

1

= i Ak · uj .

(15.131)

In this admittedly abstract expression, u stands for the external-line factor18 for the incoming X particle that has, in addition to energy and momentum, a discrete quantum number j denoting its angular momentum (for brevity we shall use the smaller word ‘spin’ throughout this section). We shall assume that j runs from 1 to N , so that there are in total N spin states : for a spin-J particle, therefore, N = 2J + 1. Similarly the final state is characterized by a discrete quantum number k alongside the continuous energy and momentum variables, and k is assumed to run from 1 to K. For instance, if 1 stands for an electron-positron state, K = 4 since there are two spin states for the electron and two for the positron. Thus, Ak denotes the total of the connected diagrams (the blob) and any external-line factors for a final state with discrete quantum number k. The total decay width Γ1 for X to go into the two-particle state 1 is given by Z 1 dΩ λ(M 2 , m2 , m02 )1/2 1 1 X Ak ·uj uj ·Ak S1 , (15.132) Γ1 = 2M N j,k (2π)2 8 M2 where m and m0 are the masses of the two particles in 1. The symmetry factor S1 equals 1 if the particles are distinguishable, and 1/2 if they are not. Ω is of course the solid angle of one of the particles in the rest frame of X. The angle- and spin-averaged transition rate is therefore Z 1 X dΩ 16πM Γ1 N M 2 Ak · uj uj · Ak = , (15.133) K j,k 4π S1 K λ1/2 with λ1/2 = λ(M 2 , m2 , m02 )1/2 . 18

This might be just a number, or a spinor, or a polarisation vector,. . . take your pick.

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475

The process X→2 is described by MX→2 =

X

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

2

= i B l · uj ,

(15.134)

where l denotes the discrete quantum numbers in the state 2. The width for the process is given by Z 1 1 X Γ2 = B l · uj uj · Bl dVn S2 , (15.135) 2M N j,l where dVn is the n-particle phase space factor going with the state 2, and S2 is the appropriate symmetry factor.

15.10.3

The rˆ ole of angular momentum conservation

Let us consider the process X→2 in some greater detail. It is easy to conceive of a final state 2 that couples only to a particle of spin J and to no other spin. Now, our important supposition : if the initial particle is at rest, and if space is isotropic so that there is no preferred direction, this does not only mean that angular momentum is conserved but also that the various 2J + 1 spin states of the X particle are to be treated on the same footing, so that each spin state must have the same decay width. This in its turn implies that the integrated-over final state must form a projection onto the pure spin-J state : X X un un (15.136) Bl B l = B(M 2 ) n

l

where n runs, of course, from 1 to N. Obviously, under the isotropy assumption B can only depend on M 2 . We find that Z X X Bl · uj uj 0 · Bl = B(M 2 ) un · uj uj 0 · un ∝ δj,j 0 , (15.137) n

l

or, in other words, XZ l

B l · uj uj 0 · Bl = 2 M Γ2 δj,j 0 .

(15.138)

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15.10.4

The unitarity bound

We now consider √ the process 1→2 by X exchange. For total scattering invariant mass s, it is given by the diagram M =

1

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

X

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

2

=

−i B l · Π · Ak . s − M 2 + iM Γ

(15.139)

Here, Π is the numerator of the X propagator : on the mass shell, therefore, we must have X Πcs=M 2 = uj uj . (15.140) n

For the total cross section we therefore have σ =

1 1 1 2λ(s, m2 , m02 )1/2 (s − M 2 )2 + m2 Γ2 K X Z (B l · Π · Ak ) (Ak · Π · Bl ) dVn S2 . ×

(15.141)

k,l

On the X mass shell, we can write, with Z 1 1 1 X (B l · uj σ = 2λ1/2 M 2 Γ2 K k,l,j,j 0 Z 1 1 1 X = (B l · uj 2λ1/2 M 2 Γ2 K k,l,j,j 0 Z 1 1 2M2Γ X dΩ = 1/2 2 2 2λ M Γ K k,j,j 0 4π

the help of Eq.(15.138), uj · Ak ) (Ak · uj 0 uj 0 · Bl ) dVn S2 uj 0 · Bl ) (Ak · uj 0 uj · Ak ) dVn S2 Ak · uj 0 uj · Ak δj,j 0 ,

(15.142)

where it must be realized that we have rewritten the integral over B-cum-A by the integral over B times the average over A. Due to angular-momentum conservation we can now write, using Eq.(15.133),    Γ1 Γ2 N 16π s σcs=M 2 = . (15.143) Γ Γ S1 K λ(s, m2 , m02 ) Now, the factor Γ2 /Γ is understandable since the X particle has only a fractional probability to decay into state 2 (there may be other decay channels available, in fact at least the decay X→1), and then symmetry between the reactions 1→2 and 2→1 requires also the presence of the factor Γ1 /Γ. We

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conclude that the cross section for the initial state 1 to go into any final state with spin J is bounded by the unitarity limit σUL =

2J + 1 16π s , S1 K λ(s, m2 , m02 )

(15.144)

where as mentioned before S1 is 1/2 for indistinguishable particles and 1 for distinguishable ones, and K is the total number of possible discrete quantum numbers for the initial state19 .

15.11

The fundamental theorem for Dirac matrices

15.11.1

Proof of the fundamental theorem

In this appendix we prove the following statement : if we have two sets of four matrices, γ µ and γˆ µ (µ = 0, 1, 2, 3) , satisfying Dirac’s anticommutation relation γ µ γ ν + γ ν γ µ = 2 g µν , γˆ µ γˆ ν + γˆ ν γˆ µ = 2 g µν ,

(15.145)

then there is a matrix S such that γˆ µ = S γ µ S −1 .

(15.146)

To this end, we first set up a basis of the Clifford algebra as follows : Γ0 = 1 , Γ1 Γ5 = γ 0 γ 1 , Γ9 = iγ 1 γ 3 , Γ13 = iγ 0 γ 2 γ 3

= γ 0 , Γ2 = iγ 1 , Γ3 = iγ 2 , Γ4 = iγ 3 , Γ6 = γ 0 γ 2 , Γ7 = γ 0 γ 3 , Γ8 = iγ 1 γ 2 , Γ10 = iγ 2 γ 3 , Γ11 = iγ 0 γ 1 γ 2 , Γ12 = iγ 0 γ 1 γ 3 , , Γ14 = γ 1 γ 2 γ 3 , Γ15 = iγ 0 γ 1 γ 2 γ 3 , (15.147)

which we denote by Γk , k = 0, 1, 2, . . . , 15 ; and using the γˆ µ we construct an ˆ k in the same way. These have a few interesting properties. analogous set Γ For example, for an initial e+ e− state we have S1 = 1, K = 4 : for an initial state of two photons S1 = 1/2, K = 4, and for an initial state of two gluons S1 = 1/2, K = 256 since gluons come with 2 possible spin states and 8 different colour states. 19

478

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

August 31, 2019 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 1 0 5 6 7 2 3 4 11 12 13 8 9 10 15 14

2 2 5 0 8 9 1 11 12 3 4 14 6 7 15 10 13

3 3 6 8 0 10 11 1 13 2 14 4 5 15 7 9 12

4 4 7 9 10 0 12 13 1 14 2 3 15 5 6 8 11

5 5 2 1 11 12 0 8 9 6 7 15 3 4 14 13 10

6 6 3 11 1 13 8 0 10 5 15 7 2 14 4 12 9

7 7 4 12 13 1 9 10 0 15 5 6 14 2 3 11 8

8 8 11 3 2 14 6 5 15 0 10 9 1 13 12 4 7

9 9 12 4 14 2 7 15 5 10 0 8 13 1 11 3 6

10 10 13 14 4 3 15 7 6 9 8 0 12 11 1 2 5

11 11 8 6 5 15 3 2 14 1 13 12 0 10 9 7 4

12 12 9 7 15 5 4 14 2 13 1 11 10 0 8 6 3

13 13 10 15 7 6 14 4 3 12 11 1 9 8 0 5 2

14 14 15 10 9 8 13 12 11 4 3 2 7 6 5 0 1

15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Table 15.3: Multiplication table for the Clifford algebra

In the first place, Γk 2 = 1 for all k. Secondly, for every pair j and k there are numbers n and cn such that Γj Γk = cn Γn ,

cn = 1, −1, i or − i.

(15.148)

From these properties it follows that simultaneously Γk Γj =

1 Γn cn

(15.149)

We can thus construct the multiplication table 15.11.120 , where the possible values of j define the rows, and those for k the columns: the corresponding entry is then the value of n. For instance, Γ6 Γ4 = Γ13 20

Kids! Don’t do this at home, since constructing this multiplication table is extremely tedious. The numbers cn are not given: they are anyhow only defined up to a sign, since we can always replace Γj by −Γj (using γ 2 γ 0 instead of γ 0 γ 2 , say) without changing the Dirac anticommutation relation.

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(in this case c13 happens to be 1). Note that in this table every row and every column contains each of the numbers from 0 to 15 precisely once. Hence, if we keep j fixed and let k run from 0 to 15, the value of n will also take on all values from 0 to 15 (although generally in a different order). Obviously, ˆ exactly the same multiplication table holds. for the set Γ We are now ready to prove the theorem. Let A be an arbitrary matrix, and define S by 15 X ˆ k A Γk . S≡ Γ (15.150) k=0

This has the desired property since ˆ j S Γj = Γ =

15 X k=0 15 X n=0

ˆj Γ ˆ k A Γk Γj Γ ˆ n A 1 Γn = S , cn Γ cn

(15.151)

in other words, ˆ j S = S Γj . Γ

(15.152)

It remains to ensure that the matrix S actually has an inverse. Since A can be chosen at will (except A = 0) this is not a problem. Let us pick another matrix B and construct 15 X ˆk . Γk B Γ (15.153) T = k=0

For this matrix we obviously have ˆj . Γj T = T Γ

(15.154)

Combining Eq.(15.152) and (15.154) we see that the product T S commutes ˆ j ). Therefore T S is proporwith Γj (and the product ST commutes with Γ tional to the unit matrix and we can adjust the elements of B such that T = S −1 . It is an interesting observation that the dimensionality of the γ µ and that of the γˆ µ does not have to be the same. In that case the matrices A and B are simply not square matrices but have different numbers of rows and columns.

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August 31, 2019

15.11.2

The charge conjugation matrix

An application of the fundamental theorem is the following. The anticommutation relation, if satisfied by the Dirac matrices γ µ , is automatically also satisfied by the matrices −(γ µ )T where T stands for the transpose. There exists, therefore, a matrix C such that γˆ µ = C γ µ C −1 = −(γ µ )T .

(15.155)

This is called the charge conjugation matrix. In the representation given in section ??, we have γˆ 0 = −γ 0 , γˆ 1 = γ 1 , γˆ 2 = −γ 2 , γˆ 3 = γ 3 ;

(15.156)

and we see that a good choice is C = C −1 = γ 0 γ 2 .

(15.157)

Since this form is not proof against change of representation, the use of the charge conjugation matrix in arguments and derivations lacks somewhat in elegance.

15.12

States of higher spin

15.12.1

The spin algebra for integer spins

In this Appendix we shall consider how to represent particles with higher spins. We shall start with integer spin. Such particles states can be represented, in the Feynman rules, as tensors of some rank r : |s, miµ1 µ2 µ3 ···µr where s stands for the total spin of the particle, and m denotes the spin along some quantization axis, for which we shal take the z direction here. That is, once we have found the correct operators of the spin algebra (Sx,y,z )µ1 µ2 ···µr ν1 ν2 ···νr and (S 2 )µ1 µ2 ···µr ν1 ν2 ···νr Then we have, by definition, (S 2 )µ1 µ2 ···µr ν1 ν2 ···νr |s, miν1 ν2 ···νr = ~2 s(s + 1) |s, miµ1 µ2 ···µr , (Sz )µ1 µ2 ···µr ν1 ν2 ···νr |s, miν1 ν2 ···νr = ~ m |s, miµ1 µ2 ···µr . (15.158)

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It is easy to see that the spin algebra is correctly constructed once we have raising and lowering operators (S± )µ1 µ2 ···µr ν1 ν2 ···νr ,

S− = (S+ )† ,

with [[S+ , S− ], S+ ] = 2~2 S+ .

(15.159)

We can then find the other algebra elements via

1 1 1 S+ + S− , S y = S+ − S− , Sz = [S+ , S− ] , 2 2i 2~ 1 S2 = S+ , S− + (Sz )2 . (15.160) 2

Sx =

We will start with particles in their rest frame21 . The spin representations are built using four unit vectors, with obvious notation, as tµ , xµ , y µ and z µ , which obey t·t = 1 , x·x = y·y = z·z = −1 , t·x = t·y = t·z = x·y = x·z = y·z = 0 . (15.161) Things will become easier if we also define
1 x± µ = √ xµ ± i y µ 2

(15.162)

so that x± · x± = 0 , x+ · x− = −1 , x± · z = x± · t = 0 .

(15.163)

Since the spin of a particle informs us about its behaviour under rotations in the three-dimensional spacelike part of Minkowski space, we always require, for particles in their rest frame, |s, miµ1 µ2 ···µr tµj = 0 , j = 1, 2, . . . , r .

(15.164)

This means that the appropriate tensors in fact contain only the three vectors x+ , x− , and z ; for instance the rank-4 tensor |s, miµ1 µ2 µ3 µ4 may contain a term x+ µ1 x− µ2 z µ3 x+ µ4 . In general, the particle’s tensor is a linear combination of such terms : which precise linear combination it is depends on s and m, and this is what we want to look into. 21

This implies that the particles are massive. For massless partices, see later on.

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August 31, 2019

15.12.2

Rank one for spin one

The simplest nontrivial case is that of a rank-1 tensor, that is, a vector. We have already considered these in Chapter 9. We can define |1, 1iµ = x+ µ , |1, 0iµ = z µ , |1, −1iµ = −x− µ ,

(15.165)

h1, 1|µ = x− µ , h1, 0|µ = z µ , h1, −1|µ = −x+ µ .

(15.166)

so that

For brevity, we shall use the easily interpretable notation |1, 1i = |+i , |1, 0i = |0i , |1, −1i = − |−i .

(15.167)

These states are properly normalized, since h1, m1 |1, m2 i = h1, m1 |µ |1, m2 iµ = − δm1 ,m2 .

(15.168)

In addition, the states are complete in the sense that X |1, λiµ h1, λ|ν = tµ tν − δ µ ν ≡ ∆µ ν .

(15.169)

λ=+,−,0

Note that ∆µα ∆αν = −∆µ ν , ∆µ µ = −3 .

(15.170)

We now proceed to set up the spin algebra. A general raising operator can always be written in the form   √ (15.171) S+ = 2~ a |+i h0| + b |0i h−| , where a and b are some complex numbers ; and so   √ ∗ ∗ S− = 2~ a |0i h+| + b |−i h0| .

(15.172)

From S+ S− S− S+

  2 2 = −2~ |a| |+i h+| + |b| |0i h0| ,   2 2 2 = −2~ |a| |0i h0| + |b| |−i h−| , 2

(15.173)

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483

we find that to get the correct form of Sz we have to take |a| = |b| = 1, since only then22   Sz = −~

|+i h+| − |−i h−|

;

furthermore, we find automatically   2 2 |+i h+| + |0i h0| + |−i h−| , S = −2~

(15.174)

(15.175)

which shows that we have here indeed a spin-one system. For reasons that will become clear later on we shall choose a = −1 and b = 1. Thus, √ √ S+ |+i = 0 , S+ |0i = 2~ |+i , S+ |−i = − 2~ |0i . (15.176) In more explicit tensorial language, we have the following matrix forms :   √ µ µ µ S+ ν = 2~ −x+ zν + z x+ ν ,   √ µ µ µ , S− ν = 2~ −z x− ν + x− zν   µ µ µ , Sz ν = ~ −x+ x− ν + x− x+ ν   µ µ 2µ 2 µ . (15.177) S ν = −2~ x+ x− ν + x− x+ ν + z zν

15.12.3

Rank-2 tensors

By taking tensor products of vectors we can build more complicated systems. Let us attempt rank-2 tensors. We can easily construct the spin algebra for this system as follows : Σj µν αβ = Sj µ α δ ν β + δ µ α Sj ν β ,

j = +, −, z ,

(15.178)

and it is easily checked that these also obey the correct commutation relations [Σ+ , Σ− ] = 2~ Σz , [Σz , Σ+ ] = ~ Σ+ ; 22

(15.179)

Do not be confused with the overall minus signs emerging here ! Remember that the states are normalized to minus unity. This is a consequence of our dealing with spacelike objects in an essentially Minkowski space.

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August 31, 2019

the operator for the total spin is of course Σ2

µν αβ

= S2

µ

µ ν µ ν µ ν ν µ 2ν α δ β +δ α S β +S+ α S− β +S− α S+ β +2Sz α Sz β

. (15.180)

There is precisely one rank-2 tensor with a spin 2~ along the z axis : it is the tensor product |2, 2iµν = |1, 1iµ |1, 1iν = x+ µ x+ ν ≡ |++i ,

(15.181)

with obvious notation. It is straightforward to check that the total spin of this object is, indeed, equal to 2~. By applying the lowering operator as given in Eq.(15.178), and normalizing, we can immediately recover the other states in the spin-2 sector : |2, 2i = |++i ,   √ |2, 1i = |+0i + |0+i / 2 ,   √ |2, 0i = − |+−i − |−+i + 2 |00i / 6 ,   √ |2, −1i = − |−0i − |0−i / 2 , |2, −2i = |−−i .

(15.182)

These five objects are totally symmetric. They are also traceless in the sense that |2, miµν gµν = 0 ; this is due to our choice for the constants a and b made above. The one object made up from |+0i and |0+i that is orthonormal to |2, 1i is |+0i − |0+i, which forms the basis of a spin-1 sector :   √ |1, 1i = |+0i − |0+i / 2 ,   √ |1, 0i = |−+i − |+−i / 2 ,   √ |1, −1i = |−0i − |0−i / 2 . (15.183) Finally, one single state is left :   √ |0, 0i = |+−i + |−+i + |00i / 3 ,

(15.184)

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485

which upon inspection is seen to have zero spin. The orthonormality of these nine states is easily checked. Some simple algebra also tells us that 2 X

|2, miµν h2, m|αβ =

1 µ 1 1 ∆ α ∆ν β + ∆µ β ∆ν α − ∆µν ∆αβ , 2 2 3

|1, miµν h1, m|αβ =

1 µ 1 ∆ α ∆ν β − ∆µ β ∆ν α , 2 2

m=−2 1 X m=−1

|0, 0iµν h0, 0|αβ =

1 µν ∆ ∆αβ , 3

(15.185)

so that there is a completeness relation of the form 2 s X X

|s, miµν hs, m|αβ = ∆µ α ∆ν β .

(15.186)

s=0 m=−s

This confirms that no states have been overlooked.

15.12.4

Rank-3 tensors

For the sake of illustration we also give the complete set of rank-3 tensorial states. These fall apart in one spin-3, two spin-2, three spin-1 and a single spin-0 sector, giving the correct total of 27 possible orthonormal states, listed below. For reasons of typography I have left out the normalizing denominators ; these can of course be trivially recovered. spin-3 : |3, 3i = |+ + +i |3, 2i = |+ + 0i + |+0+i + |0 + +i |3, 1i = 2 |+00i + 2 |0 + 0i + 2 |00+i − |+ + −i − |+ − +i − |− + +i |3, 0i = 2 |000i − |+0−i − |0 − +i − |− + 0i − |−0+i − |+ − 0i − |0 + −i |3, −1i = |+ − −i + |− + −i + |− − +i −2 |−00i − 2 |0 − 0i − 2 |00−i |3, −2i = |− − 0i + |−0−i + |0 − −i

486

August 31, 2019 |3, −3i = − |− − −i spin-2(1) : |2, 2i = |+0+i + |0 + +i − 2 |+ + 0i |2, 1i = 2 |00+i − |+ − +i − |− + +i − |+00i − |0 + 0i + 2 |+ + −i |2, 0i = |+0−i + |0 + −i − |−0+i − |0 − +i |2, −1i = 2 |00−i − |+ − −i − |− + −i − |0 − 0i − |−00i + 2 |− − +i |2, −2i = 2 |− − 0i − |0 − −i − |−0−i spin-2(2) : |2, 2i = |+0+i − |0 + +i |2, 1i = |+00i − |0 + 0i − |+ − +i + |− + +i |2, 0i = − |0 − +i + |−0+i − |+0−i + |0 + −i − 2 |+ − 0i + 2 |− + 0i |2, −1i = |+ − −i − |− + −i + |−00i − |0 − 0i |2, −2i = |0 − −i − |−0−i spin-1(1) : |1, 1i = 6 |+ + −i + 3 |0 + 0i + 3 |+00i + |+ − +i + |− + +i − 2 |00+i |1, 0i = 3 |0 + −i + 3 |+0−i + 3 |−0+i +3 |0 − +i − 2 |+ − 0i − 2 |− + 0i + 4 |000i |1, −1i = 2 |00−i − 3 |−00i − 3 |0 − 0i − |+ − −i − |− + −i − 6 |− − +i spin-1(2) : |1, 1i = |+00i − |0 + 0i + |+ − +i − |− + +i |1, 0i = |0 + −i − |+0−i + |0 − +i − |−0+i |1, −1i = |+ − −i − |− + −i + |0 − 0i − |−00i spin-1(3) : |1, 1i = |+ − +i + |− + +i + |00+i |1, 0i = |+ − 0i + |− + 0i + |000i |1, −1i = |+ − −i + |− + −i + |00−i

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spin-0 : |0, 0i = |+ − 0i + |−0+i + |0 + −i − |0 − +i − |+0−i − |− + 0i

(15.187)

Note that the spin-0 state is totally antisymmetric : obviously, this is the only possible such state in three space dimensions. We can also compute the ‘partial’ completeness relations pertaining to each spin sector. Some algebra teaches us that these are the following set of mutually orthogonal projection operators :

spin-3 :

3 X

|3, miµνρ h3, m|αβγ =

m=−3

1 6



∆µ α ∆ν β ∆ρ γ + ∆µ β ∆ν γ ∆ρ α + ∆µ γ ∆ν α ∆ρ β µ

ν

ρ

µ

ν

ρ

µ

ν

ρ



+ ∆ β ∆ α∆ γ + ∆ α∆ γ ∆ β + ∆ γ ∆ β ∆ α 
1 − ∆µν ∆ρ α ∆βγ + ∆ρ β ∆γα + ∆ρ γ ∆αβ 15
+ ∆νρ ∆µ α ∆βγ + ∆µ β ∆γα + ∆µ γ ∆αβ 
ρµ ν ν ν + ∆ ∆ α ∆βγ + ∆ β ∆γα + ∆ γ ∆αβ spin-2(1) :

2 X

|2, miµνρ h2, m|αβγ =

m=−2

1 3



µ

ν

ν

µ



∆ α ∆ β + ∆ α ∆ β ∆ρ γ  
ρ
ρ 1 µ ν ν µ µ ν ν µ ∆ β ∆ γ + ∆ β ∆ γ ∆ α + ∆ α∆ γ + ∆ α∆ γ ∆ β − 6   1 µν ρ ρ + ∆ ∆ α ∆βγ + ∆ β ∆αγ 6   1 µρ ν νρ µ + ∆ ∆ γ + ∆ ∆ γ ∆αβ 6   1 µρ ν µρ ν νρ µ νρ µ − ∆ ∆ α ∆βγ + ∆ ∆ β ∆αγ + ∆ ∆ α ∆βγ + ∆ ∆ β ∆αγ 12

488

August 31, 2019 1 − ∆µν ∆ρ γ ∆αβ 3 2 X spin-2(2) : |2, miµνρ h2, m|αβγ = m=−2

1 3



µ

ν

ν

µ



∆ α ∆ β − ∆ α ∆ β ∆ρ γ   1 µ ν ρ µ ν ρ ν µ ρ ν µ ρ + ∆ γ ∆ β ∆ α − ∆ γ ∆ α∆ β − ∆ γ ∆ β ∆ α + ∆ γ ∆ α∆ β 6   1 µρ ν µρ ν νρ µ νρ µ + ∆ ∆ α ∆βγ − ∆ ∆ β ∆αγ − ∆ ∆ α ∆βγ + ∆ ∆ β ∆αγ 4 1 X |1, miµνρ h1, m|αβγ = spin-1(1) : m=−1

1 µν ρ ∆ ∆ γ ∆αβ 15   1 µν ρ ρ − ∆ ∆ α ∆βγ + ∆ β ∆αγ 10   1 µρ ν νρ µ ∆ ∆ γ + ∆ ∆ γ ∆αβ − 10   3 µρ ν µρ ν νρ µ νρ µ ∆ ∆ α ∆βγ + ∆ ∆ β ∆αγ + ∆ ∆ α ∆βγ + ∆ ∆ β ∆αγ + 20 1 X |1, miµνρ h1, m|αβγ = spin-1(2) : m=−1

1 − 4

  µρ ν µρ ν νρ µ νρ µ ∆ ∆ α ∆βγ − ∆ ∆ β ∆αγ − ∆ ∆ α ∆βγ + ∆ ∆ β ∆αγ

spin-1(3) :

1 X

|1, miµνρ h1, m|αβγ =

m=−1

1 µν ρ ∆ ∆ γ ∆αβ 3 spin-0 : |0, 0iµνρ h0, 0|αβγ =  1 ∆µ α ∆ν β ∆ρ γ + ∆µ β ∆ν γ ∆ρ α + ∆µ γ ∆ν α ∆ρ β 6

August 31, 2019 ν

489 µ

ρ

−∆ α ∆ β ∆

γ

ν

µ

ρ

− ∆ β∆ γ∆

α

ν

µ

ρ

− ∆ γ ∆ α∆

 β

.

(15.188)

The total completeness relations is also valid : 3 s X X

|s, miµνρ hs, m|αβγ = ∆µ α ∆ν β ∆ρ γ ,

(15.189)

s=0 m=−s

provided we sum over all sectors with the same s.

15.12.5

Massless particles : surviving states

So far, we have taken our particles to be at rest, with a momentum p for which pµ = m tµ . For moving particles, we can obtain the correct states by simply performing the appropriate Lorentz boost. As already indicated, we shall take the motion of the particles to be along the z axis ; our states have been prepared for this by taking z as the spin quantization axis. The momentum of the particle will then be pµ = m tµ → pµ = p0 tµ + |~p| z µ , (15.190) and the vector z µ becomes, under the same boost z

µ

 →

|~p| m



µ

t +



p0 m



zµ .

(15.191)

The vectors x± are not affected by the boost. It is therefore sufficient to replace, in Eqns.(15.165),(15.182), (15.183),(15.184), and (15.187), z by its boosted form. Let us now consider the extreme case : that of a massless particle. We can view this as the limit p0 /m → ∞ of a massive particle. In that limit, z µ diverges badly, and we must again adopt the point of view presented in chapter 9 : the theory wil only be viable if those tensors that diverge in the massless limit decouple completely. That is, the only observable states must be those that do not diverge, i.e. those that contain x+ ’s and

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August 31, 2019

x− ’s but not any trace of a z. A quick inspection in our inventory of states reveals that only a handful of states are left : rank-1, rank-2, rank-2, rank-3,

15.12.6

spin-1 spin-2 spin-1 spin-3

: : : :

|1, 1i = |+i , |1, −1i = − |−i |2, 2i = |++i , |2, −2i = |−−i √ |1, 0i = (|+−i − |−+i)/ 2 |3, 3i = |+ + +i , |3, −3i = − |− − −i(15.192)

The Bermuda triangle

With the exception of the rank-2, spin-1 state, the so-called Kalb-Ramond state, all the surviving states have m = ±s and are totally symmetric. Is this general ? In other words, how do we know that there is no rank-31, spin-17 state that is built up from only x+ ’s and x− ’s ? We can answer this question by the following pleasing argument. Since the ladder operators Σ± transform physical states into one another, any physical state must be an eigenstate√of Σ+ Σ− or Σ− Σ+ 23 . Disregarding, for simplicity, minus signs and factors 2, the effect of Σ+ is 0 → +, − → 0, and that of Σ− is + → 0, 0 → −. We can therefore write Σ+ Σ− |+−i → Σ+ |0−i → |+−i + |00i .

(15.193)

Let us now consider a hypothetical massless-particle candidate state. It will be a linear combination of kets with lots of +’s and −’s. Among these we concentrate on three kets in particular : T1 = |· · · + + − · · ·i , T2 = |· · · + − + · · ·i , T3 = |· · · − + + · · ·i . (15.194) The rest of the content of the kets (indicated by the ellipses, and consisting of some sequences of +’s and −’s) is identical for the three kets. The candidate state contains these T ’s in some linear combination : C1 T1 + C2 T2 + C3 T3 + lots of other terms Let us now consider what happens if we let Σ+ Σ− work on these kets. T1 will turn into a lot of terms, among which we can recognize two important 23

It is of course possible that Σ+ acting on our state, say, will give zero, and then it is an eigenstate of Σ− Σ+ with eigenvalue zero. We may avoid this trivial case by choosing, instead, Σ+ Σ− , under which our state will have a nonzero eigenvalue.

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ones : T1 → |· · · + 00 · · ·i + |· · · 0 + 0 · · ·i + · · · .

(15.195)

Similarly, we find for T2 and T3 : T2 → |· · · 00 + · · ·i + |· · · + 00 · · ·i + · · · , T3 → |· · · 00 + · · ·i + |· · · 0 + 0 · · ·i + · · · .

(15.196)

We now note a few things. In the first place, a resulting ket like |· · · 0 + 0 · · ·i can only come from the T ’s (in this case, from T1 and T3 ). In the second place, our candidate state cannot contain this ket by itself, since it must be free of 0’s. In the third place, such unwanted kets must drop out because our state is an eigenstate of Σ2 . We must therefore rely on cancellations between the T ’s. In fact, we need simultaneously C1 = −C2 ,

C2 = −C3 ,

C3 = −C1 .

(15.197)

Obviously, C1,2,3 = 0 : our three T ’s do not occur at all24 ! But of course we can repeat the same argument for any other such three kets. We see that the only possibilities to have admissible massless-particle states are twofold: • Only +’s, or only −’s, occur. These are precisely the rank-s, spin-s states such as we have found, and this persists also for s > 3. Note that these states are totally symmetric — not for some deep field-theoretical reason, but because they can’t help it. • Precisely one + and one − occur. This is the Kalb-Ramond state, which now stands revealed as a lone exception.

15.12.7

Massless propagators

For massless states, the spin sums cannot be built up from objects like ∆µ α since these diverge. An often-used recipe is the following. For a massless particle of momentum pµ , define pµ = (p0 , p~) , p¯µ = (p0 , −~p) . 24

(15.198)

A three-cornered argument such as this, in which all T ’s disappear, deserves to be called a Bermuda triangle.

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August 31, 2019

Obviously, this is not a Lorentz-invariant definition, but as we shall see that is not a problem. The point is that a p¯ can be found whatever the Lorentz frame is. We can now write |+iµ h+|ν + |−iµ h−|ν = x+ µ x− ν + x− µ x+ ν   1 µ µ = p p¯ν + p¯ pν − δ µ ν ≡ ∇µ ν . p · p¯

(15.199)

In analogy to Eq.(15.170) we now have ∇µα ∇αν = −∇µ ν , ∇µ µ = −2 .

(15.200)

If, as we must promise ourselves, massless states only couple to conserved sources (on which the handlebar operation gives zero), the terms containing p¯ will always drop out. We can now write the spin sums for the surviving massless states as follows : rank-1, spin-1 : rank-2, spin-2 : rank-2, spin-1 : rank-3, spin-3 :

∇µ α ,   1 1 µ ν µ ν ∇ α ∇ β + ∇ β ∇ α − ∇µν ∇αβ , 2 2   1 ∇µ α ∇ν β − ∇µ β ∇ν α , 2  1 ∇µ α ∇ν β ∇ρ γ + ∇µ β ∇ν γ ∇ρ α + ∇µ γ ∇ν α ∇ρ β 6 µ

ν

ρ

µ

ν

ρ

µ

ν

ρ



+ ∇ β ∇ α∇ γ + ∇ α∇ γ ∇ β + ∇ γ ∇ β ∇ α 
1 ∇µν ∇ρ α ∇βγ + ∇ρ β ∇γα + ∇ρ γ ∇αβ − 12
+ ∇νρ ∇µ α ∇βγ + ∇µ β ∇γα + ∇µ γ ∇αβ 
ρµ ν ν ν + ∇ ∇ α ∇βγ + ∇ β ∇γα + ∇ γ ∇αβ (15.201) Compared to the massive case, some coefficients are different : -1/2 rather than -1/3 in the spin-2 case, and -1/12 instead of -1/15 for spin-3. This is due, of course, to the different traces of ∆ and ∇. The spin sum for the massless vector particle (rank-1, spin-1) is in fact that of the axial gauge discussed in Chapter 9, with the gauge vector r chosen to be p¯. Note that, whatever rµ , we can always move to the centre-of-mass frame of pµ and rµ , and in that frame we have precisely rµ = p¯µ .

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15.12.8

493

Spin of the Kalb-Ramond state

Concerning the Kalb-Ramond (KR) state, there may be some controversy. For a massless particle in this state, the spin along the axis of motion must, under measurement, always come out zero. It is not easy to see how such a particle can be distinguished from a scalar one. Indeed, in string theory where the KR state comes up naturally, it is considered to describe a (pseudo)scalar particle called the axion. In order to talk sensibly about the spin of the KR state it is useful to consider how it may be measured, for instance using fermions. We therefore consider the coupling of a rank-2, spin-1 state to fermions. The interaction vertex must have the properties that (a) it is an antisymmetric rank-2 tensor, and (b) it is current-conserving, in order to make sense in the massless limit. Denoting the two fermions by ψ and ψ the simplest choice appears to be ψ µνρσ pρ (A + Bγ 5 ) γσ ψ where p is the momentum of the antisymmetric tensor state, and A and B are constants. This interaction vertex vanishes trivially under the handlebar operation. For the process f¯(p1 ) f (p2 )

→

f (p3 ) f¯(p4 )

by the exchange of a KR state of mass M , we then have the amplitude M = i~ v(p1 ) µνρσ pρ (A + Bγ 5 ) γσ u(p2 ) ∆µα ∆νβ − ∆µβ ∆να × 2(s − M 2 ) × u(p3 ) αβκλ pκ (A0 + B 0 γ 5 ) γλ v(p4 ) , s = p · p , p = p1 + p2 = p 3 + p4 .

(15.202)

Because of the current conservation and the antisymmetry of the vertices, we may replace ∆µα ∆νβ − ∆µβ ∆να by 2gµα gνβ . Furthermore, since
µνρσ pρ µν κλ pκ = 2 pσ pλ − s g σλ (15.203) we have M = −2i~

1 s − M2

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August 31, 2019 
v(p1 ) A(m2 − m1 ) + B(m1 + m2 )γ 5 u(p2 ) 0

0

× u(p3 ) A (m3 − m4 ) − B (m3 + m4 )γ 
−s v(p1 ) A + Bγ 5 γ µ u(p2 ) 
0 0 5 × u(p3 ) A + B γ γµ v(p4 ) .

5




 v(p4 )

(15.204)

Here mj is the mass of momentum pj . Note that, in contrast to e.g. the case of QED, m1 = m2 or m3 = m4 is not necessary for current conservation. We can now investigate several situations. In the first place, if M 6= 0 the amplitude has a pole for some nonzero s value, which we may take as the signal of a particle. The second term in brackets in Eq.(15.204) then tells us that, indeed, a spin-1 particle has been exchanged25 . The occurrence of the first term is, then, not surprising : a similar contribution is found in e.g. the W exchange in muon decay. Secondly, we may take M = 0. In that case, the second term no longer has a pole. It can therefore not survive a truncation argument, and must not be counted as coming from any particle propagation. The first term does survive ; if we also assume flavour conservation so that m1 = m2 and m3 = m4 , the only degree of freedom that propagates is, indeed, that of a pseudoscalar.

15.12.9

Spin 1 from Dirac particles

We can, of course, also use Dirac spinors to build spinning states. As before, m is the particle mass, and pµ the particle momentum ; there is no particular advantage to take it at rest here. The spin vector is sµ and p2 = m2 , s2 = −1, (ps) = 0. We shall use the standard representation as in sec.8.1.5, and use 1 (15.205) k0µ = pµ − sµ m so that (pk0 ) = m. Thus, 1 1 u± = u(p, ±s) = √ (/p+m)u∓ (k0 ) , v± = v(p, ±s) = √ (/p−m)u± (k0 ) . 2m 2m (15.206) 25

We can measure this, for instance by looking at the angular distribution of the produced fermion-antifermion pair ; see also Appendix ??.

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We can easily check that γ 5 u+ = v− and γ 5 u− = −v+ , which is handy since it allows us to write everything using only the u’s if we want. We choose to work in a Lorentz frame where p~ k ~s : then these spinors are pure helicity states. Let us now consider the object eµ+ = √

1 8m2

v + γ µ u+ .

(15.207)

We see immediately that e+ · s ∝ v + s/u+ = v + γ 5 u+ = v + u− = 0 . (15.208) The Chisholm and reversal identities for e+ are
(15.209) e/+ = v + γ µ u+ gµ = 2 (u+ v + − v− u− ) , eµ+ = −u− γ µ v− . e+ · p ∝ v + p/u+ ∝ v + u+ = 0 ,

From the first of these we find immediately e+ · e+ = 0 ,

e+ · e∗+ = −1 .

(15.210)

Therefore, eµ+ is an object with spin 1 in the direction of p~. We can now lower the spin, using the lowering operator u+ → u− , v + → v − . This gives us the spin-0 part if the triplet : 1 (v − γ µ u+ + v + γ µ u− ) 4m
1 = −u+ γ 5 γ µ u+ + u− γ 5 γ µ u− = sµ , 4m

eµ0 =

(15.211)

exactly what you would expect for the spin-0 part. A final lowering gives us eµ− = √

1 8m2

v − γ µ u− = − (eµ+ )∗ ,

(15.212)

which yields the part with spin minus one. The spin sum is therefore X µ 1 (15.213) ej (eνj )∗ = −g µν + 2 pµ pν . m j=+,0,− So now we have constructed three spin-1 states using two spin-1/2 states. There ought to be another, spin-0 state as well. Indeed there is, somewhat trivially : it is the state −1 1 v − γ 5 γ µ u+ = u+ γ µ u+ = pµ , 2m 2m

(15.214)

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August 31, 2019

the only object with one Lorentz index orthogonal to the eµ ’s. That it has zero spin can be seen by lowering26 which turns v − γ 5 γ µ u+ into v − γ 5 γ µ u− = −u+ γ µ u− = 0.

15.12.10

Spin 3/2 particles

Adding one more Dirac spinor we can describe spin-3/2 particles. These states have one Dirac index and one Lorentz index. The highest spin resides in the state 1 uµ3/2 = √ (15.215) u+ v + γ µ u+ = u+ eµ+ . 2 8m That this state has the properties of a spin-1/2 as well as of a spin-1 particle can be seen from (/p − m) uµ3/2 = 0 ,

pµ uµ3/2 = 0 .

(15.216)

There is one more property : again using Eq.(15.209) we see that also γµ uµ3/2 = 0 .

(15.217)

The properties (15.216) and (15.217) are the so-called Rarita-Schwinger conditions. Now, repeated application of the lowering operator (and normalization) give us, successively : uµ1/2 = √ = uµ−1/2 = = uµ−3/2 =

1


u− v + γ µ u+ + u+ v − γ µ u+ + u+ v + γ µ u− 24m2  √ 1  √ u− eµ+ + 2u+ eµ0 , 3
1 √ u− v − γ µ u+ u− v + γ µ u− + u+ v − γ µ u− 24m2  √ 1  √ u+ eµ− + 2u− eµ0 , 3 1 √ (15.218) u− v − γ µ u− = u− eµ− . 2 8m

These three states also obey the Rarita-Schwinger conditions. Getting the spin sum is now less trivial than before. It helps to introduce the shorthand [αρβστ ] = uα uα γ 5 γ ρ uβ uβ γ 5 γ σ uτ uτ , 26

And, conversely, also by raising.

α, β, τ = ±

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497

We can then write 3/2 X

uµj uνj =

j=−3/2

+

1 8m2

[+ν − µ+] + [−ν + µ−]

1 24m2




[−µ + ν−] + [+µ − ν+] − 2[+µ + ν−] − 2[−µ − ν+] − 2[+µ − ν−] − 2[−µ + ν+]
+4[+µ + ν+] + 4[−µ − ν−] .

(15.219)

With some careful algebra we can turn this into 3/2 X j=−3/2

uµj uνj

  1 µ ν µν = (/p + m) −g + 2 p p m −

1 (/p + m)γ 5 γ µ (/p + m)γ 5 γ ν (/p + m) . (15.220) 2 12m

This is an interesting object ! Each of the two terms obeys the first two Rarita-Schwinger conditions separately, but for the third one (with the γµ ) we need a special combination of the two. In fact, Eq.(15.220) is the only combination with up to three p’s that obeys the Rarita-Schwinger conditions ‘from the left’ and ‘from the right’. The first term is what you would probably write down as your first guess for ‘a product of a spin-1/2 and a spin-1 state’, but that has 2 × 3 degrees of freedom : the second term kills off the two unwanted degrees of freedom. And yes, our three-Dirac world contains of course also a spin-1/2 sector : the two states  1 √ µ µ 2 u+ e− − u− e0 , , =√ 3 (15.221) µ are orthogonal to the four u ’s. Now, note that µ w1/2

 1 √ µ µ 2 u− e+ − u+ e0 =√ 3

√

µ w−1/2

1 (u− v + γ µ u+ − u+ v − γ µ u+ ) 2m
1 u− u− γ 5 γ µ u+ + u+ u+ γ 5 γ µ u+ = 2m 1 = (/p + m)γ 5 γ µ u+ , (15.222) 2m

2 u− eµ+ − u+ eµ0 =

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August 31, 2019

so, indeed X

wjµ wνj =

j=±1/2

1 (/p + m)γ 5 γ µ (/p + m)γ 5 γ ν (/p + m) . 12m2

(15.223)

The total spin sum over both sectors is therefore 3/2 X

uµj uνj

j=−3/2

+

1/2 X

wjµ wνj

j=−1/2

  1 µ ν µν = (/p + m) −g + 2 p p . m

(15.224)

Like before, we should be careful in the massless limit, in which sµ ∼ pµ blows up. Any of the Dirac-built states that contains e0 must decouple in that limit ; on the other hand, the spinors u± and v± themselves remain perfectly finite, as discussed in sec.7.4.6. We again conclude that massless spin-3/2 particles can only be seen in the two states of maximal helicity, uµ3/2 and uµ−3/2 .

15.12.11

The spin algebra for Dirac particles

In the above we have simply stated that the spinors u+ and v + correspond to positive helicity, and u+ andv − to begative helicity ; but we ought to be a bit more formal, by construviting the correct algebra. The lowering operator in this algebra, M− , must be such that M− u+ = ~ u− , v + M− = ~ v − .

(15.225)

This implies that, here, the lowering and raising operators must be M− =

~ (u− u+ − v+ v − ) , 2m

M+ = M − =

~ (u+ u− − v− v + ) . 2m (15.226)

The spin operator in the s direction is then Ms =

1 ~ (M+ M− − M− M+ ) = p/ γ 5 s/ . 2~ 2m

(15.227)

Since Ms M+ = −M+ Ms = −~M+ /2 we also have [Ms , M+ ] = ~M+ as we ought to. And the operator for total spin, finally, is M2 =

1 3~2 (M+ M− + M− M+ ) + Ms 2 = . 2 4

(15.228)

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499

So everything appears to be fine. However, looking at the spin of the various spinors, we find ~ Ms u± = ± u± , 2

~ Ms v± = ∓ v± , 2

(15.229)

so the spinor v± would appear to have the ‘wrong’ spin component. However, we employ v rather than v itself, and the transition v → v involves time reversal27 . Under time reversal, axial vectors like sµ change sign. Therefore the apparent problem disappears if you realise what the spin means for these particles. At any rate, the fact that eµ+ , say, is an honest spin-1 object justifies our spin assignments.

15.13

Generating three-particle kinematics

Here we describe how to do Monte Carlo surveys of three-particle phase space, as used in section 13.4.5. The process is that of two incoming particles with momenta pµ1,2 and masses M1,2 , and three outgoing particles with momenta µ q1,2,3 and masses m1,2,3 , respectively. This is more general than the discussion of three-body phase space in muon decay (section 8.3.2). The total energy is w, where w2 = (p1 + p2 )2 . We shall consistently work in the centre-of-mass frame, where p~1 + p~2 = ~q1 + ~q2 + ~q3 = 0. In terms of two of the energies, q10 and q20 , say, the phase space integration element is still constant : rather it is the boundaries that become very nontrivial. From (p1 + p2 − q3 )2 = (q1 + q2 )2 we find w2 − 2wq30 + m23 = 2w(q10 + q20 ) − w2 + m23 = m21 + m22 − 2q10 q20 + 2|~q1 ||~q2 | cos θ , (15.230) where θ is the angle between ~q1 and ~q2 . The allowed phase space points therefore obey 2w(q10 + q20 ) − w2 + 2q10 q20 + m23 − m21 − m22 ≤ 2|~q1 ||~q2 | . (15.231) The actual boundary, at which the inequality (15.231) becomes an equality, is 0 of multinomial type in q1,2 with powers up to (q10 q20 )2 , hence quite horrible28 . 27 28

Since v describes an outgoing antifermion, and v an incoming antifermion. The actual volume of phase space is only known for some special cases.

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August 31, 2019

What we do know are the lower and upper limits on the energies :
1 w2 + m2j − (mk + m` )2 = qjmax , {j, k, `} = {1, 2, 3} . 2w (15.232) Armed with this knowledge we can use (pseudo-)random numbers ρ, uniformly and independently distributed in the interval [0, 1], to
sample the final-state momenta29 . First, we repeat qj0 ← mj + ρ qjmax − mj for j = 1, 2 until Eq.(15.231) is satisfied : typically, one has to try only a few times. This gives us |~q1 | and |~q2 |, and cos(θ) from Eq.(15.230). We then put mj ≤ qj0 ≤

~q1 ← (0, 0, |~q1 |) , ~q2 ← (0, |~q2 | sin(θ), |~q2 | cos(θ)) , ~q3 ← −~q1 − ~q2 . This gives us the final-state momenta, sampled uniformly. For the initialstate momenta we generate a uniformly distributed solid angle by cos(η) ← −1 + 2ρ ,

φ ← 2π ρ

and the two energies by p01 =


1 w2 + M12 − M22 2w

,

p02 = w − p01

and this allows us to construct the incoming momenta : p~1 ← (|~p1 | sin(η) sin(φ), |~p1 | sin(η) cos(φ), |~p1 | cos(η)) , p~2 ← −~p1 The fact that ~q1 always points in the positive z direction is not a problem in any theory that respects angular momentum conservation, since the incoming momenta will take on all directions. You might want to have the direction of p~1 fixed instead : in that case a global rotation on all the momenta will do the trick.

15.14

The CPT theorem

In this appendix, we shall discuss the very fundamental CPT theorem30 for theories with interacting particles. This theorem deals with what happens (or 29

Everytime we make a reference to ρ it is to be understood that a new random number is picked. Therefore, a = ρ and b = ρ does not imply a = b ! 30 Also known as the CTP theorem, the TCP theorem, the TPC theorem, the PTC theorem, or the PCT theorem.

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501

ought to happen) to scattering amplitudes when we relate various physical scattering processes31 . As usual, we shall start by looking at Dirac particles.

15.14.1

Transforming spinors

In Chapter 7, we defined the standard form for the various spinors corresponding to an on-shell (anti)-particle with mass m and momentum pµ . We recapitulate them here : u± (p) = N (p) (/p + m) u∓ (k0 ) , v± (p) = N (p) (/p − m) u∓ (k0 ) , u+ (k0 ) = k/1 u− (k0 ) , u− (k0 )u− (k0 ) = ω− k/0 , p N (p) = 1/ 2(pk0 ) , k0 2 = (k0 k1 ) = 0 , k1 2 = −1 . (15.233) This is, of course, only a phase convention, where the phase choice is not explicit but implied by the choice of k0 , k1 and the complex phase of u− (k0 ). Now, let us apply γ 5 to these states. It is easy to see that γ 5 u+ (p) = v+ (p) , γ 5 u− (p) = − v− (p) , u+ (p) γ 5 = −v + (p) , u− (p) γ 5 = v − (p) .

(15.234)

In words, what this transformation does is to change an incoming, right(left)handed fermion into an outgoing, left(right)-handed antifermion (and vice versa). Thus we have (a) the interchange of particle and anti-particle (charge conjugation, C), (b) the interchange of right- and left-handedness32 (parity inversion, P), and (c) the interchange of initial and final state (time reversal T), which goes by the name of CPT transformation33 . Applied to Feynman diagrams, we can depict this as follows (where we have indicated the 31

Recall that, in these notes, we concentrate on the (perturbative)processes that are going on, that is, scattering described by diagrams and amplitudes. 32 Recall that for a particle + means right-handed, but for an antiparticle it means left-handed (cf section 8.1.5) 33 There is a slight subtlety here. An ingoing particle with three-momentum p~ is transformed into an outgoing antiparticle with the same momentum p~. Under P, momenta are inverted so that p~ becomes −~ p : but under T the velocities are again inverted. The same holds, of course, for spin vectors. It is only the fact that ”+” means right-handed for particles and left-handed for antiparticles that ensures that the net result is just a change of handedness.

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helicity) : +

→

+

→ −

+

, +

→ −

−

,

−

→

− −

.

(15.235)

As we can see, the effect of CPT on any diagram is not only to interchange initial and final states, but also to reverse the arrows on intermediate fermion lines.

15.14.2

CPT transformation on sandwiches

Let us consider the scalar current for two on-shell momenta p1,2 , with respective masses m1,2 : Jλ1 λ2 = uλ1 (p1 )uλ2 (p2 ) . (15.236) Under CPT, this scalar current behaves as follows : J++ → Jˆ++ = −v + (p1 )v+ (p2 ) , J+− → Jˆ+− = v + (p1 )v− (p2 ) .

(15.237)

At first sight, these CPT transforms look nothing like the original. Note, however, that using the standard form we can write them as traces : J++ = N (p1 )N (p2 ) Tr (ω− k/0 (/p1 + m1 )(/p2 + m2 )) , J+− = N (p1 )N (p2 ) Tr (ω− k/0 (/p1 + m1 )(/p2 + m2 )/k1 ) ,

(15.238)

Jˆ++ = − N (p1 )N (p2 ) Tr (ω− k/0 (/p1 − m1 )(/p2 − m2 )) , Jˆ+− = N (p1 )N (p2 ) Tr (ω− k/0 (/p1 − m1 )(/p2 − m2 )/k1 ) ,

(15.239)

whereas

Keeping track of which terms in these traces actually survive34 , we see that, appearances notwithstanding, Jˆλ1 λ2 = Jλ1 λ2 . 34

(15.240)

For J±± , these are the terms that contain an odd number of masses, for J±∓ those with even numbers of masses survive.

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Similar (almost trivial) trace arguments show that, under CPT, Jλ1 λ2 µ = uλ1 (p1 ) γ µ uλ2 (p2 ) → − Jλ1 λ2 µ , Jλ1 λ2 µν = uλ1 (p1 ) γ µ γ ν uλ2 (p2 ) → + Jλ1 λ2 µν , Jλ1 λ2 µνα = uλ1 (p1 ) γ µ γ ν γ α uλ2 (p2 ) → − Jλ1 λ2 µνα ,

(15.241)

and so on.

15.14.3

CPT transformation on diagrams

Consider a nontrivial but very simple diagram, for simplicity taken from the electroweak process e− (p1 )γ(q1 ) → e− (p2 )Z 0 (q2 ) : D=

(15.242)

Leaving out overall constants and denominators, this can be written as D = Aµν ¯µλZ (q2 ) λγ ν (q1 ) , Aµν = uλ2 (p2 ) ωγµ (/q + m) γν uλ1 (p1 ) , q = p1 + q1 = p2 + q2 , ω = gv + ga γ 5 ,

(15.243)

where we have indicated the handedness (helicity) of the external particles. For the polarisation vectors we take the representation given in Eq.(9.32), and for q/ we may, if we wish, use Eq.(??) to write 1 q/ = γα u+ (q)γ α u+ (q) . 2

(15.244)

Let us now see what happens if we apply CPT. In the first place, q/ → − q/ ,

(15.245)

following immediately from Eq.(15.244)35 Therefore, Aµν transforms as Aµν → −λ1 λ2 v λ1 (p2 ) ωγµ (−/q + m) γν vλ2 (p1 ) . 35

(15.246)

Antoher approach might be to find a set of timelike, positive-energy P momenta α k1,2,3,... with masses m , and a set of constants c such that = 1,2,3,... 1,2,3,... j cj kj P α q and Pj cj mj = m. Obviously, this is always possible. We can then write qP+ m = j cj (u+ (kj )u+ (kj ) + u− (kj )u− (kj )), which under CPT are transformed into / q + m. j cj (−v+ (kj )v + (kj ) − v− (kj )v − (kj )) = −/

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The arguments given in the previous section show that this evaluates again to Aµν itself. Finally, for the polarisation vectors we have, for instance, λγ ν → − λγ ν = − ¯ν−λγ ,

(15.247)

so that the CPT transform of an incoming, left(right)-handed photon can be interpreted as that of an outgoing, right(left)-handed photon with the same momentum, up to an overall minus sign. The same goes of course for λZ µ . We see that, under CPT, the amplitude M remains unchanged36 : but the interpretation is now that of the process e+ (p2 )Z 0 (q2 ) → e+ (p1 )γ(q1 ), with the understanding that left(right)-handed particles have been replaced by right(left)-handed ones. The corresponding Feynman diagram is now DCPT =

,

(15.248)

which may help you to understand the replacing of q/ by −/q : in diagrammatic terms, it comes from the fact that now q runs against the propagator’s arrow. It is now easy to see that we can perform similar operations on every conceivable Feynman diagram in our theory37 , and we shall always find that it transforms into itself. We say that our theory is CPT-invariant : if we (a) replace every external particle by its antiparticle (and vice versa), (b) interchange the initial and final states, and (c) interchange right- and lefthanded, then all amplitudes remain the same. This is the CPT theorem.

15.14.4

How to kill CPT, and what it costs

Like all such theorems, the CPT theorem can only be valid under a number of circumstances. Here, we mention the most important of these. In the first place, comparing the diagrams (15.242) and (15.248) we see that we have implicitly assumed that the vertices of the theory are insensitive to what is the ‘incoming’, and what the ‘outgoing’ particle : for instance, the 36

You might think that the fact that the two minus signs coming from the polarisation vector cancel so nicely is suspicious : but you should realize that if three external bosons were involved there would be two internal fermion propagators instead of one. 37 If push comes to shove, we can always write every vector quantity in the diagram with spinors : we then end up with a massively complicated object containing loads of (anti)spinors and their conjugates, but for the rest only fixed numbers or matrices ; for such structures, we have already proven everything that is needed.

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two vertices

µ

µ

and

_ e

e+

µ

are both assigned the value iQγ /~. More poignantly, in the electroweak sector we use the same vertex for W

U

+

W

and D

D

_

U

It is, of course, possible to let the vertex depend on the ‘orientation’ of the (sub)process : such theories, which as we see are not easily expressed diagrammatically38 , are called non-Hermitian. A non-Hermitian action would ruin CPT. In the second place, and more subtly, we have assumed that there is, at least, the very possibility of a vacuum state through which particles can move ; in the literature, this means that there is a state with lowest energy. If the spectrum of the theory is not bounded from below, CPT is ruined : but, again, it is not easy to see how any ordinary particle physics could be alive under such circumstances39 , whether CPT invariant or not. In the last place, there is the issue of Lorentz invariance. We have assumed that every vector hµ will, under CPT, turn into −hµ , and this is very important for proving the CPT invariance of amplitudes. Suppose, now, that we introduce into our theory a fixed vector40 f µ , simply a set of four universally defined41 numbers which enter nontrivially into the Feynman rules. Such a vector would, under CPT, not turn into its opposite ; but neither would it change under Lorentz transformations, it would simply remain f µ . CPT would be ruined together with Lorentz invariance. A theory violating CPT will therefore manifest itself in being Lorentz-noninvariant. You might hope to avoid this by having, built into the fabric of the universe, some physically meaningful vector quantity f µ , that does change with Lorentz transformations42 . Still, CPT would be ruined, but we must also conclude 38

At least in the way we have formulated things. In these notes, we take the existence of particles with a perturbative description for granted. 40 We speak of a ‘vector’ here in the sense that it has four components, not in the sense of its behavior under coordinate transformations : indeed, the whole point is that it doesn’t transform at all. 41 Think of having some inspiration, or a voice from heaven engraving these numbers on stone tablets. 42 Such a thing would be, for instance, the ‘momentum of the æther’. 39

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that the ‘vacuum’ state is itself simply not Lorentz invariant since there is a ‘preferred momentum’. Note that it is, in principle, possible to violate Lorentz invariance without destroying CPT. For instance we can use a fixed ‘tensor’ f µν rather than a vector f µ . Such a tensor does not change sign under CPT, exactly as it should. We can then construct theories where Lorentz invariance is violated but CPT invariance is not43 . We see that the conditions under which CPT symmetry holds are very plausible and general, but they are not unavoidable. CPT may be ruined, but we can see that by the concomitant violation of Lorentz invariance, either in the interactions of the theory or in the structure of the vacuum itself !

15.15

Mathematical Miscellanies

In this section some collected mathematical issues are discussed which are usueful in the main text, or maybe just of some interest.

15.15.1

The Gaussian doubling trick

The Gaussian integral Z∞ G=

dx exp −x2




(15.249)

−∞

is not easily computed in the standard manner. However, there is a ‘once seen, never forgotten’ way of doing it, by doubling the integral and going over to polar variables : 2

G

Z∞

2

dx dy exp − x + y

=

2





=

−∞

= 2π 0 43

dφ 0

Z∞

dr r exp −r2




Z∞

Z2π

dr r exp −r2




0

Z∞ = π

ds exp(−s) = π . (15.250) 0

As an example, we can use, for the kinetic part of a Lagrangian, the object f µν ∂µ ϕ ∂ν ϕ rather than the usual g µν ∂µ ϕ ∂ν ϕ.

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507

Thus we find   Z √ x2 dx exp − 2 = σ 2π , 2σ

Z

x2 dx exp ±i 2 2σ 



√ = σ π(1 ± i) , (15.251) 44

where the last result is obtained by analytic continuation .

15.15.2

Stirling’s approximation for n!

The following discussion will bring us right back to the start of these notes. The factorial function is defined by the integral Z∞ n! ≡ dz z n e−z (n = 0, 1, 2, 3, . . .) (15.252) 0

We can write the integrand as z n e−z = exp (n log(n) − n + n log (1 − ϕ) + nϕ) ,

(15.253)

where z = n(1 − ϕ). We recognise a path-integral-like expression : n! = nn+1 e−n

Z1

dϕ e−S(ϕ) ,

−∞

n 2 n 3 n 4 n 5 ϕ + ϕ + ϕ + ϕ + ··· , (15.254) S(ϕ) = 2 3 4 5 that is, a zero-dimensional theory with µ = n and λk = −n (k − 1)!. We find the following asymptotic result : ! X D`+1 √ n! ≈ 2π nn+1/2 e−n 1 + , (15.255) n` `≥1 where D`+1 /n` is the sum of the (` + 1)-loop vacuum diagrams for the above action, including the disconnected ones. We can make life simpler by concentrating only on the sums of the connected vacuum diagrams, C`+1 /n` , and then we write X C`+1 1 . (15.256) log(n!) ≈ (n + 1/2) log(n) − n + log(2π) + ` 2 n `≥1 44

And inspection of cos(x2 ) and sin(x2 ) which shows that these both integrate to positive numbers.

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Curiously, Ck vanishes for all odd values of k ! This is nontrivial since, for instance, the C3 terms arise from 15 three-loop diagrams. The first few nonzero terms are X C`+1 1 1 1 1 1 − ≈ + − + ` 3 5 7 n 12 n 360 n 1260 n 1680 n 1188 n9 `≥1 691 1 3617 + − + · · · (15.257) 11 13 360360 n 156 n 122400 n15 The terms alternate in sign, which is quite usual for a summable asymptotic expansion of this type. It is worth noting that the action (15.254) is actually nothing but the ‘wonderful action’ of exercise 13. There we prove that the Green’s functions are completely free of loop corrections beyond the first order. In this section, however, we deal with the vacuum diagrams ; there is therefore no contradiction. Of course no-one has actually computed C16 , say, by evaluating all 16-loop diagrams. Rather we can take the more primitive perturbative idea, by expanding the exponent of the interactions terms in powers of ϕ, dropping the odd powers of ϕ, and substituting for ϕp the Gaussian result (p − 1)!!/np/2 . The above discussion is rather intended to show how the asymptotic part of the Stirling approximation is, in fact, closely related to Feynman diagrammatics. −

15.15.3

Manipulating asymptotic series

Comparing asymptotic series In spite of the problematic convergence behaviour of asymptotic sums, we can still obtain some surprising insights. This is the field of resurgence theory, to which we do not full justice here ; but a relatively crude and simple approach is enough for our Let us consider asymptotic sums, f (x) = P P purpose. n n x f and g(x) = x g , and so on. We shall take f0 = g0 = 1. The n n n≥0 n≥0 coefficients fn and gn are growing superexponentially with n : in particular, we shall assume that fn behaves, for very large n, as fn ≈ af cnf Γ(n + βf ) (n → ∞) ,

(15.258)

and similar for g(x) and so on. Let us define some comparison between the sums : we shall say that fn =0 , n→∞ gn

f (x) ≺ g(x) if lim

fn =1 , n→∞ gn

f (x) ∼ g(x) if lim

(15.259)

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509

and f (x)  g(x) if g(x) ≺ f (x). Obviously, f (x)+g(x) ∼ f (x) if f (x)  g(x). Let us now consider the product f (x)g(x) : n X X f (x)g(x) = xn fk gn−k . (15.260) n≥0

k=0

Because the coefficients grow so fast, the sum over k is (for n → ∞) completely dominated by gn + fn , and we conclude that f (x)g(x) ∼ g(x) + f (x) .

(15.261)

For instance, we find 1/f (x) ∼ −f (x). Also, for finite p, f (x)p ∼ pf (x). Asymptotic pullbacks Consider two asymptotic sums f (x) and g(x) as above, and assume that f (x)  g(x). We want to study the asymptotic form of the pullback f (x g(x)), that is, an asymptotic sum of powers of an asymptotic sum ; at first sight this would seem hopelessly tremendous. However, we can write45 X f (x(g(x)) = fn xn g(x)n n≥0

   2 n 2 g1 + ng2 = x fn + fn−1 (ng1 ) + fn−2 2! n≥0  3   n 3 2 +fn−3 g1 + n g1 g2 + ng3 + · · · (15.262) 3! X

n

Discarding the 1/n corrections, the leading behaviour of this sum is n X X nk (15.263) f (xg(x)) ∼ xn fn−k g1 k , k! n k=0 and, for very large n, we have fn−k nk ≈ af cfn−k Γ(n + βf ), so that  k n X X 1 g1 n f (x(g(x)) ∼ fn x ∼ f (x) exp(g1 /cf ) . k! c f n k=0

(15.264)

Here, it is important that f  g to ensure that only the powers of g1 contribute in the asymptotic regime, and that the sum over k is dominated by only the lower k values. 45

Here we introduce the falling powers : ns ≡ n(n − 1) · · · (n − s + 1) = n!/(n − s)!. For asymptotically large n, we can approximate ns ≈ ns up to 1/n corrections.

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Asymptotic inversion The equation xf (x) = y can (at least formally) be inverted to give x as a function of y. We can write xf (x) = y

⇒

x = y+x(1−f (x)) = y−f1 x2 h(x) ,

h(x) =

X fn+1 n≥0

f1

xn .

(15.265) As explained in appendix 15.15.12, the Lagrange expansion is basically just the iteration of this equation, and we can (formally) write X 1  d n−1
n −f1 y 2 h(y) x = y+ n! dy n≥1  n−1 X X 1 d fm+1 m+2n n ∼ (−f1 ) y (n − 1)! dy f 1 n≥1 m≥0 XX 1 (−f1 )n−1 fm+1 mn−1 y m+n+1 ∼ − (n − 1)! n≥1 m≥0  n−1 k XX 1 f1 k fk y ∼ −y − (n − 1)! cf k≥0 n=1 ∼ −yf (y) exp(−f1 /cf ) .

(15.266)

That is, y = xf (x) = y(x) and x = x(y) are asymptotically equal up to an overall factor. Here it is important to note that for large k = m + n the sum over n is dominated by the smaller values of n (as we can see), so that the approximation h(x)n ∼ nh(x) is justified. A case study: ϕ4 theory in zero dimensions As an example we will study the ‘renormalization improvement’ discussed in section 2.4. The zero-dimensional path integral reads, after the perturbation expansion (with λ = u, and putting m = ~ = 1 for simplicity),  X  Z u 4 u k (4k + 2n)! 1 2 2n − H2n = dϕ ϕ exp − ϕ − ϕ = 2 24 24 k!(2k + n)!22k+n k≥0 X  2u k 4n ∼ − Γ(k + n) √ . (15.267) 3 π 2 k

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511

It follows that H2m  H2n if m > n. That in its turn implies G2n = H2n /H0 ∼ H2n and C2n ∼ G2n , leading to the counter-intuitive result that at very high loop order, almost all diagrams are connected ! Moreover, in low order, we have C2 = 1 − u/2 + O (u2 ) and C4 = −u + 7u2 /2 + O (u3 ) so that uˆ = u(1 − 5u/2 + O (u2 )), and uˆ(u) ∼ −H4 (u) up to an overall factor. This gives us u = uˆ(1+5ˆ u/2+O (ˆ u2 )), giving g1 = 5/2 in Eq.(15.264). In addition, Eq.(15.267) tells us that cf = −2/3, where cf is defined in Eq.(15.258). Since Hp (u)  uˆ(u) for p = 6, 8, 10, . . . we then find Hk (u(ˆ u)) ∼ Hk (ˆ u) exp(g1 /cf ) = Hk (ˆ u) e−15/4 .

(15.268)

The ‘improvement factor’ therefore has the asymptotic value exp(15/4).

15.15.4

Gamma, Digamma and Bernoulli

The Gamma function Γ(z) has a well-known integral representation : Z∞ Γ(z) =

dt tz−1 exp(−t) ,

(15.269)

0

but that works only if 0. A slightly less well-known but very useful representation, that works for almost all z, is a special limit46 : Γ(z) = lim n! nz n→∞

n Y

(k + z)−1 .

(15.270)

k=0

This shows immediately that nz n = 1 , Γ(z + 1)/Γ(z) = lim =z , n→∞ n + 1 + z n→∞ n + 1

Γ(1) = lim

(15.271)

thus proving the correctness of Eq.(15.270). Also we find Γ(m) = (m − 1)! for natural m. For z close to −m (m = 0, 1, 2, . . .) we also find very simply Γ(z) ≈ 46

Found by — who else ? — Euler.

(−)m . m! (z + m)

(15.272)

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August 31, 2019

Finally, we can work out how Γ(z) behaves for small z :


log(n!) + z log(n) −

log zΓ(z) = lim

n→∞

=

lim

n→∞

z

! log(k + z)

k=1

log(n) −

n X 1 k=1

= −zγE +

n X

X (−z)p p

p≥2

!

k

n X (−z)p X 1 + p k=1 k p p≥2

!

ζ(p) ,

(15.273)

Pn 47 where γP E = lim(− log(z)+ 1 (1/k)) is the Euler-Mascheroni constant , and ∞ ζ(p) = 1 1/k p is the Riemann zeta function48 . Using section 15.15.11, we can then write, for small z :  2  1 γE π2 Γ(z) = − γE + z + + ··· (15.274) z 2 12 The digamma function is defined as ψ(z) = Γ0 (z)/Γ(z) = lim

log(n) −

n X

1 k+z

! .

(15.275)

ψ(1/2) = − γE − log(4) .

(15.276)

n→∞

k=0

We can immediately derive ψ(1) = − γE ,

We can also find an asymptotic expansion for ψ(z) as follows, by taking an extra derivative and letting n → ∞ : 0

ψ (z) =

X k≥0

1 = (k + z)2

X

exp(−(k + z)t)

k≥0

Z∞

! dt

0

dt t 0

Z∞ =

Z∞

te

−zt

+ te

−zt

X k≥1

−kt

e

=

 dt te−zt + e−zt

t t e −1

 .

0

(15.277) 47

In short : γE = .57721566490153286060651209008240243104215933593992359881 · · ·. We have interchanged the sums over k and p in Eq.(15.273). Since the result is convergent this is actually allowed. 48

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513

The Bernoulli numbers Bm are defined as X tm t 1 1 = Bm , B0 = 1 , B1 = − , B2 = . . . t e − 1 m≥0 m! 2 6

(15.278)

Performing the integral over t and then integrating with respect to z yields X Bm 1 ψ(z) = log(z) − − z −m , (15.279) 2z m≥2 m where the constant of integration vanishes as we can check from the Stirling approximation of section 15.15.2. Finally, from the fact that t/(et − 1) + t/2 is actually symmetric in t we can see that Bm vanishes for odd m ≥ 3.

15.15.5

The Dirac delta distribution

The Kronecker delta, as introduced in chapter 0, is defined for integer arguments and reads δm,n = θ(m = n) , (15.280) so that δn,m = 0 for m 6= n ,

∞ X

δm,n = 1 .

(15.281)

n=−∞

The Dirac delta distribution49 is the continuum variant of this. Being a distribution, it is really defined in the context of integration with a test function50 . The Dirac delta is commonly denoted by (surprise ! ) δ(x) and its definition is Z∞ dx δ(x − a) f (x) = f (a) (15.282) −∞

for all test functions f (x). Viewed as some kind of function it therefore has properties analogous to those in Eq.(15.281) : Z∞ δ(x) = 0 for x 6= 0 ,

dx δ(x) = 1 .

(15.283)

−∞ 49

Colloquially, the Dirac delta function, but it is really a distribution in the sense of Schwartz. 50 A test function has compact support and is inifinitely many times differentiable : simplistically, it is a nice function.

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August 31, 2019

Applying partial integration (and assuming cavalierly that this is allowed ! ) we also find properties of its derivatives : Z∞

dx δ 0 (x − a) f (x) = −f 0 (a) ,

−∞

Z∞

dx δ 00 (x − a) f (x) = +f 00 (a) ,

−∞

(15.284) and so on. The Dirac delta can be viewed as the limit of a set of nonnegative functions with unit integral, that are increasingly narrow and more and more peaked at zero. An important result that we use extensively in the text is Z∞ dx exp(ixz) = 2π δ(z) ,

(15.285)

−∞

the proof of which forms exercise 32.

15.15.6

The principal-value distribution

Like the Dirac delta, this distribution is best defined in the context of an integral. For a test function f (x) it acts as follows : Z− Zb>0 Zb>0 f (x) f (x) 1 dx + lim ↓0 dx . dx f (x) P ≡ lim ↓0 x x x a