Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

Feynman Diagrams For PedestriansThorsten OhlInstitute for Theoretical Physics and AstrophysicsWürzburg Universityhttp://physik.uni-wuerzburg.de/ohl44th Maria Laach School for High Energy PhysicsSeptember 4-14, 2012Φ T P 2AbstractThis set of interleaved lectures and exercises will (re)introduce working experimentalparticle physicists to the techniques used for computing simple crosssections in the standard model and its extensions. The approach is deliberatelypedestrian with an emphasis on real world applications.After an introduction, gradually more and more time will be spent actuallycomputing stuff in order to build confidence and gain intuition for (new) physicssignals in cross sections.

Contents1 Introduction 11.1 Scattering Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Asymptotic States 42.1 Klein-Gordon Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Free Spin-0 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Anti Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Dirac Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 Gamma Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.6 Free Spin-1/2 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.7 Free Spin-1 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Interactions 153.1 Propagators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Feynman Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3 Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.4 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.5 Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 QED 244.1 e + e − → µ + µ − . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.2 Trace Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.3 Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.4 FORM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.5 Bhabha Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.6 FORM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 QCD 315.1 Feynman Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2 3-Jet Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 Standard Model 356.1 Propagators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356.2 Feynman Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366.3 Higgs Strahlung . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 Solutions 39Author: ohlRevision: 172Date: 2012-08-27 16:30:28 +0200 (Mon, 27 Aug 2012)

1 Introduction1.1 Scattering AmplitudesBasic principle of quantum mechanics• an accelerator prepares an initial state |in〉• that is transformed by an interaction S• and a detector measures the overlap of the resulting state with a final state |out〉.• the transition probability P is given by the absolute square of the transitionamplitude AA in→out = 〈out S in〉P in→out = |A in→out | 2• if the initial and final states |in〉 and |out〉 are not pure states, the correspondingtransition probabilities must be added (e. g. spins and flavors) or integrated(e. g. angles, energies and momenta)Task(s):1. describe |in〉 and |out〉:• pure states: completetely polarized electrons, muons, photons⇒ Dirac equation, Klein-Gordon equation, &c.• mixtures: protons, partially polarized or unpolarized electrons, muons,photons, . . .2. compute S (i. e. the part of S that contributes to 〈out S in〉)• quantum electro dynamics (QED)• quantum chromo dynamics (QCD)• standard model (SM)• “new physics”, “beyond the SM” (BSM)⇒ Feynman rules3. square A in→out and integrate P in→out• Monte Carlo1(1a)(1b)

1.2 Lorentz TransformationsBasic principle of special relativity:• the velocity of light c is the same in each inertial system.∴ the wavefronts of a spherical light wave is for every observer located at|⃗x| = ct (2)• introducing the notation x 0 = ct, this means that the solutions ofare the same in every inertial reference framex 2 0 − ⃗x 2 = 0 (2 ′ )• adding homogeneity and isotropy of space, this means thatmust be the same in every inertial reference frame.• useful notations:– 3 vectors: (covariant w. r. t. rotations)x 2 = x 2 0 − ⃗x 2 (3)⃗x = (x 1 , x 2 , x 3 ) (4)– 4 vectors: (covariant w. r. t. rotations and boosts into a moving inertial frame)x = (x 0 ;⃗x) = (x 0 ; x 1 , x 2 , x 3 ) = (x 0 ; −x 1 , −x 2 , −x 3 ) (5)• introduce Minkowski metric⎛⎞1 0 0 0g µν = g µν = ⎝0 −1 0 0 ⎠0 0 −1 0(6)0 0 0 −1to shift indices down or upx µ =3∑g µν x ν , x µ =ν=0• convenient summation conventionxp =3∑g µν x ν (7)ν=03∑x µ p µ = x µ p µ = x µ p µ = g µν x µ p ν = g µν x µ p νµ=0= x 0 p 0 −23∑x i p i = x 0 p 0 − x i p i = x 0 p 0 − ⃗x⃗p (8)i=1

• a Lorentz transformation Λ must leave xp invariant because 2xp = (x + p) 2 −x 2 − p 2 :• derivatives:for examplex µ → x ′ µ = Λ νµ x ν (mit x ′2 = x 2 ) ⇐⇒ g µµ ′ = Λ νµ Λ ν′µ ′ g νν ′ (9)∂∂x µf(x) = ∂ µ xf(x) = ∂ µ f(x) ,Problem 1. Compute the partial derivative w. r. t. xfor constant four vectors a, b and p.Problem 2. Show that∂∂x µ f(x) = ∂ µf(x) (10)∂ µ x(xp) = ∂(x ν p ν )/∂x µ = p µ (11)∂ µ e −ipx , (a∂)(b∂)e −ipx , ∂ 2 e −ipx (12)∂ µ x µ = 4(NB: ∂ µ x µ = g νµ ∂x µ /∂x ν and g νµ = δ νµ ) and compute∂ 2 e −x2 /2(13a)(13b)1.3 Schrödinger Equation• Wave functions satisfy the Schrödinger equationi̷h d Ψ(t) = HΨ(t)dt (14a)with solution (for infinitesimal time intervals)Ψ(t + δt) = e −iH·δt/̷h Ψ(t)(14b)∴ scattering amplitude (infinite time intervals)〈A in→out = 〈out S in〉 = out(t2 ) e −iH·(t 2−t 1 )/̷h in(t 1 ) 〉 (15)• Problems with this approachlimt 1 →−∞t 2 →+∞– particle production and decay has been observed, but can not be describedby wave functions (without “2nd quantization”), because probability is conserved(“unitarity”)– Schrödinger equation (14) not manifestly Lorentz covariant– free single particle equationi̷h d dt Ψ(t) = 12mis manifestly not Lorentz covariant!3( ) 2 1∇i̷hc ⃗ Ψ(t) (16)

1.4 Units• from now on, we will use units which will give us numbers with natural orderof magnitude for quantum mechanics and relativistic kinematics̷h = c = 1 . (17)• velocities and actions are dimensionless and therefore[ ]1[energy] = [momentum] = [mass] = . (18)length• in particular, our Feynman rules, will later yield cross sections in units of [energy −2 ],e. g.• the relevant conversion factors are(TeV 2 nb = GeV 2 mb) and therefore2 Asymptotic Statesσ = 4πα23E 2 (19)̷hc = 197.327 053(59) MeV fm (20)(̷hc) 2 = 0.389 379 66(23) TeV 2 nb (21)σ =• described by wave equations, that are4πα 23(E/TeV) 2 0.39 nb (19′ )1. linear: superposition principle of quantum mechanics2. relativistic: matrix elements of observables must transform under rotationsand Lorentz boosts like scalars, four vectors, tensors, &c.3. and have the correct dispersion relation: E 2 = ⃗p 2 + m 2• objects of interest– spin-0 particles: not yet unambigously observed as an elementary particle,but ATLAS and CMS have strong evidence for a state with spin 0 or 2∗ one invariant component– spin-1/2 particles: leptons, quarks∗ at least two components: spinor under rotations– spin-1 particles: gauge bosons∗ massive three components (polarizations)∗ massless two components (polarizations)4

2.1 Klein-Gordon Equation(i∂ 0 ) 2 φ(x) =[(−i⃗∂) 2 + m 2 ]φ(x) (22)• is obviously a covariant wave equation, because(□ + m2 ) φ(x) = 0 (23)• fourier transformφ(x) =∫ d 4 p(2π) 4 e−ipx ˜φ(p) , i∂ µ φ(x) =∫ d 4 p(2π) 4 e−ipx p µ ˜φ(p) , etc (24)∴ algebraic equation (p 2 − m 2) ˜φ(p) = 0 (23 ′ )p 0“mass shell”:p 0 = + √ ⃗p 2 + m 2 ,p 2 = m 2 , p 0 0|⃗p|p 0 = − √ ⃗p 2 + m 2• correct relativistic dispersion relation E = + √ ⃗p 2 + m 2• but what about the other solution E = − √ ⃗p 2 + m 2 ?2.2 Free Spin-0 Particles• general solution of the Klein-Gordon equation∫ d 4 pφ(x) =(2π) 2πΘ(p 0)δ(p 2 − m 2 ) ( φ (+) (⃗p)e −ipx + φ (−) (⃗p)e ipx) (25)4∫ ∣d 3 ⃗p ∣∣∣∣p0 (=φ(2π) 3 2p 0=√(+) (⃗p)e −ipx + φ (−) (⃗p)e ipx) (26)⃗p 2 +m∫2= ˜dp ( φ (+) (⃗p)e −ipx + φ (−) (⃗p)e ipx) (27)• conserved current ∂ 0 j 0 (x) − ⃗∇⃗j(x) = ∂ µ j µ (x) = 0 out of two solutions φ 1 and φ 2of the Klein-Gordon equation with the same mass:j µ (x) = φ ∗ 1(x)i ←→ ∂ µ φ 2 (x) = φ ∗ 1(x)[i∂ µ φ 2 (x)] − [i∂ µ φ ∗ 1(x)]φ 2 (x) (28)5

• indeed) ()∂ µ j µ (x) = ∂(φ µ ∗ 1(x)[i∂ µ φ 2 (x)] − ∂ µ [i∂ µ φ ∗ 1(x)]φ 2 (x)= i[∂ µ φ ∗ 1(x)][∂ µ φ 2 (x)] + iφ ∗ 1(x)[∂ 2 φ 2 (x)] − i[∂ 2 φ ∗ 1(x)]φ 2 (x)−i[∂ µ φ ∗ 1(x)][∂ µ φ 2 (x)] = −iφ ∗ 1(x)m 2 φ 2 (x) + im 2 φ ∗ 1(x)φ 2 (x) = 0 (29)• invariant and time independent overlap out of the conserved current∫Q =∫x 0 =td 3 ⃗x j 0 (x) =d 3 (⃗p 1(2π) 3 2p 0 2p 0φ (+)∗∫x 0 =td 3 ⃗x φ ∗ 1 (x)i←→ ∂ 0 φ 2 (x) =1(⃗p)φ (+)2(⃗p) · e ip0t i ←→ ∂ 0 e −ip0t + φ (−)∗1(−⃗p)φ (+)2(⃗p) · e −ip0t i ←→ ∂ 0 e −ip 0t+ φ (+)∗1(⃗p)φ (−)2(−⃗p) · e ip0t i ←→ ∂ 0 e ip0t + φ (−)∗1(−⃗p)φ (−)2(−⃗p) · e −ip0t i ←→ ∂ 0 e ip 0t∫ ()= ˜dp φ (+)∗1(⃗p)φ (+)2(⃗p) − φ (−)∗1(⃗p)φ (−)2(⃗p)• normalization only positive on the positive mass shell!∴ j µ (x) must not be interpreted as a probability current!• . . . anyway: the existence of the negative mass shell makes the energy unboundedfrom below and no ground state exists!∴ φ(x) must not be interpreted as Schrödinger wave function!)(30)2.3 Anti ParticlesObservations:• the positive and negative mass shells of free (i. e. not interacting) particles areindependent∴ one can simply project out the negative mass shell in this case . . .• . . . unfortunately, all local and Lorentz invariant interactions couple positiveand negative mass shell (see below)• . . . however, asymptotic states are assumed to be noninteracting∴ we can at least reinterpret the negative mass shellhowever:• the amplitude φ (+) (⃗p)e −ipx on the positive mass shell corresponds to the momemtum+⃗p, but the amplitude φ (−) (⃗p)e +ipx on the negative mass shell correspondsto the reversed momemtum −⃗p6

∴ the formalism is consistent, if all quantum numbers are reversed on the negativemass shell∵ in a stationary, i. e. time independent, state, the casescan not be distinguished.Q, ⃗p−Q, −⃗p∴ the states on the negative mass shell describe not particles with “negative energy”,but anti particles with opposite quantum numbers instead!• in stationary states, this can be taken a step further: instead of anti particlesmoving forward in time . . .• . . . one can use particles moving backward in time, without noticing a differencein the overall balance!• caveat: it is not (yet) obvious that this makes sense of interactions are switchedon . . .• . . . will be shown later.• NB: this is a calculationally convenient interpretation — there is no time travelgoing on, since we’re in a steady state• The equivalent picture with anti particles moving forward in time requires thefull machinery of quantum field theory2.4 Dirac Equation• task: find “objects” γ µ , such that(γ µ ∂ µ ) 2 = ∂ 2 (31)• because the solutions of the Dirac equation(iγ µ ∂ µ − m) ψ(x) = 0 (32)automagically satisfy the die Klein-Gordon equation as well(iγ µ ∂ µ + m) (iγ µ ∂ µ − m) ψ(x) = ( −∂ 2 − m 2) ψ(x) = 0 (33)7

• the Dirac equation is obviously linear and its solutions satisfy the proper relativisticdispersion relation.• can we construct “objects” γ µ , that satisfy (31)?• a sufficient condition is[γ µ , γ ν ] +:= γ µ γ ν +γ ν γ µ = 2g µν · 1 (34)because partial derivatives commute ∂ µ ∂ ν = ∂ ν ∂ µ .• using a useful and ubiquitous notation, the Feynman slash/a = γ µ a µ = γ µ a µ (35)this reads equivalently [/a, /b] +:= /a/b+/b/a = 2 · ab = 2 · a µ b µ2.5 Gamma Matrices• recall the Pauli matrices with the defining property]3∑[σ k , σ l = σ k σ l − σ l σ k = 2i ɛ klm σ m(σk ) †= σkm=1(36a)(36b)using the totally antisymmetric tensor ɛɛ 123 = ɛ 231 = ɛ 312 = 1, ɛ 213 = ɛ 321 = ɛ 132 = −1 (37)• concrete realisationσ 1 =( ) 0 1, σ 2 =1 0( ) ( )0 −i 1 0, σ 3 =i 0 0 −1(38)• with• in particular3∑σ k σ l = δ kl 1 + i ɛ klm σ m (39)[σ k , σ l ]+m=1= 2δ kl 1 (40)• Dirac realisation of the gamma (a. k. a. Dirac) matrices:( ) ( )1 00 σγ 0 = , γ i i=0 −1 −σ i 0(41)• there are (infinitely) many more realisations, but no smaller one8

• we can verify the anti commutation relations by explicit calculation.NB: block matrices are multiplied just like ordinary matrices with non commutingmatrix elements( ) ( ) ( ) ( )γ0 2 1 0 1 0 1 0= = = 1 (42a)0 −1 0 −1 0 1( ) ( ) ( ) ( )[γ 0 , γ i] 1 0 0 σ = + γ0 γ i + γ i γ 0 i 0 σi 1 0=0 −1 −σ i +0 −σ i 0 0 −1() ()0 1 · σ=i 0 (−σ(−1) · (−σ i +i ) · 1) 0 σ i (42b)· (−1) 0( ) ( )0 σi 0 −σi=σ i +0 −σ i = 0 (42c)0Problem 3. Verify the remaining (k, l = 1, 2, 3) anti commutation relations (34):• in the Dirac realization (41) we have obviously• however, since (γ 0 ) 2 = 1 and (γ i ) 2 = − 1∴ γ 0 must have only real eigenvalues, and∴ all γ i must have only imaginary eigenvalues∴ this must be true in all realizations.[γ k , γ l ] + = −2δ kl · 1 . (43)γ † 0 = γ 0 , γ † i = − γ i (44)• Another useful and ubiquitous notation is therefore the Dirac adjoint for matricesA = γ 0 A † γ 0 , γ µ = γ 0 γ † µγ 0 = γ µ (45)• NB: on the next page we will meet the related, but different Dirac adjoint forcolumn vectorsv = v † γ 0 (46)• Don’t mix them up!2.6 Free Spin-1/2 Particles• Ansatz:∫ψ(x) = ˜dp ( ψ (+) (p)e −ipx + ψ (−) (p)e ipx) (47)(i/∂ − m) ψ(x) = 0 ⇔{(/p − m) ψ (+) (p) = 0(/p + m) ψ (−) (p) = 0(48)9

• the adjoint solution∫¯ψ(x) = ψ(x) † γ 0 = ˜dp ( ¯ψ (+) (p)e ipx + ¯ψ (−) (p)e −ipx) (49)satisfies¯ψ(x)i ←− /∂ = i∂ µ ¯ψ(x)γ µ = i∂ µ ψ(x) † γ 0 γ µ γ 0 γ 0= i∂ µ ψ(x) † γ µ† γ 0 = (−i∂ µ γ µ ψ(x)) † γ 0 = (−i/∂ψ(x)) = −m ¯ψ(x) (50)∴(¯ψ(x) i ←− )/∂ + m = 0 , (51)or¯ψ (+) (p) (/p − m) = 0¯ψ (−) (p) (/p + m) = 0(52a)(52b)• general solution of the Dirac equationψ (+) (p) =ψ (−) (p) =2∑u k (p)b k (p)k=12∑v k (p)d k (p)with four independent basis solutions u 1 (p), u 2 (p), v 1 (p), v 2 (p) satisfying(/p − m)u k (p) = 0(/p + m)v k (p) = 0and the corresponding expansion coefficients b 1 (p), b 2 (p), d 1 (p), d 2 (p).k=1(53a)(53b)(54a)(54b)• in the rest frame of the particle, i. e. for p = (m, ⃗ 0), we have /p = mγ 0 and theDirac equation simplifies to( )m(γ 0 − 1)u k ( ⃗ 0 00) =u0 −2m · 1 k ( ⃗ 0) = 0(55a)( )m(γ 0 + 1)v k ( ⃗ 2m · 1 00) =v0 0 k ( ⃗ 0) = 0(55b)• the independent solutions in the rest frame are therefore⎛ ⎞⎛ ⎞10u 1 ( ⃗ ⎜0⎟0) = ⎝0⎠ , u 2 ( ⃗ ⎜1⎟0) = ⎝0⎠00⎛ ⎞⎛ ⎞00v 1 ( ⃗ ⎜0⎟0) = ⎝0⎠ , v 2 ( ⃗ ⎜0⎟0) = ⎝1⎠10(56a)(56b)10

• from which we can construct solutions for arbitrary on shell momenta with p 2 =m 2 u k (p) = √ /p + mp0 + m u k( ⃗ 0)(57a)v k (p) =/p − m √p0 + m v k( ⃗ 0)(57b)• since (/p+m)(/p−m) = p 2 −m 2 , these are obviously solutions of the Dirac equationfor on-shell momenta• the motivation for the not obviously covariant normalization will be apparentafter problem 4mathematical reminder:• compare the inner product of a row vector with a column vector⎛ ⎞b 1( )b 2a1 a 2 · · · a n ⎜ ⎟⎝ . ⎠ = ∑ na i b i (58)i=1b nto the outer product of a column vector with a row vector⎛ ⎞⎛⎞a 1a 1 b 1 a 1 b 2 . . . a 1 b ma 2⎜ ⎟⎝ . ⎠ ⊗ ( ) a 2 b 1 a 2 b 2 . . . a 2 b mb 1 b 2 · · · b m = ⎜⎟⎝ . . . ⎠a n a n b 1 a n b 2 . . . a n b m(59)• these are two very different operations– the inner product produces a number– the outer product produces a matrixProblem 4. Determine ū k (p) und ¯v k (p) from the definitions and show that for p 2 = m 22∑u k (p)ū k (p) = /p + m ,k=12∑v k (p)¯v k (p) = /p − m (60)k=1Problem 5. Compute (always assuming p 2 = m 2 )ū k (p)u l (p) , ¯v k (p)v l (p) , ū k (p)v l (p) , ¯v k (p)u l (p) . (61)Problem 6. Compute (always assuming p 2 = m 2 )ū k (p)γ µ u l (p) , ¯v k (p)γ µ v l (p) , ū k (⃗p)γ 0 v l (−⃗p) , ¯v k (−⃗p)γ 0 u l (⃗p) . (62)11

• this is a special case of the Gordon decomposition for arbitrary solutions of theDirac equation:ū k (p)γ µ u l (q) = ū k (p) /pγ µ + γ µ /q2mu l(q)= ū k (p) p ν(g νµ + 1[γ 2 ν, γ µ ] − ) + q ν (g µν + 1[γ 2 µ, γ ν ] − )u l (q)2m= p µ + q µ2m ūk(p)u l (q) + q ν − p ν2im ūk(p) i 2 [γ µ, γ ν ] − u l (q) (63)• there are alltogether 16 independent (anti-)hermitian 4 × 4-matrices:1 1 “scalar” (64a)γ µ 4 “vector” (64b)σ µν = i 2 [γ µ, γ ν ] −6 “tensor” (64c)γ 5 γ µ 4 “axial vector” (64d)γ 5 = iγ 0 γ 1 γ 2 γ 3 1 “pseudo scalar” (64e)• NB: the bare gamma matrices do not transform like vectors, tensors, axial vectors,or pseudo scalars!• there are additional nontrivial transformations L(Λ) that have to be applied onthe left and right, e. g.γ µ → Λµ ν L(Λ)γ ν L −1 (Λ) (65)• however sinceψ(x) → L(Λ)ψ(Λ −1 x) , ¯ψ(x) → ¯ψ(Λ −1 x)L −1 (Λ) (66)the L(Λ) compensate each other in matrix elements (a. k. a. “sandwiches”)¯ψ(x)γ µ ψ(y) → ¯ψ(Λ −1 x)L −1 (Λ)Λ νµ L(Λ)γ ν L −1 (Λ)L(Λ)ψ(Λ −1 x)= Λ νµ ¯ψ(Λ −1 x)γ ν ψ(Λ −1 x) (67)∴ and the L(Λ) can be ignored in the computation of matrix elements∴ the characterization as vector, tensor, axial vector, or pseudo scalars is meaningfulProblem 7. Compute [γ 5 , γ µ ] +Problem 8. Show the conservation of the vector current for two solutions ψ 1 (x) und ψ 2 (x) ofthe Dirac equation (32) and (51)∂ µ [ ¯ψ 1 (x)γ µ ψ 2 (x) ] = 0 . (68)12

• invariant overlap from conserved current:∫∫Q = d 3 ⃗x j 0 (x) = d 3 ⃗x ¯ψ 1 (x)γ 0 ψ 2 (x) =∫d 3 ⃗x ψ † 1 (x)ψ 2(x)x 0 =tx 0 =tx 0 =t∫(d 3 ⃗p 1=ψ (+)†(2π) 3 1(⃗p)ψ (+)2(⃗p) + ψ (−)†1(−⃗p)ψ (+)2(⃗p) · e −2ip 0t2p 0 2p 0)+ ψ (+)†1(⃗p)ψ (−)2(−⃗p) · e 2ip0t + ψ (−)†1(−⃗p)ψ (−)2(−⃗p)∫d 3 ⃗p=(2π) 3 2p 02∑ ()b † 1,k (p)b 2,k(⃗p) + d † 1,k (p)d 2,k(⃗p)k=1(69)∴ normalization positive everywhere!Problem 9. Show the Partial Conservation of the Axial Current (PCAC) for solutions ψ 1 (x)und ψ 2 (x) of the Dirac equation∂ µ [ ¯ψ 1 (x)γ µ γ 5 ψ 2 (x) ] = 2im ¯ψ 1 (x)γ 5 ψ 2 (x) . (70)Problem 10. Compute σ klGamma matrices.= i 2 [γ k, γ l ] −for k, l = 1, 2, 3 in the Dirac realization of the∴ the matrices {σ 23 , σ 31 , σ 12 } can take over the rôle of the Pauli matrices {σ 1 , σ 2 , σ 3 }and distinguish spin up from spin down in the rest frame12 σ 12u 1 ( ⃗ 0) = + 1 2 u 1( ⃗ 0) ,12 σ 12v 1 ( ⃗ 0) = + 1 2 v 1( ⃗ 0) ,12 σ 12u 2 ( ⃗ 0) = − 1 2 u 2( ⃗ 0) (71a)12 σ 12v 2 ( ⃗ 0) = − 1 2 v 2( ⃗ 0) (71b)• just as before in the case of scalar particles, we can interpret solutions on thenegative energy mass shell for spin-1/2 particles as anti particles that move inthe opposite direction in space time:– u k (p) amplitude for a particles in the initial state– ū k (p) amplitude for a particle in the final state– v k (p) amplitude for an anti particle in the final state– ¯v k (p) amplitude for an anti particle in the initial state• in addition to the other quantum numbers, we must flip the spins so that theoverall balance is maintainedQ, ⃗p, s−Q, −⃗p, −s13

2.7 Free Spin-1 Particles• combine ⃗E and ⃗B into field strength tensor⎛⎞0 −E 1 −E 2 −E 3F µν = ∂ µ A ν − ∂ ν A µ = ⎜E 1 0 −B 3 B 2⎟⎝E 2 B 3 0 −B 1⎠ (72)E 3 −B 2 B 1 0• manifestly covariant Maxwell equations• current j µ necessarily conserved: ∂ µ j µ = 0• equation of motion for the vector potential A µ• gauge invariance: F µν does not change, if• with special gauge condition ∂ µ A µ = 0: ∂ 2 A µ = j µ• more, but not most, general case∂ µ F µν = j ν , ɛ µνρσ ∂ ν F ρσ = 0 (73)(g µν ∂ 2 − ∂ µ ∂ ν )A ν = j µ (73 ′ )A µ (x) → A µ (x) − ∂ µ ω(x) (74)(g µν ∂ 2 − (1 − ξ)∂ µ ∂ ν )A ν = j µ (75)• explicit mass termor∂ µ F µν + M 2 A ν = j ν (76)(g µν (∂ 2 + M 2 ) − ∂ µ ∂ ν )A ν = j µ (76 ′ )• contraction with ∂ µM 2 ∂ ν A ν = 0 (77)• massless vector bososns have two degrees of freedom, massive ones three• polarization vectors for massless vector bosons with momentum k = (k 0 ; 0, 0, k 0 )ɛ ± = ɛ ∗ ∓ = 1 √2(0; 1, ±i, 0) (78)• propertiesand with c = (1; 0, 0, −1)∑λ=−1,+1ɛ µ λ ɛ∗ λ ′ ,µ = −δ λλ ′ɛ µ λ k µ = 0ɛ µ λ ɛν,∗ λ= −g µν + c µk ν + c ν k µck(79a)(79b)(80)• polarization vectors for massive vector bosons with momentum14

k = (k 0 ; |⃗k| sin θ cos φ, |⃗k| sin θ sin φ, |⃗k| cos θ):ɛ ± = ɛ ∗ ∓ = e∓iφ√2(0; cos θ cos φ ∓ i sin φ, cos θ sin φ ± i cos φ, − sin θ)(81a)• properties (79) and3 Interactions3.1 Propagatorsɛ 0 = k 0M (| ⃗k|/k 0 ; sin θ cos φ, sin θ sin φ, cos θ) = ɛ ∗ 0∑λ=−1,0,+1ɛ µ λ ɛν,∗ λ• Photons far from all electrical charges satisfyand we have already seen the solutions.(81b)= −g µν + k µk νM 2 (82)∂ 2 A (0)µ (x) = 0 (83)• In the presence of electrical charges, the photons couple to the electromagneticcurrent∂ 2 A µ (x) = j µ (x) = −e ¯ψ(x)γ µ ψ(x) + . . . (84)and the solutions turn out to be more complicated.• Assumption: there is a “function” D, that solves• Then(∂ 2 + m 2 )D(x, m) = −δ 4 (x) (85)∫A µ (x) = A (0)µ (x) − d 4 yD(x − y, 0)j µ (y) (86)is a solution of the inhomogeneous equation (84) for each solution of the homogeneousequation (83), since∫ [∂ 2 A µ (x) = ∂ 2 A (0)µ (x) − d 4 y]∂ 2 D(x − y, 0) j µ (y)= 0 −∫ [ ]d 4 y −δ 4 (x − y) j µ (y) = j µ (x) (87)• interpretation: the current j µ (y) acts at the space time point y as a source ofphotons, that are “propagated” by the propagator D(x − y, 0) to the space timepoint xj µ (y)D(x − y, 0)A µ (x)(86 ′ )15

• NB: retardation is built in in (86), because we integrate over the four dimensionalspace time and not over the three dimensional space at a given instant.Problem 11. Show thatS(x, m) = (i/∂ + m) D(x, m) (88)is the Feynman propagator S for Dirac particles of mass m, i. e. that(i/∂ − m) S(x, m) = δ 4 (x)if (∂ 2 + m 2 )D(x, m) = −δ 4 (x), as in (85) above.• but what is the source ??? of the Dirac field????(y)S(x − y, m)ψ(x)(89)• consider the Dirac equation with (electromagnetic) interaction(i/∂ − e /A(x) − m) ψ(x) = 0 (90)or(i/∂ − m) ψ(x) = e /A(x)ψ(x) (90 ′ )• with the formal solutionψ(x) = ψ (0) (x) +∫d 4 yS(x − y, m)e /A(y)ψ(y) (91)that can be represented graphically asS(x − y, m)ψ(x)(91 ′ )• (91) is analogous to (86), when we use the current j µ (y) = −e ¯ψ(y)γ µ ψ(y)∫A µ (x) = A (0)µ (x) − d 4 yD(x − y, 0)e ¯ψ(y)γ µ ψ(y) (92)which can be represented graphically asD(x − y, 0)A µ (x)(92 ′ )• caveat: the equations (91) and (92) are not explicit solutions, but coupled integralequations16

• that can be solved recursively by mutual series expansion∫ψ(x) = ψ (0) (x) + d 4 yS(x − y, m)e /A(y)ψ(y)∫= ψ (0) (x) + e d 4 yS(x − y, m) /A (0) (y)ψ (0) (y)(+ e∫d 2 4 yd 4 z S(x − y, m) /A (0) (y)S(y − z, m) /A (0) (z)ψ (0) (z))− S(x − y, m)γ µ ψ (0) (y)D(y − z, m) ¯ψ(z)γ µ ψ(z) + O(e 3 ) (93)• these recursive expansions become very big very fast and a graphical representationis useful:A (0)µ (y)ψ (0) (y)A (0)µ (z)ψ (0) (z)ψ(x) ψ (0) (z)S(x − y, m)S(y − z, m)ψ (0) (z)ψ(x)S(x − y, m)D(y − z, 0)(93 ′ )• remaining open questions:– does D(x − y, m) exist?– can we compute it?• fortunately, we only need to solve(∂ 2 + m 2) D(x, m) = −δ 4 (x) (85 ′ )because (most) other propagators can be obtained by taking derivatives• since the equation is translation invariant, we should use Fourier transformation• formally (with ɛ → 0+)D(x, m) =∫ d 4 p(2π) 4 e−ipx 1p 2 − m 2 + iɛ(94)(∂ 2 + m 2) D(x, m) =∫ d 4 p(2π) 4 e−ipx −p2 + m 2p 2 − m 2 + iɛ = − ∫ d 4 p(2π) 4 e−ipx = −δ 4 (x) (95)• singularities in the integral over p 0 at ± √ |⃗p| 2 + m 2 (+iɛ is a convenient shorthandfor the choice of integration contour):17

• compare1p 2 − m 2 +iɛ = 1(p 0 ) 2 − (|⃗p| 2 + m 2 ) +iɛE>0=1(p 0 ) 2 − (E−iɛ) = 12 p 0 − E+iɛImp 0Rep 0− √ |⃗p| 2 + m 2 + √ |⃗p| 2 + m 2√E=+ |⃗p| 2 +m 2=1p 0 + E−iɛ = 12E1(p 0 ) 2 − E 2 +iɛ(1p 0 − E+iɛ − 1)p 0 + E−iɛ(96)• forward in time:x 0 > 0 :limp 0 →−i∞ e−ip 0x 0→ 0the integration contour in (94) can be closed below(97a)Imp 0Rep 0• backward in time:x 0 < 0 :limp 0 →+i∞ e−ip 0x 0→ 0the integration contour in (94) can be closed above(97b)Imp 0Rep 0∴ inΦ ′ (x) =∫∫ dd 4 4 p1y D(x − y, m)Φ(y) =(2π) 4 e−ipx p 2 − m 2 +iɛ ˜Φ(p) (98)• the part of ˜Φ(p) with p 0 = + √ |⃗p| 2 + m 2 is propagated into the future• the part of ˜Φ(p) with p 0 = − √ |⃗p| 2 + m 2 is propagated into the past• Compton scattering in nonrelativistic perturbation theory contains scattering aswell as pair creation and pair annihilation contributions:t 1 t 2 +t1t 2+18t 1+t 2t 1(99)t 2

∵ intermediate states violate energy conservation and vertices can have space likedistances∴ temporal order of t 1 and t 2 depends in general on the reference frame, i. e. isundefined• Feynman’s brilliant (re-)interpretation:– particles with p 0 = + √ |⃗p| 2 + m 2 are propagated into the future– anti particles with p 0 = − √ |⃗p| 2 + m 2 and opposite charges are propagatedinto past∴ charges are conserved along the arrows in (99)!∴ the four nonrelativistic diagrams in (99) can be combined to two covariantexpressions using Feynman propagatorst 1 t 2 +11E−E 0 +iɛ E+E 0 +iɛt 1=t 21p 2 − m 2 +iɛ(100a)1E−E 0 +iɛt 1t 2+1E+E 0 +iɛt 1=t 21p 2 − m 2 +iɛ(100b)∴ the Feynman propagator allows to extend our interpretation of external, noninteractinganti particles as particles traveling backward in time to interactingparticles.Problem 12. Compute the propagator S(x, m) for Dirac particles in momentum space!• propagator for massless spin-1 particles−ig µν + i(1 − ξ) k µk νk 2 +iɛk 2 + iɛ(101)• the gauge parameter ξ ist arbitrary and must cancel in the final result• partial results can depend on ξ• the propagator for massive spin-1 particles−ig µν + i k µk νM 2k 2 − M 2 + iɛis not gauge dependent, because (76 ′ ) can be inverted, contrary to (73 ′ )(102)19

• NB: this is not what happens in the standard model, where the mass comessolely from the coupling to the Higgs sector and a gauge freedom remains.However (102) is in lowest order equivalent to the Higgs mechanism in unitaritygauge• NB: propagators and external states are universal and models differ only in theparticle content and interaction vertices!3.2 Feynman Rulesexternal spin-1/2 particles:incoming:outgoing:pp⇐⇒ · · · u(p)⇐⇒ ū(p) · · ·(103a)(103b)external spin-1/2 anti particles:incoming:outgoing:pp⇐⇒ ¯v(p) · · ·⇐⇒ · · · v(p)(103c)(103d)external spin-1 particles:incoming:outgoing:kk⇐⇒ ɛ µ (k)⇐⇒ ɛ ∗ µ(k)(103e)(103f)• internal particles and anti particles with sign of momentum relative to the arrow(i. e. charge) direction:spin-1/2:p⇐⇒i/p − m + iɛ(104a)spin-1 (m = 0):k⇐⇒ −ig µν + i(1 − ξ) k µk νk 2 +iɛk 2 + iɛ(104b)• finallyspin-0:p⇐⇒ip 2 − m 2 + iɛ(105)∵ the S-matrix always contains an uninteresting diagonal piece (no interaction) andthe global momentum conservation δ-distributionS = 1 + (2π) 4 δ 4 (p 1 + p 2 − q 1 − q 2 . . . − . . . q n )iT (106)∴ we can focus on T20

• the following Feynman rules produce an expression for iT:1. draw all diagrams using propagators and interaction vertices that connectthe initial state with the final state2. assign momenta and external wave functions accordingly3. use momentum conservation at each vertex to fix the momenta of the internallines4. follow each connected fermion line against the direction of the arrows andcollect wave functions, propagators and vertices along the way5. complete iT with the remaining wave functions, propagators and vertices6. add all diagrams with the relative signs such that that sum is anti symmetricunder the exchange of identical external (anti-)fermions (and symmetric forbosons)• in diagrams with closed loops not all momenta are fixed by momentum conservation,e. g.:p − kkpk• in this case, one must integrate over these loop momenta with∫ d 4 p(2π) 4 · · · (107)• unfortunately, there are infinitely many loop diagrams for each process• fortunately, each loop comes with additional powers of the coupling constants• in weakly interacting theories, we can expand the scattering amplitudes in thecouplings constants or the number of loops3.3 Cross Section• definition in terms of observablesσ (∆Φ) = R (∆Φ)jwhere (108)∆Φ = region of phase spaceσ (∆Φ) = cross section for scattering into ∆ΦR (∆Φ) = event rate in ∆Φj = incoming flux(109a)(109b)(109c)(109d)21

• for fixed target experiments, the incoming flux j is just the number of incomingparticles per time and area• differential cross sectionσ (∆Φ) =∫∆Φdσ(Φ)dΦ (110)dΦwith phase space element dΦ, e. g. dΩ = sin θdθdφ for 2 → 2 processes• the differential cross section can be computed from the scattering amplitude Tand the (a. k. a. “Fermi’s golden rule”)• general formula for 2 → n processesp 2 q nq 2q 1dσ =p 1|T| 2 ˜dq√1 . . .∏˜dq n4 (p 1 p 2 ) 2 − m 2 1 m2 i n (2π) 4 δ 4 (p 1 + p 2 − q 1 − . . . − q n ) (111)i!2• where we have used again˜dp =d3 ⃗p(2π) 3 2p 0∣ ∣∣∣∣p0=√⃗p 2 +m 2 (112)NB: in the old days, people use a different normalization for fermions (an additionalfactor 2m), that is only useful for heavy (slow) particles . . .• symmetry factor3.4 Kinematicsn i ={ number of identical particlesof the species i in thefinal state(113)• simplest example: 2 → 2• invariants: Mandelstam variablesp 2 q 2p 1q 1s = (p 1 + p 2 ) 2 = (q 1 + q 2 ) 2 (total energy) (114a)t = (q 1 − p 1 ) 2 = (q 2 − p 2 ) 2 (momentum transfer) (114b)u = (q 1 − p 2 ) 2 = (q 2 − p 1 ) 2(114c)22

• momentum conservation:s + t + u = p 2 1 + p 2 2 + q 2 1 + q 2 2 =4∑m 2 i (115)i=1• center of mass system (CMS) at high energiesp 1/2 = (E; 0, 0, ±E)q 1/2 = (E; ±E sin θ cos φ, ±E sin θ sin φ, ±E cos θ)(116a)(116b)s = 4E 2 , t = −2E 2 (1 − cos θ) , u = −2E 2 (1 + cos θ) (117)3.5 Phase Space• two particles in the final state:∫d 3 ⃗q 1(2π) 3 2E 1d 3 ⃗q 2(2π) 3 2E 2(2π) 4 δ 4 (q 1 + q 2 − P)= 116π 2 ∫ |⃗q1 | 2 d|q 1 |dΩ 1E 1 E 2δ(E 1 (|⃗q 1 |) + E 2 (|⃗q 1 |) − E)= 1 ∫ |⃗q1 |E 1 dE 1 dΩ 1δ(E16π 2 1 + E 2 (E 1 ) − E)= 1 ∫|⃗q 1 |d cos θE 1 E 2 16π 2 1 dφ 1E(118)• the second equality in (118) follows from E 2 = |⃗q| 2 +m 2 , which yields |⃗q|d|⃗q| = EdEindependent of the masses• in the third equality we have used that E 2 depends on E 1 through momentumconservation and dispersion relationsd(E 1 + E 2 (E 1 ) − E)dE 1= 1 + E 1 /E 2 = E/E 2 (119)• special case: high energy limit in the center of mass frame: |⃗q 1 | = |⃗q 2 | = E/2 +O(m/|⃗q 2 | 2 ).d cos θ 1 dφ 1 / (32π 2 ) (118 ′ )23

4 QED• Quantum Electro Dynamics: interacting electrons, positrons and photons (andother charged particles)• single interaction vertex for fermions:pffk, µp ′ = iQ f eγ µ (120)with the electrical charge Q f of the particle, e. g. Q e = −1 for electrons• charged bosons require an additional (“sea gull”) vertexpsk, µk ′ , ν= iQ s e(p µ + p µ),′sp ′k, µ= 2iQ 2 se 2 g µν (121)4.1 e + e − → µ + µ − iT =¯v(p 2 )v(q 2 )−ieγ ρ−ig ρσ−ieγ σ (122)(p 1 + p 2 ) 2 + iɛu(p 1 )ū(q 1 )iT = ¯v(p 2 )(−ieγ ρ −ig ρσ)u(p 1 )(p 1 + p 2 ) 2 + iɛū(q 1)(−ieγ σ )v(q 2 )∑spinsTT † = e2s= i e2s [¯v(p 2)γ ρ u(p 1 )] [ū(q 1 )γ ρ v(q 2 )] (123)e 2s [¯v(p 2)γ ρ1 u(p 1 )ū(p 1 )γ ρ2 v(p 2 )] [ū(q 1 )γ ρ 1v(q 2 )¯v(q 2 )γ ρ 2u(q 1 )]TT † = e4s 2 tr [(/p 2 − m e )γ ρ1 (/p 1 + m e )γ ρ2 ]tr [(/q 1 + m µ )γ ρ 1(/q 2 − m µ )γ ρ 2](124)= e4s 2 L ρ 2 ρ 1(p 1 , p 2 , m e )L ρ 1ρ 2(q 1 , q 2 , m µ ) (125)24

4.2 Trace Techniques• consider a matrix elementū(p)Γu(q) =• using the tensor productwe can writeū(p)Γu(q) =4∑ū k (p)Γ kl u l (q) =k,l=14∑Γ kl u l (q)ū k (p) (126)k,l=1⎛⎞u 1 (q)ū 1 (p) · · · u 1 (q)ū 4 (p)⎜u(q) ⊗ ū(p) = ⎝.. ..⎟ . ⎠ (127)u 4 (q)ū 1 (p) · · · u 4 (q)ū 4 (p)4∑Γ kl [u(q) ⊗ ū(p)] lk =k,l=1• using the trace of a matrixtr(A) =4∑(Γ[u(q) ⊗ ū(p)]) kk (128)k=14∑A kk (129)k=1we can express a matrix element equivalently as a traceū(p)Γu(p) = tr(Γ[u(p) ⊗ ū(p)]) (130)• independent of the concrete realization of the Dirac matrices, we can computetheir traces using their anti commutation relations (34) alonetr(1) = 4(131a)tr (/a/b) = 1 () (34)tr (/a/b) + tr (/b/a) = tr(1) · ab = 4 · ab (131b)2tr (/a 1 ) = tr (/a 1 /a 2 /a 3 ) = tr (/a 1 /a 2 · · · /a 2n+1 ) (γ 5γ 5 = 1)= 0 (131c)tr (/a 1 /a 2 · · · /a n ) = tr (/a n · · · /a 2 /a 1 )tr(γ 5 ) = tr(γ 5 /a) = tr(γ 5 /a/b) = tr(γ 5 /a/b/c) = 0tr(γ 5 /a/b/c/d) = 4i · ɛ(a, b, c, d)only (131d) depends on the existence of a charge conjugation matrix(131d)(131e)(131f)• also from the anti commutation relations alone, we can prove contraction formulae:γ µ /aγ µ = −2 · /aγ µ /a/b/cγ µ = −2 · /c/b/aγ µ γ µ = 4γ µ /a/bγ µ = 4 · ab(132a)(132b)(132c)(132d)25

• it’s instructive to prove at least one of the trace theorems yourselves, because ithelps to memorize the result.Problem 13. Computeusing the cyclic invariance of the tracetr (/a/b/c/d) (133)and subsequently the anti commutation relations (34).tr(ABC) = tr(BCA) (134)Problem 14. Compute the trace L µν (p, q, m) = tr [(/p + m)γ µ (/q − m)γ ν ].Problem 15. Compute L ρ2 ρ 1(p 1 , p 2 , 0)L ρ 1ρ 2(q1 , q 2 , 0) as a function of the Mandelstam variablesin the high energy limit s, −t, −u ≫ m 2 i .4.3 Cross SectionProblem 16. Compute the differential cross sectionfor e + e − → µ + µ − in the region s, −t, −u ≫ m 2 e.Problem 17. Compute the integrated cross sectionfor e + e − → µ + µ − in the region s ≫ m 2 e.dσdΩ (cos θ, E CM) (135)σ(E CM ) (136)4.4 FORM• very efficient computation using programs for symbolic manipulation, e. g. FORM.• declare variables1: vector p1, p2, q1, q2;2: symbol s, t, u, me, mq;3: indices rho1, rho2;• expressions4: local [TT*] =5: (g_(1, p2) - me*g_(1)) * g_(1, rho1)6: * (g_(1, p1) + me*g_(1)) * g_(1, rho2)7: * (g_(2, q1) + mq*g_(2)) * g_(2, rho1)8: * (g_(2, q2) - mq*g_(2)) * g_(2, rho2);26

• traces9: trace4, 1;10: trace4, 2;11: print;12: .sort;• reduction to Mandelstam variables (114)s = (p 1 + p 2 ) 2 = 2m 2 e + 2p 1 p 2t = (q 2 − p 1 ) 2 = m 2 q + m 2 e − 2q 2 p 1(137a)(137b)etc.13: id p1.p2 = 1/2 * (s - 2*me^2);14: id q1.q2 = 1/2 * (s - 2*mq^2);15: id p1.q1 = - 1/2 * (t - me^2 - mq^2);16: id p2.q2 = - 1/2 * (t - me^2 - mq^2);17: id p1.q2 = - 1/2 * (u - me^2 - mq^2);18: id p2.q1 = - 1/2 * (u - me^2 - mq^2);• human intelligence (i. e. experience, cf. (211)): the result can be written mostcompactly as a function of t und u:19: id s = - u - t + 2*me^2 + 2*mq^2;20: bracket me, mq;21: print;22: .sort;running the program:ohl@thopad2:~fdfp$ form mumu.frmFORM by J.Vermaseren,version 3.3(Mar 28 2009) Run at: Thu Jul 7 18:37:42 2011[TT*] =64*me^2*mq^2 + 32*p1.p2*mq^2 + 32*p1.q1*p2.q2 + 32*p1.q2*p2.q1 + 32*q1.q2*me^2;[TT*] =+ mq^2 * ( - 32*u - 32*t )+ mq^4 * ( 48 )+ me^2 * ( - 32*u - 32*t )+ me^2*mq^2 * ( 96 )+ me^4 * ( 48 )+ 8*u^2 + 8*t^2;0.00 sec out of 0.00 sec27

4.5 Bhabha ScatteringProblem 18. Draw all Feynman diagrams for e + e − → e + e − (“Bhabha scattering”)• do we need to add or subtract the diagrams?• more general: what’s the relative phase of the diagrams?TT † = (T t − T s )(T t − T s ) † = T t T t † − T t T s † − T s T t † + T s T s†(138)• relative signs of the diagrams from permuting the endpoints of the fermion lines• equivalently: relative signs from the number of closed fermion lines in squareddiagramsT t T t † = (−1) 2 × , T t T s † = (−1) 1 × (139a)T s T t † = (−1) 1 × , T s T s † = (−1) 2 × (139b)4.6 FORM• declaration of variables as above• expressions• T s T † s just like in e + e − → µ + µ −T s = e 2 1 s [¯v(p 2)γ ρ u(p 1 )] [ū(q 1 )γ ρ v(q 2 )] (140)T t = e 2 1 t [¯v(p 2)γ ρ v(q 2 )] [ū(q 1 )γ ρ u(p 1 )] (141)4: local [SS*] =5: (g_(1, p2) - me*g_(1)) * g_(1, rho1)6: * (g_(1, p1) + me*g_(1)) * g_(1, rho2)7: * (g_(2, q1) + me*g_(2)) * g_(2, rho1)8: * (g_(2, q2) - me*g_(2)) * g_(2, rho2);• T t T † t similar9: local [TT*] =10: (g_(1, q1) + me*g_(1)) * g_(1, rho1)11: * (g_(1, p1) + me*g_(1)) * g_(1, rho2)12: * (g_(2, p2) - me*g_(2)) * g_(2, rho1)13: * (g_(2, q2) - me*g_(2)) * g_(2, rho2);28

• the interference terms are more complicated and contain traces of eight DiracmatricesT s T t ∗ = e 4 1 st [¯v(p 2)γ ρ1 u(p 1 )] [ū(p 1 )γ ρ 2u(q 1 )] [ū(q 1 )γ ρ 1v(q 2 )] [¯v(q 2 )γ ρ2 v(p 2 )]14: local [ST*] =15: (g_(1, p2) - me*g_(1)) * g_(1, rho1)16: * (g_(1, p1) + me*g_(1)) * g_(1, rho2)17: * (g_(1, q1) + me*g_(1)) * g_(1, rho1)18: * (g_(1, q2) - me*g_(1)) * g_(1, rho2);(142)T t T s ∗ = e 4 1 st [¯v(p 2)γ ρ1 v(q 2 )] [¯v(q 2 )γ ρ 2u(q 1 )] [ū(q 1 )γ ρ 1u(p 1 )] [ū(p 1 )γ ρ2 v(p 2 )]19: local [TS*] =20: (g_(1, p2) - me*g_(1)) * g_(1, rho1)21: * (g_(1, q2) - me*g_(1)) * g_(1, rho2)22: * (g_(1, q1) + me*g_(1)) * g_(1, rho1)23: * (g_(1, p1) + me*g_(1)) * g_(1, rho2);• FORM does the traces just the same . . .24: trace4, 1;25: trace4, 2;26: .sort;(143)• reduction to Mandelstam variables again as above, but all masses are equal27: id p1.p2 = 1/2 * (s - 2*me^2);28: id q1.q2 = 1/2 * (s - 2*me^2);29: id p1.q1 = - 1/2 * (t - 2*me^2);30: id p2.q2 = - 1/2 * (t - 2*me^2);31: id p1.q2 = - 1/2 * (u - 2*me^2);32: id p2.q1 = - 1/2 * (u - 2*me^2);• human intelligence and experience: the expression for |T s | 2 is most compact asfunction of t and u33: id s = - u - t + 4*me^2;34: bracket me;35: print;36: .sort;• the expression for |T t | 2 is most compact as function of s and u29

– . . . two different reductions in the same FORM program require advancedtricks . . .• running the program:ohl@thopad2:~fdfp$ form bhabha.frmFORM by J.Vermaseren,version 3.3(Mar 28 2009) Run at: Thu Jul 7 18:37:42 2011[SS*] =+ me^2 * ( - 64*u - 64*t )+ me^4 * ( 192 )+ 8*u^2 + 8*t^2;[TT*] =+ me^2 * ( - 64*u )+ me^4 * ( 64 )+ 16*u^2 + 16*t*u + 8*t^2;[ST*] =+ me^2 * ( 64*u )+ me^4 * ( - 96 )- 8*u^2;[TS*] =+ me^2 * ( 64*u )+ me^4 * ( - 96 )- 8*u^2;0.00 sec out of 0.00 sec• NB: |T s | 2 (t, u) = |T t | 2 (s, u)• final result very symmetrical∑spinsProblem 19. Compute the differential cross section( )tTT ∗ = 8e 4 2 + u 2+ s2 + u 2+ 2u2 + O(m 2s 2 t 2 ste) (144)for Bhabha scattering in the region s, −t, −u ≫ m 2 e.dσ(cos θ) (145)dΩ30

5 QCD5.1 Feynman Rules∵ full account requires another full set of lectures∴ just a few SU(N) formulae– generators T a and totally anti symmetric structure constants f abc :(summations for c = 1, 2, . . . , (N 2 c − 1) implied).– normalization and contractions[T a , T b ] = if abc T c (146)tr (T a T b ) = 1 2 δ ab• physical degrees of freedom: quarks and gluons:(147a)T a T a = C F · 1 = N2 C − 1 · 12N C(147b)f acd f bcd = C F · δ ab(147c)k, µ, app ′ = igγ µ T a (148)• triple gluon couplings1gf a1 a 2 a 3g µ1 µ 2(k 1 µ 3− k 2 µ 3)23=+gf a1 a 2 a 3g µ2 µ 3(k 2 µ 1− k 3 µ 1)+gf a1 a 2 a 3g µ3 µ 1(k 3 µ 2− k 1 µ 2)(• quartic gluon couplings21−ig 2 f a1 a 2 bf a3 a 4 b×(g µ1 µ 3g µ4 µ 2− g µ1 µ 4g µ2 µ 3)=−ig 2 f a1 a 3 bf a4 a 2 b×(g µ1 µ 4g µ2 µ 3− g µ1 µ 2g µ3 µ 4)(3 4−ig 2 f a1 a 4 bf a2 a 3 b×(g µ1 µ 2g µ3 µ 4− g µ1 µ 3g µ4 µ 2)• Faddeev-Popov ghosts appear only in loop diagrams31

5.2 3-Jet Production• quarks are electrically chargedk, µ, app ′ = − ieQ q γ µ (and couple in the standard model (cf. (176) and (178)) to the Z 0 and W ± as well.Problem 20. Draw all Feynman diagrams without closed loops, that follow from the Feynmanrules of QED und QCD for the electroproction of 3 jetse + + e − → q + ¯q + g(at higher energies, there are additional contributions from Z 0 bosons).Write down the corresponding algebraic expression for the T-matrix.Problem 21. Verify the Ward identity for the gluonT 1∣∣ɛ∗ µ (k)=k µ+ T 2∣∣ɛ∗ µ (k)=k µ= 0 , (152)i. e. the fact that no gluons with unphysical polarization ɛ ∗ µ(k) = k µ are produced. Hint: usethe Dirac equation for the external quarks, e. g.1−/p 1 − /k − m + iɛ (/k + /p 1 + m) v(p 1 ) = −v(p 1 ) (für k 0) (153)• down the road, some additional shorthands will be usefulT n = e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 s Jµ n(q 1 , q 2 , k, ɛ) (154)with the current matrix elementJ µ (q 1 , q 2 , k, ɛ) = J µ 1 (q 1, q 2 , k, ɛ) + J µ 2 (q 1, q 2 , k, ɛ)J µ 1 (q 1, q 2 , k, ɛ) = ū(q 1)T a /ɛ ∗ (k)(/q 1 + /k + m)γ µ v(q 2 )(q 1 + k) 2 − m 2 + iɛJ µ 2 (q 1, q 2 , k, ɛ) = ū(q 1)γ µ (−/q 2 − /k + m)T a /ɛ ∗ (k)v(q 2 )(q 2 + k) 2 − m 2 + iɛ(155a)(155b)(155c)• this allows to write∑|T 1 + T 2 | 2 = e4 g 2 Q 2spins, pol.s 2 L µν (p 1 , p 2 , 0)H µν (q 1 , q 2 , k) (156)32

with the “hadronic tensor”H µν (q 1 , q 2 , k) = ∑spins,ɛ∵ with the same trick as in problem 21, we can show thatJ µ (q 1 , q 2 , k, ɛ)J ν,∗ (q 1 , q 2 , k, ɛ) (157)[q µ 1 + qµ 2 + kµ ] J µ (q 1 , q 2 , k, ɛ) = 0 (158)∴ using the center of mass momentum p = p 1 + p 2 = q 1 + q 2 + kp µ H µν (q 1 , q 2 , k) = p ν H µν (q 1 , q 2 , k) = 0 (159)• angular dependence of H µν (q 1 , q 2 , k) contains a lot of information . . .∵ . . . but the energy dependence is much simpler∴ integrate over the anglesx 1 = 2q 1 p/p 2 , x 2 = 2q 2 p/p 2 , x 3 = 2kp/p 2 (160)∫˜dq 1˜dq2 ˜dk (2π) 4 δ 4 (q 1 + q 2 + k − p) f(x 1 , x 2 , x 3 )afterwards, thre result will depend only on p and the x i .∵ from (159) we find∫( )pd ˜Ω H µν µ p ν(q 1 , q 2 , k) = − g µνp 2∴˜H(p, x 1 , x 2 ) = − 1 3= s128π 3 ∫dx 1 dx 2 f(x 1 , x 2 , 2 − x 1 − x 2 ) (161)˜H(p, x 1 , x 2 ) (162)∫d ˜Ω H µ µ(q 1 , q 2 , k) (163)• energy conservation• momentum conservation for massless particleswith equality for parallel q 1 und q 2x 1 + x 2 + x 3 = 2(q 1 + q 2 + k)pp 2 = 2 (164)x 1 + x 2 x 3 = 2 − x 1 − x 2 (165)∴x 1 + x 2 1 (165 ′ )33

x 210.80.60.40.2x 3 = 100 0.2 0.4 0.6 0.8 1x 1x 3 = 0• for massless particles∑spins, pol.∫d ˜Ω |T 1 + T 2 | 2 = 4e4 g 2 Q 2×( pµ p νss 2 (p µ 1 pν 2 + p ν 1 p µ 2 − p 1p 2 g µν )− g µν)˜H(p, x 1 , x 2 ) = 4e4 g 2 Q 2s˜H(p, x 1 , x 2 ) (166)• add phase space factord 2 σ= 1 s 1dx 1 dx 2 2s 128π 3 4∑∫d ˜Ω |T 1 + T 2 | 2 = α sα 2 Q 2spins, pol.4s˜H(p, x 1 , x 2 ) (167)Problem 22. Express the invariants q 1 q 2 , q 1 k and q 2 k by s and the x i , assuming all particlesto be massless.Problem 23. Computefor massless particles.d 2 σdx 1 dx 2(x 1 , x 2 ) (168)• start with computing H µ µ(q 1 , q 2 , k) as a function of q 1 q 2 , q 1 k and q 2 k• since (152), you may use ∑ ɛ ɛ µɛ ∗ ν = −g µν• use the contraction identities (132) before calculating the traces!34

6 Standard Model(C, T) Y Q = T 3 + Y 2leptonsν e,R ν µ,R ν τ,R (1, 1) 0 0e R ) (µ R ) (τ R )(1, 1) −2 (− 1)νµ,L ντ,L0(1, 2)e L µ L τ −1L −1(νe,Lquarks2u R c R t R (3, 1) 4/3 3d R) (s R) (b R)(3, 1) −2/3(− 1 3 2)cL tL(3, 2)3d L s L b 1/3L − 1 3(uLgauge bosonsA Z 0 W ± gHiggsΦ (1, ?) ? ?6.1 Propagators• external spin-1 particlesincoming:outgoing:kk⇐⇒ ɛ µ (k)⇐⇒ ɛ ∗ µ(k)(169a)(169b)• internal particles and anti particlesSpin-1 (m 0):k⇐⇒ −ig µν + i k µk νM 2k 2 − M 2 + iΓM(170)• polarization sum:∑pol.ɛ µ (k)ɛ ∗ ν(k) = −g µν + k µk νM 2 (171)• unitarity gauge: ∂ µ D µν = 0, other options−ig µν + i(1 − ξ) k µk νk 2 −ξM 2k 2 − M 2 + iΓMbetter for radiative corrections, but larger intermediate terms(172)• finite width Γ: (tricky for charged particles, but required)|D(p, M)| 2 ∝1(p 2 − M 2 ) 2 + Γ 2 M 2 (173)35

6.2 Feynman Rules• vector and axial vector couplingsg V = T 3 − 2Q sin 2 θ wg A = T 3(174a)(174b)• e. g. for electrons• neutral currentsfpfpZ 0fγfg V = − 1 − 4 sin2 θ w2k, µ= − ik, µ, g A = − 1 2g ( )gf2 cos θ V γ µ − g f Aγ µ γ 5w(175)(176)p ′ = − ieQ f γ µ (177)• charged currents (V ff ′ is the CKM matrix)pfk, µW −f ′ = − i √ g V ff ′τ + 1 − γ 5γ µ2 2p ′(178)• triple gauge couplingsk 1 , µ 1W −k Z 02 , µ 2 =W − k 3 , µ 3k 1 , µ 1W − γk 2 , µ 2 =W − k 3 , µ 3ie cot θ w g µ1 µ 2(k 1 µ 3− k 2 µ 3)+ie cot θ w g µ2 µ 3(k 2 µ 1− k 3 µ 1)+ie cot θ w g µ3 µ 1(k 3 µ 2− k 1 µ 2)ieg µ1 µ 2(k 1 µ 3− k 2 µ 3)+ieg µ2 µ 3(k 2 µ 1− k 3 µ 1)+ieg µ3 µ 1(k 3 µ 2− k 1 µ 2)(179)(180)36

• Yukawa couplingspp, µp, µfZ 0W ±kH= − i gm f(181)f2M Wp ′kHZ 0=p ′ , µ ′kHW ±i gM Zcos θ wg µµ ′ (182)p ′ , µ ′ = igM W g µµ ′ (183)• a lot more vertices: quartic couplings, Higgs selfcouplings, etc.6.3 Higgs StrahlungProblem 24. Compute the scattering amplitude for Higgs strahlung e + e − → ZHe + (p + )H(k)(184)e − (p − )Z 0 Z 0Z 0 (q)ignoring all terms of O(m e /M Z ), O(m e /M H ) and O(m e / √ s). You may assume that √ s ≫M 2 Z and ignore the width of the Z.Problem 25. Show that the four momentap − = (E; 0, 0, E)p + = (E; 0, 0, −E)k = (E H ; p sin θ cos φ, p sin θ sin φ, p cos θ)q = (E Z ; −p sin θ cos φ, −p sin θ sin φ, −p cos θ)(185a)(185b)(185c)(185d)withp =√λ(s, M2H, M 2 Z )2 √ s(186a)E H = s + M2 H − M2 Z2 √ sE Z = s + M2 Z − M2 H2 √ s(186b)(186c)37

andλ(a, b, c) = a 2 + b 2 + c 2 − 2ab − 2ac − 2bc (187)satisfy energy and momentum conservation and the mass shell conditions for e + e − → ZHwith m e = 0. Compute16(p + q)(p − q) . (188)Problem 26. Compute the differential cross sectionfor unpolarized Higgs strahlung e + e − → ZH.NB: use the fact that ∑ ɛ µ ɛ ∗ ν is symmetric in µ ↔ ν.dσdΩ (θ e − H) (189)Problem 27. Compute the integrated cross section for Higgs strahlung e + e − → ZH.38

7 SolutionsSolution 1.∂ µ e −ipx = −ip µ e −ipx(a∂)(b∂)e −ipx = −(ap)(bp)e −ipx∂ 2 e −ipx = −i(p∂)e −ipx = −p 2 e −ipx(190a)(190b)(190c)Solution 2.∂ µ x µ = gµ ν ∂x µ /∂x ν = gµ ν δ µ ν = g µ µ = 4 (191a)( )∂ 2 e −x2 /2 = −∂ µ x µ e −x2 /2= −x µ ∂ µ e −x2 /2 − (∂ µ x µ ) e −x2 /2= x µ x µ e −x2 /2 − g µµ e −x2 /2 = (x 2 − 4)e −x2 /2(191b)Solution 3.( ) ( )[γ k , γ l] 0 σ = k 0 σl+ −σ k 0 −σ l + (k ←→ l) =0=( )−σ k σ l 00 −σ k σ l( ) (−[σ k , σ l ] + 00 −[σ k , σ l = −2δ] kl 1 0+ 0 1+ (k ←→ l))(192a)Solution 4.ū k (p) = u † k (p)γ 0 = u † k (⃗ 0)γ 0 γ 0¯v k (p) = ¯v k ( ⃗ 0)/p − m√p0 + m/p † + m√p0 + m γ 0 = ū k ( ⃗ /p + m0) √p0 + m(193a)(193b)Using the definition and multiplying with γ 0 from the right2∑k=1u k ( ⃗ 0)ū k ( ⃗ 0) = γ 0 + 1,22∑k=1v k ( ⃗ 0)¯v k ( ⃗ 0) = γ 0 − 12(194)39

Therefore2∑k=12∑k=1u k (p)ū k (p) = (/p + m) (γ 0 + 1) (/p + m)2(p 0 + m)v k (p)¯v k (p) = (/p − m) (γ 0 − 1) (/p − m)2(p 0 + m)(195a)(195b)and (60) follows from(/p ± m)(γ 0 ± 1)(/p ± m) = /pγ 0 /p ± (mγ 0 /p + m/pγ 0 ) + m 2 γ 0 ± (/p ± m) 2= −p 2 γ 0 + 2p 0 /p ± 2p 0 m + m 2 γ 0 + 2m(/p ± m) = 2(p 0 + m)(/p ± m) . (196)Solution 5.ū k (p)u l (p) = ū k ( ⃗ 0)/p + m√p0 + m/p + m√p0 + m u l( ⃗ 0)= 2mp 0 + mūk( ⃗ 0)(/p + m)u l ( ⃗ 0)= 2mp 0 + mūk( ⃗ 0)(p 0 + m)u l ( ⃗ 0) = 2mū k ( ⃗ 0)u l ( ⃗ 0) = 2mδ kl (197)¯v k (p)v l (p) = ¯v k ( ⃗ 0)/p − m√p0 + m/p − m√p0 + m v l( ⃗ 0)= −2mp 0 + m ¯v k( ⃗ 0)(/p − m)v l ( ⃗ 0)= 2mp 0 + m ¯v k( ⃗ 0)(p 0 + m)v l ( ⃗ 0) = 2m¯v k ( ⃗ 0)v l ( ⃗ 0) = −2m · δ kl (198)ū k (p)v l (p) = ū k ( ⃗ 0)¯v k (p)u l (p) = ¯v k ( ⃗ 0)/p + m√p0 + m/p − m√p0 + m/p − m√p0 + m v l( ⃗ 0) = 0/p + m√p0 + m u l( ⃗ 0) = 0(199a)(199b)40

Solution 6.ū k (p)γ µ u l (p) = ū k ( ⃗ 0)/p + m√p0 + m γ /p + mµ √p0 + m u l( ⃗ 0)=1p 0 + mūk( ⃗ 0)2p µ (/p + m)u l ( ⃗ 0)= 2p µp 0 + mūk( ⃗ 0)(p 0 + m)u l ( ⃗ 0) = 2p µ ū k ( ⃗ 0)u l ( ⃗ 0) = 2p µ δ kl(200a)¯v k (p)γ µ v l (p) = ¯v k ( ⃗ 0)/p − m√p0 + m γ /p − mµ √p0 + m v l( ⃗ 0)=1p 0 + m ¯v k( ⃗ 0)2p µ (/p − m)v l ( ⃗ 0)= −2p µp 0 + m ¯v k( ⃗ 0)(p 0 + m)v l ( ⃗ 0) = −2p µ¯v k ( ⃗ 0)v l ( ⃗ 0) = 2p µ δ kl(200b)ū k (⃗p)γ 0 v l (−⃗p) = ū k ( ⃗ 0)/p + m√p0 + m γ /p † − m0 √p0 + m v l( ⃗ 0)= ū k ( ⃗ 0)/p + m√p0 + m/p − m√p0 + m γ 0v l ( ⃗ 0) = 0(201a)¯v k (−⃗p)γ 0 u l (⃗p) = 0(201b)Solution 7.γ 5 γ 2 = iγ 0 γ 1 γ 2 γ 3 γ 2 = −iγ 0 γ 1 γ 2 γ 2 γ 3 = iγ 0 γ 2 γ 1 γ 2 γ 3 = −iγ 2 γ 0 γ 1 γ 2 γ 3 = −γ 2 γ 5 (202)since every γ µ appears exaclty once and anti-commutes with the other three, we have for every µ[γ 5 , γ µ ] + = 0 (203)Solution 8.product rulei∂ [ µ ¯ψ 1 (x)γ µ ψ 2 (x) ] (= ¯ψ 1 (x) i ←− /∂ + i −↛ )( )∂ ψ 2 (x) = ¯ψ 1 (x) −m + m ψ 2 (x) = 0 . (204)41

Solution 9.product rule and [γ 5 , γ µ ] + = 0i∂ [ µ ¯ψ 1 (x)γ µ γ 5 ψ 2 (x) ] = ¯ψ 1 (x)(i ←− /∂ γ 5 + i −↛ )(∂ γ 5 ψ 2 (x) = ¯ψ 1 (x) i ←− /∂ γ 5 − γ 5 i −↛ )∂ ψ 2 (x)()= ¯ψ 1 (x) −mγ 5 − γ 5 m ψ 2 (x) = −2m ¯ψ 1 (x)γ 5 ψ 2 (x) . (205)Solution 10.( ) ( )0 σk 0 σl− 2iσ kl =−σ k 0 −σ l − (k ←→ l)0( )( )−[σ=k , σ l ] − 0σ0 −[σ k , σ l = −2iɛ klm m 0] − 0 σ m(206)therefore( ) σσ kl = ɛ klm m 00 σ m(207)Solution 11.(i/∂ − m) S(x, m) = ( −∂ 2 − m 2) D(x, m) = δ 4 (x) (208)Solution 12.∫ d 4 p1S(x, m) = (i/∂ + m)(2π) 4 e−ipx p 2 − m 2 +iɛ∫ d 4 ∫p /p + m d 4=(2π) 4 e−ipxp 2 − m 2 +iɛ = p 1(2π) 4 e−ipx /p − m+iɛ (209)Solution 13.tr (/a/b/c/d) = + tr (/b/c/d/a) = − tr (/b/c/a/d) + 2 · ad · tr (/b/c)= + tr (/b/a/c/d) − 2 · ac · tr (/b/d) + 2 · ad · 4 · bc= − tr (/a/b/c/d) + 2 · ab · tr (/c/d) − 2 · ac · 4 · bd + 2 · ad · 4 · bcthereforetr (/a/b/c/d) = 4 · (ab · cd − ac · bd + ad · bc) (133 ′ )42

Solution 14.L µν = tr [(/p + m)γ µ (/q − m)γ ν ] = tr [/pγ µ /qγ ν ] − m 2 tr [γ µ γ ν ]= 4 · (p µ q ν − g µν pq + p ν q µ ) − 4m 2 · g µν = 4 · (pµ q ν + p ν q µ − (pq + m 2 )g µν)(210)Solution 15.L ρ2 ρ 1L ρ 1ρ 2= 16 · (p 1,ρ2 p 2,ρ1 + p 1,ρ1 p 2,ρ2 − p 1 p 2 g ρ2 ρ 1)(q ρ 11 qρ 22+ q ρ 21 qρ 12− q 1 q 2 g ρ 1ρ 2)())= 8 · 2(p 1 q 2 )2(p 2 q 1 ) + 2(p 1 q 1 )2(p 2 q 2 ) = 8 ·(u 2 + t 2 (211)NB: cross terms cancel:(p 1,ρ2 p 2,ρ1 + p 1,ρ1 p 2,ρ2 )q 1 q 2 g ρ 1ρ 2+ p 1 p 2 g ρ2 ρ 1(q ρ 11 qρ 22+ q ρ 21 qρ 12 ) = p 1p 2 g ρ2 ρ 1q 1 q 2 g ρ 1ρ 2Solution 16.thereforet = − s 2 (1 − cos θ) , u = −s (1 + cos θ) (212)2t 2 + u 2s 2= 1 + cos2 θ2(213)dσ =therefore11 1√∏4 (p 1 p 2 ) 2 − m 2 1 m2 i n i! 4 |T|2˜dq1˜dq2 (2π) 4 δ 4 (p 1 + p 2 − q 1 − q 2 )21=4 √ 1 11 1(s/2) 2 4 |T|2 d cos θdφ =32π2 64π 2 s 4 |T|2 dΩ (214)dσdΩ = 1 ( 64π 2 s e4 1 + cos 2 θ ) = α 2 1 (1 + cos 2 θ ) (215)4sSolution 17.∫σ = dΩ dσ ∫ 1dΩ = α24s 2π d cos θ ( 1 + cos 2 θ ) =(2 α2−14s 2π + 2 )3comparewithσ =σ == 4πα23s(216)4πα 23( √ s/TeV) 2 0.39 nb (19′′ )α = e24π = 1137.0359895(61)(217)87 fb( √ s/TeV) = 8.7 pb2 ( √ (19 ′′ )s/100 GeV) 243

Solution 18.e − (p 1 )e − (q 1 )iT t =γ[, Z 0 ](218a)e + (p 2 )e − (p 1 )e + (q 2 )e − (q 1 )iT s =γ[, Z 0 ](218b)e + (p 2 )e + (q 2 )Solution 19.∵∴• finallydσdΩ = α22s(1 + cos 2 θ2t = − s ( ) θ2 (1 − cos θ) = −s sin2 (219a)2u = − s ( ) θ2 (1 + cos θ) = −s cos2 (219b)2s 2 + u 2= 1 + ( )cos4 θ2t 2 sin ( ) (220a)4 θ2u ( )2θst = −cos4 2sin ( ) (220b)2 θ+ 1 + ( ) ( ))cos4 θ2sin ( ) − 2 cos4 θ24 θsin ( )2 θ222= α24s( ) 3 + cos 2 2θ(221)1 − cos θSolution 20.¯v(p 2 )ieγ µiT 1 =u(p 1 )¯v(p 2 )iT 2 =ieγ µu(p 1 )−ig µν(p 1 + p 2 ) 2 + iɛi/q 1 + /k − m igT + iɛ a γ ρ−ig µν(p 1 + p 2 ) 2 + iɛi−/q 2 − /k − m + iɛ44igT a γ ρ−ieQγ ν−ieQγ νv(q 2 )ū(q 1 )v(q 2 )ū(q 1 )ɛ ∗ ρ(k)ɛ ∗ ρ(k)(222a)(222b)

[]iT 1 = ū(q 1 )(igT a /ɛ ∗ i(k))/q 1 + /k − m + iɛ (−ieQγµ )v(q 2 ) × (223a)−i(p 1 + p 2 ) 2 + iɛ [¯v(p 2)(ieγ µ )u(p 1 )][]iT 2 = ū(q 1 )(−ieQγ µ i)−/q 2 − /k − m + iɛ (igT a/ɛ ∗ (k))v(q 2 ) × (223b)−i(p 1 + p 2 ) 2 + iɛ [¯v(p 2)(ieγ µ )u(p 1 )]Solution 21.∣T ∣ɛ 1 ∗ µ (k)=k µ= e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 1sū(q 1)T a (/k + /q 1 − m)/q 1 + /k − m + iɛ γµ v(q 2 )on the other hand= e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 sū(q 1)T a γ µ v(q 2 ) (224)T 2∣∣ɛ∗ µ (k)=k µ= e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 sū(q 1)γ µ 1−/q 2 − /k − m + iɛ (/k + /q 2 + m)T a v(q 2 )= −e 2 Qg [¯v(p 2 )γ µ u(p 1 )] 1 sū(q 1)T a γ µ v(q 2 ) = −T 1∣∣ɛ∗ µ (k)=k µ(225)Solution 22.2q 1 q 2 = (q 1 + q 2 ) 2 = (p − k) 2 = s(1 − x 3 )2q 1 k = (q 1 + k) 2 = (p − q 2 ) 2 = s(1 − x 2 )2q 2 k = (q 2 + k) 2 = (p − q 1 ) 2 = s(1 − x 1 )(226a)(226b)(226c)Solution 23.four traces:H µν (q 1 , q 2 , k) =∑ɛ+ ∑ ɛtr[/q 1 T a /ɛ ∗ (/q 1 + /k)γ µ /q 2 γ ν (/q 1 + /k)/ɛT a ]+ ∑ (2q 1 k) 2 ɛtr[/q 1 γ µ (−/q 2 − /k)T a /ɛ ∗ /q 2 γ ν (/q 1 + /k)/ɛT a ]+ ∑ (2q 1 k)(2q 2 k)ɛtr[/q 1 T a /ɛ ∗ (/q 1 + /k)γ µ /q 2 /ɛT a (−/q 2 − /k)γ ν ](2q 1 k)(2q 2 k)tr[/q 1 γ µ (−/q 2 − /k)T a /ɛ ∗ /q 2 /ɛT a (−/q 2 − /k)γ ν ](2q 2 k) 2 (227)45

color part of the quark traces tr(T a T a ) = C F tr(1) = C F N C and polarization sumH µν tr[/q 1 (/q 1 + /k)γ µ /q 2 γ ν (/q 1 + /k)] tr[/q 1 /q 2 γ µ (/q 1 + /k)(−/q 2 − /k)γ ν ](q 1 , q 2 , k) = 2C F N c +2C(2q 1 k) 2 F N c(2q 1 k)(2q 2 k)+ 2C F N ctr[/q 1 γ µ (−/q 2 − /k)(/q 1 + /k)γ ν /q 2 ](2q 1 k)(2q 2 k)+ 2C F N ctr[/q 1 γ µ (−/q 2 − /k)/q 2 (−/q 2 − /k)γ ν ](2q 2 k) 2 (228)contractionH µ tr[/q 1 (/q 1 + /k)/q 2 (/q 1 + /k)]tr[/q 1 /q 2 ]µ(q 1 , q 2 , k) = −4C F N c +8C(2q 1 k) 2 F N c (q 1 +k)(−q 2 −k)(2q 1 k)(2q 2 k)tr[/q 1 /q 2 ]+ 8C F N c (−q 2 − k)(q 1 + k)(2q 1 k)(2q 2 k) − 4C tr[/q 1 (−/q 2 − /k)/q 2 (−/q 2 − /k)]FN c (229)(2q 2 k) 2final tracesH µ 2(q 1 k)(q 2 k) (q 1 q 2 + q 1 k + q 2 k)(q 1 q 2 ) 2(q 1 k)(q 2 k)µ(q 1 , q 2 , k) = −16C F N c −64C(2q 1 k) 2 F N c −16C F N c(2q 1 k)(2q 2 k)(2q 2 k) 21 − x 1(1 − x 3 )−8C F N c −16C F N c1 − x 2 (1 − x 1 )(1 − x 2 ) −8C 1 − x 2x 2 1FN c = −8C F N + x2 2c1 − x 1 (1 − x 1 )(1 − x 2 )(230)resultd 2 σ 4πα 2 Q 2 α s= N cdx 1 dx 2 3s 2π C x 2 1 + x2 2F(1 − x 1 )(1 − x 2 )(231)Solution 24.with p = p + + p − :giT = −i [¯v(p + )γ µ (g e V − g e2 cos θAγ 5 )u(p − )] −igµν + ip µ p ν /M 2 Z(i gM Zgw p 2 − M 2 νρ )ɛ ∗,ρ (q)Zcos θ w(232)and from current conservation [¯v(p + )γ µ (g e V − ge A γ 5)u(p − )] p µ = −2ig e A m e¯v(p + )γ 5 u(p − ) =O(m e )T = − g2 M Z 1[¯v(p2 cos 2 θ w s − M 2 + )/ɛ ∗ (q)(g e V − g e Aγ 5 )u(p − )] (233)ZSolution 25.• momentum conservation is obvious and energy conservation follows fromE H + E Z = √ s = 2E (234)46

• the Higgs mass shell follows fromE 2 H − p 2 = s2 + M 4 H + M4 Z +2sM2 H − 2sM2 Z − 2M2 H M2 Z4s− s2 + M 4 H + M4 Z −2sM2 H − 2sM2 Z − 2M2 H M2 Z4s= M 2 H , (235)and the Z mass shell analogously: E 2 Z − p2 = M 2 Z .• finally4p ∓ q = 4E(E Z ± p cos θ) = s + M 2 Z − M 2 H ±√λ(s, M 2 H , M2 Z) cos θ , (236)i. e.16(p + q)(p − q) = (s + M 2 Z − M 2 H) 2 − λ(s, M 2 H, M 2 Z) cos 2 θ= 4sM 2 Z + λ(s, M 2 H, M 2 Z) sin 2 θ (237)Solution 26.∑|T| 2 =spinsg4 M 2 Z 14 cos 4 θ w (s − M tr (/p +/ɛ ∗ (q)(g e 2Z )2 V − g e Aγ 5 )/p − (g e V + g e Aγ 5 )/ɛ(q))= g4 M 2 Z 14 cos 4 θ w (s − M 2 (ge 2 Z )2 V + g e 2 A ) tr (/p + /ɛ ∗ (q)/p − /ɛ(q))= g4 M 2 Z (ge 2 V + g e 2 A ) 14 cos 4 θ w (s − M 2 Lµν (p + , p − , 0)ɛ ∗ µ(q)ɛ ν (q) (238)Z )2polarization sum∑pol.ɛ µ (q)ɛ ∗ ν(q) = −g µν + q µq νM 2 Z(239)∑spins, pol.|T| 2 = g4 M 2 Z (ge 2 V + g e 2 A ) 1 (pµcos 4 θ+p νw (s − M 2 Z )2 − + p µ −p ν + − p + p − g µν) ×(−g µν + q )µq νM 2 Z= g4 (g e 2 V + g e 2 A ) p + p − M 2 Z + 2(p +q)(p − q)cos 4 θ w (s − M 2 Z )2= g4 (g e 2 V + g e 2 A ) 8sM 2 Z + λ(s, M2 H , M2 Z ) sin2 θ(240)8 cos 4 θ w (s − M 2 Z )247

phase space (118)dσdΩ (θ e − H) = 1 √1 λ(s, M2H, M 2 Z ) 12s 16π 2 2s 4∑|T| 2 (241)spins, pol.i. e.dσdΩ (θ e − H) =√λ(s, M2H, M 2 Z )sα 2 (g e V 2 + g e A2 )128s sin 4 θ w cos 4 θ w8sM 2 Z + λ(s, M2 H , M2 Z ) sin2 θ(s − M 2 Z )2 (242)Solution 27.with∫dΩ (a + b sin 2 θ) = 4πa + 8π 3 b (243)followsorσ =√λ(s, M2H, M 2 Z )sπα 2 (g e 2 V + g e A2 ) 12sM 2 Z + λ(s, M2 H , M2 Z ) (244)48s sin 4 θ w cos 4 θ w (s − M 2 Z )2σ =√λ(s, M2H, M 2 Z )sπα 2 (1 + (1 − 4 sin 2 θ w ) 2 )192s sin 4 θ w cos 4 θ w12sM 2 Z + λ(s, M2 H , M2 Z )(s − M 2 Z )2 (245)σ/fb25020015010050M H = 100 GeVM H = 120 GeVM H = 140 GeV0150 200 250 300 350 400 450√ s/GeV48