4442 Particle Physics Ryan Nichol Module 5 QED http://www.hep.ucl.ac.uk/~rjn/teaching/4442
Introduction • We’ve now developed a theory that incorporates all the necessities to make it useful i.e. predictive and amenable to falsification or validation by experiment. • We’ve incorporated : –Lorentz invariance (+ Special Relativity) –Quantum mechanics, spin and anti-particles (Dirac Eqn) –Gauge invariance (no dependence on “un-measurables”) –Interactions and a tractable formalism (Lagrangians) • The full formalism in Quantum Field Theory as embodied in the Standard Model has been subject to thousands of measurements never been found wrong. • But in the words of D. Rumsfeld we know there are “known unknowns”. � 2
QED Feynman Rules Vertex Factors γ ie γ µ e − e − Propagators q 2 − m 2 + i ✏ = − i ( / ( / p + m ) p + m ) − i e − e − q 2 − m 2 ✓ ◆ − i g µ ν − (1 − ⇠ ) p µ p ν = − ig µ ν µ q 2 + i ✏ ν q 2 q 2 γ External Lines U ( P ) U ( P ) e − e − V ( P ) V ( P ) e + e + ✏ ∗ µ ( P ) ✏ µ ( P ) γ γ � 3 Incoming Outgoing
QED Calculations • Will consider eµ → eµ scattering • Feynman Rules –Incoming spinors , outgoing spinors U U iq � µ –Vertices so for electrons − ie � µ − ig µ ν –Photon propagators q 2 –The usual delta functions and integrals over internal momenta –Multiply together and by i and integrate to get M • The matrix element is just a (complex) number • Calculations simplify if we sum over all spins ** ✓ s 2 + u 2 ◆ ⌦ | M | 2 ↵ = 2 e 4 t 2 � 4
Trace Algebra • There are a number of algebraic “tricks” that simplify calculation of Matrix element calculations Tr ( I ) = 4 Tr ( A + B ) ≡ Tr ( A ) + Tr ( B ) Tr ( AB....Y Z ) ≡ Tr ( ZAB...Y ) Tr ( γ µ γ ν ) = 4 g µ ν Tr (odd number γ ) = 0 Tr ( γ µ γ ν γ ρ γ σ ) = 4( g µ ν g ρσ − g µ ρ g νσ + g µ σ g νρ ) � 5
Experimental Evidence • Angular distributions in –eµ → eµ scattering –ee → µµ scattering –eq → eq scattering –The importance of spin, and helicity conservation –Interference amplitudes, A FB � • Demonstration of quark properties –spin 1/2 –fractional charge –carry colour � 6
Cross section for scattering processes • Derivation of cross section formula for e+e- → µ+µ-, e-µ- → e-µ- (**) • Correspondence between helicity and chiral operators • No chirality/handedness changing processes occur in any pure vector i.e. QCD + QED or axial vector interactions -2 10 ω φ J / ψ -3 10 ψ (2 S ) Υ ρ ′ ρ -4 Z 10 -5 σ [mb] 10 -6 10 -7 10 -8 10 2 1 10 10 � 7
Angular Distribution • QED+QCD angular distribution is symmetric due to chirality (aka helicity at high energy) and hence parity conservation of vector interactions and because we have a spin=1 boson being exchanged • Deviation from pure QED expectation is due to interference terms from Z exchange at level of ~ 10 -4 s *** • Forward-backward asymmetries : A FB ; e + e - → e + e - difference & HO corrections � 8
More Angular Distributions • Angular distributions are used to determine spins of interacting & exchanged particles –evidence that quarks are spin 1/2 (see later) –search for new physics e.g. demonstrate that spin of mythical graviton = 2 not 1 Simulation of results that we could see in 2012 if LHC finds the graviton. � 9
Evidence for Quark Properties • Spin 1/2 : angular distribution of quark jets in e+e- → qq or eq → eq • Fractionally charged • Carry colour e + e - → qq � 10
Quark jets at the LHC � 11
v’s v’s Evidence of quark charge • R = ratio of e + e - ->hadrons/muons; ratio of proton to neutron magnetic moment; charge of baryons e.g. Δ ++ 8 6 3. The ratio R as of July 1974. Ions � 12 Ions
Evidence of quark charge • R = ratio of e + e - ->hadrons/muons; ratio of proton to neutron magnetic moment; charges of baryons e.g. Δ ++ Υ 10 3 J / ψ ψ (2 S ) Z 10 2 φ ω R 10 ρ ′ 11/3 1 b-quark ρ c-quark -1 10 2 1 10 10 √ s [GeV] � 13
Evidence that quarks carry colour • Existence of Ω - particle ; R has factor of 3; there are no free quarks (confinement) � 14
Three quarks for Muster Mark! Sure he hasn't got much of a bark And sure any he has it's all beside the mark. Sir Tristram, violer d'amores, fr'over the short sea, had passen-core rearrived from North Armorica on this side the scraggy isthmus of Europe Minor to wielderfight his penisolate war: nor had topsawyer's rocks by the stream Oconee exaggerated themselse to Laurens County's gorgios while they went doublin their mumper all the time: nor avoice from afire bellowsed mishe mishe to tauftauf thuartpeatrick: not yet, though venissoon after, had a kidscad buttended a bland old isaac: not yet, though all's fair in vanessy, were sosie sesthers wroth with twone nathandjoe. Rot apeck of pa's malt had Jhem or Shen brewed by arclight and rory end to the regginbrow was to be seen ringsome on the aquaface. The fall (bababadalgharaghtakamminarronnkonnbronntonner- ronntuonnthunntrovarrhounawnskawntoohoohoordenenthur- nuk!) of a once wallstrait oldparr is retaled early in bed and later on life down through all christian minstrelsy. � 15
Evidence that gluon spin=1 • measure 3 jet events and angle of highest energy jet with respect to axis defined by the other 2 jets � 16
Photons & Polarisation • Want plane wave solutions of the form A µ = ✏ µ e ik · x � • In Lorenz gauge we have ⇤ A µ = ∂ ν ∂ ν A µ = 0 k 2 = 0 ⇒ E 2 = | ~ k | 2 � ∂ µ A µ = 0 k µ ✏ µ = 0 • The Lorenz gauge condition implies • But transformations of the form provide 0 µ = ✏ µ + � k µ ✏ an additional gauge freedom allowing us to define the Coulomb Gauge ✏ 0 = 0 ✏ · ~ ~ k = 0 � • Leaving two transverse polarisations for the photon which for a photon travelling in z-direction are: 1 ✏ µ − = 2(0 , 1 , − i, 0) ✏ µ 1 = (0 , 1 , 0 , 0) √ OR 1 ✏ µ 2 = (0 , 0 , 1 , 0) ✏ µ + = 2(0 , 1 , + i, 0) � 17 √
Massive Bosons and Polarisation Sums • For a massive spin-1 boson also have longitudinal polarisation L = 1 ✏ µ m ( p z , 0 , 0 , E ) � • The polarisation sum for real massive bosons is λ = − g µ ν + q µ q ν � X λ ✏ ν ✏ ∗ µ m 2 � • For real massless bosons this is X � λ ✏ ν λ = − g µ ν ✏ ∗ µ � • For virtual massless bosons connecting virtual fermions (higher order loops) this is λ = − g µ ν + q µ q ν X λ ✏ ν ✏ ∗ µ | ~ q | 2 � 18
Compton Scattering • Now have all the machinery necessary to calculate the cross-section for Compton scattering 0 ) + γ ( k 0 ) e − ( p ) + γ ( k ) → e − ( p � • There are two diagrams e − e − e − e − � � � � γ γ γ γ • To calculate average matrix element must = 1 1 D | M | 2 E X X | M | 2 2 2 spin pol ✓ 1 ✓ 1 " ◆ 2 # 0 ◆ p · k 0 + p · k 1 1 D | M | 2 E = 2 e 4 p · k + 2 m 2 + m 4 p · k − p · k − 0 0 p · k p · k p · k � 19
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