Experimental foundations of Quantum Chromodynamics Marcello Fanti University of Milano and INFN M. Fanti (Physics Dep., UniMi) title in footer 1/78 Introduction What we know as of today? In particle collisions we observe a wide plethora of hadrons These are understood to be bound states of quarks: mesons are qq¯, baryons are qqq —andanti-baryons areq ¯q¯q¯. Quarks (and antiquarks) carry a color charge,responsibleforbindingthemintohadrons.Colorsinteractthrough strong interaction,mediatedbygluons,whichalsocarrycolor. Quark-gluon interactions are described by a well-established gauge theory, Quantum Chromodynamics (QCD). Drawbacks: Hadrons do not carry a net color charge None of the experiments ever observed “free quarks” or “free gluons” Such experimental facts cannot be proved from first principles of QCD, although there are good hints from it ... thereforeonehastointroducea furtherpostulate: the “color confinement” the “fundamental particles” (quarks and gluons) cannot be directly studied, their properties can only be inferred ) from measurements on hadrons Despite all that, QCD works pretty well. In these slides we’ll go through the “experimental evidences” of such a theory. M. Fanti (Physics Dep., UniMi) title in footer 2/78 Bibliography Most of what follows is inspired from: F.Halzen and A.D.Martin “Quarks and leptons” — John Wiley & Sons R.Cahn and G.Goldhaber “The experimental foundations of particle physics” — Cambridge University Press D.Griffiths “Introduction to elementary particles” — Wiley-VCH Suggested reading (maybe after studying QCD): http://web.mit.edu/physics/people/faculty/docs/wilczek_nobel_lecture.pdf [Wilczec Nobel Lecture, 2004] M. Fanti (Physics Dep., UniMi) title in footer 3/78 Prequel: electron elastic scattering on pointlike fermion M. Fanti (Physics Dep., UniMi) title in footer 4/78 ef ef elastic scattering ! (“elastic” means that we have the same particles in initial state and final state) Electron scattering on pointlike fermion with charge Qf (e.g. µ)(mediatedbyaphoton) e e p1 (remember: p1 + p2 = p3 + p4) e p3 µ µ ⌫ ⌫ fermionic e.m. currents: J =¯u3γ u1 ; J =¯u4γ u2 g e f q photon propagator: µ⌫ (where q p p —orq p p ?) q2 ⌘ 1 − 3 ⌘ 2 − 4 (not important, as long as we have q2) eQf p2 p4 QED vertices: e and eQf g e2 f f matrix amplitude: = Jµ e µ⌫ eQ J⌫ = Q JµJ M e · · q2 · f · f q2 f e f ,µ ef ef ✓ ◆ ! 4 2 4 2 2 e Qf µ ⌫ e Qf µ ⌫ = (J J )(J J )⇤ = [(¯u γ u )(¯u γ u )] [(¯u γ u )(¯u γ u )]† |M| q4 e f ,µ e f ,⌫ q4 3 1 4 µ 2 3 1 4 ⌫ 2 4 2 e Qf µ ⌫ = (¯u γ u )(¯u γ u )† (¯u γ u )(¯u γ u )† q4 3 1 3 1 4 µ 2 4 ⌫ 2 h µ⌫ i h i Le Lf ,µ⌫ | {z } | {z p/ + m } p/ m For what follows, recall that averaging over spins gives uu¯ = (and likewise, vv¯ = − ) h ispin 2 h ispin 2 Also, recall that u, u¯,γare 4 1, 1 4, 4 4matrices... ⇥ ⇥ ⇥ M. Fanti (Physics Dep., UniMi) title in footer 5/78 ef ef scattering : fermionic tensors ! Few reminders: 0 0 ⌫ 0 ⌫ u¯ u†γ , γ (γ )†γ = γ µ⌫ µ ⌫ ⌘ Le =(¯u3γ u1)(u¯3γ u1)† Tr [ABC ...Z]=Tr [ZABC ...] µ ⌫ 0 =¯u3 γ u1 u1† (γ )† (u3†γ )† p/ + m p/ m ⌫ uu¯ = , vv¯ = − γ h ispin 2 h ispin 2 µ 0 ⌫ 0 µ ⌫ µ⌫ =¯u3 γ u1 u¯1 γ (γ )†γ u3 Tr [γ γ ]=4g µ ⌫ Tr [γ⇢γµγ⌫]=0 = Tr [¯u3 γ u1zu¯1 γ}|u3]{ ⇢ µ σ ⌫ ⇢µ ⌫ ⇢ µ⌫ ⇢⌫ µσ µ ⌫ Tr [γ γ γ γ ]=4(g g g g + g g ) = Tr [u3 u¯3 γ u1 u¯1 γ ] − p/ + me p/ + me Tr 3 γµ 1 γ⌫ −−−! 2 2 hispin 2 1 me me m = Tr p/ γµp/ γ⌫ + Tr p/ γµγ⌫ + Tr γµp/ γ⌫ + e Tr [γµγ⌫] 4 3 1 4 3 4 1 4 = p ph (g ⇢µg ⌫i g ⇢g µ⌫ +hg ⇢⌫g µσ)i + m2 g µ⌫h i 3,⇢ 1,σ − e = pµp⌫ + p⌫pµ p p m2 g µ⌫ 1 3 1 3 − 1 · 3 − e ... andthesameforthepointlike fermion f What if e were e+?? Need to replace u v,thisa↵ects by flipping sign in front of m ! hispin ... but m eventually enters only through m2,sotheresultdoesnotchange! What we are doing is ok for e±f e±f —andlikewisefore±f¯ e±f¯ ) ! ! M. Fanti (Physics Dep., UniMi) title in footer 6/78 ef ef scattering : matrix element ! Lµ⌫ = pµp⌫ + p⌫pµ p p m2 g µ⌫ e 1 3 1 3 − 1 · 3 − e L = p p + p p p p m2 g f ,µ⌫ 2,µ 4,⌫ 2,⌫ 4,µ− 2 · 4 − f µ⌫ Now neglect electron mass (m 0) — means that E , p~ m —butkeep m e ' 1 | 1| e f Lµ⌫ L =[pµp⌫ + p⌫pµ (p p ) g µ⌫] p p + p p p p m2 g e f ,µ⌫ 1 3 1 3 − 1 · 3 2,µ 4,⌫ 2,⌫ 4,µ − 2 · 4 − f µ⌫ =2 (p p )(p p )+(p p )(p p ) m2 (p p ) 1 · 2 3 · 4 1 · 4 ⇥2 · 3 − f 1 · 3 ⇤ ⇥ ⇤ e4Q2 e4Q2 2 = f Lµ⌫ L =2 f (p p )(p p )+(p p )(p p ) m2 (p p ) |M| q4 e f ,µ⌫ q4 1 · 2 3 · 4 1 · 4 2 · 3 − f 1 · 3 ⇥ ⇤ Using p = p + p p and neglecting p2 = p2 = m2 0: 4 1 2 − 3 1 3 e ' e4Q2 2 =2 f 2(p p )(p p )+(p p )(p p p p m2) |M| q4 1 · 2 3 · 2 1 · 3 2 · 1 − 2 · 3 − f ⇥ ⇤ M. Fanti (Physics Dep., UniMi) title in footer 7/78 ef ef scattering : matrix element in LAB frame ! LAB frame : where f is at rest at the beginning, and e moves along the z-axis outgoing( electronE’;E’sinθ,0,E’cosθ) p (E;0, 0, E);p (E 0; E 0 sin ✓, 0, E 0 cos ✓) θ 1 ⌘ 3 ⌘ incoming electron p (m ;0, 0, 0) ; p (E ; p~ ) (E;0,0,E) 2 ⌘ f 4 ⌘ 4 4 p1 p2 = Emf 2 2 · q =(p1 p3) = 2p1 p3 p1 p3 = EE 0(1 cos✓) − − · · − = 2EE 0(1 cos ✓) p p = m E 0 − − 2 · 3 f 2 2 2 2 2 2 Another useful relation: q + p = p (q + p ) = p q +2q p + m = m q +2m (E E 0)=0 2 4) 2 4) · 2 f f ) f − 2 def q ⌫ = E E 0 = (valid for any ef ef elastic scattering) − − 2mf ! 4 2 2 e Qf 2 2 =2 2m EE 0 + EE 0(1 cos ✓)(mf E mf E 0 m ) |M| q4 f − − − f recall identities : 4 2 ⇥e Q 1+cos✓ 1 cos ✓ E ⇤ E 0 1 cos ✓ ✓ f 2 − =sin2 =2 2 2mf EE 0 + − − [2EE 0(1 cos ✓)] 2 2 mf 2 2 − 1+cos✓ 2 ✓ e4Q2 ✓ q2 ✓ =cos = f m2 cos2 sin2 2 2 4EE sin4 ✓ f 2 − 2m2 2 0 2 ✓ ◆ f ✓ ◆ M. Fanti (Physics Dep., UniMi) title in footer 8/78 ef ef scattering : cross-section in LAB frame ! Compute cross-section in LAB frame (where f is at rest at the beginning) p (E;0, 0, E);p (E 0; E 0 sin ✓, 0, E 0 cos ✓) 1 ⌘ 3 ⌘ p (m ;0, 0, 0) ; p (E ; p~ ) 2 ⌘ f 4 ⌘ 4 4 Incoming speeds: v1 = c =1(relativistic electron) and v2 =0.Numberofspinstates:F3 =2and F4 =2 3 2 d p p~ d p~ d⌦ E 0 dE 0 d⌦ Phase space term for relativistic electron: 3 = | 3| | 3| = (where d⌦ d(cos ✓)dφ) 2E3 2E3 2 ⌘ Phase space term for recoiling fermion: 3 d p4 (3) δ(E4 + E 0 mf E) δ (p~3 + p~4 p~1 p~2)δ(E4 + E3 E1 E2) − − 2E − − − − −−!d3p 2(m + E E ) 4 4 f − 0 Recall Fermi’s golden rule for 1, 2 3, 4 processes:R ! 1 d 3p d 3p dσ = 3 F 4 F (2⇡)4 δ(4) (p + p p p ) 2 4(v + v )E E (2⇡)3 2E 3 (2⇡)3 2E 4 3 4 − 1 − 2 |M| 1 2 1 2 ✓ 3 ◆✓ 4 ◆ 1 1 1 2 2 (E 0 dE 0 d⌦) δ(E4 + E 0 E mf ) −−!d3p 4 Em (2⇡) m + E E − − |M| 4 f ✓ f − 0 ◆ R M. Fanti (Physics Dep., UniMi) title in footer 9/78 ef ef scattering : cross-section in LAB frame ! Now, try to elaborate dE 0 δ(E + E 0 E m ) —remember:E is also a function of E 0 4 − − f 4 2 2 2 2 2 2 2 2 For elastic ef ef , E =(p~ ) + m =(p~ p~ ) + m =(E + E 0 2EE 0 cos ✓)+m ! 4 4 f 1 − 3 f − f def δ(E 0 Eelastic0 ) Call f (E 0) = E 0 + E (E + m ) where f (E 0)=0hassomesolutionE 0 = E 0 δ(f (E 0)) = − 4 − f elastic ) f (E ) | 0 0 | 2 @(E 0 + E4 (E + mf )) @E4 1 @(E4 ) f 0(E 0) − =1+ =1+ ⌘ @E 0 @E 0 2 E4 @E 0 2E 0 2E cos ✓ m + E(1 cos ✓) =1+ − = f − 2(m + E E ) m + E E f − 0 f − 0 mf + E E 0 dE 0 δ(E4 + E 0 E mf ) − − − −−dE ! m + E(1 cos ✓) 0 f − 2 R q mf (E E 0) E and (1 cos ✓)= = − mf + E(1 cos ✓)=mf − −2EE 0 EE 0 ) − E 0 e2 Elastic di↵erential cross-section (recall that ↵ = ): 4⇡ dσ 1 E 2 = 0 2 d⌦ 4(2⇡)2m2 E |M| f ✓ ◆ ↵2Q2 E ✓ q2 ✓ = f 0 cos2 sin2 4E 2 sin4 ✓ E 2 − 2m2 2 2 ✓ ◆ f ✓ ◆ M.