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.Griths “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 =¯u3 u1 ; J =¯u4 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 u )(¯u u )] [(¯u u )(¯u u )]† |M| q4 e f ,µ e f ,⌫ q4 3 1 4 µ 2 3 1 4 ⌫ 2 4 2 e Qf µ ⌫ = (¯u u )(¯u u )† (¯u u )(¯u 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¯,are 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† , ( )† = µ⌫ µ ⌫ ⌘ Le =(¯u3 u1)(u¯3 u1)† Tr [ABC ...Z]=Tr [ZABC ...] µ ⌫ 0 =¯u3 u1 u1† ( )† (u3† )† p/ + m p/ m ⌫ uu¯ = , vv¯ = h ispin 2 h ispin 2 µ 0 ⌫ 0 µ ⌫ µ⌫ =¯u3 u1 u¯1 ( )† u3 Tr [ ]=4g µ ⌫ Tr [⇢µ⌫]=0 = Tr [¯u3 u1zu¯1 }|u3]{ ⇢ µ ⌫ ⇢µ ⌫ ⇢ µ⌫ ⇢⌫ µ µ ⌫ Tr [ ]=4(g g g g + g g ) = Tr [u3 u¯3 u1 u¯1 ] p/ + me p/ + me Tr 3 µ 1 ⌫ ! 2 2 hispin  2 1 me me m = Tr p/ µp/ ⌫ + Tr p/ µ⌫ + Tr µp/ ⌫ + e Tr [µ⌫] 4 3 1 4 3 4 1 4 = p ph (g ⇢µg ⌫i g ⇢g µ⌫ +hg ⇢⌫g µ)i + m2 g µ⌫h i 3,⇢ 1, 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 ✓)d) 2E3 2E3 2 ⌘ Phase space term for recoiling fermion: 3 d p4 (3) (E4 + E 0 mf E) (p~3 + p~4 p~1 p~2)(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 d = 3 F 4 F (2⇡)4 (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⌦) (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 (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 (E 0 Eelastic0 ) Call f (E 0) = E 0 + E (E + m ) where f (E 0)=0hassomesolutionE 0 = E 0 (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 (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⇡ d 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. Fanti (Physics Dep., UniMi) title in footer 10 / 78 Electron-proton elastic scattering

M. Fanti (Physics Dep., UniMi) title in footer 11 / 78 Recalling the “static model”

mesonic nonet The variety of discovered mesons and baryons, and their “recursive” properties, suggested that they be composite states of more fundamental particles: the quarks,thatmaycomeindi↵erentflavours (up, down, strange, . . . ) [Gell-Mann, Zweig, 1964]

baryonic octet This was a “static model” — no dynamics, no interactions, just “lego bricks”. However, it meant that hadrons are not point-like

If a proton is composite, the cross-section evaluation for fermion elastic ) scattering is expected to break down:

J⌫ =¯u ⌫u baryonic decuplet p 6 4 2 L = p p + p p p p M2 g p,µ⌫ 6 2,µ 4,⌫ 2,⌫ 4,µ 2 · 4 p µ⌫ d e4Q2 1 E ✓ q2 ✓ = f 0 cos2 sin2 d⌦ (4⇡)2 2 4 ✓ E 2 2M2 2 ep ep 6 4E sin 2 p  !  ✓ ◆ ✓ ◆

M. Fanti (Physics Dep., UniMi) title in footer 12 / 78 ep ep elastic scattering !

The fermion current of the proton must be modified into some J⌫ =¯u ⌫u p 4 2 e e The ⌫ are 4 4matricesthat“parametrizeourignorance”,yetthereare p ⇥ 1 some constraints: e p3 ⌫ Jp must be a Lorentz 4-vector and must depend only on the kinematics of the scattering q ⌫ must be built from known ingredients (i.e. -matrices and their combinations) p2 ~ p4 for soft scattering the photon wavelength = is not short enough to q~ f f probe the proton structure | | ep ep ⌫ must reduce to Dirac current for very low q2 (i.e. ⌫ ⌫) ! ) !q2 0 The parametrization must be ! ) ⌫ ⇢ ⇢ ⌫ ⌫ 2 ⌫  2 ⌫⇢ ⌫⇢ def = F1(q ) + F2(q )i q⇢ where = 2Mp 2 electric coupling magnetic coupling | {z }  = anomalous magnetic moment at rest (e.g.| p =1{z.79 for proton,} n = 1.91 for neutron) p⌫ + p⌫ 1 (recall the Gordon decomposition u¯ ⌫u =¯u 2 4 i ⌫⇢q u ) 4 2 4 2m 2m ⇢ 2  e2 Cross-section in lab frame becomes (recall ↵ def= ): 4⇡ d ↵2 E 2q2 ✓ q2 ✓ = 0 F 2(q2) F 2(q2) cos2 (F (q2)+F (q2)) sin2 d⌦ 2 4 ✓ E 1 4M2 2 2 2M2 1 2 2 ep ep 4E sin 2 p p  ! ✓ ◆ ✓ ◆ ✓ ◆ M. Fanti (Physics Dep., UniMi) title in footer 13 / 78 ep ep elastic scattering : the size of the proton !

2 2 The “form factors” F1(q ), F2(q ) are related to the extension in space of the proton: 2 3 iq~ ~r F1,2(q )= d r ⇢1,2(~r)e · — F1 for the charge distribution, F2 for the magnetic moment distribution. 2 Z 2 Low q limit: F1,2(q ) 1 !q2 0 ! iq~ ~r 1 2 Expanding e · 1+i(q~ ~r) (q~ ~r) + ' · 2 · ··· and assuming spherical symmetry (such that ⇢(~r) ⇢(r)) ⌘

(q~ ~r)2 1 F (q2) d 3r ⇢(r) 1+iq~ ~r · + =1+i q~ ~r (q~ ~r)2 1,2 ' · 2 ··· h · i2 · Z ✓ ◆ ⌦ ↵ q~ 2 r 2 Then q~ ~r =0and (q~ ~r)2 = | | for fixed q~ h · i · 3 ⌦ ↵ 2 2 2 Proof: ~r (rq, r 1, r 2)whererq is along q~ ~r q~ = q~ rq and (~r q~) = q~ rq ⌘ ? ? ⌦ ↵ ) h · i | |h i · | | 2 2 2 2 2 2 2 Due to spherical symmetry: rq =0;then rq = r 1 = r 2 ,and r = rq + r 1 + r 2 h i ? ? ⌦ ↵ ⌦ ?↵ ? ⌦ ↵ ⌦ ↵ ⌦ ↵ ⌦ ↵ ⌦ ↵ ⌦ ↵ ⌦ ↵ 1 F (q2) 1 q~ 2 r 2 + 1,2 ' 6| | ··· Here r 2 measures the “radius of the proton” ⌦ ↵ h i p

M. Fanti (Physics Dep., UniMi) title in footer 14 / 78 ep ep elastic scattering : experiment !

Elastic scattering of 188 MeV electrons from the proton and the alpha particle [R.W.McAllister and R.Hofstadter, Physical Review 1956, Vol 102]

Experimental layout Observe scattered e at several angles ✓ Measure ep ep rate at di↵erent angles !d measure (✓) ) d⌦ ep ep  ! 2 (recall: q =2EE 0(1 cos ✓)) extract F (q2) ) 1,2

QUESTION: is the scattering indeed ep ep? ! The apparatus measures the outgoing electron energy and angle : E 0 and ✓ The proton is not detected: how can we ensure that the electron beam on hydrogen ( proton) target reaction is not some ep eX ? ⌘ ! This actually happens, for large enough q2... If there is no amoving“telescope”detectsscatteredelectrons ⌫ ⌫ at di↵erent angles ✓ proton in final state, we cannot even build a Jp =¯u4 u2 — what is u4 ??? and measures their E 0 by curvature in B-field

M. Fanti (Physics Dep., UniMi) title in footer 15 / 78 ep ep elastic scattering : experiment !

QUESTION: is the scattering indeed ep ep? !

Measured E 0 vs ✓

Answer:Wesawforelastic scattering that 2 q EE 0(1 cos ✓) ⌫ E E 0 = E E 0 = ⌘ 2Mp ) Mp

Mp E E 0 = 2 ✓ ) Mp +2E sin 2 This relation can be tested experimentally, to guarantee an elastic scattering.

it is indeed elastic ep ep ) !

M. Fanti (Physics Dep., UniMi) title in footer 16 / 78 ep ep elastic scattering : experiment !

d Di↵erential cross-section vs ✓ d⌦ Theoretical curves: (a) point-like spinless proton (b) point-like Dirac proton without anomalous magnetic moment (c) point-like Dirac proton with anomalous magnetic moment

Experimental data (solid dots with error bars) don’t fit any theory prediction for point-like proton fit well a model with form factors, for 2 13 r =(0.70 0.24) 10 cm h i ± · p

M. Fanti (Physics Dep., UniMi) title in footer 17 / 78 ep ep elastic scattering : some back-of-envelope !

Using McAllister-Hofstadter results, how large is the deviation of the proton from point-like model? 2 2 2 q~ r 2 26 2 This means estimating F1,2(q ) 1 | | for r 0.5 10 cm ' 6⌦ ↵ ' · 2 2 ⌦ ↵ 2 2 2 Recalling q = 2EE 0(1 cos ✓)andq = 2M (E E 0)= 2M ⌫,andq ⌫ q~ yields: p p ⌘ | | q~ 2 = ⌫2 q2 = ⌫2 +2M ⌫ =(⌫ + M )2 M2 | | p p p In the experiment, ⌫ [0; 50 MeV], and using M =938MeV the largest e↵ect is at q~ 2 =96300MeV2 2 p | |max 22 10 Now, need to “convert” cm to MeV. Recall that ~ =6.582 10 MeV s and c =3 10 cm/s,so · · 11 1 10 1 ~c =1.975 10 MeV cm.Setting~c =1means1cm = =5.06 10 MeV . · 1.975 10 11 MeV · 2 26 2 5 2 So r 0.5 10 cm 1.25 10 MeV . · ' · ' · 2 2 ⌦ ↵ 2 q~ max r 2 So F1,2(q ) 1 | | 0.8—a20%deviationfrompoint-likeF1,2(q )=1,wellobservable! ' 6 ⌦ ↵ '

M. Fanti (Physics Dep., UniMi) title in footer 18 / 78 Electron-proton inelastic scattering and Deep Inelastic Scattering (DIS)

M. Fanti (Physics Dep., UniMi) title in footer 19 / 78 Inelastic ep scattering

2 e e For higher q ,theprotonbreaksapart:ep eX , X X1, X2,....Xn p1 ! ⌘{ } being a chunk of (several) hadrons e p X1 3 cannot define anymore a “proton current” J⌫ q ) p X q2 2 the “elastic” relation between ⌫ and ✓ is not valid anymore: ⌫ = ) 6 2Mp need to generalize the Lµ⌫L to some Lorentz-invariant form, that keeps ) e p,µ⌫ Lµ⌫ (the electron is always the same. . . ) p2 e

f Xn µ⌫ µ⌫ L Lp,µ⌫ L Wµ⌫ ep eX e ! e !

q q 1 p q p q W = W (...) g + µ ⌫ + W (...) p 2 · q p 2 · q µ⌫ 1 µ⌫ q2 2 M2 2,µ q2 µ 2,⌫ q2 ⌫ ✓ ◆ p ✓ ◆✓ ◆ W1,2(...) are functions of the kinematics at proton vertex: here two independent quantities: 2 def p2 q q and ⌫ = · = E E 0 (last step for the LAB frame, where the proton is at rest and p2 (Mp;0, 0, 0) ) Mp ⌘

d 2 ↵2 ✓ ✓ = W (⌫, q2)cos2 +2W (⌫, q2)sin2 dE d⌦ 2 4 ✓ 2 2 1 2 0 ep eX 4E sin 2  !  ✓ ◆ ✓ ◆ [see e.g. Halzen & Martin “Quarks and Leptons”]

M. Fanti (Physics Dep., UniMi) title in footer 20 / 78 Inelastic ep scattering : some kinematics

In the LAB frame where the proton is at rest: Independent L.I. quantities: p2 (Mp;0, 0, 0) and p1 p2 = MpE 2 def 2 def p2 q ⌘ · q =(p1 p3) ; ⌫ = · Mp ⌫ = E E 0 2 Invariant mass of X final state: q = 2EE 0(1 cos ✓) 2 2 2 2 EE 0(1 cos ✓) mX =(p2 + q) = Mp +2Mp⌫ + q x = M (E E ) p 0 E E 0 y = Dimensionless L.I. quantities: E

2 2 def q q Allowed region for x, y: y [0; 1] and x [0; 1] x = = Being x, y L.I., we can prove these2 in the LAB2 frame. 2p2 q 2Mp⌫ · For y this is obvious. Also obvious is x 0. Then, write m2 as def p2 q X y = · function of x, y: p2 p1 · 2 2 p1 p2 mX = Mp +2Mp · y 2(p1 p2)xy Mp · Inverse relations: E = M2 1+2y(1 x) p M ✓ p ◆ 2 q =2(p1 p2)xy 2 E · Since mX > 0foranyvalueof ,theonlywayisx 1. p1 p2 Mp  ⌫ = · y The case x =1coincideswithmX = Mp,i.e.elasticscattering. Mp As a by-product, notice that m M always X p M. Fanti (Physics Dep., UniMi) title in footer 21 / 78 Inelastic ep eX : experiment !

High-energy inelastic e-p scattering at 6 and 10 [E.D.Bloom et al., SLAC and MIT, Physical Review Letters 1969, Vol 23]

Beam energies E [7, 17] GeV (Stanford linear acceler- 2 ator) Observable scattering angle ✓ [6, 10] 2 q2 7.4 GeV ) . d 2 Measures of as a function of m —calledW dE d⌦ X here 0 for di↵erent q2 ranges: Fig(a) : 0.2 < q2 < 0.5 Fig(b) : 0.7 < q2 < 2.6 Fig(c) : 1.6 < q2 < 7.3 While at low q2 the production of excited resonances (e.g. +(1231), N+(1440), etc) are favoured, for larger q2 the observed W spectrum becomes continuum. The proton debris are more and spread.

M. Fanti (Physics Dep., UniMi) title in footer 22 / 78 Inelastic ep eX : experiment ! 2 NOTE: m W = M +2M (E E ) 2EE (1 cos ✓) is inferred from observed E 0,✓ X ⌘ p p 0 0 q PROBLEM: Final states where X is a hadronic resonance are favoured by larger cross-sections. But this requires precise values of q2,⌫. If the energy transferred from the electron (⌫)istoohightoproduceexactlytheneeded mX ,itis“convenient”toloosesomeenergyinextrae.m. radiation and fall back onto the resonance (“radiative re- turn”) HOWEVER: this “radiation” is not detectable from ob- servables E 0,✓ need ad-hoc correction. ) This can be computed from 1st principles, and also cross- checked by observing the “radiative high-W tail” of the elastic ep ep scattering. ! d 2 Fig(a): vs W before radiative correction dE 0 d⌦ d 2 Fig(b): vs W after radiative correction dE 0 d⌦ Fig(c): ratio after/before radiative correction Note that the correction enhances the resonances (as it should!)

M. Fanti (Physics Dep., UniMi) title in footer 23 / 78 Elastic vs inelastic scattering

Elastic scattering on pointlike fermion Inelastic scattering on proton q2 (E 0 is function of ✓ –recallE E 0 ⌫ = ) (E 0 and ✓ are independent) ⌘ 2mf

2 2 d ↵ 2 2 ✓ = W2(⌫, q )cos d ↵2Q2 E ✓ q2 ✓ dE d⌦ 4E 2 sin4 ✓ 2 = f 0 cos2 sin2 0 2  ✓ ◆ d⌦ 4E 2 sin4 ✓ E 2 2m2 2 ✓ 2  ✓ ◆ f ✓ ◆ +2W (⌫, q2)sin2 1 2 ✓ ◆ can be written in a similar way (proof in next slide):

2 2 2 2 2 2 d 4↵ E 0 2 d 4↵ E 0 = 4 Qf = 4 dE 0 d⌦ q ⇥ dE 0 d⌦ q ⇥ 2 2 2 ✓ q 2 ✓ q 2 2 ✓ 2 2 ✓ cos sin ⌫ + W2(⌫, q )cos +2W1(⌫, q )sin ⇥ 2 2m2 2 2m ⇥ 2 2  ✓ ◆ f ✓ ◆ ✓ f ◆  ✓ ◆ ✓ ◆ Therefore we can recover the elastic scattering on pointlike fermion from the general formula, with the replacements: q2 W (⌫, q2) Q2 ⌫ + 2 ! f · 2m ✓ f ◆ q2 q2 W (⌫, q2) Q2 ⌫ + 1 ! f · 4m2 · 2m f ✓ f ◆

M. Fanti (Physics Dep., UniMi) title in footer 24 / 78 Inelastic vs elastic scattering : proof

4↵2E 2 Let’s start with 0 : q4 2 2 ✓ using q = 2EE 0(1 cos ✓)= 4EE 0 sin we get: 2 ✓ ◆ 4E 2 4E 2 1 0 = 0 = q4 2 2 4 ✓ 2 4 ✓ 16E E 0 sin 2 4E sin 2 d 2 d For the elastic cross-section on pointlike fermion, to go from to need to integrate over dE 0: dE 0 d⌦ d⌦ (E 0 Esol0 ) Recalling that (f (E 0)) = ,beingE asolutionoff (E )=0,wehave: f (E ) sol0 0 | 0 sol0 | 2 q 2EE 0 2 ✓ dE 0 ⌫ + = dE 0 E E 0 sin 2m m 2 ✓ f ◆ ✓ f ✓ ◆◆ (E 0 Esol0 ) (E 0 Esol0 ) E 0 = dE 0 = dE 0 = dE 0 (E 0 Esol0 ) 2E 2 ✓ E E 0 E 1+ sin 1+ E mf 2 0 E 0 dE !0 E R E So the 0 term is also recovered. E

M. Fanti (Physics Dep., UniMi) title in footer 25 / 78 The “Bjorken scaling”

Is a proton indeed made of quarks? If so, for high enough q2 (i.e. small enough photon wavelength) the ep scattering should actually be an electron-quark scattering. Then, the W1, W2 functions should “collapse” back to point-like fermion scattering case: q2 W (⌫, q2) Q2 ⌫ + 2 ! q · 2m ✓ q ◆ q2 q2 W (⌫, q2) Q2 ⌫ + 1 ! q · 4m2 · 2m q ✓ q ◆ with some redefinitions, and recalling that a(ax)=(x):

2 def M q F = ⌫W = Q2 1 p 2 2 q m 2M ⌫ ✓ q p ◆ 2 2 2 def 1 M q M q F = M W = Q2 p 1 p 1 p 1 q 2 m 2M ⌫ m 2M ⌫ ✓ q ◆ ✓ p ◆ ✓ q p ◆ q2 where we spot the variable x ,soweget ⌘ 2Mp⌫ M F (x)=Q2 1 p x 2 q m ✓ q ◆ M 2 x M F (x)=Q2 p 1 p x 1 q m 2 m ✓ q ◆ ✓ q ◆ 2 2 q i.e. the new functions F1,2 depend on q ,⌫ ONLY through their combination x ⌘ 2Mp⌫ This prediction is known as the “Bjorken scaling”[D.Bjorken,PhysicalReview1969,Vol179]

M. Fanti (Physics Dep., UniMi) title in footer 26 / 78 Bjorken scaling: experimental evidence

Observed behavior of highly inelastic electron-proton scattering [M.Breidenbach et al., SLAC and MIT, Physical Review Letters 1969, Vol 23]

Beam energies E [7, 17] GeV (Stanford linear acceler- 2 ator) Observable scattering angle ✓ [6, 10] 2 q2 7.4 GeV ) . Recall: 2 2 2 d 4↵ E 0 = 4 dE 0 d⌦ q ⇥ ✓ ✓ W (⌫, q2)cos2 +2W (⌫, q2)sin2 ⇥ 2 2 1 2  ✓ ◆ ✓ ◆ at small ✓ probe W ) 2 2M ⌫ 1 Several plots of ⌫W vs ! def= p (note: ! ) 2 q2 ⌘ x at di↵erent values of E change q2,⌫ ) dependence on ! is clear (despite complicate!) (a,c) : ✓ =6,severalE ) absence of spread no dependence on q2,⌫ separately (b,d) : ✓ =10,severalE ) W1 (e) : ✓ =6, E =7GeV, R is related to (unknown) ratio W2

M. Fanti (Physics Dep., UniMi) title in footer 27 / 78 The “parton model”

The experimental evidence of Bjorken scaling (i.e. ⌫W F F (x))supportstheideathat 2 ⌘ 2 ⌘ 2 protons are made of pointlike Dirac constituents —thequarks,orthe“partons”(so named by Feynman)

M ⌫W def= F = Q2 1 p x 2 2 q m ✓ q ◆

mq 1 BUT : the -function should lock x = —whileexperimentalresultsshowbroadspectraforF2 vs x (or vs ! ). Mp ⌘ x MOREOVER (related) : what is mq ?? Yes, the quark mass... Easily defined for a “free” (non-interacting) particle, but quarks are bound inside the proton, so they are definitely interacting — they may be “virtual”, or “o↵-shell”. Let’s stick to m2 = E 2 (p~ )2 —whatweusedthroughthemath. q q q Consider a relativistic proton, moving at speed v c,withenergyE ,longitudinalmomentumpL and null transverse momentum ' p p pT =0.Insideit,aquarkwithenergyE ,longitudinalmomentumpL and negligible transverse momentum pT 0. Then p q q q ' LAB frame proton 2 2 L 2 2 2 L 2 Mp = Ep (pp ) ; mq = Eq (pq ) e pp pq def pq Proton and quark move at the same speed v = = ,thereforecallingxq = we have: Ep Eq pp pL = x pL ; E = x E ; m = E 2 (pL)2 = x E 2 (pL)2 = x M "Feynman" frame proton q q p q q p q q q q P p q p q q p p p e q This defines the “quark mass” inside the proton in terms of the fraction of the proton momentum carried by the pq pq quark: mq = xqMp = Mp.The-function therefore locks x = xq = pp pp

M. Fanti (Physics Dep., UniMi) title in footer 28 / 78 The “parton distribution functions” (PDFs)

We can now better formulate F ⌫W :theep scattering at high q2 is a combination of all possible eq scatterings, 2 ⌘ 2 for all possible quarks q,withallpossiblevaluesofxq: M 1 M Q2 1 p x dx f (x ) Q2 1 p x q m ! q q q · q m q q 0 q ✓ ◆ X Z  ✓ ◆ Here fq(xq) are the parton distribution functions (PDFs), i.e. the probability distribution that a quark q carries a fraction xq of the proton’s momentum. BEWARE: PDFs cannot be computed from theory!

1 M 1 x ⌫W F (x)= dx f (x ) Q2 1 p x = Q2 dx f (x ) 1 2 ⌘ 2 q q q · q m q q q q x q Z0  ✓ q ◆ q Z0 ✓ q ◆ X 1 X = Q2 dx f (x ) x (x x )= Q2 xf (x) q q q q q q q q q 0 q X Z X x M 2 1 M W F (x)= p F (x)= F (x) p 1 ⌘ 1 2 m 2 2x 2 ✓ q ◆

In summary:

2 F2(x) W2(⌫, q )= 2xF (x)=F (x)=x Q2 f (x) ⌫ 1 2 q q F (x) F (x) q 2 1 2 X W1(⌫, q )= = Mp 2xMp

(the relation F2(x)=2xF1(x) is known as Callan-Gross relation)

M. Fanti (Physics Dep., UniMi) title in footer 29 / 78 ep vs eq cross-sections: “factorization”

Cross-section of eq eq “elastic” scattering: !

2 2 2 2 2 d 4↵ E 0 2 2 ✓ q 2 ✓ q = 4 Qq cos 1 2 tan ⌫ + dE 0 d⌦ eq eq q 2 · 2mq 2 2mq ✓ ◆ ! ✓ ◆  ✓ ◆ ✓ ◆ 2 2 2 2 4↵ E 0 ✓ Mp ⌫ q ✓ 1 x Helpers: 2 2 2 2 = 4 Qq cos 1+ 2 tan 1 def mq def q q 2 · " mq Mp 2Mp⌫ 2 # ⌫ xq x = ; x = ✓ ◆ ✓ ◆ ✓ ◆ q M 2M ⌫ 2 2 P p 4↵ E 0 2 2 ✓ 1 x 2 ✓ q2 1 x = 4 Qq cos + 2 tan xq(x xq) ⌫ + = 1 q 2 · ⌫ xq Mp 2 2m ⌫ x ✓ ◆  ✓ ◆ ✓ q ◆ ✓ q ◆ 4↵2E 2 ✓ x tan2 ✓ = 0 Q2 cos2 + 2 (x x ) q4 q 2 · ⌫ M q ✓ ◆ " p # Cross-section of ep eX DIS: !

2 2 2 d 4↵ E 0 2 ✓ 2 ✓ = 4 cos W2 +2W1 tan dE 0 d⌦ ep eX q 2 · 2 ✓ ◆ ! ✓ ◆  ✓ ◆ 4↵2E 2 ✓ x tan2 ✓ F = 0 cos2 + 2 2 4 Helpers: q 2 · "⌫ Mp # x ✓ ◆ F2 F2 2 2 2 ✓ W = ;2W = 4↵ E 0 ✓ x tan 2 1 = cos2 + 2 Q2f (x) ⌫ xMp q4 2 · ⌫ M q q " p # q ✓ ◆ X d 2 = dx f (x ) q q q dE d⌦ q 0 eq eq X Z ✓ ◆ !

M. Fanti (Physics Dep., UniMi) title in footer 30 / 78 “Hadronic” vs “partonic” cross-section : factorization

We got hadronic cross-section partonic cross-section d 2 d 2 = dx f (x ) dE d⌦ q q q dE d⌦ z 0 }| ep eX{ q z 0 }| eq eq{ ✓ ◆ ! X Z ✓ ◆ ! i.e. the ep “hadronic cross-section”canbefactorized as the eq “partonic cross-section”timesthePDFs

Reminder: the PDFs are unknown ... BUT:

partonic cross-sections can be evaluated the PDFs can be probed hadronic cross-sections can be measured )

part proc1 computation had measurement ) proc1 9 9 Ideally, once PDFs are “known” (so > > part > > computation > > to say...) they can be used, to- proc2 > fit PDFs > had measurement ) > ) > proc2 > > predict had gether with other partonic cross- = > ) procX > sections, to predict other hadronic part computation = proc3 > had measurement > cross-sections proc3 ) > > > > > > part ;> > procX computation > > > M. Fanti (Physics Dep., UniMi) title in footer ;> 31 / 78 Can we extract nucleon’s PDFs ?

2 Recall: F2(x) can be measured, and F2(x)=x Qq fq(x) —howtoextractallfq’s from F2? q X 2 1 Simplest nucleon model: proton u, u, d and neutron u, d, d ,withQ = ; Q = ⌘{ } ⌘{ } u 3 d 3 (p) (n) [NOTE: fu (x)isthePDFofboth u-quarks inside the proton — likewise, fd (x)isthePDFofboth d-quarks inside the neutron]

(p) 2 (p) 4 (p) 1 (p) but isospin flips p n and u d F2 (x)=x q=u,d Qq fq (x)=x 9fu (x)+9fd (x) $ $ h i f (n) = f (p) f ) P u d d (n) 2 (n) 4 (n) 1 (n) (n) (p) ⌘ F (x)=x Q fq (x)=x fu (x)+ f (x) ) f = fu f 2 q=u,d q 9 9 d ( d ⌘ u h i From ep , en inelasticP scattering measure F (p)(x) , F (n)(x) extract f (x) , f (x) — DOES THIS WORK??? 2 2 ) u d Average momentum fraction carried by u, d inside p: x (p) = dx x f (x) h iu,d u,d Z If simply p u, u, d we’d expect x (p) + x (p) =1.But: from measurements we have: ⌘{ } h iu h id (p) (p) (p) (p) dx F (x)=4 dx x f (x)+1 dx x f (x)=4 x + 1 x dx F2 (x) 0.18 2 9 u 9 d 9 h iu 9 h id ' R (n) R (n) R R (p) (p) dx F (x)=4 dx x f (x)+1 dx x f (x)=4 x + 1 x dx F2 (x) 0.12 2 9 d 9 u 9 h id 9 h iu ' (p) (p) R Rsolving we get x R(p) 0.36 and Rx 0.18 —i.e. x (p) + x 0.54 ) h iu ' h id ' h iu h id ' NO, DOESN’T WORK! Where is the rest of proton’s momentum? ) M. Fanti (Physics Dep., UniMi) title in footer 32 / 78 Summary

From ep scattering we learnt: 2 13 elastic scattering: protons are not pointlike ( r 0.7 10 cm) h i' · deep inelastic scattering (DIS): protons are madep of pointlike charged spin-1/2 constituents: the partons (Bjorken scaling) electron-proton cross-sections can be described in terms of combinations of electron-parton cross-sections: d 2 d 2 = dx f (x ) dE d⌦ q q q dE d⌦ 0 ep eX q 0 eq eq ✓ ◆ ! X Z ✓ ◆ ! the fq(xq) are the parton distribution functions (PDFs) — to be determined experimentally assuming p u, u, d , n u, d, d ,fromep, en DIS one can try to measure f (x), f (x) ⌘{ } ⌘{ } u d however this does explain only half the momentum of the nucleon! ⇠ Some speculations The parton model is successful, but apparently there are more partons than just the “valence quarks” u, u, d { } There must be partons not sensitive to e.m. probes electrically neutral. ) What keeps partons together inside the proton? Cannot be e.m. interaction (the proton is +ve charged, would repel apart). A new interaction, “stronger” than e.m., with its mediators? If there are interactions between “valence quarks”, the interaction energy may possibly materialize into more fermion-antifermion pairs, just like electron-positron pairs in the Dirac theory we may need to consider also ) quark-antiquark pairs “from the (Dirac) sea”.

M. Fanti (Physics Dep., UniMi) title in footer 33 / 78 Quark pair-production + in e e collisions

M. Fanti (Physics Dep., UniMi) title in footer 34 / 78 + The e e f f¯ process !

e f

p1 p3 µ µ ⌫ ⌫ fermionic e.m. currents: J =¯v2 u1 ; J =¯u3 v4 g e f e eQ photon propagator: µ⌫ (where q p + p ) q f q2 ⌘ 1 2 p2 p4 QED vertices: e and eQf g e2 matrix amplitude: = e2Q Jµ µ⌫ J⌫ = Q JµJ M f e q2 f q2 f e f ,µ e+ f¯ ✓ ◆

+ e e f f¯ ! 4 2 4 2 2 e Qf µ ⌫ e Qf µ ⌫ = (J J )(J J )⇤ = [(¯v u )(¯u v )] [(¯v u )(¯u v )]† |M| q4 e f ,µ e f ,⌫ q4 2 1 3 µ 4 2 1 3 ⌫ 4 4 2 e Qf µ ⌫ = (¯v u )(¯v u )† (¯u v )(¯u v )† q4 2 1 2 1 3 µ 4 3 ⌫ 4 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¯,are 4 1, 1 4, 4 4matrices... ⇥ ⇥ ⇥

M. Fanti (Physics Dep., UniMi) title in footer 35 / 78 + The e e f f¯ process : fermionic tensors !

Few reminders:

0 0 ⌫ 0 ⌫ u¯ u† , ( )† = ⌘ p/ + m p/ m µ⌫ µ ⌫ uu¯ spin = , vv¯ spin = Le =(¯v2 u1)(¯v2 u1)† h i 2 h i 2 µ ⌫ 0 =¯v2 u1 u1† ( )† (v2† )† Tr [ABC ...Z]=Tr [ZABC ...] µ ⌫ µ⌫ ⌫ Tr [ ]=4g ⇢ µ ⌫ µ 0 ⌫ 0 Tr [ ]=0 =¯v2 u1 u¯1 ( )† v2 Tr [⇢µ⌫]=4(g ⇢µg ⌫ g ⇢g µ⌫ + g ⇢⌫g µ) µ ⌫ = Tr [¯v2 u1zu¯1 }|v2]{ µ ⌫ = Tr [v2 v¯2 u1 u¯1 ] p/ me p/ + me Tr 2 µ 1 ⌫ ! 2 2 hispin  2 1 me me m = Tr p/ µp/ ⌫ + Tr p/ µ⌫ Tr µp/ ⌫ e Tr [µ⌫] 4 2 1 4 3 4 1 4 = p ph (g ⇢µg ⌫i g ⇢g µ⌫ +hg ⇢⌫g µ)i m2 g µ⌫h i 2,⇢ 1, e = pµp⌫ + p⌫pµ p p + m2 g µ⌫ 1 2 1 2 1 · 2 e ... andsimilarlyfor Lµ⌫ —startfromLµ⌫,replace1 3, 2 4, and µ ⌫ f e ! ! $

Lµ⌫ = pµp⌫ + p⌫pµ p p + m2 g µ⌫ f 3 4 3 4 3 · 4 f

M. Fanti (Physics Dep., UniMi) title in footer 36 / 78 + The e e f f¯ process : matrix element !

Lµ⌫ L = pµp⌫ + p⌫pµ p p + m2 g µ⌫ p p + p p p p + m2 g e f ,µ⌫ 1 2 1 2 1 · 3 e 3,µ 4,⌫ 3,⌫ 4,µ 2 · 4 f µ⌫ =2 (p p )(p p )+(p p )(p p )+m2 (p p )+m2 (p p )+4m2m2 ⇥ 2 · 3 1 · 4 2 · 4 1 · 3⇤⇥ f 1 · 2 e 3 · 4 e f ⇤

4⇥ 2 ⇤ ) e Q θ 2 f µ⌫ f cos = Le Lf ,µ⌫ E |M| q4 β , 0 , outgoingθ sin E β + ; Now consider e e collisions in center-of-mass frame, with electron energy E incoming e- ( incoming e+ E me (i.e. neglect me)andoutgoingfermionatangle✓,withenergyE (E ; 0 , 0 , E) (E ; 0 , 0 , -E) ) θ and momentum E: f cos E β

, 0 , - outgoingθ sin E 2 β p1 p2 =2E ; - E · 2 ( p1 p3 = E (1 cos ✓) · 2 p1 p4 = E (1 + cos ✓) · 2 p1 ( E ;0, 0 , E ) p2 p3 = E (1 + cos ✓) ⌘ · 2 p2 ( E ;0, 0 , E ) p2 p4 = E (1 cos ✓) ⌘ · 2 2 p3 ( E ; E sin ✓, 0 ,E cos ✓ ) p3 p4 = E (1 + ) ⌘ 2 · 2 2 p4 ( E ; E sin ✓, 0 , E cos ✓ ) q =(p1 + p2) =4E ⌘ e4Q2 2 = f 2 E 4(1 + cos ✓)2 + E 4(1 cos ✓)2 +2E 2m2 |M| 16 E 4 f e4Q2 ⇥ ⇤ = f 1+2 cos2 ✓ +(1 2) 4 M. Fanti (Physics Dep., UniMi) ⇥ title in footer ⇤ 37 / 78 + The e e f f¯ process : cross-section ! Recall Fermi’s golden rule for 2 2processes: ! d = F F 2 d(cos ✓) 128⇡ E 2 3 4|M| 2 (where we use s =4E and where F3,4 are the degrees of freedom of outgoing f , f¯ —hereF3 = F4 =2duetospin states)

=4 4 2 d e Qf 2 2 2 = 2 F3F4 1+ cos ✓ +(1 ) d(cos ✓) + ¯ 128⇡ E 4 ✓ ◆e e f f ! e4Q2 z}|{ ⇥ ⇤ = f 1+2 cos2 ✓ +(1 2) 128⇡ E 2 ⇥ ⇤ In the relativistic limit, E m , 1, f ! 4 2 d e Qf 2 = 2 (1 + cos ✓) d(cos ✓) e+e f f¯ 128⇡ E ✓ ◆ ! 1 1 (note the energy dependency as and the angular dependency 1+cos2 ✓). / E 2 / s / Total cross-section ( d(cos✓)): Z 4 2 2 2 e Qf 4⇡↵ Qf e+e f f¯ = = ! 48⇡ E 2 3 s

M. Fanti (Physics Dep., UniMi) title in footer 38 / 78 + Do we observe e e qq¯ reactions? !

Example: 2 4⇡↵ d 3 2 + + = ; = + + (1 + cos ✓) e e µ µ e e µ µ ! 3 s d⌦ e+e µ+µ 16⇡ ! · ✓ ◆ ! 2 1 If quarks exist, as point-like fermions with fractional charge Q =+ , Q = ,weshouldobservesimilarreactions, u 3 d 3 as:

1 4⇡↵2 e+e dd¯ = ! 9 3 s d 3 2 = + (1 + cos ✓) 4 4⇡↵2 e e qq¯ d⌦ e+e qq¯ 16⇡ ! · e+e uu¯ = ✓ ◆ ! ! 9 3 s

How would we expect to “observe” quarks? ) Will they behave as expected as function of cos ✓ and s ? )

M. Fanti (Physics Dep., UniMi) title in footer 39 / 78 + JADE: an experiment at e e collider

+ JADE experiment was run in 1979–86 at the PETRA e e collider in DESY (Hamburg) (center-of-mass energy ps =2Ebeam up to 30 GeV)

central tracker in magnetic field allows detection of charged particles and measurement of their momenta lead-glass calorimeter allows detection of e.m. showers produced by e± and

M. Fanti (Physics Dep., UniMi) title in footer 40 / 78 Interlude: particle identification by ionization energy loss

Charged particles crossing material (e.g. a gas) loose energy by collisions with electrons. The mean energy loss per p material depth depends on the particle speed — actually, on = . m

2 2 Z Bethe-Block formula works for 0.1 . . 1000 K =4⇡NAr mec e A 2 2 2 2 2 2 2 dE Q 1 2mec Wmax 2 () 2mec = K ln Wmax = 2 dx 2 2 2 2 1+2me/m +(me/m) ⌧  I = mean excit. energy (eV) [see pdg.lbl.gov “Passage of particles through matter”] I

dE p Measuring allows a measurement of (especially at low ) dx ⌘ m ⌧ Measuring also p (with curvature in magnetic field) allows knowledge of m particle identification ) dE 2 1 Due to Q2,fractionalcharges(e.g. , )wouldbeidentified dx / ±3 ± 3 by very⌧ low energy loss.

1 2 ALAS, particles with Q = , were never observed, sorry . . . ) ±3 ±3

M. Fanti (Physics Dep., UniMi) title in footer 41 / 78 + Hadronic jets at e e collisions

There are events with several hadrons produced. Hadrons happen to be not isotropic in the event, rather they cluster around two opposite directions: the “jets”

Define a 3 3tensor:S ↵ def= p↵p (↵, x, y, z ) ⇥ i i 2{ } i tracks 2{X } 3ortogonaleigenvectorsnˆ1, nˆ2, nˆ3 and 3 real eigenvalues 1 2 3. Some) properties:  

= Tr [S]= S ↵↵ = p~ 2 k | i | k ↵ i X 1 X 1 X p~ 2 1  3 k ⌘ 3 | i | k i 1 X 1 X p~ 2 3 3 k ⌘ 3 | i | Xk Xi ↵ ↵ ↵ ↵ 2 [dijet event from JADE] k =ˆnk (Snˆk )= n S n = n p p n = (p~i nˆk ) · k k k i i k · X↵, Xi X↵, Xi direction nˆ maximizes sum of squared tracks’ projected momenta choose it as jets’ axis ) 3 ) 2 nˆ3 1,2 are sums of squared tracks’ transverse momenta wrtn ˆ3: 1 + 2 = pi? Xi ⇣ ⌘

M. Fanti (Physics Dep., UniMi) title in footer 42 / 78 Sphericity and jets’ evidence

2 nˆ3 3 + 3 i pi? Define sphericity : S def= 1 2 2 ⌘ 2 ⇣ p~ 2⌘ k k P i | i | Properties: 0 S 1;eventswithsharpjetshaveS 0, while perfectly isotropic events have S 1.   P P ! ! + Evidence for jet structure in hadron production by e e annihilation [G.Hanson et at, LBL and SLAC, Physical Review Letters 1975, Vol. 35] (data taken at several energies : ps E =2E ) ⌘ cm beam

Sphericity decreases at higher collision energies (a) ps =3GeV (b) ps =6.2 GeV (c) ps =7.4 GeV At low energy observation compatible with both isotropic and jet models At higher energy, isotropic model (dashed) predicts increasing S,notcompatiblewithdata Jet model (solid) predicts decreasing S,confirmedby data

M. Fanti (Physics Dep., UniMi) title in footer 43 / 78 cos ✓ distribution of jets axis

Measuring the angle between jet axis (ˆn3)andincoming + e e beams: d find the expected behaviour 1+cos2 ✓ ) d⌦ /

jets appear to be the manifestation of quarks’ pair-production )

+ + take e e hadrons events as manifestation of e e qq¯ process ) ! !

M. Fanti (Physics Dep., UniMi) title in footer 44 / 78 Rµ and the number of “colors”

psqq¯ expected Rµ observed Rµ + def e e hadrons Define Rµ = ! . ¯ 2 + + 2 ms uu¯, dd, ss¯ 2 e e µ µ 3 ⇡ ! 2 2 4⇡↵ Qf 2 ¯ 10 From e+e f f¯ = we’d expect Rµ = Qq 2 mc uu¯, dd, ss¯, cc¯ ??? ! 3 s 9 q 11 11 X 2 m uu¯, dd¯, ss¯, cc¯, bb¯ b 9 3

obviously, each qq¯-pair is pro- ) duced 3 times more abundant than expected! this suggest the existence of ) an internal degree of freedom for quarks, that may take 3 values.

Quarks may exist in 3 “colors” red, green, blue ) —andlikewiseanti-quarks may exist in 3 “anti-colors” anti-red, anti-green, anti-blue

M. Fanti (Physics Dep., UniMi) title in footer 45 / 78 Hadrons are colorless

Apostulateisthatcolor cannot be observed free.

Therefore, when quarks (and antiquarks) are produced, more of them are created in order to form color-less aggregates: Mesons are qq¯ bound states. Their color composition may be some linear combination of rr¯, gg¯, bb¯,suchtogive zero net color. Baryons are qqq bound states. Each quark must have a di↵erent color state, like rgb,butinsuchawaythatthe three colors “neutralize”. The latter seems not intuitive. Luckily, there is a group SU(3) with commuting rules that ensures that a totally anti-symmetric combination satisfies the requirement: 1 (rgb + gbr + brg grb rbg bgr) p6 By the way, the introduction of “colors” was already proposed [W.Greenberg, 1964] to explain the “++ puzzle”. This resonance, with mass 1232 MeV,isauuu composition with spin 3/2, and being the lightest of a series must have zero orbital angular momenta. Therefore, the orbital wave function is entirely symmetric under quark exchanges. Likewise is the spin state ( to give +3/2). How """ can three identical fermions have a symmetric wave function? This would violate Pauli’s principle! The solution came postulating the existence of a further degree of freedom — the color — that antisymmetrizes the wave function. obs exp That seemed quite an artifact at the time, but the completely independent observation Rµ =3Rµ ,longlater,revivedit!

M. Fanti (Physics Dep., UniMi) title in footer 46 / 78 Quantum Chromodynamics (QCD)

M. Fanti (Physics Dep., UniMi) title in footer 47 / 78 Where are we?

Hadrons exist in multiplets, suggesting they are composites of some more fundamental particles — the “quarks”; — what binds quarks together is unknown Bjorken scaling in deep inelastic scattering proved that nucleons are made of point-like Dirac particles — the “partons” (partons and quarks are identified to be the same objects) — they do not account for the full momentum of the nucleon,though + production of free qq¯ is not observed in e e collisions; —collimatedjetsofhadronsareproduced,whoseangulardistributionfollowsthatexpectedforqq¯ production — why quarks are not directly observable? + e e jets cross-sections is 3 larger than expected, suggesting that quarks have one more internal degree of ! ⇠ ⇥ freedom — the color, taking 3 values the SU(3) group provides a perfect way to combine qq¯ and qqq into colorless composite states.

Conjecture: quarks exist in the simplest representation of SU(3) — the triplet. ) SU(3) is an exact symmetry — quarks masses and interactions do not change if we permute our definition of colors r,g,b. Making the SU(3) symmetry local generates a new interaction (recall Yang-Mills gauge theories) SU(3) has 8 generators 8gaugebosons,the“gluons” ) gluons interact with quarks through gs qqg vertices, (gs being the coupling) SU(3) is non-abelian gluons self-interact through g ggg and g 2 gggg vertices ) s s

M. Fanti (Physics Dep., UniMi) title in footer 48 / 78 SU(3) and QCD

R Quarks are fermions carrying a “color charge” described by 3 degrees of freedom: red, green, blue: q ⌘ 0 G 1 B Gauge invariance is expressed by SU(3) group:unitary3 3complexmatriceswithdet=1 @ A ⇥ 8groupparameters(1) ) 8gaugefields,the gluons ) q g g g

g g

q g g g

quark-gluon interaction triple gluon coupling quadruple gluon coupling To our knowledge, gluons are massless (as gauge fields should be)

1 SU(n)hasn2 1freeparameters.Proof:SU(n)matriceshaven2 complex coecients 2n2 real parameters; unitarity imposes n2 ) constraints (n diagonal real constraints and n(n 1)/2o↵-diagonalcomplexconstraints)plusthedet=1constraint(unitarymatrices always have det =1). | |

M. Fanti (Physics Dep., UniMi) title in footer 49 / 78 SU(3) group and its representations

gs 8 Any element U(✓~) ei 2 p=1 ✓pp must be unitary with det(U(✓~)) = 1 ⌘ each generator mustP be hermitian and traceless ) p What characterizes SU(3) is its set of commuting rules: [a; b]=2ifabc c There are several “representations” of the abstract group, where group elements are represented by d d matrices on ⇥ a d-dimensional space. Arepresentationisaclosedvectorspacewrtallgroupoperators,i.e.wrtallthegenerators.Thereisalwaysan operator commuting with all generators: 2 def= —tryyourselftheproofthatindeed[2; ]=0 a p p a 8 p X Arepresentationisavectorspacewhereallvectorshavethesameeigenvalueof2 ) The smallest representation of SU(3) must be dimension-3, with 3 3matrices. ⇥ Vectors inside a representation are further classified according to their eigenvalues wrt some other p operators that commute with each other and with 2 —sothattheyallcanbesimultaneouslydiagonalized.Theobviousthingisto seek diagonal matrices. Since commuting rules are the same for all representations, we consider 3 3matrices.With ⇥ the constraint to be traceless, we have 2 independent ones — traditionally named 3 and 8. eigenvectors in any representation can be mapped onto a plane )

As a comparison, consider the SU(2) group for the spin — or the isospin. There are 3 generators J1, J2, J3 and its smallest

representation is dimension-2, made of 2 2matrices—thePaulimatrices1, 2, 3.Therepresentationsaredefinedbyeigenvalues 2 2 2 2 ⇥ of J J + J + J .Eachrepresentationisfurthermappedontoaline,bytheeigenvaluesoftheonlydiagonalgenerator,J3. ⌘ 1 2 3

M. Fanti (Physics Dep., UniMi) title in footer 50 / 78 The 3 representation — quarks

There are two smallest representations of SU(3), both with dimension-3, labelled 3 and 3.¯

In representation 3, vectors are linear combinations of r, g, b,anddescribequarks. 1 Generators are written as r = 0 0 1 010 0 i 0 100 001 0 = 100 = i 00 = 0 10 = 000 @ 0 A 1 0 1 2 0 1 3 0 1 4 0 1 000 000 000 100 g = 1 0 1 @ 00 iA @ 000A @ 00 0A @ 10A 0 0 1 = 00 0 = 001 = 00 i = 01 0 @ 0 A 5 0 1 6 0 1 7 0 1 8 p 0 1 i 00 010 0 i 0 3 00 2 b = 0 0 1 @ A @ A @ A @ A 1 [Gell-Mann matrices](normalizedsuchthatTr [ab]=2ab) @ A λ Most fabc vanish, apart for 8

f123 =1 1 g r f = f = f = f = f = f = 147 246 257 345 516 637 2 p3 f = f = λ 458 678 2 3 and their permutations, with appropriate sign

In the (3; 8)plane,theeigenvectorsof3 are as in figure. b

M. Fanti (Physics Dep., UniMi) title in footer 51 / 78 The 3¯ representation — anti-quarks

i(Et p~ ~x) +i(Et p~ ~x) [Recall: fermions = ue · ;anti-fermions = ve · anti-particles imply complex conjugation] ) In representation 3¯, vectors are linear combinations of r¯, g¯, b¯,anddescribeanti-quarks. 1 Generators are obtained as ¯ = ⇤ k k r¯ = 0 0 1 0 10 0 i 0 100 00 1 0 ¯ = 100 ¯ = i 00 ¯ = 010 ¯ = 00 0 @ 0 A 1 0 1 2 0 1 3 0 1 4 0 1 000 000 000 10 0 g¯ = 1 0 1 @ 00 i A @000A @ 00 0A @ 100A 0 1 ¯ = 00 0 ¯ = 001 ¯ = 00 i ¯ = 0 10 @ 0 A 5 0 1 6 0 1 7 0 1 8 p 0 1 ¯ i 00 0 10 0 i 0 3 002 b = 0 0 1 @ A @ A @ A @ A 1 This way, [¯a; ¯b]=2ifabc ¯c and Tr ¯a¯b =2ab are preserved. @ A ⇥ ⇤ λ 8 ¯ In the (3; 8)plane,theeigenvectorsof3 and 3 are as in figure. b g r Anti-quarks have opposite (3; 8)eigenvalues(i.e.quantumnumbers)than corresponding quarks λ 3 g These are two independent representations: r, g, b and r¯, g¯, b¯ live in di↵erent r vector spaces, not connected by any SU(3) operator. b

M. Fanti (Physics Dep., UniMi) title in footer 52 / 78 The 8 representations — gluons

Another representation is given by the gluons.Theytransformintoeachotherwithagroupofmatricesthatisa representation of SU(3). The dimension is obviously 8 — it’s called 8

Proof: For ✓~ uniform in space-time, gluons transform as G µ = g f ✓ G µ a s abc b c First, note that these transforms are not complex — they can’t be, as gluon fields are real µ gs µ gs 8 8 Yet, let’s write it as G = i ✓b (2ifabc ) G and compare with infinitesimal U = 1 + i ✓b —the stands for dim-8 representation. a 2 c 2 b 8 8 We can identify fabc with matrix elements of b:precisely[b]ac = 2ifbac . Do they fulfill SU(3) requirements? 8 8 8 It’s easy to check that are hermitian :[( )†] = [ ] ⇤ =( 2if )⇤ =( 2i)⇤(f )⇤ =2i( f )=[ ] b b ca b ac bac bac bca b ca and traceless : Tr[8]= [8] = 2if =0 b a b aa a aba Now the commutators! We want to check whether [8; 8]=2if 8 P P a b abc c First, using the operator identity [A;[B; C]] + [B;[C; A]] + [C;[A; B]] = 0 for generic SU(3) operators a, b, c and using SU(3) commutation rules we find fadr frbc + facr frdb + fabr frcd =0—tryit!It’sjustalgebra...Then: [ ; ] =[ ] [ ] [ ] [ ] = 4(f f f f )= 4(f f + f f ) a b cd a cr b rd b cr a rd acr brd bcr ard acr rdb adr rbc =4f f =2if ( 2if )=2if [8] abr rcd abr rcd abr r cd

Note:Thisisactuallyageneralpropertyofallgaugetheories:gauge bosons transform into each other as a representation of the symmetry group —the“adjointrepresentation”.

Gluons belong to a color representation gluons carry color ) [Compare with QED: photons do not that’s why they interact with each other (ggg and gggg vertices) ) carry electric charge] when a gluon interacts with a quark, it changes quark’s color )

M. Fanti (Physics Dep., UniMi) title in footer 53 / 78 Building composite representations: 3 3¯ and 3 3 3 ⌦ ⌦ ⌦ We know that in nature there are several aggregates of quarks and antiquarks: mesons (qq¯ states) and baryons (qqq)—andofcourseanti-baryons(¯qq¯q¯) can they be described in terms of SU(3) representations? ) λ 8

Group theory tells us that 3 3¯ = 8 1 —meaningthataqq¯ system can gb rb ⌦ sit in a SU(3) octet, or in a SU(3) singlet. The exagone shows 6 obvious color-anticolor combinations at the vertices, rb¯, rg¯, bg¯, br¯, gr¯, gb¯,plus3combinationsofrr¯, gr rg λ gg¯, bb¯ in the centre. One of them belongs to color singlet: 3 1 rr¯+ gg¯ + bb¯ p3 br bg —thisisthecolorstateofmesons .

Likewise, 3 3 3 = 10 8 8 1 —aqqq system can stay in a decuplet, or in one of two independent octets, ⌦ ⌦ or again in a singlet. [You may build triangles centred over triangles 3 times now, but it gets even less clear how to separate the multiplets! Let’s leave this to groups experts. . . ] Baryons are in the singlet representation 1 (rgb + gbr + brg grb rbg bgr) p6

M. Fanti (Physics Dep., UniMi) title in footer 54 / 78 QCD e↵ective potential at short distances

r In QCD, fermions have a color state c g together with normal helicity and particle/antiparticle (u, v)states, ⌘ 0 1 b @ A u, v have indexes in spinor space, c has indexes in color space; quarks: uc -matrices and -matrices commute, because they act, respectively, on u, v components anti-quarks: vc and on c-components

g f1 f3 f1 f3 =[¯u (eµ) u ] µ⌫ [¯v (e⌫) v ] QED 3 1 q2 2 4 g M · · e S [¯u µu ] g [¯v ⌫v ] = e2 3 1 µ⌫ 2 4 · q2 g a µ gµ⌫ab b ⌫ = u¯ c† g u c v¯ c† g v c MQCD 3 3 S 2 1 1 · q2 · 2 2 S 2 4 4 e gS  ✓ ◆ µ  ✓ ⌫ ◆ 2 1 [¯u3 u1] gµ⌫ [¯v2 v4] = g c†ac1 c†ac4 f¯ f¯ f¯ f¯ S 4 3 2 · q2 2 4 2 4 ⇣ ⌘⇣ ⌘ QED scattering QCD scattering fQCD | {z } 2 2 ↵em ↵S e gS e↵ective potential for f f¯: VQED = VQCD = fQCD —(↵em = and ↵S = ) r )↵ r ↵ 4⇡ 4⇡ likewise e↵ective potential for ↵ : V =+ em V =+f S QED r ) QCD QCD r

M. Fanti (Physics Dep., UniMi) title in footer 55 / 78 QCD e↵ective potential at short distances

configuration qq¯ singlet qq¯ octet qq sextet qq triplet (symmetric) (antisymmetric)

4 1 1 2 f QCD 3 6 3 3

color factor fQCD: (see Grith for computation) ↵ ↵ f S +f S QCD r QCD r 4 ↵ 1 ↵ 1 ↵ 2 ↵ potential S S S S 3 r 6 r 3 r 3 r

attractive repulsive repulsive attractive colorless qq¯ singlets form bound states colored qq¯ octets repel apart colorless qqq singlet states are completely antisymmetric each qq pair inside binds ) colored qqq decuplets are completely symmetric each qq pair inside repel apart ) colored qqq octet may have symmetric or antisymmetric qq pairs inside — intermediate situation All this does not prove, but supports, the stability of colorless combinations wrt colored ones.

M. Fanti (Physics Dep., UniMi) title in footer 56 / 78 Running coupling in QED

Going to higher perturbative orders (i.e. adding loops) modifies the cross-section computation. All e↵ects can be “reabsorbed” in a redefinition of the coupling e: e0 is the “bare charge” (unphysical), what we measure is e. Creating virtual loops is more favoured for larger transferred squared momentum, Q2 def= q2 —note:hereQ is NOT acharge!Sinceq2 < 0always,it’sconvenienttointroduceQ2 > 0. The result is that the physical charge depends on Q2: e e(Q2). ⌘

2 2 ↵(µ ) ↵(Q )= 2 1 ↵(µ ) ln Q2 3⇡ µ2 ⇣ ⌘

[credits: Halzen-Martin] This is the running of the coupling constant :onceknownthevalue↵(µ2) at a given energy scale µ2,itsvalue↵(Q2) at a given energy scale Q2 can be predicted. 1 The well-known value of the fine structure constant, ↵ , corresponds to a scale µ2 m2 (electron mass 0 ' 137 ' e squared). 2 2 2 ↵0 Q ↵(Q )growswithQ , very slowly — eventually it diverges, when ln 2 =1(“Landau pole”) 3⇡ me 3⇡ ✓ ◆ 2 2↵ 277 This happens at transferred energies Q = me e 0 10 GeV —wecansleepwithoutfears! 5 ' ~c 19 (Planck’s scale MPl = 10 GeV)[yet,it’suncomfortablehavingadivergingprobability...]p r G ⇡ M. Fanti (Physics Dep., UniMi) title in footer 57 / 78 Running coupling in QCD

Similar situation in QCD — but there are also gluon loops:

2 2 ↵S (µ ) ↵S (Q )= 2 1+↵S (µ ) (11n 2n )ln Q2 12⇡ C F µ2 ⇣ ⌘ nC =numberofcolors=3 n =numberofflavours:u, d, c, s, t, b =6 F ) [credits: Halzen-Martin] (for “low” Q2 top-loop highly suppressed n =5) ) F anyway, 11n 2n > 0 ↵ (Q2)decreasesasQ2 increases,and↵ 0 as Q2 ) C F ) S S ! !1 This is the so-called “asymptotic freedom” [Nobel Prize 2004 to Gross, Politzer, Wilczek]

In high-energy collisions ↵S (and gS ) is small enough to allow perturbative expansion, and therefore Feynman rules.

On the opposite, at low energies ↵S increases, QCD becomes non-perturbative. 12⇡ There is a scale ⇤ at which ↵ diverges: ln(⇤2 )= ln(µ2) QCD S QCD (11n 2n )↵ (µ2) C F S 2 2 12⇡ Then ↵S (Q )canbeexpressedasfunctionof⇤QCD: ↵S (Q )= (11n 2n )ln Q2 C F ⇤2 QCD ✓ ◆

M. Fanti (Physics Dep., UniMi) title in footer 58 / 78 Running coupling in QCD — theory vs experiments

2 12⇡ ↵S (Q )= (11n 2n )ln Q2 C F ⇤2 QCD ✓ ◆

⇤QCD is not predicted by the theory: it must be deter- 2 mined experimentally, from measurements of ↵S vs Q . ⇤ 0.2 GeV — i.e. masses of light hadrons! ) QCD ' ⇡ clearly at energy scales hadron masses, ↵ blows up ) ⇡ S cannot describe hadronic bound states perturbatively )

[http://arxiv.org/abs/hep-ex/0211012v1]

M. Fanti (Physics Dep., UniMi) title in footer 59 / 78 The “color string” model

[D.Perkins “Introduction to High Energy Physics” Addison-Wesley] 1 3 5 Experimentally, several hadrons with a spin hierarchy (J =0, 1, 2,... or J = , , ,...)exhibitasimplerelation 2 2 2 2 2 between spin and mass: J = k0m + const,withk0 0.93 GeV (experimental) ' The color string model accounts for this relation. It assumes that the potential energy between quarks at distance r behaves as U(r)=kr.

For a meson (qq¯)assumethat: quarks’ masses are negligible c the qq¯ pair is separated by a distance 2r0;itrotatestogeneratetheangular momentum; the rotating speed of the quarks is c (= 1) (!) v q dU there is a color string connecting q andq ¯,withenergydensityu = = k;italso dr r r rotates with the quark pair; a piece of string between r and r + dr rotates with 0 r r v = c = ; r0 r0 all the “mass” and the angular momentum come from the kinetic energy of the rotating string; a string snippet has “mass” dm = dU = kdr,kineticenergy dm kdr d = = ,andangularmomentum q 2 2 2 E 1 v /c 1 (r/r0) dm kdr r 2 dL = p vr =p 1 v 2/c2 1 (r/r )2 r0 0 p p

M. Fanti (Physics Dep., UniMi) title in footer 60 / 78 The “color string” model

Integration gives: r0 r0 kdr M =2 d =2 = ⇡ kr0 2 0 E 0 1 (r/r ) Z Z 0 r0 r0 kdr r 2 ⇡ J =2 dL =2 p = kr2 2 0 0 0 1 (r/r ) r0 2 Z Z 0 1 J = M2 (eliminatingp r between 1st and 2nd eq) ) 2⇡ k 0 2 2 2 1 Therefore the relation J M is found. Replacing k0 0.93 GeV we get k 0.17 GeV 0.86 GeV fm .(for / ' 1 ' ' the latter, recall that setting c = ~ =1wegetGeV =5.06 fm ).

Just for fun: what is the newtonian force between a qq¯ pair? 10 dU 1.6 10 J F = = k 0.86 · 15 14 000 N —abouttheweightofamediumcar!Now,youseewhythiswascalled“strongforce”? dr ' 10 m '

M. Fanti (Physics Dep., UniMi) title in footer 61 / 78 The “color confinement” concept

E↵ective QCD qq¯ potential:

4 ↵ V = S + kr qq¯ 3 r long-distance short-distance | {z } | {z } QCD does not prove color confinement.

However, there are strong hints for it from the theory: (1) ↵ increases indefinitely at scales pQ2 hadron masses S ⇡ quarks inside hadrons are unlikely to separate ) (2) the “color string” model supports a potential V (r) kr ' can grow indefinitely with distance ) quarks inside hadrons are unlikely to separate (again!) )

M. Fanti (Physics Dep., UniMi) title in footer 62 / 78 Fragmentation and hadronization concepts

When quarks are produced at high energy, they travel apart. Their potential energy increases linearly as the color string stretches. At some point it is energetically more convenient to break the string by creating another qq¯ pair, and so on. . . This process is called fragmentation —oralso parton shower. The process goes on, converting part of the initial energy into quarks’ masses and kinetic energies, until eventually many colorless quarks aggregates (i.e. the hadrons) are formed. This phase is called hadronization. There are no more color strings between two hadrons. In the end color strings exist only inside each hadron. Typical masses of light hadrons (for which quarks’ masses are negligible) are . 0.5 GeV,suggestingthattheirsizebe 1 0.5 fm —recallk 1 GeV fm . . ⇡

M. Fanti (Physics Dep., UniMi) title in footer 63 / 78 Experimental evidence of gluons

M. Fanti (Physics Dep., UniMi) title in footer 64 / 78 + Gluon “observation”: 3 jets in e e collisions

+ If gluons exist, they should be produced in e e interactions:

e q e q e q

g g

e+ q¯ e+ q¯ e+ q¯ + + + e e qq¯ (NLO) e e qqg¯ e e qqg¯ ! ! !

As for quarks, the gluon would not be visible alone (color confinement, again!) we expect to observe it as a 3rd jet )

M. Fanti (Physics Dep., UniMi) title in footer 65 / 78 + Gluon “observation”: 3 jets in e e collisions

3jets 3 colored “objects” that hadronize )

+ Since the initial state (e e)iselectricallyneutraland colorless, the 3 final “objects” must also form a color singlet, with overall null electric charge (as for the initial + state e e)

Q: Why can’t they be 3 quarks (e.g. u, d, d)inafullyantisym- metric color configuration? A: because of angular momentum conservation...initial state has integer J,finalstatewouldbehalf-integerdueto3fermions,each carrying a 1 contribution due to its spin. ±2

[3-jet event from JADE]

M. Fanti (Physics Dep., UniMi) title in footer 66 / 78 The JADE algorithm for jets clustering

Now we have 3 jets we need a better way to define their kinematic properties, i.e. which observed particles belong ) to each jet (eigenvectors of sphericity tensor are not enough!)

Naively, a jet is a chunk of collimated par- ticles emerging from a parton fragmenta- tion. we need to define a way to group ) particles such to represent the underlying hard process (quarks and gluons), without being too sensitive to soft subprocesses (fragmentation, hadronization) CAVEAT: never hope for a perfect parton jet match — partons are colored, $ jets aren’t! Ideally, group together particles that are close in phase space,e.gmeaningtheyhavesmall invariant mass Recall: m2 =(E + E )2 (p~ + p~ )2 2E E (1 cos ↵ ),being↵ the angle between p~ ,p~ . jk j k j k ' j k jk jk j k m2 2 Define “distance” y = jk = E E (1 cos ↵ ) ( Q2 ( E )2 ( p~ )2 “total visible invariant mass” ) ) jk Q2 Q2 j k jk ⌘ j j Xj Xj (0) start with a list of “objects” given by the “particles”, each having 4-momentum p (E ; p~ ) j ⌘ j j 2 (1) compute y = E E (1 cos ↵ ) for all object pairs (jk) jk Q2 j k jk (2) choose pair with smallest yjk :ifyjk < ycut group the two constituents to make one new object with 4-momentum pj + pk ,remove objects j, k from the list, then reiterate from (1);

(3) if smallest yjk is > ycut,stopiteration:each“object”leftinthelistisconsideredajet

M. Fanti (Physics Dep., UniMi) title in footer 67 / 78 + Kinematics of 3 jets in e e collisions

Variable of interest: pT

pT

here defined as the transverse momentum component of the quark that radiated the gluon, with respect to the direction of the other quark In practice: choose jet with highest p~ as the “quark jet | | that did not radiate the gluon”. Either of the two remaining jets would have the same transverse-p~-component with respect to the former one. Continuous curves: Experimental points: Observation consistent with emission of gauge boson, QCD predictions at di↵erent : ps =12GeV with spin-1 and couplings as from QCD center-of-mass energies : ps [27.4, 31.6] GeV • 2 ps E : ps [35.0, 36.6] GeV ⌘ cm ⇥ 2 [credits: Halzen-Martin]

M. Fanti (Physics Dep., UniMi) title in footer 68 / 78 Scaling violation in D I S

M. Fanti (Physics Dep., UniMi) title in footer 69 / 78 E↵ect of gluons in the ep DIS: scaling violation

Reminder: in DIS the idea that the ep scattering could be described as an elastic eq scattering against a point- like spin-1/2 quark brought to the Bjorken scaling,which received experimental confirmation.

BUT: if QCD works, then the scattered quark could emit a gluon

no more only an “elastic eq eq scattering”, ) ! rather and more often an “inelastic eq eqg scattering” ! Bjorken conjecture would break ) (especially at large q2 when probability of gluon emission | | increases)

Even more: e-gluon scattering becomes possible (via g qq¯ splitting) !

M. Fanti (Physics Dep., UniMi) title in footer 70 / 78 E↵ect of gluons in the ep DIS: scaling violation

Experiments probing ep scattering at higher energies E —thusathigherQ2 — showed a clear dependence F F (x, Q2). 2 ⌘ 2 [Bjorken scaling implied F F (x)] 2 ⌘ 2 Remember: Q2 def= q2 =2(p p )xy and 0 y 1 1 · 2   On fixed target, (p p )=EM Q2 < 2EM x 1 · 2 p ) p

QUESTION:whydidformerexperimentsobserve F F (x) —independentofQ2 — 2 ⌘ 2 thus confirming Bjorken scaling ?

Former SLAC experiments probed x [ 0.03; 1] at energies E 2 ⇠ 2 [7; 17] GeV and small angles ✓ =6, 10 2 2 Q From Q =2EE 0(1 cos ✓)andx = 2Mp(E E 0) M x ✓ Q2 =2E 2(1 cos ✓) p 4EM sin2 x ) E + M x ' p 2 p ✓ ◆ at SLAC Q2 0.3 GeV2 ) . ... That’swhytheydidn’tobservethescalingviolation!

[credits: Halzen-Martin]

M. Fanti (Physics Dep., UniMi) title in footer 71 / 78 2 Probing at higher Q : e±p colliders

Recall: Q2 =2(p p )xy — x [0; 1] and y [0; 1] 1 · 2 2 2 2 At fixed target Q =2EMpxy

In a e±p collider, where both e± and p are relativistic, p (E ;0, 0, E )andp (E ;0, 0, E ) 1 ⌘ 1 1 2 ⌘ 2 2 (p p )=2E E Q2 =4E E xy ) 1 · 2 1 2 ) 1 2

HERA at DESY (1992–2007) was a e±p collider with e± beam at energy E1 =27.5 GeV p beam at energy E2 =460– 920 GeV could probe up to Q2 105 GeV2 ) ⇡

M. Fanti (Physics Dep., UniMi) title in footer 72 / 78 PDFs — scaling

2 Evolution of F2 with Q reflects evolution of PDFs: F (x, Q2) f (x, Q2) 2 ) a

Consider the parton splitting functions, Pab(z), Q probability that parton a comes from collinear emission from parton b,withp = zp p p a b call x def= a and y def= b x = zy pp pp ) a x Then, the PDFs evolve according to the Dokshitzer-Gribov-Lipatov-Altarelli-Parisi equation (DGLAP): Q

a 2 2 1 x=zy dfa(x, Q ) ↵S (Q ) dy x 2 b y 2 Pab fb y, Q d ln Q ' 2⇡ x y y P b Z ✓ ◆ ab (z) X

4 1+z2 P (z) Examples: L.O. parton splitting functions qq ' 3 1 z  4 1+(1 z)2 P (z) gq ' 3 z  1 P (z) z2 +(1 z)2 qg ' 2 P (z) z P (z) z P (z) z P (z) z ⇥1 z ⇤ z qq gq qg gg P (z) 6 + z(1 z)+ gg ' z 1 z 

M. Fanti (Physics Dep., UniMi) title in footer 73 / 78 PDFs at higher Q2

With DGLAP equations, the PDFs at high Q2 can be LHC parton kinematics 9 inferred from lower Q2 measurements 10 x = (M/14 TeV) exp(±y) 1,2 fixed target HERA 8 ! 10 Q = M M = 10 TeV HERA LHC ! 107 Few words about LHC 2 6 The (x, Q ) map @ LHC is more tricky... 10 M = 1 TeV

proton-proton collisions 5 )

2 10 2PDFs,2x-variables, x , x ) 1 2 104 M = 100 GeV (GeV proton, E proton, E 2 (collision) Q !parton, (x E;0,0,+x E) parton,(x E;0,0, x E) 3 1 1 2 2 10

proton-proton collision energy Ecm = ps =2E y = 6 4 2 0 2 4 6 2 parton-parton collision energy Q = x1x2 s 102 · M = 10 GeV kinematics of collision: Eˆ =(x1 + x2)E, pˆ =(x1 x2)E ˆ k 1 E +ˆp 1 x 1 fixed 1 10 HERA rapidity of collision: Y = ln k = ln target 2 Eˆ pˆ ! 2 x2 k ✓ ◆ 0 2 2 10 Q +Y Q Y -7 -6 -5 -4 -3 -2 -1 0 x = e ; x = e 10 10 10 10 10 10 10 10 1 s 2 s r r x

M. Fanti (Physics Dep., UniMi) title in footer 74 / 78 PDFs — some features

PDFs are extracted from global fits to several deep-inelastic-scattering (DIS) experiments

MSTW 2008 LO PDFs (68% C.L.)

) 1.2 ) 1.2 2 2 Behaviour vs x: Q2 = 10 GeV2 Q2 = 104 GeV2

xf(x,Q 1 xf(x,Q 1 At large x the valence quarks (u, d) g/10 g/10 dominate, with fu 2fd ; 0.8 0.8 ⇡ At low x gluons dominate by far; also qq¯-pairs from the sea are possi- 0.6 0.6 b,b u u ble, heavier quarks are less probable. d 0.4 0.4 c,c d Behaviour vs Q2: c,c s,s 0.2 s,s 0.2 d u d u With larger Q2 sea (anti-)quarks be- ¯ 0 0 come probable also at higher x; bb- -3 -3 10-4 10 10-2 10-1 1 10-4 10 10-2 10-1 1 pair also appear despite their large x x mass.

[http://mstwpdf.hepforge.org/]

M. Fanti (Physics Dep., UniMi) title in footer 75 / 78 PDFs — some features

MSTW 2008 LO PDFs (68% C.L.) PDFs must satisfy some normalization rules:

) 1.2 ) 1.2 2 2

Q2 = 10 GeV2 Q2 = 104 GeV2

xf(x,Q 1 xf(x,Q 1 Nu =2+Nu¯ ; Nd =1+Nd¯ ; Nq = Nq¯ q / u, d g/10 g/10 8 2{ } 0.8 0.8 1 0.6 0.6 b,b dx [f (x) f (x)] = 2 u u u u¯ 0 Z 1 d 0.4 0.4 c,c d dx [f (x) f ¯(x)] = 1 d d c,c s,s 0 0.2 s,s 0.2 Z 1 u d d u dx [f (x) f (x)] = 0 ( q = s, c, b ) q q¯ 0 0 0 -3 -3 Z 10-4 10 10-2 10-1 1 10-4 10 10-2 10-1 1 x x Also, momentum conservation implies that: The proton is much more complex than a uud system! 1 dx xf (x)+xf (x) =1 2 q g 3 0 ¯ ¯ 3“valencequarks”uud Z q=u,u¯,dX,d,s,s¯,c,c¯,b,b several low-x gluons 4 5 (emitted-reabsorbed by quarks) several qq¯-pairs “from the sea” Do you remember that “valence quarks” uud carry only 1 (via g qq¯ g processes) 2 of proton momentum? ! ! ⇠ ... incontinuoustransformation The rest is carried by gluons

M. Fanti (Physics Dep., UniMi) title in footer 76 / 78 PDFs — some features

MSTW 2008 LO PDFs (68% C.L.)

) 1.2 ) 1.2 2 2

Q2 = 10 GeV2 Q2 = 104 GeV2

xf(x,Q 1 xf(x,Q 1 g/10 g/10 0.8 0.8

0.6 0.6 b,b u u

d 0.4 0.4 c,c d

c,c s,s 0.2 s,s 0.2 d u d u

0 0 -3 -3 10-4 10 10-2 10-1 1 10-4 10 10-2 10-1 1 x x MSTW 2008 NLO PDFs (68% C.L.)

) 1.2 ) 1.2 2 2 E↵ect of perturbative order: Q2 = 10 GeV2 Q2 = 104 GeV2

xf(x,Q 1 xf(x,Q 1 g/10 g/10 From top to bottom: the hard scattering (partonic cross-sections) are eval- 0.8 0.8 uated respectively at leading order (LO), next-to-leading order (NLO) and 0.6 u 0.6 b,b u next-to-next-to-leading order (NNLO). Consequently, the PDFs also evolve. d 0.4 d 0.4 c,c

c,c s,s 0.2 s,s d 0.2 d u u

0 0 -3 -3 10-4 10 10-2 10-1 1 10-4 10 10-2 10-1 1 [http://mstwpdf.hepforge.org/] x x MSTW 2008 NNLO PDFs (68% C.L.)

) 1.2 ) 1.2 2 2

Q2 = 10 GeV2 Q2 = 104 GeV2

xf(x,Q 1 xf(x,Q 1 g/10 g/10 0.8 0.8

0.6 u 0.6 b,b u

d 0.4 d 0.4 c,c

c,c s,s 0.2 s,s d 0.2 d u u

0 0 -3 -3 10-4 10 10-2 10-1 1 10-4 10 10-2 10-1 1 x x

M. Fanti (Physics Dep., UniMi) title in footer 77 / 78 Summary

+ Quarks belong to triplets, r, g, b —mainproofbeingR at e e colliders { } µ Quarks dynamics is symmetric wrt color exchange SU(3) global symmetry ) Making SU(3) local gauge interaction theory, QCD,with8mediatorfields,thegluons ) Color is not observable. Quarks are present only inside hadrons: 1 mesons are qq¯ bound states in a color singlet: rr¯+ gg¯ + bb¯ p3 1 baryons are qqq bound states in a color singlet: (rgb + gbr +brg grb rbg bgr) p6 When quarks are hit or produced at high energy, they behave as free (“asymptotic freedom”) and travel apart. ⇠ Then, at larger distance, more quarks/antiquarks materialize until all particles are colorless hadrons. jets of hadrons ) The “color confinement” cannot be proved from QCD Lagrangian, but there are strong hints: qq¯ qq short-distance QCD e↵ective potentials, VQCD, VQCD,suggestthatcolor-singletaggregatesarestable, other combinations repel apart 2 running of ↵S show large coupling at smaller Q larger distance ) ↵ increase of QCD e↵ective potential at large distance: V = f S + kr QCD QCD r Gluons, as quarks, carry color, therefore cannot be directly observed. Best experimental evidences of gluons — and therefore of QCD — are + the observation of 3-jet events in e e collision the success of Altarelli-Parisi equations

M. Fanti (Physics Dep., UniMi) title in footer 78 / 78