Experimental foundations of Quantum Chromodynamicsfanti/Particelle3/01-QCD.pdfExperimental foundations of Quantum Chromodynamics Marcello Fanti University of Milano and INFN M. Fanti

Experimental foundations

of Quantum Chromodynamics

Marcello Fanti

University of Milano and INFN

M. Fanti (Physics Dep., UniMi) title in footer 1 / 76

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 — and anti-baryons are qqq.

Quarks (and antiquarks) carry a color charge, responsible for binding them into hadrons. Colors interact through

strong interaction, mediated by gluons, which also carry color.

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

. . . therefore one has to introduce a further postulate: 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.


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

[Winczec Nobel Lecture, 2004]


http://web.mit.edu/physics/people/faculty/docs/wilczek_nobel_lecture.pdf

Prequel: electron scattering

on pointlike fermion


ef → ef scattering

Electron scattering on pointlike fermion with charge Qf (e.g. µ) (mediated by a photon)

p2

p1

p3

q

p4

f

e

f

e

e

eQf

ef → ef

fermionic e.m. currents: Jµe = u3γµu1 ; Jνf = u4γ

νu2

photon propagator:gµνq2

(where q ≡ p1 − p3)

QED vertices: e and eQf

matrix amplitude: M = e2Qf Jµe

(gµνq2

)Jνf =

e2

q2Qf Jµe Jf ,µ

|M|2 =e4Q2

f

q4(Jµe Jf ,µ) (Jνe Jf ,ν)∗ =

e4Q2f

q4[(u3γ

µu1) (u4γµu2)] [(u3γνu1) (u4γνu2)]†

=e4Q2

f

q4

[(u3γ

µu1) (u3γνu1)†

]︸︷︷︸

Lµνe

[(u4γµu2) (u4γνu2)†

]︸︷︷︸

Lf ,µν

For what follows, recall that averaging over spins gives 〈uu〉spin =/p + m

2(and likewise, 〈v v〉spin =

/p −m

2)

Also, recall that u, u, γ are 4× 1, 1× 4, 4× 4 matrices. . .


ef → ef scattering : fermionic tensors

Few reminders:

u ≡ u†γ0 , γ0(γν)†γ0 = γν

〈uu〉spin =/p + m

2, 〈v v〉spin =

/p −m

2

Tr [ABC . . .Z ] = Tr [ZABC . . . ]Tr [γµγν] = 4gµν

Tr [γργµγν] = 0Tr [γργµγσγν] = 4 (g ρµgσν − g ρσgµν + g ρνgµσ)

Lµνe = (u3γµu1) (u3γ

νu1)†

= u3 γµ u1 u†1 (γν)† (u†3γ

0)†

= u3 γµ u1 u1

γν︷︸︸︷γ0(γν)†γ0 u3

= Tr [u3 γµ u1 u1 γ

ν u3]

= Tr [u3 u3 γµ u1 u1 γ

ν]

−−−−→〈〉spin

Tr

[/p3

+ me

2γµ/p1

+ me

2γν]

=1

4Tr[/p3γµ/p1

γν]

+me

4Tr[/p3γµγν

]+

me

4Tr[γµ/p1

γν]

+m2

e

4Tr [γµγν]

= p3,ρp1,σ (gρµgσν − gρσgµν + gρνgµσ) + m2e gµν

= pµ1 pν3 + pν1 pµ3 −(

p1 · p3 −m2e

)gµν

. . . and the same for the pointlike fermion f

What if e were e+?? Need to replace u → v , this affects 〈〉spin by flipping sign in front of m

. . . but m eventually enters only through m2, so the result does not change!

⇒ What we are doing is ok for e±f → e±f — and likewise for e±f → e±f


ef → ef scattering : matrix element

Lµνe = pµ1 pν3 + pν1 pµ3 −(

p1 · p3 −m2e

)gµν

Lf ,µν = p2,µp4,ν + p2,νp4,µ −(

p2 · p4 −m2f

)gµν

Now neglect electron mass (me ' 0) — means that E1, |~p1| me — but keep mf

Lµνe Lf ,µν = [pµ1 pν3 + pν1 pµ3 − (p1 · p3) gµν][p2,µp4,ν + p2,νp4,µ −

(p2 · p4 −m2

f

)gµν]

= 2[

(p1 · p2)(p3 · p4) + (p1 · p4)(p2 · p3) − m2f (p1 · p3)

]|M|2 =

e4Q2f

q4Lµνe Lf ,µν = 2

e4Q2f

q4

[(p1 · p2)(p3 · p4) + (p1 · p4)(p2 · p3) − m2

f (p1 · p3)]

Using p4 = p1 + p2 − p3 and neglecting p21 = p2

3 = m2e ' 0 :

|M|2 = 2e4Q2

f

q4

[2(p1 · p2)(p3 · p2) + (p1 · p3)(p2 · p1 − p2 · p3 −m2

f )]


ef → ef scattering : matrix element in LAB frame

LAB frame : where f is at rest at the beginning

p1 ≡ (E ; 0, 0,E ) ; p3 ≡ (E ′; E ′ sin θ, 0,E ′ cos θ)

p2 ≡ (mf ; 0, 0, 0) ; p4 ≡ (E4; ~p4)θ

(E;0,0,E)

incoming electron

(E’;E’sinθ,0,E’cosθ)outg

oing el

ectro

n

p1 · p2 = Emf

p1 · p3 = EE ′(1− cosθ)

p2 · p3 = mf E ′

q2 = (p1 − p3)2 = − 2p1 · p3

= −2EE ′(1− cos θ)

Another useful relation: q + p2 = p4⇒ (q + p2)2 = p24⇒ q2 + 2q · p2 + m2

f = m2f⇒ q2 + 2mf (E − E ′) = 0

νdef= E − E ′ = − q2

2mf

(. . . and recall1− cos θ

2= sin2 θ

2and

1 + cos θ

2= cos2 θ

2)

|M|2 = 2e4Q2

f

q4

[2m2

f EE ′ + EE ′(1− cos θ)(mf E −mf E ′ −m2f )]

= 2e4Q2

f

[2EE ′(1− cos θ)]22m2

f EE ′[

1 + cos θ

2+

1− cos θ

2

E − E ′

mf

]=

e4Q2f

4EE ′ sin4(θ2

)m2f

[cos2

(θ

2

)− q2

2m2f

sin2

(θ

2

)]M. Fanti (Physics Dep., UniMi) title in footer 8 / 76

ef → ef scattering : cross-section in LAB frame

Compute cross-section in LAB frame (where f is at rest at the beginning)

p1 ≡ (E ; 0, 0,E ) ; p3 ≡ (E ′; E ′ sin θ, 0,E ′ cos θ)

p2 ≡ (mf ; 0, 0, 0) ; p4 ≡ (E4; ~p4)

Incoming speeds: v1 = c = 1 (relativistic electron) and v2 = 0. Number of spin states: F3 = 2 and F4 = 2

Phase space term for relativistic electron:d 3p3

2E3=|~p3|2 d |~p3| dΩ

2E3=

E ′ dE ′ dΩ

2(where dΩ ≡ d(cos θ)dφ)

Phase space term for recoiling fermion:d 3p4

2E4δ(3)(~p3 + ~p4 − ~p1 − ~p2)δ(E4 + E3 − E1 − E2) −−−→∫

d3p4

δ(E4 + E ′ −mf − E )

2(mf + E − E ′)

Recall Fermi’s golden rule for 1, 2→ 3, 4 processes:

dσ =1

4(v1 + v2)E1E2

(d 3p3

(2π)3 2E3F3

)(d 3p4

(2π)3 2E4F4

)(2π)4 δ(4) (p3 + p4 − p1 − p2) |M|2

−−−→∫d3p4

1

4 E mf

1

(2π)2(E ′ dE ′ dΩ)

(1

mf + E − E ′

)δ(E4 + E ′ − E −mf ) |M|2


ef → ef scattering : cross-section in LAB frame

Now, try to elaborate dE ′ δ(E4 + E ′ − E −mf ) — remember: E4 is also a function of E ′

For elastic ef → ef , E 24 = (~p4)2 + m2

f = (~p1 − ~p3)2 + m2f = (E 2 + E ′

2 − 2EE ′ cos θ) + m2f

Call f (E ′)def= E ′ + E4 − (E + mf ) where f (E ′) = 0 has some solution E ′ = E ′elastic ⇒ δ(f (E ′)) =

δ(E ′ − E ′elastic)

|f ′(E ′)|

f ′(E ′) ≡ ∂(E ′ + E4 − (E + mf ))

∂E ′= 1 +

∂E4

∂E ′= 1 +

1

2 E4

∂(E 24 )

∂E ′

= 1 +2E ′ − 2E cos θ

2(mf + E − E ′)=

mf + E (1− cos θ)

mf + E − E ′

dE ′ δ(E4 + E ′ − E −mf ) −−−→∫dE ′

mf + E − E ′

mf + E (1− cos θ)

and (1− cos θ) = − q2

2EE ′=

mf (E − E ′)

EE ′⇒ mf + E (1− cos θ) = mf

E

E ′

Elastic differential cross-section (recall that α =e2

4π):

dσ

dΩ=

1

4(2π)2m2f

(E ′

E

)2

|M|2

=α2Q2

f

4E 2 sin4(θ2

) E ′

E

[cos2

(θ

2

)− q2

2m2f

sin2

(θ

2

)]


Electron-proton

elastic scattering


Recalling the “static model”

mesonic nonet

baryonic octet

baryonic decuplet

The variety of discovered mesons and baryons, and their “recursive” properties,

suggested that they be composite states of more fundamental particles: the

quarks, that may come in different flavours (up, down, strange, . . . )

[Gell-Mann, Zweig, 1964]

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:

Jνp 6= u4γνu2

Lp,µν 6= p2,µp4,ν + p2,νp4,µ −(

p2 · p4 −M2p

)gµν[

dσ

dΩ

]ep→ep

6= e4Q2f

(4π)2

1

4E 2 sin4 θ2

E ′

E

[cos2

(θ

2

)− q2

2M2p

sin2

(θ

2

)]


ep → ep elastic scattering

The fermion current of the proton must be modified into some Jνp = u4Γνu2

The Γν are 4 × 4 matrices that “parametrize our ignorance”, yet there are

some constraints:

Jνp must be a Lorentz 4-vector and must depend only on the kinematics

of the scattering

Γν must be built with known ingredients (i.e. γ-matrices and their

combinations)

for soft scattering the photon wavelength λ =~|~q|

is not short enough to

probe the proton structure

⇒ Γν must reduce to Dirac current for very low q2 (i.e. Γν −−−−−→q2→0

γν)

p2

p1

p3

q

p4

f

e

f

e

e

ep → ep

⇒ The parametrization must be

Γν = F1(q2)γν︸︷︷︸electric coupling

+κ

2MpF2(q2)iσνρqρ︸︷︷︸

magnetic coupling

where σνρdef=γνγρ − γργν

2

κ = anomalous magnetic moment at rest (e.g. κp = 1.79 for proton, κn = −1.91 for neutron)

(recall the Gordon decomposition u4γνu2 = u4

[pν2 + pν4

2m− i

1

2mσνρqρ

]u2 )

Cross-section in lab frame becomes (recall αdef=

e2

4π):[

dσ

dΩ

]ep→ep

=α2

4E 2 sin4(θ2

) E ′

E

[(F 2

1 (q2)− κ2q2

4M2p

F 22 (q2)

)cos2

(θ

2

)− q2

2M2p

(F1(q2) + κF2(q2)) sin2

(θ

2

)]M. Fanti (Physics Dep., UniMi) title in footer 13 / 76

ep → ep elastic scattering : the size of the proton

The “form factors” F1(q2), F2(q2) are related to the extension in space of the proton, by F1,2 =

∫d 3r ρ1,2(~r)ei~q·~r —

F1 for the charge distribution, F2 for the magnetic moment distribution.

Low q2 limit: F1,2(q2) −−−−−→q2→0

1

Assuming spherical symmetry (ρ(~r) ≡ ρ(r)) and expanding ei~q·~r in power series:

F1,2(q2) '∫

d 3r ρ(r)

(1 + i~q · ~r − (~q · ~r)2

2+ · · ·

)= 1 − 1

6|~q|2

⟨r 2⟩

+ · · ·

Here√〈r 2〉 measures the “radius of the proton”

Is the scattering indeed ep → ep?The apparatus measures the outgoing electron energy and angle : E ′ and θ

The proton is not detected: how can we ensure that the reaction is not some ep → eX ?

This actually happens, for large enough −q2. . . If there is no proton in final state, we cannot even build a

Jνp = u4Γνu2 — what is u4 ???

Answer: We saw for elastic scattering that ν ≡ E − E ′ = − q2

2Mp⇒ E − E ′ =

EE ′(1− cos θ)

Mp

⇒ E ′ =Mp E

Mp + 2 E sin2(θ2

)This relation can be tested experimentally, to guarantee an elastic scattering.


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

Measured E ′ vs θ

⇒ it is indeed elastic ep → ep


ep → ep elastic scattering : experiment

Differential cross-sectiondσdΩ

vs θ

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√〈r 2〉 = (0.70± 0.24) · 10−13 cm


ep → ep elastic scattering : some back-of-envelope

Using McAllister-Hofstadter results, how large is the deviation of the proton from point-like model?

This means estimating F1,2(q2) ' 1−|~q|2

⟨r 2⟩

6for⟨

r 2⟩' 0.5 · 10−26 cm2

Recalling q2 = −2EE ′(1− cos θ) and q2 = −2Mp(E − E ′) = −2Mpν, and q2 ≡ ν2 − |~q|2 yields:

|~q|2 = ν2 − q2 = ν2 + 2Mpν = (ν + Mp)2 −M2p

In the experiment, ν ∈ [0; 50 MeV], and using Mp = 938 MeV the largest effect is at |~q|2max = 96300 MeV2

Now, need to “convert” cm to MeV. Recall that ~ = 6.582 · 10−22 MeV s and c = 3 · 1010 cm/s, so

~c = 1.975 · 10−11 MeV cm. Setting ~c = 1 means 1 cm =1

1.975 · 10−11 MeV= 5.06 · 1010 MeV−1.

So⟨

r 2⟩' 0.5 · 10−26 cm2 ' 1.25 · 10−5 MeV−2.

So F1,2(q2) ' 1−|~q|2max

⟨r 2⟩

6' 0.8 — a 20% deviation from point-like F1,2(q2) = 1, well observable!


Electron-proton

inelastic scattering

and Deep Inelastic Scattering

(DIS)


Inelastic ep scattering

p2

p1

p3

q

f

e

Xn

X2

X1

e

e

ep → eX

For higher −q2, the proton breaks apart: ep → eX , X ≡ X1,X2, . . . .Xnbeing a chunk of (several) hadrons

⇒ cannot define anymore a “proton current” Jνp

⇒ the “elastic” relation between ν and θ is not valid anymore: ν 6= − q2

2Mp

⇒ need to generalize the Lµνe Lp,µν to some Lorentz-invariant form, that keeps

Lµνe (the electron is always the same. . . )

Lµνe Lp,µν −→ Lµνe Wµν

Wµν = W1(. . . )

(−gµν +

qµqνq2

)+ W2(. . . )

1

M2p

(p2,µ −

p2 · qq2

qµ

)(p2,ν −

p2 · qq2

qν

)W1,2(. . . ) are functions of the kinematics at proton vertex: here two independent quantities:

q2 and νdef=

p2 · qMp

= E − E ′ (last step for the LAB frame, where the proton is at rest and p2 ≡ (Mp; 0, 0, 0) )

[d 2σ

dE ′ dΩ

]ep→eX

=α2

4E 2 sin4(θ2

) [W2(ν, q2) cos2

(θ

2

)+ 2 W1(ν, q2) sin2

(θ

2

)][see e.g. Halzen & Martin “Quarks and Leptons”]


Inelastic ep scattering : some kinematics

Independent L.I. quantities:

q2 def= (p1 − p3)2 ; ν

def=

p2 · qMp

Invariant mass of X final state:

m2X = (p2 + q)2 = M2

p + 2Mpν + q2

Dimensionless L.I. quantities:

xdef=−q2

2p2 · q=−q2

2Mpν

ydef=

p2 · qp2 · p1

Inverse relations:

−q2 = 2(p1 · p2)xy

ν =p1 · p2

Mpy

In the LAB frame where the proton is at rest:

p2 ≡ (Mp; 0, 0, 0) and p1 · p2 = MpE

ν = E − E ′

q2 = −2EE ′(1− cos θ)

x =EE ′(1− cos θ)

Mp(E − E ′)

y =E − E ′

E

Allowed region for x , y : y ∈ [0; 1] and x ∈ [0; 1]Being x , y L.I., we can prove these in the LAB frame.For y this is obvious. Also obvious is x ≥ 0. Then, write m2

X asfunction of x , y :

m2X = M2

p + 2Mpp1 · p2

Mpy − 2(p1 · p2)xy

= M2p

(1 + 2y(1− x)

E

Mp

)Since m2

X > 0 for any value ofE

Mp, the only way is x ≤ 1.

The case x = 1 coincides with mX = Mp, i.e. elastic scattering.

As a by-product, notice that mX ≥ Mp always


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-

ator)

Observable scattering angle θ ∈ [6, 10]

⇒ −q2 . 7.4 GeV

Measures ofd 2σ

dE ′ dΩas a function of mX — called W

here

for different −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.


Inelastic ep → eX : experiment

NOTE: mX ≡ W =√

M2p + 2Mp(E − E ′)− 2EE ′(1− cos θ) is inferred from observed E ′, θ

PROBLEM: Final states where X is a hadronic resonance

are favoured by larger cross-sections. But this requires

precise values of q2, ν. If the energy transferred from the

electron (ν) is too high to produce exactly the needed

mX , it is “convenient” to loose some energy in extra e.m.

radiation and fall back onto the resonance (“radiative re-

turn”)

HOWEVER: this “radiation” is not detectable from ob-

servables E ′, θ ⇒ 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.

Fig(a):d 2σ

dE ′ dΩvs W before radiative correction

Fig(b):d 2σ

dE ′ dΩvs W after radiative correction

Fig(c): ratio after/before radiative correction

Note that the correction enhances the resonances (as it

should!)


Elastic vs inelastic scattering

Elastic scattering on pointlike fermion

(E ′ is function of θ – recall E − E ′ ≡ ν = − q2

2mf)

Inelastic scattering on proton

(E ′ and θ are independent)

dσ

dΩ=

α2Q2f

4E 2 sin4(θ2

) E ′E

[cos2

(θ

2

)− q2

2m2f

sin2

(θ

2

)] d2σ

dE ′ dΩ=

α2

4E 2 sin4(θ2

) [W2(ν, q2) cos2

(θ

2

)+ 2W1(ν, q2) sin2

(θ

2

)]

can be written in a similar way (proof in next slide):

d2σ

dE ′ dΩ=

4α2E ′2

q4Q2

f ×

×[

cos2

(θ

2

)− q2

2m2f

sin2

(θ

2

)]δ

(ν +

q2

2mf

)d2σ

dE ′ dΩ=

4α2E ′2

q4×

×[W2(ν, q2) cos2

(θ

2

)+ 2W1(ν, q2) sin2

(θ

2

)]Therefore we can recover the elastic scattering on pointlike fermion from the general formula, with the replacements:

W2(ν, q2) −→ Q2f · δ

(ν +

q2

2mf

)W1(ν, q2) −→ Q2

f ·−q2

4m2f

· δ(ν +

q2

2mf

)


Inelastic vs elastic scattering : proof

Let’s start with4α2E ′2

q4:

using q2 = −2EE ′(1− cos θ) = −4EE ′ sin2

(θ

2

)we get:

4E ′2

q4=

4E ′2

16E 2E ′2 sin4(θ2

) =1

4E 2 sin4(θ2

)For the elastic cross-section on pointlike fermion, to go from

d2σ

dE ′ dΩto

dσ

dΩneed to integrate over dE ′:

Recalling that δ(f (E ′)) =δ(E ′ − E ′sol)

|f ′(E ′sol)|, being E ′sol a solution of f (E ′) = 0, we have:

dE ′ δ

(ν +

q2

2mf

)= dE ′ δ

(E − E ′ − 2EE ′

mfsin2

(θ

2

))= dE ′

δ(E ′ − E ′sol)∣∣∣1 + 2Emf

sin2(θ2

)∣∣∣ = dE ′δ(E ′ − E ′sol)∣∣1 + E−E ′

E ′

∣∣ = dE ′ δ(E ′ − E ′sol)E ′

E

−−−→∫dE ′

E ′

E

So theE ′

Eterm is also recovered.


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:

W2(ν, q2) −→ Q2q · δ

(ν +

q2

2mq

)W1(ν, q2) −→ Q2

q ·−q2

4m2q

· δ(ν +

q2

2mq

)with some redefinitions, and recalling that aδ(ax) = δ(x):

F2def= νW2 = Q2

q δ

(1− Mp

mq

−q2

2Mpν

)F1

def= MpW1 = Q2

q

1

2

(Mp

mq

)2 ( −q2

2Mpν

)δ

(1− Mp

mq

−q2

2Mpν

)

where we spot the variable x ≡ −q2

2Mpν, so we get

F2(x) = Q2q δ

(1− Mp

mqx

)F1(x) = Q2

q

(Mp

mq

)2x

2δ

(1− Mp

mqx

)i.e. the new functions F1,2 depend on −q2, ν ONLY through their combination x ≡ −q2

2MpνThis prediction is known as the “Bjorken scaling” [D.Bjorken, Physical Review 1969, Vol 179]


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-

ator)

Observable scattering angle θ ∈ [6, 10]

⇒ −q2 . 7.4 GeV

Recall:

d2σ

dE ′ dΩ=

4α2E ′2

q4×

×[W2(ν, q2) cos2

(θ

2

)+ 2W1(ν, q2) sin2

(θ

2

)]⇒ at small θ probe W2

Several plots of νW2 vs ωdef=

2Mpν

−q2(note: ω ≡ 1

x)

at different values of E ⇒ change −q2, ν

⇒ dependence on ω is clear (despite complicate!)

absence of spread⇒ no dependence on −q2, ν separately

(a,c) : θ = 6, several E(b,d) : θ = 10, several E

(e) : θ = 6, E = 7 GeV, R is related to (unknown)W1

W2ratio


The “parton model”

The experimental evidence of Bjorken scaling (i.e. νW2 ≡ F2 ≡ F2(x)) supports the idea that

protons are made of pointlike Dirac constituents — the quarks, or the “partons” (so named by Feynman)

νW2def= F2 = Q2

q δ

(1− Mp

mqx

)BUT : in this equation νW2 ≡ F2 is a δ-function, that should lock x =

mq

Mp.

On the opposite, experimental results show broad spectra for νW2 vs x (or vs ω ≡ 1

x).

MOREOVER (related) : what is mq ??

Yes, the quark mass. . . But we easily define masses for “free” (non-interacting) particles, whereas quarks are bound inside the proton, so

they are definitely interacting — they may well be “virtual”, or “off-shell”.

Consider a very relativistic proton, with energy Ep, longitudinal momentum PLp and null transverse momentum PT

p = 0. Inside it, a quarkwith energy Eq, longitudinal momentum PL

q and negligible transverse momentum PTq ' 0. Then

M2p = E 2

p − (PLp )2 ; m2

q = E 2q − (PL

q )2

Proton and quark move at the same speed v =Pp

Ep=

Pq

Eq, therefore calling xq

def=

Eq

Epwe have:

PLq = xq P

Lp ; Eq = xq Ep ; mq =

√E 2q − (PL

q )2 = xq

√E 2P − (PL

p )2 = xq Mp

This defines the “quark mass” inside the proton in terms of the fraction of the proton momentum carried by the

quark: mq = xqMp =Pq

PpMp. The δ-function therefore locks x = xq =

Pq

PpM. Fanti (Physics Dep., UniMi) title in footer 27 / 76

The “parton distribution functions” (PDFs)

We can now better formulate F2 ≡ νW2: the ep scattering at high −q2 is a combination of all possible eq scatterings,

for all possible quarks q, with all possible values of xq.

Question: what values picks xq? Any possible values in [0; 1], with some probability distribution fq(xq) — these are

called parton distribution functions (PDFs) . . . and are unknown

νW2 ≡ F2(x) =∑q

∫ 1

0

dxq fq(xq) ·[Q2

q δ

(1− Mp

mqx

) ]=∑q

Q2q

∫ 1

0

dxq fq(xq)δ

(1− x

xq

)=∑q

Q2q

∫ 1

0

dxq fq(xq) xq δ(x − xq) =∑q

Q2q x fq(x)

MpW1 ≡ F1(x) =x

2

(Mp

mq

)2

F2(x) =1

2xF2(x)

In summary:

2xF1(x) = F2(x) = x∑q

Q2q fq(x)

W2(ν, q2) =F2(x)

ν

W1(ν, q2) =F1(x)

Mp=

F2(x)

2x Mp

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


ep vs eq cross-sections: “factorization”

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

(d2σ

dE ′ dΩ

)eq→eq

=4α2E ′2

q4Q2

q cos2

(θ

2

)·[

1− q2

2m2q

tan2

(θ

2

)]δ

(ν +

q2

2mq

)=

4α2E ′2

q4Q2

q cos2

(θ

2

)·

[1 +

M2p

m2q

ν

Mp

−q2

2Mpνtan2

(θ

2

)]1

νδ

(1− x

xq

)=

4α2E ′2

q4Q2

q cos2

(θ

2

)·[

1

ν+

x

x2q Mp

tan2

(θ

2

)]xqδ(x − xq)

=4α2E ′2

q4Q2

q cos2

(θ

2

)·

[x

ν+

tan2(θ2

)Mp

]δ(x − xq)

Helpers:

xqdef=

mq

MP; x

def=−q2

2Mpν

δ

(ν +

q2

2mq

)=

1

νδ

(1− x

xq

)

Cross-section of ep → eX DIS:

(d2σ

dE ′ dΩ

)ep→eX

=4α2E ′2

q4cos2

(θ

2

)·[W2 + 2W1 tan2

(θ

2

)]=

4α2E ′2

q4cos2

(θ

2

)·

[x

ν+

tan2(θ2

)Mp

]F2

x

=4α2E ′2

q4cos2

(θ

2

)·

[x

ν+

tan2(θ2

)Mp

] ∑q

Q2q fq(x)

=∑q

∫dxq fq(xq)

(d2σ

dE ′ dΩ

)eq→eq

Helpers:

W2 =F2

ν; 2W1 =

F2

x Mp


“Hadronic” vs “partonic” cross-section : factorization

We got

hadronic cross-section︷︸︸︷(d 2σ

dE ′ dΩ

)ep→eX

=∑q

∫dxq fq(xq)

partonic cross-section︷︸︸︷(d 2σ

dE ′ dΩ

)eq→eq

i.e. the ep “hadronic cross-section” can be factorized as the eq “partonic cross-section” times the PDFs

Reminder: the PDFs are unknown . . . BUT:

partonic cross-sections can be evaluated

hadronic cross-sections can be measured⇒ the PDFs can be probed

Ideally, once PDFs are “known” (so

to say. . . ) they can be used, to-

gether with other partonic cross-

sections, to predict other hadronic

cross-sections

σpartproc1 computation

σhadproc1 measurement

⇒



⇒



⇒

⇒ fit PDFs

σpartprocX computation

⇒ predict σhadprocX


Can we extract nucleon’s PDFs ?

Recall: F2(x) can be measured, and F2(x) = x∑q

Q2q fq(x) — how to extract all fq’s from F2?

Simplest nucleon model: proton≡ u, u, d and neutron≡ u, d , d, with Qu =2

3; Qd = −1

3NOTE: f

(p)u (x) describes PDF of both u-quarks inside the proton — likewise, f

(n)d (x) describes PDF of both d-quarks inside the proto

⇒F

(p)2 (x) = x

∑q=u,d Q2

q f(p)q (x) = x

[49f

(p)u (x) + 1

9f(p)d (x)

]F

(n)2 (x) = x

∑q=u,d Q2

q f(n)q (x) = x

[49f

(n)u (x) + 1

9f(n)d (x)

]but isospin flips p ↔ n and u ↔ d

⇒

f

(n)u = f

(p)d ≡ fd

f(n)d = f

(p)u ≡ fu

From ep , en inelastic scattering measure F(p)2 (x) , F

(n)2 (x) ⇒ extract fu(x) , fd(x) — DOES THIS WORK???

Average momentum fraction carried by u, d inside p: 〈x〉(p)u,d =

∫dx x fu,d(x)

If simply p ≡ u, u, d we’d expect 〈x〉(p)u + 〈x〉(p)

d = 1. But:∫dx F

(p)2 (x) = 4

9

∫dx x fu(x) + 1

9

∫dx x fd(x) = 4

9 〈x〉(p)u + 1

9 〈x〉(p)d∫

dx F(n)2 (x) = 4

9

∫dx x fd(x) + 1

9

∫dx x fu(x) = 4

9 〈x〉(p)d + 1

9 〈x〉(p)u

from measurements we have:∫dx F

(p)2 (x) ' 0.18

∫dx F

(n)2 (x) ' 0.12

⇒ solving we get 〈x〉(p)u ' 0.36 and 〈x〉(p)

d ' 0.18 — i.e. 〈x〉(p)u + 〈x〉(p)

d ' 0.54

⇒ NO, DOESN’T WORK! Where is the rest of proton’s momentum?M. Fanti (Physics Dep., UniMi) title in footer 31 / 76

Summary

From ep scattering we learnt:

elastic scattering: protons are not pointlike (√〈r 2〉 ' 0.7 · 10−13 cm)

deep inelastic scattering (DIS): protons are made 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 2σ

dE ′ dΩ

)ep→eX

=∑q

∫dxq fq(xq)

(d 2σ

dE ′ dΩ

)eq→eq

the fq(xq) are the parton distribution functions (PDFs) — to be determined experimentally

assuming p ≡ u, u, d, n ≡ u, d , d, from ep, en DIS one can try to measure fu(x), fd(x)

however this does explain only ∼ half the momentum of the nucleon!

Some speculationsThe parton model is successful, but apparently there are more partons than just the “valence quarks” u, u, dThere 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”.


Quark pair-production

in e+e− collisions


The e+e−→ f f process

p1

p4p2

q

p3

e+

e−

f

f

e eQf

e+e− → f f

fermionic e.m. currents: Jµe = v2γµu1 ; Jνf = u3γ

νv4

photon propagator:gµνq2

(where q ≡ p1 + p2)

QED vertices: e and eQf

matrix amplitude: M = e2Qf Jµe

(gµνq2

)Jνf =

e2

q2Qf Jµe Jf ,µ

|M|2 =e4Q2

f

q4(Jµe Jf ,µ) (Jνe Jf ,ν)∗ =

e4Q2f

q4[(v2γ

µu1) (u3γµv4)] [(v2γνu1) (u3γνv4)]†

=e4Q2

f

q4

[(v2γ

µu1) (v2γνu1)†

]︸︷︷︸

Lµνe

[(u3γµv4) (u3γνv4)†

]︸︷︷︸

Lf ,µν

For what follows, recall that averaging over spins gives 〈uu〉spin =/p + m

2(and likewise, 〈v v〉spin =

/p −m

2)

Also, recall that u, u, γ are 4× 1, 1× 4, 4× 4 matrices. . .


The e+e−→ f f process : fermionic tensors

Few reminders:

u ≡ u†γ0 , γ0(γν)†γ0 = γν

〈uu〉spin =/p + m

2, 〈v v〉spin =

/p −m

2

Tr [ABC . . .Z ] = Tr [ZABC . . . ]Tr [γµγν] = 4gµν

Tr [γργµγν] = 0Tr [γργµγσγν] = 4 (g ρµgσν − g ρσgµν + g ρνgµσ)

Lµνe = (v2γµu1) (v2γ

νu1)†

= v2 γµ u1 u†1 (γν)† (v †2γ

0)†

= v2 γµ u1 u1

γν︷︸︸︷γ0(γν)†γ0 v2

= Tr [v2 γµ u1 u1 γ

ν v2]

= Tr [v2 v2 γµ u1 u1 γ

ν]

−−−−→〈〉spin

Tr

[/p2−me

2γµ/p1

+ me

2γν]

=1

4Tr[/p2γµ/p1

γν]

+me

4Tr[/p3γµγν

]− me

4Tr[γµ/p1

γν]− m2

e

4Tr [γµγν]

= p2,ρp1,σ (gρµgσν − gρσgµν + gρνgµσ) − m2e gµν

= pµ1 pν2 + pν1 pµ2 −(

p1 · p2 + m2e

)gµν

. . . and similarly for Lµνf — start from Lµνe , replace 1→ 3, 2→ 4, and µ↔ ν

Lµνf = pµ3 pν4 + pν3 pµ4 −(

p3 · p4 + m2f

)gµν


The e+e−→ f f process : matrix element

Lµνe Lf ,µν =[pµ1 pν2 + pν1 pµ2 −

(p1 · p3 + m2

e

)gµν] [

p3,µp4,ν + p3,νp4,µ −(

p2 · p4 + m2f

)gµν]

= 2[

(p2 · p3)(p1 · p4) + (p2 · p4)(p1 · p3) + m2f (p1 · p2) + m2

e (p3 · p4) + 4m2em2

f

]|M|2 =

e4Q2f

q4Lµνe Lf ,µν

Now consider e+e− collisions in center-of-mass frame, with electron energy

E me (i.e. neglect me) and outgoing fermion at angle θ, with energy E

and momentum βE :

incoming e-

(E ; 0 , 0 , E)

incoming e+

(E ; 0 , 0 , -E)

outg

oing

f

(E ;

βE sin

θ ,

0 , β

E cos

θ)

outg

oing

f

(E ;

-βE sin

θ ,

0 , -

βE cos

θ)

p1 · p2 = 2E 2

p1 · p3 = E 2 (1− β cos θ)

p1 · p4 = E 2 (1 + β cos θ)

p2 · p3 = E 2 (1 + β cos θ)

p2 · p4 = E 2 (1− β cos θ)

p3 · p4 = E 2 (1 + β2)

q2 = (p1 + p2)2 = 4E 2

p1 ≡ ( E ; 0 , 0 , E )

p2 ≡ ( E ; 0 , 0 , − E )

p3 ≡ ( E ; βE sin θ , 0 , βE cos θ )

p4 ≡ ( E ; − βE sin θ , 0 , − βE cos θ )

|M|2 =e4Q2

f

16 E 42[E 4(1 + β cos θ)2 + E 4(1− β cos θ)2 + 2E 2m2

f

]=

e4Q2f

4

[1 + β2 cos2 θ + (1− β2)

]M. Fanti (Physics Dep., UniMi) title in footer 36 / 76

The e+e−→ f f process : cross-section

Recall Fermi’s golden rule for 2→ 2 processes:

dσ

d(cos θ)=

β

128π E 2F3F4|M|2

(where we use s = 4E 2 and where F3,4 are the degrees of freedom of outgoing f , f — here F3 = F4 = 2 due to spin

states)

(dσ

d(cos θ)

)e+e−→f f

=β

128π E 2

=4︷︸︸︷F3F4

e4Q2f

4

[1 + β2 cos2 θ + (1− β2)

]=

e4Q2f

128π E 2β[1 + β2 cos2 θ + (1− β2)

]In the relativistic limit, E mf , β → 1,(

dσ

d(cos θ)

)e+e−→f f

=e4Q2

f

128π E 2(1 + cos2 θ)

(note the energy dependency as ∝ 1

E 2∝ 1

sand the angular dependency ∝ 1 + cos2 θ).

Total cross-section (

∫d(cosθ)):

σe+e−→f f =e4Q2

f

48π E 2=

4π α2Q2f

3 s


Do we observe e+e−→ qq reactions?

Example:

σe+e−→µ+µ− =4π α2

3 s;

(dσ

dΩ

)e+e−→µ+µ−

=3

16πσe+e−→µ+µ− · (1 + cos2 θ)

If quarks exist, as point-like fermions with fractional charge Qu = +2

3, Qd = −1

3, we should observe similar reactions,

as:

σe+e−→dd =1

9

4π α2

3 s

σe+e−→uu =4

9

4π α2

3 s

(dσ

dΩ

)e+e−→qq

=3

16πσe+e−→qq · (1 + cos2 θ)

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


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√

s = 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 γ


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

material depth depends on the particle speed — actually, on βγ =p

m.

Bethe-Block formula works for 0.1 . βγ . 1000⟨−dE

dx

⟩= K

Q2

β2

[1

2ln

2mec2β2γ2Wmax

I2− β2 − δ(βγ)

2

][see pdg.lbl.gov “Passage of particles through matter”]

K = 4πNAr2emec

2Z

A

Wmax =2mec

2β2γ2

1 + 2γme/m + (me/m)2

I = mean excit. energy (eV)

Measuring

⟨−dE

dx

⟩allows a measurement of βγ ≡ p

m(especially at low βγ)

Measuring also p (with curvature in magnetic field) allows knowledge of m

⇒ particle identification

Due to

⟨−dE

dx

⟩∝ Q2, fractional charges (e.g. ±2

3, ± 1

3) would be identified

by very low energy loss.

⇒ ALAS, particles with Q = ±1

3,±2

3were never observed, sorry . . .


pdg.lbl.gov

Hadronic jets at e+e− collisions

[dijet event from JADE]

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× 3 tensor: Sαβ def=

∑i∈tracks

pαi pβi (α, β ∈ x , y , z)

⇒ 3 ortogonal eigenvectors n1, n2, n3 and 3 real eigenvalues λ1 ≤ λ2 ≤ λ3.Some properties:

∑k

λk = Tr [S ] =∑α

Sαα =∑i

|~pi |2

λ1 ≤1

3

∑k

λk ≡1

3

∑i

|~pi |2

λ3 ≥1

3

∑k

λk ≡1

3

∑i

|~pi |2

λk = nk · (Snk) =∑α,β

nαk Sαβnβk =

∑i

∑α,β

nαk pαi p

βi n

βk =

∑i

(~pi · nk)2

⇒ direction n3 maximizes sum of squared tracks’ projected momenta ⇒ choose it as jets’ axis

λ1, λ2 are sums of squared tracks’ transverse momenta wrt n3: λ1 + λ2 =∑i

(p⊥n3i

)2


Sphericity and jets’ evidence

Define sphericity : Sdef=

3

2

λ1 + λ2∑k λk

≡ 3

2

∑i

(p⊥n3i

)2∑i |~pi |2

Properties: 0 ≤ S ≤ 1; events with sharp jets have S −→ 0, while perfectly isotropic events have S −→ 1.

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 :√

s ≡ Ecm = 2Ebeam)

Sphericity decreases at higher collision energies

(a)√

s = 3 GeV

(b)√

s = 6.2 GeV

(c)√

s = 7.4 GeV

At low energy observation compatible with both

isotropic and jet models

At higher energy, isotropic model (dashed) predicts

increasing S , not compatible with data

Jet model (solid) predicts decreasing S , confirmed by

data


cos θ distribution of jets axis

Measuring the angle between jet axis (n3) and incoming

e+e− beams:

⇒ find the expected behaviourdσ

dΩ∝ 1 + cos2 θ

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

⇒ take e+e− → hadrons events as manifestation of e+e− → qq process


Rµ and the number of “colors”

Define Rµdef=σe+e−→hadrons

σe+e−→µ+µ−.

From σe+e−→f f =4π α2Q2

f

3 swe’d expect Rµ =

∑q

Q2q

√s qq expected Rµ observed Rµ

2ms uu, dd , ss2

3≈ 2

2mc uu, dd , ss, cc10

9???

2mb uu, dd , ss, cc , bb11

9

11

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

— and likewise anti-quarks may exist in 3 “anti-colors” anti-red, anti-green, anti-blue


Hadrons are colorless

A postulate is that color 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 r r , gg , bb, such to give

zero net color.

Baryons are qqq bound states. Each quark must have a different color state, like rgb, but in such a way that the

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√6

(rgb + gbr + brg − grb − rbg − bgr)

By the way, the introduction of “colors” was already proposed [W.Greenberg, 1964] to explain the “∆++ puzzle”. This resonance, withmass 1232 MeV, is a uuu 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). Howcan 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.

That seemed quite an artifact at the time, but the completely independent observation Robsµ = 3Rexp

µ , long later, revived it!


Quantum Chromodynamics

( Q C D )


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;

— collimated jets of hadrons are produced, whose angular distribution follows that expected for qq 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 ⇒ 8 gauge bosons, the “gluons”

gluons interact with quarks through gs qqg vertices, (gs being the coupling)

SU(3) is non-abelian ⇒ gluons self-interact through gs ggg and g 2s gggg vertices


SU(3) and QCD

Quarks are fermions carrying a “color charge” described by 3 degrees of freedom: red, green, blue: ψq ≡

ψR

ψG

ψB

Gauge invariance is expressed by SU(3) group: unitary 3× 3 complex matrices with det=1

⇒ 8 group parameters (1)

⇒ 8 gauge fields, the gluons

gq

q

quark-gluon interaction

gg

g

triple gluon coupling

g

g

g

g

quadruple gluon coupling

To our knowledge, gluons are massless (as gauge fields should be)

1 SU(n) has n2 − 1 free parameters. Proof: SU(n) matrices have n2 complex coefficients ⇒ 2n2 real parameters; unitarity imposes n2

constraints (n diagonal real constraints and n(n − 1)/2 off-diagonal complex constraints) plus the det=1 constraint (unitary matricesalways have |det|=1).


SU(3) group and its representations

Any element U(~θ) ≡ eigs2

∑8p=1 θpλp must be unitary with det(U(~θ)) = 1

⇒ each generator λp must be hermitian and traceless

What characterizes SU(3) is its set of commuting rules: [λa;λb] = 2i fabc λc

There are several “representations” of the abstract group, where group elements are represented by d × d matrices on

a d -dimensional space.

A representation is a closed vector space wrt all group operators, i.e. wrt all the generators. There is always an

operator commuting with all generators: λ2 def=∑p

λpλp — try yourself the proof that indeed [λ2;λa] = 0 ∀ a

⇒ A representation is a vector space where all vectors have the same eigenvalue of λ2

The smallest representation of SU(3) must be dimension-3, with 3× 3 matrices.

Vectors inside a representation are further classified according to their eigenvalues wrt some other λp operators that

commute with each other and with λ2 — so that they all can be simultaneously diagonalized. The obvious thing is to

seek diagonal matrices. Since commuting rules are the same for all representations, we consider 3× 3 matrices. With

the constraint to be traceless, we have 2 independent ones — traditionally named λ3 and λ8.

⇒ 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× 2 matrices — the Pauli matrices σ1, σ2, σ3. The representations are defined by eigenvalues

of J2 ≡ J21 + J2

2 + J23. Each representation is further mapped onto a line, by the eigenvalues of the only diagonal generator, J3.


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, and describe quarks.Generators are written as

λ1 =

0 1 01 0 00 0 0

λ2 =

0 −i 0i 0 00 0 0

λ3 =

1 0 00 −1 00 0 0

λ4 =

0 0 10 0 01 0 0

λ5 =

0 0 −i0 0 0i 0 0

λ6 =

0 0 00 0 10 1 0

λ7 =

0 0 00 0 −i0 i 0

λ8 =1√3

1 0 00 1 00 0 −2

[Gell-Mann matrices] (normalized such that Tr [λaλb] = 2δab)

r =

100

g =

010

b =

001

Most fabc vanish, apart for

f123 = 1

f147 = f246 = f257 = f345 = f516 = f637 =1

2

f458 = f678 =

√3

2

and their permutations, with appropriate sign

In the (λ3;λ8) plane, the eigenvectors of 3 are as in figure.

rg

b

λ3

λ8


The 3 representation — anti-quarks

In representation 3, vectors are linear combinations of r , g , b, and describe anti-quarks.Generators are obtained as λk = − λ∗k

λ1 =

0 −1 0−1 0 0

0 0 0

λ2 =

0 −i 0i 0 00 0 0

λ3 =

−1 0 00 1 00 0 0

λ4 =

0 0 −10 0 0−1 0 0

λ5 =

0 0 −i0 0 0i 0 0

λ6 =

0 0 00 0 −10 −1 0

λ7 =

0 0 00 0 −i0 i 0

λ8 =1√3

−1 0 00 −1 00 0 2

This way, [λa; λb] = 2i fabc λc and Tr

[λaλb

]= 2δab are preserved.

r =

100

g =

010

b =

001

In the (λ3;λ8) plane, the eigenvectors of 3 and 3 are as in figure.

Anti-quarks have opposite (λ3;λ8) eigenvalues (i.e. quantum numbers) than

corresponding quarks

These are two independent representations: r , g , b and r , g , b live in different

vector spaces, not connected by any SU(3) operator.

rg

b

r g

b

λ3

λ8


The 8 representations — gluons

Another representation is given by the gluons. They transform into each other with a group of matrices that is a

representation of SU(3). The dimension is obviously 8 — it’s called 8

Proof:For ~θ uniform in space-time, gluons transform as δGµ

a = −gsfabcθbGµc

First, note that these transforms are not complex — they can’t be, as gluon fields are real

Yet, let’s write it as δGµa = i

gs2θb (2ifabc)Gµ

c and compare with infinitesimal U = 1 + igs2θbλ

8b — the 8 stands for dim-8 representation.

We can identify fabc with matrix elements of λ8b: precisely [λ8

b]ac = −2ifbac . Do they fulfill SU(3) requirements?It’s easy to check that λb are hermitian : [(λ8

b)†]ca =([λ8

b]ac)∗

= (−2ifbac)∗ = (−2i)∗(fbac)∗ = 2i(−fbca) = [λ8b]ca

and traceless : Tr[λ8b] =

∑a[λ8

b]aa =∑

a 2ifaba = 0Now the commutators! We want to check whether [λ8

a;λ8b] = 2i fabc λ

8c

First, using the operator identity [A; [B ;C ]] + [B ; [C ;A]] + [C ; [A;B]] = 0 for generic SU(3) operators λa, λb, λc and using SU(3)commutation rules we find fadr frbc + facr frdb + fabr frcd = 0 — try it! It’s just algebra. . . Then:

[λa;λb]cd = [λa]cr [λb]rd − [λb]cr [λa]rd = − 4 (facr fbrd − fbcr fard) = − 4 (facr frdb + fadr frbc)

= 4 fabr frcd = 2i fabr (−2ifrcd) = 2i fabr [λ8r ]cd

Note: This is actually a general property of all gauge theories: gauge bosons transform into each other as a representation of the

symmetry group — the “adjoint representation”.

Gluons belong to a color representation ⇒ gluons carry color

⇒ that’s why they interact with each other (ggg and gggg vertices)

⇒ when a gluon interacts with a quark, it changes quark’s color

[Compare with QED: photons do not

carry electric charge]


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) — and of course anti-baryons (qqq)

⇒ can they be described in terms of SU(3) representations?

Group theory tells us that 3 ⊗ 3 = 8 ⊕ 1 — meaning that a qq system can

sit in a SU(3) octet, or in a SU(3) singlet.

The exagone shows 6 obvious color-anticolor combinations at the

vertices, r b, r g , bg , br , g r , gb, plus 3 combinations of r r ,

gg , bb in the centre. One of them belongs to color singlet:1√3

(r r + gg + bb

)— this is the color state of mesons.

r

g

br g

b

λ3

λ8

gr

b

b r

g

Likewise, 3⊗ 3⊗ 3 = 10⊕ 8⊕ 8⊕ 1 — a qqq 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 maths experts. . . ]

Baryons are in the singlet representation1√6



QCD effective potential at short distances

In QCD, fermions have a color state c ≡

r

g

b

together with normal helicity and particle/antiparticle (u, v) states,

quarks: uc

anti-quarks: vc

u, v have indexes in spinor space, c has indexes in color space;

γ-matrices and λ-matrices commute, because they act, respectively, on u, v components

and on c-components

γ

f2

f1

f4

f3

e

e

QED scattering

g

f2

f1

f4

f3

gS

gS

QCD scattering

MQED = [u3 (eγµ) u1] · gµνq2· [v2 (eγν) v4]

= e2 · [u3γµu1] gµν [v2γ

νv4]

q2

MQCD =

[u3c

†3

(gSλa2γµ)u1c1

]· gµνδab

q2·[v2c†2

(gSλb2γν)v4c4

]= g 2

S

1

4

(c†3λac1

)(c†2λac4

)︸︷︷︸

fQCD

·[u3γµu1] gµν [v2γ

νv4]

q2

effective potential for f f : VQED = −αem

r⇒ VQCD = −fQCD

αS

r— (αem =

e2

4πand αS =

g 2S

4π)

likewise effective potential for ff : VQED = +αem

r⇒ VQCD = +fQCD

αS

r


QCD effective potential at short distances

color factor fQCD :

(see Griffith for computation)

configuration qq singlet qq octet qq sextet qq triplet

(symmetric) (antisymmetric)

fQCD4

3−1

6

1

3−2

3

−fQCDαS

r+fQCD

αS

r

potential −4

3

αS

r

1

6

αS

r

1

3

αS

r−2

3

αS

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.


Running coupling in QED

Going to higher perturbative orders (i.e. adding loops) modifies the cross-section computation. All effects 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: here Q is NOT

a charge! Since q2 < 0 always, it’s convenient to introduce Q2 > 0.

The result is that the physical charge depends on Q2: e ≡ e(Q2).

[credits: Halzen-Martin]

α(Q2) =α(µ2)

1− α(µ2)3π ln

(Q2

µ2

)

This is the running of the coupling constant : once known the value α(µ2) at a given energy scale µ2, its value α(Q2)

at a given energy scale Q2 can be predicted.

The well-known value of the fine structure constant, α0 '1

137, corresponds to a scale µ2 ' m2

e (electron mass

squared).

α(Q2) grows with Q2, very slowly — eventually it diverges, whenα0

3πln

(Q2

m2e

)= 1 (“Landau pole”)

This happens at transferred energies√

Q2 = me e3π

2α0 ' 10277 GeV — we can sleep without fears!

(Planck’s scale MPl =

√~c5

G≈ 1019 GeV) [yet, it’s uncomfortable having a diverging probability. . . ]


Running coupling in QCD

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


αS(Q2) =αS(µ2)

1 + αS (µ2)12π (11nC − 2nF ) ln

(Q2

µ2

)nC = number of colors = 3

nF = number of flavours: u, d , c, s, t, b ⇒ = 6

(for “low” Q2 top-loop highly suppressed ⇒ nF = 5)

⇒ anyway, 11nC − 2nF > 0 ⇒ αS(Q2) decreases as Q2 increases, and αS → 0 as Q2 →∞

This is the so-called “asymptotic freedom” [Nobel Prize 2004 to Gross, Politzer, Wilczek]

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

On the opposite, at low energies αS increases, QCD becomes non-perturbative.

There is a scale ΛQCD at which αS diverges: ln(Λ2QCD) = ln(µ2)− 12π

(11nC − 2nF )αS(µ2)

Then αS(Q2) can be expressed as function of ΛQCD : αS(Q2) =12π

(11nC − 2nF ) ln

(Q2

Λ2QCD

)M. Fanti (Physics Dep., UniMi) title in footer 57 / 76

Running coupling in QCD — theory vs experiments

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

αS(Q2) =12π

(11nC − 2nF ) ln

(Q2

Λ2QCD

)

ΛQCD is not predicted by the theory: it must be deter-

mined experimentally, from measurements of αS vs Q2.

⇒ ΛQCD ' 0.2 GeV — i.e. ≈ masses of light hadrons!

⇒ clearly at energy scales ≈ hadron masses, αS blows up

⇒ cannot describe hadronic bound states perturbatively


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

The “color string” model

[D.Perkins “Introduction to High Energy Physics” Addison-Wesley]

Experimentally, several hadrons with a spin hierarchy (J = 0, 1, 2, . . . or J =1

2,

3

2,

5

2, . . . ) exhibit a simple relation

between spin and mass: J = k ′m2 + const, with k ′ ' 0.93 GeV−2 (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) assume that:

quarks’ masses are negligible

the qq pair is separated by a distance 2r0; it rotates to generate the angularmomentum; the rotating speed of the quarks is c (= 1) (!)

there is a color string connecting q and q, with energy density u =dU

dr= k ; it also

rotates with the quark pair; a piece of string between r and r + dr rotates with

v =r

r0c =

r

r0;

all the “mass” and the angular momentum come from the kinetic energy of the

rotating string; a string snippet has “mass” dm = dU = k dr , kinetic energy

dE =dm√

1− v 2/c2=

k dr√1− (r/r0)2

, and angular momentum

dL =dm√

1− v 2/c2vr =

k dr√1− (r/r0)2

r 2

r0

rr0

v

c

q

q


The “color string” model

Integration gives:

M = 2

∫ r0

0

dE = 2

∫ r0

0

k dr√1− (r/r0)2

= π k r0

J = 2

∫ r0

0

dL = 2

∫ r0

0

k dr√1− (r/r0)2

r 2

r0=

π

2k r 2

0

⇒ J =1

2π kM2 (eliminating r0 between 1st and 2nd eq)

Therefore the relation J ∝ M2 is found. Replacing k ′ ' 0.93 GeV−2 we get k ' 0.17 GeV2 ' 0.86 GeV fm−1. (for

the latter, recall that setting c = ~ = 1 we get GeV = 5.06 fm−1).

Just for fun: what is the newtonian force between a qq pair?

F =dU

dr= k ' 0.86

1.6 · 10−10 J

10−15 m' 14 000 N — about the weight of a medium car! Now, you see why this was called “strong force”?


The “color confinement” concept

Effective QCD qq potential:

Vqq = − 4

3

αS

r︸︷︷︸short-distance

+ kr︸︷︷︸long-distance

QCD does not prove color confinement.

However, there are strong hints for it from the theory:

(1) αS increases indefinitely at scales√

Q2 ≈ hadron masses

⇒ 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!)


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 — or also

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, suggesting that their size be

. 0.5 fm — recall k ≈ 1 GeV fm−1.


Experimental evidences of gluons


Gluon “observation”: 3 jets in e+e− collisions

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

e+

e−

q

q

e+e− → qq (NLO)

e+

e−

q

g

q

e+e− → qqg

e+

e−

q

g

q

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


Gluon “observation”: 3 jets in e+e− collisions

[3-jet event from JADE]

3 jets ⇒ 3 colored “objects” that hadronize

Since the initial state (e+e−) is electrically neutral and

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) in a fully antisym-

metric color configuration?

A: because of angular momentum conservation. . . initial state has

integer J , final state would be half-integer due to 3 fermions, each

carrying a ±12 contribution due to its spin.


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 groupparticles such to represent the underlyinghard process (quarks and gluons), withoutbeing 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.g meaning they have small invariant mass

Recall: m2jk = (Ej + Ek)2 − (~pj + ~pk)2 ' 2EjEk(1− cosαjk), being αjk the angle between ~pj , ~pk .

⇒ Define “distance” yjk =m2

jk

Q2=

2

Q2EjEk(1− cosαjk) ( Q2 ≡ (

∑j

Ej)2 − (

∑j

~pj)2 “total visible invariant mass” )

(0) start with a list of “objects” given by the “particles”, each having 4-momentum pj ≡ (Ej ; ~pj)

(1) compute yjk =2

Q2EjEk(1− cosαjk) for all object pairs (jk)

(2) choose pair with smallest yjk : if yjk < ycut group the two constituents to make one new object with 4-momentum pj + pk , removeobjects j , k from the list, then reiterate from (1);

(3) if smallest yjk is > ycut , stop iteration: each “object” left in the list is considered a jet


Kinematics of 3 jets in e+e− collisions

Continuous curves:QCD predictions at differentcenter-of-mass energies√s ≡ Ecm

Experimental points: :

√s = 12 GeV

• :√s ∈ [27.4, 31.6] GeV

× :√s ∈ [35.0, 36.6] GeV


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.

Observation consistent with emission of gauge boson,

with spin-1 and couplings as from QCD


Effect 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)


Effect of gluons in the ep DIS: scaling violation


Remember: Q2 def= −q2 = 2(p1 · p2)xy and 0 ≤ y ≤ 1

On fixed target, (p1 · p2) = EMp ⇒ Q2 < 2EMp x

Experiments probing ep scattering at higher energies E

— thus at higher Q2 —

showed a clear dependence F2 ≡ F2(x ,Q2).

QUESTION: why did former experiments observe

F2 ≡ F2(x) — independent of Q2 —

thus confirming Bjorken scaling ?

Former SLAC experiments probed x ∈ [∼ 0.03; 1] at energies E ∈[7; 17] GeV and small angles θ = 6, 10

From Q2 = 2EE ′(1− cos θ) and x =Q2

2Mp(E − E ′)

⇒ Q2 = 2E 2(1− cos θ)Mpx

E + Mpx' 4EMp sin2

(θ

2

)x

⇒ at SLAC Q2 . 0.3 GeV2

. . . That’s why the didn’t observe the scaling violation!


Probing at higher Q2: e±p colliders

Recall: Q2 = 2(p1 · p2)xy — x ∈ [0; 1] and y ∈ [0; 1]

At fixed target Q2 = 2EMpxy

In a e±p collider, where both e± and p are relativistic,

p1 ≡ (E1; 0, 0,E1) and p2 ≡ (E2; 0, 0,−E2)

⇒ (p1 · p2) = 2E1E2 ⇒ Q2 = 4E1E2 xy

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


PDFs — scaling

Evolution of F2 with Q2 reflects evolution of PDFs:

F2(x ,Q2) ⇒ fa(x ,Q2)

We need to consider the parton splitting functions, Pab(z),

⇒ probability that a parton a carrying fraction x of proton momentum comes

from a collinear emission from a parton b, carrying a fraction y > x of proton’s

momentum.

⇒ call z =x

ythe fraction of momentum of b carried away by a

zP (z)qqzP (z)gq

zP (z)qgzP (z)gg

Pqq(z) ' 4

3

[1 + z2

1− z

]Pgq(z) ' 4

3

[1 + (1− z)2

z

]Pqg(z) ' 1

2

[z2 + (1− z)2

]Pgg(z) ' 6

[1− z

z+ z(1− z) +

z

1− z

]

Then, the PDFs evolve according to the Dokshitzer-Gribov-Lipatov-Altarelli-Parisi equation (DGLAP):

d fa(x ,Q2)

d ln Q2' αS(Q2)

2π

∑b

∫ 1

x

dy

yPab

(x

y

)fb(

y ,Q2)


PDFs at higher Q2

With DGLAP equations, the PDFs at high Q2 can be

inferred from lower Q2 measurements

fixed target → HERA

HERA → LHC

Few words about LHCThe (x ,Q2) map @ LHC is more tricky. . .

proton-proton collisions

⇒ 2 PDFs, 2 x-variables, x1, x2

proton, E−−−−−−−−−−−−→parton, (x1E ;0,0,+x1E )

(collision)proton, E←−−−−−−−−−−−−

parton, (x2E ;0,0,−x2E )

proton-proton collision energy Ecm =√s = 2E

parton-parton collision energy Q2 = x1x2 · skinematics of collision: E = (x1 + x2)E , p‖ = (x1 − x2)E

rapidity of collision: Y =1

2ln

(E + p‖

E − p‖

)=

1

2ln

(x1

x2

)

x1 =

√Q2

se+Y ; x2 =

√Q2

se−Y 10

-710

-610

-510

-410

-310

-210

-110

010

0

101

102

103

104

105

106

107

108

109

fixed

targetHERA

x1,2

= (M/14 TeV) exp(±y)

Q = M

LHC parton kinematics

M = 10 GeV

M = 100 GeV

M = 1 TeV

M = 10 TeV

66y = 40 224

Q2

(G

eV

2)

x


PDFs — some features

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

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

uss,

cc,

2 = 10 GeV2Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

u

ss,

cc,

bb,

2 GeV4 = 102Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

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

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

Behaviour vs x :

At large x the valence quarks (u, d)

dominate, with fu ≈ 2fd ;

At low x gluons dominate by far;

also qq-pairs from the sea are possi-

ble, heavier quarks are less probable.

Behaviour vs Q2:

With larger Q2 sea (anti-)quarks be-

come probable also at higher x ; bb-

pair also appear despite their large

mass.


http://mstwpdf.hepforge.org/


x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

uss,

cc,

2 = 10 GeV2Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

u

ss,

cc,

bb,

2 GeV4 = 102Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2


The proton is much more complex than a uud system!

3 “valence quarks” uud

several low-x gluons

(emitted-reabsorbed by quarks)

several qq-pairs “from the sea”

(via g → qq → g processes)

. . . in continuous transformation

PDFs must satisfy some normalization rules:

Nu = 2+Nu ; Nd = 1+Nd ; Nq = Nq ∀q /∈ u, d

∫ 1

0

dx [fu(x)− fu(x)] = 2∫ 1

0

dx [fd(x)− fd(x)] = 1∫ 1

0

dx [fq(x)− fq(x)] = 0 ( q = s, c , b )

Also, momentum conservation implies that:∫ 1

0

dx

∑q=u,u,d ,d ,s,s,c ,c ,b,b

xfq(x) + xfg(x)

= 1

Do you remember that “valence quarks” uud carry only

∼ 12 of proton momentum?

The rest is carried by gluons



x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

uss,

cc,

2 = 10 GeV2Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

u

ss,

cc,

bb,

2 GeV4 = 102Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2


x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

uss,

cc,

2 = 10 GeV2Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

u

ss,

cc,

bb,

2 GeV4 = 102Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

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

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

uss,

cc,

2 = 10 GeV2Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

g/10

d

d

u

u

ss,

cc,

bb,

2 GeV4 = 102Q

x

-410

-310

-210

-110 1

)2

xf(

x,Q

0

0.2

0.4

0.6

0.8

1

1.2

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

Effect of perturbative order:

From top to bottom: the hard scattering (partonic cross-sections) are eval-

uated respectively at leading order (LO), next-to-leading order (NLO) and

next-to-next-to-leading order (NNLO). Consequently, the PDFs also evolve.

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


http://mstwpdf.hepforge.org/

Summary

Quarks belong to triplets, r , g , b — main proof being Rµ at e+e− colliders

Quarks dynamics is symmetric wrt color exchange ⇒ SU(3) global symmetry

Making SU(3) local ⇒ gauge interaction theory, QCD, with 8 mediator fields, the gluons

Color is not observable. Quarks are present only inside hadrons:

mesons are qq bound states in a color singlet:1√3

(r r + gg + bb

)baryons are qqq bound states in a color singlet:

1√6


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:

short-distance QCD effective potentials, V qqQCD , V qq

QCD , suggest that color-singlet aggregates are stable,

other combinations repel apart

running of αS show large coupling at smaller Q2 ⇒ larger distance

increase of QCD effective potential at large distance: VQCD = −fQCDαS

r+ kr

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


Documents

Experimental foundations of Quantum Chromodynamicsfanti/Particelle3/01-QCD.pdfExperimental foundations of Quantum Chromodynamics Marcello Fanti University of Milano and INFN M. Fanti