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Home , Meson, Omega meson, Vector meson

The Parton Model

• • Pseudoscalar JP = 0– • Vector JP = 1– Mesons • P + • J = 3/2 P + • J = ½ Baryons • Resonances • Resonance Detecon • Discovery of the ω meson • Dalitz Plots Quark Model

Beginning of 60s: Regularies in the spectra indicated that they are formed by 3 (valence) : u, d, s and their anparcles u, d, s • Quarks are not observed as free parcles “symmetry driven” model in 1964; 10 more years were needed to develop a dynamical model

• Beginning of 70s: deep-inelasc scaering of off (Rutherford-like experiment): result: inside the there are light, quasi-free, point-like called partons (our quarks!)

• Their effecve is: m ~1/3 of the mass, with a momentum p: ~1/R (~1 fm)

• For heavy quarks the non-relavisc approximaon holds: m>>1/R

• Regularies in the hadron spectra were accounted for by the quark model ( mulplets) Classificaon of Hadrons Classificaon of Hadrons πN Scaering Quark Model d u s =0 s = 1 s Quarks Antiquarks

s = -1 s Q = 2/3 s =0 u d

Q = -1/3 Q = -2/3 Q = 1/3

• Parcles with same , and charge conjugaon symmetry described as mulplet - Different Iz and Y • Raising and lowering operators to navigate around the mulplet • Gell-Mann and Zweig: Paerns of mulplets explained if all hadrons were made of quarks

Q Q • Model originally developed using group theory alone Q - No need for physical quarks Q - Fact that quark charges non-integer suggested Q perhaps they were not real parcles Mesons

Mesons are built with quark/an-quark pairs. If we assume only u, d, s quarks (with their an-quarks) we could have families with 32=9 elements each (nonets). One has then:

J=1 spin triplet states (↑↑, 1/√2 (↑↓+↓↑), ↓↓) with J=1 and Jz=1, 0, -1 J=0 spin singlet states (1/√2 (↑↓-↓↑)) with J=0 S K*0 K*+ +1 e.g. the JP=1- meson nonet

- 0 + ρ ρ φ ρ I3 vector mesons: JPC=1-- -1 -1/2 ω +1/2 +1 (as the )

*- K K*0 -1 Mesons Pseudoscalar JP= 0– Mesons

Quark plus anquark with spins ½ + -½ = 0 (↑ ↓) (→ 9 combinaons)

⇥0 :(uu¯ dd¯)/⇥2 :(uu¯ + dd¯ 2ss¯)/⇥6 :(uu¯ + dd¯+ ss¯)/⇥3

1 Pseudoscalar JP= 0– Mesons Vector JP= 1– Mesons

Quark plus anquark with spins ½ + ½ = 1 (↑ ↑) (→ 9 combinaons)

0 ρ = 1/√2 (dd – uu) ω = 1/√2 (dd + uu) ϕ = ss Vector Mesons JP= 1– Vector Mesons Higher Masses

States shown so far have no between the qq or the qqq

Higher mass states are obtained by having orbital angular momentum between the qq or the qqq → J = 2, 3, . . .5/2, 7/2 . . .

Meson Masses Decays Vector meson decays into pairs: one more proof of the hadron quark composion and charge assignment The leptonic paral width Γ(e+e-) is proporonal to the square of the quark charges (Rutherford): 16πα 2Q2 Γ(V → l+l− ) = ψ(0) 2 Amplitude squared for the two quarks interacng with the photon in 2 one point of the space-me =1/volume of the meson MV 2 2 with Q = ΣaiQi (square of the mean quark charge, with ai amplitude coefficients)

The formula is derived taking into account a (1/q2)2 propagator term, the phase space € for 2-body decay (q2) and the coupling of the photon to the quarks in the meson: € q l + √α Q √α

1/q2 q l + 1 2 1 2 1 2 1 e.g. for the ρ : (uu − dd) one has: Q = [ ( − (− )] = 2 2 3 3 2 1 2 1 2 1 2 1 (uu + dd) Q = [ ( + (− )] = ω : 2 2 3 3 18 1 1 € € Q2 = (− )2 = φ : ss 3 9 € €

€ € Vector Meson Decays

Indeed, the measured leptonic widths for the various vector mesons are quite different:

Γe + e − (ρ) = 6.8 ± 0.3

Γe + e − (ω) = 0.6 ± 0.02

Γe + e − (φ) =1.37 ± 0.05 But their differences are completely understood within the quark meson charge assignment and composion:

€ Γe + e − (ρ) 2 =13.6 ± 0.6 Σai Qi

Γe + e − (ω) 2 =10.8 ± 0.4 Σai Qi

Γe + e − (φ) 2 =12.3± 0.5 Σai Qi

€ The Quark Parton Model Experimentally hadron states classified by mass, spin and parity and associated into families

P + Baryons with J = 3/2

Mass I = – 3/2 –1 – 1/2 0 +1/2 +1 +3/2 Strangeness 3 (MeV/c2) – 0 + ++ 1230 0 I = 3/2 Δ Δ Δ Δ

I = 1 Σ*– Σ*0 Σ*+ 1380 -1

I = ½ Ξ*– Ξ*0 1530 -2

I = 0 Ω– 1670 -3 P + J = 3/2 Baryons

These can all be explained by a basic set of 3 different spin ½ 'quarks' u, d, s combined in sets of 3 i.e. qqq with their spins aligned to give:

½ + ½ + ½ = 3/2 (↑ ↑ ↑)

2 with mu ≈ md and ms ≈ mu + 150 MeV/c

Quark B J I I3 S Q u ⅓ ½ħ ½ +½ 0 ⅔e d ⅓ ½ħ ½ -½ 0 -⅓e s ⅓ ½ħ 0 0 -1 -⅓e P + J = 3/2 Baryons

d → u

d → s

Not known at the time

These are strongly decaying resonances P + J = 3/2 Baryons Regularies in the hadron spectra: baryon JP = 3/2+ decuplet

I3 = -3/2 I3 = 0 I3 = +3/2 S Λ- Λ0 (ddu) Λ+ (duu) Λ++ (uuu) S = 0 (ddd) I = 3/2 Q = +2

Σ- (dds) Σ0 (dus) Σ+ (uus) I S = -1 3 I = 1 Q = +1

- S = -2 Ξ (dss) Ξ0 (uss) I = 1/2 Q = 0

S = -3 Ω- (sss) Q = - 1 I = 0

Q B + S Y = I + = I + | e | 3 2 3 2

€ P + J = 1/2 Baryons 3 quarks with spins ½ + -½ + ½ = ½ (↑ ↓ ↑)

0 Λ = 1/√2 (ud – du) s 0 Σ = 1/√2 (ud + du) s

Lightest baryons Decay weakly (except which is stable) P + J = 1/2 Baryons Why are there no uuu, ddd or sss here?

With u, d and s quarks there are 3 × 3 × 3 = 27 combinaons From symmetry arguments these can be grouped into 1 (Singlet) + 8 (Octet) + 8 (Octet) + 10 (Decuplet) The 10 are the JP = 3/2+ shown first and one of the 8 are these JP = 1/2 +

Baryons are fermions (spin 1/2, 3/2, . . .) so the overall wavefuncon must be ansymmetric (A)

The space × spin × flavour part must be symmetric (S)

With angular momentum ℓ = 0 the space part ~ (-1) ℓ = +1 so we want

spin × flavour to be symmetric Resonances There are over 100 known hadrons – most are 'Resonances' i.e. Excited States with the same quark content but higher internal angular momentum

p, n → N*, Δ K0, K± → K* Λ, Σ → Λ*, Σ* π0, π± → ρ, ω

Resonances are formed and decay via the Strong Interacon

Lifemes very short ~ 10-22 to 10-23 s By Uncertainty Principle:

Width Γ~ 66 MeV

At only travel 3 × 108 × 10-23 ~ 3 × 10-15 m → leave no tracks in detectors Resonance Detecon Detect their presence via stable (or longer lived) parcles into which they decay or are produced

+ ++ + π + p → Δ (1232) → π + p π – + p → Δ0(1232) → π – + p

Beam momentum Equivalent CM Energy 1.0 1.25 1.75 2.0 Discovery of the ω Meson

Note: ω (omega) meson not Ω– baryon

Take bubble chamber pictures of p annihilaon - look at events with exactly 5 produced

+ + – – 0 p + p → π + π + π + π + π Plot of all possible 3π combinaons:

+ + – π + π + π – – + π + π + π All have net → + + 0 π + π + π No structure seen – – 0 π + π + π The only neutral combinaon + – 0 π + π + π → Sharp peak at M = 790 MeV/c2, Γ = 12 MeV/c2 Discovery of the ω Meson

Peak Dalitz Plots

– + – In π + p → π + π + n there could be intermediate short lived states: long lived Delta Δ ≡ excited (Nπ) rho ρ ≡ excited (ππ)

(a) ρ0 (= ππ resonance) is (b) Δ (= Nπ resonance) is produced via π exchange produced via ρ exchange

How to tell if either (or both) happen? → Dalitz Plot Dick Dalitz Dalitz Plots

Measure many thousand 'events' (= collisions of this type) and plot result on 2-D histogram

Intermediate states (ρ0, Δ+) cause clustering of points at specific values of M² giving 'bands' on Dalitz Plot

Spread of M² (width of band) gives measure of width of resonance (somemes increased or masked by experimental resoluon) Dalitz Plots

Note: Stascal treatment – cannot idenfy specific events

So many hadrons discovered like this no longer considered 'fundamental' Discovery of the Ω-

Discovery of the Ω- (aer being postulated to complete the decuplet) 1964, Brookhaven Lab.

Ω- (sss) uud Combinaons

For uud Can have flavour symmetric (uud + udu + duu) and spin symmetric (↑↑↑ + ↑↑↑ + ↑↑↑) 3 Same quantum → J = /2 numbers so can form linear or combinaons flavour ansymmetric (uud) It's symmetric if it stays the same if and spin ansymmetric you swap any two flavours or spins (↑↑↓) → J = ½

+ So there is a uud state in both the J = 3/2 decuplet (Δ ) and the J = ½ octet (p) uuu Combinaons

For uuu (or ddd or sss) Can have only have flavour symmetric (uuu) and spin symmetric (↑↑↑)

→ J = 3/2

No flavour ansymmetric and spin ansymmetric (↑↑↓) possible

++ So there is only a uuu state in the J = 3/2 decuplet (Δ ) and not in the J = ½ octet Construcng Baryon States Baryon Masses