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Home , Quark model

1 MODEL OF

1 Quark model of hadrons

1.1 Symmetries and evidence for

1 Hadrons as bound states of quarks – point-like - 2 objects, charged (‘coloured’) under the strong force.

Baryons as qqq combinations. as qq¯ combinations. 1 number = 3 (N(q) − N(¯q)) and its conservation.

You are not expected to know all of the names of the in the baryon and multiplets, but you are expected to know that such multiplets exist, and to be able to interpret them if presented with them. You should also know the quark contents of the simple light and mesons (and their anti-particles):

p = (uud) n = (udd) π0 = (mixture of uu¯ and dd¯) π+ = (ud¯) K0 = (ds¯) K+ = (us¯). and be able to work out others with some hints.

P 1 + Lowest-lying baryons as L = 0 and J = 2 states of qqq. P 3 + Excited versions have L = 0 and J = 2 . Pauli exclusion and non existence of uuu, ddd, sss states in lowest lying multiplet.

Lowest-lying mesons as qq¯0 states with L = 0 and J P = 0−. First excited levels (particles) with same quark content have L = 0 and J P = 1−. Ability to explain contents of these multiplets in terms of quarks.

J/Ψ as a of cc¯ Υ (upsilon) as a bound state of b¯b. Realization that these are hydrogenic-like states with suitable reduced mass (c.f. ), and subject to the strong force, so with energy levels paramaterized by αs rather than αEM

Top is very heavy and decays before it can form hadrons, so no top hadrons exist.

You should (for example) be able to ⇒ draw quark-flow diagrams for production and decay of resonances; ⇒ determine if a reaction can proceed via the strong force (no flavour changes, only strongly-interacting particles participate); ⇒ be able to apply Breit-Wigner formula to extract/interpret properties of new resonances (N.B. CM energy needed – first year relativity).

Isospin was touched on in the problem set but is non-examinable.

1 1.2 as bound states of quarks 2 THE

1.2 Nucleons as bound states of quarks p = uud, n = udd

You should be able to draw quark-flow and Feynman diagrams at the quark level for reactions involving and , e.g. β± decay, capture, solar reactions

1.3 Deep inelastic scattering

Scattering of a very high-momentum ( 1 GeV) probe from some target (usually p or n).

Probe has small wavelength and so scatters from what appear to be “temporarily free quarks” inside the hadron, rather than from hadron as a whole. Subsequently outgoing quarks cannot exist as free objects so generate ‘jets’ of hadrons along their direction of motion.

Probe may be or charged lepton (usually e±).

You should be able to draw Feynman diagrams at the quark level for charged- current (W ±) and neutral-current (Z0 and/or γ as appropriate). Vertex factors and relativistic propagators 1/ P · P − m2 for virtual particles

You should be able to ⇒ work out the energy-momentum of the ZMF, and of the virtual/intermediate ⇒ indicate the vertex couplings and calculate the propagator factor(s). (Note, need basic relativistic kinematics again here. Life is easier with 4-vectors.)

2 The Standard Model

2.1 Quark and lepton families

Members of the families (a.k.a. generations): quarks: (u, d), (c, s), (t, b); charged 1 : e, µ, τ and their associated . Spin= 2 . Point-like nature. Some idea of mass. Quantum numbers (, lepton number, , , bottomness, , colour charge [non-examinable: ]).

Fundamental interactions: ⇒ electromagnetic charges and couplings of each to (vertex factor of Qge where Q is the charge in units of charge); ⇒ strong charges and couplings of each quark to the (vertex factor of gs for quarks only) ;

2 2.2 , confinement 2 THE STANDARD MODEL

± 0 ⇒ vertices involving W and Z with vertex factors ≈ gw (W couplings involving quarks of different generations are in fact suppressed). Feynman diagrams involving each vertex.

Flavour mixing: flavour changes only occur in quarks via W ± boson (weak eigen- states 6= mass/energy eigenstates). Inter-family mixing (e.g. usW or bcW ver- tices) suppressed relative to mixing within families (e.g. udW or csW vertices). No flavour-changing neutral currents.

Flavour mixing: flavour changes only occur in leptons via neutrino oscillations (mass/energy eigenstates 6= flavour/weak eigenstates).

2.2 Strong interaction, confinement

Force transmitted by massless, coloured . Force is strong since αs = gs/(4π) ≈ 1. Gluons carry colour ⇒ gluon self-interactions.

As quarks pulled apart potential energy continues to increase. If quarks pulled far apart, the energy in the field is transformed into new particles. qq¯ pairs are pulled out of the vacuum, allowing the formation of multiple colour-neutral states (mesons and baryons). Result is jets of hadrons in direction of original quark and/or anti- quark momenta. (The sprays or jets are created because of the relativistic headlight effect). Net result: quarks can’t be found free in nature – they are confined within colour- neutral hadrons of size ∼ fm.

2.3

Mediated by W ± and Z0 . Couple to quarks and leptons with vertex factors ∼ gw. Universality of lepton couplings. For quarks W couplings suppressed for inter-generational W qq0 vertices.

Weak boson vertices violate : amplitudes not the same after the parity oper- ation P: x → −x. Helicity is projection of spin in momentum direction h = S·p/|p|. We showed in the examples that the helicity eigen-states change sign under a parity operation. W ± bosons couple preferentially to negative-helicity quarks and leptons, and to positive- helicity anti-quarks and anti-leptons. In the relativistic limit the W ± bosons couple only to those helicities.1 The reactions rates are very, very nearly the same after the combined operations of parity (x → −x) and charge conjugation (replacement of particles with their corresponding anti-particles). The combined operation is known as CP. 1Non-examinable: in the relativistic limit the helicity of a spin-half coincides with a property of a spin-half particle which is called its chirality or ‘handedness’.

3 2.4 Production and decay of the W and Z bosons2 THE STANDARD MODEL

2.4 Production and decay of the

Production in hadron and lepton collisions. (Relativistic) calculations involving energy, momentum, CM energy. Experimental identification of final states involving leptons, jets and missing energy & momentum from neutrinos.

Width of the Z0:

+ − + − + − Γ(Z) = Γ(qq,¯ q ∈ u, d, s, c, b) + Γe e + Γµ µ + Γµ µ + Nν × Γ(νν¯).

Can measure full width from width of Breit-Wigner. Can calculate the partial widths to the visible final states from their rates. With a calculation of Γ(νν¯) we can then 0 infer Nν , the number of light neutrinos that couple to the Z . Find Nν = 3 to high precision. Good evidence for only three generations of leptons (and so we presume quarks).

2.5 Neutrino oscillations

Individual lepton flavour [for any individual ` ∈ {e, µ , τ}, the lepton flavour number − + ± is L` = N(` ) − N(` ) + N(ν`) − N(¯ν`)] conserved at W vertices, however over long distances neutrinos found to change flavour. Experiments consistent with quantum oscillations. Calculation for two-state system (handout). Results in lepton flavour violation (but not violation of total lepton number: L = Le + Lµ + Lτ . Examples of solar and atmospheric neutrino oscillations. Implication is that neutrinos must have (small, sub-eV) mass.

4 3 EXAMPLE: B1 2006 QUESTION 6

3 Example: B1 2006 question 6

Link to 2006 B1 paper

The following list gives some combinations of quarks. State which combinations occur in nature, and in each such case give an example.

q, qq, qqq, qqqq, qq, qqq, qqqq, qqqqq, qqq, qqq

The D+ meson is the lowest-mass charmed meson with a mass of 1869 MeV/c2. The (D+)∗ meson has a mass of 2010 MeV/c2 and has spin-parity J P = 1−. Give the spin and parity of the D+ and explain how the D+ and (D+)∗ mesons fit into the quark model.

+ ∗0 + The D may decay into K µ νµ. What is the maximum possible momentum the could have in a decay in which the D+ has total energy of 1885 MeV? Enumerate the likely decay modes of the (D+)∗ meson and draw a Feynman diagram or quark flow diagram for each. ∗0 [The mass of K is 892 MeV/c2.]

Of the combinations given, the candidates that might exist in nature must be colourless. The quark–antiquark combinations that can be made colourless are:

quarks example quarks example hadron qqq uud proton qq¯ ud¯ π+ meson q¯q¯q¯ u¯d¯d¯ anti- qqqqq¯ – –

Only the baryons (qqq), mesons (qq¯) and the anti-baryons (q¯q¯q¯), are observed in nature. (The baryon–meson ‘’ qqqqq¯ has not been confirmed to exist.)

The D+ is the lowest lying charmed meson. It must therefore be a ground-state combination of a charmed quark with a first-generation quark. To get the positive charge the charm must be in the form of c rather than c¯ and the anti-quark must be a d¯; so it has quark content cd¯.

All the lowest-lying mesons for light quark content have J = S = L = 0 and negative parity, so we would expect the same to be true for the D+.

J P = 0−

The (D+)∗ is an excited state of the D+. Again, we know that the light-quark mesons tend to have first excited states with L = 0 and J = S = 1, and negative

5 3 EXAMPLE: B1 2006 QUESTION 6 parity, and so we expect that this is also true for the (D+)∗. So this excited state is also a bound state of a c and a d¯, but with S = 1 rather than S = 0.

We are asked for the maximum energy of the muon in in the decay

+ ∗0 + D → K¯ + µ + νµ.

In the ZMF the mass-energy of the D+ is shared out between the decay products. The maximum energy for the muon occurs when the neutrino has energy (and hence momentum) tending to zero in the ZMF2. This means that in the ZMF, as far as the momenta are concerned, we have an effective 2-body decay. Placing the x-axis along the direction of the K meson we then have the following (E, p) four-vectors (written in natural units with c = 1 for convenience):

PD = (MD, 0)

Pν = (0, 0)

PK = (EK , pK , 0, 0)

Pµ = (Eµ, −pK , 0, 0) , where we have used momentum conservation to set the x-momentum of the muon pµ = −pK .

Since all these particles are real (rather than virtual) we require P · P = m2 for both the K and the µ:

2 2 2 EK − pK = MK (1) 2 2 2 Eµ − pK = Mµ (2)

We can use energy conservation EK + Eµ = MD to eliminate EK from (1), and then eliminate pK using (2) to obtain:

2 2 max MD − MK Eµ = 2MD

Decay modes of (D+)∗?

Possible modes in order of typical rates are strong, electromagnetic, weak. Take each in turn.

Strong No flavour change allowed. Final state must include charmed hadron. Lightest is D+, so we’re interested in a reaction of the type:

∗ D+ → D+ + X

2This statement about the maximum of the three-body decay occurring at zero neutrino mo- mentum requires the knowledge from the example done in relativity course. There one treats the combination of the K and ν as a composite particle with P = PK + Pν . One then maximizes Eµ with respect to the invariant mass of this composite. This calculation is no longer on the syllabus for the subatomic part of the course. The result of this calculation is that we arrive at an effective two-body decay case. The kinematics of the two-body case you should all be able to do, since it only requires first-year level relativity, which you are expected to be able to use.

6 3 EXAMPLE: B1 2006 QUESTION 6

Where X is some other meson, which for no flavour change must have vacuum flavour numbers. This would be allowed by flavour conservation laws for e.g. X = π0 which has no net flavour. However let us look at energy conservation. The mass difference between the (D+)∗ and the D+ is only 2010 − 1885 = 125 MeV. This is less than the mass of the π0 so this reaction cannot proceed. No other strong interaction seems possible either – since the mass difference is less than the mass of any meson.

Electromagnetic Again no flavour changes, so we’re interested in reactions of the type: ∗ D+ → D+ + γ Unlike our attempt at a strong decay above, this is energetically allowed since the photon is massless. The reaction is allowed by all the conservation laws, so will proceed, and – in the absence of a competing strong interaction process – is likely to dominate. Example Feynman diagram is two straight lines with appropriate arrows for the c and d¯ (representing the (D+)∗ on the ‘in’ side and the D+ on the ‘out’ side) and an outgoing photon line attached to one of those quarks and to the ‘out’ side.

Weak via W propagator Flavour changes allowed at W vertices. Charm is a heavy quark so we expect it to decay via c → W +q0 vertices. Possibilities for q0 are s (same generation = unsuppressed) or d (different generation = suppressed). The second vertex of the virtual W + can connect to leptons or to hadrons. In the case + of leptons the final state will include ` + ν`. In the case of quarks the options are W + → ud¯ or W + → us¯ (inter-generation vertex).

Pulling together the final state quarks into mesons, we get final state possibilities that should include:

∗ D+ → K+ + K¯ 0 0 + → K¯ + ` + ν` (` = e, µ, τ) → K¯ 0 + π+ 0 + → π + ` + ν` (` = e, µ, τ)

[The large mass of the (D+)∗ meson makes multi-meson final states energetically allowable, and indeed the PDG suggests that such states are common.]

7 4 EXAMPLE: B1 2007 Q7

4 Example: B1 2007 Q7

Link to 2007 B1 paper

A neutrino of energy Eν is incident on a stationary neutron. Draw a Feynman diagram at the quark level for a charged-current interaction in which a charged lepton (τ) and a proton are produced. Indicate clearly the type of neutrino involved in this interaction. Compute the threshold energy Eν for this interaction. Above this threshold, the cross section for producing tau leptons in- creases slowly as a function of energy. One of the factors in the cross section is the phase-space factor ρ = d√N/dEτ given by Fermi’s Golden Rule. Show that ρ is proportional to Eτ just above threshold, where Eτ is the energy of the tau lepton. Discuss whether any of the other factors in the cross section has a strong energy dependence near the threshold. The fluxes of neutrinos with energies around 1 GeV are measured in an underground detector and compared with simulations of the production of neutrinos in the atmosphere. The measured flux of upward going muon-type neutrinos (i.e. the sum of νµ + νµ) is a factor of two lower than expected from the simulations, whereas the measured fluxes of downward going muon-type neutrinos and both downward and upward going electron-type neutrinos is the same as in the simulations. Give an interpretation of this difference in terms of neutrino properties.

We are told that a neutrino hits a neutron generating a proton and a tau lepton. We therefore have ν + n → p + τ By charge conservation the charge on the tau lepton must be negative. By conser- vation of lepton number we therefore know that the neutrino must be a ντ . The Feynman diagram has uud set of quarks on the ‘in’ side (to represent the pro- ton), along with a ντ . On the ‘out’ side there is a udd set of quarks (collectively a neutron). The intermediate W boson changes the u quark to a d quark at one vertex and the neutrino to the τ at the other.

dN The phase space factor in Fermi’s Golden Rule is where Ef is the final energy. dEf For two-body kinematics, if we know the momentum of one outgoing particle, then the other is fully specified by momentum conservation. Hence the total density of states is that of one of the outgoing particles,

d3p 4πp2dp dN = = (3) (2π)3 (2π)3

dN To obtain we shall need a relationship between dp and dEf . In general we dEf need to relativistic kinematics for questions involving creation and of particles. However just above threshold, the velocity of the outgoing tau lepton and

8 4 EXAMPLE: B1 2007 Q7 the proton must be small in the ZMF (since at threshold they would be at rest). Thus p2 p2 Eτ ≈ mτ + Ep ≈ mp + (4) 2mτ 2mp where p is the magnitude of the momentum of either particle in the ZMF. The term needed for FGR involves the total final state energy is Ef = Eτ + Ep, and is dN dN dp = dEf dp dEf

dN 4πp2 dp The term we can obtain from (3) giving 3 . The term we can get by dp (2π) dEf doing an implicit differential of the terms in (4) giving

dEf d p p = (Eτ + Ep) = + ∝ p dp dp mτ mp So looking at the dependence on p we find dN 1 ∝ p2 × ∝ p (5) dEf p √ From (4) we have that p = 2mτ Tτ where Tτ = Eτ − mτ is the kinetic energy of the tau lepton. Substituting this into (5) we obtain

dEf p ∝ T dp τ which was presumably the result that the examiners expected us to obtain. dEf [The result given in the question: dp ∝ Eτ appears to be incorrect).]

We are asked if any of the other factors in FGR will have strong energy dependence near the threshold. The other terms to consider are: the vertex factors (which have no energy dependence at all) and the propagator factor 1 1 2 = 2 2 2 P · P − mW E − p − mW Near threshold energies and momenta are of order GeV, so are are small compared −2 to mW . The propagator term will then be very close to mW , which also has no energy dependence. So we expect the density of states term to dominate the energy dependence of the cross section.

Measure flux of upward-going muon-type neutrinos νµ +ν ¯µ is factor of two lower than simulations. Downward-going neutrinos are the same as simulations. Why?

Important points in interpretation

• upward coming neutrinos travel further than downward coming neutrinos • over long distances neutrinos are observed to oscillate between flavours

2 2  • probability of oscillation ∝ sin const × (∆mν ) L/E

9 4 EXAMPLE: B1 2007 Q7

2 • observe oscillations ⇒ (∆mν ) 6= 0, so neutrinos must have mass • also has the consequence that neutrino flavour is not conserved

• [Not expected to remember: form of oscillation for atmospheric neutrinos is νµ ↔ ντ ]

As a of interest, let’s find out if we can see the ντ via the charged-current reaction we considered in the first part of the question. . . . What would be the ZMF energy of a collision of a Eν = pν ≈ GeV energy neutrino with a proton?

2 2 2 2 2 ECM = Etot − ptot = (Eν + mp) − (Eν ) 2 2 = 2Eν mn + mn ≈ 3 GeV

So ECM ≈ 1.7 GeV < (mp + mτ ). There’s insufficient energy available for a to undergo the charged current interaction.

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