Particle Physics - Solution Sheet 1

D1 Composite models of quarks and leptons have been proposed using “preons”, which are spin 1/2 fermions with three colours r, g, b. They have antipreons with equal and opposite quantum numbers. Consider these two simple models:

(a) The two “rishon” model with T (charge +1/3e) and V (charge 0) Harari, Phys.Lett. B86:83 (1979) (b) The three preon model with α, δ (both +1/3e) and β (-2/3e) Dugne et al., Europhys.Lett.57:188 (2002) hep-ph/0208135

For each model discuss how the quarks and leptons can be constructed from the preons.

(a) The lowest states of Hariri’s rishons give the ﬁrst generation of quarks and leptons:

e+ = T T T , e− = T¯T¯T¯ νe = VVV ,ν ¯e = V¯ V¯ V¯ u = TTV/TVT/VTT ,u ¯ = T¯T¯V/¯ T¯V¯ T/¯ V¯ T¯T¯ d¯= TVV/VTV/VVT , d = T¯V¯ V/¯ V¯ T¯V/¯ V¯ V¯ T¯

The second and third generations are imagined to be dynamical excitations of the ﬁrst generation. (b) The three preon model gives all three generations of quarks and leptons, provided that the two like-charge preons α and δ have diﬀerent masses:

+ − e = β¯β¯δ¯, e = ββδ, νe = αβδ,ν ¯e =α ¯β¯δ¯ u = αβ¯δ¯,u ¯ =αβδ ¯ , d = ββ¯δ¯, d¯= ββδ¯ + − µ = ααδ, µ =α ¯α¯δ¯, νµ = β¯α¯δ¯,ν ¯µ = βαδ c = αα¯β¯,c ¯ =ααβ ¯ , s = αα¯δ¯,s ¯ =ααδ ¯ + − τ = β¯α¯β¯, τ = βαβ, ντ = ααβ,ν ¯τ =α ¯α¯β¯ t = δβ¯δ¯, t¯ = δβδ¯ , b = βα¯β¯, ¯b = βαβ¯

There are some additional states that have not yet been seen:

A heavy charged lepton κ+ = δαδ Two more neutrinos νκ = δβδ, νλ = δαβ A heavy quark doublet h = δα¯β¯, g = δα¯δ¯

1 An exotic -4/3e quark X = βα¯δ¯!

The vector bosons could be preon-antipreon composites:

W + = αβ¯, W − =αβ ¯ , Z0 = αα¯

Again, there are additional Z states that have not yet been seen:

Z′ = ββ,¯ δδ,¯ αδ,¯ δα¯

What things might you hope to explain with a preon model?

- Fermion mass hierarchy - W,Z,g,γ might also be composite states - Electroweak prediction sin2 θ 1/4 W ≈ - Baryon and Lepton number violation, but B L conservation - Links to GUTS, SUSY, Supergravity, String Theory− - Structure of CKM matrix - Neutrino mixing (Dugne et al)

What problems are there with preon models?

- Dynamical mechanism between preons - Diﬀerent preon masses - Pointlike nature of quarks and leptons - Non-observation of proton decay - Flavour-changing neutral currents µ eγ, s dγ etc. - Colour states and quark conﬁnement→ → - Lack of other states, VV T¯ etc. (Harari) - Predictions of additional states (Dugne et al)

D2 Weak charged currents are described by γµ(1 γ5), whereas electro- magnetic currents are just γµ. −

(a) Explain why the weak charged current only couples to left-handed fermions or right-handed antifermions.

5 The operator PL = (1 γ ) is the left-handed helicity projection. This means that the− only spinor available for u is u2, i.e. a fermion must be left-handed in the weak charged current.

2 By operating from the left with P = γ0 we are doing the equivalent of charge conjugation, if CP is conserved. By matrix multiplica- tion it can be shown that:

γ0(1 γ5)=(1+ γ5)γ0 γ0γµ = γµγ0 − − Hence the parity-inverted weak charged current can be written as:

PJ µ = u¯(1 + γ5)γµγ0u − 5 The operator PR = (1+ γ ) was shown to be the right-handed helicity projection operator. Acting with the operator to the right on the ﬁnal state antifermion, the only spinor available is v1, and an antifermion must be right-handed in the weak charged current.

(b) Why does the weak neutral current have coefficients cV and cA? What is the significance of the values cV = cA = 1 for neutrinos? Why do other fermions have different values of cV and cA?

Due to electroweak symmetry breaking the Z does not have purely left-handed couplings. It acquires a small right-handed coupling from the mixing with the photon. The left-handed couplings are proportional to c + c , which gives a (1 γ5) term, and the V A − right-handed couplings are proportional to cV cA, which gives a (1 + γ5 term. −

Neutrinos are essentially massless, so they are purly left-handed (antineutrinoes are purely right-handed). Hence they only have left-handed Z couplings, so cV = cA for neutrinos.

The other fermion couplings are modiﬁed by an additional vector coupling proportional to the fermion charge which comes from the mixing of the photon. In detail:

c = g + g = 1/2 2Q sin2 θ c = g g = 1/2 V L R ± − W A L − R ± where the + sign refers to u-type and the sign to d-type fermions. −

3 Standard Problems * denotes a harder problem, which you are nevertheless encouraged to try!

S1 Draw the lowest order Feynman diagrams for these interactions:

electron muon scattering: • µ− µ− √α

√α ¡ e− e− electron positron annihilation • into a muon pair: + e+ µ

√α √α

γ ¡ e− µ− electron electron scattering: • e− e− √α

√α ¡ e− e− Plus diagram with interchange of the identical outgoing electrons. electron positron scattering: • e+ e− e+ e+ √α

√α √α γ γ

¡ √α ¡ e− e+ e− e−

4 S2 Describe qualitatively how the angular distribution of Bhabha scattering e+e− e+e− diﬀers from e+e− µ+µ−. → → The s-channel diagram describing e+e− µ+µ− gives an angular distribution (1 + cos2 θ), where θ is the CM→ scattering angle. This peaks slightly in the forward and background directions and is symmetric about θ = 90◦.

For Bhabha scattering there is an additional t-channel diagram. The matrix element squared just from this diagram has a 1/t2 dependence. Since t (1 cos θ) there is a pole at θ = 0 where the cross-section becomes∝ inﬁnite− for small angle scattering. Note that there is no equivalent pole in the backward direction at θ = 180◦. To get the complete distribution it is also necessary to include the interference term between the s and t-channel amplitudes. This has a 1/ts dependence which also gives a forward peak.

Bhabha scattering is characterised by a very large cross-section for forward scattering. This peak is absent in µ-pair production.

5 S3 The normalisation condition for Dirac spinors is u†u =2E. Show that the normalisation constant is N = √E + m.

pz px−ipy 1 0 E+m E+m  0   1   px+ipy   −pz  1 2 2 + 1 + u = pz u = px−ipy v = E m v = E m +      E m   E+m       +     1   0   px ipy   −pz       E+m   E+m   0   1                  The products of these spinors with their conjugates are:

p2 p2 + p2) u†u =1+ z + x y (E + m)2 (E + m)2

(E + m)2 + p2 u†u = (E + m)2 Using the relativistic equation E2 = p2 + m2 this simpliﬁes to:

(2E2 +2Em) 2E u†u = = (E + m)2 E + m

For this to be equal to the required normalisation condition 2E, each spinor must have an additional normalisation constant: 1 1 = N √E + m

S4 For the Dirac spinors u1 and u2 show that the lower components are smaller than the upper ones by a factor of β = v/c for a relativistic particle. What happens in the non-relativistic limit?

The momentum direction can be taken to be pz for simplicity. For a relativistic particle we have E m, and the momentum depen- dent part of each spinor reduces to≫ a single component proportional to β = p/E.

In the non-relativistic limit β 1 the lower components can be ig- ≪ nored. In the extreme relativistic limit β 1 the two components are the same size. →

6 S5 * The operators: 1 1 P = (1 γ5) P = (1 + γ5) L 2 − R 2 are said to project out the left-handed and right-handed helicity states of particles and antiparticles. 0 1 γ5 iγ0γ1γ2γ3 = ≡ 1 0 ! and the projection operators can be written as 4x4 matrices: 1 1 1 1 P = P = L 1− 1 R 1 1 − ! ! or written out in full: 1 0 1 0 1010 − 0 1 0 1 0101 P =   P =   L 10 1− 0 R 1010  −     0 1 0 1   0101       −    (a) In the ultrarelativistic limit β 1 show that the helicity projection operators have the following→ properties:

1 1 1 2 2 2 PLu =0 PLv = v PLu = u PLv =0

1 1 2 2 2 1 PRu = u PRv = v PRu =0 PRv =0

At high energies pz = E and the spinors become: 1 0 1 0 1 0 2 1 2 0 1 1 u =   u =   v =   v =  −  1 0 1 0          0   1   0   1             −      Multiplying each of these by the projection operators gives the required results. If we allow for β < 1: 1 β 0 1 −0 1 1+ β P u1 =   P u2 =   L 2 (1 β) L 2 0      − −     0   (1 + β)     −  For ultrarelativistic particles the helicity operators project out the spin states as left-handed (u2, v1) or right-handed (u1, v2). Note

7 that for massless fermions (neutrinos), the right-handed states are not physical, and only the left-handed neutrino states exist. Similarily only right-handed antineutrino states exist. For β < 1 the projection operators do not fully project onto the helicity states, since there is a small component of the opposite helicity, proportional to 1 β. − (b) Show that the operators satisfy the following relations:

2 2 PL = PL PR = PR PL + PR =1 PLPR =0

Starting from the matrix expressions for the projection operators, and performing some more 4x4 matrix multiplications and addi- tions, all the above results can be veriﬁed up to an overall normalisation factor 1/2 that we have ignored in most of this question.

The last result shows that the projection operators are orthogonal to each other. They project out two orthogonal components of the spinors. The ﬁrst two results say that applying the same projection operator twice is the same as applying it once. The third result says that the sum of the two projections must give back the original two-component spinor.

S6 The decay width for muon decay is described by:

G2 m5 Γ(µ− e−ν¯ ν )= F µ → e µ 192π3

Determine the value of GF from muon decay.

We need the muon lifetime and mass:

τ =2.197 10−6s m =0.1057GeV µ × µ We also need to insert anh ¯ =6.582 10−25GeV s into the decay width −2 × to get GF in GeV .

192π3h¯ G2 = F 2.197 10−6(0.1057)5 × G2 =1.352 10−10GeV −4 G =1.163 10−5GeV −2 F × F × which agrees with the correct answer 1.166 10−5 to within the round- ing errors of this calculation. ×

8 What is the dimensionless weak coupling constant g and how does it compare with e?

8 2 8 −5 2 g = GF M = 1.166 10 (80.4) =0.42 √2 × W √2 × × e = √4πα = 4π/137 = 0.30 In natural units, the dimensionlessq coupling constant g is (1) and similar in size to e! O

S7 Explain how the following measurements demonstrate lepton universality between weak couplings:

m = 105.65869 0.000009MeV τ =2.19703 0.00004µs µ ± µ ± m = 1776.99 0.26MeV τ = 290.6 1.0fs τ ± τ ± BF (µ eν¯ ν ) = 100% → e µ BF (τ eν¯ ν )=17.85 0.05% BF (τ µν¯ ν )=17.36 0.05% → e τ ± → µ τ ± If lepton universality is satisﬁed the decay widths are proportional to the ﬁfth power of the lepton mass:

Γ(τ e) m5 → = τ Γ(µ e) m5 → µ The branching fractions (BF) are related to the decay widths and life- times: BF (τ e) τ Γ(τ e) → = τ → BF (µ e) τ Γ(µ e) → µ → Putting these together:

BF (τ e) τ m5 → = τ τ BF (µ e) τ m5 → µ µ 290.6 10−15 (1.777)5 0.1784 = × × 2.197 10−6 (0.10566)5 × × which agrees to within the errors of the input measurements.

Note that the accuracy of this check is limited by the 0.3% accuracy in the knowledge of the τ lifetime.

9 The τµ BF is slightly lower because the mass of the muon cannot be neglected. When a correction is made to the ﬁnal state phase space it can be shown that the τ µ rate is also comsistent with lepton universality. →