Particle Physics: Problem Sheet 5 Weak, Electroweak and LHC Physics

2010 — Subatomic: Particle Physics 1

Particle Physics: Problem Sheet 5 Weak, electroweak and LHC Physics

1. Draw a quark level Feynman diagram for the decay K+ → π+π0. This + is a weak decay. K has strange quark number, Ns = −1. The ﬁnal state has no strange quarks so Ns = 0. The number of strange quarks can only change when a W boson is exchanged.

2. Write down all possible decay modes of the W − boson into quarks and leptons. What is the strength of each of these vertices?

Decay Strength − − W → e ν¯e gW − − W → τ ν¯τ gW − − W → µ ν¯µ gW − W → ¯us gW Vus − W → ¯ud gW Vud − W → ¯ub gW Vub − W → ¯cd gW Vcd − W → ¯cs gW Vcs − W → ¯cb gW Vcb − W → ¯t d gW Vtd − W → ¯t s gW Vts − W → ¯t b gW Vtb

3. By drawing the lowest order Feynman diagrams, show that both charged (W ± exchange) and neutral current (Z0 exchange) contribute to neutrino − − electron scattering: νe + e → νe + e . 2010 — Subatomic: Particle Physics 2

Charged current means exchange of the W ± boson, neutral current means exchange of the Z0 boson.

Because we can draw both diagrams, whilst obeying all the conservation laws, it means that both are able to happen. When we observe an electron-neutrino scattering event we can never tell, on an event-by-event basis, whether the Z-boson or the W -boson was responsilbe. However by measuring the total cross section and comparing with predictions, we can show that both charged and neutral currents are involved.

4. Muons decay with a lifetime of 2.2 µs.

(a) What boson is responsible for the decay of the muon? It’s the W − boson. (b) What are the ﬁnal decay products? They have to be lighter than the muon. The only quark or charged lepton lighter than a muon is an electron. So we have µ− → e−X, where X are some other particles to make sure the quantum numbers are conserved. In particular − − Le and Lµ is conserved. Therefore we get: µ → e ν¯eνµ. (c) Draw a Feynman diagram illustrating this decay. (d) What is the matrix element for this decay? How does this relate to the Fermi constant: √ 2 g2 G = w ? F 2 8 mW 2010 — Subatomic: Particle Physics 3

2 gw M ∝ 2 2 mW − q

2 q can’t be any larger than mµ which is a lot smaller than mW . So the matrix

element, M is proportional to GF .

5. Draw the lowest order Feynman diagram for the decay of a tau into an − − − − electron and into a muon: τ → e ν¯eντ and τ → µ ν¯µντ , respectively. Calculate the matrix element for each of these decays. Compare this to the matrix element for muon decay, in Q4 above.

2 − − gw M(τ → e ν¯eντ ) ∝ 2 2 mW − q 2 − − gw M(τ → µ ν¯µντ ) ∝ 2 2 mW − q

Same as for muon decay above!

6. Recall that the rate of a process, such as the decay width, Γ, for a given process is given by Fermi’s Golden rule: 2π Γ = |M|2ρ h¯ where ρ is the phase space factor. For this question we will use dimen- sional analysis to understand ρ. The decay width Γ is measured in units of energy. Considering muon and tau decays, what is the (energy) dimensions of the Fermi constant, 5 GF ? Therefore, show that the dimension of ρ must be [E] . −2 2 −4 The dimension of GF is [E] , and therefore the dimension of |M| is [E] . There- fore to obtain a dimension of [E] for Γ, we need rho to have a dimension of [E]5 The only energy scale for muon or tau decay is the mass of the decaying lepton (mµ in Q4; mτ in Q5). Use this to calculate the ratio of following decays: − − Γ(µ → e ν¯eνµ) − − Γ(τ → e ν¯τ νµ) How does this relate to the lifetime of the tau lepton? (The lifetime of the muon is given in Q4, and the masses of the tau and muon are on the particle handout.)

− − 2 5 5 5 Γ(µ → e ν¯eνµ) GF mµ mµ 105.7 −7 − − = 2 5 = 5 = = 7.45 × 10 Γ(τ → e ν¯τ νµ) GF mτ mτ 1777 This is not equal to the ratio of the branching ratios, as there are more possible decay modes for the τ - but not for the muon. But it shows that the lifetime for the tau is at least 7.45 × 10−7 times smaller than for the muon - even though they decay by the same force! 2010 — Subatomic: Particle Physics 4

0 7. Estimate the total decay width, ΓZ , and the lifetime of the Z boson using:

• The measurement of the total hadronic width: Γ(Z → hadrons) = 1744.4 MeV • The measurement of the width into charged leptons: Γ(Z → `+`−) = 84.0 MeV. (` ≡ {e, µ, τ }). • The prediction for the width into one ﬂavour of neutrino: Γ(νν¯) = 167.0 MeV.

Compare with the observed resonance width, ΓZ = 2495.2 ± 2.3 MeV. The Z-boson can decay as:

Z0 → e+e− Z0 → µ+µ− Z0 → τ +τ − Z0 → u¯u Z0 → dd¯ Z0 → s¯s Z0 → c¯c Z0 → bb¯ 0 0 0 Z → νe ν¯e Z → νµ ν¯µ Z → ντ ν¯τ

0 It can’t decay as Z → t¯t, because mZ < mt/2. The hadronic width accounts for the decays into quarks. The total width is:

+ − + − + − ΓZ = Γ(Z → qq¯) + Γ(Z → e e ) + Γ(Z → µ µ ) + Γ(Z → τ τ ) + Nν Γ(Z → νν¯) = 1744.4 + 84.0 + 84.0 + 84.0 + 3 · 167.0 = 2497.4 MeV

where Nν = 3 is the number of neutrino ﬂavours. This is 2.2 MeV from the observed width, or just over one standard deviation! A beautiful validation of the Standard Model. Notice here that we cannot directly measure the neutrino cross section, as neutrinos don’t interact in the detector. Therefore this validate our assumption that there are only three kinds of neutrinos that the Z-boson can decay into. −22 The lifetime of the Z-boson is τZ =h/ ¯ ΓZ . Usingh ¯ = 6.58 × 10 MeV s gives −25 τZ = 2.63 × 10 s. What are the branching ratios of each of the decay modes? The branching ratio are the ratio of the individual widths to the total widths:

Γ(Z → X) BR(Z → X) = ΓZ • BR(Z0 → `+`−) = 3.36% • BR(Z0 → hadrons) = 69.9% • The branching ratio for all neutrino flavours is: BR(Z0 → νν¯) = 20.0%. This total is what we can measure, not into the ratio into particular neutrino flavours, as we don’t detect the neutrino directly, so we certainly can’t tell which flavour they were! 2010 — Subatomic: Particle Physics 5

8. The key predications of electroweak theory is that the couplings and the masses of the bosons are related as:

2 2 2 e = gW sin θW mZ = mW / cos θW

At high energies, the measured values of the electromagnetic coupling 2 constant, the “weak mixing angle” and the Z-boson mass are αEM(q = 2 2 2 2 2 mZ ) = 1/128, sin θW (q = mZ ) = 0.23120 and mZ = 91.188 GeV/c .

Calculate the weak coupling constant, αW , and predicted mass of the W -boson in the electroweak model, and compare this to the measured 2 mass of mW = 80.413 GeV/c .

2 2 gW e α 1 1 αW = = 2 = 2 = = 0.03379 ∼ 4π 4π sin θW sin θW 128 · 0.23120 29.6 We can see here the weak force is actually stronger than the electromagnetic force 2 (αW = 1/29.6 > α = 1/128) when the momentum transferred by the boson (q ) is around the mass of the Z-boson. At lower energies the weak force appears weak due to mass of the W and Z bosons - not because the strength of the weak interaction is intrinsically weak. The mass of the W boson is given by: √ p 2 2 mW = mZ cos θW = mZ 1 − sin θW = 91.188 × 1 − 0.23120 = 80.00 GeV/c

A 0.5% difference between the measured value. The mass of the W boson is measured to ±0.023 GeV/c2 so this is actually a significant difference. The resolution between these two values lies in the fact that the W -boson receives extra, small con- tributions due to its interactions with other particles, in particular its interaction with top quark, and to a lesser extent, with the Higgs boson. This is why we have constraints on the Higgs boson mass - the Higgs boson causes a small correction to many quantities. 9. Higgs at the LHC At LHC the Higgs boson, H can be produced in association with a W or Z boson, e.g. pp → W H, pp → ZH. (a) Draw the Feynman diagram for this process. (Select the correct one from page 10 of lecture 10.) What partons are involved in the hard scatter?

The partons involved in the hard scatter are a quark (from one proton) and an anti-quark (from the other proton). 2010 — Subatomic: Particle Physics 6

√ (b) What is the minimum eﬀective collision energy, sˆ needed to create 2 a real Higgs boson of mass, mH = 120 GeV/c in association with √a Z boson. What are typical values of Feynman-x for this value of sˆ? √ The minimum eﬀective collision energy is, sˆ = 211 GeV, the energy required to make a Higgs boson (120 GeV/c2) and a Z-boson (91 GeV/c2). √ sˆ √ 211 √ = x x = = 0.015 s 1 2 14000

If we assume the Feynman x of both partons is roughly the same, x1 ∼ x2, we get a typical value of x = 0.015. (c) Looking at the Higgs branching ratio plot on Lecture 10, page 7, what 2 is the most likely decays of a Higgs boson with mH = 120 GeV/c ? It’s H → bb,¯ Higgs decaying into a pair of b-quarks. (d) What are the possible final states for the process pp → ZH where 2 mH = 120 GeV/c ? What would this look like in the ATLAS detector? Let’s take the case that the Higgs decays into a b-quark and anti-b-quark, 2 which would be the most likely scenario for mH = 120 GeV/c . The Z-boson can decay into a pair of anything: either quark–anti-quark pair, a charged lepton–anti-lepton pair (e+e−, µ+µ−, τ +τ −) or a neutrino–anti-neutrino pair. So the possible final states are: (a) bbq¯q,where¯ q is any quark flavour, except top. (b) bb¯`+`−, where ` is e, µ, τ. (c) bb¯νν¯ In the detector quarks appear as jets. In lecture 5, slide 19 we saw the neutrinos leave a signature momentum imbalance or ”missing momentum”. We didn’t talk very much about tau leptons in the detector, but in lecture 7, slide 7, we saw the tau decays primarily into hadrons plus a neutrino, so actually the main signature of a tau decay is also a jet, plus a small amount of ”missing momentum” due to the neutrinos. But it is often possible to distinguish between a quark jet and a tau jet. So the signatures would be: (a) 4 jets (b) (i) 2 jets, 2 electrons (ii) 2 jets, 2 muons (iii) 2 quark jets, 2 tau jets, small missing momentum (c) 2 jets, large missing momentum.

10. Feynman diagrams In each of the following decays try to draw a quark level Feynman diagram and determine which interaction is responsible.

D?+ → D0π+ Σ0 → Λγ + 0 + − − D → K π τ → ρ ντ K∗0 → K+π− π0 → γγ 2010 — Subatomic: Particle Physics 7

Hints: The photon only interacts electromagnetically and the neutrino only weak. First establish the relevant quark and lepton flavour quantum numbers of the initial and final states. While quark and lepton flavours are conserved at each vertex for the strong and electromagnetic processes, the weak charged current is flavour changing. Always draw the simplest possible diagrams.

For the Σ0 decay the photon can come from any of the quarks. (Although it’s more likely to come from the up quark, as the coupling is proportional to the charge, and the up quark has the largest charge.) √ The π0 is (u¯u − dd)¯ / 2. For cases like this, just choose one of them to draw. Sometime both combination, u¯uand dd,¯ work, sometimes only one of them. The π0 must decay into 2 photons, not just one, in order to conserve four-momentum! Do your answers agree with the observed lifetimes? We can work out which force is responsible, by looking at the boson involved.

Decay Force Lifetime (s) D?+ → D0π+ strong 1.3 × 10−23 0 D+ → K π+ weak 1.04 × 10−11 K∗0 → K+π− strong 6.8 × 10−21 Σ0 → Λγ electromagnetic 7.4 × 10−20 − − −13 τ → ρ ντ weak 2.90 × 10 π0 → γγ electromagnetic 8.4 × 10−17

From lecture 2, typical lifetimes for decays are: 2010 — Subatomic: Particle Physics 8

• Strong: 10−20 − 10−23 s • Electromagnetic: 10−20 − 10−16 s • Weak: 10−13 − 10+3 s

So, yes, the lifetimes do agree!