Solutions Manual for Introduction to the Standard Model and Beyond Quantum Field Theory, Symmetries

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Intro to the Standard Model and Beyond: Quantum Field Theory, Symmetries, and Phenomenology Solutions to Exercises

Stuart Raby

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1

Poincare Invariance

Exercises 1.1 A particle with mass m decays into two particles with masses, m1 , m2 , and four momenta, p1 , p2 , respectively (see Fig. 1.1). Find the energy and momenta of particles 1 and 2 in the rest frame of the decaying particle.

Solution

m p2, m2

t

Fig. 1.1

p1, m1

Particle with mass m at rest decays into two particles with the mass and momenta as shown.

We have p⃗1 + p⃗2 = 0 and we define p⃗ ≡ p⃗1 = −p⃗2 . In addition, the four momenta satisfy (m, ⃗0) = P1 + P2 . We then have √ 1 p| λ(m2 , m21 , m22 ) = m|⃗ (1.1) 2 √ 2 2 2 1 (m + m1 − m2 ) E1 = m21 + p⃗2 = 2 m √ 2 2 2 1 (m − m 1 + m2 ) E2 = m22 + p⃗2 = .

2 m

1.2 (a) What is the minimum energy per proton for two beams of protons with equal and opposite momenta to produce a pair of top quarks (mt ∼ 175 GeV )? (see Fig. 1.2) (b) What is the minimum energy required in the same problem if a beam of protons hits a fixed hydrogen target?

Solution, part (a) √ The center of momentum energy, s = (E1 + E2 ) = 2E1 , since p⃗1 + p⃗2 = 0.

But we also have s = (Pt + Pt̄ )2 = (Et + Et̄ )2 , since p⃗t + p⃗t̄ = 0.. Thus the minimum value of s is given by smin = 4m2t . Or √ (1.2) Emin = smin /2 = mt ≈ 175 GeV. Solution, part (b) 1

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Poincare Invariance

2

Pt

P2

P1

t

Fig. 1.2

Pt̄ Energy momentum conservation gives us P1 + P2 = Pt + Pt̄ . ⃗ In this case, we have P1 = (E1lab , p⃗lab 1 ), P2 = (mp , 0). We still have s = (P1 + P2 )2 = (2m2p + 2mp E1lab ).

Thus E1lab =

(1.3)

(s−2m2p ) 2mp . We finally have

(E1lab )min =

(smin − 2m2p ) (2m2t − m2p ) 2mt = ≃ mt ≈ 350mt 2mp mp mp

(1.4)

(E1lab )min ≈ 350 × 175 GeV ∼ 65 T eV.

(1.5)

or

This energy is huge. This is the reason that colliders are so important !

1.3 Using the group property U (Λ)U (Λ′ )U −1 (Λ) = U (ΛΛ′ Λ−1 ) where U (Λ) = µν

i

e 2 αµν M̂ and Λµν ≃ gµν + αµν + O(α2 ), show that [ ] ( ) M̂ µν , M̂ ρσ = i M̂ µρ g νσ + M̂ νσ g µρ − M̂ νρ g µσ − M̂ µσ g νρ .

(1.6)

Solution We use the group relation U (Λ)U (Λ′ )U −1 (Λ) = U (ΛΛ′ Λ−1 )

(1.7)

where Λµν ∼ gµν + αµν + O(α2 ) and U (Λ) ∼ I + 2i αµν M̂ µν + · · · . We then have i i ′ i i (I + αµν M̂ µν )(I + ασρ M̂ σρ )(I − αµν M̂ µν ) ∼ (I + α̃σρ M̂ σρ ) (1.8) 2 2 2 2 where α̃σρ is given by 2 Λσ α Λ′α Λ−1 βρ ≡ gσρ + α̃σρ + O(α̃ ) β

(1.9)

′ β ∼ (gσ α + ασ α )(gα β + αα )(gβρ − αβρ ) β ′ α ′ ∼ gσρ + ασρ + ασ ααρ − ασ′ αβρ + O(α2 ).

Hence ′ ′ α̃σρ = ασρ + ασ α ααρ − ασ′ ααρ + O(α2 ). α

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(1.10)

Exercises

3

Consider the term O(αα′ , we have ] 1[ i α ′ ′ − αµν M̂ µν , ασρ M̂ σρ = (ασ α ααρ − ασ′ ααρ )M̂ σρ .

4 2

(1.11)

[ ] α ′ ′ αµν ασρ M̂ µν , M̂ σρ = −2i(ασ α ααρ − ασ′ ααρ )M̂ σρ (1.12) ( ) ′ = −iαµν ασρ M̂ µρ g νσ + M̂ νσ g µρ − M̂ νρ g µσ − M̂ µσ g νρ .

The last step was necessary to have the correct symmetry for the indices.

Hence [ ] ( ) M̂ µν , M̂ ρσ = i M̂ µρ g νσ + M̂ νσ g µρ − M̂ νρ g µσ − M̂ µσ g νρ .

(1.13)

1.4 B and B̄ = anti-B mesons are produced in the process e+ e− → B B̄ at a total

√ center of momentum energy s = 90 GeV (see Fig. 1.3). Given mB = mB̄ = 5277.6 M eV and τB = τB̄ = 11.8 × 10−13 s, (mB = B mass, τB = B lifetime) calculate the distance a B travels in its lifetime.

Solution

B

e+

t

Fig. 1.3

e−

B̄ Energy momentum conservation gives us Pe+ + Pe− = PB + PB̄ .

2 In the center of momentum system we have s = (PB + PB̄ )2 = 4EB . Hence EB = 45 GeV . In this boosted frame the B lifetime is

τB = τB (rest)

EB mB

(1.14)

and the distance the B travels, dB , is √ pB EB 2 pB τB = τB (rest) = ( ) − 1 τB (rest) d B = vB τ B = EB mB mB √ 2 − m2 . Finally, we find where in the last step we used pB = EB B dB ≃ .3 cm.

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(1.15)

(1.16)