Intro to the Standard Model and Beyond: Quantum Field Theory, Symmetries, and Phenomenology Solutions to Exercises
Stuart Raby
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
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
(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.
(1.15)
(1.16)
Poincare Invariance
4
1.5 The Thomson cross-section for light scattering on electrons is given by σT = 8πα2 3m2e
where α ≃ 1/137 and me ≃ 12 M eV is the electron mass. Calculate σT and give your answer in cm2 . Solution Plugging in the given numbers we find σT in energy units, σT ≃ 1800 GeV −2 .
(1.17)
Using the approximate formula for ~c or 200 M eV ∼ 1/(10−13 cm) or GeV −1 ∼ .2 × 10−13 cm, we find σT ≃ 7 × 10−25 cm2 ≃ 700 mb where 1 mb = 10−27 cm2 .
(1.18)
2
Quantum Mechanics
Exercises 2.1 Two particle phase space -
√ (1) Given p = p′1 + p′2 with p2 = s, p = ( s, ⃗o), p′1 = (E1′ , p⃗1 ′ ) and define p′ dΩ p′ = |⃗ p ′ |, show that dLips(s; p′1 , p′2 ) = 16π 2 s1/2 where dΩ = d cos θdϕ and √ 2 2 λ(s, m1 , m2 ) p′ = . The three vector p⃗ ′ is defined in Fig. 2.1. 4s Solution We begin with the definitions dLips(s, p′1 , p′2 ) ≡ (2π)4 δ 4 (p − p′1 − p′2 ) and
d3 p⃗2 ′ d3 p⃗1 ′ ′ 3 (2π) 2E1 (2π)3 2E2′
√ δ 4 (p − p′1 − p′2 ) ≡ δ( s − E1′ − E2′ )δ 3 (⃗ p − p⃗1 ′ − p⃗2 ′ ).
(2.1)
(2.2)
In the CMS, we have p⃗ = 0 and p2′ + p⃗1′ ) = 1 d3 p⃗2′ δ(⃗
(2.3)
in the sense of performing the integral with an arbitrary function of p⃗2′ , i.e. ∫ d3 p⃗2′ δ(⃗ p2′ + p⃗1′ )f (⃗ p2′ ) = f (−⃗ p1′ ). (2.4) Thus we have (2π)−2 √ d3 p⃗ ′ (2.5) δ( s − E1′ − E2′ ) ′ ′ 4 E1 E2 √ √ p2′ and E1′ = (⃗ where we define p⃗ ′ ≡ p⃗1′ = −⃗ p ′ )2 + m21 , E2′ = (⃗ p ′ )2 + m22 . 2 Now d3 p⃗ ′ = p′ dp′ dΩ and let E ≡ E1′ + E2′ , then ( ′ ) p p′ p′ E ′ dE = + dp = dp′ . (2.6) E1′ E2′ E1′ E2′ dLips(s, p′1 , p′2 ) =
Hence dLips(s, p′1 , p′2 ) =
dE ′ 1 p′ (2π)−2 √ √ dΩ. δ( s − E) p dΩ = 4 E 16π 2 s
(2.7)
(2) Consider two body scattering 1 + 2 → 3 + 4 with p1 + p2 = p3 + p4 . In the CMS, let p1 = (E, p⃗), 5
p = |⃗ p| and p3 = (E ′ , p⃗ ′ ),
p′ = |⃗ p ′|
Quantum Mechanics
6
with p⃗ · p⃗ ′ = pp′ cos θ. Define the Lorentz invariants, s = (p1 + p2 )2 , t = ∑4 (p3 −p1 )2 , u = (p1 −p4 )2 with s+t+u ≡ i=1 m2i . Show that the differential ′
1 p cross-section is given by dσ = 4s p
T (s,Ω) 4π
2
dΩ = 4pπ2 s
T (s, t) 4π
2
dt. Where
Ω = {θ, ϕ} and the second equality assumes T is independent of ϕ.
z p~ ′
θ
y φ x
t
Fig. 2.1
Definition of the polar angles, θ, ϕ. Solution Given (Eqn. 5.25 in the book), we have 1 |T |2 dLips(s, p3 , p4 ). dσ = √ 2 λ(s, m21 , m22 )
(2.8)
Using the result of the previous problem and the identity (Eqn. 4.33 in the book) √ √ λ(s, m21 , m22 ) = 2 s p, we find dσ =
p′ T 2 | | dΩ. 4sp 4π
(2.9)
note that the transition amplitude T For the second part of the problem, ∫ dσ is a Lorentz scalar, since σ = dΩ dΩ is a Lorentz scalar. Thus T must be a function of Lorentz scalar quantities made up of the momenta in the problem or T = T (s, t). Here we have assumed that there are no other vectors in the problem, i.e. spin zero particles. In the CM system we have p⃗1 + p⃗2 = 0 = p⃗3 + p⃗4 with p = |⃗ p1 |, p′ = |⃗ p3 | and t = p23 + p21 − 2p3 · p1 = m23 + m21 − 2(E3 E1 − p⃗3 · p⃗1 ) = m21 + m23 − 2(E3 E1 − pp′ cos θ).
(2.10)
Exercises
7
√ √ In addition, we have E1 = p2 + m21 , E3 = p′ 2 + m23 . √ Note, E1 + E2 = s = E3 + E4 , thus given s, p and p′ are fixed. Hence dt = 2pp′ d cos θ.
(2.11)
Then we have p′ T 2 | | dΩ 4sp 4π π T (s, t) 2 = | dt | 4sp2 4π ∫ 2π where we have used dΩ = d cos θdϕ and 0 dϕ = 2π. dσ =
(2.12)
2.2 Three body phase space - Show that in the CMS system, where p⃗1 +⃗p2 +⃗p3 = 0,
dLips(s; p1 , p2 , p3 ) = 81 (2π)−5 dΩ′ dE1 dE2 where Ω′ is all the angles in Ω1 , Ω2 , except for θ12 defined by p⃗1 · p⃗2 = p1 p2 cos θ12 . Thus three body phase space is constant on the energy surface, E1 , E2 . Solution dLips(s; p1 , p2 , p3 ) (2.13) √ d3 p ⃗2 d3 p ⃗3 d3 p ⃗1 3 = (2π) δ( s − E1 − E2 − E3 )δ (⃗ p1 + p⃗2 + p⃗3 ) (2π3 )2E1 (2π3 )2E2 (2π3 )2E3 4
−5
= (2π)8
√ δ( s−E1 −E2 −E3 ) 3 d p⃗1 d3 p⃗2 E1 E2 E3
with p⃗3 = −(⃗ p1 + p⃗2 ). Using the following identities, d3 p⃗1 = p21 dp1 dΩ1 , d3 p⃗2 = p22 dp2 dΩ2 , √ √ E3 = (⃗ p1 + p⃗2 )2 + m23 = (p21 + p22 + 2p1 p2 cos θ12 + m23 , (2.14) E1 dE1 = p1 dp1 , E2 dE2 = p2 dp2 and finally dΩ1 dΩ2 ≡ dΩ′ d cos θ12
(2.15)
we obtain (2π)−5 √ p1 p2 δ( s − E1 − E2 − E3 ) d cos θ12 dE1 dE2 dΩ′ . 8 E3 (2.16) √ Note, however, that dE3 = pE1 p32 d cos θ12 . Therefore, δ( s−E1 −E2 −E3 ) pE1 p32 d cos θ12 = √ δ( s − E1 − E2 − E3 )dE3 = 1. Hence, we finally have dLips(s; p1 , p2 , p3 ) =
dLips(s; p1 , p2 , p3 ) =
(2π)−5 dE1 dE2 dΩ′ . 8
(2.17)
2.3 Now calculate two body phase space in the Laboratory frame described by the incoming 4 momenta - P = (m, 0), K = (E, ⃗k). Calculate dLips(s, K ′ , P ′ ) ′ with the outgoing 4 momenta given by- P ′ = (Ep′ , p⃗ ′ ) and K ′ = (E ′ , ⃗k ) and ′ in the Lab frame the scattering angle between ⃗k and ⃗k is given by θ such ′ that ⃗k · ⃗k = kk ′ cos(θ) (see Fig. 2.2).
Quantum Mechanics
8
~k′, m3 θ ~k, m1
t
Fig. 2.2
m2 ~p′, m4
Definition of the scattering angle, θ.
Solution The initial and final four momenta are given by Pi = (m2 + E, ⃗k)
(2.18)
Pf = (Ep′ + E ′ , p⃗ ′ + ⃗k ′ )
√ √ with Ep′ = (⃗ p ′ )2 + m24 , E ′ = (⃗k ′ )2 + m23 . The two body phase space in the laboratory frame is given by dLips(s, K ′ , P ′ ) = (2π)4 δ 4 (pi − pf )
d3⃗k ′ d3 p⃗ ′ (2π)3 2E ′ (2π)3 2Ep′
(2.19)
p ′ |. where s = Pi2 = m22 + m21 + 2m2 E and we define k = |⃗k|, k ′ = |⃗k ′ |, p′ = |⃗ We then have dLips(s, K ′ , P ′ ) = δ(m2 + E − Ep′ − E ′ ) Define the function f (k ′ , cos θ) ≡ m2 + E −
(k ′ )2 dk ′ dΩ . 16π 2 E ′ Ep′
√ √ (⃗ p ′ )2 + m24 − (⃗k ′ )2 + m23
(2.20)
(2.21)
with p⃗ ′ ≡ ⃗k − ⃗k ′ , (p′ )2 = k 2 + (k ′ )2 − 2⃗k · ⃗k ′ and ⃗k · ⃗k ′ = kk ′ cos θ. Thus (p′ )2 = k 2 + (k ′ )2 − 2kk ′ cos θ. We can now evaluate |
∂f k′ k ′ − k cos θ | = + . ∂k ′ Ep′ E′
(2.22)
Finally we obtain dLips(s, K ′ , P ′ ) = =
(k ′ )2 dΩ ∂f 16π 2 E ′ Ep′ | ∂k ′| (k ′ )2 dΩ 16π 2 [E ′ (k ′ − k cos θ) + Ep′ k ′ ]
with f (k ′ , cos θ) = 0 defining the boundary conditions on phase space.
(2.23)
3
Introduction to Field Theory
Exercises 3.1 Redefine the Hamiltonian -
) ( Ep † Ĥ = Ep a (⃗ p)a(⃗ p) + I (3.1) 2 ( ) ∫ 3 Ep where we’ve added an infinite c-number constant (2π)d 3p⃗2Ep I which we 2 shall see is interpreted as the zero-point vacuum energy and ∫
d3 p⃗ (2π)3 2Ep
I ≡ (2π)3 2Ep δ 3 (⃗0). Define
)1/2 ( ) Ep P (⃗ p) ≡ −i a(⃗ p) − a† (−⃗ p) 2 )1/2 ( ( ) 1 a(⃗ p) + a† (−⃗ p) . Q(⃗ p) ≡ 2Ep
(3.2)
(
Note P † (⃗ p) = P (−⃗ p), Q† (⃗ p) = Q(−⃗ p). Show that ( ) ∫ Ep2 d3 p⃗ 1 2 2 Ĥ = |P (⃗ p)| + |Q(⃗ p)| . (2π)3 2Ep 2 2
(3.3) (3.4)
(3.5)
Thus Ĥ is a ’sum’ over an infinite set of harmonic oscillators where P (⃗ p) is the canonical momentum and Q(⃗ p) is the coordinate. Solution Ep † (a (⃗ p) − a(−⃗ p))(a(⃗ p) − a† (−⃗ p)) (3.6) 2 Ep † [a (⃗ p)a(⃗ p) + a(−⃗ p)a† (−⃗ p) − a† (⃗ p)a† (−⃗ p) − a(−⃗ p)a(⃗ p)]. = 2
|P (⃗ p)|2 =
Ep † (a (⃗ p) + a(−⃗ p))(a(⃗ p) + a† (−⃗ p)) (3.7) 2 Ep † [a (⃗ p)a(⃗ p) + a(−⃗ p)a† (−⃗ p) + a† (⃗ p)a† (−⃗ p) + a(−⃗ p)a(⃗ p)]. = 2
Ep2 |Q(⃗ p)|2 =
9