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

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

Introduction to Field Theory

10

Hence Ep2 1 Ep † |P (⃗ p)|2 + |Q(⃗ p)|2 = [a (⃗ p)a(⃗ p) + a(−⃗ p)a† (−⃗ p)].

2 2 2 Finally, the integral ∫

3

d p⃗ (2π)3 2Ep

(

Ep2

1 |P (⃗ p)|2 + |Q(⃗ p)|2 2 2

)

∫ = ∫

(3.8)

(3.9) ) d p⃗ Ep ( † a (⃗ p)a(⃗ p) + a(−⃗ p)a† (−⃗ p) 3 (2π) 2Ep 2 3

) d3 p⃗ Ep ( † a (⃗ p)a(⃗ p) + a(⃗ p)a† (⃗ p) 3 (2π) 2Ep 2 ( ) ∫ Ep d3 p⃗ † † p), a (⃗ p)a(⃗ p) + p)] [a(⃗ = Ep a (⃗ (2π)3 2Ep 2 ( ) ∫ d3 p⃗ Ep † E a I = (⃗ p )a(⃗ p ) + (3.10) p (2π)3 2Ep 2 =

where [a(⃗ p), a† (⃗ p)] = (2π)3 2Ep δ 3 (0) ≡ I.

3.2 Given

∫ ϕ(x) ≡

(a) Show that 1 2

[ ip·x † ] d3 p⃗ e a (⃗ p) + e−ip·x a(⃗ p) .

3 (2π) 2Ep

∫

∫ d3 ⃗x ϕ̇(x)2 |t=0 =

d3 p⃗ 1 2 |P (⃗ p)| (2π)3 2Ep 2

0 where ϕ̇(x) ≡ ∂ϕ ∂t and x = t. (b) Show that [( ) ] ∫ ∫ 2 Ep2 1 d3 p⃗ 2 ⃗ d3 ⃗x ∇ϕ + m2 ϕ2 |t=0 = |Q(⃗ p)| .

3 2 (2π) 2Ep 2

Thus

(

∫ Ĥ =

d3 ⃗x

Solution (Part a)

[ ]) 1 1 ( ⃗ )2 ϕ̇(x)2 + ∇ϕ + m2 ϕ2 .

2 2

(3.11)

(3.12)

(3.13)

(3.14)

(3.15)

∫

d3 p⃗ iEp (a† (⃗ p)e−i⃗p·⃗x − a(⃗ p)ei⃗p·⃗x ) (2π)3 2Ep ∫ d3 p⃗ = iEp (a† (−⃗ p) − a(⃗ p))ei⃗p·⃗x (2π)3 2Ep ∫ d3 p⃗ √ = 2Ep P (⃗ p)ei⃗p·⃗x .

(2π)3 2Ep

ϕ̇(x)|t=0 =

(3.16)

(3.17)

Exercises

11

Thus 1 2

∫

1 d ⃗x ϕ̇(x) |t=0 = 2 ∫ × 3

2

∫

∫ 3

d ⃗x

d3 p⃗ √ P (⃗ p)ei⃗p·⃗x (2π)3 2Ep

′ d3 p⃗ ′ √ P (⃗ p ′ )ei⃗p ·⃗x 3 (2π) 2Ep′ ∫ 1 d3 p⃗ = |P (⃗ p)|2 2 (2π)3 2Ep

(3.18)

(3.19)

(Part b) ∫

( ) d3 p⃗ −i⃗ pa† (⃗ p)e−i⃗p·⃗x + i⃗ pa(⃗ p)ei⃗p·⃗x ) 3 (2π) 2Ep ∫ ( ) d3 p⃗ = i⃗ p a† (−⃗ p) + a(⃗ p) ei⃗p·⃗x 3 (2π) 2Ep ∫ √ d3 p⃗ = i⃗ p 2Ep Q(⃗ p)ei⃗p·⃗x .

3 (2π) 2Ep

⃗ x)|t=0 = ∇ϕ(⃗

We also have

∫ ϕ(⃗x)|t=0 =

d3 p⃗ √ 2Ep Q(⃗ p)ei⃗p·⃗x .

(2π)3 2Ep

(3.20)

(3.21)

Thus [( ) ] ∫ ∫ ∫ ∫ 2 1 d3 p⃗ d3 p⃗ ′ 1 ⃗ d3 ⃗x ∇ϕ + m2 ϕ2 |t=0 = d3 ⃗x (3.22) 2 2 (2π)3 2Ep (2π)3 2Ep′ √ √ ′ × (−⃗ pp⃗ ′ + m2 ) 2Ep Q(⃗ p) 2Ep′ Q(⃗ p ′ )ei(⃗p+⃗p )·⃗x ∫ 1 d3 p⃗ p)|2 . = E 2 |Q(⃗ (3.23) 2 (2π)3 2Ep p

3.3 Define Πϕ (⃗x, t) ≡ ϕ̇(⃗x, t) ≡ ϕ̇(x).

(3.24)

Show that (using the commutation relations of the creation and annihilation operators) [ϕ(⃗x, t), Πϕ (⃗y , t)] = iδ 3 (⃗x − ⃗y ).

(3.25)

Solution ∫ ϕ̇(x) =

d3 p⃗ iEp (a† (⃗ p)eip·x − a(⃗ p)e−ip·x ).

(2π)3 2Ep

(3.26)

Introduction to Field Theory

12

Then

] (3.27) ϕ(⃗x, t), ϕ̇(⃗y , t) [∫ ] [ ] ∫ 3 3 ′ ′ ′ d p ⃗ d p ⃗ ip·x † † = a (⃗ p) + e−ip·x a(⃗ p) , (2π) p ′ )eip ·y − a(⃗ p ′ )e−ip ·y ] 3 2E ′ iEp′ [a (⃗ (2π)3 2Ep e p ([ ] [ ]) ∫ ∫ d3 p⃗ ′ ′ ′ d3 p ⃗ −ip·x † ′ ip a(⃗ p), a (⃗ p )e ·y − eip·x a† (⃗ p), a(⃗ p ′ )e−ip ·y ] = (2π)3 2Ep (2π)3 2E ′ iEp′ e p ) ( ∫ d3 p⃗ i⃗p·(⃗x−⃗y) ∫ d3 p ⃗ = (2π)3 2Ep iEp ei⃗p·(⃗x−⃗y) + e−i⃗p·(⃗x−⃗y) = i (2π) = iδ 3 (⃗x − ⃗y ) (3.28) 3e with y = (⃗y , t).

[

4

Complex Scalar Field

Exercises 4.1 Given

∫ ϕ(x) =

[ ip·x † ] d3 p⃗ e b (⃗ p) + e−ip·x a(⃗ p) .

(2π)3 2Ep

Show (using the commutation relations [ ] a(⃗ p ′ ), a† (⃗ p) = (2π)3 2Ep δ 3 (⃗ p − p⃗ ′ ) [ ] b(⃗ p ′ ), b† (⃗ p) = (2π)3 2Ep δ 3 (⃗ p − p⃗ ′ ),

(4.1)

(4.2)

with all others zero) that [ϕ(x), Πϕ (y)] |x0 =y0 = iδ 3 (⃗x − ⃗y )

(4.3)

∂ϕ∗ δL = ϕ̇∗ (x) ≡ . ∂t δ ϕ̇(x)

(4.4)

where Πϕ (x) ≡

Solution First we have ∫ [ ] d3 p⃗ ′ ∗ −ip′ ·y ′ ip′ ·y † ′ ′) e ϕ̇(y) = (−iE b(⃗ p ) − e a (⃗ p ) . p (2π)3 2Ep′ Then [ϕ(x), Πϕ (y)] |

(4.5)

∫

x0 =y 0

∫ ([ ] d3 p⃗ ′ d3 p⃗ ip·x † −ip′ ·y ′ ′) (−iE e b (⃗ p ), e b(⃗ p ) = p (2π)3 2Ep (2π)3 2Ep′ [ ]) ′ − e−ip·x a(⃗ p), eip ·y a† (⃗ p ′) (4.6) ∫ ∫ ( [ ] ′ d3 p⃗ ′ d3 p⃗ (iEp′ ) eip·x−ip ·y b(⃗ p ′ ), b† (⃗ p) = 3 3 (2π) 2Ep (2π) 2Ep′ [ ]) −ip·x+ip′ ·y +e a(⃗ p), a† (⃗ p ′) ∫ ∫ ) ( d3 p⃗ d3 p⃗ ′ −i⃗ p·⃗ x+i⃗ p ′ ·⃗ y i⃗ p·⃗ x−i⃗ p ′ ·⃗ y ′) = (iE e + e p (2π)3 2Ep (2π)3 2Ep′ ×(2π)3 2Ep δ 3 (⃗ p − p⃗ ′ ) ∫ ) ( d3 p⃗ −i⃗ p·(⃗ x−⃗ y) i⃗ p·(⃗ x−⃗ y) = iδ 3 ((⃗x − ⃗y ). = (iE ) e + e p (2π)3 2Ep

13

Complex Scalar Field

14

4.2 Given

∫ d3 ⃗xj 0

(4.7)

← → j 0 ≡ iϕ∗ ∂ t ϕ,

(4.8)

Q̂ = where

show that

[

] Q̂, ϕ = −ϕ.

(4.9)

Solution ∫ Q= Thus

∫

∗

d3 ⃗xi(ϕ∗ Πϕ∗ − Πϕ ϕ).

d ⃗xi(ϕ ϕ̇ − ϕ˙∗ ϕ) ≡ 3

[ ] ∫ Q̂, ϕ = d3 ⃗xi [−Πϕ (⃗x), ϕ(⃗y )] ϕ(⃗x) ∫ = d3 ⃗xi(iδ 3 (⃗x − ⃗y ))ϕ(⃗x) = −ϕ(⃗y ).

(4.10)

(4.11)

4.3 Prove ϕ′ ≡ eiαQ̂ ϕ e−iαQ̂ = e−iα ϕ.

(4.12)

[ ] dϕ′ (α) ≡ ieiαQ̂ Q̂, ϕ e−iαQ̂ dα = −iϕ′ (α).

(4.13)

Solution

(4.14)

Therefore, upon integrating we have ϕ′ ≡ e−iα ϕ.

(4.15)

4.4 Show that the following equation, which implies that T −1 ≡ T †

T +T = T T + = 1

T ϕ(⃗x, t) T −1 = η ϕ† (⃗x, −t) †

|η| = 1

(4.16)

′

≡ η ϕ (⃗x, t ), is a consequence of the transformation law, T a(⃗ p) T −1 = η a† (−⃗ p) T b(⃗ p) T Solution

−1

∗

†

= η b (−⃗ p).

(4.17)

Exercises

15

∫ ϕ(x) =

[ ip·x † ] d3 p⃗ e b (⃗ p) + e−ip·x a(⃗ p) .

(2π)3 2Ep

(4.18)

T a(⃗ p) T −1 = η a† (−⃗ p)

(4.19)

Then using T b(⃗ p) T we find

−1

∗

†

= η b (−⃗ p)

∫

] [ ip·x d3 p⃗ e T b† (⃗ p) T −1 + e−ip·x T a(⃗ p) T −1 3 (2π) 2Ep ∫ [ ip·x ] d3 p⃗ =η e b(−⃗ p) + e−ip·x a† (−⃗ p) 3 (2π) 2Ep ∫ [ ] d3 p⃗ −ip·x′ ip·x′ † =η e b(⃗ p ) + e a (⃗ p ) (2π)3 2Ep

T ϕ(⃗x, t) T −1 =

= η ϕ† (⃗x, t′ ) where x′ = (⃗x, −t) and t′ = −t.

(4.20)

5

Spin 1/2 Particles

Exercises 5.1 Given the anti-commutation relations, {ψα (⃗x, t), ψβ† (⃗y , t)} = δ 3 (⃗x − ⃗y ) δαβ

(5.1)

{ψα (⃗x, t), ψβ (⃗y , t)} = 0, define

∫ QΓ ≡

d3 ⃗xψ † (⃗x, t)Γψ(⃗x, t)

(5.2)

where ψ † Γψ ≡ ψα† Γαβ ψβ and Γαβ may contain a differential operator acting on ⃗x. Show that [QΓ , QΓ′ ] = Q[Γ, Γ′ ] .

(5.3)

Solution We can use the identity [AB, CD] ≡ A {B, C} D − {A, C} BD − C {A, D} B + CA {D, B} .

(5.4)

We have [QΓ , QΓ′ ] ] ∫ = d3 ⃗xψ † (⃗x, t)Γψ(⃗x, t), d3 ⃗x ′ ψ † (⃗x ′ , t)Γ′ ψ(⃗x ′ , t) ( { } ∫ ∫ = d3 ⃗x d3 ⃗x ′ ψ † (⃗x, t) Γψ(⃗x, t), ψ † (⃗x ′ , t) Γ′ ψ(⃗x ′ , t) { } ) −ψ † (⃗x ′ , t) ψ † (⃗x, t), Γ′ ψ(⃗x ′ , t) Γψ(⃗x, t) ∫ = d3 ⃗x ψ † (⃗x, t) [Γ, Γ′ ] ψ(⃗x, t) = Q[Γ, Γ′ ] . [∫

5.2 Given

∫ H0 ≡

with

d3 ⃗x ψ † ΓH ψ

(5.5)

(5.6)

( ) ⃗ + βm . ΓH ≡ −i⃗ α·∇

(5.7)

[H0 , ψα (⃗x, t)] = −iψ̇α (⃗x, t)

(5.8)

Show that

16

Exercises

17

reproduces the Dirac equation.

Solution We can use the identity [AB, C] = A {B, C} − {A, C} B. We then have

(5.9)

∫

[ ] d3 ⃗x ′ ψ † (⃗x ′ , t)ΓH ψ(⃗x ′ , t), ψα (⃗x, t) ∫ { } = − d3 ⃗x ′ ψ † (⃗x ′ , t), ψα (⃗x, t) ΓH ψ(⃗x ′ , t) ( ) ⃗ + βm ψ(⃗x, t) = −iψ̇α (⃗x, t) = − −i⃗ α·∇

[H0 , ψα (⃗x, t)] =

where the last equality follows from ∫ ⃗ (−i⃗ α · ∇ + β m)ψ(⃗x, t) =

(5.10)

d3 p⃗ (⃗ α · p⃗ + β m)S(⃗ p, t)ei⃗p·⃗x (2π)3 2Ep

≡ iψ̇.

(5.11)

5.3 a) Given the Dirac Lagrangian with external sources coupled to fermions L = ψ̄(i∂/ − m)ψ + η̄ ψ + ψ̄ η

(5.12)

where η, η̄ are fermionic classical sources, we obtain the equations of motion (i∂/x − m)ψ(x) = −η(x).

Find the Green’s function which solves this equation, ∫ ψ(x) = − d4 ySF (x − y) η(y).

Solution (a) We start with (i∂/x − m)ψ(x) = −

(5.13)

(5.14)

∫ d4 y(i∂/x − m)SF (x − y) η(y).

Let’s define the Fourier transform of the Green’s function, we have ∫ d4 p −ip·(x−y) SF (x − y) ≡ e S̃(p).

(2π)4 Then ∫ ∫ ∫ − d4 y(i∂/x −m)SF (x−y) η(y) = − d4 y

(5.15)

(5.16)

d4 p −ip·(x−y) (p η(y). /−m)S̃(p)e (2π)4

We then choose p 1 /+m = 2 (p − m) p − m2 /

(5.17)

(i∂/x − m)ψ(x) = −η(x).

(5.18)

S̃(p) = and obtain

Spin 1/2 Particles

18

b) Given the free Dirac field, ∫ ∑ d3 p⃗ (b(⃗ p, s, t) u(⃗ p, s) e−ip·x + d† (⃗ p, s, t) v(⃗ p, s)eip·x ), ψα (⃗x, t) = 3 (2π) 2Ep s (5.19) show that ( ) i[SF (x − y)]αβ ≡ ⟨0|T ψα (x)ψ̄β (y) |0⟩ (5.20) ( ) ∫ 4 p d p −ip·(x−y) /+m =i e 4 2 (2π) p − m2 + iϵ αβ where ( ) T ψα (x)ψ̄β (y) ≡ θ(x0 − y 0 )ψα (x)ψ̄β (y) − θ(y 0 − x0 )ψ̄β (y)ψα (x).

(5.21)

Note, the minus sign is for fermions.

Solution (b) We have ( ) i[SF (x − y)]αβ ≡ ⟨0|T ψα (x)ψ̄β (y) |0⟩ ∫ ∑ d3 p⃗ = (uα (⃗ p, s)ūβ (⃗ p, s)) e−ip·(x−y) θ(x0 − y0 ) (2π)3 2Ep s ∫ ∑ d3 p⃗ (vα (⃗ p, s)v̄β (⃗ p, s)) eip·(x−y) θ(y0 − x (5.22) − 0) (2π)3 2Ep s where the minus sign is from the definition of the time-ordered product and we have used the identities ⟨0|b(⃗ p, s)b† (⃗ p ′ , s′ )|0⟩ = (2π)3 2Ep δ 3 (⃗ p − (⃗ p ′ )δss′ †

′

′

(5.23)

′

⟨0|d(⃗ p, s)d (⃗ p , s )|0⟩ = (2π) 2Ep δ (⃗ p − (⃗ p )δss′ .

3

3

Now using ∑

u(⃗ p, s)ū(⃗ p, s) = p /+m

(5.24)

v(⃗ p, s)v̄(⃗ p, s) = p / − m.

(5.25)

s

∑ s

we have

∫

{ } d3 p⃗ −ip·(x−y) ip·(x−y) (p θ(x0 − y0 ) + (−p θ(y0 − x0 ) / + m)αβ e / + m)αβ e 3 (2π) 2Ep ∫ { } d3 p⃗ −ip·(x−y) ip·(x−y) / e θ(x − y ) + e θ(y − x ) = (i∂ x + m)αβ 0 0 0 0 (2π)3 2Ep ∫ { } d3 p⃗ −ip·(x−y) ip·(x−y) e δ(x − y ) − e δ(x − y ) .

(5.26) −i(γ0 )αβ 0 0 0 0 (2π)3 2Ep

i[SF (x − y)]αβ =

Note, the term in the last line vanishes since we can change variables p⃗ → −⃗ p

19

Exercises

in the second term. Finally, we have

( ∫ ) d4 p e−ip·(x−y) i[SF (x − y)]αβ = (i∂/x + m)αβ i (2π)4 (p2 − m2 + iϵ) ( ) ∫ p d4 p −ip·(x−y) /+m =i e .

(2π)4 p2 − m2 + iϵ

(5.27)

6

QED

Exercises 6.1 Calculate the scattering amplitude for the Bhabha scattering process, (S − 1) = (2π)4 δ 4 (p1 + p2 − p3 − p4 ) × T

(6.1)

where the transition amplitude is given by T = [−v̄(⃗ p1 , s1 )γ µ u(⃗ p2 , s2 ) ū(⃗ p4 , s4 )γ ν v(⃗ p3 , s3 )] (6.2) 1 (ie)2 ]} { contraction (1) - s channel} × [−igµν (p1 + p2 )2 + iϵ {[+ū(⃗ p4 , s4 )γ ν u(⃗ p2 , s2 ) v̄(⃗ p1 , s1 )γ µ v(⃗ p3 , s3 )] (6.3) 1 × [−igµν (ie)2 ] { contraction (2) - t channel }, (p1 − p3 )2 + iϵ starting from the lSZ formalism (as in the example in Eqn. 12.36 in the book), ∫ (S − 1)p′ q′ ;pq ≡ i4

(6.4) → → − → d4 x1 . . . d4 x4 eiq ·x1 eip ·x2 ū(p ′ , s′ ) Kx1 Dx2 ′

←

′

←

− ×(ψ0 T(ϕ(x1 ) ψ(x2 ) ψ̄(x3 ) ϕ† (x4 )) ψ0 ) Kx4 Dx3 u(→ p , s) e−iq·x4 e−ip·x3 , then using the Gell-Mann-Low formula and Wick’s theorem to first derive the amplitude in position space. Then do all the space-time integrals to obtain the amplitude in momentum space.

Solution ∫ (S − 1)p′ q′ ;pq ≡ i4

→ → − − d4 x1 . . . d4 x4 eip4 ·x4 eip3 ·x3 (v̄(→ p1 , s1 ) Dx1 ) (ū(→ p4 , s4 ) Dx4 )

×(ψ0 T (ψ̄(x3 ) ψ(x1 ) ψ(x4 ) ψ̄(x2 ))ψ0 ) ← ← → → × (D u(− p , s )) (D v(− p , s ))e−ip1 ·x1 e−ip2 ·x2 . x2

20

2

2

x3

3

3

(6.5)

Exercises

21

∫ Now to second order in the perturbation (−i d4 zeψ̄(z)A(z)Qψ(z), / where Q = −1), i.e. using the Gell-Mann-Low theorem, we have ∫ (S − 1)p′ q′ ;pq ≡ i4

→ → → → d4 x1 . . . d4 x4 eip4 ·x4 eip3 ·x3 (v̄(− p1 , s1 ) Dx1 ) (ū(− p4 , s4 ) Dx4 )

⟨0|T (ψ̄(x3 ) ψ(x1 ) ψ(x4 ) ψ̄(x2 ))ψ̄(z1 ) ∫ 2 / / ⟨0|[1 + d4 z2 (ie) 2 ⟨0|ψ̄(z1 )A(z1 )ψ(z1 )ψ̄(z2 )A(z2 )ψ(z2 )|0⟩] [ ] ∫ ∫ 2 (ie) 4 4 / 1 )ψ(z1 )ψ̄(z2 )A(z / 2 )ψ(z2 ) |0⟩ × 1 + d z1 d z2 A(z 2 ← ← → → × (D u(− p , s )) (D v(− p , s ))e−ip1 ·x1 e−ip2 ·x2 .

×

∫

x2

d4 z1

2

2

x3

3

3

Note, all terms linear in the perturbation vanish due to Wick’s theorem with an odd number of photon fields. Also, as discussed in the text, the term in the denominator, when re-expanded in e2 , cancels the vacuum bubble term in the numerator. Finally, all disconnected diagrams vanish due to having two kinetic operators acting on one Green’s function. Therefore we are left with the following contribution to connected Green’s functions, ∫ (S − 1)p′ q′ ;pq = i4

∫ d4 x1 . . . d4 x4

d4 z1 d4 z2

(6.6)

→ → (ie)2 ip4 ·x4 ip3 ·x3 − − e e (v̄(→ p1 , s1 ) Dx1 ) (ū(→ p4 , s4 ) Dx4 ) 2 ( ) / 1 )ψ(z1 )ψ̄(z2 )A(z / 2 )ψ(z2 ) |0⟩ ×⟨0|T ψ̄(x3 ) ψ(x1 ) ψ(x4 ) ψ̄(x2 )ψ̄(z1 )A(z

×

← ← → → × (Dx2 u(− p2 , s2 )) (Dx3 v(− p3 , s3 ))e−ip1 ·x1 e−ip2 ·x2 .

Then given the two inequivalent contractions (Eqn. 13.33 in the book),

Aµ (z1 )Aν (z2 )×   

(6.7)

ψ̄(x2 )ψ(x1 )ψ(x4 )ψ̄(x3 )ψ̄(z1 )γ µ ψ(z1 )ψ̄(z2 )γ ν ψ(z2 )

(1)

+ } µ

ν

ψ̄(x2 )ψ(x1 )ψ(x4 )ψ̄(x3 )ψ̄(z1 )γ ψ(z1 )ψ̄(z2 )γ ψ(z2 ) (2)

,

QED

22

we have

∫

∫

(S − 1)p′ q′ ;pq = i4

d4 x1 . . . d4 x4

d4 z1 d4 z2 (ie)2

(6.8)

→ → → − × eip4 ·x4 eip3 ·x3 (v̄(− p1 , s1 ) Dx1 )α1 (ū(→ p4 , s4 ) Dx4 )α4 (−igµν ∆F (z1 − z2 ))

× {−[iSF (x4 − z2 )γ ν iSF (z2 − x3 )]α4 α3 [iSF (x1 − z1 )γ µ iSF (z1 − x2 )]α1 α2 + [iSF (x1 − z1 )γ µ iSF (z1 − x3 )]α1 α3 [iSF (x4 − z2 )γ ν iSF (z2 − x2 )]α4 α2 } ← ← → → × (D u(− p , s )) (D v(− p , s )) e−ip1 ·x1 e−ip2 ·x2 . x2

2

2

α2

x3

3

3

α3

Upon applying the kinetic operators we find ∫ ∫ 4 4 4 ′ ′ (S − 1)p q ;pq = i d x1 . . . d x4 d4 z1 d4 z2 (ie)2

(6.9)

→ → × eip4 ·x4 eip3 ·x3 (v̄(− p1 , s1 )α1 (ū(− p4 , s4 )α4 (−igµν ∆F (z1 − z2 )) { ν µ 4 × −[γ ]α4 α3 [γ ]α1 α2 δ (x1 − z1 )δ 4 (x4 − z2 )δ 4 (x2 − z1 )δ 4 (x3 − z2 ) } + [γ µ ]α1 α3 [γ ν ]α4 α2 δ 4 (x1 − z1 )δ 4 (x4 − z2 )δ 4 (x2 − z2 )δ 4 (x3 − z1 ) → → × (u(− p , s )) (v(− p , s )) e−ip1 ·x1 e−ip2 ·x2 .

2

2

α2

3

3

α3

∫ (S − 1)p′ q′ ;pq = i

4

d4 z1 d4 z2 (ie)2

(6.10)

→ − × (v̄(− p1 , s1 )α1 (ū(→ p4 , s4 )α4 (−igµν ∆F (z1 − z2 )) { ν µ × −[γ ]α4 α3 [γ ]α1 α2 eip4 ·z2 eip3 ·z2 e−ip1 ·z1 e−ip2 ·z1 } + [γ µ ]α1 α3 [γ ν ]α4 α2 eip4 ·z2 eip3 ·z1 e−ip1 ·z1 e−ip2 ·z2 → → × (u(− p , s )) (v(− p , s )) .

2

2

α2

3

3

α3

∫ (S − 1)p′ q′ ;pq = i

4

d4 z1 d4 z2 (ie)2

(6.11)

∫ d4 p eip·(z1 −z2 ) − → × (v̄(→ p1 , s1 )α1 (ū(− p4 , s4 )α4 (−igµν (2π)4 p2 + iϵ { ν µ ip4 ·z2 ip3 ·z2 −ip1 ·z1 −ip2 ·z1 × −[γ ]α4 α3 [γ ]α1 α2 e e e e } µ ν ip4 ·z2 ip3 ·z1 −ip1 ·z1 −ip2 ·z2 + [γ ]α1 α3 [γ ]α4 α2 e e e e → → × (u(− p , s )) (v(− p , s )) .

2

2

α2

3

3

α3

∫ (S − 1)p′ q′ ;pq = i4 (2π)4

d4 p (ie)2

(6.12)

−igµν → − × (v̄(− p1 , s1 )α1 (ū(→ p4 , s4 )α4 2 p + iϵ { × −[γ ν ]α4 α3 [γ µ ]α1 α2 δ 4 (p − p1 − p2 )δ 4 (p − p3 − p4 ) } + [γ µ ]α1 α3 [γ ν ]α4 α2 δ 4 (p − p1 + p3 )δ 4 (p + p2 − p4 ) → → × (u(− p , s )) (v(− p , s )) .

2

2

α2

3

3

α3

Exercises

23

(S − 1)p′ q′ ;pq = (2π)4 δ 4 (p1 + p2 − p3 − p4 ) (ie)2 → → × (v̄(− p1 , s1 )α1 (ū(− p4 , s4 )α4 { −igµν × −[γ ν ]α4 α3 [γ µ ]α1 α2 (p1 + p2 )2 + iϵ } −igµν µ ν + [γ ]α1 α3 [γ ]α4 α2 (p1 − p3 )2 + iϵ → − → − × (u(p2 , s2 ))α2 (v(p3 , s3 ))α3 .

(6.13)

(S − 1)p′ q′ ;pq = (2π)4 δ 4 (p1 + p2 − p3 − p4 ) (6.14) { −igµν (ie)2 → → → − × −(ū(− p4 , s4 )γ ν v(− p3 , s3 ))(v̄(− p1 , s1 )γ µ u(→ p2 , s2 )) (p1 + p2 )2 + iϵ } −igµν (ie)2 → → → → + (v̄(− p1 , s1 )γ µ v(− p3 , s3 ))(ū(− p4 , s4 )γ ν u(− p2 , s2 )) . (p1 − p3 )2 + iϵ

6.2 Calculate the differential cross-section for Bhabha scattering (in the limit Ei ≫ me ; i.e. neglect the electron mass) for unpolarized initial states and assuming that only the momentum of the final states are measured.

Solution ∑

|T |2 =

spins

(6.15) {[ ] 1 µ e ū(p4 , s4 )γ u(p2 , s2 )v̄(p1 , s1 )γµ v(p3 , s3 ) t s1 ,s2 ,s3 ,s4 [ ] } 1 − v̄(p1 , s1 )γ µ u(p2 , s2 )ū(p4 , s4 )γµ v(p3 , s3 ) (2) s ] {[ 1 (3) × ū(p2 , s2 )γ ν u(p4 , s4 )v̄(p3 , s3 )γν v(p1 , s1 ) t [ ] } 1 ν − ū(p2 , s2 )γ v(p1 , s1 )v̄(p3 , s3 )γν u(p4 , s4 ) (4) . s 4

∑

(1)

There are four terms given by (1) × (3) =

) ( ) e4 ( µ ν Tr (p /1 − me )γµ (p /3 − me )γν . /4 + me )γ (p /2 + me )γ Tr (p 2 t (6.16)

QED

24

In the relativistic limit, we can neglect me . We thus have ) ( ) e4 ( µ ν Tr (p (6.17) /4 )γ (p /2 )γ Tr (p /1 )γµ (p /3 )γν 2 t 16e4 = 2 (pµ4 pν2 + pν4 pµ2 − g µν p4 · p2 ) (p1µ p3ν + p1ν p3µ − gµν p1 · p3 ) t 16e4 = 2 (2p1 · p4 p2 · p3 + 2p3 · p4 p1 · p2 ) . t

(1) × (3) =

) ( ) e4 ( µ ν Tr ( p )γ ( p )γ Tr ( p )γ ( p )γ / / / / µ ν 1 2 4 3 s2 4 32e = 2 (p1 · p4 p2 · p3 + p2 · p4 p1 · p3 ) . s

(2) × (4) =

(6.18)

(1) × (4) = −

) e4 ( µ ν Tr (p /4 )γ (p /2 )γ (p /1 )γµ (p /3 )γν . st

(6.19)

(2) × (3) = −

) e4 ( µ ν Tr (p /1 )γ (p /2 )γ (p /4 )γµ (p /3 )γν . st

(6.20)

In order to evaluate the trace in the latter two terms, let’s consider the quantity ν µ ν µ ν γµp /2 γ p /1 γµ = −γ p / 2 γ γµ p /1 + 2γ p /2 γ p1µ

(6.21)

where we’ve used the identity {γµ , γν } = 2gµν .

(6.22)

We can use this identity over and over until we have γ µ γµ = 4 to the left. We finally obtain ν ν ν ν ν γµp /2 γ p /1 γµ = −2p /2 γ p /1 − 2γ p /2 p /1 + 2p /1 p /2 γ ≡ −2p /1 γ p /2 .

(6.23)

Using this identity we can now evaluate the term (1) × (4), we have (1) × (4) = −

( ) ) e4 ( ν Tr (p /4 ) −2p /1 γ p / 2 (p /3 )γν . st

(6.24)

Using similar techniques we find γν p /2 p /3 γν = 2(p /2 p /3 + p /3 p /2 ) ≡ 4p2 · p3 .

(6.25)

Hence (1) × (4) =

32e4 p4 · p1 p2 · p3 . st

(6.26)

(2) × (3) =

32e4 p1 · p4 p2 · p3 . st

(6.27)

Similarly

Exercises

25

Finally, putting it all together, we have { ∑ 1 |T |2 = 32e4 2 (p1 · p4 p2 · p3 + p3 · p4 p1 · p2 ) t

(6.28)

spins

1 + 2 (p1 · p4 p2 · p3 + p2 · p4 p1 · p3 ) s } 2 + p4 · p1 p2 · p3 . st We then use the relations, u = (p1 − p4 )2 ≈ −2p1 · p4 = −2p2 · p3

(6.29)

s = (p1 + p2 ) ≈ 2p1 · p2 = 2p3 · p4 2

t = (p1 − p3 )2 ≈ −2p1 · p3 = −2p2 · p4 .

We then obtain ∑ spins

{ |T | = 8e 2

4

] ] 2u2 1 [ 2 1 [ 2 2 2 u + s + u + t + t2 s2 st

} .

(6.30)

The differential cross-section is given by ∑ 1 1 dσ(e+ e− → e+ e− ) = √ dLips(s, p3 , p4 ) |T |2 2 λ(s, m2e , m2e ) 4 spins

(6.31)

where the factor of 14 comes from averaging over initial spins. ′

We have, in the CMS, dLips(s, p3 , p4 ) = 16πp2 s1/2 dΩ with p′ ≡ |⃗ p3 |, p ≡ |⃗ p1 |, √ √ p ≡ p′ and λ(s, m2e , m2e ) ≡ 2 sp. We thus obtain { 2 } e4 u + s2 u 2 + t2 2u2 dσ(e+ e− → e+ e− ) = + + dΩ (6.32) 32π 2 s t2 s2 st and the CMS we have E = E1 = E2 = E3 = E4 , s = 4E 2 , t ≈ −4E 2 sin2 θ2 , u ≈ −4E 2 cos2 θ2 . { } 4 θ 2 cos dσ(e+ e− → e+ e− ) α2 (1 + cos4 θ2 ) θ θ 2 = + cos4 + sin4 − dΩ 2s 2 2 sin4 θ2 sin2 θ2 } { α2 (1 + cos4 θ2 ) 1 + cos2 θ 2 cos4 θ2 + . (6.33) ≡ − 2s 2 sin4 θ2 sin2 θ2 Note, for unpolarized e+ e− → µ+ µ− scattering, only the s-channel process occurs and we have ( for energies above the muon mass) dσ(e+ e− → µ+ µ− ) α2 = (1 + cos2 θ). dΩ 4s

(6.34)

QED

26

6.3 Calculate the differential cross-section for an incoming electron with negative helicity and positron with positive helicity to to produce a µ− with negative + − + helicity and a µ+ with positive helicity, i.e (e− −1/2 e+1/2 → µ−1/2 µ+1/2 ). Com+ − + pare the result for a similar scattering process with (e− −1/2 e+1/2 → µ−1/2 µ−1/2 ).

Do not neglect the muon mass! Explain the results in terms of angular momentum and/or chirality conservation.

Solution In general we have the differential cross-section for the polarized process (e− (pe , se )e+ (pē , sē ) → µ− (pµ , sµ )µ+ (pµ̄ , sµ̄ )) is given by dσ(e− (pe , se )e+ (pē , sē ) → µ− (pµ , sµ )µ+ (pµ̄ , sµ̄ )) =

dσ(e− (−1/2)e+ (+1/2) → µ− (−1/2)µ+ (sµ̄ )) =

p′ |T (pe , se ; pē , sē ; pµ , sµ ; pµ̄ , sµ̄ )|2 dΩ.

64π 2 sp (6.35)

p′ |T (pe , −1/2; pē , +1/2; pµ , −1/2; pµ̄ , sµ̄ )|2 dΩ.

64π 2 sp (6.36)

e2 ū(⃗ pµ , −1/2)γ µ v(pµ̄ , sµ̄ ) v̄(⃗ pē , +1/2)γµ u(⃗ pe , −1/2). s (6.37) We’ll evaluate the amplitude in the CMS with p⃗e + p⃗ē = 0 and p̂e ≡ p̂. Then we have T (pe , −1/2; pē , +1/2; pµ , −1/2; pµ̄ , sµ̄ ) ∼

v̄(−p̂, sē ) =

( ) √ −2sē p ∗ E + me N ∗ χ−p̂ (−sē ) − χ∗p̂ (−sē ) .

E + me

(6.38)

Thus (6.39) ( ) ( ) 2sē p χp̂ (−1/2) ∗ ∗ ∗ v̄(−p̂, sē )γµ u(⃗ pe , −1/2) = −(E + me )N χ−p̂ (−sē ) χ (−sē ) γµ −p E + me −p̂ E+me χp̂ (−1/2) ( ) ( ) 2sē p χp̂ (−1/2) = −(E + me )N ∗ χ∗p̂ (sē ) χ∗p̂ (sē ) γµ . −p E + me E+me χp̂ (−1/2)

Now let me = 0, sē = 1/2, p̂ = ẑ such that θ = ϕ = 0 and p = E. We then

Exercises

27

have



 0  1   v̄(−pẑ, 1/2)γµ u(pẑ, −1/2) = −iEN ∗ (1 0 1 0)γµ  (6.40)  0  −1  0 µ = 0, 3  ( ) = −2iEN ∗ 0 µ = 1, 2  (1 0) σ i 1   0 µ = 0, 3 = −2iEN ∗ 1 µ=1 .  −i µ=2

Now consider the second half of the calculation, ū(⃗ pµ , −1/2)(γ 1 − iγ 2 )v(−⃗ pµ , sµ̄ ) (6.41) ) ) ( −2sµ̄ pµ )( ( 1 2 χ (s ) 0 σ − iσ p̂ µ̄ p µ µ E+mµ χ∗p̂µ (−1/2) = Nµ (E + mµ ) χ∗p̂µ (−1/2) E+m µ −(σ 1 − iσ 2 ) 0 χp̂µ (sµ̄ ) ( )( 2 ) (2sµ̄ )p = Nµ (E + mµ ) χ∗p̂µ (−1/2) (σ 1 − iσ 2 ) χp̂µ (sµ̄ ) 1 + (E+mµµ)2 where we used the identities σ 2 χ∗−p̂ (s) = N χ−p̂ (−s),

and

χ−p̂ (−s) = χp̂ (s).

(6.42)

Now using 1 + (2sµ̄ )

p2µ E − mµ = 1 + (2sµ̄ ) 2 (E + mµ ) E + mµ { 2E sµ̄ = +1/2 E+mµ = .

2mµ sµ̄ = −1/2 E+mµ

(6.43)

Thus we have T (pe , −1/2; pē , +1/2; pµ , −1/2; pµ̄ , sµ̄ ) (6.44) { ( ) 1 2 1 sµ̄ = +1/2 ) = −2iNe∗ Nµ e2 χ∗p̂µ (−1/2) (σ −iσ χp̂µ (sµ̄ ) 2 mµ /E sµ̄ = −1/2 { i(1 + cos θ)e−iϕ sµ̄ = +1/2 = −iNe∗ Nµ e2 . mµ −iϕ sµ̄ = −1/2 E sin θe Finally, we obtain in the case sµ̄ = +1/2: πα2 p′ dσ − (e (−1/2)e+ (+1/2) → µ− (−1/2)µ+ (+1/2)) = (1 + cos θ)2 . (6.45) dθ 2sp In Fig. 6.1 we show a graphical representation of the scattering process e− (−1/2)e+ (+1/2) → µ− (−1/2)µ+ (+1/2).

Let’s analyze the result.

QED

28

µ− θ e+

e−

t

Fig. 6.1

µ+

e− (−1/2)e+ (+1/2) → µ− (−1/2)µ+ (+1/2).

• Note, chirality is conserved at both vertices, thus the amplitude is nonvanishing for E ≫ mµ .

• The differential cross-section vanishes in the backward direction (θ = π) by angular momentum conservation.

And in the case sµ̄ = −1/2: πα2 p′ mµ 2 2 dσ − (e (−1/2)e+ (+1/2) → µ− (−1/2)µ+ (−1/2)) = ( ) sin θ. dθ 2sp E (6.46) In Fig. 6.2 we show a graphical representation of the scattering process e− (−1/2)e+ (+1/2) → µ− (−1/2)µ+ (+1/2).

Let’s analyze the result. µ− θ e+

e−

t

Fig. 6.2

µ+

e− (−1/2)e+ (+1/2) → µ− (−1/2)µ+ (+1/2).

• Note, chirality is not conserved at the muon vertex, thus the amplitude vanishes for E ≫ mµ .

• Angular momentum conservation is satisfied for θ = π2 .

7

Charge Conjugation and Parity

Exercises 7.1 Calculate the transformation of the following operators under C and P using CAµ (x)C† = −Aµ (x), †

†

Cψ(x)C = ηc γ2 ψ (x),

PAµ (x)P† = PAµ (x′ ) †

0

(7.1)

′

Pψ(x)P = ηp γ ψ(x )

where x′ = (−⃗x, t) ≡ Px. ψ̄Γψ

(7.2)

Γ = {I, γ5 , γµ γ5 , γµ , Σµν , γ5 Σµν },

(7.3)

with

and 2 Fµν = Fµν Fαβ g µα g νβ , F F̃ =

1 Fµν Fαβ ϵµναβ .

2

(7.4)

Solution Consider all operators as being normal ordered.

Cψ̄ψC† = ψγ2† γ0 γ2 ψ † = −ψγ0 ψ † = ψ † γ0T ψ = ψ̄ψ

γ2† = −γ2

(7.5)

Cψ̄γ5 ψC = ψγ2† γ0 γ5 γ2 ψ † = ψγ0 γ5 ψ † = −ψ † (γ0 γ5 )T ψ = ψ̄γ5 ψ (γ0 γ5 )T = −γ0 γ5 Cψ̄γµ γ5 ψC† = −ψγ2 γ0 γµ γ5 γ2 ψ † = −ψγ0 γµ∗ γ5 ψ † = ψ † (γ0 γµ∗ γ5 )T ψ = ψ̄γµ γ5 ψ(7.6) γ2 γµ γ2 ≡ γµ∗ (γ0 γµ∗ γ5 )T = γ5 㵆 γ0 Cψ̄γµ ψC† = −ψγ2 γ0 γµ γ2 ψ † = ψγ0 γµ∗ ψ † = −ψ † (γ0 γµ∗ )T ψ = −ψ̄γµ ψ (γ0 γµ∗ )T = γ0 γµ Cψ̄Σµν ψC† = −ψγ2 γ0 Σµν γ2 ψ † = ψγ0 Σ∗µν ψ † = −ψ † (γ0 Σ∗µν )T ψ = −ψ̄Σµν ψ (γ0 Σ∗µν )T = γ0 Σµν Cψ̄γ5 Σµν ψC† = −ψγ2 γ0 γ5 Σµν γ2 ψ † = ψγ0 γ5 Σ∗µν ψ † = −ψ † (γ0 γ5 Σ∗µν )T ψ = ψ̄γ5 Σµν ψ (γ0 γ5 Σ∗µν )T = −γ0 γ5 Σµν . †

29

Charge Conjugation and Parity

30

Pψ̄(x)ψ(x)P† = ψ(x′ )† γ0 γ0 γ0 ψ(x′ ) = ψ(x′ )† γ0 ψ(x′ ) = ψ̄(x′ )ψ(x′ ) †

′ †

′

′

(7.7)

′

Pψ̄(x)γ5 ψ(x)P = ψ(x ) γ0 γ0 γ5 γ0 ψ(x ) = −ψ̄(x )γ5 ψ(x ) Pψ̄(x)γµ γ5 ψ(x)P† = ψ(x′ )† γµ γ5 γ0 ψ(x′ ) = −ψ̄(x′ )㵆 γ5 ψ(x′ ) = −P ψ̄(x′ )γµ γ5 ψ(x′ ) 㵆 ≡ P γµ = (γ0 , −γi ) Pψ̄(x)γµ ψ(x)P† = ψ(x′ )† γµ γ0 ψ(x′ ) = ψ̄(x′ )㵆 ψ(x′ ) = P ψ̄(x′ )γµ ψ(x′ ) Pψ̄(x)Σµν ψ(x)P† = ψ(x′ )† Σµν γ0 ψ(x′ ) = ψ̄(x′ )Σ†µν ψ(x′ ) ≡ P ψ̄(x′ )Σµν ψ(x′ ) P Σµν = (Σ00 , −Σ0i , Σij ) †

′ †

′

Pψ̄(x)γ5 Σµν ψ(x)P = ψ(x ) γ5 Σµν γ0 ψ(x ) = −ψ̄(x′ )γ5 Σ†µν ψ(x′ ) ≡ −P ψ̄(x′ )γ5 Σµν ψ(x′ ).

CFµν (x)C† = −Fµν (x), †

Fµν (x) = ∂µ Aν (x) − ∂ν Aµ (x)

′

(7.8)

∂µ = P ∂µ′ .

PFµν (x)P = P Fµν (x ), Thus

CFµν (x)F µν (x)C† = Fµν (x)F µν (x) µνρσ

CFµν (x)Fρσ (x)ϵ

†

C = Fµν (x)Fρσ (x)ϵ

(7.9) µνρσ

PFµν (x)F µν (x)P† = Fµν (x′ )F µν (x′ ) PFµν (x)Fρσ (x)ϵ

µνρσ

†

′

′

(7.10)

P = −Fµν (x )Fρσ (x )ϵ

µνρσ

.

7.2 The magnetic dipole interaction is given by Fµν ψ̄Σµν ψ.

(7.11)

Show that this interaction term is invariant under C and P. Determine whether the electric dipole interaction given by Fµν ψ̄γ5 Σµν ψ

(7.12)

is invariant under C, P or CP. Solution CFµν (x)ψ̄(x)Σµν ψ(x)C† = CFµν (x)CC† ψ̄(x)Σµν ψ(x)C† = Fµν (x)ψ̄(x)Σµν ψ(x) PFµν (x)ψ̄(x)Σµν ψ(x)P† = PFµν (x)PP† ψ̄(x)Σµν ψ(x)P† = Fµν (x′ )ψ̄(x′ )Σµν ψ(x′ ) (7.13) †

†

†

CFµν (x)ψ̄(x)γ5 Σ ψ(x)C = CFµν (x)CC ψ̄(x)γ5 Σ ψ(x)C = −Fµν (x)ψ̄(x)γ5 Σ ψ(x) µν

µν

µν

PFµν (x)ψ̄(x)γ5 Σµν ψ(x)P† = PFµν (x)PP† ψ̄(x)γ5 Σµν ψ(x)P† = −Fµν (x′ )ψ̄(x′ )Σµν ψ(x′ ).

Hence Fµν (x)ψ̄(x)Σµν ψ(x)

(7.14)

Exercises

31

is invariant under C and P, while Fµν (x)ψ̄(x)γ5 Σµν ψ(x)

(7.15)

is not invariant under either C or P.

7.3 Given Cb† (⃗ p, s)C† = ηc∗ d† (⃗ p, s) †

†

(7.16)

†

Cd (⃗ p, s)C = ηc b (⃗ p, s).

Let ηc = 1 and show that ] [ ] [ ∑ ) π ∑( † † † d (⃗ b (⃗ p, s)b(⃗ p, s) ×exp p, s)b(⃗ p, s) − b (⃗ p, s)d(⃗ p, s) .

C ≡ exp iπ 2 s s (7.17) Hint: Define the transformation [ ] ∑( ) † † Mθ ≡ exp θ d (⃗ p, s)b(⃗ p, s) − b (⃗ p, s)d(⃗ p, s) (7.18) s

where θ is a continuous real parameter. Show via differentiation with respect to θ that Mθ b† (⃗ p, s)Mθ† = cos θb† (⃗ p, s) + sin θd† (⃗ p, s)

(7.19)

Mθ d (⃗ p, s)Mθ† = − sin θb† (⃗ p, s) + cos θd† (⃗ p, s). †

Note, also that in the expressions above ∑ ∑∫ d3 p⃗ . ≡ (2π)3 2Ep s s

(7.20)

Solution [ ] ∑ ∂ † † † † † (d b − b d), b Mθ† Mθ b (s)Mθ = Mθ (7.21) ∂θ s′ { } { } = d† (s′ ) b(s′ ), b† (s) − b† (s), d† (s′ ) b(s) = Mθ d† (s)Mθ† .

Hence ∂ Mθ b† (⃗ p, s)Mθ† = Mθ d† (⃗ p, s)Mθ† . ∂θ

(7.22)

∂ Mθ d† (⃗ p, s)Mθ† = −Mθ b† (⃗ p, s)Mθ† . ∂θ

(7.23)

Similarly

Thus ∂2 Mθ b† (⃗ p, s)Mθ† = −Mθ b† (⃗ p, s)Mθ† .

(7.24) ∂θ2 We can continue this process ad infinitum. This provides each term in a Taylor

Charge Conjugation and Parity

32

series expansion of the function Mθ b† (s)Mθ† about the point θ = 0. We obtain the result in Eqn. 7.19. Therefore we can now evaluate the result at θ = π2 .

We have M π2 b† (⃗ p, s)M †π = d† (⃗ p, s) 2

(7.25)

M π2 d† (⃗ p, s)M †π = −b† (⃗ p, s).

2

This is almost the transformation we want, but Cd† C† = b† and not −b† .

We thus need an additional rotation which takes b† to −b† , while leaving d† untouched. Consider the operator ∑ b† (⃗ exp[iα p, s)b(⃗ p, s)] (7.26) s

and show in a way similar to the previous analysis that ∑ ∑ exp[iα b† (⃗ p, s)b(⃗ p, s)] b† exp[−iα b† (⃗ p, s)b(⃗ p, s)] ≡ eiα b† . s

Hence exp[iπ

∑ s

(7.27)

s

b† (⃗ p, s)b(⃗ p, s)] b† exp[−iπ

∑

b† (⃗ p, s)b(⃗ p, s)] ≡ −b† .

(7.28)

s

Finally, putting it all together we have [ ] [ ] ∑ ) π ∑( † † † C ≡ exp iπ b (⃗ p, s)b(⃗ p, s) ×exp d (⃗ p, s)b(⃗ p, s) − b (⃗ p, s)d(⃗ p, s) .

2 s s (7.29)

8

Global Symmetries

Exercises 8.1 Calculate the differential cross-sections for the processes (a) e+ e− → µ+ µ− and (b) e+ e− → π + π − .

Assume unpolarized initial states and sum over the final spins. Explain the difference in the angular dependence θ (measured between the incoming electron and outgoing positively charged particle) for the two processes and the difference in the threshold behavior, i.e. the momentum dependence of the differential cross-sections near threshold. [Hint: both are consequences of angular momentum conservation.]

Solution (a) In the calculation of Bhabha scattering (Chapter 13), we noted that part of the solution corresponding to the s-channel process alone, gives the differential cross-section for the process e+ e− → µ+ µ− . The result presented there in exercise 13.2 was, however, only valid for energies much greater than the muon mass.

e+ e−

−→

µ+ µ−

p1 p2

p3 p4

s1 s2

s3 s4

(8.1)

The scattering amplitude is given by

−igµν (ie)2 → → → → T = −(ū(− p4 , s4 )γ ν v(− p3 , s3 ))(v̄(− p1 , s1 )γ µ u(− p2 , s2 )) (p1 + p2 )2 + iϵ 33

(8.2)

Global Symmetries

34

and ∑

|T |2 =

spins

) ( ) e4 ( µ ν Tr (p /1 )γ (p /2 )γ Tr (p /4 + mµ )γµ (p /3 − mµ )γν 2 s

) 32e4 ( p1 · p4 p2 · p3 + p2 · p4 p1 · p3 + p1 · p2 m2µ s2 ( ) m4µ 8e4 2 2 2 = 2 u + t + mµ s − . s 2 ∑ At threshold s = 4m2µ , u = −m2µ = t and s |T |2 ̸= 0.

The differential cross-section is given by ∑ dσ(e+ e− → µ+ µ− ) 1 1 = √ dLips(s, p3 , p4 ) |T |2 .

2 2 dΩ 2 λ(s, me , me ) 4

(8.3)

=

(8.4)

spins

In the high energy limit, s ≫ µ2 , we have dσ(e+ e− → µ+ µ− ) α2 u2 + t2 = ( ) (8.5) dΩ 2s s2 ) α2 α2 ( = (1 + cos θ)2 + (1 − cos θ)2 = (1 + cos2 θ).

8s 4s For the helicity understanding why the first term peaks at θ = 0 (see Fig. 8.1), while the second term peaks at θ = π (see Fig. 8.2). The helicity amplitude µ− θ e+

e−

t

Fig. 8.1

µ+

This contribution peaks at θ = 0.

µ− θ e+

e−

t

Fig. 8.2

µ+

This contribution peaks at θ = π. peaks at θ = 0, π, since chirality/helicity is conserved at each vertex.

Exercises

35

(b) e+ e−

−→

π+ π−

(8.6)

p1 p2

p3 p4

s1 s2

s3 = 0 s4 = 0

The scattering amplitude is given by T (e+ e− → π + π − ) =

e2 v̄(p1 )γµ u(p2 )(p3 − p4 )µ . s

(8.7)

and ∑

|T |2 =

s

) 4e4 ( 2p1 · (p3 − p4 ) p2 · (p3 − p4 ) − p1 · p2 (p3 − p4 )2 s2

(8.8)

) 4e4 ( −8(⃗ p · p⃗ ′ )2 + 2|⃗ p ′ |2 s 2 s 8e4 |⃗ p ′ |2 sin2 θ. = s

=

∑ Note s |T |2 = 0 at threshold. Since the pions have spin 0, helicity conservation requires that the outgoing state has orbital angular momentum with L = 1 and the amplitude peaks at θ = ± π2 (see Fig. 8.3). The extra factor of π− θ e+

e−

t

Fig. 8.3

π+

This contribution peaks at θ = ± π2 . |⃗ p ′ |2 ≡ (p′ )2 is due to the L = 1 final state. dσ α2 p′ (p′ )2 = sin2 θ. dΩ 2sp s

(8.9)

9

Testing Isospin and G Parity

Exercises 9.1 Define the generator of isospin 1 Ii = 2

∫

† d3 ⃗xψN (x)τi ψN (x)

where ψN (x), defined in Eqn. 9.2 ψN (x) =

(

ψp (x) ψn (x)

(9.1)

) ,

(9.2)

is an iso-doublet field containing two four component fields, ψp , ψn , representing the proton and neutron, respectively; τi , i = 1, 2, 3 are Pauli matrices acting in isospin space. (a) Show that [Ii , Ij ] = iϵijk Ik

(9.3)

and (b) [Ii , ψN (x)] = −

τi ψN (x).

2

(9.4)

(c) Define rotation operators ⃗ ≡ exp(−iI⃗ · θ) ⃗ UI (θ)

(9.5)

where θi are three real parameters. Show that ⃗ N (x)U † (θ) ⃗ = uI ψN (x) UI (θ)ψ I

(9.6)

where ⃗ ⃗ = exp(i ⃗τ · θ) (9.7) uI (θ) 2 is the representation of the isospin rotation group in the two dimensional ⃗ and θ̂ = θ/α ⃗ space. [Hint: Let θ⃗ = αθ̂ where α = |θ| and differentiate with respect to α.] (d) Assume [Ii , Hstrong ] = 0.

36

(9.8)

Exercises

37

Show that this implies that the proton and neutron, which are eigenstates of Hstrong , have equal mass.

Solution (a) In exercise 9.1 you showed that given the charge operator ∫ QΓ ≡ d3 ⃗xψ † (⃗x, t)Γψ(⃗x, t) (9.9) it satisfies the commutation relations [QΓ , QΓ′ ] = Q[Γ, Γ′ ] .

(9.10)

In this case the fermionic field is ψ → ψN and Γ → τi /2. Hence we have [ ] ∫ ∫ 1 1 1 † † 3 [Ii , Ij ] = d ⃗xψN (x) τi , τj ψN (x) = iϵijk d3 ⃗xψN (x) τk ψN (x) = iϵijk Ik .

2 2 2 (9.11) (b) [ ] ∫ 1 † [Ii , ψN (x)b ] = d3 ⃗x ′ ψN (x′ ) τi ψN (x′ ), ψN (x)b (9.12) 2 ∫ { }1 1 † = − d3 ⃗x ′ ψN (x′ )a , ψN (x)b τi ψN (x′ )a = − τi ψN (x)b .

2 2 (c) Let ⃗ ≡ exp(−iI⃗ · θ) ⃗ = exp(−iαI0 ) UI (θ)

(9.13)

where I0 = I⃗ · θ̂. Then ∂ UI (α)ψN (x)UI† (α) = UI (α) (−i [I0 , ψN (x)]) UI† (α). ∂α

(9.14)

But [I0 , ψN (x)] = θ̂i [Ii , ψN (x)] = −

τ0 ψN (x) 2

(9.15)

where τ0 ≡ θ̂i τi . Therefore ∂ τ0 UI (α)ψN (x)UI† (α) = i UI (α)ψN (x)UI† (α). ∂α 2 Thus consider the Taylor series expansion f (α) ≡ UI (α)ψN (x)UI† (α) = f (0) + αf ′ (0) + =

∞ ∑ (iα τ0 )n 2

n=0 τ ⃗ ⃗ iθ·

=e

2

n!

α2 ′′ f (0) + · · · 2

(9.16)

(9.17)

τ0

ψN (x) = eiα 2 ψN (x)

⃗ N (x). ψN (x) = uI (θ)ψ

(d) mp = ⟨p|Hstrong |p⟩ = ⟨p|Hstrong I+ |n⟩ = ⟨p|I+ Hstrong |n⟩ = ⟨n|Hstrong |n⟩ = mn .

(9.18)

Testing Isospin and G Parity

38

9.2 The pions, π ± , π 0 , transform as a vector under isospin, where I3 |π ± ⟩ = ±|π ± ⟩, √ I ± |π 0 ⟩ = ∓ 2|π ± ⟩,

I3 |π 0 ⟩ = 0

√ I |π ⟩ = 0, I |π ⟩ = − 2|π 0 ⟩ +

−

+

(9.19)

+

and 1 π ± ≡ √ (π1 ± iπ2 ), π 0 ≡ π3 , 2

(9.20)

I ± ≡ (I1 ± iI2 ).

(9.21)

and πi are real scalar fields and I ± are raising and lowering operators.

The pion-Nucleon scattering cross-section, at a laboratory pion energy ∼ 300 M eV , is dominated by the resonance, ∆(1232) with isospin 3/2, i.e. π + N → ∆(1232) → π + N . Use conservation of isospin, [Ii , S] = 0 (where S is the scattering operator) and the Wigner-Eckart theorem to show that the ratio of cross-sections, σ(π + p → π + p) : σ(π − p → π 0 n) : σ(π − p → π − p) = 9 : 2 : 1.

(9.22)

Solution Using the notation for isospin states |I, I3 ⟩ we have the Clebsch-Gordan decomposition |π + p⟩ = |3/2, 3/2⟩

(9.23)

−

|π p⟩ = |3/2, −1/2⟩(3/2, −1/2|1, −1; 1/2, 1/2) + |1/2, −1/2⟩(1/2, −1/2|1, −1; 1/2, 1/2) |π 0 n⟩ = |3/2, −1/2⟩(3/2, −1/2|1, 0; 1/2, −1/2) + |1/2, −1/2⟩(1/2, −1/2|1, 0; 1/2, −1/2) Now consider the scattering amplitudes ⟨π − p|S|π − p⟩, ⟨π + p|S|π + p⟩, ⟨π 0 n|S|π − p⟩.

(9.24)

We assume that [Ii , Hstrong ] = 0 and thus [Ii , S] = 0. Thus [I3 , S] = 0 = [I⃗2 , S] implies that the scattering operator, S, cannot change the value of I or I3 , i.e. ⟨I ′ , I3′ |S|I, I3 ⟩ ≡ SI,I3 δII ′ δI3 ,I3′ .

(9.25)

Moreover, SI,I3 ≡ SI is independent of the ’magnetic’ quantum number, I3 , since [Ii , S] = 0. This is a special case of the Wigner-Eckart theorem for a scalar operator.

Thus ⟨π + p|S|π + p⟩ ≡ S3/2 −

−

(9.26)

⟨π p|S|π p⟩ ≡ S3/2 |(3/2, −1/2|1, −1; 1/2, 1/2)| + S1/2 |(1/2, −1/2|1, −1; 1/2, 1/2)|2 2

⟨π 0 n|S|π − p⟩ ≡ S3/2 (3/2, −1/2|1, 0; 1/2, −1/2)∗ (3/2, −1/2|1, −1; 1/2, 1/2) + S1/2 (1/2, −1/2|1, 0; 1/2, −1/2)∗ (1/2, −1/2|1, −1; 1/2, 1/2) (9.27)

Exercises

39

where S3/2 , S1/2 implicitly depend on the center of momentum energy squared, s = (pπ + pN )2 . √ At s = 1232 MeV, we have S3/2 ≫ S1/2 ( if ∆(1232) is a resonance with √ I = 3/2). Thus at s = 1232 MeV, we have σ(π + p → π + p) = ΦP S |S3/2 |2 −

−

σ(π p → π p) = ΦP S |S3/2 | |(3/2, −1/2|1, −1; 1/2, 1/2)| 2

(9.28) 4

σ(π − p → π 0 n) = ΦP S |S3/2 |2 |(3/2, −1/2|1, 0; 1/2, −1/2)|2 |(3/2, −1/2|1, −1; 1/2, 1/2)|2 where ΦP S is the result of integrating over phase space, summing over spins, etc. But √ √ (3/2, −1/2|1, −1; 1/2, 1/2) = 1/ 3, (3/2, −1/2|1, 0; 1/2, −1/2) = 2/3.

(9.29) We thus find the result σ(π + p → π + p) : σ(π − p → π 0 n) : σ(π − p → π − p) = 9 : 2 : 1.

(9.30)

These relations are fairly well satisfied by the experimental data. Reading off √ the Particle Data Tables (Tanabashi et al., 2018) by eye, we see at s ∼ 1232 MeV, σ(π + p → π + p) ∼ 200mb, σ(π − p → π − p) ∼ 28mb, σ(π − p)total ∼ 75mb.

(9.31)

10

Evidence for New Particles, Quantum Numbers and Interactions Exercises 10.1 Consider the decay K → 2π. (a) What is the orbital angular momentum of the pions? (b) Use Bose statistics to identify the allowed isospin states of the two pions, i.e I = 0, 1, 2 ? (c) If in weak decays, only |∆I| ≤ 1 is allowed, is the process K + → π + π 0 permitted? (d) Given, τK + = 1.24 × 10−8 s, τK 0 = 0.892 × 10−10 s,

BR(K + → π + π 0 ) = (21.17 ± 0.16)% (10.1) BR(K 0 → π + π − ) = (68.61 ± 0.28)%.

Calculate the ratio of the |∆I| = 3/2 to ∆I = 1/2 amplitudes for these processes.

Solution (a) The orbital angular momentum of the pions is L = 0, since the kaon has spin 0. (b) The total wavefunction of the pions must be symmetric, therefore it can only be I = 0, 2. (c) Since the final state must be symmetric under isospin, it must be I = 2.

Therefore ∆I = 3/2 would be forbidden, assuming |∆I| ≤ 1. (d) The amplitude for the process K + → π + π 0 is given by 3/2

(2, +1|HW |1/2, +1/2).

(10.2)

While the amplitude for the process K 0 → π + π − is given by 1 1 3/2 1/2 √ (2, 0|HW |1/2, −1/2) + √ (0, 0|HW |1/2, −1/2) 6 3

(10.3)

where 3/2

1/2

HW = HW + HW .

40

(10.4)

Exercises

41

The decay rates for the two processes satisfy BR(K 0 → π + π − ) ≈ 7.69 × 109 s−1 τK 0 BR(K + → π + π 0 ) ≫ Γ(K + →π+ π0 ) = = 1.71 × 107 s−1 . τK + Γ(K 0 →π+ π− ) =

(10.5)

Thus in the process K 0 → π + π − , ∆I = 1/2 dominates ! In order to get an approximation for the ratio of ∆I = 3/2 : ∆I = 1/2 we can ignore the contribution of ∆I = 3/2 to the process K 0 → π + π − . We then have 1 (∆ = 1/2)2 3

(10.6)

Γ(K + →π+ π0 ) = (∆I = 3/2)2 .

(10.7)

Γ(K 0 →π+ π− ) = and

Hence ∆I = 3/2 = ∆I = 1/2

(

Γ(K + →π+ π0 ) 3Γ(K 0 →π+ π− )

)1/2

( =

1.71 × 107 3 × 7.69 × 109

)1/2 =

1 .

36

(10.8)

10.2 The scattering process K + p → K 0 π + p is measured in the laboratory with a lab K + bean with energy EK + = 3 GeV incident on a hydrogen bubble chamber.

The K 0 decays into π + π − . The energy and momenta of the final states are measured and a cartoon of the observed differential cross-section [in the CMS frame]

d2 σ dEp dEK 0

(10.9)

is plotted in Fig. 10.1 (Fig. 21.5 in the book). This is known as a Dalitz plot,

t

Fig. 10.1

This is a Dalitz plot of the differential cross-section for the process K + p → K 0 π + p. where each point represents an event and the dark bands represent clusters

Evidence for New Particles, Quantum Numbers and Interactions

42

of points. From this plot we learn that there are two resonances produced K + p → K ∗+ p followed by the decay K ∗+ → K 0 π + and K + p → K 0 ∆++ followed by the decay ∆++ → π + p.

Using the location of the bands given in the figure, find the mass of the ∆++ and K ∗+ . Note, mp = 0.938 GeV , mK + = 0.494 GeV , mK 0 = 0.498 GeV and mπ+ = 0.140 GeV .

Solution From the Dalitz plot one finds two resonances. Recall in the CMS the phase space is a constant. Thus the Dalitz plot measures the square of the scattering amplitude. The center of mass energy squared, s = (PK + + Pp )2 = m2K + + m2p + 2mp EK + = 6.752 GeV2 and

√ s = 2.598 GeV.

(10.10)

(10.11)

In two body scattering the energy of the outgoing particles are determined in terms of the CMS energy and the masses √ of the outgoing particles 1 and 2.

Using the result of Chapter 4, 2ps1/2 = λ(s, m21 , m22 ), we can show that the energy of particle 1 is given by s − m22 + m21 √ .

2 s

(10.12)

1 s − m2∆++ + m2K 0 √ 2 s

(10.13)

E1 = In the CMS EK 0 =

on resonance. From the Dalitz plot we have EK 0 = 1.0549 GeV.

(10.14)

√ m2∆ = −2 sEK 0 + s + m2K 0 = 1.232 GeV.

(10.15)

We then find that

The second resonance give Ep =

1 s − m2K ∗ + m2p √ 2 s

(10.16)

on resonance and from the Dalitz plot we have Ep = 1.3154 GeV.

(10.17)

√ m2K ∗ = −2 sEp + s + m2p = 0.892 GeV.

(10.18)

We then find that

The total amplitude can be written in terms of the sum of two Breit-Wigner amplitudes.

11

Representation Theory for SU (2)

Exercises 11.1 Find the generators of SU (2) in the I = 3/2 representation, as we discussed with (Ti )t3 t′ ≡ ⟨t3 |Ti |t′3 ⟩, 3/2

(11.1)

3

for i = 1, 2, 3, i.e. write out the three 4 × 4 matrices.

Show (without actually commuting the matrices) that these matrices satisfy the Lie algebra [ ] (Ti )3/2 , (Tj )3/2 = iϵijk (Tk )3/2 .

(11.2) Solution √ 3 3 5 3 T+ | , t3 ⟩ = × − t3 (t3 + 1) | , t3 + 1⟩ 2 2 2 2 √ 3 3 5 3 T− | , t3 ⟩ = × − t3 (t3 − 1) | , t3 − 1⟩ 2 2 2 2 Thus we have

√ 3 3 15 ⟨ , t3 + 1|T+ | , t3 ⟩ = − t3 (t3 + 1) 2 2 4 √ 3 3 15 ⟨ , t3 − 1|T− | , t3 ⟩ = − t3 (t3 − 1) 2 2 4

Therefore we find



0  0 3/2 T+ =   0 0 in the basis

(

√ 3 0 0 0

 0 0 2 √0   0 3  0 0

3/2 1/2 −1/2

−3/2

)

.

(11.3)

(11.4)

(11.5)

(11.6)

We also have 3/2

T− 43

3/2 †

= T+

.

(11.7)

Representation Theory for SU (2)

44

Finally, 3/2

T1 and

[

=

1 3/2 1 3/2 3/2 3/2 3/2 (T + T− ), T2 = (T+ − T− ) 2 + 2i

 3/2 0 0 0  0 1/2 0 0  3/2 .

T3 =   0 0 −1/2 0  0 0 0 −3/2

(11.8)



(11.9)

The matrices satisfy the Lie algebra ) ] ∑( 3 3 ′′ 3 ′′ 3 ′ 3 3 ′′ 3 ′′ 3 ′ 3/2 3/2 = ⟨ , t |T | , t ⟩⟨ , t |T | , t ⟩ − ⟨ , t |T | , t ⟩⟨ , t |T | , t ⟩ Ti , Tj 3 i j 3 j i 2 2 3 2 3 2 3 2 2 3 2 3 2 3 t3 t′3 ′′ t3

3 3 3 3 3/2 ≡ ⟨ , t3 | [Ti , Tj ] | , t′3 ⟩ = iϵijk ⟨ , t3 |Tk || , t′3 ⟩ = iϵijk (Tk )t3 t′ 3 2 2 2 2 where we’ve used the fact ∑ 3 3 | , t′′3 ⟩⟨ , t′′3 | ≡ I. (11.10) 2 2 ′′ t3

12

SU (3) Symmetry

Exercises 12.1 Given the Gell-Mann matrices, {λa }, The generators of SU (3) in the triplet representation, Ta ≡ λ2a , satisfy the Lie algebra of SU (3), i.e.

[Ta , Tb ] = ifabc Tc

(12.1)

where fabc are the structure constants (see Table 23.1 in the book). (a) Show that the generators, T̄a ≡ −TaT ,

(12.2)

satisfy the same algebra. This is the conjugate representation, 3̄. (b) Use the Jacobi identity, [[Ta , Tb ] , Tc ] + [[Tb , Tc ] , Ta ] + [[Tc , Ta ] , Tb ] ≡ 0,

(12.3)

to show that the eight 8 × 8 matrices, (Fa )bc ≡ −ifabc , a, b, c = 1 . . . 8

(12.4)

satisfy the Lie algebra of SU (3). This is the octet ( or adjoint ) representation. (c) (i) Given the highest weight state, |t3 , y⟩ with t3 = 3/2, y = 1, use raising and lowering operators to complete the multiplet. (ii) Draw the weight diagram for this representation. Assuming Q = T3 + Y2 , identify the charge of each state.

Solution (a) Given [Ta , Tb ] = ifabc Tc ,

(12.5)

take the transpose of both sides. We have [ T T] Tb , Ta = ifabc TcT .

(12.6)

Thus

45

[ ] −TaT , −TbT = ifabc (−TcT ).

(12.7)

SU (3) Symmetry

46

(b) The Jacobi identity, [[Ta , Tb ] , Tc ] + [[Tb , Tc ] , Ta ] + [[Tc , Ta ] , Tb ] ≡ 0

(12.8)

[ifabd Td , Tc ] + [ifbcd Td , Ta ] + [ifcad Td , Tb ] ≡ 0

(12.9)

implies −[fabd fdce + fbcd fdae + fcad fdbe ] = 0 (Fa )bd (Fe )dc − (Fe )bd (Fa )dc − ifaed (Fd )bc = 0 and finally [(Fa ), (Fe )]bc = ifaed (Fd )bc .

(12.10)

(c) (i) √ 1 3 T− | , 1⟩ = 3| , 1⟩ 2 2 1 1 T− | , 1⟩ = 2| − , 1⟩ 2 2 √ 3 1 T− | − , 1⟩ = 3| − , 1⟩ 2 2 3 3 1 3 3 3 V3 | , 1⟩ = ( Y + T3 )| , 1⟩ = | , 1⟩ 2 4 2 2 2 2 √ 3 V− | , 1⟩ = 3|1, 0⟩ 2 √ T− |1, 0⟩ = 2|0, 0⟩ √ T− |0, 0⟩ = 2| − 1, 0⟩ 1 V− |1, 0⟩ = 2| , −1⟩ 2 1 1 T− | , −1⟩ = | − , −1⟩ 2 2 √ 1 V− | , −1⟩ = 3|0, −2⟩ 2 (ii) See Fig. 12.1

(12.11)

Exercises

47

Y ∆−

∆0

J P = 32

∆+

+

∆++

1 Σ− 0 − 23

−1

∗

Σ0 − 12 Ξ−

1 Ξ0

∗

Ω−

Fig. 12.1

Σ+ 1 2

0

−1

t

∗

+

Baryon decuplet with spin, parity, J P = 32 .

∗

∗

3 2

References

Tanabashi, M., et al. 2018. Review of Particle Physics. Phys. Rev., D98(3), 030001.

48