Introduction to Quantum Field Theory Classical Mechanics to Gauge Field Theories Anthony G. Williams

Solutions Manual for Teachers Introduction to Quantum Field Theory: Classical Mechanics to Gauge Field Theories ETHAN N. CARRAGHER AND ANTHONY G. WILLIAMS

SOLUTION MANUAL

Contents

Preface to Solutions Manual

iii

page vi

1 Lorentz and Poincaré Invariance 1.1 Problem 1 1.2 Problem 2 1.3 Problem 3 1.4 Problem 4 1.5 Problem 5 1.6 Problem 6 1.7 Problem 7 1.8 Problem 8 1.9 Problem 9 1.10 Problem 10 1.11 Problem 11 1.12 Problem 12 1.13 Problem 13 1.14 Problem 14 1.15 Problem 15 1.16 Problem 16

1 1 2 3 5 5 7 8 8 9 10 11 12 13 13 14 15

2 Classical Mechanics 2.1 Problem 1 2.2 Problem 2 2.3 Problem 3 2.4 Problem 4 2.5 Problem 5 2.6 Problem 6 2.7 Problem 7 2.8 Problem 8 2.9 Problem 9 2.10 Problem 10 2.11 Problem 11 2.12 Problem 12 2.13 Problem 13 2.14 Problem 14 2.15 Problem 15

18 18 19 20 22 24 29 30 34 35 36 37 40 42 43 45

Contents

iv

3 Relativistic classical fields 3.1 Problem 1 3.2 Problem 2 3.3 Problem 3 3.4 Problem 4 3.5 Problem 5 3.6 Problem 6 3.7 Problem 7 3.8 Problem 8

48 48 52 53 57 60 61 61 64

4 Relativistic Quantum Mechanics 4.1 Problem 1 4.2 Problem 2 4.3 Problem 3 4.4 Problem 4 4.5 Problem 5 4.6 Problem 6 4.7 Problem 7 4.8 Problem 8 4.9 Problem 9 4.10 Problem 10 4.11 Problem 11 4.12 Problem 12

68 68 69 72 75 76 77 79 81 82 86 88 89

5 Introduction to Particle Physics 5.1 Problem 1 5.2 Problem 2 5.3 Problem 3 5.4 Problem 4 5.5 Problem 5 5.6 Problem 6 5.7 Problem 7 5.8 Problem 8 5.9 Problem 9 5.10 Problem 10 5.11 Problem 11 5.12 Problem 12

91 91 92 94 95 97 98 100 102 103 104 108 110

6 Formulation of Quantum Field Theory 6.1 Problem 1 6.2 Problem 2 6.3 Problem 3 6.4 Problem 4 6.5 Problem 5

117 117 118 120 121 123

Contents

v

6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13

Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13

127 130 133 135 147 150 151 154

7 Interacting Quantum Field Theories 7.1 Problem 1 7.2 Problem 2 7.3 Problem 3 7.4 Problem 4 7.5 Problem 5 7.6 Problem 6 7.7 Problem 7 7.8 Problem 8 7.9 Problem 9 7.10 Problem 10

163 163 164 164 165 166 167 168 170 171 173

8 Symmetries and Renormalization 8.1 Problem 1 8.2 Problem 2 8.3 Problem 3 8.4 Problem 4 8.5 Problem 5 8.6 Problem 6 8.7 Problem 7 8.8 Problem 8

176 176 177 179 180 181 182 189 191

9 Gauge Field Theories 9.1 Problem 1 9.2 Problem 2 9.3 Problem 3 9.4 Problem 4 9.5 Problem 5 9.6 Problem 6 9.7 Problem 7 9.8 Problem 8 9.9 Problem 9

192 192 194 198 200 202 205 208 220 227

References

231

Preface to Solutions Manual

This solutions manual has been provided to assist teachers who adopt the textbook as a teaching resource for their classes. It is not intended for broad distribution as that would defeat the purpose of the problem sets at the end of each chapter. Some of the problems are challenging for students and such problems could be assigned to a small group or as small projects for individual students.

As can be seen from the length of this manual, it took some time and effort for us to prepare and typeset these solutions and it is almost certain that we have not caught every error. So we ask the reader’s forgiveness for any mistakes found and we would be very grateful if these could be reported at the website below. In that way corrections can be made and an appropriate acknowledgement given.

Ethan N. Carragher and Anthony G. Williams Adelaide, May 28, 2022

Numbering of equations Two-part equation numbers, such as Eq. (1.42) and Eq. (9.51), refer to equations in this Solutions Manual. Three-part equation numbers, such as Eq. (3.1.16) and Eq. (9.4.23), refer to equations in the book.

Corrections to this book As is the case for the book itself, the current list of corrections for this Solutions Manual along with the names of those who suggested them can be found at: www.cambridge.org/WilliamsQFT. It would be greatly appreciated if anyone finding additional errors in the book itself or in this Solutions Manual could please report them using the relevant corrections link provided on this website.

vi

1

Lorentz and Poincaré Invariance 1.1 Problem 1 Problem: A muon is a more massive version of an electron and has a mass of 105.7 MeV/c2 . The dominant decay mode of the muon is to an electron, an electron antineutrino and a muon neutrino, µ− → e− + ν̄e + νµ . If we have N (t) unstable particles at time t then the fraction of particles decaying per unit time is a constant, i.e., we have dN/N = −(1/τ )dt for some constant τ . This gives dN/dt = −(1/τ )N , which has the solution N (t) = N0 e−t/τ where we have N0 unstable particles at t = 0. The fraction of particles decaying in the interval t to t + dt −t/τ is −dN/N dt. So the mean lifetime (or lifetime) 0 = (−dN/dt)dt/NR 0 = (1/τ )e R∞ ∞ −x is 0 t (−dN/dt) dt/N0 = τ 0 xe dx = τ , where x = t/τ . The half-life, t1/2 , is the time taken for half the particles to decay, e−t1/2 /τ = 21 , which means that t1/2 = τ ln 2. The decay rate, Γ, is defined as the probability per unit time that a particle will decay, i.e., Γ = (−dN/dt)/N = 1/τ is the inverse mean lifetime. A muon at rest has a lifetime of τ = 2.197 × 10−6 s. Cosmic rays are high-energy particles that have traveled enormous distances from outside our solar system. Primary cosmic rays are particles that have been accelerated by some extreme astrophysical event and secondary cosmic rays are those resulting from collisions of primary cosmic rays with interstellar gas or with our atmosphere. Most cosmic rays reaching our atmosphere will be stable particles such as photons, neutrinos, electrons, protons and stable atomic nuclei (mostly helium nuclei). Muon cosmic rays therefore are secondary cosmic rays produced when primary or secondary cosmic rays collide with out atmosphere. A typical height in the atmosphere for the production of cosmic ray muons is ∼ 15 km. What is the minimum velocity that this muon be produced with in order that it have a 50% chance of reaching the surface of the Earth before decaying?

Solution: A stationary observer on Earth will see the time experienced by a muon traveling at speed v to be dilated by a factor of γ = (1 − v 2 /c2 )−1/2 compared to their own.

So according to the observer, the half-life of the muon will be γ t1/2 . If the muon travels at a speed that allows it to traverse the L = 15 km of the atmosphere in this time, it will therefore have a 50% chance of reaching the Earth’s surface. That

1

Lorentz and Poincaré Invariance

2

is, the speed of the muon must satisfy q L = γ t1/2

v=

2

1 − vc2 L

(1.1)

t1/2

to have a 50% chance of reaching the surface before decaying. This can be rearranged to give L . v=q 2 t21/2 + Lc2

(1.2)

Now from the given information, the half-life of the muon is t1/2 = τ ln 2 = (2.197 × 10−6 s) ln 2 = 1.523 × 10−6 s,

(1.3)

so the necessary speed is v=q

15 × 103 m

3

2

(15×10 m) (1.523 × 10−6 s)2 + (3×10 8 ms−1 )2

= 299, 653, 700 ms−1 = 0.9995c.

(1.4)

An inertial observer traveling with the muon will see the height of the atmosphere contracted by a factor of γ, which at this speed gives a height of r L v2 = 1− 2L γ c p = 1 − 0.99952 (15 × 103 m) = 456 m.

(1.5)

In other words, from the muon’s point of view, it only needs to travel 456 m to reach the Earth’s surface.

1.2 Problem 2 Problem: The diameter of our Milky Way spiral galaxy is approximately 100,000 - 180,000 light years and our solar system is approximately 25,000 light years from the center of our galaxy. Recalling the effects of time dilation, approximately how fast would you have to travel to reach the center of the galaxy in your lifetime? Estimate how much energy would it take to accelerate your body to this speed.

Solution:

3

Problem 3

From the solution to Problem 1.1, the speed needed for an observer to travel a proper distance L in a time T is L v=q .

2 T 2 + Lc2

(1.6)

So for a human with a lifetime of, say, T = 80 years, to travel to the center of the Milky Way L = 25,000 light years = 25,000c years away, a speed of 25,000c years v=p (80 years)2 + (25,000 years)2 = 0.999995c

(1.7) (1.8)

would be required. The kinetic energy possessed by a human of mass m = 70 kg at this speed is Ekin = (γ − 1)mc2

1 = √ − 1 (70 kg)(3 × 108 ms−1 ) 1 − 0.9999952 = 2 × 1021 J.

(1.9)

For reference, it would take humanity slightly more than 3 years to consume this much energy at current rates.

1.3 Problem 3 Problem: Consider two events that occur at the same spatial point in the frame of some inertial observer O. Explain why the two events occur in the same temporal order in every inertial frame connected to it by a Lorentz transformation that does not invert time. Show that the time separation between the two events is a minimum in the frame of O. (Hint: Consider Figs. 1.2 and 1.5.)

Solution: Since the two events E1 and E2 occur at the same spatial point in frame O, the displacement vector E2 − E1 between them will point entirely along the time axis in this frame, in the positive direction, say. Under any given Lorentz transformation, displacement vectors will be moved along hyperbolae in a spacetime diagram, as illustrated in Fig. 1.1. Further, Lorentz transformations that do not invert time will only move displacement vectors along branches of those hyperbolae. Our vector of interest E2 − E1 lies on a positive-time branch, so under an orthochronus Lorentz transformation it will remain on that positive-time branch. That is, the temporal order of the events remains the same. And the point on such a positive-time branch that has the smallest time component is the one that lies on the time axis, as E2 −E1

Lorentz and Poincaré Invariance

4

iBK2 1 1 2

bT +2

2 2

1

t

Figure 1.1

A two-dimensional of spacetime displacement with displacements 6B;m`2representation k, >vT2`#QH b Q7 7` K2@BMp `B M+2X UTXR8V represented by vectors. Vectors labeled by the same number (1 or 2) are related by Lorentz transformations. y

o0

does in frame O. So under an orthochronus Lorentz transformation, E2 − E1 will be moved to a point on the spacetime diagram with a time component at least as o separation between the events is large as that in frame O, showing that `0 the time minimized in this frame. ` in frame O as To see this algebraically, label the two events 00

o

T

✓E1 = (ct1 , x) , 00 E2✓= (ct2 , x)T`,

(1.10) (1.11)

where t2 > t1 . Then under a Lorentz transformation Λ, the events go to E1 → E1′ = ΛE1 = Λ(ct1 , x)T ,

x

(1.12)

6B;m`2 j, UTXRNV

E2 → E2′ = ΛE2 = Λ(ct2 , x)T .

(1.13)

The difference between theseyevents is 0

y

T

E2′ − E1′ = Λ (c(t2 − t1 ), 0) .

(1.14)

ois Λ00 c(t2 − t1 ), which is at In particular, the temporal component of this difference least as great as c(t2 −t1 ), because Lorentz transformations that do not reverse time satisfy Λ00 ≥ 1. Again we conclude the time separation ` is minimized in frame O. x0 ✓ x 6B;m`2 9, UTXRNV

Problem 4

5

1.4 Problem 4 Problem: Consider any two events that occur at the same time in the frame of an inertial observer O. Show that by considering any Lorentz transformation there is no limit to the possible time separation of the two events and that the smallest spatial separation of the two events occurs in the frame of O. Solution: Label the two events in frame O as E1 = (ct, x1 )T ,

(1.15)

T

(1.16)

E2 = (ct, x2 ) .

Assume without loss of generality that the events are separated in the x-direction by an amount |∆x| = ̸ 0, and consider a boost in the x-direction with speed v. The difference in the boosted temporal coordinates will be given by Eq. (1.2.89),

v v (1.17) c∆t′ = γ c∆t − ∆x = − γ∆x, c c and so the magnitude of the time separation in the boosted coordinates will be v |∆x|.

(1.18) |∆t′ | = q 2 c2 1 − vc2 The factor |∆x| is a non-zero constant, so this expression approaches infinity as v → c. Hence, there is no limit to the possible time separation between the two events.

Next, the spatial separations between the events along each direction in the boosted frame, ∆x′ = γ (∆x − v∆t) = γ∆x,

(1.19)

′

(1.20)

′

∆z = ∆z,

(1.21)

imply a total spatial separation of p |∆x′ | = (γ∆x)2 + (∆y)2 + (∆z)2 p ≥ (∆x)2 + (∆y)2 + (∆z)2 = |∆x|.

(1.22)

∆y = ∆y,

(1.23)

Hence, the smallest spatial separation between the two events occurs in frame O.

1.5 Problem 5 Problem:

6

Lorentz and Poincaré Invariance

y

γ1

θ/2

x

θ/2

t

Figure 1.2

γ2

Coordinate system being used for Problem 1.5, where γ1 and γ2 label the two beams of light.

Two narrow light beams intersect at angle θ, where θ is the angle between the outgoing beams. The beams intersect head on when θ = 180◦ . Using the addition of velocities formula in Eq. (1.2.126) show that for any angle θ there is always an inertial frame in which the beams intersect head on.

Solution: Set up the coordinate system such that the beams of light lie in the x − y plane, with the origin at their intersection and the x axis bisecting the angle θ made by the beams, as in Fig. 1.2. Then the velocity vectors of the beams will be T

θ θ , c sin ,0 , 2 2

T θ θ u2 = c cos , −c sin ,0 .

2 2

u1 =

c cos

(1.24) (1.25)

We will boost with speed v in the +x-direction, hoping to find a speed such that the beams are directed entirely in the y ′ -direction and moving oppositely. The velocities in the x′ -direction, given by Eq. (1.2.126) in the text, must therefore vanish: u′x =

ux − v =0 1 − ucx2v

=⇒

v = ux = c cos

(1.26)

θ .

2

(1.27)

If cos(θ/2) = 1 then the beams intersect head on in the original frame, and it is not necessary to perform a boost. If not, then Eq. 1.27 shows that v < c and therefore our proposed boost is valid. The velocities in the y ′ -direction are u′y =

1 uy γ 1 − ucx2v

Problem 6

7

y0

y

` θ `x

t

Figure 1.3

boost `y

`0

`0y = `y

θ0

x

`0x = `γx

x0

Diagram showing the length contraction of the ruler in a boosted frame. uy =q 2 1 − vc2

±c sin θ2 =q

1 − cos2 θ2 = ±c

(1.28)

where the ± refers to the first and the second beam, respectively. We see that the beams are oppositely directed in the y ′ -direction and therefore intersect head on in this frame.

1.6 Problem 6 Problem: A ruler of rest length ℓ is at rest in the frame of inertial observer O and is at an angle θ with respect to the +x-direction. Now consider an inertial observer O′ in an identical inertial frame except that it has been boosted by speed v in the +xdirection. What is the length ℓ′ of the ruler and what is the angle θ′ with respect to the +x′ -direction that will be measured by O′ ?

Solution: As a result of Eq. (1.2.89), the lengths measured by observer O′ will be contracted compared to the lengths in frame O by a factor of γ in the x′ -direction, and remain the same in the perpendicular directions, as illustrated in Fig. 1.3. If the ruler lies in the x − y plane, O′ will therefore measure the lengths in the x′ and y ′ directions to be r ℓx v2 ′ ℓx = = 1 − 2 ℓ cos θ, (1.29) γ c ℓ′y = ℓy = ℓ sin θ,

(1.30)

Lorentz and Poincaré Invariance

8

and therefore the total length to be ′

ℓ =

q

r (ℓ′x )2 + (ℓ′y )2 = ℓ

1−

v2 cos2 θ. c2

(1.31)

The angle θ′ in the boosted frame will satisfy ℓ′y ℓy tan θ′ ≡ ′ = γ ≡ γ tan θ ℓx ℓx



 tan θ . =⇒ θ′ = tan−1  q 2 1 − vc2

(1.32)

1.7 Problem 7 Problem: If a spaceship approaches earth at 1.5×108 ms−1 and emits a microwave frequency of 10 GHz, what frequency will an observer on Earth detect?

Solution: The ratio of the frequencies in the source frame and observer frame is given by Eq. (1.2.123) in the text, s 1+β fs = .

(1.33) fo 1−β Since the ship is moving towards Earth at a speed of 1.5 × 108 ms−1 = c/2, we write v = −c/2 and so β = v/c = −1/2. An observer on Earth will then detect a frequency of s 1−β fo = fs (1.34) 1+β s 1 + 21 = (10 GHz) (1.35) 1 − 21 = 17.3 GHz.

(1.36)

Note that fo > fs , so the beam has been blueshifted as expected when the source is moving towards the observer.

1.8 Problem 8 Problem: An inertial observer observes two spaceship moving directly toward one another.

She measures one to be traveling at 0.7 c in her inertial frame and the other at 0.9

Problem 9

9

y γ θ x v

t

Figure 1.4

A particular ray of light emitted by the moving light source in frame O. c in the opposite direction. What is the magnitude of the relative velocity that each spaceship measures the other to have?

Solution: Take the +x-direction to be the direction of motion of the spaceship moving at 0.9c in the frame of the inertial observer. Boost by v = 0.9c in this direction to go to the rest frame of the spaceship. In this frame, the other spaceship will have a speed u′x =

ux − v , 1 − ucx2v

(1.37)

where ux = −0.7c is the speed of that spaceship in the original frame along the +x-direction, by Eq. (1.2.126). Substituting in the given values, we find u′x =

−0.7c − 0.9c

1 − (−0.7c)(0.9c) c2

= −0.9816c.

(1.38)

That is, each spaceship measures the other to be moving towards them at 0.98c.

1.9 Problem 9 Problem: A light source moves with constant velocity v in the frame of inertial observer O. The source radiates isotropically in its rest frame. Show that in the inertial frame of O the light is concentrated in the direction of motion of the source, where half of the photons lie in a cone of semi-angle θ, where cos θ = v/c.

Solution:

Lorentz and Poincaré Invariance

10

In frame O, assume without loss of generality the source of light is moving in the +x-direction at speed v. Consider the ray of light emitted by the source in the x − y plane making an angle θ with the x-axis, shown in Fig. 1.4. It will have velocity vector u = (c cos θ, c sin θ, 0)T .

(1.39)

Boost by speed v in the +x-direction to move to the rest frame of the light source.

Here, the x′ -component of the ray’s velocity vector will be given by Eq. (1.2.126): ux − v 1 − ucx2v c cos θ − v = 1 − vc cos θ

u′x =

≡ c cos θ′ ,

(1.40)

which implies cos θ =

v v + 1 − cos θ cos θ′ . c c

(1.41)

Now the source emits photons uniformly in all directions in this boosted frame, so half of the photons it emits will lie in the region −90◦ < θ′ < +90◦ . The boundary of this region satisfies cos θ′ = 0, which corresponds in the original frame to cos θ = v/c

by Eq. 1.41. Since the boost transformation is continuous, the interior of the original region (satisfying cos θ′ > 0) will be mapped to the interior of the boosted region (satisfying cos θ > v/c). That is, half of the emitted photons will lie in a cone of semi-angle θ = cos−1 (v/c) < 90◦ , showing that the light is concentrated in the direction of motion of the source.

1.10 Problem 10 Problem: Write down the Lorentz transformation rule for an arbitrary (3, 2) tensor Aµνρστ .

Hence show that any double contraction of this tensor leads to a contravariant vector, i.e., to a (1, 0) tensor.

Solution: Under a Lorentz transformation Λ, a (3, 2) tensor Aµνρστ will transform as ′ ′ ′

′

′

Aµνρστ → A′µνρστ = Λµµ′ Λν ν ′ Λρρ′ Aµ ν ρ σ′ τ ′ (Λ−1 )σ σ (Λ−1 )τ τ .

Define T µ = Aµνρνρ . Then using Eq. 1.42, T µ will transform as T µ → T ′µ = A′µνρνρ

′ ′ ′

′

′

= Λµµ′ Λν ν ′ Λρρ′ Aµ ν ρ σ′ τ ′ (Λ−1 )σν (Λ−1 )τ ρ

(1.42)

Problem 11

11

h ih i ′ ′ ′ ′ ′ = (Λ−1 )σν Λν ν ′ (Λ−1 )τ ρ Λρρ′ Λµµ′ Aµ ν ρ σ′ τ ′ ′

′ ′ ′

′

= δ σν ′ δ τ ρ′ Λµµ′ Aµ ν ρ σ′ τ ′ ′ ′ ′

= Λµµ′ Aµ ν ρ ν ′ ρ′ ′

= Λµµ′ T µ .

(1.43)

This is the transformation rule for a (1, 0) tensor, so T µ is a (1, 0) tensor. Of course, the same steps apply if any other pairs of indices were chosen in the double contraction of Aµνρστ to form T µ - our choice here is equivalent to any other choice under a change of basis. Therefore, any double contraction of Aµνρστ produces a (1, 0) tensor.

1.11 Problem 11 Problem: A rocket of initial rest mass M0 has a propulsion system that accelerates it by converting matter into light with negligible heat loss and directs the light in a collimated beam behind it. The propulsion system is turned on for some period of time during which the speed of the rocket is boosted by speed v. Use energy and momentum p conservation to show that the final rest mass of the rocket is given by Mv = M0 (c − v)/(c + v).

Solution: In the frame where the rocket is initially at rest, the total momentum is p = 0 and the total energy is the rest energy of the rocket, E = M0 c2 . Once the propulsion is completed, the rocket will have momentum of magnitude γMv v and energy γMv c2 where γ = (1 − v 2 /c2 )−1/2 , while the emitted light will have some momentum pγ directed oppositely to that of the rocket, and an energy pγ c. Momentum conservation then imposes the condition γMv v − pγ = |p| = 0, =⇒

pγ = γMv v,

(1.44)

and energy conservation the condition γMv c2 + pγ c = E = M0 c2 .

Putting this information together, γMv c2 + γMv vc = M0 c2

v = M0 γMv 1 + c q =⇒

Mv = M0

2

1 − vc2

1 + vc

(1.45)

Lorentz and Poincaré Invariance

12

q = M0

1 − vc

1 + vc s

1 − vc 1 + vc

r

c−v , c+v

= M0 = M0

1 + vc

(1.46)

as required.

1.12 Problem 12 Problem: Two identical spaceships approach an inertial observer O at equal speeds but from opposite directions. A second observer O′ traveling on one of the spaceships measures the length of the other spaceship to be only 60% of its length at rest. How fast is each space ship traveling with respect to O? Solution: In frame O, say that each spaceship has speed v and that observer O′ is moving in the +x-direction, so the other spaceship has velocity ux = −v in this direction.

The rest frame of O′ is obtained by a boost of speed v along the +x-direction, so in this frame the speed of the other spaceship will be given by the velocity addition formula Eq. (1.2.126), u′x =

2v ux − v 2 , ux v = − 1 − c2 1 + vc2

(1.47)

Now O′ measures the other spaceship to be contracted by a factor 1 = γ

s

′ 2 ux = 0.6, 1− c

(1.48)

so its speed must be |u′x | = 0.8c. With the previous information, this implies 2v 2 = 0.8c, 1 + vc2

or

c 0 = c2 − 2.5cv + v 2 = v − (v − 2c) .

2

(1.49)

Discounting the superluminal solution, it must be that the spaceships are traveling at speed v = c/2 with respect to O.

Problem 13

13

1.13 Problem 13 Problem: An observer on earth observes the hydrogen-β line of a distant galaxy shifted from its laboratory measured value of 434 nm to a value of 510 nm. How fast was the galaxy receding from the Earth at the time the light was emitted? [You may neglect the effects of the cosmological redshift due to the expansion of the universe during the time of flight of the photons].

Solution: The relation between the frequency fs in the source frame (the frequency that would be measured in a laboratory) and the frequency fo in the observer’s frame on Earth is given by Eq. (1.2.123), s 1−β fs .

(1.50) fo = 1+β Since c = f λ, the analogous equation between wavelengths is s 1+β λo = λs .

1−β

(1.51)

With this, we get

λo λs

2 =

c+v , c−v

(1.52)

and rearranging for the relative speed between the source and the observer, 2 λo −1 λs c.

(1.53) v = 2 λo + 1 λs For the case at hand we have λo /λs = (510 nm)/(434 nm) = 1.175, meaning the galaxy is receding from Earth at a speed of v=

1.1752 − 1 c = 0.16c.

1.1752 + 1

(1.54)

1.14 Problem 14 Problem: An inertial observer O is midway between two sources of light at rest in her frame with the sources 2 km apart. Each source emits a flash of light that reach O simultaneously. Another inertial observer O′ is moving parallel to the line that

Lorentz and Poincaré Invariance

14

passes through the light sources and inertial observer O. In her frame O′ measures the time of each flash and finds that they are spaced 10 ns apart in time. What is the speed of O′ with respect to O? Solution: Say without loss of generality that the sources of light lie on the x-axis in the frame of observer O. Then according to O, the events at which the flashes originated are separated in time by an amount ∆t = 0 and in space by an amount ∆x = 2 km.

The temporal separation of the events in the frame of O′ , who is boosted by speed v in the x-direction, is given by the Lorentz transformation rule Eq. (1.2.89),

v v c∆t′ = γ c∆t − ∆x = − q ∆x.

(1.55) 2 c c 1 − vc2 Rearranging for the speed, we get v=p

c∆t′ (∆x)2 + (c∆t′ )2

c.

(1.56)

Now we are told that ∆t′ = 10 ns (and so c∆t′ = 3 m), meaning observer O′ must be traveling at a speed of v=p

3m (2 × 103 m)2 + (3 m)2

c = 0.0015c = 4.5 × 105 ms−1

(1.57)

with respect to observer O.

1.15 Problem 15 Problem: A particle with integer or half-integer spin s has 2s+1 values of spin with respect to any arbitrary spin quantization axis n in the particle’s rest frame. If parity is a good symmetry, a spin eigenstate for a massless particle is a helicity eigenstate with helicity eigenvalues ±s. If parity is not conserved then a single helicity eigenstate is possible with eigenvalue either +s or −s. As succinctly as you can, summarize the key reasons for this result.

Solution: Particles are classified according to the representations of the Poincaré group under which they transform. For the classification of the internal states of the particle, i.e. the intrinsic spin, it helps to consider the little group of Poincaré transformations that leave the four-momentum of the particle invariant. For massive particles, we can always consider the rest frame with zero momentum, in which case the little group corresponds to the group SO(3) of 3D rotations. We have seen that the representations of SO(3) are labeled according to an integer or half-integer s,

15

Problem 16

and are 2s + 1 dimensional. This corresponds to the 2s + 1 possible spin states for a massive particle of spin s.

For massless particles, on the other hand, there is no frame in which the momentum vanishes. The little group is instead the group SE(2) of translations in a plane orthogonal to the momentum vector, and rotations in that plane. For this group, requiring a finite number of helicity degrees of freedom in fact forces there to be only a single possible helicity λ, and the opposite helicity for the antiparticle. This is because the ladder operators that raise or lower the helicity do not give rise to a terminating tower of helicity states (there is no analog to the maximal spin ms = s state that occurs in the massive case), unless there is only a single valid state that they both annihilate. Now under a parity transformation, helicity reverses sign: Shelicity = Sspin · np → −Sspin · np = −Shelicity .

(1.58)

So if a theory is parity invariant, it must be possible for a massless particle to have both helicity +λ and −λ. However, if parity is not a good symmetry, then negating the helicity of a massless particle may not result in a valid particle state, and only a single helicity would be possible.

1.16 Problem 16 Problem: The Poincaré group is a subgroup of a larger Lie group, the conformal group C(1, 3). This fifteen-parameter group consists of the ten-parameter subgroup of Poincaré transformations, the one-parameter subgroup of scale transformations (dilatations), and the four-parameter subgroup of special conformal transformations (SCT). The generators of the unitary representation of the group are: P µ = iℏ ∂ µ (translations); M µν = iℏ (xµ ∂ ν −xν ∂ µ ) = xµ P ν − xν P µ (Lorentz transformations); D = iℏ xµ ∂µ = x · P (dilatations); and K µ = iℏ 2xµ x·∂ −x2 ∂ µ = 2xµ D − x2 P µ (SCT). Using [P µ , xν ] = iℏg µν we have already seen that P µ and M µν satisfy the Lie algebra of the Poincaré group. Show that the remaining commutators of the Lie algebra of the conformal group are: [D, P µ ] = −iℏ P µ , [D, K µ ] = iℏ K µ , [D, M µν ] = 0, [K µ , P ν ] = −iℏ 2(M µν +g µν D), [K µ , K ν ] = 0 and [K µ , M ρσ ] = iℏ (g µρ K σ −g µσ K ρ ).

Solution: The following derivations will make extensive use of the commutator identities [AB, C] = A[B, C] + [A, C]B,

(1.59)

[A, BC] = [A, B]C + B[A, C],

(1.60)

which may readily be verified by expansion of both sides of the equations.

Lorentz and Poincaré Invariance

16

(i) Using the first identity, we have [D, P µ ] = [xν Pν , P µ ] = xν [Pν , P µ ] + [xν , P µ ]Pν = 0 − iℏg νµ Pν = −iℏP µ .

(1.61) (ii) For this part it helps to first calculate [D, xµ ] = [xν Pν , xµ ] = xν [Pν , xµ ] + [xν , xµ ]Pν = iℏg νµ xν + 0 = iℏxµ (1.62) and [D, x2 ] = [D, xµ xµ ] = [D, xµ ]xµ + xµ [D, xµ ] = iℏxµ xµ + iℏxµ xµ = 2iℏx2 .

(1.63) Note that a commutator containing a covariant index is just the covariant version of that same commutator with a contravariant index, and vice versa, because the metric components are constant and thus commute with all operators. Now we use these findings to calculate [D, K µ ] = [D, 2xµ D − x2 P µ ]

= 2[D, xµ ]D − [D, x2 ]P µ − x2 [D, P µ ]

= 2iℏxµ D − 2iℏx2 P µ + iℏx2 P µ

= iℏ(2xµ D − x2 P µ ) = iℏK µ .

(1.64)

(iii) Next we see [D, xµ P ν ] = [D, xµ ]P ν + xµ [D, P ν ] = iℏxµ P ν − iℏxµ P ν = 0,

(1.65)

and so [D, M µν ] = [D, xµ P ν − xν P µ ] = 0.

(1.66)

(iv) First, [x2 , P µ ] = xν [xν , P µ ] + [xν , P µ ]xν = −iℏxν g νµ − iℏδνµ xν = −2iℏxµ . (1.67) Hence, [K µ , P ν ] = [2xµ D − x2 P µ , P ν ]

= 2xµ [D, P ν ] + 2[xµ , P ν ]D − x2 [P µ , P ν ] − [x2 , P ν ]P µ

= −2iℏxµ P ν − 2iℏg µν D − 0 + 2iℏxν P µ

= −2iℏ(xµ P ν − xν P µ + g µν D)

= −2iℏ(M µν + g µν D)

(1.68)

(v) Using our earlier results, we have [K µ , xν ] = [2xµ D − x2 P µ , xν ] = 2xµ [D, xν ] − x2 [P µ , xν ] = 2iℏxµ xν − iℏx2 g µν

(1.69)

Problem 16

17

and [K µ , x2 ] = [K µ , xν ]xν + xν [K µ , xν ] = 2iℏxµ xν xν − iℏx2 g µν xν + 2iℏxν xµ xν − iℏxν x2 δνµ

= 2iℏx2 xµ ,

(1.70)

since components of the position vector commute amongst themselves. It follows that [K µ , K ν ] = [K µ , 2xν D − x2 P ν ]

= 2[K µ , xν ]D + 2xν [K µ , D] − [K µ , x2 ]P ν − x2 [K µ , P ν ]

= 4iℏxµ xν D − 2iℏx2 g µν D − 2iℏxν K µ − 2iℏx2 xµ P ν + 2iℏx2 (M µν + g µν D)

= 2iℏ(2xµ xν D − xν K µ − x2 xµ P ν + x2 M µν )

= 0.

(1.71)

(vi) Finally, let us calculate [xµ , M ρσ ] = [xµ , xρ P σ − xσ P ρ ]

= xρ [xµ , P σ ] − xσ [xµ , P ρ ]

=⇒ =⇒

µ

xµ [x , M

ρσ

2

ρσ

[x , M

= −iℏxρ g µσ + iℏxσ g µρ ,

(1.72)

] = 0.

(1.73)

ρ σ

σ ρ

] = −iℏx x + iℏx x = 0

Then [K µ , M ρσ ] = [2xµ D − x2 P µ , M ρσ ]

= 2[xµ , M ρσ ]D + 2xµ [D, M ρσ ] − [x2 , M ρσ ]P µ − x2 [P µ , M ρσ ]

= 2iℏ(xσ g µρ − xρ g µσ )D + 0 − 0 − iℏx2 (g µρ P σ − g µσ P ρ )

= iℏ g µρ (2xσ D − x2 P σ ) − g µσ (2xρ D − x2 P ρ ) = iℏ(g µρ K σ − g µσ K ρ ),

(1.74)

given we already know P µ and M µν satisfy the Lie algebra of the Poincaré group, Eq. (1.2.167).

2

Classical Mechanics 2.1 Problem 1 Problem: A cylinder of mass M , moment of inertia I about the cylindrical axis and radius R rolls on a horizontal surface without slipping: (a) Express the no-slip constraint in differential form. Is it holonomic? Explain. (b) Write down a Lagrangian for the system.

Solution: (a) To not slip, the distance traveled by the center of mass of the cylinder must equal the distance rolled along the circumference. So if a particular point on the circumference turns through a differential angle dθ while the center of the cylinder moves a horizontal distance dx, the no-slip constraint is given by dx = Rdθ,

(2.1)

x − Rθ − constant = 0.

(2.2)

or equivalently

This is the form of a holonomic constraint since there is no dependence on the generalized velocities ẋ and θ̇. (b) The cylinder, being a rigid body, has kinetic energy equal to the sum of its translational kinetic energy and the rotational kinetic energy about its center of mass: 1 1 (2.3) T = M ẋ2 + I θ̇2 .

2 2 Assuming the cylinder is constrained to always be touching the ground, its potential energy will be constant and can therefore be regarded as zero. So an appropriate Lagrangian for the system is L=T −V =

1 1 M ẋ2 + I θ̇2 .

2 2

(2.4)

Lagrange multipliers will be needed to enforce the constant-height constraint in the equations of motion.

18

Problem 2

19

x2 m x1 − `

t

Figure 2.1

M

x=0 Coordinates used to analyse the sliding blocks of Problem 2.2. The natural length of the spring is denoted by ℓ.

2.2 Problem 2 Problem: A small block of mass m slides without friction on top of a larger block of mass M and is attached to a pin in this block by a massless spring with spring constant k. All motion is in one dimension. The larger block slides without friction on a flat table. Construct a Lagrangian for the system and obtain the equations of motion.

Identify any conserved quantities. Relate these to the symmetries of the system.

Solution: Let us use the coordinates shown in Fig. 2.1: x1 is the horizontal coordinate of the pin in the larger block plus the natural length of the spring, and x2 is the horizontal coordinate of the smaller block, so that the end of the spring is displaced from its equilibrium position by an amount (x2 − x1 ). Then the total kinetic energy of the system is simply T =

1 1 M ẋ21 + mẋ22 , 2 2

(2.5)

while the potential energy is that stored in the spring: V =

1 k(x2 − x1 )2 .

2

(2.6)

In total, the Lagrangian for the system is L=T −V =

1 1 1 M ẋ21 + mẋ22 − k(x2 − x1 )2 .

2 2 2

(2.7)

The equation of motion for the coordinate x1 is d ∂L ∂L − dt ∂ ẋ1 ∂x1 d = (M ẋ1 ) − k(x2 − x1 ) dt = M ẍ1 − k(x2 − x1 ),

0=

(2.8)

Classical Mechanics

20

and that for x2 is d ∂L ∂L − dt ∂ ẋ2 ∂x2 d = (mẋ2 ) + k(x2 − x1 ) dt = mẍ2 + k(x2 − x1 ).

0=

(2.9)

Notice that summing Eq. 2.8 and Eq. 2.9 yields the equation d (M ẋ1 + mẋ2 ) = 0, dt

(2.10)

which is simply a statement of the conservation of momentum. This is to be expected because momentum is the conserved quantity related to spatial translations x1 → x1 + s and x2 → x2 + s, which are clearly a symmetry of our Lagrangian. Further, we note that the system is natural and the Lagrangian is time independent, so the total energy E = T + V should be conserved too. This is simple to check, using the above equations of motion:

dE d 1 1 1 2 2 2 = M ẋ1 + mẋ2 + k(x2 − x1 ) dt dt 2 2 2 = M ẋ1 ẍ1 + mẋ2 ẍ2 + k(x2 − x1 )(ẋ2 − ẋ1 )

= ẋ1 [M ẍ1 − k(x2 − x1 )] + ẋ2 [mẍ2 + k(x2 − x1 )] = 0,

(2.11)

as expected.

2.3 Problem 3 Problem: A uniform circular wire of radius R is forced to rotate about a fixed vertical diameter at constant angular velocity ω. A bead of mass m experiences gravity, is smoothly threaded on the wire and can slide without friction. Is the system holonomic? Construct a Lagrangian for the system. Show that there is just one solution for the off-axis motion of the bead where it does not oscillate. Does the system conserve energy?

Solution: Let us use the cartesian coordinate system of Fig. 2.2 with origin at the center of the wire circle. When the bead is at an angle θ from the bottom of the wire, its position can be written as x = (R sin θ cos(ωt), R sin θ sin(ωt), −R cos θ)T ,

(2.12)

Problem 3

21

z

y

x θ

t

Figure 2.2

R

ω Coordinates used to analyse the rotating wire and bead system of Problem 2.3. given an appropriate time coordinate. Note that this implies a constraint of the form x · x − R2 = x2 + y 2 + z 2 − R2 = 0,

(2.13)

which does not depend on the generalized velocities and is thus holonomic. The kinetic energy of the bead is 1 m|ẋ|2 , 2 while the potential energy is entirely gravitational: T =

V = mgz = −mgR cos θ.

(2.14)

(2.15)

Now the velocity of the bead can be decomposed into perpendicular components, one tangential to the wire circle (of magnitude Rθ̇), and the other tangential to the circle of radius R sin θ swept out by the bead (of magnitude (R sin θ)ω). It follows that the speed of the bead satisfies |ẋ|2 = R2 θ̇2 + R2 ω 2 sin2 θ,

(2.16)

as can be verified explicitly from Eq. 2.12. Hence, the Lagrangian of the system is given by 1 mR2 (θ̇2 + ω 2 sin2 θ) + mgR cos θ.

2 The equation of motion is therefore L=T −V =

d ∂L ∂L − dt ∂ θ̇ ∂θ 2 = mR θ̈ − mR2 ω 2 sin θ cos θ + mgR sin θ.

(2.17)

0=

(2.18)

Classical Mechanics

22

Now suppose the bead is off-axis (so sin θ ̸= 0) and does not oscillate (so θ̈ = 0).

Then the equation of motion dictates −mR2 ω 2 sin θ cos θ + mgR = 0.

(2.19)

Therefore there is only one solution, cos θ = g/(Rω 2 ), for off-axis motion of the bead with no oscillations.

It might be thought that energy is conserved in this system because the Lagrangian has no explicit time dependence. However, this is incorrect because the system is not natural (the kinetic energy is not homogeneous in θ̇), so the Hamiltonian is not equal to the energy of the system. In fact, it is conceptually clear that energy E = T + V is generically not conserved in this system: the wire is driven by an external force so that it rotates with a constant angular speed ω independently of the motion of the bead, so the energy associated to this rotation, supplied by the external driving system, does not need to balance the other forms of energy of the bead. Checking this algebraically:

dE d 1 2 2 2 2 = mR (θ̇ + ω sin θ) − mgR cos θ dt dt 2 = mR2 θ̇θ̈ + mR2 θ̇ω 2 sin θ cos θ + mgRθ̇ sin θ = 2mR2 θ̇ω 2 sin θ cos θ

(by the equation of motion)

̸= 0,

(2.20)

as was reasoned above.

2.4 Problem 4 Problem: A bead of mass m is threaded without friction on a massless wire hoop of radius R which is forced to oscillate vertically in a fixed vertical plane at angular frequency ω and with amplitude a. Is the system holonomic? Construct a Hamiltonian for the system and obtain Hamilton’s equations. Does the system conserve energy?

Solution: Let us use a cartesian coordinate system with origin at the average position of the center of the wire hoop, as in Fig. 2.3. Then the position of the center can be parameterized as xcenter = (0, a sin ωt)T .

(2.21)

The position of the bead, x = (x, y)T , is constrained to lie on a circle of radius R around this point: |x − xcenter |2 = R2 ,

(2.22)

Problem 4

23

y

a a

x θ

t

Figure 2.3

R

Coordinates used to analyse the oscillating hoop and bead system of Problem 2.4. and therefore the coordinates obey the holonomic constraint x2 + (y − a sin ωt)2 − R2 = 0.

(2.23)

Now whether the system is natural depends on the form of the kinetic energy T = mẋ2 /2. To calculate this, we first parameterize the coordinates by the angle θ of the bead from the bottom of the hoop as x = xcenter + (R sin θ, −R cos θ)T = (R sin θ, a sin ωt − R cos θ)T ,

(2.24)

ẋ = (Rθ̇ cos θ, aω cos ωt + Rθ̇ sin θ)T ,

(2.25)

leading to

and therefore ẋ2 = R2 θ̇2 cos2 θ + a2 ω 2 cos2 ωt + 2Raω θ̇ sin θ cos ωt + R2 θ̇2 sin2 θ = R2 θ̇2 + a2 ω 2 cos2 ωt + 2Raω θ̇ sin θ cos ωt.

(2.26)

This expression is not homogeneous in θ̇: it contains both θ̇ and θ̇2 terms. The kinetic energy is therefore not homogeneous in θ̇, and so the system is not natural.

To construct a Hamiltonian for the system we must first write a Lagrangian. We already have an expression for the kinetic energy, and the potential energy of the bead is entirely gravitational: V = mgy = mg(a sin ωt − R cos θ), so our Lagrangian is L=T −V =

1 1 mR2 θ̇2 + ma2 ω 2 cos2 ωt + mRaω θ̇ sin θ cos ωt 2 2

(2.27)

Classical Mechanics

24

− mg(a sin ωt − R cos θ).

(2.28)

Now we must calculate the conjugate momentum ∂L = mR2 θ̇ + mRaω sin θ cos ωt ∂ θ̇ in order to finally construct the Hamiltonian pθ :=

(2.29)

H := pθ θ̇ − L

1 1 = mR2 θ̇2 + mRaω θ̇ sin θ cos ωt − mR2 θ̇2 − ma2 ω 2 cos2 ωt 2 2 − mRaω θ̇ sin θ cos ωt + mg(a sin ωt − R cos θ) 1 1 = mR2 θ̇2 − ma2 ω 2 cos2 ωt + mg(a sin ωt − R cos θ).

(2.30) 2 2 Importantly, the Hamiltonian is a function of position and momentum only, and not of the generalized velocities. This means we must use Eq. 2.29 to substitute out the generalized velocity θ̇ and write the Hamiltonian instead as H=

1 (pθ − mRaω sin θ cos ωt)2 − ma2 ω 2 cos2 ωt + mg(a sin ωt − R cos θ).

2 2mR 2 (2.31)

Hamilton’s equations of motion are θ̇ =

∂H pθ − mRaω sin θ cos ωt , = ∂pθ mR2

(2.32)

which is simply a restatement of Eq. 2.29, and ∂H (pθ − mRaω sin θ cos ωt) = (−mRaω cos θ cos ωt) − mgR sin θ. (2.33) ∂θ mR2 A rather involved calculation would show that the total energy E = T + V is not conserved over time when these equations of motion are satisfied. This does not follow from the explicit time dependence of the Lagrangian, because the system is not natural and so the Hamiltonian is not equal to the total energy. However, it is clear on physical grounds that the energy will generically not be conserved: imagine the bead sitting at the bottom of the hoop, driven at a very high frequency ω with a very small amplitude a. The kinetic energy of the bead would vary wildly, while the potential energy would be almost constant, so the total energy would be changing over time. The energy is not conserved because the system is not isolated; the external system driving the oscillation is supplying extra energy to the bead. ṗθ = −

2.5 Problem 5 Problem: Two point masses m and M are connected by a massless rod of length ℓ and placed

Problem 5

25

y x

M

b

θ ` x

m

t

Figure 2.4

Diagram showing the generalized coordinates x and θ for the constrained rod and spring system of Problem 2.5. on a horizontal table. The mass m is also connected to a fixed point P on the table by a spring with spring constant k and zero natural length; it can slide without friction in any direction on the table. The mass M is constrained to slide along a straight line a perpendicular distance b < ℓ from P . Set up a Lagrangian for the system and obtain the Euler-Lagrange equations. In the limit of small oscillations, solve for the characteristic angular frequencies and normal modes. Explain whether or not energy is conserved?

Solution: Let us use a cartesian coordinate system with origin located at P , as in Fig. 2.4.

The position of mass M will then be written xM = (x, b)T .

(2.34)

If θ is the angle the rod makes from the vertical (going clockwise), the mass m at the end of the rod will be located at the point xm = xM + (−ℓ sin θ, −ℓ cos θ)T = (x − ℓ sin θ, b − ℓ cos θ)T .

(2.35)

Our two generalized coordinates here are the horizontal position x of the mass M , and the angle θ of the rod.

The velocities of the two masses are ẋM = (ẋ, 0)T , ẋm = (ẋ − ℓθ̇ cos θ, ℓθ̇ sin θ)T , so the kinetic energy of the system is T =

1 1 M ẋ2M + mẋ2m 2 2

(2.36)

Classical Mechanics

26

1 1 M ẋ2 + m(ẋ2 − 2ẋℓθ̇ cos θ + ℓ2 θ̇2 cos2 θ + ℓ2 θ̇2 sin2 θ) 2 2 1 1 = M ẋ2 + m(ẋ2 − 2ẋℓθ̇ cos θ + ℓ2 θ̇2 ).

2 2 =

(2.37)

The potential energy of the system is that stored in the spring, which is extended by a length |xm |: 1 2 kx 2 m

1 = k (x − ℓ sin θ)2 + (b − ℓ cos θ)2 2 1 ∼ = k(x2 − 2xl sin θ − 2bl cos θ), 2

V =

(2.38)

where we have dropped inconsequential constant terms in the last line. In total, an appropriate Lagrangian for the system is L=T −V =

1 1 1 M ẋ2 + m(ẋ2 − 2ẋℓθ̇ cos θ + ℓ2 θ̇2 ) − k(x2 − 2xl sin θ − 2bl cos θ).

2 2 2 (2.39)

The Euler-Lagrange equation for the coordinate x is d ∂L ∂L − dt ∂ ẋ ∂x d = (M ẋ + mẋ − mℓθ̇ cos θ) − (−kx + kℓ sin θ) dt = (M + m)ẍ − mℓθ̈ cos θ + mℓθ̇2 sin θ + k(x − ℓ sin θ),

0=

(2.40)

and that for the coordinate θ is d ∂L ∂L − dt ∂ θ̇ ∂θ d = (mℓ2 θ̇ − mẋℓ cos θ) − (mẋℓθ̇ sin θ + kxℓ cos θ − kbℓ sin θ) dt = mℓ2 θ̈ − mẍℓ cos θ + kℓ(b sin θ − x cos θ).

0=

(2.41)

Now to find the behavior of small oscillations, we must first identify the stable equilibrium points of the system, about which oscillations can take place. At such points, the gradient of the potential energy must vanish: ∂V

kx − kℓ sin θ 0 ∂x = = .

(2.42) ∂V −kxℓ cos θ + kbℓ sin θ 0 ∂θ There are two classes of solutions to these equations: 1) 2)

x = 0, p x = ± ℓ2 − b2 ,

sin θ = 0, x sin θ = , ℓ

and and

cos θ = ±1; b cos θ = . ℓ

(2.43) (2.44)

Since b < ℓ, both of these solutions are valid. As for which of these is a stable

Problem 5

27

equilibrium, we look to the matrix of second derivatives !

∂2V ∂2V k −kℓ cos θ 2 ∂x ∂x∂θ K(x, θ) ≡ = . ∂2V ∂2V −kℓ cos θ kxℓ sin θ + kbℓ cos θ ∂θ∂x ∂θ 2

(2.45)

Stability occurs if this matrix is positive-definite, which, since k > 0, will be the case if and only if det K(x, θ) > 0. Now in the first class of solutions we have cos θ = ±1, so the second derivative matrix will have the form

k −kℓ(±1) K= , (2.46) −kℓ(±1) kbℓ(±1) and therefore a determinant of k 2 ℓ(±b − ℓ) < 0, so this solution does not represent a stable equilibrium point. For the second class of solutions, where cos θ = b/ℓ, the second derivative matrix is

k −kb K= .

(2.47) −kb kℓ2 The determinant here is k 2 (ℓ2 − b2 ) > 0, so this solution does represent a stable equilibrium setup. To summarize, the only stable equilibria in this system are those √ configurations where x = ± ℓ2 − b2 , sin θ = x/ℓ, and cos θ = b/ℓ. Let us denote the coordinates at such a point by x0 and θ0 , so then we can expand our coordinates in terms of small deviations x̃ and θ̃ about that point as x =: x0 + x̃, θ =: θ0 + θ̃.

(2.48)

Small oscillations are found by expanding the Lagrangian to second order in x̃ and θ̃, and their derivatives. In fact we already know the second order expansion of the potential energy: the only important terms are the quadratic terms, dictated by the matrix K in Eq. 2.47. So only the kinetic energy needs to be expanded further. To do so, we note the Taylor expansion cos θ = cos(θ0 + θ̃) = cos θ0 − θ̃ sin θ0 + O(θ̃2 ) =

b + O(θ̃) ℓ

(2.49)

allows us to expand the kinetic energy as 1 1 M ẋ2 + m(ẋ2 − 2ẋℓθ̇ cos θ + ℓ2 θ̇2 ) 2 2

1 1 b 2 2 2 2 = M ẋ + m ẋ − 2ẋℓθ̇ + ℓ θ̇ + O(θ3 ) 2 2 ℓ

1 ẋ M + m −mb + O(θ3 ). = (ẋ θ̇) −mb mℓ2 θ̇ 2

T =

(2.50)

In an abuse of notation we have replaced the symbols x̃ and θ̃ by simply x and θ from the second line onwards. We recognize the matrix

M + m −mb 0 M = , (2.51) −mb mℓ2

Classical Mechanics

28

that dictates the frequency ω of small oscillations through the relation 0 = det(K − ω 2 M 0 )

k − ω 2 (M + m) −kb + ω 2 mb = det −kb + ω 2 mb kℓ2 − ω 2 mℓ2

= ω 4 [M mℓ2 + m2 (ℓ2 − b2 )] − ω 2 k[M ℓ2 + 2m(ℓ2 − b2 )] + k 2 (ℓ2 − b2 )

k 2 =m ω − [M ℓ2 + m(ℓ2 − b2 )]ω 2 − k(ℓ2 − b2 ) .

(2.52) m

The two solutions are r ω1 =

k , m

s ω2 =

k(ℓ2 − b2 ) .

M ℓ2 + m(ℓ2 − b2 )

(2.53)

Now the normal modes v1 and v2 corresponding to these frequencies are those vectors that satisfy (K − ω12 M 0 )v1 = 0,

(K − ω22 M 0 )v2 = 0.

(2.54)

If the top component of one of these vector equations is satisfied then the bottom 2 component will be satisfied too, since det(K − ω1,2 M 0 ) = 0. So let us solve the top equation for the first normal mode:

M +m v11 = 0, =⇒ v11 = 0.

(2.55) k−k m Really we only need to solve for v1,2 up to a scalar multiple C1,2 , so we can write

0 v1 = C1 .

(2.56) 1 As for the normal mode v2 corresponding to ω2 , we get the equation (K − ω22 M 0 )v2 = 0,

k(ℓ2 −b2 )mb kb − 1 2 2 2 M ℓ +m(ℓ −b ) ℓ2 v2 = , = =⇒ 2 2 2 −b )(M +m) v2 b k − k(ℓ M ℓ2 +m(ℓ2 −b2 ) !

2 and therefore

v2 = C2

ℓ b

1

.

The normal mode oscillations are then simply

x(t) x(t) = v1 cos(ω1 t + ϕ1 ) and = v2 cos(ω2 t + ϕ2 ), θ(t) θ(t) for some arbitrary constants ϕ1 and ϕ2 .

(2.57)

(2.58)

Problem 6

29

2.6 Problem 6 Problem: A mass-m particle moves along the x-axis subject to the potential energy V (x) = λ[(x2 − a2 )2 + 2ax3 ], where λ and a are positive constants: (a) Find all equilibrium points and determine which are stable. (b) Find the angular frequency of small oscillations about the point with the lowest energy.

Solution: (a) Equilibrium points are the stationary points of the potential, i.e. the points that satisfy

∂V = λ 2(x2 − a2 ) · 2x + 6ax2 = 0 ∂x =⇒ 2x(2x − a)(x + 2a) = 0.

(2.59)

Hence, there are three stationary points: x = 0, x = a/2, and x = −2a. Whether these are stable depend on the second derivative of the potential:  2  −4λa for x = 0 ∂2V 2 2 2 = λ(12x − 4a + 12ax) = (2.60) 5λa for x = a2  ∂x2 2 20λa for x = −2a.

The first of these is negative, so x = 0 is an unstable equilibrium point, while the others are positive, so x = a/2 and x = −2a are stable equilibrium points. (b) By the second derivative test, the point with the lowest energy is either x = a/2 or x = −2a. Direct calculation reveals

13 4 λa > −7λa4 = V (−2a), (2.61) 16 so x = −2a is the point with the lowest energy. To find the frequency of small oscillations about this point, let us change coordinates to η := x + 2a.

Equilibrium occurs at η = 0. The kinetic energy is simply V (a/2) =

1 1 mẋ2 = mη̇ 2 , 2 2 while Taylor expansion of the potential gives T =

(2.62)

∂V 1 ∂2V (x + 2a) + (x + 2a)2 + O((x + 2a)3 ) ∂x x=−2a 2 ∂x2 x=−2a 1 = −7λa4 + 0 + 20λa2 η 2 + O(η 3 ).

(2.63) 2 Using the general procedure outlined in Section 2.3, we recognize the matrices

V (x) = V (−2a) +

Classical Mechanics

30

M 0 = [m] and K = [20λa2 ] from the above equations. The frequency ω of small oscillations about η = 0 then satisfies det(K − ω 2 M 0 ) = 0,

(2.64)

giving r ω=

20λa2 . m

(2.65)

In this simple case where the system is one-dimensional, it is easy to obtain this same result by recognizing the Lagrangian 1 1 mη̇ 2 − 20λa2 η 2 + O(η 3 ), (2.66) 2 2 is, for small enough η, the Lagrangian of a simple harmonic oscillator with the above frequency.

L=T −V =

2.7 Problem 7 Problem: A uniform thin rod of length ℓ and mass m is suspended from fixed points A and B by two identical massless springs with zero natural length and spring constant k. At equilibrium (as shown in the figure), the angle between the springs and the vertical is ϕ and the length of the springs is a. All motion is restricted to this vertical plane. (a) Construct a Lagrangian for the system and obtain the Euler-Lagrange equations. (b) Find the normal modes of small oscillations of the system about equilibrium and their angular frequencies. (c) Construct the Hamiltonian, obtain Hamilton’s equations and show equivalence to the Euler-Lagrange equations.

Solution: (a) Let us use cartesian coordinates x = (x, y)T to denote the position of the center of the rod, with origin such that equilibrium occurs at (x, y) = (0, 0), and also let θ be the angle the rod makes with the horizontal. These will be our generalized coordinates, depicted in Fig. 2.5. With these coordinates the kinetic energy takes on a simple form, for it is the sum of the translational kinetic energy of the rod and the rotational kinetic energy about its center of mass: 1 1 T = m(ẋ2 + ẏ 2 ) + I θ̇2 , (2.67) 2 2

Problem 7

31

y A

B `/2 xL

t

Figure 2.5

x

`/2

xR

θ x

Diagram showing our generalized coordinates x and θ for the system of Problem 2.7. where I = mℓ2 /12 is the moment of inertia of the rod about its center of mass. The potential energy of the system is more involved. It is the sum of the potential energy stored in the each spring, and also the gravitational potential energy of the rod. We need to find the length of each spring in terms of x, y, θ, and the given constants. Now the left endpoint of the rod will be at position T

ℓ ℓ , (2.68) xL = x − cos θ, y − sin θ 2 2 and the right at position xR =

T ℓ ℓ x + cos θ, y + sin θ .

2 2

(2.69)

Meanwhile, the positions of the fixed endpoints of the left and right springs (A and B, say) are

T ℓ xA = − − a sin ϕ, a cos ϕ , 2

T ℓ xB = + a sin ϕ, a cos ϕ , (2.70) 2 as may be ascertained from the geometry of the equilibrium position. The potential energy will then be 1 1 k|xL − xA |2 + k|xR − xB |2 + mgy 2 2

2 2 1 ℓ ℓ ℓ = k x − cos θ + + a sin ϕ + y − sin θ − a cos ϕ 2 2 2 2

2 2 ! ℓ ℓ ℓ + x + cos θ − − a sin ϕ + y + sin θ − a cos ϕ + mgy 2 2 2

V =

(2.71) The identity (a + b)2 + (a − b)2 = 2(a2 + b2 ) is useful here, ultimately leading

Classical Mechanics

32

to V =

1 ℓ 2 k 2x2 + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ + mgy, 2 2

(2.72)

where inconsequential constant terms have been discarded. In total, our Lagrangian for the system is

1 1 2 2 2 2 L = T − V = m ẋ + ẏ + ℓ θ̇ 2 12

1 ℓ 2 2 − k 2x + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ − mgy.

2 2 (2.73) The Euler-Lagrange equations for the x coordinate are d ∂L ∂L − dt ∂ ẋ ∂x = mẍ + 2kx,

0=

(2.74)

for the y coordinate are 0=

d ∂L ∂L − dt ∂ ẏ ∂y

= mÿ + 2k(y − a cos ϕ) + mg, and for the θ coordinate are d ∂L ∂L − 0= dt ∂ θ̇ ∂θ

1 ℓ 2 = mℓ θ̈ + kℓ + a sin ϕ sin θ.

12 2

(2.75)

(2.76)

(b) Let us find the equilibrium points of this system, where the gradient of the potential energy vanishes:  ∂V      2kx 0 ∂x  ∂V  =  2k(y − a cos ϕ) + mg  =  0  .

(2.77) ∂y

ℓ ∂V kℓ + a sin ϕ sin θ 0 2 ∂θ Now by definition of our coordinates, (x, y, θ) = (0, 0, 0) is an equilibrium point.

The middle equation then amounts to a relation between given constants: −2ka cos ϕ + mg = 0.

(2.78)

We see that the equilibrium points are those such that x = 0, y = 0, and sin θ = 0. Whether they are stable depends on the matrix of second derivatives  ∂2V    ∂2V ∂2V 2k 0 0 ∂x2 ∂x∂y ∂x∂θ 2 2  ∂2V  ∂ V .

K(x, y, θ) ≡  ∂y∂x ∂∂yV2 =  0 2k 0 ∂y∂θ 

ℓ ∂2V ∂2V ∂2V 0 0 kℓ 2 + a sin ϕ cos θ 2 ∂θ∂x

∂θ∂y

∂θ

(2.79)

Problem 7

33

At an equilibrium point we have sin θ = 0 and so cos θ = ±1. Now for stability, K needs to be positive-definite, which in this case means all entries on the diagonal are positive. This requires cos θ = +1, so in fact the only stable equilibrium point is (x, y, θ) = (0, 0, 0). Furthermore, we see from Eq. 2.79 that our Lagrangian is already diagonal up to second order in our coordinates:

1 1 1 1 L= mẋ2 − 2kx2 + mẏ 2 − 2ky 2 2 2 2 2

1 1 1 ℓ + mℓ2 θ̇2 − kℓ + a sin ϕ θ2 + constant + O(θ3 ).

(2.80) 2 12 2 2 That is, it is the Lagrangian of three non-interacting harmonic oscillators of frequencies s r r

kℓ 2ℓ + a sin ϕ 2k 2k , ωy = , and ωθ = , (2.81) ωx = 1 2 m m 12 mℓ so the three normal modes of oscillation are     x(t) cos(ωx t + ϕx )  y(t)  = Cx  , 0 θ(t) 0     x(t) 0  y(t)  = Cy  cos(ωy t + ϕy )  , θ(t) 0     x(t) 0 . and  y(t)  = Cθ  0 θ(t) cos(ωθ t + ϕθ )

(2.82)

(c) To construct the Hamiltonian, we must calculate the conjugate momenta ∂L = mẋ, ∂ ẋ ∂L py := = mẏ, and ∂ ẏ ∂L 1 mℓ2 θ̇. pθ := = 12 ∂ θ̇

px :=

(2.83)

Then the Hamiltonian is H := px ẋ + py ẏ + pθ θ̇ − L

1 2 2 1 2 2 = m ẋ + ẏ + ℓ θ̇ 2 12

1 ℓ 2 2 + k 2x + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ + mgy.

2 2

(2.84)

To make sure this is a function of position and momentum coordinates only,

Classical Mechanics

34

we substitute out the generalized velocities using Eq. 2.83, yielding H=

p2y p2x p2 + + 6 θ2 2m 2m mℓ

ℓ 1 2 2 + k 2x + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ + mgy.

2 2

(2.85)

Now half of Hamilton’s equations of motion just reproduce the relations in Eq. 2.83: ∂H px = , ∂px m ∂H py ẏ = = , ∂py m pθ ∂H = 12 2 , θ̇ = ∂pθ mℓ

ẋ =

(2.86)

while the other half of the equations are ∂H = −2kx, ∂x ∂H p˙y = − = −2k(y − a cos ϕ) − mg, ∂y

∂H ℓ p˙θ = − = −kℓ + a sin ϕ sin θ. ∂θ 2

p˙x = −

(2.87)

Clearly, these contain the same content as the Euler-Lagrange equations Eqs. (2.74), (2.75), and (2.76).

2.8 Problem 8 Problem: 2 4 Show that the Lagrangian L = q1 q̇2 − q2 q̇1 − a q1 − q2 is invariant under the transformations

q1 1 + s −s q1 → q2 s 1−s q2 and construct the corresponding conserved quantity.

Solution: Under the given transformation, we have q1 q̇2 − q2 q̇1 → ((1 + s)q1 − sq2 )(sq̇1 + (1 − s)q̇2 )

− (sq1 + (1 − s)q2 )((1 + s)q̇1 − sq̇2 )

= (1 + s)sq1 q̇1 + (1 − s2 )q1 q̇2 − s2 q2 q̇1 − s(1 − s)q2 q̇2

− s(1 + s)q1 q̇1 + s2 q1 q̇2 − (1 − s2 )q2 q̇1 + s(1 − s)q2 q̇2

Problem 9

35

= q1 q̇2 − q2 q̇1 ,

(2.88)

and q1 − q2 → (1 + s)q1 − sq2 − (sq1 + (1 − s)q2 ) = q1 − q2 .

(2.89)

It follows that the Lagrangian is invariant under these transformations. Now the identity transformation is at the point s = 0, so the conserved quantity corresponding to this symmetry is given by Eq. (2.2.13) in the text, C =η·

∂L , ∂ η̇

(2.90)

where d ds

q1′ q2′

d = ds

1 + s −s s 1−s

η :=

s=0

q1 q2

= s=0

q1 − q2 q1 − q2

.

(2.91)

Explicitly, C = (q1 − q2 )[2(q1 q̇2 − q2 q̇1 )(−q2 )] + (q1 − q2 )[2(q1 q̇2 − q2 q̇1 )(q1 )] = 2(q1 q̇2 − q2 q̇1 )(q1 − q2 )2 .

(2.92)

It is straightforward but tedious to check that C is indeed conserved, i.e. dC/dt = 0, when the equations of motion are satisfied.

2.9 Problem 9 Problem: A system with a single degree of freedom is governed by the Lagrangian L = βt 1 e ( 2 mq̇ 2 − 21 kq 2 ): (a) Write down the equation of motion. What sort of motion does it describe?; (b) Show that the explicit time dependence of L can be removed by introducing a new coordinate variable Q = eβt/2 q; and (c) Construct the first integral whose conservation is implied by this property.

Solution: (a) The equation of motion is d ∂L ∂L − dt ∂ q̇ ∂q d βt = (e mq̇) − (−eβt kq) dt

0=

Classical Mechanics

36

= βeβt mq̇ + eβt mq̈ + eβt kq =⇒

0 = mq̈ + mβ q̇ + kq.

(2.93)

This is recognized as the equation of motion of a damped harmonic oscillator. (b) If we define Q = eβt/2 q, then we have Q̇ =

β βt/2 e q + eβt/2 q̇, 2

q̇ = e

−βt/2

(2.94)

or

β Q̇ − Q .

2

In these new coordinates, the Lagrangian becomes !

2 β 1 −βt 2 1 −βt βt me Q̇ − Q − ke Q L=e 2 2 2

1 1 1 mβ 2 2 = mQ̇ − mβ Q̇Q − k− Q2 .

2 2 2 4

(2.95)

(2.96)

The explicit time dependence has vanished. (c) As the Lagrangian in Eq. 2.96 has no explicit time dependence, the energy function ∂L −L h := Q̇ ∂ Q̇

1 1 1 1 mβ 2 2 = Q̇ mQ̇ + mβQ − mQ̇ + mβ Q̇Q + k− Q2 2 2 2 2 4

1 mβ 2 1 k− Q2 (2.97) = mQ̇2 + 2 2 4 will be conserved over time. Note, however, that since the system is not natural, h is not equal to the energy of the system. The energy of a damped harmonic oscillator will not be conserved over time.

2.10 Problem 10 Problem: A particle of mass m moves in one dimension in a potential V (q) = −Cqe−λq where C and λ are positive constants. Write down the Lagrangian and find the point of stable equilibrium. In the small-oscillation approximation find the angular frequency of the normal mode.

Solution:

Problem 11

37

With the given potential, the Lagrangian is 1 2 1 mq̇ − V = mq̇ 2 + Cqe−λq .

(2.98) 2 2 Points of stable equilibrium are local minima of the potential energy. These satisfy L=

dV = −Ce−λq + Cλqe−λq = 0 dq

(2.99)

d2 V = Cλe−λq + Cλe−λq − Cλ2 qe−λq > 0. dq 2

(2.100)

and

There is only one solution to Eq. 2.99, given by q = λ−1 . At this point the second derivative Eq. 2.100 is equal to Cλ/e > 0, so this is indeed a point of stable equilibrium.

Just as in Problem 2.6, we solve for the small perturbations about this minimum by expanding the Lagrangian in terms of η := q − λ−1 . The kinetic energy is T =

1 2 1 mq̇ = mη̇ 2 , 2 2

(2.101)

while the potential is

∂V 1 ∂2V V λ−1 + η = V λ−1 + η 2 + O(η 3 ) η+ ∂q q=λ−1 2 ∂x2 q=λ−1 =−

C 1 Cλ 2 +0+ η + O(η 3 ). λe 2 e

(2.102)

We recognize the matrices M 0 = [m] and K = [Cλ/e] that dictate the frequency ω of small oscillations about η = 0 through the relation det(K − ω 2 M 0 ) = 0,

(2.103)

giving r ω=

Cλ . em

(2.104)

2.11 Problem 11 Problem: The Lagrangian of a particle of mass m and charge e in a uniform, static magnetic field B is given by L(x, ẋ) = 12 mẋ2 + (e/2m)L · B, where L = x × mẋ is the mechanical angular momentum: (a) Show that the Hamiltonian is H(x, p) = p̄2 /2m, where p̄ = p − (e/2)B × x; (b) Obtain the Poisson bracket result {p̄i , p̄j } = e ϵijk Bk ; and (c) Show that p̄ · B is conserved using Poisson brackets; and (d) Are components of p̄ perpendicular to B conserved?

Classical Mechanics

38

Solution: (a) In index notation, the Lagrangian is given by L=

1 e mẋi ẋi + εijk xi ẋj Bk , 2 2

(2.105)

where repeated indices are summed over and we do not distinguish between upper and lower indices. To find the Hamiltonian, we must first compute the conjugate momenta pn :=

∂L 1 1 e = mδin ẋi + mẋi δin + εijk xi δjn Bk ∂ ẋn 2 2 2 1 e 1 = mẋn + mẋn − εikn xi Bk 2 2 2 e = mẋn − (x × B)n , 2

(2.106)

or in vector notation, e p = mẋ − (x × B).

2

(2.107)

Then the Hamiltonian is H := p · ẋ − L e 1 e = mẋ2 − (x × B) · ẋ − mẋ2 − (x × ẋ) · B 2 2 2 1 e e 2 = mẋ − (ẋ × x) · B − (x × ẋ) · B 2 2 2 1 2 = mẋ , 2

(2.108)

where we have used the cyclic property of the scalar triple product in going from the second to the third line, and the antisymmetry of the cross product between the third and the fourth. But the Hamiltonian must be expressed in terms of position and momentum only, so we use Eq. 2.107 to substitute out the velocity and obtain

2 p + 2e (x × B) p̄2 1 ≡ , (2.109) H= m 2 m 2m where p̄ := p + 2e (x × B) = p − 2e (B × x). (b) Using the bilinearity of Poisson brackets, we can write e e {p̄i , p̄j } = {pi − εikl Bk xl , pj − εjmn Bm xn } 2 2 e e = {pi , pj } − εjmn Bm {pi , xn } − εikl Bk {xl , pj } 2 2 e2 + εikl εjmn Bk Bm {xl , xn }, 4

(2.110)

Problem 11

39

because the components of B are constant. Now we use the fundamental Poisson brackets Eq. (2.4.28), {xi , pj } = δij ,

{xi , xj } = 0 = {pi , pj },

and the antisymmetry of the brackets to simplify e e {p̄i , p̄j } = 0 − εjmn Bm (−δin ) − εikl Bk δlj + 0 2 2 e = (εjmi Bm − εikj Bk ) 2 = e εijk Bk .

(2.111)

(2.112)

Here, the total antisymmetry of εijk was used and we relabeled the dummy index m → k. (c) The time derivative of p̄ · B is given, from Eq. (2.4.24), by ∂(p̄ · B) d(p̄ · B) = {p̄ · B, H} + dt ∂t = {p̄ · B, H}.

(2.113)

Going back to index notation and using the product rule of the Poisson bracket, we calculate Bi d(p̄ · B) = {p̄i , p̄j p̄j } dt 2m Bi = ({p̄i , p̄j }p̄j + p̄j {p̄i , p̄j }) 2m Bi p̄j = {p̄i , p̄j } m Bi p̄j = e εijk Bk m e = p̄ · (B × B) m = 0.

(2.114) Thus, p̄ · B is conserved. (d) Those components of p̄ perpendicular to B are given by p̄⊥ i = p̄i −

(p̄ · B) Bi , B2

(2.115)

as may readily be verified by observing that p̄⊥ · B = 0. These evolve in time as ∂ p̄⊥ dp̄⊥ i i = {p̄⊥ i , H} + dt ∂t = {p̄⊥ i , H} Bi = {p̄i , H} − 2 {p̄ · B, H} B

Classical Mechanics

40

e εijk p̄j Bk − 0 m e = (p̄ × B)i , m =

(2.116)

using the results of Part (c). Generically this time derivative is non-zero, showing the components of p̄ perpendicular to B are not necessarily conserved, as might reasonably be expected from the Lorentz force law. In general, the perpendicular components are all conserved only when all of them vanish.

2.12 Problem 12 Problem: 2

(a) Show that the relativistic Lagrangian L = −(1/γ)mc2 eU/c reduces to the Newtonian Lagrangian for a particle of mass m in a gravitational potential U in the slow motion, weak field limit v ≪ c, |U | ≪ c2 . (b) Obtain the Euler-Lagrange equations of motion, and show that they can be written in the relativistic form dV µ /dτ = (g µν − (1/c2 )V µ V ν )∂ν U , where V is the four-velocity and τ is the proper time of the particle. Verify that not all four component equations are independent.

Solution: (a) We find the slow motion, weak field limit of the Lagrangian by Taylor expanding the various factors in terms of v/c and U/c2 , and keeping only the first-order terms: L=−

mc2 U/c2 e γ r

v 2 U/c2 e c2

4 2 1 v2 v U U 2 = −mc 1 − +O 1+ 2 +O 2 c2 c4 c c4 2 1 v U = −mc2 + mv 2 − mU + O .

2 c4 = −mc2

1−

(2.117)

Ignoring the higher-order terms, and also the constant −mc2 , which does not affect the equations of motion, we have L=

1 mv 2 − mU = T − V, 2

(2.118)

where V := mU is the Newtonian gravitational potential energy. This is the standard Newtonian Lagrangian for a particle of mass m in a gravitational field.

Problem 12

41

(b) To find the equations of motion, it is first helpful to calculate ! r

d 1 γv i d 1 2v i v2 q = − . = = − 1 − 2 dv i γ dv i c2 c2 c2 2 1 − vc2

(2.119)

Then the equations of motion are d ∂L ∂L − i i dt ∂v ∂x

mc2 U/c2 1 ∂U d −γv i U/c2 2 e − − = −mc e dt c2 γ c2 ∂xi

2 2 d 0=γ γv i eU/c + eU/c ∂i U. (2.120) dt 0=

=⇒

Recall the 4-velocity is given by V µ = (γc, γv), so γv i = V i . Also recall that γ≡

dt , dτ

so

γ

d d = . dt dτ

With these, the equations of motion Eq. 2.120 become 2 d i U/c2 0= V e + (∂i U )eU/c dτ 2 1 dU 2 dV i U/c2 = e + V i eU/c 2 + (∂i U )eU/c dτ c dτ V i dU dV i + 2 + ∂i U. =⇒ 0= dτ c dτ

(2.121)

(2.122)

To simplify further, note that dU ∂U dxν = = (∂ν U )V ν , dτ ∂xν dτ

(2.123)

∂i U = −(−δ iν )∂ν U = −g iν ∂ν U,

(2.124)

and that

where g µν is the Minkowski metric tensor. We are following the convention that Greek letters denote spacetime indices, while Latin letters denote only spatial indices. With these identities, Eq. 2.122 reduces to

dV i 1 i ν iν 0= − g − 2V V ∂ν U. (2.125) dτ c The tensor on the right-hand side of this equation evidently has vanishing spatial components in all inertial frames. The only tensor with this property is the zero tensor, so the above equation must also remain true if we replace i → µ. We arrive at the desired equations of motion

dV µ 1 µ ν µν = g − 2V V ∂ν U. (2.126) dτ c Now we deduced the µ = 0 equation from the three spatial equations, so it

Classical Mechanics

42

is clear that not all of these equations are independent. To see this explicitly, consider the fact that V µ Vµ is the constant c2 . This implies 0=

1 d dV 0 dV i dV µ (V µ Vµ ) = Vµ = V0 + Vi , 2 dτ dτ dτ dτ

(2.127)

leading to dV 0 Vi dV i =− , dτ V0 dτ

(2.128)

and therefore showing the equations are not independent. Checking that we do indeed arrive at the desired temporal equation, we substitute in the known spatial equations and use the fact that there is no numerical difference between covariant and contravariant temporal components of a tensor:

Vi 1 i ν dV 0 iν ∂ν U =− g − 2V V dτ γc c

1 1 i i ν =− Vi ∂ U − 2 Vi V V ∂ν U γc c

1 1 ν 0 i ν =− V ∂ν U − V ∂0 U − 2 Vi V V ∂ν U γc c

1 1 = ∂0 U − 1 − 2 Vi V i V ν ∂ν U γc c

1 1 = ∂0U − 1 − 2 (Vµ V µ − V0 V 0 ) V ν ∂ν U γc c

1 1 = g 0ν ∂ν U − 1 − 2 (c2 − γcV 0 ) V ν ∂ν U γc c

1 (2.129) = g 0ν − 2 V 0 V ν ∂ν U, c which is the desired form.

2.13 Problem 13 Problem: Consider a differentiable function V (q) of a generalized coordinate q. Consider the Lagrangian L = (1/12)m2 q̇ 4 + mq̇ 2 V (q) − V 2 (q). Show that this system is equivalent to a much simpler Lagrangian for a particle moving in a potential.

Solution: Two Lagrangians are equivalent if their equations of motion have the same solutions. The equation of motion of the given Lagrangian is 0=

d ∂L ∂L − dt ∂ q̇ ∂q

Problem 14

43

d 1 2 3 dV dV m q̇ + 2mq̇V − mq̇ 2 − 2V dt 3 dq dq

dV dq dV dV = m2 q̇ 2 q̈ + 2mq̈V + 2mq̇ − mq̇ 2 − 2V dq dt dq dq

1 2 dV mq̇ + V , = 2 mq̈ + dq 2

=

(2.130)

where we have made extensive use of the chain rule. For the above equation to be satisfied, at least one of the following equations must hold: dV = 0, dq

(2.131)

1 2 mq̇ + V = 0.

2

(2.132)

mq̈ +

Suppose the second equation holds. Then also the first equation holds, for

d 1 2 mq̇ + V = 0 dq 2 dV dq̇ dt + =0 mq̇ dt dq dq dV =⇒ mq̈ + = 0.

(2.133) dq In other words, the motion of the system is entirely dictated by the equation mq̈ = −

dV . dq

(2.134)

But this is the equation of motion of the more familiar Lagrangian L′ :=

1 2 mq̇ − V (q) 2

(2.135)

for a particle moving in a potential V , and therefore the two systems are equivalent.

2.14 Problem 14 Problem: (a) A particle with rest mass M phas kinetic energy T . Show that the magnitude of its momentum is p = (1/c) T (T + 2M c2 ). (b) A particle A decays at rest into a lighter particle B and a photon γ. Use fourvector methods to determine the energies of B and γ in terms of the rest masses mA and mB of A and B. Solution:

Classical Mechanics

44

(a) The magnitude of the particle’s momentum is given by p = γM v ≡ q

Mv 2

1 − vc2

,

(2.136)

while the particle’s kinetic energy is equal to the total energy, minus the rest mass energy: T = γM c2 − M c2 = (γ − 1)M c2 .

(2.137)

Performing some algebra, we find

T (T + 2M c2 ) = (γ − 1)M c2 (γ + 1)M c2 = (γ 2 − 1)M 2 c4

1 = 1 − 2 γ 2 M 2 c4 γ

v2 = 1− 1− 2 γ 2 M 2 c4 c = γ 2 M 2 v 2 c2 = p 2 c2 p T (T + 2M c2 ) , p= c

=⇒

(2.138)

as desired. (b) Denote the four-momenta of the three particles by pA , pB , and pγ . By the conservation of four-momentum, we have pA = pB + pγ .

(2.139)

Consider now the quantity p2γ = (pA − pB )2

= p2A − 2pA · pB + p2B .

(2.140)

Recall the following properties of a particle’s four-momentum vector conveyed

by Eq. (2.6.4) in the text: (i) The temporal component is equal to the particle’s energy, divided by c. (ii) The length of the vector is equal to the particle’s mass, multiplied by c.

The above equation therefore reduces to 0 = m2A c2 − 2 (mA c)

EB c

+ m2B c2 ,

(2.141)

where we have used the fact that in the rest frame of particle A, its fourmomentum is pA = (EA /c, 0)T = (mA c, 0)T . Rearranging, we find the energy of particle B EB =

m2A + m2B 2 c .

2mA

(2.142)

Problem 15

45

γ1 y

θ1 e+

t

Figure 2.6

e−

x θ2

γ2

Geometry of the e+ + e− → 2γ process in the rest frame of the electron. Of course, the electron and positron will not exist at the same times as the two photons.

Similarly, the energy of the photon is found through the following calculation: p2B = (pA − pγ )2

= p2A − 2pA · pγ + p2γ

Eγ 2 2 2 2 +0 mB c = mA c − 2 (mA c) c m2 − m2B 2 =⇒ Eγ = A c .

2mA

(2.143)

Note that for this decay process to be physical we must have Eγ > 0, which enforces mB < mA .

2.15 Problem 15 Problem: A relativistic positron e+ with kinetic energy T annihilates a stationary electron − e and produces two photons e+ + e− → γ + γ: (a) At what angles are the photons emitted when they have equal energies?. (b) Is it possible for the photons to have unequal energies? Explain.

Solution: (a) We will use the conservation of four-momentum to answer this question. In the rest frame of the electron, the four-momenta of the initial particles are T T

M c2 E e− , 0 = , 0 , pe− = c c

Classical Mechanics

46

pe+ =

Ee+ , p, 0, 0 c

T =

T + M c2 , c

p

T (T + 2M c2 ) , 0, 0 c

!T ,

(2.144)

where we have taken the +x-direction to be the direction of motion of the positron, and we have used the result of Problem 2.14 (a) to express the momentum of the positron in terms of its kinetic energy. The total four-momentum is then !T p T (T + 2M c2 ) T + 2M c2 , , 0, 0 , (2.145) ptot = pe− + pe+ = c c and this must be equal to the total four-momentum of the two photons. Now it is easy to see that the three-momenta of the photons and of the positron all lie in a single plane: the momentum vectors of the two photons define a plane, and the momentum of the positron in the direction normal to this plane must vanish in order for momentum to be conserved. For convenience, we shall take this to be the x − y plane, as illustrated in Fig. 2.6. Say the two photons make respective angles θ1 and θ2 to the x-axis. Then under the assumption the photons have equal energies Eγ , their four-momenta will be Eγ T (1, cos θ1 , sin θ1 , 0) , c Eγ T pγ2 = (1, cos θ2 , sin θ2 , 0) . c

pγ1 =

(2.146)

(The magnitude of a photon’s three-momentum is always Eγ /c). Therefore, the total four-momentum may equivalently be written ptot = pγ1 + pγ2 =

Eγ T (2, cos θ1 + cosθ2 , sin θ1 + sin θ2 , 0) . c

(2.147)

Comparing with Eq. 2.145, we get the following relations: 2Eγ = T + 2M c2 , p Eγ (cos θ1 + cosθ2 ) = T (T + 2M c2 ), sin θ1 + sin θ2 = 0.

(2.148) (2.149) (2.150)

The last of these tells us θ1 = −θ2 ≡ θ. Then combining this with the first two, we get p 2Eγ cos θ = (T + 2M c2 ) cos θ = T (T + 2M c2 ).

(2.151) Hence, the photons make angles −1

θ = cos

r

T T + 2M c2

!

on either side of the line of motion of the positron.

(2.152)

47

Problem 15

(b) Consider now the collision in the center-of-mass frame of the electron and positron, so the magnitude of the total three-momentum is zero. To conserve momentum, the photons must therefore be directed oppositely with equal energies:

T Eγ pγ1 = ,p , c

T Eγ pγ2 = , −p .

(2.153) c Clearly, any direction p is consistent with the requirement of energy and momentum conservation, and is therefore allowed. Now to return to our original frame in part (a), let us boost by some speed v in the x-direction so that the electron becomes stationary. Four-momenta transform as tensors, so in this boosted frame the temporal components of the photons’ four-momenta will be

Eγ′ 1 Eγ v =γ − px , c c c

Eγ′ 2 Eγ v =γ − (−px ) .

(2.154) c c c These energies are not equal in general, so it is possible for the photons to have unequal energies in the stationary frame of the electron.

3

Relativistic classical fields 3.1 Problem 1 Problem: √ We define a free complex scalar field as ϕ(x) = (1/ 2)[ϕ1 (x) + iϕ2 (x)] ∈ C in terms of two real free scalar fields ϕ1 (x), ϕ2 (x) ∈ R and the Lagrangian density as the sum of the two free Lagrangian densities. (a) Show that the Lagrangian density in terms of ϕ(x) and ϕ∗ (x) is as given in Eq. (3.2.48), L = (∂µ ϕ)∗ (∂ µ ϕ) − m2 ϕ∗ ϕ. Using the Euler-Lagrange equations for ϕ1 and ϕ2 show that the complex scalar field and its complex conjugate satisfy (∂µ ∂ µ + m2 )ϕ(x) = (∂µ ∂ µ + m2 )ϕ∗ (x) = 0.

R (b) Using the action S[ϕ] = d4 x L and the Wirtinger calculus results in Sec. A.2.5, show that these equations of motion follow from Hamilton’s Principle, δS[ϕ]/δϕ∗ (x) = δS[ϕ]/δϕ(x) = 0. (c) Using Eq. (3.1.15) evaluate the conjugate momenta π(x) = ∂L/∂(∂0 ϕ)(x) and π ∗ (x) = ∂L/∂(∂0 ϕ∗ )(x). Use these R results Rto construct the Hamiltonian using Eq. (3.1.16) and show that H = d3 x H = d3 x [π ∗ (x)π(x) + ∇ϕ∗ (x) · ∇ϕ(x) + m2 ϕ∗ (x)ϕ(x)]. (d) Obtain Hamilton’s equations using Eq. (3.1.21) in terms of ϕ and ϕ∗ and show that these reproduce the Klein-Gordon equations for ϕ and ϕ∗ . (e) The complex scalar field Lagrangian in terms of ϕ1 and ϕ2 has a well-defined Legendre transform since the Hessian matrix ML is positive definite. This follows from the straightforward generalization of Eq. (3.1.48), (ML )ij (x, y)|x0 =y0 =

δ2 L = δij δ 3 (x − y) , δ ϕ̇i (x)δ ϕ̇j (y) x0 =y0

(3.1)

for i, j = 1, 2. Show that the Wirtinger calculus version of the nonsingular Hessian matrix has the four elements δ 2 L/δ ϕ̇(x)δ ϕ̇(y) δ 2 L/δ ϕ̇(x)δ ϕ̇∗ (y)

Solution: 48

x0 =y 0 x0 =y 0

= δ 2 L/δ ϕ̇∗ (x)δ ϕ̇∗ (y) = δ 2 L/δ ϕ̇∗ (x)δ ϕ̇(y)

x0 =y 0

x0 =y 0

= 0,

= δ 3 (x − y) .

(3.2)

49

Problem 1

(a) For two free real scalar fields ϕ1 and ϕ2 of mass m, the Lagrangian density is

1 1 1 1 L= (∂µ ϕ1 )(∂ µ ϕ1 ) − m2 ϕ21 + (∂µ ϕ2 )(∂ µ ϕ2 ) − m2 ϕ22 2 2 2 2 1 1 = [(∂µ ϕ1 )(∂ µ ϕ1 ) + (∂µ ϕ2 )(∂ µ ϕ2 )] − m2 (ϕ21 + ϕ22 ).

(3.3) 2 2 √ √ Now with the definitions ϕ := (ϕ1 + iϕ2 )/ 2 and ϕ∗ := (ϕ1 − iϕ2 )/ 2, we note the identity

ϕ1 − iϕ2 ϕ1 + iϕ2 1 ∗ √ √ ϕ ϕ= = (ϕ21 + ϕ22 ), (3.4) 2 2 2 and similarly for the derivatives,

µ

∂µ ϕ1 − i∂µ ϕ2 ∂ ϕ1 + i∂ µ ϕ2 √ √ (∂µ ϕ)∗ (∂ µ ϕ) = 2 2 1 = [(∂µ ϕ1 )(∂ µ ϕ1 ) + (∂µ ϕ2 )(∂ µ ϕ2 )] .

2

(3.5)

From these it follows that the Lagrangian density can equivalently be written as L = (∂µ ϕ)∗ (∂ µ ϕ) − m2 ϕ∗ ϕ.

(3.6)

Looking back to the first expression of the Lagrangian density, and using the identities ∂(∂µ ϕi ) = δ ν µ δ ji , ∂(∂ν ϕj )

∂(∂ µ ϕi ) = g µν δ ji , ∂(∂ν ϕj )

(3.7)

the Euler-Lagrange equation for the field ϕ1 is seen to be

∂L ∂L 0 = ∂ν − ∂(∂ν ϕ1 ) ∂ϕ1

1 ν µ = ∂ν (δ µ ∂ ϕ1 + ∂µ ϕ1 g µν ) − (−m2 ϕ1 ) 2 = ∂ν ∂ ν ϕ1 + m2 ϕ1 = (∂ 2 + m2 )ϕ1 .

(3.8)

The equation for ϕ2 is then obviously (∂ 2 + m2 )ϕ2 = 0, because the Lagrangian density is symmetric with respect to ϕ1 and ϕ2 . Now we can combine these two equations to see

1 0 = √ (∂ 2 + m2 )ϕ1 ± i(∂ 2 + m2 )ϕ2 2 ϕ1 ± iϕ2 = (∂ 2 + m2 ) √ , 2 or in other words, (∂ 2 + m2 )ϕ = 0 and (∂ 2 + m2 )ϕ∗ = 0.

(3.9)

Relativistic classical fields

50

(b) Hamilton’s Principle that the equations of motion will be satisfied when δS = 0 under variations of ϕ and of ϕ∗ will lead to the Euler-Lagrange equations of motion, where ϕ and ϕ∗ are the fields:

∂L ∂L ∂L ∂L .

(3.10) 0 = ∂ν − and 0 = ∂ν − ∂(∂ν ϕ) ∂ϕ ∂(∂ν ϕ∗ ) ∂ϕ∗ The Wirtinger calculus results tell us that these derivatives, with ϕ and ϕ∗ defined as above, act in the expected way as long as ϕ and ϕ∗ are thought of as completely independent fields. So the first of these Euler-Lagrange equations, for example, gives by Eq. 3.6

∂L ∂L 0 = ∂ν − ∂(∂ν ϕ) ∂ϕ = ∂ν (∂ µ ϕ)∗ δ νµ − (−m2 ϕ∗ )

= (∂ 2 + m2 )ϕ∗ ,

(3.11)

and the second gives

0 = ∂ν

∂L ∂(∂ν ϕ∗ )

−

∂L ∂ϕ∗

= ∂ν (∂ ν ϕ) − (−m2 ϕ) = (∂ 2 + m2 )ϕ,

(3.12)

where we have used the fact that (∂µ ϕ∗ ) = (∂µ ϕ)∗ . We see that we have derived the equations of motion from Part (a) by using Hamilton’s Principle, with ϕ and ϕ∗ as the variational fields. (c) Using Eq. 3.6, we calculate π=

∂L ∂L = ∂ 0 ϕ∗ , and π ∗ = = ∂ 0 ϕ. ∂(∂0 ϕ) ∂(∂0 ϕ∗ )

(3.13)

Note that since ∂ 0 = ∂0 , we can also write π = ∂0 ϕ∗ and π ∗ = ∂0 ϕ. The Hamiltonian density is then by definition H := π∂0 ϕ + π ∗ ∂0 ϕ∗ − L

= ππ ∗ + π ∗ π − g µν (∂µ ϕ)∗ (∂ν ϕ) − m2 ϕ∗ ϕ

= 2ππ ∗ − g 00 (∂0 ϕ)∗ (∂0 ϕ) − g ij (∂i ϕ)∗ (∂j ϕ) + m2 ϕ∗ ϕ = 2ππ ∗ − (∂0 ϕ)∗ (∂0 ϕ) + δ ij (∂i ϕ)∗ (∂j ϕ) + m2 ϕ∗ ϕ = 2ππ ∗ − ππ ∗ + δ ij (∇i ϕ)∗ (∇j ϕ) + m2 ϕ∗ ϕ = ππ ∗ + (∇ϕ∗ ) · (∇ϕ) + m2 ϕ∗ ϕ.

(3.14)

Here we have used the fact that ∂µ = (∂0 , ∇)T . Or in terms of the full Hamiltonian, Z Z

3 H = d x H(x) = d3 x π(x)π ∗ (x) + (∇ϕ∗ (x)) · (∇ϕ(x)) + m2 ϕ∗ (x)ϕ(x) .

(3.15)

Problem 1

51

(d) With the Hamiltonian density Eq. 3.14, the Hamilton’s equations that are relevant to us are ∂H = π ∗ , and ∂π ∂H ∂H ∂0 π = − +∇· = −m2 ϕ∗ + ∇ · (∇ϕ∗ ), ∂ϕ ∂(∇ϕ) ∂0 ϕ =

(3.16)

and of course the conjugate counterparts ∂H = π, and ∂π ∗ ∂H ∂H ∂0 π ∗ = − ∗ + ∇ · = −m2 ϕ + ∇ · (∇ϕ). ∂ϕ ∂(∇ϕ∗ ) ∂0 ϕ∗ =

(3.17)

Combining these two sets of equations yields ∂0 ∂0 ϕ = ∂0 π ∗ = −m2 ϕ + ∇ · (∇ϕ)

=⇒

(∂0 ∂ 0 − ∇2 + m2 )ϕ = 0, (3.18)

and similarly ∂0 ∂0 ϕ∗ = ∂0 π = −m2 ϕ∗ + ∇ · (∇ϕ∗ )

=⇒

(∂0 ∂ 0 − ∇2 + m2 )ϕ∗ = 0.

(3.19)

These are simply the Klein-Gordon equations (∂ 2 +m2 )ϕ = 0 and (∂ 2 +m2 )ϕ∗ = 0 that we derived previously. (e) For a functional of the form F [σ] := the property that

R

dx f (σ(x)), functional derivatives have

∂f (σ(x)) δF = . δσ(x) ∂σ(x)

(3.20)

Importantly, this holds as long as F is not a functional of the derivatives of σ.

Applied to the case at hand, we can identify F → L, f → L, and σ → ϕ̇(∗) ≡ ∂0 ϕ(∗) . Then, for example, δL ∂L(y) = δ ϕ̇(y) ∂ ϕ̇(y)

∂ ∗ = ϕ̇ (y) ϕ̇(y) + ∂i ϕ∗ (y) ∂ i ϕ(y) − m2 ϕ∗ (y)ϕ(y) ∂ ϕ̇(y) = ϕ̇∗ (y).

(3.21)

It follows that δ2 L δ ϕ̇∗ (y) δ2 L δ ϕ̇∗ (y) = = 0, and = = δ 4 (x − y). δ ϕ̇(x)ϕ̇(y) δ ϕ̇(x) δ ϕ̇∗ (x)ϕ̇(y) δ ϕ̇∗ (x) Similarly, we have δL ∂L(y) = ∗ δ ϕ̇ (y) ∂ ϕ̇∗ (y)

(3.22)

Relativistic classical fields

52

=

∂ ϕ̇∗ (y) ϕ̇(y) + ∂i ϕ∗ (y) ∂ i ϕ(y) − m2 ϕ∗ (y)ϕ(y) ∂ ϕ̇∗ (y)

= ϕ̇(y),

(3.23)

and so δ2 L δ ϕ̇(y) δ2 L δ ϕ̇(y) = = δ 4 (x − y), and = = 0. (3.24) ∗ ∗ ∗ δ ϕ̇(x)ϕ̇ (y) δ ϕ̇(x) δ ϕ̇ (x)ϕ̇ (y) δ ϕ̇∗ (x) Restricted to equal-time spacetime points (x0 = y 0 ), these reduce to the desired equations with factors of δ 3 (x − y) (the common infinite factors of δ(x0 − y 0 ) = δ(0) should be understood to be implicit).

3.2 Problem 2 Problem: Let ϕ be a Lorentz-scalar field.

(a) Show that the quantities Tµν = ∂µ ϕ∂ν ϕ − 12 gµν (∂ϕ)2 − λ2 ϕ2 with λ ∈ R are components of a Lorentz (0, 2) tensor field. (b) Show that ∂µ T µν = 0 if ϕ(x) obeys the wave equation (∂ 2 + κ2 )ϕ = 0.

Solution: (a) First, we note that since ϕ is a Lorentz-scalar field, it transforms under a Lorentz transformation Λ as ϕ(x) → ϕ′ (x′ ) = ϕ(x).

(3.25)

Obviously ϕ2 will also be a Lorentz-scalar, transforming as ϕ2 (x) → (ϕ′ )2 (x′ ) = ϕ2 (x). From Eq. (1.2.134), the derivatives of ϕ will form a Lorentz four-vector, ′

∂µ ϕ(x) → ∂µ′ ϕ′ (x′ ) = (Λ−1 )µ µ ∂µ′ ϕ(x),

(3.26)

from which it follows that (∂ϕ)2 ≡ (∂µ ϕ)(∂ µ ϕ) is also a Lorentz-scalar. Then knowing that gµν is a (0, 2) tensor, we see that Tµν will transform as ′ Tµν (x) → Tµν (x′ ) ′

′

= (Λ−1 )µ µ ∂µ′ ϕ(x) (Λ−1 )ν ν ∂ν ′ ϕ(x)

′ ′ 1 − (Λ−1 )µ µ (Λ−1 )ν ν gµ′ ν ′ {(∂ϕ)2 (x) − λ2 ϕ2 (x)} 2

′ ′ 1 = (Λ−1 )µ µ (Λ−1 )ν ν ∂µ′ ϕ(x) ∂ν ′ ϕ(x) − gµ′ ν ′ {(∂ϕ)2 (x) − λ2 ϕ2 (x)} 2 ′

′

= (Λ−1 )µ µ (Λ−1 )ν ν Tµ′ ν ′ (x).

(3.27)

This is the transformation rule for a (0, 2) tensor, so Tµν are the components

Problem 3

53

of a (0, 2) tensor. (b) By direct calculation we have

1 ∂µ T µν = ∂µ ((∂ µ ϕ)(∂ ν ϕ)) − g µν ∂µ (∂ϕ)2 − λ2 ϕ2 2

1 2 ν µ = (∂ ϕ)(∂ ϕ) + (∂ ϕ)(∂µ ∂ ν ϕ) − ∂ ν (∂µ ϕ)(∂µ ϕ) − λ2 ϕ2 2

1 2 ν µ ν = (∂ ϕ)(∂ ϕ) + (∂ ϕ)(∂ ∂µ ϕ) − 2(∂ µ ϕ)(∂ ν ∂µ ϕ) − 2λ2 ϕ∂ ν ϕ 2

= ∂ 2 ϕ + λ2 ϕ ∂ ν ϕ.

(3.28) We see that ∂µ T µν will vanish if ϕ obeys the wave equation (∂ 2 + λ2 )ϕ = 0.

3.3 Problem 3 Problem: Consider n real equal-mass scalar fields ϕ1 , . . . , ϕn with a quartic interaction of the form Lint = −(λ/4!)(ϕ21 + · · · + ϕ2n )2 then

Pn Pn L = i=1 Lfree + Lint = i=1 12 ∂µ ϕi ∂ µ ϕi − 21 m2 ϕ2i − (λ/4!)(ϕ21 + · · · + ϕ2n )2 i = 12 ∂µ ϕT ∂ µ ϕ − 21 m2 ϕT ϕ − (λ/4!)(ϕT ϕ)2 .

(3.29)

(a) Obtain the Euler-Lagrange equations for this theory; (b) Explain why this theory is invariant under O(n) transformations. Derive the conserved Noether current and explain why it corresponds to the SO(n) subgroup and not O(n); and (c) For n complex equal-mass scalar fields write a corresponding Lagrangian density invariant under U (n). Obtain the Euler-Lagrange equations. Construct the conserved Noether current.

Solution: (a) Our fields are all independent, so we have ∂ϕi = δij , ∂ϕj

and

∂(∂µ ϕi ) = δ αµ δij . ∂(∂α ϕj )

(3.30)

Then since ∂µ ϕi ∂ µ ϕi = g µν ∂µ ϕi ∂ν ϕi , it follows that ∂ (∂µ ϕi ∂ν ϕi ) = g µν [(δ αµ δij )∂ν ϕi + ∂µ ϕi (δ αν δij )] ∂(∂α ϕj ) = 2∂ α ϕi δij .

With these, we calculate the Euler-Lagrange equations of motion as

∂L ∂L 0 = ∂α − ∂(∂α ϕj ) ∂ϕj

(3.31)

Relativistic classical fields

54

= ∂α

n X

! α

∂ ϕi δij

i=1

−

n X λ −m ϕi δij − 2(ϕ21 + · · · + ϕ2n ) 2ϕi δij 4! i=1 i=1 2

n X

!

λ 2 (ϕ + · · · + ϕ2n )ϕj .

(3.32) 3! 1 For each field ϕj this is simply the free equation of motion, with an added interaction term. = ∂ 2 ϕj + m2 ϕj +

(b) Inspection of Eq. 3.29 reveals the Lagrangian density, rather than depending on the individual field values, is actually a function of only the “lengths” in field space q p (3.33) |ϕ| := ϕT ϕ = ϕ21 + · · · + ϕ2n , and similarly for the derivatives. Any transformation of the fields that preserves these lengths is therefore a symmetry of the theory. But the group of lengthpreserving transformations acting on n elements is precisely O(n), so this theory is invariant under O(n). Algebraically, O(n) consists of those n×n matrices A with the property AT A = I, as seen in Section 2.1.1. Then under such a transformation ϕ → Aϕ,

∂µ ϕ → ∂µ (Aϕ) = A ∂µ ϕ,

(3.34)

the Lagrangian density transforms as 1 λ 1 (A ∂µ ϕ)T (A ∂ µ ϕ) − m2 (Aϕ)T (Aϕ) − [(Aϕ)T (Aϕ)]2 2 2 4!

1 1 λ = ∂µ ϕT (AT A)∂ µ ϕ − m2 ϕT (AT A)ϕ − [ϕT (AT A)ϕ]2 2 2 4!

1 2 T λ T 2 1 T µ = ∂µ ϕ ∂ ϕ − m ϕ ϕ − (ϕ ϕ) 2 2 4! = L.

L→

(3.35)

That is, the Lagrangian density is invariant under O(n) transformations, as was reasoned earlier.

This is an internal symmetry, so the associated conserved Noether current is given by Eq. (3.2.18) from the text, ⃗ j µ = ⃗π µ · Φ,

(3.36)

⃗ is the vector of field changes under an infinitesimal O(n) transformawhere Φ tion, and ⃗π is the vector of conjugate momenta

T ∂L ∂L T µ ⃗π := , ... , = (∂ µ ϕ1 , . . . , ∂ µ ϕn ) .

(3.37) ∂(∂µ ϕ1 ) ∂(∂µ ϕn ) ⃗ i.e. we must determine the behavior of ϕ under an infinitesIt remains to find Φ, imal O(n) transformation. Now we know from Section 2.1.1 that the matrices of O(n) are grouped into those that satisfy det(A) = +1, which constitute

Problem 3

55

the group SO(n), and those that satisfy det(A) = −1. Infinitesimal transformations must have determinants infinitesimally close to det(I) = +1, so any infinitesimal transformation of O(n) must lie in SO(n). The conserved current coming from an O(n) symmetry is therefore the same as that from an SO(n) symmetry1 .

Now SO(n) is the group of rotations of n-dimensional space. A rotation is specified by a plane of rotation, and a rotation angle. The number of linearly independent planes in n-dimensional space is equal to the number of ways of choosing 2 dimensions out of n, which is

n(n − 1) n = .

(3.38) 2 2 We expect this many degrees of freedom in parameterizing arbitrary SO(n) transformations, along with the freedom afforded by a rotation angle. Infinitesimal SO(n) transformations A can then be parameterized as A(⃗ α) = I + iαa Ma ,

(3.39)

where αa (a = 1, . . . , n(n − 1)/2) are a set of infinitesimal rotation angles, and Ma are as-yet unspecified transformations that each rotate about an independent plane. Such matrices Ma are known as generators of SO(n). To make sure that A is indeed an element of SO(n), we must impose AT A ≈ I + iαa (Ma + MaT ) = I,

(3.40)

giving the condition MaT = −Ma . In other words, the generators must be antisymmetric. The vector space of antisymmetric n×n matrices indeed has dimension n(n − 1)/2, so everything will be consistent if we take the generators to be any basis of the antisymmetric matrices. Now under this transformation, our fields transform as ϕ → Aϕ = ϕ + iαa Ma ϕ,

or ϕi → ϕi + iαa

n X (Ma )ij ϕj .

(3.41)

j=1

There is actually an independent symmetry transformation here for each Ma , so there will be a conserved current associated to each of these, which we also index by a. Our vectors of infinitesimal field changes are then given by  T n n X X d(Aϕ) ⃗ a := = i (Ma )1j ϕj , . . . , i (Ma )nj ϕj  , (3.42) Φ dαa |⃗α=⃗0 j=1 j=1 1

More generally, the conserved current is associated to the component of the symmetry group that is continuously connected to the identity element. For O(n) this connected component is SO(n), because on any continuous path in O(n) the determinant map is continuous, so it cannot jump from +1 to −1 at any point. Since det(I) = +1, the only elements of O(n) continuously connected to I are those with determinant +1, i.e. those in SO(n).

Relativistic classical fields

56

allowing us to finally arrive at the conserved currents, ⃗a = i jaµ = ⃗π µ · Φ

n X n X (∂ µ ϕi )(Ma )ij ϕj = i(∂ µ ϕ)T Ma ϕ.

(3.43)

i=1 j=1

It might seem strange that we have not specified the matrices Ma exactly, but in fact any choice of an antisymmetric basis will lead to current conservation under the equations of motion. (c) If we now have a collection of n complex scalar fields ϕ1 , . . . , ϕn , we can generalize our previous result Eq. 3.6 to write an appropriate Lagrangian density with interactions as L=

n X i=1

Lfree + Lint i

n X

λ = (∂µ ϕi )∗ (∂ µ ϕi ) − m2 ϕ∗i ϕi − (ϕ∗1 ϕ1 + · · · + ϕ∗n ϕn )2 4! i=1

= (∂µ ϕ)† (∂ µ ϕ) − m2 ϕ† ϕ −

λ † 2 (ϕ ϕ) .

4!

(3.44)

Now the group U (n) is similar to O(n), only it preserves terms of the form ϕ† ϕ instead of ϕT ϕ; complex matrices U of U (n) are those that satisfy U † U = I. Clearly our Lagrangian density is invariant under U (n), then, for much the same reason as in Part (b): L → (U ∂µ ϕ)† (U ∂ µ ϕ) − m2 (U ϕ)† (U ϕ) − = ∂µ ϕ† (U † U )∂ µ ϕ − m2 ϕ† (U † U )ϕ − = ∂µ ϕ† ∂ µ ϕ − m2 ϕ† ϕ −

λ [(U ϕ)† (U ϕ)]2 4!

λ † † [ϕ (U U )ϕ]2 4!

λ † 2 (ϕ ϕ) 4!

= L.

(3.45)

We construct the corresponding conserved current(s) in the same way as above. First we note that our vector of conjugate momenta is now

T ∂L ∂L ∂L ∂L µ , , ... , , ⃗π := ∂(∂µ ϕ1 ) ∂(∂µ ϕ∗1 ) ∂(∂µ ϕn ) ∂(∂µ ϕ∗n ) T

= (∂ µ ϕ∗1 , ∂ µ ϕ1 , . . . , ∂ µ ϕ∗n , ∂ µ ϕn ) ,

(3.46)

because we need to treat ϕi and ϕ∗i as independent fields, as per Problem 3.1. ⃗ of infinitesimal field transformations must also include Similarly, the vector Φ those of the conjugate fields. For infinitesimal unitary transformations U (⃗ α) = I + iαa Ha ,

(3.47)

Problem 4

57

where Ha are the generators of U (n), we require U † U ≈ I + iαa (Ha − Ha† ) = I.

(3.48)

Since α ⃗ is an arbitrary infinitesimal, this gives the condition Ha† = Ha . That is, the generators of U (n) must be hermitian. Any basis of the n×n hermitian matrices will be suitable for our purposes. Hermitian matrices must have a symmetric real component and an antisymmetric imaginary component, so in total there are [n(n + 1)/2] + [n(n − 1)/2] = n2 real degrees of freedom available. Our index a therefore ranges from 1 to n2 . Under this infinitesimal transformation our fields transform as ϕ → U ϕ = ϕ + iαa Ha ϕ,

ϕ† → (U ϕ)† = ϕ† − iαa ϕ† Ha† = ϕ† − iαa ϕ† Ha ,

(3.49)

or in other words, ϕi → ϕi + iα

a

ϕ∗i → ϕ∗i − iαa

n X

(Ha )ij ϕj ,

j=1 n X

ϕ∗j (Ha )ji .

(3.50)

j=1

Our vectors of infinitesimal field transformations are then   n n n n X X X X ⃗ a = i ϕ∗j (Ha )jn  .

(Ha )nj ϕj , −i ϕ∗j (Ha )j1 , . . . , i Φ (Ha )1j ϕj , −i j=1

j=1

j=1

j=1

(3.51) These give conserved currents ⃗a jaµ = ⃗π µ · Φ n X µ ∗ =i (∂ ϕ1 )(Ha )1j ϕj − (∂ µ ϕ1 )ϕ∗j (Ha )j1 + · · · j=1

+(∂ µ ϕ∗n )(Ha )nj ϕj − (∂ µ ϕn )ϕ∗j (Ha )jn n n X X µ ∗

(∂ ϕi )(Ha )ij ϕj − (∂ µ ϕi )ϕ∗j (Ha )ji =i i=1 j=1

= i (∂ µ ϕ)† Ha ϕ − ϕ† Ha (∂ µ ϕ) .

(3.52)

Since U (n) is connected, these n2 currents are indeed associated with U (n) symmetry, and not a smaller symmetry group.

3.4 Problem 4 Problem:

Relativistic classical fields

58

A Lagrangian density with two complex scalar fields, ϕ and χ, is given by L = ∂µ ϕ∂ µ ϕ∗ + ∂µ χ∂ µ χ∗ − m2ϕ |ϕ|2 − m2χ |χ|2 − λϕ |ϕ|4 − λχ |χ|4 − g[χ∗ ϕ2 + χ(ϕ∗ )2 ],

(3.53)

where mϕ , mχ , λϕ , λχ , g are real constants. (a) Using the transformations ϕ → eiα ϕ and χ → eiβ χ choose α, β such that L is invariant. Construct the Noether current j µ ; and (b) use the Euler-Lagrange equations to verify that ∂µ j µ (x) = 0.

Solution: (a) Under the phase transformations ϕ → eiα ϕ and χ → eiβ χ, the first line of Eq. 3.53 is clearly invariant because all of its terms depend only on the magnitudes |ϕ|, |χ|, |∂ϕ|, |∂χ| of the fields and their derivatives. Only the term in the second line is not necessarily invariant, for it transforms as χ∗ ϕ2 + χ(ϕ∗ )2 → (eiβ χ)∗ (eiα ϕ)2 + (eiβ χ)(eiα ϕ∗ )2

= ei(2α−β) χ∗ ϕ2 + e−i(2α−β) χ(ϕ∗ )2 .

(3.54)

Evidently, it is only when ei(2α−β) = 1 that this term is invariant, as would happen if we fixed β = 2α for example. In this case the fields and conjugate fields transform as ϕ → eiα ϕ ≈ ϕ + iαϕ,

χ → ei2α χ ≈ χ + 2iαχ,

ϕ∗ → e−iα ϕ∗ ≈ ϕ∗ − iαϕ∗ ,

χ∗ → e−2iα χ∗ ≈ χ∗ − 2iαχ∗

(3.55)

for infinitesimal α. This tells us that the vector of changes in the fields under this symmetry is Φ = (iϕ, −iϕ∗ , 2iχ, −2iχ∗ )T . Because the symmetry is internal, we know from Eq. (3.2.18) in the text that the associated conserved Noether current is given by ⃗ j µ = ⃗π µ · Φ,

(3.56)

where ⃗π is the vector of conjugate momenta

T ∂L ∂L ∂L ∂L ⃗π µ : = , , , ∂(∂µ ϕ) ∂(∂µ ϕ∗ ) ∂(∂µ χ) ∂(∂µ χ∗ ) T

= (∂ µ ϕ∗ , ∂ µ ϕ, ∂ µ χ, ∂ µ χ∗ ) .

(3.57)

We therefore have the current ⃗ = i(ϕ∂ µ ϕ∗ − ϕ∗ ∂ µ ϕ) + 2i(χ∂ µ χ∗ − χ∗ ∂ µ χ). j µ = ⃗π µ · Φ

(3.58)

Note that if we had instead used β as our infinitesimal transformation parameter and fixed α = β/2, then the current we would have calculated would be scaled by a factor of 1/2 compared to the one above. This scaling would have no effect on the conserved nature of the current.

Problem 4

59

(b) Noting that |ϕ|2 = ϕ∗ ϕ and |ϕ|4 = (ϕ∗ )2 (ϕ)2 , and the same for χ, the EulerLagrange equations can easily be found by using the Wirtinger calculus. We have in particular

∂L ∂L 0 = ∂µ − ∂(∂µ ϕ) ∂ϕ

µ ∗ 2 ∗ = ∂µ ∂ ϕ − −mϕ ϕ − 2λ(ϕ∗ )2 ϕ − 2gχ∗ ϕ , (3.59) and

0 = ∂µ

∂L ∂(∂µ χ)

−

∂L ∂χ

= ∂µ ∂ µ χ∗ − −m2χ χ∗ − 2λ(χ∗ )2 χ − g(ϕ∗ )2 ,

(3.60)

and since the Lagrangian is real-valued, the equations for the conjugate fields are simply the conjugates of these equations:

0 = ∂µ ∂ µ ϕ − −m2ϕ ϕ − 2λϕ∗ ϕ2 − 2gχϕ∗ ,

0 = ∂µ ∂ µ χ − −m2χ χ − 2λχ2 χ∗ − gϕ2 .

(3.61) These results will allow us to show the current Eq. (3.58) is indeed conserved, for we calculate ∂µ j µ = i(∂µ ϕ∂ µ ϕ∗ + ϕ∂µ ∂ µ ϕ∗ − ∂µ ϕ∗ ∂ µ ϕ − ϕ∗ ∂µ ∂ µ ϕ)

+ 2i(∂µ χ∂ µ χ∗ + χ∂µ ∂ µ χ∗ − ∂µ χ∗ ∂ µ χ − χ∗ ∂µ ∂ µ χ)

= i(ϕ∂µ ∂ µ ϕ∗ − ϕ∗ ∂µ ∂ µ ϕ) + 2i(χ∂µ ∂ µ χ∗ − χ∗ ∂µ ∂ µ χ).

(3.62)

By the equations of motion, the first term here is i(ϕ∂µ ∂ µ ϕ∗ − ϕ∗ ∂µ ∂ µ ϕ) = iϕ −m2ϕ ϕ∗ − 2λ(ϕ∗ )2 ϕ − 2gχ∗ ϕ

− iϕ∗ −m2ϕ ϕ − 2λϕ∗ ϕ2 − 2gχϕ∗

= 2igχ(ϕ∗ )2 − 2igχ∗ ϕ2 ,

(3.63)

and similarly the second term is 2i(χ∂µ ∂ µ χ∗ − χ∗ ∂µ ∂ µ χ) = 2iχ −m2χ χ∗ − 2λ(χ∗ )2 χ − g(ϕ∗ )2

− 2iχ∗ −m2χ χ − 2λχ2 χ∗ − gϕ2 = 2igχ∗ ϕ2 − 2igχ(ϕ∗ )2 .

(3.64)

Together, these give ∂µ j µ = 2igχ(ϕ∗ )2 − 2igχ∗ ϕ2 + 2igχ∗ ϕ2 − 2igχ(ϕ∗ )2 = 0,

(3.65)

so the current is conserved as expected. Note that the factor of 2 multiplying the χ terms in the current, which was so essential for all of the cancellations in ∂µ j µ , comes directly from the relation β = 2α in accordance with the symmetry.

Relativistic classical fields

60

3.5 Problem 5 Problem: Let F µν be the electromagnetic tensor and consider the symmetric stress-energy tensor defined in Eq. (3.3.9), T̄ µν = −F µτ Fντ + 14 δ µν F στ Fστ . In the presence of an external conserved four-vector current density j µ show that ∂µ T̄ µν = 0 becomes ∂µ T̄ µν = (1/c)jµ F µν .

Solution: First of all, we can raise an index on T̄ µν to get 1 1 T̄ µν = g νρ T̄ µρ = −g νρ F µτ Fρτ + g νρ δ µρ F στ Fστ = −F µτ F ντ + g µν F στ Fστ .

4 4 (3.66) Now Maxwell’s equation’s can be expressed in Lorentz covariant form as ∂µ F µν =

1 ν j , c

(3.67)

together with the Jacobi identity ∂ ν F αβ + ∂ α F βν + ∂ β F να = 0.

(3.68)

With these, we calculate 1 ∂µ T̄ µν = −∂µ (F µτ F ντ ) + g µν ∂µ (F στ Fστ ) 4 1 = − (∂µ F µτ ) F ντ − F µτ (∂µ F ντ ) + ∂ ν (Fστ F στ ) 4

1 τ 1 =− j F ντ − Fµτ (∂ µ F ντ ) + Fστ (∂ ν F στ ) c 2 1 1 = − jτ F ντ − Fµτ (∂ µ F ντ ) + Fστ (−∂ σ F τ ν − ∂ τ F νσ ) , c 2 and now recognizing we can relabel dummy indices,

1 1 σ τ ν 1 τ νσ νµ µν σ ντ ∂µ T̄ = − jµ F − Fστ ∂ F + ∂ F + ∂ F c 2 2

1 1 1 ∂ σ F ντ + ∂ τ F νσ , = jµ F µν − Fστ c 2 2

(3.69)

(3.70)

where we have also used the antisymmetry of F µν . Now the term in brackets is symmetric in σ and τ , so its contraction with the antisymmetric Fστ vanishes. We are left with ∂µ Gµν = as desired.

1 jµ F µν , c

(3.71)

Problem 6

61

3.6 Problem 6 Problem: Let Aµ be the electromagnetic four-vector potential and define the four-vector Kµ = ϵµνρσ Aν ∂ ρ Aσ . Show that ∂µ K µ = −(2/c)E · B. (Hint: consider F̃ µν .)

Solution: By direct calculation, we have ∂µ K µ = ϵµνρσ (∂µ Aν )(∂ρ Aσ ) + ϵµνρσ Aν (∂µ ∂ρ Aσ ).

(3.72)

Now partial derivatives commute, so ∂µ ∂ρ Aσ is symmetric in µ and ρ. Its contraction with the antisymmetric tensor ϵµνρσ therefore vanishes, giving ∂µ K µ = ϵµνρσ (∂µ Aν )(∂ρ Aσ ).

(3.73)

We can further exploit the antisymmetry of ϵµνρσ and the arbitrariness of dummy index labels to write 1 ∂µ K µ = (ϵµνρσ − ϵνµρσ − ϵµνσρ + ϵνµσρ ) (∂µ Aν )(∂ρ Aσ ) 4 1 µνρσ = ϵ [(∂µ Aν )(∂ρ Aσ ) − (∂ν Aµ )(∂ρ Aσ ) 4 −(∂µ Aν )(∂σ Aρ ) + (∂ν Aµ )(∂σ Aρ )] 1 = ϵµνρσ (∂µ Aν − ∂ν Aµ )(∂ρ Aσ − ∂σ Aρ ) 4 1 µνρσ = ϵ Fµν Fρσ 4 1 = Fµν F̃ µν , (3.74) 2 using the definitions of the field strength and dual field strength tensors Fµν and F̃ µν . But we have seen in Eq. (2.7.26) that the contraction of these two tensors evaluates to −4(E · B)/c , so in total 2 ∂µ K µ = − E · B. c

(3.75)

3.7 Problem 7 Problem: A general real scalar Lagrangian density at most quadratic in the four-vector potential is L = − 12 (∂µ Aν ∂ µ Aν + ρ ∂µ Aν ∂ ν Aµ ) + λ Aµ Aµ , where ρ, λ ∈ R: (a) Obtain the Euler-Lagrange equations for this system. (b) Assume plane wave solutions with the general form Aµ (x) = Cϵµ (k)e−ik·x and express the Euler-Lagrange equation in terms of ϵµ (k) and k µ .

Relativistic classical fields

62

(c) For k 2 ̸= 0 we can define transverse and longitudinal projectors as T µν ≡ g µν − (k µ k ν /k 2 ) and Lµν ≡ (k µ k ν /k 2 ). Verify the properties T µρ Tρν = T µν , Lµρ Lρν = Lµν and T µν + Lµν = δ µν . Define ϵµℓ ≡ Lµν ϵν and ϵµt ≡ T µν ϵν and so verify that k·ϵℓ = k·ϵ, k·ϵt = 0 and ϵµt + ϵµℓ = ϵµ . (d) Show that there is a plane wave solution to the Euler-Lagrange equations with k 2 = m2ℓ ≡ 2λ/(1 + ρ) that is longitudinal ϵµ = ϵµℓ . (e) Similarly show that there is a plane wave solution with with k 2 = m2t ≡ 2λ that is transverse. (f) We say that the ℓ and t modes propagate with masses mℓ and mt respectively.

Choosing ρ = −1 gives mℓ → ∞ so that the ℓ modes cannot propagate, leaving only the t modes. Show that in this limit we recover the Proca Lagrangian density for a massive vector field given in Eq. (6.5.1) with mt becoming the mass of the vector field.

Solution: (a) Using the identities ∂Aν ∂(∂µ Aν ) = δ αµ δ βν , = δ βν and ∂Aβ ∂(∂α Aβ )

(3.76)

we get for the given Lagrangian density the equations of motion

∂L ∂L 0 = ∂α − ∂(∂α Aβ ) ∂Aβ

1 α β µ ν µ ν α β α β ν µ µ ν α β = ∂α − δ µδ ν ∂ A + ∂ A δ µδ ν + ρ δ µδ ν ∂ A + ρ ∂ A δ ν δ µ 2 − λ(δ βµ Aµ + Aµ δ βµ )

= −∂α (∂ α Aβ + ρ ∂ β Aα ) − 2λAβ ,

(3.77)

or in a more conventional form, ∂ 2 Aµ + ρ ∂ µ (∂ · A) + 2λAµ = 0.

(3.78)

(b) If we assume a plane wave ansatz Aµ (x) = Cϵµ (k)e−ik·x , the derivatives of the four-vector potential will be

α ∂ν Aµ = Cϵµ ∂ν e−ikα x = Cϵµ e−ik·x ∂ν (−ikα xα ) = −ikν Cϵµ e−ik·x = −ikν Aµ ,

(3.79)

∂ µ (∂ · A) = −ikν ∂ µ Aν = (−ikν )(−ik µ Aν ) = −k ν (k · A),

(3.80)

from which it follows

Problem 7

63

and ∂ 2 Aµ = ∂ ν ∂ν Aµ = −ikν ∂ ν Aµ = (−ikν )(−ik ν Aµ ) = −k 2 Aµ .

(3.81)

The Euler-Lagrange equations found in Part (a) then reduce to −k 2 Aµ − ρk ν (k · ϵ) + 2λAµ = 0

(2λ − k 2 )ϵµ − ρk µ (k · ϵ) = 0,

=⇒

(3.82)

having factored out Ce−ik·x , which is nowhere vanishing. (c) To avoid redundant calculations, we shall prove the operator identities in the reverse order given. First, we show T µν = gνα T µα = gνα g µα − gνα kµ kν k2 µ = δ ν − Lµν

kµ kα k2

= δ µν −

=⇒

T µν + Lµν = δ µν .

(3.83)

Next, k µ k ρ kρ kν k 2 k µ kν k µ kν = = = Lµν , k2 k2 k4 k2 and finally, using the above, Lµρ Lρν =

(3.84)

T µρ Tρν = (g µρ − Lµρ ) (gρν − Lρν )

= δ µν − Lµν − Lµν + Lµρ Lρν = δ µν − Lµν = T µν .

(3.85)

The required dot products are straightforward to evaluate: kµ kν k2 ν ϵ = k ϵν = k · ϵ, ν k2 k2 k · ϵt = kµ T µν ϵν = kµ (g µν − Lµν )ϵν = k ν ϵν − kµ Lµν ϵν = 0,

k · ϵℓ = kµ Lµν ϵν = kµ

(3.86) (3.87)

and hidden in the second line here is the last required identity ϵµℓ + ϵµt = (T µν + Lµν )ϵν = g µν ϵν = ϵµ .

(3.88)

(d) If we substitute ϵ → ϵℓ and k 2 → m2ℓ in the equation of motion Eq. 3.82, we can simplify using the results of Part (c): 0 = (2λ − m2ℓ )ϵµℓ − ρk µ (k · ϵℓ ) kµ kν = (2λ − m2ℓ ) 2 ϵν − ρk µ (k · ϵ) mℓ

2λ − 1 − ρ kµ . = (k · ϵ) m2ℓ

(3.89)

64

Relativistic classical fields

Discounting the case where k · ϵ = 0, it must be that

2λ 2λ − 1 − ρ = 0, i.e. m2ℓ = . m2ℓ 1+ρ

(3.90)

Therefore, there is a valid longitudinal solution to the Euler-Lagrange equations with the given mass. (e) Similarly, if we substitute ϵ → ϵt and k 2 → m2t in the equation of motion Eq. 3.82, the results of part (c) allow us to simplify considerably: 0 = (2λ − m2t )ϵµt − ρk µ (k · ϵt ) = (2λ − m2t )ϵµt ,

(3.91)

so discounting the case where ϵµt = 0, it must be that m2t = 2λ. Hence, there is a valid transverse solution to the Euler-Lagrange equations with this mass. (f) With ρ = −1, the given Lagrangian density is

1 L = − (∂µ Aν ∂ µ Aν − ∂µ Aν ∂ ν Aµ ) + λAµ Aµ 2 1 m2 = − (∂µ Aν )(∂ µ Aν − ∂ ν Aµ ) + t Aµ Aµ 2 2 1 1 2 µν µ = − (∂µ Aν )F + mt Aµ A , 2 2

(3.92)

where we have substituted λ → m2t /2 from part (e). Now since the field strength tensor F µν is antisymmetric, we can use the same trick as in Problem 3.6 to write 1 1 L = − (∂µ Aν )(F µν − F νµ ) + m2t Aµ Aµ 4 2 1 1 2 µν = − (∂µ Aν − ∂ν Aµ )F + mt Aµ Aµ 4 2 1 1 2 µν µ = − Fµν F + mt Aµ A , (3.93) 4 2 by relabeling dummy indices to arrive at the second line. This is exactly the Proca Lagrangian density for a free massive vector field, where mt is the mass of the vector field (cf. Eq. (6.5.1)).

3.8 Problem 8 Problem: Scalar electrodynamics is the gauge invariant minimal coupling of a charged scalar field and the electromagnetic field. It has the Lagrangian density L = − 14 Fµν F µν + (Dµ ϕ)∗ (Dµ ϕ) − m2 ϕϕ∗ , where the covariant derivative is defined as as Dµ ≡ ∂µ + iqAµ and where q is the electric charge of the scalar particle:

Problem 8

65

(a) The system is invariant under global phase transformations ϕ → e−iα ϕ. Use this to derive the Noether current that replaces Eq. (3.2.49), j µ = iqc [ϕ∗ (Dµ ϕ)−(Dµ ϕ)∗ ϕ] = iqc [ϕ∗ (∂ µ ϕ)−(∂ µ ϕ∗ )ϕ+2iqAµ ϕ∗ ϕ] (b) Show that the Euler-Lagrange equations can be written as ∂µ F µν = j ν /c and (Dµ Dµ + m2 )ϕ = 0; and (c) Show that Aµ (x) → Aµ (x) + ∂ µ α(x), ϕ(x) → e−iqα(x) ϕ(x) leaves the system invariant. This is an example of U (1) gauge invariance.

Solution: (a) The given symmetry transformation is internal, so the corresponding conserved current is given by ⃗ j µ = ⃗π µ · Φ,

(3.94)

⃗ is the vector of field changes under an infinitesimal symmetry transwhere Φ formation, and ⃗π is the vector of conjugate momenta

T ∂L ∂L µ ⃗π := , .

(3.95) ∂(∂µ ϕ) ∂(∂µ ϕ∗ ) Now with the covariant derivatives written out explicitly, the Lagrangian density is 1 L = − Fµν F µν + (∂µ ϕ∗ − iqAµ ϕ∗ )(∂ µ ϕ + iqAµ ϕ) − m2 ϕϕ∗ , 4

(3.96)

and so its derivatives are ⃗π µ = (∂ µ ϕ∗ − iqAµ ϕ∗ , ∂ µ ϕ + iqAµ ϕ)T T

= ((Dµ ϕ)∗ , Dµ ϕ) .

(3.97)

⃗ Under the given symmetry transformation, the Now all we need to find is Φ. two fields transform as ϕ → e−iα ϕ = ϕ − iαϕ + O(α2 ), ∗

ϕ → (e

−iα

∗

∗

∗

(3.98) 2

ϕ) = ϕ + iαϕ + O(α ).

(3.99)

From these, we see that the vector of field changes under an infinitesimal sym⃗ = (−iϕ, +iϕ∗ )T . Hence, the conserved current is metry is Φ j µ = (Dµ ϕ)∗ (−iϕ) + (Dµ ϕ)(iϕ∗ ) = i [ϕ∗ Dµ ϕ − ϕ(Dµ ϕ)∗ ]

= i [ϕ∗ (∂ µ ϕ + iqAµ ϕ) − ϕ(∂ µ ϕ∗ − iqAµ ϕ∗ )]

= i [ϕ∗ ∂ µ ϕ − ϕ∂ µ ϕ∗ + 2iqAµ ϕ∗ ϕ] .

(3.100)

66

Relativistic classical fields

Of course, any constant multiple of this current will also be conserved, so we may rescale j µ → qc j µ to arrive at the normalization given in the question, j µ = iqc [ϕ∗ ∂ µ ϕ − ϕ∂ µ ϕ∗ + 2iqAµ ϕ∗ ϕ] . (b) For the equations of motion of the field Aµ , we note the result

∂L ∂ 1 αβ = − Fαβ F = −F µν ∂(∂µ Aν ) ∂(∂µ Aν ) 4

(3.101)

(3.102)

from Eq. (3.3.5) of the text. The Euler-Lagrange equations then reduce to

∂L ∂L 0 = ∂µ − ∂(∂µ Aν ) ∂Aν

= −∂µ F µν − iqϕ∂ ν ϕ∗ − iqϕ∗ ∂ µ ϕ + 2q 2 Aν ϕ∗ ϕ

1 ν µν = −∂µ F − − j , (3.103) c or ∂µ F µν = j ν /c. And for the scalar field, the equation of motion is

∂L ∂L 0 = ∂µ − ∂(∂µ ϕ∗ ) ∂ϕ∗

= ∂µ (Dµ ϕ) − −iqAµ (∂µ ϕ + iqAµ ϕ) − m2 ϕ = ∂µ (Dµ ϕ) + iqAµ Dµ ϕ + m2 ϕ

= (∂µ + iqAµ )Dµ ϕ + m2 ϕ = Dµ Dµ ϕ + m2 ϕ = (Dµ Dµ + m2 )ϕ.

(3.104)

Of course, the equation of motion of the conjugate field is simply the conjugate of this equation. (c) Under a local phase transformation, the field strength tensor transforms as Fµν = ∂µ Aν − ∂ν Aµ → ∂µ (Aν + ∂ν α) − ∂ν (Aµ + ∂µ α)

= (∂µ Aν − ∂ν Aµ ) + (∂µ ∂ν α − ∂ν ∂µ α)

= Fµν ,

(3.105)

because partial derivatives commute. That is, the field strength tensor is invariant under a local phase transformation. Next, we look at the covariant derivative: Dµ ϕ = ∂µ ϕ + iqAµ ϕ → ∂µ (e−iqα ϕ) + iq(Aµ + ∂µ α)(e−iqα ϕ) = −iq(∂µ α)e−iqα ϕ + e−iqα ∂µ ϕ

+ iqAµ e−iqα ϕ + iq(∂µ α)(e−iqα ϕ)

= e−iqα (∂µ ϕ + iqAµ ϕ) = e−iqα Dµ ϕ.

(3.106)

67

Problem 8 We see that Dµ ϕ transforms in the same way as ϕ. It follows that (Dµ ϕ)∗ (Dµ ϕ) will transform in the same way as the mass term ϕ∗ ϕ. Now the mass term is invariant under the local transformation: ϕ∗ ϕ → (e−iqα ϕ)∗ (e−iqα ϕ) = e+iqα ϕ∗ e−iqα ϕ = ϕ∗ ϕ,

(3.107)

and therefore all terms in the Lagrangian density are unchanged under a local phase transformation. In other words, such a transformation leaves the whole system invariant.

4

Relativistic Quantum Mechanics 4.1 Problem 1 Problem: Show that the symmetric n-boson basis states in Eq. (4.1.149) are normalized, i.e., show ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ = 1.

Solution: The symmetric n-boson basis states are defined by rQ X m nm ! |bσ(i1 ) bσ(i2 ) · · · bσ(in ) ⟩, |bi1 · · · bin ; S⟩ = n! σ

(4.1)

where the sum is over all distinct permutations σ of the elements i1 , i2 , · · · , in , and nm is the number of times bm occurs in the list bi1 bi2 · · · bin . The squared norm of such a state is then !2 rQ XX n ! m m ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ = ⟨bτ (i1 ) · · · bτ (in ) |bσ(i1 ) · · · bσ(in ) ⟩ n! σ τ Q nm ! X X = m δτ (i1 )σ(i1 ) δτ (i2 )σ(i2 ) · · · δτ (in )σ(in ) , n! σ τ (4.2) where we have used Eq. (4.1.149) from the text. Note that the terms in this sum vanish unless σ(im ) = τ (im ) for all m = 1, · · · , n, i.e. unless σ and τ are the same permutation, in which case the terms are unity. So the double sum reduces to Q nm ! X ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ = m 1.

(4.3) n! σ We see this sum is counting the number of distinct permutations of our symbols bi1 , bi2 , · · · , bin . The number of permutations of n distinct elements is n!, but in fact this overcounts the number of permutations we have by a factor of nm ! for each bm in our list, because permuting these identical symbols does not result in a different ordering. That is to say, X σ

68

n! , m nm !

1= Q

(4.4)

Problem 2

69

so in total, Q ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ =

m nm ! Q n!

n!

m nm !

= 1.

(4.5)

The states are correctly normalized.

4.2 Problem 2 Problem: For any vector operator V̂ in quantum mechanics we have [V̂ i , Jˆj ] = iℏϵijk V̂ k from Eq. (4.1.114). It will be helpful to also recall Eq. (1.2.64). (a) Use these results to show that ⟨jb mb |V̂z |ja ma ⟩ = 0 unless ma = mb . (b) Define V̂± ≡ V̂x ± iV̂y in the usual way and show that ⟨jb mb |V̂± |ja ma ⟩ = 0 unless mb = ma ± 1. (c) Evaluate [Jˆ+ , V̂+ ] and use this result for the special case j = ja = jb to obtain the dependence of ⟨j (m + 1)|V̂+ |jm⟩ on m; and (d) Extending this result show that ⟨jmb |V̂|jma ⟩ = Cj ⟨jmb |Ĵ|jma ⟩, where Cj is a constant independent of mb and ma .

Solution: (a) From [V̂ i , Jˆj ] = iℏϵijk V̂ k , we have that [V̂z , Jˆz ] = 0, or V̂z Jˆz = Jˆz V̂z . Now since Jˆz |jm⟩ = ℏm|jm⟩, and since Jˆz is self-adjoint, we can calculate ℏma ⟨jb mb |V̂z |ja ma ⟩ = ⟨jb mb |V̂z Jˆz |ja ma ⟩ = ⟨jb mb |Jˆz V̂z |ja ma ⟩

= ℏmb ⟨jb mb |V̂z |ja ma ⟩

(4.6)

Hence, if ma ̸= mb , we must have ⟨jb mb |V̂z |ja ma ⟩ = 0. (b) Using the linearity of the commutator, we have [V̂± , Jˆz ] = [V̂x , Jˆz ] ± i[V̂y , Jˆz ] = −iℏV̂y ± i(iℏV̂x ) = ∓ℏ(V̂x ± iV̂y ) = ∓ℏV̂± .

(4.7)

This allows us to perform a similar calculation to that in Part (a): ℏma ⟨jb mb |V̂± |ja ma ⟩ = ⟨jb mb |V̂± Jˆz |ja ma ⟩ = ⟨jb mb |(Jˆz V̂± ∓ ℏV̂± )|ja ma ⟩ = ℏ(mb ∓ 1)⟨jb mb |V̂± |ja ma ⟩.

(4.8)

Relativistic Quantum Mechanics

70

Hence, if mb ̸= ma ± 1, we must have ⟨jb mb |V̂± |ja ma ⟩. (c) With our known commutation relations, noting that [V̂ i , Jˆj ] = −[Jˆj , V̂ i ], we have [Jˆ+ , V̂+ ] := [Jˆx + iJˆy , V̂x + iV̂y ] = [Jˆx , V̂x ] + i[Jˆx , V̂y ] + i[Jˆy , V̂x ] − [Jˆy , V̂y ] = 0 + i(iℏV̂z ) + i(−iℏV̂z ) − 0 = 0,

To continue, we must know that p Jˆ± |jm⟩ = ℏ j(j + 1) − m(m ± 1)|j (m ± 1)⟩. † Then since Jˆ+ = Jˆx + iJˆy = (Jˆx − iJˆy )† = Jˆ− , it follows that †

p ⟨jm|Jˆ+ = Jˆ− |jm⟩ = ℏ j(j + 1) − m(m − 1) ⟨j (m − 1)|,

(4.9)

(4.10)

(4.11)

and so 1 ⟨j (m + 1)|V̂+ Jˆ+ |j (m − 1)⟩ ⟨j (m + 1)|V̂+ |jm⟩ = p ℏ j(j + 1) − m(m − 1) 1 = p ⟨j (m + 1)|Jˆ+ V̂+ |j (m − 1)⟩ ℏ j(j + 1) − m(m − 1) s j(j + 1) − m(m + 1) = ⟨j m|V̂+ |j (m − 1)⟩.

(4.12) j(j + 1) − m(m − 1) The inner product on the right-hand side here is of the same form as the one on the left-hand side, only with m → m − 1. We may therefore continually repeat this process on the right-hand side until m reaches its minimum possible value, m = −j. Fortunately the product telescopes, leaving us with s j(j + 1) − m(m + 1) ⟨j (m + 1)|V̂+ |jm⟩ = ⟨j (−j + 1)|V̂+ |j (−j)⟩. j(j + 1) − (−j)(−j + 1) (4.13) We may gather the terms that depend on j, but not on m, into ⟨j (−j + 1)|V̂+ |j (−j)⟩ Cj := p , j(j + 1) − (−j)(−j + 1) to see that the m-dependence of our expression is given by p ⟨j (m + 1)|V̂+ |jm⟩ = Cj j(j + 1) − m(m + 1).

(4.14)

(4.15)

(d) Note that the above result can be written as ⟨j (m + 1)|V̂+ |jm⟩ = Cj ⟨j (m + 1)|Jˆ+ |jm⟩,

(4.16)

Problem 2

71

and therefore more generally ⟨j mb |V̂+ |jma ⟩ = Cj ⟨j mb |Jˆ+ |jma ⟩

(4.17)

because both sides vanish whenever mb ̸= ma + 1, by Part (b). In order to arrive at the analogous equation for V̂z and Jˆz , we must relate V̂z to V̂+ , for this is used in the definition of Cj . The only relations at our disposal are [V̂ i , Jˆj ] = iℏϵijk V̂ k , which tell us 1 [V̂x , Jˆy ] iℏ i V̂z = [V̂y , Jˆx ] ℏ

=⇒

V̂z =

=⇒

1 1 [V̂x + iV̂y , Jˆy ] = [V̂+ , Jˆy ], iℏ iℏ 1 1 V̂z = [V̂x + iV̂y , Jˆx ] = [V̂+ , Jˆx ], ℏ ℏ V̂z =

(4.18) (4.19)

and therefore 1 V̂z = 2ℏ

1 [V̂+ , Jˆx ] + [V̂+ , Jˆy ] i

=

1 [V̂+ , Jˆ− ].

2ℏ

(4.20)

Using this, we have 1 ⟨j mb |(V̂+ Jˆ− − Jˆ− V̂+ )|jma ⟩ 2ℏ 1 p ℏ j(j + 1) − ma (ma − 1)⟨j mb |V̂+ |j (ma − 1)⟩ = 2ℏ

p −ℏ j(j + 1) − mb (mb + 1)⟨j (mb + 1)|V̂+ |j ma )⟩ C j p = j(j + 1) − ma (ma − 1)⟨j mb |Jˆ+ |j (ma − 1)⟩ 2

p − j(j + 1) − mb (mb + 1)⟨j (mb + 1)|Jˆ+ |j ma )⟩ p C j p j(j + 1) − ma (ma − 1) j(j + 1) − ma (ma − 1)δma mb = 2

p p − j(j + 1) − mb (mb + 1) j(j + 1) − ma (ma + 1)δma mb

⟨j mb |V̂z |jma ⟩ =

Cj [j(j + 1) − ma (ma − 1) − j(j + 1) + ma (ma + 1)] δma mb 2 = Cj ma δma mb = Cj ⟨j mb |Jˆz |jma ⟩.

(4.21) =

For compactness, we used δma mb to denote 1 if ma = mb , and 0 otherwise.

Finally, we need to derive the analogous relation for V̂− and Jˆ− , for then the result follows. Using the same technique, we have V̂x = [Jˆy , V̂z ]/(iℏ) and V̂y = −[Jˆx , V̂z ]/(iℏ), so V̂− = V̂x − iV̂y =

1 ˆ 1 [Jx − iJˆy , V̂z ] = [Jˆ− , V̂z ]. ℏ ℏ

Hence, ⟨j mb |V̂− |jma ⟩ =

1 ⟨j mb |(Jˆ− V̂z − V̂z Jˆ− )|jma ⟩ ℏ

(4.22)

Relativistic Quantum Mechanics

72

1 p ℏ j(j + 1) − mb (mb + 1)⟨j (mb + 1)|V̂z |j ma ⟩ ℏ

p −ℏ j(j + 1) − ma (ma − 1)⟨j mb |V̂z |j (ma − 1)⟩ p = Cj j(j + 1) − mb (mb + 1)⟨j (mb + 1)|Jˆz |j ma ⟩

p − j(j + 1) − ma (ma − 1)⟨j mb |Jˆz |j (ma − 1)⟩ p = Cj j(j + 1) − mb (mb + 1)ma δma (mb +1)

p − j(j + 1) − ma (ma − 1)mb δ(ma −1) mb p j(j + 1) − ma (ma − 1)ma = Cj

p − j(j + 1) − ma (ma − 1)(ma − 1) δ(ma −1) mb p = Cj j(j + 1) − ma (ma − 1)δ(ma −1) mb = Cj ⟨j mb |Jˆ− |jma ⟩.

(4.23) =

Combining these results, we see

1 ⟨j mb | V̂+ + V̂− |jma ⟩ 2

Cj ⟨j mb | Jˆ+ + Jˆ− |jma ⟩ = 2 = Cj ⟨j mb |Jˆx |jma ⟩,

⟨j mb |V̂x |jma ⟩ =

(4.24)

and similarly

1 ⟨j mb | V̂+ − V̂− |jma ⟩ 2i

Cj = ⟨j mb | Jˆ+ − Jˆ− |jma ⟩ 2i = Cj ⟨j mb |Jˆy |jma ⟩.

⟨j mb |V̂y |jma ⟩ =

(4.25)

Therefore in total, ⟨j mb |V̂|jma ⟩ = Cj ⟨j mb |Ĵ|jma ⟩, where Cj is defined in Eq. 4.14.

4.3 Problem 3 Problem: A scalar particle with mass m and charge q interacts with a time-independent electromagnetic potential given by qA0 = 0 for z < 0 and qA0 = V with V > 0 for z ≥ 0.

A particle plane wave is moving in the +z-direction with energy E > 0. Calculate the reflection (R) and transmission (T ) coefficients, where these are the √ probability of reflection and transmission respectively, R + T = 1.

Define p ≡ E 2 −m2 and p 2 2 K ≡ (E −V ) −m .

73

Problem 3

(a) Show that if |E − V | < m, then there is no transmission, R = 1 and T = 0. (b) Show that if E−V > m, then R = |(p−K)2 /(p+K)2 | < 1 and T = 1−R and there is partial transmission and partial reflection. (c) Show that if E−V < −m, then R = |(p+K)2 /(p−K)2 | > 1 and T = 1 − R, which corresponds to a negative current being transmitted and more positive current reflected than was incident. This is the Klein paradox and shows that |V | > 2m can lead to particle-antiparticle pair creation. This illustrates the breakdown of relativistic quantum mechanics in the presence of strong fields. (Hint: Use continuity of the wavefunction and its first derivative.)

Solution: In the presence of an electromagnetic field, the behavior of a scalar particle ϕ is governed by the modified Klein-Gordon equation

mc 2 q q ϕ = 0, (4.26) ∂ µ + i Aµ ∂µ + i Aµ + ℏc ℏc ℏ from Eq. (4.3.57) in the text. In our case, this reduces to  h i  0 = ∂ µ ∂µ + mc 2 ϕ for z < 0, ℏ i h

2  0 = ∂ 0 + i V ∂0 + i V − ∇2 + mc ϕ for z ≥ 0. ℏc ℏc ℏ

(4.27)

In the region z < 0 we have the usual Klein-Gordon equation, which has the familiar plane wave solutions iEt ipz ipz (z < 0), (4.28) ϕ(x) = Ae ℏ + Be− ℏ e− ℏ where A and B are arbitrary constants, and 1p 2 p := E − m2 c4 . c

(4.29)

The term proportional to A represents a wave moving in the +z-direction, i.e. the incident wave, while the term proportional to B represents a wave moving in the −z-direction, i.e. the reflected wave.

Now to solve the equation in the region z ≥ 0, let us assume a solution of the form ϕ(x) = Ceiκz/ℏ e−iEt/ℏ , and find conditions for C and κ. Substituting this in, the equation becomes 2 2

mc 2 E V iκ 0 = −i +i ϕ(x) − ϕ(x) + ϕ(x), (4.30) ℏc ℏc ℏ ℏ which is easily seen to be solved by 1p κ=± (E − V )2 − m2 c4 ≡ ±K c

(4.31)

for arbitrary C. Here we run into a subtle point that lies at the heart of this exercise: are we to take the positive or negative solution for κ? The answer lies in considering

Relativistic Quantum Mechanics

74

the group velocity of this wave; on physical grounds it must point in the positive zdirection, because this solution represents the transmitted wave. The group velocity is found from Eq. (4.2.21): vg =

dE Kc êz = ± c êz , dκ E−V

(4.32)

and therefore we must take the positive solution when E − V > 0, and the negative solution when E − V < 0. That is,

+K if E − V > 0, κ= (4.33) −K if E − V < 0.

This will be important later.

In total, our wavefunction is iEt ipz ( ipz ϕ− (x) ≡ Ae ℏ + Be− ℏ e− ℏ ϕ(x) = iEt iκz ϕ+ (x) ≡ Ce ℏ e− ℏ

for z < 0 for z ≥ 0.

(4.34)

Since this wavefunction is non-normalizable, in order to find the reflection and transmission coefficients we need only find the relative probabilities of reflection and transmission, given by the strengths of the reflected and transmitted current densities relative to the incident current density. The current densities associated to each of the wave components are found using Eq. (4.3.24): j µ = iqc[ϕ∗ ∂ µ ϕ − (∂ µ ϕ)∗ ϕ].

(4.35)

The vector component of the current for the incident wave, for example, is h ipz ipz iEt iEt jA = −iqc (Ae ℏ e− ℏ )∗ ∇(Ae ℏ e− ℏ ) i ipz ipz iEt iEt −(∇(Ae ℏ e− ℏ ))∗ (Ae ℏ e− ℏ ) qc = + (p + p∗ )|A|2 êz ℏ 2qc p|A|2 êz ≡ jA êz , (4.36) =+ ℏ and similarly 2qc p|B|2 êz ≡ jB êz , (4.37) ℏ qc jC = + (κ + κ∗ )|C|2 êz ≡ jC êz .

(4.38) ℏ Note that p is always real for an on-shell particle, which allowed us to simplify the expressions for jA and jB , but the same is not necessarily true of κ. The reflection and transmission coefficients are therefore jB = −

R≡

−jB |B|2 = , jA |A|2

T ≡

jC |C|2 κ + κ∗ = . jA |A|2 2p

(4.39)

Now we must impose some relations between A, B, and C so that we may calculate these coefficients. Since the Klein-Gordon equation is second order, we have two

Problem 4

75

conditions to impose: ϕ(x) must be continuous everywhere, and ∂µ ϕ(x) must be continuous everywhere. If we consider the boundary plane z = 0, these conditions respectively require ϕ− (x)|z=0 = ϕ+ (x)|z=0

=⇒

A + B = C,

(4.40)

∂z ϕ− (x)|z=0 = ∂z ϕ+ (x)|z=0

=⇒

p(A − B) = κC.

(4.41)

and p(A − (C − A)) = κC,

(4.42)

|C|2 κ + κ∗ 2p (κ + κ∗ ). = 2 |A| 2p |p + κ|2

(4.43)

Together, these imply p(A − B) = κ(A + B) allowing us to conclude R=

|B|2 |p − κ|2 = , |A|2 |p + κ|2

T =

Notice that since p is real, it is always the case that R+T =

(p − κ)(p − κ∗ ) + 2p(κ + κ∗ ) (p + κ)(p + κ∗ ) = = 1.

2 |p + κ| |p + κ|2

(4.44)

(a) If we assume that |E − V | < mc2 , then from its definition κ will be purely imaginary. As a result, κ + κ∗ = κ + (−κ) = 0, and so T = 0: there is no transmission. Furthermore since R + T = 1, we must have R = 1. (b) If we assume that E − V > mc2 , then κ will be real. From our considerations above, we must have in this case κ = K. Then our coefficients reduce to R=

(p − K)2 , (p + K)2

T =

4pK . (p + K)2

(4.45)

Now clearly R > 0 and T > 0, so since R + T = 1 it must be that R < 1 and T < 1: there is partial transmission and partial reflection. (c) Finally, if we assume that E − V < −mc2 , then (E − V )2 > m2 c4 and once again κ is real, though now we must have κ = −K. Then κ∗ = −K, so we may write R=

(p + K)2 , (p − K)2

T =−

4pK . (p − K)2

(4.46)

Notice that T < 0. Then since R + T = 1, we must have R > 1 in this regime.

4.4 Problem 4 Problem: Using the properties of the Dirac γ-matrices verify that the matrices ( 12 σ µν ) satisfy the Lie algebra of the Lorentz transformations in Eq. (1.2.179).

Relativistic Quantum Mechanics

76

Solution: We wish to find the commutation relations among the matrices M µν := 21 σ µν = i µ ν 4 [γ , γ ]. For calculating these, note that the anticommutation relations of the γ matrices imply

−γ ν γ µ for µ ̸= ν µ ν γ γ = (4.47) ±I for µ = ν, which in turn implies [γ µ γ ν , A] = −[γ ν γ µ , A] for all values of µ and ν and any matrix A. This simplifies our calculation greatly, for then i µ ν i [γ γ , M ρσ ] − [γ ν γ µ , M ρσ ] 4 4 i µ ν ρσ = [γ γ , M ] 2 i2 = [γ µ γ ν , γ ρ γ σ ] by the same logic.

4 Then using the commutator identities [M µν , M ρσ ] =

[A, BC] = [A, B]C + B[A, C]

and

[AB, C] = A{B, C} − {A, C}B,

(4.48)

(4.49) (4.50)

we get 1 [M µν , M ρσ ] = − ([γ µ γ ν , γ ρ ]γ σ + γ ρ [γ µ γ ν , γ σ ]) 4 1 µ ν ρ σ = − (γ {γ , γ }γ − {γ µ , γ ρ }γ ν γ σ + γ ρ γ µ {γ ν , γ σ } − γ ρ {γ µ , γ σ }γ ν ) 4 1 = − (g νρ γ µ γ σ − g µρ γ ν γ σ + g νσ γ ρ γ µ − g µσ γ ρ γ ν ) .

(4.51) 2 Now we note that 1 γ µ γ ν = ([γ µ , γ ν ] + {γ µ , γ ν }) 2 = −2iM µν + g µν , (4.52) and so, using the symmetry of g µν and antisymmetry of M µν , 1 [M µν , M ρσ ] = − (g νρ (−2iM µσ + g µσ ) − g µρ (−2iM νσ + g νσ ) 2 +g νσ (−2iM ρµ + g ρµ ) − g µσ (−2iM ρν + g ρν )) = i (g νρ M µσ − g µρ M νσ − g νσ M µρ + g µσ M νρ ) .

(4.53)

This is exactly the Lie algebra of the Lorentz transformations, as was to be shown.

4.5 Problem 5 Problem:

Problem 6

77

Explain why the hermiticity of γ 0 and γ 5 and the antihermiticity of γ are preserved in any representation unitarily equivalent to the Dirac representation.

Solution: Suppose H+ is a hermitian matrix such as γ 0 or γ 5 , and suppose H− is an antihermitian matrix such as γ i (working in the Dirac representation). Then we † can write H± = ±H± . In any representation unitarily equivalent to the Dirac representation, these matrices will be related to those in the original representation through a similarity transformation H̄± ≡ U H± U −1 = U H± U † ,

(4.54)

for some unitary matrix U . We see that † H̄± = (U H± U † )†

† † = (U † )† H± U

= U (±H± )U † = ±H̄± .

(4.55)

In other words, H̄+ is hermitian and H̄− is antihermitian; (anti)hermiticity is preserved in any representation unitarily equivalent to the Dirac representation. Note that the unitarity is important here because it allowed us to express the similarity transformation in terms of U † rather than U −1 , enabling the later manipulations.

4.6 Problem 6 Problem: Provide detailed proofs of the trace identities: (a) traces of products of γ matrices (with no γ µ† ) are representation independent (same for γ µ and γ̄ µ ≡ Sγ µ S −1 ). (b) tr (I) = 4, tr (γ µ γ ν ) = 4g µν , tr (γ µ ) = tr (odd # of γ µ′ s) = tr (σ µν ) = 0, tr (γ µ γ ν γ ρ γ σ ) = 4 (g µν g ρσ −g µρ g νσ +g µσ g νρ ).

(c) tr γ 5 γ µ γ ν = tr γ 5 (odd # of γ µ′ s) = 0 and tr γ 5 γ µ γ ν γ ρ γ σ = −4iϵµνρσ .

Solution: (a) Moving from one representation of γ matrices to another amounts to a change of basis. Since the trace of a matrix is the sum of its eigenvalues, it does not depend on the basis in which the matrix is expressed. Therefore, the trace of any product of γ matrices is representation independent.

To see this algebraically, suppose we have any collection of γ matrices γ µi ,

Relativistic Quantum Mechanics

78

i = 1, . . . , N , in a certain representation. In any other representation, these matrices will be of the form γ̄ µi = Sγ µi S −1 for some invertible matrix S. Then

tr (γ̄ µ1 γ̄ µ2 · · · γ̄ µN ) = tr Sγ µ1 S −1 Sγ µ2 S −1 · · · Sγ µN S −1

= tr Sγ µ1 γ µ2 · · · γ µN S −1

= tr γ µ1 γ µ2 · · · γ µN S −1 S = tr (γ µ1 γ µ2 · · · γ µN ) ,

(4.56)

where we have used the cyclic property of the trace1 in the third step. So we see the trace of any collection of γ matrices is independent of representation. (b)

(i) The 4×4 identity matrix is I = diag(1, 1, 1, 1), so tr(I) = 1 + 1 + 1 + 1 = 4. (ii) Using the anticommutation relations {γ µ , γ ν } = 2g µν I and the linearity and cyclic properties of the trace, we have 1 [tr (γ µ γ ν ) + tr (γ ν γ µ )] 2 1 = tr ({γ µ , γ ν }) 2 1 = 2g µν tr(I) 2 = 4g µν .

tr (γ µ γ ν ) =

(4.57)

(iii) Consider an odd number of γ matrices γ µi , i = 1, . . . , 2N + 1. Since γ 5 anticommutes with each of these matrices, we have γ 5 γ µ1 γ µ2 · · · γ µ2N +1 = (−1)2N +1 γ µ1 γ µ2 · · · γ µ2N +1 γ 5 .

(4.58)

Then since (γ 5 )2 = I, using the cyclic property of the trace we calculate

tr (γ µ1 γ µ2 · · · γ µ2N +1 ) = tr γ 5 γ 5 γ µ1 γ µ2 · · · γ µ2N +1

= (−1)2N +1 tr γ 5 γ µ1 γ µ2 · · · γ µ2N +1 γ 5 = −tr (γ µ1 γ µ2 · · · γ µ2N +1 ) .

(4.59)

We conclude the trace of any odd number of γ matrices must vanish. Note this includes the case for a single γ matrix. (iv) It follows from the cyclic property of the trace that the trace of any commutator vanishes (at least for finite-dimensional matrices). So then since σ µν = i[γ µ , γ ν ]/2, we must have tr(σ µν ) = 0. (v) Using the anticommutation relations along with our previous results, we calculate tr (γ µ γ ν γ ρ γ σ ) = 2g µν tr (γ ρ γ σ ) − tr (γ ν γ µ γ ρ γ σ )

= 2g µν · 4g ρσ − 2g µρ tr (γ ν γ σ ) + tr (γ ν γ ρ γ µ γ σ )

= 8g µν g ρσ − 2g µρ · 4g νσ + 2g µσ tr (γ ν γ ρ ) − tr (γ ν γ ρ γ σ γ µ )

1

This is the easily-proven property that tr(AB) = tr(BA) for any square matrices A and B of the same size.

Problem 7

79

= 8g µν g ρσ − 8g µρ g νσ + 2g µσ · 4g νρ − tr (γ µ γ ν γ ρ γ σ ) , (4.60) having used the cyclic property of the trace in the last step. Solving for tr(γ µ γ ν γ ρ γ σ ) yields the desired relation tr (γ µ γ ν γ ρ γ σ ) = 4 (g µν g ρσ − g µρ g νσ + g µσ g νρ ) .

(4.61)

Obviously, this approach generalizes to any even number of γ matrices. (c)

(i) Since γ 5 = iγ 0 γ 1 γ 2 γ 3 , any odd number of gamma matrices times γ 5 will be the product of an odd number of gamma matrices, and this has a vanishing trace by an earlier result. (ii) Now consider tr(γ 5 γ µ γ ν ). Let γ ρ be distinct from both γ µ and γ ν . Then γ ρ appears exactly once in the product γ 5 γ µ γ ν = iγ 0 γ 1 γ 2 γ 3 γ µ γ ν , and it anticommutes with the other five matrices in the product, meaning that γ ρ γ 5 γ µ γ ν = (−1)5 γ 5 γ µ γ ν γ ρ . Since (γ ρ )2 = ±I depending on ρ, we may calculate

tr γ 5 γ µ γ ν = ±tr γ ρ γ ρ γ 5 γ µ γ ν

= ±tr −γ ρ γ 5 γ µ γ ν γ ρ

= −tr γ 5 γ µ γ ν , (4.62) using the cyclic property of the trace in the last step. This implies the result tr(γ 5 γ µ γ ν ) = 0. (iii) Finally, consider the quantity tr(γ 5 γ µ γ ν γ ρ γ σ ). If any two of γ µ , γ ν , γ ρ , and γ σ are equal, then they can be brought together by some number of commutations with the other γ matrices. Since (γ µ )2 ∝ I, this leaves us with the trace of the product of γ 5 with only two other γ matrices. But this vanishes by the previous result, so we see that all of the γ matrices in the product must be distinct for the trace to be non-zero. Distinct γ matrices all anticommute, so this means tr(γ 5 γ µ γ ν γ ρ γ σ ) must be proportional to the antisymmetric tensor with four indices, ϵµνρσ . We can evaluate the constant of proportionality by considering the special case

tr γ 5 γ 0 γ 1 γ 2 γ 3 = −i tr γ 5 γ 5 = −i tr(I) = −4i = −4iϵ0123 , (4.63) implying that tr(γ 5 γ µ γ ν γ ρ γ σ ) = −4iϵµνρσ .

4.7 Problem 7 Problem: Prove the following identities: (a) γµ γν γρ = gµν γρ − gµρ γν + gνρ γµ + iϵµνρσ γ σ γ 5 .

Relativistic Quantum Mechanics

80

(b) γ µ γµ = 4, γ µ γ 5 γµ = −4γ 5 , γ µ ̸ kγµ = −2̸ k, γ µ γ 5 ̸ kγµ = 2γ 5 ̸ k, γ µ ̸ k̸ ℓγµ = 4k · ℓ, γ µ ̸ k̸ ℓ̸ pγµ = −2̸ p̸ ℓ̸ k, γ µ ̸ k̸ ℓ̸ p̸ qγµ = 2 (̸ q̸ k̸ ℓ̸ p + ̸ p̸ ℓ̸ k̸ q).

Solution: (a) We begin by performing a series of commutations: γµ γν γρ = 2gµν γρ − γν γµ γρ

= 2gµν γρ − 2gµρ γν + γν γρ γµ

= 2gµν γρ − 2gµρ γν + 2gνρ γµ − γρ γν γµ ,

(4.64)

giving 1 (γµ γν γρ + γρ γν γµ ) = gµν γρ − gµρ γν + gνρ γµ .

2 The required identity therefore follows as long as

(4.65)

1 (γµ γν γρ + γρ γν γµ ) = γµ γν γρ − iϵµνρσ γ σ γ 5 .

(4.66) 2 It is easy to see that this relation holds if any two of µ, ν, and ρ are equal, because the square of any γ matrix is proportional to the identity and therefore commutes with everything, and also ϵµνρσ will vanish. It remains to check the cases where µ, ν, and ρ are all distinct. Since distinct γ matrices anticommute, in these cases the above condition simplifies to γµ γν γρ = iϵµνρσ γ σ γ 5 .

(4.67)

Both sides here are completely antisymmetric, so it suffices to check the four cases (µ, ν, ρ) = (0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3). This is a tedious but rather straightforward application of the identities γ0 = γ 0 , γi = −γ i , (γ 0 )2 = I, (γ i )2 = −I, and the the anticommutivity of distinct γ matrices: iϵ012σ γ σ γ 5 = −ϵ0123 γ 3 γ 0 γ 1 γ 2 γ 3 = (−1)3 γ 0 γ 1 γ 2 γ 3 γ 3 = γ0 γ1 γ2 , σ 5

2 0 1 2 3

σ 5

1 0 1 2 3

σ 5

0 0 1 2 3

2 0 1 2 2 3

iϵ013σ γ γ = −ϵ0132 γ γ γ γ γ = −(−1) γ γ γ γ γ = γ0 γ1 γ3 , 1 0 1 1 2 3

iϵ023σ γ γ = −ϵ0231 γ γ γ γ γ = (−1) γ γ γ γ γ = γ0 γ2 γ3 , 1 2 3

iϵ123σ γ γ = −ϵ1230 γ γ γ γ γ = −γ γ γ = γ1 γ2 γ3 .

(4.68) (4.69) (4.70) (4.71)

We have now verified all distinct cases of our relation, proving that γµ γν γρ = gµν γρ − gµρ γν + gνρ γµ + iϵµνρσ γ σ γ 5 ,

(4.72)

as required. (b)

(i) With a slick application of the anticommutation relations, we have γ µ γµ = 1 µ ν µν µ 2 {γ , γ }gµν = g gµν I = δ µ I = 4I. Clearly, generalizing to d-dimensional µ spacetime will give γ γµ = dI. (ii) Since γ 5 anticommutes with all other γ matrices, we have from above γ µ γ 5 γµ = −γ µ γµ γ 5 = −4Iγ 5 = −4γ 5 .

Problem 8

81

(iii) Another application of the anticommutation relations gives γ µ ̸ kγµ = γ µ kν γ ν γµ = kν (2g µν − γ ν γ µ )γµ = 2̸ k − ̸ kγ µ γµ = 2̸ k − 4̸ k = −2̸ k. (iv) Similarly, γ µ γ 5 ̸ kγµ = −γ 5 γ µ ̸ kγµ = 2γ 5 ̸ k. (v) The same technique yields γ µ ̸ k̸ ℓγµ = kν ℓρ γ µ γ ν γρ γµ = kν ℓρ (2g µν I − γ ν γ µ )(2gρµ I − γµ γρ ) = (2k µ I − ̸ kγ µ )(2ℓµ I − γµ ̸ ℓ) = 4k · ℓI − 4̸ k̸ ℓI + ̸ kγ µ γµ ̸ ℓ = 4k · ℓ, using the first result. (vi) Going further, γ µ ̸ k̸ ℓ̸ pγµ = (2k µ − ̸ kγ µ )̸ ℓ̸ pγµ = 2̸ ℓ̸ p̸ k − ̸ k(4ℓ · p), from the previous result. Now ̸ ℓ̸ p = ℓµ pν γ µ γ ν = ℓµ pν (2g µν I − γ ν γ µ ) = 2ℓ · p − ̸ p̸ ℓ, so this reduces to γ µ ̸ k̸ ℓ̸ pγµ = −2̸ p̸ ℓ̸ k as desired. (vii) Finally, γ µ ̸ k̸ ℓ̸ p̸ qγµ = γ µ ̸ k̸ ℓ̸ p(2qµ −γµ ̸ q) = 2̸ q̸ k̸ ℓ̸ p−(−2̸ p̸ ℓ̸ k)̸ q = 2(̸ q̸ k̸ ℓ̸ p+ ̸ p̸ ℓ̸ k̸ q), again using the previous result.

4.8 Problem 8 Problem: Prove each of the identities: γ µ σ ρσ γµ = 0, σ µν σµν = 12, γ µ σ ρσ γ λ γµ = 2γ λ σ ρσ , µ λ ρσ γ γ σ γµ = 2σ ρσ γ λ , σ µν γ ρ σµν = 0, σ µν γ ρ γ σ σµν = 4 (4g ρσ −γ ρ γ σ ) and σ µν σ ρσ σµν = −4σ ρσ .

Solution: The following identities concern the quantities σ µν := 2i [γ µ , γ ν ]. For proving them, it helps to note the implications of the identities in Problem 4.5: γ µ ̸ kγµ = −2̸ k µ

γ ̸ k̸ ℓγµ = 4k · ℓ µ

γ ̸ k̸ ℓ̸ pγµ = −2̸ p̸ ℓ̸ k

=⇒ =⇒ =⇒

γ µ γ ν γµ = −2γ ν , µ ρ σ

(4.73)

ρσ

γ γ γ γµ = 4g , µ ρ σ λ

(4.74) λ σ ρ

γ γ γ γ γµ = −2γ γ γ .

(4.75)

(a) It follows from Eq. 4.74 that γ µ (γ ρ γ σ − γ σ γ ρ )γµ = 4(g ρσ − g σρ ) = 0, and therefore γ µ σ ρσ γµ = 0. (b) With a judicious relabeling of dummy indices, we can write σ µν σµν = (i2 /4)(γ µ γ ν − γ ν γ µ )(γµ γν − γν γµ ) = (−1/2)(γ µ γ ν γµ γν − γ µ γ ν γν γµ ).

(4.76)

Using Eq. 4.73 along with the result γ µ γµ = 4I, this reduces to σ µν σµν = (−1/2)(−8I − (4I)(4I)) = 12I. (c) From Eq. 4.75, we have γ µ σ ρσ γ λ γµ = (i/2)(γ µ γ ρ γ σ γ λ γµ − γ µ γ σ γ ρ γ λ γµ ) = −i(γ λ γ σ γ ρ − γ λ γ ρ γ σ ) = 2γ λ σ ρσ ,

(4.77)

Relativistic Quantum Mechanics

82

as required. (d) Similarly, γ µ γ λ σ ρσ γµ = (i/2)(γ µ γ λ γ ρ γ σ γµ − γ µ γ λ γ σ γ ρ γµ ) = −i(γ σ γ ρ γ λ − γ ρ γ σ γ λ ) = 2σ ρσ γ λ ,

(4.78)

as required. (e) The easiest way to calculate σ µν γ ρ σµν is to consider the form it must have. It is a linear combination of products of five γ matrices, with only γ ρ left uncontracted. These products can be simplified, of course, by repeated application of the anticommutation relations, which would introduce factors of g µν into the expression and nothing else. The only terms that might exist in the expression, then, are proportional to either g µν gµν γ ρ ∝ γ ρ , g µν γµ γν γ ρ ∝ γ ρ or g ρν γν = γ ρ .

That is, σ µν γ ρ σµν = kγ ρ for some constant k. Now multiplying by γρ from the right gives σ µν γ ρ σµν γρ = 4kI. The left-hand side vanishes by the result in Part (a), so we get k = 0 and therefore σ µν γ ρ σµν = 0. (f) Again by relabeling dummy indices, we can calculate σ µν γ ρ γ σ σµν = (i2 /4)2[γ µ γ ν γ ρ γ σ γµ γν − γ µ γ ν γ ρ γ σ γν γµ ] = (−1/2)[(−2γ σ γ ρ γ ν )γν − γ µ (4g ρσ )γµ ] = (−1/2)[−2γ σ γ ρ (4I) − 4g ρσ (4I)] = 4(2g ρσ + γ σ γ ρ ),

(4.79)

where we have used Eqs. 4.74 and 4.75. Now using the anticommutation relations, we arrive at the desired identity σ µν γ ρ γ σ σµν = 4(4g ρσ − γ ρ γ σ ). (g) Finally, we calculate with the previous result σ µν σ ρσ σµν = (i/2)[σ µν γ ρ γ σ σµν − σ µν γ σ γ ρ σµν ]

= 2i[(4g ρσ − γ ρ γ σ ) − (4g σρ − γ σ γ ρ )] = −2i[γ ρ , γ σ ] = −4σ ρσ ,

as required.

4.9 Problem 9 Problem:

(4.80)

Problem 9

83

Consider the sixteen matrices {Γ1 , Γ2 , . . . , Γ16 } ≡ {I, γ µ , σ µν |µ<ν , γ 5 γ µ , iγ 5 }.

Show that (a) γ 0 (ΓI )† γ 0 = ΓI in any representation unitarily equivalent to the Dirac representation; (b) Γ2I = ±1; (c) ΓI ΓJ = ±ΓJ ΓI ; (d) ΓJ ΓK = aJK L ΓL for some L (no sum implied) with aJK L = ±1, ±i; (e) Γ−1 I ΓJ ΓI = −ΓJ for J ̸= 1 for at least one ΓI ; (f) tr(ΓJ ) = 0 for J ̸= 1; (g) tr(Γ−1 Γ ) = tr(ΓI Γ−1 = 4δ ; J ) P16I J−1 P16 IJ −1 (h) I=1 (ΓI )ij (ΓI )kℓ = I=1 (ΓI )ij (ΓI )kℓ = 4δiℓ δkj ; P 16 (i) if I=1 cI ΓI = 0 then we must have all cI = 0 and hence the ΓJ are sixteen linearly independent matrices; and (j) using the result of part (i), explain why any complex 4×4 matrix can be written P16 as A = I=1 aI ΓI with aI ≡ 41 tr(Γ−1 I A). So these sixteen matrices form a basis for the set of all 4 × 4 complex matrices.

Solution: Throughout this question, it helps to keep in mind the following facts: distinct gamma matrices anticommute, and (γ 0 )2 = −(γ i )2 = I in any representation unitarily equivalent to the Dirac representation. For convenience, we write all of the Γ matrices here explicitly, grouped into the scalar, vector, tensor, axial-vector, and pseudo-scalar combinations of γ matrices: Γ1 : I, Γ2 , . . . Γ5 : γ 0 , γ 1 , γ 2 , γ 3 , Γ6 , . . . Γ11 : iγ 0 γ 1 , iγ 0 γ 2 , iγ 0 γ 3 , iγ 1 γ 2 , iγ 1 γ 3 , iγ 2 γ 3 , Γ12 , . . . Γ15 : −iγ 1 γ 2 γ 3 , −iγ 0 γ 2 γ 3 , iγ 0 γ 1 γ 3 , −iγ 0 γ 1 γ 2 , Γ16 : −γ 0 γ 1 γ 2 γ 3 .

(4.81)

Note that we have used γ 5 = iγ 0 γ 1 γ 2 γ 3 , and σ µν = 2i [γ µ , γ ν ] = iγ µ γ ν for µ ̸= ν. (a) Clearly for Γ1 we have γ 0 Γ†1 γ 0 = γ 0 γ 0 = I = Γ1 . The same equation holds for Γ2 , . . . Γ5 as a consequence of Eq. (4.4.56). We also see that γ 0 (iγ µ γ ν )† γ 0 = −iγ 0 γ ν † γ µ † γ 0 = −iγ 0 γ ν † γ 0 γ 0 γ µ † γ 0 = −iγ ν γ µ = iγ µ γ ν for µ ̸= ν, so the equation holds further for Γ6 , . . . Γ11 . It is easy to see the same technique can be used for Γ12 , . . . Γ16 to show that γ 0 (ΓI )† γ 0 = ΓI in general. This will hold in any representation unitarily equivalent to the Dirac representation as a result of the generalization of Eq. (4.4.58). (b) It is clear that Γ2I = ±I for I = 1, . . . , 5. Now for γ µ , γ ν distinct, we have (γ µ γ ν )2 = γ µ γ ν γ µ γ ν = −γ µ γ µ γ ν γ ν = ±I, so this also holds for I = 6, . . . , 11.

The same technique works for any product of distinct gamma matrices (with

84

Relativistic Quantum Mechanics coefficients that square to ±1) such as Γ12 , . . . Γ16 , so Γ2I = ±I in general. (c) ΓI ΓJ = ±ΓJ ΓI trivially holds for I = J, so it suffices to check this relation for I < J. The case I = 1 is also trivial, because Γ1 ΓJ = ΓJ = ΓJ Γ1 . All other cases amount to the statement that any product of gamma matrices is ±1 times the product of any permutation of those same matrices, but this is a straightforward consequence of all distinct gamma matrices anticommuting.

To illustrate, consider Γ14 Γ15 = (iγ 0 γ 1 γ 3 )(−γ 0 γ 1 γ 2 γ 3 ) = (−1)3 (iγ 0 γ 1 )(−γ 0 γ 1 γ 2 γ 3 )γ 3 = (−1)3 (−1)3 (iγ 0 )(−γ 0 γ 1 γ 2 γ 3 )γ 1 γ 3 = (−1)3 (−1)3 (−1)3 (−γ 0 γ 1 γ 2 γ 3 )(iγ 0 γ 1 γ 3 ) = (−1)3 (−1)3 (−1)3 Γ15 Γ14 .

(4.82)

Hence, ΓI ΓJ = ±ΓJ ΓI for all I, J. (d) The Γ matrices consist of all of the distinct products of the γ matrices. That is, any product of γ matrices can be simplified by commuting the distinct matrices past each other (picking up factors of −1) and using (γ µ )2 = ±I until there are only distinct γ matrices in the product, which will match ΓL for some L up to a phase factor of ±1 or ±i. Since ΓJ ΓK is a product of gamma matrices, it too is subject to this logic (except if J = 1 or K = 1, but then the result is trivial), so ΓJ ΓK = aJK L ΓL for some L, with aJK L = ±1, ±i. (e) We can rewrite the desired relation Γ−1 I ΓJ ΓI = −ΓJ in perhaps a more familiar form ΓJ ΓI = −ΓI ΓJ . This is obviously symmetric in I and J, so it suffices to check this relation holds for values of I and J that collectively exhaust the elements in the set {2, . . . , 16}. Now since γ 5 anticommutes with every product of an odd number of γ matrices (see Eq. 4.58), we see the relation holds for J = 16 and I = 2, . . . , 5, and also I = 12, . . . , 15. The remaining cases to check are J = 6, . . . , 11, where ΓJ = iγ µ γ ν for some µ ̸= ν. If we take I such that ΓI = γ µ , then we see ΓJ ΓI = iγ µ γ ν γ µ = −iγ µ γ µ γ ν = −ΓI ΓJ , so the relation also holds for all such J for some values of I. This proves that for any J ̸= 1, there is some I such that Γ−1 I ΓJ ΓI = −ΓJ . (f) Taking the trace of the above result, we see for J ̸= 1 −1 tr(ΓJ ) = tr(−Γ−1 I ΓJ ΓI ) = −tr(ΓJ ΓI ΓI ) = −tr(ΓJ ),

(4.83)

having used the cyclic property of the trace. This implies tr(ΓJ ) = 0 for J ̸= 1, as desired. (g) If I = J, then clearly tr(Γ−1 I ΓJ ) = tr(I) = 4. Let us now suppose that I ̸= J. From Part (b), we know that Γ−1 I = ±ΓI for all I. Then from Part (d), we get

Problem 9

85

L Γ−1 I ΓJ = ±ΓI ΓJ = ±aIJ ΓL for some L. Now if L = 1 then ΓI ΓJ , a product of γ matrices in which at least one γ µ appears an odd number of times, would be proportional to the identity. This cannot be the case because the simplification process outlined in Part (d) will never be able to remove such a γ µ from the product completely, meaning the product must be proportional to some ΓL where L ̸= 1, and such a ΓL cannot be proportional to the identity as a result of L Part (f). All that is to say, Γ−1 I ΓJ = ±aIJ ΓL for some L ̸= 1, so from Part (f) −1 L we get tr(Γ−1 I ΓJ ) = ±aIJ tr(ΓL ) = 0. This shows that tr(ΓI ΓJ ) = 4δIJ , and the I ↔ J symmetry on the right-hand side here, along with the cyclic property of the trace, further imply tr(ΓI Γ−1 J ) = 4δIJ , completing the result.

(h) Here we assume the result of Part (j), that any 4×4 complex matrix A can be written as 16

A=

1X tr(Γ−1 I A)ΓI .

4

(4.84)

I=1

Expressed in component form, this is 16

Akℓ =

4

4

1 X X X −1 (ΓI )ij Aji (ΓI )kℓ .

4 i=1 j=1

(4.85)

I=1

But also by definition Akℓ =

4 X 4 X

δiℓ δkj Aji ,

(4.86)

i=1 j=1

and since these equations hold for any set of complex numbers Akℓ , it must be that 16 X

(Γ−1 I )ij (ΓI )kℓ = 4δiℓ δkj .

(4.87)

I=1

Furthermore, we can exploit the (i, ℓ) ↔ (k, j) symmetry on the right-hand side of this equation to also write 16 X

(ΓI )ij (Γ−1 I )kℓ = 4δkj δiℓ ,

(4.88)

I=1

as was to be shown. (i) Suppose we have 0=

16 X

cI ΓI

(4.89)

I=1

for some coefficients cI . We multiply from the left by Γ−1 J for any J and take

Relativistic Quantum Mechanics

86

the trace, finding 0=

16 X

cI tr Γ−1 J ΓI

I=1

=

16 X

4cI δIJ

I=1

= 4cJ

(4.90)

by Part (g). We see that we must have cI = 0 for I = 1, . . . , 16. Hence, there is no non-trivial linear combination of Γ matrices that vanishes: these sixteen matrices are linearly independent. (j) The space of complex 4×4 matrices is 42 = 16 (complex-) dimensional, so any set of sixteen linearly independent complex matrices, such as the Γ matrices, form a basis. Any complex 4×4 matrix A, for example, can then be written as A=

16 X

aI ΓI

(4.91)

I=1

for some complex coefficients aI . We can use the same technique as in Part (i) to calculate for any J tr(Γ−1 J A) =

16 X

aI tr(Γ−1 J ΓI ) =

I=1

16 X

4aI δIJ = 4aJ .

(4.92)

I=1

That is, the coefficients are given by aI =

1 tr(Γ−1 I A), 4

(4.93)

as was to be shown.

4.10 Problem 10 Problem: (a) Prove that if ψ(x) is a solution of the Dirac equation, Eq. (4.4.59), then it is also a solution of the Klein-Gordon equation, Eq. (4.3.12). (b) Note that ∂µ and ∂ν commute, [∂µ , ∂ν ]ψ = 0, whereas Dµ and Dν do not. Evaluate [Dµ , Dν ]ψ = Dµ (Dν ψ) − Dν (Dµ ψ) and so prove [Dµ , Dν ] = i(q/ℏc)Fµν . (c) Prove that if ψ(x) is a solution of the Dirac equation in the presence of an electromagnetic field, of a modified Klein Eq. (4.4.61), that it is also a solution

Gordon equation, Dµ Dµ + (q/2ℏc)σ µν Fµν + (mc/ℏ)2 ψ = 0.

Solution:

87

Problem 10 (a) Let us assume that ψ is a solution of the Dirac equation, (iℏγ ν ∂ν − mc)ψ = 0.

Act on this equation from the left by the operator (iℏγ µ ∂µ + mc) to obtain 0 = (iℏγ µ ∂µ + mc)(iℏγ ν ∂ν − mc)ψ = (−ℏ2 γ µ γ ν ∂µ ∂ν − m2 c2 )ψ.

(4.94)

Now ∂µ ∂ν is symmetric in the indices µ and ν, so this contraction picks out only the symmetric part of γ µ γ ν = 12 ({γ µ , γ ν } + [γ µ , γ ν ]):

mc 2 1 0 = −ℏ2 {γ µ , γ ν }∂µ ∂ν + ψ 2 ℏ

mc 2 = −ℏ2 g µν ∂µ ∂ν + ψ ℏ

mc 2 ψ, (4.95) =⇒ 0 = ∂2 + ℏ hence, ψ is also a solution of the Klein-Gordon equation. Conversely, let us now assume that ψ is a solution of the Klein-Gordon equation, but not necessarily of the Dirac equation. We want to show that there is at least one such ψ that is in fact not a solution of the Dirac equation. With the same logic as above, we may factorize the Klein-Gordon equation as

mc 2 2 0= ∂ + ψ ℏ 1 (4.96) = − 2 (iℏγ µ ∂µ − mc)(iℏγ ν ∂ν + mc)ψ, ℏ and therefore any solution of the equation (iℏγ ν ∂ν +mc)ψ = 0 will be a suitable candidate. Now if a ψ from this class of solutions also satisfies the Dirac equation (iℏγ ν ∂ν − mc)ψ = 0, then by taking the difference of the two equations we get 2mcψ = 0. In other words, any non-trivial solution of (iℏγ ν ∂ν + mc)ψ = 0 will be a solution of the Klein-Gordon equation, but not of the Dirac equation.

Many such non-trivial solutions exist (simply take the known solutions of the Dirac equation and substitute m → −m), and therefore a solution of the KleinGordon equation need not be a solution of the Dirac equation. (b) By direct calculation, taking care to use the product rule where necessary, [Dµ , Dν ]ψ = Dµ (Dν ψ) − Dν (Dµ ψ)

q q q q = ∂µ + i Aµ ∂ν ψ + i Aν ψ − ∂ν + i Aν ∂µ ψ + i Aµ ψ ℏc ℏc ℏc ℏc q 2 q = ∂µ ∂ν ψ + i [(∂µ Aν )ψ + Aν ∂µ ψ + Aµ ∂ν ψ] + i Aµ Aν ψ ℏc ℏc q 2 q [(∂ν Aµ )ψ + Aµ ∂ν ψ + Aν ∂µ ψ] + i Aν Aµ ψ − ∂ν ∂µ ψ − i ℏc ℏc q = i (∂µ Aν − ∂ν Aµ )ψ ℏc q = i Fµν ψ.

(4.97) ℏc

Relativistic Quantum Mechanics

88

q That is, [Dµ , Dν ] = i ℏc Fµν . Note that we have assumed that Aµ and Aν commute, as is the case for the electromagnetic field. Generalizing to nonabelian gauge fields yields the same formula, as long as Fµν is redefined to include the [Aµ , Aν ] term above.

(c) Now suppose ψ is a solution of the modified Dirac equation (iℏγ ν Dν −mc)ψ = 0.

Using the same technique as above, we operate with −(iℏγ µ Dµ + mc)/ℏ2 from the left to obtain a Klein-Gordon-like equation: 1 0 = − 2 (iℏγ µ Dµ + mc)(iℏγ ν Dν − mc)ψ ℏ mc 2 ψ. = γ µ γ ν Dµ Dν + ℏ

(4.98)

The trick here is to decompose γ µ γ ν and Dµ Dν into their symmetric and antisymmetric components, and then use the known (anti)commutation relations to simplify. Contractions between symmetric components and antisymmetric components vanish, so we get

mc 2 1 1 0= ({γ µ , γ ν } + [γ µ , γ ν ]) · ({Dµ , Dν } + [Dµ , Dν ]) + ψ 2 2 ℏ

mc 2 1 µ ν 1 = {γ , γ }{Dµ , Dν } + [γ µ , γ ν ][Dµ , Dν ] + ψ 4 4 ℏ

mc 2 i q 1 µν g {Dµ , Dν } + − σ µν i Fµν + = ψ 2 2 ℏc ℏ

mc 2 q µν σ Fµν + ψ, (4.99) =⇒ 0 = Dµ Dµ + 2ℏc ℏ as desired.

4.11 Problem 11 Problem: Using the standard representation of the Pauli spin matrices show that complex conjugation with a rotation of π about the y-axis reverses the orientation of the spin, i.e., show that (iσ 2 )σ ∗ (iσ 2 )† = −σ.

Solution: In the standard representation, the Pauli spin matrices are given by

0 1 0 −i 1 0 1 2 3 σ = , σ = , σ = .

1 0 i 0 0 −1

(4.100)

Problem 12

89

Arriving at the desired result is then an exercise in matrix multiplication:

0 1 0 1 0 −1 0 −1 (iσ 2 )σ 1 ∗ (iσ 2 )† = = = −σ 1 , (4.101) −1 0 1 0 1 0 −1 0

0 1 0 i 0 −1 0 i (iσ 2 )σ 2 ∗ (iσ 2 )† = = = −σ 2 , (4.102) −1 0 −i 0 1 0 −i 0

0 1 1 0 0 −1 −1 0 2 3∗ 2 † (iσ )σ (iσ ) = = = −σ 3 .

(4.103) −1 0 0 −1 1 0 0 1 In total, (iσ 2 )σ ∗ (iσ 2 )† = −σ.

4.12 Problem 12 Problem: Prove directly from the definition of M (Λ) and M (Λ) in Eq. (4.6.14) that iσ 2 χ∗ transforms with M (Λ) and −iσ 2 ϕ∗ transforms with M (Λ) and hence verify that the charge conjugate of the Dirac spinor in the chiral representation in Eq. 4.6.44 transforms as a Dirac spinor.

Solution: The matrices M (Λ) and M (Λ) are defined for a restricted Lorentz transformation Λ that is parameterized by the rotation and boost parameters (αn, ηnb ), as in the main text, by M (Λ) = exp[(−iαn − ηnb ) · σ/2],

(4.104)

M (Λ) = exp[(−iαn + ηnb ) · σ/2].

(4.105)

These are the matrices under which the Weyl spinor components of a Dirac spinor (in the chiral representation) transform. That is, for a Dirac spinor

ϕ ψ= (4.106) χ in the chiral representation, the components transform under the Lorentz transformation Λ as ϕ → M (Λ)ϕ,

(4.107)

χ → M (Λ)χ.

(4.108)

Therefore, the quantities Φ := −iσ 2 ϕ∗ and X := iσ 2 χ∗ transform as Φ → −iσ 2 (M (Λ)ϕ)∗ = (iσ 2 )[M (Λ)]∗ (iσ 2 )−1 Φ,

X→

2

∗

2

∗

2 −1

iσ (M (Λ)χ) = (iσ )[M (Λ)] (iσ )

X.

Now since (iσ 2 )(iσ 2 )−1 = I, it follows that (iσ 2 )[M (Λ)]∗ (iσ 2 )−1 = (iσ 2 ) exp[(+iαn − ηnb ) · σ ∗ /2](iσ 2 )−1

(4.109) (4.110)

Relativistic Quantum Mechanics

90

= exp[(+iαn − ηnb ) · ((iσ 2 )σ ∗ (iσ 2 )−1 )/2],

(4.111)

because the two expressions have the same Taylor expansions, and similarly (iσ 2 )[M (Λ)]∗ (iσ 2 )−1 = (iσ 2 ) exp[(+iαn + ηnb ) · σ ∗ /2](iσ 2 )−1

= exp[(+iαn + ηnb ) · ((iσ 2 )σ ∗ (iσ 2 )−1 )/2].

To calculate these, we note that

0 (iσ 2 )−1 = −1

−1 1 0 = 0 1

−1 = (iσ 2 )† , 0

(4.112)

(4.113)

and therefore from the result in Problem 4.4, we have (iσ 2 )[M (Λ)]∗ (Λ)(iσ 2 )−1 = exp[(+iαn − ηnb ) · ((iσ 2 )σ ∗ (iσ 2 )† )/2] = exp[(+iαn − ηnb ) · (−σ)/2] = M (Λ),

2

∗

2 −1

(iσ )[M (Λ)] (iσ )

2

∗

(4.114) 2 †

= exp[(+iαn + ηnb ) · ((iσ )σ (iσ ) )/2] = exp[(+iαn + ηnb ) · (−σ)/2] = M (Λ).

(4.115)

In total, we see that Φ → M (Λ)Φ and X → M (Λ)X. That is, −iσ 2 ϕ∗ transforms with the matrix M (Λ), and iσ 2 χ∗ transforms with M (Λ). But X and Φ are merely the components of the charge conjugate of the Dirac spinor ψ, 2 ∗ iσ χ X 0 ∗ ψC ≡ Cγ ψ = = , (4.116) −iσ 2 ϕ∗ Φ so we see ψC has the transformation properties of a Dirac spinor, Eq. 4.108.

5

Introduction to Particle Physics 5.1 Problem 1 Problem: Show that the area of two other of the six nontrivial unitarity triangles of the CKM matrix are equal to half of the Jarlskog invariant, J. Solution: We have from Eq. (5.2.15) the defining expression for the Jarlskog invariant X ∗ Im(Vij Vkℓ Viℓ∗ Vkj )=J ϵikm ϵjℓn , (5.1) m,n

where no sum is implied on the left-hand side. A proof has been given in the text that the area of the unitarity triangle defined by ∗ Vcd Vcb∗ + Vtd Vtb∗ + Vud Vub =0

(5.2)

is equal to J/2. The proof carries over verbatim to all of the five other unitarity triangles, ∗ ∗ Vud Vus + Vcd Vcs + Vtd Vts∗ = 0, ∗ Vus Vub + Vcs Vcb∗ + Vts Vtb∗ = 0, ∗ ∗ Vud Vcd + Vus Vcs + Vub Vcb∗ = 0,

Vud Vtd∗ + Vus Vts∗ + Vub Vtb∗ = 0, Vcd Vtd∗ + Vcs Vts∗ + Vcb Vtb∗ = 0,

(5.3)

with the appropriate substitution of index labels (under the identifications u, d ∼ 1, c, s ∼ 2, and t, b ∼ 3). For example, consider the last of these, ∗ ∗ ∗ V21 V31 + V22 V32 + V23 V33 = 0.

(5.4)

∗ This represents a triangle in the complex plane with vertices at the points V21 V31 , ∗ ∗ ∗ ∗ ∗ V21 V31 + V22 V32 , and V21 V31 + V22 V32 + V23 V33 = 0. If we rescale the plane by a ∗ factor V21 V31 , then the vertices will instead lie at the points 0, 1, and ∗ ∗ ∗ V21 V31 + V22 V32 V23 V33 = − ∗ ∗ V21 V31 V21 V31

(5.5)

by virtue of the defining equation of the triangle. This rescaled triangle has a base 91

92

Introduction to Particle Physics

length of b = 1 and a height of ∗ V23 V33 h = Im − ∗ V21 V31

,

(5.6)

and therefore a signed area

∗ 1 1 V23 V33 . bh = Im − ∗ 2 2 V21 V31

(5.7)

Since areas scale as distance squared, the area A of the unscaled triangle must be

∗ ∗ ∗ 2 1 1 V23 V33 V23 V33 |V21 V31 | ∗ 21 ∗ ∗ = = − Im (V23 V33 A = |V21 V31 | Im − Im − V21 V31 ) , ∗ ∗ 2 V21 V31 2 V21 V31 2 (5.8) having used the identity |z|2 = zz ∗ for complex numbers z. A more direct way to arrive at this result would be to find the area as half of the determinant of the matrix (⃗ p1 , p⃗2 ), where p⃗1 and p⃗2 are the vector coordinates of the two non-zero triangle vertices. In any case, we recognize this expression for the area appears in the definition of the Jarlskog invariant, 1 X 1 1 1 ∗ ∗ V21 ) = J ϵ23m ϵ31n = Jϵ231 ϵ312 = J(+1)(+1), (5.9) A = − Im (V23 V31 V33 2 2 m,n 2 2 and so A = J/2 as claimed. That this proof works for all such triangles indicates that J is well-defined.

5.2 Problem 2 Problem: In some circumstances the contribution of one neutrino flavor can be approximately neglected: e.g., in νµ → ντ atmospheric oscillations the role played by the νe is very small. Then the PMNS matrix reduces to a single 2 × 2 matrix of the Cabibbo form with a single mixing angle, θ. Show that in this case (in natural units) we have Pℓ,ℓ′ ;ℓ̸=ℓ′ = sin2 (2θ) sin2 (∆m2ℓℓ′ L/4E).

Solution: In the ultrarelativistic limit that applies to neutrinos, we have seen from Eq. (5.2.34) that the neutrino mixing probabilities are given by !

X ∆m2jk L ∗ ∗ Pℓℓ′ = Uℓj Uℓ′ j Uℓk Uℓ′ k exp −i , (5.10) 2E j,k

where U is the neutrino mixing matrix and the sums are over all neutrino flavours

Problem 2

93

that contribute to the mixing. Here we are only considering the case of mixing between two neutrino flavours, so the mixing matrix has the Cabibbo form

U11 U12 cos θ sin θ U= = (5.11) U21 U22 − sin θ cos θ for some mixing angle θ. Substituting these matrix values into the above expression will lead to the desired result, but we will take a slightly more sophisticated approach. Notice that U is a purely real matrix, so we can manipulate the probability expression into ! ! # " # " 2 X X m L m2j L k T T Ujℓ Ukℓ Uℓk exp +i Pℓℓ′ = Uℓj exp −i ′ ′ 2E 2E j k

=

X j

Uℓj exp −i

≡ (V V T )ℓℓ′

2

m2j L 2E

2

!

T Ujℓ ′

,

(5.12)

where Vℓj := exp(−im2j L/4E)Uℓj . Explicitly, we have    m2 m2 m2 m2 1L 2L 1L 1L −i 4E −i 4E −i 4E −i 4E e cos θ e sin θ  e cos θ −e sin θ VVT = m2 m2 m2 m2 1L 2L 2L 2L −i 4E −i 4E −i 4E −i 4E −e sin θ e cos θ e sin θ e cos θ   2 2 2 m1 L m2 L m2 L m2 1L 2 −i 2E 2 −i 2E −i 2E −i 2E e cos θ + e sin θ sin θ cos θ(e −e ) . (5.13) = m2 m2 m2 m2 1L 1L 2L 2L 2 −i 2E −i 2E −i 2E 2 −i 2E −e ) e sin θ + e cos θ sin θ cos θ(e This matrix is symmetric, so to find Pℓℓ′ ;ℓ̸=ℓ′ we need only consider the absolute square of one of the off-diagonal elements:

2 m2′ L m2 ℓL ℓ Pℓℓ′ ;ℓ̸=ℓ′ = sinθ cosθ e−i 2E − e−i 2E

m2′ L m2′ L m2 m2 ℓL ℓ ℓL ℓ 2 = (sinθ cosθ) e−i 2E − e−i 2E e+i 2E − e+i 2E 2

∆m2 ′ L ∆m2′ L sin(2θ) ℓℓ ℓ ℓ −i 2E −i 2E = 1−e −e +1 2

∆m2 ′ L ∆m2 ′ L sin2 (2θ) ℓℓ ℓℓ = 2 − e−i 2E − e+i 2E 4 2   ∆mℓℓ′ L 1 − cos 2E 2  = sin (2θ)  2

∆m2ℓℓ′ L 2 = sin (2θ) sin , 4E

as desired.

(5.14)

94

Introduction to Particle Physics

5.3 Problem 3 Problem: Show that the set of permutations of n symbols Sn satisfies the properties of a group. Recall that the four properties of a group are closure, associativity, existence of the identity and the existence of inverses.

Solution: A permutation of n symbols is an operation that changes the ordering of those symbols. Sn is the set of all such permutations together with the operation ◦ of composing those permutations successively. We may think of a permutation σ ∈ Sn as a map σ : {1, 2, . . . , n} → {1, 2, . . . , n} i 7→ σ(i),

(5.15)

signifying that σ maps the ith symbol into the position σ(i) for all i = 1, . . . , n. Now σ cannot map any two symbols to the same position by definition of a permutation, so it is injective and therefore also bijective because its domain and codomain are the same finite sets. Conversely, any bijection σ : {1, 2, . . . , n} → {1, 2, . . . , n} is a permutation in Sn . We can also adopt the permutation notation of Section 5.3.1 to write

1 2 ··· n σ= , (5.16) σ(1) σ(2) · · · σ(n) where we note the ordering of the columns does not matter. Knowing this, we can now prove the group properties hold for Sn . (i) Consider the actions on n symbols induced by the two permutations σ, τ ∈ Sn .

Under σ, the ith element moves to position σ(i), and subsequently to position τ (σ(i)) under the action of τ . In other words, successively composing τ with σ amounts to ordinary function composition: (τ ◦ σ)(i) = τ (σ(i)).

(5.17)

The composition of two bijections is again a bijection, so τ ◦ σ is again a permutation of n elements, showing the set Sn is closed under the composition operation. (ii) For σ, τ, ρ ∈ Sn we have [(ρ ◦ τ ) ◦ σ](i) ≡ (ρ ◦ τ )(σ(i)) ≡ ρ(τ (σ(i))) ≡ ρ(τ ◦ σ(i)) ≡ [ρ ◦ (τ ◦ σ)](i), (5.18) implying (ρ ◦ τ ) ◦ σ = ρ ◦ (τ ◦ σ), i.e., the composition of permutations is associative.

Problem 4

95

(iii) The identity element is simply the permutation that takes all elements to themselves, e(i) = i, or

1 2 ··· n e= , (5.19) 1 2 ··· n as σ ◦ e(i) = σ(i) = e ◦ σ(i) for all σ ∈ Sn . (iv) Bijective functions have unique inverses that are also bijective, so for any σ ∈ Sn there is an inverse permutation σ −1 ∈ Sn . More conceptually, the inverse of σ should map σ(i) to i, and so can be written explicitly as

σ(1) σ(2) · · · σ(n) −1 σ = .

(5.20) 1 2 ··· n It is clear this expression is well-defined because {σ(1), σ(2), . . . , σ(n)} is simply a reordering of {1, 2, . . . , n}. Then by direct computation we see

σ(1) σ(2) · · · σ(n) 1 2 ··· n −1 σ σ= 1 2 ··· n σ(1) σ(2) · · · σ(n)

1 2 ··· n = 1 2 ··· n = e,

(5.21)

and similarly σσ −1 =

1 σ(1)

2 σ(2)

σ(1) σ(1)

σ(2) σ(2)

1 2 1 2

··· ···

= =

= e,

··· ···

··· ···

n n

n σ(1) σ(n) 1

σ(n) σ(n)

σ(2) 2

··· ···

σ(n) n

(5.22)

so σ −1 is indeed the inverse permutation of σ. This proves that Sn fulfils the definition of a group.

5.4 Problem 4 Problem: Draw the weight diagram for the (6, 3)D multiplet of SU (3) and indicate the multiplicities of each layer. Use this to calculate the number of states in the irrep/multiplet. Draw the corresponding Young diagram and use the factors over hooks rule to calculate the the number of dimensions of the irrep/multiplet and verify that the two results agree.

Introduction to Particle Physics

96

n = 6 steps m = 3 steps

4

4

3

2

1

n = 6 steps

t

Figure 5.1

m = 3 steps Weight diagram for the (6, 3)D multiplet of SU (3), with multiplicities of the points in each layer shown.

Table 5.1 Number of points in each layer of the weight diagram Figure 5.1, and their multiplicities.

Layer

Number of points

Multiplicity

1 2 3 4 5

27 21 15 9 1

1 2 3 4 4

Solution: From the procedure outlined in the text, we know the weight diagram for the (6, 3)D multiplet of SU (3) consists of all points in a regular triangular (isometric) grid that are bounded by an irregular hexagon with alternating sidelengths consisting of 6 + 1 and 3 + 1 lattice points, or 6 and 3 steps between points respectively, as shown in Figure 5.1. Each hexagonal or triangular layer in this grid, working from the outside inwards, is connected by a dashed line indicating all points on that line have the same multiplicity. The multiplicities increase from 1 in the outermost layer to 4 in the first triangular layer, at which point they remain 4 for all further inner layers. For convenience, the number of points in each layer and their multiplicities are listed in Table 5.1.

The number of states in the multiplet is simply the number of points in the weight diagram, with multiplicities included. We may therefore calculate the number of

Problem 5

97

states N from Table 5.1 as N = (27 × 1) + (21 × 2) + (15 × 3) + (9 × 4) + (1 × 4) = 154.

(5.23)

This number should be equal to the dimensionality of the representation. Let us check this by calculating the dimension directly with the “factors over hooks” rule.

The notation (6, 3)D signifies a Young diagram with 3 boxes in its last row and 6 + 3 = 9 boxes in the row above that:

. This diagram has a

factor and hook F : 3 4 5 6 7 8 9 10 11 2 3 4

=⇒

H : 10 9 8 6 5 4 3 2 1 3 2 1

=⇒

11! × 4! = 12!, 2!

10!

H= × 3!, 7 F =

(5.24)

and therefore denotes the representation of dimension F/H = 12 × 11 × 7/3! = 154, matching the number of states N calculated from the weight diagram as expected.

5.5 Problem 5 Problem: Draw the Young diagram for the (1, 0, 1)D irrep/multiplet of SU (4). Draw all standard Young tableaux for this irrep/multiplet and construct the corresponding Young symmetrizers.

Solution: The notation λ = (1, 0, 1)D indicates a Young diagram with 1 box in its last row, 0 + 1 = 1 box in the row above that, and 1 + 0 + 1 = 2 boxes in the top row: λ = (1, 0, 1)D

=⇒

.

(5.25)

A standard Young tableaux for this diagram is one in which the boxes are filled with the numbers 1 − 4, with the numbers increasing along the top row and increasing down the first column. This implies the top-left box must be labeled 1 in any standard Young tableaux, while 2 can be put in either box next to it. It is not difficult to exhaust all possible standard Young tableaux working in this way; there are only three distinct possibilities: 1 2 λ1 = 3 , 4

1 3 λ2 = 2 , 4

1 4 λ3 = 2 .

3

(5.26)

The corresponding Young symmetrizers are those combinations of permutations that completely antisymmetrize the symbols in each column, and completely symmetrize the symbols in each row. The symmetric combination of the two symbols

98

Introduction to Particle Physics

(1, 2), for example, is (1, 2) + (2, 1), which is obtained by operating upon (1, 2) with the combination of permutations (e) + (12). And the antisymmetric combination of the three symbols (1, 3, 4), for example, is (1, 3, 4) − (3, 1, 4) − (4, 3, 1) − (1, 4, 3) + (3, 4, 1) + (4, 1, 3), which is obtained by operating upon (1, 3, 4) with the combination of permutations (e) − (13) − (14) − (34) + (134) + (143). So we see the Young symmetrizer corresponding to λ1 is Ŷ1 = [(e) − (13) − (14) − (34) + (134) + (143)][(e) + (12)] = (e) − (13) − (14) − (34) + (134) + (143)

(12) − (13)(12) − (14)(12) − (34)(12) + (134)(12) + (143)(12)

= (e) − (13) − (14) − (34) + (134) + (143)

(12) − (123) − (124) − (34)(12) + (1234) + (1243).

(5.27)

The symmetrizers for the other two Young tableaux can be found similarly, but it is quicker to realize they can also be obtained by permuting the symbols in the above result in accordance with the diagrams, Ŷ2 = (e) − (12) − (14) − (24) + (124) + (142)

(13) − (132) − (134) − (24)(13) + (1324) + (1342),

Ŷ3 = (e) − (12) − (13) − (23) + (123) + (132)

(14) − (142) − (143) − (23)(14) + (1423) + (1432),

(5.28)

because the labels are arbitrary.

5.6 Problem 6 Problem: Prove the SU (2) results in Eq. (5.3.28): 2 ⊗ 2 = 3 ⊕ 1; 3 ⊗ 3 = 5 ⊕ 3 ⊕ 1; and 1 +j2 2j1 +1 ⊗ 2j2 +1 = ⊕jJ=|j (2J+1).

1 −j2 | Solution: Note that the first and second results are simply special cases of the last, but we shall prove them separately to make the general process clear. (a) In SU (2), the 2 representation is denoted by the Young diagram process outlined in the text, we have ⊗ a =

a ⊕

a

.

. Using the

(5.29)

Both of the diagrams on the right-hand side give admissible sequences, allowing us to conclude ⊗

=

⊕

.

(5.30)

Problem 6

99

The right-most diagram here is the singlet 1 representation of SU (2) while is the representation of dimension F/H = (2 × 3)/(2 × 1) = 3, showing that 2 ⊗ 2 = 3 ⊕ 1. (b) Knowing that the 3 representation of SU (2) is given by the Young diagram , we calculate ⊗ a a =

a a ⊕

a

a ⊕

a a

⊕ a a

.

(5.31)

Now the right-most diagram here has more than N = 2 rows, so it is not a valid Young diagram for SU (2) and should be discarded. Similarly, the columns of length two can be ignored because they do not affect the dimensions of their diagrams. In total, we get ⊗ The diagram

⊕

=

⊕

.

(5.32)

has a factor and hook F : 2 3 4 5

=⇒

F = 5!,

H: 4 3 2 1

=⇒

H = 4!,

(5.33)

and so is the representation of dimension F/H = 5!/4! = 5, showing that 3 ⊗ 3 = 5 ⊕ 3 ⊕ 1. (c) The above calculation clearly generalizes to the case of the diagram n

z

}| { ···

(5.34)

to show it is the representation of dimension F/H = (n + 1)!/n! = n + 1.

The 2j + 1 representation is therefore given by the single-row diagram with 2j boxes. Now suppose for simplicity (and without loss of generality) that j2 ≤ j1 , so j1 − j2 = |j1 − j2 |. Then we have 2j

2j1

z

2j2 }| { z }|2 { M ··· ⊗ a a ··· a = k=0

2j1 +2j2 −k

z

}| { ··· ··· a a ··· a , a a ··· a

(5.35)

where the bottom row has k boxes. Here we have used the fact that the only diagrams that are valid are those containing at most two rows. Note that our condition j2 ≤ j1 ensures the second row is always shorter than the first in this sum, as it must be. Now the k columns of length two can be ignored, reducing the sum to 2j1

z

2j

2j2 z }| { z }|2 { M ··· ⊗ a a ··· a = k=0

(2j1 +2j2 −k)−k

···

}| { a a ··· a .

(5.36)

Introduction to Particle Physics

100

The representations in this sum have dimension 2j1 +2j2 −2k +1, by the above.

Defining J = j1 + j2 − k, we are therefore summing over the representations 2J + 1 for J between j1 + j2 − 2j2 = |j1 − j2 | and j1 + j2 − 0, so in total (2j1 + 1) ⊗ (2j2 + 1) =

jM 1 +j2

(2J + 1),

(5.37)

J=|j1 −j2 |

as required.

5.7 Problem 7 Problem: Prove the SU (3) results in Eq. (5.3.29): 3⊗3 = 8⊕1; 3⊗3 = 6⊕3; 8⊗3 = 15⊕6⊕3; and 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1.

Solution: (a) In SU (3) the 3 representation is denoted by the Young diagram , and so the conjugate 3 representation is the one-column diagram with N − 1 = 2 rows, .

Their product is found from

a ⊗ = ⊗ b a ⊕ b a =

a b ⊕ b | {z }

a ⊕

inadmissible

=⇒

⊗

=

⊕

b a | {z }

⊕ a b

inadmissible

.

(5.38)

Of course, the same result could be found in a simpler way via ⊗ a =

a ⊕

.

(5.39)

a

Either way, the right-most diagram here is the trivial 1 representation of SU (3), while the other diagram has a factor and hook F : 3 4 2

=⇒

F = 3 × 4 × 2 = 24,

H: 3 1 1

=⇒

H = 3,

and so has dimension F/H = 24/3 = 8. In total, 3 ⊗ 3 = 8 ⊕ 1.

(5.40)

Problem 7

101

(b) We know from Problem 5.6 that ⊗

⊕

=

,

(5.41)

since that calculation works in SU (N ) for any N ≥ 2. The diagram has F/H = (3 × 4)/(2 × 1) = 6, so in SU (3) this diagrammatic relation translates to 3 ⊗ 3 = 6 ⊕ 3. (c) We can calculate 8 ⊗ 3 with our known diagrams a ⊕

⊗ a =

a

⊕

.

(5.42)

a

The column of length three can be ignored because it does not affect the dimension of the diagram, so we are left with ⊗

=

Now is simply the conjugate of diagram has

⊕

⊕

.

(5.43)

∼ = 6 in SU (3), while the other unfamiliar

F : 3 4 5 2

=⇒

F = 3 × 4 × 5 × 2 = 120,

H: 4 2 1 1

=⇒

H = 4 × 2 = 8,

(5.44)

and therefore dimension F/H = 120/8 = 15. In total, we have 8⊗3 = 15⊕6⊕3. (d) Using the result of Part (b), we have

⊗ ⊗ a = ⊕ ⊗ a

= ⊗ a ⊕ ⊗ a 

= ⊕ a ⊕ a a

 ⊕

,

(5.45)

.

(5.46)

a

and so ⊗

⊗

⊕

=

⊕

We are familiar with all of these diagrams except F : 3 4 5

=⇒

H: 3 2 1

=⇒

⊕

, which has

F = 3 × 4 × 5 = 60, H = 3 × 2 = 6,

(5.47)

and therefore dimension F/H = 60/6 = 10. Our diagrammatic equation then translates to 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1, as desired.

Introduction to Particle Physics

102

5.8 Problem 8 Problem: Prove Eq. (5.3.38) for SU (4): 4 ⊗ 4 = 15 ⊕ 1.

Solution: In SU (4) the 4 representation is denoted by the Young diagram , and so the conjugate 4 representation is the one-column diagram with N − 1 = 3 rows, . We

could then calculate 4 ⊗ 4 as

a ⊗ b = ⊗ b a ⊕ a c c   = 

a b ⊕ b | {z }

a ⊕

inadmissible

 ∼ = =

b a | {z }

⊕

inadmissible

a b | {z }

 ⊕a  ⊗ c b

inadmissible

 a ⊕

b

a ⊗ c b

a c ⊕ b | {z }

inadmissible

⊕

c a b | {z }

a

a

a b c | {z }

⊕ b c

inadmissible

⊕

inadmissible

∼ = b c



a c b | {z }

⊕

inadmissible

a b c | {z }

⊕ a b c

inadmissible

⊕ a , b c

(5.48)

implying ⊗

⊕

=

.

(5.49)

Of course, the same result could be obtained in a much simpler way via a ⊗ a =

⊕

.

(5.50)

a Now the right-most diagram is simply the trivial 1 representation in SU (4), while

Problem 9

103

the other diagram has 4 5 F : 3 2

=⇒

F = 4 × 5 × 3 × 2 = 120,

4 1 H: 2 1

=⇒

H = 4 × 2 = 8,

(5.51)

and therefore dimension F/H = 120/8 = 15. So we see 4 ⊗ 4 = 15 ⊕ 1, as desired.

5.9 Problem 9 Problem: Prove Eq. (5.3.41) for SU (6): 6 ⊗ 6 ⊗ 6 = 56 ⊕ 70 ⊕ 70 ⊕ 20. Explain using Young diagrams why the 56 multiplet is symmetric, why the 70 multiplets have mixed symmetry and why the 20 is antisymmetric.

Solution: In SU (6) the 6 representation is denoted by the Young diagram therefore make use of Eq 5.46, ⊗

⊗

=

⊕

⊕

⊕

,

. We can

(5.52)

to calculate 6 ⊗ 6 ⊗ 6, as the calculation that led to this holds in SU (N ) for any N ≥ 2. The three distinct diagrams here have factors and hooks

F : 6 7 8 =⇒ F = 6 × 7 × 8 = 336 =⇒ F/H = 56, H: 3 2 1 =⇒ H = 3 × 2 = 6 F :

6 7 5

  =⇒ F = 6 × 7 × 5 = 210  

H:

3 1 1

=⇒ H = 3

F :

6 5 4

    =⇒ F = 6 × 5 × 4 = 120    

H:

3 2 1

      

=⇒ H = 3 × 2 = 6

=⇒

F/H = 70,

=⇒

F/H = 20.

  

(5.53)

So translated into the language of SU (6), we have 6 ⊗ 6 ⊗ 6 = 56 ⊕ 70 ⊕ 70 ⊕ 20, as desired. Now every Young diagram admits one or multiple Young tableaux, which correspond to Young symmetrizer operators that completely symmetrize the labels

Introduction to Particle Physics

104

in each row of the tableaux and antisymmetrize the labels in each column. A Young diagram that is a single row of boxes, then, such as the 56 of SU (6), denotes a symmetric multiplet, while a Young diagram that is a single column of boxes, such as the 20 of SU (6), denotes an antisymmetric multiplet. The 70 of SU (6), having both a non-trivial row and a non-trivial column structure in its Young diagram, corresponds to a multiplet that is symmetrized in some indices and antisymmetrized in others, i.e. a multiplet with mixed symmetry.

5.10 Problem 10 Problem: For SU (3)flavor recall that we defined |i1 ⟩ = |u⟩, |i2 ⟩ = |d⟩, |i3 ⟩ = |s⟩ and |ī1 ⟩ = ¯ |ī3 ⟩ = |s̄⟩. From Eq. (5.3.15) the raised basis states for the meson |ū⟩, |ī2 ⟩ = |d⟩, states in 3 ⊗ 3 are |ik (ī)ℓ1 ℓ2 ⟩ ≡ 12 ϵℓ1 ℓ2 ℓ3 |ik īℓ3 ⟩, where the Young symmetrizers Ŷ act on the labels k, ℓ1 , ℓ2 , e.g., (12)|ik (ī)ℓ1 ℓ2 ⟩ = |iℓ1 (ī)kℓ2 ⟩ and (123)|ik (ī)ℓ1 ℓ2 ⟩ = |iℓ3 (ī)kℓ1 ⟩ and so on. The singlet Young diagram is and the octet Young diagram is . The corresponding Young symmetrizers are given in Eq. (5.3.22): (a) Show that applying the singlet Young symmetrizer Ŷ = (e)−(12)−(13)−(23)+ (123)+(132) to the raised basis states gives a vanishing result unless k, ℓ1 , ℓ2 are all different. When they are all different show that Ŷ projects out the onedimensional space with normalized basis vector √13 |uū + dd¯ + ss̄⟩ = |η1 ⟩, which recovers the result in Eq. (5.3.16) for this case. (b) Consider the octet Young symmetrizer Ŷ = (e)+(12)−(13)−(123). Show that ¯ applying this when k, ℓ1 , ℓ2 are all different leads to the states 21 |uū − dd⟩, 1 ¯− ss̄⟩ and 1 |ss̄ − uū⟩, from which we can construct two orthonormal states |d d 2 2 ¯ = |π 0 ⟩ and √1 |uū + dd¯ − 2ss̄⟩ = |η8 ⟩. Since |ik (ī)ℓ ℓ ⟩ chosen to be √1 |uū − dd⟩ 2

6

1 2

is antisymmetric in ℓ1 and ℓ2 then ℓ1 ̸= ℓ2 and there are only six remaining independent possibilities k = ℓ1 ̸= ℓ2 , i.e., kℓ1 ℓ2 = 112, 113, 221, 223, 331, 332.

Show that applying the symmetrizer Ŷ to these leads to the orthonormal states ¯ = |π + ⟩, |ds̄⟩ = |K 0 ⟩, |dū⟩ = |π − ⟩, |sd⟩ ¯ = |K 0 ⟩, |sū⟩ = |K − ⟩ |us̄⟩ = |K + ⟩, |ud⟩ respectively, where we assume J P = 0− here and so arrive at the pseudoscalar meson octet. Assuming J P = 1− gives the vector octet. (c) Show that the second Young symmetrizer, Ŷ = (e) + (13) − (12) − (132), in Eq. (5.3.22) leads to the same states and hence to the same irrep/multiplet.

This is due to the antisymmetry in ℓ1 and ℓ2 for the mesons. The baryon octets 8 and 8′ in the Clebsch-Gordan series for 3 ⊗ 3 ⊗ 3 are different. However, for baryons we need to use the SU (6)spin−flavor formalism because of the need to satisfy Eq. (5.3.39).

Solution:

Problem 10

105

We are given that the basis states are 1 ℓ1 ℓ2 ℓ′ ϵ |ik īℓ′ ⟩.

2 For convenience, let us relabel (k, ℓ1 , ℓ2 ) → (ℓ1 , ℓ2 , ℓ3 ) and write instead |ik (ī)ℓ1 ℓ2 ⟩ =

(5.54)

1 ℓ′ ℓ2 ℓ3 (5.55) ϵ |iℓ1 īℓ′ ⟩, 2 so that the permutation operator labels match with the symbols they permute.

Note that this definition tells us |iℓ1 (ī)ℓ2 ℓ3 ⟩ is antisymmetric in ℓ2 and ℓ3 . |iℓ1 (ī)ℓ2 ℓ3 ⟩ =

(a) Acting with the given singlet Young symmetrizer on the state yields Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩ = [(e) − (12) − (13) − (23) + (123) + (132)] |iℓ1 (ī)ℓ2 ℓ3 ⟩ = |iℓ1 (ī)ℓ2 ℓ3 ⟩ − |iℓ2 (ī)ℓ1 ℓ3 ⟩ − |iℓ3 (ī)ℓ2 ℓ1 ⟩ − |iℓ1 (ī)ℓ3 ℓ2 ⟩ + |iℓ2 (ī)ℓ3 ℓ1 ⟩ + |iℓ3 (ī)ℓ1 ℓ2 ⟩

= 2 (|iℓ1 (ī)ℓ2 ℓ3 ⟩ + |iℓ2 (ī)ℓ3 ℓ1 ⟩ + |iℓ3 (ī)ℓ1 ℓ2 ⟩) ,

(5.56)

having used the antisymmetry mentioned above. This antisymmetry carries over to the symmetrized state, for we get Ŷ |iℓ1 (ī)ℓ3 ℓ2 ⟩ = 2 (|iℓ1 (ī)ℓ3 ℓ2 ⟩ + |iℓ3 (ī)ℓ2 ℓ1 ⟩ + |iℓ2 (ī)ℓ1 ℓ3 ⟩)

= 2 (−|iℓ1 (ī)ℓ2 ℓ3 ⟩ − |iℓ3 (ī)ℓ1 ℓ2 ⟩ − |iℓ2 (ī)ℓ3 ℓ1 ⟩) = −Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩.

(5.57)

simply from swapping the labels ℓ2 ↔ ℓ3 , by Eq. 5.56. A similar calculation with ℓ1 ↔ ℓ2 yields Ŷ |iℓ2 (ī)ℓ1 ℓ3 ⟩ = 2 (|iℓ2 (ī)ℓ1 ℓ3 ⟩ + |iℓ1 (ī)ℓ3 ℓ2 ⟩ + |iℓ3 (ī)ℓ2 ℓ1 ⟩)

= 2 (−|iℓ2 (ī)ℓ3 ℓ1 ⟩ − |iℓ1 (ī)ℓ2 ℓ3 ⟩ − |iℓ3 (ī)ℓ1 ℓ2 ⟩) = −Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩.

(5.58)

Further exploiting these two antisymmetries, we see Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩ = −Ŷ |iℓ2 (ī)ℓ1 ℓ3 ⟩ = Ŷ |iℓ2 (ī)ℓ3 ℓ1 ⟩ = −Ŷ |iℓ3 (ī)ℓ2 ℓ1 ⟩ = Ŷ |iℓ3 (ī)ℓ1 ℓ2 ⟩, (5.59) i.e. Ŷ |iℓ1 (ī)ℓ3 ℓ2 ⟩ is totally antisymmetric in its indices, and therefore vanishes unless ℓ1 , ℓ2 , and ℓ3 are all different. If we suppose now ℓ1 , ℓ2 , and ℓ3 are indeed distinct, then Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩ = 2 (|iℓ1 (ī)ℓ2 ℓ3 ⟩ + |iℓ2 (ī)ℓ3 ℓ1 ⟩ + |iℓ3 (ī)ℓ1 ℓ2 ⟩) ′

′

′

= ϵℓ ℓ2 ℓ3 |iℓ1 īℓ′ ⟩ + ϵℓ ℓ3 ℓ1 |iℓ2 īℓ′ ⟩ + ϵℓ ℓ1 ℓ2 |iℓ3 īℓ′ ⟩

= ϵℓ1 ℓ2 ℓ3 |iℓ1 īℓ1 ⟩ + ϵℓ2 ℓ3 ℓ1 |iℓ2 īℓ2 ⟩ + ϵℓ3 ℓ1 ℓ2 |iℓ3 īℓ3 ⟩ (no sum) = ϵℓ1 ℓ2 ℓ3 (|iℓ1 īℓ1 ⟩ + |iℓ2 īℓ2 ⟩ + |iℓ3 īℓ3 ⟩)

¯ + |ss̄⟩ = ϵℓ1 ℓ2 ℓ3 |uū⟩ + |dd⟩ ≡ ϵℓ1 ℓ2 ℓ3 |uū + dd¯ + ss̄⟩.

(5.60)

Introduction to Particle Physics

106

¯ We see that Ŷ projects out the one-dimensional space spanned by |uū+dd+ss̄⟩, which has the normalized basis vector 1 √ |uū + dd¯ + ss̄⟩ ≡ |η1 ⟩, (5.61) 3 ¯ |ss̄⟩} is an orthonormal set of states. given {|uū⟩, |dd⟩, (b) With the given Young octet symmetrizer we have Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩ = [(e) + (12) − (13) − (123)] |iℓ1 (ī)ℓ2 ℓ3 ⟩

= |iℓ1 (ī)ℓ2 ℓ3 ⟩ + |iℓ2 (ī)ℓ1 ℓ3 ⟩ − |iℓ3 (ī)ℓ2 ℓ1 ⟩ − |iℓ3 (ī)ℓ1 ℓ2 ⟩ = |iℓ1 (ī)ℓ2 ℓ3 ⟩ + |iℓ2 (ī)ℓ1 ℓ3 ⟩ ,

(5.62)

by the antisymmetry of the last two indices. So then if ℓ1 , ℓ2 , and ℓ3 are all different, Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩ = |iℓ1 (ī)ℓ2 ℓ3 ⟩ + |iℓ2 (ī)ℓ1 ℓ3 ⟩ 1 ′ 1 ′ = ϵℓ ℓ2 ℓ3 |iℓ1 īℓ′ ⟩ + ϵℓ ℓ1 ℓ3 |iℓ2 īℓ′ ⟩ 2 2 1 1 ℓ1 ℓ2 ℓ3 |iℓ1 īℓ1 ⟩ + ϵℓ2 ℓ1 ℓ3 |iℓ2 īℓ2 ⟩ = ϵ 2 2 1 ℓ1 ℓ2 ℓ3 = ϵ (|iℓ1 īℓ1 ⟩ − |iℓ2 īℓ2 ⟩) .

2

(5.63)

Hence,

1 1 ¯ = 1 |uū − dd⟩, ¯ (|i1 ī1 ⟩ − |i2 ī2 ⟩) = |uū⟩ − |dd⟩ 2 2 2

1 1 ¯ − |ss̄⟩ = 1 |dd¯ − ss̄⟩, |dd⟩ Ŷ |i2 (ī)31 ⟩ = (|i2 ī2 ⟩ − |i3 ī3 ⟩) = 2 2 2 1 1 1 Ŷ |i3 (ī)12 ⟩ = (|i3 ī3 ⟩ − |i1 ī1 ⟩) = (|ss̄⟩ − |uū⟩) = |ss̄ − uū⟩.

(5.64) 2 2 2 ¯ + |dd¯ − ss̄⟩ + |ss̄ − uū⟩ = 0, and so these three states are Note that |uū − dd⟩ Ŷ |i1 (ī)23 ⟩ =

not linearly independent - they only span a plane. Since

¯ − dd⟩ ¯ = ⟨uū|uū⟩ − ⟨uū|dd⟩ ¯ − ⟨dd|uū⟩ ¯ ¯ d⟩ ¯ ⟨uū − dd|uū + ⟨dd|d = 12 − 0 − 0 + 12 = 2,

(5.65)

a suitable normalized basis vector for this plane is 1 ¯ = |π 0 ⟩. √ |uū − dd⟩ 2

(5.66)

Using the Gram-Schmidt process, we can calculate an orthogonal state as |dd¯ − ss̄⟩ −

¯ d¯ − ss̄⟩ |uū − dd⟩ ¯ ⟨uū − dd|d 1 ¯ √ √ = |dd¯ − ss̄⟩ + |uū − dd⟩ 2 2 2 1 = |uū + dd¯ − 2ss̄⟩, (5.67) 2

Problem 10

107

and since ⟨uū + dd¯ − 2ss̄|uū + dd¯ − 2ss̄⟩ = 12 + 12 + 22 = 6, we find the orthonormal state 1 √ |uū + dd¯ − 2ss̄⟩ = |η8 ⟩.

6

(5.68)

This completes our analysis of the case where ℓ1 , ℓ2 , and ℓ3 are distinct. Now let us consider the second case, where ℓ1 = ℓ2 ≡ ℓ ̸= ℓ3 . From Eq. 5.62, we get now Ŷ |iℓ (ī)ℓℓ3 ⟩ = |iℓ (ī)ℓℓ3 ⟩ + |iℓ (ī)ℓℓ3 ⟩ = 2|iℓ (ī)ℓℓ3 ⟩ ′

= ϵℓ ℓℓ3 |iℓ īℓ′ ⟩,

(5.69)

leading to the set of states Ŷ |i1 (ī)12 ⟩ = ϵ312 |i1 ī3 ⟩ = |us̄⟩ = |K + ⟩, ¯ = −|π + ⟩, Ŷ |i1 (ī)13 ⟩ = ϵ213 |i1 ī2 ⟩ = −|ud⟩ Ŷ |i2 (ī)21 ⟩ = ϵ321 |i2 ī3 ⟩ = −|ds̄⟩ = −|K 0 ⟩, Ŷ |i2 (ī)23 ⟩ = ϵ123 |i2 ī1 ⟩ = |dū⟩ = |π − ⟩, 0

¯ = |K ⟩, Ŷ |i3 (ī)31 ⟩ = ϵ231 |i3 ī2 ⟩ = |sd⟩

Ŷ |i3 (ī)32 ⟩ = ϵ132 |i3 ī1 ⟩ = −|sū⟩ = −|K − ⟩,

(5.70)

which complete the pseudoscalar meson octet. (c) Finally, the second Young octet symmetrizer gives Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩ = [(e) + (13) − (12) − (132)] |iℓ1 (ī)ℓ2 ℓ3 ⟩

= |iℓ1 (ī)ℓ2 ℓ3 ⟩ + |iℓ3 (ī)ℓ2 ℓ1 ⟩ − |iℓ2 (ī)ℓ1 ℓ3 ⟩ − |iℓ2 (ī)ℓ3 ℓ1 ⟩ = |iℓ1 (ī)ℓ2 ℓ3 ⟩ + |iℓ3 (ī)ℓ2 ℓ1 ⟩

= |iℓ1 (ī)ℓ2 ℓ3 ⟩ − |iℓ3 (ī)ℓ1 ℓ2 ⟩ ,

(5.71)

by the antisymmetry of the last two indices. Note that if we swap ℓ2 ↔ ℓ3 , we get Ŷ |iℓ1 (ī)ℓ3 ℓ2 ⟩ = |iℓ1 (ī)ℓ3 ℓ2 ⟩ − |iℓ2 (ī)ℓ1 ℓ3 ⟩

= −|iℓ1 (ī)ℓ2 ℓ3 ⟩ − |iℓ2 (ī)ℓ1 ℓ3 ⟩,

(5.72)

which is the same as the action of the first octet Young symmetrizer on the original state, Eq. 5.62, up to an unimportant phase factor of −1. The set of all states obtained by considering Ŷ |iℓ1 (ī)ℓ2 ℓ3 ⟩ for all possible triplets (ℓ1 , ℓ2 , ℓ3 ) is therefore the same as that found in Part (b), i.e. we get again the pseudoscalar meson octet.

Introduction to Particle Physics

108

5.11 Problem 11 Problem: A beam of π − mesons is directed at a proton target and produces a variety of final states. For each of the final states listed below propose a reaction that leads to the final state. Draw a simple labeled diagram showing quark, lepton and boson lines as done for example in Fig. 5.5. Choose a reaction that you expect to be one of the most important ways for this final state to be produced keeping in mind the relative strengths of the strong, electromagnetic and weak interactions. Assume that the π − beam has sufficient energy for each reaction to be kinematically allowed: (a) ∆++ resonance; (b) Z 0 boson; (c) electron neutrino, νe , without an anti-neutrino; (d) K − meson; and (e) anti-neutrino and a charmed baryon.

Solution:

(a) The quark content of the ∆++ resonance is uuu, while the combined quark content of π − +p is ūd+uud. A π − +p initial state, then, can only form a ∆++ resonance if a u quark is created. This could happen, for example, through a weak interaction d → u+W − , or as part of pair production d → d+X → d+uū, where X represents a gluon, Z 0 boson, or photon (there is no particular reason for using a d quark here as opposed to a u quark). Because of the relative strengths of these interactions, we expect pair production through the strong interaction to be the dominant pathway. The total process is then π − +p → ∆++ +π − +π − : Introduction to Particle Physics

110

ū ⇡

d

ū d

⇡

d ⇡ ū

d p u

u

u

u

++

u (b) To produce a Z 0 boson, obviously at least one weak interaction vertex is needed.

0 One vertex aisZalso sufficient, as the boson can be producedvertex by theisannihila0 (b) To produce boson, obviously at Zleast one weak interaction needed.

0 0 tion u + ū !

Z , leading to the reaction process ⇡ + p !

Z + n: 0 One vertex is also sufficient, as the Z boson can be produced by the annihila− 0 tion u + ū → Z 0 , leading ū to the reaction process π + p → Z + n: ⇡ d Z0

u p u d

d u n d

(c) Electron neutrinos can only be produced in a weak interaction, either charged (W + ! ⌫e + e+ ) or neutral (Z 0 ! ⌫e + ⌫¯e ). A single charged weak interaction is therefore the dominant pathway for producing an electron neutrino without

d d

⇡

d p ud p u u u

109

d ⇡ ūd ⇡ ū u uu

++

++ uu u 0 (b) To produce a Z boson, obviously at least one weak interaction vertex is needed.

0 Problem 0 11 (b) One To produce a Zalso boson, obviously atZleast one can weakbeinteraction needed. vertex is sufficient, as the boson produced vertex by the isannihila0 One uvertex is Z also sufficient, as the Z 0 boson can ⇡be + produced the annihilation + ū ! , leading to the reaction process p ! Z 0 by + n: tion u + ū ! Z 0 , leading to the reaction process ⇡ + p ! Z 0 + n: ū ⇡ dū Z0 ⇡ d Z0

u d u p u ud n p u u n d d d d (c) Electron neutrinos can only be produced in a weak interaction, either charged (c) neutrinos can a weak interaction, either charged + + (c) Electron Electron neutrinos can only only be be0produced produced¯ein in either charged (W ). aA weak singleinteraction, charged weak interaction + ! ⌫e + e+ ) or neutral (Z 0 ! ⌫e + ⌫ (W → ν + e ) or neutral (Z → ν + ν̄ ).

A single charged weak interaction + e + e+ ) or neutral (Z 0 ! ⌫e + ⌫ e ). A single charged weak interaction (W ! ⌫ ¯ e the dominant pathway for e producing e is therefore an electron neutrino without is the is therefore therefore the dominant dominant pathway pathway for for producing producing an an electron electron neutrino neutrino without without its anti-neutrino: its anti-neutrino: its anti-neutrino: ū ⇡ dū ⇡ ū d ⇡ ū d ⇡ d d ud n d u n d p ud d p u u e+ u W+ e+ ⌫e W+ ⌫e + The overall process is ⇡ + p ! ⇡ + n + e + ⌫e . There was no choice in − + The overall is is π −⇡to+the p p→u!πquark ν⌫ee. . There was no The overall + ⇡ ++nhere; n++ee++it++could There was no choice choice coupling theprocess Wprocess boson not be produced in in a + + coupling the W the quark it could not be produced in a coupling the from W boson boson toquark the uubecause quark here; single vertex a ū or to d of charge conservation. single single vertex vertex from from aa ūū or or dd quark quark because because of charge conservation.

111

(d) The quark content of the K meson is ūs, and therefore an s quark would need Introduction to Particle Physics (d) The The quark content ofinteraction. the K K− meson meson isasūs, (d) content the s quark need to bequark created in thisof Justis inand Parttherefore (a), this an could occurwould through a to be be created created in in this this interaction. interaction. Just Just as as in Part (a), this could occur through a to charged occur through through strong strong interaction interaction charged weak weak interaction, interaction, but but is is more more likely likely to to occur pair production: pair production:

⇡

ū d u p u d

ū K s s̄ K+ u d u n d

As we have drawn it, the total interaction here is ⇡ + p ! K + K + + n.

The extraneous s̄ quark could also have paired instead with a d quark, giving instead ⇡ + p ! ūs + s̄d + uud = K + K 0 + p. (e) Just as in Part (c), at least one weak interaction is necessary to produce an anti-neutrino. Fortunately, this can at the same time lead to a c quark in the final state via d ! W + c ! e + ⌫¯e + c. A c quark could also be created as part of pair production, but this would be a less likely pathway because of the extra energy needed to create the additional c̄. Our suggested reaction pathway is then

⇡

ū K s

ū d

s̄ K+ u

u p u d Introduction to Particle Physics

110

d u n d

− − + AsAs wewe have havedrawn drawnit,it,the thetotal totalinteraction interactionhere hereisisπ⇡ ++pp→ !K K + +K K+ + + n. n.

The extraneous s̄ quark could also have paired instead with a d quark, giving The extraneous s̄ quark could also have paired instead with a d quark, giving − − 0 instead π instead ⇡ + + pp → ! ūs ūs + + s̄d s̄d + + uud uud = =K K + +K K0 + + p. p.

(e) Just Just as (e) as in in Part Part (c), (c), at at least least one one weak weak interaction interaction is is necessary necessary to to produce produce an an anti-neutrino.

Fortunately, this can at the same time lead to a c quark anti-neutrino. Fortunately, this can at the same time lead to a c quark in in the the − − final state final state via via d d→ !W W + + cc → ! ee + + ν̄⌫¯ee + + c. c. A A cc quark quark could could also also be be created created as as part of part of pair pair production, production, but but this this would would be be aa less less likely likely pathway pathway because because of of the the extra energy extra energy needed needed to to create create the the additional additional c̄. c̄. Our Our suggested suggested reaction reaction pathway pathway is then is then

⇡

ū 0 ⇡ u

ū d

d u ⌃+ c

u

c

p u d W

e ⌫¯e

++ The charmed The charmedbaryon baryonhere hereis isudc udc==Σ⌃ sothe thetotal totalreaction reaction isis π⇡− + + pp → ! c c, ,so 0 − + Σc + π ⌃ ⇡ + e + ν̄⌫¯e .

Problem 12 12 5.12 Problem Problem: Problem: Consider the following reactions. Which reactions are allowed? If allowed, explain what the main reaction mechanism is using a labeled diagram showing quark, lepton and boson lines. Explain whether it occurs through the strong and/or electromagnetic and/or weak interactions? If not allowed, why is it forbidden? (Hint: For the quark flavor content of hadrons, see above.) (a) p + p → K + + K + + Ξ0 ; (b) Σ0 → Λ0 + γ; (c) π − + p → π 0 + K − + (Σ̄)+ ; (d) Λ0 → p+µ− + ν̄µ ; (e) π + +n → K̄ 0 +K + ; (f) K 0 +n → K − +p; (g) Σ+ → p+π 0 ; (h) νe + e+ + n → p + π 0 ; (i) p + π + + π − → ∆+ + π 0 ; (j) Λ0 → K − + p; (k) Σ− → n + π − ; (l) π 0 + νe → π + + e− .

Solution: For convenience, we list the quark content of the hadrons in Tables 5.2 and 5.3.

Throughout this problem it is helpful to keep in mind that no process in the Standard Model can alter the baryon number (the number of quarks minus the number of

Problem 12

111

antiquarks), allowing us to conclude many of the processes are impossible. Baryons have a baryon number of +1, antibaryons a baryon number of −1, and mesons a baryon number of 0. It should also be understood that gluons couple equally strongly to all quarks, so the specific quark in a hadron that a gluon couples to in the following diagrams is arbitrary. Any W ± bosons could have their order of emission/absorption swapped and charge negated, and the end result would also be a valid diagram.

Table 5.2 Quark content of the mesons appearing in this problem.

Table 5.3 Quark content of the baryons appearing in this problem.

Meson

Baryon

−

π π0 π+ K− K0 0 K + K

Quark content

Quark content

+

dū uū, dd¯ ud¯ sū ds̄ sd¯ us̄

p, ∆ n Σ− Λ0 , Σ0 Σ+ (Σ̄)+ Ξ0

uud udd dds uds uus ¯ d¯ds̄ uss

(a) The baryon number of the state p + p is 1 + 1 = +2, while the baryon number of the state K + + K + + Ξ0 is 0 + 0 + 1 = +1. Since baryon number is conserved, there is no process that can take one state to the other. (b) The Σ0 baryon is an excited version of Λ0 , so the reaction Σ0 → Λ0 + γ is kinematically allowed. It occurs through an entirely electromagnetic process:

u ⌃

0

d s

u d ⇤0 s

The photon could be emitted by any of the quarks, but it is most likely to come µ from the u quark because of its greater charge. (c) The reaction π − + p → π 0 + K − + (Σ̄)+ is forbidden because of baryon number conservation (0 + 1 → 0 + 0 − 1), as in Part (a).

s

W

⌫¯µ

⇤0 u d

u u p d

s̄ K0

ū W

K

Introduction to Particle Physics

112

u 0

d than a proton and muon combined, so there (d) The Λ0 baryon has more⌃mass s u + ν̄µ . Because there is a net is nothing forbidding the reaction Λ0 → p + µ− change in quark flavour in this process, it must occur through a charged weak d ⇤0 interaction: s

µ

W

s

⌫¯µ

⇤0 u d

u u

u p

⌃0 d s

d

u

d ⇤0 ū s + 0 + (e) The reaction π 0+ n → K̄ + K is forbidden because of baryon K number conK servation (0 + 1 → 0 + 0), as in Part (a). W µs d

s̄

u

(f) After some thought, it becomes apparent that the reaction K 0 + n → K − + p can occur through several pathways. ForWexample:

u d W s p n u ⌫¯µu proton (1) The neutron may emit a W − boson and turn into the directly;

⇤0 u d u d (i) and the remaining reaction occurs entirely u pthrough charged weak interactions, such as in d

d

Q`

s̄ K

ū

0

W

d

u

K s

W u u p

d n u d Q`

k

d

or

k

Problem 12

113

d d K 00 d KK0 s̄ s̄s̄

ū ūū W W W

d d n du n u n u d dd

Q` or Q`Q`

s ss K K ū K ūū

W W W

u u uu p uu pp d dd

W W W u uu

d K 00 dd s̄ K K 0 s̄ s̄

s s K ūs K K ūū

W W W

u d u d d n u uu p n u u pc n ud du p ; d d d d 0 (ii) or the K particle annihilates itself after a weak interaction: W W Wu s s d u s 0 s K u K0 d s ūs K K0 ds̄ K s̄ ū K W s̄ ū W W u d u d n du uu p n u u p n ud du p d d d d . h?Bb `2 +iBQM T i?r v rQmH/ HbQ #2 p HB/ B7 i?2 ;HmQM r2`2 `2TH +2/ #v h?Bb `2 +iBQM T i?r v rQmH/ HbQ #2 p HB/ B7 i?2 ;HmQM r2`2 `2TH +2/ #v This reaction pathway would be2H2+i`QK ;M2iB+ valid if the gluon were replaced Z #QbQM Q` T?QiQM#mirQmH/ i?2b2 r2 F M/p HB/ BMi2` +iBQMb `2 #v H2bb h?Bb `2 +iBQM T i?r v HbQalso #2 B7 i?2 ;HmQM r2`2 `2TH +2/ Z #QbQM Q` T?QiQM#mi i?2b2 r2 F M/ 2H2+i`QK ;M2iB+ BMi2` +iBQMb `2 H2bb by a Z boson or #mi photon, but these weak and electromagnetic interaci?2 bi`QM; BMi2` +iBQM ZbB;MB}+ Mi #QbQM Q` i? M T?QiQMi?2b2 r2 F M/b?QrMX 2H2+i`QK ;M2iB+ BMi2` +iBQMb `2 H2bb bB;MB}+ Mi i? M i?2 bi`QM; BMi2` +iBQM b?QrMX tions are lessi?2 significant than the strong bB;MB}+ Mi i? M bi`QM; BMi2` +iBQM b?QrMXinteraction shown.

(2) Another possibility is that a gluon produces a uū pair that contributes to both the proton and K − , such as in

K0

d s̄

W ū

u jj j

W+ d n u

⌃

+

s ū

K

u u p

d

d

u u

u u p

s

u

d

W

u ū

⌫e

u

+

ū

.

⇡0

⇡0

Introduction to Particle Physics

114

Since all of these pathways involve four weak interaction vertices, they all cons d to the amplitudeWof this process. tribute roughly equally K0 K s̄ ū ū (g) The Σ+ resonance has a greater u the proton and π 0 combined, so the W + mass than + 0 decay Σ → p + π is kinematically allowed and there u is nothing to forbid this d process. The transition from Σ+ to p requires a net flavour change of s → d, n u u p so at least two charged weak interaction vertices are necessary in the reaction. d The neutral pion can then be created from a gluon viadthe strong interaction:

⌃

+

u u s

u u p u

d

W

u ū

⇡0

⌫e could be replaced by a Z boson u Just as above the gluon or photon, but the ⇡0 + strong interaction shown ed is the most significant of the ūsthree. Note that the π 0 W K0 K W is technically a superposition of +uū and dd¯ states. We have s̄ ū only included the uū ū ¯ component to simplifydthe diagram; the udd component ucan clearly be produced W+ through the same process. n dd du p Another possibility is that the W − boson does not form a loop and instead u u decays into d + ū, nandu a W + boson then completes thep Σ+ → p transition ¯ The decay products collectively d before decaying into uud+ d. form the π 0 with no need for the strong ⇡ + ¯interaction, but since this pathway requires many weak du significantly suppressed. uu interaction vertices it is ⇡0 + u ⌃ ūu p d s (h) A charged weak interaction is necessary to remove the neutrino in the reaction d ⇡ u ν + e+ + n → p + π 0ū. This can at the same time remove the positron, and e

convert the neutron into a proton via W νe + e+ + d →u W + + d → u. A π 0 u u 0 meson can then be created through pair production, most ⇡likely via the strong ū + p u u interaction: d d ⌫e u ⇡0 ū e+ W+

d n d u

u d p u

9 (i) The process p + π +++ u π − → ∆+ + π 0 simply requires removing a dd¯ pair from ⇡ + the initial state, and exciting the proton to become a ∆ d¯ u resonance. This can ⇡0 easily be achieved through annihilation via the strong interaction: ū d ⇡ ū u

u

p u

u

d

d

+

W

u ū

⌫e

u

e+

ū

⇡0

⇡0

W+ 115

d n d u ⇡+

u Problem 12

u d¯

u ū

d ū

⇡

d p u

u

u

p u

u

d

d

⇡0

+

(j) Consider the rest frame of a particular Λ0 resonance, and assume this is an inertial frame. The energy of the resonance is simply its rest mass energy, 1115 MeV. This is less than the combined rest mass energies of a proton (938 MeV) and a K − meson (493 MeV), so the decay Λ0 → K − +p would be forbidden by energy conservation. All inertial observers would then agree this decay is forbidden. (k) The mass of the Σ− resonance is greater9 than the combined masses of a neutron and π − meson, so the reaction Σ− → n + π − is kinematically allowed. Since there is a net change of quark flavour s → d in this process, two charged weak interaction vertices are necessary. This could either be done by changing the s quark into a d quark directly,

d d n ud d n

d ⌃

d sd

⌃

d s

u ū

u

d ū

W

⇡

u ⇡ d or by changing the s into a u that contributes to the neutron, and having the d d W π − be a direct product of the W − : ⌃ d d n sd ud ⌃ d d n s W du ⇡ ū W d u u ⇡ ū + ⇡0 ⇡ ū d¯ u the amplitude for this Both of these diagramsuwould contribute significantly to ⇡0 ⇡+ process.

W+ ū d¯ ⌫e e W+ ⌫e

e

d ⌃

u

d s ū

u

d

W

116

⌃

⇡

d

d

d Introduction to Particle Physics s

d n u

(l) A charged weak interaction is needed for the reaction π 0 + νe → π + + e− , as W d this is the only way for the neutrino to disappear. The weak ⇡ boson could then ū 0 + go on to convert the π into a π directly:

⇡0

u

u ⇡+ d¯

ū W+ ⌫e

e

The dd¯ component of π 0 could annihilate into a gluon (or Z boson or photon) that then gets absorbed by another quark.

8

6

Formulation of Quantum Field Theory 6.1 Problem 1 Problem: Substitute Eq. (6.2.14) for ϕ̂s (x) and π̂s (x) into Eq. (6.2.16) and hence show that the expressions given for the annihilation and creation operators, âp and â†p , are correct.

Solution: From Eqs. (6.2.14) we have the following expressions for ϕ̂s (x) and π̂s (x): Z

1 d3 p p âp eip·x + â†p e−ip·x , ϕ̂s (x) = 3 (2π) 2ωp r Z 3

d p ωp π̂s (x) = (−i) âp eip·x − â†p e−ip·x .

(6.1) (2π)3 2 Putting these into the expressions in Eqs. (6.2.16), we calculate Z h i 1 d3 x p e−ip·x ωp ϕ̂s (x) + iπ̂s (x) 2ωp " Z Z

ωp 1 d3 p′ † −ip′ ·x 3 −ip·x ip′ ·x ′e p e = d xp e â + â ′ p p (2π)3 2ωp 2ωp′ r

ωp′ † −ip′ ·x ip′ ·x âp′ e + − âp′ e 2 r Z Z

1 d3 p′ ωp † −i(p′ +p)·x 3 i(p′ −p)·x ′ = d x â e + â e ′ p p 2 (2π)3 ωp′ r

ωp′ † −i(p′ +p)·x i(p′ −p)·x ′ âp e − âp′ e + ωp

Z r

1 d3 p′ ωp † 3 3 ′ 3 3 ′ ′ (2π) δ (p − p) + â ′ (2π) δ (p + p) = â p p 2 (2π)3 ωp′ r

ωp′ + âp′ (2π)3 δ 3 (p′ − p) − â†p′ (2π)3 δ 3 (p′ + p) ωp r

r r r 1 ωp ωp † ωp ω−p † = âp + â−p + âp − â−p 2 ωp ω−p ωp ωp = âp ,

117

(6.2)

Formulation of Quantum Field Theory

118

having used ω−p = Z

p p | − p|2 + m2 = |p|2 + m2 = ωp , and similarly,

h i 1 d3 x p eip·x ωp ϕ̂s (x) − iπ̂s (x) 2ωp " Z Z

′ ′ 1 d3 p′ ω 3 ip·x p p = d xp e âp′ eip ·x + â†p′ e−ip ·x 3 (2π) 2ωp 2ωp′ r

′ ′ ωp′ âp′ eip ·x − â†p′ e−ip ·x − 2 r Z

3 ′ Z 1 d p ωp † −i(p′ −p)·x 3 i(p′ +p)·x ′e = d x â + â e ′ p p 2 (2π)3 ωp′ r

′ ′ ωp′ âp′ ei(p +p)·x − â†p′ e−i(p −p)·x − ωp r Z

3 ′ 1 ωp d p = âp′ (2π)3 δ 3 (p′ + p) + â†p′ (2π)3 δ 3 (p′ − p) 3 2 (2π) ωp′ r

ωp′ † 3 3 ′ 3 3 ′ âp′ (2π) δ (p + p) − âp′ (2π) δ (p − p) − ωp r

r r r ωp ωp † ω−p ωp † 1 = â−p + â − â−p + â 2 ω−p ωp p ωp ωp p = â†p ,

(6.3)

showing that Eqs. (6.2.16) contain the correct expressions for âp and â†p .

6.2 Problem 2 Problem: Show that the commutation relations of âp and â†p in Eq. (6.2.17) reproduce each of the Schrödinger picture canonical commutation relations in Eq. (6.2.12).

Solution: Eq. (6.2.17) gives the commutation relations [âp , âp′ ] = [â†p , â†p′ ] = 0,

[âp , â†p′ ] = −[â†p′ , âp ] = (2π)3 δ 3 (p − p′ ).

(6.4)

Using Eqs. (6.2.14),

d3 p 1 p âp eip·x + â†p e−ip·x , 3 (2π) 2ωp r Z 3

d p ωp π̂s (x) = (−i) âp eip·x − â†p e−ip·x , 3 (2π) 2 Z

ϕ̂s (x) =

(6.5)

Problem 2

119

we can then calculate Z i h ′ ′ 1 d3 p d3 p′ 1 p p [ϕ̂s (x), ϕ̂s (y)] = âp eip·x + â†p e−ip·x , âp′ eip ·y + â†p′ e−ip ·y 3 3 (2π) (2π) 2ωp 2ωp′ Z

3 ′ 3 ′ ′ 1 1 d p d p = [âp , âp′ ]eip·x eip ·y + [âp , â†p′ ]eip·x e−ip ·y , √ 3 3 ′ 2 (2π) (2π) ωp ωp

′ ′ +[â†p , âp′ ]e−ip·x eip ·y + [â†p , â†p′ ]e−ip·x e−ip ·y 1 = 2

′ 1 d3 p d3 p′ 0 + (2π)3 δ 3 (p − p′ )eip·x e−ip ·y √ (2π)3 (2π)3 ωp ωp′

Z

′

−(2π)3 δ 3 (p′ − p)e−ip·x eip ·y + 0 =

1 2

1 d3 p eip·(x−y) − e−ip·(x−y) √ 3 (2π) ωp ωp

Z

= 0,

(6.6)

because the integrand is odd in p. Similarly, r r Z i d3 p d3 p′ ωp ωp′ h † −ip′ ·y 2 ip·x † −ip·x ip′ ·y ′e [π̂s (x), π̂s (y)] = (−i) â e − â e , â − â e ′ p p p p (2π)3 (2π)3 2 2 Z

3 3 ′ ′ ′ 1 d p d p √ = ωp ωp′ [âp , âp′ ]eip·x eip ·y − [âp , â†p′ ]eip·x e−ip ·y , 3 3 2 (2π) (2π)

′ ′ −[â†p , âp′ ]e−ip·x eip ·y + [â†p , â†p′ ]e−ip·x e−ip ·y Z

′ 1 d3 p d3 p′ √ ′ = ω ω 0 − (2π)3 δ 3 (p + p′ )eip·x e−ip ·y p p 3 3 2 (2π) (2π)

′ −(2π)3 δ 3 (p′ − p)e−ip·x eip ·y + 0 Z

1 d3 p √ −ip·(x−y) ip·(x−y) = ω ω e − e p p 2 (2π)3 = 0,

(6.7)

again because the integrand is odd in p. Finally, we calculate r Z i ′ ′ 1 ωp′ h d3 p d3 p′ p (−i) âp eip·x + â†p e−ip·x , âp′ eip ·y − â†p′ e−ip ·y [ϕ̂s (x), π̂s (y)] = 3 3 (2π) (2π) 2 2ωp Z

3 3 ′ r ′ ′ −i d p d p ωp′ = [âp , âp′ ]eip·x eip ·y − [âp , â†p′ ]eip·x e−ip ·y , 3 3 2 (2π) (2π) ωp

′ ′ +[â†p , âp′ ]e−ip·x eip ·y + [â†p , â†p′ ]e−ip·x e−ip ·y −i = 2

Z

d3 p d3 p′ (2π)3 (2π)3

r

′ ωp′ 0 − (2π)3 δ 3 (p − p′ )eip·x e−ip ·y ωp

Formulation of Quantum Field Theory

120

′

−(2π)3 δ 3 (p′ − p)e−ip·x eip ·y + 0

Z r

−i d3 p ωp ip·(x−y) −e − e−ip·(x−y) 3 2 (2π) ωp

−i 1 = −(2π)3 δ 3 (x − y) − (2π)3 δ 3 (x − y) 3 2 (2π) =

= iδ 3 (x − y),

(6.8)

confirming all of the Schrödinger picture canonical commutation relations in Eq. (6.2.12).

6.3 Problem 3 Problem: Obtain the result for the total three-momentum operator P̂ for a free scalar field in terms of annihilation and creation operators as given in Eq. (6.2.34).

Solution: We can rewrite our expressions Z

1 d3 p p ϕ̂s (x) = âp eip·x + â†p e−ip·x , 3 (2π) 2ωp r Z 3

ωp d p (−i) âp eip·x − â†p e−ip·x π̂s (x) = 3 (2π) 2

(6.9)

in a more convenient form by making the change of variables p → −p in the second term of each integrand (which preserves the integration measure and range), yielding Z

d3 p 1 † ip·x p ϕ̂s (x) = e â + â p −p , (2π)3 2ωp r Z

ωp ip·x d3 p † (−i) e â − â .

(6.10) π̂s (x) = p −p (2π)3 2 The spatial derivative of our free scalar field is then Z

d3 p 1 p ∇ϕ̂s (x) = ip eip·x âp + â†−p , 3 (2π) 2ωp so the momentum operator is found as Z P̂ ≡ − d3 x π̂s (x)∇ϕ̂s (x) r Z Z

d3 p ωp ip·x = − d3 x (−i) e âp − â†−p 3 (2π) 2 Z

′ d3 p′ 1 p × ip′ eip ·x âp′ + â†−p′ 3 (2π) 2ωp′

(6.11)

Problem 4

121

=−

1 2

Z

d3 p d3 p′ (2π)3 (2π)3

r

ωp ′ p ωp′

Z

′ d3 x ei(p+p )·x âp − â†−p âp′ + â†−p′ , (6.12)

as given in Eq. (6.2.34). We fill in the remaining steps of the derivation: Z Z r

1 d3 p d3 p′ ωp ′ † † 3 i(p+p′ )·x ′ + â P̂ = − â − â â p d x e ′ p p −p −p 2 (2π)3 (2π)3 ωp′ Z r

1 ωp ′ d3 p d3 p′ † † 3 3 ′ ′ =− p (2π) δ (p + p ) â − â â + â ′ p p −p −p 2 (2π)3 (2π)3 ωp′ Z r

d3 p ωp 1 (−p) âp − â†−p â−p + â†p =− 3 2 (2π) ω−p Z

3 d p 1 † † † † p â â + â â − â â − â â = p −p p −p p p −p −p 2 (2π)3 Z

1 d3 p † † = p â â − â â , (6.13) p −p p −p 2 (2π)3 having recognized that p â−p âp and p â†−p â†p are odd in p because â−p âp = âp â−p and â†−p â†p = â†p â†−p , so those terms integrate to zero. Now we again make the substitution p → −p in the second term of the integrand, taking care to remember the factor of p that will change its sign, to get Z

1 d3 p P̂ = p âp â†p + â†p âp 3 2 (2π)

Z d3 p 1 † † = p â â + [â , â ] .

(6.14) p p p p (2π)3 2 Using the commutation relations we see p [âp , â†p ] = p (2π)3 δ 3 (0) is odd in p, so this too integrates to zero. We are left with Z Z d3 p d3 p † P̂ = p â â ≡ p N̂p , (6.15) p p (2π)3 (2π)3 as was to be shown.

6.4 Problem 4 Problem: Explain why the expressions for the n-boson identity operator Iˆn in Eqs. (6.2.63) and (6.2.64) are identical and verify that Iˆn |p1 · · · pn ⟩ = |p1 · · · pn ⟩ and Iˆn |p1 · · · pn ; So ⟩ = |p1 · · · pn ; So ⟩ as required.

Solution: Eq. (6.2.63) gives the identity operator in terms of the ordered momentum states

Formulation of Quantum Field Theory

122

|p1 · · · pn ; So ⟩ as

Z d3 p1 d3 pn Iˆn = · · · |p1 · · · pn ; So ⟩⟨p1 · · · pn ; So |.

2Ep1 (2π)3 2Epn (2π)3 ord(p)

(6.16)

The notation here signifies that the integration is only over the range in momentum space that satisfies the ordering relation imposed on the momenta, e.g. the range that satisfies p11 < p12 < · · · < p1n . But because the labels are arbitrary, this would give the same result as integrating over the range p12 < p11 < · · · < p1n , or indeed any other permutation of the momenta. There are n! total permutations, and collectively their ranges cover the entire momentum space1 , so we have the result

X d3 p1 d3 pn d3 pn d3 p1 ··· = ··· n!

2Ep1 (2π)3 2Epn (2π)3 ord(p) 2Ep1 (2π)3 2Epn (2π)3 ord(p) ord∈Sn

=

=⇒

d3 p1 d3 pn · · · 2Ep1 (2π)3 2Epn (2π)3

d3 pn 1 d3 p1 d3 pn d3 p1 · · · = ··· , (6.17) 3 3 3 2Ep1 (2π) 2Epn (2π) ord(p) n! 2Ep1 (2π) 2Epn (2π)3

where there is no ordering in the last expression. The identity operator can therefore be written in terms of states |p1 · · · pn ⟩ with symmetrized momenta as Z 1 d3 p1 d3 pn ˆ In = · · · |p1 · · · pn ⟩⟨p1 · · · pn |, (6.18) n!

2Ep1 (2π)3 2Epn (2π)3 as claimed in Eq. (6.2.64). To verify these are indeed the correct expressions for the identity operator, let us calculate their action on an arbitrary state. We will make use of the normalisations given in Eqs. (6.2.63) and (6.2.64), ⟨p′1 · · · p′n ; So |p1 · · · pn ; So ⟩ = ⟨p′1 · · · p′n |p1 · · · pn ⟩ =

n Y

2Ep′k (2π)3 δ 3 (pk − p′k ),

k=1 " n Y

k=1

# 3

2Ep′k (2π)

n X Y σ∈Sn i=1

δ 3 (pi − p′σ(i) ).

(6.19) (6.20)

First we consider the action on an ordered state |p1 · · · pn ; So ⟩: Iˆn |p1 · · · pn ; So ⟩

Z d3 p′1 d3 p′n = · · · |p′ · · · p′n ; So ⟩⟨p′1 · · · p′n ; So |p1 · · · pn ; So ⟩ 2Ep′1 (2π)3 2Ep′n (2π)3 ord(p′ ) 1

Z n Y d3 p′1 d3 p′n ′ ′ = · · · |p · · · p ; S ⟩ 2Ep′k (2π)3 δ 3 (pk − p′k ) o 1 n 2Ep′1 (2π)3 2Ep′n (2π)3 ord(p′ ) k=1 Z 3 ′

′ ′ 3 ′ 3 ′ = d p1 · · · d pn ord(p′ ) |p1 · · · pn ; So ⟩ δ (p1 − p1 ) · · · δ 3 (pn − p′n ) 1

Except for planes where any of the momenta are equal, but these constitute a set of measure zero and so can be ignored.

Problem 5

123

= |p1 · · · pn ; So ⟩.

(6.21)

The last step is valid because (p1 , . . . , pn ) is ordered by definition, so it lies within the integration range and therefore the delta functions can be satisfied. This shows that Iˆn is the identity operator. And to double check the expression in the symmetric basis, we act on the state |p1 · · · pn ⟩ with symmetrized momenta: Iˆn |p1 · · · pn ⟩ Z d3 pn 1 d3 p1 ··· |p′ · · · p′n ⟩⟨p′1 · · · p′n |p1 · · · pn ⟩ = 3 n!

2Ep1 (2π) 2Epn (2π)3 1 " n # Z n Y X Y d3 pn 1 d3 p1 ′ ′ 3 ′ (2π) · · · |p = · · · p ⟩ 2E δ 3 (pi − p′σ(i) ) p n k n!

2Ep1 (2π)3 2Epn (2π)3 1 σ∈Sn i=1 k=1 Z 1 X = d3 p′1 · · · d3 p′n |p′1 · · · p′n ⟩ δ 3 (p1 − p′σ(1) ) · · · δ 3 (pn − p′σ(n) ) n! σ∈Sn 1 X = |pσ−1 (1) · · · pσ−1 (n) ⟩ n! σ∈Sn 1 X = |p1 · · · pn ⟩ n! σ∈Sn

= |p1 · · · pn ⟩.

(6.22)

Here the last steps follow because the ordering of the momenta in the symmetrized state is immaterial and there are n! terms in the sum over the permutations. This completes our proof that we have the correct expressions for the identity operator Iˆn .

6.5 Problem 5 Problem: Use Eq. (6.2.129) and symmetry to explain why P̂ and M̂ ij do not need normalordering. Show that for a scalar boson field

Z d3 p † ij i ∂ j ∂ L̂ = −i â p −p âp and hence (2π)3 p ∂pj ∂pi Z d3 p † k ∂ L̂i = 21 ϵijk L̂jk = iϵijk â p âp , (6.23) (2π)3 p ∂pj where L̂µν is the coordinate component of M̂ µν and L̂ is the orbital angular momentum operator. Show that the orbital angular momentum of a single boson at rest is zero, i.e., L̂|p = 0⟩ = 0. (Hint: Consider integration by parts to move the momentum derivative.)

Solution:

124

Formulation of Quantum Field Theory

We can combine Eqs. (6.2.21) and (6.2.34) to see that the (non-normal-ordered) four-momentum operator for a scalar field is given by

Z 1 d3 p µ † µ † p âp âp + [âp , âp ] P̂nno = (2π)3 2

Z 3 1 d p µ † = p N̂p + [âp , âp ] , (6.24) (2π)3 2 where N̂p is the number operator for particles of momentum p. We note that µ [âp , â†p ] = (2π)3 δ 3 (0) is just an (infinite) constant operator, so the action of P̂nno on the vacuum is

Z Z 1 d3 p µ d3 p µ µ † p N̂ + p |0⟩.

(6.25) P̂nno [â , â |0⟩ = ] |0⟩ ∝ p p p 3 (2π) 2 (2π)3 By symmetry considerations, this vanishes for µ = i, but is non-zero for µ = 0, p because pi is odd in p while p0 = p2 + m2 is even in p. Since the purpose of normal-ordering an operator is to ensure it annihilates the vacuum state, we see 0 i that P̂nno needs to be normal-ordered while P̂nno does not (or put another way, i P̂nno is already in normal-ordered form). This is also the case for the stress-energy tensor T̂ 0i as seen in Eq. (6.2.127): Z 1 d3 p i 0i 0i p = : T̂ 0i : , (6.26) T̂ = : T̂ : + 2 (2π)3 while T̂ 00 = Ĥ needs normal-ordering for the same reason as P̂ 0 = Ĥ. This carries over to M̂ µν by Eq. (6.2.127): Z

M̂ µν = d3 x xµ : T̂ 0ν : −xν : T̂ 0µ : , (6.27) so we see M̂ 0i needs normal-ordering but M̂ ij = L̂ij does not.

Now for a scalar boson field, the stress-energy tensor is given by Eq. (6.2.119), T̂ µν = π̂ µ ∂ ν ϕ̂ − g µν L, and so the coordinate component of M̂ µν is Z

L̂ij = M̂ ij = d3 x xi π̂ 0 ∂ j ϕ̂ − xj π̂ 0 ∂ i ϕ̂ ,

(6.28)

(6.29)

by Eq. (6.27). Here π̂ 0 ≡ ∂L/∂(∂0 ϕ̂) is the conjugate momentum field. We know from Problem 6.3 that we can write our fields as Z

d3 p ipj † ip·x p , ∂ j ϕ̂(x) = â + â p −p e (2π)3 2ωp r Z

d3 p ωp † 0 π̂ (x) = eip·x , (6.30) (−i) â − â p −p (2π)3 2

125

Problem 5 and so the first term in L̂ij , for example, is Z

d3 x xi π̂ 0 ∂ j ϕ̂ Z Z r

′ 1 ωp′ i j d3 p d3 p′ = d3 x x p âp′ − â†−p′ âp + â†−p ei(p+p )·x 3 3 2 (2π) (2π) ωp Z r

Z ′ 1 d3 p d3 p′ ωp′ j † † = p âp′ − â−p′ âp + â−p d3 x xi ei(p+p )·x 2 (2π)3 (2π)3 ωp Z r

1 ∂ Z ′ d3 p d3 p′ 1 ωp′ j † † ′ − â â â + â d3 x ei(p+p )·x p = ′ p p −p −p 3 3 2 (2π) (2π) ωp i ∂pi Z

3 ′ r 3 ′ i ωp j d p d p ∂ =− p âp′ − â†−p′ âp + â†−p (2π)3 δ 3 (p + p′ ).

2 (2π)3 (2π)3 ωp ∂pi (6.31) At this stage we can integrate by parts and pick up the surface term pi =+∞ r Z

d2 p d3 p′ ωp′ j i † † 3 ′ ′ p âp − â−p′ âp + â−p δ (p + p ) − . (6.32) 2 (2π)3 (2π)3 ωp pi =−∞ This vanishes, of course, because the integral only ranges over finite values of p′ , so δ 3 (p + p′ ) = 0 at all points on the boundary. This leaves us with Z Z

i d3 p d3 p′ d3 x xi π̂ 0 ∂ j ϕ̂ = (2π)3 δ 3 (p + p′ ) 2 (2π)3 (2π)3 r

ωp′ j ∂ × p âp′ − â†−p′ âp + â†−p . ∂pi ωp (6.33) The derivative must be taken before the delta function can be applied. Here we can save ourselves some trouble: since L̂ij is antisymmetric in i and j, we know that any terms above that are symmetric in i and j will cancel out contribute nothing to L̂ij . Among these terms are r ωp′ ∂ pj ∝ pj pi (6.34) ∂pi ωp p (because ωp = p2 + m2 ), and also ∂pj /∂pi = g ij . Therefore, the only contribution to L̂ij comes from the derivative acting on the creation and annihilation operators.

So in total we have Z

L̂ij = d3 x xi π̂ 0 ∂ j ϕ̂ − xj π̂ 0 ∂ i ϕ̂ Z r

i d3 p d3 p′ ωp′ 3 3 ′ = (2π) δ (p + p ) âp′ − â†−p′ 3 3 2 (2π) (2π) ωp

j ∂ i ∂ × p −p âp + â†−p ∂pi ∂pj

Z

3 i ∂ i d p † † j ∂ =− .

(6.35) â − â p − p â + â −p p p −p 2 (2π)3 ∂pj ∂pi

Formulation of Quantum Field Theory

126

Let us examine each of the terms in this expression in detail. This will require use of the commutation relations [âp , âp′ ] = [â†p , â†p′ ] = 0,

[âp , â†p′ ] = (2π)3 δ 3 (p − p′ ),

and the derivatives of these relations, #

" †

∂âp † ∂âp † ∂ 3 ∂âp , âp′ = , âp′ = 0, , âp′ = (2π)3 δ (p − p′ ). ∂pi ∂pi ∂pi ∂pi Note in particular that this implies

∂ 3 ∂âp † , â = (2π)3 δ (0) = 0. ∂pi p ∂pi

(6.36)

(6.37)

(6.38)

The first term in Eq. (6.35), for example, can then be calculated as

Z Z ∂pi ∂âp ∂ d3 p i d3 p i i ∂â−p p â = p â â − â â − p â −p −p p −p p p (2π)3 ∂pj (2π)3 ∂pj ∂pj ∂pj

Z ∂â−p d3 p −g ij â−p âp − pi âp = (2π)3 ∂pj

Z 3 d p ∂âp ij i = −g â−p âp − (−p )â−p (2π)3 ∂(−pj ) Z Z 3 3 ∂âp d p ij d p i 1 p â−p g â−p âp .

(6.39) =⇒ =− 3 (2π) ∂pj 2 (2π)3 In the above calculation, the total derivative term vanished on the boundary2 , the annihilation operator was commuted past the derivative of the annihilation operator by virtue of Eq. (6.37), and a change of variables p → −p was performed in the final term. We see that the first term is therefore symmetric in i and j, and therefore drops out of the antisymmetric L̂ij . Identical reasoning can be applied to the case with two creation operators to see that Z † d3 p i † ∂â−p p âp (6.40) (2π)3 ∂pj also does not contribute to L̂ij . Finally, a similar calculation shows

Z Z ∂pi d3 p ∂ d3 p i ∂â†p i † † i ∂âp † p â = p â â − â â − p â p p p p p (2π)3 ∂pj (2π)3 ∂pj ∂pj ∂pj p

Z d3 p ij † i † ∂âp i ∂âp † = −g â â − p â − p , â p p p (2π)3 ∂pj ∂pi p

Z d3 p ∂âp = −g ij âp â†p − pi â†p , (6.41) (2π)3 ∂pj having used Eq. (6.38). The first term here, again, is symmetric in i and j and does 2

By this we mean that the matrix elements of the boundary term between any two physical states will vanish, because there are no physical states with infinite momentum.

Problem 6

127

not contribute to L̂ij . Putting all of these relations into Eq. (6.35), we arrive at the final result

Z d3 p † ij i ∂ j ∂ L̂ = −i â p −p âp .

(6.42) (2π)3 p ∂pj ∂pi The orbital angular momentum operator is therefore

Z 1 d3 p † 1 k ∂ j ∂ â p − p âp L̂i ≡ ϵijk L̂jk = iϵijk 2 2 (2π)3 p ∂pj ∂pk

Z 1 d3 p † k ∂ k ∂ = iϵijk â p + p âp 2 (2π)3 p ∂pj ∂pj Z d3 p † k ∂ â p = iϵijk âp , (2π)3 p ∂pj

(6.43)

due to the antisymmetry of ϵijk in j and k. With this expression, we can calculate the orbital angular momentum of the state of a single boson with zero linear momentum, √ (6.44) |p = 0⟩ ≡ 2ω0 â†0 |0⟩.

Explicitly, √ L̂i |p = 0⟩ = iϵijk 2ω0

d3 p k † ∂âp † p âp â |0⟩ (2π)3 ∂pj 0 Z

√ d3 p k † ∂ † = iϵijk 2ω0 p â â â |0⟩ p 0 p (2π)3 ∂pj Z

√ d3 p k † ∂ † p âp = iϵijk 2ω0 â0 âp |0⟩ + [âp , â†0 ]|0⟩ 3 (2π) ∂pj Z 3

√ d p k † ∂ = iϵijk 2ω0 p âp 0 + (2π)3 δ 3 (p)|0⟩ 3 (2π) ∂pj " # Z ∂pk † 3 √ ∂â†p 3 ∂ ijk 3 k † 3 k = iϵ 2ω0 d p p âp δ (p) − â δ (p) − p δ (p) |0⟩. ∂pj ∂pj p ∂pj Z

(6.45) Now here the total derivative term vanishes on the boundary, ϵijk ∂pk /∂pj = ϵijk g jk is a contraction between symmetric and antisymmetric tensors and therefore vanishes, and the delta function sets the last term to zero. So in total the state has zero orbital angular momentum, L̂i |p = 0⟩ = 0, as expected.

6.6 Problem 6 Problem:

(6.46)

128

Formulation of Quantum Field Theory

In Eq. (6.2.168) we showed, using its vacuum expectation value form, that DF (x− y) satisfied (∂x2 + m2 )DF (x − y) = −iδ 4 (x − y). Using their vacuum expectation value forms, similarly show that this is true for each of Dret (x − y), Dadv (x − y) and D̄F (x − y).

Solution: The vacuum expectation value forms of the different propagators are given in Eq. (6.2.163) and Eq. (6.2.164): Dret (x − y) = θ(x0 − y 0 )⟨0|[ϕ̂(x), ϕ̂(y)]|0⟩,

Dadv (x − y) = −θ(y 0 − x0 )⟨0|[ϕ̂(x), ϕ̂(y)]|0⟩, 0

(6.47)

0

0

0

D̄F (x − y) = −θ(x − y )⟨0|ϕ̂(y)ϕ̂(x)|0⟩ − θ(y − x )⟨0|ϕ̂(x)ϕ̂(y)|0⟩.

We shall look at each of these in turn. (i) Let us take the first derivative of the retarded propagator with respect to the coordinate xµ : ∂µ(x) Dret (x − y) = θ(x0 − y 0 )∂µ(x) ⟨0|[ϕ̂(x), ϕ̂(y)]|0⟩ h i + ∂µ(x) θ(x0 − y 0 ) ⟨0|[ϕ̂(x), ϕ̂(y)]|0⟩ = θ(x0 − y 0 )⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩

+ δ 0µ δ(x0 − y 0 )⟨0|[ϕ̂(x), ϕ̂(y)]|0⟩

= θ(x0 − y 0 )⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩.

(6.48)

The second term in the second line vanishes because the delta function can be used to set x0 = y 0 , and the resulting equal-time commutator is zero by the fundamental commutation relations. A second derivative proceeds similarly, µ µ ∂(x) ∂µ(x) Dret (x − y) = θ(x0 − y 0 )∂(x) ⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩ h i µ + ∂(x) θ(x0 − y 0 ) ⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩ µ = θ(x0 − y 0 )⟨0|[∂(x) ∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩

+ δ µ0 δ(x0 − y 0 )⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩

= θ(x0 − y 0 )⟨0|[−m2 ϕ̂(x), ϕ̂(y)]|0⟩ (x)

+ δ(x0 − y 0 )⟨0|[∂0 ϕ̂(x), ϕ̂(y)]|0⟩

= −m2 θ(x0 − y 0 )⟨0|[ϕ̂(x), ϕ̂(y)]|0⟩ + δ(x0 − y 0 )⟨0|[π̂(x), ϕ̂(y)]|0⟩

= −m2 Dret (x − y) − iδ(x0 − y 0 )δ 3 (x − y).

(6.49)

In the third line we made use of the equation of motion (∂ µ ∂µ + m2 )ϕ̂(x) = 0, the fourth line substituted the definition π̂(x) = ∂0 ϕ̂(x), and the final line used the equal-time commutation relation [π̂(x), ϕ̂(y)] = −iδ 3 (x − y) of

Problem 6

129

2 Eq. (6.2.81). A simple rearrangement gives (∂(x) + m2 )Dret (x − y) = −iδ 4 (x − y), as desired.

(ii) The calculation for the advanced propagator is almost exactly the same as the above, so we shall leave out some intermediate steps for brevity. The only (x) (x) difference we need to note here is that ∂µ θ(y 0 − x0 ) = −∂µ θ(x0 − y 0 ). The first derivative is then ∂µ(x) Dadv (x − y) = −θ(y 0 − x0 )⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩

+ δ 0µ δ(x0 − y 0 )⟨0|[ϕ̂(x), ϕ̂(y)]|0⟩

= −θ(y 0 − x0 )⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩,

(6.50)

and so the second is µ µ ∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩ ∂µ(x) Dadv (x − y) = −θ(y 0 − x0 )⟨0|[∂(x) ∂(x)

+ δ µ0 δ(x0 − y 0 )⟨0|[∂µ(x) ϕ̂(x), ϕ̂(y)]|0⟩

= −θ(y 0 − x0 )⟨0|[−m2 ϕ̂(x), ϕ̂(y)]|0⟩ + δ(x0 − y 0 )⟨0|[π̂(x), ϕ̂(y)]|0⟩

= −m2 Dadv (x − y) − iδ(x0 − y 0 )δ 3 (x − y).

(6.51)

2 This rearranges to give (∂(x) + m2 )Dadv (x − y) = −iδ 4 (x − y), as desired.

(iii) Finally we have the anti-time-ordered propagator D̄F (x − y). Using the same method as above, we calculate the first derivative ∂µ(x) D̄F (x − y)

= −θ(x0 − y 0 )⟨0|ϕ̂(y)∂µ(x) ϕ̂(x)|0⟩ − δ 0µ δ(x0 − y 0 )⟨0|ϕ̂(y)∂µ(x) ϕ̂(x)|0⟩

− θ(y 0 − x0 )⟨0|∂µ(x) ϕ̂(x)ϕ̂(y)|0⟩ + δ 0µ δ(x0 − y 0 )⟨0|∂µ(x) ϕ̂(x)ϕ̂(y)|0⟩

= −θ(x0 − y 0 )⟨0|ϕ̂(y)∂µ(x) ϕ̂(x)|0⟩ − θ(y 0 − x0 )⟨0|∂µ(x) ϕ̂(x)ϕ̂(y)|0⟩ + δ(x0 − y 0 )⟨0|[π̂(x), ϕ̂(y)]|0⟩

= −θ(x0 − y 0 )⟨0|ϕ̂(y)∂µ(x) ϕ̂(x)|0⟩ − θ(y 0 − x0 )⟨0|∂µ(x) ϕ̂(x)ϕ̂(y)|0⟩, (6.52) and therefore the second derivative µ ∂(x) ∂µ(x) D̄F (x − y)

µ = −θ(x0 − y 0 )⟨0|ϕ̂(y)∂(x) ∂µ(x) ϕ̂(x)|0⟩ − δ µ0 δ(x0 − y 0 )⟨0|ϕ̂(y)∂µ(x) ϕ̂(x)|0⟩

µ − θ(y 0 − x0 )⟨0|∂(x) ∂µ(x) ϕ̂(x)ϕ̂(y)|0⟩ + δ µ0 δ(x0 − y 0 )⟨0|∂µ(x) ϕ̂(x)ϕ̂(y)|0⟩

= −θ(x0 − y 0 )⟨0|ϕ̂(y)(−m2 )ϕ̂(x)|0⟩ + δ(x0 − y 0 )⟨0|[π̂(x), ϕ̂(y)]|0⟩ − θ(y 0 − x0 )⟨0|(−m2 )ϕ̂(x)ϕ̂(y)|0⟩

= −m2 D̄F (x − y) − iδ(x0 − y 0 )δ 3 (x − y), 2 once again giving (∂(x) + m2 )D̄F (x − y) = −iδ 4 (x − y), as desired.

(6.53)

Formulation of Quantum Field Theory

130

6.7 Problem 7 Problem: Prove that (a) the only nonzero commutation relations for the annihilation and creation operators of a charged scalar field (fˆp , fˆp† , ĝp , ĝp† ) are given by Eq. (6.2.180), i.e., [fˆp , fˆp†′ ] = [ĝp , ĝp† ′ ] = (2π)3 δ 3 (p − p′ ), and (b) the only nonzero equal-time canonical commutation relations for the complex field operator ϕ̂ and its canonical momentum density operator π̂ are given by Eq. (6.2.181), i.e., [ϕ̂(t, x), π̂(t, x′ )] = [ϕ̂† (t, x), π̂ † (t, x′ )] = iδ 3 (x − x′ ).

Solution: (a) The annihilation and creation operators of a charged scalar field are given by Eq. (6.2.178), √ √ fˆp† ≡ (1/ 2)[â†1p − iâ†2p ], fˆp ≡ (1/ 2)[â1p + iâ2p ], √ √ ĝp† ≡ (1/ 2)[â†1p + iâ†2p ], (6.54) ĝp ≡ (1/ 2)[â1p − iâ2p ], where (â1p , â†1p ) and (â2p , â†2p ) are two sets of mutually commuting creation and annihilation operators with the usual commutation relations [âp , âp′ ] = [â†p , â†p′ ] = 0,

[âp , â†p′ ] = −[â†p′ , âp ] = (2π)3 δ 3 (p − p′ ). (6.55)

Since all annihilation operators â commute regardless of their momenta, it is immediately apparent that [fˆp , fˆp′ ] = [ĝp , ĝp′ ] = [fˆp , ĝp′ ] = 0,

(6.56)

[fˆp† , fˆp†′ ] = [ĝp† , ĝp† ′ ] = [fˆp† , ĝp† ′ ] = 0

(6.57)

and similarly

because all creation operators ↠commute regardless of their momenta. There are two further vanishing commutators that are non-trivial to show: 1 [â1p + iâ2p , â†1p + iâ†2p ] 2

1 = [â1p , â†1p ] + i2 [â2p , â†2p ] 2

1 = (2π)3 δ 3 (p − p′ ) − (2π)3 δ 3 (p − p′ ) 2 = 0,

[fˆp , ĝp† ′ ] =

and 1 † [â − iâ†2p , â1p − iâ2p ] 2 1p

1 † = [â1p , â1p ] + i2 [â†2p , â2p ] 2

[fˆp† , ĝp′ ] =

(6.58)

Problem 7

131

1 −(2π)3 δ 3 (p − p′ ) + (2π)3 δ 3 (p − p′ ) 2 = 0.

=

(6.59)

The second relation here actually follows from the first because [fˆp† , ĝp′ ] = −[fˆp , ĝ † ′ ]† , as can be seen by expanding both sides. This leaves two distinct p

commutators to calculate: 1 [fˆp , fˆp†′ ] = [â1p + iâ2p , â†1p − iâ†2p ] 2

1 [â1p , â†1p ] − i2 [â2p , â†2p ] = 2

1 = (2π)3 δ 3 (p − p′ ) + (2π)3 δ 3 (p − p′ ) 2 = (2π)3 δ 3 (p − p′ ),

(6.60)

and 1 [â1p − iâ2p , â†1p + iâ†2p ] 2

1 [â1p , â†1p ] − i2 [â2p , â†2p ] = 2

1 = (2π)3 δ 3 (p − p′ ) + (2π)3 δ 3 (p − p′ ) 2 = (2π)3 δ 3 (p − p′ ),

[ĝp , ĝp† ′ ] =

(6.61)

proving the result. (b) The complex fields and their canonical conjugate momentum densities can be written in terms of the above operators as Z

1 ˆ −ip·x d3 p † ip·x p ϕ̂(x) = f e + ĝ e , p p (2π)3 2Ep Z

1 ˆ† ip·x d3 p −ip·x † p f e + ĝ e , ϕ̂ (x) = p p (2π)3 2Ep r Z

d3 p Ep ˆ† ip·x −ip·x f e − ĝ e , π̂(x) = i p p (2π)3 2 r Z

d3 p Ep ˆ −ip·x † ip·x π̂(x)† = −i f e − ĝ e , (6.62) p p (2π)3 2

by Eq. (6.2.177) in the text. Here it is to be understood that p0 = Ep ≡ p p2 + m2 in all cases. Now from the results in Part (a), we know that any two such fields can only fail to commute if their expansions contain fp and fp† , or gp and gp† operators collectively. It is then immediately apparent that [ϕ̂(x), ϕ̂(x′ )] = [ϕ̂† (x), ϕ̂† (x′ )] = [π̂(x), π̂(x′ )] = [π̂ † (x), π̂ † (x′ )] = 0,

(6.63)

and also [ϕ̂(x), π̂ † (x′ )] = [ϕ̂† (x), π̂(x′ )] = 0.

(6.64)

Formulation of Quantum Field Theory

132

The other commutators must be calculated on a case-by-case basis. To begin, †

′

Z

1 d3 p d3 p′ p 3 3 (2π) (2π) 2 Ep Ep′ h i ′ ′ ′ ′ × fˆp e−ip·x + ĝp† eip·x , fˆp†′ eip ·x + ĝp′ e−ip ·x

Z

1 d3 p d3 p′ p (2π)3 (2π)3 2 Ep Ep′ h i

′ ′ ′ ′ × fˆp , fˆp†′ e−i(p·x−p ·x ) + ĝp† , ĝp′ ei(p·x−p ·x )

[ϕ̂(t, x), ϕ̂ (t, x )] =

=

d3 p d3 p′ (2π)3 δ 3 (p − p′ ) −i(p·x−p′ ·x′ ) i(p·x−p′ ·x′ ) p e − e (2π)3 (2π)3 2 Ep Ep′ Z

3 d p 1 −ip·(x−x′ ) ip·(x−x′ ) p = e − e (2π)3 2 Ep Ep Z

d3 p 1 ip·(x−x′ ) −ip·(x−x′ ) = e − e (2π)3 2Ep Z

=

= 0,

(6.65)

because the integrand is odd in p. Note that because we are calculating equaltime commutators, we have x0 − x′0 = 0 and so p · (x − x′ ) = −p · (x − x′ ). A similar calculation yields p Ep Ep′ d3 p d3 p′ [π̂(t, x), π̂ (t, x )] = 3 3 (2π) (2π) 2 i h ′ ′ ′ ′ × fˆp† eip·x − ĝp e−ip·x , fˆp′ e−ip ·x − ĝp† ′ eip ·x p Z Ep Ep′ d3 p d3 p′ = (2π)3 (2π)3 2 h i h i

′ ′ ′ ′ × fˆp† , fˆp′ ei(p·x−p ·x ) + ĝp , ĝp† ′ e−i(p·x−p ·x ) p Z Ep Ep′ d3 p d3 p′ (2π)3 δ 3 (p − p′ ) = 3 3 (2π) (2π) 2

′ ′ ′ ′ × −ei(p·x−p ·x ) + e−i(p·x−p ·x ) p Z

Ep Ep ip·(x−x′ ) d3 p −ip·(x−x′ ) −e + e = (2π)3 2 Z

3 d p Ep −ip·(x−x′ ) ip·(x−x′ ) = −e + e (2π)3 2 †

′

Z

= 0,

(6.66)

133

Problem 8

again because the integrand is odd in p. Finally, s Z 3 ′ 3 i d p Ep′ d p [ϕ̂(t, x), π̂(t, x′ )] = 2 (2π)3 (2π)3 Ep h i ′ ′ ′ ′ × fˆp e−ip·x + ĝp† eip·x , fˆp†′ eip ·x − ĝp′ e−ip ·x s Z d3 p d3 p′ Ep′ i = 3 3 2 (2π) (2π) Ep h i

′ ′ ′ ′ × fˆp , fˆp†′ e−i(p·x−p ·x ) − ĝp† , ĝp′ ei(p·x−p ·x ) s Z Ep′ d3 p i = (2π)3 δ 3 (p − p′ ) 3 2 (2π) Ep

′ ′ ′ ′ × e−i(p·x−p ·x ) + ei(p·x−p ·x ) s Z

′ d3 p Ep −ip·(x−x′ ) i e + eip·(x−x ) = 3 2 (2π) Ep Z

d3 p i ip·(x−x′ ) −ip·(x−x′ ) e + e = 2 (2π)3

i 3 = δ (x − x′ ) + δ 3 (x − x′ ) 2 = i δ 3 (x − x′ ).

(6.67) Taking the hermitian conjugate of this equation shows [ϕ̂(t, x), π̂(t, x′ )]† = [π̂ † (t, x′ ), ϕ̂† (t, x)] = −i δ 3 (x−x′ ). In total, we have seen that the only non-zero equal-time canonical commutation relations between a complex field operator and its canonical momentum density operator are given by [ϕ̂(t, x), π̂(t, x′ )] = [ϕ̂† (t, x), π̂ † (t, x′ )] = i δ 3 (x − x′ ).

6.8 Problem 8 Problem: Prove Eq. (6.3.197) for writing the fermion creation and annihilation operators in terms of the field operators; Solution: From Eq. (6.3.194) we have the expression for the fermion field operator Z i d3 p 1 Xh s s −ip·x ˆs† v s (p)eip·x p ψ̂(x) = b̂ u (p)e + d p p (2π)3 2Ep s

Formulation of Quantum Field Theory

134

≡

Z

i d3 p 1 Xh s s† ˆ p b̂ u (x) + d v (x) , p ps p ps (2π)3 2Ep s

(6.68)

and hence i 1 X h s † s† ip·x d3 p p u (p) b̂p e + v s (p)† dˆsp e−ip·x 3 (2π) 2Ep s Z i d3 p 1 Xh † s† † ˆs p ≡ u (x) b̂ + v (x) d ps ps p p .

(2π)3 2Ep s

ψ̂(x)† =

Z

(6.69)

We note here that these spinors are normalized such that ′

′

′

us† (p)us (p) = v s† (p)v s (p) = 2Ep δ ss

(6.70)

in natural units, by Eq. (4.4.125), and also obey the orthogonality condition ′

′

us† (p0 , p) v s (p0 , −p) = v s† (p0 , p) us (p0 , −p) = 0,

(6.71)

by Eq. (4.4.121). With these, we can perform some intermediate calculations for future convenience: Z Z ′ ′ d3 x u†ps (x)up′ s′ (x) ≡ d3 x us† (p)us (p′ )ei(p−p )·x ′

′0

0

= us† (p)us (p′ )ei(p −p )t (2π)3 δ 3 (p − p′ ) ′

= us† (p)us (p)e0 (2π)3 δ 3 (p − p′ ) ′

= 2Ep (2π)3 δ ss δ 3 (p − p′ ),

(6.72)

having noted that if p′ = p then p′ = p more generally. Continuing, Z Z ′ ′ d3 x u†ps (x)vp′ s′ (x) ≡ d3 x us† (p)v s (p′ )ei(p+p )·x ′

′0

0

= us† (p)v s (p′ )ei(p +p )t (2π)3 δ 3 (p + p′ ) ′

0

= us† (p0 , p)v s (p0 , −p)e2ip t (2π)3 δ 3 (p + p′ )

Z

= 0, Z ′ ′ 3 † d x vps (x)up′ s′ (x) ≡ d3 x v s† (p)us (p′ )e−i(p+p )·x ′

(6.73)

′0

0

= v s† (p)us (p′ )e−i(p +p )t (2π)3 δ 3 (p + p′ ) ′

0

= v s† (p0 , p)us (p0 , −p)e−2ip t (2π)3 δ 3 (p + p′ ) = 0,

and finally Z

† d x vps (x)vp′ s′ (x) ≡ 3

Z

′

(6.74)

′

d3 x v s† (p)v s (p′ )e−i(p−p )·x ′

0

′0

= v s† (p)v s (p′ )e−i(p −p )t (2π)3 δ 3 (p − p′ ) ′

= v s† (p)v s (p)e0 (2π)3 δ 3 (p − p′ ) ′

= 2Ep (2π)3 δ ss δ 3 (p − p′ ).

(6.75)

Problem 9

135

Using these results, we quickly calculate Z Z Z i Xh ′ 1 d3 p′ 3 † 3 † s† † ˆs′ ′ s′ (x) b̂ ′ + vp′ s′ (x) d p d x ψ̂ (x)ups (x) = d x u ′ ups (x) p p p (2π)3 2Ep′ ′ s Z i Xh ′ ′ 1 d3 p′ p 2Ep (2π)3 δ ss δ 3 (p − p′ )b̂sp′† + 0 = 3 (2π) 2Ep′ s′ i h 1 =p 2Ep b̂s† p 2Ep p = 2Ep b̂s† (6.76) p , Z Z Z h i 3 ′ X ′ 1 d p † s′ ˆs † p d3 x u†ps (x)ψ̂(x) = d3 x u (x) b̂ ′ up′ s′ (x) + dp′ vp′ s′ (x) ps p (2π)3 2Ep′ ′ s Z i Xh 1 d3 p′ 3 ss′ 3 ′ s′ p 2E (2π) δ δ (p − p ) b̂ = ′ + 0 p p (2π)3 2Ep′ ′ s i 1 h s =p 2Ep b̂p 2Ep p = 2Ep b̂sp , (6.77) Z Z Z h ′ i 3 ′ X ′ 1 d p s† † s † ˆ ′ ′ ′ ′ p v (x) b̂ u (x) + d v (x) d3 x vps (x)ψ̂(x) = d3 x ′ ′ p ps p ps (2π)3 2Ep′ ′ ps s Z i Xh ′ d3 p′ 1 3 ss′ 3 ′ ˆs † p = 0 + 2E (2π) δ δ (p − p ) d ′ p p (2π)3 2Ep′ ′ s i 1 h =p 2Ep dˆs† p 2Ep p = 2Ep dˆs† (6.78) p , Z Z Z h i 3 ′ X ′ ′ 1 d p p d3 x ψ̂ † (x)vps (x) = d3 x up′ s′ (x)† b̂sp′† + vp′ s′ (x)† dˆsp′ vps (x) (2π)3 2Ep′ ′ s Z h i 3 ′ X ′ ′ 1 d p p 0 + 2Ep (2π)3 δ ss δ 3 (p − p′ )dˆsp′ = 3 (2π) 2Ep′ s′ h i 1 =p 2Ep dˆsp 2Ep p (6.79) = 2Ep dˆs , p

confirming Eq. (6.3.197).

6.9 Problem 9 Problem: Show that the fermion operators P̂ µ and M̂ µν are the generators of the Poincaré

Formulation of Quantum Field Theory

136

group by proving Eqs. (6.3.214), (6.3.212) and (6.3.213). Show that P̂ µ and M̂ µν are hermitian. (Hint: First prove before normal-ordering and then prove that normalordering only involves adding a real constant. While each step is relatively straightforward, this proof requires care and effort.)

Solution: From Eq. (6.3.198) we have the expressions Z P̂ µ = d3 x : iψ̂ † (x)∂ µ ψ̂(x) : Z ≡ d3 x : P̂ µ : , Z M̂ ρσ = d3 x : iψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + [γ ρ , γ σ ]/4)ψ̂(x) : Z ≡ d3 x : iψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + Σρσ )ψ̂(x) : Z ≡ d3 x : M̂ρσ : ,

(6.80)

(6.81)

for our candidate generators of the Poincaré group. The time coordinates of all operators and spacetime points in this solution should be understood to be equal to the same fixed value. We wish to show that P̂ µ and M̂ ρσ are hermitian and satisfy the Lie algebra of the Poincaré group. The solution to the problem is rather unwieldy, so for convenience we separate it into distinct sections. (a) Let us first prove these operators are hermitian. To do so, we note that together Eq. (6.3.80) and Eq. (6.3.203) imply that normal-ordering a product of two fermion creation or annihilation operators introduces at most a real infinite constant, e.g. ′

′

′

′

′

′

s† s† s s s† s s† s s† ss 3 : b̂sp b̂p δ (p − p′ ). ′ : ≡ −b̂p′ b̂p = b̂p b̂p′ − {b̂p , b̂p′ } = b̂p b̂p′ − δ

(6.82)

Therefore, P̂ µ and M̂ ρσ differ from their non-normal-ordered forms (call them µ ρσ P̂nno and M̂nno ) by at most a complex infinite constant, µ ˆ P̂ µ = P̂nno + cP I, ρσ ˆ M̂ ρσ = M̂nno + cM I,

(6.83)

depending on the coefficients in their expressions. We shall see in Part (a) (iii) that in fact these constants are real, so that µ† µ† ˆ P̂ µ† = P̂nno + c∗P Iˆ† = P̂nno + cP I, ρσ† ρσ† ˆ M̂ ρσ† = M̂nno + c∗M Iˆ† = M̂nno + cM I.

(6.84)

µ† µ ρσ† ρσ Hence, if P̂nno = P̂nno and M̂nno = M̂nno , then P̂ µ† = P̂ µ and M̂ ρσ† = M̂ ρσ as well. This will be the method by which we prove the hermiticity of the generators.

Problem 9

137

(i) We begin by considering the hermiticity of the non-normal-ordered momentum operator, Z

† µ† P̂nno = d3 x iψ̂ † (x)∂ µ ψ̂(x) Z

† = d3 x (−i) ∂ µ ψ̂(x) ψ̂(x) Z

= d3 x (−i) ∂ µ ψ̂ † (x) ψ̂(x) Z

i h (6.85) = d3 x (−i) ∂ µ ψ̂ † (x)ψ̂(x) − ψ̂ † (x)∂ µ ψ̂(x) .

Here we must consider two different cases. If µ = i, then the integral over i xi yields a factor of [ψ̂ † (x)ψ̂(x)]xxi =+∞ =−∞ in the first term, which vanishes for fields that vanish at infinity. Otherwise, we have µ = 0. In this case we cleverly recognize that ψ̂ † ψ̂ = ψ̂ † γ 0 γ 0 ψ̂ = ψ̄ˆγ 0 ψ̂ = ĵ 0 , so we have the term ∂ 0 ĵ 0 = ∂0 ĵ 0 = −∂i ĵ i = −∂i (ψ̄ˆγ i ψ̂) by current conservation Eq. (6.3.207). The divergence theorem then converts this to a surface term, which vanishes for fields that vanish at infinity. So in all cases, we get Z µ† µ P̂nno = d3 x iψ̂ † (x)∂ µ ψ̂(x) = P̂nno (6.86) µ is hermitian. and so P̂nno ρσ , but here we must (ii) The same sort of manipulations can be done for M̂nno ρσ be careful. We wish to show that the specific operator M̂nno is hermitian.

We do not want to show that the matrix M̂nno is hermitian. That is to say, σρ ∗ ρσ ∗ ρσ ) . It ) , not (M̂nno is by definition (M̂nno the hermitian conjugate of M̂nno is also crucial to keep track of spinor indices here, which have so far been kept implicit. With these clarifications understood, we calculate ρσ† ρσ ∗ M̂nno ≡ (M̂nno ) Z

∗ = d3 x iψ̂α† (δαβ xρ ∂ σ − δαβ xσ ∂ ρ + Σρσ ) ψ̂ β αβ Z

∗ † = (−i) d3 x xρ ψ̂α ∂ σ ψ̂α† − xσ ψ̂α ∂ ρ ψ̂α† + ψ̂α (Σρσ αβ ) ψ̂β Z

= (−i) d3 x ∂ σ (xρ ψ̂α† ψ̂α ) − ψ̂α† (∂ σ xρ )ψ̂α − xρ ψ̂α† ∂ σ ψ̂α

−∂ ρ (xσ ψ̂α† ψ̂α ) + ψ̂α† (∂ ρ xσ )ψ̂α + xσ ψ̂α† ∂ ρ ψ̂α ∗ +ψ̂α† (Σρσ βα ) ψ̂β

(6.87)

Here we have made use of the fact that there is no difference between ψ̂α∗ and ψ̂α† , and in the last term we have relabeled α ↔ β. The second terms

Formulation of Quantum Field Theory

138

in the first two lines cancel, since ∂ σ xρ = g στ ∂τ xρ = g στ δ ρτ = g σρ = g ρσ = ∂ ρ xσ . So we see Z

ρσ† † ρσ M̂nno = i d3 x xρ ψ̂α† ∂ σ ψ̂α − xσ ψ̂α† ∂ ρ ψ̂α + ψ̂α† Σρσ αβ ψ̂β − ψ̂α Σαβ ψ̂β ∗ +∂ ρ (xσ ψ̂α† ψ̂α ) − ∂ σ (xρ ψ̂α† ψ̂α ) − ψ̂α† (Σρσ βα ) ψ̂β ρσ = M̂nno +i

Z

d3 x ∂ ρ (xσ ψ̂α† ψ̂α ) − ∂ σ (xρ ψ̂α† ψ̂α ) i h ρσ ∗ −ψ̂α† Σρσ ψ̂β . αβ + (Σβα )

(6.88)

Our goal, therefore, is to show that this last integral vanishes. Without σρ loss of generality we may take σ = j, since M̂nno is antisymmetric in ρ and σ. Then the second term in the integral integrates to zero provided xρ ψ̂α† ψ̂α vanishes sufficiently quickly as xj → ±∞. This gives us Z

h i ρj ∗ ρj† ρj M̂nno = M̂nno + i d3 x ∂ ρ (xj ψ̂α† ψ̂α ) − ψ̂α† Σρj ψ̂β . αβ + (Σβα ) (6.89) We can only evaluate this integral on a case-by-case basis, as in Part (a) (i).

First let us consider the case where ρ is spatial, ρ = i. Then the first term ij ∗ integrates to zero and all that remains is the term ψ̂α† [Σij αβ + (Σβα ) ]ψ̂β .

In matrix notation this is written as ψ̂ † [Σij + (Σij )† ]ψ̂, where the dagger acts on the spinor indices, not the Lorentz indices. Now from the definition Σij ≡ [γ i , γ j ]/4 we see (Σij )† =

1 1 1 i j† [γ , γ ] = [(γ j )† , (γ i )† ] = [−γ j , −γ i ] = Σji = −Σij .

4 4 4 (6.90)

These steps simply follow from the definition of the commutator and the properties of the γ-matrices, (γ 0 )† = γ 0 and (γ i )† = −γ i . Therefore ij ij† = M̂nno . (Σij )† + Σij = 0, and so ultimately M̂nno The last remaining case, where ρ = 0, is more subtle. Here the first term in the integrand of Eq. (6.89) becomes ∂ 0 (xj ψ̂α† ψ̂α ) = xj ∂0 ĵ 0 = −xj ∂i ĵ i = ĵ j − ∂i (xj ĵ i ). This last total derivative will give a vanishing boundary contribution provided xj ĵ i vanishes as xi → ±∞, so we are left with Z

h i 0j ∗ 0j† 0j M̂nno = M̂nno + i d3 x ĵ j − ψ̂α† Σ0j ψ̂β αβ + (Σβα ) Z

0j = M̂nno + i d3 x ψ̄ˆγ j ψ̂ − ψ̂ † Σ0j + (Σ0j )† ψ̂ Z

0j = M̂nno + i d3 x ψ̂ † γ 0 γ j ψ̂ − ψ̂ † Σ0j + (Σ0j )† ψ̂ Z

0j = M̂nno + i d3 x ψ̂ † γ 0 γ j − Σ0j − (Σ0j )† ψ̂.

(6.91)

Problem 9

139

Calculating the matrix in brackets, 1 1 γ 0 γ j − Σ0j − (Σ0j )† = γ 0 γ j − [γ 0 , γ j ] − [γ 0 , γ j ]† 4 4 1 0 j 1 j † 0 † 0 j = γ γ − 2γ γ − [(γ ) , (γ ) ] 4 4 1 0 j 1 j 0 = γ γ − [−γ , γ ] 2 4 1 0 j 1 j 0 = γ γ + 2γ γ 2 4 = 0,

(6.92)

having used the fact that distinct γ-matrices anticommute. We see that 0j† 0j i0† i0 M̂nno = M̂nno , and therefore also M̂nno = M̂nno by antisymmetry. Comij† ij ρσ† ρσ bined with the previous result M̂nno = M̂nno , we get that M̂nno = M̂nno for all values of ρ and σ.

It is helpful to note here that these calculations imply the general formula (Σµν )† = −Σµν + (δ µ0 δ νk − δ ν0 δ µk )γ 0 γ k ,

(6.93)

which will be useful later. (iii) We have found that the non-normal-ordered forms of the candidate generators are hermitian. Now all that is left is to show that the normal-ordered forms differ from these by a real constant, and hence are hermitian as well.

We can do this most efficiently by noting that each of our generators is of the form Z d3 x : ψ̂ † Oψ̂ : (6.94) for some differential operator O, so we are interested in calculating the general quantity Z n o ˆ ∆O := d3 x : ψ̂ † Oψ̂ : −ψ̂ † Oψ̂ (6.95) and showing that it is real for our specific operators. The demonstration rests on the expression for the fermion field Eq. (6.3.194), Z

1 X s s d3 p −ip·x ˆs† v s (p)eip·x p b̂ u (p)e + d ψ̂(x) = p p (2π)3 2Ep s Z

d3 p 1 X s ˆs† vps (x) , p b̂ u (x) + d (6.96) ≡ ps p p (2π)3 2Ep s which gives Z Z d3 p d3 p′ 1 ˆ O = d3 x p ∆ (2π)3 (2π)3 2 Ep Ep′

Formulation of Quantum Field Theory

140

×

Xh ′ † ˆs v † O b̂s′′ us′ p′ + dˆs ′† vs′ p′ : : b̂s† u + d p p sp p sp p s,s′

′ i ′ † s ˆs † ˆs † − b̂s† . p usp + dp vsp O b̂p′ us′ p′ + dp′ vs′ p′ (6.97) Now most of the terms in the first line here are already normal-ordered, ′ ′ s′ s† s′ s† ˆs † s† ˆs † ˆs s′ ˆs s′ since : b̂s† p b̂p′ := b̂p b̂p′ , : b̂p dp′ := b̂p dp′ , and : dp b̂p′ := dp b̂p′ . Many terms therefore cancel, leaving only Z Z h i X ′ ′ 1 d3 p d3 p′ † ˆ O = d3 x p ∆ vsp Ovs′ p′ : dˆsp dˆsp′† : −dˆsp dˆsp′† 3 3 (2π) (2π) 2 Ep Ep′ ′ s,s Z Z h i 3 ′ 3 X ′ ′ 1 d p d p † ˆs ′† dˆs − dˆs dˆs ′† ′ p′ −d p v = d3 x Ov s sp p p p p (2π)3 (2π)3 2 Ep Ep′ ′ s,s Z Z 3 3 ′ X

d p 1 d p † p vsp Ovs′ p′ δss′ (2π)3 δ 3 (p − p′ ) = − d3 x 3 3 (2π) (2π) 2 Ep Ep′ ′ s,s Z Z 3 X d p 1 v † Ovsp .

(6.98) = − d3 x (2π)3 2Ep s sp Here we have simply used the canonical anticommutation relations. The operator O will need to be specified to progress further.

Taking O = i∂ µ will give us the case for P̂ µ . A simple calculation then gives Z Z d3 p 1 X s† µ µ 3 P̂ − P̂nno = − d x v (p)e−ip·x i∂ µ v s (p)eip·x (2π)3 2Ep s Z Z d3 p 1 X s† v (p)i2 pµ v s (p) = − d3 x (2π)3 2Ep s Z Z d3 p X µ = d3 x p , (6.99) (2π)3 s as follows from Eq. (4.4.125), v s† (p)v s (p) = 2Ep . This is obviously a real constant operator, so it follows that P̂ µ is hermitian3 .

Next, we take O = i(xρ ∂ σ − xσ ∂ ρ + Σρσ ) to obtain the case for M̂ ρσ . As above, the derivatives will translate into momentum factors, so that Z Z d3 p 1 X s† ρσ ρσ 3 M̂ − M̂nno = d x v (p)(xρ pσ − xρ pσ − iΣρσ )v s (p) (2π)3 2Ep s 3

Of course this is already clear from the text, Z d3 p X µ s† s ˆs P̂ µ = p (b̂p b̂p + dˆs† p dp ).

(2π)3 s

(6.100)

s† s s † ˆs† ˆs † ˆs† ˆs This is obviously hermitian since (b̂s† p b̂p ) = b̂p b̂p and (dp dp ) = dp dp , but this approach is µν not conducive to showing that M̂ is hermitian.

Problem 9

141

Z =

d3 x

Z

d3 p X ρ σ (x p − xρ pσ ) − iv s† (p)Σρσ v s (p) .

3 (2π) s (6.101)

The first two terms always integrate to zero, since either they are odd in one of the integration variables xi or pi , or they simply cancel each other.

Hence, Z Z d3 p X s† ρσ ρσ 3 iv (p)Σρσ v s (p).

(6.102) M̂ − M̂nno = − d x (2π)3 s If this operator is hermitian, then M̂ ρσ will be hermitian. Checking this: [iv s† (p)Σρσ v s (p)]† = −iv s† (p)Σρσ† v s (p).

(6.103)

We will need to proceed by considering the various component combinations individually. For example, since Σij† = −Σij , the above operator is hermitian in this case and so M̂ ij† = M̂ ij . The only other case that needs to be considered is ρ, σ = 0, j. Here we can make use of Eq. (4.4.135) to calculate4 Z Z d3 p X s† 0j v (p)Σ0j v s (p) M̂ 0j − M̂nno = −i d3 x (2π)3 s Z Z d3 p X 1 s† = −i d3 x v (p)γ 0 γ j v s (p) (2π)3 s 2 Z Z d3 p X j = −i d3 x p (2π)3 s = 0,

(6.104)

because the integrand is odd in pj . This is a real constant, and so M̂ 0j† = M̂ 0j . This completes the proof that M̂ ρσ† = M̂ ρσ for all values of ρ and σ. (b) So much for hermiticity. Now we must show that P̂ and M̂ ρσ satisfy the Lie algebra of the Poincaré group. For this we may neglect the normal-ordering in P̂ and M̂ ρσ , because this only introduces a constant and it is evident that the Lie algebra relations are preserved under such a constant shift.

To begin, we note a peculiar fact: any derivative of the integrand of either generator integrates to zero, e.g.

Z Z d3 x ∂ ν : P̂ µ : = d3 x ∂ ν P̂ µ = 0.

(6.105) (The derivative of any operator is equal to the derivative of its normal-ordered form, since they differ by at most a constant). The reasoning for this is as i =+∞ follows. If ν = i, then the integral over xi yields a factor of [P̂ µ ]xxi =−∞ , which 4

Incidentally, this calculation shows that actually M̂ 0i does not need normal ordering after all, making the reasoning in Problem 6.5 moot.

Formulation of Quantum Field Theory

142

vanishes for fields that vanish at infinity. If instead ν = 0 then the derivative can be taken outside of the integral and we get ∂ 0 P̂ µ , which is zero because P̂ µ is the conserved quantity arising from the spacetime translational of R 3 ν invariance ρσ the Dirac Lagrangian. Identical considerations apply to d x ∂ M̂ .

This solution will make heavy use of the equal-time anticommutation relations Eq. (6.3.94), {ψ̂α (x), ψ̂β (y)} = {ψ̂α† (x), ψ̂β† (y)} = 0,

{ψ̂α (x), ψ̂β† (y)} = δαβ δ 3 (x − y), (6.106)

in conjunction with the identities [A, BC] = {A, B}C − B{A, C}, [AB, C] = A[B, C] + [A, C]B.

(6.107) (6.108)

The first of these gives, for example, Z µ [ψ̂α (x), P̂ µ ] = d3 y [ψ̂α (x), iψ̂β† (y)∂(y) ψ̂β (y)] Z

µ µ = i d3 y {ψ̂α (x), ψ̂β† (y)}∂(y) ψ̂β (y) − ψ̂β† (y)∂(y) {ψ̂α (x), ψ̂β (y)} Z

µ = i d3 y δαβ δ 3 (x − y)∂(y) ψ̂β (y) − 0 = i∂ µ ψ̂α (x).

(6.109)

Here we have explicitly shown the spinor indices, but it is easier to keep these implicit and write [ψ̂(x), P̂ µ ] = i∂ µ ψ̂(x). Keeping track of the spinor indices is, however, necessary in arriving at this equation. A similar calculation reveals Z σ d3 y [ψ̂α (x), ψ̂ † (y)y ρ ∂(y) ψ̂(y)] Z

σ σ = d3 y y ρ {ψ̂α (x), ψ̂β† (y)}∂(y) ψ̂β (y) − y ρ ψ̂β† (y)∂(y) {ψ̂α (x), ψ̂β (y)} Z

σ = d3 y y ρ δαβ δ 3 (x − y)∂(y) ψ̂β (y) − 0 = xρ ∂ σ ψ̂α (x),

(6.110)

and Z

d3 y[ψ̂α (x), ψ̂ † (y)Σρσ ψ̂(y)] Z

† ρσ = d3 y {ψ̂α (x), ψ̂β† (y)}Σρσ ψ̂ (y) − ψ̂ (y)Σ { ψ̂ (x), ψ̂ (y)} γ α γ βγ β βγ Z

= d3 y δαβ δ 3 (x − y)Σρσ βγ ψ̂γ (y) − 0 = Σρσ αγ ψ̂γ (x).

(6.111)

Problem 9

143

These results combine to give Z [ψ̂(x), M̂ ρσ ] = i d3 y[ψ̂(x), ψ̂ † (y)(y ρ ∂ σ − y σ ∂ ρ + Σρσ )ψ̂(y)] h i = i xρ ∂ σ ψ̂(x) − xσ ∂ ρ ψ̂(x) + Σρσ ψ̂(x) , (6.112) having made the spinor components implicit. We have now proven both commutation relations of Eq. (6.3.212). It proves useful to derive the commutators with the hermitian conjugate fields using the identity [A† , B † ] = −[A, B]† : [ψ̂ † (x), P̂ µ ] = −[ψ̂(x), P̂ µ ]† = −(i∂ µ ψ̂(x))† = i∂ µ ψ̂ † (x), †

[ψ̂ (x), M̂

ρσ

(6.113)

ρσ †

] = −[ψ̂(x), M̂ ] h i † = − i xρ ∂ σ ψ̂(x) − xσ ∂ ρ ψ̂(x) + Σρσ ψ̂(x) h i = i xρ ∂ σ ψ̂ † (x) − xσ ∂ ρ ψ̂ † (x) + ψ̂ † (x)(Σρσ )† ,

(6.114)

since P̂ µ and M̂ ρσ are hermitian. Note that the hermitian conjugation acts only on the spinor indices, not the Lorentz indices. Now we are in a position to calculate the commutators of the generators.

Z [P̂ µ , P̂ ν ] = d3 x [iψ̂ † (x)∂ µ ψ̂(x), P̂ ν ] Z

= i d3 x [ψ̂ † (x), P̂ ν ]∂ µ ψ̂(x) + ψ̂ † (x)∂ µ [ψ̂(x), P̂ ν ] Z

= i d3 x i∂ ν ψ̂ † (x)∂ µ ψ̂(x) + iψ̂ † (x)∂ µ ∂ ν ψ̂(x) Z

= i d3 x ∂ ν iψ̂ † (x)∂ µ ψ̂(x) = 0.

(6.115)

This last step follows from Eq. (6.105). We were able to pass P̂ ν through the derivative because all of its derivatives vanish, by the same reasoning. This proves the first set of Poincaré Lie algebra relations. Next, Z [P̂ µ , M̂ ρσ ] = d3 x [P̂ µ , iψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + Σρσ )ψ̂(x)] Z

= i d3 x [P̂ µ , ψ̂ † (x)](xρ ∂ σ − xσ ∂ ρ + Σρσ )ψ̂(x)

+ψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + Σρσ )[P̂ µ , ψ̂(x)] Z

= i d3 x −i∂ µ ψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + Σρσ )ψ̂(x)

−iψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + Σρσ )∂ µ ψ̂(x) Z

= i d3 x −i∂ µ ψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + Σρσ )ψ̂(x)

Formulation of Quantum Field Theory

144

h i −iψ̂ † (x)∂ µ (xρ ∂ σ − xσ ∂ ρ + Σρσ )ψ̂(x)

+iψ̂ † (x)(∂ µ xρ ∂ σ − ∂ µ xσ ∂ ρ )ψ̂(x) Z h

i = i d3 x −i∂ µ ψ̂ † (x)(xρ ∂ σ − xσ ∂ ρ + Σρσ )ψ̂(x)

+iψ̂ † (x)(g µρ ∂ σ − g µσ ∂ ρ )ψ̂(x) .

(6.116) Here we have used the fact that ∂ µ xρ = g µν ∂ν xρ = g µν δ ρν = g µρ . Now the first term in the last line is ∂ µ M̂ρσ , so it integrates to zero for the same reason as Eq. (6.105). This leaves us with Z [P̂ µ , M̂ ρσ ] = i d3 x iψ̂ † (x)(g µρ ∂ σ − g µσ ∂ ρ )ψ̂(x) Z Z = ig µρ d3 x iψ̂ † (x)∂ σ ψ̂(x) − ig µσ d3 x iψ̂ † (x)∂ ρ ψ̂(x)

= i g µρ P̂ σ − g µσ P̂ ρ , (6.117) the second set of the Lie algebra relations. Finally, we calculate the commutator Z µν ρσ [M̂ , M̂ ] = d3 x [iψ̂ † (x)(xµ ∂ ν − xν ∂ µ + Σµν )ψ̂(x), M̂ ρσ ] Z n = i d3 x ψ̂ † (xµ ∂ ν − xν ∂ µ + Σµν )[ψ̂, M̂ ρσ ] o +[ψ̂ † , M̂ ρσ ](xµ ∂ ν − xν ∂ µ + Σµν )ψ̂ Z

= i d3 x iψ̂ † (xµ ∂ ν − xν ∂ µ + Σµν )(xρ ∂ σ ψ̂ − xσ ∂ ρ ψ̂ + Σρσ ψ̂)

+i(xρ ∂ σ ψ̂ † − xσ ∂ ρ ψ̂ † + ψ̂ † Σρσ† )(xµ ∂ ν − xν ∂ µ + Σµν )ψ̂ Z n

= i d3 x iψ̂ † xµ g νρ ∂ σ ψ̂ + xρ ∂ ν ∂ σ ψ̂ − g νσ ∂ ρ ψ̂ − xσ ∂ ν ∂ ρ ψ̂ + Σρσ ∂ ν ψ̂

− xν g µρ ∂ σ ψ̂ + xρ ∂ µ ∂ σ ψ̂ − g µσ ∂ ρ ψ̂ − xσ ∂ µ ∂ ρ ψ̂ + Σρσ ∂ µ ψ̂

+ Σµν xρ ∂ σ ψ̂ − xσ ∂ ρ ψ̂ + Σρσ ψ̂ h i − ∂ σ xρ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ h i + ∂ ρ xσ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂

o +Σρσ† xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ Z +i

n h

i d3 x i ∂ σ ψ̂ † xρ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ h

io −∂ ρ ψ̂ † xσ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂

(6.118)

Problem 9

145

This last equality is simply an expansion of the previous line, having used the product rule along with the identity ∂ ν xρ = g νρ . Many terms here cancel after using Eq. (6.93) to deal with the Σρσ† factor, ultimately leaving Z

n [M̂ µν , M̂ ρσ ] = i d3 x iψ̂ † xµ g νρ ∂ σ ψ̂ − g νσ ∂ ρ ψ̂ − xν g µρ ∂ σ ψ̂ − g µσ ∂ ρ ψ̂

+ xσ g µρ ∂ ν ψ̂ − g νρ ∂ µ ψ̂ − xρ g µσ ∂ ν ψ̂ − g νσ ∂ µ ψ̂ +Σµν Σρσ ψ̂ − Σρσ Σµν ψ̂ Z +i

o

n h i d3 x i ∂ σ ψ̂ † xρ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂

i h − ∂ ρ ψ̂ † xσ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂

o +ψ̂ † (δ ρ0 δ σk − δ σ0 δ ρk )γ 0 γ k xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ .

(6.119)

For convenience, let us collect this second integral into a single symbol, µνρσ .

We will prove later that in fact µνρσ = 0. But first let us focus on the full expression for the commutator, which can be rearranged to give Z n

µν ρσ [M̂ , M̂ ] = i d3 x iψ̂ † g νρ xµ ∂ σ ψ̂ − xσ ∂ µ ψ̂ + Σµσ ψ̂ − Σµσ ψ̂

− g µρ xν ∂ σ ψ̂ − xσ ∂ ν ψ̂ + Σνσ ψ̂ − Σνσ ψ̂

− g νσ xµ ∂ ρ ψ̂ − xρ ∂ µ ψ̂ + Σµρ ψ̂ − Σµρ ψ̂

+ g µσ xν ∂ ρ ψ̂ − xρ ∂ ν ψ̂ + Σνρ ψ̂ − Σνρ ψ̂ o +Σµν Σρσ ψ̂ − Σρσ Σµν ψ̂ + µνρσ

= i g νρ M̂ µσ − g µρ M̂ νσ − g νσ M̂ µρ + g µσ M̂ νρ Z − d3 x ψ̂ † (g µρ Σνσ − g νρ Σµσ − g µσ Σνρ + g νσ Σµρ +Σµν Σρσ − Σρσ Σµν ) ψ̂ + µνρσ .

(6.120)

It is possible to show that the matrix in the integrand here vanishes. Indeed, consider the case where µ, ν, ρ, σ are all distinct. Then because distinct γmatrices anticommute, Σµν Σρσ =

1 1 µ ν ρ σ γ γ γ γ = (−1)4 γ ρ γ σ γ µ γ ν = Σρσ Σµν , 4 4

(6.121)

and so the last two terms cancel. All of the metric components will also be zero because the metric is diagonal, so the entire integrand vanishes. All other cases

Formulation of Quantum Field Theory

146

have at least two Lorentz indices equal. Because of the antisymmetry present, we may suppose without loss of generality that µ = ρ, and that µ ̸= ν and µ ̸= σ. The matrix in the integrand then reduces to 1 1 g µµ Σνσ + Σµν Σµσ − Σµσ Σµν = g µµ Σνσ + γ µ γ ν γ µ γ σ − γ µ γ σ γ µ γ ν 4 4 1 µ µ ν σ 1 µ µ σ ν µµ νσ =g Σ − γ γ γ γ + γ γ γ γ 4 4 1 µ 2 ν σ µµ νσ = g Σ − (γ ) [γ , γ ] 4 = g µµ Σνσ − (γ µ )2 Σνσ = 0,

(no sum)

(6.122)

as follows from (γ 0 )2 = −(γ i )2 = I = g 00 I = −g ii I. This proves that the integrand vanishes in all cases, leaving us with

[M̂ µν , M̂ ρσ ] = i g νρ M̂ µσ − g µρ M̂ νσ − g νσ M̂ µρ + g µσ M̂ νρ + µνρσ . (6.123) We will therefore have the desired result once we show µνρσ = 0. To remind, this symbol is defined as Z n h

i µνρσ ≡ i d3 x i ∂ σ ψ̂ † xρ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ h

i − ∂ ρ ψ̂ † xσ xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ o +ψ̂ † (δ ρ0 δ σk − δ σ0 δ ρk )γ 0 γ k (xµ ∂ ν − xν ∂ µ + Σµν ) ψ̂ .

(6.124) If both ρ and σ are spatial, then the first two terms integrate to zero given sufficient boundary conditions, and the last term simply vanishes. Hence, µνij = 0.

Because of antisymmetry, it is sufficient to now only consider the case where ρ, σ = 0, j. After discarding the boundary term, here we are left with Z n h i µν0j  ≡ i d3 x i −xj ∂ 0 ψ̂ † xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂

o +ψ̂ † γ 0 γ j xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ .

(6.125)

Now we must once again appeal to current conservation. As known from Eqs. (3.2.96), (3.2.97), and (6.3.129), M̂ ρσ is the conserved charge associated to the conserved current operator Jˆµρσ , Z ρσ M̂ = d3 x Jˆ0ρσ , (6.126) where Jˆµρσ ≡ xρ T̂ µσ − xσ T̂ µρ +

∂L ∂(∂µ ψ̂)

Σµν ψ̂

Problem 10

147

= iψ̄ˆγ µ xρ ∂ σ ψ̂ − xσ ∂ ρ ψ̂ + Σµν ψ̂ .

(6.127)

(This matches the formula we have been using, since ψ̄ˆγ 0 = ψ̂ † γ 0 γ 0 = ψ̂ † ). The first term in µν0j is therefore xj ∂ 0 Jˆ0µν = −xj ∂i Jˆiµν = Jˆjµν −∂i (xj Jˆiµν ) by the current conservation equation. The last total derivative integrates to zero given the fields vanish at infinity, so we have Z o

n µν0j  = i d3 x i −Jˆjµν + ψ̂ † γ 0 γ j xµ ∂ ν ψ̂ − xν ∂ µ ψ̂ + Σµν ψ̂ = 0.

(6.128) Together these results imply µνρσ = 0, and we get the third set of Lie algebra relations

[M̂ µν , M̂ ρσ ] = i g νρ M̂ µσ − g µρ M̂ νσ − g νσ M̂ µρ + g µσ M̂ νρ , (6.129) as desired.

6.10 Problem 10 Problem: In the covariant quantization approach: 1 (a) Prove Eq. (6.4.148); i.e., the addition of − 2ξ (∂µ Aµ )2 to the Maxwell langrangian gives the equations of motion □Aν − (1 − 1ξ )∂ ν (∂µ Aµ ) = 0. (b) Complete the details of the sketch proof and show how to obtain the covariant ˙ canonical commutation relations for µ and µ in Eq. (6.4.152).

Solution: (a) We are considering the modified Maxwell Lagrangian density 1 1 Lξ = − Fµν F µν − (∂µ Aµ )2 , 4 2ξ

(6.130)

where F µν = ∂ µ Aν − ∂ ν Aµ . Since a field and its derivatives are independent variables, we see that ∂Lξ /∂Aν = 0 and so the Euler-Lagrange equations reduce to

∂Lξ ∂Lξ ∂Lξ 0= − ∂µ = −∂µ .

(6.131) ∂Aν ∂(∂µ Aν ) ∂(∂µ Aν ) Now from Eq. (3.3.5) we know ∂ ∂(∂µ Aν )

1 αβ − Fαβ F = −F µν , 4

(6.132)

148

Formulation of Quantum Field Theory

so all that remains is calculating the derivative of the second term,

∂ 1 ∂ 1 2 − (∂ · A) = − 2 (∂ · A) (∂ · A) ∂(∂µ Aν ) 2ξ 2ξ ∂(∂µ Aν )

1 ∂ = − (∂ · A) g αβ ∂α Aβ ξ ∂(∂µ Aν ) 1 = − (∂ · A) g αβ δ µα δ νβ ξ 1 = − (∂ · A) g µν . ξ Hence, the equations of motion are

∂Lξ 1 µν µν 0 = −∂µ = −∂µ −F − (∂ · A) g ∂(∂µ Aν ) ξ 1 = ∂µ (∂ µ Aν − ∂ ν Aµ ) + g µν ∂µ (∂ · A) ξ 1 = □Aν − ∂ ν ∂µ Aµ + ∂ ν (∂ · A) ξ

1 = □Aν − 1 − ∂ ν (∂µ Aµ ), ξ

(6.133)

(6.134)

having made use of the fact that partial derivatives commute. (b) The sketch proof in the text below Eq. (6.4.152) tells us to consider the commutation relations between the spatial and temporal components of µ and π̂ ν separately to arrive at Eq. (6.4.152), where the canonical momenta are given by 1 ˙ 1 π̂ 0 = − ∂ρ Âρ = − (Â0 + ∂j Âj ), ξ ξ

˙ π̂ i = −Âi + ∂ i Â0 .

(6.135)

This requires knowing the equal-time commutators of Eq. (6.4.151), [µ (x), Âν (y)]|x0 =y0 = [π̂ µ (x), π̂ ν (y)]|x0 =y0 = 0, [µ (x), π̂ ν (y)]|x0 =y0 = ig µν δ 3 (x − y).

(6.136)

Taking spatial derivatives of the first line in Equation (6.136) leads to the useful relations [∂i µ (x), Âν (y)]|x0 =y0 = [∂i µ (x), ∂j Âν (y)]|x0 =y0 = 0, [∂i π̂ µ (x), π̂ ν (y)]|x0 =y0 = [∂i π̂ µ (x), ∂j π̂ ν (y)]|x0 =y0 = 0.

(6.137)

In what follows, all commutators should be understood to be between operators at equal times. (i) With Equations (6.136) and (6.137) we can calculate :0

1 ˙ 1 j  (x), Â0 (y)] [π̂ 0 (x), Â0 (y)] = −ig 00 δ 3 (x−y) = − [Â0 (x), Â0 (y)]− [∂j ξ ξ

Problem 10

149

˙ =⇒ [Â0 (x), Â0 (y)] = ig 00 [1 + (ξ − 1)g00 ] δ 3 (x − y),

(6.138)

having made use of the fact that g00 = g 00 = 1. A similar calculation yields :0

˙ 0 [π̂ i (x), Â0 (y)] = 0 = −[Âi (x), Â0 (y)] + [∂i  (x), Â0 (y)]

˙ =⇒ [Âi (x), Â0 (y)] = 0 = ig i0 [1 + (ξ − 1)gi0 ] δ 3 (x − y), (6.139) since g i0 = gi0 = 0. Together, Equations (6.138) and (6.139) can be ˙ written as [µ (x), Â0 (y)] = ig µ0 [1 + (ξ − 1)gµ0 ] δ 3 (x − y). Continuing in the same manner, :0

1 1 ˙ i j [π̂ 0 (x), Âj (y)] = 0 = − [Â0 (x), Âj (y)] − [∂i  (x),  (y)] ξ ξ ˙ =⇒ [Â0 (x), Âj (y)] = 0 = ig 0j [1 + (ξ − 1)g00 ] δ 3 (x − y), (6.140) and finally, :0

˙ i 0 [π̂ i (x), Âj (y)] = −ig ij δ 3 (x − y) = −[Âi (x), Âj (y)] + [∂ Â (x), Âj (y)]

˙i j ij 3 =⇒ [ (x),  (y)] = ig [1 + (ξ − 1)gi0 ] δ (x − y), (6.141) again since gi0 = 0. Together, Equations (6.140) and (6.141) can be writ˙ ten as [µ (x), Âj (y)] = ig µj [1 + (ξ − 1)gµ0 ] δ 3 (x − y), which combined with the previous result yields ˙ [µ (x), Âν (y)]|x0 =y0 = ig µν [1 + (ξ − 1)gµ0 ] δ 3 (x − y),

(6.142)

exactly the second line of Eq. (6.4.152). ˙ (ii) From Equation (6.142) we see that [µ (x), Âν (y)] = 0 if µ ̸= ν, and ˙ therefore by taking a spatial derivative [µ (x), ∂ j Âν (y)] = 0 unless µ = ν.

So we can calculate 1 ˙0 [Â (x) + ∂i Âi (x), Â0 (y) + ∂j Âj (y)] ξ2 1 ˙ ˙ = 2 [Â0 (x), Â0 (y)], (6.143) ξ

[π̂ 0 (x), π̂ 0 (y)] = 0 =

since all other commutators in the sum vanish by the above. Similarly, ˙ ˙ [π̂ i (x), π̂ j (y)] = 0 = [−Âi (x) + ∂ i Â0 (x), −Âj (y) + ∂ j Â0 (y)] ˙ ˙ = [Âi (x), Âj (y)].

(6.144) Together these relations complete the first line of Eq. (6.4.152).

Formulation of Quantum Field Theory

150

(iii) Finally, let us calculate 1 ˙ ˙ [π̂ 0 (x), π̂j (y)] = 0 = − [Â0 (x) + ∂i Âi (x), −Âj (y) + ∂j Â0 (y)] ξ 1 ˙0 ˙ ˙ [Â (x), Âj (y)] − [Â0 (x), ∂j Â0 (y)] =+ ξ

=⇒

: 0

˙ i 0 + [∂i Âi (x), Âj (y)] − [∂i  (x), ∂j  (y)]

˙0 ˙ ˙0 ˙ 0 i [ (x), Âj (y)] = [ (x), ∂j  (y)] − [∂i  (x), Âj (y)].

(6.145)

Now by lowering an index in Equation (6.142), we see that ˙ ˙ [Âi (x), Âj (y)] = gjν [Âi (x), Âν (y)] = −igjν g iν [1 + (ξ − 1)gj0 ] δ 3 (x − y) = −iδ ij δ 3 (x − y), (6.146) and therefore ∂ ˙ [∂i Âi (x), Âj (y)] = −iδ ij ∂i δ 3 (x − y) = −i j δ 3 (x − y). ∂x

(6.147)

In the same way, ˙ [Â0 (x), Â0 (y)] = ig 00 [1 + (ξ − 1)g00 ] δ 3 (x − y) ∂ ∂ ˙ =⇒ [Â0 (x), ∂j Â0 (y)] = iξ j δ 3 (x − y) = −iξ j δ 3 (x − y). (6.148) ∂y ∂x Substituting these results into Equation (6.145), we get ∂ ∂ ˙ ˙ [Â0 (x), Âj (y)]|x0 =y0 = −iξ j δ 3 (x − y) + i j δ 3 (x − y) ∂x ∂x ∂ = i(1 − ξ) j δ 3 (x − y), (6.149) ∂x completing the last line of Eq. (6.4.152).

6.11 Problem 11 Problem: Generalize the arguments for the neutral vector field and obtain some of the key results for the charged vector field case in analogy with the extension of the neutral scalar field to the charged scalar field. (Hint: A line-by-line derivation is not necessary, but explain the rationale for the summary of results.)

Solution: Let µ (x) denote the charged vector field and µ† (x) be the hermitian

Problem 12

151

conjugate field. In analogy with Eq. (6.2.177) and Eq. (6.5.45) we write Z 3 i 1 P3 h µ d k ˆλ −ik·x + ϵµ (k, λ)∗ ĝ λ† eik·x . √ µ (x) = λ=1 ϵ (k, λ)fk e k 3 (2π) 2ωk

(6.150)

where in place of Eq. (6.2.180) and Eq. (6.5.54) we have all commutators of annihilation and creation operators vanishing except for ′

′

[fˆkλ , fˆkλ′ † ] = [ĝkλ , ĝkλ′ † ] = (2π)3 δλλ′ δ 3 (k − k′ ) .

(6.151)

The classical Lagrangian with sources is ∗ F µν + m2 A∗µ Aµ − jµ∗ Aµ − j µ A∗µ .

L = − 12 Fµν

(6.152)

where F µν = ∂ µ Aν − ∂ν Aµ in the usual way. Note the factor of two difference with respect to the neutral vector boson case, which is similar to the case of the charged scalar field. In place of Eq. (6.2.184) we have R R d3 p P3 P3 ˆλ† ˆλ (6.153) N̂f ≡ d3 p λ=1 N̂fλp ≡ (2π) 3 λ=1 fp fp , R R 3 P3 P 3 3 d p λ λ† λ ≡ (2π) N̂g ≡ d p λ=1 N̂gp 3 λ=1 ĝp ĝp .

It is not difficult to see that we will have Z Z P3 P3 λ λ Ĥ = d3 k λ=1 Ek (N̂fλk + N̂gk ) , P̂ = d3 k λ=1 k(N̂fλk + N̂gk ), Z P3 λ Q̂ = N̂f − N̂g = d3 k λ=1 (N̂fλk − N̂gk ).

(6.154) (6.155)

Similalry, in place of Eq. (6.2.198) and Eq. (6.5.56) we will have for the noncovariant canonical quantization procedure that ⟨0|T µ (x)Âν† (y)|0⟩NC = ∆µν V (x − y) µ

ν

µ†

(6.156) ν†

⟨0|T Â (x)Â (y)|0⟩NC = ⟨0|T Â (x)Â (y)|0⟩NC = 0 .

(6.157)

In the covariant canonical quantization procedure the above vev’s have no ‘NC’ label µν and we arrive at the Proca Feynman propagator ∆µν F (x − y) replacing ∆V (x − y) in the above equations.

6.12 Problem 12 Problem: For a massive vector boson field: (a) Prove Eq. (6.5.51), which is the result that Â0 (x) = −∇ · Ê(x)/m2 for a free massive vector boson field. (b) Prove Eq. (6.5.54), which are the commutation relations for the annihilation and creation operators.

Formulation of Quantum Field Theory

152

Solution: (a) We are given the electric field operator in Eq. (6.5.48), r Z 3 i d3 k ωk X h λ −ik·x ∗ λ† ik·x i ϵ̃(k, λ)â , e − ϵ̃(k, λ) â e Ê(x) = k k (2π)3 2

(6.158)

λ=1

where ϵ̃(k, λ) ≡ ϵ(k, λ) − k

k · ϵ(k, λ) . ωk2

(6.159)

The divergence of the electric field operator is then calculated to be r Z 3 i d3 k ωk X h ik·x i −ik · ϵ̃(k, λ)âλk e−ik·x − ik · ϵ̃(k, λ)∗ âλ† . ∇ · Ê(x) = k e 3 (2π) 2 λ=1

(6.160) Here we unravel definitions to see that 2

2 ωk − k2 k · ϵ(k, λ) m k · ϵ̃(k, λ) = = ϵ0 (k, λ), ωk ωk ωk

(6.161)

having noted that k ·ϵ(k, λ) = 0 implies ϵ0 (k, λ) = k·ϵ(k, λ)/ωk . Our divergence therefore simplifies to r Z 3 i ωk X m2 h 0 d3 k λ −ik·x 0 ∗ λ† ik·x ∇ · Ê(x) = ϵ (k, λ)â e + ϵ (k, λ) â e , k k (2π)3 2 ωk λ=1

(6.162) giving ∇ · Ê(x) = m2

Z

3 i d3 k 1 Xh 0 λ −ik·x 0 ∗ λ† ik·x √ ϵ (k, λ)â e + ϵ (k, λ) â e k k (2π)3 2ωk λ=1

0

= −Â (x),

(6.163)

by Eq. (6.5.45), as was to be shown. (b) As stated in the text, our expressions for the fields  and Ê can be inverted to give expressions for the creation and annihilation operators, Z i e−ik·x h λ ωk ϵ̃(k, λ)∗ · Â(x) − iϵ(k, λ)∗ · Ê(x) , âk = d3 x √ 2ωk Z i ik·x h e âλ† d3 x √ ωk ϵ̃(k, λ) · Â(x) + iϵ(k, λ) · Ê(x) , (6.164) k = 2ωk from Eq. (6.5.53). Calculating the creation and annihilation operator commutator relations therefore requires knowing the equal-time commutators of the  and Ê fields, given in Eq. (6.5.41) as [Ê i (x), Ê j (y)]|x0 =y0 = [Âi (x), Âj (y)]|x0 =y0 = 0,

Problem 12

153

[Âi (x), Ê j (y)]|x0 =y0 = −iδ ij δ 3 (x − y).

(6.165)

By direct substitution we then have Z Z −ik·x −ik′ ·x′ h e λ λ′ 3 3 ′ e √ âk , âk′ ] = d x d x √ ωk ϵ̃i (k, λ)∗ Âi (x) − iϵi (k, λ)∗ Ê i (x), 2ωk 2ωk′ i ωk′ ϵ̃j (k ′ , λ′ )∗ Âj (x′ ) − iϵj (k ′ , λ′ )∗ Ê j (x′ ) =−

i 2

Z

d3 x

Z

′

d3 x′

′

e−i(k·x+k ·x ) h ωk ϵ̃i (k, λ)∗ ϵj (k ′ , λ′ )∗ [Âi (x), Ê j (x′ )] √ ωk ωk′ i +ωk′ ϵi (k, λ)∗ ϵ̃j (k ′ , λ′ )∗ [Ê i (x), Âj (x′ )]

i =− 2

Z

3

d x

Z

3 ′ e

d x

−i(k·x+k′ ·x′ )

√

ωk ωk′

ωk ϵ̃i (k, λ)∗ ϵj (k ′ , λ′ )∗ [−iδ ij δ 3 (x − x′ )]

+ωk′ ϵi (k, λ)∗ ϵ̃j (k ′ , λ′ )∗ iδ ij δ 3 (x − x′ )

1 = 2 =

′

e−i(k+k )·x d x √ [−ωk ϵ̃(k, λ)∗ · ϵ(k ′ , λ′ )∗ + ωk′ ϵ(k, λ)∗ · ϵ̃(k ′ , λ′ )∗ ] ωk ωk′

Z

3

1 (2π)3 δ 3 (k + k′ ) [−ωk ϵ̃(k, λ)∗ · ϵ(k ′ , λ′ )∗ + ωk′ ϵ(k, λ)∗ · ϵ̃(k ′ , λ′ )∗ ] . √ 2 ωk ωk′ (6.166)

The delta function here allows us to set k′ = −k in all other terms. In this case we have by definition ϵ(k ′ , 1) = (1, 0, 0)T = ϵ(k, 1), ϵ(k ′ , 2) = (0, 1, 0)T = ϵ(k, 2), ϵ(k ′ , 3) = (0, 0, ω−k /m)T = (0, 0, ωk /m)T = ϵ(k, 3),

(6.167)

or more compactly, ϵ(k ′ , λ′ ) = ϵ(k, λ′ ) for λ′ = 1, 2, 3. It then follows from Equation (6.159) that also ϵ̃(k ′ , λ′ ) = ϵ̃(k, λ′ ) for λ′ = 1, 2, 3. So going back to our commutator expression, we get for λ, λ′ = 1, 2, 3 ′

(2π)3 δ 3 (k + k′ ) ∗ [ωk′ ϵ(k, λ) · ϵ̃(k ′ , λ′ ) − ωk ϵ̃(k, λ) · ϵ(k ′ , λ′ )]|k′ =−k √ 2 ωk ωk′ 1 ∗ = (2π)3 δ 3 (k + k′ ) [ϵ(k, λ) · ϵ̃(k, λ′ ) − ϵ̃(k, λ) · ϵ(k, λ′ )] 2

1 k · ϵ(k, λ′ ) = (2π)3 δ 3 (k + k′ ) ϵ(k, λ) · ϵ(k, λ′ ) − k 2 ωk2

∗ k · ϵ(k, λ) ′ − ϵ(k, λ) − k · ϵ(k, λ ) ωk2

[âλk , âλk′ ] =

= 0,

(6.168)

as was to be shown. Taking the hermitian conjugate of this equation immediλ′ † ately leads to the second desired result [âλ† k , âk′ ] = 0. The final commutator

Formulation of Quantum Field Theory

154

is calculated in a similar way: Z Z −ik·x ik′ ·x′ h e λ′ † λ 3 3 ′ e √ [âk , âk′ ] = d x d x √ ωk ϵ̃i (k, λ)∗ Âi (x) − iϵi (k, λ)∗ Ê i (x), 2ωk 2ωk′ i ωk′ ϵ̃j (k ′ , λ′ )Âj (x′ ) + iϵj (k ′ , λ′ )Ê j (x′ ) =

i 2

Z

d3 x

Z

′

d3 x′

′

e−i(k·x−k ·x ) h ωk ϵ̃i (k, λ)∗ ϵj (k ′ , λ′ )[Âi (x), Ê j (x′ )] √ ωk ωk′ i −ωk′ ϵi (k, λ)∗ ϵ̃j (k ′ , λ′ )[Ê i (x), Âj (x′ )]

=

i 2

Z

d3 x

Z

′

d3 x′

′

e−i(k·x−k ·x ) ωk ϵ̃i (k, λ)∗ ϵj (k ′ , λ′ )[−iδ ij δ 3 (x − x′ )] √ ωk ωk′

−ωk′ ϵi (k, λ)∗ ϵ̃j (k ′ , λ′ ) iδ ij δ 3 (x − x′ ) ′

e−i(k−k )·x d3 x √ [ωk ϵ̃(k, λ)∗ · ϵ(k ′ , λ′ ) + ωk′ ϵ(k, λ)∗ · ϵ̃(k ′ , λ′ )] ωk ωk′

=

1 2

=

1 (2π)3 δ 3 (k − k′ ) [ωk ϵ̃(k, λ)∗ · ϵ(k ′ , λ′ ) + ωk′ ϵ(k, λ)∗ · ϵ̃(k ′ , λ′ )] . √ 2 ωk ωk′

Z

(6.169)

The delta function now allows us to set k′ = k in the other terms, and therefore more generally k ′ = k. It follows from Eq. (6.5.52) that in this case ϵ̃(k ′ , λ′ )∗ · ϵ(k, λ) = ϵ̃(k, λ′ )∗ · ϵ(k, λ) = δλλ′ ,

(6.170)

1 (2π)3 δ 3 (k − k′ ) [δλλ′ + δλλ′ ] = (2π)3 δλλ′ δ 3 (k − k′ ), 2

(6.171)

so we get ′

[âλk , âλk′ † ] =

completing the result.

6.13 Problem 13 Problem: (a) Define {V1 , V2 , . . . , V16 } ≡ 12 {I, γ 0 , iγ j , iσ 0j , σ jk |j<k , iγ 5 γ 0 , γ 5 γ j , γ 5 }. Show that these sixteen matrices are hermitian, VJ = VJ† ; (b) Using the γ-matrix anticommutation relations show that: (i) VJ2 = 14 I; (ii) VJ VK = ±VK VJ ; and (iii) VK VJ VK = − 14 VJ for every J ̸= 1 for at least one K (K = 16 for J = 2, . . . , 5 and 12, . . . , 15; K is any of 2, . . . , 5 for J = 16; and K is at least one of 2, . . . , 5 for J = 6, . . . , 11). (c) Show for every VJ and VK that VJ VK = 21 cJKL VL where cJKL = ±1, ±i for only one VL and where no sum over L is implied. Note that for fixed J each

155

Problem 13

K leads to a different L and for fixed K each J leads to a different L. This is referred to as the Rearrangement Lemma. (d) By considering tr(VK VJ VK ) explain why trVJ = 0 for every J ̸= 1; (e) Explain why trace identities lead to ‘orthonormality’, tr(VJ† VK ) = tr(VJ VK ) = δJK ; (f) Show that cJKL = 2tr(VJ VK VL ); (g) Define the 16 vectors (XJ )r ≡ (VJ )ij with r ≡ j + 4(i − 1) = 1, . . . , 16. Hence P show that tr(VJ† VK ) = r (XJ )∗r (XK )r = XJ† XK = δJK so that the XJ are 16 orthonormal sixteen component complex vectors and hence are an orthonormal basis of C16 . Show that the completeness relation for this orthonormal basis P P P (XJ )r (XJ∗ )s = δrs can be written as J (VJ )ij (VJ∗ )kℓ = J (VJ )ij (VJ† )ℓk = J P J (VJ )ij (VJ )ℓk = δik δjℓ . (h) Any 4 × 4 complex matrix can be regarded as a vector in C16 . Completeness of the XJ then means completeness of the matrices VJ and so for any matrix A P we can write A ≡ J cJ VJ . Show that cJ = tr(VK A). (i) Explain why a matrix that commutes with all four γ-matrices must commute with all matrices and so must be a multiple of the identity. This is sometimes referred to Schur’s Lemma.

P P (j) Show (VJ )ij (VK )kℓ = m,n (VJ )im (VK )kn δmj δnℓ = L (VJ VL )iℓ (VK VL )kj using completeness. Then show using the Rearrangement Lemma that (VJ )ij (VK )kℓ = P (1/4) L cJLM cKLN (VM )iℓ (VN )kj . Explain why we can write for some coefficients CJKM N ∈ C, P (VJ )ij (VK )kℓ ≡ M,N CJKM N (VM )iℓ (VN )kj , (6.172) which is referred to as the general form of the Fierz identity. Explain why CJKM N = 0 if J = K unless M = N and vice versa. (k) Using this result and orthonormality show that CJKM N = tr(VM VJ VN VK ). (l) Rewrite the sixteen basis matrices as 2VJ → ΓaA , where a labels the appropriate combination of spacetime indices and where A = S, V, T, A, P denote scalar, vector, tensor, axial vector, and pseudoscalar respectively. Specifically we define Γ1S ≡ I, Γ1V , . . . , Γ4V ≡ γ µ , Γ1T , . . . , Γ6T ≡ σ µν |µ<ν , Γ1A , . . . , Γ4A ≡ iγ 5 γ ν , and Γ1P ≡ γ 5 . Use the Einstein summation convention for the collective space-time index a. Summations over the index A will always be explicitly written. We define ΓAa ≡ gab ΓbA , where gab represents the appropriate number of gµν for the particular A. For example, we have ΓS1 ≡ I, ΓV 1 , . . . , ΓV 4 ≡ γµ , ΓT 1 , . . . , ΓT 6 ≡ σµν |µ<ν , ΓA1 , . . . , ΓA4 ≡ iγ 5 γµ and ΓP 1 ≡ γ 5 . With this notation show that the above results can be rewritten as: P tr(ΓaA ΓBb ) = 4δAB δba ; A ≡ A cAa ΓaA for some cAa ∈ C; P P a a δ ; A (ΓA )ij (ΓAa )kℓ = A (ΓAa )ij (ΓA )kℓ = 4δ iℓ kj 1 a b c a b abc abc tr Γ Γ Γ ; ΓA ΓB = ρABC Γ with ρ = Cc A B C ABC 4

P abcd (ΓaA )ij ΓbB kl = C,D CABCD (ΓCc )iℓ (ΓDd )kj (Fierz transformation); 1 abcd abcd CABCD = 16 tr(ΓcC ΓaA ΓdD ΓbB ), where CABCD vanishes if (a, A) = (b, B) unless (c, C) = (d, D) and vice versa.

Formulation of Quantum Field Theory

156

(m) The standard Fierz transformation has (a, A) = (b, B) and hence (c, C) = (d, D) and is a special case. Recall that we sum over repeated spacetime indices. Show P 1 that (ΓaA )ij (ΓAa )kl = C CAC (ΓcC )iℓ (ΓCc )kj with CAB = 16N tr ΓBb ΓaA ΓbB ΓAa B and NB ≡ 1, 4, 6, 4, 1 for B = S, V, T, A, P respectively. (n) The contraction identities can be written as ΓaA ΓbB ΓAa ≡ fAB ΓbB , which defines the coefficients fAB . Show that CAB = 14 fAB , which can be written in matrix form as   1 1 1 1 1  4 −2 0 2 −4   1 1  (6.173) C= f=  6 0 −2 0 6  , 4 4 4 2 0 −2 −4  1 −1 1 −1 1 where the row (A) and column (B) indices occur in the order S, V, T, A, P .

Explicitly verify at least six of these matrix elements. (o) Adopt the notation of Itzykson and Zuber (1980) to write the standard Fierz identities in a more familiar form. Consider four arbitrary spinors ψ1 , . . . , ψ4 and five Lorentz scalar quadrilinear spinor combinations, s(1, 2; 3, 4) ≡ (ψ̄1 ΓS ψ2 )(ψ̄3 ΓS ψ4 ) = (ψ̄1 ψ2 )(ψ̄3 ψ4 )

v(1, 2; 3, 4) ≡ (ψ̄1 ΓaV ψ2 )(ψ̄3 ΓV a ψ4 ) = (ψ̄1 γ µ ψ2 )(ψ̄3 γµ ψ4 )

t(1, 2; 3, 4) ≡ (ψ̄1 ΓaT ψ2 )(ψ̄3 ΓT a ψ4 ) = (ψ̄1 σ µν |µ<ν ψ2 )(ψ̄3 σµν |µ<ν ψ4 ) = 12 (ψ̄1 σ µν ψ2 )(ψ̄3 σµν ψ4 )

a(1, 2; 3, 4) ≡ (ψ̄1 ΓaA ψ2 )(ψ̄3 ΓAa ψ4 ) = (ψ̄1 (iγ 5 γ µ )ψ2 )(ψ̄3 (iγ 5 γµ )ψ4 ) = (ψ̄1 γ 5 γ µ ψ2 )(ψ̄3 γµ γ 5 ψ4 )

p(1, 2; 3, 4) ≡ (ψ̄1 ΓP ψ2 )(ψ̄3 ΓP ψ4 ) = (ψ̄1 γ 5 ψ2 )(ψ̄3 γ 5 ψ4 ) .

(6.174)

Define Ψ to be the column vector with elements ΨA being s, v, t, a, p for A = S, V, T, A, P respectively, i.e., Ψ(1, 2; 3, 4) ≡ [s, v, t, a, p] (1, 2; 3, 4). Show that the standard Fierz identity can be compactly summarized as Ψ(1, 4; 3, 2) = CΨ(1, 2; 3, 4)

(6.175)

for arbitrary spinors ψ1 , ψ2 , ψ3 , ψ4 . So the action of the matrix C is to exchange the second and fourth spinors (or equivalently the first and third). We have Ψ(1, 2; 3, 4) = CΨ(1, 4; 3; 2) = C 2 Ψ(1; 2; 3, 4) and so C 2 = I. Explicitly verify that this is true of the matrix C. Note that if instead of spinors we have anticommuting fermion operators then the required exchange of two fermion operators in the re-ordering gives an overall minus sign, i.e., Ψ̂(1, 2; 3, 4) = −C Ψ̂(1, 4; 3, 2), where the caret on the Ψ denotes that it is an operator quadilinear made up of four fermion operators.

Solution:

Problem 13

157

(a) The factors of i are chosen to ensure that the VI for I = 1, 2, . . . , 16 are hermitian in the Dirac representation and therefore in any representation unitarily equivalent to it. It is straightforward to verify on a case by case basis. There can only be 16 linearly independent hermitian 4 × 4 matrices, (4 × 4 complex matrix has contains 32 real numbers and being hermitian imposes 16 conditions leaving only 16 independent real numbers to specify. So any hermitian 4 × 4 matrix can be written as some real linear combination of these sixteen matrices (b) We can rewrite these VI as 2V1 ≡ I

2V2 , . . . , 2V5 ≡ γ 0 ,

2V6 , . . . , 2V11 ≡ γ 0 γ 1 ,

iγ 1 ,

2V12 , . . . , 2V15 ≡ γ 1 γ 2 γ 3 ,

iγ 2 ,

γ0γ2,

iγ 3 γ0γ3,

iγ 0 γ 2 γ 3 ,

2V16 ≡ iγ 0 γ 1 γ 2 γ 3 .

i(γ 1 γ 2 ),

i(γ 1 γ 3 ),

−iγ 0 γ 1 γ 3 ,

iγ 0 γ 1 γ 2

i(γ 2 γ 3 ) (6.176)

Using the fact that distinct γ-matrices anticommute and that (γ 0 )2 = −(γ i )2 = I we observe that VI VJ = ±VJ VI and also that (VI )2 ∝ I/4 up to a factor of ±i or ±1. Since the VI were chosen to be hermitian it follows that the constant of proportionality for (VI )2 = VI VI† is +1, since det(VI VI† ) ≥ 0. Hence we have the results (VI )2 = +I/4

and VI VJ = ±VJ VI .

(6.177)

We can readily verify that for every VJ with J ̸= 1 there exists at least one VI such that VI VJ VI = −VJ /4 ,

(6.178)

where: I = 16 for J = 2, . . . , 5 and 12, . . . , 15; I is any of 2, . . . , 5 for J = 16; and I = 2, . . . , 5 for J = 6, . . . , 11 where in this case the γ-matrix in the VI must correspond to at least one of the γ-matrices in the VJ as defined in Eq. (6.176) above. Note that this result relies only on the anticommutation relations of the γ-matrices and is therefore true in any representation of them. (c) Since Eq. (6.176) contains every possible product of the γ-matrices and since products of products of the γ-matrices can always be reduced using Eq. (A.3.3) to one of these up to a phase factor of ±1 or ±i then we have the result that for every VJ and VK VJ VK = cJKL VL /2

where

cJKL = ±1, ±i

(6.179)

for some VL and where no sum over L is implied. For a given VJ each choice of VK leads to a different VL , since if it did not we would have VJ VK ∝ VJ VM and acting from the left with VJ would give VK ∝ VM which is not possible if K ̸= M . Similarly we can argue that for a given VK each choice of VJ leads to a different VL as we see by acting on an assumed VJ VK ∝ VM VK from the right with VK . In particular, if K = J then from Eq. (6.177) we have that L = 1 and so if J ̸= K then L ̸= 1.

158

Formulation of Quantum Field Theory

(d) Taking the trace of both sides of Eq. (6.178) and using Eq. (6.177) it follows that tr(VI VJ VI ) = tr(VJ VI2 ) = tr(VJ )/4 = −tr(VJ )/4 for J ̸= 1 and so tr(VJ ) = 0

for

J ̸= 1 .

(6.180)

(e) It also follows from Eq. (6.179) that for J ̸= K we have L ̸= 1 and tr(VJ VK ) ∝ tr(VL ) = 0 and since tr(VJ2 ) = tr(I)/4 = 1 then tr(VI† VJ ) = tr(VI VJ ) = δIJ

(6.181)

and so the sixteen VI form an hermitian orthonormal basis in any representation unitarily equivalent to the Dirac representation. Then any representation, γ̄ µ , of the γ-matrices that is similar to the Dirac representation will lead to an orthonormal-like basis of V̄I matrices. (f) tr(VJ VK VL ) = 21 cJKL′ tr(VL′ VL ) = 12 cJKL′ δL′ L = 12 cJKL . (g) Firstly we observe that P4 P4 δJK = tr(VJ† VK ) = i,j=1 (VJ† )ij (VK )ji = i,j=1 (VJ )∗ji (VK )ji P16 = r=1 (XJ )∗r (XK )r = XJ† XK .

(6.182) Since there are sixteen orthonormal sixteen component vectors XJ , then they P16 span C16 and so satisfy completeness, J=1 (XJ )r (XJ )∗s = δrs . It then follows that with r = j + 4(i − 1) and s = ℓ + 4(k − 1) we have P P P δrs = δik δjℓ = J (XJ )r (XJ )∗s = J (VJ )ij (VJ )∗kℓ = J (VJ )ij (VJ† )ℓk P = J (VJ )ij (VJ )ℓk .

(6.183) (h) Any 4 × 4 complex matrix A can be written as the vector XA where (XA )r = P Aij and we can always expand any vector XA as XA = J cJ XJ , where P P ∗ cJ = (X ) .

This can be rewritten as A = (X ) c VJ and cJ = A r J J r r J P † † (V ) A = tr(V A) = tr(V A) since the V are hermitian.

J J J r J ji ij (i) If the matrix X commutes with all γ-matrices then it will commute with all products of γ-matrices and hence with all sixteen of the VJ . But since any complex 4 × 4 matrix A can be written as a linear superposition of the VJ , then [X, A] for every matrix A. This means that we must have X = cI with c ∈ C. (j) We see that P P (VJ )ij (VK )kℓ = m,n (VJ )im (VK )kn δmj δnℓ = L,m,n (VJ )im (VK )kn (VL )mℓ (VL )nj P P = L (VJ VL )iℓ (VK VL )kj = 41 L cJLM cKLN (VM )iℓ (VN )kj (6.184) using completeness and the Rearrangement Lemma. Recall that for fixed J each L leads to a different M in cJLM and similarly for fixed K each L leads to a different N in cKLN . For fixed J and K and fixed M and N it follows that there is at most one L for which cJLM cKLN is nonvanishing. For nonvanishing combinations of J, K, M, N define CJKM N ≡ (1/4)cJLM cKLN and for all other

Problem 13

159

combinations define CJKM N ≡ 0. Note that if J = K then M = N . We have shown the general Fierz identity, P (VJ )ij (VK )kℓ ≡ M,N CJKM N (VM )iℓ (VN )kj .

(6.185) (k) The proof follows if we exercise some care with indices, P tr(VM VJ VN VK ) = i,j,k,ℓ (VM )ℓi (VJ )ij (VN )jk (VK )kℓ (6.186) P P = i,j,k,ℓ (VM )ℓi (VN )jk M ′ ,N ′ CJKM ′ N ′ (VM ′ )iℓ (VN ′ )kj P = M ′ ,N ′ CJKM ′ N ′ tr(VM VM ′ )tr(VN VN ′ ) P = M ′ ,N ′ CJKM ′ N ′ δM M ′ δN N ′ = CJKM N . (l) It is easily verified from the results for the VJ that tr(ΓaA ΓBb ) = 4δAB δba .

(6.187)

Since we know that the ΓaA must form a basis, then for an arbitrary matrix P A we have A = A cAa ΓaA with the orthonormal-like relation in Eq. (6.187) giving the result (1/4)tr(ΓAa A) = cAa . Thus for any 4 × 4 matrix A we have X X 1 (6.188) A= cAa ΓaA = ΓaA tr(ΓAa A) .

4 A

A

In terms of matrix elements this has the form 1X a Aij = (ΓA )ij (ΓAa )kℓ Aℓk 4

(6.189)

A

and since the matrix elements of A are arbitrary then it follows that we must we have the closure-like relation X X (ΓaA )ij (ΓAa )kℓ = (ΓAa )ij (ΓaA )kℓ = 4δiℓ δkj .

(6.190) A

A

Since with this notation we essentially have J → a, A then in terms of this basis we can write the Rearrangement Lemma in the form ΓaA ΓbB = ρabc ABC ΓCc ,

(6.191)

where for any given pair of combined indices Aa and Bb there is only one combined index Cc for which ρ does not vanish. Acting from the right with the matrix ΓcC with no sum over c implied and taking the trace gives

1 a b c ρabc (6.192) ABC = 4 tr ΓA ΓB ΓC , where we have made use of Eq. (6.187) to arrive at this result. The cyclic properties of the trace imply that ρabc ABC is symmetric under cyclic permutations of the combined indices Aa, Bb and Cc. Similarly it must be the case that the Fierz transformation of Eq. (6.184) can be written as

P abcd (ΓaA )ij ΓbB kl = C,D CABCD (ΓCc )iℓ (ΓDd )kj (6.193)

160

Formulation of Quantum Field Theory abcd for some coefficients CABCD . Using Eq. (6.187) and following similar steps that gave the result for CJKM N it easily follows that abcd CABCD =

1 tr(ΓcC ΓaA ΓdD ΓbB ) , 16

(6.194)

abcd where CABCD vanishes if a, A = b, B unless c, C = d, D. (m) The standard Fierz transformation is the special case of the general Fierz transformation, P (ΓaA )ij (ΓAa )kl = m,n (ΓaA )im (ΓAa )kn δmj δnℓ

P P = m,n (ΓaA )im (ΓAa )kn B ΓbB mℓ (ΓBb )nj

P = B ΓaA ΓbB iℓ (ΓAa ΓBb )kj

P c = B ρabc ABC ρABCabc (ΓCc )iℓ (ΓC )kj P P = C CAC (ΓCc )iℓ (ΓcC )kj = C CAC (ΓcC )iℓ (ΓCc )kj , (6.195)

where we used the Rearrangement Lemma and where CAC ≡ ρabc ABC ρABCabc .

Note that CAC depends only on the indices A and C since the indices a, b, c are summed over and since for fixed A, C there is only one B for which the ρ does not vanish. We can act on both sides of Eq. (6.195) with (ΓDd )ℓi ΓdD jk , sum over each of i, j, k, ℓ and use the orthonormality condition in Eq. (6.187) to give X

tr ΓDd ΓaA ΓdD ΓAa = CAC tr (ΓcC ΓDd ) tr ΓCc ΓdD C

= CAD 16δdd = 16ND CAD ,

(6.196)

where NA ≡ 1, 4, 6, 4, 1 for A = S, V, T, A, P respectively. We have then

1 tr ΓBb ΓaA ΓbB ΓAa .

(6.197) CAB = 16NB (n) In order to evaluate the coefficients CAB it will be convenient to make use of the various contraction identities given in Eq. (A.3.32). We can summarize the relevant contraction identities using the notation ΓaA ΓbB ΓAa ≡ fAB ΓbB ,

(6.198)

which defines the coefficients fAB . We know that contraction identities of this form must exist because the different ΓbB transform uniquely under spacetime transformations. Since ΓaA ΓbB ΓAa and ΓbB transform identically then the uniqueness property implies that they must be proportional. We can now evaluate the CAB in terms of the fAB using Eq. (6.197) and Eq. (6.198) and remembering that a repeated index b is summed over

1 1 1 fAB tr ΓBb ΓbB = fAB (4NB ) = fAB .

(6.199) CAB = 16NB 16NB 4 It remains then to use the contraction identities to evaluate the fAB . We see that: IΓbB I = ΓbB gives fSB = 1 for all B;

161

Problem 13 γ µ Iγµ = 4I gives fV S = 4; γ µ γ ν γµ = −2γ ν gives fV V = −2; γ µ σ ρσ γµ = 0 gives fV T = 0; γ µ (iγ 5 γ ν )γµ = 2(iγ 5 γ ν ) gives fV A = 2; γ µ γ 5 γµ = −4γ 5 gives fV P = −4; σ µν |µ<ν Iσµν |µ<ν = (1/2)σ µν Iσµν = 6 gives fT S = 6; (1/2)σ µν γ ρ σµν = 0 gives fT V = 0; (1/2)σ µν σ ρσ σµν = −2σ ρσ gives fT T = −2; (1/2)σ µν (iγ 5 γ ρ )σµν = 0 gives fT A = 0; (1/2)σ µν γ 5 σµν = 6 gives fT P = 6; (iγ 5 γ µ )I(iγ 5 γµ ) = γ 5 γ µ Iγµ γ 5 = 4I gives fAS = 4; γ 5 γ µ γ ν γµ γ 5 = −2γ 5 γ ν γ 5 = 2γ ν gives fAV = 2; γ 5 γ µ σ ρσ γµ γ 5 = 0 gives fAT = 0; γ 5 γ µ (iγ 5 γ ν )γµ γ 5 = 2γ 5 (iγ 5 γ ν )γ 5 = −2(iγ 5 γ ν ) gives fAA = −2; γ 5 γ µ γ 5 γµ γ 5 = −4γ 5 gives fAP = −4; γ 5 Iγ 5 = I gives fP S = 1; γ 5 V ν γ 5 = −V ν gives fP V = −1; γ 5 σ ρσ γ 5 = σ ρσ gives fP T = 1; γ 5 (iγ 5 γ ν )γ 5 = −(iγ 5 γ ν ) gives fP A = −1; and γ 5 γ 5 γ 5 = γ 5 gives fP P = 1; We can summarize these results as   1 1 1 1 1  4 −2 0 2 −4   1 1  C= f=  6 0 −2 0 6  , 4 4 4 2 0 −2 −4  1 −1 1 −1 1

(6.200)

where each of the row (A) and column (B) indices occur in the order S, V, T, A, P . (o) Consider the four arbitrary spinors ψ1 , . . . , ψ4 and consider the five Lorentz scalar quadrilinear spinor combinations s(1, 2; 3, 4) ≡ (ψ̄1 ΓS ψ2 )(ψ̄3 ΓS ψ4 ) = (ψ̄1 ψ2 )(ψ̄3 ψ4 )

v(1, 2; 3, 4) ≡ (ψ̄1 ΓaV ψ2 )(ψ̄3 ΓV a ψ4 ) = (ψ̄1 γ µ ψ2 )(ψ̄3 γµ ψ4 )

t(1, 2; 3, 4) ≡ (ψ̄1 ΓaT ψ2 )(ψ̄3 ΓT a ψ4 ) = (ψ̄1 σ µν |µ<ν ψ2 )(ψ̄3 σµν |µ<ν ψ4 ) = 21 (ψ̄1 σ µν ψ2 )(ψ̄3 σµν ψ4 )

a(1, 2; 3, 4) ≡ (ψ̄1 ΓaA ψ2 )(ψ̄3 ΓAa ψ4 ) = (ψ̄1 (iγ 5 γ µ )ψ2 )(ψ̄3 (iγ 5 γµ )ψ4 ) = (ψ̄1 γ 5 γ µ ψ2 )(ψ̄3 γµ γ 5 ψ4 )

p(1, 2; 3, 4) ≡ (ψ̄1 ΓP ψ2 )(ψ̄3 ΓP ψ4 ) = (ψ̄1 γ 5 ψ2 )(ψ̄3 γ 5 ψ4 ) .

(6.201)

The above quadrilinears are composed of two spinor bilinears with the same Lorentz tensor structure and are combined so as to contract to a Lorentz scalar.

Note that these definitions of s, v, t, a, p agree with those in Itzykson and Zuber (1980) and we have now proven their result. To summarize this in a compact

162

Formulation of Quantum Field Theory

notation let us define Ψ to be the column vector with elements ΨA being the Lorentz scalar combination s, v, t, a, p for A = S, V, T, A, P respectively, i.e.,   s  v     (6.202) Ψ(1, 2; 3, 4) ≡   t  (1, 2; 3, 4) .  a  p Then we can summarize the standard Fierz identity as Ψ(1, 4; 3, 2) = CΨ(1, 2; 3, 4)

(6.203)

for arbitrary spinors ψ1 , ψ2 , ψ3 , ψ4 , which is a compact form of the identity given in Itzykson and Zuber (1980). So the action of the matrix C is to exchange the second and fourth spinors (or equivalently the first and third) and we see that Ψ(1, 2; 3, 4) = CΨ(1, 4; 3; 2) = C 2 Ψ(1; 2; 3, 4). Hence we must have C 2 = I which is easily verified explicitly from Eq. (6.173).

Let g be the column vector consisting of constants gs , gv , gt , ga , gp then we can write (gs s + gv v + gt t + ga a + gp p)(1, 2; 3, 4) = g T Ψ .

(6.204)

The equation ĝ T Ψ(1, 2; 3, 4) = g T Ψ(1, 4; 3, 2) will be satisfied when the constants ĝ and g have the relationship ĝ T = g T C or equivalently ĝ = C T g.

Writing this explicitly in terms of the elements gives      ĝs 1 4 6 4 1 gs  ĝv   1 −2   0 2 −1       gv   ĝt  = 1  1   (6.205) 0 −2 0 1   gt   .    ĝa  4  1 2 0 −2 −1   ga  ĝp 1 −4 6 −4 1 gp This agrees with the matrix given in Cheng and Li (1984) after we take account of the fact that they choose a different normalization convention for t and a, where tCL = 2t and aCL = −a. It is useful to observe how to transform the matrix C for different normalization conventions of the spinor bilinears s, v, t, a, p.

For example, let us define the Cheng and Li normalization as ΨCL A = rA ΨA , CL which gives CAB = (rA /rB )CAB where rA = 2, −1 for A = T, A respecP tively and rA = 1 for A = S, V, P . [Proof: ΨCL = rA ΨA = rA B CAB ΨB = A P P CL CL CL B (rA /rB )CAB ΨB ≡ B CAB ΨB .]

7

Interacting Quantum Field Theories 7.1 Problem 1 Problem: Consider the proton-proton collisions in the Large Hadron Collider as discussed in Sec. 7.3.1. Assume that: improved beam optics gave an improved luminosity reduction factor Fred = 0.88 and a transverse beam size of 15 µm, average time spacing between bunches of 24 ns, and the initial protons per bunch increased to 1.3× 1011 . Calculate the following: (a) the number of proton bunches in each beam, (b) the peak luminosity, (c) the peak rate of detectable events, (d) the peak number of detectable events per bunch crossing and (e) the approximate peak observed rate for diphoton events coming from Higgs decays?

Solution: (a) We are told that the average time spacing between proton bunches is ∆t = 24 × 10−9 s, so the average spatial separation between the bunches will be approximately ∆x = c∆t = (3 × 108 ms−1 )(24 × 10−9 s) = 7.2 m because the protons in the collider travel at almost exactly the speed of light (as shown in the text, protons with energy 6.5 TeV travel at a speed of v = 0.99999999c). A circular tunnel of length L = 26.7 × 103 m would then be able to accommodate Nbunch = L/∆x = (26.7 × 103 m)/(7.2 m) = 3, 708 such bunches. (b) With a luminosity reduction factor of Fred = 0.88, transverse beam sizes of σx = σy = 15 µm = 1.5×10−3 cm at the intersection point, a collision frequency of fcoll = 1/∆t = 1/(24×10−9 s) = 4.17×107 s−1 , and Np = 1.3×1011 protons per bunch, we calculate from Eq. (7.3.23) a maximum luminosity of L=

Np2 fcoll Fred (1.3 × 1011 )2 (4.17 × 107 s−1 )(0.88) = 4πσx σy 4π(1.5 × 10−3 cm)2 = 2.19 × 1034 cm−2 s−1

(7.1)

in the upgraded collider. (c) It is given in the text that the total cross-section of detectable events from 163

164

Interacting Quantum Field Theories 6.5 TeV proton collisions is σdet ∼ = σinel = 58 × 10−27 cm, so the number of detectable events per second will be Rdet = Lσdet = (2.19 × 1034 cm−2 s−1 )(58 × 10−27 cm) = 1.27 × 109 s−1 . (d) We have seen that the number of bunch crossings per second is fcoll = 4.17 × 107 s−1 , so the number of detectable collisions per bunch crossing is Ndet/bunch ≡ Rdet /fcoll = (1.27 × 109 s−1 )/(4.17 × 107 s−1 ) = 30.5. (e) It is given in the text that the total cross-section for Higgs production and subsequent decay into two photons in the ATLAS detector is σfid (pp → h + X, h → γγ) = 65 × 10−39 cm2 , so the detectable rate for such diphoton Higgs decays in ATLAS at the peak luminosity is R(pp → h + X, h → γγ) = Lσfid (pp → h + X, h → γγ)

= (2.19 × 1034 cm−2 s−1 )(65 × 10−39 cm2 ) = 1.4 × 10−3 s−1 .

(7.2)

This is more than double the rate that would occur with the original collider specifications.

7.2 Problem 2 Problem: Sketch three examples of connected four-point (n = 4) Feynman diagrams in ϕ3 theory contributing to the left-hand side of Eq. (7.5.37). Show that each diagram can be uniquely identified as a contribution to the right-hand side.

Solution: Three illustrative examples are:

7.3 Problem 3 Problem: Sketch three examples of connected two-point (n = 2) Feynman diagrams with ten vertices in ϕ4 theory that contribute to DF (p). Show that these can be uniquely identified with contributions to the right-hand side of Eq. (7.5.49).

Problem 4

165

Solution: Three examples are:

7.4 Problem 4 Problem: Consider a Yukawa theory with the Lagrangian density L = 12 ∂µ ϕ∂ µ ϕ − 12 m2 ϕ2 + ψ̄(i̸ ∂ − M )ψ − g ψ̄(1 + irγ 5 )ϕψ ,

(7.3)

where g, r ∈ R, ϕ is a scalar field and ψ is the field for fermion f . (a) Does this theory respect parity? Explain. (b) Write down the Feynman rules for this theory. (c) Calculate the tree-level invariant amplitude for f f¯ → f f¯. (d) Sketch but do not evaluate all distinct Feynman diagrams contributing to the invariant amplitude for ϕϕ → ϕϕ at lowest nonvanishing order in g.

Solution: (a) Since ϕ is a scalar, rather than a pseudoscalar, then g ψ̄ϕψ is partity invariant but rg ψ̄γ 5 ϕψ is maximally parity violating. So if r ̸= 0 then parity is violated; (b) The Feynman rules for external fermion and scalar states are as given in Eq. (7.6.3). The fermion and scalar propagators are given in Eq. (7.6.4). There are two vertices as given in Eq. (7.6.15) with values iL = −ig and iL = grγ 5 . (c) The tree level amplitude for f f¯ → f f¯ is O(g 2 ) and is a generalization of Eq. (7.6.37) to include both vertices. There will be four copies of each of the two diagrams with the relative minus sign between them: (i) a copy where both vertices are −ig and conserve parity; (ii) a copy where both vertices are grγ 5 and preserve parity; (iii) two copies where the vertices are mixed with one −ig and one grγ 5 leading to parity violation. So the tree-level amplitude for f f¯ → f f¯ has eight contributing diagrams and so has eight terms. The amplitude is then given by ( ′ ) ′ ′ ′ ūs (p′ )us (p)v̄ r(k)v r (k ′ ) ūs (p′ )v r (k ′ )v̄ r(k)us (p) 2 iMf f¯→f f¯ = i(−ig) − t−m2ϕ s−m2ϕ

Interacting Quantum Field Theories

166

(

) ′ ′ ′ ′ ūs (p′ )γ 5 us (p)v̄ r(k)γ 5 v r (k ′ ) ūs (p′ )γ 5 v r (k ′ )v̄ r(k)γ 5 us (p) + i(gr) − t−m2ϕ s−m2ϕ ( ′ ) ′ ′ ′ ūs (p′ )γ 5 us (p)v̄ r(k)v r (k ′ ) ūs (p′ )γ 5 v r (k ′ )v̄ r(k)us (p) + i(−ig)(gr) − t−m2ϕ s−m2ϕ ( ′ ) ′ ′ ′ ūs (p′ )us (p)v̄ r(k)γ 5 v r (k ′ ) ūs (p′ )v r (k ′ )v̄ r(k)γ 5 us (p) − + i(−ig)(gr) . t−m2ϕ s−m2ϕ 2

(7.4) (d) The lowest order diagrams contributing to ϕϕ → ϕϕ contribute at O(g 2 ) and are the simple t-channel exchange of a fermion, where there are again four possible combinations of vertices: two of (−ig) with parity conservation; two of (rgγ 5 ) with parity conservation; and two with one of each of (−ig) and (grγ 5 ) that violate pairty. So there are four t-channel diagrams.

7.5 Problem 5 Problem: By drawing on the key steps in the derivation of the LSZ formalism for scalars and fermions outline the key steps in the derivation of the LSZ formalism for charged scalar bosons. There is no need to build explicit wavepackets to satisfy the asymptotic condition, which can be assumed.

Solution: Learning from both the neutral scalar and fermion cases, we arrive at the ‘midway’ case relevant for a charged scalar as follows: ⟨Ω|T ϕ̂(x)ϕ̂† (y)|Ω⟩1−particle = |⟨Ω|ϕ̂(0)|p = 0⟩|2 D0 (m2 ; x − y) , Z = |⟨Ω|ϕ̂(0)|p = 0⟩|2 = |⟨Ω|ϕ̂(0)|p⟩|2

and

(7.5) (7.6)

⟨Ω|T ϕ̂(x)ϕ̂† (y)|Ω⟩1−particle = ZD0 (m ; x−y) = Z⟨Ω|T ϕ̂as (x)ϕ̂†as (y)|Ω⟩ .

2

|p⟩ ≡

p

† 2Ep fˆasp |Ω⟩ ,

⟨Ω|ϕ̂as (x)|p⟩ = e−ip·x , ⟨p|ϕ̂†as (x)|Ω⟩ = eip·x ,

The asymptotic condition is that √ ⟨β|ϕ̂(x)|α⟩ = Z⟨β|ϕ̂as (x)|α⟩ ,

|p̄⟩ ≡

p

† 2Ep ĝasp |Ω⟩ .

⟨Ω|ϕ̂†as (x)|p̄¯⟩ = e−ip̄·x , ⟨p̄|ϕ̂as (x)|Ω⟩ = e

⟨β|ϕ̂† (x)|α⟩ =

√

ip̄·x

(7.7) (7.8)

.

Z⟨β|ϕ̂†as (x)|α⟩ ,

The in and out creation operators satisfy p p † 2Ek fˆkin |Ω⟩ = |k in⟩ and ⟨Ω| 2Ek fˆkout = ⟨k out| , p p † 2Ek ĝkin |Ω⟩ = |k̄ in⟩ and ⟨Ω| 2Ek ĝkout = ⟨k̄ out| .

(7.9)

(7.10) (7.11)

Problem 6

167

↔ R † 2Ek fˆask = −2i d3 x e−ik·x ∂ 0 ϕ̂in (x)|x0 =−T , p p ↔ R † † 2Ek fˆkout = 2Ek Ŝ † fˆask Ŝ = −2i d3 x e−ik·x ∂ 0 ϕ̂out (x)|x0 =T , p p ↔ R † † 2Ek ĝkin = 2Ek ĝask = −2i d3 x e−ik·x ∂ 0 ϕ̂†in (x)|x0 =−T , p p ↔ R † † 2Ek ĝkout = 2Ek Ŝ † ĝask Ŝ = −2i d3 x e−ik·x ∂ 0 ϕ̂†out (x)|x0 =T .

p

† 2Ek fˆkin =

p

(7.12)

(7.13) (7.14)

Let |i⟩ = |k1 , · · · , km , k̄1 · · · k̄m̄ ⟩ and ⟨f | = ⟨p̄n̄ , · · · , p̄1 , pn , · · · , p1 |, where the ki and the k̄j are the incoming particle and antiparticle momenta respectively and the pi and the p̄j are the outgoing particle and antiparticle momenta respectively.

Then we have hnn+n̄ o nm+ oi Y Ym̄ √ Sf i = Z m+n+m̄+n̄ DF−1 (pj ) G(m+n;m̄+n̄) DF−1 (ki ) (7.15) =

√

j=1

i=1

m̄+n̄) Z m+n+m̄+n̄ G(m+n; amp

on−shell

on−shell

,

(7.16)

where G(m+n;m̄+n̄) ≡ G(−k1 , · · · , −km , p̄1 , · · · , p̄n̄ ; p1 , · · · , pn , −k̄1 , · · · , −k̄m̄ ) , and where with all momenta outgoing we have R Qn Qn′ 4 iqj ·yj 4 ipi ·xi G(p1 , · · · , pn ; q1 , · · · , qn′ ) ≡ i=1 d xi e j=1 d yj e

(7.17)

(7.18)

× G(x1 , · · · , xn ; y1 , · · · , yn′ ) ,

G(x1 , · · · , xn ; y1 , · · · , yn′ ) ≡ ⟨Ω|T ϕ̂(y1 ) · · · ϕ̂(yn′ )ϕ̂† (x1 ) · · · ϕ̂† (xn )|Ω⟩ .

(7.19)

7.6 Problem 6 Problem: Write down the version of the photon LSZ formula of Eq. (7.5.125) that is appropriate for massive vector Proca bosons. Justify your answer by describing the key steps needed to arrive at this result.

Solution: As stated in the text the LSZ formalism for the Proca vector boson is similar to that for the photon, except that we need not be concerned with gauge invariance, there is a nonzero mass and there are three polarizations rather than two. Following this observation we obtain Eq. (7.5.109) with p2 + iϵ → p2 − m2 + iϵ for each momentum, with the three polarizations, and with Z3 → ZB for some appropriate ZB . Using k · ϵ(p, λ) from Eq. (6.5.36) and the massive vector boson Feynman propagator DFµν in Eq. (6.5.72) we again find Eq. (7.5.125) with the corresponding amputated covariant Green’s function. √ Understanding that there are three polarizations and that Ek = k2 + m2 we find in analogy with the photon case that p ′ |kλ⟩ ≡ |k, λ⟩ ≡ 2Ek âλ† ⟨kλ|k′ λ′ ⟩ = 2Ek δ λλ (2π)3 δ 3 (k−k′ ) , kas |Ω⟩ ,

Interacting Quantum Field Theories

168

⟨Ω|Â(x)|kλ⟩ =

p

ZB ⟨Ω|Âas (x)|kλ⟩ =

p ZB ϵ(k, λ)e−ik·x ,

(7.20)

where λ is the polarization index. Eq. (6.4.21) becomes √

↔ R 2ωk âλk = 2i d3 x eik·x ∂ 0 ϵ(k, λ)∗ · Âas (x) , R 3 −ik·x ↔ √ 2ωk âλ† ∂ 0 ϵ(k, λ)· Âas (x) . k = −2i d x e

(7.21)

Followomg the arguments for the photon case we then find p ν1 ···µm Sf i = (− ZB )m+n ϵν1 (p1 , κ1 )∗ · · · ϵνn (pn , κn )∗ G(m+n) amp (p1 , · · ·, −km ) = (−1)

m+n

× ϵµ1 (k1 , λ1 ) · · · ϵµm (km , λm ) , ∗

ϵν1 (p1 , κ1 )

(7.22)

∗

ν1 ···µm · · · ϵνn (pn , κn ) G(m+n) r,amp (p1 , · · ·, −km )

× ϵµ1 (k1 , λ1 ) · · · ϵµm (km , λm ) ,

where G(m+n) (p1 , · · · , pn ; −k1 , · · · , −km )ν1 ···νn µ1 ···µm has the same definition as the photon case.

7.7 Problem 7 Problem: Consider a theory with three interacting scalar fields Φ, ϕa and ϕb with masses M , m and m, respectively, with 3m > M > 2m and with the Lagrangian density L = 21 ∂µ Φ∂ µ Φ − 12 M 2 Φ2 + 21 ∂µ ϕa ∂ µ ϕa − 12 m2 ϕ2a + 21 ∂µ ϕb ∂ µ ϕb − 12 m2 ϕ2b − (λ/4!)Φ4 + 12 gaa Φϕ2a + gab Φϕa ϕb .

(7.23)

(a) What decays of the heavy scalar Φ are possible in a tree-level calculation?

Sketch the lowest order Feynman diagram for the process Φ → ϕb ϕb .

(b) Can ϕa or ϕb decay? Explain.

(c) Evaluate the ratio of partial decay rates, Γ(Φ → ϕa ϕb )/Γ(Φ → ϕa ϕa ), at tree level. (d) Evaluate the total tree-level decay rate for a Φ at rest, Γ(Φ → anything).

Solution: (a) At tree level only the decays Φ → ϕa ϕa and Φ → ϕa ϕb are kinematically possible and correspond to the interaction vertices Φϕ2a and Φϕa ϕb respectively.

Since M < 3m only decays to two particles are possible. The decay Φ → ϕb ϕb is kinematically allowed but can only occur at one loop or higher. The one-loop diagram that contributes is

Problem 7

169

b

a Φ

Φ a

b

(b) Neither ϕa nor ϕb can decay since they are the lightest particles and their masses are equal. Their decays are kinematically forbidden. (c) Using Eq. (7.3.74) the differential decay rates for Φ → ϕa ϕa and Φ → ϕa ϕb are |MΦ→ϕa ϕa |2 dΠLIPS ϕa ϕa , 2M |MΦ→ϕa ϕb |2 dΓΦ→ϕa ϕb = dΠLIPS ϕa ϕb , 2M

dΓΦ→ϕa ϕa =

(7.24) (7.25)

where from Eq. (7.3.41) we have P2 Q2 dΠLIPS ≡ (2π)4 δ 4 (pΦ − j=1 pj ) k=1 [d3 pk /(2π)3 2Epk ] (7.26) ab p with Epk = p2k + m2 and with the ϕa and ϕb bosons having four-momenta pµ1 and pµ2 respectively. Note that ϕa and ϕb have the same masses and so it follows that dΠLIPS = 12 dΠLIPS aa ab , because the final state ϕa ϕa contains two to avoid identical particles and so we must include a factor of 1/2! in dΠLIPS aa overcounting. So in this case at tree level we find dΓΦ→ϕa ϕb 2|MΦ→ϕa ϕb |2 2g 2 = = 2ab .

2 dΓΦ→ϕa ϕa |MΦ→ϕa ϕa | gaa

(7.27)

The right-hand side comes from the tree-level Feynman rules noting that the interactions iLΦϕa ϕb = igab Φϕa ϕb and iLΦϕa ϕa = i 12 gaa Φϕa ϕa lead to the 2 vertices igab and igaa respectively. This then gives |MΦ→ϕa ϕb |2 = gab and 2 2 |MΦ→ϕa ϕa | = gaa . (d) At tree level only two decay channels are available as stated above, so 2 2 gab + 21 gaa dΠLIPS (7.28) Φ→ϕa ϕb 2M 1 2 2 P2 Q2 g + g = ab 2 aa (2π)4 δ 4 (pΦ − j=1 pj ) k=1 [d3 pk /(2π)3 2Epk ] 2M

dΓΦ→anything = dΓΦ→ϕa ϕb + dΓΦ→ϕa ϕa =

where we have used Eq. (7.3.74). The total decay rate at tree-level for a Φ boson at rest is then 2 Z 2 gab + 21 gaa dΩp1 |p1 | √ (7.29) Γtree = Φ→anything 2M 16π 2 s p1 +p2 =0 g2 + 1 g2 1 p = ab 2 aa (1/4) − (m2 /M 2 ) , (7.30) 2M 4π √ where we have we have used that p1 = −p2 and s = M , we have adapted Eq. (7.3.79) to carry out the two-bodypphase-space integration p and have also used that M = Ep1 + Ep2 = 2Ep1 = 2 p21 + m2 ⇒ |p1 | = (M 2 /4) − m2 .

Interacting Quantum Field Theories

170

7.8 Problem 8 Problem: Consider a physical system described by the Lagrangian density L = 12 [∂µ ϕ∂ µ ϕ+∂µ χ∂ µ χ−m2ϕ ϕ2 −m2χ χ2 ]− 12 gχ2 ϕ−(λ/4!)χ4 ,

(7.31)

where ϕ, χ, mϕ , mχ , g and λ are real. (a) Calculate dσ/d cos θ in the centre of momentum (CM) frame at tree-level for χϕ → χϕ in terms of CM energy and scattering angle. (b) Repeat this calculation for χχ → χχ at tree-level in the CM frame. Express your answer in terms of the Mandelstam variables s, t and u.

Solution: (a) The only tree-level diagram contributing to χϕ → χϕ is χ

χ χ

φ

φ

The only relevant interaction is iLχ2 ϕ = −i 12 gχ2 ϕ leading to the χ2 ϕ vertex −ig, since if a field appears n times in an interaction this leads to a factor of n!.

Using the Feynman rules the invariant amplitude from this Feynman diagram is iM = (−ig)2 (i/s − m2χ ) and so |M|2 = g 4 /(s − m2χ )2 . Then from Eq. (7.3.82) we have

dσ |M|2 LIPS 1 pcm g4 f = dΠ2 = |M|2 = , (7.32) cm 2 2 dΩ cm 4F 64π s pi 64π s(s − m2χ )2 since the initial and final state masses are the same and so pcm = pcm i f by energy conservation. Then using Eq. (7.3.83) we arrive at

1 pcm g4 dσ f 2 = |M| = .

(7.33) d(cos θ) cm 32πs pcm 32πs(s − m2χ ) i

2 Since s = Ecm then this differential cross-section is independent of the scattering angle θ as well as the azimuthal angle ϕ in the CM frame as we would expect for an s-channel scattering process, i.e., a scalar mediator in the CM frame carries no directional information. (b) There are four tree-level diagrams contributing to χχ → χχ, which are χ

χ

χ

φ

χ

χ

χ

φ χ

χ

χ

χ

χ

χ

χ

χ

φ χ

χ

Problem 9

171

The first three diagrams are the s, t and u channel exchanges. Since we attach momentum labels p1 , . . . , p4 to each external point, the t and u channels are distinct because they involve different combinations of momenta. We must also remember to divide the phase space by 2! since we have two identical χ particles in the final state. The relevant interactions are iLχ2 ϕ = −i 21 gχ2 ϕ and iLχ4 = −i(λ/4!)χ4 leading to the vertices −ig and −iλ respectively. Using the Feynman rules the invariant amplitude resulting from adding these four Feynman diagrams is " # i i i 2 iM = (−ig) + + + (−iλ) .

(7.34) s − m2ϕ t − m2ϕ u − m2ϕ Generalizing the above arguments and remembering to multiply by 12 because of two χ particles in the final state, we then have

dσ 1 1 pcm f |M|2 (7.35) = d(cos θ) cm 2 32πs pcm i " # 2 1 1 1 1 2 g = + + +λ .

64πs s − m2ϕ t − m2ϕ u − m2ϕ Since from

Eq. (7.3.85) we see that t depends on θ and similarly for u, then dσ will depend on θ but not ϕ in this case. d(cos θ) cm

7.9 Problem 9 Problem: The tree-level invariant amplitude for the equivalent of Compton scattering in scalar QED was calculated in Eq. (7.6.54). Use this to calculate the unpolarized differential cross-section dσ/dΩ in the fixed-target frame (initial ϕ at rest).

P [Hint: Use κ ϵµ (k, κ)ϵµ (k, κ)∗ → −g µν as in Eq. (7.6.83)].

Solution: We being by defining Mνµ (k ′ , k) (suppressing the dependence on p′ ,p for notational convenience only), from Eq. (7.6.54) such that ( ) ′ ′ (2p−k ′ )ν (2p′−k)µ 2 ν ′ ′ ∗ (2p+k)µ (2p +k )ν iM = i(−ie) ϵ (k , κ ) + −2gµν ϵµ (k, κ) s − m2ϕ u − m2ϕ ≡ ϵν (k ′ , κ′ )∗ Mνµ (k ′ , k)ϵµ (k, κ) ,

(7.36)

where we will choose qc = e. As for the fermion Compton scattering case we have p2 = p′2 = m2ϕ and k 2 = k ′2 = 0, and we have for the denominators that s − m2ϕ = (p + k)2 − m2ϕ = 2p · k and u − m2ϕ = (p − k ′ )2 − m2ϕ = −2p · k ′ .

Then forming |M|2 and averaging over initial photon polarization κ and summing over final photon polarization κ′ leads to P P 1 1 2 ν ′ ′ ∗ µ ′ σ ′ ′ ρ ∗ ′ ∗ κ′,κ |M| = 2 κ′,κ ϵ (k , κ ) ϵ (k, κ)Mνµ (k , k)ϵ (k , κ )ϵ (k, κ) Mσρ (k , k) 2

Interacting Quantum Field Theories

172

= 12 (−g νσ )(−g ρµ )Mνµ (k ′, k)Mσρ (k ′, k)∗ = 21 Mνµ (k ′, k)Mνµ (k ′, k)∗ ≡ |M|2 , (7.37) where

(2p+k)µ (2p′+k ′ )ν (2p−k ′ )ν (2p′−k)µ −2gµν (7.38) + 2 2p · k −2p · k ′

(2p+k)µ (2p′+k ′ )ν (2p−k ′ )ν (2p′−k)µ −2g µν + × 2p · k −2p · k ′

(2p+k)2 (2p′+k ′ )2 (2p−k ′ )2 (2p′−k)2 1 4 = 2e + +16 (2p · k)2 (2p · k ′ )2 [(2p+k) · (2p′ −k)][(2p′ +k ′ ) · (2p−k ′ )] (2p+k) · (2p′+k ′ ) −2 −4 (2p · k)(2p · k ′ ) 2p · k

′ ′ (2p−k ) · (2p −k) . +4 2p · k ′

|M|2 = 1 e4

Recall that the Mandelstam variables are: s = (p + k)2 = (p′ + k ′ )2 = m2ϕ + 2p · k = m2ϕ + 2p′ · k ′ ; t = (p′ − p)2 = (k ′ − k)2 = 2m2ϕ − 2p′ · p = −2k ′ · k; u = (k ′ − p)2 = (p′ − k)2 = m2ϕ − 2p · k ′ = m2ϕ − 2p′ · k.

So we have p · k = p′ · k ′ , p · k ′ = p′ · k and k ′ · k = p′ · p − m2ϕ . Also note that p′ − k = p + k − k ′ − k = p − k ′ . It then follows that in the fixed target frame with pµ = (mϕ , 0) we have p · k = mϕ ω; p · k ′ = mϕ ω ′ ; (2p + k)2 = (2p′ + k ′ )2 = 4(m2ϕ + p · k) = 4mϕ (mϕ + ω) ; (2p − k ′ )2 = (2p′ − k)2 = 4(m2ϕ − p · k ′ ) = 4mϕ (mϕ − ω ′ ) ; (2p + k) · (2p′ − k) = (2p′ + k ′ ) · (2p − k ′ ) = 4m2ϕ + 2p · (k − k ′ ) = 4m2ϕ + 2mϕ (ω − ω ′ ); (2p′ + k ′ ) · (2p − k ′ ) = 4m2ϕ + 2mϕ (ω − ω ′ ); (2p + k) · (2p′ + k ′ ) = 4m2ϕ + 4p · k + k · k ′ = 4m2ϕ + 4mϕ ω + ωω ′ (1 − cos θ); (2p − k ′ ) · (2p′ − k) = 4m2ϕ − 4p · k ′ + k · k ′ = 4m2ϕ − 4mϕ ω ′ + ωω ′ (1 − cos θ). These results then lead to

8[mϕ + 21 (ω − ω ′ )]2 4(mϕ + ω)2 4(mϕ − ω ′ )2 + +16− |M|2 = 12 e4 ω2 ω ′2 ωω ′ ) 8m2ϕ + 8mϕ ω + 2ωω ′ (1 − cos θ) 8m2ϕ − 8mϕ ω ′ + 2ωω ′ (1 − cos θ) − + mϕ ω mϕ ω ′ ( )

8m2ϕ 1 1 1 1 2 1 4 = 2 e 4mϕ + +8+8mϕ − − , (7.39) ω 2 ω ′2 ω ω′ ωω ′ where we have used Eq. (7.6.95) in the form (1/ω ′ )−(1/ω) = (1−cos θ)/mϕ ,

(7.40)

which follows from four-momentum conservation. Then we combine the resulting form with Eq. (7.3.94) to obtain the final form of interest. As for the fermion Compton scattering case we identify m1 = m3 = mγ = 0, m2 = m4 = mϕ ,

Problem 10

173

pµ1 = k µ , pµ2 = pµ , pµ3 = k ′µ and pµ4 = p′µ . Now write ω ≡ k 0 = |k| = 1/λ and ω ′ ≡ k ′0 = |k′ | = 1/λ′ . This then gives the form analogous to the fermion Compton scattering case, 2

1 1 ω′ dσ 1 ω′ 2 = |M| |M|2 , = dΩ 64π 2 ωmϕ mϕ + ω(1 − cos θ) 64π 2 m2ϕ ω 2

(7.41)

where |M|2 is now the form above. To get the right-hand side we again use Eq. (7.6.95).

7.10 Problem 10 Problem: Consider a gauge theory of charged scalars and charged fermions interacting with photons, which is a combined form of QED and scalar QED, L = LDirac + LKG + LMaxwell + LQED + LsQED . int int

(7.42)

(a) Calculate the invariant amplitude for the process e− e+ → ϕϕ∗ to lowest order in α, where the ϕ charge is qc = +|e|. (b) Calculate dσ/dΩ in the CM frame. Compare with the result for e+ e− → µ+ µ− in Eq. (7.6.71).

Solution: It is relatively straightforward to adapt the approach used to obtain Eq. (7.6.71) for e+ e− → µ+ µ− using the appropriate Feynman rules. (a) At tree level and O(α) only one Feynman diagram contributes. We have e−

p

p0

φ+

e+

k

k0

φ−

iMe− e+ →ϕ+ ϕ− =

−igµν = [(−ie)(p − k ) ] [v̄ r (k)(+ie)γ ν us (p)] s ′

= −ie2

′ µ

v̄ r (k)(̸ p′ − ̸ k ′ )us (p) , s

(7.43)

where s = (p + k)2 = (p′ + k ′ )2 and e2 = 4πα. (b) The final state scalars have no polarizations to sum over, but averaging over the initial state spins we have |M|2 = e4

1 X |v̄ r (k)(̸ p′ − ̸ k ′ )us (p)|2 4 r,s s2

(7.44)

Interacting Quantum Field Theories

174

= (e4 /s2 )(p′ − k ′ )µ (p′ − k ′ )ν 41 4

2

′

′ µ

′

P

r,s tr[γµ (̸ p + me )γν (̸ k − me )]

′ ν

= (e /s )(p − k ) (p − k ) {4(pµ kν +pν kµ −gµν p·k)−4m2e gµν } = 4(e4 /s2 ){2[(p′ −k ′ )·p][(p′ −k ′ )·k]−(p′ −k ′ )2 (p·k+m2e )} ,

where we have used the Casimir trick in Eq. (7.6.61) and then the result in Eq. (7.6.66). The Mandelstam variables are s = 2m2e +2p·k = 2m2ϕ +2p′·k ′ ,

t = m2e +m2ϕ −2p·p′ = m2e +m2ϕ −2k·k ′ ,

u = m2e +m2ϕ −2p·k ′ = m2e +m2ϕ −2k·p′ ,

(7.45)

which means that p · k ′ = p′ · k, k · k ′ = p · p′ and p′ · k ′ = m2e − m2ϕ + p · k.

The unpolarized differential cross-section in the CM frame is then obtained from ′ ′ Eq. (7.3.82), where in the CM frame we have pcm = |p| = |k|, pcm i f = |p | = |k |, √ 1 1 0 ′0 0 ′0 ′ ′ ′ p = p = k = k = 2 Ecm = 2 s ≡ E, p = −k, p = −k and p · p = cm − |p′ ||p| cos θ = pcm f pi cos θ. Note that here θ is the angle between the initial e and + the final ϕ in the CM frame. Comparing to the left-hand side of Fig. 7.11 we have p1 = p, p2 = k, p3 = p′ and p4 = k ′ and the result for e− e+ → ϕ+ ϕ− is then

1 pcm dσ f = |M|2 (7.46) dΩ cm 64π 2 s pcm i α2 |p′ | {2[(p′ −k ′ )·p][(p′ −k ′ )·k]−(p′ −k ′ )2 (p·k+m2e )} s3 |p| α2 |p′ | α2 |p′ |3 = 3 −8(p′ · p)2 + 8|p′ |2 E 2 = 1 − (|p|2 /E 2 ) cos2 θ s |p| 8E 4 |p| α2 βϕ3 = 1 − (|p|2 /E 2 ) cos2 θ , 8E 2 βe

=

where in the CM frame we have used s = 4E 2 , p′ − k ′ = (0, 2p′ ), (p′ − k ′ ) · p = −2p′ · p = −2|p′ ||p| cos θ, (p′ − k ′ ) · k = +2p′ · p = 2|p′ ||p| cos θ, (p′ − k ′ )2 = 2 2 2 −4|p′ |2 ,p p · k = E 2 + |p| + |p|2 + m2e = 2E 2 . Recall that p and (p · k + me ) = E p γ = 1/q 1 − β 2 = 1/q1 − (v/c)2 and soq β = 1 − (1/γ 2 ), which means that |p′ | =

E 2 − m2ϕ = E

1 − (m2ϕ /E 2 ) = E

1 − (1/γϕ2 ) = Eβϕ . Similarly we have

used |p| = Eβe . There is no need to include an additional factor of 12 in the phase space here since ϕ+ and ϕ− are distinguishable particles. To compare with the e+ e− → µ+ µ− case in Eq. (7.6.71) we note that ϕ± → µ∓ . The results are similar but the θ-dependence and other details are slightly different as could be expected.

In the ultralrelativistic limit, E ≫ me , mϕ , we have |p| → E and we find

dσ α2 α2 = {1 − cos2 θ} = sin2 θ , (7.47) 2 dΩ cm 8E 8E 2 which peaks at ±π/2 and vanishes at forward and backward angles, 0 and π. This can be compared with the ultrarelativistic case for e+ e− → µ+ µ− with

α2 1 dσ = {1 + cos2 θ} , (7.48) dΩ cm 8E 2 2

175

Problem 10

which never vanishes and peaks at forward and backward angles.

8

Symmetries and Renormalization 8.1 Problem 1 Problem: Use the equal-time canonical commutation and anticommutation relations to prove Eq. (8.3.22). Then for either a boson or fermion operator X̂(y) generalize Eq. (6.2.166) and show that ∂µx ⟨0|T ĵ µ (x)X̂(y)|0⟩ = δ(x0 −y 0 )⟨0|[ĵ 0 (x), X̂(y)]|0⟩.

Then further generalize this result to verify the QED Ward identity in Eq. (8.3.21).

Solution: Recall that we have the equal-time commutation relations between independent fields [µ (t, x), ψ̂(t, x′ )] = [µ (t, x), ψ̄ˆ(t, x′ )] = 0 (8.1) and the fermion equal-time anticommutation relations in Eq. (6.3.94), {ψ̂α (x), ψ̂β† (y)} = δαβ δ 3 (x − y) ,

{ψ̂α (x), ψ̂β (y)} = {ψ̂β† (x), ψ̂α† (y)} = 0 .

(8.2)

We note in passing, but do not need here, the covariant formulation of the equaltime commutation relations for the photon in Eq. (6.4.122), ˙ ˙ [µ (x), Âν (y)]x0 =y0 = [µ (x), Âν (y)]x0 =y0 = 0 ˙ [µ (x), Âν (y)]x0 =y0 = −ig µν δ 3 (x − y) .

and (8.3)

Since ĵ 0 (x) = qc ψ̄ˆ(x)γ 0 ψ̂(x) it is clear from Eq. (8.1) that [ĵ 0 (t, x), µ (t, y)] = 0.

Using an obvious shorthand we see from Eq. (8.2) that we have [ψ̂α† (t, x)ψ̂β (t, x), ψ̂γ (t, y)] = [ψ̂α† ψ̂β , ψ̂γ ] = ψ̂α† ψ̂β ψ̂γ − ψ̂γ ψ̂α† ψ̂β

(8.4)

= −ψ̂α† ψ̂γ ψ̂β − ψ̂γ ψ̂α† ψ̂β = −{ψ̂α† , ψ̂γ }ψ̂β = −{ψ̂α† (t, x), ψ̂γ (t, y)}ψ̂β (t, x) = −δαγ δ 3 (x − y)ψ̂β (t, x)

and similarly [ψ̂α† (t, x)ψ̂β (t, x), ψ̂γ† (t, y)] = [ψ̂α† ψ̂β , ψ̂γ† ] = ψ̂α† ψ̂β ψ̂γ† − ψ̂γ† ψ̂α† ψ̂β

(8.5)

= ψ̂α† ψ̂β ψ̂γ† + ψ̂α† ψ̂γ† ψ̂β = ψ̂α† {ψ̂β , ψ̂γ† } = ψ̂α† (t, x){ψ̂β (t, x), ψ̂γ† (t, y)} = δβγ δ 3 (x − y)ψ̂α† (t, x) .

Using ĵ 0 (x) = qc ψ̂ † (x)ψ̂(x) we then find that [ĵ 0 (x), µ (x′ )] = 0 , 176

(8.6)

Problem 2

177

[ĵ 0 (x), ψ̂γ (x′ )]δ(x0−x′0 ) = −qc ψ̂γ (x)δ 4 (x−x′ ), [ĵ 0 (x), ψ̄ˆ (x′ )]δ(x0−x′0 ) = q ψ̄ˆ (x)δ 4 (x−x′ ) . γ

c

γ

We next need the generalization of Eq. (6.2.166) when fermion and photon field operators are present. Begin by observing that ĵ µ (x) behaves as a boson operator would in a time-ordered product since it is a fermion bilinear. Then for either a boson or fermion operator X̂(y) we have ∂µx ⟨Ω|T ĵ µ (x)X̂(y)|Ω⟩ = ∂µx ⟨Ω|θ(x0 − y 0 )ĵ µ (x)X̂(y) + θ(y 0 − x0 )X̂(y)ĵ µ (x)|Ω⟩ = θ(x0 − y 0 )⟨Ω|∂µx ĵ µ (x)X̂(y)|Ω⟩ + θ(y 0 − x0 )⟨0|X̂(y)∂µx ĵ µ (x)|0⟩ + δ 0µ δ(x0 − y 0 )⟨Ω|[ĵ µ (x), X̂(y)]|Ω⟩

(8.7)

= ⟨Ω|T ∂µx ĵ µ (x)X̂(y)|Ω⟩ + δ 0µ δ(x0 − y 0 )⟨Ω|[ĵ µ (x), X̂(y)]|Ω⟩ = δ(x0 − y 0 )⟨Ω|[ĵ 0 (x), X̂(y)]|Ω⟩ ,

Next we observe that if we replace X̂(y) with X̂1 (y1 ) · · · X̂k (yk ), then we will only get contributions when x0 = yi0 for one of the Xi (yi ). Having more than two time coordinates equal is a set of measure zero and so we need not consider such a situation. So we only need consider x0 ≃ yi0 for one i at a time. Then it follows that we can hold all times fixed with a finite separation from x0 except that we let yi0 vary infinitesimally around x0 . Then in that case T ĵ µ (x)X̂1 (y1 ) · · · X̂k (yk ) = · · · T [ĵ µ (x)X̂i (yi )] · · · , where the dots represent the remaining operators timeordered with their times fixed. It is then clear that Pk ∂µx ⟨Ω|T ĵ µ (x)X̂1 (y1 ) · · · X̂k (yk )|Ω⟩ = i=1 ∂µx ⟨Ω| · · · [T ĵ µ (x)X̂i (yi )] · · · |Ω⟩ (8.8) Pk 0 0 0 = i=1 δ(x − yi )⟨Ω| · · · [ĵ (x), X̂i (yi )] · · · |Ω⟩ , Combining this result with Eq. (8.6) it then readily follows that ∂µx ⟨Ω|T ĵ µ (x)ψ̂(x1 )ψ̄ˆ(y1 ) · · · ψ̂(xn )ψ̄ˆ(yn )µ1 (z1 ) · · · µm (zm )|Ω⟩ Pn = qc i=1 {−δ 4 (x−xi ) + δ 4 (x−yi )} × ⟨Ω|T ψ̂(x )ψ̄ˆ(y ) · · · ψ̂(x )ψ̄ˆ(y )µ1 (z ) · · · µm (z )|Ω⟩ , 1

1

n

n

1

(8.9)

m

which is Eq. (8.3.21).

8.2 Problem 2 Problem: Using the Feynman rules for a ϕψ̄ψ Yukawa theory, write down the one loop contribution to the ϕ4 proper vertex. Show that this one-loop contribution is divergent and explain why we must include a ϕ4 vertex in Yukawa theory.

Solution: The one-loop contribution to the ϕ4 vertex in Yukawa theory has a fermion loop. Representing the fermion loop as a clockwise loop, there are six

Symmetries and Renormalization

178

topologically distinct ways to attach the external scalar legs. Labeling each external leg from 1 to 4, these six diagrams can be represented as 1234, 1243, 1324, 1342, 1423 and 1432 as shown: 1

2

1

2

1

3

1

3

1

4

1

4

4

3

3

4

4

2

2

4

3

2

2

3

Let us assign the external legs 1, 2, 3 and 4 the outgoing momenta p1 , p2 , p3 and p4 = −(p1 + p2 + p3 ), respectively. Then the contribution from the first diagram to the 1PI four-point vertex is p1

p2 `

` − p2

` + p1 p4

= (−1)(−ig)4

Z

d4 ℓ (2π)4

(8.10)

p3 ` − p2 − p3

×

tr[i(̸ ℓ+̸ p1 −m)i(̸ ℓ−̸ p2 −̸ p3 −m)i(̸ ℓ−̸ p2 −m)i(̸ ℓ−m)] , 2 [(ℓ+p1 ) −m2 +iϵ][(ℓ−p2 −p3 )2 −m2 +iϵ][(ℓ−p2 )2 −m2 +iϵ][ℓ2 −m2 +iϵ]

where the (−1) is due to the fermion loop, where each vertex gives a factor (−ig) and where as always the trace is ordered from right to left in the fermion arrow direction. The most divergent part of this amplitude has the form Z 4 d ℓ −g 2 tr[̸ ℓ̸ ℓ̸ ℓ̸ ℓ] 4 2 2 (2π) [(ℓ+p1 ) −m +iϵ][(ℓ−p2 −p3 )2 −m2 +iϵ][(ℓ−p2 )2 −m2 +iϵ][ℓ2 −m2 +iϵ] Z Λ2 2 2 2 Z Λ2 2 π dℓ ℓ 4ℓ4 dℓ 1 g4 4 = −g ∼− ln Λ2 , (8.11) = −g 4 4 8 7 2 2 (2π) ℓ + O(ℓ ) (2π) ℓ +O(ℓ) (2π)2 which we see is a logarithmic divergence. Each of the six diagrams has the same logarithmic divergence and so the total divergence is −6g 4 ln Λ2 /(2π)4 . In order to renormalize this divergence we need to introduce a ϕ4 interaction so that the corresponding counterterm can compensate for this divergence as well as the divergence in the ϕ4 proper vertex coming from the ϕ4 interactions themselves. In short, Yukawa theory must also have a ϕ4 interaction in order that the theory be remormalizable.

This divergence was not present in the four-photon one-loop diagram in QED shown in Eq. (8.4.16), since the Ward-Takahashi identity ensured that the logarithm divergence was absent and so the loop integral was finite.

Problem 3

179

8.3 Problem 3 Problem: Prove Eq. (8.7.36).

Solution: We have from Eq. (8.7.34) by definition N µ ≡ [−y̸ q+z̸ p]γ µ [(1−y)̸ q+z̸ p]−2m(1−2y)q µ −4mzpµ +m2 γ µ .

(8.12)

We need to obtain N0µ (p, q), where this is a simplified version of N µ (p, q) that gives the same result when evaluated between plane wave states, ′

ūs (p′ )N µ (p, q)us (p) (8.13)

s′ ′ µ µ µ 2 µ s = ū (p ) [−y̸ q+z̸ p]γ [(1−y)̸ q+z̸ p]−2m(1−2y)q −4mzp +m γ u (p)

′ = ūs (p′ ) [−y̸ q+z(̸ p ′ − ̸ q)]γ µ [(1−y)̸ q+z̸ p]−2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) [−y̸ q+z(m − ̸ q)]γ µ [(1−y)̸ q+zm]−2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) [(x−1)̸ q+zm]γ µ [(1−y)̸ q+zm]−2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (x−1)(1−y)̸ qγ µ ̸ q+zm[(x−1)̸ qγ µ +(1−y)γ µ ̸ q]+z 2 m2 γ µ −2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (x−1)(1−y)[2q µ ̸ q − q 2 γ µ ]+zm[(x−1)̸ qγ µ +(1−y)γ µ ̸ q]+z 2 m2 γ µ

−2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (1−x)(1−y)q 2 γ µ +zm[(x−1)(̸ p ′ −̸ p)γ µ +(1−y)γ µ (̸ p ′ −̸ p)]+z 2 m2 γ µ

−2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (1−x)(1−y)q 2 γ µ +zm[(x−1)(m−̸ p)γ µ +(1−y)γ µ (̸ p ′ −m)]+z 2 m2 γ µ

−2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (1−x)(1−y)q 2 γ µ−zm2 [(1−x)+(1−y)]γ µ+zm[(1−x)̸ pγ µ +(1−y)γ µ ̸ p ′ ] +z 2 m2 γ µ −2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (1−x)(1−y)q 2 γ µ −zm2 [1+z]γ µ +zm[(1−x)(2pµ −γ µ ̸ p)

+(1−y)(2p′µ −̸ p ′ γ µ )]+z 2 m2 γ µ −2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (1−x)(1−y)q 2 γ µ−zm2 [1+z]γ µ+zm[(1−x)(2pµ −γ µ m) +(1−y)(2p′µ − mγ µ )]+z 2 m2 γ µ −2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (1−x)(1−y)q 2 γ µ −zm2 [1+z]γ µ −zm2 [1+z]γ µ

+2zm[(1−x)pµ +(1−y)p′µ ]+z 2 m2 γ µ −2m(1−2y)q µ −4mzpµ +m2 γ µ us (p)

′ = ūs (p′ ) (1−x)(1−y)q 2 γ µ +m2 [1−2z−z 2 ]γ µ +2zm[(1−x)pµ +(1−y)p′µ ]−2m(1−2y)q µ −4mzpµ } us (p)

′ = ūs (p′ ) γ µ [(1−x)(1−y)q 2 +(1−2z−z 2 )m2 ]+(p′ +p)µ mz(z−1) +q µ [m(z−2)(x−y)]} us (p) ′

≡ ūs (p′ )N0µ (p, q)us (p),

Symmetries and Renormalization

180

′

′

′

where we have used a̸ γ µ = 2aµ −γ µa̸ , ūs (p′ )̸ p ′ us (p) = ūs (p′ )̸ pus (p) = mus (p′ )us (p) ′ and ūs (p′ )̸ qus (p) = 0 and where in the last step we have used 2zm[(1−x)pµ +(1−y)p′µ ]−2m(1−2y)q µ −4mzpµ ′µ

µ

′µ

µ

µ

(8.14) µ

= zm[(1−x)({p +p }−q )+(1−y)({p +p }+q )]−2m(1−2y)q ′µ

µ

µ

(8.15)

µ

− 2mz({p +p }−q )

= (p′µ +pµ )[2mz−mz(x+y)−2mz]+q µ [zm{(1−y)−(1−x)}−2m(1−2y)+2mz] = (p′µ +pµ )mz(z−1)+q µ m[z{(1−y)−(1−x)}−2(1−2y)+2z] = (p′µ +pµ )mz(z−1)+q µ m[z(x−y)−2+ 4y+2(1−x−y)] = (p′µ +pµ )mz(z−1)+q µ m(z−2)(x−y).

8.4 Problem 4 Problem: Prove the results in Eq. (8.7.53) in the chiral representation or any other representation that you care to use.

Solution: We will work in the chiral representation. It then follows that i hp p p ′ ′ √ ūs (p′ )γ 0 us (p) = us † (p′ )us (p) = ϕ†s′ p′ · σ p · σ + p′ · σ p · σ ϕs (8.16)

σ · p′ σ · p σ · p σ · p′ 1− = mϕ†s′ 1− 1+ + 1+ ϕs +O(p2 ) 2m 2m 2m 2m = 2mϕ†s′ ϕs +O(p2 ) = 2mδs′ s +O(p2 ),

h p p p i √ −σ i 0 s † s′ ′ i s s′ † ′ ′ ·σσ i p·σ+ ′ ·σσ i p·σ ϕ ū (p )γ u (p) = u (p ) p p u (p) = ϕ − ′ s s 0 σi

σ · p σ · p′ σ · p σ · p′ σ i 1− σ i 1+ ϕs + 1+ = mϕ†s′ − 1− 2m 2m 2m 2m

= ϕ†s′ σ · p′ σ i +σ i σ · p ϕs +O(p2 )

= ϕ†s′ (p′ + p)i − iϵijk (p′ − p)j σ k ϕs +O(p2 ) ′

(p +p)i iϵijk q j σ k = 2mϕ†s′ − ϕs +O(p2 ), 2m 2m

′ ′ ′ 0 σj ūs (p′ )σ 0j qj us (p) = qj us † (p′ )γ 0 σ 0j us (p) = qj us † (p′ ) us (p) −σ j 0 hp i p p √ = qj ϕ†s′ p′ · σσ j p · σ − p′ · σσ j p · σ ϕs

σ · p′ σ · p σ · p′ σ · p † j j = qj mϕs′ 1− σ 1+ − 1+ σ 1− ϕs 2m 2m 2m 2m = qj mϕ†s′ [O(p)] us (p) = O(p2 ), s′

′

ij

s

ū (p )σ qj u (p) = qj u

s′ †

′

0 ij s

(p )γ σ u (p) = qj ϵ

ijk s′ †

u

′

(p )

0 σk

σk 0

us (p)

Problem 5

181

i hp p p √ p′ · σσ k p · σ + p′ · σσ k p · σ ϕs

σ · p′ σ · p′ σ · p σ · p † ijk k k = qj ϵ mϕs′ 1− + 1+ ϕs σ 1+ σ 1− 2m 2m 2m 2m

= qj ϵijk mϕ†s′ 2σ k +O(p) us (p) = 2mϕ†s′ ϵijk qj σ k us (p)+O(p2 )

= 2mϕ†s′ −ϵijk q j σ k ϕs +O(p2 ), = qj ϵijk ϕ†s′

where we have used the chiral representaion of the γ-matrices given in Sec. A.3 and σ i σ j = δ ij + iϵijk σ k . Note that O(p) means either of O(p) or O(p′ ) and O(p2 ) means either O(p2 ) or O(p′2 ) or O(p′ · p).

8.5 Problem 5 Problem: Complete the detailed steps to arrive at the scattering of a charged fermion from a static electromagnetic field given in Eq. (8.7.48).

Solution: As explained in the text, for the scattering of a single charged particle from a static potential we know that energy is conserved and so we can extract an overall energy-conserving δ-function. Then we can define M′f i in terms of Tf i such that Tf i = (2π)δ(Ef −Ei )M′f i . At first-order we have the Born approximation where the interaction Lagrangian density is Lint = −j µ Aµ from Eqs. (3.3.4) and (7.6.22), where j µ = qc ψ̄γ µ ψ. The momentum space Feynman rules tell us how to manage the fermion part of the interaction and we need to use the momentum space form of Aµ , which is the four-dimensional Fourier transform Aµ (q). For a static Aµ (x) we have independence of x0 = t and so can define Aµ (q) using 0 Aext µ (q) ≡ 2πδ(q )õ (q).

(8.17)

This provides the needed energy-conserving δ-function, δ(q 0 ) = δ(Ef − Ei ), in Tf i .

So at tree level we then find ′

Tf i = (2π)δ(Ef −Ei )M′f i = −qc ūs (p′ )γ µus (p) Aext µ (q)

s′ ′ µ s = 2πδ(Ef − Ei ) −qc ū (p )γ u (p) õ (q) ,

(8.18)

which is just Eq. (8.7.51), Tf i = (2π)δ(Ef −Ei )M′f i (8.19)

µν ′ iσ qν F2 (q 2 ) us (p) Aext = −qc ūs (p′ ) γ µ F1 (q 2 )+ µ (q) , 2m

′ iσ µν qν = (2π)δ(Ef −Ei ) −qc ūs (p′ ) γ µ F1 (q 2 )+ F2 (q 2 ) us (p) õ (q) 2m with F1 (q 2 ) → 1 and F2 (q 2 ) → 0 as q 2 → 0. For a single interaction with the external field we can include higher order terms in the Feynman diagrams, such

Symmetries and Renormalization

182

as the one-loop calculation in Eq. (8.7.32) and what follows, which leads to higher loop effects in F1 (q 2 ) and F2 (q 2 ). From Eq. (8.19) we then identify

′ iσ µν qν M′f i = −qc ūs (p′ ) γ µ F1 (q 2 )+ F2 (q 2 ) us (p) õ (q) .

(8.20) 2m Returning to Eq. (7.3.32) we see that the initial state contains only a single particle and so one factor of 2Ep V is absent in the denominator. The external field does not change and so is irrelevant in this. Adapting the previous arguments for squaring the invariant amplitude with a δ-function we arrive at a new form for the numerator of Eq. (7.3.38), which is ∆T 2πδ(Ef − Ei )|M′f i |2 , which means that we still arrive at a factor ∆T /V . There is a subtlety here that every particle scatters at least slightly from an electromagnetic field and so we should really frame the derivation in terms of the differential cross-section. In place of Eq. (7.3.30) we have dP = (∆T /V )vdσ. So the resulting differential cross section is as claimed dσ =

(2π)δ(Ep′ −Ep )|M′f i |2 d3 p′ , (2π)3 2Ep′ |v|2Ep

(8.21)

where p = γmv and i → f means p → p′ .

8.6 Problem 6 Problem: Consider the pseudoscalar Yukawa theory in Eq. (7.6.11) with a ϕ4 interaction, λ 4 L = 21 (∂µ ϕ∂ µ ϕ−m2ϕ ϕ2 )+ ψ̄(i̸ ∂ − m)ψ− 4! ϕ − igϕψ̄γ 5 ψ .

(8.22)

[Hint: This problem requires some care and effort.] (a) Calculate the one-loop self-energy for the scalar and for the fermion and write down the M S counterterms; (b) Calculate the M S one-loop β-functions for both couplings, βλ (g, λ) and βg (g, λ).

Solution: We begin by combining the discussions on Sec. 8.7.1 and Sec. 8.7.2 using dimensional regularization in the M S-scheme. Then Eq. (8.23) generalizes to L = 21 ∂µ ϕ0 ∂ µ ϕ0 − 12 m2ϕ0 ϕ20 + ψ̄0 (i̸ ∂ − m0 )ψ0 − λ4!0 ϕ40 − ig0 ϕ0 ψ̄0 γ 5 ψ0 1 2

µ

1 2

= Zϕ ∂µ ϕ∂ ϕ− Zϕ mϕ0 ϕ

2

+Z2 ψ̄(i̸ ∂ − m0 )ψ−(λ0 /4!)Zϕ2 ϕ4 − iZ2

(8.23) p

Zϕ g0 ϕψ̄γ 5 ψ

= 12 Zϕ ∂µ ϕ∂ µ ϕ− 12 Zϕ mϕ0 ϕ2 +Z2 ψ̄(i̸ ∂ − m0 )ψ−(µϵ λ/4!)Z1λ ϕ4 − iZ1g µϵ/2 gϕψ̄γ 5 ψ

= 21 (∂µ ϕ∂ µ ϕ−m2ϕ ϕ2 )+ ψ̄(i̸ ∂ − m)ψ−(µϵ λ/4!)ϕ4 − iµϵ/2 gϕψ̄γ 5 ψ n o + 21 δz ∂µ ϕ∂ µ ϕ− 21 δmϕ ϕ2 +iδ2 ψ̸̄ ∂ψ − δm ψ̄ψ−(µϵ δ1λ /4!)ϕ4 −iδ1g µϵ/2 gϕψ̄γ 5 ψ ,

where ϕ0 =

p

Zϕ ϕ, ψ0 =

√

Z2 ψ, Z2

p Zϕ g0 = Z1g µϵ/2 g, Zϕ2 λ0 = Z1λ µϵ λ, Zϕ m2ϕ0 =

Problem 6

183

p m2ϕ + δmϕ , Z2 m0 = m + δm, Zϕ2 λ0 = µϵ λ + µϵ δ1λ , and Z2 Zϕ g0 = µϵ/2 g(1 + δ1g ).

2 2 It follows that m0 = Zm m with Zm ≡ (1 + δm m )/Z2 , mϕ0 ≡ Zmϕ mϕ with Zmϕ ≡ δ

(1 + mmϕ 2 )/Zϕ , Z1g = 1 + δ1g and Z1λ = 1 + (δ1λ /λ). Then we have for example,

w

ϕ

z

Z1g (µ, ϵ) p µϵ/2 g(µ), Z2 (µ, ϵ) Zϕ (µ, ϵ) y Z (µ, ϵ) x 1λ λ0 (ϵ) ≡ Zλ (µ, ϵ)µϵ λ(µ) = 2 µϵ λ(µ).

Zϕ (µ, ϵ) g0 (ϵ) ≡ Zg (µ, G ϵ)µc ϵ/2 g(µ) =

(8.24) (8.25)

Note that comparing with Eq. (8.23) we have defined δλ → δ1λ , since here we prefer to later define δλ to mean something else. (a) There are two>1_1y one-loop contributions to the scalar self-energy and a countert>1_1y erm,

w

z Gc

x

w

Gc −iΠ(p )= x

y

z 2

=

+

+

,

(8.26)

y

where we keep only the O(λ) part of Eq. (8.7.7). We have already calculated >1_1y the first diagram on the right-hand side in Eq. (8.6.19), with the result that " ! # im2ϕ λ 2 m2ϕ ` −γ = +ln(4πe )−ln +O(ϵ) .

(8.27) 32π 2 ϵ µ2 e p p >1_1R 5 Using the vertex in Eq. (7.6.15), iLYuk int = gγ , the second diagram gives a result >1_1R very similar to that in Eq. (8.7.18),

`+p Z d d ℓ tr γ 5 i(̸ p+̸ ℓ+m)γ 5 i(̸ ℓ+m) ϵ/2 2 9 `+p = (−1)(µ g) (8.28) `+p (2π)d [(p+ℓ)2 −m2 +iϵ][ℓ2 −m2 +iϵ] ` Z d >1_1R `+p 4[−(p+ℓ)·ℓ+m2 ] d ℓ ϵ 2 =µ g ` (2π)d [(p+ℓ)2`−m2 +iϵ][ℓ2 −m2 +iϵ] Z 1 Z d m2 −(p+ℓ)·ℓ d ℓ `+p = 4µϵ g 2 >1_1k dx d 2 2 2 2 ` ` Z0 Z (2π) {ℓ + 2xp·ℓ+xp −m +iϵ} `+p 1 m2 +x(1−x)p2 −ℓ′2 p= 4µϵ g 2 pdx dd ℓ′ >1_1k ` 9 (2π)d {ℓ′2 +x(1−x)p2 −m2 +iϵ}2 0 Z 1 Z d ′ ` d ℓ 2m2 −∆−ℓ′2 = 4µϵ g 2 dx d ′2 2 (2π) {ℓ −∆+iϵ} 0 >1_1k )

Z 1 ( (2m2 −∆)Γ(2− d2 ) d2 Γ(1 − d2 ) i ϵ 2 9 = 4µ g dx + 1−(d/2) , (4π)d/2 0 ∆2−(d/2) ∆ Z 1 n −1 o Γ(2− d2 ) 1 = 4ig 2 dx (2m2 −∆)+ d2 1− d2 ∆ (4π)d/2 (∆/µ2 )2−(d/2) 0 Z 1 n o −1 = 4ig 2 dx (2m2 −∆) + d2 1− d2 ∆ 0

8

Symmetries and Renormalization

184

1 2 −γ 2 +ln(4πe )−ln(∆/µ )+O(ϵ) (4π)2 ϵ Z 1 n o ϵ 2 g2 ϵ dx − 1+ m + 3+ x(1−x)p2 +O(ϵ2 ) = 4i 2 (4π) 0 2 2

2 × +ln(4πe−γ )−ln(∆/µ2 )+O(ϵ) ϵ

2 Z 1

g −m2 +3x(1−x)p2 dx = 4i 2 (4π) 0

2 × +ln(4πe−γ )−ln(∆/µ2 ) −m2 +x(1−x)p2 +O(ϵ), ϵ ×

>1_1y

where we have defined ℓ′ = ℓ + xp and ∆ ≡ m2 − x(1 − x)p2 , where we have used analogous steps to those used in Eq. (8.7.18) and where we have used Eq. (A.6.17). From Eq. (8.7.16) the counterterm is = i(p2 δzϕ −δmϕ ).

(8.29)

In the M S-scheme we use the counterterm to cancel just the 2/ϵ divergences, whereas in the M S-scheme it cancels terms proportional to [(2/ϵ) + ln(4πe−γ )].

So isolating these contribution in Eq. (8.28) we have

Z 1

g2 2 2 2 −γ 4i dx −m +3x(1−x)p +ln(4πe ) (8.30) (4π)2 0 ϵ

g2 p2 2 = i 2 −m2 + +ln(4πe−γ ) , 4π 2 ϵ R1 since 0 dx x(1 − x) = 16 . So in the M S-scheme we have arrived at ( )

2 2 2 λm g2 2 g m 2 ϕ −γ −γ `+p δzϕ = − 2 +ln(4πe ) , δmϕ = − +ln(4πe ) . (8.31) 8π ϵ 32π 2 4π 2 ϵ

9

>1_1R

`+p

The fermion self-energy up to one-loop has only one diagram plus the counterterm,`

dd ℓ γ 5 i(̸ p+̸ ℓ+m)γ 5 i ϵ/2 2 = p(µ g) p 1 2 p0 = + pi pn i(̸ pδ2 −δm ) p i+1 2 1 i(2π) 1 d (p+ℓ) 2p−m +iϵ ℓn2 −m ϕ +iϵ Z d ̸ p+̸ ki+1ℓ−m k=1 µϵ g 2 d ℓ ki kn + i(̸ pδ2 −δm ) (2π)d [(p+ℓ)2 −m2 +iϵ][ℓ2 −m2ϕ +iϵ] Z 1 Z d d ℓ ̸ p+̸ ℓ−m = µϵ g 2 dx + i(̸ pδ2 −δm ) d 2 2 (2π) {ℓ +2xp·ℓ+xp −xm2 +(1−x)m2ϕ +iϵ}2 0 q Z 1 Z d ′ p1 pi ℓ pi +q pi+1 +q d (1−x)̸ p+̸ ℓ ′ −mp0 +q ϵ 2pi 1 =µ g dx + i(̸ pδ2 −δm ) (2π)d {ℓ′2 +x(1 − x)p2 −xm2 −(1−x)m2ϕ +iϵ}2 0 Z 1 Zki d ′ ki+1 k1 k d ℓ (1−x)̸ p−m n = µϵ g 2 dx + i(̸ pδ2 −δm ) (2π)d {ℓ′2 +x(1 − x)p2 −xm2 −(1−x)m2ϕ +iϵ}2 0

Gc

w

z

p

y

p

(8.32)

Z

x

>1_1k

+ `

− iΣ(p) =

Problem 6

185

{m−(1−x)̸ p}Γ(2− d2 ) + i(̸ pδ2 −δm ) 2 {xm +(1−x)m2ϕ −x(1−x)p2 }2−(d/2) 0

Z 1 Γ(2− d2 ) 1 + i(̸ pδ2 −δm ) = −ig 2 dx {m−(1−x)̸ p} d/2 (4π) (∆Σ /µ2 )2−(d/2) 0

Z 1 2 g2 ∆Σ −γ dx{m−(1−x)̸ p} +ln(4πe )−ln 2 +O(ϵ) +i(̸ pδ2 −δm ) , = −i (4π)2 0 ϵ µ

= −µϵ g 2

i (4π)d/2

Z 1 dx

where we have defined ℓ′ = ℓ + xp and ∆Σ ≡ xm2 +(1−x)m2ϕ −x(1−x)p2 . In the M S-scheme we then find

g2 2 g2 m 2 −γ −γ δ2 = − +ln(4πe ) , δ = − +ln(4πe ) , m 32π 2 ϵ 16π 2 ϵ R1 since 0 dx(1 − x) = 12 . (b) To extract the β functioms we also need the one-loop counterterms for the vertices in the M S-scheme. We do not need to calculate the full renormalized vertices at one-loop. It is sufficient to identify the divergent (2/ϵ) parts of the one-loop vertices and we can then easily obtain the δ1g and δλ in either the M S- or M S-schemes. The divergent contributions can be obtained by simply ignoring all masses and external momenta relative to the loop momentum ℓ in the one-loop integrals.

The renormalized Yukawa proper (1PI) vertex is analogous to that in QED. >1_1k >1_1k Recall that in dimensional regularization the tree-level vertex is µϵ/2 gγ 5 . In analogy with QED we have in place of Eq. (8.7.32),

q+` q+` (µϵ/2 g)Λ2 (p′ , p) =

Z

= (µϵ/2 g)3

qq

pp0 0 `` pp +

``

pp11

pp22

pp44

pp33

(8.33)

pp

d

i γ 5 i(̸ q+̸ ℓ+m)γ 5 i(̸ ℓ+m)γ 5 d ℓ + µϵ/2 γ 5 δ1g (2π)d (ℓ−p)2 −m2ϕ +iϵ [(ℓ+q)2 −m2 +iϵ][ℓ2 −m2 +iϵ]

Z d d ℓ {i(̸ q+̸ ℓ−m)(̸ ℓ+m)γ 5 } = (µϵ/2 g)3 + µϵ/2 γ 5 δ1g d 2 (2π) [(ℓ−p) −m2ϕ +iϵ][ℓ′2 −m2 +iϵ][ℓ2 −m2 +iϵ] Z 1 Z d {· · · } d ℓ + µϵ/2 γ 5 δ1g g = 2(µϵ/2 g)3 [dxdydz δ(1−x−y−z)] d (2π) [xA1 +yA2 +zA3 ]3 0 Z 1 Z d d k i[(k 2 +O(k 0 )] = 2(µϵ/2 g)3 γ 5 [· · · ] + µϵ/2 γ 5 δ1g g d [k 2 − ∆ + iϵ]3 (2π) 0 ( ) Z 1 i(−1)3 d Γ(2− d2 ) ϵ/2 3 5 i +finite terms +µϵ/2 γ 5 δ1g g = −2(µ g) γ [· · · ] (4π)d/2 2 2∆2−(d/2) 0

2−(d/2) Z 1 Γ(2− d2 ) µ2 d ϵ/2 3 5 +finite terms +µϵ/2 γ 5 δ1g g = −2µ g γ [· · · ] d/2 ∆/µ2 4 (4π) 0

Z 1 1 d 2 ϵ/2 3 5 = −2µ g γ [· · · ] +finite terms +finite terms +µϵ/2 γ 5 δ1g g (4π)2 ϵ 4 0

Symmetries and Renormalization

186

Z 1 2 g2 +finite terms +µϵ/2 γ 5 δ1g g = µϵ/2 gγ 5 − [· · · ] 2 8π ϵ 0

2 g 2 = µϵ/2 gγ 5 − +finite terms+δ 1g , 16π 2 ϵ where in analogy with the QED case we have k = ℓ + yq − zp and ∆ ≡ −2yzq·p−y(1−y)q 2 −z(1−z)p2 +(1−z)m2 +zm2ϕ , and where we have used Eqs. (A.6.5), (A.6.17) and R1 R1 R 1 R 1−x R1 [· · · ] = 0 dxdydz δ(1−x−y−z) = 0 dx 0 dy = 0 dx(1−x) = 21 .

0 ϵ

(8.34)

(8.35)

2

Note that µ = exp(ϵ ln µ) = 1 + ϵ ln µ + O(ϵ ). So in the M S-scheme we have arrived at

g2 2 −γ δ1g = + ln(4πe ) + O(ϵ).

(8.36) 16π 2 ϵ It remains to calculate δλ . At tree level on dimensional regularization the ϕ4 vertex is >1_1R −iµϵ λ arising from Eqs. (7.6.10) and (8.6.14).w There arez two contri>1_1k butions to the 1PI (proper) ϕ4 vertex as one-loop. Including the counterterm, Gc −iδλ , we have for the one-loop contribution q+`

`+p x

p0

`+p `

(−iµϵ λ)Γ2 (p2`) =p q

`

p

+

p1

p2

p4

p3

+

y

.

(8.37)

`

The first diagram on the right-hand side was evaluated in Eq. (8.7.12) and as we saw in>1_1k the following discussion there are three topologically distinct ways to attach the four external lines to the loop. We then have a divergent contribution to δλ from this scalar-loop diagram of δ1λ (scalar loop) = (λ2 /32π 2 )(6ϵ) = (3λ2 /32π 2 )(2/ϵ).

(8.38)

We only need retain the most divergent part of the second diagram to evaluate its contribution to δλ , q+` q

p0 ` p

`

p

p1

p2

p4

p3

(8.39)

Z d 9 d ℓ tr[γ 5 i(̸ ℓ+· · · )γ 5 i(̸ ℓ+· · · )γ 5 i(̸ ℓ+· · · )γ 5 i(̸ ℓ+· · · )] = (−1)(µϵ/2 g)4 (2π)d [(ℓ+· · · )2 −· · · ][(ℓ+· · · )2 −· · · ][(ℓ+· · · )2 −· · · ][ℓ2 −· · · ] Z 1 Z d d ℓ 4(ℓ2 )2 +· · · = −6(µϵ/2 g)4 dx1 · · · dx4 δ(1−x1 −· · ·−x4 ) d 2 (2π) {ℓ +2X ·ℓ+· · · )}4 0 Z 1 Z d ′ d ℓ (ℓ′2 )2 +O(ℓ′2 ) = −24(µϵ/2 g)4 dx1 · · · dx4 δ(1−x1 −· · ·−x4 ) (2π)d {ℓ′2 − ∆ + iϵ}4 0 Z 1 Γ(2 − d2 ) + O(ϵ) i = −24µϵ g 4 dx1 · · · dx4 δ(1−x1 −· · ·−x4 ) (4π)d/2 (∆/µ2 )2−(d/2) 0

e

187

Problem 6

Z 1 2 i +finite terms = −24µϵ g 4 dx1 · · · dx4 δ(1−x1 −· · ·−x4 ) (4π)2 ϵ 0

g4 2 = −i 2 +finite terms , 4π ϵ

(8.40)

where we have (−1) from the fermion loop, where we have used Eqs. (A.5.6), where we do not need to evaluate the O(ℓ0 ) term X µ or the O(ℓ′0 ) term ∆ here, where we have done a change of variable ℓ′ = ℓ + X, where we have used Eqs. (A.6.7) and (A.6.17), and where we have used R1 R1 R 1−x R 1−x −x dx1 · · · dx4 δ(1−x1 −· · ·−x4 ) = 0 dx1 0 1 dx2 0 1 2 dx3 1 (8.41) 0 R1 R 1−x1 R1 R 1−x1 1−x = 0 dx1 0 dx2 (1 − x1 − x2 ) = 0 dx1 0 [x − 2(1 − x1 ) − 21 x22 ]0 1 R1 R 1−x1 R1 R 1−x1 = 0 dx1 0 dx2 (1 − x1 − x2 ) = 0 dx1 0 [(1 − x1 )2 − 21 (1 − x1 )2 ] R 1 = 21 0 dx1 [1 − 2x1 + 12 x21 ] = 12 [x1 − x21 + 13 x31 ]10 = 16 .

It is useful to note that there is a shortcut for the above calculation, which arises from the observation that to obtain the divergent behavior it is sufficient to set all external momenta to zero. Then we have Z d d ℓ tr[γ 5 i(̸ ℓ+m)γ 5 i(̸ ℓ+m)γ 5 i(̸ ℓ+m)γ 5 i(̸ ℓ+m)] = (−1)(µϵ/2 g)4 (8.42) (2π)d [ℓ2 −m2 +iϵ]4 Z d d ℓ tr[(−̸ ℓ+m)(̸ ℓ+m)(−̸ ℓ+m)(̸ ℓ+m)] ϵ/2 4 = (−1)(µ g) (2π)d [ℓ2 −m2 +iϵ]4 Z d Z d 2 2 2 1 d ℓ d ℓ 4(ℓ − m ) ϵ/2 4 ϵ/2 4 = −4(µ g) = (−1)(µ g) (2π)d [ℓ2 −m2 +iϵ]4 (2π)d [ℓ2 −m2 +iϵ]2 2 2−(d/2) 1 µ d = −4iµϵ g 4 Γ(2 − ) 2 m2 (4π)d/2

2

2 m g4 2 g4 −γ +ln(4π )−ln +O(ϵ) = −i +finite terms , = −4i (4π)2 ϵ µ2 4π 2 ϵ where we have used Eqs. (A.6.4) and (A.6.17). So this confirms the result.

There are six topologically distinct ways to assign the four external momenta to vertices on the loop and each of these will have the same divergent behavior.

We then have a divergent contribution to δλ from this fermion-loop diagram of δλ (fermion loop) = −

3g 4 2 .

2π 2 ϵ

So in the M S-scheme we have arrived at

3g 4 3λ2 2 −γ δλ = − 2 + + ln(4πe ) .

2π 32π 2 ϵ

(8.43)

(8.44)

We now have what we need to construct the two β-functions from the relevant form of the Callan-Symanzik equation. From Eq. (8.24) and the analog of Eq. (8.7.65) we have

Symmetries and Renormalization

188

g0 (ϵ) ≡ Zg (µ, ϵ)µϵ/2 g(µ),

q Zg (µ, ϵ) ≡ Z1g (µ, ϵ)/Z2 (µ, ϵ) Zϕ (µ, ϵ) = 1 + δ1g − δ2 − 12 δzϕ 2

g g2 g2 2 −γ = 1+ + + + ln(4πe ) 16π 2 32π 2 16π 2 ϵ

5g 2 2 + ln(4πe−γ ) . = 1+ 32π 2 ϵ

(8.45) (8.46)

(8.47)

The analog of Eq. (8.7.65) is then

∂ ∂ ϵ µ ∂Zg µ dg g0 = µ [Zg µϵ/2 g] = µϵ/2 Zg g + + ∂µ ∂µ 2 Zg ∂µ g dµ dg ϵ µ ∂Zg ⇒ βg (g, λ) ≡ µ = − g−g . dµ 2 Zg ∂µ

0=µ

(8.48)

Note that Zg = 1 + O(g 2 ) and so to leading order in g 2 we have from the above equation that µ(∂g/dµ) = − 2ϵ g + O(g 3 ) and so g

µ ∂Zg dg 1 ∂Zg dg µ ∂Zg =g =g µ Zg ∂µ Zg ∂g dµ Zg ∂g dµ

ϵ 1 10g 2 −γ 3 − =g + ln(4πe ) g+O(g ) 1+O(g 2 ) 32π 2 ϵ 2 5g 3 =− +O(g 5 )+O(ϵ).

16π 2

(8.49)

Note that it might seem strange to write Zg = 1 + O(g 2 ) when the corefficient of the g 2 terms contains 2/ϵ. But recall that infinities are canceled order by order in perturbation theory in a renormalizable theory and so we do not need to concern ourselves with tracking infinities that occur at higher order in g.

Then we have ϵ 5g 3 5g 3 ϵ→0 βg (g, λ) = − g+ +O(g 5 )+O(ϵ) −−−→ +O(g 5 ), (8.50) 2 2 16π 16π 2 which is independent of λ.

The corresponding calculation for βλ (g, λ) is a little more complicated but proceeds in the analogous way. From Eq. (8.25) we have Z1λ (µ, ϵ) ϵ µ λ(µ), (8.51) Zϕ2 (µ, ϵ)

δλ 3g 4 3λ g2 2 −γ Zλ (µ, ϵ) = 1 + − 2δzϕ = 1+ − + + 2 + ln(4πe ) . λ 2λπ 2 32π 2 4π ϵ (8.52) λ0 (ϵ) ≡ Zλ (µ, ϵ)µϵ λ(µ) =

The analog of Eq. (8.7.65) in this case is

∂ µ ∂Zλ µ dλ ∂ ϵ ϵ + 0 = µ λ0 = µ [Zλ µ λ] = µ Zλ λ ϵ+ ∂µ ∂µ Zλ ∂µ λ dµ

(8.53)

Problem 7

189

⇒

βλ (g, λ) ≡ µ

dλ µ ∂Zλ = −ϵλ−λ . dµ Zλ ∂µ

We have Zλ = 1 + O(g 2 ) + O(λ), where we have seen from above that one loop corrections are order O(g 2 ) and O(λ). So to one-loop order we have from the above equation that µ(∂λ/dµ) = −ϵλ + O(g 3 ) + O(λg 2 ) and so

1 ∂Zλ dg ∂Zλ dλ µ ∂Zλ =λ µ + µ (8.54) λ Zλ ∂µ Zλ ∂g dµ ∂λ dµ

1 g ϵ 6g 3 =λ + − g+O(g 3 ) − 2 2 2 1 + O(g ) + O(λ) λπ 2π 2

4

3 3g 3 2 + −ϵλ + O(g ) + O(λg ) + λ2 2π 2 32π 2

2 × + ln(4πe−γ ) .

(8.55) ϵ Then we have

g 3 6g 3 3g 4 βλ (g, λ) −−−→ −λ − 2 + 2 (−g) + + (−2λ) λπ 2π λ2 2π 2 32π 2 6g 4 λg 2 3g 4 3λ2 3g 4 λg 2 3λ2 =− 2 + 2+ 2 + = − + + . π 2π π 16π 2 π 2 2π 2 16π 2

ϵ→0

(8.56) (8.57)

which depends on g 2 and λ.

8.7 Problem 7 Problem: Calculate the one-loop contribution to the γγγ 1PI vertex in QED. Show that it vanishes as expected from Furry’s theorem and so needs no renormalization.

(Hint: Eq. (A.3.29) may prove useful.)

Solution: The calcuation is very similar to that for the one-loop vacuum polarization of the photon except there is one additional vertex and one additional fermion propagator. We will not bother with dimensional regularization or counterterms here, since the two contributions will be shown to cancel exactly and so this is not necessary. Consider one of the two one-loop contributions, µ q1 ` q3 = −q1 − q2 ρ

` + q1

(8.58)

q2 ` + q1 + q2

ν

Symmetries and Renormalization

190

Z d tr[γ ρ i(̸ ℓ+̸ q 1 +̸ q 2 +m)γ ν i(̸ ℓ+̸ q 1 +m)γ µ i(̸ ℓ+m)] d ℓ = (−1)(−iqc )3 , d (2π) [(ℓ+q1 +q2 )2 −m2 +iϵ][(ℓ+q1 )2 −m2 +iϵ][ℓ2 −m2 +iϵ] where the fermion loop gives the factor (−1) and the spinor trace is ordered right to left in the direction of the fermion arrow. The only other one-loop contribution is the fermion loop with the opposite direction. In order to have the convenience of having the same denominator in the loop integral we simply reverse all of the fermion momenta to give µ q1 −(` + q1)

−` q3 = −q1 − q2

(8.59)

q2 −(` + q1 + q2)

ρ

Z

= (−1)(−iqc )3

ν

dd ℓ tr[γ µ i(−̸ ℓ−̸ q 1 +m)γ ν i(−̸ ℓ−̸ q 1 −̸ q 2 +m)γ ρ i(−̸ ℓ+m)] .

(2π)d [(ℓ+q1 +q2 )2 −m2 +iϵ][(ℓ+q1 )2 −m2 +iϵ][ℓ2 −m2 +iϵ]

So if the sum of the two traces vanishes then the one-loop contribution vanishes.

We know from Eq. (A.3.24) that the trace of an odd number of γ-matrices vanishes.

So O(m) and O(m3 ) terms vanish in the traces, which leaves O(m0 ) and O(m2 ) contributions. This means that we can reverse the sign of m in the second trace and extract a factor (−1)3 = −1 by changing the sign of the three propagator numerators. Using this observation we find that tr[γ ρ i(̸ ℓ+̸ q 1 +̸ q 2 +m)γ ν i(̸ ℓ+̸ q 1 +m)γ µ i(̸ ℓ+m)] µ

ν

(8.60)

ρ

+ tr[γ i(−̸ ℓ−̸ q 1 +m)γ i(−̸ ℓ−̸ q 1 −̸ q 2 +m)γ i(−̸ ℓ+m)]

= −i {tr[γ ρ (̸ ℓ+̸ q 1 +̸ q 2 +m)γ ν (̸ ℓ+̸ q 1 +m)γ µ (̸ ℓ+m)]

−tr[γ µ (̸ ℓ+̸ q 1 +m)γ ν (̸ ℓ+̸ q 1 +̸ q 2 +m)γ ρ (̸ ℓ+m)]}

= −i {tr[γ ρ (̸ ℓ+̸ q 1 +̸ q 2 +m)γ ν (̸ ℓ+̸ q 1 +m)γ µ (̸ ℓ+m)]

−tr[(̸ ℓ+m)γ µ (̸ ℓ+̸ q 1 +m)γ ν (̸ ℓ+̸ q 1 +̸ q 2 +m)γ ρ ]}

= −i {tr[γ ρ (̸ ℓ+̸ q 1 +̸ q 2 +m)γ ν (̸ ℓ+̸ q 1 +m)γ µ (̸ ℓ+m)] = 0,

−tr[γ ρ (̸ ℓ+̸ q 1 +̸ q 2 +m)γ ν (̸ ℓ+̸ q 1 +m)γ µ (̸ ℓ+m)]}

where in the third step we used the cyclic property of the trace to move (̸ ℓ + m) in the second trace and in the fourth step we used Eq. (A.3.29) that states that the trace is invariant under a reversal of the order of all γ-matrices.

Note that if we consider any odd number of vertices then we will have an equal odd number of propagators. Again it follows that all odd powers of the mass vanish since that implies we have an odd number of γ-matrices in the trace. Then only even powers of the mass survive and so we can again reverse the sign of m in the second trace. We then extract an odd number of minus signs from the odd number

Problem 8

191

of propagators, which means that the second trace acquires an overall minus sign.

Then we can use the cyclic property of the trace and again reverse all γ-matrices in the second trace. The two resulting traces are then equal and hence cancel. So the loop together with its opposite orientation vanishes for any odd number of photon vertices in QED. This is then a direct proof of Furry’s theorem starting from the momentum-space Feynman rules. Furthermore, with an even number of vertices and propagators it again follows that only even powers of m survive in the traces.

So in the second trace we pull out an even number of minus signs and the rest of the tricks still apply. So rather than the loops in opposite directions canceling, for an even number of vertices the two resulting traces are equal and add together as noted in the proof of Furry’s theorem.

8.8 Problem 8 Problem : In the discussion of the Casimir effect we placed the parallel plates in the middle of the box as shown in Fig. 8.4. Show that the Casimir effect depends only on the plate separation a and not on plate location as L, A → ∞.

Solution: Let us divide the length L as L = (ℓ − 21 a) + a + (L − ℓ − 12 a), which then gives before UV regularization that ℓ−(a/2)

δE = [E0a + E0

(L−ℓ)−(a/2)

+ E0

] − E0L ,

where for any very large ℓ we have from Eq. (8.9.5) that p R∞ R∞ E0ℓ → (ℓ/π)(Aℏc/2π) 0 kdk 0 dkz k 2 +kz2 .

(8.61)

(8.62)

The key observation is simply that for large ℓ we have E0ℓ ∝ ℓ. We see that as ℓ, L → ∞ we have p R∞ R∞ ℓ−(a/2) (L−ℓ)−(a/2) E0 + E0 −E0L → −(a/π)(Aℏc/2π) 0 kdk 0 dkz k 2 +kz2 , (8.63) which is independent of ℓ. We also showed that for finite a R∞ P∞ E0a = (Aℏ/2π) 0 kdk n=(0)1 ωna (k 2 ),

(8.64)

and so we again find the last line of Eq. (8.9.5), i h (L−ℓ)−(a/2) ℓ−(a/2) δE = E0a + E0 + E0 − E0L (8.65) nP o √ √ R R

R ∞ ∞ ∞ ∞ = (Aℏcπ 2 /4a3 ) du u+n2 − 0 dn 0 du u+n2 . n=(0)1 0 The remainder of the arguments go through exactly as before and so we again find δE = −Aℏcπ 2 /720a3

⇒

F/A = −d(δE/A)/da = ℏcπ 2 /240a4 .

(8.66)

9

Gauge Field Theories 9.1 Problem 1 Problem: (a) Show that Noether’s theorem for global SU (3)color symmetry in QCD leads abc b c a a to the conserved color current R 3 ajµ = ψ̄i γµ Tij ψj − f Aν Fµν and hence that a a dQ /dt = 0 where Q = d x j0 in the usual way. a (b) Show that the quark component of the color current jqµ ≡ ψ̄i γµ Tija ψj satisfies µ a µbc c D jqµ = 0 and hence that ∂ jqµ ̸= 0. Explain why QED-like Ward-Takahashi and Ward identities do not apply in QCD. Solution: (a) The Lagrangian of QCD with a single quark field ψ is 1 a µν / − m)ψ L = − Fµν Fa + ψ̄(iD 4 1 a µν = − Fµν Fa + ψ̄(iγ µ ∂µ ψ + gγ µ Aaµ Ta ψ − mψ), 4

(9.1)

where Ta (a = 1, . . . , 8) are generators of SU (3) in the fundamental representation. Under an infinitesimal SU (3)color transformation parameterized by αa , we know that the quark field, being in the fundamental representation, transforms as ψ → ψ + iαa T a ψ,

(9.2)

while the gauge fields transform as 1 Acν → Acν + ∂ν αc − f cab αa Abν , g

(9.3)

by Eq. (9.1.4) with a slight change of indices. Note in particular that δψ = iT a ψ, δαa δAcν = −f cab Abν = −f abc Abν , δαa

(9.4)

by antisymmetry of the structure constants f abc . Now since SU (3)color is an 192

Problem 1

193

internal symmetry of this Lagrangian, we have conserved currents given by Eq. (3.2.18), jaµ = −

∂L δψ ∂L δAcν − , a ∂(∂µ ψ) δα ∂(∂µ Acν ) δαa

(9.5)

including an overall factor of −1 for future convenience. Eq. (3.3.5) in the text carries over to the nonabelian case, so it is easy to see from Eq. (9.1) that our currents evaluate to jaµ = −(ψ̄iγ µ )(iTa ψ) − (−Fcµν )(−f abc Abν ) = ψ̄γ µ Ta ψ − f abc Abν Fcµν .

(9.6)

c With the colour indices explicit, this is jµa = ψ̄i γµ Tija ψj −f abc Abν Fµν , as desired. µ These currents are guaranteed to satisfy ∂µ ja = 0 when the equations of motion are satisfied, by Noether’s theorem. Written another way,

1 ∂ja0 + ∇ · ja = 0. c ∂t R Hence, if we define Qa := d3 x j0a , then Z Z Z d ∂j a dQa = d3x j0a = d3x 0 = −c d3x ∇ · ja = 0, dt dt ∂t

(9.7)

(9.8)

since Gauss’ divergence theorem turns this into a surface integral on the boundary of space, which vanishes for a physical current. (b) Isolating the quark component of the current, jqµc ≡ ψ̄γ µ T c ψ, we have Dµbc jqµc ≡ (∂µ δ bc − gf bca Aaµ )jqµc

= ∂µ (δ bc ψ̄γ µ T c ψ) − gf bca ψ̄γ µ Aaµ T c ψ

= (∂µ ψ̄)γ µ T b ψ + ψ̄γ µ T b (∂µ ψ) − gf abc ψ̄γ µ Aaµ T c ψ,

(9.9)

having used the antisymmetry of f abc again. To proceed, we need to use the equations of motion. The Euler-Lagrange equation for ψ is

∂L ∂L 0 = ∂µ = ∂µ (iψ̄γ µ ) − (g ψ̄γ µ Aaµ T a − mψ̄), (9.10) − ∂(∂µ ψ) ∂ψ implying (∂µ ψ̄)γ µ = −iψ̄(gγ µ Aaµ T a − m). A similar calculation can be done for the ψ̄ field provided we write the Lagrangian in terms of ∂µ ψ̄ rather than ∂µ ψ.

A simple integration by parts allows us to replace ψ̄(∂µ ψ) → −(∂µ ψ̄)ψ in the Lagrangian, since the four-divergence term has no physical effect. Explicitly, 1 a µν L = − Fµν Fa − i(∂µ ψ̄)γ µ ψ + ψ̄(gγ µ Aaµ Ta − m)ψ.

4 The Euler-Lagrange equation then gives

∂L ∂L 0 = ∂µ − = ∂µ (−iγ µ ψ) − (gγ µ Aaµ T a − m)ψ, ∂(∂µ ψ̄) ∂ ψ̄

(9.11)

(9.12)

Gauge Field Theories

194

or γ µ ∂µ ψ = i(gγ µ Aaµ T a − m)ψ. With these equations of motion, our covariant derivative becomes Dµbc jqµc = −iψ̄(gγ µ Aaµ T a − mψ̄)T b ψ + ψ̄T b i(gγ µ Aaµ T a − m)ψ − gf abc ψ̄γ µ Aaµ T c ψ = −ig ψ̄γ µ Aaµ [T a , T b ]ψ − gf abc ψ̄γ µ Aaµ T c ψ

= −ig ψ̄γ µ Aaµ (if abc T c )ψ − gf abc ψ̄γ µ Aaµ T c ψ = 0,

(9.13)

by definition of the structure constants, [T a , T b ] = if abc T c . Hence, ∂µ jqµa ≡ Dµab jqµb + gf abc Acµ jqµb = gf abc Acµ jqµb ̸= 0.

The QED Ward-Takahashi and subsequent Ward identities are derived from the conservation of the fermion charged current. Since the fermion color-charge current in QCD is not conserved then they will not apply to QCD.

9.2 Problem 2 Problem: Calculate the M S counterterm for quark-gluon vertex in an arbitrary covariant gauge and hence obtain the result for δ1 given in Eq. (9.2.28).

Solution: Shortcuts: To reduce the tedium of these calculations it is useful to note that we can isolate the divergence using 2 2−(d/2)

Z d µ i 1 i 2 d ℓ ϵ d Γ(2 − 2 ) = +· · · , (9.14) µ = (2π)d [ℓ2 − ∆]2 ∆ (4π)2 ϵ (4π)d/2 2−(d/2)

µν Z d ℓµ ℓν −i(−1)3 g µν Γ(2 − d2 ) µ2 i 2 g d ℓ ϵ µ = = +· · · .

(2π)d [ℓ2 − ∆]3 2 ∆ (4π)2 ϵ 4 (4π)d/2 2 The full 1PI quark-gluon vertex with dimensional regularization has the form ϵ/2 iµϵ/2 gΓaµ gγ µ Tija . Using ij and the tree-level vertex from the Feynman rules is iµ notation similar to that used for QED we then have

ϵ/2 4 iµϵ/2 gΓaµ g γ µ Tija + Λaµ (9.15) ij = iµ ij + O(g ) 2 where we define iµϵ/2 gΛaµ ij to be the O(g ) one-loop contribution. This contribution is

iµϵ/2 gΛau ij =

+

+

where the third diagram is the the counterterm iµϵ/2 gγ µ Tija δ1 .

,

Problem 2

195

Let us begin with the first diagram on the right-hand side, which is the QCD analog of the one-loop contribution to the QED vertex in Eq. (8.7.32).

p0 ρ, c

`+q ′ (−iν ϵ/2 g)Λaµ ij (p , p)

diagram 1

i

µ, a

=

q

`−p, δ bc ν, b

` p

j

The difference from QED is the color matrices, remembering that we order fermion propagators right to left in the direction of the arrow gives a color-factor (T c T a T b )δ bc = T b T a T b = T b T b T a + T b [T a , T b ] = CF T a + if abc T b T c a

1 2

= CF T + f

abc

b

c

a

1 2

[T , T ] = CF T − f

abc bcd

f

d

(9.16)

T = (CF − CA )T a .

1 2

So in place of Eq. (8.7.32) we then have in any Rξ gauge, ′ (iµϵ/2 g)Λaµ ij (p , p)

diagram 1

≡ (CF − 12 CA )T a (iµϵ/2 g)Λµij (p′ , p)

diagram 1

, (9.17)

where (iµϵ/2 g)Λµij (p′ , p)

diagram 1

=

(9.18)

Z d i[−g +(1−ξ) (ℓ−p)ν (ℓ−p)ρ ] νρ γ ρ i(̸ ℓ+̸ q+m)γ µ i(̸ ℓ+m)γ ν d ℓ (ℓ−p)2 . = (iµϵ/2 g)3 d 2 (2π) (ℓ−p) +iϵ [(ℓ + q)2 −m2 +iϵ][ℓ2 −m2 +iϵ] Recall from Eq. (6.4.154) that the 1/(ℓ − p)2 factor in the (1−ξ) term is shorthand for 1/(ℓ − p)2 + iϵ. To obtain the divergent part of the diagram it is sufficient to take the various limits p/ℓ, q/ℓ, mℓ → 0 in the numerator and to set all external momenta to zero in the denominator. Then we have (iµϵ/2 g)Λµij (p′ , p) diagram 1

Z d d ℓ (1−ξ)ℓ2 γ µ ℓ2 +· · · γν ̸ ℓγ µ ̸ ℓγ ν +· · · ϵ/2 3 = (µ g) − 2 (2π)d [ℓ2 +iϵ][ℓ2 −m2 +iϵ]2 [ℓ +iϵ]2 [ℓ2 −m2 +iϵ]2

Z d ϵ µ −2(1 − 2 )̸ ℓγ ̸ ℓ+· · · d ℓ (1−ξ)γ µ +· · · ϵ/2 3 = (µ g) − 2 (2π)d [ℓ2 +iϵ][ℓ2 −m2 +iϵ]2 [ℓ −m2 +iϵ]2 Z 1 Z d d k −2̸ kγ µ ̸ k+· · · = (µϵ/2 g)3 2 dxdydz δ(1−x−y−z) (2π)d [k 2 −∆]3 0

Z 1 Z d ′ d k (1−ξ)γ µ +· · · − dxdy δ(1−x−y) (2π)d [k ′2 − ∆′ ]2 0 Z 1 Z d d k kρ kσ +· · · ϵ/2 3 = (µ g) 2 dxdydz δ(1−x−y−z) (−2γ ρ γ µ γ σ ) 2 d (2π) [k −∆]3 0

(9.19)

Gauge Field Theories

196

Z d ′ Z 1 d k (1−ξ)γ µ +· · · dxdy δ(1−x−y) (2π)d [k ′2 − ∆′ ]2 0

Z 1 i 2 gρσ = µϵ/2 g 3 +· · · 2 dxdydz δ(1−x−y−z)2γ ρ γ µ γ σ (4π)2 ϵ 4 0

Z 1 + dxdy δ(1−x−y)(1−ξ)γ µ + · · · −

0

Z 1 g2 2 gρσ +· · · 2 dxdydz δ(1−x−y−z)(−2γ ρ γ µ γ σ ) 2 (4π) ϵ 4 0

Z 1 − dxdy δ(1−x−y)(1−ξ)γ µ + · · ·

= (iµϵ/2 g)

0

Z 1 2 g2 = (iµ gγ ) +· · · 2 dxdydz δ(1−x−y−z) (4π)2 ϵ 0

Z 1 − dxdy δ(1−x−y)(1−ξ) + · · · 0

g2 2 ϵ/2 µ = (iµ gγ ) +· · · [1−(1−ξ)].

(4π)2 ϵ ϵ/2

µ

For the second diagram we have p0 p0 −` ′ (−iµϵ/2 g)Λaµ ij (p , p)

diagram 2

i ρ, c

µ, a

=

.

`

q ν, b

p−` p

j

Remembering to order the fermions right to left in the direction of the fermion arrow we have the color factors 1 i i f abc T c T b = 21 f abc [T c , T b ] = f abc if cbd T d = − f abc f dbc T d = − CA T a . (9.20) 2 2 2 We will need to choose the momenta consistent for the three-gluon vertex consistent with the choice f abc according to the Feynman rules for QCD. In order to keep track of spacetime indices we use a λ between ν and µ and a τ between µ and ρ. So in place of Eq. (8.7.32) we then have in any Rξ gauge, ′ (iµϵ/2 g)Λaµ ij (p , p)

diagram 2

i ≡ − CA T a (iµϵ/2 g)Λµij (p′ , p) , 2 diagram 2

(9.21)

where (iµϵ/2 g)Λµij (p′ , p)

diagram 2 Z d

= (iµϵ/2 g)2 (µϵ/2 g)

(9.22)

d ℓ [g µλ (p′ +ℓ−2p)τ +g λτ (p′ +p−2ℓ)µ +g τ µ (p+ℓ−2p′ )λ ] (2π)d

Problem 2

197

×

(p′−ℓ)τ (p′−ℓ)ρ ν (p−ℓ)λ ] γ ρ i(̸ ℓ+m)γ ν i[−gνλ +(1−ξ) (p−ℓ) ] [(p′−ℓ)2 (p−ℓ)2 . ′ 2 2 2 2 [(p −ℓ) +iϵ] [ℓ −m +iϵ] [(p−ℓ) +iϵ]

i[−gτ ρ +(1−ξ)

We again set all external momenta to zero and retain only leading order in ℓ in the numerator and find (iµϵ/2 g)Λµij (p′ , p) diagram 2 Z d d ℓ [g µλ ℓτ −g λτ 2ℓµ +g τ µ ℓλ ] = i(µϵ/2 g)3 (2π)d

(9.23)

ℓ ℓ

[−gτ ρ +(1−ξ) τℓ2 ρ ] [−gνλ +(1−ξ) ℓνℓ2ℓλ ] γ ρ ̸ ℓγ ν [ℓ2 +iϵ] [ℓ2 −m2 +iϵ] [ℓ2 +iϵ] Z d d ℓ = i(µϵ/2 g)3 [g µλ ℓτ −g λτ 2ℓµ +g τ µ ℓλ ] (2π)d γτ ̸ ℓγλ −(1−ξ)(ℓτ γλ + γτ ℓλ )+(1 − ξ)2 ̸ ℓ(ℓτ ℓλ /ℓ2 ) × [ℓ2 +iϵ]2 [ℓ2−m2 +iϵ] Z d d ℓ 2[ℓ2 γ µ + (2 − ϵ)ℓµ ̸ ℓ−(1−ξ)(ℓ2 γ µ −ℓµ ̸ ℓ) = i(µϵ/2 g)3 (2π)d [ℓ2 +iϵ]2 [ℓ2−m2 +iϵ] Z d d ℓ 2γ µ [1−(1−ξ)]ℓ2 +2γν [(2−ϵ)+(1−ξ)]ℓµ ℓν = i(µϵ/2 g)3 (2π)d [ℓ2 +iϵ]2 [ℓ2−m2 +iϵ] Z d 1 µ 2 d ℓ 3γ ℓ [1− 2 (1−ξ)] + O(ϵ) = i(µϵ/2 g)3 (2π)d [ℓ2 +iϵ]2 [ℓ2−m2 +iϵ] Z d 3γ µ [1− 12 (1−ξ)] d k + O(ϵ) = i(µϵ/2 g)3 (2π)d [ℓ2 +iϵ][ℓ2 −m2 +iϵ]2 Z 1 Z d d k 3γ µ [1− 21 (1−ξ)] = i(µϵ/2 g)3 2 dxdy δ(1−x−y) + ··· , (2π)d [k 2 −∆+iϵ]2 0 Z 1 Z d d k 6[1− 21 (1−ξ)] = (iµϵ/2 gγ µ )3 µϵ g 2 dxdy δ(1−x−y) + ··· (2π)d [k 2 −∆+iϵ]2 0

Z 1 i 2 ϵ/2 µ 3 2 1 = (iµ gγ ) g dxdy δ(1−x−y)6[1− 2 (1−ξ)] +· · · (4π)2 ϵ 0

g2 2 = (iµϵ/2 gγ µ )3 +· · · (3i)[1− 21 (1−ξ)], (4π)2 ϵ ×

R R where we have used dd ℓ ℓµ ℓν f (ℓ2 ) = dd ℓ d1 ℓ2 g µν f (ℓ2 ) from Eq. (A.6.9), d = 4−ϵ, R1 R1 the first line of Eq. (9.14) and 0 dxdy δ(1 − x − y) = 0 dx (1 − x) = 21 .

So putting everything together we have found for Eq. (9.16) that

g2 2 iµϵ/2 gΛau = (CF − 21 CA )T a (iµϵ/2 gγ µ ) +· · · [1−(1−ξ)] (9.24) (4π)2 ϵ

i g2 2 + − CA T a (iµϵ/2 gγ µ )3 +· · · (3i)[1− 21 (1−ξ)] 2 (4π)2 ϵ + iµϵ/2 gγ µ T a δ1

Gauge Field Theories

198

2 g2 +ln(4πe−γ ) ϵ 2(4π)2 1 × {2(CF − 2 CA )[1−(1−ξ)]+3CA [1 − 12 (1 − ξ)]}+δ1 ) + · · · ,

= iµϵ/2 gγ µ T a

which leads to the result quoted in Eq. (9.2.28),

δ1 =

2 g2 [−2CF −2CA +2(1−ξ)CF + 21 (1−ξ)CA ] . +ln(4πe−γ ) ϵ 2(4π)2

(9.25)

9.3 Problem 3 Problem: Draw the diagrams contributing to the O(g 2 ) (one-loop) corrections to the tree-level ghost-gluon, three-gluon and four-gluon vertices. Write down the loop integrals for the ghost-gluon vertex.

Solution: There are three distinct diagrams contributing O(g 2 ) corrections to the ghostgluon vertex,

and five diagrams contributing to the three-gluon vertex,

The three-gluon quark loop diagram vanishes for the same reason it does in QED, where it led to Furry’s theorem. While this loop contribution vanishes, the threegluon vertex does not vanish and so there is no gluon equivalent of Furry’s theorem for odd numbers of gluons.

Problem 3

199

The one-loop contributions to the four-gluon vertex are

,

where all permutations of the external lines are to be understood in the usual way, where the third diagram has a symmetry factor of 21 arising from symmetry under a horizontal flip of the loop and where the fourth diagram has a (−1) from the fermion loop as usual.

We need to insert appropriate factors of the renormalization scale for dimensional regularization as we discussed in Sec. 8.6.2, where from their kinetic terms we note that the gluon and ghost fields scale like photons and scalar bosons, [Aaµ ] = [ca ] = M(d−2)/2 , while as on QED the quarks have dimension [ψ] = M(d−1)/2 . So we have [g]quark−gluon = Mϵ/2 as in QED and so we make the replacement g → µϵ/2 g for the quark-gluon vertex as done when computing the QCD β-function in Sec. 9.2.3.

As a check on other vertices in the QCD Lagrangian density in Eq. (9.2.12) we see that in the ghost-gluon, three-gluon and four-gluon vertices gAaµ must have the same dimension as ∂ µ , which is the same as the requirement for the quark-gluon vertex and so we can replace g → µϵ/2 g everywhere in the QCD Feynman rules.

The tree-level vertex is given by (−µϵ/2 g)f abc p′µ and the corresponding counterterm is then (−µϵ/2 g)f abc p′µ δ1c . Noting that q = p′ − p the first of the two one-loop contributions to the ghost-gluon vertex is evaluated as p0 `+q

a ρ, h

f

d

µ, b q

`−p

e `

g

λ, i p

ϵ/2

= (−µ

Z

3

g)

c

dd ℓ ahf ′ρ df dbe [f p ]δ [f (ℓ + q)µ ]δ eg [f gic ℓλ ]δ hi (2π)d

(9.26)

Gauge Field Theories

200

(ℓ−p))λ (ℓ−p)ρ

×

i[−gλρ +(1−ξ) (ℓ−p)2 i i 2 2 2 2 [(ℓ + q) − mc +iϵ] [ℓ − mc +iϵ] (ℓ−p)2 +iϵ

]

,

where we have temporarily added an infinitesimal ghost mass mc to keep the integrals well-defined during the intermediate steps. We could also have added a gluon mass like we have done for the photon but doing both is not needed here. The second diagram is evaluated in a way similar to the second diagram for the quark-gluon vertex. We have p0 p0 −`

σ, g

a h

ρ, d q

λ, e p−`

(9.27)

`, δ ih

µ, b i

τ, f

c

p

Z d d ℓ = (µϵ/2 g)(−µϵ/2 g)2 (2π)d × {f agh p′σ }{f bed [g µλ (p′ +ℓ−2p)+g λρ (p′ +p−2ℓ)+g ρµ (p+ℓ−2p′ )]}{f if c ℓτ }δ ih ×

′ (ℓ−p)σ (ℓ−p)ρ )λ (ℓ−p′′ )τ ] i[−gλτ +(1−ξ) (ℓ−p(ℓ−p) ] 2 i (ℓ−p)2 . (ℓ−p)2 +iϵ (ℓ−p′ )2 +iϵ [ℓ2 − m2c +iϵ]

i[−gσρ +(1−ξ)

9.4 Problem 4 Problem: Define the BRST transformation of some field ϕ as Qϕ, where under the BRST transformation ϕ → ϕ + δϕ with δϕ ≡ θQϕ. For example, from Eq. (9.2.42) we see that QAaµ = (1/g)Dµab cb . Show that the BRST variations are nilpotent, which is the statement that Q2 ϕ = 0 when acting on all fields ϕ.

Solution: The BRST transformations of each type of field are given in Eqs. 9.2.42 and (9.2.45), ψ → ψ + iθca T a ψ, 1 ca → ca − θf abc cb cc , 2

1 Aaµ → Aaµ + θDµab cb , g 1 c̄a → c̄a − θ∂ µ Aaµ . ξg

(9.28)

Recall that ca and c̄a are Grassmann-valued ghost fields, and so anticommute

Problem 4

201

amongst themselves and also the Grassmann variable θ. From these transformations we see 1 1 1 Qψ = ica T a ψ, QAaµ = Dµab cb , Qca = − f abc cb cc , Qc̄a = − ∂ µ Aaµ . (9.29) g 2 ξg Under another BRST transformation, Qψ, for example, will transform as

1 abc b c a Qψ → i c − θf c c T a ψ + iθcd T d ψ 2 1 1 = ica T a ψ − ca T a θcd T d ψ − iθf abc cb cc T a ψ + iθf abc cb cc T a θcd T d ψ 2 2 1 a d a d abc b c a = Qψ + θc c T T ψ − iθf c c T ψ + 0 2 ≡ Qψ + θQ2 ψ.

(9.30) In the third step we have used that ca θ = −θca and θ2 = 0. Hence, exploiting antisymmetry further and relabeling dummy indices, 1 Q2 ψ = cb cc T b T c ψ − if abc cb cc T a ψ 2 1 1 = (cb cc − cc cb )T b T c ψ − if abc cb cc T a ψ 2 2 1 1 = cb cc (T b T c − T c T b )ψ − if abc cb cc T a ψ 2 2 1 1 = cb cc (if bca T a )ψ − if abc cb cc T a ψ 2 2 = 0,

(9.31)

since f bca = f abc . This proves the result for fermion fields. For gauge fields, the result has already been shown in the text: by Eq. (9.2.46), QAaµ =

1 1 ab b Dµ c → Dµab cb = QAaµ ≡ QAaµ + θQ2 Aaµ , g g

and so Q2 Aaµ = 0. For the ghost fields, we have

1 abc b 1 bde d e 1 cf g f g c a Qc → − f c − θf c c c − θf c c 2 2 2 1 1 1 = − f abc cb cc + θf abc f bde cd ce cc + f abc f cf g cb θcf cg + 0 2 4 4 1 abc bde d e c 1 a abc cf g b f g a = Qc + θf f c c c − Qc f f c c c + 0 4 4 ≡ Qca + θQ2 ca .

(9.32)

(9.33)

Since cd ce cc = (−1)2 cc cd ce , we can then write 1 acb cde b d e 1 abc cde b d e f f c c c − f f c c c 4 4 1 acb cde b d e = f f c c c 2

Q2 ca =

(9.34)

Gauge Field Theories

202

having relabeled b ↔ c in the first term and f → d, g → e in the second. The last step follows from f acb = −f abc . By anticommutativity this is equal to 1 acb cde 1 b d e f f (c c c + cd ce cb + ce cb cd ) 2 3 1 = (f acb f cde + f ace f cbd + f acd f ceb )cb cd ce 6 =0

Q2 ca =

(9.35)

because of the Jacobi identity Eq. (A.7.7). Finally, the BRST variation of the antighost field transforms as

1 ab b 1 µ a a Qc̄ → − ∂ Aµ + θDµ c ξg g 1 1 = − ∂ µ Aaµ − 2 θ∂ µ Dµab cb ξg ξg ≡ Qc̄a + θQ2 c̄a ,

(9.36)

and therefore Q2 c̄a = −(1/ξg 2 )∂ µ Dµab cb . Here we must turn to the equations of motion to see that this vanishes. From the Lagrangian Eq. (9.2.41) 1 a µν 1 / − m)ψ − (∂ µ Aaµ )2 − c̄a ∂ µ Dµab cb , L = − Fµν Fa + ψ̄(iD 4 2ξ

(9.37)

it is easy to see that the Euler-Lagrange equation for c̄a is simply ∂ µ Dµab cb = 0, and therefore Q2 c̄a = 0 on-shell. We have now shown that the BRST transformations are nilpotent, as required.

9.5 Problem 5 Problem: Starting from Eq. (9.4.2), complete the derivation of the Lagrangian densities Lgauge and LHiggs in terms of the physical fields (Wµ± , Zµ , Aµ and h) given in Eqs. (9.4.14) and (9.4.15) respectively.

Solution: We are given the electroweak Lagrangian Eq. (9.4.2), 1 1 a 2 L = − (Wµν ) − (Bµν )2 + (Dµ Φ)† (Dµ Φ) + m2 Φ† Φ − λ(Φ† Φ)2 , 4 4

(9.38)

in terms of the gauge fields Wµ1,2,3 and Bµ . The physical fields Wµ± , Zµ , Aµ are related to these by Wµ1 =

Wµ+ + Wµ− √ , 2

iWµ+ − iWµ− √ , 2

(9.39)

Bµ = cos θw Aµ − sin θw Zµ .

(9.40)

Wµ2 =

and also Eq. (9.4.8), Wµ3 = sin θw Aµ + cos θw Zµ ,

Problem 5

203

The abelian field strength is given by Bµν = ∂µ Bν − ∂ν Bµ , while the non-abelian a field strengths are Wµν = ∂µ Wνa − ∂ν Wµa + gf abc Wµb Wνc by Eq. (9.1.5). For SU (2) gauge fields with generators T a = σ a /2, we know from Eq. (A.3.2) that the structure constants are f abc = ϵabc , so explicitly, 1 Wµν = ∂µ Wν1 − ∂ν Wµ1 + g(Wµ2 Wν3 − Wµ3 Wν2 ), 2 Wµν = ∂µ Wν2 − ∂ν Wµ2 + g(Wµ3 Wν1 − Wµ1 Wν3 ), 3 Wµν = ∂µ Wν3 − ∂ν Wµ3 + g(Wµ1 Wν2 − Wµ2 Wν1 ).

(9.41)

Notice that the first two derivative terms in these field strength expressions are linear in the gauge fields. Substituting the physical fields into these expressions then yields

1 + − 1 + Wµν + ig (Wµ+ − Wµ− )(sin θw Aν + cos θw Zν ) Wµν = √ Wµν 2

−(sin θw Aµ + cos θw Zµ )(Wν+ − Wν− ) ,

i + − 2 − Wµν − ig (sin θw Aµ + cos θw Zµ )(Wν+ + Wν− ) Wµν = √ Wµν 2

−(Wµ+ + Wµ− )(sin θw Aν + cos θw Zν ) , 3 Wµν = sin θw Fµν + cos θw Zµν + ig(Wµ− Wν+ − Wµ+ Wν− ),

Bµν = cos θw Fµν − sin θw Zµν .

(9.42)

± ≡ ∂µ Wν± − ∂ν Wµ± for convenience. Some simple but tedious We have defined Wµν

algebra then reveals

1 2 Wµν W1µν + Wµν W2µν + + = 2Wµν W µν− + 4igWµν (sin θw Aµ + cos θw Z µ )W ν− − − 4igWµν (sin θw Aµ + cos θw Z µ )W ν+

− 4g 2 (sin θw Aµ + cos θw Z µ )(sin θw Aν + cos θw Z ν )Wµ+ Wν−

+ 4g 2 (sin θw Aµ + cos θw Z µ )(sin θw Aµ + cos θw Zµ )Wν+ W ν− .

(9.43)

Similarly for the other field strengths, 3 Wµν W3µν + Bµν B µν

= Fµν F µν + Zµν Z µν − g 2 [(Wµ− Wν+ − Wµ+ Wν− )(W µ− W ν+ − W µ+ W ν− )] + 2ig(Wµ− Wν+ − Wµ+ Wν− )(sin θw F µν + cos θw Z µν )

= Fµν F µν + Zµν Z µν − 2g 2 [Wµ+ W µ+ Wν− W ν− − (Wµ+ W µ− )2 ] + 4ig(Wµ− Wν+ − Wµ+ Wν− )(sin θw ∂ µ Aν + cos θw ∂ µ Z ν ).

(9.44)

The last step follows by antisymmetry of F µν , Z µν . To bring these to the form used in the text, we must use the relation for the electric charge e = g sin θw of Eq. (9.4.11). Then g 2 = e2 / sin2 θw , g cos θw = e cot θw , g 2 sin2 θw = e2 , g 2 cos2 θw = e2 cot2 θw , and g 2 sin θw cos θw = e2 cot θw . Expanding everything in the gauge kinetic terms explicitly therefore yields 1 1 a 2 − (Wµν ) − (Bµν )2 4 4

Gauge Field Theories

204

1 1 1 = − Fµν F µν − Zµν Z µν − (∂µ Wν+ − ∂ν Wµ+ )(∂ µ W ν− − ∂ ν W µ− ) 4 4 2 µ ν + − + ie cot θw [∂ Z (Wµ Wν − Wµ− Wν+ ) − Z µ (∂µ Wν+ − ∂ν Wµ+ )W ν− + Z µ (∂µ Wν− − ∂ν Wµ− )W ν+ ]

+ ie[∂ µ Aν (Wµ+ Wν− − Wµ− Wν+ ) − Aµ (∂µ Wν+ − ∂ν Wµ+ )W ν− + Aµ (∂µ Wν− − ∂ν Wµ− )W ν+ ]

e2 [Wµ+ W µ+ Wν− W ν− − (Wµ+ W µ− )2 ] 2 sin2 θw + e2 cot2 θw [Z µ Wµ+ Z ν Wν− − Zµ Z µ Wν+ W ν− ] +

+ e2 [Aµ Wµ+ Aν Wν− − Aµ Aµ Wν+ W ν− ]

+ e2 cot θw [Aµ Wµ+ Z ν Wν− + Aµ Wµ− Z ν Wν+ − 2Aµ Z µ Wν+ W ν− ]

(9.45)

Careful inspection reveals this is identical to Eq. (9.4.14), except for the missing gauge boson mass terms that come from the Higgs sector.

The Higgs sector is easiest to analyze in the unitary gauge, where

1 0 .

(9.46) Φ= √ 2 v+h In this gauge the covariant derivative is ig ig ′ Dµ Φ = ∂µ Φ + igWµa T a Φ + ig ′ Y Bµ Φ = ∂µ Φ + Wµa σ a Φ + Bµ Φ 2 2 #

" 0 ig ′ Bµ 0 ig Wµ1 − iWµ2 Wµ3 + = ∂µ + v+h √ −Wµ3 0 Bµ 2 Wµ1 + iWµ2 2 2

2 1 1 i v + h g(Wµ − iWµ ) 0 √ =√ + −gWµ3 + g ′ Bµ ∂ h 2 2 2 µ √

+ 1 i v+h g 0 p 2Wµ √ =√ + .

(9.47) − g 2 + g ′2 Zµ 2 2 ∂µ h 2 p p The last step follows from Eq. (9.4.9): cos θw ≡ g/ g 2 + g ′2 and sin θw ≡ g ′ / g 2 + g ′2 .

Hence, the Higgs kinetic term is 1 g 2 (v + h)2 (∂µ h)(∂ µ h) + (Wµ+ )† (W µ+ ) 2 4 (g 2 + g ′2 )(v + h)2 + (Zµ )† (Z µ ) 8 2

1 g (g 2 + g ′2 ) = (∂µ h)(∂ µ h) + (v + h)2 Wµ− W µ+ + Zµ Z µ .

2 4 8 (9.48) p We can make some final replacements using mw = gv/2 and mZ = g 2 + g ′2 v/2 to get 2

mW − µ+ m2Z 1 † µ µ 2 µ (Dµ Φ) (D Φ) = (∂µ h)(∂ h) + (v + h) Wµ W + 2 Zµ Z 2 v2 2v (Dµ Φ)† (Dµ Φ) =

Problem 6

205

=

1 m2 2 1 (∂µ h)(∂ µ h) + 1 + h + 2 h2 m2W Wµ− W µ+ + Z Zµ Z µ .

2 v v 2 (9.49)

The terms with no dependence on the Higgs field h are the gauge boson mass terms, which when included in Eq. (9.45) complete the derivation of Eq. (9.4.14). We shall ignore these from now on.

All that remains now is to expand the Higgs potential terms. In the unitary √ gauge, using the Higgs minimum condition v = m/ λ of Eq. (9.4.5), the potential terms become

2 4 m2 v + h v+h 2 † † 2 2 √ √ − 2 m Φ Φ − λ(Φ Φ) = m v 2 2 2 2 2 m v m m2 = (9.50) − m2 h2 − h3 − 2 h4 .

4 v 4v The first term is constant and so can be ignored. Matching with the conventionally normalized Higgs mass term − 21 m2h h2 , we see that m2 ≡ 21 m2h . We can make a final replacement using the relation mW = gv/2 to get 1 gm2h 3 g 2 m2h 4 m2 Φ† Φ − λ(Φ† Φ)2 = − m2h h2 − h . h − 2 4mW 32m2W Together, Eqs. (9.49) and (9.51) form the Higgs part of the Lagrangian,

2 2 1 gm2h 1 g mh h4 LHiggs = (∂µ h)(∂ µ h) − m2h h2 − h3 − 2 2 4mW 32m2W

2 1 m2 + h + 2 h2 m2W Wµ− W µ+ + Z Zµ Z µ , v v 2

(9.51)

(9.52)

in agreement with Eq. (9.4.15).

9.6 Problem 6 Problem: Show that the fermion kinetic contribution Lkin−ferm in Eq. (9.4.20) can be rewritten in terms of the physical vector boson fields as Eq. (9.4.22) with the µ± µ electroweak currents JW , JZµ and Jem given in Eq. (9.4.23).

Solution: Eq. (9.4.20) gives the fermion kinetic contribution to the Lagrangian as a

a

/ T a + ig ′ YL B)L / + iQ̄(∂/ + ig W / T a + ig ′ YQ B)Q / Lkin−ferm = iL̄(∂/ + ig W ′ ′ / R + id¯R (∂/ + ig Yd B)d / R + iēR (∂/ + ig ′ Ye B)e / R + iūR (∂/ + ig Yu B)u / R + iν̄R ∂ν / + Q̄i∂Q / + ūR i∂u / R + d¯R i∂d / R + ēR i∂e / R + ν̄R i∂ν / R = L̄i∂L − gWµa (L̄γ µ T a L + Q̄γ µ T a Q)

(9.53)

Gauge Field Theories

206

− g ′ Bµ (L̄γ µ YL L + Q̄γ µ YQ Q + ūR γ µ Yu uR + d¯R γ µ Yd dR + ēR γ µ Ye eR ).

Here we are using shorthand matrix notation to include all particle generations, e.g. uR ≡ (u1R , u2R , u3R )T = (uR , cR , tR )T . Note the right-handed neutrinos νR do not couple to anything and so can be ignored. To write this in terms of the physical gauge fields Wµ± , Zµ , Aµ , we use Eq. (9.4.8) Wµ+ + Wµ− √ , 2 iWµ+ − iWµ− √ Wµ2 = , 2 Wµ1 =

Wµ3 = sin θw Aµ + cos θw Zµ , Bµ = cos θw Aµ − sin θw Zµ .

(9.54)

Then Wµ+ + Wµ− 1 iWµ+ − iWµ− 2 √ √ T + T + (sin θw Aµ + cos θw Zµ )T 3 2 2 Wµ− Wµ+ = √ (Tµ1 + iTµ2 ) + √ (Tµ1 − iTµ2 ) + (sin θw Aµ + cos θw Zµ )T 3 2 2

1

Wµ+ 0 1 Wµ− 0 0 0 = √ + √ + (sin θw Aµ + cos θw Zµ ) 2 , 0 0 1 0 0 − 12 2 2 (9.55)

Wµa T a =

from the explicit form of the SU (2) generators in the fundamental representation T a = 12 σ a . So since L = (νL , eL )T , we get that

Wµ+ eL Wµ− sin θw Aµ + cos θw Zµ νL 0 a a √ √ Wµ T L = + + −eL 0 νL 2 2 2 Wµ+ Wµ− =⇒ Wµa L̄γ µ T a L = √ ν̄L γ µ eL + √ ēL γ µ νL 2 2 + (sin θw Aµ + cos θw Zµ )( 12 ν̄L γ µ νL − 12 ēL γ µ eL ).

(9.56)

The left-handed quark terms Q = (uL , dL )T are similar, giving in total Wµa (L̄γ µ T a L + Q̄γ µ T a Q) (9.57)

1 1 = Wµ+ √ (ν̄L γ µ eL + ūL γ µ dL ) + Wµ− √ ēL γ µ νL + d¯L γ µ uL 2 2 + (sin θw Aµ + cos θw Zµ )( 21 ν̄L γ µ νL − 12 ēL γ µ eL + 21 ūL γ µ uL − 12 d¯L γ µ dL ) µ+ µ− ≡ Wµ+ JW + Wµ− JW

+ (sin θw Aµ + cos θw Zµ )( 21 ν̄L γ µ νL − 12 ēL γ µ eL + 21 ūL γ µ uL − 12 d¯L γ µ dL ).

µ± The charged weak currents JW are exactly those of Eq. (9.4.23). Finally we must consider the hypercharge terms in Eq. (9.54). The hypercharges of each field are YL = −1/2, YQ = +1/6, Yu = +2/3, Yd = −1/3, and Ye = −1, so

Bµ (L̄γ µ YL L + Q̄γ µ YQ Q + ūR γ µ Yu uR + d¯R γ µ Yd dR + ēR γ µ Ye eR )

1 1 1 1 = (cos θw Aµ − sin θw Zµ ) − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL 2 2 6 6

Problem 6

207

2 1 + ūR γ µ uR − d¯R γ µ dR − ēR γ µ eR .

3 3

(9.58)

Hence, Eq. (9.54) expands to / + Q̄i∂Q / + ūR i∂u / R + d¯R i∂d / R + ēR i∂e / R + ν̄R i∂ν / R Lkin−ferm = L̄i∂L µ+ µ− − g[Wµ+ JW + Wµ− JW ]

− g cos θw Zµ ( 21 ν̄L γ µ νL − 12 ēL γ µ eL + 12 ūL γ µ uL − 21 d¯L γ µ dL ) − g sin θw Aµ ( 12 ν̄L γ µ νL − 12 ēL γ µ eL + 12 ūL γ µ uL − 12 d¯L γ µ dL )

1 1 1 1 ′ + g sin θw Zµ − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL 2 2 6 6

2 1 + ūR γ µ uR − d¯R γ µ dR − ēR γ µ eR 3 3

1 1 1 1 − g ′ cos θw Aµ − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL 2 2 6 6

2 1¯ µ µ µ + ūR γ uR − dR γ dR − ēR γ eR 3 3 / + Q̄i∂Q / + ūR i∂u / R + d¯R i∂d / R + ēR i∂e / R + ν̄R i∂ν / R = L̄i∂L µ+ µ− − g[Wµ+ JW + Wµ− JW ]

1 1 1¯ µ 1 µ µ µ ν̄L γ νL − ēL γ eL + ūL γ uL − dL γ dL − gZµ cos θw 2 2 2 2

′ 1 1 1 g 1 − sin θw − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL g 2 2 6 6

1¯ µ 2 µ µ + ūR γ uR − dR γ dR − ēR γ eR 3 3

1 1 1 1 − g sin θw Aµ ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL − d¯L γ µ dL 2 2 2 2

′ g cos θw 1 1 1 1 + − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL g sin θw 2 2 6 6

2 1 + ūR γ µ uR − d¯R γ µ dR − ēR γ µ eR 3 3 ¯ / + Q̄i∂Q / + ūR i∂u / R + dR i∂d / R + ēR i∂e / R + ν̄R i∂ν / R ≡ L̄i∂L µ+ µ− µ − g[Wµ+ JW + Wµ− JW ] − gZµ JZµ − g sin θw Aµ Jem .

(9.59)

It remains to show that these currents match those in Eq. (9.4.23). Let us begin µ with Jem . First, we note that Eq. (9.4.9) implies g ′ cos θw /g sin θw = 1, and we know from Eq. (9.4.11) that e = g sin θw . Therefore

1 1 1 1 µ g sin θw Aµ Jem = eAµ ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL − d¯L γ µ dL 2 2 2 2 1 1 1 1 − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL 2 2 6 6

Gauge Field Theories

208

2 1 + ūR γ µ uR − d¯R γ µ dR − ēR γ µ eR 3 3

2 1 = eAµ −ēL γ µ eL + ūL γ µ uL − d¯L γ µ dL 3 3

2 1¯ µ µ µ −ēR γ eR + ūR γ uR − dR γ dR 3 3

2 1 ¯ µd . = eAµ −ēγ µ e + ūγ µ u − dγ 3 3

(9.60)

The last step follows from e.g. e = eL + eR together with eL eR = eR eL = 0. This matches the electromagnetic current in Eq. (9.4.23). Finally we consider the weak neutral current again using Eq. (9.4.9),

1 1 1 1 1¯ µ µ 2 µ µ µ JZ = cos θw ν̄L γ νL − ēL γ eL + ūL γ uL − dL γ dL cos θw 2 2 2 2

1 1 1 g′ 1 − cos θw sin θw − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL g 2 2 6 6

2 1 + ūR γ µ uR − d¯R γ µ dR − ēR γ µ eR 3 3

1 1 1¯ µ 1 1 2 µ µ µ ν̄L γ νL − ēL γ eL + ūL γ uL − dL γ dL (1 − sin θw ) = cos θw 2 2 2 2

1 1 1 1 − sin2 θw − ν̄L γ µ νL − ēL γ µ eL + ūL γ µ uL + d¯L γ µ dL 2 2 6 6

2 1 + ūR γ µ uR − d¯R γ µ dR − ēR γ µ eR 3 3

1 1 1 = νL + ēL γ µ − + sin2 θw eL + ēR γ µ sin2 θw eR ν̄L γ µ cos θw 2 2

1 2 2 + ūL γ µ − sin2 θw uL + ūR γ µ − sin2 θw uR 2 3 3

1 1 1 +d¯L γ µ − + sin2 θw dL + d¯R γ µ sin2 θw dR , (9.61) 2 3 3 confirming Eq. (9.4.23).

9.7 Problem 7 Problem: (a) Use the definition of the Fermi coupling constant in Eq. (5.2.7) and the lowmomentum limit contribution to the scattering e− ν̄e → µ− ν̄µ to √ of the2 tree-level 2 show that GF / 2 √ = g /8MW = 1/2v 2 . Since we know that GF ≃ 1.167 × 10−5 −2 GeV then v = ( 2GF )−1/2 ≃ 246 GeV.

Problem 7

209

(b) Calculate the total cross-section for e+ e− → Zh at tree level in the CM frame (the electron-Higgs coupling is very small and can be neglected). (c) Estimate the total cross-section for e+ e− → hadrons at tree-level in the CM frame, including both the intermediate γ and Z contributions to M and their interference in the total cross-section, σ(e+ e− → hadrons).

Solution: (a) From Eq. (5.2.7) we know that the relevant interaction Lagrangian term for the reaction e− ν̄e → µ− ν̄µ in the Fermi theory is √

LFermi = i2 2GF (ψ̄νe γ α 12 (1 − γ 5 )ψe )(ψ̄µ γα 12 (1 − γ 5 )ψνµ ) int

+(ψ̄νµ γ α 12 (1 − γ 5 )ψµ )(ψ̄e γα 21 (1 − γ 5 )ψνe ) √ ≡ i2 2GF [(ν̄e γ α PL e)(µ̄γα PL νµ ) + (ν̄µ γ α PL µ)(ēγα PL νe )] . (9.62) The second equality simply rewrites the first in the more familiar notation of the Standard Model. This tells us that the tree-level interaction vertex is e−

p

µ−

p0

√ = i2 2GF v̄(k)γ α PL u(p)ū(p′ )γα PL v(k ′ ). k ν̄e

k0

ν̄µ

(9.63) On the other hand, in the electroweak theory we know from Eqs. (9.4.22) and (9.4.23) that the relevant interactions arise from the terms

g Lweak = − √ Wα+ (ν̄eL γ α eL + ν̄µL γ α µL ) + Wα− (ēL γ α νeL + µ̄L γ α νµL ) int 2

g + = − √ Wα (ν̄e γ α PL e + ν̄µ γ α PL µ) + Wα− (ēγ α PL νe + µ̄γ α PL νµ ) , 2 (9.64) which give the tree-level amplitude e−

p

p0

µ−

k0

ν̄µ

W−

ν̄e

k

qα qβ

2 αβ −i g − 2 g mW = −√ v̄(k)γα PL u(p) ū(p′ )γβ PL v(k ′ ) 2 − m2 q 2 W

Gauge Field Theories

210

=i

g2 v̄(k)γ α PL u(p)ū(p′ )γα PL v(k ′ ) + O 2m2W

q2 m4W

,

(9.65)

where q ≡ p + k. In the low-momentum limit q → 0, the higher-order terms can be neglected and we can match the two amplitudes to get √ g2 g2 GF v2 √ = 2 2GF = =⇒ = , (9.66) 2 2 2mW 8mW 2 2 as required. (b) As we are neglecting the small electron-Higgs coupling, the only tree-level diagram for the reaction e+ e− → Zh is e−

p

p0

Z

k0

h

Z

e+

k

We see from Eq. (9.4.15) that the hZZ vertex carries a factor of 2im2Z /v, while Eq. (9.4.23) shows that the e+ e− Z vertex factor is (−ig/ cos θw )γ µ [(− 21 + sin2 θw )PL + (sin2 θw )PR ] = (−ig/2 cos θw )γ µ [2 sin2 θw − 21 + 12 γ 5 ], since PR,L = 1 (1 ± γ 5 ). It is conventional to write the vector and axial vector couplings 2 as ve = 2 sin2 θw − 21 and ae = − 12 . Again defining q = p + k, the tree-level amplitude is therefore

qµ qν 2im2Z µ ′ ′ ∗ −i gµν − m2Z −igγ ν [ve − ae γ 5 ] s iM = ϵ (p , λ ) v̄ r (k) u (p) 2 2 v q − mZ 2 cos θw q q

µ ν 2m3Z µ ′ ′ ∗ gµν − m2Z r = −i 2 ϵ (p , λ ) v̄ (k)γ ν [ve − ae γ 5 ]us (p), v q 2 − m2Z

(9.67)

having simplified using Eq. (9.4.9). By Eq. (7.6.60), the conjugate amplitude is q q

ρ σ 2m3Z ρ ′ ′ gρσ − m2Z s −iM = i 2 ϵ (p , λ ) 2 ū (p)γ σ [ve − ae γ 5 ]v r (k), v q − m2Z

∗

(9.68)

and so

4m6 |M|2 = 4Z ϵµ (p′ , λ′ )∗ ϵρ (p′ , λ′ ) v

q q

µ ν gµν − m 2 Z

qρ qσ gρσ − m 2

(q 2 − m2Z )2

Z

× v̄ r (k)γ ν [ve − ae γ 5 ]us (p)ūs (p)γ σ [ve − ae γ 5 ]v r (k).

(9.69)

Averaging over the initial polarizations r, s and summing over the final state polarization λ′ , we get the unpolarized squared amplitude X1X |M|2 = |M|2 4 r,s ′ λ

211

Problem 7

qµ qν qρ qσ gρσ − m 2 1 X 4m6Z µ ′ ′ ∗ ρ ′ ′ gµν − m2Z Z ϵ (p , λ ) ϵ (p , λ ) = 4 ′ v4 (q 2 − m2Z )2 λ

σ 5 × tr γ ν [ve − ae γ 5 ](p k − me ) . / + me )γ [ve − ae γ ](/

(9.70)

The second line follows from the Casimir trick Eq. (7.6.61), since γ 0 [γ µ (ve + γ 5 )]† γ 0 = γ 0 (ve + γ 5† )γ 0 γ 0 γ µ† γ 0 = (ve − γ 5 )γ µ = γ µ (ve + γ 5 ) by Eq. (A.3.11).

To proceed, we can use Eq. (6.5.38) to evaluate the polarization sum:

X qρ qσ qµ qν gρσ − 2 (9.71) ϵµ (p′ , λ′ )∗ ϵρ (p′ , λ′ ) gµν − 2 mZ mZ λ′

p′µ p′ρ qµ qν qρ qσ = −g µρ + g − g − µν ρσ m2Z m2Z m2Z

q ρ qν p′ p′ρ (p′ · q)p′ρ qν qρ qσ = −δ ρν + 2 + ν 2 − g − ρσ mZ mZ m4Z m2Z

p′ p′ (p′ · q)(p′σ qν + p′ν qσ ) q2 (p′ · q)2 qν qσ = −gνσ + ν 2σ − + 2 − + . mZ m4Z m2Z m4Z m2Z Next, the trace in Eq. (9.70) can be evaluated by repeated application of Eq. (A.3.30). Specifically, since the trace of an odd number of γ-matrices vanishes, and also the trace of γ 5 times an odd number of γ-matrices (or two γ-matrices) vanishes, we can expand

σ 5 tr γ ν [ve − ae γ 5 ](p k − me ) / + me )γ [ve − ae γ ](/ 2 ν σ σ 5 = tr ve γ (p k − me ) + a2e γ ν γ 5 (p k − me ) / + me )γ (/ / + me )γ γ (/ σ σ 5 −ve ae γ ν γ 5 (p k − me ) − ve ae γ ν (p k − me ) / + me )γ (/ / + me )γ γ (/

σ 2 2 ν σ 2 ν 5 σ 5 2 2 ν 5 σ 5 = tr ve2 γ ν p /γ k/ − ve me γ γ + ae γ γ p /γ γ k/ − ae me γ γ γ γ

σ 2 ν 5 σ σ 5 2 ν σ −ve ae γ ν γ 5 p γ γ γ − ve ae γ ν p γ γ γ5 /γ k/ + ve ae me /γ γ k/ + ve ae me

σ 2 2 ν σ 2 2 ν σ 2 2 ν σ = tr ve2 γ ν p /γ k/ − ve me γ γ + (−1) ae γ p /γ k/ − (−1)ae me γ γ σ 3 5 ν σ −(−1)ve ae γ 5 γ ν p /γ k/ − (−1) ve ae γ γ p /γ k/

= (ve2 + a2e )pα kβ tr γ ν γ α γ σ γ β − (ve2 − a2e )m2e tr {γ ν γ σ }

+ 2ve ae pα kβ tr γ 5 γ ν γ α γ σ γ β .

(9.72)

In the third step we have used that γ 5 anticommutes with every other γ-matrix, and (γ 5 )2 = I, from Eq. (A.3.5). These traces in the last line are given in Eqs. (A.3.24) and (A.3.25),

σ 5 tr γ ν [ve + γ 5 ](p k − me ) / + me )γ [ve + γ ](/ = (ve2 + a2e )pα kβ 4(g να g σβ − g νσ g αβ + g νβ g ασ ) − (ve2 − a2e )m2e (4g νσ ) + 2ve ae pα kβ (−4iϵνασβ )

= 4(ve2 + a2e )(pν k σ + k ν pσ ) − 4[ve2 (p · k + m2e ) + a2e (p · k − m2e )]g νσ

Gauge Field Theories

212

− 8ive ae ϵνασβ pα kβ .

(9.73)

Note that the last term here will vanish upon contraction with every term in Eq. (9.72) because it is antisymmetric in ν, σ. Substituting Eqs. (9.72) and (9.73) into Eq. (9.70) then gives |M|2 =

4m6Z 1 v 4 (q 2 − m2Z )2

(p · p′ )(k · p′ ) × 2(ve2 + a2e ) −p · k + m2Z (p′ · q) [(p · q)(k · p′ ) + (p · p′ )(k · q)] − m4Z

q2 (p′ · q)2 (p · q)(k · q) + 2− 2 + mZ m4Z m2Z

(p′ )2 − ve2 (p · k + m2e ) + a2e (p · k − m2e ) −4 + 2 mZ

2 ′ 2 2 ′ 2 2(p · q) q (p · q) q − + 2− 2 + m4Z mZ m4Z m2Z

(9.74)

In the centre-of-momentum frame we have pµ = (Ee , p) and k µ = (Ee , −p), so q µ ≡ pµ + k µ = (2Ee , 0) ≡ (Ecm , 0). The outgoing momenta are p′µ = (EZ , p′ ) and k ′µ = (Eh , −p′ ), subject to Ecm = Eh + EZ . As usual we have p2 = k 2 = m2e , (p′ )2 = m2Z , and (k ′ )2 = m2h . Simple algebra then shows Ecm (Eh − EZ ) = Eh2 − EZ2 = (m2h + |p′ |2 ) − (m2Z + |p′ |2 ) = m2h − m2Z . Then 2 Ecm (Eh − (Ecm − Eh )) = m2h − m2Z =⇒ Ecm Eh = 21 (Ecm − m2Z + m2h ) and 1 2 2 2 similarly Ecm EZ = 2 (Ecm + mZ − mh ). The quantities appearing in the above equation are therefore 2 q 2 = Ecm ,

p·q =k·q =

p·k =p·q−p·p=

1 2 E , 2 cm

1 2 E − m2e , 2 cm

1 2 (E + m2Z − m2h ), 2 cm 1 2 q · k ′ = Ecm Eh = (Ecm − m2Z + m2h ), 2 1 1 2 p · p′ = Ecm EZ − p · p′ = (Ecm + m2Z − m2h ) − |p||p′ | cos θ, 2 4 1 1 2 k · p′ = Ecm EZ + p · p′ = (Ecm + m2Z − m2h ) + |p||p′ | cos θ.

2 4 q · p′ = Ecm EZ =

(9.75)

Here θ is defined to be the angle between the incoming e− and the outgoing Z boson. The unpolarized squared amplitude is then |M|2 =

8m6Z 1 2 − m2 )2 v 4 (Ecm Z

Problem 7

213

1 2 (E 2 + m2Z − m2h )2 |p|2 |p′ |2 cos2 θ × (ve2 + a2e ) Ecm + m2e + cm − 4 16m2Z m2Z

(E 2 + m2Z − m2h )2 + a2e m2e −3 − cm 2m4Z 2

2 (Ecm + m2Z − m2h )2 Ecm E2 + .

(9.76) + 2 − cm m2Z 4m4Z m2Z The differential cross-section in the centre-of-momentum frame is calculated from Eq. (7.3.82),

pcm 1 dσ 1 |p′ | f 2 = |M| |M|2 .

(9.77) = 2 pcm 2 |p| dΩ cm 64π 2 Ecm 64π 2 Ecm i R The total cross-section is then of course σ(e+ e− → Zh) = dΩ (dσ/dΩ) = R1 2π −1 dcos θ (dσ/dΩ), so we can write σ(e+ e− → Zh) =

1 1 |p′ | 16m6Z (9.78) 2 |p| 2 − m 2 )2 32πEcm v 4 (Ecm Z

2 + m2Z − m2h )2 1 2 (Ecm |p|2 |p′ |2 2 2 2 × (ve + ae ) Ecm + me + − 4 16m2Z 3m2Z

2 2 2 2 (E + mZ − mh ) + a2e m2e −3 − cm 2m4Z 2

2 2 + m2Z − m2h )2 Ecm (Ecm Ecm . + 2− 2 + mZ 4m4Z m2Z

Explicitly the momenta are r

1 2 E − m2e , 4 cm s q 2 + m2 − m2 )2 1 (Ecm ′ Z h 2 2 − m2Z , |p | = EZ − mZ = 2 4 Ecm |p| =

p

Ee2 − m2e =

(9.79)

but we shall not substitute these into the above expression.

We can obtain a simplified expression for the cross-section by taking the reasonable limit Ecm ≫ me . Then |p| = Ecm /2, and the cross-section becomes σ(e+ e− → Zh)|Ecm ≫me

2 2 m6Z (Ecm + m2Z − m2h )2 Ecm |p′ |2 |p′ | 1 2 2 2 = − (ve + ae ) Ecm + 3 (E 2 − m2 )2 πv 4 Ecm 4 16m2Z 12m2Z cm Z

2 2 2 2 2 2 2 2 2 2 2 ae me (Ecm + mZ − mh ) Ecm (Ecm + mZ − mh ) 2 + 2 − + 2− 2 + Ecm . mZ 2m2Z mZ 4m4Z

The term proportional to m2e in the second line is not set to zero because it is normalized by m2Z and so cannot be compared with Ecm .

Going further, we can take the ultra-high energy limit Ecm → ∞. This is equivalent to setting any mass scale divided by Ecm to zero, taking care to note that v = (2 cos θw /g)mZ . In this regime the momenta are |p| = |p′ | = Ecm /2.

Gauge Field Theories

214

Since only the terms with the highest powers of Ecm are relevant, the crosssection reduces to σ(e+ e− → Zh)|Ecm →∞

1 g 4 m2Z 1 (9.80) 2 4 2πEcm 16 cos4 θw Ecm

4

4

4 4 Ecm Ecm a2e m2e Ecm Ecm 2 × (ve2 + a2e ) − + − + E cm 16m2Z 48m2Z m2Z 2m2Z 4m2Z

ve2 + a2e g4 1 a2e m2e 1 a2e m2e = + − 4 2 64π cos4 θw 2 mZ Ecm 12 m2Z

8 sin4 θw − 4 sin2 θw + 1 1 m2e 1 m2e 1 g4 + 2 − . = 64π cos4 θw 2 m4Z Ecm 24 4 m2Z

=

Interestingly, this tends towards a non-zero value as Ecm → ∞. Ultimately this is because of the qµ qν /m2Z term in the massive vector boson propagator. (c) We wish to calculate the cross-section for e+ e− → hadrons, i.e. e+ e− → qi q̄i for all quark types qi , including all tree-level photon and Z boson contributions.

As in Part (b), we will first need to know the vector and axial vector couplings of the quarks. From Eq. (9.4.23) we see that the qi q̄i Z vertex factor for up-type (electric charge +2/3) quarks is −ig µ γ cos θw

1 2 2 − sin2 θw PL + − sin2 θw PR 2 3 3

1 4 1 −ig γµ − sin2 θw − γ 5 = 2 cos θw 2 3 2 h i −ig (u) (u) ≡ γ µ vi − ai γ 5 , 2 cos θw

(9.81)

while for the the down-type (electric charge −1/3) quarks the vertex factor is −ig µ γ cos θw

1 1 1 − + sin2 θw PL + sin2 θw PR 2 3 3

−ig 1 2 1 = γ µ − + sin2 θw + γ 5 2 cos θw 2 3 2 h i −ig (d) (d) ≡ γ µ vi − a i γ 5 .

2 cos θw

(9.82)

It is then straightforward to evaluate the only tree-level diagrams that con-

Problem 7

215

tribute to this process: e−

qi

p0

p γ

p

k0

q̄i

p0

qi

k0

q̄i

Z

+

k

e+

e−

k

e+

−igµν r v̄ (k)(ieγ ν )us (p) (9.83) q2

qµ qν −i gµν − m µ 5 2 −igγ ν [ve − ae γ 5 ] s Z s′ ′ −igγ [vi − ai γ ] r ′ ′ + ū (p ) v̄ r (k) v (k ) u (p).

2 2 2 cos θw q − mZ 2 cos θw ′

′

= ūs (p′ )(−iqi γ µ )v r (k ′ )

Here we are using qi to denote the electric charge of the quark qi . To avoid 2 confusion we will write Ecm in place of the squared momentum q 2 , as in Part (b).

Note that since the spinors are on-shell, we can use the equations of motion Eq. (A.3.36), p k = /u(p) = mu(p), ū(p)p / = mū(p) and k/v(k) = −mv(k), v̄(k)/ ′ ′ −mv̄(k) to simplify the qµ qν term, where q = p + k = p + k by momentum conservation: ū(p′ )γ µ [vi − ai γ 5 ]v(k ′ )qµ ′

′ 5 ′ = ū(p′ )(p / + k/ )[vi − ai γ ]v(k )

′

′ 5 ′ ′ 5 = ū(p′ )p k v(k ′ ) / [vi − ai γ ]v(k ) + ū(p )[vi + ai γ ]/

= mi ū(p′ )[vi − ai γ 5 ]v(k ′ ) − mi ū(p′ )[vi + ai γ 5 ]v(k ′ ) = −2mi ai ū(p′ )γ 5 v(k ′ ),

(9.84)

where mi is the mass of quark qi , and similarly qν v̄(k)γ ν [ve − ae γ 5 ]u(p)

5 = v̄(k)(p / + k/)[ve − ae γ ]u(p)

= v̄(k)/ k [ve − ae γ 5 ]u(p) + v̄(k)[ve + ae γ 5 ]p /u(p)

= −me v̄(k)[ve − ae γ 5 ]u(p) + me v̄(k)[ve + ae γ 5 ]u(p) = 2me ae v̄(k)γ 5 u(p).

Then the invariant amplitude simplifies to " ′ ′ ūs (p′ )γ µ v r (k ′ )v̄ r (k)γµ us (p) iM = −i qi e 2 Ecm ′

(9.85)

(9.86) ′

g2 ūs (p′ )γ µ [vi − ai γ 5 ]v r (k ′ )v̄ r (k)γµ [ve − ae γ 5 ]us (p) 2 2 − m2 4 cos θw Ecm Z # ′ ′ g 2 mi me ai ae ūs (p′ )γ 5 v r (k ′ )v̄ r (k)γ 5 us (p) − 2 .

2 − m2 cos θw m2Z Ecm Z

−

Gauge Field Theories

216

By Eq (7.6.60) the conjugate amplitude is then " ′ ′ v̄ r (k ′ )γ α us (p′ )ūs (p)γα v r (k) ∗ −iM = +i qi e 2 Ecm ′

(9.87) ′

g2 v̄ r (k ′ )γ α [vi − ai γ 5 ]us (p′ )ūs (p)γα [ve − ae γ 5 ]v r (k) − 2 − m2 4 cos2 θw Ecm Z # ′ ′ r ′ 5 s ′ s 2 mi me ai ae v̄ (k )γ u (p )ū (p)γ 5 v r (k) g . − 2 2 − m2 cos θw m2Z Ecm Z For the unpolarized squared amplitude we average over the initial spins r, s and sum over the final state spins r′ , s′ , |M|2 =

X1X r ′ ,s′

4 r,s

MM∗ .

(9.88)

These sums will produce traces by the Casimir trick Eq. (7.6.61), e.g.

XX ′ ′ ′ ′ [ūs (p′ )γ µ v r (k ′ )v̄ r (k)γµ us (p)][v̄ r (k ′ )γ α us (p′ )ūs (p)γα v r (k)] r,s r ′ ,s′

=

X r,s

′

′ r s s r tr{γ µ (/ k − mi )γ α (p / + mi )}[v̄ (k)γµ u (p)][ū (p)γα v (k)] ′

′ = tr{γ µ (/ k − mi )γ α (p k − me )}. / + mi )} tr{γµ (p / + me )γα (/

(9.89)

Doing this for every term in the product we get qi2 e2 ′ ′ tr{γ µ (/ k − mi )γ α (p k − me )} / + mi )} tr{γµ (p / + me )γα (/ 4 4Ecm 1 qi eg 2 ′ ′ tr{γ µ (/ k − mi )γ α [vi − ai γ 5 ](p − / + mi )} 2 2 2 16 cos θw Ecm (Ecm − m2Z )

|M|2 =

5 × tr{γµ (p k − me )} / + me )γα [ve − ae γ ](/

−

qi eg 2 mi me ai ae ′ ′ tr{γ µ (/ k − mi )γ 5 (p / + mi )} 2 m2 (E 2 − m2 ) 4 cos2 θw Ecm cm Z Z 5 × tr{γµ (p k − me )} / + me )γ (/

−

qi eg 2 1 ′ ′ tr{γ µ [vi − ai γ 5 ](/ k − mi )γ α (p / + mi )} 2 (E 2 − m2 ) 16 cos2 θw Ecm cm Z × tr{γµ [ve − ae γ 5 ](p k − me )} / + me )γα (/

+

1 g4 ′ ′ tr{γ µ [vi − ai γ 5 ](/ k − mi )γ α [vi − ai γ 5 ](p / + mi )} 4 2 64 cos θw (Ecm − m2Z )2 5 × tr{γµ [ve − ae γ 5 ](p k − me )} / + me )γα [ve − ae γ ](/

+

mi me ai ae g4 ′ ′ tr{γ µ [vi − ai γ 5 ](/ k − mi )γ 5 (p / + mi )} 2 − m 2 )2 16 cos4 θw m2Z (Ecm Z 5 × tr{γµ [ve − ae γ 5 ](p k − me )} / + me )γ (/

Problem 7

217

−

qi eg 2 mi me ai ae ′ ′ tr{γ 5 (/ k − mi )γ α (p / + mi )} 2 2 2 − m2 ) 4 cos θw Ecm m2Z (Ecm Z × tr{γ 5 (p k − me )} / + me )γα (/

+

mi me ai ae g4 ′ ′ tr{γ 5 (/ k − mi )γ α [vi − ai γ 5 ](p / + mi )} 2 − m 2 )2 16 cos4 θw m2Z (Ecm Z 5 × tr{γ 5 (p k − me )} / + me )γα [ve − ae γ ](/

+

g4 m2i m2e a2i a2e ′ ′ tr{γ 5 (/ k − mi )γ 5 (p / + mi )} 2 − m2 )2 4 cos4 θw m4Z (Ecm Z 5 × tr{γ 5 (p k − me )} / + me )γ (/

(9.90)

These traces have several distinct forms, which can all be evaluated using the rules in Sec. A.3. First, we have seen in Part (b) that µ α α µ µα 2 µα tr{γ µ (/ k − m)γ α (p / + m)} = 4(p k + p k − g p · k) − 4m g .

(9.91)

Then because the trace of γ 5 times one, two, or three γ-matrices vanishes, we also have tr{γ µ (/ k − m)γ 5 (p / + m)} = 0, 5

(9.92)

α

tr{γ (/ k − m)γ (p / + m)} = 0,

(9.93)

and µ α 5 tr{γ µ (/ k − m)γ α γ 5 (p / + m)} = tr{γ k/γ γ p /}

= −kρ pσ tr{γ 5 γ µ γ ρ γ α γ σ } = 4ikρ pσ ϵµρασ ,

µ 5

α

µ 5

α

tr{γ γ (/ k − m)γ (p / + m)} = tr{γ γ k/γ p /} = 4ikρ pσ ϵ

(9.94) µρασ

,

(9.95)

and further since γ 5 anticommutes with all other γ-matrices and (γ 5 )2 = I, µ 5 5 tr{γ µ γ 5 (/ k − m)γ α γ 5 (p k − m)(−γ α )(p / + m)} = tr{γ γ γ (−/ / + m)}

= tr{γ µ (/ k + m)γ α (p / + m)}

= 4(pµ k α + pα k µ − g µα p · k) + 4m2 g µα , (9.96)

µ tr{γ µ γ 5 (/ k − m)γ 5 (p k − m)(p / + m)} = tr{γ (−/ / + m)}

µ µ = m tr{−γ µ k/ − γ µ p /} = −4m(k + p ), (9.97)

tr{γ 5 (/ k − m)γ α γ 5 (p k + m)γ α (p / + m)} = tr{(/ / + m)} 5

5

α α = m tr{/ kγ α + γ α p /} = 4m(k + p ),

(9.98)

2 2 = −tr{/ kp / + m } = −4p · k − 4m .

(9.99)

tr{γ (/ k − m)γ (p k − m)(p / + m)} = tr{(−/ / + m)}

With these identities the unpolarized squared amplitude reduces to |M|2 =

8qi2 e2 ′ (p · p)(k ′ · k) + (p′ · k)(k ′ · p) + (p′ · k ′ )m2e + (p · k)m2i + 2m2i m2e 4 Ecm

Gauge Field Theories

218

2qi eg 2 1 {2vi ve [(p′ · p)(k ′ · k) + (p′ · k)(k ′ · p) 2 2 2 cos θw Ecm (Ecm − m2Z )

+(p′ · k ′ )m2e + (p · k)m2i + 2m2i m2e − ai ae kρ′ p′σ pτ k ω ϵµρασ ϵµτ αω 2 1 g4 (vi + a2i )(ve2 + a2e ) [(p′ · p)(k ′ · k) + (p′ · k)(k ′ · p)] + 2 − m2 )2 2 cos4 θw (Ecm Z

−

− (vi2 + a2i )(a2e − ve2 )(p′ · k ′ )m2e − (a2i − vi2 )(ve2 + a2e )(p · k)m2i

+2(a2i − vi2 )(a2e − ve2 )m2i m2e − 2vi ai ve ae kρ′ p′σ pτ k ω ϵµρασ ϵµτ αω ′ ′

2g 4 m2i m2e a2i a2e + (p · k + m2i )(p · k + m2e ) − m2Z (p′ + k ′ ) · (p + k) 4 2 4 2 2 cos θw mZ (Ecm − mZ ) (9.100) Beware that the orderings of the τ, ω indices in the middle lines are important.

These terms can be simplified using Eq. (A.2.3) as kρ′ p′σ pτ k ω ϵµρασ ϵµτ αω = 2kρ′ p′σ pτ k ω (δ ρω δ στ − δ ρτ δ σω ) = 2[(k ′ · k)(p′ · p) − (k ′ · p)(p′ · k)].

Now the kinematics in this reaction are identical to those in Part (b) provided we replace mZ , mh → mi . We can therefore read off from Eq. (9.75) that 2 (p′ + k ′ ) · (p + k) = Ecm , 1 2 − m2e , p · k = Ecm 2 1 2 p′ · k ′ = Ecm − m2i , 2 1 2 p · p′ = k · k ′ = Ecm − |p||p′ | cos θ, 4 1 2 k · p′ = p · k ′ = Ecm + |p||p′ | cos θ.

(9.101) 4 With these we get

4 E 2 (m2 + m2e ) 8q 2 e2 Ecm + cm i + 2|p|2 |p′ |2 cos2 θ |M|2 = i4 Ecm 8 2

4

2 4qi eg 2 1 Ecm Ecm (m2i + m2e ) 2 ′ 2 2 − v v + + 2|p| |p | cos θ i e 2 (E 2 − m2 ) cos2 θw Ecm 8 2 cm Z 2 +ai ae Ecm |p||p′ | cos θ

4 g4 1 Ecm 2 ′ 2 2 2 2 2 2 + + 2|p| |p | cos θ (vi + ai )(ve + ae ) 2 − m2 )2 2 cos4 θw (Ecm 8 Z

E2 − cm (vi2 + a2i )(a2e − ve2 )m2e + (a2i − vi2 )(ve2 + a2e )m2i 2 2 +4(2a2i a2e − a2i ve2 − a2e vi2 )m2i m2e + 4vi ai ve ae Ecm |p||p′ | cos θ

2g 4 m2i m2e a2i a2e 1 2 2 Ecm − m2Z Ecm .

(9.102) − 4 4 2 − m2 )2 cos θw mZ (Ecm 4 Z

As we have it now, this is the amplitude for e+ e− → qi q̄i , for a single color of quark qi . To get the amplitude for e+ e− → hadrons, we should multiply by the number of colors Nc = 3 and sum over the quark flavors i.

Problem 7

219

We calculate the total cross section in the centre-of-momentum frame using Eq. (7.3.82) as in Part (b), σ(e+ e− → hadrons)

X Z 1 X Nc |p′ | Z 1 dσ = Nc dcos θ dcos θ |M|2 2π = 2 |p| dΩ 32πE −1 −1 cm cm i i X Nc |p′ | = 2 |p| 32πEcm i 4

2 16qi2 e2 Ecm Ecm (m2i + m2e ) 2|p|2 |p′ |2 × + + 4 Ecm 8 2 3

2 4 2 8qi eg vi ve Ecm Ecm (m2i + m2e ) 2|p|2 |p′ |2 − + + 2 (E 2 − m2 ) cos2 θw Ecm 8 2 3 cm Z

4 4 2 ′ 2 1 Ecm g 2|p| |p | 2 2 2 2 (vi + ai )(ve + ae ) + + 2 − m2 )2 cos4 θw (Ecm 8 3 Z 2

E − cm (vi2 + a2i )(a2e − ve2 )m2e + (a2i − vi2 )(ve2 + a2e )m2i 2 +4(2a2i a2e − a2i ve2 − a2e vi2 )m2i m2e

1 2 4g 4 m2i m2e a2i a2e 2 2 E − mZ Ecm .

(9.103) − 4 2 − m2 )2 cos θw m4Z (Ecm 4 cm Z If we are to take the ultra-high energy limit Ecm → ∞ then only the highest powers of Ecm matter and |p| = |p′ | = Ecm /2, so σ(e+ e− → hadrons)|Ecm →∞

4 4 4 X Nc 16q 2 e2 E 4 Ecm 8qi eg 2 vi ve Ecm Ecm cm i + − + = 2 4 4 32πEcm Ecm 8 24 cos2 θw Ecm 8 24 i

4 4 g 4 (vi2 + a2i )(ve2 + a2e ) Ecm Ecm 4g 4 m2i m2e a2i a2e 1 4 + 4 + E − 4 4 cos θw Ecm 8 24 cos4 θw m4Z Ecm 4 cm

X Nc 4qi eg 2 vi ve 8qi2 e2 − = 2 32πEcm 3 3 cos2 θw i

g4 (vi2 + a2i )(ve2 + a2e ) m2i m2e a2i a2e + 4 − . (9.104) cos θw 6 m4Z As a check, consider the first of these terms which is due solely to photon exchange, X i

X q2 X q 2 4πα2 Nc 8qi2 e2 e4 i i = N = N c c 2 2 12πE 2 2 3E 2 32πEcm 3 e e cm cm i i X q2

(9.105)

i σ(e+ e− → µ+ µ− )|Ecm →∞ 2 e i "

2 # 2 2 −1 =3 3 +3 σ(e+ e− → µ+ µ− )|Ecm →∞ .

3 3

= Nc

Gauge Field Theories

220

We see that this contribution is equal to the ultrarelativistic cross-section for P e+ e− → µ+ µ− found in Eq. (7.6.73), modified by a factor of Nc i qi2 /e2 as was reasoned in Eq. (9.2.40).

9.8 Problem 8 Problem: (a) Calculate to lowest order the decay rate h → bb̄; and (b) Calculate the decay rate for h → gg. (Hint: This problem requires a one-loop calculation and some care and effort.) (c) Draw the lowest order diagrams contributing to the decay h → γγ.

Solution: (a) The Higgs boson directly couples to the bottom quark and so the decay h → bb̄ can occur at tree-level. From Eq. (9.4.30) we see that the bb̄h vertex factor is −imb /v, where mb is the mass of the bottom quark. The amplitude for the decay is therefore p

b

q

= iM =

h p0

−imb s ū (p)v r (p′ ), v

(9.106)

b̄

so the unpolarized squared amplitude is X X m2 b [ūs (p)v r (p′ )][v̄ r (p′ )us (p)] |M|2 = MM∗ = 2 v r,s r,s m2b ′ tr[(p / − mb )(p / + mb )] v2

m2 ′ 2 = 2b tr[p /p /] − mb tr[I] v

4m2b ′ (9.107) = 2 p · p − m2b v by the Casimir trick Eq. (7.6.61) together with Eq. (A.3.24). Momentum conservation tells us q 2 = (p′ + p)2 = p′2 + 2p′ · p + p2 , or p′ · p = 12 (q 2 − p′2 − p2 ) = 1 (m2h − 2m2b ). This gives 2 =

2m2b m2h − 4m2b .

(9.108) 2 v So far this is only the squared amplitude for the Higgs to decay to a single |M|2 =

221

Problem 8

color of bottom quark. Since there are Nc = 3 different color-anticolor pairs of quarks the Higgs could decay into, we should multiply by the number of colors Nc = 3 to get the full squared amplitude for h → bb̄.

The decay rate for this decay is then obtained from Eq. (7.3.72), Z d3 p d3 p′ (2π)4 δ 4 (q − p − p′ ) Nc |M|2 Γ(h → bb̄) = , (9.109) (2π)3 (2π)3 4Ep Ep′ 2Eh p p p where Ep = m2b + p2 , Ep′ = m2b + p′2 , and Eh = m2h + q2 . Let us evaluate this in the Higgs rest frame, so q µ = (mh , 0) and Eh = mh . Then Z d3 p d3 p′ (2π)4 δ(mh − Ep − Ep′ )δ 3 (−p − p′ ) Nc |M|2 Γ(h → bb̄) = (2π)3 (2π)3 4Ep Ep′ 2mh Z 3 2 d p δ(mh − Ep − E−p ) Nc |M| = (2π)2 4Ep E−p 2mh p

Z d3 p δ(mh − 2 m2b + p2 ) Nc m2b m2h − 4m2b = .

(9.110) (2π)2 4(m2b + p2 ) v 2 mh This is most easily evaluated in spherical coordinates, reusing the label p for the radial coordinate p ≡ |p|, p

Z Z ∞ dp p2 δ(mh − 2 m2b + p2 ) Nc m2b m2h − 4m2b Γ(h → bb̄) = dΩ3 2 4(m2b + p2 ) v 2 mh 0 (2π)

p Z ∞ dp p2 δ(mh − 2 m2b + p2 ) Nc m2b m2h − 4m2b . (9.111) = 4π 2 4(m2b + p2 ) v 2 mh 0 (2π) p Here the Dirac delta function vanishes unless p = 21 m2h − 4m2b , which is simply the statement that energy is conserved. From Eq. (A.2.8) we then get that p q δ(p − 12 m2h − 4m2b ) 2 2 δ(mh − 2 mb + p ) = p d 2 2 dp (mh − 2 mb + p ) p q m2b + p2 δ(p − 21 m2h − 4m2b ), = (9.112) 2p ignoring the unphysical negative solution for p. The decay rate then reduces to p

Z ∞ dp p δ(p − 21 m2h − 4m2b ) Nc m2b m2h − 4m2b p Γ(h → bb̄) = 8π v 2 mh m2b + p2 0 p

m2h − 4m2b 1 Nc m2b m2h − 4m2b = 1 16π mh v 2 mh 2 3/2 Nc m2b m2h − 4m2b = .

(9.113) 8πv 2 m2h Notice this is positive, and therefore represents a possible decay, only when mh > 2mb , as should be expected.

Gauge Field Theories

222

(b) In contrast to the case of the bottom quark, the Higgs boson does not directly couple to the gluon and so the decay h → gg can only occur at loop level. The most significant diagrams contributing to this process are those with a top quark loop because the top quark is the heaviest elementary particle and therefore has the strongest coupling to the Higgs boson. For now we will assume a single quark with mass m and sum over quark flavors f at the end of the calculation.

From Eq. (9.4.30), the quark-Higgs vertex is given by −imδ ij /v where m is the quark mass and i, j are color indices. The q q̄g vertex has the vertex igγ µ Tija .

There are two topologically distinct diagrams with the quark loop reversed in direction. We label the loop momentum as −ℓ in the second diagram so that the propagator denominators have the same form. The total amplitude is therefore p `+p q h

i

j

n

k

m

`−p0

µ, a

+

` ν, b

l p0

p

−imδ jk v

−(`+p) q j h

i

k −(`−p0 )

µ, a

n m

−` ν, b

l p0

a b ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ (igTni )δ ij δ kl (igTlm )δ mn (9.114)

Z ′ tr γ µ i(/ℓ + p d4 ℓ / + m)i(/ℓ − p / + m)γ ν i(/ℓ + m) × (2π)4 [(ℓ + p)2 − m2 + iϵ][(ℓ − p′ )2 − m2 + iϵ][ℓ2 − m2 + iϵ]

−imδ jk b a + (−1) ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ (igTml )δ lk δ ij (igTin )δ nm (9.115) v

Z ′ tr γ ν i(−/ℓ + p d4 ℓ / + m)i(−/ℓ − p / + m)γ µ i(−/ℓ + m) × (2π)4 [(ℓ + p)2 − m2 + iϵ][(ℓ − p′ )2 − m2 + iϵ][ℓ2 − m2 + iϵ]

mg 2 1 ab =− δ ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ (9.116) v 2

Z ′ d4 ℓ tr γ µ (/ℓ + p / +m)(/ℓ − p / +m)γ ν (/ℓ +m) +(µ, p ↔ ν, p′ ; ℓ → −ℓ) × (2π)4 [(ℓ + p)2 − m2 + iϵ][(ℓ − p′ )2 − m2 + iϵ][ℓ2 − m2 + iϵ]

= (−1)

where we have used tr[T a T b ] = tr[T b T a ] = 21 δ ab from Eq. (A.7.26), where the fermion loop gives a factor of (−1) and where the spinor traces are ordered right to left in the direction of the fermion arrow. Now since the trace of an odd number of γ-matrices vanishes, only the terms O(m) and O(m3 ) survive the trace. The numerator of the integrand then reduces to

′ ν ′ tr γ µ (/ℓ + p (9.117) / + m)(/ℓ − p / + m)γ (/l + m) + (µ, p ↔ ν, p ; ℓ → −ℓ)

′ ν µ ν µ ′ ν 3 µ ν = mtr[γ µ (/ℓ − p / )γ /ℓ +γ (/ℓ + p /)γ /ℓ +γ (/ℓ + p /)(/ℓ − p / )γ ]+m tr[γ γ ] + (µ, p ↔ ν, p′ ; ℓ → −ℓ)

= m δαµ (ℓ−p′ )β δγν ℓδ + δαµ (ℓ+p)β δγν ℓδ + δαµ (ℓ+p)β (ℓ−p′ )γ δδν tr[γ α γ β γ γ γ δ ]

Problem 8

223

+4m3 g µν + (µ, p ↔ ν, p′ ; ℓ → −ℓ)

= 4m δαµ (ℓ−p′ )β δγν ℓδ + δαµ (ℓ+p)β δγν ℓδ + δαµ (ℓ+p)β (ℓ−p′ )γ δδν

× (g αβ g γδ − g αγ g βδ + g αδ g βγ )+m2 g µν + (µ, p ↔ ν, p′ ; ℓ → −ℓ)

= 4m (ℓ−p′ )µ ℓν −g µν (ℓ−p′ )·ℓ+ℓµ (ℓ−p′ )ν +(ℓ+p)µ ℓν −g µν (ℓ+p)·ℓ+ℓµ (ℓ+p)ν

+(ℓ+p)µ (ℓ−p′ )ν −(ℓ+p)ν (ℓ−p′ )µ +g µν (ℓ+p)·(ℓ−p′ )+m2 g µν +(µ, p ↔ ν, p′ ; ℓ → −ℓ),

= 4m 4ℓµ ℓν + pν p′µ −pµ p′ν +2pµ ℓν −2p′ν ℓµ +g µν {m2 −ℓ2 −p·p′ } +(µ, p ↔ ν, p′ ; ℓ → −ℓ),

= 8m 4ℓµ ℓν + pν p′µ −pµ p′ν +2pµ ℓν −2p′ν ℓµ +g µν {m2 −ℓ2 −p·p′ } , where we have used tr[γ α γ β γ γ γ δ ] = 4(g αβ g γδ − g αγ g βδ + g αδ g βγ ). The gluon polarization vectors satisfy pµ ϵµ (p, λ)∗ = 0 and p′ν ϵν (p′ , λ′ )∗ = 0. Therefore only the ℓµ ℓν , pν p′µ and g µν terms will survive. Conservation of momentum tells us m2h = q 2 = (p + p′ )2 = p2 + 2p · p′ + p′2 = 2p · p′ , or p · p′ = 21 m2h . We have since gluons are massless p2 = p′2 = 0. The invariant amplitude Eq. (9.114) becomes 2 2 4g m δ ab ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ (9.118) iM = − v Z 4ℓµ ℓν +pν p′µ +g µν [m2 −ℓ2 − 21 m2h ] d4 ℓ × .

(2π)4 [(ℓ+p)2 −m2 +iϵ][(ℓ−p′ )2 −m2 +iϵ][ℓ2 −m2 +iϵ] Using the Feynman parameterization in Eq. (A.5.6) we have 1 [(ℓ+p)2 −m2 +iϵ][(ℓ−p′ )2 −m2 +iϵ][ℓ2 −m2 +iϵ]

(9.119)

Z 1

= Γ(3)

dxdydz δ(1−x−y−z) 0

1 {x[(ℓ+p)2 −m2 +iϵ] + y[(ℓ−p′ )2 −m2 +iϵ]+z[ℓ2 −m2 +iϵ]}3 Z 1 Z 1−x 1 =2 dx dy 2 {ℓ +2(xp−yp′ )·ℓ−m2 +iϵ]}3 0 0 Z 1 Z 1−x 1 =2 dx , dy ′2 [ℓ −∆+iϵ]3 0 0 ×

where we have defined ℓ′ ≡ ℓ + (xp − yp′ ) and ∆ ≡ (xp − yp′ )2 + m2 = −2xyp · p′ + m2 = m2 − xym2h . The numerator in Eq. (9.118) when expressed in terms of ℓ′ and contracted with ϵµ (p, λ)∗ ϵµ (p′ , λ′ )∗ becomes ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ {4ℓµ ℓν +pν p′µ +g µν [m2 −ℓ2 − 21 m2h ]} ∗

′

′ ∗

′µ ′ν

ν ′µ

→ ϵµ (p, λ) ϵν (p , λ ) {4ℓ ℓ +(1−4xy)p p +g µν [m2 −ℓ′2 −(xp−yp′ )2 − 21 m2h ]}

→ ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ {(4ℓ′µ ℓ′ν − g µν ℓ′2 )+(1−4xy)pν p′µ

(9.120)

224

Gauge Field Theories +g µν [m2 +(xy− 12 )m2h ]}, where all terms involving a contraction of pµ and/or p′ν with their respective polarization vectors vanish, terms odd in ℓ′ are omitted since they vanish in the integration after the change of variables and again using p2 = p′2 = 0.

Using Eqs. (9.119) and (9.120) in Eq. (9.118) and then applying Eq. (A.6.4) leads to Z 1 Z 1−x Z d ′ 8m2 g 2 ab d ℓ dy (9.121) iM = − δ ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ dx d v (2π) 0 0 {(4ℓ′µ ℓ′ν −g µν ℓ′2 )+(1−4xy)pν p′µ +g µν [m2 +(xy− 21 )m2h ]} × [ℓ′2 − ∆ + iϵ]3 Z 1 Z 1−x Z d ′ d ℓ 8m2 g 2 ab ∗ ′ ′ ∗ dy dx =− δ ϵµ (p, λ) ϵν (p , λ ) d v (2π) 0 0 1 µν 4 ′2 ν ′µ µν 2 {g ( d − 1)ℓ +(1−4xy)p p +g [m +(xy− 2 )m2h ]} × , [ℓ′2 − ∆ + iϵ]3 where we have used Eq. (A.6.9). The numerator consists of an ℓ′ -independent component and an ℓ′ -dependent component. Let us start with the latter, which requires the evaluation of

2−(d/2)

Z d ′ ( d4 −1)ℓ′2 i d ℓ 4 d Γ(2− d2 ) 1 −1 (9.122) = (2π)d [ℓ′2 −∆+iϵ]3 d 2 ∆ (4π)d/2 2

4 id 1 2 ϵ i 2 i = −1 −ln ∆+ln(4πe−γ )+O(ϵ) = +O(ϵ) = , d 4 (4π)2 ϵ 4 (4π)2 ϵ 32π 2 where we have used Eqs. (A.6.5) and (A.6.17). The ℓ′ -independent numerator term gives a finite contribution and using Eq. (A.6.4) we have 3−(d/2) Z d ′ d ℓ 1 −i Γ(3− d2 ) 1 −i 1 = = . (9.123) (2π)d [ℓ′2 −∆+iϵ]3 2 ∆ 32π 2 ∆ (4π)d/2 Then putting these results together we find that Eq. (9.121) becomes Z 1 Z 1−x 8m2 g 2 ab ∗ ′ ′ ∗ iM = − δ ϵµ (p, λ) ϵν (p , λ ) dx dy (9.124) v 0 0

µν i (1−4xy)pν p′µ +g µν [m2 +(xy− 12 )m2h ] ig − × 32π 2 32π 2 ∆ Z 1 Z 1−x 2 2 8m g ab =− δ ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ dx dy v 0 0 i g µν ∆−(1−4xy)pν p′µ −g µν [m2 +(xy− 12 )m2h ] × 32π 2 ∆ Z 1 Z 1−x 8m2 g 2 ab ∗ ′ =− δ ϵµ (p, λ) ϵν (p , λ′ )∗ dx dy v 0 0 i g µν ( 12 −2xy)m2h −(1−4xy)pν p′µ × 32π 2 m2 − xym2h

Problem 8

225

Z 1−x Z 1 i 8m2 g 2 ab 2 µν ν ′µ 1 dy dx δ ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ m g −p p v 32π 2 2 h 0 0 1 1−4xy × 2 m 1 − xy(mh /m)2

ig 2 = − 2 δ ab ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ 21 m2h g µν −pν p′µ I(mh /m), 4π v where for a quark of flavor f and mass mf we define

Z 1 Z 1−x mh (1 − 4xy) If .

(9.125) ≡ dx dy 2 mf 1−xy(m h /mf ) 0 0 =−

Then summing over quark flavors in the quark loop the invariant amplitude is given by

X iαs ab mh ν ′µ ∗ ′ ′ ∗ 1 2 µν iM = − If δ ϵµ (p, λ) ϵν (p , λ ) 2 mh g −p p . (9.126) πv mf f

We need to sum over final state polarizations and colors (a, b = 1, . . . , 8). Since we do not have a Ward identity for QCD then we need to use the full expression for the sum over polarizations given in Eq. (9.127). Since the gluons are on-shell we can use k 2 = 0 and so P µ k µ nν +nµ k ν kµ kν µν ν ∗ + . ϵ (k, λ)ϵ (k, λ) = −g − λ (k·n)2 k·n With a little effort one can show that only the g µν contribution survives for on-shell gluons in an unpolarized final state, which we had to expect since the physical decay rate cannot depend on the arbitrary direction nµ . We then have for the unpolarized modulus-squared of the invariant amplitude P P |M|2 = a,b λ,λ′ |M|2 (9.127) 2 X X α = 2s2 δ ab δ ab ϵµ (p, λ)∗ ϵν (p′ , λ′ )∗ ϵα (p, λ)ϵβ (p′ , λ′ ) π v ′ a,b

×

1 2

λ,λ

m2h g µν −pν p′µ

8α2 = 2 s2 (−gµα )(−gνβ ) π v

1 2

m2h g αβ −pβ p′α

X

If

f

1 m2h g µν −pν p′µ 2

mh mf

2

X 1 m2h g αβ −pβ p′α If 2

f

mh mf

2

2

=

X 8αs2 4( 12 m2h )2 −m2h p · p′ If 2 2 π v f

mh mf

2

=

4m4h αs2 π2 v2

X f

If

mh mf

=

4m4h ααs2 πm2W sin2 θw

X f

If

mh mf

2 , (9.128)

where we have used mW = gv/2 and e = g sin θw so that v = 2mW sin θw /e,

Gauge Field Theories

226

P where a,b δ ab δ ab = 8, and where since gluons are massless then like photons P we can effectively replace λ ϵµ (p, λ)ϵν (p, λ)∗ → −gµν by Eq. (9.127).

The decay rate is given by Eq. (7.3.72), including a symmetry factor of 21 because the gluons in the final state are identical, Z 1 d3 p d3 p ′ (2π)4 δ 4 (q − p − p′ )|M|2 Γ(h → gg) = , (9.129) 3 3 2 (2π) 2Ep (2π) 2Ep′ 2Eh where q µ is the Higgs four-momentum. It is easiest to work in the CM frame where q µ = (mh , 0) and p = −p′ . Since gluons are massless we can write p ≡ |p| = Ep and p′ ≡ |p′ | = Ep′ and so p′ = p = 21 mh . Since |M|2 has no angular dependence then the phase space integral becomes Z Z 3 1 d3 p d3 p ′ 1 1 d p 4 4 ′ (2π)δ(mh − 2p) (2π) δ (q − p − p ) = 2 (2π)3 2Ep (2π)3 2Ep′ 2 m2h (2π)3 Z Z 1 1 4π 1 1 3 1 1 d p [ dp p2 12 δ(p− 12 mh ) = = , δ(p − m )] = h 2 2 2 2 2 2 2mh (2π) 2mh (2π) 16π where we have used Eq. (A.2.8) for the δ-function. So for a Higgs boson at rest the decay rate for h → gg is 1 |M|2 |M|2 m3 α 2 X Γ(h → gg)rest = = = h3 s2 If 16π 2mh 32πmh 8π v f

X ααs2 m3h If = 2 2 2 8π mW sin θw f

mh mf

mh mf

2

(9.130)

2 .

(9.131)

From the expression in Eq. (9.125) for If (mm /mf ), we see that the top quark loop will give the largest contributon as we might expect since the Higgs-quark coupling is proportional to the quark mass. We also note that xy ≤ 12 and so If develops a singularity when 2mf < mh , which is when the loop momenta contain the f f¯ production threshold. To remove the h → f f¯ production contribution to If we need to take the principal part of If for each quark flavor with 2mf < mh . (c) The lowest-order contributions to h → γγ involve: (i) a sum over all quark (u, d, s, c, t) and massive lepton (e− , µ− , τ − ) one-loop contributions in the first diagram and (ii) the two W -loop contributions in the second and third diagrams. This can be deduced from examining the available interaction vertices in Eqs. (9.4.14), (9.4.15), (9.4.22) and (9.4.30).

W+

W+ h

W+

h W

h

+

W+

(9.132)

Problem 9

227

9.9 Problem 9 Problem: Consider an SU (3) nonabelian gauge theory with coupling to eight real scalar fields in the adjoint representation, ϕa for a = 1, . . . , 8. The covariant derivative is then Dµab ϕb = ∂µ + gf abc Abµ ϕc . Define Φ = ϕa T a , where T a are the eight SU (3) generators. Write down the Lagrangian density for the theory. Assume that some potential is added that causes Φ to have a vacuum expectation value, Φ0 = ⟨Ω|Φ̂|Ω⟩ ̸= 0. Since Φ0 is a traceless hermitian matrix we can diagonalize it.

Assume that after doing so we found Φ0 → ΦD 0 = vD with v > 0 and with D being the diagonal matrix D = diag( 12 , 12 , −1). By referring to Eq. (5.3.33) explain why this corresponds to to the symmetry breaking SU (3) → SU (2) × U (1) with SU (3) and SU (2) in their defining 3 × 3 and 2 × 2 representations respectively. Explain why the “hypercharge operator” Y must satisfy Y ∝ D. How many gauge bosons acquire a mass and how many remain massless? Calculate the mass of the massive gauge bosons at tree level.

Solution: An SU (3) gauge theory with real scalar fields ϕa (a = 1, . . . , 8) of equal mass m has a Lagrangian L=

1 1 a µν 1 ab b (D ϕ )(Dµ ac ϕc ) − m2 ϕa ϕa − Fµν Fa , 2 µ 2 4

(9.133)

where Dµab ϕb ≡ ∂µ ϕa + gf abc Abµ ϕc for fields in the adjoint representation. Defining Φ = ϕa T a where T a are the generators of SU (3) in the fundamental representation, note that Dµ Φ ≡ (Dµ ϕ)a T a = (Dµab ϕb )T a = (∂µ ϕa + gf abc Abµ ϕc )T a .

(9.134)

Then using the general relation tr[T a T b ] = 12 δ ab of Eq. (A.7.26), we have tr[Φ2 ] = tr[(ϕa T a )(ϕb T b )] =

1 1 a b ab ϕ ϕ δ = ϕ a ϕa 2 2

(9.135)

and similarly tr[(Dµ Φ)(Dµ Φ)] = tr[(Dµab ϕb )T a (Dµ dc ϕc )T d ] 1 = (Dµab ϕb )(Dµ dc ϕc )δ ad 2 1 = (Dµab ϕb )(Dµ ac ϕc ), 2

(9.136)

and so the Lagrangian can be written as 1 a µν L = tr[(Dµ Φ)(Dµ Φ)] − m2 tr[Φ2 ] − Fµν Fa .

4

(9.137)

Now let us suppose that Φ acquires a vacuum expectation value ⟨Ω|Φ|Ω⟩ ≡ Φ0 =

228

Gauge Field Theories

vD ≡ v diag( 12 , 12 , −1) for some v > 0. We can write this in the notation of Eq. (5.3.33),

1 0 2 I2 (9.138) Φ0 = v 0 − 11 I1 to see that we have broken SU (3) → SU (2)×SU (1)×U (1) with SU (3) and SU (2) in their defining representations. But since SU (1) is the trivial group with one element, this is equivalent to the breaking SU (3) → SU (2) × U (1). To be more explicit, a symmetry is broken if the vacuum expectation value is not invariant under the symmetry transformation. To see how Φ transforms under the symmetry, consider its covariant derivative Dµ Φ = (∂µ ϕa + gf abc Abµ ϕc )T a = ∂µ Φ − igAbµ ϕc (if bca T a ) = ∂µ Φ − igAbµ ϕc [T b , T c ] = ∂µ Φ − igAbµ [T b , Φ]. a

(9.139) a

The symmetry generated by T will therefore be broken if and only if [T , Φ0 ] ̸= 0.

It is easy to see from the expressions for the generators in Eq. (A.7.27) that only T 1 , T 2 , T 3 , and T 8 commute with Φ0 . The first three of these generate an SU (2) subgroup while T 8 generates a U (1) subgroup, so again we arrive at the unbroken symmetry group SU (2) × U (1). Having dim SU (3) = 32 − 1 = 8 symmetries, of which dim SU (2) × U (1) = (22 − 1) + 1 = 4 are unbroken, we expect there to be 8 − 4 = 4 massive gauge bosons and 4 gauge bosons that remain massless.

Since hypercharge corresponds to an unbroken U (1) subgroup, which is abelian, its operator Y must commute with every unbroken generator. It follows from some quick computations with the matrices T 1 , T 2 , T 3 , and T 8 that Y must have the block diagonal form

AI2 0 Y = .

(9.140) 0 BI1 Furthermore, since Y is in the Lie algebra of SU (3), it must be traceless, ensuring 2A + B = 0. In total then, we have

1 I 0 , (9.141) Y = −B 2 2 0 − 11 I1 and therefore Y ∝ D. The masses of the gauge bosons arise from the covariant derivative acting on √ Φ0 .

To calculate this it helps to recognize from Eq. (A.7.27) that Φ0 = vD = v√ 3T 8 , i.e. that the vacuum expectation values of the original fields ϕa are ϕa0 = v 3δ a8 . √ √ Then Dµab ϕb0 = gf abc Abµ ϕc0 = 3gvf abc Abµ δ c8 = 3gvf ab8 Abµ , so 1 ab b (D ϕ )(Dµ ac ϕc0 ) 2 µ 0 3 = g 2 v 2 f ab8 f ac8 Abµ Aµ c .

2

tr[(Dµ Φ0 )(Dµ Φ0 )] =

(9.142)

229

Problem 9 From Eq. (A.7.28) we know that the relevant structure constants are f 458 = f 678 = √ 3/2 with all others vanishing, so the summation explicitly becomes √ !2

3 2 2 3 µ tr[(Dµ Φ0 )(D Φ0 )] = g v A4µ Aµ 4 + A5µ Aµ 5 + A6µ Aµ 6 + A7µ Aµ 7 .

2 2 (9.143) Recall that the conventionally normalized gauge boson mass terms (in the mass 2 2 Aµ Aµ . Matching to the above, we see 12 Mgauge = eigenbasis) are of the form + 12 Mgauge 2 8 4 5 6 9g v /8, or Mgauge = 3gv/2. Our massive gauge bosons are Aµ , Aµ , Aµ , and A7µ , as would be expected from the broken generators.

References

Itzykson, C., and Zuber, J.B. 1980. Quantum Field Theory. New York: McGrawHill.

231