ALL 9 CHAPTER COVERED
SOLUTIONS MANUAL
, Contents
Preface to Solutions Manual page vi
1 Lorentz and Poincaré Invariance 1
1.1 Problem 1 1
1.2 Problem 2 2
1.3 Problem 3 3
1.4 Problem 4 5
1.5 Problem 5 5
1.6 Problem 6 7
1.7 Problem 7 8
1.8 Problem 8 8
1.9 Problem 9 9
1.10 Problem 10 10
1.11 Problem 11 11
1.12 Problem 12 12
1.13 Problem 13 13
1.14 Problem 14 13
1.15 Problem 15 14
1.16 Problem 16 15
2 Classical Mechanics 18
2.1 Problem 1 18
2.2 Problem 2 19
2.3 Problem 3 20
2.4 Problem 4 22
2.5 Problem 5 24
2.6 Problem 6 29
2.7 Problem 7 30
2.8 Problem 8 34
2.9 Problem 9 35
2.10 Problem 10 36
2.11 Problem 11 37
2.12 Problem 12 40
2.13 Problem 13 42
2.14 Problem 14 43
2.15 Problem 15 45
iii
,iv Contents
3 Relativistic classical fields 48
3.1 Problem 1 48
3.2 Problem 2 52
3.3 Problem 3 53
3.4 Problem 4 57
3.5 Problem 5 60
3.6 Problem 6 61
3.7 Problem 7 61
3.8 Problem 8 64
4 Relativistic Quantum Mechanics 68
4.1 Problem 1 68
4.2 Problem 2 69
4.3 Problem 3 72
4.4 Problem 4 75
4.5 Problem 5 76
4.6 Problem 6 77
4.7 Problem 7 79
4.8 Problem 8 81
4.9 Problem 9 82
4.10 Problem 10 86
4.11 Problem 11 88
4.12 Problem 12 89
5 Introduction to Particle Physics 91
5.1 Problem 1 91
5.2 Problem 2 92
5.3 Problem 3 94
5.4 Problem 4 95
5.5 Problem 5 97
5.6 Problem 6 98
5.7 Problem 7 100
5.8 Problem 8 102
5.9 Problem 9 103
5.10 Problem 10 104
5.11 Problem 11 108
5.12 Problem 12 110
6 Formulation of Quantum Field Theory 117
6.1 Problem 1 117
6.2 Problem 2 118
6.3 Problem 3 120
6.4 Problem 4 121
6.5 Problem 5 123
,v Contents
6.6 Problem 6 127
6.7 Problem 7 130
6.8 Problem 8 133
6.9 Problem 9 135
6.10 Problem 10 147
6.11 Problem 11 150
6.12 Problem 12 151
6.13 Problem 13 154
7 Interacting Quantum Field Theories 163
7.1 Problem 1 163
7.2 Problem 2 164
7.3 Problem 3 164
7.4 Problem 4 165
7.5 Problem 5 166
7.6 Problem 6 167
7.7 Problem 7 168
7.8 Problem 8 170
7.9 Problem 9 171
7.10 Problem 10 173
8 Symmetries and Renormalization 176
8.1 Problem 1 176
8.2 Problem 2 177
8.3 Problem 3 179
8.4 Problem 4 180
8.5 Problem 5 181
8.6 Problem 6 182
8.7 Problem 7 189
8.8 Problem 8 191
9 Gauge Field Theories 192
9.1 Problem 1 192
9.2 Problem 2 194
9.3 Problem 3 198
9.4 Problem 4 200
9.5 Problem 5 202
9.6 Problem 6 205
9.7 Problem 7 208
9.8 Problem 8 220
9.9 Problem 9 227
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) un-
stable 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 un-
stable particles at t = 0. The fraction of particles decaying in the interval t to t + dt
−t/τ
is −dN/N
R∞ 0 = (−dN/dt)dt/NR 0 = (1/τ )e dt. So the mean lifetime (or lifetime)
∞ −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 par-
ticles that have traveled enormous distances from outside our solar system. Primary
cosmic rays are particles that have been accelerated by some extreme astrophysi-
cal event and secondary cosmic rays are those resulting from collisions of primary
cosmic rays with interstellar gas or with our atmosphere. Most cosmic rays reach-
ing our atmosphere will be stable particles such as photons, neutrinos, electrons,
protons and stable atomic nuclei (mostly helium nuclei). Muon cosmic rays there-
fore 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
,2 Lorentz and Poincaré Invariance
is, the speed of the muon must satisfy
q
v2
L 1− c2 L
v= = (1.1)
γ t1/2 t1/2
to have a 50% chance of reaching the surface before decaying. This can be rearranged
to give
L
v=q . (1.2)
L2
t21/2 + c2
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
15 × 103 m
v=q
(15×103 m)2
(1.523 × 10−6 s)2 + (3×108 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 . (1.6)
L2
T2 + c2
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 (1.7)
(80 years)2 + (25,000 years)2
= 0.999995c (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
, 4 Lorentz and Poincaré Invariance
iBK2
1
1
2 2 bT
+2
2
1
t
Figure 1.1 6B;m`2representation
A two-dimensional k, >vT2`#QH
b Q7 7`
K2@BMp
`B
M+2X
of spacetime UTXR8V
displacement with displacements
represented by vectors. Vectors labeled by the same number (1 or 2) are related by
Lorentz transformations.
o
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
large as that in frame O, showing that o separation between the events is
` the time
minimized in this frame.
` in frame O as
To see this algebraically, label the two events
o
T
E1 = (ct1 , x) , (1.10)
E2 = (ct2 , x) `,
T
(1.11)
where t2 > t1 . Then under a Lorentz transformation Λ, the events go to
E1 → E1′ = ΛE1 = Λ(ct1 , x)T , (1.12)
E2 → 6B;m`2
= ΛEj,
E2′ 2 =UTXRNV
Λ(ct2 , x) . T
(1.13)
The difference between these events is
T
E2′ − E1′ = Λ (c(t2 − t1 ), 0) . (1.14)
In particular, the temporal component of this differenceois Λ00 c(t2 − t1 ), which is at
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.
6B;m`2 9, UTXRNV
SOLUTIONS MANUAL
, Contents
Preface to Solutions Manual page vi
1 Lorentz and Poincaré Invariance 1
1.1 Problem 1 1
1.2 Problem 2 2
1.3 Problem 3 3
1.4 Problem 4 5
1.5 Problem 5 5
1.6 Problem 6 7
1.7 Problem 7 8
1.8 Problem 8 8
1.9 Problem 9 9
1.10 Problem 10 10
1.11 Problem 11 11
1.12 Problem 12 12
1.13 Problem 13 13
1.14 Problem 14 13
1.15 Problem 15 14
1.16 Problem 16 15
2 Classical Mechanics 18
2.1 Problem 1 18
2.2 Problem 2 19
2.3 Problem 3 20
2.4 Problem 4 22
2.5 Problem 5 24
2.6 Problem 6 29
2.7 Problem 7 30
2.8 Problem 8 34
2.9 Problem 9 35
2.10 Problem 10 36
2.11 Problem 11 37
2.12 Problem 12 40
2.13 Problem 13 42
2.14 Problem 14 43
2.15 Problem 15 45
iii
,iv Contents
3 Relativistic classical fields 48
3.1 Problem 1 48
3.2 Problem 2 52
3.3 Problem 3 53
3.4 Problem 4 57
3.5 Problem 5 60
3.6 Problem 6 61
3.7 Problem 7 61
3.8 Problem 8 64
4 Relativistic Quantum Mechanics 68
4.1 Problem 1 68
4.2 Problem 2 69
4.3 Problem 3 72
4.4 Problem 4 75
4.5 Problem 5 76
4.6 Problem 6 77
4.7 Problem 7 79
4.8 Problem 8 81
4.9 Problem 9 82
4.10 Problem 10 86
4.11 Problem 11 88
4.12 Problem 12 89
5 Introduction to Particle Physics 91
5.1 Problem 1 91
5.2 Problem 2 92
5.3 Problem 3 94
5.4 Problem 4 95
5.5 Problem 5 97
5.6 Problem 6 98
5.7 Problem 7 100
5.8 Problem 8 102
5.9 Problem 9 103
5.10 Problem 10 104
5.11 Problem 11 108
5.12 Problem 12 110
6 Formulation of Quantum Field Theory 117
6.1 Problem 1 117
6.2 Problem 2 118
6.3 Problem 3 120
6.4 Problem 4 121
6.5 Problem 5 123
,v Contents
6.6 Problem 6 127
6.7 Problem 7 130
6.8 Problem 8 133
6.9 Problem 9 135
6.10 Problem 10 147
6.11 Problem 11 150
6.12 Problem 12 151
6.13 Problem 13 154
7 Interacting Quantum Field Theories 163
7.1 Problem 1 163
7.2 Problem 2 164
7.3 Problem 3 164
7.4 Problem 4 165
7.5 Problem 5 166
7.6 Problem 6 167
7.7 Problem 7 168
7.8 Problem 8 170
7.9 Problem 9 171
7.10 Problem 10 173
8 Symmetries and Renormalization 176
8.1 Problem 1 176
8.2 Problem 2 177
8.3 Problem 3 179
8.4 Problem 4 180
8.5 Problem 5 181
8.6 Problem 6 182
8.7 Problem 7 189
8.8 Problem 8 191
9 Gauge Field Theories 192
9.1 Problem 1 192
9.2 Problem 2 194
9.3 Problem 3 198
9.4 Problem 4 200
9.5 Problem 5 202
9.6 Problem 6 205
9.7 Problem 7 208
9.8 Problem 8 220
9.9 Problem 9 227
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) un-
stable 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 un-
stable particles at t = 0. The fraction of particles decaying in the interval t to t + dt
−t/τ
is −dN/N
R∞ 0 = (−dN/dt)dt/NR 0 = (1/τ )e dt. So the mean lifetime (or lifetime)
∞ −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 par-
ticles that have traveled enormous distances from outside our solar system. Primary
cosmic rays are particles that have been accelerated by some extreme astrophysi-
cal event and secondary cosmic rays are those resulting from collisions of primary
cosmic rays with interstellar gas or with our atmosphere. Most cosmic rays reach-
ing our atmosphere will be stable particles such as photons, neutrinos, electrons,
protons and stable atomic nuclei (mostly helium nuclei). Muon cosmic rays there-
fore 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
,2 Lorentz and Poincaré Invariance
is, the speed of the muon must satisfy
q
v2
L 1− c2 L
v= = (1.1)
γ t1/2 t1/2
to have a 50% chance of reaching the surface before decaying. This can be rearranged
to give
L
v=q . (1.2)
L2
t21/2 + c2
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
15 × 103 m
v=q
(15×103 m)2
(1.523 × 10−6 s)2 + (3×108 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 . (1.6)
L2
T2 + c2
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 (1.7)
(80 years)2 + (25,000 years)2
= 0.999995c (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
, 4 Lorentz and Poincaré Invariance
iBK2
1
1
2 2 bT
+2
2
1
t
Figure 1.1 6B;m`2representation
A two-dimensional k, >vT2`#QH
b Q7 7`
K2@BMp
`B
M+2X
of spacetime UTXR8V
displacement with displacements
represented by vectors. Vectors labeled by the same number (1 or 2) are related by
Lorentz transformations.
o
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
large as that in frame O, showing that o separation between the events is
` the time
minimized in this frame.
` in frame O as
To see this algebraically, label the two events
o
T
E1 = (ct1 , x) , (1.10)
E2 = (ct2 , x) `,
T
(1.11)
where t2 > t1 . Then under a Lorentz transformation Λ, the events go to
E1 → E1′ = ΛE1 = Λ(ct1 , x)T , (1.12)
E2 → 6B;m`2
= ΛEj,
E2′ 2 =UTXRNV
Λ(ct2 , x) . T
(1.13)
The difference between these events is
T
E2′ − E1′ = Λ (c(t2 − t1 ), 0) . (1.14)
In particular, the temporal component of this differenceois Λ00 c(t2 − t1 ), which is at
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.
6B;m`2 9, UTXRNV
