Foundations Of Nuclear And Particle Physics Instructor Solution Manual Solutions T William Donnelly

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Foundations of Nuclear
and Particle Physics:
Solutions to Exercises
T. W. DONNELLY
MASSACHUSETTS INSTITUTE OF TECHNOLOGY, CAMBRIDGE, MA
J. A. FORMAGGIO
B. R. HOLSTEIN
UNIVERSITY OF MASSACHUSETTS, AMHERST, MA
R. G. MILNER
B. SURROW
TEMPLE UNIVERSITY, PHILADELPHIA, PA

Preface
The exercises and their solutions here were developed by us as a central pedagogical
tool in the use of the book by the student of nuclear and particle physics. We
have found that the typical graduate student will find them challenging and their
solution will demand a mature level of understanding. To preserve the effectiveness
of these exercises, the solutions manual should be retained by the instructor and
not be distributed to the students. We welcome all suggestions for improving the
solutions.
April 2019 T.W. Donnelly, B.R. Holstein, R.G. Milner
iii

Contents
1 Introduction page 1
2 Symmetries 5
3 Building Hadrons from Quarks 23
4 The Standard Model 40
5 QCD and Confinement 54
6 Chiral Symmetry and QCD 66
7 Elastic Electron Scattering from the Nucleon 77
8 Elastic Electron Scattering from the Nucleon 108
9 Hadron Structure via Lepton-Nucleon Scattering 118
10 High-Energy QCD 125
11 The Nucleon-Nucleon Interaction 134
12 The Structure and Properties of Few-Body Nuclei 147
13 Overview of Many-Body Nuclei 157
14 Models of Many-Body Nuclei 175
15 Electron Scattering from Discrete States 204
16 Electroexcitation of High-Lying Excitations of the Nucleus 244
17 Beta Decay 267
18 Neutrino Physics 278
iv

v Contents
19 The Physics of Relativistic Heavy-Ions 293
20 Astrophysics 307
21 Beyond the Standard Model Physics 319
22 Useful information 320
23 Quantum Theory 321
References 331

1 Introduction
1.1 US Energy Production
In 2011, the United States required 3,856 billion kW-hours of electricity. About
20% of this power was generated by ∼100 nuclear fission reactors. About 67% was
produced by the burning of fossil fuels, which accounted for about one-third of all
greenhouse gas emissions in the U.S. The remaining 13% was generated using other
renewable energy resources. Consider the scenario where all the fossil fuel power
stations are replaced by new 1-GW nuclear fission reactors. How many such reac-
tors would be needed?
Exercise 1.1 Solution: US Energy Production
3.856 × 106
GW-hours × 0.67 ×
years
8766 hours
×
Reactor
GW
= 295 Reactors
1.2 Geothermal Heating
It is estimated that 20 TW of heating in the earth is due to radioactive decay: 8
TW from 238
U decay, 8 TW from 232
Th decay, and 4 TW from 40
K decay. Estimate
the total amount of 238
U, 232
Th, and 40
K present in the Earth in order to produce
such heating.
Exercise 1.2 Solution: Geothermal Heating
The number of decays in unit time dt is equal to N(1 − e−dt/τ
) ≈ Ndt/τ. That
means the total mass of an isotope can be found from:
Mtot. =
τP
Edecay
× Matom
1) 238
U
i) P = 8 × 1012
W = 4.99 × 1025
MeV/s
ii) τ = 6.446 × 109
years = 1.64 × 1017
s
iii) Edecay = 4.267 MeV
iv) Matom = 238 amu = 3.95 × 10−25
kg
Mtot. = 9.39 × 1017
kg
2) 232
Th
1

2 Introduction
i) P = 8 × 1012
W = 4.99 × 1025
MeV/s
ii) τ = 2.027 × 101
0 years = 6.39 × 1017
s
iv) Matom = 232 amu = 3.85 × 10−25
kg
Mtot. = 3.00 × 1018
kg
3) 40
K
i) P = 4 × 1012
W = 2.50 × 1025
MeV/s
ii) τ = 1.805 × 109
years = 5.70 × 1016
s
iv) Matom = 40 amu = 6.64 × 10−26
kg
Mtot. = 7.23 × 1016
kg
1.3 Radioactive Thermoelectric Generators
A useful form of power for space missions which travel far from the sun is a ra-
dioactive thermoelectric generator (RTG). Such devices were first suggested by the
science fiction writer Arthur C. Clarke in 1945. An RTG uses a thermocouple to
convert the heat released by the decay of a radioactive material into electricity by
the Seebeck effect. The two Voyager spacecraft have been powered since 1977 by
RTGs using 238
Pu. Assuming a mass of 5 kg of 238
Pu, estimate the heat produced
and the electrical power delivered. (Do not forget to include the ∼ 5% thermocouple
efficiency.)
Exercise 1.3 Solution: Radioactive Thermoelectric Generators
Let’s first look at the instantaneous power produced. We know that:
P(t) =
N(t)Edecay
τ
=
N(0)e−t/τ
Edecay
τ
238
Pu has a 5.593 MeV α decay with a lifetime of τ = 126.5 years (3.99×109
s). 5 kg
of 238
Pu is 1.27 × 1025
atoms. At t = 0 years (1977), the thermal power produced
was 1.77 × 1016
MeV/s = 2.84 kW. At t = 38 years (2015), the power was down to
2.10 kW. A typical RTG efficiency is about 5%, so we can estimate that Voyager’s
power budget has been between 140 and 100 W.
We can integrate the thermal power to get the total heat produced.
Q =
Z t
0
P(t)dt = N(0)Edecay
h
1 − e−t/τ
i
For t = 38 years, we get a total heat Q = 2.95 × 1012
J.
1.4 Fission versus Fusion
Energy can be produced by either nuclear fission or nuclear fusion.

3 Introduction
a) Consider the fission of 235
U into 117
Sn and 118
Sn, respectively. Using the mass
information from a Table of Isotopes, calculate (i) the energy released per
fission and (ii) the energy released per atomic mass of fuel.
b) Consider the deuteron-triton fusion reaction
2
H + 3
H → 4
He + n .
Using the mass information from the Periodic Table of the Isotopes, calculate
(i) the energy released per fusion and (ii) the energy released per atomic mass
unit of fuel.
Exercise 1.4 Solution: Fission versus Fusion
Mass defect of 235
U: 40.9218 MeV
Mass defect of 117
Sn: −90.3977 MeV
Mass defect of 118
Sn: −91.6528 MeV
. Per fission:
E = (∆M235U − ∆M117Sn − ∆M118Sn) = 222.97 MeV
. Per amu:
E = 222.97 MeV/atom ×
atom
235 amu
= 0.949 MeV/amu
Mass defect of 2
H: 13.1357 MeV
Mass defect of 3
H: 14.9498 MeV
Mass defect of 4
He: 2.4249 MeV
Mass defect of n: 8.0713 MeV
. Per fission:
E = (∆M2H + ∆M3H − ∆M4He − ∆Mn) = 17.59 MeV
. Per amu:
E = 17.59 MeV/fission ×
fission
5 amu
= 3.518 MeV/amu
1.5 Absorption Lengths
A flux of particles is incident upon a thick layer of absorbing material. Find the
absorption length, the distance after which the particle intensity is reduced by a
factor of 1/e ∼ 37% (the absorption length) for each of the following cases:
a) when the particles are thermal neutrons (i.e., neutrons having thermal energies),
the absorber is cadmium, and the cross section is 24,500 barns,
b) when the particles are 2 MeV photons, the absorber is lead, and the cross section
is 15.7 barns per atom,

4 Introduction
c) when the particles are anti-neutrinos from a reactor, the absorber is the Earth,
and the cross section is 10−19
barns per atomic electron.
Exercise 1.5 Solution: Absorption Lengths
The beam intensity will fall exponentially, according to the differential equation:
dN
dx
= −
N
λ
where λ is the absorbtion length. The beam is reduced by factor of 1/e when x = λ.
The reduction of beam intensity is proportional to the number of interactions per
distance, i.e. σn, where σ is the interaction cross section and n is the number density
in the absorber.
λ = −
N
dN
dx
=
1
σn
In each part of this problem, one must find a plausible value for n and solve for λ.
1. Neutrons in Cadmium:
The density of cadmium is 8.65 g/cm3
. The atomic weight of cadmium is 112 amu
= 1.86 × 10−22
g, making n = 4.65 × 1022
cm−3
.
λ =
1
[4.65 × 1022 cm−3] [24500 b]
×
b
10−24 cm
= 8.78 × 10−4
cm = 8.78µm
2. Photons in Lead:
The density of lead is 11.3 g/cm3
. The atomic weight of lead is 207 amu =
3.44 × 10−22
g, making n = 3.28 × 1022
cm−3
.
λ =
1
[3.28 × 1022 cm−3] [15.7 b]
×
b
10−24 cm
= 1.94 cm
3. Anti-neutrinos through the Earth:
The average density of the earth is 5.51 g/cm3
. The four most abundant elements
on earth, oxygen, magnesium, silicon, and iron, make up more than 93% of
the earth’s mass. A mass-weighted average of their values of e per amu comes
to 0.488 e/amu = 2.94 ×1023
e/g. The electron density of the earth is thus
1.51 × 1024
e/cm3
.
λ =
1
[1.51 × 1024 cm−3] [10−19 b]
×
b
10−24 cm
= 6.62 × 1018
cm ≈ 7 light-years

2 Symmetries
2.1 Noether’s Theorem
The mathematician Emmy Noether proved a very important theorem to physics
which states that for any invariance of the classical action under a continuous field
transformation there exists a classical charge Q which is independent of time and
which is connected to a conserved current, ∂µJµ
= 0. That is, the existence of a
a symmetry requires the validity of a corresponding conservation law. Well known
examples in classical physics are the invariance of the equations of motion under
spatial translation, time translation, and rotation, which lead to conservation of
momentum, energy, and angular momentum respectively. The purpose of this ex-
ercise is to prove this theorem.
We begin with an action
S =
Z
d4
xL(φ, ∂µφ)
and consider an infinitesimal transformation of the field
φ → φ0
= φ + f(φ) , .
where f(φ) is some function of the field.
a) Find the equation of motion for a constant value of the infinitesimal and show
that one finds the Euler-Lagrange equation
δL
δφ
− ∂µ
δL
δ∂µφ
= 0 .
b) Calculate the change of the action under this field transformation in the case
that = (x) and show that
S → S0
= S +
Z
d4
x∂µjµ
,
where
jµ
=
δL
δ∂µφ
f(φ) .
c) Now integrate by parts and show that if the action is invariant we require that
∂µjµ
= 0, up to a total derivative.
5

6 Symmetries
d) Integrate the equation
∂j0
∂t
= −∇ · j = 0
over all space, and show that for a local “charge” distribution we have
dQ
dt
= 0 where Q =
Z
dxj0
,
which proves Noether’s theorem.
Exercise 2.1 Solution: Noether’s Theorem
a) We have
0 = δS =
Z
d4
x

δL
δφ
δφ +
δL
δ∂µφ
δ∂µφ

=
Z
d4
xf(φ)

δL
δφ
− ∂µ
δL
δ∂µφ

so
δL
δφ
− ∂µ
δL
δ∂µφ
= 0
b) Now
0 = δS =
Z
d4
x

δLδφ +
δL
δ∂µφ
δ∂µφ

=
Z
d4
x

f(φ)
δL
δφ
+ (f(φ)∂µ + ∂µf(φ))
δL
δ∂µφ

=
Z
d4
xf(φ)

δL
δφ
− ∂µ
δL
δ∂µφ

+ ∂µf(φ)
δL
δ∂µφ
c) Then, using, result of a), we have
0 =
Z
d4
x∂µf(φ)
δL
δ∂µφ
=
Z
d4
x∂µ
jµ
with
jµ = f(φ)
δL
δ∂µφ
d) Since
0 =
δρ
δt
+ ∇ · j
we have
d
dt
Z
d3
xρ = −
Z
d3
x∇ · j =
Z
dA · j
R→∞
−→ 0
2.2 Rotation Matrices and Finite Rotations

7 Symmetries
The effect on a spin eigenstate |S, Szi under a rotation by angle χ about an axis n̂
is given by
|S, Szi
0
= RS
(n̂, χ) |S, Szi ,
where
RS
(n̂, χ) = exp(−iχS · n̂) ,
and where S are the (2S + 1) × (2S + 1) component spin matrices constructed from
the relations
Sz |S, mi = m |S, mi
(Sx ± iSy) |S, mi =
p
(S ∓ m)(S ± m + 1) |S, m ± 1i .
a) Evaluate the rotation matrix for a spin-1
2 system and show that
R
1
2 (n̂, χ) = exp(−iχS
1
2 · n̂) = cos
χ
2
− iσ · n̂ sin
χ
2
,
where σ are the Pauli matrices.
b) Calculate the rotated spin-1
2 state |1/2, mi
0
for the initial state |1/2, mi for m =
±1
2 using n̂ = êz and rotation angle χ and demonstrate that
|1/2, mi
0
= exp(−imχ) |1/2, mi .
c) Verify the commutation relations for the representations of the spin-1 operators
in Eq. (2.41),
[Sz, S±] = ±S±
[S+, S−] = 2Sz .
d) Now, evaluate the rotation matrix for a spin-1 system and show that
R1
(n̂, χ) = exp(−iχS1
· n̂) = 1 − (S1
· n̂)2
(1 − cos χ) − iS1
· n̂ sin χ ,
where S1
are the 3 × 3 spin matrices constructed in the text.
e) Calculate the rotated spin state |1, mi
0
for the initial state |1, mi for the cases
m = 1, 0, −1 using n̂ = êz and rotation angle χ and demonstrate that
|1, mi
0
= exp(−imχ) |1, mi .
Exercise 2.2 Solution: Rotation Matrices and Finite Rotations
a) We have σ · n̂2
= 1 so
exp −i
χ
2
σ · n̂ =
∞
X
n=0
1
(2n)!

−
χ
2
2n
+ iσ · n̂
∞
X
n=0
1
(2n + 1)!

−
χ
2
2n+1
= cos
χ
2
− iσ · n̂ sin
χ
2

8 Symmetries
b) Pick n̂ = ẑ so
R
1
2 (χ, n̂)|
1
2
, ±
1
2
=

cos
χ
2
− iσz sin
χ
2

|
1
2
, ±
1
2

=

cos
χ
2
∓ i sin
χ
2

|
1
2
, ±
1
2
= exp

∓i
χ
2

|
1
2
, ±
1
2

c) Have
SzS+ =


1 0 0
0 0 0
0 0 −1




0
√
2 0
0 0
√
2
0 0 0

 =


0
√
2 0
0 0 0
0 0 0


and
SzS+ =


0
√
2 0
0 0
√
2
0 0 0




1 0 0
0 0 0
0 0 −1

 =


0 0 0
0 0 −
√
2
0 0 0


so
[Sz, S+] = SzS+ − S+Sz =


0
√
2 0
0 0
√
2
0 0 0

 = S+
Also, we have
S+S− =


0
√
2 0
0 0
√
2
0 0 0




0 0 0
√
2 0 0
0
√
2 0

 =


2 0 0
0 2 0
0 0 0


and
S−S+ =


0 0 0
√
2 0 0
0
√
2 0




0
√
2 0
0 0
√
2
0 0 0

 =


0 0 0
0 2 0
0 0 2


so
[S+, S−] = S+S− − S−S+ =


2 0 0
0 0 0
0 0 −2

 = 2Sz
d) Since (S · n̂)
3
= S · n̂ we have
exp −iχS · n̂ = 1 + (S · n̂)
2
∞
X
n=1
1
(2n)!
(−χ)
2n
+ iS · n̂
∞
X
n=0
1
(2n + 1)!
(−χ)
2n+1
= 1 + (S · n̂)
2
(cos χ − 1) − iS · n̂ sin χ
e) Pick n̂ = ẑ so
R1
(χ, n̂)|1, m = 1 − S2
z (1 − cos χ) − iSz sin χ

|1, m

9 Symmetries
= 1 − m2
(1 − cos χ) − im sin χ

|1, m =


(cos χ − i sin χ)|11 m = 1
|10 m = 0
(cos χ + i sin χ)|1, −1 m = −1


= exp(−imχ)|1, m
2.3 Symmetry and Dipole Moments
Since the only three-vector associated with an elementary particle in its rest frame
is the spin, any dipole moment of the particle must be along the spin direction. For
example, in the case of a spin-1
2 system such as the nucleon, one writes
m = gm
e
2m
S ,
where m is the magnetic dipole moment, gm is the gyromagnetic ratio, and e/2m
is called the Bohr magneton. When such a dipole is placed in a magnetic field B,
the corresponding interaction energy is
UM = −m · B .
a) Analyze this interaction from the point of view of parity and time reversal, and
demonstrate that UM is even under both.
Now suppose that we define an analogous electric dipole moment p in terms a
gyroelectric ratio ge
p = ge
e
2m
S .
When placed in an external electric field E there will exist an interaction energy
UE = −p · E .
b) Analyze this interaction from the point of view of parity and time reversal and
demonstrate that UE is odd under both.
Thus, an elementary particle cannot possess an electric dipole moment if parity
and/or time reversal are conserved.
c) We know that atoms and molecules can have large static electric dipole moments.
Explain why this fact does this not indicate a violation of parity and time re-
versal invariance.
Exercise 2.3 Solution: Symmetry and Dipole Moments
a) Note that vector potential Aµ is a four-vector so transforms like xµ. That
is, under parity Aµ
P
−→ Aµ while under time reversal Aµ
T
−→ −Aµ
. Thus B =

10 Symmetries
∇ × A
P
−→ B and B = ∇ × A
T
−→ −B. On the other hand, spin is a pseudo-vector
so S =
P
−→ S and S =
T
−→ −S. Thus we have
S · B
P
−→ S · B and S · B
T
−→ S · B
b) On the other hand for the electric field we have E = ∂
∂t A − ∇A0
P
−→ −E and
E =
T
−→ E, while for spin S =
P
−→ S and S =
T
−→ −S. Thus we have
S · E
P
−→ −S · E and S · E
T
−→ −S · E
c) For a non-elementary system such as a molecule, there exist many nearly de-
generate states so that a field reversal has virtually no change in the energy of the
system.
2.4 Spin Coupling
Consider a state with two spin-1
2 particles having spinors χ
ms1
1/2 (1) and χ
ms2
1/2 (2),
where the particles are labeled 1 and 2 and the spin projections along some axis
of quantization have the values msi
= ±1
2 . If the spin angular momenta of the two
particles are coupled to total spin S with projection MS one has the state
AMS
S (1, 2) ≡

χ1/2(1) ⊗ χ1/2(2)
MS
S
.
What values can S and MS have? By explicit evaluation of the Clebsch-Gordan
coefficients or 3-j symbols (see [Edm74]) show that the familiar answers are obtained
(see [Sch55]). Using the properties of the 3-j symbols prove upon interchange of the
coordinates of the two particles that one has
AMS
S (2, 1) = (−)S+1
AMS
S (1, 2) .
These results can trivially be extended to the case of two isospin-1
2 particles, for
instance two nucleons, where one has isospinors ξ
mt1
1/2 (1) and ξ
mt2
1/2 (2) with mti = ±1
2 ,
one has the coupled state
BMT
T (1, 2) ≡

ξ1/2(1) ⊗ ξ1/2(2)
MT
T
,
and obtains the symmetry under interchange
BMT
T (2, 1) = (−)T +1
BMT
T (1, 2) .
What happens upon coordinate interchange for a system of two nucleons (NN)
where one has both spin and isospin?
Exercise 2.4 Solution: Spin Coupling

11 Symmetries
One begins by writing out the coupling of the two spin-spinors in detail:
AMS
S (1, 2) =
X
ms1
ms2
h1/2 ms1
1/2 ms2
| 1/2 1/2 SMSi χ
ms1
1/2 (1)χ
ms2
1/2 (2)
= (−)
MS
[S]
X
ms1
ms2

1/2 1/2 S
ms1
ms2
−MS

χ
ms1
1/2 (1)χ
ms2
1/2 (2),
where as usual [S] ≡
√
2S + 1. Clearly, using either the Clebsch-Gordan coefficients
or the 3-j symbols one sees that the only values of the total spin quantum numbers
are S = MS = 0 and S = 1 with MS = −1, 0, +1. The former is easily written out
explicitly using the fact (see Edmonds [Edm74]) that

1/2 1/2 0
ms1
ms2
0

=
(−)
1/2−ms1
√
2
δms2
,−ms1
yielding
A0
0(1, 2) =
1
√
2
h
χ
+1/2
1/2 (1)χ
−1/2
1/2 (2) − χ
−1/2
1/2 (1)χ
+1/2
1/2 (2)
i
.
For the S = 1, MS = 0 case one has from Edmonds

1/2 1/2 1
ms1
ms2
0

=
r
2
3
(−)
1/2−ms1
ms1
δms2
,−ms1
yielding
A0
1(1, 2) =
1
√
2
h
χ
+1/2
1/2 (1)χ
−1/2
1/2 (2) + χ
−1/2
1/2 (1)χ
+1/2
1/2 (2)
i
.
For the remaining cases one has

1/2 1/2 1
1/2 1/2 −1

=

1/2 1/2 1
−1/2 −1/2 +1

= −
1
√
3
,
again from Edmonds, yielding
A+1
1 (1, 2) = χ
+1/2
1/2 (1)χ
+1/2
1/2 (2)
A−1
1 (1, 2) = χ
−1/2
1/2 (1)χ
−1/2
1/2 (2).
These are all the expected familiar results from standard discussions of quantum
mechanics.
The 3-j symbols have the symmetry property under exchange of the magnetic
quantum numbers of the two spinors:

1/2 1/2 S
ms2 ms1 −MS

= (−)
S+1

1/2 1/2 S
ms1 ms2 −MS

.
Upon exchange of the particle coordinates 1 ↔ 2 in the basic expression for the

12 Symmetries
coupled spin-spinor, one has
AMS
S (2, 1) = (−)
MS
[S]
X
ms1
ms2

1/2 1/2 S
ms1 ms2 −MS

χ
ms1
1/2 (2)χ
ms2
1/2 (1)
= (−)
MS
[S]
X
µs1 µs2

1/2 1/2 S
µs2 µs1 −MS

χ
µs1
1/2(1)χ
µs2
1/2(2)
= (−)
S+1
× (−)
MS
[S]
X
µs1 µs2

1/2 1/2 S
µs1
µs2
−MS

χ
µs1
1/2(1)χ
µs2
1/2(2)
= (−)
S+1
AMS
S (1, 2),
where in the second line the replacements ms1 → µs2 and ms2 → µs1 have been
made, and in the third line the symmetry relation above has been employed. Note
that in the second line the spin-spinors have been written in the opposite order:
since they operate in different spaces (namely, those for particles 1 and 2), they
commute.
Clearly all of the above goes through for isospin and, specifically, one has the
cases T = MT = 0 and T = 1 with MT = −1, 0, +1 having analogous expressions
to those above. The exchange symmetry for isospin is immediate,
BMT
T (2, 1) = (−)
T +1
AMT
T (1, 2),
and exchange of the coordinates for systems having both spin and isospin yields
AMS
S (2, 1)BMT
T (2, 1) = (−)
S+T
AMS
S (1, 2)AMT
T (1, 2).
2.5 L-S Coupling and Central Potentials
Consider a nonrelativistic system of two particles, which interact via a potential. In
general such an interaction can depend on spin and isospin as well as the particle
separation. This problem deals with how to deal with such a potential.
If we represent the spatial coordinates of the two particles by r1 and r2, then one
can change variables to a center-of-mass variable R = (r1 + r2)/2 and a relative
coordinate r = r1−r2. The two-particle spatial wavefunction may then be written as
a product of a plane wave for the center-of-mass (cm) part times a function involving
the relative coordinate, Φ(r). If the state has good orbital angular momentum L,
with projection ML, where L is an integer and −L ≤ ML ≤ L, then the spatial
wavefunction can be written
ΦML
L (r) =RL(r)Y ML
L (Ωr) ,
where RL(r), the radial wavefunction depends only on r = |r| and the dependence
on the polar and azimuthal angles Ωr = (θ, φ) specifying the direction of r is
captured in the spherical harmonic Y ML
L (Ωr). For a general discussion of these

13 Symmetries
basic ideas see one of the standard books on quantum mechanics, such as [Sch55];
see also [Edm74].
a) Prove that under a parity transformation, namely under inversion of coordinates
r → −r one has Y ML
L (Ωr) → (−)L
Y ML
L (Ωr). One may now form the total
wavefunction for an NN system in a given partial wave (i.e., with good orbital
angular momentum),
ΨMJ ;MT
(LS)J;T (1, 2) = R(LS)J;T (r) [YL(Ωr) ⊗ AS(1, 2)]
MJ
J BMT
T (1, 2) ,
where here for simplicity the cm plane wave has been omitted. The total state
is presumed to have orbital and spin angular momenta coupled to total angular
momentum J with projection MJ . What happens upon interchange of all of
the coordinates of the two nucleons? Since the Pauli exclusion principle states
that the total wavefunction must be antisymmetric, what quantum numbers
are allowed for the NN system?
b) Suppose that one has a spin-isospin dependent central potential of the form
VC(r) = V1(r) + V2(r)σ(1) · σ(2) + V3(r)τ(1) · τ(2)
+ V4(r)σ(1) · σ(2)τ(1) · τ(2) ,
where the factors σ(i) and τ(i) are the familiar Pauli matrices discussed in this
chapter. For a state of the type introduced in the previous exercise evaluate
the spin and isospin matrix elements of the potential.
c) If a so-called spin-orbit potential of the form
VLS(r) = {V5(r) + V6(r) (τ(1) · τ(2))} (L · S)
were to be added to the central potential in the previous Exercise, what would
be the form of the orbital-spin-isospin matrix elements?
d) One might also want to add a so-called tensor potential of the form
VT (r) = {V7(r) + V8(r) (τ(1) · τ(2))} S12(1, 2),
involving the tensor operator
S12(1, 2) = 3(r · σ(1))(r · σ(2))/r2
− σ(1) · σ(2) .
Find the result of applying the tensor operator to a singlet spin state, i.e., a
state having S = 0. For a potential that contains a tensor term one finds that
the orbital angular momentum is not a good quantum number and that states
with differing values of L must mix; which states are these? Do you know a
relatively familiar example of this mixing? Can you think of other types of
potentials not included in the types discussed here in Exercises 2.4 and 2.5?
Exercise 2.5 Solution: L-S Coupling and Central Potentials
a) Inversion of coordinates r → −r corresponds to making the transformation

14 Symmetries
(r, θ, ϕ) → (r, π − θ, π + ϕ) under which cos θ → − cos θ, sin θ → + sin θ and
eim(π+ϕ)
= (−)
m
eimϕ
. Employing this with expressions for the spherical harmonics
(see Edmonds [Edm74]) gives the desired result: Y ML
L (Ωr) → (−)L
Y ML
L (Ωr). In
Exercise 2.4 we saw that interchange of coordinates for an NN system having both
spin and isospin yielded the phase (−)
S+T
and hence interchange of coordinates for
space-spin-isospin NN wavefunctions yields the total symmetry (−)
L+S+T
. Since
the total wavefunction must be antisymmetric, this implies that L + S + T = odd.
So, for example, one has that 3
S1 and 3
D1,2,3 partial waves have isospin T = 0,
while 1
S0 and 3
P0,1,2 partial waves have isospin T = 1 (here the notation is, as
usual, 2S+1
LJ with J = L + S). All of these developments are used in Chapter 11.
b) Starting with the spin operators, one has
S(1, 2) = s(1) + s(2),
where as usual s(i) = σ(i)/2. Thus
S2
(1, 2) = s2
(1) + s2
(2) + 2s(1) · s(2)
having eigenvalues S(S + 1) for S2
(1, 2) and 1/2 × 3/2 = 3/4 for s2
(1) and s2
(2).
Using this and rewriting the result in terms of Pauli matrix operators one has
σ(1) · σ(2) = 2S(S + 1) − 3 ≡ aS
=

−3 S = 0 spin singlet
+1 S = 1 spin triplet
The isospin case is proven in an analogous way, yielding
τ(1) · τ(2) = 2T(T + 1) − 3 ≡ bT
=

−3 T = 0 isospin singlet
+1 T = 1 isospin triplet
which then immediately leads to the desired matrix elements of the potential
VC(r) = V1(r) + aSV2(r) + bT aV3(r)+aSbT V4(r).
c) Noting that
J = L + S
one can form
J2
= L2
+ S2
+ 2L · S
having eigenvalues J(J + 1), L(L + 1) and S(S + 1) for J, L and S, respectively,
which leads to the relationship
L · S =
1
2
{J(J + 1) − L(L + 1) − S(S + 1)} ≡ cLSJ
=







0 S = 0 J = L
− (L + 1) S = 1 J = L − 1
−1 S = 1 J = L
+L S = 1 J = L + 1

15 Symmetries
This would lead to the additional terms
VLS(r) = {V5(r) + bT V6(r)} cLSJ .
d) The tensor potential involves somewhat more work. We start by writing the
tensor operator
S12(1, 2) = 3(r · σ(1))(r · σ(2))/r2
− σ(1) · σ(2)
in terms of contributions that operate only in coordinate space or only in spin space,
begining by decomposing the vectors involved in terms of spherical tensors
σ(i) =
X
m=0,±1
σm(i)e∗
m
r =
r
4π
3
r
X
m=0,±1
Y m
1 (Ωr)e∗
m.
Let us define tensor products in the two spaces:
ΣM
K (1, 2) ≡ [σ(1) ⊗ σ(2)]
MK
K
=
X
µµ0
h1µ1µ0
| 11KMKi σµ(1)σµ0 (2)
= (−)
MK
[K]
X
µµ0

1 1 K
µ µ0
−MK

σµ(1)σµ0 (2)
ZML
L (Ωr) ≡
4π
3
[Y1(Ωr) ⊗ Y1(Ωr)]
ML
L
=
4π
3
(−)
ML
[L]
X
µµ0

1 1 K
µ µ0
−MK

Y µ
1 (Ωr)Y µ0
1 (Ωr)
with inverses
σm(1)σm0 (2) =
X
KMK
(−)
MK
[K]

1 1 K
m m0
−MK

ΣMK
K (1, 2)
Y m
1 (Ωr)Y m0
1 (Ωr) =
3
4π
X
LML
(−)
ML
[L]

1 1 L
m m0
−ML

ZML
L (Ωr),
which are easily derived using the orthogonality properties of the 3-j symbols; as
usual [x] ≡
√
2x + 1. By using Eq. (4.6.5) in Edmonds [Edm74] one has that
ZML
L (Ωr) =
√
4π

1 1 L
0 0 0

Y ML
L (Ωr),
where the 3-j symbol (the so-called parity 3-j symbol) must have the sum of its
entries even, implying that only L = 0 and L = 2 enter. By using the fact that
σ(1) · σ(2) =
X
m
(−)
m
σm(1)σ−m(2)

16 Symmetries
one has immediately that
σ(1) · σ(2) = −
√
3Σ0
0(1, 2),
and by direct evaluation that
Z0
0 (Ωr) = −
1
√
3
.
Now the remainder of the tensor operator involves
3(r · σ(1))(r · σ(2))/r2
= 3
X
mm0
(−)
m+m0 4π
3
Y m
1 (Ωr)Y −m0
1 (Ωr)
×σ−m(1)σm0 (2)
= 3
X
KMK
(−)
MK
ZMK
K (Ωr)ΣMK
K (1, 2)
= Z0
0 (Ωr)Σ0
0(1, 2)
+3
X
MK
(−)
MK
ZMK
2 (Ωr)Σ−MK
2 (1, 2)
= −
√
3Σ0
0(1, 2)
+3
X
MK
(−)
MK
ZMK
2 (Ωr)Σ−MK
2 (1, 2),
where again orthogonality of the 3-j symbols has been employed, together with
other relations from above. The first term here cancels with σ(1) · σ(2) (in fact, by
design: this is why it is called the tensor operator). One is left with
S12(1, 2) = 3
X
MK
(−)
MK
ZMK
2 (Ωr)Σ−MK
2 (1, 2)
involving the scalar product of two rank-2 tensor operators. The matrix elements
of ZMK
2 (Ωr) involve doing the integral over the solid angle of the product of three
spherical harmonics,
Z
dΩY
ML0
L0 (Ωr)∗
Y MK
2 (Ωr)Y ML
L (Ωr),
which by Eq. (4.6.3) of Edmonds [Edm74] is proportional to another parity 3-j
symbol

L0
2 L
0 0 0

,
and, since then L + L0
must be even, L and L0
must both be either even or odd
for mixing to occur. Also, using the Wigner-Eckart theorem for the spin matrix
elements,
D
S0
MS0 Σ−MK
2 (1, 2) SMS
E
= (−)
S0
−MS0

S0
2 S
−MS0 −MK MS

× hS0
kΣ2(1, 2)k Si ,

17 Symmetries
tells us that only when both S0
and S are equal to 1 (triplet) can mixing occur.
Thus in the sequence of allowed NN partial waves, 3
S1, 3
P0,1,2, 3
D1,2,3, 3
F2,3,4, · · · ,
only mixings of the type 3
S1− 3
D1, 3
P2−3
F2, · · · can mix; the other triplet states
are sometimes called isolated triplets; and the singlet states do not mix, as we have
seen. A familiar example of this (see Chapter 11 of the text) is the deuteron whose
non-relativistic ground state involves both 3
S1 and 3
D1 configurations mixed via
the tensor force.
Finally, it is also possible to have additional types of forces. For instance, using
only the facts that the charge is conserved and that all terms must be invariant
under interchange of identical particles would allow additional contributions pro-
portional to τ3(1)τ3(2) and τ3(1) + τ3(2). For low-energy NN scattering one does
not need such contributions and finds that the NN force is charge independent. If
one assumes the force to be only charge symmetric, then terms of the τ3(1)τ3(2)
type are also needed.
One may also have contributions that are quadratic in orbital angular momentum,
for instance terms involving L2
, (σ(1) · σ(2)) L2
and (L · S)
2
times 1 or τ(1) · τ(2).
2.6 Single-Particle Wavefunctions
As we shall see later in the book, it is useful to employ a basis of single-particle
wavefunctions when discussing the nuclear many-body problem. One starts with
some mean-field potential in which the individual nucleons move and, upon coupling
the orbital and spin angular momenta to form the total angular momentum, forms
the single-particle wavefunctions
ψms;mt
(`s)j;t (r) = R(`s)j;t(r) [Y` (Ωr) ⊗ χs]
mj
j ξmt
t ,
where s = t = 1/2 and here the spatial coordinate is with respect to the origin of
the potential. What values are allowed for the orbital and total angular momentum
quantum numbers ` and j? If the potential has a spin-orbit term such as in Exercise
2.5 c), what does one expect for the splitting between states with the same orbital
angular momentum, but different values of j?
Exercise 2.6 Solution: Single-Particle Wavefunctions
We use a relationship that is very similar to that in Exercise 2.5 c), namely,
starting with
j = ` + s
and forming
j2
= `2
+ s2
+ 2` · s
with eigenvalues j(j +1), `(`+1) and s(s+1) = 3/4, for j2
, `2
and s2
, respectively,

18 Symmetries
one has
` · s =
1
2
{j(j + 1) − `(` + 1) − 3/4} ≡ α`j
=
1
2
×

− (` + 1) j = ` + 1/2
+` j = ` − 1/2
and hence a spin-orbit potential that has the form Vso (r) ` · s would require evalu-
ating the radial integral of Vso (r) and multiplying by α`j given above. The phys-
ical situation is that the integral comes out to be positive and so the so-called
“stretched” states with j = ` + 1/2 are found to be lower in energy, whereas the
so-called “jack-knifed” states with j = ` − 1/2 lie higher in energy.
2.7 SU(6) Symmetry
Employing the rules given in the text for determining the dimensions of represen-
tations of SU(N) groups using the Young tableaux, consider the direct product of
three copies of the fundamental in SU(6), 6,
6 ⊗ 6 ⊗ 6
(in the text the cases of SU(2) and SU(3) were both discussed). Determine the
resulting direct sum of the representations that emerge and find their dimensions.
Exercise 2.7 Solution: SU(6) Symmetry
We proceed as in the text, first taking the direct product of two copies of the
fundamental representation which yields the result shown in Fig. 2.3, and then
taking the direct product with a third fundamental representation we obtain the
result in Fig. 2.7. Finally, using the hook rules on the individual Young tableaux,
inserting the factors N, etc., and adding the hooks, we obtain the dimensions as
discussed in the text. The results are shown in the accompanying figures. For case
A we add the factors N, N + 1 and N + 2), which yields dimension
DA =
N (N + 1) (N + 2)
2 · 3
,
namely, 10 as in the text for SU(3) and 56 in the present case of SU(6). Similarly,
for tableaux B1,2 in the second figure, upon inserting the factors N − 1, N and
N + 1, yielding these as numerator factors for the three hooks involved, we find
dimension
DB1,2 =
(N − 1) N (N + 1)
3
,
namely, 8 as in the text for SU(3) and 70 in the case of SU(6). Finally, the remaining
Young tableau has factors N, N − 1 and N − 2, as shown in the third figure, and
the three hooks as illustrated, yielding
DC =
N (N − 1) (N − 2)
2 · 3
,

19 Symmetries
A
N
N+1
N+2
3 2 1
t
Fig. 2.1 Applying the hook rules for Young tableau A.
namely, 1 for SU(3) and 20 in the case of SU(6).
2.8 Spontaneous Symmetry Breaking in Classical Mechanics
Consider a frictionless bead of mass m free to slide on a hoop of radius R, which is
rotating about a vertical axis with angular velocity ω.
a) Show that the potential energy is given by
V (θ) = −
1
2
mω2
R2
sin2
θ − mgR cos θ ,

20 Symmetries
B
N N+1
N-1
3 1
1
1,2
t
Fig. 2.2 Applying the hook rules for Young tableaux B1,2.
where θ is the angle of the bead as measured from the bottom of the hoop.
b) Show that for ω2
g
R the shape of the potential has a single minimum, so that
the position of stable equilibrium is at θ = 0.
c) Show that for ω2
g
R the shape of the potential has two minima, so that there
exist two positions of stable equilibrium at
θ = ± cos−1 g
ω2R
.

21 Symmetries
C
N
N-2
N-1
3
2
1
t
Fig. 2.3 Applying the hook rules for Young tableau C.
Exercise 2.8 Solution: Spontaneous Symmetry Breaking in Classical Me-
chanics
a) Kinetic energy of rotating bead is T = 1
2 (R2
θ̇2
+ R2
ω2
sin2
θ) while potential
energy is V = −mgR cos θ so Lagrangian is
L = T − V =
1
2
R2
θ̇2
+
1
2
mR2
ω2
sin2
θ + mgR cos θ

22 Symmetries
so effective potential is
Veff (θ) = −
1
2
mR2
ω2
sin2
θ − mgR cos θ
Find minimum via
0 =
dVeff
dθ
= mgR sin θ − mR2
ω2
sin θ cos θ = mR sin θ(Rω2
cos θ − g)
b) If Rω2
g then factor in parentheses is negative definite so equilibrium must
be at θ = 0.
c) If Rω2
g then two new equilibrium points at θ = ± cos−1 g
Rω2
Note can check stability via
d2
Veff
dθ2
= mR cos θ(Rω2
cos θ − g) − mR2
ω2
sin2
θ
= mR2
ω2
cos 2θ − mgR cos θ
Then if Rω2
g curvature is negative at θ = 0 and positive at θ = π so θ = 0 is
stable equilibrium point. However, if Rω2
g then curvature is positive at both
θ = 0, π so these are unstable points, but negative at θ = ± cos−1 g
ω2R so these are
stable.

3 Building Hadrons from Quarks
3.1 Isospin Symmetry and K → 2π Decay
The dominant decay modes of the K mesons are the two-pion channels
K0
→ π+
π−
, K0
→ π0
π0
, K±
→ π±
π0
.
a) Using the fact that pions are spinless and are therefore bosons, show that the
final two-pion state resulting from the decay of a kaon must have either isospin
zero or two.
b) Since ūd carries I = 1 while s̄u carries I = 1
2 the strangeness-changing nonlep-
tonic Hamiltonian
Hw ∼
G
√
2
s̄uūd
involves a linear combination of I = 1
2 and I = 3
2 components, leading to
K → 2π decay amplitudes a1 and a3. Show that the decay amplitude can be
written as
A(K+
→ π+
π0
) =
3
2
√
2
a3
A(K0
→ π+
π−
) = −a1 +
1
2
a3
A(K0
→ π0
π0
) = a1 + a3 .
c) Compare this parametrization with the experimental lifetimes for the following
channels
τK0→π+π− ' τK0→π0π0 τK±→π±π0 .
What do you conclude about the amplitudes a1 and a3?
Exercise 3.1 Solution: Isospin Symmetry and K → 2π Decay
a) Bose symmetry requires that spatial × spin × isospin piece of the wavefunc-
tion be symmetric under interchange. In the case of the two-pion state there is
no spin and the spatial component must be an S-wave, which is symmetric under
interchange. Hence, the isospin piece must also be symmetric. Now in general, the
product of two isotriplet states reads (I1 = 1) × (I2 = 1) → I = 0, 1, 2, where the
states with I = 0, 2 are symmetric and the state I = 1 is antisymmetric. Hence,
only the states with I = 0, 2 are permitted.
23

b) We have, by isotopic spin symmetry
πa
πb
|Hw|Kn
= ã1 1, a; 1, b|0, 0; 11 0, 0;
1
2
1
2
|
1
2
,
1
2
;
1
2
, n
+ã3 1, a; 1, b|2, a + b; 11 2, a + b;
1
2
1
2
|
3
2
,
1
2
;
1
2
, n
Here
1, +; 1, −|0, 0; 11 =
r
1
3
= − 1, 0; 1, 0|0, 0; 11
1, 1; 1, 0|2, 1; 11 =
r
1
2
0, 0;
1
2
1
2
|
1
2
,
1
2
;
1
2
, −
1
2
=
r
1
2
1, +; 1, −|2, 0; 11 =
r
1
6
=
1
2
1, 0; 1, 0|2, 0; 11
2, 0;
3
2
1
2
|
3
2
,
1
2
;
1
2
, −
1
2
=
r
1
2
=
r
2
3
2, 1;
3
2
1
2
|
3
2
,
1
2
;
1
2
,
1
2

Then
π+
π+
|Hw|K0
=
r
1
6
ã1 +
r
1
12
ã3
π0
π0
|Hw|K0
= −
r
1
6
ã1 + 2
r
1
12
ã3
π+
π0
|Hw|K+
=
√
3
2
√
2
ã3
Result given in the problem results from redefinitions ã1 ≡ −
√
6a1 and ã3 ≡
√
3a3.
c) Since τK+→π+π0 τK0→π+π− ' τK0→π0π0 we must have |a3| |a1|.
3.2 Isospin Invariance
Just as in the case of angular momentum invariance where the Hamiltonian is
required to be a scalar under arbitrary rotations, isospin invariance requires that
the Hamiltonian be a scalar under rotations in isospin space. Spin and isospin
invariance have another similarity in that, since the commutation relations of spin
and isospin operators have the same structure
[Si, Sj] = iijkSk , [Ii, Ij] = iijkIk ,

the irreducible representations must be the same — particles must lie in represen-
tations labeled by |I, I3i with I = 0, 1
2 , 1, 3
2 , . . . and −I ≤ I3 ≤ I.
In the case of the pion-nucleon interaction, the interaction Lagrangian must be of
the form
Lint = gπNN N̄τ · πγ5N ,
where τ are the Pauli isospin matrices, π is the isovector pion field and the presence
of γ5 is required by the feature that the pion is a pseudoscalar particle.
a) Since the nucleon has isospin 1
2 while the pion has isospin 1, demonstrate that
under an infinitesimal isospin rotation by angle δχ
N → N +
i
2
δχ · τN
π → π − δχ × π .
b) From these transformation properties show that the interaction Lagrangian given
above is invariant:
Lint
rot
−→ Lint .
c) Using the representations
π+
= −
r
1
2
(π1 + iπ2), π0
= π3, π−
=
r
1
2
(π1 − iπ2)
of the pions in terms of their Cartesian components demonstrate that the
isospin invariance of Lint requires the pion couplings to nucleons to have the
form
r
1
2
g(npπ−
) = −
r
1
2
g(pnπ+
) = g(ppπ0
) = −g(nnπ0
) = gπNN .
d) Compare the results of c) with the requirements of the Wigner-Eckart theorem
g(NaNbπc) =

1
2
ma1mc|
1
2
1
1
2
mb

hN||g||πNi ,
where 1
2 ma1mc|1
2 11
2 mb is a Clebsch-Gordan coefficient (see Chapter 2).
Of course, unlike angular momentum invariance which is exact, these isospin pre-
dictions are expected to be broken at the percent level due to the small differences
between the n, p and π+
, π0
, π−
masses.
Exercise 3.2 Solution: Isospin Invariance
a) We have under rotation
N → (1 + iτ · δχ) N

Then
πi → πi + ijkδχjπk
b) Then
N̄τ · πN → N̄ (1 − iτ · δχ) τ · (π − δχ × π) (1 + iτ · δχ) N
= N̄τ · πN + N̄ (i[τ · π, τ · δχ] + τ · δχ × π) N
= N̄τ · πN + N̄τ · i2
π × δχ + π × δχ

N = N̄τ · πN
c) Since
τ · π = π0
τ3 − π+
r
1
2
τ− + π−
r
1
2
τ+
we have
gπNN = g(p̄pπ0
) = −g(n̄nπ0
) = −
r
1
2
g(p̄nπ−
) =
r
1
2
g(n̄pπ+
)
d) Note that
−
r
1
3
=
1
2
,
1
2
; 1, 0|
1
2
,
1
2
;
1
2
1 = −
1
2
, −
1
2
; 1, 0|
1
2
, −
1
2
;
1
2
1
=
r
1
2

1
2
, −
1
2
; 1, 1|
1
2
,
1
2
;
1
2
1 = −
r
1
2

1
2
,
1
2
; 1, −1| −
1
2
,
1
2
;
1
2
1
so completely agrees.
3.3 SU(3) Invariance
The near degeneracy of the u, d quarks as well as the n, p and π+
, π0
, π−
masses
indicate that isospin should be a quite good symmetry, with breaking at the percent
level. In the case of SU(3) the symmetry is still useful, with breaking, however, at
the ∼20% level, as suggested by the difference between the N, ∆ and ρ, K∗
masses.
Since the lightest pseudoscalar mesons and spin-1
2 baryons are both members of
SU(3) octet representations, it is interesting to see what the implications of SU(3)
symmetry are both for the masses and the couplings of these particles. In order to
study these predictions it is useful to represent these octets in the form
M =
8
X
j=1
λjφj =





q
1
2 π0
+
q
1
6 η8 π+
K+
π−
−
q
1
2 π0
+
q
1
6 η8 K0
K−
K̄0
−2
q
1
6 η8





B =
8
X
j=1
λjBj =





q
1
2 Σ0
+
q
1
6 Λ0
Σ+
p
Σ−
−
q
1
2 Σ0
+
q
1
6 Λ0
n
Ξ−
Ξ0
−2
q
1
6 Λ0






where λj are the Gell-Mann matrices introduced in Chapter 2. Then an SU(3)
rotation is given by
M0
= UMU†
, B0
= UBU†
with U = exp (
i
2
8
X
j=1
αjλj) ,
and the simplest possible SU(3) invariants are constructed using the trace
Tr(B̄B),
1
2
Tr(B̄{B, M}),
1
2
Tr(B̄[B, M]) .
That such forms are invariant is clear from the properties of the trace
Tr(ABC) → Tr(UAU†
UBU†
UCU†
) = Tr(ABC) .
Thus the Lagrangian describing the pseudoscalar meson-baryon couplings should
have the form
L =
X
abc
gabcMaB̄bγ5Bc
with
gabc =
1
4
DTr(λaλb, λc) +
1
4
FTr(λa[λb, λc]) = Ddabc + iFfabc,
where dabc and fabc are the SU(3) structure constants and D and F are the corre-
sponding couplings.
a) Use these results to write out the various possible meson-baryon couplings in
terms of their SU(3) decompositions, e.g.,
g(pp̄π0
) =
1
2
(D + F)
g(pp̄η0
) = −
1
2
√
3
(D − 3F) .
b) From the experimental results
gpp̄π0 ∼
= 13.5 and gpp̄η0 ∼
= −3.9
determine the SU(3) couplings D, F.
To the extent that the mass operator can be written in terms of an SU(3) singlet
plus octet forms, we can then fit the pseudoscalar and 1
2
+
baryon masses.
c) Show that the masses can then be written in the form
M2
ij = M2
0 δij + M2
Ddij8 + iM2
F fij8 mesons
Bij = B0
δij + BDdij8 + iBF fij8 baryons .
d) Determine the constants M2
0 , M2
D, M2
F for mesons and B0, BD, BF for baryons
from experimental data.

e) Show that the mass predictions obey the Gell-Mann-Okubo sum rules
1
2
(mN + mΞ) =
1
4
(3mΛ + mΣ)
1
2
(m2
K+ + m2
K0 ) =
1
4
(3m2
η + m2
π)
and compare with experiment.
Exercise 3.3 Solution: SU(3) Invariance
a) Strong couplings are
gπ0pp =
1
2
Dd4−i5,4+i5,3 −
i
2
Ff4−i5,4+i5,3 =
1
2
D +
1
2
F
gη0pp =
1
2
Dd4−i5,4+i5,8 −
i
2
Ff4−i5,4+i5,8 = −
1
2
√
3
D +
√
3
2
F
gπ0Ξ−Ξ− =
1
2
Dd4+i5,4−i5,3 −
i
2
Ff4+i5,4−i5,3 =
1
2
D −
1
2
F
gη0Ξ−Ξ− =
1
2
Dd4+i5,4−i5,8 −
i
2
Ff4+i5,4−i5,8 = −
1
2
√
3
D −
√
3
2
F
gπ0Σ+Σ+ =
1
2
Dd1−i2,1+i2,3 −
i
2
Ff1−i2,1+i2,3 = F
gη0Σ+Σ+ =
1
2
Dd1−i2,1+i2,8 −
i
2
Ff1−i2,1+i2,8 =
1
√
3
D
b) We have
D + F = 27.2
D − 3F = 13.5
so, subtracting, have 4F = 40.7, or F = 10.2 and D = 27.2 − 10.2 = 17.0.
c) We have in general
mB = B0TrB̄B + BDTr{B̄, B}λ8 + iBF Tr[B̄, B]λ8
m2
P = M2
0 TrM̄M + M2
DTr{M̄, M}λ8 + iM2
F Tr[M̄, M]λ8
so
M2
ij = M2
0 δij + M2
Ddij8 + iM2
F fij8
B2
ij = B0δij + BDdij8 + iBF fij8

d) Calculate masses via
m2
K+ =
1
2
M2
0 δ4−i5,4+i5+
1
2
M2
Dd4−i5,4+i5,8+
i
2
M2
F f4−i5,4+i5,8 = M2
0 −
1
2
√
3
M2
D−
√
3
2
M2
F
m2
π+ =
1
2
M2
0 δ1−i2,1+i2 +
1
2
M2
Dd1−i2,1+i2,8 +
i
2
M2
F f1−i2,1+i2,8 = M2
0 +
1
√
3
M2
D
m2
K0 =
1
2
M2
0 δ6−i7,6+i7+
1
2
MDd6−i7,6+i7,8+
i
2
M2
F f6−i7,6+i7,8 = M2
0 −
1
2
√
3
M2
D−
√
3
2
M2
F
m2
η = M2
0 δ8,8 + M2
Dd8,8,8 + M2
F f8,8,8 = M2
0 −
1
√
3
M2
D
Then
M2
0 =
1
2
(m2
π + m2
η)
M2
F =
1
√
3
(m2
K − m2
π)
M2
D =
√
3(m2
η − m2
π)
mp =
1
2
B0δ4−i5,4+i5 +
1
2
BDd4−i5,4+i5,8 +
i
2
BF f4−i5,4+i5,8 = B0 −
1
2
√
3
BD −
√
3
2
BF
mΣ+ =
1
2
B0δ1−i2,1+i2 +
1
2
BDd1−i2,1+i2,8 +
i
2
BF f1−i2,1+i2,8 = B0 +
1
√
3
BD
mΞ+ =
1
2
B0δ4+i5,4−i5+
1
2
BDd4+i5,4−i5,8+
i
2
BF f4+i5,4−i5,8 = B0−
1
2
√
3
BD +
√
3
2
BF
mΛ = B0δ8,8 + BDd8,8,8 + f8,8,8 = B0 −
1
√
3
BD
Then
B0 =
1
2
(mΣ + mΛ)
BF =
1
√
3
(mΞ − mN )
BD =
√
3(mΛ − mN )
e) We have then
1
2
(mN + mΞ) = B0 −
1
2
√
3
BD =
1
4
(3mΛ + mΣ)
1
2
(m2
K+ + m2
K0 ) = M2
0 −
1
2
√
3
M2
D =
1
4
(3m2
η + m2
π)

3.4 SU(4) Representations
Just as in the case of SU(3) symmetry, where one looks at rotations among u, d, s
quarks, one can look at SU(4) symmetry which involves rotations among u, d, s and
c (charm) quarks. Of course, since the c quark is very much heavier than the oth-
ers, SU(4) should be strongly broken as a dynamical symmetry, but nevertheless its
irreducible representations in the meson and baryon sectors should be useful in un-
derstanding the shape of the particle spectrum, and that is the goal of this exercise.
In the case of SU(2) we know that the irreducible representations can be given
in terms of a line along which the particles having the same total isospin but
different values of I3 are given. In the case of SU(3) we require a two dimensional
picture in which I3 is given along the x-axis and hypercharge Y , defined through
the relationship Q = I3 + Y/2, is plotted along the y-axis, leading to the familiar
octet and decuplet representations. For SU(4) we will require a three-dimensional
plot with charm plotted along the z-axis and a stacking of SU(3) representaions
a) In the case of the pseudoscalar mesons, show that we expect a 15-dimensional
representation, and plot the result in terms of the quark content in a three-
dimensional plot along I3, Y, C axes.
b) Analyze this 15-dimensional representation in terms of its SU(3) content and
show that we have {15}SU(4) = (8 + 3 + 3̄ + 1)SU(3)
c) In the case of the spin-1
2 baryons show that we expect a 20-dimensional represen-
tation, and plot the results in terms of the quark content in a three-dimensional
plot along I3, Y, C axes.
d) Analyze this 20-dimensional representation in terms of its SU(3) content and
show that we have {20}SU(4) = (8 + 6 + 3 + 3̄)SU(3)
e) Compare these representations with what has been observed experimentally.
Exercise 3.4 Solution: SU(4) Representations
a) The pseudoscalar mesons have the structure
q̄i
qj −
1
4
δi
jq̄k
qk
with i, j = 1, 2, 3, 4 so the dimensionality of this representation is d = 4×4−1 =
15. A picture of this reprentation in C, Y, I space can be found in Figure 15.1a
of the Quark Model review by C. Amsler, T. DeGrand, and B. Krusche in the
Particle Data Group Review of Particle Properties.
b) The SU(3) representations which make up the 15-dimensional representation of

SU(4) are
q̄i
qj −
1
3
δi
jq̄k
qk, + q̄i
q4 + q̄4
qi +
4
X
k=1
q̄k
qk
where i, j = 1, 2, 3 which have the dimensionality 8 + 3 + 3̄ + 1.
c) The baryons have the structure qiqjqkjk`m
where i, j, k = 1, 2, 3, 4,which has the
dimensionality d = 4 × 4×3
2 − 2 × 2 = 20. A picture of this representation in
C, Y, I space can be found in Figure 15.4a of the Quark Model review by C.
Amsler, T. DeGrand, and B. Krusche in the Particle Data Group Review of
Particle Properties.
d) The SU(3) representations which make up the 20-dimensional representation of
SU(4) are
qiqjq`j`n
−
1
3
δn
i kj`
qkqjq` + q4qiqj + q4qiqjijk
+ q4q4qi
where i, j = 1, 2, 3 which have the dimensionality 8 + 6 + 3̄ + 3.
e) The latest PDG summary lists, in the meson case, C = ±1 mesons D±
(1870), D̄0
, D0
(1865), D±
S (1969)
corresponding to the c̄ quark coupled to u, d, s quarks and the c quark cou-
pled to the ū, ¯
d, s̄ quarks. These are the 3 and 3̄ pieces of the 15-dimensional
SU(4) representation. The singlet SU(3) representation is associated with the
ηC(2984).
In the case of the baryons the singly charmed particles making up the sex-
tet are the Σ0,+,++
C (2455), the Ξ0,+
C (2577), and the Ω0
C(2696); the anti-triplet
is the Λ+
C(2287) and the Ξ+,0
C (2469); the triplet is the Ξ++,+
CC (3620) and the yet
unseen Ω+
CC
3.5 Symmetry and Weak Nonleptonic Λ Decay
The Λ(1115) has spin-1
2 and a lifetime of about 200 ps. Its primary decay mode are
into a proton and a negatively charged pion or to a neutron and a neutral pion.
The ∼10−10
second lifetime and the fact that strangeness is changed by one unit
indicates that this decay is due to the weak interaction. Hence the decay amplitude
will possess a parity-violating as well as a parity-conserving component.
a) Show that angular momentum conservation requires that the transition ampli-
tude must be either S-wave or P-wave so that the decay amplitude must have
the form
M = ASχ†
pχΛ + AP χ†
pσ · p̂πχΛ ,
where σ is a Pauli spin matrix and p̂π is a unit vector in the direction of the
outgoing pion.
b) Evaluate the decay distribution from an unpolarized Λ and show that the distri-
bution is isotropic.

c) Evaluate the decay distribution for the case of decay from a Λ with polarization
Pn̂, and show that it has the form
dΓ
dΩ
∼ 1 + A1Pn̂ · p̂π
with
A1 =
2ReA∗
SAP
|AS|2 + |AP |2
.
d) Show that a nonzero value of A1 implies that pions tend to be emitted either
parallel or antiparallel to the Λ spin direction according to whether A1 is
positive or negative and show, using mirror arguments, that the existence of
such an asymmetry requires a violation of parity invariance. Compare with the
explicit form for the asymmetry in terms of AS and AP and comment.
e) Evaluate the polarization of the final-state nucleon in the direction n̂ × p̂π for
the case of decay from a Λ with polarization Pn̂ and show that it is of the
form
P N · n̂ × p̂π
|n̂ × p̂π|
= P
2ImA∗
SAP
|AS|2 + |AP |2 + 2PReA∗
SAP n̂ · p̂π
|n̂ × p̂π| .
f) Analyze this result from the point of view of simple time reversal invariance
arguments and show that one might have expected this polarization to vanish.
g) What does the fact that the experimental value for this polarization is nonzero
say about the weak decay amplitudes?
Exercise 3.5 Solution: Symmetry and Weak Nonleptonic Λ Decay
a) Since initial state has spin 1
2 , this must be the total angular momentum. Since
the nucleon is spin 1
2 the orbital angular momentum can only be 0 or 1, with
amplitudes
M = ASχ†
pχΛ + AP χ†
pσ · p̂πχΛ
b) Find decay rate via
1
2
X
sp,sΛ
|M|2
=
1
2
Tr|AS + AP σ · p̂π|2
= |AS|2
+ |AP |2
c) Find
1
2
Tr(AS + AP σ · p̂π)(1 + Pσ · n̂)(A∗
S + A∗
P σ · p̂π)
= |AS|2
+ |AP |2
+ 2ReA∗
SAP Pn̂ · p̂π
d) Define asymmetry via number of pions emitted into the hemisphere parallel
to the polarization to those emitted into the opposite hemispherel. Find then
A1 =

#(n̂ · p̂π 0) − #(n̂ · p̂π 0)
#(n̂ · p̂π 0) + #(n̂ · p̂π 0)

=
2ReA∗
SAP
|AS|2 + |AP |2

Since spin stays invariant under a parity change while momentum reverses, the
correlation n̂ · p̂π is odd under parity. Hence it requires an interference of the
opposite parity amplitudes AS and AP .
e) Calculate final state nucleon polarization via
P p =
Trσ(AS + AP σ · p̂π)(1 + Pσ · n̂)(A∗
S + A∗
P σ · p̂π)
Tr(AS + AP σ · p̂π)(1 + Pσ · n̂)(A∗
S + A∗
P σ · p̂π)
=
2iImASA∗
P n̂ × p̂π
|AS|2 + |AP |2 + 2ReA∗
SAP Pn̂ · p̂π
f) Under time reversal both spin and momentum reverse sign so a triple correla-
tion such as sp · p̂π × sΛ is odd and should vanish.
g) Since ImA∗
SAP 6= 0, it says that the relative phase between the amplitudes is
nonvanishing. By the Fermi-Watson theorem then the pπ−
scattering phase in the
S- and P-channels must be different and nonvanishing.
3.6 Mesonic States
Consider the simple mesonic states which can be constructed from binding a quark
and antiquark.
a) Write down all of the qq̄ states in the pseudoscalar and vector nonets having good
G-parity.
b) Using Eq. (3.34), obtain the masses of the pseudoscalar and vector nonets shown
in Fig. 3.2, and compare with the physical values (note: the source for the
masses is [PDG14], with which the reader should become familiar).
Exercise 3.6 Solution: Mesonic States
a) For the low-lying non-strange pseudoscalar mesons we have
π+
=
1
√
2
ud + du

π0
=
1
2

dd − |uui

+ dd − |uui

π−
= −
1
√
2
(|udi + |dui) ,
which all have negative G-parity, as discussed in the text, while the non-strange
vector mesons discussed there are
ρ+
=
1
√
2
ud − du

ρ0
=
1
2

dd − |uui

− dd − |uui

ρ−
=
1
√
2
(|udi − |dui)

Table 3.1 Pseudoscalar nonet
Pseudoscalar Computed Value Measured Value % Difference
meson (MeV) (MeV)
π±
139.0 139.57018 +0.4
π0
139.0 134.9766 -2.9
K±
486.8 493.677 +1.4
K0
, K
0
486.8 497.614 +2.2
η 772.0 547.862 -29
η0
772.0 957.78 +24
with
Mmeson
0 = m1 + m2
from Eq. (3.31). Equation (3.34) provides a model for the interaction,
Mmeson
int =
ξ
4m1m2
×

−3 pseudoscalar
+1 vector
,
where in the text the following values were proposed for the constituent quark
masses and the spin-spin interaction strength:
mu ' md ≡ m = 308 MeV
ms ≡ m0
= 483 MeV
ξ = 4m2
× 159 MeV.
This results in the values in the table given here, where they are compared with
the measured values [PDG14], first for the pseudoscalar nonet of mesons (see Fig.
3.2 in the text). Except for the η and η0
where mixing occurs (see the discussions in
Chapter 3 of the text), the constituent quark model with this simple approach does
a very good job of accounting for the measured masses of the low-lying pseudoscalar
nonet. For the vector nonet (whose weight diagram is also shown in Fig. 3.2) we
find the results in the table and again the model works very well except for the ω
and φ where again mixing occurs.
3.7 Baryonic States
Consider the simple baryonic states which can be constructed from binding three
quarks.
a) Using the proton and neutron states given in Eqs. (3.60) and (3.61) together with
the charge operator, verify that the correct charges are obtained.
b) Using the magnetic dipole operator given in Eq. (3.71) with the baryon octet
states given in the text, obtain predictions for the magnetic moments. Compare
these results with the physical values given in [PDG14].

Table 3.2 Vector nonet
Vector Computed Value Measured Value % Difference
meson (MeV) (MeV)
ρ±
775.0 775.26 +0.03
ρ0
775.0 775.26 +0.03
K∗±
892.4 891.66 (895.5) -0.08 (+0.3)
K∗0
, K
∗0
892.4 895.81 +0.4
ω 1030.7 782.65 -24
φ 1030.7 1019.461 -1.1
c) Using the mass formulas given in the text, obtain predictions for the masses of
both the baryon octet and decuplet (see Fig. 3.4). Compare these results with
the physical values given in [PDG14].
Exercise 3.7 Solution: Baryonic States
a) The one-body charge operator is given by
Q =
X
i
q (i)
with charges q (i), i = 1, 2, 3 equal to +2/3 and −1/3 for u- and d-quarks, respec-
tively. As discussed in the text, since the states involve symmetric configurations,
one can set i = 1 and multiply by three in computing
Qp = hp ↑ |Q| p ↑i = 3 hp ↑ |q (1)| p ↑i .
Upon substituting Eq. (3.60) for the proton state in terms of the six contributions
involving two u quarks and one d quark one gets nine terms to add; these are
detailed below:
Qp =
3
18

2 × 2 ×
2
3
+ 2 × 2 ×
2
3
+ 2 × 2 ×

−
1
3

+

1 × 1 ×
2
3
+ 1 × 1 ×
2
3
+ 1 × 1 ×

−
1
3

× 2

= 1,
as expected. For the neutron one interchanges the factors 2/3 and −1/3 above and
gets Qn = 0, also as expected.
b) The one-body magnetic dipole operator is handled the same way as the charge
operator in part a):
µ =
X
i
µ (i) σ(i),
where µ (i) = q (i)·e/2m(i), where q (i) is the quark charge as in part a) with +2/3

Table 3.3 Baryon octet magnetic moments
State Model Computed Value Measured Value Difference
(µN ) (µN ) (µN )
p µp = 1
3
(4µu − µd) 2.59 2.792847356 0.21
n µn = 1
3
(4µd − µu) -1.73 -1.9130427 -0.19
Σ+
µΣ+ = 1
3
(4µu − µs) 2.49 2.458 -0.04
Σ0
µΣ0 = 1
3
(2µu + 2µd − µs) 0.77 – –
Σ−
µΣ− = 1
3
(4µd − µs) -0.96 -1.160 -0.20
Ξ0
µΞ0 = 1
3
(4µs − µu) -1.35 -1.250 -0.10
Ξ−
µΞ− = 1
3
(4µs − µd) -0.49 -0.6507 -0.16
Λ0
µΛ0 = µs -0.58 -0.613 -0.03
for u and −1/3 for d and s quarks, and m(i) is the quark mass, taken here to be
mu ' md ≡ m = 363 MeV for u and d quarks and ms ≡ m0
= 538 MeV for s
quarks, as discussed in the text. One has for any of the baryons
µ ≡ hB ↑ |µ3| B ↑i = 3µ3 (1) hB ↑ |σ3(1)| B ↑i ,
again using the symmetry of the states. For the states in the baryon octet, starting
with the proton, one has
µp = 3µ3 (1) hp ↑ |σ3(1)| p ↑i ,
again with nine terms as in part a) which are detailed below:
µp =
3
18
{2 × 2 × µu + 2 × 2 × µu + 2 × 2 × (−µd)
+1 × 1 × µu + 1 × 1 × µu + 1 × 1 × µd
+1 × 1 × (−µu) + 1 × 1 × (−µu) + 1 × 1 × µd}
=
1
3
(4µu − µd) ,
namely the result in Eq. (3.74) in the text. The neutron is easily obtained from this
by making the interchange µu ↔ µd:
µn =
1
3
(4µd − µu) ,
the result in Eq. (3.75) in the text. The other members of the baryon octet are han-
dled similarly using the configurations given in Chapter 3. The results for the entire
multiplet are listed in the accompanying table together with the measured values
from [PDG14]. Clearly the trends are good, although the differences presented in
the last column show offsets of up to roughly 0.2µN .
c) The baryon mass formula was derived in Chapter 3 (see the weight diagrams
in Fig. 3.4 for the baryon octet and decuplet). One has
Mbaryon
= Mbaryon
0 + Mbaryon
int

with
Mbaryon
0 = m1 + m2 + m3,
where, as in part b) we use the quark masses m and m0
defined there and in the
text. For the baryon decuplet from Eq. (3.80) one has simply
Mbaryon [10]
= Mbaryon
0

1 +
ξ0
4m1m2m3

with ξ0
= 4m2
× 50 MeV, yielding
M
baryon [10]
∆ = 3m

1 +
ξ0
4m3

M
baryon [10]
Σ∗ = (2m + m0
)

1 +
ξ0
4m2m0

M
baryon [10]
Ξ∗ = (m + 2m0
)

1 +
ξ0
4mm02

M
baryon [10]
Ω = 3m0

1 +
ξ0
4m03

.
The baryon octet is found to be somewhat more complicated (see Eqs. (3.81, 3.84,
3.85 and 3.86) in the text):
M
baryon [8]
N = 3m −
3ξ0
4m2
M
baryon [8]
Σ = 2m + m0
+
1
4

1
m2
−
1
mm0

ξ0
M
baryon [8]
Λ = 2m + m0
−
3ξ0
4m2
M
baryon [8]
Ξ = m + 2m0 1
4

1
m02
−
1
mm0

ξ0
.
From these we have the results given in the accompanying table for the baryon
decuplet — remarkably good agreement with the measured values. As noted in the
text, the constituent quark model was used to predict the mass and properties of the
Ω−
; it was subsequently found experimentally just as predicted, a major triumph
for the model. For the low-lying baryon octet we find the results in the table, and
again excellent agreement with the measured values is observed.

Table 3.4 Baryon decuplet masses
State Computed Value (MeV) Measured Value (MeV) % Deviation
∆ 1239.0 1232 -0.6
Σ∗+
1381.5 1382.80 +0.1
Σ∗0
1381.5 1383.7 +0.2
Σ∗−
1381.5 1387.2 +0.4
Ξ∗0
1529.2 1531.80 +0.2
Ξ∗−
1529.2 1535.0 +0.4
Ω−
1682.3 1672.45 -0.6
Table 3.5 Baryon octet masses
State Computed Value (MeV) Measured Value (MeV) % Deviation
p 939.0 938.272046 -0.08
n 939.0 939.565379 +0.06
Σ+
1179.1 1189.37 +0.9
Σ0
1179.1 1192.642 +1.2
Σ−
1179.1 1197.449 +1.6
Λ0
1114.0 1115.683 +0.2
Ξ0
1326.8 1314.86 -0.9
Ξ−
1326.8 1321.71 -0.4

4 The Standard Model
4.1 Z0
Decays
The Z0
boson decays via the weak neutral current to a pair of fundamental fermions:
ūu, ¯
dd, ...ν̄τ ντ
a) Show that the decay amplitude of the Z0
boson to a pair of fundamental spin-1
2
fermions is given by
Amp = i
g2
cos θW
µ
Zū(p1)γµ
h
(If
3 + 2 sin2
θW Qf ) + If
3 γ5
i
v(p2) ,
where If
3 is the SU(2)L eigenvalue of the fermion, Qf is the fermion charge,
and θW is the weak mixing angle.
b) Calculate the decay width for each quark and lepton in the approximation that
mf MZ0 .
c) Compare your results with the results from the particle data book [PDG14],
Γ(ēe) = Γ(µ̄µ) = Γ(τ̄τ) ' 84 MeV
Γ(ūu) = Γ(c̄c) ' 300 MeV
Γ( ¯
dd) = Γ(s̄s)) = Γ(b̄b) ' 380 MeV
d) Calculate the width due to (unseen) neutrino modes.
Exercise 4.1 Solution: Z0
Decays
Consider Z0
decay to fermion pair.
a) From Eq. 4.38 we have
L =
g
2 cos θW
µ
Zū(p1)γµ
h
If
3 (1 − γ5) − 2 sin2
θW Qf
i
v(p2)
≡
g
2 cos θW
µ
Zū(p1)γµ
h
gf
V − gf
Aγ5
i
v(p2)
where gf
V = If
3 − 2 sin2
θW Qf and gf
A = If
3
b) Calculate the differential decay rate
dΓ( ¯
ff) =
Nc
2mZ
Z
d3
p1
(2π)32E1
d3
p2
(2π)32E2
(2π)4
δ4
(PZ − p1 − p2)

g
2 cos θW
2
40

×
1
3

−ηµν
+
(p1 + p2)µ
(p1 + p2)ν
m2
Z

Trγµ(gf
V − gf
Aγ5) 6p2γν(gf
V − gf
Aγ5) 6p1
= Nc
dΩ
96π2
g2
cos2 θW
gf2
V + gf2
A
2mZ
!
−ηµν
+
(p1 + p2)µ
(p1 + p2)ν
m2
Z

(p1µp2ν + p1νp2µ − ηµνp1 · p2)
= Nc
dΩ
96π2
g2
cos2 θW

g2
V + g2
A
2mZ

2p1 · p2 =
g2
mZ(gf2
V + gf2
A )
6 cos2 θW
Using g2
2 cos2 θW
= 2
√
2GF m2
Z we have
Γ( ¯
ff) = Nc
GF m3
Z
6
√
2π
(gf2
V + gf2
A ) ' (gf2
V + gf2
A ) × 330 MeV
c) Now look at various channels:
i) e, µ, τ: Nc = 1; gV = −1
2 (1 − 4 sin2
θW ); gA = 1
2 . Then
Γ(ēe) = Γ(µ̄µ) = Γ(τ̄τ) ' 83 MeV 1 + (1 − 4 sin2
θW )2

' 83 MeV
ii) u, c: Nc = 3; gV = 1
2 − 4
3 sin2
θW ; gA = 1
2 . Then
Γ(ūu) = Γ(c̄c) =' 250 MeV

1 + (1 −
8
3
sin2
θW )2

' 290 MeV
iii) d, s, b: Nc = 3; gV = −1
2 + 2
3 sin2
θW ; gA = −1
2 . Then
Γ( ¯
dd) = Γ(s̄s) = Γ(b̄b) ' 250 MeV

1 + (1 −
4
3
sin2
θW )2

' 370 MeV
d) For neutrino modes we have Nc = 1; gV = gA = 1
2 . Then width is
Γ(ν̄eνe) = Γ(ν̄µνµ) = Γ(ν̄τ ντ ) = 83 MeV(1 + 1) = 166 MeV
so the total invisible width is
Γ(inv.) = 3 × 166 Mev = 498 MeV
4.2 Yang-Mills Theory
The Dirac Lagrangian for a spin-1
2 particle of mass m
L[ψ(x)] = ψ̄(x) [i 6∂ − m] ψ(x)
is invariant under the U(1) global phase transformation
ψ(x) → ψ0
(x) = exp(iβ)ψ(x) .
a) Demonstrate this invariance.

It is also well known that the Dirac Lagrangian for a spin-1
2 particle of mass m and
charge e can be made invariant under a local U(1) phase transformation
ψ(x) → ψ0
(x) = exp(ieβ(x))ψ(x) ≡ Uψ(x).
b) Demonstrate that the Lagrangian L[ψ(x)] is not invariant under a local U(1)
transformation.
The invariance can be accomplished if we introduce an additional dynamical vector
field Aµ(x) with the transformation property
Aµ(x) → UAµ(x)U−1
−
i
e
U∂µU−1
= Aµ(x) − ∂µβ(x)
and modify the Lagrangian to become
L[ψ(x), Aµ(x)] = ψ̄(x) [i 6D − m] ψ(x) −
1
4
Fµν(x)Fµν
(x),
where
Dµ = ∂µ + ieAµ(x)
is the covariant derivative and has the property
Dµψ(x) → D0
µψ0
(x) = UDµψ(x).
c) Verify this transformation property of the covariant derivative.
d) Demonstrate the invariance of L[ψ(x), Aµ(x)] under a local U(1) transformation.
Yang and Mills showed how to generalize this local invariance to the case of non-
Abelian transformations. We consider the case of SU(2) with a doublet of spin-1
2
fields
N(x) =

ψ1(x)
ψ2(x)

which transforms as
N(x) → N0
(x) = exp(ig
1
2
τ · β(x))N(x) ≡ UN(x)
under a local SU(2) gauge transformation, where τ are the Pauli isospin matrices
and satisfy the commutation relations
[τi, τj] = 2iijkτk.
e) Demonstrate that the Lagrangian L[N(x)] is not invariant under a local SU(2)
gauge transformation.
Now define a triplet of vector fields Aµ(x) which transform as
1
2
τ · Aµ(x) → U
1
2
τ · Aµ(x)U−1
− iU∂µU−1
and a covariant derivative
Dµ = ∂µ + ig
1
2
τ · A(x).

f) Demonstrate that the covariant derivative has the property
DµN(x) → D0
µN0
(x) = UDµψ(x) , .
where
U = exp(i
1
2
τ · β(x))
g) For an infinitesimal δβ show that
Aµ(x) → Aµ(x) − ∂µδβ(x) + gδβ(x) × Aµ(x).
h) Demonstrate that the Lagrangian
L[N(x), A(x)](x) = N̄(x) [i 6D − m] N(x) −
1
4
F µν(x) · F µν
(x)
is invariant under a local SU(2) gauge transformation provided that F µν(x) is
defined via
[Dµ, Dν]N(x) = −ig
1
2
τ · F µν(x)N(x)
i.e.,
F µν(x) = ∂µAν(x) − ∂νAµ(x) − gAµ(x) × Aν(x)
and has the transformation property
1
2
τ · F µν → U
1
2
τ · F µνU−1
.
Exercise 4.2 Solution: Yang-Mills Theory
a) If
ψ(x) → ψ0
(x) = Uψ(x)
with U = exp(iβ) then
ψ̄(x) → ψ̄0
(x) = ψ̄(x)U−1
with U−1
= U†
= exp(−iβ) so
L = ψ̄(x)(i 6∂ − m)ψ(x) → ψ̄(x)U−1
(i 6∂ − m)Uψ(x) = L
b) If β → eβ(x) then
L = ψ̄(x)(i 6∂ − m)ψ(x) → ψ̄(x)U−1
(i 6∂ − m)Uψ(x)
= L − eψ̄(x)U−1
(6∂β(x))Uψ(x) = L − eψ̄(x)(6∂β(x))ψ(x)
c) Define iDµ = i∂µ − eAµ so
iDµψ(x) = (i∂µ − eAµ(x))ψ(x) → [i∂µ − eAµ(x) + e∂µβ(x)) Uψ(x)
= U (i∂µ − eAµ − e∂µβ(x) + e∂µβ(x)) ψ(x) = UiDµψ(x)

d) Then, since
Fµν → Fµν − e∂µ∂νβ(x) + e∂ν∂µβ(x) = Fmuν
we have
L = ψ̄(x)(i 6D−m)ψ(x)−
1
4
FµνFµν
→ ψ̄(x)U−1
U(i 6D−m)ψ(x)−
1
4
FµνFµν
= L
e) We find
N̄(i 6D − m)N → N̄U−1
(i 6D − m)UN = N̄(i 6D − m)N − gN̄
1
2
τ · ∂µβN
f) Then for covariant derivative
iDµN(x) = (i∂µ − gτ · Aµ(x))N(x) → (i∂µ − gτ · Aµ(x) + τ · ∂µβ) UN(x)
= U

i∂µ − g
1
2
τ · Aµ(x) +
1
2
τ · ∂µβ −
1
2
τ · ∂µβ

N(x) = UiDµN(x)
g) Since
1
2
τ · Aµ → U
1
2
τ · AµU−1
− iU∂µU−1
we have
1
2
τ · Aµ →
1
2
τ · Aµ + ig[
1
2
τ · δβ,
1
2
τ · Aµ] +
1
2
gτ · ∂µδβ
or
Aµ → Aµ − gδβ × Aµ + g∂µδβ
h) First look at
−
1
4
F µν · F µν
= −
1
2
Tr
1
2
τ · F µν ·
1
2
τ · F µν
→ −
1
2
TrU
1
2
τ · F µνU−1
· U
1
2
τ · F µνU−1
= −
1
4
F µν · F µν
Then
L = N̄(i 6D − m)N −
1
4
F µν · F µν
→ N̄U−1
U(i 6D − m)N −
1
4
F µν · F µν
= L
Check:
[Dµ, Dν]N = −ig∂µ
1
2
τ · Aν + ig∂ν
1
2
τ · Aµ + ig2 1
2
τ · Aµ × Aν = −igF µν
with
F µν = ∂µAν − ∂νAµ − gAµ × Aν
Also, for infinitesimal transformation
1
2
τ · F µν → U
1
2
τ · F µνU−1
=
1
2
τ · F µν − ig[
1
2
τ · δβ,
1
2
τ · F µν]
=
1
2
τ · F µν + g
1
2
τ · δβ × F µν

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205
Then came the pancake, and the final drop. But in the
end the plane received little damage nor were its
occupants much thrown about. The carriage holding the
wheels, torn loose in front when the wheels scraped the
upper edge of the Coliseum’s tiers of seats, was still
firmly fastened at the rear. Thus, the wheels hung
slantwise. Had Frank, ignorant of what had occurred,
attempted the usual landing, the results would have
been disastrous. But by pancaking and dropping, the
wheels were pushed up against the bottom of the plane
and held firmly in place, instead of being torn entirely
from the fastenings.
The result was that the plane, although racketted about
a bit, suffered no more than in a bumpy landing, and
came to rest without burying nose or wings in the sand
as had been feared would be the case.
All climbed stiffly out, and the next minute Frank and
Bob were hugging each other like a couple of kids, and
thumping each other on the back with terrific whacks.
In the meantime, Roy Stone and Amrath stood aside,
and it was not until Frank and he had pummelled each
other to their mutual satisfaction that Bob turned to the
aviator.
“Haven’t had much chance for personal conversation
with you up there in the plane, Stone,” he said, as he
wrung the other’s hand. “But I want to tell you—Oh,
shucks, what’s the use? I can’t sling language much.
Only, I will say I never got more benefit out of a fight in
my life than out of that one with you in the cave back in
Old Mexico.”
Roy Stone grinned through the sun-wrinkles about his
eyes. He knew Bob’s reference was to the affray

206
207
between the two parties in the lonely mountains of Old
Sonora, when the boys were striving to rescue Mr.
Hampton from the hands of the Mexican rebels. At that
time, as recorded in “The Radio Boys on the Mexican
Border,” Stone had been in the rebel forces. But later he
changed his allegiance, and warm, indeed, had been the
friendship between him and the boys, particularly
between him and Bob, who had been his own individual
opponent in the fight in the cave.
“You like fighting so much,” said Stone, “that it’s a
wonder you consented to let us take you away from the
Coliseum back there in Athensi.”
Bob shook his head and threw up his hands.
“A fellow can get too much of any good thing,” he said.
“Well, let’s snap into it and go back to this place where
our friends are fighting. Maybe we can help a little. But
first I’m going to leave this hardware here.”
Whereupon he stripped off the various pieces of heavy
armor and tossed them into the pit of the airplane,
standing revealed in nothing but a G string—a superb
figure who caused Amrath, for one, to draw in a breath
of admiration.
“Monsieur would have been a hard man to beat in the
Sacrificial Games,” he said in French.
“Aw, forget it,” said Bob. “Come on. Got to give Jack and
Mr. Hampton a hand.”

208
CHAPTER XXVI.
THE REVOLUTIONISTS SUCCEED.
In advance of the mounted re-inforcements from the
other pass, which still were some distance in the rear,
the four adventurers entered the Great Road and
started at a trot up the gradual ascent, Bob in the lead.
“Don’t hear any firing yet, do you?” he shouted over his
shoulder to the others. “You fellows have got revolvers,
but I’m going to hop ahead and root for one in the
luggage.”
Frank had explained about the grove where their own
party was encamped and where the radio had been set
up. It was here Bob intended to look for his automatic,
which he had not taken with him when departing from
the distant oasis on that memorable ostrich hunt.
“Not much use this, unless at close quarters,” he called,
waving a short, heavy sword of hard wood—a dummy
weapon which he had been using against a trainer when
rescued from the Coliseum. “Might brain a man with it,
but that’s all.”
With a farewell wave of the wooden sword, Bob’s naked
figure drew away from the others. It was late afternoon,
and the Great Road already lay in the shadow cast by

209
the western wall of the pass. Hot though it was, the
relief from the heat of the desert was instantaneous,
and the others felt it at once and began to increase
their speed.
As they passed abreast of the grove, Bob emerged,
flourishing his automatic, the dummy sword left behind.
As he fell in beside them he cried with a grin:
“Well, I’m all dressed up now.”
Despite their labored breathing, the others could not
restrain a laugh at the ridiculous idea of a naked man
considering himself dressed with a revolver.
After all, their services were not needed. When they
arrived at the barricade, they found the defenders still
awaiting the attack which had failed to materialize.
Jack’s earlier arrival with Roy Stone’s message that he
intended to drop gas bombs in the midst of the
Janissaries had given them the solution of the mystery,
and the explanation of the fliers regarding the damage
wrought was greeted with delight.
The little band had suffered slightly by comparison with
the terrible execution they had worked among the
Janissaries at the tunnel exit of the subterranean river.
Yet their losses had been severe enough. Lieutenant
Horeb and one of his men had been killed; Akmet, two
other Arabs, and three revolutionists had suffered
dangerous, though not fatal, injuries, and not one had
escaped without some slight wound.
To the boys the fact that Mr. Hampton, praised by all for
covering the retreat with his repeater, had come
through safely with no more than a flesh wound in the
calf of his right leg, was a matter for the greatest

210
thankfulness. As the three of them foregathered with
Mr. Hampton and Roy Stone, a little to one side of the
main group, the thought occurred to all that they had
reason, indeed, for gratitude at having passed
practically unscathed through their numerous and
deadly perils.
Mr. Hampton, who was not given to outward religious
manifestations, said simply:
“Almighty Providence has looked after us all, fellows,
and we mustn’t forget to give thanks.”
And for a moment, each bowed his head and voiced the
thankfulness in his heart in his own way.
A clatter of approaching hoofs rang in the road, and up
dashed the score of hard-riding horsemen from the
other pass, for whom Jepthah had despatched the
messenger.
A condensed account of events was given their leader, a
lean hard-bitten man older than the majority of the
young revolutionists whom, the boys later learned was
Maspah, a nobleman whose gorge had risen at the
terrible punishment meted out by the Oligarchy to those
earlier exiles who had shown kindness to Professor
Souchard and aided his return to civilization, and who
forthwith had fled to join the little outlaw bands which
finally concentrated at Korakum under Captain
Amanassar and launched the revolution.
His eyes gleamed when he was told of the
demoralization wrought among the Janissaries by the
dropping of the gas bombs. While waiting the arrival of
the footmen, peasants armed with bows and arrows and

211
numbering 200, he had the breach re-opened to admit
the passage of his horsemen.
In the meantime, too, scouts were sent ahead with
glasses furnished by Amrath and Mr. Hampton, who had
worn his in a case slung over his shoulder, to mount into
the tops of a grove of date palms just beyond the
mouth of the pass and inspect the valley. They returned
presently with word that in the distance, where the gas
bombs had fallen, the Great Road was still littered with
men, but that to the left of this spot, in the cleared
space in front of the ruins of the ancient temple, where
the revolutionists had been accustomed to hold their
meetings, officers were re-assembling the scattered
Janissaries not struck down by the gas. A considerable
number, perhaps four or five hundred, were collecting.
Lieutenant Maspah looked thoughtful.
“They will be better armed than we,” he said. “Yet we
have thirty horsemen, which gives us a big advantage
and if we strike at once we shall have the advantage of
surprise, while if we delay they will recover from their
demoralization. Ah, here come the footmen,” he added.
“I shall attack at once.”
Only four of the camels of the Hampton party had been
brought in, the others having lumbered away to their
grazing grounds in a distant portion of the valley when
their masters had been wounded. Akmet and his two
companions had been carried to the barricade on the
camels of their comrades. But from mounting these four
camels, Ali and his remaining Arabs could not be
dissuaded. Their blood was up and they wanted a hand
in the last phase of the battle.

212
213
This left no mounts for the boys and Roy Stone, which
caused Bob, who wanted to “take a crack” at the bloody
rascals, as he expressed it, to grumble exceedingly. Mr.
Hampton, however, was pleased that it should be so, as
he felt the lives of all had been risked sufficiently.
Besides, he had undertaken to look after the wounded,
who as yet lay on the roadside in the shadow of the
western wall, and he needed aid to transport them to
the shade of their own camp in the grove where, with
medical instruments and drugs, he could make shift to
probe wounds, extract bullets, bandage and do his best
to ease pain.
“The four of you,” he said to his son and Frank, Bob and
Roy, “can do vastly more good helping me than out
there in Korakum. We need litters to move these fellows
to the grove, so hurry back, cut down some of those
young trees coming up in the brush, and then return.
Make your best speed, too. I’ll go along and get out my
supplies and have everything ready to do what I can
when you bring me the wounded.”
An hour later, word arrived by messenger sent back by
Amrath, who knew Mr. Hampton would be anxious to
hear the result of the battle, that the Janissaries had put
up only a feeble resistance in their demoralized state
and that, after being badly cut up by the horsemen,
they had surrendered. A little later Ali and his Arabs
returned, unwounded, swaggering a bit, and gave them
a lurid account of the fight.

214
CHAPTER XXVII.
ATHENSI FALLS.
After all these events culminating in the rescue of Bob
and the disastrous rout of the Janissaries at Korakum,
Mr. Hampton decided instead of returning to civilization
without having accomplished his main objective—
namely, the exploration of the ruins of Korakum and the
gaining of entrance to Athensi—to stay and await the
result of the revolution.
The Korakum expedition had been timed by the
Oligarchs to coincide with an attack in force launched
through the mountains against Captain Amanassar’s
main body of revolutionists in the field. There, too, the
Janissaries had been unsuccessful. Though not beaten
so decisively as at Korakum, they had been unable to
penetrate the strong position held by the rebels and,
sullen and alarmed at the unexpected strength of the
opposition, they had fallen back to the shelter of the
walls of Athensi.
In their retreat they carried off all the livestock for miles
from the country between Captain Amanassar and the
city, stripping the poor peasants of everything, and
herding the young men into the city while leaving the
children and the old people to live as best they might.

215
Mr. Hampton made a trip to Captain Amanassar’s camp,
into which the stricken country people from the
devastated districts were making their way, and on his
return reported many pitiable sights. The rebel leader’s
assurance that the fall of Athensi, in view of the two
disasters to the arms of its defenders, was inevitable,
caused the American to decide to stay.
He was moved by more than an explorer’s interest,
moreover. Deeply stirred by the ideals of these young
Athensians, sons of a semi-savage race dating from the
dawn of time, who were resolved to redeem their
country from the rule of the Oligarchs who so long had
held it in thrall, he felt that his engineering experience
would be valuable in the final siege of the city and that
later his knowledge of world affairs would be worth
much to Captain Amanassar when the latter and his
compatriots came to the point of opening
communication with the outside world.
Week by week the lines about Athensi grew tighter, with
every sally of the Janissaries repulsed. Reports from
friends within the city, where the revolutionists had
many adherents, continued to reach the rebel camp,
and all were to the effect that famine was beginning to
raise its head amid the crowded population.
That great numbers of his countrymen should be
starved to death or die of plague, for sickness also
broke out in Athensi, was not Captain Amanassar’s
object. On several occasions, he made overtures to the
Oligarchs looking to the surrender of the city on terms
which would spare their lives, but these were all
rejected. The rulers of the priest clan could not bring
themselves to a realization that at last the power they
had exercised through uncounted centuries was

216
seriously threatened, and seemed bent on involving all
in ruin rather than continue to live shorn of power. To
storm Athensi was an impossibility for Captain
Amanassar’s numerous but ill-equipped army, and
apparently the only thing to do was to play a waiting
game.
Such a course, however, was repugnant to the rebel
leader, whose heart bled for the miseries of the cooped-
up population, and he sought by every known method
of appeal to prevail on those residents who managed to
steal out of Athensi and reach his camp, to bring about
an uprising in the city which would open its gates to his
forces.
At length, when the miseries of the city reached a point
too great to be borne any longer, his arguments
prevailed. A half dozen of his stoutest-hearted aides
entered Athensi with a drove of lean cattle, announcing
boldly they had been burned out by the rebels and
came to the city for shelter. They disappeared amid the
city warrens after being admitted at the great gate, and
then scattered to rouse the city to fever pitch.
That night the Janissaries, going to change guard on
the walls, were attacked as they passed through the
streets, and were driven back to the shelter of the Inner
City. The guard at the great gate was surprised and
overcome, and the gates opened to admit a force of
picked warriors from the rebel ranks, who had stolen up
under cover of darkness.
The Janissaries posted on the walls in the vicinity of the
gate were overcome, although fighting desperately, and
before help could reach them from other parts of the

217
walls, the main force of the rebels, which had moved up
by forced marches, entered the city.
Many of the Janissaries were cut down as they fell back
to the Inner City, where their heartless comrades
refused to open the gates to admit them lest the rebels
also force their way in.
Dawn found Athensi in the possession of Captain
Amanassar’s forces, with the Inner City beleaguered on
every side, and its fall only a matter of time. Three
weeks it managed to hold out and then its defenders
weak from hunger, were forced to seek unconditional
surrender. The Oligarchs were imprisoned to stand trial
later for their crimes, and the surviving Janissaries were
disarmed and, although their lives were spared, they
were put to work as state peons repairing the ravaged
countryside.
Bob, Jack, Frank and Roy Stone followed the first wave
of the attack into Athensi in a company of 200 rebels
commanded by Jepthah. At Bob’s special request, this
group made its way through the tumultuous streets to
the Coliseum. It was a moonless night, and the great
amphitheater lay dark and mysterious outside the walls
of the Inner City.
Around those walls raged a furious battle but in the
Coliseum itself, which the Janissaries had no idea of
defending, all was silent. That is, until the rebels with
Bob at their head, clad again in the gladiator’s armor he
had worn on being rescued, entered the arena with
their wavering torches.
The tumult of the desperate fighting within the city was
reduced to a murmur down there, on the sand, at the

218
base of those towering tiers of seats. Yet here, too, it
had penetrated and the poor captives, locked in their
quarters for the night, and awaiting the coming of the
Sacrificial Games, now only a week away, were awake
and moving about restlessly.
As the light of the torches fell through the massive bars
of the great door set in the solid stone of the wall, and
penetrated the interior of the single great room where
all the alien gladiators were quartered and where Bob,
too, had lived, the poor fellows crowded forward.
They did not know what the tumult in the city and now
the arrival of this armed force portended, but Bob was
easily recognizable in his armor and made friendly signs
indicating he had come to release them. At the same
time, men armed with stout axes and wrenching bars
attacked the gate. It was stubborn and resisted all
assaults a long time but eventually gave way, and then
the slaves threw themselves at Bob’s feet and tried to
kiss his hands. To these men, most of whom were
Negroes, although a few Berbers and Tuaregs were in
the number, Bob’s sudden rescue by airplane had
appeared as a miracle. And now his return to release
them had an even greater effect on their primitive
intelligences.
While this was going on Jepthah headed another party
which broke down a similar gate on the other side of
the arena, behind which were confined the young
Athensians destined to fight the slaves in the Sacrificial
Games. To one or two of them he was known, and
when he spread the word of the success of the
revolution the joy of these young fellows, snatched from
their families by the Oligarchs to go to death, knew no
bounds.

220
CHAPTER XXVIII.
CONCLUSION.
After the final capitulation of the Oligarchy, Mr. Hampton
and the members of his party went to live in quarters
assigned them in one of the palaces of the Inner City. It
was an age-worn stone structure of immensely thick
walls, two stories in height, and covering five acres of
ground. In it were hundreds of rooms and apartments,
sumptuously appointed with many luxuries.
“It’s all right, this business of living in a palace,” said
Bob, one day. “Just the same I for one can never
accustom myself to living in a tomb. And that’s what
this seems like, with its old stone walls and courts and
secret passages, and what not.”
With this opinion, Jack and Frank were in hearty
agreement. Likewise Roy Stone, who after repairing his
airplane had flown it to the plain outside Athensi where
it rested now with just sufficient fuel to carry him out of
the desert when the time came to depart. Departure,
however, he kept putting off from time to time at the
insistence of his friends.
Ali and his Arabs continued to stay with Mr. Hampton,
the wounded members of the party now fully restored

221
to health.
In the Great Library of Athensi, the biggest building in
extent within the Inner City, were found as the
revolutionists had predicted many thousands of
manuscripts or papyrus rolls written in the ancient
mother tongue of Atlantis of which Athensian was a
corruption. Few of the young nobles among the
revolutionists ever had been within the library before, as
the ancients of the Oligarchs had guarded it jealously.
They were even more eager than Mr. Hampton to
browse, if that word can properly be employed in this
connection. But when they came to examining the rolls,
they found that it was only with difficulty they could
here and there decipher a word.
However, the similarity of languages was such that in
time the mother tongue could be learned and the
treasured knowledge of this most ancient of libraries in
the oldest living city on earth, could be unlocked and
given to the world. To the task of learning the language
and of putting the library in order, Captain Amanassar
who had been elected President of the new republican
government, assigned Amonasis, Amrath and two dozen
assistants, comprising the best educated of the young
revolutionists. Eagerly, they began their task.
At length, with a story that later was to astound not
only the scientific world but all civilization, Mr. Hampton,
finding his advice no longer was required, decided to
depart. They had been absent from home five months.
Bob and Frank were overdue for their Senior year at
Yale. Mr. Hampton was to be the unofficial
representative of the Athensian government to the
United States, and was to pave the way for official

222
representatives to be sent to the various world capitals
by making public his account of events.
In addition, he was to interest capitalists in developing
the resources of the country, and in building a railroad
linking up Athensi with the Cape-to-Cairo Railroad.
He promised to return the following year, estimating it
would require that length of time at least to perform his
various commissions. On his return, the boys planned to
accompany him and to build a great radio station at
Athensi, which would put the mountain people in touch
with all the world.
True to his promise, they did return the following year,
carrying to Athensi a great caravan of supplies for the
erection of a completely equipped radio sending and
receiving station. These supplies were taken up the
Niger by boat and finally across the desert by camel.
But after finishing the erection of the station, the three
Radio Boys set out on an exploring expedition through
the heart of Africa in the interests of a new motion
picture producing corporation among the backers of
which were both Mr. Temple and Mr. Hampton. And the
adventures which befell them upon this 5,000 mile
journey through jungle wilds and in coming into contact
with savage men and beasts, were numerous and
varied.
All will be duly chronicled in “The Radio Boys in Darkest
Africa.” Until then, let us bid them good-bye.
THE END.

223
The Radio Boys Series
BY GERALD BRECKENRIDGE
A new series of copyright titles for boys of all ages.
Cloth Bound, with Attractive Cover Designs
PRICE, 65 CENTS EACH
THE RADIO BOYS ON THE MEXICAN BORDER
THE RADIO BOYS ON SECRET SERVICE DUTY
THE RADIO BOYS WITH THE REVENUE GUARDS
THE RADIO BOYS’ SEARCH FOR THE INCA’S TREASURE

224
THE RADIO BOYS RESCUE THE LOST ALASKA
EXPEDITION
The Boy Troopers Series
BY CLAIR W. HAYES
Author of the Famous “Boy Allies” Series.
The adventures of two boys with the Pennsylvania State
Police.
All Copyrighted Titles.
Cloth Bound, with Attractive Cover Designs.

225
PRICE, 65 CENTS EACH.
THE BOY TROOPERS ON THE TRAIL
THE BOY TROOPERS IN THE NORTHWEST
THE BOY TROOPERS ON STRIKE DUTY
THE BOY TROOPERS AMONG THE WILD
MOUNTAINEERS
The Golden Boys Series
BY L. P. WYMAN, PH.D.
Dean of Pennsylvania Military College.

226
A new series of instructive copyright stories for boys of
High School Age.
Handsome Cloth Binding.
THE GOLDEN BOYS AND THEIR NEW ELECTRIC CELL
THE GOLDEN BOYS AT THE FORTRESS
THE GOLDEN BOYS IN THE MAINE WOODS
THE GOLDEN BOYS WITH THE LUMBER JACKS
THE GOLDEN BOYS ON THE RIVER DRIVE
The Ranger Boys Series
BY CLAUDE H. LA BELLE

227
A new series of copyright titles telling of the adventures
of three boys with the Forest Rangers in the state of
Maine.
Handsome Cloth Binding.
THE RANGER BOYS TO THE RESCUE
THE RANGER BOYS FIND THE HERMIT
THE RANGER BOYS AND THE BORDER SMUGGLERS
THE RANGER BOYS OUTWIT THE TIMBER THIEVES
THE RANGER BOYS AND THEIR REWARD

The Boy Allies
(Registered in the United States Patent Office)
With the Navy
BY ENSIGN ROBERT L. DRAKE
For Boys 12 to 16 Years.
All Cloth Bound Copyright Titles
Frank Chadwick and Jack Templeton, young American
lads, meet each other in an unusual way soon after the
declaration of war. Circumstances place them on board
the British cruiser, “The Sylph,” and from there on, they

228
share adventures with the sailors of the Allies. Ensign
Robert L. Drake, the author, is an experienced naval
officer, and he describes admirably the many exciting
adventures of the two boys.
THE BOY ALLIES ON THE NORTH SEA PATROL; or,
Striking the First Blow at the German Fleet.
THE BOY ALLIES UNDER TWO FLAGS; or, Sweeping the
Enemy from the Sea.
THE BOY ALLIES WITH THE FLYING SQUADRON; or,
The Naval Raiders of the Great War.
THE BOY ALLIES WITH THE TERROR OF THE SEAS; or,
The Last Shot of Submarine D-16.
THE BOY ALLIES UNDER THE SEA; or, The Vanishing
Submarine.
THE BOY ALLIES IN THE BALTIC; or, Through Fields of
Ice to Aid the Czar.
THE BOY ALLIES AT JUTLAND; or, The Greatest Naval
Battle of History.
THE BOY ALLIES WITH UNCLE SAM’S CRUISERS; or,
Convoying the American Army Across the Atlantic.
THE BOY ALLIES WITH THE SUBMARINE D-32; or, The
Fall of the Russian Empire.
THE BOY ALLIES WITH THE VICTORIOUS FLEETS; or,
The Fall of the German Navy.
The Boy Allies
(Registered in the United States Patent Office)
With the Army
BY CLAIR W. HAYES

In this series we follow the fortunes of two American
lads unable to leave Europe after war is declared. They
meet the soldiers of the Allies, and decide to cast their
lot with them. Their experiences and escapes are many,
and furnish plenty of good, healthy action that every
boy loves.
THE BOY ALLIES AT LIEGE; or, Through Lines of Steel.
THE BOY ALLIES ON THE FIRING LINE; or, Twelve Days
Battle Along the Marne.
THE BOY ALLIES WITH THE COSSACKS; or, A Wild Dash
Over the Carpathians.
THE BOY ALLIES IN THE TRENCHES; or, Midst Shot and
Shell Along the Aisne.

229
THE BOY ALLIES IN GREAT PERIL; or, With the Italian
Army in the Alps.
THE BOY ALLIES IN THE BALKAN CAMPAIGN; or, The
Struggle to Save a Nation.
THE BOY ALLIES ON THE SOMME; or, Courage and
Bravery Rewarded.
THE BOY ALLIES AT VERDUN; or, Saving France from
the Enemy.
THE BOY ALLIES UNDER THE STARS AND STRIPES; or,
Leading the American Troops to the Firing Line.
THE BOY ALLIES WITH HAIG IN FLANDERS; or, The
Fighting Canadians of Vimy Ridge.
THE BOY ALLIES WITH PERSHING IN FRANCE; or, Over
the Top at Chateau Thierry.
THE BOY ALLIES WITH THE GREAT ADVANCE; or,
Driving the Enemy Through France and Belgium.
THE BOY ALLIES WITH MARSHAL FOCH; or, The Closing
Days of the Great World War.
The Boy Scouts Series
BY HERBERT CARTER

For Boys 12 to 16 Years
New Stories of Camp Life
THE BOY SCOUTS’ FIRST CAMPFIRE; or, Scouting with
the Silver Fox Patrol.
THE BOY SCOUTS IN THE BLUE RIDGE; or, Marooned
Among the Moonshiners.
THE BOY SCOUTS ON THE TRAIL; or, Scouting through
the Big Game Country.
THE BOY SCOUTS IN THE MAINE WOODS; or, The New
Test for the Silver Fox Patrol.

230
THE BOY SCOUTS THROUGH THE BIG TIMBER; or, The
Search for the Lost Tenderfoot.
THE BOY SCOUTS IN THE ROCKIES; or, The Secret of
the Hidden Silver Mine.
THE BOY SCOUTS ON STURGEON ISLAND; or,
Marooned Among the Game-Fish Poachers.
THE BOY SCOUTS DOWN IN DIXIE; or, The Strange
Secret of Alligator Swamp.
THE BOY SCOUTS AT THE BATTLE OF SARATOGA; A
story of Burgoyne’s Defeat in 1777.
THE BOY SCOUTS ALONG THE SUSQUEHANNA; or, The
Silver Fox Patrol Caught in a Flood.
THE BOY SCOUTS ON WAR TRAILS IN BELGIUM; or,
Caught Between Hostile Armies.
THE BOY SCOUTS AFOOT IN FRANCE; or, With The Red
Cross Corps at the Marne.
The Jack Lorimer Series
BY WINN STANDISH

CAPTAIN JACK LORIMER; or, The Young Athlete of
Millvale High.
Jack Lorimer is a fine example of the all-around
American high-school boys. His fondness for clean,
honest sport of all kinds will strike a chord of
sympathy among athletic youths.
JACK LORIMER’S CHAMPIONS; or, Sports on Land and
Lake.

231
There is a lively story woven in with the athletic
achievements, which are all right, since the book
has been O. K’d. by Chadwick, the Nestor of
American Sporting journalism.
JACK LORIMER’S HOLIDAYS; or, Millvale High in Camp.
It would be well not to put this book into a boy’s
hands until the chores are finished, otherwise they
might be neglected.
JACK LORIMER’S SUBSTITUTE; or, The Acting Captain of
the Team.
On the sporting side, this book takes up football,
wrestling, and tobogganing. There is a good deal of
fun in this book and plenty of action.
JACK LORIMER, FRESHMAN; or, From Millvale High to
Exmouth.
Jack and some friends he makes crowd
innumerable happenings into an exciting freshman
year at one of the leading Eastern colleges. The
book is typical of the American college boy’s life,
and there is a lively story, interwoven with feats on
the gridiron, hockey, basketball and other clean
honest sports for which Jack Lorimer stands.

The Girl Scouts Series
BY EDITH LAVELL
A new copyright series of Girl Scouts stories by an
author of wide experience in Scouts’ craft, as Director of
Girl Scouts of Philadelphia.
Clothbound, with Attractive Color Designs.
THE GIRL SCOUTS AT MISS ALLEN’S SCHOOL
THE GIRL SCOUTS AT CAMP
THE GIRL SCOUTS’ GOOD TURN
THE GIRL SCOUTS’ CANOE TRIP

232
THE GIRL SCOUTS’ RIVALS
Marjorie Dean College Series
BY PAULINE LESTER.
Author of the Famous Marjorie Dean High School Series.
Those who have read the Marjorie Dean High School
Series will be eager to read this new series, as Marjorie
Dean continues to be the heroine in these stories.
All Clothbound. Copyright Titles.

233
MARJORIE DEAN, COLLEGE FRESHMAN
MARJORIE DEAN, COLLEGE SOPHOMORE
MARJORIE DEAN, COLLEGE JUNIOR
MARJORIE DEAN, COLLEGE SENIOR
Marjorie Dean High School Series
BY PAULINE LESTER
Author of the Famous Marjorie Dean College Series
These are clean, wholesome stories that will be of great
interest to all girls of high school age.

234
MARJORIE DEAN, HIGH SCHOOL FRESHMAN
MARJORIE DEAN, HIGH SCHOOL SOPHOMORE
MARJORIE DEAN, HIGH SCHOOL JUNIOR
MARJORIE DEAN, HIGH SCHOOL SENIOR
The Camp Fire Girls Series
By HILDEGARD G. FREY
A Series of Outdoor Stories for Girls 12 to 16 Years.

THE CAMP FIRE GIRLS IN THE MAINE WOODS; or, The
Winnebagos go Camping.
THE CAMP FIRE GIRLS AT SCHOOL; or, The Wohelo
Weavers.
THE CAMP FIRE GIRLS AT ONOWAY HOUSE; or, The
Magic Garden.
THE CAMP FIRE GIRLS GO MOTORING; or, Along the
Road That Leads the Way.
THE CAMP FIRE GIRLS’ LARKS AND PRANKS; or, The
House of the Open Door.
THE CAMP FIRE GIRLS ON ELLEN’S ISLE; or, The Trail of
the Seven Cedars.
THE CAMP FIRE GIRLS ON THE OPEN ROAD; or, Glorify
Work.
THE CAMP FIRE GIRLS DO THEIR BIT; or, Over the Top
with the Winnebagos.
THE CAMP FIRE GIRLS SOLVE A MYSTERY; or, The
Christmas Adventure at Carver House.
THE CAMP FIRE GIRLS AT CAMP KEEWAYDIN; or, Down
Paddles.
For sale by all booksellers, or sent postpaid on receipt of
price by the Publishers
A. L. BURT COMPANY
114-120 EAST 23rd STREET NEW YORK

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