Assignment II
Solutions
• 5.2M A clever experimentalist figures out how to collide two Ω particles of opposite charge into a resonance (ie. a short-lived bound state) called an omegaball.
(a) What are the possible spins of an omegaball in its lowest-energy state?
(b) The experimentalist then figures out how to make a supply of 10,000 omegaballs of lowest energy by repeat-edly colliding Ω+ and Ω− together with opposite spins along thez-axis. What must these spin components be if more highest-spin omegaballs are made than any other kind? Find approximately how many of each spin are made in the sample of 10,000.
(c) How does your answer to (b) change if the highest-spin omegaballs are fewest in number in the sample?
Solution
(a) Two Ω’s each have spins 3/2 so the total spin of the resonance can have the values between their sum and their difference, or
3,2,1,0
where there is no orbital angular momentum since the resonance is in its lowest-energy state.
(b) We must decompose a bound state of two Ω’s with opposite spin components on the z-axis. From the Clebsch tables with 3/2⊗3/2 we must have a total z-component of 0, which means we must use the middle part of the table. The highest-spin omegaball will be in a|30istate. The highest probability for this state to occur is 9/20 or 45%, which is when the z-components are 1/2 and−1/2. In this case about 4500 of these kind of omegaballs are made, and about 2500 each of the |00iand|20i kinds are made. Only 500|10iomegaballs are made.
(c) If the highest spin omegaballs are fewest in number then the z-components are 3/2 and−3/2, with about 500 of these being made. We still have 2500 each of the |00i and |20i kinds and 4500 of the |10i kind of omegaballs.
• 5.7E In 1932 the decay of the neutron into a proton and an electron was observed. What conservation laws did this decay violate, if any?
Solution
• 6.2E According to the Standard Model, which of the following reactions are allowed and which are forbidden? State the reasons why if not. If allowed, state what interaction is responsible for the process.
(a) ¯n−→p¯+e++νe (b) γ+Z0−→νe+π0
(c) µ++τ−−→γ+γ+γ (d) Z0−→νµ¯ +νµ
(e) D0−→K−+ρ+ (f) e−+ ¯νe−→t¯+b
Solution
Reaction Allowed? Violation/Interaction
¯
n−→p¯+e++ν Yes Weak interaction (due to flavour change)
γ+Z0−→ν
e+π0 No Violates
angular momentum lepton number
conservation
µ++τ− −→γ+γ+γ No Violates
muon number tau number
conservation
Z0−→νµ¯ +νµ Yes Weak interaction (Z0 only couples via weak force)
D0−→K−+ρ+ Yes Strong interaction (due toρ+ in final state)
e−+ ¯νe−→¯t+b Yes Weak interaction (due to flavour change)
• 6.7H Time reversal interchanges initial and final states, so that ifT|χi=|χit then
T(hϑ|χi) = thϑ|χit=hχ|ϑi= (hϑ|χi)∗
where the first equality is due to interchange of initial and final states and the second equality is the property of quantum-mechanical amplitudes.
(a) Show that the above relation impliesTc=c∗Twherec is any complex number.
(b) Given thatT˜J=−˜JTwhere˜Jis the angular momentum operator, show that
|χ1i ≡ −T
1 2,−
1 2
and |χ2i ≡T
1 2,
1 2
form a spin-1/2 doublet in the time-reversed system.
(c) What isT2 for a state|φiwith an odd number of spin-1/2 particles? What is
T2 for a state|ϕi with an
even number of spin-1/2 particles?
(d) Suppose the state|φiin (c) is an eigenstate of the Hamiltonian. What is the degree of the degeneracy of this state? (i.e. how many distinct states are there at a given energy level?) This degeneracy is calledKramers degeneracy.
(e) Is Kramers degeneracy preserved in an electric field? Is it preserved in a magnetic field?
(a) Let|χi=c|ζi=|cζi, wherecis any complex number. Then by the above relation
T(chϑ|ζi) =T(hϑ|cζi) = (hϑ|cζi)∗=c∗(hϑ|ζi)∗=c∗T(hϑ|ζi)
Since this must hold for arbitrary states|χiand|ϑiwe haveTc=c∗T.
(b) SinceT˜J=−˜JT, then we have
TJz = −JzT and TJ±=−J∓T
sinceTJ± = T(Jx±iJy) =−(JxT±i∗JyT) =−(JxT∓iJyT) =−J∓T
Hence we have
Jz|χ1i ≡ −JzT
1 2,−
1 2
=TJz
1 2,−
1 2
=−1
2T 1 2,−
1 2
= 1 2|χ1i
Jz|χ2i ≡ JzT 1 2, 1 2
=−TJz
1 2, 1 2
=−1
2T 1 2, 1 2
=−1
2|χ2i
and also
J+|χ1i ≡ −J+T
1 2,−
1 2
=−TJ−
1 2,−
1 2
= 0
J−|χ1i ≡ −J−T
1 2,−
1 2
=TJ+
1 2,−
1 2 =T 1 2, 1 2
=|χ2i
J+|χ2i ≡ J+T 1 2, 1 2
=−TJ−
1 2, 1 2
=−T
1 2,−
1 2
=|χ1i
J−|χ2i ≡ J−T 1 2, 1 2
=−TJ+
1 2, 1 2 = 0 Hence
|χ1i is a state of j= 1 2, jz=
1
2 =⇒ |χ1i=κ
1 0
|χ2i is a state ofj= 1
2, jz=− 1
2 =⇒ |χ2i=κ
0 1
whereκ is a complex phase (i.e. κ∗κ= 1).
(c) We have
T2 1 2, 1 2
= T|χ2i=T κ 1 2,−
1 2
=κ∗T
1 2,−
1 2
=−κ∗|χ1i=−κ∗κ 1 2, 1 2 =− 1 2, 1 2 T2 1 2,−
1 2
= −T|χ1i=−T κ 1 2, 1 2
=−κ∗T 1 2, 1 2
=−κ∗|χ2i=−κ∗κ
1 2,−
1 2 =− 1 2,−
1 2
So forT2acting on a single spin-state|ηi,
T2|ηi=− |ηi. Hence onN spin states
T2|η1,...,ηNi= (−1)N|η1,...,ηNi=
|η1,...,ηNi N= even
(d) SupposeH|φi=E|φi. E is a real number, sinceH is Hermitian. Then since [H,T] = 0 we obtain
H(T|φi) =T(H|φi) =T(E|φi) =ET|φi
and soT|φiand|φiare both degenerate.
Are they distinct? If they weren’t distinct, then we’d haveT|φi=ξ|φi, whereξ∗ξ= 1. For a single spin state
|φi,T2|φi=− |φi. However we also have
T2|φi=Tξ|φi=ξ∗T|φi=ξ∗ξ|φi=|φi, which is a contradiction.
HenceT|φiand|φiare also both distinct.
(e) In an electric fieldE~, and a magnetic fieldB~
TE~ =E~T andTB~ =−B~T
Now let
|φi = T|χi=⇒ |φ↑i=−T|χ↓i and |φ↓i=T|χ↑i
from (d) =⇒
|φ↑i and |χ↑i |φ↓i and |χ↓i
are degenerate
For a state|χiwith an electric dipole moment
~ d
and magnetic dipole moment|~µ|we have
~
σ·E~|χ↑i = +
~ d
|χ↑i and
~
σ·E~|χ↓i=−
~ d
|χ↓i
~σ·B~|χ↑i = +|~µ| |χ↑i and ~σ·B~|χ↓i=− |~µ| |χ↓i
Calculating the dipole moments of|φi, we get (recall from (b) that T~σ=−~σT)
~
σ·E~|φ↑i=−~σ·E~T|χ↓i=T~σ·E~|χ↓i=−T
~ d
|χ↓i=
~ d
|φ↑i
~
σ·E~|φ↓i=~σ·E~T|χ↑i=−T
~
σ·E~|χ↑i=−T
~ d
|χ↑i=−
~ d
|φ↓i
degenerate with |χi
~
σ·B~|φ↑i=−~σ·B~T|χ↓i=−T~σ·B~|χ↓i= +T|~µ| |χ↓i=− |~µ| |φ↑i
~
σ·B~|φ↓i=~σ·B~T|χ↑i=T~σ·B~|χ↑i= +T|~µ| |χ↑i=|~µ| |φ↓i
not degenerate with |χi
So Kramers degeneracy is preserved in an electric field, but not preserved in a magnetic field.
• 9.3E Show that the resonance function
R(ω) = Ω 2Γ2
(ω2−Ω2)2+ Ω2Γ2
can be approximated by the Breit-Wigner function nearω= Ω.
Taking the series expansion ofR(ω) nearω= Ω gives
R(ω) = 1 + dR
dω
ω=Ω
(ω−Ω) + 1 2!
d2R dω2
ω=Ω
(ω−Ω)2
= 1− 4
Γ2(ω−Ω) 2
to leading order in (ω−Ω). However the Breit-Wigner function is
Γ 2
2
(ω−Ω)2+ Γ 2
2 =
1
1 +Γ42(ω−Ω)
2 = 1− 4
Γ2(ω−Ω) 2
and so the two expressions are equivalent to leading order in (ω−Ω).
• 9.5M (a) Show that
q
(p1·p2) 2
−p2
1p22= (E1+E2)|~p1|/c in the CM frame.
(b) Compute this quantity in the lab frame.
Solution (a) In the CM frame pµ1 = (E1, ~p) and pµ2 = (E2,−~p), so (p1·p2) =E1E2+|~p| 2
, and p21=E12−|~p| 2
,
p2
2=E22− |~p| 2
. Hence
(p1·p2) 2
−p21p22 = E1E2+|~p| 22
−E12− |~p|2 E22− |~p|2
= |~p|2 E12+E22+ 2E1E2
= |~p1| 2
(E1+E2) 2
where I have put~p=~p1 in the last line. To get the units right we need something with units of momentum2 on both sides of this equation, which means that the energies have to be divided byc. Hence
q
(p1·p2) 2
−p2
1p22= (E1+E2)|~p1|/c
(b) Taking particle #1 to be the moving particle, in the lab frame pµ1 = (E, ~p) and pµ2 = (m2,0), so (p1·p2) =Em2and p12=m21, p22=m22. Hence
(p1·p2) 2
−p21p22 = (Em2) 2
− m21
m22
= m22 E 2
−m21
= m22|~p|2
and so (putting in the factor ofc to get the units right):
q
(p1·p2) 2
−p2
• 10.1H ConsiderABBtheory in which theA is massless.
(a) Find the differential cross-section in the lab-frame for the reaction
B+A−→B+A
What is the angular dependence of the cross-section at high energies?
(b) Find the differential cross-section in the CMS for the reaction
A+A−→B+B
Solution
The diagram is given in the figure below.
1
'
p
1
p
2
p
2'
p
A
B
B
A
q
B
+
1
'
p
1
p
p
22
'
p
A
B
B
A
q
B
From the rules ofABBtheory we have
iM =
Z d4q
(2π)4(2π) 8
[δ(p10+p20−q)δ(q−p1−p2) +δ(p20+q−p1)δ(p 0
1−q−p2)]
(ig)2i q2−m2
B
= −i(2π)4δ(p10+p20−p2−p1)
"
g2
(p1+p2)2−M2
+ g
2
(p10−p2)2−M2
#
where I have set mB =M. In the lab frame the general scenario is given in the figure below.
1
p
1
p!
( )
,
0
2
!
M
p
=
!
A
B
B
A
Hence
pµ2 = (M,0) p1µ= (|~p|, ~p) = (E, ~p) p10µ= (|~p0|, ~p0) = (E0, ~p0) (p1+p2)
2
−M2 = (p1)2+ (p2)2+ 2p1·p2−M2=M2+ 0 + 2M E−M2= 2M E
(p20) 2
= (p2+p1−p10) 2
⇒p10·p1=M(E−E0)
⇒ EE0(1−cosθ) =M(E−E0)⇒E0 = M E
M +E(1−cosθ)
(p10−p2) 2
−M2 = (p10) 2
+ (p2)2−2p10·p2−M2=−2M E0
⇒ M= g
2
2M E − g2
2M E0 =−
g2
2M EE0(E−E
0) =− g2
2M2(1−cosθ)
and so the cross-section is
dσ dΩ
2-body-LAB
=
}
8π
2 S |M|2|~p0 1|
2
M2|~p1|[|~p10|(E1+M2c2)− |~p1|E10cosθ]
=
}
8π
2 |M|2E02
M E[E0(E+M)−EE0cosθ]
=
}
8π
2 E02
(M E)2
g2
2M2(1−cosθ)
2
=
}g2
16πM2
2 (1−cosθ)2
[M +E(1−cosθ)]2
=
}g2
8πM2csin 2θ
2
2
1
M c2+ 2Esin2θ
2
2
using the above relations and putting in the correct factors of c in the last line. These factors follow upon recognizing that g has units of momentum and }c has units of energy×length. We see from this that the differential cross-section is independent of angle at high energies.
(b) Find the differential cross-section in the CMS for the reaction
A+A−→B+B
Now the diagrams are
1
'
p
1
p
p
22
'
p
B
B
A
A
q
B
+
1
'
p
1
p
p
22
'
p
A
A
B
B
and we have
iM =
Z d4q
(2π)4
h
(2π)8δ(p10+q−p1)δ(p20−q−p2) + (2π)8δ(p20+q−p1)δ(p10−q−p2)i (ig) 2
i q2−m2
B
= −i(2π)4δ(p10+p20−p2−p1)
"
g2
(p20−p2)2−M2 +
g2
(p10−p2)2−M2
#
where againmB =M. In the CMS frame
pµ1 = (|~p|, ~p) = (E, ~p) pµ2 = (|~p|,−~p) = (E,−~p) where |~p|=E =E0 p20µ= (E0,−~p0) p10µ = (E0, ~p0)
⇒ M=− g
2
2E(E− |~p0|cosθ)−
g2
2E(E+|~p0|cosθ) =−
g2
(E2− |~p0|2cos2θ)
and so, putting in the correct factors ofc
dσ dΩ =
}c
8π
2
S |M|2
(E1+E2) 2
|~p0| |~p|
= 1
2!
}c
8π
2 1
(2E)2
r
1−M
2c4
E2
g2c2
(E2− |~p0|2
c2cos2θ)
!2
= 1
2
}c3g2
16πE
2
q
1−M2c4 E2
(E2sin2
θ+M2c4cos2θ)2
At high energies the cross-section falls off as 1/E6; the minimum energy for this process to take place is
E > M c2.
• 11.3M (a) Write the Dirac equation in Hamiltonian form by isolating the time-derivative of the spinorψon the left-hand side of the equation. The HamiltonianH will be the operator on the right-hand side of the equation. What isH?
(b) Find the commutator ofH with the orbital angular momentum operator~L.
(c) Find the commutator ofH with the spin angular momentum operator S~.
(d) Find the commutator ofH with the total angular momentum operatorJ~=L~+S~.
(e) Show that all spinors are eigenstates ofS~·S~ =S2. Since S2 =
}2s(s+ 1), what is the value of s for a
Dirac spinor?
Solution
(a) The Dirac equation is
(iγµ∂µ−m)ψ= 0 or
iγ0∂ ∂t+i~γ·
~ ∇ −m
Multiplying byγ0and putting the time-derivative on the left-hand side gives
i∂ ∂tψ=γ
0−i~γ·∇~ +mψ
which upon comparing to the Schroedinger-type formi∂t∂ψ=Hψ gives
H =γ0(~γ·p~+mc)c
recalling that~p=−i~γ·∇~ and putting in the correct factors ofc so thatH has units of energy.
(b) The orbital angular momentum operator~L=~r×~pso
[H, Lj] =
γ0(γkpk+mc)c, εjilripl
= γ0(γkc)εjil[pk, ri]pl
= −i~cγ0γkεjklpl
= −i~cγ0(~γ×p~)j
Note that the commutator is nonzero and so orbital angular momentum is not conserved.
(c) The spin angular momentum operatorS~= } 2Σ =~ }2
~ σ 0 0 ~σ
, so
[H, Sj] = } 2
γ0(γkpk+mc)c,Σj
= }c 2 pk
γ0γk,Σj
= }c 2 pk
0 σk σk 0
,
σj 0
0 σj
= }c 2 pk
0 [σk, σj]
[σk, σj] 0
= }c 2 pk
0 2iεkjlσl
2iεkjlσl 0
= i}cγ0pkεkjlγl
= +i~cγ0(~γ×~p)j
(d) SinceJ~=~L+S~ we easily see that
[H, Jj] = [H, Lj+Sj] = [H, Lj] + [H, Sj] =−i~cγ0(~γ×~p)j+i~cγ
0(~γ×~p)
j= 0
(e) ComputingS~·S~ =S2, we find
S2=} 2
4
~ σ·~σ 0 0 ~σ·~σ
= } 2
4
3I 0 0 3I
= 3} 2
4 I
Since S2 is proportional to the identity matrix it commutes with H. Setting S2 =
}2s(s+ 1) = 3}
2
• 12.5E Write the equation
(iγµDµ−m)ψ= 0
in Hamiltonian form by isolating the time-derivative of the spinorψon the left-hand side of the equation. The HamiltonianH will be the operator on the right-hand side of the equation. What isH?
Solution
We need to break this equation up into temporal and spatial parts. WritingAµ=φ, ~Ayields
(iγµDµ−m)ψ = iγµ(∂µψ+ieAµψ)−mψ= 0
⇒iγ0∂ψ ∂t =
h
−i~γ·∇ψ~ +eφγ0ψ−e~γ·Aψ~ +mψi
or
i∂ψ ∂t =
h
−iγ0~γ·∇ψ~ +eφψ−eγ0~γ·Aψ~ +mγ0ψi
Comparing to the Schroedinger-type formi∂t∂ψ=Hψgives
H =γ0~γ·p~−e ~A+eφγ0+m
• 12.8M Consider a theory of one complex scalar particle with wavefunctionϕand two spin-12 particlesψandχ, each of which couples to the photon via the equations
(iγµDµ−mψ)ψ = gϕ∗χ
(iγµDµ−mχ)χ = gϕψ DµDµϕ−m2ϕϕ = gψχ¯
wherem2
ϕ , mψ, andmχ are the respective masses of theϕ,ψandχparticles.
(a) Find the most general local phase transformation of ϕ,ψandχ that leaves this system gauge-covariant.
(b) What is the relationship between the charges ofϕ,ψand χ?
(c) In Maxwell’s equations
∂µFµν=Jv
what is the currentJv for this theory?
Solution
(a) Since the gauge field must transform asA0µ=Aµ−1e∂µα, whereeis the charge of the field, we must adjust
the value ofefor each wavefunction so that these equations are gauge-covariant. This is most easily done by rescalingα→eαfor each wavefunction, so that
ψ0(x) = eieψα(x)ψ(x) χ0(x) =eieχα(x)χ(x) and ϕ0(x) =eieϕα(x)ϕ(x)
A0µ = Aµ−∂µα
is the gauge transformation, and the gauge-covariant derivatives are
The transformation of the above equations is now
eieψα(x)(iγµDµ−mψ)ψ = gei(eχ−eϕ)α(x)ϕ∗χ
eieχα(x)(iγµDµ−mχ)χ = gei(eψ+eϕ)α(x)ϕψ
eieϕα(x) DµD
µϕ−m2ϕϕ
= gei(eχ−eψ)α(x)ψχ¯
which respectively mean
eχ−eϕ = eψ
eϕ+eψ = eχ eχ−eψ = eϕ
(b) The equations in (a) are redundant and have the solution
eϕ+eψ=eχ
Henceχ must have the same charge as the sum of theϕandψcharges.
(c) The current is given by the sum of the Dirac currents forψandχ and the Klein-Gordon current forϕ:
Jµ = ieψψγ¯ µψ+ieχχγ¯ µχ+ieϕ ϕ∗Dµϕ−bϕ(Dµϕ)
∗
= ieψψγ¯ µψ+i(eϕ+eψ) ¯χγµχ+ieϕ ϕ∗Dµϕ−bϕ(Dµϕ)∗
• 13.1H Consider a theory in which the electron couples to a massless pseudoscalar particle (the φ) with the vertex
=
ig
!
5
(a) Draw the lowest-order Feynman diagrams for the processe++e− −→2φ.
(b) Compute to lowest order the differential cross-section for the process in (a) in the CMS frame.
Solution
1
p
p
22
'
p
1
'
p
+
1
p
p
22
'
p
1
'
p
(b) The diagram on the left yields
Z d4q (2π)4
"
¯
v(p2)igγ5
i /q+m
q2−m2 igγ5u(p1)
#
(2π)4δ(p01+q−p1) (2π)4δ(p02−q−p2)
whereas the one on the right yields
Z d4q (2π)4
"
¯
v(p2)igγ5
i /q+m
q2−m2 igγ5u(p1)
#
(2π)4δ(p02+q−p1) (2π)4δ(p01−q−p2)
Since there is no interchange of fermions, these diagrams add together. Integrating over q and removing the overall delta-function yields the lowest order matrix element
−iM = (−i) (ig)2
"
¯
v(p2) /p1−p/ 0 1−m (p1−p01)
2
−m2u(p1) + ¯v(p2)
/ p
1−/p 0 2−m (p1−p02)
2
−m2u(p1)
#
= ig2
"
¯
v(p2) −/p 0 1 (p1−p01)
2
−m2u(p1) + ¯v(p2)
−/p0
2 (p1−p02)
2
−m2u(p1)
#
where in the second line theγ5matrix was anticommuted through, and the third line follows because
/
p1−mu(p1) = 0 . Note that the denominators simplify
(p1−p01) 2
−m2 = (p1)2+ (p01)2−2 (p1·p01)−m
2=−2 (p 1·p01) (p1−p02)
2
−m2 = (p1)2+ (p02) 2
−2 (p1·p02)−m2=−2 (p1·p02)
because theφis massless (and sop02
1 =p022= 0). This gives
−iM=ig2v¯(p2)
/p0
1 2p1·p01
+ /p 0 2 2p1·p02
The Casimir trick implies
|M|2 = 1 4
1 2
2
g4tr
"
/ p2−m
/p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
/ p1+m
/p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
# = g 4 16tr / p
2−m
/p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
/ p
1+m
/p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
= g 4 16tr / p2
/p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
/ p1
/p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
−g
4
16m 2tr
/p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
/ p0
1 (p1·p01)
+ /p 0 2 (p1·p02)
= 4g 4
16
"
2 (p2·p01) (p1·p01)−(p2·p1) (p01·p01)
(p1·p01)
2 +
2 (p2·p02) (p1·p02)−(p2·p1) (p02·p02)
(p1·p02) 2
+2(p2·p 0
2) (p1·p01)−(p2·p1) (p01·p02) + (p2·p01) (p1·p02)−m2(p01·p02) (p1·p01) (p1·p02)
= g
4
4
4(p2·p 0 1) (p1·p01)
+ 4(p2·p 0 2) (p1·p02)
−2 (p01·p02)
(p2·p1) +m2 (p1·p01) (p1·p02)
again usingp0 2
1 =p022= 0. Now simplify in the CMS:
~
p1 = ~p=−~p2 and ~p10 =~p0 =−~p20 also ~p·~p0 =|~p0| |~p|cosθ=E0|~p|cosθ
pµ1 = (E, ~p) p2µ= (E,−~p) p10µ= (E0, ~p0) p20µ= (E0,− |~p0|)
whereE0=E by conservation of energy. This gives
(p2·p01) (p1·p01)
= EE
0+E0|~p|cosθ
EE0−E0|~p|cosθ =
E+|~p|cosθ E− |~p|cosθ
(p2·p02) (p1·p02)
= EE
0−E0|~p|cosθ
EE0+E0|~p|cosθ =
E− |~p|cosθ E+|~p|cosθ
(p2·p1) +m2 (p1·p01) (p1·p02)
(p01·p02) =
E02+|~p0|2 hE2+|~p|2 +m2i
(EE0+E0|~p|cosθ) (EE0−E0|~p|cosθ) =
4E2
E2− |~p|2cos2θ
and so
|M|2 = g4
E+|~p|cosθ E− |~p|cosθ +
E− |~p|cosθ E+|~p|cosθ −
2E2
E2− |~p|2cos2θ
= g4
2|~p|2cosθ2
E2− |~p|2cos2θ
and so the CMS cross-section is
dσ dΩ =
}c
8π
2 S|M|2
(E1+E2)2
|~p0| |~p|
= 1
2!
}cg2
8π
2" √E2−m2cosθ2 2E E2sin2θ+m2cos2θ
#
• 13.7 Find all diagrams of ordere4 for the processes below. Label them but do not calculate them.
(a)e++γ−→e++γ
(b)γ+γ−→e++e−
Solution
(a) The lowest order diagrams are
k
k! p!
p
k
k!
p!
p
+
and the next-order ones are
Plus 8 more
diagrams with
the initial &
final photon
legs crossed
k
p! k!
p
k!
p
k!
p
k!
p
k!
p
k!
p
k!
p
k!
p p!
k
p!
k
p!
k
p!
k
p!
k
p!
k p!
k
1
p!
1
k
1
p!
1
k
2
p!
2
k
+
2
p!
2
k
and the next-order ones are
Plus eight more with the incoming photon legs crossed
1
k
1
p!
1
k
1
p!
1
k
1
p!
1
k
1
p!
1
k
1
p!
1
k
1
p!
1
k
1
p!
1
k
1
p!
2
p!
2
k
2
k
2
p! p!2 p!2
2
p!
2
p!
2
p!
2
p!
2
k k2 k2 k2
2
k
2
