Greiner W., Schrammm S., Stein E. Quantum Chromodynamics
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36 2. Review of Relativistic Field Theory
Making use of the final result (2.101) and taking into account that the square of
momentum functions, δ
4
, can be expressed as
(2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)
2
= (2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)(2π)
4
δ
4
(0)
= (2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)VT ,
(2.109)
we obtain
P
(1)
fi
=(2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)(N
1
N
2
N
3
N
4
)
2
|
F
fi
|
2
. (2.110)
This transition rate is still proportional to the flux of incoming particles and the
density of target particles. In order to eliminate this dependence we have to di-
vide by these two quantities. The beam particle flux (projectile quantities carry
index 1, here, for example, π
+
particles) is defined as the number of incoming
particles per unit area that can reach the target (K
+
particles, index 2) per unit
time. The velocity of such beam particles is v =v
1
−v
2
. The densities are nor-
malized to 2E/V particles per unit volume (see (2.107)) and therefore the flux
factor is
|
j
1
|
=
|
v
|
2E
1
V
. (2.111)
For the density of target particles we obtain
2
=
2E
2
V
. (2.112)
If the target particles are at rest, (2.112) assumes the value 2m
2
/V . All together
(2.110) has to be multiplied by the factor
1
|
v
|
V
2E
1
V
2E
2
. (2.113)
In order to derive the cross section, one has to sum over all two-particle final
states. If the volume V contains a particle, this yields an integration over the two-
particle phase space with the voulume element
V
(2π)
3
d
3
p
3
V
(2π)
3
d
3
p
4
. (2.114)
According to the normalization employed above, the phase space factor is
V
(2π)
3
d
3
p
3
2E
3
V
(2π)
3
d
3
p
4
2E
4
, (2.115)
which then yields the following cross section:
dσ = P
fi
V
2
2E
1
2E
2
|
v
|
V
(2π)
3
d
3
p
3
2E
3
V
(2π)
3
d
3
p
4
2E
4
(2.116)
=
|
F
|
2
2E
1
2E
2
|
v
|
(2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)
d
3
p
3
(2π)
3
2E
3
d
3
p
4
(2π)
3
2E
4
.
2.2 Scalar Quantum Electrodynamics 37
Since N
i
=1/
√
V , all factors V cancel and the cross section does not depend
on the normalization volume. The Lorentz invariant form of the flux factor (see
Exercise 2.2) is
E
1
E
2
|
v
|
=
( p
1
· p
2
)
2
−m
2
1
m
2
2
. (2.117)
This identity holds only for collinear collisions. For noncollinear collisions only
the relativistically invariant expression on the right-hand side of (2.117) remains
valid. We therefore introduce the Lorentz-invariant phase-space factor
dLips(s; p
3
, p
4
)
=(2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)
1
(2π)
3
d
3
p
3
2E
3
1
(2π)
3
d
3
p
4
2E
4
, (2.118)
where
s = ( p
1
+ p
2
)
2
(2.119)
denotes one of the so-called Mandelstam variables. Equation (2.116) then
assumes the form
dσ =
| F |
2
4
( p
1
· p
2
)
2
−m
2
1
m
2
2
dLips(s; p
3
, p
4
). (2.120)
EXERCISE
2.2 The Flux Factor
Problem. Verify that in the center-of-mass system as well as in the laboratory
system the flux factor 4E
1
E
2
|v| is given by the invariant expression
4E
1
E
2
|v|=4
( p
1
p
2
)
2
−m
2
1
m
2
2
.
Solution. In the center-of-mass system the momenta of the projectile and target
point in opposite directions, i.e.,
E
1
=
m
2
1
+ p
2
, p
1
=+p ,
E
2
=
m
2
2
+ p
2
, p
2
=−p . (1)
38 2. Review of Relativistic Field Theory
Exercise 2.2
Therefore p
1
p
2
= E
1
E
2
+ p
2
, and we obtain
( p
1
p
2
)
2
−m
2
1
m
2
2
= (E
1
E
2
+ p
2
)
2
−m
2
1
m
2
2
= E
2
1
(E
2
2
−m
2
2
) +2E
1
E
2
p
2
+ p
4
+(E
2
1
−m
2
1
)m
2
2
= p
2
(E
2
1
+2E
1
E
2
+E
2
2
)
= p
2
(E
1
+E
2
)
2
= (E
1
E
2
)
2
p
1
E
1
−
p
2
E
2
2
, (2)
which immediately leads to
4
( p
1
p
2
)
2
−m
2
1
m
2
2
=4E
1
E
2
p
1
E
1
−
p
2
E
2
=4E
1
E
2
|
v
1
−v
2
|
. (3)
In the laboratory system the target particle (particle 2) is at rest and we have
E
1
=
m
1
√
1 −v
2
, p
1
=
m
1
v
√
1 −v
2
E
2
= m
2
, p
2
=0 . (4)
Then the scalar product p
1
· p
2
is simply equal to m
1
m
2
/
√
1 −v
2
and we obtain
4
( p
1
· p
2
)
2
−m
2
1
m
2
2
= 4m
1
m
2
!
1
1 −v
2
−1
= 4
m
1
m
2
√
1 −v
2
|v|
= 4E
1
E
2
|v| . (5)
EXERCISE
2.3 The Mandelstam Variable s
Problem. Introduce the Mandelstam variable s = ( p
1
+ p
2
)
2
and show that
4
( p
1
· p
2
)
2
−m
2
1
m
2
2
=
s −(m
1
+m
2
)
2
s −(m
1
−m
2
)
2
holds.
Solution. The Mandelstam variable s provides an invariant measure for the en-
ergy of the particles participating in the reaction. In the center-of-mass system,
where one has p
1
+ p
2
= 0,
√
s is equal to the sum of all particle energies:
s = ( p
1
+ p
2
)
2
= (E
1
+E
1
, p
1
+ p
2
)
2
=(E
1
+E
2
)
2
.
2.2 Scalar Quantum Electrodynamics 39
In general
s = ( p
1
+ p
2
)
2
= p
2
1
+2 p
1
· p
2
+ p
2
2
=m
2
1
+2 p
1
· p
2
+m
2
2
,
i.e.,
2 p
1
· p
2
=s −m
2
1
−m
2
2
. (1)
The flux factor 4
( p
1
· p
2
)
2
−m
2
1
m
2
2
then assumes the form
(2 p
1
· p
2
−2m
1
m
2
)(2 p
1
· p
2
+2m
1
m
2
)
=
s −(m
1
+m
2
)
2
s −(m
1
−m
2
)
2
.
EXERCISE
2.4 The Lorentz-Invariant Phase-Space Factor
Problem. Show that the phase-space factor
dLips(s; p
3
, p
4
)
=(2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)
1
(2π)
3
d
3
p
3
2E
3
1
(2π)
3
d
3
p
4
2E
4
(1)
is Lorentz invariant.
Solution. The phase-space factor is apparently invariant under spacial rotations.
We must therefore investigate its behavior under proper Lorentz transformations,
which are induced by the matrix Λ(v
b
) with a boost velocity v
b
. Owing to ro-
tational invariance one can, without loss of generality, put v
b
parallel to the
z axis. This considerably simplifies the dependence of the particle coordinates
in the system at rest p
n
(n = 1, 2, 3, 4) on the new coordinates p
n
in the moving
reference system. In general this dependence is
p
n
=Λ(v
b
) p
n
, (2)
with the inversion
p
n
=Λ
−1
(v
b
) p
n
=Λ(−v
b
) p
n
. (3)
Exercise 2.3
40 2. Review of Relativistic Field Theory
Exercise 2.4
For the differentials d p
x
n
,dp
y
n
,dp
z
n
we obtain
γ = 1/
1 −v
2
b
:
dp
x
n
=dp
x
n
,
dp
y
n
=dp
y
n
,
dp
z
n
=γ
dp
z
n
+|v
b
|dE
n
=γ dp
z
n
1 +|v
b
|
p
z
n
E
n
=dp
z
n
E
n
E
n
. (4)
Here we have employed the relations
E
n
≡ p
0
n
=
p
2
n
+m
2
n
1
2
= γ(E
n
+|v
b
|p
z
n
), (5)
which verifies the last step in equation (4), and
dE
n
dp
z
n
=
d
dp
z
n
!
m
2
+
p
x
n
2
+
p
y
n
2
+
p
z
n
2
=
p
z
n
E
n
. (6)
For the volume element d
3
p
n
= dp
x
n
dp
y
n
dp
z
n
we therefore have
d
3
p
n
E
n
=
d
3
p
n
E
n
. (7)
It remains to prove that the four-delta function is a Lorentz scalar. By definition
we have
d
4
pδ
4
( p) = 1(8)
in any reference frame. Owing to the properties of the matrix Λ(v) the volume
element is a Lorentz scalar too:
d
4
p =
∂ p
µ
∂ p
ν
d
4
p
=|det(Λ(−v
b
))|d
4
p
= d
4
p
. (9)
2.2 Scalar Quantum Electrodynamics 41
EXAMPLE
2.5 π
+
π
+
and π
+
π
−
Scattering
As an example of the scattering of identical particles we study π
+
π
+
scattering.
The main modification compared with the π
+
K
+
scattering discussed above is
the symmetrization of initial and final states. The total scattering amplitude has
to be symmetric under exchange of the incoming or outgoing identical bosons.
The direct graph of π
+
π
+
scattering is depicted in Fig. 2.8.
Substituting the final state p
3
for p
4
and vice versa leads to the corresponding
exchange graph (Fig. 2.9). The complete scattering amplitude, is then
S
(1)
fi
(π
+
π
+
) = S
(1)
fi
(direct) +S
(1)
fi
(exchange).
Employing (2.101), S
(1)
fi
assumes the form
S
(1)
fi
(π
+
π
+
) =−i(2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)N
1
N
2
N
3
N
4
×
−e
2
( p
1
+ p
3
)
µ
( p
2
+ p
4
)
µ
( p
2
− p
4
)
2
+
−e
2
( p
1
+ p
4
)
µ
( p
2
+ p
3
)
µ
( p
2
− p
3
)
2
≡−i(2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)
×N
1
N
2
N
3
N
4
F
π
+
π
+
( p
1
p
2
; p
3
p
4
). (1)
Again F
π
+
π
+
denotes the invariant scattering amplitude. All further steps can be
performed in complete analogy to π
+
K
+
scattering.
A similar argument holds for π
+
π
−
scattering, if we take the antiparticle
interpretation into account (cf. the discussion connected with (2.17)–(2.20)).
Fig. 2.9. The exchange am-
plitude of π
+
π
+
scattering
in the one-photon exchange
approximation
Fig. 2.10. The interrela-
tion between π
+
π
+
and
π
+
π
−
scattering according
to the Feynman reinterpre-
tation of outgoing particles
with four-momentum p as
incoming antiparticles with
four-momentum −p,and
vice versa
Fig. 2.8. A Feynman dia-
gram for the direct ampli-
tude of π
+
π
+
scattering in
the one-photon exchange ap-
proximation
42 2. Review of Relativistic Field Theory
Fig. 2.11. The one-photon
exchange amplitude for
π
+
π
−
scattering in detail.
Both the physical picture
(left) and the Feynman rein-
terpretation of outgoing par-
ticles with four-momentum
p as incoming antiparticles
with four-momentum −p
(right) are shown for the
direct graph
Fig. 2.12. The Feynman pic-
ture of the direct graph in
Fig. 2.11 (left) is equiva-
lent to the exchange graph
of π
+
π
−
scattering in the
physical picture (right)
Fig. 2.13. The detailed ex-
change graph for π
+
π
−
scattering. Obviously this
amplitude can also be under-
stood as one-photon π
+
π
−
annihilation connected with
π
+
π
−
pair creation
We can therefore make the identification shown in Fig. 2.10. Owing to the
antiparticle concept disussed earlier we can interpret an incoming π
−
with four-
momentum p
b
as an outgoing π
+
with four-momentum −p
b
and an outgoing π
−
with four-momentum p
d
as an incoming π
+
with four-momentum −p
d
. Hence
we can immediately write down the invariant scattering amplitude for π
+
π
−
scattering:
F
π
+
π
−
( p
a
, p
b
; p
c
, p
d
) = F
π
+
π
+
( p
a
, −p
d
; p
c
, −p
b
), (2)
which explicitly is
F
π
+
π
−
( p
a
p
b
; p
c
p
d
) =
−e
2
( p
a
+ p
c
)
µ
(−p
d
− p
b
)
µ
(−p
d
+ p
b
)
2
+
−e
2
( p
a
− p
b
)
µ
(+p
c
− p
d
)
µ
( p
a
+ p
b
)
2
. (3)
A graphical representation of the two contributions more detailed than that given
above is shown in Figs. 2.11 and 2.12. The physical and the Feynman pictures of
the direct and the exchange graphs are given side by side. The exchange graph is
best understood in the Feynman picture where the outgoing particles are simply
exchanged. As a consequence, in the physical picture, the outgoing particle and
the incoming antiparticle are exchanged. The difference between the direct parts
of the amplitudes for π
+
π
−
scattering (or π
+
K
+
scattering) and π
+
π
+
scatter-
ing is only that of sign if the corresponding four-momenta are considered equal
(because of the different masses of K
+
and π
−
they are in general not equal).
The different signs of these two processes correspond to attractive and repulsive
interaction, respectively.
2.2 Scalar Quantum Electrodynamics 43
EXERCISE
2.6 The Cross Section for Pion–Kaon Scattering
Problem. Derive the explicit form of the differential cross section for electro-
magnetic π
+
K
+
scattering in the center-of-momentum system (cm system).
Solution. According to (2.120) the cross section for π
+
K
+
scattering is
dσ =
|F|
2
4
( p
1
· p
2
)
2
−m
2
1
m
2
2
dLips(s; p
3
p
4
), (1)
where
F =−
e
2
q
2
( p
1
+ p
3
) ·( p
2
+ p
4
),
q
µ
=( p
3
− p
1
)
µ
=−( p
4
− p
2
)
µ
, (2)
and
dLips(s; p
3
p
4
)
= (2π)
4
δ
4
( p
3
+ p
4
− p
1
− p
2
)
1
(2π)
3
d
3
p
3
2E
3
1
(2π)
3
d
3
p
4
2E
4
(3)
denote the invariant scattering amplitude and the Lorentz-invariant phase-space
factor, respectively. The cm system is defined by
p
1
+ p
2
= p
3
+ p
4
= 0 . (4)
In this system the scattering process is described by the scattering angle θ
cms
(see
Fig. 2.14). Now we transform the four-momenta to the cm system,
p
µ
1
=(E
1
, p), p
µ
2
=(E
2
, −p),
p
µ
3
= (E
3
, p
), p
µ
4
=(E
4
, −p
), (5)
which leads to the total energy
E
cms
= E
1
+E
2
= E
3
+E
4
=
p
2
+m
2
1
+
p
2
+m
2
2
=
p
2
+m
2
1
+
p
2
+m
2
2
=
p
2
+m
2
1
+
p
2
+m
2
2
. (6)
Here and in the following we denote the absolute value of the spatial momentum
by
|p|=|p
|≡p . (7)
Fig. 2.14. π
+
K
+
scattering
in the cm system
44 2. Review of Relativistic Field Theory
Exercise 2.6
By integrating the invariant phase-space factor d Lips(s; p
3
p
4
) over the spatial
momenta d
3
p
4
we obtain
d
3
p
4
E
4
δ
4
( p
3
+ p
4
− p
1
− p
2
) =
1
E
4
δ(E
3
+E
4
−E
1
−E
2
). (8)
The right-hand side of (8) contains E
4
as well as
|
p
4
|
. These two variables,
however, are not independent of each other: they are connected by
p
4
= p
1
+ p
2
− p
3
, E
4
=
p
2
4
+m
2
2
,(m
2
= m
4
). (9)
Next we transform d
3
p
3
into spherical coordinates
d
3
p
3
= p
2
3
dp
3
dΩ, (10)
where p
3
=
|
p
3
|
and dΩ denotes the spherical angle into which the π
+
is
scattered. Because of
E
2
3
= p
2
3
+m
2
1
(11)
we have
E
3
dE
3
= p
3
dp
3
(12)
and
dLips(s; p
3
p
4
) =
1
(4π)
2
δ(E
3
+E
4
−E
1
−E
2
)
p
3
dE
3
E
4
dΩ. (13)
This formula is valid in any Lorentz system. Now we go into the cm system by
making use of the relations
E
2
3
= p
2
+m
2
1
, E
2
4
= p
2
+m
2
2
, (14)
and
E
3
dE
3
= E
4
dE
4
= pdp . (15)
Introducing the free variable
E
= E
3
+E
4
, (16)
we also have, according to (15),
dE
=
p
E
3
dp +
p
E
4
dp =
E
E
3
E
4
pdp =
E
E
4
dE
3
. (17)
Now (13) assumes the form
dLips(s; p
3
p
4
)
cms
=
1
(4π)
2
δ(E
cms
−E
)
p
E
dE
dΩ. (18)
2.2 Scalar Quantum Electrodynamics 45
An integration over E
then yields
dLips(s; p
3
p
4
)
cms
=
1
(4π)
2
p
E
cms
dΩ (19)
for the two-particle phase-space factor in the cm system. Finally the flux fac-
tor has to be rewritten in terms of cm variables. With the help of (5) and (6) we
immediately find that
( p
1
· p
2
)
2
−m
2
1
m
2
2
=
(E
1
E
2
+ p
2
)
2
−(E
2
1
− p
2
)(E
2
2
− p
2
)
= pE
cms
. (20)
The result for the differential cross section is then
dσ
dΩ
cms
=
1
(8πE
cms
)
2
|F|
2
. (21)
By introducing the second Mandelstam variable t the result (21) can be brought
into an invariant form. t is simply defined as the square of the four-momentum
transfer:
t =q
2
= ( p
1
− p
3
)
2
=( p
2
− p
4
)
2
. (22)
In the cm system we consequently have
t = (0, p − p
)
2
=−( p − p
)
2
=−
p
2
−2 p · p
+ p
2
=−2p
2
(
1 −cos θ
cms
)
, (23)
which leads to
dt = 2p
2
dcos
θ
cms
. (24)
Since the cross section for spinless particles is cylindrically symmetric around
the beam axis, i.e.
dΩ
cms
= 2πdcos
θ
cms
, (25)
we obtain the relation
d
dt
=
π
p
2
d
dΩ
cms
. (26)
Therefore the two-particle cross section is, in invariant form
dσ
dt
=
1
64π
1
( pE
cms
)
2
|F|
2
=
1
64π
|F|
2
( p
1
· p
2
)
2
−m
2
1
m
2
2
, (27)
Exercise 2.6