7Quantum chromodynamics (QCD)
III The Standard Model
7.3 e
+
e
−
→ hadrons
Doing QCD computations is very hard
TM
. Partly, this is due to the problem of
confinement. Confinement in QCD means it is impossible to observe free quarks.
When we collide quarks together, we can potentially produce single quarks or
anti-quarks. Then because of confinement, jets of quarks, anti-quarks and gluons
would be produced and combine to form colour-singlet states. This process is
known as hadronization.
Confinement and hadronization are not very well understood, and these
happen in the non-perturbative regime of QCD. We will not attempt to try to
understand it. Thus, to do computations, we will first ignore hadronization,
which admittedly isn’t a very good idea. We then try to parametrize the
hadronization part, and then see if we can go anywhere.
Practically, our experiments often happen at very high energy scales. At
these energy scales,
α
S
is small, and we can expect perturbation theory to work.
We now begin by ignoring hadronization, and try to compute the amplitudes
for the interaction
e
+
e
−
→ q¯q.
The leading process is
e
−
e
+
q
¯q
p
1
p
2
k
2
k
1
γ
We let
q = k
1
+ k
2
= p
1
+ p
2
,
and
Q
be the quark charge. Repeating computations as before, and neglecting
fermion masses, we find that
M = (−ie)
2
Q¯u
q
(k
1
)γ
µ
v
q
(k
2
)
−ig
µν
q
2
¯v
e
(p
2
)γ
ν
u
e
(p
1
).
We average over initial spins and sum over final states. Then we have
1
4
X
spins
|M|
2
=
e
4
Q
2
4q
4
Tr(
/
k
1
γ
µ
/
k
2
γ
ν
) Tr(
/
p
1
γ
µ
/
p
2
γ
ν
)
=
8e
4
Q
2
q
4
(p
1
· k
1
)(p
2
· k
2
) + (p
1
· k
2
)(p
2
· k
1
)
= e
4
Q
2
(1 + cos
2
θ),
where
θ
is the angle between
k
1
and
p
1
, and we are working in the COM frame
of the e
+
and e
−
.
Now what we actually want to do is to work out the cross section. We have
dσ =
1
|v
1
− v
2
|
1
4p
0
1
p
0
2
d
3
k
1
(2π)
3
2k
0
1
d
3
k
2
(2π)
3
2k
0
2
(2π)
4
δ
(4)
(q −k
1
− k
2
) ×
1
4
X
spins
|M|
2
.
We first take care of the
|v
1
− v
2
|
factor. Since
m
= 0 in our approximation,
they travel at the speed of light, and we have
|v
1
− v
2
|
= 2. Also, we note that
k
0
1
= k
0
2
≡ |k| due to working in the center of mass frame.
Using these, and plugging in our expression for |M |, we have
dσ =
e
4
Q
2
2
·
1
16
·
d
3
k
1
d
3
k
2
(2π)
2
|k|
4
δ
(4)
(q −k
1
− k
2
)(1 + cos
2
θ).
This is all, officially, inside an integral, and if we are only interested in what
directions things fly out, we can integrate over the momentum part. We let dΩ
denote the solid angle, and then
d
3
k
1
= |k|
2
d|k| dΩ.
Then, integrating out some delta functions, we obtain
dσ
dΩ
=
Z
d|k|
e
4
Q
2
8π
2
q
4
1
2
δ
p
q
2
2
− |k|
!
(1 + cos
2
θ)
=
α
2
Q
2
4q
2
(1 + cos
2
θ),
where, as usual
α =
e
2
4π
.
We can integrate over all solid angle to obtain
σ(e
+
e
−
→ q¯q) =
4πα
2
3q
2
Q
2
.
We can compare this to
e
+
e
−
→ µ
+
µ
−
, which is exactly the same, except we
put Q = 1 because that’s the charge of a muon.
But this isn’t it. We want to include the effects of hadronization. Thus, we
want to consider the decay of
e
+
e
−
into any possible hadronic final state. For
any final state X, the invariant amplitude is given by
M
X
=
e
2
q
2
hX|J
µ
h
|0i¯v
e
(p
2
)γ
ν
u
e
(p
1
),
where
J
µ
h
=
X
f
Q
f
¯q
f
γ
µ
q
f
,
and Q
f
is the quark charge. Then the total cross section is
σ(e
+
e
−
→ hadrons) =
1
8p
0
1
p
0
2
X
X
1
4
X
spins
(2π)
4
δ
(4)
(q −p
X
)|M
X
|
2
.
We can’t compute perturbatively the hadronic bit, because it is non-perturbative
physics. So we are going to parameterize it in some way. We introduce a hadronic
spectral density
ρ
µν
h
(q) = (2π)
3
X
X,p
X
δ
(4)
(q − p
X
) h0|J
µ
h
|XihX|J
ν
h
|0i.
We now do some dodgy maths. By current conservation, we have
q
µ
ρ
µν
= 0.
Also, we know
X
has positive energy. Then Lorentz covariance forces
ρ
h
to take
the form
ρ
µν
h
(q) = (−g
µν
q
2
+ q
µ
q
ν
)Θ(q
0
)ρ
h
(q
2
).
We can then plug this into the cross-section formula, and doing annoying
computations, we find
σ =
1
8p
0
1
p
0
2
(2π)e
4
4q
4
4(p
1µ
p
2ν
− p
1
· p
2
g
µν
+ p
1ν
p
2µ
)(−g
µν
q
2
+ q
µ
q
ν
)ρ
h
(q
2
)
=
16π
3
α
2
q
2
ρ
h
(q
2
).
Of course, we can’t compute this
ρ
h
directly. However, if we are lazy, we can
consider only quark-antiquark final states
X
. It turns out this is a reasonably
good approximation. Then we obtain something similar to what we had at the
beginning of the section. We will be less lazy and include the quark masses this
time. Then we have
ρ
µν
h
(q
2
) = N
c
X
f
Q
2
f
Z
d
3
k
1
(2π)
3
2k
0
1
Z
d
3
k
2
(2π)
3
2k
0
2
(2π)
3
δ
(4)
(q − k
1
− k
2
)
× Tr[(
/
k
1
+ m
f
)γ
µ
(
/
k
2
− m
f
)γ
ν
]|
k
2
1
=k
2
2
=m
2
f
,
where
N
c
is the number of colours and
m
f
is the quark
q
f
mass. We consider
the quantity
I
µν
=
Z
d
3
k
1
k
0
1
Z
d
3
k
2
k
0
2
δ
(4)
(q − k
1
− k
2
)k
µ
1
k
ν
2
k
2
1
=k
2
2
=m
2
f
.
We can argue that we can write
I
µν
= A(q
2
)q
µ
q
ν
+ B(q
2
)g
µν
.
We contract this with
g
µν
and
q
µ
q
ν
(separately) to obtain equations for
A, B
.
We also use
q
2
= (k
1
+ k
2
)
2
= 2m
2
f
+ 2k
1
· k
2
.
We then find that
ρ
h
(q
2
) =
N
c
12π
2
X
f
Q
2
f
Θ(q
2
− 4m
2
f
)
1 −
4m
2
f
q
2
!
1/2
q
2
+ 2m
2
f
q
2
.
That’s it! We can now plug this into the equation we had for the cross-section.
It’s still rather messy. If all
m
f
→
0, then this simplifies very nicely, and we find
that
ρ
h
(q
2
) =
N
c
12π
2
X
f
Q
2
f
.
Then after some hard work, we find that
σ
LO
(e
+
e
−
→ hadrons) = N
c
4πα
2
3q
2
X
f
Q
2
f
,
where LO denotes “leading order”.
An experimentally interesting quantity is the following ratio:
R =
σ(e
+
e
−
→ hadrons)
σ(e
+
e
−
→ µ
+
µ
−
)
.
Then we find that
R
LO
= N
c
X
f
Q
2
f
=
2
3
N
c
when u, d, s are active
10
9
N
c
when u, d, s, c are active
11
9
N
c
when u, d, s, c, b are active
In particular, we expect “jumps” as we go between the quark masses. Of course,
it is not going to be a sharp jump, but some continuous transition.
We’ve been working with tree level diagrams so far. The one-loop diagrams
are UV finite but have IR divergences, where the loop momenta
→
0. The
diagrams include
e
−
e
+
q
¯q
γ
e
−
e
+
q
¯q
γ
e
−
e
+
q
¯q
γ
However, it turns out the IR divergence is cancelled by tree level
e
+
e
−
→ q¯qg
such as
e
−
e
+
q
¯q
g
γ