7Quantum chromodynamics (QCD)
III The Standard Model
7.2 Renormalization
We now spend some time talking about renormalization of QCD. We didn’t
talk about renormalization when we did electroweak theory, because the effect
is much less pronounced in that case. Renormalization is treated much more
thoroughly in the Advanced Quantum Field Theory course, as well as Statistical
Field Theory. Thus, we will just briefly mention the key ideas and results for
QCD.
QCD has a coupling constant, which we shall call
g
. The idea of renormal-
ization is that this coupling constant should depend on the energy scale
µ
we
are working with. We write this as
g
(
µ
). However, this dependence on
µ
is not
arbitrary. The physics we obtain should not depend on the renormalization point
µ
we chose. This imposes some restrictions on how
g
(
µ
) depends on
µ
, and this
allows us to define and compute the quantity.
β(g(µ)) = µ
d
dµ
g(µ).
The β-function for non-abelian gauge theories typically looks like
β(g) = −
β
0
g
3
16π
2
+ O(g
5
)
for some
β
. For an
SU
(
N
) gauge theory coupled to fermions
{f}
(both left- and
right-handed), up to one-loop order,we have
β
0
=
11
3
N −
4
3
X
f
T
f
,
where
T
f
is the Dynkin index of the representations of the fermion
f
. For the
fundamental representation, which is all we are going to care about, we have
T
f
=
1
2
.
In our model of QCD, we have 6 quarks. So
β
0
= 11 − 4 = 7.
So we find that the β-function is always negative!
This isn’t actually quite it. The number of “active” quarks depends on the
energy scale. At energies
m
top
≈ 173 GeV
, then the top quark is no longer
active, and
n
f
= 5. At energies
∼ 100 MeV
, we are left with three quarks, and
then
n
f
= 3. Matching the
β
functions between these regimes requires a bit of
care, and we will not go into that. But in any case, the
β
function is always
negative.
Often, we are not interested in the constant
g
itself, but the strong coupling
α
S
=
g
2
4π
.
It is an easy application of the chain rule to find that, to lowest order,
µ
dα
S
dµ
= −
β
0
2π
α
2
S
.
We now integrate this equation, and see what we get. We have
Z
α
S
(µ)
α
S
(µ
0
)
dα
S
α
2
S
= −
β
0
2π
Z
µ
µ
0
dµ
µ
.
So we find
α
S
(µ) =
2π
β
0
1
log(µ/µ
0
) +
2π
β
0
α
S
(µ
0
)
.
There is an energy scale where
α
S
diverges, which we shall call Λ
QCD
. This is
given by
log Λ
QCD
= log µ
0
−
2π
β
0
α
S
(µ
0
)
.
In terms of Λ
QCD
, we can write α
S
(µ) as
α
S
(µ) =
2π
β
0
log(µ/Λ
QCD
)
.
Note that in the way we defined it,
β
0
is positive. So we see that
α
S
decreases
with increasing
µ
. This is called asymptotic freedom. Thus, the divergence occurs
at low
µ
. This is rather different from, say, QED, which is the other way round.
Another important point to get out is that we haven’t included any mass
term yet, and so we do not have a natural “energy scale” given by the masses.
Thus,
L
QCD
is scale invariant, but quantization has led to a characteristic scale
Λ
QCD
. This is called dimensional transmutation.
What is this scale? This depends on what regularization and renormalization
scheme we are using, and tends to be
Λ
QCD
∼ 200-500MeV.
We can think of this as approximately the scale of the border between perturbative
and non-perturbative physics. Note that non-perturbative means we are in low
energies, because that is when the coupling is strong!
Of course, we have to be careful, because these results were obtained by only
looking up to one-loop, and so we cannot expect it to make sense at low energy
scales.