6Non-abelian gauge theory
III Advanced Quantum Field Theory
6.7 Renormalization of Yang–Mills theory
To conclude the course, we compute the
β
-function of Yang–Mills theory. We
are mostly interested in the case of
SU
(
N
). While a lot of the derivations (or
lack of) we do are general, every now and then, we use the assumption that
G = SU(N), and that the fermions live in the fundamental representation.
Usually, to do perturbation theory, we expand around a vacuum, where all
fields take value 0. However, this statement doesn’t make much sense for gauge
theory, because
A
µ
is only defined up to gauge transformations, and it doesn’t
make sense to “set it to zero”. Instead, what we do is that we fix any classical
solution
A
0
µ
to the Yang–Mills equation, and take this as the “vacuum” for
A
µ
.
This is known as the background field. We write
A
µ
= A
0
µ
+ a
µ
.
Whenever we write ∇
µ
, we will be using the background field, so that
∇
µ
= ∂
µ
+ A
0
µ
.
We can compute
F
µν
= ∂
µ
A
ν
− ∂
ν
A
µ
+ [A
µ
, A
ν
]
= F
0
µν
+ ∂
µ
a
ν
− ∂
ν
a
µ
+ [a
µ
, a
ν
] + [A
0
µ
, a
ν
] + [a
µ
, A
0
ν
]
= F
0
µν
+ ∇
µ
a
ν
− ∇
ν
a
µ
+ [a
µ
, a
ν
].
Thus, if we compute the partition function, we would expect to obtain
something of the form
Z ≡ e
−S
eff
[A]
= exp
−
1
2g
2
Y M
Z
(F
0
µν
, F
0,µν
) d
d
x
(something).
A priori, the “something” will be a function of
A
, and also the energy scale
µ
.
Then since the result shouldn’t actually depend on
µ
, this allows us to compute
the β-function.
We will work in Feynman gauge, where we pick ξ = 1. So we have
S[a, ¯c, c,
¯
ψ, ψ] =
Z
d
d
x
1
4g
2
(F
0
µν
+ ∇
[µ
a
ν]
+ [a
µ
, a
ν
])
2
−
1
2g
2
(∂
µ
A
µ
+ ∂
µ
a
µ
)
2
− ¯c∂
µ
∇
µ
c − ¯c∂
µ
a
µ
c +
¯
ψ(
/
∇ + m)ψ +
¯
ψ
/
aψ
.
This allows us to rewrite the original action in terms of the new field
a
µ
. We
will only compute the β-function up to 1-loop, and it turns out this implies we
don’t need to know about the whole action. We claim that we only need to know
about quadratic terms.
Let
L
be the number of loops,
V
the number of vertices, and
P
the number
of propagators. Then, restricting to connected diagrams, Euler’s theorem says
P − V = L − 1.
By restricting to 1-loop diagrams, this means we only care about diagrams with
P = V .
For each node
i
, we let
n
q,i
be the number of “quantum” legs coming out of
the vertex, i.e. we do not count background fields. Then since each propagator
connects two vertices, we must have
2P =
X
vertices i
n
q,i
.
Also, almost by definition, we have
V =
X
vertices i
1.
So this implies we only care about fields with
0 =
X
(n
q,i
− 2).
It can be argued that since
A
0
µ
satisfies the Yang–Mills equations, we can ignore
all linear terms. So this means it suffices to restrict to the quadratic terms.
Restricting to the quadratic terms, for each of
c, ψ, a
, we have a term that
looks like, say
Z
Dc D¯c e
R
d
d
x ¯c∆c
for some operator ∆. Then the path integral will give
det
∆. If the field is a
boson, then we obtain
1
det
instead, but ultimately, the goal is to figure out what
this ∆ is.
Note that each particle comes with a representation of
SO
(
d
) (or rather, the
spin group) and the gauge group
G
. For our purposes, all we need to know about
the representations of
SO
(
d
) is the spin, which may be 0 (trivial),
1
2
(spinor) or
1 (vector). We will refer to the representation of
G
as “
R
”, and the spin as
j
.
We then define the operator
∆
R,j
= ∇
2
+ 2
1
2
F
a
µν
J
µν
(j)
t
a
R
,
where
{t
a
R
}
are the images of the generators of
g
in the representation, and
J
µν
(j)
are the generators of so(d) in the spin j representation. In particular,
J
µν
(0)
= 0, J
µν
(
1
2
)
= S
µν
=
1
4
[γ
µ
, γ
ν
], J
µν
(1)
= i(δ
ρ
µ
δ
σ
ν
− δ
ρ
ν
δ
σ
µ
).
For simplicity, we will assume the fermion masses
m
= 0. Then we claim that
(up to a constant factor), the ∆ of the
c
,
ψ
and
a
fields are just ∆
adj,0
,
q
∆
R,
1
2
and ∆
adj,1
respectively. This is just a computation, which we shall omit.
Thus, if there are n many fermions, then we find that we have
Z = exp
−
1
2g
2
Y M
Z
(F
0
µν
, F
0,µν
) d
d
x
(det ∆
adj,0
)(det ∆
R,
1
2
)
n/2
(det ∆
adj,1
)
1/2
.
We are going to view these extra terms as being quantum corrections to the
effective action of
A
µ
, and we will look at the corrections to the coupling
g
2
Y M
they induce. Thus, we are ultimately interested in the logarithm of these extra
terms.
To proceed, write write out ∆ explicitly:
∆
R,j
= −∂
2
+ (∂
µ
A
a
µ
+ A
a
µ
∂
µ
)t
a
(R)
| {z }
∆
(1)
+ A
µ,a
A
b
µ
t
a
R
t
b
R
| {z }
∆
(2)
+ 2
1
2
F
a
µν
J
µν
(j)
t
a
(R)
| {z }
∆
(J)
.
Then we can write the logarithm as
log det ∆
R,j
= log det(−∂
2
+ ∆
(1)
+ ∆
(2)
+ ∆
(J)
)
= log det(−∂
2
) + tr log(1 − (∂
2
)
−1
(∆
(1)
+ ∆
(2)
+ ∆
(J)
)).
Again, the first term is a constant, and we will ignore it. We want to know the
correction to the coupling
1
4g
2
Y M
. Since everything is covariant with respect to
the background field
A
0
, it is enough to just compute the quadratic terms in
A
µ
, because all the other terms are required to behave accordingly.
We now see what quadratic terms we obtain in the series expansion of
log
.
In the case of
G
=
SU
(
N
), We have
tr t
a
=
tr J
µν
= 0. So certain cross terms
vanish, and the only quadratic terms are
log det ∆
R,j
∼ tr(−∂
−2
∆
(2)
)
| {z }
(a)
+
1
2
tr(∂
−2
∆
(1)
∂
−2
∆
(1)
)
| {z }
(b)
+
1
2
tr(∂
−2
∆
(J)
∂
−2
∆
(J)
)
| {z }
(c)
.
(a) =
Z
d
d
k
(2π)
d
A
a
µ
(k)A
b
ν
(−k)
Z
d
d
p
(2π)
d
tr
R
(t
a
t
b
)
p
2
δ
µν
d(j)
(b) =
Z
d
d
k
(2π)
d
d
d
p
(2π)
d
(k + 2p)
µ
(k + 2p)
ν
tr(t
a
t
b
)A
a
µ
(k)A
b
ν
(k)
p
2
(p + k)
2
(c) =
Z
d
d
k
(2π)
d
d
d
p
(2π)
d
A
a
µ
(k)A
b
ν
(−k)(k
2
δ
µν
− k
µ
k
ν
)
1
p
2
(p + k)
2
C(j) tr(t
a
t
b
).
where
d
(
j
) is the number of spin components of the field and
C
(
j
) is some
constant depending on j. Explicitly, they are given by
scalar Dirac 4-vector
d(j) 1 4 4
C(j) 0 1 2
In terms of Feynman diagrams, we can interpret (
a
) as being given by the
loop diagram
k
A
a
µ
A
b
ν
p
while (b) and (c) are given by diagrams that look like
k
A
a
µ
A
b
ν
We now define the quantity C(R) by
tr
R
(t
a
t
b
) = C(R)δ
ab
,
where the trace is to be taken in the representation
R
. For
G
=
SU
(
N
), we have
C(adj) = N, C(fund) =
1
2
.
Then, evaluating all those integrals, we find that, in dimensional regularization,
we have
log ∆
R,j
=
Γ(2 −
d
2
)
4
Z
d
d
x µ
4−d
(−∂
2
)
d−4
2
F
a
µν
F
a,µν
·
C(R)
(4π)
2
d(j)
3
− 4C(j)
.
Thus, combining all the pieces, we obtain
S
eff
[A] =
1
4g
2
Y M
Z
d
d
x F
a
µν
F
a,µν
−
Γ
2 −
d
2
4
Z
d
d
x µ
4−d
(−∂
2
)
d−4
2
F
a
µν
F
µν
a
×
1
2
C
adj,1
− C
adj,0
−
n
2
C
R,1/2
,
where
C
R,j
=
C(R)
(4π)
2
d(j)
3
− 4C(j)
.
Explicitly, we have
C
R,j
=
C(R)
(4π)
2
×
1
3
scalars
−
8
3
Dirac
−
20
3
vectors
.
As always, the Γ function diverges as we take
d →
4, and we need to remove
the divergence using counterterms. In the
MS
scheme with scale
µ
(with
g
2
Y M
=
µ
4−d
g
2
(
µ
)), we are left with logarithmic dependence on
µ
2
. The independence
of µ gives the condition
µ
d
dµ
1
g
2
(µ)
+ log
µ
2
something
1
2
C
adj,1
− C
adj,0
−
n
2
C
R,
1
2
= 0.
So
−
2
g
3
(µ)
β(g) + 2
1
2
C
adj,1
− C
adj,0
−
n
2
C
R,1
= 0.
In other words, we find
β(g) = g
3
(µ)
1
2
C
adj,1
− C
adj,0
−
n
2
C
R,1
= −
g
3
(µ)
(4π)
2
11
3
C(adj) −
4n
3
C(R)
= −
g
3
(4π)
2
11
3
N −
2n
3
.
Thus, for
n
sufficiently small, the
β
-function is negative! Hence, at least for
small values of
g
, where we can trust perturbation theory, this coupling now
increases as Λ → 0, and decreases as Λ → ∞.
The first consequence is that this theory has a sensible continuum limit! The
coupling is large at low energies, and after a long story, this is supposed to lead
to the confinement of quarks.
In fact, in 1973, Gross–Coleman showed that the only non-trivial QFT’s in
d
= 4 with a continuum limit are non-abelian gauge theories. The proof was by
exhaustion! They just considered the most general kind of QFT we can have,
and then computed the
β
-functions etc., and figured the only ones that existed
were non-abelian gauge theories.