4Wilsonian renormalization
III Advanced Quantum Field Theory
4.1 Background setting
We are now going to study renormalization. Most of the time, we will assume we
are talking about a real scalar field ϕ, but the ideas and results are completely
general.
Supposed we did some experiments, obtained some results, and figured that
we are probably working with a quantum field theory described by some action.
But we didn’t test our theory to arbitrarily high energies. We don’t know how
“real physics” looks like at high energy scales. So we can’t really write down a
theory that we can expect is valid to arbitrarily high energies.
However, we have previously seen that “integrating out” high energy particles
has the same effect as just modifying the coupling constants of our theory.
Similarly, even with a single fixed field
ϕ
, we can integrate out the high energy
modes of the field
ϕ
, and obtain an effective theory. Suppose we integrate out
all modes with k
2
≥ Λ
0
, and obtain an effective action
S
Λ
0
[ϕ] =
Z
M
d
d
x
"
1
2
(∂ϕ)
2
+
X
i
g
i
O
i
(ϕ, ∂ϕ)
#
.
This, by definition, means the partition function of the theory is now given by
Z =
Z
C
∞
(M)
≤Λ
0
Dϕ e
−S
Λ
0
[ϕ]
,
where
C
∞
(
M
)
≤Λ
0
denotes the space of all functions on
M
consisting of sums
(integrals) of eigenmodes of the Laplacian with eigenvalues
≤
Λ
0
(in “layman”
terms, these are fields with momentum
k
2
≤
Λ
0
). This effective action can
answer questions about “low energy physics”, at scales
<
Λ
0
, which we can use
to test our theory against experiments.
Note that in the case of a compact universe, the Laplacian has discrete
eigenvalues. For example, if we work on a flat torus (equivalently,
R
n
with
periodic boundary conditions), then the possible eigenvalues of the Laplacian lie
on a lattice. Then after imposing a cutoff, there are only finitely many energy
modes, and we have successfully regularized the theory into something that
makes mathematical sense.
In the case of a non-compact universe, this doesn’t happen. But still, in
perturbation theory, this theory will give finite answers, not infinite ones. The
loop integrals will have a finite cut-off, and will thus give finite answers. Of
course, we are not saying that summing all Feynman diagrams will give a finite
answer — this doesn’t happen even for 0-dimensional QFTs. (This isn’t exactly
true. There are further subtleties due to “infrared divergences”. We’ll mostly
ignore these, as they are not really problems related to renormalization)
Either way, we have managed to find ourselves a theory that we are reasonably
confident in, and gives us the correct predictions in our experiments.
Now suppose 10 years later, we got the money and built a bigger accelerator.
We can then test our theories at higher energy scales. We can then try to write
down an effective action at this new energy scale. Of course, it will be a different
action, since we have changed the energy scale. However, the two actions are
not unrelated! Indeed, they must give the same answers for our “low energy”
experiments. Thus, we would like to understand how the action changes when
we change this energy scale.
In general, the coupling constants are a function of the energy scale Λ
0
, and
we will write them as
g
i
(Λ
0
). The most general (local) action can be written as
S
Λ
0
[ϕ] =
Z
M
d
d
x
"
1
2
(∂ϕ)
2
+
X
i
g
i
(Λ
0
)Λ
d−d
i
0
O
i
(ϕ, ∂ϕ)
#
,
where
O
(
ϕ, ∂ϕ
) are monomials in fields and derivatives, and
d
i
the mass di-
mension [
O
i
]. Note that this expression assumes that the kinetic term does not
depend on Λ
0
. This is generally not the case, and we will address this issue later.
We inserted the factor of Λ
d−d
i
0
such that the coupling constants
g
i
(Λ
0
) are
dimensionless. This is useful, as we are going to use dimensional analysis a
lot. However, this has the slight disadvantage that even without the effects of
integrating out fields, the coupling constants
g
i
must change as we change the
energy scale Λ
0
.
To do this, we need to actually figure out the mass dimension of our operators
O
i
. Thus, we need to figure out the dimensions of
ϕ
and
∂
. We know that
S
itself is dimensionless, since we want to stick it into an exponential. Thus, any
term appearing in the integrand must have mass dimension
d
(as the measure
has mass dimension −d).
By looking at the kinetic term in the Lagrangian, we deduce that we must
have
[(∂ϕ)
2
] = d.
as we have to eventually integrate it over space.
Also, we know that [∂
µ
] = 1. Thus, we must have
Proposition.
[∂
µ
] = 1, [ϕ] =
d − 2
2
.