1QFT in zero dimensions

1.4 An effective theory
We now play with another toy theory. Suppose we have two scalar fields
φ, χ R
,
and consider the action
S(φ, χ) =
m
2
2
φ
2
+
M
2
2
χ
2
+
λ
4
φ
2
χ
2
.
For convenience, we will set ~ = 1. We have Feynman rules
1/m
2
1/M
2
with a vertex
λ
We can use these to compute correlation functions and expectation values. For
example, we might want to compute
log(Z/Z
0
).
We have a diagram that looks like
log
Z(m
2
, λ)
Z(m
2
, 0)
+ + + + ···
λ
4m
2
M
2
+
λ
2
16m
4
M
4
+
λ
2
16m
4
M
4
+
λ
2
8m
4
M
4
+ ···
We can also try to compute
hφ
2
i
. To do so, we consider diagrams with two
vertices connected to a solid line:
The relevant diagrams are
hφ
2
i + + + + + ···
1
m
2
+
λ
2m
4
M
4
+
λ
2
4m
6
M
4
+
λ
2
2m
6
M
4
+
λ
2
4m
6
M
4
+ ···
Let’s arrive at this result in a different way. Suppose we think of
χ
as “heavy”,
so we cannot access it directly using our experimental equipment. In particular,
if we’re only interested in the correlation functions that depend only on
φ
, then
we could try to “integrate out” χ first.
Suppose we have a function f(φ), and consider the integral
Z
R
2
dφ dχ f(φ)e
S(φ,χ)/~
=
Z
R
dφ
f(φ)
Z
R
dχ e
S(φ,χ)/~
.
We define the effective action for φ, S
eff
(φ) by
S
eff
(φ) = ~ log
Z
R
dχ e
S(φ,χ)/~
.
Then the above integral becomes
Z
R
dφ f(φ)e
S
eff
(φ)/~
.
So doing any computation with
φ
only is equivalent to pretending
χ
doesn’t
exist, and using this effective potential instead.
In general, the integral is very difficult to compute, and we can only find
an asymptotic series for the effective action. However, in this very particular
example, we have chosen
S
such that it is only quadratic in
χ
, and one can
compute to get
Z
R
dχ e
S(φ,χ)/~
= e
m
2
φ
2
/2~
s
2π~
M
2
+ λφ
2
/2
.
Therefore we have
S
eff
(φ) =
m
2
2
φ
2
+
~
2
log
1 +
λφ
2
2M
2
+
~
2
log
M
2
2π~
=
m
2
2
+
~λ
4M
2
φ
2
~λ
2
16M
4
φ
4
+
~λ
3
48M
6
φ
6
+ ··· +
~
2
log
M
2
2π~
=
m
2
eff
2
φ
2
+
λ
4
4!
φ
4
+
λ
6
6!
φ
6
+ ··· +
~
2
log
M
2
2π~
,
where
λ
2k
= (1)
k+1
~(2k)!
2
k+1
k!
λ
k
M
2k
.
We see that once we integrated out
χ
, we have generated an infinite series of
new interactions for φ in S
eff
(φ). Moreover, the mass also shifted as
m
2
7→ m
2
eff
= m
2
+
~λ
2M
4
.
It is important to notice that the new vertices generated are quantum effects.
They vanish as
~
0 (they happen to be linear in
~
here, but that is just a
coincidence). They are also suppressed by powers of
1
M
2
. So if
χ
is very very
heavy, we might think these new couplings have very tiny effects.
This is a very useful technique. If the universe is very complicated with
particles at very high energies, then it would be rather hopeless to try to account
for all of the effects of the particles we cannot see. In fact, it is impossible to
know about these very high energy particles, as their energy are too high to
reach. All we can know about them is their incarnation in terms of these induced
couplings on lower energy fields.
A few things to note:
Z
2
× Z
2
symmetry, given by (
φ, χ
)
7→
(
±φ, ±χ
).
This symmetry is preserved, and we do not generate any vertices with odd
powers of φ.
The field S
eff
(φ) also contains a field independent term
~
2
log
M
2
2π~
.
This plays no role in correlation functions
hf
(
φ
)
i
. Typically, we are just
going to drop it. However, this is one of the biggest problem in physics.
This term contributes to the cosmological constant, and the scales that are
relevant to this term are much larger than the actual observed cosmological
constant. Thus, to obtain the observed cosmological constant, we must
have some magic cancelling of these cosmological constants.
In this case, passing to the effective action produced a lot of new inter-
actions. However, if we already had a complicated theory, e.g. when we
started off with an effective action, and then decided to integrate out
more things, then instead of introducing new interactions, the effect of this
becomes shifting the coupling constants, just as we shifted the mass term.
In general, how can we compute the effective potential? We are still computing
an integral of the form
Z
R
dχ e
S(φ,χ)/~
,
so our previous technique of using Feynman diagrams should still work. We will
treat
φ
as a constant in the theory. So instead of having a vertex that looks like
λ
we drop all the φ lines and are left with
λφ
2
But actually, this would be rather confusing. So instead what we do is that
we still draw the solid
φ
lines, but whenever such a line appears, we have to
terminate the line immediately, and this contributes a factor of φ:
λφ
2
For our accounting purposes, we will say the internal vertex contributes a factor
of
λ
2
(since the automorphism group of this vertex has order 2, and the action
λ
4
), and each terminating blue vertex contributes a factor of φ.
Since we have a “constant” term
m
2
2
φ
2
as well, we need to add one more
diagram to account for that. We allow a single edge of the form
m
2
With these ingredients, we can compute the effective potential as follows:
S
eff
(φ) + + + + ···
m
2
2
φ
2
+
λ
4M
2
φ
2
+
λ
2
φ
4
16M
4
+
λ
3
φ
6
48M
6
+ ···
These are just the terms we’ve previously found. There are, again, a few things
to note
The diagram expansion is pretty straightforward in this case, because we
started with a rather simple interacting theory. More complicated examples
will have diagram expansions that are actually interesting.
We only sum over connected diagrams, as
S
eff
is the logarithm of the
integral.
We see that the new/shifted couplings in
S
eff
(
φ
) are generated by loops of
χ fields.
When we computed the effective action directly, we had a “cosmological
constant” term
~
2
log
M
2
2π~
, but this doesn’t appear in the Feynman
diagram calculations. This is expected, since when we developed our
Feynman rules, what it computed for us was things of the form
log
(
Z/Z
0
),
and the Z
0
term is that “cosmological constant”.
Example. We can use the effective potential to compute hφ
2
i:
hφ
2
i + + ···
1
m
2
eff
+
λ
2m
6
eff
+ ···
We can see that this agrees with our earlier calculation correct to order λ
2
.
At this moment, this is not incredibly impressive, since it takes a lot of work
to compute
S
eff
(
φ
). But the computation of
S
eff
(
φ
) can be reused to compute
any correlation involving
φ
. So if we do many computations with
φ
, then we we
can save time.
But the point is not that this saves work. The point is that this is what we
do when we do physics. We can never know if there is some really high energy
field we can’t reach with our experimental apparatus. Thus, we can only assume
that the actions we discover experimentally are the effective actions coming from
integrating out some high energy fields.