1QFT in zero dimensions

1.3 Feynman diagrams
We now try to understand the terms in the asymptotic series in terms of Feynman
diagrams. The easy bits are the powers of
~λ
m
4
n
.
This combination is essentially fixed by dimensional analysis. What we really
want to understand are the combinatorial factors, given by
1
(4!)
n
n!
×
(4n)!
4
n
(2n)!
.
To understand this, we can write the path integral in a different way. We have
Z(~, m, λ) =
Z
dφ exp
1
~
m
2
2
φ
2
+
λ
4!
φ
4

=
Z
dφ
X
n=0
1
n!
λ
4!~
n
φ
4n
exp
m
2
2~
φ
2
X
n=0
1
(4!)
n
n!
λ
~
n
hφ
4n
i
free
Z
free
.
Thus, we can apply our previous Wick’s theorem to compute this! We now see
that there is a factor of
1
(4!)
n
n!
coming from expanding the exponential, and
the factor of
(4n)!
4
n
(2n)!
came from Wick’s theorem it is the number of ways to
pair up 4
n
vertices together. There will be a factor of
~
2k
coming from Wick’s
theorem when evaluating hφ
4n
i, so the expansion is left with a ~
k
factor.
Let’s try to compute the order
λ
term. By Wick’s theorem, we should
consider diagrams with 4 external vertices, joined by straight lines. There are
three of these:
But we want to think of these as interaction vertices of valency 4. So instead of
drawing them this way, we “group” the four vertices together. For example, the
first one becomes
Similarly, at order λ
2
, we have a diagram
Note that in the way we are counting diagrams in Wick’s theorem means we
consider the following diagrams to be distinct:
However, topologically, these are the same diagrams, and we don’t want to count
them separately. Combinatorially, these diagrams can be obtained from each
other by permuting the outgoing edges at each vertex, or by permuting the
vertices. In other words, in the “expanded view”, when we group our vertices as
we are allowed to permute the vertices within each block, or permute the blocks
themselves. We let
D
n
be the set of all graphs, and
G
n
be the group consisting
of these “allowed” permutations. This
G
n
is a semi-direct product (
S
4
)
n
o S
n
.
Then by some combinatorics,
|D
n
| =
(4n)!
4
n
(2n)!
, |G
n
| = (4!)
n
n!.
Recall that
|D
n
|
is the number Wick’s theorem gives us, and we now see that
1
|G
n
|
happens to be the factor we obtained from expanding the exponential. Of
course, this isn’t a coincidence we chose to put
λ
4!
λ
4
in front
of the φ
4
term so that this works out.
We can now write the asymptotic series as
Z(m
2
, λ)
Z(m
2
, 0)
N
X
n=0
|D
n
|
|G
n
|
~λ
m
4
n
.
It turns out a bit of pure mathematics will allow us to express this in terms
of the graphs up to topological equivalence. By construction, two graphs are
topologically equivalent if they can be related to each other via
G
n
. In other
words, the graphs correspond to the set of orbits
O
n
of
D
n
under
G
n
. For
n
= 1,
there is only one graph up to topological equivalence, so
|O
1
|
= 1. It is not hard
to see that |O
2
| = 3.
In general, each graph Γ has some automorphisms that fix the graph. For
example, in the following graph:
3 4
21
we see that the graph is preserved by the permutations (1 3), (2 4) and (1 2)(3 4).
In general, the automorphism group
Aut
(Γ) is the subgroup of
G
n
consisting of
all permutations that preserve the graph Γ. In this case,
Aut
(Γ) is generated by
the above three permutations, and |Aut(Γ)| = 8.
Recall from IA Groups that the orbit-stabilizer theorem tells us
|D
n
|
|G
n
|
=
X
Γ∈O
n
1
|Aut(Γ)|
.
Thus, if we write
O =
[
nN
O
n
,
then we have
Z(m
2
, λ)
Z(m
2
, 0)
X
Γ∈O
1
|Aut(Γ)|
~λ
m
4
n
,
where the
n
appearing in the exponent is the number of vertices in the graph Γ.
The last loose end to tie up is the factor of
~λ
m
4
n
appearing in the sum.
While this is a pretty straightforward thing to come up with in this case, for
more complicated fields with more interactions, we need a better way of figuring
out what factor to put there.
If we look back at how we derived this factor, we had a factor of
λ
~
coming from each interaction vertex, whereas when computing the correlation
functions
hφ
4n
i
, every edge contributes a factor of
~m
2
(since, in the language
we expressed Wick’s theorem, we had
M
=
m
2
). Thus, we can imagine this as
saying we have a propagator
~/m
2
and a vertex
λ/~
Note that the negative sign appears because we have
e
S
in “Euclidean” QFT.
In Minkowski spacetime, we have a factor of i instead.
After all this work, we can now expand the partition function as
Z(m
2
, λ)
Z(m
2
, 0)
+ + + + + ···
1 +
λ~
m
4
1
8
+
λ
2
~
2
m
8
1
48
+
λ
2
~
2
m
8
1
16
+
λ
2
~
2
m
8
1
128
+ ···
In generally, if we have a theory with several fields with propagators of value
1
P
i
, and many different interactions with coupling constants λ
α
, then
Z({λ
α
})
Z({0})
X
Γ∈O
1
|Aut Γ|
Q
α
λ
|v
α
(Γ)|
α
Q
i
|P
i
|
|e
i
(Γ)|
~
E(Γ)V (Γ)
,
where
e
i
(Γ) is the number of edges of type i in Γ;
v
α
(Γ) is the number of vertices of type α in Γ;
V (Γ) is the number of vertices in total; and
E(Γ) the total number of edges.
There are a few comments we can make. Usually, counting all graphs is rather
tedious. Instead, we can consider the quantity
log(Z/Z
0
).
This is the sum of all connected graphs. For example, we can see that the term for
the two-figure-of-eight is just the square of the term for a single figure-of-eight
times the factor of
1
2!
, and one can convince oneself that this generalizes to
arbitrary combinations.
In this case, we can simplify the exponent of
~
. In general, by Euler’s theorem,
we have
E V = L C,
where
L
is the number of loops in the graph, and
C
is the number of connected
components. In the case of
log
(
Z/Z
0
), we have
C
= 1. Then the factor of
~
is
just ~
L1
. Thus, the more loops we have, the less it contributes to the sum.
It is not difficult to see that if we want to compute correlation functions, say
hφ
2
i
, then we should consider graphs with two external vertices. Note that in
this case, when computing the combinatorial factors, the automorphisms of a
graph are required to fix the external vertices. To see this mathematically, note
that the factors in
1
|G
n
|
=
1
(4!)
n
n!
came from Taylor expanding
exp
and the coefficient
1
4!
of
φ
4
. We do not get
these kinds of factors from φ
2
when computing
Z
dφ φ
2
e
S[φ]
.
Thus, we should consider the group of automorphisms that fix the external
vertices to get the right factor of
1
|G
n
|
.
Finally, note that Feynman diagrams don’t let us compute exact answers
in general. Even if we had some magical means to compute all terms in the
expansions using Feynman diagrams, if we try to sum all of them up, it will
diverge! We only have an asymptotic series, not a Taylor series. However, it
turns out in some special theories, the leading term is the exact answer! The tail
terms somehow manage to all cancel each other out. For example, this happens
for supersymmetry theories, but we will not go into details.