2QFT in one dimension (i.e. QM)
III Advanced Quantum Field Theory
2.2 Feynman rules
Consider a theory with a single field x : S
1
→ R, and action
S[x] =
Z
S
1
dt
1
2
˙x
2
+
1
2
m
2
x
2
+
λ
4!
x
4
.
We pick
S
1
as our universe so that we don’t have to worry about boundary
conditions. Then the path integral for the partition function is
Z
Z
0
=
1
Z
0
Z
S
1
Dx x(t
1
) ···x(t
n
) e
−S[x]
∼
1
Z
0
N
X
n=0
Z
n
Y
i=1
dt
i
Z
S
1
Dx e
−S
free
[x]
λ
n
(4!)
n
n!
n
Y
i=1
x(t
i
)
4
=
N
X
n=0
Z
n
Y
i=1
dt
i
λ
n
(4!)
n
n!
*
n
Y
i=1
x(t
i
)
4
+
free
So we have again reduced the problem to computing the correlators of the free
theory.
Instead of trying to compute these correlators directly, we instead move to
momentum space. We write
x(t) =
X
k∈Z
x
k
e
−ikt
.
For the sake of brevity (or rather, laziness of the author), we shall omit all
factors of 2π. Using orthogonality relations, we have
S[x] =
X
k∈Z
1
2
(k
2
+ m
2
)|x
k
|
2
+
X
k
1
,k
2
,k
3
,k
4
∈Z
δ(k
1
+ k
2
+ k
3
+ k
4
)
λ
4!
x
k
1
x
k
2
x
k
3
x
k
4
.
Note that here the
δ
is the “discrete” version, so it is 1 if the argument vanishes,
and 0 otherwise.
Thus, we may equivalently represent
Z
Z
0
∼
N
X
n=0
X
{k
(i)
j
}
λ
n
4!
n
n!
n
Y
i=1
δ(k
(i)
1
+ k
(i)
2
+ k
(i)
3
+ k
(i)
4
)
*
n
Y
i=1
4
Y
j=1
x
k
(i)
j
+
free
.
This time, we are summing over momentum-space correlators. But in momentum
space, the free part of the action just looks like countably many free, decoupled
0-dimensional fields! Moreover, each correlator involves only finitely many of
these fields. So we can reuse our results for the 0-dimensional field, i.e. we can
compute these correlators using Feynman diagrams! This time, the propagators
have value
1
k
2
+m
2
.
If we are working over a non-compact space, then we have a Fourier transform
instead of a Fourier series, and we get
Z
Z
0
∼
N
X
n=0
Z
n
Y
i=1
4
Y
j=1
dk
(i)
j
λ
n
4!
n
n!
n
Y
i=1
δ(k
(i)
1
+k
(i)
2
+k
(i)
3
+k
(i)
4
)
*
n
Y
i=1
4
Y
j=1
x(k
(i)
j
)
+
free
.