3Interacting fields
III Quantum Field Theory
3.3 Wick’s theorem
Now the above process was still sort-of fine, because we were considering first-
order terms only. If we considered higher-order terms, we would have complicated
expressions involving a mess of creation and annihilation operators, and we would
have to commute them past each other to act on the states.
Our objective is thus to write a time-ordered product
T φ
(
x
1
)
φ
(
x
2
)
···φ
(
x
n
)
as a sum of normal-ordered things, since it is easy to read off the action of a
normal-ordered operator on a collection of states. Indeed, we know
hf|:O: |ii
is non-zero if the creation and annihilation operators
:O:
match up exactly with
those used to create |fi and |ii from the vacuum.
We start with a definition:
Definition (Contraction). The contraction of two fields φ, ψ is defined to be
φψ = T (φψ) −:φψ: .
More generally, if we have a string of operators, we can contract just some of
them:
···φ(x
1
) ···φ(x
2
) ··· ,
by replacing the two fields with the contraction.
In general, the contraction will be a c-function, i.e. a number. So where we
decide to put the result of the contraction wouldn’t matter.
We now compute the contraction of our favorite scalar fields:
Proposition. Let φ be a real scalar field. Then
φ(x
1
)φ(x
2
) = ∆
F
(x
1
− x
2
).
Proof. We write φ = φ
+
+ φ
−
, where
φ
+
(x) =
Z
d
3
p
(2π)
3
1
p
2E
p
a
p
e
−ip·x
, φ
−
(x) =
Z
d
3
p
(2π)
3
1
p
2E
p
a
†
p
e
ip·x
.
Then we get normal order if all the φ
−
appear before the φ
+
.
When x
0
> y
0
, we have
T φ(x)φ(y) = φ(x)φ(y)
= (φ
+
(x) + φ
−
(x))(φ
+
(y) + φ
−
(y))
= φ
+
(x)φ
+
(y) + φ
−
(x)φ
+
(y) + [φ
+
(x), φ
−
(y)]
+ φ
−
(y)φ
+
(x) + φ
−
(x)φ
−
(y).
So we have
T φ(x)φ(y) = :φ(x)φ(y): + D(x − y).
By symmetry, for y
0
> x
0
, we have
T φ(x)φ(y) = :φ(x)φ(y): + D(y − x).
Recalling the definition of the Feynman propagator, we have
T φ(x)φ(y) = :φ(x)φ(y): + ∆
F
(x − y).
Similarly, we have
Proposition. For a complex scalar field, we have
ψ(x)ψ
†
(y) = ∆
F
(x − y) = ψ
†
(y)ψ(x),
whereas
ψ(x)ψ(y) = 0 = ψ
†
(x)ψ
†
(y).
It turns out these computations are all we need to figure out how to turn a
time-ordered product into a sum of normal-ordered products.
Theorem
(Wick’s theorem)
.
For any collection of fields
φ
1
=
φ
1
(
x
1
)
, φ
2
=
φ
2
(x
2
), ···, we have
T (φ
1
φ
2
···φ
n
) = :φ
1
φ
2
···φ
n
: + all possible contractions
Example. We have
T (φ
1
φ
2
φ
3
φ
4
) = :φ
1
φ
2
φ
3
φ
4
: + φ
1
φ
2
:φ
3
φ
4
: + φ
1
φ
3
:φ
2
φ
4
: + ··· + φ
1
φ
2
φ
3
φ
4
+ ···
Proof sketch. We will provide a rough proof sketch.
By definition this is true for
n
= 2. Suppose this is true for
φ
2
···φ
n
, and
now add φ
1
with
x
0
1
> x
0
k
for all k ∈ {2, ··· , n}.
Then we have
T (φ
1
···φ
n
) = (φ
+
1
+ φ
−
1
)(:φ
2
···φ
n
: + other contractions).
The
φ
−
term stays where it is, as that already gives normal ordering, and the
φ
+
1
term has to make its way past the
φ
−
k
operators. So we can write the RHS
as a normal-ordered product. Each time we move past the
φ
−
k
, we pick up a
contraction φ
1
φ
k
.
Example.
We now consider the more complicated problem of nucleon scattering:
ψψ → ψψ.
So we are considering interactions of the form
p
1
p
2
p
0
1
p
0
2
something happens
Then we have initial and final states
|ii =
p
4E
p
1
E
p
2
b
†
p
1
b
†
p
2
|0i
|fi =
q
4E
p
0
1
E
p
0
2
b
†
p
0
1
b
†
p
0
2
|0i
We look at the order
g
2
term in
hf|
(
S − 1
)
|ii
. We remove that
1
as we are not
interested in the case with no scattering, and we look at order
g
2
because there
is no order g term.
The second order term is given by
(−ig)
2
2
Z
d
4
x
1
d
4
x
2
T
ψ
†
(x
1
)ψ(x
1
)φ(x
1
)ψ
†
(x
2
)ψ(x
2
)φ(x
2
)
.
Now we can use Wick’s theorem to write the time-ordered product as a sum of
normal-ordered products. The annihilation and creation operators don’t end up
killing the vacuum only if we annihilate two
b
p
and then construct two new
b
p
.
This only happens in the term
:ψ
†
(x
1
)ψ(x
1
)ψ
†
(x
2
)ψ(x
2
): φ(x
1
)φ(x
2
).
We know the contraction is given by the Feynman propagator ∆
F
(
x
1
− x
2
). So
we now want to compute
hp
0
1
, p
0
2
|:ψ
†
(x
1
)ψ(x
1
)ψ
†
(x
2
)ψ(x
2
): |p
1
, p
2
i
The only relevant term in the normal-ordered product is
ZZZZ
d
3
k
1
d
3
k
2
d
3
k
3
d
3
k
4
(2π)
12
p
16E
k
1
E
k
2
E
k
3
E
k
4
b
†
k
1
b
†
k
2
b
k
3
b
k
4
e
ik
1
·x
1
+ik
2
·x
2
−ik
3
·x
1
−ik
4
·x
2
.
Now letting this act on hp
0
1
, p
0
2
| and |p
1
, p
2
i would then give
e
ix
1
·(p
0
1
−p
1
)+ix
2
·(p
0
2
−p
2
)
+ e
ix
1
(p
0
2
−p
1
)+ix
1
(p
0
1
−p
1
)
, (∗)
plus what we get by swapping
x
1
and
x
2
by a routine calculation (swap the
b
-operators in the
ψ
with those that create
|ii, |fi
, insert the delta functions as
commutators, then integrate).
What we really want is to integrate the Hamiltonian. So we want to integrate
the result of the above computation with respect to all spacetime to obtain
(−ig)
2
2
Z
d
4
x
1
d
4
x
2
(∗) ∆
F
(x
1
− x
2
)
=
(−ig)
2
2
Z
d
4
x
1
d
4
x
2
(∗)
Z
d
4
k
(2π)
4
e
ik·(x
1
−x
2
)
k
2
− m
2
+ iε
= (−ig)
2
Z
d
4
k
(2π)
4
i(2π)
8
k
2
− m
2
+ iε
δ
4
(p
0
1
− p
1
+ k)δ
4
(p
0
2
− p
2
− k) + δ
4
(p
0
2
− p
1
+ k)δ
4
(p
0
1
− p
2
− k)
= (−ig)
2
(2π)
4
i
(p
1
− p
0
1
)
2
− m
2
+
i
(p
1
− p
0
2
)
2
− m
2
δ
4
(p
1
+ p
2
− p
0
1
− p
0
2
).
What we see again is a δ-function that enforces momentum conservation.