graphically by lines connecting two vertices:
by a line connecting
x
i
and
x
j
and oriented from
x
j
to
x
i
;
by a dashed, undirected
line joining
x
i
to
x
j
.
The reason why the latter line is not oriented is due to the
symmetry of the photon propagator
D
F
μν
(
x
i
−
x
j
)
with respect to an exchange of
x
i
and
x
j
:
D
F
μν
(
x
i
−
x
j
)
=
D
F
μν
(
x
j
−
x
i
).
Each of these internal lines describes
a
virtual
particle propagating between the two vertices. By virtual particle we mean
a particle whose momentum does not satisfy the
onshell
condition:
p
2
−
m
2
=
0
for the electron,
k
2
=
0 for the photon,
k
μ
being the photon 4momentum.
Thenormalorderedterms,besidesthecontractions,willalsocontainuncontracted
field operators, which we shall refer to as
free
. These operators will be represented by
lines extending from the vertex in which they are computed to infinity: The line rep
resenting
ψ(
x
)
will originate at infinity and end in
x
. In the matrix element between
initial and final state, it will contribute only if the initial state contains a free electron
or if the final state contains a free positron. In these two cases
ψ(
x
)
will destroy
the incoming electron or create the outgoing positron in
x
, respectively. In the first
case, if the electron state is

p
e
−
,
r
=
c
†
(
p
e
−
,
r
)

0
, ψ(
x
)
would give the following
nonvanishing contribution to the amplitude
15
:
0

ψ(
x
)

p
e
−
,
r
=
d
q
√
2
m
2
s
=
1
u
(
q
,
s
)
0

c
(
q
,
s
)
c
†
(
p
e
−
,
r
)

0
e
−
iq
·
x
=
d
q
√
2
m
2
s
=
1
u
(
q
,
s
)
0
[
c
(
q
,
s
),
c
†
(
p
e
−
,
r
)
]
−

0
e
−
e
−
iq
·
x
=
√
2
mu
(
p
e
−
,
r
)
e
−
ip
e
−
·
x
.
(12.132)
Similarly, in the second case, if the final positron state is

p
e
+
,
r
=
d
†
(
p
e
+
,
r
)

0 ,
the field
ψ(
x
)
will contribute the following nonvanishing quantity:
p
e
+
,
r

ψ(
x
)

0
=
√
2
m
v(
p
e
+
,
r
)
e
−
ip
e
+
·
x
.
(12.133)
By the same token we can show that a free
ψ(
x
)
operator contributes only to those
processeswithanincomingpositron,whichwillbedestroyedin
x
,orwithanoutgoing
electron, which will be created in
x
by the same operator. It will be represented by
15
Recall
that,
in
the
light
of
our
comments
below
eq.
(12.76),
we
have
replaced
everywhere the normalization volume
V
with 1/(2
E
), so that, for instance
[
c
(
p
,
r
),
c
†
(
q
,
s
)
]
−
=
(
2
π)
3
2
E
δ
3
(
p
−
q
)
. Note that, had we kept the normalization volumes, the calculation below would
yield 0

ψ(
x
)

p
,
r
=
m
E
p
V
e
−
ip
·
x
, contributing a factor 1
/
√
V
to the amplitude, as anticipated in
our discussion below eq. (12.76).