3Discrete symmetries
III The Standard Model
3.5 S-matrix
We consider how S-matrices transform. Recall that S is defined by
hp
1
, p
2
, ···|S |k
A
, k
B
, ···i =
out
hp
1
, p
2
, ···|k
A
, k
b
, ···i
in
= lim
T →∞
hp
1
, p
2
, ···|e
−iH2T
|k
A
, k
B
, ···i
We can write
S = T exp
−i
Z
∞
−∞
V (t) dt
,
where T denotes the time-ordered integral, and
V (t) = −
Z
d
3
x L
I
(x),
and L
I
(x) is the interaction part of the Lagrangian.
Example. In QED, we have
L
I
= −e
¯
ψ(x)γ
µ
A
µ
(x)ψ(x).
We can draw a table of how things transform:
P C T
L
I
(x) L
I
(x
P
) L
I
(x) L
I
(x
T
)
V (t) V (t) V (t) V (−t)
S S S ??
A bit more care is to figure out how the
S
matrix transforms when we reverse
time, as we have a time-ordering in the integral. To figure out, we explicitly
write out the time-ordered integral. We have
S =
∞
X
n=0
(−i)
n
Z
∞
−∞
dt
1
Z
t
1
−∞
dt
2
···
Z
t
n−1
−∞
dt
n
V (t
1
)V (t
2
) ···V (t
n
).
Then we have
S
T
=
ˆ
T S
ˆ
T
−1
=
X
n
(+i)
n
Z
∞
−∞
dt
1
Z
t
1
−∞
dt
2
···
Z
t
n−1
−∞
dt
n
V (−t
1
)V (−t
2
) ···V (−t
n
)
We now put τ
i
= −t
n+1−2
, and then we can write this as
=
X
n
(+i)
n
Z
∞
−∞
dt
1
Z
−τ
n
−∞
···
Z
−τ
2
−∞
dt
n
V (τ
n
)V (τ
n−1
) ···V (τ
1
)
=
X
n
(+i)
n
Z
∞
−∞
dτ
n
Z
∞
τ
n
dτ
n−1
···
Z
∞
τ
2
dτ
1
V (τ
n
)V (τ
n−1
) ···V (τ
1
).
We now notice that
Z
∞
−∞
dτ
n
Z
∞
τ
n
dτ
n−1
=
Z
∞
−∞
dτ
n−1
Z
τ
n−1
−∞
dτ
n
.
We can see this by drawing the picture
τ
n
τ
n−1
τ
n−1
= τ
n
So we find that
S
T
=
X
n
(+i)
n
Z
∞
−∞
dτ
1
Z
τ
1
−∞
dτ
2
···
Z
τ
n−1
−∞
dτ
n
V (τ
n
)V (τ
n−1
) ···V (τ
1
).
We can then see that this is equal to S
†
.
What does this tell us? Consider states |ηi and |ξi with
|η
T
i =
ˆ
T |ηi
|ξ
T
i =
ˆ
T |ξi
The Dirac bra-ket notation isn’t very useful when we have anti-linear operators.
So we will write inner products explicitly. We have
(η
T
, Sξ
T
) = (
ˆ
T η, S
ˆ
T ξ)
= (
ˆ
T η, S
†
T
ˆ
T ξ)
= (
ˆ
T η,
ˆ
T S
†
, ξ)
= (η, S
†
ξ)
∗
= (ξ, Sη)
where we used the fact that
ˆ
T is anti-unitary. So the conclusion is
hη
T
|S |ξ
T
i = hξ|S |ηi.
So if
ˆ
T L
I
(x)
ˆ
T
−1
= L
I
(x
T
),
then the
S
-matrix elements are equal for time-reversed processes, where the
initial and final states are swapped.