λλ
′
(2
π
)
3
δ
(3)
(
vector
p
−
vector
q
)
λ,λ
′
= 1
,
2
,
3
(6.46)
But for the timelike annihilation and creation operators, we have
[
a
0
vector
p
,a
0
†
vector
q
] =
−
(2
π
)
3
δ
(3)
(
vector
p
−
vector
q
)
(6.47)
This is very odd! To see just how strange this is, we take the Lorentz invariant vacuum
|
0
)
defined by
a
λ
vector
p
|
0
)
= 0
(6.48)
Then we can create one-particle states in the usual way,
|
vector
p,λ
)
=
a
λ
†
vector
p
|
0
)
(6.49)
For spacelike polarization states,
λ
= 1
,
2
,
3, all seems well.
But for the timelike
polarization
λ
= 0, the state
|
vector
p,
0
)
has negative norm,
(
vectorp,
0
|
vector
q,
0
)
=
(
0
|
a
0
vector
p
a
0
†
vector
q
|
0
)
=
−
(2
π
)
3
δ
(3)
(
vector
p
−
vector
q
)
(6.50)
Wtf? That’s very very strange. A Hilbert space with negative norm means negative
probabilities which makes no sense at all. We can trace this negative norm back to the
wrong sign of the kinetic term for
A
0
in our original Lagrangian:
L
= +
1
2
˙
vector
A
2
−
1
2
˙
A
2
0
+
...
.
At this point we should remember our constraint equation,
∂
μ
A
μ
= 0, which, until
now, we’ve not imposed on our theory. This is going to come to our rescue. We will see
that it will remove the timelike, negative norm states, and cut the physical polarizations
down to two. We work in the Heisenberg picture, so that
∂
μ
A
μ
= 0
(6.51)
makes sense as an operator equation. Then we could try implementing the constraint
in the quantum theory in a number of different ways.
Let’s look at a number of
increasingly weak ways to do this
•
We could ask that
∂
μ
A
μ
= 0 is imposed as an equation on operators. But this
can’t possibly work because the commutation relations (6.39) won’t be obeyed
for
π
0
=
−
∂
μ
A
μ
. We need some weaker condition.
– 133 –
•
We could try to impose the condition on the Hilbert space instead of directly
on the operators. After all, that’s where the trouble lies! We could imagine that
there’s some way to split the Hilbert space up into good states
|
Ψ
)
and bad states
that somehow decouple from the system. With luck, our bad states will include
the weird negative norm states that we’re so disgusted by. But how can we define
the good states? One idea is to impose
∂
μ
A
μ
|
Ψ
)
= 0
(6.52)
on all good, physical states
|
Ψ
)
. But this can’t work either! Again, the condition
is too strong. For example, suppose we decompose
A
μ
(
x
) =
A
+
μ
(
x
) +
A
−
μ
(
x
) with
A
+
μ
(
x
) =
integraldisplay
d
3
p
(2
π
)
3
1
radicalbig
2
|
vector
p
|
3
summationdisplay
λ
=0
ǫ
λ
μ
a
λ
vector
p
e
−
ip
·
x
A
−
μ
(
x
) =
integraldisplay
d
3
p
(2
π
)
3
1
radicalbig
2
|
vector
p
|
3
summationdisplay
λ
=0
ǫ
λ
μ
a
λ
†
vector
p
e
+
ip
·
x
(6.53)
Then, on the vacuum
A
+
μ
|
0
)
= 0 automatically, but
∂
μ
A
−
μ
|
0
) negationslash
= 0. So not even
the vacuum is a physical state if we use (6.52) as our constraint
•
Our final attempt will be the correct one. In order to keep the vacuum as a good
physical state, we can ask that physical states
|
Ψ
)
are defined by
∂
μ
A
+
μ
|
Ψ
)
= 0
(6.54)
This ensures that
(
Ψ
′
|
∂
μ
A
μ
|
Ψ
)
= 0
(6.55)
so that the operator
∂
μ
A
μ
has vanishing matrix elements between physical states.
