382
G. Barnich and M. Bonte
Since
{Ω Q , K ξ } =
d 3 x
∂ k A k π 0 + A 0
−∂ i π i + j 0
+ iPρ + i∂ i ¯
C∂ i η −
1
2
ξπ 0 π 0
,
(61)
the gauge fixed Hamiltonian contains in particular the correct source term. When
using the decomposition π i = π i
T +
1
Δ ∂ i ∂ j π j , it follows that
1
2
d
3 x π
i π i =
1
2
d
3 x π
i
T π
T
i −
1
2
d
3 x ∂ j π
j 1
Δ
∂ k π
k .
(62)
The last term can be written as
−
1
2
d
3 x ∂ j π
j 1
Δ
∂ k π
k
=
Ω
Q , −
1
2
d
3 x P
1
Δ
∂ i π
i
+ j
0
−
1
2
d
3 x j
0 1
Δ
j
0 ,
(63)
so that
H ξ = H
ph
+
Ω
Q , ˜
K
Q
ξ
,
(64)
H
ph
=
1
2
d
3 x
π
i
T π
T
i − A
T
j ΔA
j
T − j
0 1
Δ
j
0
,
(65)
˜
K
Q
ξ = K ξ −
1
2
d
3 x P
1
Δ
∂ i π
i
+ j
0
.
(66)
In Feynman gauge ξ = 1, when expressed in terms of modes, we have
ˆ
H ξ =1 = ˆ
H
phys
+
ˆ
Ω
Q , ˆ ˜
K
Q
ξ =1
,
(67)
ˆ
H
phys
=
d
3 k ω(
k)
ˆ
a
†
a (
k) ˆ
a
a (
k) + q(
k)
2
,
(68)
ˆ ˜
K
Q
ξ =1 =
d
3 k ω(
k)
ˆ ¯
c
† (
k)
ˆ
b(
k) +
ω(
k)
2
+
ˆ
b
† (
k) +
ω(
k)
2
ˆ ¯
c(
k)
, (69)
and
ˆ
Ω
Q , ˆ ˜
K
Q
ξ =1
=
d
3 k ω(
k)
ˆ
a
† (
k)b(
k) + ˆ
b
† (
k) ˆ
a(
k) + ˆ ¯
c
† (
k) ˆ
c(
k) + ˆ
c
† (
k) ˆ ¯
c(
k)
+
d
3 k ω(
k)q(
k)[ ˆ
a 0 (
k) + ˆ
a
†
0 (
k)],
(70)