Soft Degrees of Freedom, Gibbons–Hawking Contribution and Entropy from. . .
383
where the first line is proportional to the number operator for unphysical oscillators,
while the second line contains the correct source term. Since ˆ
a † ˆ
b + ˆ
b † ˆ
a + q ˆ
a 0 +
q ˆ
a
†
0 = ˆ
a
†
3 ˆ
a 3 − ( ˆ
a
†
0 − q)( ˆ
a 0 − q) + q 2 , it follows that |0 Q is an eigenstate of this
gauge fixed Hamiltonian
ˆ
H ξ =1 |0
Q
=
d
3 k ω(
k)q(
k)
2
|0
Q .
(71)
In the context of BRST quantization, one may modify the gauge fixing fermion
and remove the source dependent term therein, that is to say, one may replace ˜
K
Q
ξ
by
˜
K ξ = K ξ −
1
2
d
3 x P
1
Δ
∂ i π
i ,
(72)
ˆ ˜
K ξ =1 =
d
3 k ω(
k)
ˆ ¯
c
† (
k) ˆ
b(
k) + ˆ
b
† (
k) ˆ ¯
c(
k)
,
(73)
ˆ
Ω
Q , ˆ ˜
K ξ =1
=
d
3 k ω(
k)
ˆ
a
Q† (
k)b(
k) + ˆ
b
† (
k) ˆ
a
Q (
k)
+ ˆ ¯
c
† (
k) ˆ
c(
k) + ˆ
c
† (
k) ˆ ¯
c(
k)
,
(74)
since this modifies the ghost number 0 part of the Hamiltonian by terms that are
proportional to the constraints. It now follows that |0 Q is an eigenstate of the new
gauge fixed Hamiltonian ˆ
H
ξ =1 ,
ˆ
H
ξ =1 = ˆ
H
ph
+
ˆ
Ω
Q , ˆ ˜
K ξ =1
,
ˆ
H
ξ =1 |0
Q
=
d
3 k ω(
k)q(
k)
2
|0
Q ,
(75)
with the same eigenvalue than |0 Q is of ˆ
H ξ =1 .
Note also that the difference between e
d 3 k
q 2 (
k)
2 |0 Q and |0 Q is BRST exact.
Indeed,
e
d 3 k
q 2 (
k)
2 |0
Q
− |0
Q
=
e
−
d 3 k
q(
k)
2 ˆ
a Q† (
k)
− ˆ
1
|0
Q .
(76)
The result follows from the fact that −
d 3 k
q(
k)
2 ˆ
a Q† (
k) is BRST exact,
−
d
3 k
q(
k)
2
ˆ
a
†Q (
k) = [ ˆ
K, ˆ
Ω
Q
],
ˆ
K = −
d
3 k
q(
k)
2
ˆ
c
† (
k),
(77)