384
G. Barnich and M. Bonte
and that the difference of the exponential of a BRST exact operator minus the unit
operator is a BRST exact operator,
e
[ ˆ
K, ˆ
Ω Q ]
− ˆ
1 =
ˆ
L, ˆ
Ω
Q
,
(78)
for some operator ˆ
L (see, e.g. [4], exercise 14.3 for the proof), so that
e
d 3 k
q 2 (
k)
2 |0
= |0
Q
− ˆ
Ω
Q ˆ
L|0
Q ,
(79)
since |0 Q is BRST closed.
Some additional comments on [6] are in order.
(i) In the computation (2.8), an obvious infrared regularization is understood since
the Fourier transform of k −2 is
1
4πr only when using such a regulator,
1
(2π) 3
d
3 k
1
k 2 + μ 2 e
i
k·· x
=
1
4πr
e
−μr ,
with the desired result obtained when μ → 0 + .
(ii) Equation (2.7) is not correct. Starting from
∂ i A
i
=
i
(2π) 3/2
d
3
k
ω(
k)
2
a(
k) + 2b(
k)
2
e
i
k·· x
− c.c.
(80)
one finds instead of (2.7) that
Q
i ˆ
A
i
|0
Q
=
i
(2π) 3/2
d
3
k
ω(
k)
2
1
2
q(
k)e
i
k·· x
− c.c.
= 0.
(81)
Indeed, the two terms cancel since both ω(
k) and q(
k) are even under
k →
−
k. There is no explanation needed for the difference of a factor 2 between
(2.5) and (2.6) because A μ is not a gauge invariant quantity, as opposed to
π
and
∇ ×
A whose associated expectation values are correctly given in (2.8)
and (2.9). Note however that the Hamiltonian ˆ
H
ξ =1 gives rise to the usual
oscillating behavior for all oscillators in the Heisenberg picture, except for
ˆ
a(
k), ˆ
a † (
k) which evolve according to
ˆ
a
Q (t,
k) ≡ ˆ
a(t,
k) − q(
k) = e
−iω(
k)t
ˆ
a
Q (
k),
(82)
and its complex conjugate.
(iii) In order to make contact with the original [5] and subsequent work, note that
the new vacuum corresponds to the old one “dressed” by