7.4 Particle Detection—A “Nonideal” Measurement
195
Similar considerations could be applied also to f (t) ≡ ≡ 1 ⊗ ϕ| exp(−it (H A +
H B ))| 1 ⊗ ϕ; the spectrum of H A from (7.4.1) acting on the Hilbert space H vac of
the used representation consists of a single eigenvalue {0}, and of absolutely continuous part consisting of the interval [−2, +2] ⊂ R, which can be seen from the Sect. 7.3,
and from [36]. So the function f (t) = = 1 | exp(−it H A )| 1 · ·ϕ| exp(−it H B )|ϕ =
f 1 (t)g(t) has the Fourier image ˆ
f (u) = (2π)
−
1
2 ˆ
f 1 ∗ ˆ
g(u), which with the help of
(7.4.63) and (7.4.68f) gives
ˆ
f (u) =
1
√
2π
ˆ
f 1 ∗ ˆ
g(u) =
1
√
2π
dτ ˆ
f 1 (τ ) ˆ
g(u − τ )
(7.4.69)
=
1
2π
2
−2
dτ
4 − τ 2
τ − u
2
θ(τ − u)
4π
dφ | ˆ
ϕ(
√
τ − u, φ)|
2
.
Remember that ϕ ∈ D(R
3
), hence its Fourier image ˆ
ϕ ∈ S(R
3
) is an entire
analytic function of three complex variables, so that the function p →
4π dφ
| ˆ
ϕ( p, φ)|
2
> 0, a.e. for p > 0. Then (7.4.69) implies that ˆ
f (u) = 0 for u > 2, and
ˆ
f (u) > 0 for almost all u < 2.
For checking finally the conditions of the positivity of w(ψ) from its expression
(7.4.60), we have to check under which conditions it is | ˆ
F
0
+ (u)|θ(2 − u) > 0, u ∈
S ⊂ R, for some S of positive Lebesgue measure.
Let us assume that ˆ
F
0
+ (u) ≡ 0 in some nonzero interval: u ∈ I ⊂ R. The function ˆ
F
0
+ (u) ≡
1
√
2π
+∞
0
e
−itu
ϕ| exp(−it H B )|ψ dt, cf. (7.4.6a), can be continued
to a function analytic in the lower complex half plane Im u < 0 and continuous
on the real axis R. The identical vanishing of this function on an interval I ⊂ R
would imply (with the help of the Schwarz Reflection Principle) its analyticity on
I , and consequent vanishing everywhere in the analyticity domain, hence also on
the whole real axis (i.e. vanishing also on the boundary of the analyticity domain).
The identical vanishing ˆ
F
0
+ (u) ≡ 0, ∀u ∈ R, would imply, however, the identical
vanishing ϕ| exp(−it H B )|ψ ≡ 0, which contradicts (7.4.24). This proves that, for
γ
2
> 0 satisfying (7.4.25), it is ψ|W γ |ψ > 0, iff ψ satisfies (7.4.24). Since the condition (7.4.24) does not depend on the parameter γ, the subspace of H B consisting
of those vectors ψ for which it is ψ|W γ |ψ = 0 does not depend on γ, hence also
its orthogonal complement H W ⊂ H B is independent of γ, cf. Lemma 7.4.4.
It remains to show that, at least for some values of γ ∈ R, it is W
2
γ = W γ , i.e. that
the positive operator W γ is not a projector. For any nonzero orthogonal projector
P ∈ L(H) there exists a subspace PH ≡ H P ⊂ H such that for any normalized
vector ψ ∈ H P it is ψ|P|ψ = 1, and for all vectors ψ from its orthogonal complement: ψ ∈ H
⊥
P := H H P , it is ψ|P|ψ = 0. If an operator W γ would be a nonzero
projector, for all the normalized vectors ψ ∈ H W it would be ψ|W γ |ψ = 1. Such
a ψ would necessarily satisfy (7.4.24), and then ψ|W γ |ψ > 0 for any γ satisfying
(7.4.25).
For any given normalized ψ satisfying (7.4.24), the numerical function γ
2
→
ψ|W γ |ψ expressed in (7.4.60) is continuous and monotonically increasing in
195
Similar considerations could be applied also to f (t) ≡ ≡ 1 ⊗ ϕ| exp(−it (H A +
H B ))| 1 ⊗ ϕ; the spectrum of H A from (7.4.1) acting on the Hilbert space H vac of
the used representation consists of a single eigenvalue {0}, and of absolutely continuous part consisting of the interval [−2, +2] ⊂ R, which can be seen from the Sect. 7.3,
and from [36]. So the function f (t) = = 1 | exp(−it H A )| 1 · ·ϕ| exp(−it H B )|ϕ =
f 1 (t)g(t) has the Fourier image ˆ
f (u) = (2π)
−
1
2 ˆ
f 1 ∗ ˆ
g(u), which with the help of
(7.4.63) and (7.4.68f) gives
ˆ
f (u) =
1
√
2π
ˆ
f 1 ∗ ˆ
g(u) =
1
√
2π
dτ ˆ
f 1 (τ ) ˆ
g(u − τ )
(7.4.69)
=
1
2π
2
−2
dτ
4 − τ 2
τ − u
2
θ(τ − u)
4π
dφ | ˆ
ϕ(
√
τ − u, φ)|
2
.
Remember that ϕ ∈ D(R
3
), hence its Fourier image ˆ
ϕ ∈ S(R
3
) is an entire
analytic function of three complex variables, so that the function p →
4π dφ
| ˆ
ϕ( p, φ)|
2
> 0, a.e. for p > 0. Then (7.4.69) implies that ˆ
f (u) = 0 for u > 2, and
ˆ
f (u) > 0 for almost all u < 2.
For checking finally the conditions of the positivity of w(ψ) from its expression
(7.4.60), we have to check under which conditions it is | ˆ
F
0
+ (u)|θ(2 − u) > 0, u ∈
S ⊂ R, for some S of positive Lebesgue measure.
Let us assume that ˆ
F
0
+ (u) ≡ 0 in some nonzero interval: u ∈ I ⊂ R. The function ˆ
F
0
+ (u) ≡
1
√
2π
+∞
0
e
−itu
ϕ| exp(−it H B )|ψ dt, cf. (7.4.6a), can be continued
to a function analytic in the lower complex half plane Im u < 0 and continuous
on the real axis R. The identical vanishing of this function on an interval I ⊂ R
would imply (with the help of the Schwarz Reflection Principle) its analyticity on
I , and consequent vanishing everywhere in the analyticity domain, hence also on
the whole real axis (i.e. vanishing also on the boundary of the analyticity domain).
The identical vanishing ˆ
F
0
+ (u) ≡ 0, ∀u ∈ R, would imply, however, the identical
vanishing ϕ| exp(−it H B )|ψ ≡ 0, which contradicts (7.4.24). This proves that, for
γ
2
> 0 satisfying (7.4.25), it is ψ|W γ |ψ > 0, iff ψ satisfies (7.4.24). Since the condition (7.4.24) does not depend on the parameter γ, the subspace of H B consisting
of those vectors ψ for which it is ψ|W γ |ψ = 0 does not depend on γ, hence also
its orthogonal complement H W ⊂ H B is independent of γ, cf. Lemma 7.4.4.
It remains to show that, at least for some values of γ ∈ R, it is W
2
γ = W γ , i.e. that
the positive operator W γ is not a projector. For any nonzero orthogonal projector
P ∈ L(H) there exists a subspace PH ≡ H P ⊂ H such that for any normalized
vector ψ ∈ H P it is ψ|P|ψ = 1, and for all vectors ψ from its orthogonal complement: ψ ∈ H
⊥
P := H H P , it is ψ|P|ψ = 0. If an operator W γ would be a nonzero
projector, for all the normalized vectors ψ ∈ H W it would be ψ|W γ |ψ = 1. Such
a ψ would necessarily satisfy (7.4.24), and then ψ|W γ |ψ > 0 for any γ satisfying
(7.4.25).
For any given normalized ψ satisfying (7.4.24), the numerical function γ
2
→
ψ|W γ |ψ expressed in (7.4.60) is continuous and monotonically increasing in