7.4 Particle Detection—A “Nonideal” Measurement
193
We can rewrite this expression to the forms
f 1 (t) =
1
2π
2
−2
e
itu
4 − u 2 du
=
1
2π
+∞
−∞
e
itu
θ(2 − |u|)
4 − u 2 du
=
F
−1
1
√
2π
θ(2 − |u|)
4 − u 2
(t),
(7.4.62)
hence, we obtain from (7.4.62) the wanted Fourier image immediately:
ˆ
f 1 (u) ≡ F( f 1 )(u) =
1
√
2π
θ(2 − |u|)
4 − u 2 .
(7.4.63)
The expression (7.4.61) of f 1 leads, in agreement with its definition (7.4.7), to the
estimates
| f 1 (t)| ≤
4
π
1
0
dx| cos(2t x)|
1 − x 2 ≤
4
π
1
0
dx
1 − x 2
=
4
π
π
2
0
dα cos
2
α = 1,
(7.4.64)
where we used the change of the integration variable x := sin α, the identity
sin
2
α + cos
2
α ≡ 1, and the symmetry properties of the goniometric functions. Since
both functions f 1 , g are continuous, g(t) = (ϕ, exp(−it H B )ϕ), ϕ ∈ D(R
3
) ⇒ the
Fourier image ˆ
ϕ ∈ S(R
3
) is an entire analytic function of three complex variables
[262, Theorem IX.12], the function t → g(t) = 0 (a.e. for t ∈ R) according to
Lemma 7.4.4, and the continuous function f 1 (t) is not constant, hence the function | f 1 (t)| < 1 on certain intervals of R, the estimate for L
1 -norms gives:
f 1 ≡ ≡ f 1 · g 1 < g 1 ,
(7.4.65)
hence we have here obtained the sharp inequality. From the definition of the Fourier
transformation it is seen that the following trivial inequality is valid for any function
h ∈ L
1
(R):
ˆ
h ∞ ≤
1
√
2π
h 1 .
(7.4.66)
These considerations give an estimate for the denominator in (7.4.60) by
2πγ
2
ˆ
g + ˆ
f + ∞ ≤ 2πγ
2
ˆ
g + ∞ ˆ
f + ∞ ≤ γ
2
g + 1 f + 1
=
γ
2
4
g 1 f 1 <
γ
2
4
g
2
1 .
(7.4.67)