Page 23 - Phase Space Optics Fundamentals and Applications

4 Chapter One

where we have assumed that the stochastic process is temporally sta-
tionary. After Fourier transforming the mutual coherence function
˜ (r 1 , r 2 , ), we get the mutual power spectrum 15,16 or cross-spectral
density: 17

(r 1 , r 2 , ) = ˜ (r 1 , r 2 , ) exp(i2 ) d =: (r 1 , r 2 ) (1.7)
The basic property 16,17 of (r 1 , r 2 ) is that it is a nonnegative definite
Hermitian function of r 1 and r 2 , i.e.,

∗
(r 1 , r 2 ) = (r 2 , r 1 ) and g(r 1 ) (r 1 , r 2 )g (r 2 ) dr 1 dr 2 ≥ 0
∗
(1.8)
for any function g(r). The input-output relationship can now be for-
mulated in the temporal-frequency domain as

∗
o (r 1 , r 2 ) = h(r 1 , 1 ) i ( 1 , 2 ) h (r 2 , 2 ) d 1 d 2 (1.9)
which expression replaces Eq. (1.5). Note that in the completely co-
∗
herent case, for which (r 1 , r 2 ) takes the product form f (r 1 ) f (r 2 ), the
coherence is preserved and Eq. (1.9) reduces to Eq. (1.5).

1.2.3 Some Basic Examples
of Optical Signals
Importantbasicexamples of coherentsignals,asthey appearinaplane
z = constant, are as follows:

1. An impulse in that plane at position r o , f (r) = (r−r o ). In optical
terms, the impulse corresponds to a point source.
2. The crossing with that plane of a plane wave with spatial
t
frequency q , f (r) = exp(i2 q r). The plane wave example
o o
shows us how we should interpret the spatial-frequency vec-
tor q . We assume that the wavelength of the light equals o ,
o
in which case the length of the wave vector k equals 2 / o .If
t
we express the wave vector in the form k = [k x ,k y ,k z ] , then
t
t
2 q = 2 [q x ,q y ] = [k x ,k y ] is simply the transversal part of k,
o
that is, its projection onto the plane z = constant. Furthermore, if
the angle between the wave vector k and the z axis equals , then
the length of the spatial-frequency vector q equals sin / o .
o
3. The crossing with that plane of a spherical wave (in the parax-
t
ial approximation), f (r) = exp(i r Hr), whose curvature is