Page 444 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 425

at frequency 2 lead to the generation of an electromagnetic wave at 2. Crystals such as quarts have no center
of symmetry and, therefore, they have a nonzero (2) coefficient. In the case of optical glass fibers, SiO 2
(silica) is a symmetric molecule and (2) is zero. Therefore, second harmonic generation does not normally
occur in optical fibers. The third-order susceptibility (3) is responsible for third harmonic generation and the
Kerr effect.
Suppose the incident electromagnetic field has only E and H components. For a centrally symmetric
y
x
dielectric material, the tensor equation (10.41) can be simplified to obtain
(1) (3) 3
P = E + E , (10.42)
x 0 xx x 0 xxxx x
(3) (3)
where xxxx is a component of the fourth-rank tensor . Suppose the incident optical field is a monochro-
matic wave,
E = E exp (−it). (10.43)
x 0
3
To find E , as pointed out in Section 1.6.2, we should first take the real part of E ,
x
x
1 ∗
Re[E ]= [E exp (−it)+ E exp (it)], (10.44)
x 0 0
2
1 3 ∗3
3
{Re[E ]} = {E exp (−3it)+ E exp (3it)
x
0
0
8
2
∗
+ 3|E | [E exp (−it)+ E exp (it)]}. (10.45)
0
0
0
From Eqs. (10.42) and (10.45), we find that the incident field oscillating at frequency leads to a component
of polarization oscillating at frequency 3, which is responsible for third harmonic generation. The electro-
magnetic wave at frequency 3 becomes significant only when special phase-matching techniques are used.
Otherwise, the component of polarization at frequency 3 can be ignored. Hence, we ignore the first two
terms on the right-hand side of Eq. (10.45).
Let the polarization at frequency be
P = P exp (−it), (10.46)
x 0
1 ∗
Re[P ]= [P exp (−it)+ P exp (it)]. (10.47)
x 0 0
2
From Eq. (10.42), we have
(1) (3) 3
Re[P ]= Re[E ]+ Re[E ] , (10.48)
x
0 xxxx
0 xx
x
x
where the imaginary parts of the susceptibility are ignored. Substituting Eqs. (10.44) and (10.45) into
Eq. (10.48), collecting the terms that are proportional to exp (−it), and comparing it with Eq. (10.47), we
obtain ( 2 )
3 E |
|
(1) (3)
P = + | 0| E = E , (10.49)
0 0 xx xxxx 0 0 eff 0
4
where is the effective susceptibility that includes both linear and nonlinear susceptibilities. From Eq.
eff
(1)
2
(10.29), in the absence of nonlinearity, we have n = 1 + . Now, we modify it as
xx
2
0
2 (1) 3|E | (3)
n = 1 + = 1 + + (10.50)
eff xx xxxx
4
2
(3)
0
2
= n + 3|E | xxxx , (10.51)
0 4