Nonlinear Fiber Optics - 4 ed. Agrawal

192 Chapter 6. Polarization Effects
it is better to write Eqs. (6.3.1) and (6.3.2) in terms of linearly polarized components
using Eq. (6.1.14). The resulting equations are
dĀ x
dz − i 2 (Δβ)Ā x = 2iγ (|Ā x | 2 + 2 )
3 3 |Ā y | 2 Ā x + iγ 3 Ā∗xĀ2 y, (6.3.14)
dĀ y
dz + i 2 (Δβ)Ā y = 2iγ
3
(|Ā y | 2 + 2 3 |Ā x | 2 )
Ā y + iγ 3 Ā∗ yĀ2 x. (6.3.15)
These equations can also be obtained from Eqs. (6.1.11) and (6.1.12).
At this point, we introduce the four real variables known as the Stokes parameters
and defined as
S 0 = |Ā x | 2 + |Ā y | 2 , S 1 = |Ā x | 2 −|Ā y | 2 ,
S 2 = 2Re(Ā ∗ xĀy), S 3 = 2Im(Ā ∗ xĀy),
(6.3.16)
and rewrite Eqs. (6.3.14) and (6.3.15) in terms of them. After considerable algebra, we
obtain
dS 0
dz = 0, dS 1
dz = 2γ
3 S 2S 3 , (6.3.17)
dS 2
dz = −(Δβ)S 3 − 2γ
3 S dS 3
1S 3 ,
dz =(Δβ)S 2. (6.3.18)
It can be easily verified from Eq. (6.3.16) that S0 2 = S2 1 + S2 2 + S2 3 . As S 0 is independent
of z from Eq. (6.3.17), the Stokes vector S with components S 1 , S 2 , and S 3
moves on the surface of a sphere of radius S 0 as the CW light propagates inside the
fiber. This sphere is known as the Poincaré sphere and provides a visual description of
the polarization state. In fact, Eqs. (6.3.17) and (6.3.18) can be written in the form of a
single vector equation as [33]
dS
= W × S, (6.3.19)
dz
where the vector W = W L + W NL such that
W L =(Δβ,0,0), W NL =(0,0,−2γS 3 /3). (6.3.20)
Equation (6.3.19) includes linear as well as nonlinear birefringence. It describes evolution
of the polarization state of a CW optical field within the fiber under quite general
conditions.
Figure 6.6 shows motion of the Stokes vector on the Poincaré sphere in several
different cases. In the low-power case, nonlinear effects can be neglected by setting
γ = 0. As W NL = 0 in that case, the Stokes vector rotates around the S 1 axis with an
angular velocity Δβ (upper left sphere in Figure 6.6). This rotation is equivalent to the
periodic solution given in Eq. (6.3.3) obtained earlier. If the Stokes vector is initially
oriented along the S 1 axis, it remains fixed. This can also be seen from the steady-state
(z-invariant) solution of Eqs. (6.3.17) and (6.3.18) because (S 0 ,0,0) and (−S 0 ,0,0)
represent their fixed points. These two locations of the Stokes vector correspond to the
linearly polarized incident light oriented along the slow and fast axes, respectively.