Feb 17, 2009 ... Nonlinear Optics is a collection of new optical phenomena brought about ... of
Boyd. In the later part we shall discuss nonlinear effects in optical ...
Nonlinear Optics Yaron Silberberg February 17, 2009
Chapter 1
Introduction Nonlinear Optics is a collection of new optical phenomena brought about by the availability of laser light. Its the high intensity and (sometimes) the monochromaticity of lasers that leads to most nonlinear effects, although in retrospect, some nonlinear effects can be produced by other sources. In this course we shall introduce the basic tools of nonlinear optics and review the most useful effects. In the introductory part we shall follow the text of Boyd. In the later part we shall discuss nonlinear effects in optical fibers, where the book of Agrawal provides more detailed account.
1.1
Sources:
1. Boyd, R. W. ”Nonlinear Optics”, Academic Press, (1992) 2. Agrawal, G. ”Nonlinear Fiber Optics”, Academic Press (1994) 3. Shen, Y.R. ”Principles of Nonlinear Optics”, John Wiley & Sons, New York (1984) 4. Yariv, A. ”Quantum Electronics”, John Wiley & Sons, New York (1975) 5. Siegman, A. ”Lasers”, University Science Books, 6. Allan & Eberly ”Optical Resonances and Two Level Atoms”, Dover
1
Chapter 2
Linear Optics 2.1
Atoms as Harmonic oscillators
Linear optical phenomena in dielectrics are explained by assigning n and α. The source of n and α in a given material system should be described using q.m., but most of the optics can use the classical Lorentz model, assumeing an electron bound to heavy nucleus is described as an harmonic oscillator. d2 x + mω02 x = eE(t) (2.1) dt2 A classical oscillating dipole decays. From the rate of energy loss we derive a decay time: m
τ rad =
6π²mc3 e2 ω02
(2.2)
For light, ω ' 4 × 1014 , we get τ ' 10−8 sec. This is pure radiative decay. In practice there will be nonradiative decays. In general 1/τ = 1/τrad + 1/τnr d2 x 1 dx e + + ω02 x = E(t) (2.3) dt2 τ dt m The microscopic polarization is pxi = exi . Macroscopically Px = N px for dilute collection of coherently oscillating dipoles. Note that macroscopic polarization may decay by dephasing. QM source of dipoles
2.2
Susceptibility
Assuming a monochromatic field ³ ´ ˜x e−iωt + c.c. Ex (t) = E
2
(2.4)
N e2 1 ˜x P˜x = N epx = E m ω02 − ω 2 − iω/τ
(2.5)
In a linear, isotropic medium: ˜ P˜ (ω) = χ(ω) ˜ E(ω)
(2.6)
˜ =E ˜ + 4π P˜ = (1 + 4π χ) ˜ = ²˜E ˜ D ˜E
χ(ω) ˜ = χ(ω) ˜ =
N e2 P˜ 1 = ˜ m ω02 − ω 2 − iωγ E
N e2 1 N e2 (ω0 − ω) + iγ/2 = 2ωm ω0 − ω − iγ/2 2ωm (ω0 − ω)2 + γ 2 /4 χ = χ0 + iχ00
Note: A full QM calculation for a two level atom in the linear limit (low power, no repopulation of level) gives similar result with e2 /2mω → d2ge /¯h.
2.3
QM derivation of susceptibility H = H0 − µ · E
(2.7)
with µ = −er(t). Using first-order perturbation theory, i.e. ψ = |g > +af (t)|f >, we find af (t) =
1 i¯h
Z
t
µf g E(τ )eiωf g τ dτ =
− inf
1 µf g ei(ωf g −ω0 )t ¯ ωf g − ω0 + iγf g h
(2.8)
Dipole moment µ2f g µ2f g 1 { + }E0 eiω0 t (2.9) ¯h ωf g − ω0 − iγf g ωf g + ω0 − iγf g P The macroscopic polarization P = N < p >, and in general Pi = χij Ej (ω) where µign µjng µign µjng NX { + } (2.10) χij (ω) = ¯h n ωng − ω0 − iγng ωng + ω0 − iγng < p >=< ψ|µ|ψ >=
Define Oscillator Strength: fng =
2mωng |µng |2 3¯he2 3
(2.11)
which obey the sum rule
P n
χ(ω) '
fng = 1, and then X
fng {
n
2 ωng
N e2 /m } − ω 2 − 2iωγng
(2.12)
Compared with the classical expression, the QM treatment introduces many resonances, each with its own ”oscillator strength”.
2.4
Refractive index and Absorption Coefficient
The basic wave equation in one dimension ∂ 2 E(z, t) 1 ∂ 2 E(z, t) 4π ∂ 2 P (z, t) − = ∂z 2 c2 ∂t2 c2 ∂t2 For a monochromatic wave at ω ˜ 4πω 2 ˜ ∂2E ω2 ˜ E = − + χ ˜E ∂z 2 c2 c2 E = Aeikz k2 = (
nω α ω 2 n2 ω ω2 + i )2 ≈ 2 + iαn = 2 [1 + 4π(χ0 + iχ00 )] c 2 c c c n=
2.5
p
1 + 4πχ0 ,
α = 4πk0
χ00 n
Problems
1. Derive the Kramers-Kronig relation from causality. Start with the definition Z ∞ P (t) = χ(τ )E(t − τ )dτ 0
to get the K-K- relations: Z 00 0 1 χ (ω ) 0 0 χ (ω) = ℘ dω π ω0 − ω
1 χ (ω) = − ℘ π
Z
00
χ0 (ω 0 ) 0 dω ω0 − ω
Hint: Find F [Θ(t)], where Θ(t) is the step function, and use convolution theorem. 2. Find the inpulse response χ(τ ) for the Lorentz model.
4
Chapter 3
Nonlinear Interactions 3.1
SHG as anharmonic oscillator
d2 x 1 dx e + + ω02 x + ax2 = (Ee−iwt + c.c.) (3.1) dt2 τ dt m Note implies asymmetry between x and -x. Will happen only in noncentrosymmetric materials (crystals). ¡ ¢ x = x1 e−iωt + x2 e−2iωt + c.c.
(3.2)
e 1 ˜ E. m ω02 − ω 2 − iω/τ
(3.3)
x(1) (ω) = µ (2)
x
(2ω) = −a
eE m
¶2
1 1 (ω02 − ω 2 − iω/τ )2 ω02 − 4ω 2 − i2ω/τ
(3.4)
Since P (2ω) = exN ≡ χ(2) (2ω)E 2 (ω) we get ”Miller’s Rule”: χ(2) (2ω) =
ma 2 [χ(ω)] χ(2ω) e3 N 2
(3.5)
Estimate of magnitude: assume that when x ' 3A the anharmonicity is ω02 x = ax2 then for ω0 ' 10−16 one gets χ(2) ' 10−12 m/V (or 310− 8 esu.
3.2
Nonlinear Susceptibility
Linear optics, in vector terms, Z P(r, t) = χ(ρ, τ ) · E(r − ρ, t − τ )dτ dρ
5
(3.6)
we will assume local response. Z Pi (t) =
χij (τ )Ej (t − τ )dτ,
(3.7)
In general, for any nonlinear polarization dependence on fields, PR Pi (t) = χij (τ )Ej (t − τ )dτ + PR j + χijk (τ1 , τ2 )Ej (t − τ1 )Ek (t − τ2 )dτ1 dτ2 + jk R P + χijkl (τ1 , τ2 , τ3 )Ej (t − τ1 )Ek (t − τ2 )El (t − τ3 )dτ1 dτ2 dτ3 + . . .
(3.8)
jkl
P PR R Pi (ω) = χij (ω)Ej (ω) + χijk (ω; ω1 , ω2 )Ej (ω1 )Ek (ω2 )δ(ω1 + ω2 − ω)dω1 dω2 + jk P jR + χijkl (ω; ω1 , ω2 , ω3 )Ej (ω1 )Ek (ω2 )El (ω3 )δ(ω1 + ω2 + ω3 − ω)dω1 dω2 dω3 + . . . jkl
(3.9)
3.3
Symmetry Considerations
Real Fields, Polarizations: Ej (−ω) = Ej∗ (ω). Intrinsic Permutation: Each combination of polarization and frequencies contribute several times to the resulting polarization. For example, in second order, χijk (ω3 ; ω1 , ω2 ) and χikj (ω3 ; ω2 , ω1 ) always appear together, and the convention is to take the two terms equal. For example, the sum frequency polarization induced by fields at two frequencies: X Pi (ω3 = ω1 + ω2 ) = [ χijk (ω3 ; ω1 , ω2 )Ej (ω1 )Ek (ω2 ) + (3.10) jk
+
X
χijk (ω3 ; ω2 , ω1 )Ej (ω2 )Ek (ω1 )]
jk
Pi (ω3 = ω1 + ω2 ) = 2
X
χijk (ω3 ; ω1 , ω2 )Ej (ω1 )Ek (ω2 )
(3.11)
jk
for x-polarized fields: Pi (ω3 = ω1 + ω2 ) = 2²0 χixx (ω3 ; ω1 , ω2 )Ex (ω1 )Ex (ω2 ). However, if only one field present the SHG polarization X Pi (ω3 = 2ω1 ) = χijk (ω3 ; ω1 , ω1 )Ej (ω1 )Ek (ω1 ) (3.12) jk
or, for x-polarized fields, Pi (ω3 = 2ω1 ) = ²0 χixx (ω3 ; ω1 , ω1 )Ex (ω1 )Ex (ω1 ). Note the difference by a factor of 2; It represents the number of distinct permutations of frequencies contributing to the polarization. This reflects the fact that if we 6
have √ two distinct √ fields to sum, E(ω1 ) + E(ω2 ), as ω1 − > ω2 we should ”assign” 1/ 2E(ω) + 1/ 2E(ω). For electronic nonlinearities in the transparency regime, P (t) = ²0 χ(2) E(t)E(t), that is the polarization follows the field instantaneously and is therefor dispersionless, and then χijk = χikj . Similarly, χijkl is symmetric under permutations of (jkl). It is common to use for χ(2) the contracted notation: 12 χijk = dim (ω) with m defined m jk
1 2 11 22
3 33
4 23, 32
5 6 13, 31 12, 21
(3.13)
From transparency (energy conservation) it can be shown that all permutations are equal. These are ’Kleinman symmetries’. It brings χ(2) from 27 to 10 terms, χ(3) from 81 to 15. In that case, for example d12 ≡ d122 = d212 ≡ d26 ; similarly d21 = d16 , d14 = d25 = d36 , d31 = d15 , d32 = d24 , d34 = d23 and d13 = d35 . The crystal symmetry group will cause many of these terms to vanish. (see tables in Yariv or Boyd). For KDP (KH2 PO4 ) d14 = d25 ≈ d36 = 5 · 10−13 m/V, all other terms vanish. For a fixed geometry, it is convenient to work with a scalar relationship P (ω3 ) = 4def f E(ω1 )E(ω2 )
(3.14)
for sum freq. generation, and similarly P (2ω) = 2def f E(ω)2 , where def f should be obtained from the geometry. χijk (ω; ω1 , ω2 ) = 0 in any centrosymmetric crystal. In isotropic materials, χijkl (ω; ω1 , ω2 , ω3 ) has only 3 independent terms of its 81 terms. There: xxxx = yyyy = zzzz xxyy = xxzz = yyxx = yyzz = zzxx = zzyy xyxy = yxyx = xzxz = zxzx = yzyz = zyzy xyyx = xzzx = yxxy = yzzy = zxxz = zyyz xxxx = xxyy + xyxy + xyyx
3.4
(3.15)
QM Derivation of the susceptibility
See Boyd, Chap. 3.
3.5
Problems
1. Optical rectification. Use the Lorentz model to find the strength of the optically induced dc field. ¡ ¢ 2. Lorentz model with two frequencies. Start with E = 12 E1 e−iω1 t + E2 e−iω2 t + c.c. Find expressions for the sum and difference frequency polarizations.
7
3. Prove that χxxxx = χxxyy + χxyxy + χxyyx . Find Px induced by Ex once in the princilpe coordinate system, and once in a system (x0 , y 0 ) rotated by 45◦ to the principle system.
8
Chapter 4
Second Order Effects 4.1
Slowly varying amplitude approximation
For a plane wave propagating along z: ∇2 E(z, t) −
1 ∂ 2 E(z, t) 4π ∂ 2 P (z, t) = 2 2 2 c ∂t c ∂t2
(4.1)
we will write P = χ(1) E + P N L . We will also assume a solution ³ ´ E = A(z)ei(kz−ωt) + c.c. ,
³ ´ P N L = P N L (z)ei(kz−ωt) + c.c.
(4.2)
real χ(1) : no absorption ∇2 A(z) + i2k
∂A(z) n2 ω 2 4πω 2 − k 2 A(z) + 2 A(z) = − 2 P N L (z) ∂z c c
SVA approximation: A changes slowly on a wavelength scale, so k ∂A(z) ∂t . Then : ∂A(z) 2πω N L =i P (z) ∂z nc Generalization: if wave changes in transverse dimension: i 2 2πω N L ∂A(x, y, z) − ∇ A(x, y, z) = i P (x, y, z) ∂z 2k ⊥ nc
4.2
(4.3) ∂ 2 A(z) ∂z 2
Is. However, as we know the true P (E) most of the time there is no real reason to expand.
5.4 5.4.1
Two Wave Interaction Two waves in a Kerr medium
Two waves interfere to form an intensity grating. Assume now general frequencies E = A1 (z) exp i(k1 · r−ω1 t) + A2 (z) exp i(k2 · r − ω2 t) + c.c I
=
A1 A∗1 + A2 A∗2 + A1 A∗2 exp i[(k1 −k2 )·r − (ω1 − ω2 )t] + +A∗1 A2
exp −i[(k1 −k2 )·r − (ω1 − ω2 )t]
18
(5.14) (5.15)
this grating has the right periodicity to reflect one wave into the other. If the index follow the intensity instantaneously, i.e. nN L = n2 I, then the SVA gives
dA1 dz dA2 dz
dA = ik0 n2 IA dz
(5.16)
=
ik0 n2 (I1 + 2I2 )A1
(5.17)
=
ik0 n2 (2I1 + I2 )A2
Intensities do not change: dIi d(Ai A∗i ) = =0 dz dz
(5.18)
but each of the waves aquired nonlinear phase shift: θi = k0 n2 (Ii + 2I3−i )
(5.19)
which is different for waves with unequal intensities. Even though the grating is at the Bragg periodicity, its phase is such that no energy is trasfered. In photorefractive materials there is a phase shift between the intensity grating and the index grating, that leads to energy transfer. Assume now a slugish material where τ
dnN L + nN L = n2 I dt
which is solved by nN L
n2 = τ
Z
t
0
I(t0 )e(t −t)/τ dt0
(5.20)
(5.21)
−∞
or nN L
= A1 A∗1 + A2 A∗2 +
dA1 dz dA2 dz
= =
1 1 − Ωt] A1 A∗2 exp i[q · r − Ωt] + A∗ A2 exp −i[q · r(5.22) 1 − iΩτ 1 + iΩτ 1
1 I2 A1 1 − iΩτ 1 ik0 n2 (I1 + I2 )A2 + ik0 n2 I1 A2 1 + iΩτ ik0 n2 (I1 + I2 )A1 + ik0 n2
(5.23)
The two waves can now exchange energy: dI1 dz dI2 dz
Ωτ I2 I1 1 + Ω2 τ 2 Ωτ = 2k0 n2 I1 I2 1 + Ω2 τ 2 =
−2k0 n2
19
(5.24)
5.4.2
Two Waves in a Saturable Absorber
Two waves propagate at small angles ±θ to the z axis, enter a medium with saturable absorber with initial small signal absorption α0 L. (a) Find the intensity of the emerging waves as a function of their input intensity I1 and I2 . (b) Find the intensity of the first and second order diffracted waves. E(ω) = A1 (z) exp(ik1 r) + A2 (z) exp(ik2 r) + c.c
(5.25)
I = A1 A∗1 + A2 A∗2 + A1 A∗2 exp[i(k1 − k2 ) · r] + A∗1 A2 exp[−i(k1 − k2 ) · r] (5.26) µ ¶ α0 I I2 dA1 =− 1− + 2 − ... E termswith exp(ik1 r) (5.27) dz 2 Is Is use I in units of Is, dA1 α0 α0 = − A1 + (I1 + 2I2 )A1 dz 2 2
(5.28)
It might be thought that strong enough intensity will result in gain. However, this results hold only for α0
