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Nonlinear Optics Suggested Reading Fundamentals of Photonics – Saleh and Teich 1991 and 2006 Nonlinear Optics – R.W. Royd Principles of Nonlinear Optics – Butcher and D. Cotter Applied Nonlinear Optics – Zernike and Midwinter Handbook of Nonlinear optics – Sutherland 2003 Great web sites – used lots of figures/slides from these http://www.ph.surrey.ac.uk/intranet/undergraduate/3mol http://www.physics.gatech.edu/gcuo/UltrafastOptics/index.html What are nonlinear optical effects? Example: Sending infrared light into a crystal yielded this display of green light (second-harmonic generation): Nonlinear optics allows us to change the colour (frequency) of a light beam, to change its shape in space and time, and to create ultrashort laser pulses, the shortest events ever made by Man. NL is key element for optical data processing First demonstration of second-harmonic generation P.A. Franken (1961) The second-harmonic beam was very weak because the process was not phase-matched. First demonstration of second-harmonic generation The actual published results… Why don't we see nonlinear optical effects in our daily life? 1. Intensities of daily life are too weak. Sunlight 102 W/m2 Nonlinear effects ~ 1013 W/m2 2. Normal light sources are incoherent. 3. Some NLO effects require specific crystal symmetries 4. “Phase-matching” is required, and it doesn't usually happen on its own. What are examples of NLO effects? Linear optics: shine light of a given frequency on a material and will find light of same frequency at output with perhaps some absorption NLO Examples: the medium adds and subtracts light SHG Medium ω (IR) 2ω (Green) ω1 SHG Medium ω1+ω2 ω2 THG Medium ω (IR) 3ω (UV) Can get sum 1064 nm + 532 nm => 355nm and difference 500nm-700nm=1500nm Difference-frequency generation ω1 ω2 = ω3 − ω1 Green=UV-Blue ω3 ω1 ω1 ω3 ω2 Optical Parametric Amplification (OPA) Input high power signal at high frequency ω3 and a weaker signal at ω1 can achieve amplification of the signal at ω1 and generate signal at ω2 And Frequency Sum ω2 = ω3 + ω1 Non co-linear – third order Spatial effects: Third order example: two different input beams, whose frequencies can be different. So in addition to generating the third harmonic of each input beam, the medium will generate interesting sum and difference frequencies with spatial separation. ω2 THG medium Signal #1 2ω1 +ω2 2ω2 +ω1 ω1 Signal #2 Self-diffraction … but the frequencies don’t have to be different to generate new optical fields propagating in different directions… Signal #1 ω Nonlinear medium ω ω Signal #2 ω Self-focusing Consequence of OPTICAL KERR EFFECT Collimated beam of light propagating through a NLO medium is brought to a focus: the light causes the medium to act like a lens Optical computing and optical data processing Nonlinear absorption Two-photon absorption detectors Saturable Absorbers Optical limiting Nonlinearity key element for optical switches and optical bistability Optical logic gates, flip-flops Phase conjugation Reflection of a plane wave from an ordinary and phase conjugate mirror Reflection of a spherical wave from an ordinary and phase conjugate mirror Phase conjugate mirror: Light reflected behaves as if time reversal occurring Questions Many different types of NLO effects... Why do nonlinear-optical effects occur? How can we describe them? How can we use them? Second Order Effects: Second-harmonic generation Sum- and difference-frequency generation Autocorrelation Third Order Non-linear Effects: Frequency generation, Nonlinear refractive index, Selfphase modulation, Phase conjugation… Consequences and Applications Nonlinear Response Generic example: describe response with a polynomial 2 3 4 R = ζ 1S + ζ 2 S + ζ 3 S + ζ 4 S + ... Example: R = ζ 1S + ζ 3 S 3 S = V cos ωt R = ζ 1V cos ωt + ζ 3V 3 cos3 ωt cos3 ωt = 34 cos ωt + 14 cos 3ωt R = (ζ 1V + 34 ζ 3V 3 ) cos ωt + 14 ζ 3V 3 cos 3ωt Small cubic nonlinearity gives rise to modified response at ω and generates a new frequency component at 3ω Representation of Nonlinearity Linear R = ξ1S S R R S s Non-Linear R S R = ξ1S − ξ 2 S 2 S R The Fourier components R The same frequency as the stimulus Double the frequency of the stimulus A DC component Describing light interacting with matter r r r r r ∂B ∇× E = − Maxwell’s equations ∇ ⋅ E = 0 ∂t in a medium: r r r r r r ∂D ∇⋅B = 0 ∇ × B = µ0 J + µ0 ∂t These equations reduce to the wave equation: 2 2 1 E P ∂ ∂ 2 ∇ E − 2 2 = µ0 2 c ∂t ∂t The “Inhomogeneous Wave Equation” induced polarization, P, is what contains the effect of the medium on the EM wave propagation And the form of P determines the solution Linear optics For low light intensity, the polarization is proportional to the incident field: c = ε 0 χX Optical polarization of dielectric crystals – mostly due to outer loosely bound valence electrons displaced by the optical electric field. Separation of charges gives rise to a dipole moment P(t)=-Nex(t) Polarization = dipole moment per unit volume Polarization is alternating with the same frequency as the applied E field. Electron oscillates about the equilibrium position – oscillating dipole is a source of EM radiation. + - E Solving the wave equation in the presence of linear induced polarization For low irradiances, the polarization is proportional to the incident field: c = ε 0 χX In this simple (and most common) case, the wave equation becomes: ∂2E 1 ∂2E ∂2P − 2 2 = µ0 2 2 ∂z c ∂t ∂t ∂2E 1+ χ ∂2E 1 − 2 = 0; ε 0 µ 0 = 2 2 2 ∂z c ∂t c 1 1+ χ c = ; n = = 1+ χ 2 2 v c v This equation has the solution: X ( z, t ) ∝ E cos(ωt − k z) 0 where ω = v k and v = c /n and n = (1+χ)1/2 The induced polarization, at the same frequency as the incident field and only changes the refractive index. FAMILIAR If only the polarization contained other frequencies… Lorentz Model Lorentz model – analogous to a mass on a spring Electron of mass, m, and charge, e, is attached to the ion by a spring. r r ∑ F = ma Frestoring + Fdamping + Fapplied = ma dU Frestoring = − dx 2 2 3 4 1 1 1 U = 2 mω0 x + 3 mζ 2 x + 4 mζ 3 x + ... Cont… U = 12 mω02 x 2 + 13 mζ 2 x 3 + 14 mζ 3 x 4 + ... Frestoring P(t)=-Nex(t) x(t) is small, harmonic potential regime 2 restoring 0 F dU =− = −mω02 x − mζ 2 x 2 − mζ 3 x 3 − ... dx = − mω x = −kx Happens when applied field <<< internal field Internal field due to nucleus ~1011 V/m Sunlight for example ~1 kV/m Consequently, for small E, the linear approximation is very accurate (1) P (t ) = ε 0 χ E (t ) Nonlinear optics and anharmonic oscillators For field strength > 1 kV/m (i.e. lasers), x(t) is sufficiently large that the potential of the electron or nucleus (in an atom/molecule) is not a simple harmonic potential. In an anharmonic potential: Polarization expanded as a power series in E to give: r r r r (1) (2) 2 ( 3) 3 P = ε 0 χ E + χ E + χ E + ... ( χ(2) = 2nd order susceptibility χ(3) = 3rd order susceptibility In order for the series to converge: χ(3)E3<< χ(2)E2<< χ(1)E Access for optical intensities I ~1013 W/m2 ) Nonlinear Optics Regime Response (polarization) is nonlinear with respect to the stimulus (applied electric field) When polarization term in the inhomogeneous wave equation has higher order terms it can now drive solutions at new frequencies 2 2 1 ∂ ∂ E P 2 ∇ E − 2 2 = µ0 2 ∂t c ∂t i.e. Nonlinear optics is what happens when the polarization contains higher-order (nonlinear!) terms Generating nonlinear effects r r ( 2) 2 P = ε0χ E Example the term can generate light at many other frequencies, for example for a single input field: Describing the input field: [ E = E0 cos(kz − ωt ) = 12 E0 ei ( kz −ωt ) + e − i ( kz −ωt ) ] E = 12 E0 eikz e −iωt + 12 E0 e −ikz eiωt E = E1e −iωt + E1 eiωt * atom or molecule Generated fields from polarization term P = ε 0 χ ( 2) E 2 P = ε0χ ( 2) [E e 2 − i 2 ωt 1 * 2 i 2 ωt 1 +E e ]+ ε χ Second harmonic term 0 ( 2) * 1 2 E1 E input photons D.C. component different colour! emitted photons More than one field For second order can have up to two input fields interacting E1 = E 01e iω1t + E 01 e −iω1t * Input fields E 2 = E 02 e iω2t + E 02 e −iω2t * ( P = ε 0 χ ( 2 ) E 2 = ε 0 χ ( 2 ) (E1 + E 2 ) = ε 0 χ ( 2) E1 + E 2 + 2 E1 E 2 2 P= ε0χ ( 2) ε0χ ( 2) [E [E 2 i 2ω1t 01 e 2 i 2ω 2 t 02 [ [E e *2 ] ]+ + E 01 e − 2iω1t + *2 2 Second Second + E 02 e − 2iω2t 2ε 0 χ ( 2 ) * i (ω1 −ω 2 ) t −i (ω1 −ω 2 ) t E e + E E e 01 02 01 02 * * * 2ε 0 χ ( 2 ) E 01 E 01 + 2ε 0 χ ( 2 ) E 02 E 02 * D.C. ) harmonic of first frequency harmonic of second frequency ] ]+ 2ε 0 χ ( 2 ) E 01 E 02 e i (ω1 +ω2 )t + E 01 E 02 e −i (ω1 +ω2 ) t + * 2 Frequency Frequency sum difference components for each field What is the form of the higher order susceptibility terms? r r ∑ F = ma Frestoring + Fdamping + Fapplied = ma dx Fdamping = −2mγ dt Fapplied = −eE (t ) Susceptibility is crucial in determining the strength of the nonlinear signal Back to the Lorentz Model U = 12 mω02 x 2 + 13 mζ 2 x 3 + 14 mζ 3 x 4 + ... dU Frestoring = − = −mω02 x − mζ 2 x 2 − mζ 3 x 3 − ... dx d 2x dx −e 2 2 3 + 2γ + ω0 x + ζ 2 x + ζ 3 x + .. = E (t ) 2 dt dt m Cont.. Solving for x, no longer a linear solution so we can look for a solution in the form of a power series Use same form for field as before x = x1 + x2 + x3 + ... xl = al E l E = E (ω )e −iωt + c.c.; use E = E (ω )e −iωt x = a1 E (ω )e −iωt + a2 E 2 (ω )e −i 2ωt + ... Equate Fourier components on both sides of our equation, i.e. same order of E on both sides of the equation −e d 2x dx 2 2 3 + 2γ + ω0 x + ζ 2 x + ζ 3 x + .. = E (t ) 2 dt dt m d 2 x1 dx1 −e 2 + 2γ + ω0 x1 = E (t ) (1) Harmonic oscillator equation 2 dt dt m d 2 x2 dx2 2 2 + 2 γ + ω x + ζ x (2) Has first anharmonic term 0 2 2 1 =0 2 dt dt Solving equation (1) 2 d x1 dx1 −e 2 + 2γ + ω0 x1 = E (t ) (1) 2 dt dt m x1 = a1 E (ω )e − iωt dx1 = −iωa1 E (ω )e −iωt dt d 2 x1 2 − iωt ( ) i a E e = − ω ω 1 2 dt −e a1 E (ω )e ω − 2iγω − ω = E (ω )e −iωt m − eE (ω )e −iωt ⇒ x1 = m ω02 − 2iγω − ω 2 − iωt [ [ 2 0 2 ] ] Cont… The polarization: P = − Nex = − Ne∑ xl = ε 0 ∑ χ ( l ) E l = ∑ Pl l P1 = ε 0 χ (1) E (ω )e −iωt χ (1) l Ne 2 E (ω )e − iωt = − Nex1 = m ω02 − 2iγω − ω 2 Ne 2 = 2 2 mε 0 ω0 − 2iγω − ω [ l [ ] ] The Linear Susceptibility Linear susceptibility χ (1) Ne 2 = ε 0 m ω02 − 2iγω − ω 2 [ ] ω0 Optical Frequency, ω c n= = v µ0ε ε = = εr = ε0 µ0ε 0 (1 + χ ) (1) Refraction and Absorption The refractive index is a complex quantity n = 1+ χ (1) n = n0 − iκ ) ( α (ω ) κ = Im( 1 + χ ) = 4π n0 = Re 1 + χ (1) (1) Dispersion, frequency dependent speed of propagation i.e. proportional to the material absorption If we are far from the absorption, the imaginary parts are negligible. In the linear case the dipoles and the polarization oscillate at the same frequency as the incident field. Solving for the second order susceptibility d 2 x2 dx2 2 2 + 2γ + ω0 x2 + ζ 2 x1 = 0 (2) 2 dt dt x1 = a1 E (ω )e −iωt 2 2 2 x1 = a1 E (ω )e − i 2 ωt x2 = a2 E 2 (ω )e −i 2ωt dx2 = −i 2ωa2 E 2 (ω )e −i 2ωt dt 2 d x2 2 2 − i 2ω t ω ω = − i 4 a E ( ) e 1 dt 2 Cont… d 2 x2 dx2 2 2 + 2γ + ω0 x2 + ζ 2 x1 = 0 (2) 2 dt dt Fill into 2 a2 E (ω )e − i 2 ωt [ω − 4iγω − 4ω ] = −ξ a 2 2 0 2 1 2 2 E (ω )e x2 = a2 E 2 (ω )e −i 2ωt − ξ 2 a1 E 2 (ω )e −i 2ωt x2 = 2 ω0 − 4iγω − 4ω 2 2 [ ] −e a1 = 2 m ω0 − 2iγω − ω 2 [ ] − ξ 2 E (ω )e −e x2 = 2 2 2 2 ω0 − 4iγω − 4ω m ω0 − 2iγω − ω 2 [ −i 2ωt ] [ ] 2 − i 2ωt Χ(2) P2 = − Nex2 = ε 0 χ ( 2 ) E 2 2 − ξ 2 E (ω )e −e x2 = 2 2 2 2 ω0 − 4iγω − 4ω m ω0 − 2iγω − ω 3 Ne ξ2 ( 2) χ = 2 2 2 2 ε 0 m ω0 − 4iγω − 4ω ω0 − 2iγω − ω 2 2 [ − i 2 ωt ] [ [ χ (1) Ne 2 = 2 ε 0 m ω0 − 2iγω − ω 2 [ ] ][ ] 2 ] 3 Ne ξ 2 (1) ( 2) (1) (1) χ (2ω ) = χ ( ω ) χ ( ω ) χ (2ω ) 2 ε 0m Strength of the polarization, which is the source term for fields at new frequencies, depends on linear susceptibility at applied field frequency and the generated field frequency Many interacting fields r ( 2) r r (2) P = ∑∑ ε 0 χ (ωn ,ωm ) E(ωn ) E(ωm ) e − i (ωn +ωm ) t n χ (2) ( ω n ,ω m ) = m 2 0 ε mξ ( 2) 2 3 N e χ ((ω1) ) χ ((ω1) ) χ ((ω1) +ω n m n m) The 2nd order: interaction of two fields producing a third Result: all frequencies (ωn+ ωm) for all possible values of n,m For just two fields n,m = ±1,±2 Will get all the terms we had before: DC component Second harmonic generation Frequency sum and frequency difference terms Third order susceptibility Continuing our analysis of the anharmonic oscillator, for 1 driving field we will find χ ( 3) = 3 0 ε mξ ( 3) 3 4 N e (1) (ω ) (1) (ω ) (1) (ω ) χ χ χ χ (1) ( 3ω ) Third order: 3 input fields producing a fourth Nonlinear polarization will find (ωn+ ωm+ ωp) for all possible values of n,m,p = ±1,±2, ±3 r ( 3) r r r − i (ω n +ω m +ω p ) t ( 3) P = ∑∑∑ ε 0 χ (ωn ,ωm ,ω p ) E(ωn ) E(ωm ) E(ω p ) e n χ ( 3) ( ω n ,ω m ,ω p ) m = p 3 0 ε mξ ( 3) 3 4 N e χ ((ω1) ) χ ((ω1) ) χ ((ω1) ) χ ((ω1) +ω n m p n m +ω p ) Third harmonic generation and many frequency sum/frequency difference terms.