Model
SHG. A monochromatic plane-wave pump at ω and its second harmonic at 2ω travel collinearly along z through a lossless crystal. The fields are Ej = Re{Aj(z) ei(kjz − ωjt)}, with Aj the peak amplitude. The tool integrates the intensity-normalised amplitudes aj = √(njε₀c/2) Aj, so |aj|² = Ij exactly:
da₁/dz = iκ s(z) a₂ a₁* eiΔkz, da₂/dz = iκ s(z) a₁² e−iΔkz, κ = ω deff √(2/(n₁²n₂ε₀c³)), Δk = k₂ − 2k₁.
With the same κ in both equations, d(I₁ + I₂)/dz = 0 identically. Pump depletion therefore conserves energy by construction, and photon fluxes satisfy Manley–Rowe: N₁ + 2N₂ = const (two pump photons make one SH photon). s(z) = ±1 is the sign of d(z): always +1 in a bulk crystal, a square wave of period Λ in a poled one. The solver is fixed-step RK4 (core.rk4Step). A poled crystal is integrated domain by domain so that no step crosses a sign flip.
Kerr. A pulse envelope A(z, T) in watts½ obeys the nonlinear Schrödinger equation ∂A/∂z = −(iβ₂/2)∂²A/∂T² + iγ|A|²A, with T = t − z/vg. The symmetric split-step method applies half a dispersion step exactly in the Fourier domain, then the Kerr phase exp(iγ|A|²h) exactly, then the other half dispersion step. Its global error is O(h²).
- I₁, I₂
- pump and SH intensities (W/m²); η = I₂(L)/I₁(0)
- deff
- effective nonlinear coefficient (pm/V) for the chosen polarisations and angles
- κ, Γ
- coupling (m⁻¹(W/m²)^−½) and gain coefficient Γ = κ√I₁(0) (m⁻¹)
- Δk, Lc
- wave-vector mismatch 2ω(n₂ − n₁)/c; coherence length π/|Δk|
- θ, θpm, ρ
- angle between k and the optic axis; phase-matching angle; walk-off angle of the e-wave
- Λ
- poling period; first-order QPM needs Λ = 2π/Δk = 2Lc
- β₂, γ
- group-velocity dispersion (ps²/km) and Kerr coefficient γ = n₂ω₀/(cAeff) (W⁻¹km⁻¹)
- T₀, P₀
- pulse half-width (sech: TFWHM = 1.763 T₀; Gaussian: 1.665 T₀) and peak power
- LD, LNL, N
- T₀²/|β₂|, 1/(γP₀), soliton order √(LD/LNL)
Assumptions and validity: infinite plane waves (no diffraction, no walk-off), CW or quasi-CW pump for SHG (no group-velocity mismatch), no absorption, Kleinman symmetry, and the slowly varying envelope approximation (|d²A/dz²| ≪ k|dA/dz|). The NLSE keeps β₂ and instantaneous Kerr response only: no β₃, Raman or self-steepening. The fibre is lossless and single-mode, with Aeff absorbed into γ.
Derivation: coupled equations, sinc², tanh², Manley–Rowe and QPM
Driven wave equation. With PNL(t) = 2ε₀deffE(t)², the 2ω part of E² is ½A₁²e2i(k₁z−ωt) + c.c., so the peak polarisation phasor is P2ω = ε₀deffA₁²e2ik₁z. Inserting E₂ into ∇²E − (n²/c²)∂²E/∂t² = μ₀∂²PNL/∂t² and dropping d²A₂/dz² (slowly varying envelope) gives dA₂/dz = (iω₂/(2n₂cε₀))P2ωe−ik₂z = i(ωdeff/(n₂c))A₁²e−iΔkz. The ω part of E² couples A₂A₁*, which gives dA₁/dz = i(ωdeff/(n₁c))A₂A₁*eiΔkz.
Normalisation. Substituting Aj = aj√(2/(njε₀c)) gives the symmetric form above with κ = ωdeff√(2/(n₁²n₂ε₀c³)). Then d(|a₁|² + |a₂|²)/dz = 2Re[iκX] + 2Re[iκX*] = 0 with X = a₁*²a₂eiΔkz, which is energy conservation. Since Nj = Ij/(ħωj) and ω₂ = 2ω₁, this is the same statement as N₁ + 2N₂ = const (Manley–Rowe).
Undepleted limit. If a₁ ≈ a₁(0), then a₂(L) = iκa₁(0)² ∫₀L e−iΔkzdz, so |a₂|² = κ²I₁(0)²L² sinc²(ΔkL/2): η = (ΓL)² sinc²(ΔkL/2). This formula predicts η > 1 when ΓL > 1, which signals that its assumption has failed.
Depleted, Δk = 0. Take a₁ real and a₂ = iu. Then da₁/dz = −κua₁ and du/dz = κa₁², with a₁² + u² = I₁(0). This gives du/dz = κ(I₁(0) − u²), so u = √I₁(0) tanh(Γz) and η = tanh²(ΓL) (Armstrong, Bloembergen, Ducuing and Pershan, 1962).
Birefringent phase matching. In a negative uniaxial crystal (ne < no), an extraordinary SH wave at angle θ sees 1/ne(θ)² = cos²θ/no² + sin²θ/ne². Setting ne(2ω, θ) = no(ω) gives sin²θpm = [no(ω)⁻² − no(2ω)⁻²]/[ne(2ω)⁻² − no(2ω)⁻²]. Near θpm, Δk ≈ (∂Δk/∂θ)δθ, so the sinc² in ΔkL sets an angular acceptance δθFWHM = 5.566/(L|∂Δk/∂θ|).
Quasi-phase matching. Write s(z) = Σm odd (4/(πm)) sin(2πmz/Λ). The m = 1 term contains (2/π)ei2πz/Λ/i. When 2π/Λ = Δk it cancels e−iΔkz, so on average the crystal behaves like a phase-matched one with deff → (2/π)deff, and the efficiency is multiplied by (2/π)² ≈ 0.405. Within each domain the SH still grows along an arc, which is the staircase visible in the growth plot.
Solitons. In normalised units U = A/√P₀, ξ = z/LD and τ = T/T₀, the NLSE becomes iUξ = (sgn β₂/2)Uττ − N²|U|²U. For β₂ < 0 and N = 1, U = sech(τ)eiξ/2 is an exact solution. The chirp from dispersion is cancelled by the SPM chirp at every z.