Frequency_doubling_with_imperfect_phase_matching.gif


Summary

Description
English: Numerical solution of equations 2.7.10 and 2.7.11 on Boyd's "Nonlinear optics". If the phase matching is not perfect having your crystal too long can be detrimental, as the energy flows back into the pump.
Date
Source https://twitter.com/j_bertolotti/status/1196849637904322561
Author Jacopo Bertolotti
Permission
( Reusing this file )
https://twitter.com/j_bertolotti/status/1030470604418428929

Mathematica 11.0 code

d = 0.05; c = 1; \[Omega]1 = 1; \[Omega]2 = 2 \[Omega]1; n1 = 1; n2 = 1.1;
k1 = (n1 \[Omega]1)/c; k2 = (n2 \[Omega]2)/c; \[CapitalDelta]k =  2 k1 - k2;
K1 = (2 I \[Omega]1^2 d)/(k1 c^2); K2 = (I \[Omega]2^2 d)/(k2 c^2); 
sol = NDSolve[{A1'[z] == K1 Conjugate[A1[z]] A2[z] E^(-I \[CapitalDelta]k z), A2'[z] == K2 (A1[z])^2 E^(I \[CapitalDelta]k z), A2[0] == 0, A1[0] == 1}, {A1[z], A2[z]}, {z, 0, 50}]

t1 = Re[Evaluate[(A1[z] /. sol) /. {z -> 20}] E^(I (k1 (z) - \[Omega]1 t))];
t2 = Re[(Evaluate[A2[z] /. sol /. {z -> 20}]) E^(I (k2 (z) - \[Omega]2 t))];
p1 = Table[
   Show[
    Plot[{Re[ E^(I (k1 z - \[Omega]1 t))], 0}, {z, -10, 0}, PlotStyle -> {Purple, Orange},  Axes -> False], Plot[{Re[Evaluate[A1[z] /. sol] E^(I (k1 z - \[Omega]1 t))], Re[Evaluate[A2[z] /. sol] E^(I (k2 z - \[Omega]2 t))]}, {z, 0, 20}, PlotPoints -> 40, PlotStyle -> {Purple, Orange}, PlotLegends -> LineLegend[{"\[Omega]=\!\(\*SubscriptBox[\(\[Omega]\), \(1\)]\)", "\[Omega]=2\!\(\*SubscriptBox[\(\[Omega]\), \(1\)]\)"}]],
    Plot[ t1, {z, 20, 30}, PlotStyle -> {Purple}], Plot[ t2, {z, 20, 30}, PlotStyle -> {Orange}], PlotRange -> All, Prolog -> {LightGray, Polygon[{{0, 1}, {0, -1}, {20, -1}, {20, 1}, {0, 1}}]}, 
    PlotLabel -> "\!\(\*SuperscriptBox[\(\[Chi]\), \((2)\)]\) crystal", LabelStyle -> {Black, Bold}
    ]
   , {t, 0, (2 \[Pi])/\[Omega]1, (2 \[Pi])/\[Omega]1 1/20}];
ListAnimate[p1]

Licensing

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Captions

Schematic of frequency doubling (SHG) with imperfect phase matching.

19 November 2019

image/gif