P. Piot, PHYS 630 – Fall 2008Phase matching bandwidthIΔkPhase-matching only works exactly for one wavelength, say λ0.Since ultrashort pulses have lots of bandwidth, achievingapproximate phase-matching for all frequencies is a big issue.The range of wavelengths (or frequencies) that achieve approximatephase-matching is the phase-matching bandwidth.[ ]4( ) ( ) ( / 2)k n n!" " ""# = $0!02!WavelengthRefractive index! ne! no2 2( ) ( / ) sinc ( / 2)sigI L L k L!" #Recall that the intensity out of anSHG crystal of length L is:where:( ) )/ 2 (n n! !"2!!P. Piot, PHYS 630 – Fall 2008Phase matching bandwidth!k (") =4#"n(") $ n("/ 2)[ ]0 0 0 00 04( ) 1 ( ) ( ) ( / 2) ( / 2)2k n n n n! "# "## # "# # # ## #$ %$ %& &' = ( + ( () *) *+ ,+ ,because, when the input wavelength changes by δλ, the second-harmonic wavelength changes by only δλ/2.The phase-mismatch is:Assuming the process is phase-matched at λ0, let’s see what thephase-mismatch will be at λ = λ0 + δλx xBut the process is phase-matched at λ0 0 004 1( ) ( ) ( / 2)2k n n! "## # ##$ %& &' = () *+ ,to first orderin δλP. Piot, PHYS 630 – Fall 2008Phase matching bandwidthThe sinc2 curve will decrease by afactor of 2 when Δk L/2 = ± 1.39.So solving for the wavelengthrange that yields |Δk | < 2.78/Lyields the phase-matchingbandwidth.010 020.44 /( ) ( / 2)FWHMLn n!"!! !=# #$0 004 12.78 / ( ) ( / 2) 2.78 /2L n n L! "## ##$ %& &' < ' <( )* +IΔkFWHM2.78/L-2.78/Lsinc2(ΔkL/2)P. Piot, PHYS 630 – Fall 2008Phase matching bandwidth examplesBBO KDPThe phase-matching bandwidth is usually too small, but it increases asthe crystal gets thinner or the dispersion decreases (i.e., thewavelength approaches ~1.5 microns for typical media).The theory breaks down, however, when the bandwidthapproaches the wavelength.P. Piot, PHYS 630 – Fall 2008Group velocity mismatchInside the crystal the two different wavelengths have different groupvelocities.Define the Group-VelocityMismatch (GVM):0 01 1v ( / 2) v ( )g gGVM! !" #CrystalAs the pulse enters the crystal:As the pulseleaves the crystal:Second harmonic createdjust as pulse enters crystal(overlaps the input pulse)Second harmonic pulse lagsbehind input pulse due to GVMP. Piot, PHYS 630 – Fall 2008Group velocity mismatch0/ ( )v ( )1 ( )( )gc nnn!!!!!="#0 0 0 00 00 0 0 0( / 2) / 2 ( )1 ( / 2) 1 ( )( / 2) ( )n nn nc n c n! ! ! !! !! !" # " #$ $= % % %& ' & '( ) ( )00 001( ) ( / 2)2GVM n nc!! !" #$ $= %& '( )Calculating GVM:01 ( )1 ( )v ( ) ( )gnnc n! !!! !" #$= %& '( )So:0 01 1v ( / 2) v ( )g gGVM! !" #But we only care about GVM when n(λ0/2) = n(λ0)P. Piot, PHYS 630 – Fall 2008Effect of group velocity mismatchAssuming that a very short pulseenters the crystal, the length of the ,SH pulse, δt, will be determined bythe difference in light-travel timesthrough the crystal:!t =Lvg("0/ 2)#Lvg("0)= L GVMCrystalL GVM <<!pWe always try to satisfy:P. Piot, PHYS 630 – Fall 2008Effect of group velocity mismatchL /LDSecond-harmonic pulse shape for different crystal lengths:It’s best to use a very thin crystal. Sub-100-micron crystals are common.! LD"#pGVMInputpulseshapeLD is the crystallength thatdoubles thepulse length.P. Piot, PHYS 630 – Fall 2008Effect of group velocity mismatchP. Piot, PHYS 630 – Fall 2008Effect of group velocity mismatchLet’s compute the second-harmonic bandwidth due to GVM.Take the SH pulse to have a Gaussian intensity, for which δt δν = 0.44.Rewriting in terms of the wavelength, δt δλ = δt δν [dν/dλ]–1 = 0.44 [dν/dλ]–1 = 0.44 λ2/c0So the bandwidth is:010 020.44 /( ) ( / 2)FWHMLn n!"!! !#$ $%Calculating the bandwidth by considering the GVM yields the sameresult as the phase-matching bandwidth!2 20 0 0 00.44 / 0.44 /FWHMc ct L GVM! !"!"# =00 001( ) ( / 2)2GVM n nc!! !" #$ $= %& '( )P. Piot, PHYS 630 – Fall 2008Difference frequency generationω1ω1ω3ω2 = ω3 − ω1Parametric Down-Conversion(Difference-frequency generation)Optical ParametricOscillation (OPO)ω3ω2"signal""idler"By convention:ωsignal > ωidlerω1ω3ω2Optical ParametricAmplification (OPA)ω1ω1ω3ω2Optical ParametricGeneration (OPG)Difference-frequency generation takes many useful forms.mirror