Ikkinchi garmonika generatsiyasi
Ikkinchi garmonikaning hosil boʻlishi nochiziqli optikaning asosiy elementlaridan biri boʻlib, lazer nurlanishlarini hosil qilishda muhim rol oʻynaydi. Avval kvadratik nochiziqli muhitda amplitudasi sekin oʻzgaruvchi toʻlqinlar uchun tenglamalarni keltirib chiqaramiz. Buning uchun Makswell tenglamalarini yozamiz: rot E − 1 c ∂ B ∂ t ,       div B = 0 {\displaystyle {\textrm {rot}}{\textbf {E}}-{\dfrac {1}{c}}{\dfrac {\partial B}{\partial t}},\ \ \ {\textrm {div}}{\textbf {B}}=0} rot H = 1 c ∂ D ∂ t ,       div D = 0 {\displaystyle {\text{rot}}{\textbf {H}}={\dfrac {1}{c}}{\dfrac {\partial D}{\partial t}},\ \ \ {\text{div}}{\textbf {D}}=0} Bu tenglamalarning birinchisidan rotor olib uchinchi tenglama yordamida undan magnit maydonini yoʻqotamiz. Natijada elektr maydon uchun quyidagi tenglamalar sistemasini olamiz: { Δ E + grad   div E = 1 c 2 ∂ 2 D ∂ t 2 div D = 0 {\displaystyle {\begin{cases}\Delta {\textbf {E}}+{\text{grad}}\ {\text{div}}{\textbf {E}}={\dfrac {1}{c^{2}}}{\dfrac {\partial ^{2}D}{\partial t^{2}}}\\{\text{div}}{\textbf {D}}=0\end{cases}}} Bu yerda muhit dielektrik boʻlgani uchun magnit singdiruvchanlik μ = 1 {\displaystyle \mu =1} deb olindi. Toʻlqinning elektr maydonini Furye integraliga yoyamiz: E α ( r , t ) = 1 2 π ∫ − ∞ ∞ E α ( r , ω ) e i ω t d ω {\displaystyle E_{\alpha }({\textbf {r}},t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }E_{\alpha }({\textbf {r}},\omega )e^{i\omega t}d\omega } Buni yuqoridagi ikkita tenglamalarga qoʻyib quyidagini hosil qilamiz: div ( ε ( ω ) E ( r , ω ) ) = − 4 π div P n c h ( r , ω ) {\displaystyle {\text{div}}(\varepsilon (\omega ){\textbf {E}}(r,\omega ))=-4\pi {\text{div}}{\textbf {P}}^{nch}(r,\omega )} Δ E ( r ω ) − grad ( div E ( r , ω ) ) + ω 2 ε ( ω ) c 2 E ( r , ω ) = 4 π ω 2 c 2 P n c h ( r , ω ) {\displaystyle \Delta {\textbf {E}}(r\omega )-{\text{grad}}({\text{div}}{\textbf {E}}(r,\omega ))+{\dfrac {\omega ^{2}\varepsilon (\omega )}{c^{2}}}{\textbf {E}}(r,\omega )={\dfrac {4\pi \omega ^{2}}{c^{2}}}{\textbf {P}}^{nch}(r,\omega )} Chastotasi ω {\displaystyle \omega } va toʻlqin vektori k boʻlgan lazer hosil qilayotgan yorugʻlik dastasining muhitga taʼsirini koʻrib chiqamiz. Asosiy garmonikadagi nurlanish intensivligini pasayishini inobatga olmaslik uchun ikkinchi garmonikaning intensivligini kichik deb faraz qilamiz. Qutblanish vektorining nochiziqli qismini va elektr maydonni quyidagi koʻrinishda yozamiz: E ( z , t ) = e ^ E ( ω 1 ) e i ( k 1 z − ω 1 t ) + e ^ E ∗ ( ω 1 ) e − i ( k 1 z − ω 1 t ) {\displaystyle {\textbf {E}}(z,t)={\hat {e}}{\textbf {E}}(\omega _{1})e^{i(k_{1}z-\omega _{1}t)}+{\hat {e}}{\textbf {E}}^{*}(\omega _{1})e^{-i(k_{1}z-\omega _{1}t)}} P α n c h ( z , t ) = E 2 ( ω 1 ) ∑ β , γ χ α β γ e ^ β e ^ γ e 2 i ( k 1 z − ω 1 t ) + k . q . … {\displaystyle P_{\alpha }^{nch}(z,t)=E^{2}(\omega _{1})\sum \limits _{\beta ,\gamma }\chi _{\alpha \beta \gamma }{\hat {e}}_{\beta }{\hat {e}}_{\gamma }e^{2i(k_{1}z-\omega _{1}t)}+k.q.\ldots } Bu yerda e ^ {\displaystyle {\hat {e}}} toʻlqinning qutblanish yoʻnalishidagi birlik vektor. Izotrop dielektrikda toʻlqin z {\displaystyle z} oʻqi boʻylab tarqalganda, elektr maydon ( x y {\displaystyle xy} ) tekisligida yotadi. Maʼlumki, tushayotgan toʻlqin koʻndalang boʻlishiga qaramasdan, muhit qutblanish vektorining koʻndalang va boʻylama tashkil etuvchilari paydo boʻladi: P n c h = z ^ P ∥ n c h + P ⊥ n c h {\displaystyle {\textbf {P}}^{nch}={\hat {z}}{\textbf {P}}_{\parallel }^{nch}+{\textbf {P}}_{\perp }^{nch}} Shu sababli ω 2 = 2 ω 1 {\displaystyle \omega _{2}=2\omega _{1}} chastotali ikkinchi garmonika maydonining koʻndalang va parallel tashkil etuvchilari boʻladi, yaʼni E ( z , ω 2 ) = z ^ E ∥ ( z , ω 2 ) + E ⊥ ( z , ω 2 ) {\displaystyle {\textbf {E}}(z,\omega _{2})={\hat {z}}E_{\parallel }(z,\omega _{2})+E_{\perp }(z,\omega _{2})} Bularni inobatga olsak, boshlangʻich toʻlqin tenglamalari quyidagi koʻrinishni qabul qiladi: ∂ 2 E ⊥ ∂ z 2 + ω 2 2 c 2 ( ε ⊥ ( ω 2 ) E ⊥ + 4 π P ⊥ n c h ) + ω 2 2 c 2 ( ε ∥ ( ω 2 ) E ∥ + 4 π P ∥ n c h ) {\displaystyle {\dfrac {\partial ^{2}E_{\perp }}{\partial z^{2}}}+{\dfrac {\omega _{2}^{2}}{c^{2}}}\left(\varepsilon _{\perp }(\omega _{2}){\textbf {E}}_{\perp }+4\pi {\textbf {P}}_{\perp }^{nch}\right)+{\dfrac {\omega _{2}^{2}}{c^{2}}}\left(\varepsilon _{\parallel }(\omega _{2}){\textbf {E}}_{\parallel }+4\pi {\textbf {P}}_{\parallel }^{nch}\right)} ∂ ∂ z ( ε ∥ ( ω 2 ) E ∥ + 4 π P ∥ n c h ) {\displaystyle {\dfrac {\partial }{\partial z}}\left(\varepsilon _{\parallel }(\omega _{2})E_{\parallel }+4\pi P_{\parallel }^{nch}\right)} Bu yerda ε ∥ {\displaystyle \varepsilon _{\parallel }} , ε ⊥ {\displaystyle \varepsilon _{\perp }} optik oʻqqa parallel va perpendikulyar dielektrik singdiruvchanliklardir. Yuqoridagilarga asosan tenglamani parallel va perpendikulyar tashkil etuvchilarga ajratamiz: ( ∂ 2 ∂ z 2 + k 2 2 ) E ⊥ = − 4 π ω 2 2 c 2 P ⊥ ( n c h ) {\displaystyle \left({\dfrac {\partial ^{2}}{\partial z^{2}}}+k_{2}^{2}\right){\textbf {E}}_{\perp }=-{\dfrac {4\pi \omega _{2}^{2}}{c^{2}}}{\textbf {P}}_{\perp }^{(nch)}} ε ∥ ( ω 2 ) E ∥ = − 4 π P ∥ ( n c h ) {\displaystyle \varepsilon _{\parallel }(\omega _{2})E_{\parallel }=-4\pi P_{\parallel }^{(nch)}} Yuqoridagi tenglamalarni haqiqiy va mavhum qismlarga ajratib, ρ 1 , 2 {\displaystyle \rho _{1,2}} va φ 1 , 2 {\displaystyle \varphi _{1,2}} uchun toʻrtta tenglamalar sistemasi hosil qilamiz: { ∂ ρ 2 ∂ ξ = − ρ 1 2 sin ⁡ ( 2 φ 1 − φ 2 ) ∂ ρ 1 ∂ ξ = ρ 1 ρ 2 sin ⁡ ( 2 φ 1 − φ 2 ) ρ 2 ∂ φ 2 ∂ ξ = ρ 1 2 cos ⁡ ( 2 φ 1 − φ 2 ) ∂ φ 1 ∂ ξ = ρ 2 cos ⁡ ( 2 φ 1 − φ 2 ) {\displaystyle {\begin{cases}{\dfrac {\partial \rho _{2}}{\partial \xi }}=-\rho _{1}^{2}\sin(2\varphi _{1}-\varphi _{2})\\{\dfrac {\partial \rho _{1}}{\partial \xi }}=\rho _{1}\rho _{2}\sin(2\varphi _{1}-\varphi _{2})\\\rho _{2}{\dfrac {\partial \varphi _{2}}{\partial \xi }}=\rho _{1}^{2}\cos(2\varphi _{1}-\varphi _{2})\\{\dfrac {\partial \varphi _{1}}{\partial \xi }}=\rho _{2}\cos(2\varphi _{1}-\varphi _{2})\end{cases}}} Nisbiy faza ψ = 2 φ 1 − φ 2 {\displaystyle \psi =2\varphi _{1}-\varphi _{2}} kiritib, yuqoridagi toʻrtta tenglamani uchta tenglamaga keltirish mumkin: { ∂ ρ 2 ∂ ξ = − ρ 1 2 sin ⁡ ( ψ ) ∂ ρ 1 ∂ ξ = ρ 1 ρ 2 sin ⁡ ( ψ ) ∂ ψ ∂ ξ = ( 2 ρ 2 − ρ 1 2 ρ 2 ) cos ⁡ ψ {\displaystyle {\begin{cases}{\dfrac {\partial \rho _{2}}{\partial \xi }}=-\rho _{1}^{2}\sin(\psi )\\{\dfrac {\partial \rho _{1}}{\partial \xi }}=\rho _{1}\rho _{2}\sin(\psi )\\{\dfrac {\partial \psi }{\partial \xi }}=\left(2\rho _{2}-{\dfrac {\rho _{1}^{2}}{\rho _{2}}}\right)\cos \psi \\\end{cases}}} Bu tenglamalar yordamida quyidagi tenglikni yozish mumkin: tg ψ ∂ ψ ∂ ξ − 2 ρ 1 ∂ ρ 1 ∂ ξ − 1 ρ 2 ∂ ρ 2 ∂ ξ = 0 {\displaystyle {\text{tg}}\psi {\dfrac {\partial \psi }{\partial \xi }}-{\dfrac {2}{\rho _{1}}}{\dfrac {\partial \rho _{1}}{\partial \xi }}-{\dfrac {1}{\rho _{2}}}{\partial \rho _{2}}{\partial \xi }=0} yoki ∂ ∂ ξ [ ln ⁡ ( ρ 1 2 ρ 2 cos ⁡ ψ ) ] = 0 {\displaystyle {\dfrac {\partial }{\partial \xi }}\left[\ln \left(\rho _{1}^{2}\rho _{2}\cos \psi \right)\right]=0} Demak, bu yerda qavs ichidagi kattalik ρ 1 2 ρ 2 cos ⁡ ψ = c o n s t {\displaystyle \rho _{1}^{2}\rho _{2}\cos \psi =const} saqlanuvchi kattalik, yaʼni harakat integrali ekan. Shunday qilib, hamma ξ {\displaystyle \xi } lar uchun ψ = ± π / 2 {\displaystyle \psi =\pm \pi /2} boʻlishi mumkinligi kelib chiqadi. Muhitda toʻlqin tarqalishi davomida ikkinchi garmonikaning intensivligi oʻsib borishini inobatga olsak, bu ikkita holdan minus ishorani tanlash kerak. U holda ∂ ρ 2 ∂ ξ = ρ 1 2 ,       ∂ ρ 1 ∂ ξ = − ρ 1 ρ 2 {\displaystyle {\dfrac {\partial \rho _{2}}{\partial \xi }}=\rho _{1}^{2},\ \ \ {\dfrac {\partial \rho _{1}}{\partial \xi }}=-\rho _{1}\rho _{2}} Bundan quyidagini topamiz: ρ 2 ∂ ρ 2 ∂ ξ + ρ 1 ∂ ρ 1 ∂ ξ = 0 {\displaystyle \rho _{2}{\dfrac {\partial \rho _{2}}{\partial \xi }}+\rho _{1}{\dfrac {\partial \rho _{1}}{\partial \xi }}=0} yoki ρ 1 2 + ρ 2 2 = 1 {\displaystyle \rho _{1}^{2}+\rho _{2}^{2}=1} Bu tenglik energiyaning saqlanish qonunini anglatadi. Yuqoridagi shart yordamida sistemaning birinchisidan ρ 1 {\displaystyle \rho _{1}} ni yoʻqotib tenglamani integrallaymiz va ρ 1 , 2 {\displaystyle \rho _{1,2}} uchun quyidagi ifodalarni topamiz: ρ 2 = th ξ ,       ρ 1 = sech ξ {\displaystyle \rho _{2}={\text{th}}\xi ,\ \ \ \rho _{1}={\text{sech}}\xi } Endi boshlangʻich funksiya va oʻzgaruvchilarga qaytamiz: E ( z , ω 2 ) = i E ( ω 1 ) th ( z L c ) {\displaystyle E(z,\omega _{2})=iE(\omega _{1}){\text{th}}\left({\frac {z}{L_{c}}}\right)} E ( z , ω 1 ) = E ( ω 1 ) sech ( z L c ) ,     φ 1 = 0 ,     φ 2 = π 2 {\displaystyle E(z,\omega _{1})=E(\omega _{1}){\text{sech}}\left({\frac {z}{L_{c}}}\right),\ \ \varphi _{1}=0,\ \ \varphi _{2}={\frac {\pi }{2}}}