Light field in frequency domain In linear system, the frequency remains the same. Therefore,
E ( r , ω ) = E ( r , ω ) ⋅ e i φ ( r , ω ) ⋅ e ^ ( r , ω ) \textbf{E}\left(r,\omega\right)=E\left(\textbf r,\omega\right) \cdot \mathrm{e}^{\mathrm{i}\varphi\left(\textbf r,\omega\right)} \cdot \hat{\textbf e}\left(\textbf r,\omega\right) E ( r , ω ) = E ( r , ω ) ⋅ e i φ ( r , ω ) ⋅ e ^ ( r , ω ) In this article, the bold character stand for vector in x, y, z, for example r = ( x , y , z ) \textbf{r} = (x, y, z) r = ( x , y , z ) .
The three components stand for amplitude, phase, and polarization, separately.
The relation of k , n , ε k, n, \varepsilon k , n , ε Here φ ( r , ω ) = ∫ 0 r k ⋅ d r − φ 0 \varphi\left(\textbf r,\omega\right) = \int_{0}^{\textbf r}\textbf k\cdot\mathrm{d}\textbf r-\varphi_{0} φ ( r , ω ) = ∫ 0 r k ⋅ d r − φ 0 , k = 2 π / λ \textbf k = 2 \pi / \lambda k = 2 π / λ represents the wave number, and importantly, the direction of k \textbf k k is the direction of light wave. Here k = 2 π n / λ = n k 0 = ε / ε 0 ⋅ k 0 k = 2\pi n/ \lambda = n k_0 = \sqrt{\varepsilon/\varepsilon_0} \cdot k_0 k = 2 πn / λ = n k 0 = ε / ε 0 ⋅ k 0 , which related to the index of the material.
Maxwell Eq In frequency domain, Maxwell Eqs are:
{ ∇ × E = − ∂ B ∂ t ∇ × H = ∂ D ∂ t + J ∇ ∙ D = ρ ∇ ∙ B = 0 \begin{cases}\nabla\times\boldsymbol{E}=-\frac{\partial\boldsymbol{B}}{\partial t}\\\nabla\times\boldsymbol{H}=\frac{\partial\boldsymbol{D}}{\partial\boldsymbol{t}}+\boldsymbol{J}\\\nabla\bullet\boldsymbol{D}=\rho\\\nabla\bullet\boldsymbol{B}=0\end{cases} ⎩ ⎨ ⎧ ∇ × E = − ∂ t ∂ B ∇ × H = ∂ t ∂ D + J ∇ ∙ D = ρ ∇ ∙ B = 0 In most real optical field, there is neither conduction current nor space charge. Therefore, J = 0 , ρ = 0 \textbf{J} = 0, \rho=0 J = 0 , ρ = 0
In addition, we have
D = ε 0 E + P B = μ 0 H + M \begin{aligned}\boldsymbol{D}=&\varepsilon_0\boldsymbol{E}+\boldsymbol{P}\\\boldsymbol{B}=&\mu_0\boldsymbol{H}+\boldsymbol{M}\end{aligned} D = B = ε 0 E + P μ 0 H + M In optical field, medium is Non-magnetic, so M = 0 ， B = μ 0 H \textbf{M}= 0， \textbf{B}=\mu_0\textbf{H} M = 0 ， B = μ 0 H . For D = ε 0 E + P \boldsymbol{D}=\varepsilon_0\boldsymbol{E}+\boldsymbol{P} D = ε 0 E + P , it is complex. When E \textbf E E is independent of time, P ∝ E \textbf{P} \propto \textbf{E} P ∝ E . However, when it’s not, P ∝ E \textbf{P} \propto \textbf{E} P ∝ E should be convoluted. So, P = ε 0 ∫ − ∞ + ∞ x ( 1 ) ( t − t 1 ) E ( r , t 1 ) d t 1 \boldsymbol{P}=\varepsilon_0\int_{-\infty}^{+\infty}x^{(1)}\left(t-t_1\right)\boldsymbol{E}\left(\boldsymbol{r},t_1\right)\mathrm{d}t_1 P = ε 0 ∫ − ∞ + ∞ x ( 1 ) ( t − t 1 ) E ( r , t 1 ) d t 1 which indicates that the change of P \textbf{P} P lags behind E \textbf{E} E by some time. And we all know that the frequency of light is high, so it should be considered.
D = ε 0 E + P = ε 0 { E + ∫ − ∞ + ∞ x ( 1 ) ( t − t 1 ) ∙ E ( r , t 1 ) d t 1 } \boldsymbol{D}=\varepsilon_0\boldsymbol{E}+\boldsymbol{P}=\varepsilon_0\left\{\boldsymbol{E}+\int_{-\infty}^{+\infty}x^{(1)}(t-t_1)\bullet\boldsymbol{E}(\boldsymbol{r},t_1)\mathrm{d}t_1\right\} D = ε 0 E + P = ε 0 { E + ∫ − ∞ + ∞ x ( 1 ) ( t − t 1 ) ∙ E ( r , t 1 ) d t 1 } With Fourier Transformation, in frequency domain, we can simplify it to
D ( ω ) = ε ( ω ) E ( ω ) \textbf D\left(\omega\right)=\varepsilon\left(\omega\right)\textbf E\left(\omega\right) D ( ω ) = ε ( ω ) E ( ω ) ε c ( ω ) \varepsilon_c\left(\omega\right) ε c ( ω ) is always a complex number, the real part is phase shift and the imaginary part is loss.
Now, we can deduce the Maxwell eqs in frequency domain
∇ × E = i ω μ 0 H ∇ × H = − i ω D ∇ ∙ D = 0 ∇ ∙ H = 0 \begin{aligned}&\nabla\times\boldsymbol{E}=\mathrm{i}\omega\mu_0\boldsymbol{H} \\&\nabla\times\boldsymbol{H}=-\mathrm{i}\omega\boldsymbol{D} \\&\nabla\bullet{\boldsymbol{D}}=0 \\&\nabla\bullet{\boldsymbol{H}}=0\end{aligned} ∇ × E = i ω μ 0 H ∇ × H = − i ω D ∇ ∙ D = 0 ∇ ∙ H = 0 As we have D ( ω ) = ε ( ω ) E ( ω ) \textbf D\left(\omega\right)=\varepsilon\left(\omega\right)\textbf E\left(\omega\right) D ( ω ) = ε ( ω ) E ( ω ) , we can get ∇ ∙ D = ∇ ∙ ( ε E ) = ∇ ε ⋅ E + ε ∇ ⋅ E = 0 \nabla\bullet{\boldsymbol{D}}=\nabla\bullet({{\varepsilon E}}) = \nabla \varepsilon \cdot \boldsymbol E+ \varepsilon \nabla \cdot \boldsymbol E = 0 ∇ ∙ D = ∇ ∙ ( εE ) = ∇ ε ⋅ E + ε ∇ ⋅ E = 0 , so we can get
∇ ⋅ E = − ∇ ε ε ⋅ E \nabla\boldsymbol{\cdot}\boldsymbol{E}=\frac{-\nabla\varepsilon}\varepsilon\boldsymbol{\cdot}\boldsymbol{E} ∇ ⋅ E = ε − ∇ ε ⋅ E which have a definite physical meaning, that the inhomogeneity of the distribution of the medium leads to the uneven distribution of polarization charges, which are manifested as active fields .
Finally, we deduce the the Maxwell eqs in Linear Time Invariant:
∇ × E = i ω μ 0 H ∇ × H = − i ω ε E ∇ ∙ E = − ∇ ε ε ⋅ E ∇ ∙ H = 0 \begin{aligned}&\nabla\times\boldsymbol{E}=\mathrm{i}\omega\mu_0\boldsymbol{H} \\&\nabla\times\boldsymbol{H}=-\mathrm{i}\omega\boldsymbol\varepsilon\boldsymbol{E} \\&\nabla\bullet{\boldsymbol{E}}=\frac{-\nabla\boldsymbol\varepsilon}{\boldsymbol\varepsilon}{\boldsymbol{\cdot}\boldsymbol{E}}
\\&\nabla\bullet{\boldsymbol{H}}=0\end{aligned} ∇ × E = i ω μ 0 H ∇ × H = − i ω ε E ∇ ∙ E = ε − ∇ ε ⋅ E ∇ ∙ H = 0 Helmholtz Equation With
∇ × ( ∇ × E ) = ∇ ( ∇ ∙ E ) − ∇ 2 E \nabla\times(\nabla\times \textbf E)=\nabla(\nabla\bullet \textbf E)-\nabla^2\textbf E ∇ × ( ∇ × E ) = ∇ ( ∇ ∙ E ) − ∇ 2 E and also for H \textbf H H , we can deduce Helmholtz equation that
{ ∇ 2 E + k 0 2 n 2 E + ∇ ( E ∙ ∇ ε ε ) = 0 ∇ 2 H + k 0 2 n 2 H + ∇ ε ε × ( ∇ × H ) = 0 \begin{cases}\nabla^2\boldsymbol{E}+k_0^2n^2\boldsymbol{E}+\nabla\left(\boldsymbol{E}\bullet\frac{\nabla\varepsilon}\varepsilon\right)=0\\\\\nabla^2\boldsymbol{H}+k_0^2n^2\boldsymbol{H}+\frac{\nabla\varepsilon}\varepsilon\times(\nabla\times\boldsymbol{H})=0\end{cases} ⎩ ⎨ ⎧ ∇ 2 E + k 0 2 n 2 E + ∇ ( E ∙ ε ∇ ε ) = 0 ∇ 2 H + k 0 2 n 2 H + ε ∇ ε × ( ∇ × H ) = 0 Light modes in Waveguide First the index should be constant along the z direction.
ε ( x , y , z ) = ε ( x , y ) k = β z ^ \varepsilon(x, y, z) = \varepsilon(x, y)\\\textbf k = \beta \hat z ε ( x , y , z ) = ε ( x , y ) k = β z ^ So we can deduce that (neglect ω \omega ω )
E ( r ) = E ( x , y ) e i β z e ^ ( x , y ) H ( r ) = H ( x , y ) e i β z h ^ ( x , y ) \textbf{E}\left(\textbf{r}\right)=E\left(x,y\right){e}^{i\beta z}\boldsymbol{\hat{e}}\left(x,y\right)
\\\textbf{H}\left(\textbf{r}\right)=H\left(x,y\right){e}^{i\beta z}\boldsymbol{\hat{h}}\left(x,y\right) E ( r ) = E ( x , y ) e i β z e ^ ( x , y ) H ( r ) = H ( x , y ) e i β z h ^ ( x , y ) The amplitude and polarization related to x , y x, y x , y , and the phase related to z z z .
Let e ( x , y ) = E ( x , y ) e ^ ( x , y ) , h ( x , y ) = H ( x , y ) h ^ ( x , y ) {{\textbf e}}\left(x,y\right) = E\left(x,y\right)\boldsymbol{\hat{e}}\left(x,y\right) , {{\textbf h}}\left(x,y\right) = H\left(x,y\right)\boldsymbol{\hat{h}}\left(x,y\right) e ( x , y ) = E ( x , y ) e ^ ( x , y ) , h ( x , y ) = H ( x , y ) h ^ ( x , y ) , which are the mode fields, which represent the distribution of the optical fields E ,H along the cross-section.
Through Helmholtz equation and decompose the model field horizontally and vertically. We can know for a specific boundary condition, the Helmholtz equation have infinite discrete eigen-solution . （skip the detailed process here
( E n H n ) = ( e n h n ) ( x , y ) e i β n z , n = 1 , 2 , 3 , ⋅ ⋅ ⋅ \binom{\boldsymbol{E_n}}{\boldsymbol{H_n}}=\binom{\boldsymbol{e}_n}{\boldsymbol{h}_n}\left(x,y\right)\mathrm{e}^{i\beta_nz},\quad n=1,2,3,\cdotp\cdotp\cdotp ( H n E n ) = ( h n e n ) ( x , y ) e i β n z , n = 1 , 2 , 3 ,⋅⋅⋅ And each eigen-solution is a optical mode. But what is this mean in physics.
The physical meaning of mode and its properties The most important property of the optical mode lies that the optical mode is a stable distribution along the z-direction, which means when a mode is transmitted along the longitudinal direction, its field distribution form is unchanged . However, for other ( E , H ) (\boldsymbol E, \boldsymbol H) ( E , H ) , the field will change when it transmits. The total distribution of optical field is the linear combination of modes, ( E H ) = ∑ n ( a n e n b n h n ) ( x , y ) e i β n z \binom{\boldsymbol{E}}{\boldsymbol{H}}=\sum_n\binom{a_n\boldsymbol{e}_n}{b_n\boldsymbol{h}_n}\left(x,y\right)\mathrm{e}^{i\beta_nz} ( H E ) = n ∑ ( b n h n a n e n ) ( x , y ) e i β n z Since the limit of boundary condition, not all modes can exist in waveguide. And in real application, in fact, we always want only one mode can exist. In transmission process, the basic physics quantity is transmission constant β \beta β . The real part is the phase shift in transmission and the imaginary part stands for the loss in transmission. Solution in x-y plane Now we have (t means x-y plane
( E t E z H t H z ) = ( e t e z h t h z ) e i β z \begin{pmatrix}\boldsymbol{E}_\mathrm{t}\\\\\boldsymbol{E}_z\\\\\boldsymbol{H}_\mathrm{t}\\\\\boldsymbol{H}_z\end{pmatrix}=\begin{pmatrix}\boldsymbol{e}_\mathrm{t}\\\\\boldsymbol{e}_z\\\\\boldsymbol{h}_\mathrm{t}\\\\\boldsymbol{h}_z\end{pmatrix}\mathrm{e}^\mathrm{i\beta z} E t E z H t H z = e t e z h t h z e i β z We have ∇ = ∇ t + z ^ ∂ ∂ z \nabla=\nabla_{\mathrm{t}}+\hat{z}\:\frac{\partial}{\partial z} ∇ = ∇ t + z ^ ∂ z ∂ , so in x-y plane, we can write Helmholtz eq (In each area, we have ∇ ε = 0 \nabla \varepsilon = 0 ∇ ε = 0
∇ 2 E ( x , y , z ) = ∇ t 2 [ E ( x , y ) e i β z ] + ∂ 2 ∂ z 2 [ E ( x , y ) e i β z ] = [ ∇ t 2 E ( x , y ) ] e i β z − β 2 E ( x , y ) e i β z \begin{aligned} \nabla^{2}\boldsymbol{E}\left(x,y,z\right)& =\nabla_{\mathrm{t}}^{2}\Big[E\left(x\:,y\:\right)\mathrm{e}^{\mathrm{i}\beta z}\Big]+\frac{\partial^{2}}{\partial z^{2}}\Big[E\left(x\:,y\:\right)\mathrm{e}^{\mathrm{i}\beta z}\Big] \\ &=\left[\nabla_{\mathrm{t}}^{2}\mathbf{E}\left(x\:,y\:\right)\right]\mathrm{e}^{\mathrm{i}\beta z}-\beta^{2}\mathbf{E}\left(x\:,y\:\right)\mathrm{e}^{\mathrm{i}\beta z} \end{aligned} ∇ 2 E ( x , y , z ) = ∇ t 2 [ E ( x , y ) e i β z ] + ∂ z 2 ∂ 2 [ E ( x , y ) e i β z ] = [ ∇ t 2 E ( x , y ) ] e i β z − β 2 E ( x , y ) e i β z Therefore, it Helmholtz eq can be simplified as
∇ t 2 E ( x , y ) + ( k 0 2 n 2 − β 2 ) E ( x , y ) = 0 ∇ t 2 H ( x , y ) + ( k 0 2 n 2 − β 2 ) H ( x , y ) = 0 \nabla_{\mathrm{t}}^{2}\boldsymbol{E}(x,y)+(k_{0}^{2}n^{2}-\beta^{2})\boldsymbol{E}(x,y) = 0\\
\nabla_{\mathrm{t}}^{2}\boldsymbol{H}(x,y)+(k_{0}^{2}n^{2}-\beta^{2})\boldsymbol{H}(x,y) = 0 ∇ t 2 E ( x , y ) + ( k 0 2 n 2 − β 2 ) E ( x , y ) = 0 ∇ t 2 H ( x , y ) + ( k 0 2 n 2 − β 2 ) H ( x , y ) = 0 TEM E z = H z = 0 E_z = H_z = 0 E z = H z = 0 , only if ω 2 μ 0 ε − β 2 = 0 \omega^2\mu_0\varepsilon-\beta^2=0 ω 2 μ 0 ε − β 2 = 0 . The β \beta β is constant for one mode, but the ε \varepsilon ε will change for different place.
Therefore, it exists only in a infinite uniform large medium.
TE H z = 0 H_z = 0 H z = 0
TM E z = 0 E_z = 0 E z = 0
Waveguide plane waveguide In this situation,
E ( x , y ) → E ( x ) H ( x , y ) → H ( x ) \mathbf E(x,y) \rightarrow \mathbf E(x) \\\mathbf H(x,y) \rightarrow \mathbf H(x) E ( x , y ) → E ( x ) H ( x , y ) → H ( x ) We have
x ^ × ∂ E t ( x ) ∂ x = i ω μ 0 H z ( x ) x ^ × ∂ H t ( x ) ∂ x = − i ω ε 0 E z ( x ) x ^ × ∂ E z ( x ) ∂ x + i β z ^ × E t ( x ) = i ω μ 0 H t ( x ) x ^ × ∂ H z ( x ) ∂ x + i β z ^ × H t ( x ) = − i ω ε E t ( x ) \begin{aligned}&\hat{\boldsymbol{x}}\times\frac{\partial\boldsymbol{E}_\mathrm{t}(x)}{\partial x}=\mathrm{i}\omega\mu_0\boldsymbol{H}_z(x)\\&\hat{\boldsymbol{x}}\times\frac{\partial\boldsymbol{H}_\mathrm{t}(x)}{\partial x}=-\mathrm{i}\omega\varepsilon_0\boldsymbol{E}_z(x)\\&\hat{\boldsymbol{x}}\times\frac{\partial\boldsymbol{E}_z(x)}{\partial x}+\mathrm{i}\beta\hat{\boldsymbol{z}}\times\boldsymbol{E}_\mathrm{t}(x)=\mathrm{i}\omega\mu_0\boldsymbol{H}_\mathrm{t}(x)\\&\hat{\boldsymbol{x}}\times\frac{\partial\boldsymbol{H}_z(x)}{\partial x}+\mathrm{i}\beta\hat{\boldsymbol{z}}\times\boldsymbol{H}_\mathrm{t}(x)=-\mathrm{i}\omega\varepsilon\boldsymbol{E}_\mathrm{t}(x)\end{aligned} x ^ × ∂ x ∂ E t ( x ) = i ω μ 0 H z ( x ) x ^ × ∂ x ∂ H t ( x ) = − i ω ε 0 E z ( x ) x ^ × ∂ x ∂ E z ( x ) + i β z ^ × E t ( x ) = i ω μ 0 H t ( x ) x ^ × ∂ x ∂ H z ( x ) + i β z ^ × H t ( x ) = − i ω ε E t ( x ) For TE mode, H z = 0 H_z = 0 H z = 0 .
H x = − β ω μ 0 E y H z = − i ω μ 0 d E y d x \begin{aligned}H_x&=-\frac\beta{\omega\mu_0}E_y\\H_z&=-\frac{\mathrm{i}}{\omega\mu_0}\frac{\mathrm{d}E_y}{\mathrm{d}x}\end{aligned} H x H z = − ω μ 0 β E y = − ω μ 0 i d x d E y So, we only need to know E y E_y E y to get all the components.
Step-index waveguide It is the simple plane waveguide as below, a a a is the thickness of the waveguide
n ( x ) = { n 1 , ∣ x ∣ < a / 2 n 2 , ∣ x ∣ > a / 2 n\left(x\right)=\begin{cases}n_1,&\mid x\mid<a/2\\n_2,&\mid x\mid>a/2&\end{cases} n ( x ) = { n 1 , n 2 , ∣ x ∣< a /2 ∣ x ∣> a /2 In each area, we have ∇ ε = 0 \nabla \varepsilon = 0 ∇ ε = 0 . So the Helmholtz equation simplify as
d 2 E y d x 2 + ( k 0 2 n 2 − β 2 ) E y = 0 \frac{\mathrm{d}^2E_y}{\mathrm{d}x^2}+(k_0^2n^2-\beta^2)E_y=0 d x 2 d 2 E y + ( k 0 2 n 2 − β 2 ) E y = 0 With the boundary condition that the filed in core is oscillatory, and in the cladding is decay.
E y ∣ x − > ± ∞ = 0 E_y|_{x->\pm\infty} = 0 E y ∣ x − > ± ∞ = 0 The solution should be
E y = A 1 cos ( κ x ) at core E y = A 1 cos ( 2 κ a ) e q ( x − a ) at bottom cladding E y = A 1 cos ( 2 κ a ) e − q ( x + a ) at bottom cladding \begin{aligned}&E_{_y}= A_1\cos(\kappa x) &\text{ at core}\\
&E_y = A_1 \cos(2\kappa a)e^{ q(x-a)} &\text{ at bottom cladding}\\
&E_y = A_1 \cos(2\kappa a)e^{ -q(x+a)} &\text{ at bottom cladding}\end{aligned} E y = A 1 cos ( κ x ) E y = A 1 cos ( 2 κa ) e q ( x − a ) E y = A 1 cos ( 2 κa ) e − q ( x + a ) at core at bottom cladding at bottom cladding with
κ = ( k 0 2 n 1 2 − β 2 ) 1 2 q = ( β 2 − k 0 2 n 2 2 ) 1 2 tan ( κ a ) = q κ \begin{gathered} \kappa=(k_0^2n_1^2-\beta^2)^{\frac12} \\ q=(\beta^2-k_0^2n_2^2)^{\frac12} \\\tan(\kappa a)=\frac{q}{\kappa}\end{gathered} κ = ( k 0 2 n 1 2 − β 2 ) 2 1 q = ( β 2 − k 0 2 n 2 2 ) 2 1 tan ( κa ) = κ q So
κ a = m π + arctan ( q / κ ) \kappa a = m\pi + \arctan(q/\kappa) κa = mπ + arctan ( q / κ ) the m = 0 , 1 , 2 , 3 , . . . m=0,1,2,3,... m = 0 , 1 , 2 , 3 , ... here is the mode number.
Define effective index n e f f = β / k 0 n_{eff} = \beta / k_0 n e ff = β / k 0 ,
( n 1 2 − N 2 ) 1 2 k 0 a = m π + arctan [ ( N 2 − n 2 2 n 1 2 − N 2 ) 1 2 ] (n_1^2-N^2)^{\frac12}k_0 a=m\pi+\arctan\left[\left(\frac{N^2-n_2^2}{n_1^2-N^2}\right)^{\frac12}\right] ( n 1 2 − N 2 ) 2 1 k 0 a = mπ + arctan [ ( n 1 2 − N 2 N 2 − n 2 2 ) 2 1 ] When N = n 2 N = n_2 N = n 2 , it reaches its cut-off width. And the light cannot propogate in the waveguide if a < w c u t o f f ( m ) a < w_{cutoff}(m) a < w c u t o ff ( m ) .
Multi-layers waveguide (5 symmetric layers) Transfer Matrix Still we have
d 2 E y d x 2 + ( k 0 2 n j 2 − β 2 ) E y = 0 \frac{\mathrm{d}^2E_y}{\mathrm{d}x^2}+(k_0^2n_j^2-\beta^2)E_y=0 d x 2 d 2 E y + ( k 0 2 n j 2 − β 2 ) E y = 0 Define the transmission coefficient for each layer as
U j = k 0 2 n j 2 − β 2 U_j = \sqrt{k_0^2n_j^2-\beta^2} U j = k 0 2 n j 2 − β 2 The solution should be
E y j = a j sin ( U j x ) + b j cos ( U j x ) E_{yj}=a_j\sin(U_jx)+b_j\cos(U_jx) E y j = a j sin ( U j x ) + b j cos ( U j x ) i ω μ 0 h z = U j [ a j cos ( U j x ) − b j sin ( U j x ) ] {\mathrm{i}\omega\mu_0}h_z=U_j[a_j\cos(U_jx)-b_j\sin(U_jx)] i ω μ 0 h z = U j [ a j cos ( U j x ) − b j sin ( U j x )] Define Transfer Matrix as
M j ( x j − x j − 1 ) = [ cos [ U j ( x j − x j − 1 ) ] 1 U j sin [ U j ( x j − x j − 1 ) ] − U j sin [ U j ( x j − x j − 1 ) ] cos [ U j ( x j − x j − 1 ) ] ] \boldsymbol{M}_j(x_j-x_{j-1})=\begin{bmatrix}\cos[U_j(x_j-x_{j-1})]&\frac{1}{U_j}\sin[U_j(x_j-x_{j-1})]\\\\-U_j\sin[U_j(x_j-x_{j-1})]&\cos[U_j(x_j-x_{j-1})]\end{bmatrix} M j ( x j − x j − 1 ) = cos [ U j ( x j − x j − 1 )] − U j sin [ U j ( x j − x j − 1 )] U j 1 sin [ U j ( x j − x j − 1 )] cos [ U j ( x j − x j − 1 )] [ e y i ω μ 0 h z ] x = x j = M j ( x j − x j − 1 ) [ e y i ω μ 0 h z ] x = x j − 1 \begin{bmatrix}e_y\\\\\mathrm{i}\omega\mu_0h_z\end{bmatrix}_{x=x_j}=\boldsymbol{M}_j(x_j-x_{j-1})\begin{bmatrix}e_y\\\\\mathrm{i}\omega\mu_0h_z\end{bmatrix}_{x=x_{j-1}} e y i ω μ 0 h z x = x j = M j ( x j − x j − 1 ) e y i ω μ 0 h z x = x j − 1 Let’s consider the boundary condition:
For the left side, x < − x 2 x < -x_2 x < − x 2 ,
E y = b l ⋅ e x p ( U l x ) E_y = b_l \cdot exp(U_lx) E y = b l ⋅ e x p ( U l x ) For the right side, x > x 2 x > x_2 x > x 2 ,
E y = b r ⋅ e x p ( − U r x ) E_y = b_r \cdot exp(-U_rx) E y = b r ⋅ e x p ( − U r x ) 5 layers Solution For 5 layers
M − 1 ( x − 1 − x − 2 ) = [ cos ( U − 1 h 2 ) 1 U − 1 sin ( U − 1 h 2 ) − U − 1 sin ( U − 1 h 2 ) cos ( U − 1 h 2 ) ] \boldsymbol{M}_{-1}(x_{-1}-x_{-2})=\begin{bmatrix}\cos(U_{-1}h_2)&\frac{1}{U_{-1}}\sin(U_{-1}h_2)\\\\-U_{-1}\sin(U_{-1}h_2)&\cos(U_{-1}h_2)\end{bmatrix} M − 1 ( x − 1 − x − 2 ) = cos ( U − 1 h 2 ) − U − 1 sin ( U − 1 h 2 ) U − 1 1 sin ( U − 1 h 2 ) cos ( U − 1 h 2 ) M 1 ( x 1 − x − 1 ) = [ cos ( 2 U 1 h 1 ) 1 U 1 sin ( 2 U 1 h 1 ) − U 1 sin ( 2 U 1 h 1 ) cos ( 2 U 1 h 1 ) ] \boldsymbol{M}_1(x_1-x_{-1})=\begin{bmatrix}\cos(2U_1h_1)&\frac{1}{U_1}\sin(2U_1h_1)\\\\-U_1\sin(2U_1h_1)&\cos(2U_1h_1)\end{bmatrix} M 1 ( x 1 − x − 1 ) = cos ( 2 U 1 h 1 ) − U 1 sin ( 2 U 1 h 1 ) U 1 1 sin ( 2 U 1 h 1 ) cos ( 2 U 1 h 1 ) M 2 ( x 2 − x 1 ) = [ cos ( U 2 h 2 ) 1 U 2 sin ( U 2 h 2 ) − U 2 sin ( U 2 h 2 ) cos ( U 2 h 2 ) ] \boldsymbol{M}_2(x_2-x_1)=\begin{bmatrix}\cos(U_2h_2)&\frac{1}{U_2}\sin(U_2h_2)\\\\-U_2\sin(U_2h_2)&\cos(U_2h_2)\end{bmatrix} M 2 ( x 2 − x 1 ) = cos ( U 2 h 2 ) − U 2 sin ( U 2 h 2 ) U 2 1 sin ( U 2 h 2 ) cos ( U 2 h 2 )
Parabolic medium refractive index Parabolic medium is defined as below:
n 2 ( x ) = n 1 2 − ( n 1 2 − n 2 2 ) ( x a ) 2 = n 1 2 [ 1 − 2 Δ ( x a ) 2 ] n^2(x)=n_1^2-(n_1^2-n_2^2)\left(\frac xa\right)^2 = n_1^2[1-2\Delta(\frac{x}{a})^2] n 2 ( x ) = n 1 2 − ( n 1 2 − n 2 2 ) ( a x ) 2 = n 1 2 [ 1 − 2Δ ( a x ) 2 ] For TE mode, H z = 0 H_z = 0 H z = 0 . So we have
d 2 E y d x 2 + [ k 0 2 n 2 ( x ) − β 2 ] E y = 0 \frac{\mathrm{d}^2E_y}{\mathrm{d}x^2}+\left[k_0^2n^2(x)-\beta^2\right]E_y=0 d x 2 d 2 E y + [ k 0 2 n 2 ( x ) − β 2 ] E y = 0 Substitute n ( x ) n(x) n ( x )
d 2 E y d x 2 + [ ( k 0 2 n 1 2 − β 2 ) − 2 k 0 2 n 1 2 Δ ( x a ) 2 ] E y = 0 \begin{aligned} &\frac{\mathrm{d}^2E_y}{\mathrm{d}x^2}+\left[(k_0^2n_1^2-\beta^2)-2k_0^2n_1^2\Delta\left(\frac xa\right)^2\right]E_y=0\end{aligned} d x 2 d 2 E y + [ ( k 0 2 n 1 2 − β 2 ) − 2 k 0 2 n 1 2 Δ ( a x ) 2 ] E y = 0 Let
w 0 2 = a 2 V = a k 0 n 1 ( 2 Δ ) 1 / 2 ξ = x w 0 w_{0}^{2} =\frac{a^2}V=\frac a{k_0n_1(2\Delta)^{1/2}} \\ \xi= \frac x{w_0} \\ w 0 2 = V a 2 = k 0 n 1 ( 2Δ ) 1/2 a ξ = w 0 x We deduce
d 2 E y d ξ 2 + ( λ − ξ 2 ) E y = 0 λ = ( k 0 2 n 1 2 − β 2 ) w 0 2 \frac{\mathrm{d}^2E_y}{\mathrm{d}\xi^2}+(\lambda-\xi^2)E_y=0 \\ \lambda=(k_0^2n_1^2-\beta^2)w_0^2 d ξ 2 d 2 E y + ( λ − ξ 2 ) E y = 0 λ = ( k 0 2 n 1 2 − β 2 ) w 0 2 , which is the same as the Schrödinger Equation for One-Dimensional Harmonic Oscillator.
The eigenvalue is λ = 2 m + 1 , m = 0 , 1 , 2 , . . . \lambda = 2m+1, m=0,1,2,... λ = 2 m + 1 , m = 0 , 1 , 2 , ...
The solution is
P 2 = 1 − ( 2 m + 1 V ) , P = n e f f 2 − n 2 2 n 1 2 − n 2 2 V = k 0 n 1 ( 2 Δ ) 1 / 2 a P^2 = 1-(\frac{2m+1}{V}),
\\P = \frac{n_{eff}^2-n_2^2}{n_1^2-n_2^2}
\\V = k_0 n_1 (2\Delta)^{1/2}a P 2 = 1 − ( V 2 m + 1 ) , P = n 1 2 − n 2 2 n e ff 2 − n 2 2 V = k 0 n 1 ( 2Δ ) 1/2 a The field distribution is as below, dash line for m = 0 m=0 m = 0 , dot line for m = 1 m=1 m = 1 , and the solid line for m = 2 m=2 m = 2
Symbol List P \textbf{P} P
E \textbf{E} E
References [1] 导波光学，清华大学出版社，ISBN：9787302531609
