# Two-Mode Coupling

This is a continuation from the previous tutorial - coupled-mode theory.

In most applications, we are interested in the coupling between two modes. This includes coupling between two modes in the same waveguide, such as that in a periodic waveguide, or coupling between two parallel waveguides, such as that in a directional coupler. For coupling between two modes, the coupled-mode equations can be written in a simple form that can be solved analytically. In this tutorial, we consider the general formulation and general solutions for this important case of two-mode coupling. The characteristics of specific couplers are discussed in later tutorials.

We have shown that both coupling among modes in the same waveguide and coupling among multiple waveguides can be described by coupled-mode equations of the same form as given in (33) and (39) [refer to the coupled-mode theory tutorial]. The only difference is that the coupling coefficients in (39) for multiple-waveguide coupling are defined differently from those in (33) for single-waveguide mode coupling. This is convenient because general solutions of the coupled-mode equations can be applied to both cases. For a particular problem, we only have to calculate the coupling coefficients specific to the problem under consideration.

For two-mode coupling either in a single waveguide or between two separate waveguides, the field expansion in (25) and (26) [refer to the coupled-mode theory tutorial] consists of only two modes with amplitudes A and B. Thus, coupled-mode equations of the form given in (33) or (39) simply reduces to the following two coupled equations:

$\tag{45}\pm\frac{\text{d}A}{\text{d}z}=\text{i}\kappa_{aa}A+\text{i}\kappa_{ab}B\text{e}^{\text{i}(\beta_b-\beta_a)z}$$\tag{46}\pm\frac{\text{d}B}{\text{d}z}=\text{i}\kappa_{bb}B+\text{i}\kappa_{ba}A\text{e}^{\text{i}(\beta_a-\beta_b)z}$

For coupling in a single waveguide, the coupling coefficients in these equations are simply given by (34) [refer to the coupled-mode theory tutorial] in the case of an isotropic waveguide or by (36) in the case of an anisotropic waveguide. According to (37), we also have $$\kappa_{ab}=\kappa_{ba}^*$$ if the waveguide is lossless.

For coupling between two waveguides, the coupling coefficients are given by (40) [refer to the coupled-mode theory tutorial], which can be expressed explicitly as

\tag{47}\begin{align}\kappa_{aa}=\frac{\tilde{\kappa}_{aa}-c_{ab}\tilde{\kappa}_{ba}/c_{bb}}{1-c_{ab}c_{ba}/c_{aa}c_{bb}},\qquad\kappa_{ab}=\frac{\tilde{\kappa}_{ab}-c_{ab}\tilde{\kappa}_{bb}/c_{bb}}{1-c_{ab}c_{ba}/c_{aa}c_{bb}},\\\kappa_{ba}=\frac{\tilde{\kappa}_{ba}-c_{ba}\tilde{\kappa}_{aa}/c_{aa}}{1-c_{ab}c_{ba}/c_{aa}c_{bb}},\qquad\kappa_{bb}=\frac{\tilde{\kappa}_{bb}-c_{ba}\tilde{\kappa}_{ab}/c_{aa}}{1-c_{ab}c_{ba}/c_{aa}c_{bb}}.\end{align}

As discussed earlier and expressed in (43) and (44) [refer to the coupled-mode theory tutorial], in general, $$\kappa_{ab}\ne\kappa_{ba}^*$$ for coupling between two waveguides.

There is a self-coupling term in each of the coupled equations (45) and (46). These terms are caused by the fact that normal modes see an index profile in the perturbed waveguide different from that of the original waveguide where the modes are defined. They can be removed from these equations by expressing the normal-mode expansion coefficient as follows:

$\tag{48}A(z)=\tilde{A}(z)\exp\left[\pm\text{i}\displaystyle\int\limits_0^z\kappa_{aa}(z)\text{d}z\right]$$\tag{49}B(z)=\tilde{B}(z)\exp\left[\pm\text{i}\displaystyle\int\limits_0^z\kappa_{bb}(z)\text{d}z\right]$

where a plus or minus sign is chosen for a forward-propagating or backward-propagating mode, respectively.

Before transforming (45) and (46) into two coupled equations in terms of $$\tilde{A}$$ and $$\tilde{B}$$ to remove the self-coupling terms, we have to consider the fact that all of the coupling coefficients can be a function of z because $$\Delta\boldsymbol{\epsilon}$$ can be a function of z but the integration in (36) and (42) is carried out only over x and y [refer to the coupled-mode theory tutorial]. In case $$\kappa_{ab}(z)$$ and $$\kappa_{ba}(z)$$ are arbitrary functions of z, the coupled-mode equations in (45) and (46) cannot be solved analytically. In this situation, there is no need to simplify the coupled-mode equations further because they can only be solved numerically. However, for waveguide structures of practical interest that are designed for two-mode coupling, $$\Delta\boldsymbol{\epsilon}$$ is either independent of z or is a periodic function of z. Then, the coupling coefficients are either constant or periodic in z. In either case, (45) and (46) can be reduce to the following general form:

$\tag{50}\pm\frac{\text{d}\tilde{A}}{\text{d}z}=\text{i}\kappa_{ab}\tilde{B}\text{e}^{\text{i}2\delta z}$$\tag{51}\pm\frac{\text{d}\tilde{B}}{\text{d}z}=\text{i}\kappa_{ba}\tilde{A}\text{e}^{-\text{i}2\delta z}$

in terms of $$\tilde{A}$$ and $$\tilde{B}$$ with $$\kappa_{ab}$$ and $$\kappa_{ba}$$ in these two equations being constants independent of z. The parameter $$2\delta$$ is the phase mismatch between the two modes being coupled. Phase-matched coupling with $$\delta=0$$ between two modes is always symmetric with $$\kappa_{ab}=\kappa_{ba}^*$$ irrespective of whether these two modes belong to the same waveguide or two different waveguides.

The general form of (50) and (51) applies to both cases of constant and periodic perturbations, but the details of the parameters in these two equations vary.

1. Constant perturbation.  In this case, $$\Delta\boldsymbol{\epsilon}$$ is not a function of z. Then all of the coupling coefficients $$\kappa_{aa}$$, $$\kappa_{bb}$$, $$\kappa_{ab}$$, and $$\kappa_{ba}$$ are constant that are independent of z. We then find that

$\tag{52}A(z)=\tilde{A}(z)\text{e}^{\pm\text{i}\kappa_{aa}z}\qquad\text{and}\qquad B(z)=\tilde{B}(z)\text{e}^{\pm\text{i}\kappa_{bb}z}$

and

$\tag{53}2\delta=(\beta_b\pm\kappa_{bb})-(\beta_a\pm\kappa_{aa})$

The choice of sign in each ± here is consistent with that in (48) and (49) discussed above. The physical meaning of the self-coupling coefficients is a change in the propagating constant of each normal mode. While the propagation constants of the normal modes in the original waveguide are $$\beta_a$$ and $$\beta_b$$, their values are changed because of the perturbation on the waveguide. These modes now propagate with the modified propagation constants $$\beta_a\pm\kappa_{aa}$$ and $$\beta_b\pm\kappa_{bb}$$, respectively, which take into account the effect of the perturbation. In addition, they couple to each other through $$\kappa_{ab}$$ and $$\kappa_{ba}$$.

2. Periodic perturbation.  In this case, $$\Delta\boldsymbol{\epsilon}$$ is a periodic function of z and so are the coupling coefficients $$\kappa_{aa}(z)$$, $$\kappa_{bb}(z)$$, $$\kappa_{ab}(z)$$, and $$\kappa_{ba}(z)$$. The periodic perturbation has a period $$\Lambda$$ and a wavenumber

$\tag{54}K=\frac{2\pi}{\Lambda}$

The coupling coefficients $$\kappa_{ab}(z)$$ and $$\kappa_{ba}(z)$$, being periodic in z with a periodicity $$\Lambda$$, can be expanded in a Fourier series with constant coefficients $$\kappa_{ab}(q)$$ and $$\kappa_{ba}(q)$$ and a phase factor $$qK$$, where $$q$$ is an integer. Because $$\kappa_{aa}(z)$$ and $$\kappa_{bb}(z)$$ are periodic in z, we find that

$\tag{55}\left|\displaystyle\int\limits_0^z\kappa_{aa}(z)\text{d}z\right|\ll Kz\qquad\text{and}\qquad\left|\displaystyle\int\limits_0^z\kappa_{bb}(z)\text{d}z\right|\ll Kz$

Therefore, the contribution to the phase-mismatch parameter $$2\delta$$ by $$\kappa_{aa}$$ and $$\kappa_{bb}$$ is negligible compared to the contribution by $$qK$$. As a result, we find that the coupled-mode equations in the case of periodic perturbation can also be expressed in the form of (50) and (51) but with $$\kappa_{ab}=\kappa_{ab}(q)$$ and $$\kappa_{ba}=\kappa_{ba}(q)$$ being constants that are independent of z and

$\tag{56}2\delta=\Delta\beta+qK=\beta_b-\beta_a+qK$

where $$\Delta\beta=\beta_b-\beta_a$$ and $$q$$ is an integer that minimizes the value of $$\delta$$.

With these general considerations, (50) and (51) represent the most general coupled equations for two-mode coupling in waveguide structures of practical interest. They can be solved analytically and their solutions apply to many different two-mode coupling problems.

Codirectional coupling

First, we consider the coupling of two modes propagating in the same direction, say the forward direction in $$z$$, over a length $$l$$, as is shown in figure 3 below. In this case, $$\beta_a>0$$ and $$\beta_b>0$$. The coupled equations are

$\tag{57}\frac{\text{d}\tilde{A}}{\text{d}z}=\text{i}\kappa_{ab}\tilde{B}\text{e}^{\text{i}2\delta z}$$\tag{58}\frac{\text{d}\tilde{B}}{\text{d}z}=\text{i}\kappa_{ba}\tilde{A}\text{e}^{-\text{i}2\delta z}$

These equations for codirectional coupling are generally solved as an initial-value problem with the initial values of $$\tilde{A}(z_0)$$ and $$\tilde{B}(z_0)$$ given at $$z=z_0$$ to find the values of $$\tilde{A}(z)$$ and $$\tilde{B}(z)$$ at any other location $$z$$. The general solution can be expressed in the following matrix form:

$\tag{59}\begin{bmatrix}\tilde{A}(z)\\\tilde{B}(z)\end{bmatrix}=\mathbf{F}(z;z_0)\begin{bmatrix}\tilde{A}(z_0)\\\tilde{B}(z_0)\end{bmatrix}$

where the forward-coupling matrix $$\mathbf{F}(z;z_0)$$ relates the field amplitudes at the location $$z_0$$ to those at the location $$z$$. It has the form

\tag{60}\begin{align}\mathbf{F}(z;z_0)&=\\&\begin{bmatrix}\frac{\beta_c\cos\beta_c(z-z_0)-\text{i}\delta\sin\beta_c(z-z_0)}{\beta_c}\text{e}^{\text{i}\delta(z-z_0)}&\frac{\text{i}\kappa_{ab}}{\beta_c}\sin\beta_c(z-z_0)\text{e}^{\text{i}\delta(z+z_0)}\\\frac{\text{i}\kappa_{ba}}{\beta_c}\sin\beta_c(z-z_0)\text{e}^{-\text{i}\delta(z+z_0)}&\frac{\beta_c\cos\beta_c(z-z_0)+\text{i}\delta\sin\beta_c(z-z_0)}{\beta_c}\text{e}^{-\text{i}\delta(z-z_0)}\end{bmatrix}\end{align}

where

$\tag{61}\beta_c=(\kappa_{ab}\kappa_{ba}+\delta^2)^{1/2}$

We consider a simple case when power is launched only into mode $$a$$ at $$z = 0$$. Then the initial values are $$\tilde{A}(0)\ne0$$ and $$\tilde{B}(0)=0$$. By applying these conditions to (59) and taking $$z_0=0$$ in (60), we find that

$\tag{62}\tilde{A}(z)=\tilde{A}(0)\left(\cos\beta_c z-\frac{\text{i}\delta}{\beta_c}\sin\beta_c z\right)\text{e}^{\text{i}\delta z}$$\tag{63}\tilde{B}(z)=\tilde{A}(0)\left(\frac{\text{i}\kappa_{ba}}{\beta_c}\sin\beta_c z\right)\text{e}^{-\text{i}\delta z}$

The power in the two modes varies with $$z$$ as follows:

$\tag{64}\frac{P_a(z)}{P_a(0)}=\left|\frac{A(z)}{A(0)}\right|^2=\left|\frac{\tilde{A}(z)}{\tilde{A}(0)}\right|^2=\frac{\kappa_{ab}\kappa_{ba}}{\beta_c^2}\cos^2\beta_c{z}+\frac{\delta^2}{\beta_c^2}$$\tag{65}\frac{P_b(z)}{P_a(0)}=\left|\frac{B(z)}{A(0)}\right|^2=\left|\frac{\tilde{B}(z)}{\tilde{A}(0)}\right|^2=\frac{|\kappa_{ba}|^2}{\beta_c^2}\sin^2\beta_c{z}$

The coupling efficiency for a length $$l$$ is

$\tag{66}\eta=\frac{P_b(l)}{P_a(0)}=\frac{|\kappa_{ba}|^2}{\beta_c^2}\sin^2\beta_c{l}$

Thus, power is exchanged periodically between two modes with a coupling length

$\tag{67}l_c=\frac{\pi}{2\beta_c}$

where maximum power transfer occurs. Figure 4 below shows the periodic power exchange between the two coupled modes as a function of $$z$$. As can be seen from figure 4, complete power transfer can occur only in the phase-matched condition when $$\delta=0$$.

We now consider the coupling of two modes propagating in opposite directions over a length $$l$$, as is shown in figure 5 below where mode $$a$$ is forward propagating and mode $$b$$ is backward propagating.

In this case, $$\beta_a\gt0$$ and $$\beta_b\lt0$$. Thus, the coupled equations are

$\tag{68}\frac{\text{d}\tilde{A}}{\text{d}z}=\text{i}\kappa_{ab}\tilde{B}\text{e}^{\text{i}2\delta{z}}$$\tag{69}-\frac{\text{d}\tilde{B}}{\text{d}z}=\text{i}\kappa_{ba}\tilde{A}\text{e}^{-\text{i}2\delta{z}}$

These equations for contradirectional coupling are generally solved as a boundary-value problem with the boundary values of $$\tilde{A}(0)$$ at one end and $$\tilde{B}(l)$$ at the other end to find the values of $$\tilde{A}(z)$$ and $$\tilde{B}(z)$$ at any location $$z$$ between the two ends. The general solution can be expressed in the following matrix form:

$\tag{70}\begin{bmatrix}\tilde{A}(z)\\\tilde{B}(z)\end{bmatrix}=\mathbf{R}(z;0,l)\begin{bmatrix}\tilde{A}(0)\\\tilde{B}(l)\end{bmatrix}$

where the reverse-coupling matrix $$\mathbf{R}(z;0,l)$$ relates the filed amplitudes $$\tilde{A}(0)$$ at $$z=0$$ and $$\tilde{B}(l)$$ at $$z=l$$ to those at location $$z$$. It has the following form:

$\tag{71}\mathbf{R}(z;0,l)=\begin{bmatrix}\frac{\alpha_c\cosh\alpha_c(l-z)+\text{i}\delta\sinh\alpha_c(l-z)}{\alpha_c\cosh\alpha_cl+\text{i}\delta\sinh\alpha_cl}\text{e}^{\text{i}\delta{z}}&\frac{\text{i}\kappa_{ab}\sinh\alpha_cz}{\alpha_c\cosh\alpha_cl+\text{i}\delta\sinh\alpha_cl}\text{e}^{\text{i}\delta(z+l)}\\\frac{\text{i}\kappa_{ba}\sinh\alpha_c(l-z)}{\alpha_c\cosh\alpha_cl+\text{i}\delta\sinh\alpha_cl}\text{e}^{-\text{i}\delta{z}}&\frac{\alpha_c\cosh\alpha_cz+\text{i}\delta\sinh\alpha_cz}{\alpha_c\cosh\alpha_cl+\text{i}\delta\sinh\alpha_cl}\text{e}^{-\text{i}\delta(z-l)}\end{bmatrix}$

where

$\tag{72}\alpha_c=(\kappa_{ab}\kappa_{ba}-\delta^2)^{1/2}$

we consider a simple case when power is launched only into mode $$a$$ at $$z=0$$. Then the boundary values are $$\tilde{A}(0)\ne0$$ and $$\tilde{B}(l)=0$$. By applying these conditions to (70), we find that

$\tag{73}\tilde{A}(z)=\tilde{A}(0)\frac{\alpha_c\cosh\alpha_c(l-z)+\text{i}\delta\sinh\alpha_c(l-z)}{\alpha_c\cosh\alpha_cl+\text{i}\delta\sinh\alpha_cl}\text{e}^{\text{i}\delta{z}}$$\tag{74}\tilde{B}(z)=\tilde{A}(0)\frac{\text{i}\kappa_{ba}\sinh\alpha_c(l-z)}{\alpha_c\cosh\alpha_cl+\text{i}\delta\sinh\alpha_cl}\text{e}^{-\text{i}\delta{z}}$

The power in the two contradirectionally coupled modes varies with $$z$$ as follows:

$\tag{75}\frac{P_a(z)}{P_a(0)}=\left|\frac{A(z)}{A(0)}\right|^2=\left|\frac{\tilde{A}(z)}{\tilde{A}(0)}\right|^2=\frac{\cosh^2\alpha_c(l-z)-\delta^2/\kappa_{ab}\kappa_{ba}}{\cosh^2\alpha_cl-\delta^2/\kappa_{ab}\kappa_{ba}}$$\tag{76}\frac{P_b(z)}{P_a(0)}=\left|\frac{B(z)}{A(0)}\right|^2=\left|\frac{\tilde{B}(z)}{\tilde{A}(0)}\right|^2=\frac{\kappa_{ba}^*}{\kappa_{ab}}\frac{\sinh^2\alpha_c(l-z)}{\cosh^2\alpha_cl-\delta^2/\kappa_{ab}\kappa_{ba}}$

Because mode $$b$$ is propagating backward with no input at $$z=l$$ but an output at $$z=0$$, the coupling efficiency for a length $$l$$ is

$\tag{77}\eta=\frac{P_b(0)}{P_a(0)}=\frac{\kappa_{ba}^*}{\kappa_{ab}}\frac{\sinh^2\alpha_cl}{\cosh^2\alpha_cl-\delta^2/\kappa_{ab}\kappa_{ba}}$

Figure 6 below shows the power exchange between the two contradirectionally coupled modes as a function of $$z$$. As can be seen from figure 6, complete power transfer occurs as $$l\rightarrow\infty$$ if $$\delta^2\lt\kappa_{ab}\kappa_{ba}$$.

In the case when $$\tilde{A}(0)\ne0$$ and $$\tilde{B}(l)=0$$, as considered above, contradirectional coupling can be viewed as reflection of the field amplitude $$\tilde{A}(0)$$ at $$z=0$$ with a reflection coefficient

$\tag{78}r=|\text{r}|\text{e}^{\text{i}\varphi_\text{DBR}}=\frac{\tilde{B}(0)}{\tilde{A}(0)}=\frac{\text{i}\kappa_{ba}\sinh\alpha_cl}{\alpha_c\cosh\alpha_cl+\text{i}\delta\sinh\alpha_cl}$

The reflectivity is $$R=|\text{r}|^2=\eta$$ as is given in (77). The phase shift is

$\tag{79}\varphi_{\text{DBR}}=\varphi_\text{B}-\tan^{-1}\left(\frac{\delta}{\alpha_c}\tanh\alpha_cl\right)$

Conservation of Power

Conservation of power requires that in a lossless waveguide structure the net power flowing across any cross section of the waveguide be a constant independent of the longitudinal location of the cross section. For codirectional coupling with the power initially launched into only one mode so that $$P_a(0)\ne0$$ but $$P_b(0)=0$$, this requirement suggests that the sum of power in the two waveguides, $$P_a(z)+P_b(z)$$, is a constant because the power in the two modes flows in the same direction. For contradirectional coupling with the power launched into only one mode so that $$P_a(0)\ne0$$ and $$P_b(l)=0$$, this requirement suggests that $$P_a(z)-P_b(z)$$ is a constant because the power in mode $$b$$ flows in the backward direction while that in mode $$a$$ flows in the forward direction. These conclusions are correct for mode coupling in a single waveguide, but they do not generally hold for coupling between different waveguides.

It can be seen from (64) and (65) that $$P_a(z)+P_b(z)$$ is not a constant for codirectional coupling unless $$\kappa_{ab}=\kappa_{ba}^*$$. Similarly, from (75) and (76), it is also found that $$P_a(z)-P_b(z)$$ is not a constant for contradirectional coupling when $$\kappa_{ab}\ne\kappa_{ba}^*$$. It seems that the total power is not conserved in a lossless waveguide structure in the case of asymmetric coupling with $$\kappa_{ab}\ne\kappa_{ba}^*$$. A close examination reveals that because $$c_{ab}\ne0$$ [refer to the coupled-mode theory tutorial] in this case of asymmetric coupling, the two modes being coupled are not orthogonal to each other. Therefore, the total power flow cannot be fully accounted for by gathering the power in each individual mode as if the modes were orthogonal to each other. Indeed, by taking the total electric field and the total magnetic field expanded as (25) and (26), respectively, [refer to the coupled-mode theory tutorial], for two modes to calculate the power of the entire structure, we find that the total power flow as a function of space is

\tag{80}\begin{align}P(z)&=c_{aa}|A(z)|^2+c_{bb}|B(z)|^2+2\text{Re}[c_{ab}A^*(z)B(z)\text{e}^{\text{i}\Delta\beta{z}}]\\&=c_{aa}P_a(z)+c_{bb}P_b(z)+P_{ab}(z)\end{align}

where $$P_{ab}(z)=2\text{Re}[c_{ab}A^*(z)B(z)\text{e}^{\text{i}\Delta\beta{z}}]$$ can be considered as the power residing between the two nonorthogonal modes of the two different waveguides. As defined in the coupled-mode theory tutorial, $$c_{\nu\nu}=1$$ if mode $$\nu$$ is forward propagating and $$c_{\nu\nu}=-1$$ if mode $$\nu$$ is backward propagating. It can be shown, using (62) and (63) for the case of codirectional coupling and using (73) and (74) for the case of contradirectional coupling, that $$P(z)$$ is a constant independent of $$z$$ no matter whether $$\kappa_{ab}=\kappa_{ba}^*$$ or $$\kappa_{ab}\ne\kappa_{ba}^*$$. Therefore, conservation of power holds as expected.

When $$P_{ab}(z)=0$$, it can be shown simply by applying conservation of power that $$\kappa_{ab}=\kappa_{ba}^*$$; hence the coupling is symmetric. Conversely, if the coupling is symmetric, $$P_{ab}(z)$$ always vanishes even when mode $$a$$ and mode $$b$$ are not orthogonal to each other. Two conclusions can thus be made:

1. When $$c_{ab}=0$$, mode $$a$$ and mode $$b$$ are orthogonal to each other. Then $$P_{ab}(z)=0$$ and $$\kappa_{ab}=\kappa_{ba}^*$$ even when $$\delta\ne0$$ so that the two waveguide modes are not phase matched.
2. When the two modes are phase matched, $$\delta=0$$. In this case, $$P_{ab}(z)=0$$ and $$\kappa_{ab}=\kappa_{ba}^*$$ even when mode $$a$$ and mode $$b$$ are not orthogonal to each other with $$c_{ab}\ne0$$.

Consequently, coupling between two modes $$a$$ and $$b$$ is symmetric with $$\kappa_{ab}=\kappa_{ba}^*$$ if these two modes are orthogonal to each other or if they are phase matched.

Phase Matching

As can be seen from figure 4 and 6 above, power transfer is most efficient when $$\delta=0$$. The parameter $$\delta$$ is a measure of phase mismatch between the two modes being coupled. For the simple case when $$2\delta=\Delta\beta=\beta_b-\beta_a$$, the phase-matching condition $$\delta=0$$ is achieved when $$\beta_a=\beta_b$$. Then, the two modes have the same phase velocity and are synchronized. In case $$\delta$$ includes a contribution from additional structure, such as a periodic grating, phase matching of the two modes being coupled can be accomplished by matching the difference $$\Delta\beta=\beta_b-\beta_a$$ with a grating phase factor to make $$\delta=0$$. When considering phase matching between two modes, it is important always to include all sources of contribution to the phase-mismatch parameter $$\delta$$.

Phase-matched coupling is always symmetric, meaning that $$\kappa_{ab}=\kappa_{ba}^*$$ whenever $$\delta=0$$. This statement is true even when $$c_{ab}\ne0$$ and $$\beta_a\ne\beta_b$$. However, symmetric coupling does not necessarily imply a phase-matched condition. Therefore, it is also possible to have $$\kappa_{ab}=\kappa_{ba}^*$$ while $$\delta\ne0$$. The most obvious example of this situation is the coupling between two phase-mismatched modes in the same waveguide.

When perfect phase matching is accomplished, we can take

$\tag{81}\kappa=\kappa_{ab}=\kappa_{ba}^*\qquad\text{with}\qquad\kappa=|\kappa|\text{e}^{\text{i}\varphi}$

Because $$\delta=0$$, we find that

$\tag{82}\beta_c=\alpha_c=|\kappa|$

With these relations under the condition of perfect phase matching, the matrix $$\mathbf{F}(z;z_0)$$ for codirectional coupling is reduced to the simple form

$\tag{83}\mathbf{F}_{\text{PM}}(z;z_0)=\begin{bmatrix}\cos|\kappa|(z-z_0)&\text{ie}^{\text{i}\varphi}\sin|\kappa|(z-z_0)\\\text{ie}^{-\text{i}\varphi}\sin|\kappa|(z-z_0)&\cos|\kappa|(z-z_0)\end{bmatrix}$

and the matrix $$\mathbf{R}(z;0,l)$$ for contradirectional coupling is reduce to

$\tag{84}\mathbf{R}_{\text{PM}}(z;0,l)=\begin{bmatrix}\frac{\cosh|\kappa|(l-z)}{\cosh|\kappa|l}&\text{ie}^{\text{i}\varphi}\frac{\sinh|\kappa|z}{\cosh|\kappa|l}\\\text{ie}^{-\text{i}\varphi}\frac{\sinh|\kappa|(l-z)}{\cosh|\kappa|l}&\frac{\cosh|\kappa|z}{\cos|\kappa|l}\end{bmatrix}$

For codirectional coupling with perfect phase matching, the coupling efficiency is

$\tag{85}\eta_{\text{PM}}=\sin^2|\kappa|l$

and the coupling length is

$\tag{86}l_c^{\text{PM}}=\frac{\pi}{2|\kappa|}$

By choosing the interaction length to be $$l=l_c^{\text{PM}}$$, or any odd multiple of $$l_c^{\text{PM}}$$, 100% power transfer from one mode to the other with $$\eta_{\text{PM}}=1$$ can be accomplished.

Example 1

A phase-matched codirectional coupler has a coupling length of $$l_c^{\text{PM}}=1 \text{mm}$$ for a 100% coupling efficiency. What is the coupling coefficient of the coupler? For the same coupling coefficient, what is the length of a 3-dB codirectional coupler that has a 50% coupling efficiency?

From (86), we find that the coupling coefficient has a value

$|\kappa|=\frac{\pi}{2l_c^{\text{PM}}}=1.57\text{ mm}^{-1}$

From (85), we find that $$\eta_\text{PM}=1/2$$ when $$|\kappa|l=\pi/4$$. Therefore, the length of the 3-dB codirectional coupler is simply

$l_\text{3dB}=\frac{\pi}{4|\kappa|}=\frac{l_c^{\text{PM}}}{2}=0.5\text{ mm}$

A 3-dB codirectional coupler can be made by cutting the length of a 100% codirectional coupler in half. This statement is true even if the 100% coupler has a length longer than $$l_c^{\text{PM}}$$ at any odd integral multiple of $$l_c^{\text{PM}}$$.

For contradirectional coupling with perfect phase matching, the coupling efficiency is

$\tag{87}\eta_{\text{PM}}=\tanh^2|\kappa|l$

For an interaction length of $$l=l_c^{\text{PM}}$$ defined in (86), this gives a coupling efficiency of $$\eta_\text{PM}\approx84\%$$. Although complete power transfer with 100% efficiency cannot be accomplished in the case of contradirectional coupling, most power is transferred in a length comparable to the coupling length of codirectional coupling if phase matching is accomplished.

Example 2

A phase-matched contradirectional coupler has the same coupling coefficient as that of the codirectional coupler in example 1. What is the length of the contradirectional coupler for a 99% coupling efficiency? What is the length of a 3-dB contradirectional coupler with a 50% coupling efficiency?

A contradirectional coupler only approaches 100% efficiency asymptotically. From (87), we find that $$\eta_{\text{PM}}=99\%$$ when $$|\kappa|l=3\approx0.96\pi$$. Therefore, the length of the 99% contradirectional coupler with $$|\kappa|=1.57\text{ mm}^{-1}$$ as found in example 1 is

$l=\frac{3}{|\kappa|}=1.91\text{ mm}$

which is almost twice the length of the 100% codirectional coupler of the same coupling coefficient. We also find from (87) that $$\eta_\text{PM}=0.5$$ when $$|\kappa|l=0.88\approx0.28\pi$$. The length of the 3-dB contradirectional coupler is thus

$l_\text{3dB}=\frac{0.88}{|\kappa|}=0.56\text{ mm}$

which again is longer than the 3-dB codirectional coupler of the same coupling coefficient found in example 1. We also see that, unlike codirectional couplers, a 3-dB contradirectional coupler cannot be made by cutting in half a contradirectional coupler of nearly completely coupling at 99% efficiency.

In the presence of phase mismatch, symmetric coupling with $$\kappa_{ab}=\kappa_{ba}^*$$ is also true for coupling between two modes in the same waveguide but is not necessarily true for coupling between two different waveguides. Nevertheless, to illustrate the effect of phase mismatch on the coupling efficiency between two modes, we consider the simple case that $$\kappa=\kappa_{ab}=\kappa_{ba}^*$$, as expressed in (81). Then the coupling efficiency obtained in (66) for codirectional coupled modes can be written as

$\tag{88}\eta=\frac{1}{1+|\delta/\kappa|^2}\sin^2(|\kappa|l\sqrt{1+|\delta/\kappa|^2})$

The maximum efficiency is

$\tag{89}\eta_\text{max}=\frac{1}{1+|\delta/\kappa|^2}$

at a coupling length of

$\tag{90}l_c=\frac{l_c^{\text{PM}}}{\sqrt{1+|\delta/\kappa|^2}}$

The maximum coupling efficiency is clearly less than unity when $$\delta\ne0$$. As shown in figure 7(a) below, both $$l_c$$ and $$\eta_\text{max}$$ decreases as $$|\delta/\kappa|$$ increases. If the interaction length is fixed at $$l=l_c^{\text{PM}}$$, the efficiency also drops quickly as $$|\delta/\kappa|$$ increases, as shown in figure 7(b) below.

For contradirectional coupled modes, the coupling efficiency can also be expressed in terms of $$|\kappa|l$$ and $$|\delta/\kappa|$$:

$\tag{91}\eta=\frac{\sinh^2(|\kappa|l\sqrt{1-|\delta/\kappa|^2})}{\cosh^2(|\kappa|l\sqrt{1-|\delta/\kappa|^2})-|\delta/\kappa|^2}$

The coupling efficiency also decreases as phase mismatch increases, as shown in figure 8 below.

Example 3

Find the coupling efficiencies of codirectional and contradirectional couplers when the phase mismatch has the same magnitude as the coupling coefficient.

For a codirectional coupler with $$|\delta/\kappa|=1$$, we find from (88) that

$\tag{92}\eta=\frac{1}{2}\sin^2(\sqrt{2}|\kappa|l)$

For a contradirectional coupler with $$|\delta/\kappa|=1$$, we find from (91) that

$\tag{93}\eta=\frac{|\kappa|^2l^2}{1+|\kappa|^2l^2}$

It is interesting to see that when the phase mismatch has the same magnitude as the coupling coefficient, a codirectional coupler can only have a maximum coupling efficiency of 50% but a contradirectional coupler can still have an efficiency higher than 50% if $$|\kappa|l\gt1$$. However, the coupling efficiency of a contradirectional coupler varies with $$|\kappa|l$$ sinusoidally when $$|\delta/\kappa|\gt1$$ rather than monotonically as it does when $$|\delta/\kappa|\lt1$$.

In summary, to accomplish efficient coupling between two waveguide modes, the following three parameters have to be considered:

1. Coupling coefficient. The coupling coefficient $$\kappa$$ has to exist and be large enough.
2. Phase matching. The phase mismatch has to be minimized so that $$|\delta/\kappa|$$ is made as small as possible. Ideally, perfect phase matching with $$\delta=0$$ is desired.
3. Interaction length. For codirectional coupling, the length has to be properly chosen as the efficiency oscillates with interaction length. An overly long length is neither required nor beneficial. For contradirectional coupling, the length has to be sufficiently long but does not have to critically chosen as the efficiency increases monotonically with interaction length. A very long length is not necessary, either.

The next part continues with the grating waveguide couplers tutorial.