# Wave Equations for Optical Waveguides

This is a continuation from the Optical Waveguide Field Equations tutorial.

The field equations obtained in the optical waveguide field equations tutorial establish the relations among the field components. In general, it is only necessary to find $$\mathcal{E}_z$$ and $$\mathcal{H}_z$$. Then all other components can be calculated by simply using (16) - (19). The common approach to finding $$\mathcal{E}_z$$ and $$\mathcal{H}_z$$ is to solve the wave equations together with boundary conditions. In this tutorial, we examine the form of the wave equations for waveguides.

To obtain the wave equations, we need the other two Maxwell's equations in addition to (8) and (9). For the case of a linear, isotropic waveguide with a spatially dependent ε(x, y) discussed here, they can be written as

$\tag{21}\pmb{\nabla}\cdot(\epsilon\mathbf{E})=0$

$\tag{22}\pmb{\nabla}\cdot\mathbf{H}=0$

Note that (21) implies that

$\tag{23}\pmb{\nabla}\cdot\mathbf{E}=-\frac{\pmb{\nabla}\epsilon}{\epsilon}\cdot\mathbf{E}$

which does not vanish in general because ε(xy) is spatially dependent. Using the four Maxwell's equations in (8), (9), (21), and (22), together with (23) and the vector identity $$\pmb{\nabla}\times\pmb{\nabla}\times=\pmb{\nabla}\pmb{\nabla}\cdot-\nabla^2$$, we have

$\tag{24}\nabla^2\mathbf{E} + k^2\mathbf{E} = - \pmb{\nabla}\left(\frac{\pmb{\nabla}\epsilon}{\epsilon}\cdot\mathbf{E}\right)$

$\tag{25}\nabla^2\mathbf{H} + k^2\mathbf{H} = - \frac{\pmb{\nabla}\epsilon}{\epsilon}\times\pmb{\nabla}\times\mathbf{H}$

It can be seen that the three components Ex, Ey, and Ez for the electric field are generally coupled together because ∇ε ≠ 0 in a waveguide. For the same reason, HxHy, and Hz are also coupled. This fact indicates that the vectorial characteristics of a mode field in a waveguide are strongly dependent on the geometry and index profile of the waveguide.

If the terms on the right-hand sides of (24) and (25) vanish, then the field components are decoupled. This condition exists only in certain special cases. For example, in the case of a TE mode,

$\tag{26}\pmb{\nabla}\epsilon\bot\mathbf{E}\qquad\text{so that}\qquad\pmb{\nabla}\epsilon\cdot\mathbf{E}=0$

As a consequence, each component of the electric field of a TE mode satisfies a homogeneous scalar differential equation. The magnetic field components of a TE mode are still coupled because the right-hand term of (25) does not vanish.

The index profile of a step-index waveguide is piecewise constant. We can write a homogeneous wave equation separately for each region of constant ε because ∇ε = 0 within each region. By taking E and H in the form of (1) and (2), respectively [refer to the waveguide modes tutorial], and using (24) and (25) with ∇ε = 0 for each region of constant ε, we obtain

$\tag{27}\frac{\partial^2\mathcal{E}_z}{\partial x^2} + \frac{\partial^2\mathcal{E}_z}{\partial y^2} + (k_i^2-\beta^2)\mathcal{E}_z = 0$

$\tag{28}\frac{\partial^2\mathcal{H}_z}{\partial x^2} + \frac{\partial^2\mathcal{H}_z}{\partial y^2} + (k_i^2-\beta^2)\mathcal{H}_z = 0$

where

$\tag{29}k_i^2=\omega^2\mu_0\epsilon_i=n_i^2\frac{\omega^2}{c^2}$

is a constant for region i, which has a constant index of refraction ni.

A homogeneous equation in the same form can be written for each of the other four field components, $$\mathcal{E}_x,\mathcal{E}_y,\mathcal{H}_x,\text{and}\,\mathcal{H}_y$$. However, it is not necessary to solve the wave equations for all field components because the transverse field components can be found from $$\mathcal{E}_z$$ and $$\mathcal{H}_z$$ using the relations in (16) - (19) [refer to the field equations tutorial]. Therefore, the mode field pattern can be obtained by solving only (27) and (28) for each region of constant index and by requiring the fields to satisfy the boundary conditions at the interfaces between neighboring regions. Clearly, this approach does not work for graded-index waveguides because (27) and (28) are not valid for such waveguides.

Wave Equations for Planar Waveguides

Homogeneous wave equations exist for planar waveguides of any index profile n(x). For a planar waveguide, the modes are either TE or TM.  Furthermore, ∂/∂y = 0 because the index profile is independent of the y coordinate. The wave equations are substantially simplified.

1. TE modes. For any TE mode of a planar waveguide, $$\mathcal{E}_z=0$$. It can be seen from (16) - (19) that [refer to the field equations tutorial] $$\mathcal{E}_x=\mathcal{H}_y=0$$ as well because $$\partial\mathcal{H}_z/\partial y=0$$. The only non-vanishing field components are $$\mathcal{H}_x,\mathcal{E}_y,\text{and}\,\mathcal{H}_z$$. Because there is only one non-vanishing electric field component $$\mathcal{E}_y$$, the wave equation for $$\mathcal{E}_y$$ is naturally decoupled from the other field components. Therefore, we have

$\tag{30}\frac{\partial^2\mathcal{E}_y}{\partial x^2}+(k^2-\beta^2)\mathcal{E}_y=0$

where

$\tag{31}k^2=\omega^2\mu_0\epsilon(x)=\frac{\omega^2}{c^2}n^2(x)$

Using (10) and (12) [refer to the field equations tutorial], the other two non-vanishing field components can be obtained from $$\mathcal{E}_y$$:

$\tag{32}\mathcal{H}_x=-\frac{\beta}{\omega\mu_0}\mathcal{E}_y$

$\tag{33}\mathcal{H}_z=\frac{1}{\text{i}\omega\mu_0}\frac{\partial\mathcal{E}_y}{\partial x}$

2. TM modes. For any TM mode of a planar waveguide, $$\mathcal{H}_z=0$$. Then, $$\mathcal{H}_x=\mathcal{E}_y=0$$ because $$\partial\mathcal{E}_z/\partial y=0$$. The only non-vanishing field components are $$\mathcal{E}_x,\mathcal{H}_y,\text{and}\,\mathcal{E}_z$$. In this case, there is only one non-vanishing magnetic field component $$\mathcal{H}_y$$, and the wave equation for $$\mathcal{H}_y$$ is naturally decoupled from the other field components. From (25), we have

$\tag{34}\frac{\partial^2\mathcal{H}_y}{\partial x^2}+(k^2-\beta^2)\mathcal{H}_y=\frac{1}{\epsilon}\frac{\text{d}\epsilon}{\text{d}x}\frac{\partial\mathcal{H}_y}{\partial x}$

where k2 = k2(x) is the same as that given by (31). The other two non-vanishing field components can be obtained from $$\mathcal{H}_y$$:

$\tag{35}\mathcal{E}_x=\frac{\beta}{\omega\epsilon}\mathcal{H}_y$

$\tag{36}\mathcal{E}_z=-\frac{1}{\text{i}\omega\epsilon}\frac{\partial\mathcal{H}_y}{\partial x}$

In the case of planar waveguide, it is convenient to solve for the unique transverse field component first: $$\mathcal{E}_y$$ for a TE mode and $$\mathcal{H}_y$$ for a TM mode. The other field components, including the longitudinal component, then follow directly.

Although there is only one non-vanishing longitudinal field component for each type of mode in a planar waveguide, it is coupled to a transverse field component. For example, $$\mathcal{H}_z$$ of a TE mode is coupled to $$\mathcal{H}_x$$ and is not described by a simple equation of the form (30).

Power and Orthogonality

Except for evanescent fields, the energy of the fields in a waveguide flows only in the longitudinal direction. The intensity of a waveguide mode ν is thus given by

$\tag{37}\mathbf{I}_\nu=\bar{\mathbf{S}}_\nu\cdot\hat{z}=(\mathbf{S}_\nu+\mathbf{S}_\nu^*)\cdot\hat{z}=(\mathbf{\mathcal{E}}_\nu\times\mathbf{\mathcal{H}}_\nu^*+\mathbf{\mathcal{E}}_\nu^*\times\mathbf{\mathcal{H}}_\nu)\cdot\hat{z}$

which is a function of x and y. The power, , of the mode is obtained by integrating Iν(x, y) over the entire transverse cross section of the waveguide. It can be seen that the longitudinal components, $$\mathcal{E}_z$$ and $$\mathcal{H}_z$$, of the mode fields do not contribute to the mode intensity or the mode power.

For TE and TM modes, the power obtained by integrating Iν(xy) given in (37) can be transformed into other forms. It can be shown, using (10) and (11) [refer to the field equations tutorial], that the power of a TE mode is simply given by

$\tag{38}P_{\text{TE}}=\frac{2\beta}{\omega\mu_0}\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}|\pmb{\mathcal{E}}|^2\text{d}x\text{d}y$

By using (13) and (14) [refer to the field equations tutorial], the power of a TM mode can be expressed as

$\tag{39}P_{\text{TM}}=\frac{2\beta}{\omega}\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\frac{1}{\epsilon(x,y)}|\pmb{\mathcal{H}}|^2\text{d}x\text{d}y$

In a lossless isotropic waveguide, the mode fields have the following orthogonality relation:

$\tag{40}\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}(\pmb{\mathcal{E}}_\nu\times\pmb{\mathcal{H}}_\mu^*+\pmb{\mathcal{E}}_\mu^*\times\pmb{\mathcal{H}}_\nu)\cdot\hat{z}\text{d}x\text{d}y=\pm P_\nu\delta_{\nu\mu}$

The mode fields can be normalized to have the following orthonormality relation:

$\tag{41}\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}(\hat{\pmb{\mathcal{E}}}_\nu\times\hat{\pmb{\mathcal{H}}}_\mu^*+\hat{\pmb{\mathcal{E}}}_\mu^*\times\hat{\pmb{\mathcal{H}}}_\nu)\cdot\hat{z}\text{d}x\text{d}y=\pm\delta_{\nu\mu}$

where the plus sign is for forward-propagating modes while the minus sign is for backward-propagating modes.

The electric and magnetic field patterns of a particular mode ν are represented by the normalized mode field distribution $$\hat{\pmb{\mathcal{E}}}_\nu(x,y)$$ and $$\hat{\pmb{\mathcal{H}}}_\nu(x,y)$$, respectively.

Here δνμ is the Kronecker delta function for discrete modes. For a nonplanar waveguide, ν = mn and μ = m'n'; hence δνμ = δmm' δnn'. For a planar waveguide, ν = m and μ = m', and δνμ = δmm' .

For continuous modes, δνμ has to be replaced by the Dirac delta function δ(ν - μ). For a nonplanar waveguide, δ(ν - μ) = δ(a - a')δ(b - b') for ν = ab and μ = a'b'. The Dirac delta function is defined as

\tag{42}\delta(\nu-\mu)=\left\lbrace\begin{align}&0,\qquad\;\;\;\nu\ne\mu\\&\infty,\qquad\;\nu=\mu\end{align}\right.

and

$\tag{43}\displaystyle\int\limits_{-\infty}^{\infty}\delta(\nu)\text{d}\nu=1$

For TE modes, the orthonormality relation in (41) can be transformed to

$\tag{44}\frac{2\beta_\nu}{\omega\mu_0}\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\hat{\pmb{\mathcal{E}}}_\nu\cdot\hat{\pmb{\mathcal{E}}}_\mu^*\text{d}x\text{d}y=\delta_{\nu\mu}$

For TM modes, we have

$\tag{45}\frac{2\beta_\nu}{\omega}\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\frac{1}{\epsilon(x,y)}\hat{\pmb{\mathcal{H}}}_\nu\cdot\hat{\pmb{\mathcal{H}}}_\mu^*\text{d}x\text{d}y=\delta_{\nu\mu}$

The orthogonality relation in (40), or the orthonormality relation in (41), indicates that power cannot be transferred between different modes in a linear, lossless waveguide.

For anisotropic or lossy waveguides, (40) and (41) do not apply, neither do (44) and (45). The orthogonality conditions for modes of such waveguides have other forms.

The next part continues with the step-index planar waveguides tutorial.