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# The Transfer Matrix for a Boundary

1. Introduction

Let’s apply the material from Making Friends with Electromagnetic Boundary Conditions to calculate a transfer matrix. We will consider a TE wave travelling between materials with parameters ${\mu_1,\epsilon_1}$ and ${\mu_2,\epsilon_2}$: There are a few things to note:

• This is a TE wave, so the electric field ${\mathbf{E}}$ is transverse to the plane of incidence. In this case we chose to let ${\mathbf{E}}$ point into the page, which then determines the direction of the magnetic field since ${(\mathbf{k},\mathbf{E},\mathbf{B})}$ form a right-handed coordinate system.
• The total wavevector ${\mathbf{k}}$ is made up of a component ${k_z}$ in the ${z}$-direction and ${k_x}$ in the ${x}$ direction.
• The field on each side is made up of a forward-propagating wave ( ${a_1,c_1}$) and a backward propagating wave ( ${b_1,d_1}$). On the left we have $\displaystyle E=\left(a_1e^{i(k_{1z}z-\omega t)}+b_1e^{i(-k_{1z}z-\omega t)}\right)e^{ik_xx}, \ \ \ \ \ (1)$

and similarly for the right.

We want ${(a_1,b_1)}$ to represent the electric field in the first medium, but do we want them to represent ${\mathbf{E}}$ or ${\mathbf{D}}$? To decide, let’s take a look at the boundary conditions: $\displaystyle \begin{gathered} \hat{\mathbf{n}}\cdot\left(\mathbf{B}_2-\mathbf{B}_1\right)=0;\; \hat{\mathbf{n}}\times\left(\mathbf{E}_1-\mathbf{E}_2\right)=0, \\ \hat{\mathbf{n}}\cdot\left(\mathbf{D}_2-\mathbf{D}_1\right)=0;\;\hat{\mathbf{n}}\times\left(\mathbf{H}_2-\mathbf{H}_1\right)=0. \end{gathered} \ \ \ \ \ (2)$

The electric field is perpendicular to the boundary, so ${\hat{\mathbf{n}}\cdot(\mathbf{D}_2-\mathbf{D}_1)=0}$. On the other hand since ${\mathbf{E}}$ is perpendicular to ${\hat{\mathbf{n}}}$, you can convince yourself that the other boundary condition for the electric field gives ${\mathbf{E}_1=\mathbf{E}_2}$. Thus the electric field is continuous across the boundary, so we want ${(a_1,b_1)}$ to represent the electric field. In this case if ${a_1,b_1}$ are the fields just before the boundary and ${c_1,d_1}$ the fields just after, we have \displaystyle \begin{aligned} \mathbf{E}_1 &= \mathbf{E}_2, \\ a_1+b_1&=c_1+d_1. \end{aligned} \ \ \ \ \ (3)

Note that we have chosen coordinates such that the exponentials in Eq. (1) are one right at the boundary.

Aside: If we had a TM wave we would choose ${(a_1,b_1)}$ to represent ${\mathbf{H}}$ rather than ${\mathbf{B}}$, since ${\mathbf{H}}$ is what would be continuous across the boundary and we would again have ${a_1+b_1=c_1+d_1}$. Note that this is just a convenience; you could use ${\mathbf{B}}$ if you really wanted to, and using ${\mathbf{B}=\mu \mathbf{H}}$ the boundary condition would then become ${(a_1+b_1)/\mu_1=(c_1+d_1)/\mu_2}$.

We need one more relation between the ${a_1,b_1,c_1,d_1}$, and we have two two boundary conditions we haven’t used so far: $\displaystyle \hat{\mathbf{n}}\cdot(\mathbf{B}_2-\mathbf{B}_1)=0;\;\hat{\mathbf{n}}\times(\mathbf{H}_2-\mathbf{H}_1)=0. \ \ \ \ \ (4)$

These are in terms of the magnetic field, which we want to relate to the electric field. This comes from Maxwell’s equation: $\displaystyle \nabla\times\mathbf{E}=-\partial_t\mathbf{B}=i\omega\mathbf{B}. \ \ \ \ \ (5)$

What would happen if we took the ${\mathbf{B}}$ equation? Since ${\mathbf{B}}$ is being dotted with ${\hat{\mathbf{n}}}$, we see that the ${z}$-component of ${\mathbf{B}}$ is continuous across the boundary. This tells us that ${(\nabla\times\mathbf{E})_z}$ is continuous, which is: $\displaystyle (\nabla\times\mathbf{E})_z=\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}=\frac{\partial E_y}{\partial x}, \ \ \ \ \ (6)$

since only ${E_y}$ is nonzero. Meanwhile the ${\mathbf{H}}$ equation gives us a relation for the ${x}$-component of ${\mathbf{H}}$, and $\displaystyle (\nabla\times\mathbf{E})_x=\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}=-\frac{\partial E_y}{\partial z}. \ \ \ \ \ (7)$

The wave is propagating in the ${z}$ direction, so this is what we want.

Using ${\mathbf{B}=\mu\mathbf{H}}$, Eq. (7) tells us that $\displaystyle \left.\frac{1}{\mu_1}\frac{\partial E_y}{\partial z}\right\rvert_{Left}=\left.\frac{1}{\mu_2}\frac{\partial E_y}{\partial z}\right\rvert_{Right}. \ \ \ \ \ (8)$

Looking at \cref{eq:TotalEField} the left hand side is $\displaystyle \frac{ik_{1z}}{\mu_1}\left(a_1-b_1\right), \ \ \ \ \ (9)$

while the right hand side will be $\displaystyle \frac{ik_{2z}}{\mu_2}\left(c_1-d_1\right). \ \ \ \ \ (10)$

Thus this boundary condition is: $\displaystyle a_1-b_1=\frac{\mu_1}{\mu_2}\frac{k_{2z}}{k_{1z}}(c_1-d_1). \ \ \ \ \ (11)$

The last thing is to write ${k_{2z}/k_{1z}}$ in terms of the parameters of the material. We know that ${|k|=\omega\sqrt{\mu\epsilon}}$, and ${k_x^2+k_z^2=k^2}$. Thus \displaystyle \begin{aligned} k_{1z} &= \sqrt{\omega^2\mu_1\epsilon_1-k_x^2}, \\ k_{2z} &= \sqrt{\omega^2\mu_2\epsilon_2-k_x^2}. \\ \end{aligned} \ \ \ \ \ (12)

Eq. (11) thus gives us our second boundary equation: $\displaystyle a_1-b_1=\frac{\mu_1}{\mu_2}\sqrt{\frac{\omega^2\mu_2\epsilon_2-k_x^2}{\omega^2\mu_1\epsilon_1-k_x^2}}(c_1-d_1). \ \ \ \ \ (13)$

This is in terms of the frequency of the wave ${\omega}$, the parameters ${\mu_i,\epsilon_i}$ of the material, as well as ${k_x}$, which is a parameter telling us the angle of the input wave. For normal incidence we have ${k_x=0}$.

2. References

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