Theory of the transfer matrix method

On the left of the interface the electric field is given by

\[E_{1}=E^{+}_{1} e^{-j k_1 z}+E^{-}_{1} e^{j k_1 z} \label{efield1}\] and on the right hand side of the interface the electric field is given by \[E_{2}=E^{+}_{2} e^{-j k_2 z}+E^{-}_{2} e^{j k_2 z} \label{efield2}\]

Maxwel’s equations give us the relationship between the electric and magnetic fields for a plane wave.

\[\nabla \times E=-j\omega \mu H\] which simplifies to: \[\frac{\partial E} {\partial z}=-j\omega \mu H \label{maxwel}\]

Applying equation ?? to equations ??-??, we can get the magnetic field on the left of the interface \[-j \mu \omega H^{y}_{1}=-j k_1 E^{+}_{1} e^{-j k_1 z}+j k_1 E^{-}_{1} e^{j k_1 z}\] and on the right of the interface \[-j \mu \omega H^{y}_{2}=-j k_2 E^{+}_{2} e^{-j k_2 z}+j k_2 E^{-}_{2} e^{j k_2 z}.\]

Tidying up gives, \[H^{y}_{1}=\frac{k}{\omega \mu}E^{+}_{1} e^{-j k_1 z}-\frac{k}{\omega \mu} E^{-}_{1} e^{j k_1 z}\]

\[H^{y}_{2}=\frac{k}{\omega \mu}E^{+}_{2} e^{-j k_2 z}-\frac{k}{\omega \mu} E^{-}_{2} e^{j k_2 z}\]

Boundary conditions

We now apply the electric and magnetic boundary conditions \[\mathbf{n} \times (\mathbf{E_2}-\mathbf{E_1})=0\]

\[\mathbf{n} \times (\mathbf{H_2}-\mathbf{H_1})=0\]

We let the interface be at z=0, which gives, \[(E_{2}^{+}+E_{2}^{-})-(E_{1}^{+}+E_{1}^{-})=0 \label{electric_boundary}\] and \[\frac{k_1}{\omega \mu}(E_{2}^{+}-E_{2}^{-})-(E_{1}^{+}-E_{1}^{-})\frac{k_2}{\omega \mu}=0\] . The wavevector is given by \[k=\frac{2 \omega }{\lambda}=\frac{\omega n}{c}\] . We can therefore write the magnetic boundary condition as \[n_2 (E_{2}^{+}-E_{2}^{-}) - n_1 (E_{1}^{+}-E_{1}^{-})=0 \label{mag_boundary}\]

Forward propagating wave

Rearrange equation, ?? to give,

\[E_{1}^{-} = E_{1}^{+}-\frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})\] Inserting in equation ??, gives \[E_{2}^{+}+E_{2}^{-}=E_{1}^{+}+E_{1}^{+}-\frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})\]

\[2E_{1}^{+}=E_{2}^{+}+E_{2}^{-}+\frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})\]

\[2E_{1}^{+}\frac{n_1}{n_1+n_2}=E_{2}^{+}+E_{2}^{-}\frac{n_1-n_2}{n_1+n_2}\]

Backwards propagating wave

Rearrange equation, ?? to give,

\[E_{1}^{+}=E_{1}^{-} +\frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})\]

Inserting in equation ??, gives \[E_{2}^{+}+E_{2}^{-}=E_{1}^{-} +\frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})+E_{1}^{-}\]

\[2E_{1}^{-}=E_{2}^{+}+E_{2}^{-}- \frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})\]

\[2E_{1}^{-}\frac{n_1}{n_1+n_2}=E_{2}^{+}\frac{n_1-n_2}{n_1+n_2}+E_{2}^{-}\] Which is the same result as obtained in .

These equations become:

\[E_{1}^{-}t_{12}=E_{2}^{+}r_{12}+E_{2}^{-}\]

and \[E_{1}^{+}t_{12}=E_{2}^{+}+E_{2}^{-}r_{12}\]

Accounting for propagation we can write. Note the change in sign between and this work, this is because of how I have defined my wave equation. \[E_{1}^{+}t_{12}=E_{2}^{+}e^{\zeta_2 d_1}+E_{2}^{-}r_{12}e^{-\zeta_2 d_1}\] and

\[E_{1}^{-}t_{12}=E_{2}^{+}r_{12}e^{\zeta_2 d_1}+E_{2}^{-}e^{-\zeta_2 d_1}\]

where \[\zeta=\frac{2\pi}{\lambda} \bar{n}\]

For a device with non reflecting back contacts:

\(\begin{pmatrix} e^{\zeta d} & 0 & 0 & 0 &r_{01}e^{-\zeta d} & 0 & 0 & 0 \\ -t_{12} & e^{\zeta d} & 0 & 0 &0 & r_{12}e^{-\zeta d} & 0 & 0 \\ 0 & -t_{23} & e^{\zeta d} & 0 &0 & 0 & r_{23}e^{-\zeta d} & 0 \\ 0 & 0 & -t_{34} & e^{\zeta d} & 0 & 0 & 0 & r_{34}e^{-\zeta d} \\ 0 &r_{12}e^{\zeta d_1} & 0 & 0 & -t_{12} &e^{-\zeta d} & 0 & 0 \\ 0 &0 & r_{23}e^{\zeta d} & 0 & 0 &-t_{23} & e^{-\zeta d} & 0 \\ 0 &0 & 0 & r_{34}e^{\zeta d} & 0 &0 & -t_{34} & e^{-\zeta d} \\ 0 &0 & 0 & 0 & 0 &0 & 0 & -t_{45} \\ \end{pmatrix} \begin{pmatrix} E_{1}^{+} \\ E_{2}^{+} \\ E_{3}^{+} \\ E_{4}^{+} \\ E_{1}^{-} \\ E_{2}^{-} \\ E_{3}^{-} \\ E_{4}^{-} \\ \end{pmatrix} = \begin{pmatrix} t_{01}E_{external} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}\)

Refractive index and absorption

\[E(z,t)=Re(E_0 e^{j(-kz+\omega t)})= Re(E_0 e^{j(\frac{-2 \pi (n+j\kappa)}{\lambda}z + \omega t)})=e^{\frac{2\pi\kappa z}{\lambda}}Re(E_0 e^{\frac{j(-2 \pi (n+j\kappa)}{\lambda}z +\omega t})\] And because the intensity is proportional to the square of the electric field the absorption coefficient becomes

\[e^{-\alpha x}=e^{\frac{2\pi\kappa z}{\lambda}}\]

\[\alpha=-\frac{4\pi\kappa}{\lambda_0}\]