Transfer Matrix method is used to calculate the reflection and transmission behaviour of multilayer stacks. Multilayer stacks are used in Dielectric mirrors, Anti reflective coatings, High Reflective Dielectric mirrors etc. Here we will derive this method step-by-step using one layer sandwiched between air (n1) and substrate (n3).

In this system, we have three mediums - air, coating or layer, substrate. There are two interfaces - Interface 12 (between air and coating) and Interface 23 (between coating and substrate). We will assume that the substrate is thick enough such that there is not back reflection.

Normal Incidence

Assume light is incident normally on an interface. For a deeper understanding of what happens at this interface, refer Fresnel Coefficients article.

Medium 1 (air), Medium 2(layer), Medium 3(substrate) each has two waves (eg reflected and transmitted). If the thickness is along z, then + denotes fields in +z direction while - represents field in -z direction.

At interface 12:

\[ E_1^+ + E_1^- = E_2^+ + E_2^- \]

\[ H_1^+ - H_1^- = H_2^+ - H_2^- \]

Using the relation between H and E:

\[ H = \frac{n}{\mu_0 c} E \], we bring the fields in medium 1 in terms of medium 2. There is a reason why we choose to write medium 1 on the left side (instead of 2 which is the resultant medium).

\[ E_1^+ = \frac{1}{2}\left(1+\frac{n_2}{n_1}\right)E_2^+ + \frac{1}{2}\left(1-\frac{n_2}{n_1}\right)E_2^- \]

\[ E_1^- = \frac{1}{2}\left(1-\frac{n_2}{n_1}\right)E_2^+ + \frac{1}{2}\left(1+\frac{n_2}{n_1}\right)E_2^- \]

Combining these two, we get:

\[ \begin{pmatrix} E_1^+ \\ E_1^- \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1+\frac{n_2}{n_1} & 1-\frac{n_2}{n_1} \\ 1-\frac{n_2}{n_1} & 1+\frac{n_2}{n_1} \end{pmatrix}\begin{pmatrix} E_2^+ \\ E_2^- \end{pmatrix} \]

Now lets factor out the 1/n1 divide and multiply by n1+n2, we start gettign familiar expressions of fresnel coefficients r and t for this interface

\[ \begin{pmatrix} E_1^+ \\ E_1^- \end{pmatrix} = \frac{n_1+n_2}{2n_1}\begin{pmatrix} 1 & \frac{n_1-n_2}{n_1+n_2} \\ \frac{n_1-n_2}{n_1+n_2} & 1 \end{pmatrix}\begin{pmatrix} E_2^+ \\ E_2^- \end{pmatrix} \]

Replacing the expressions of r and t:

\[ \begin{pmatrix} E_1^+ \\ E_1^- \end{pmatrix} = \frac{1}{t_{12}}\begin{pmatrix} 1 & r_{12} \\ r_{12} & 1 \end{pmatrix}\begin{pmatrix} E_2^+ \\ E_2^- \end{pmatrix} \]

We call this matrix interface matrix and is denoted by:

\[ I_{12} = \frac{1}{t_{12}}\begin{pmatrix} 1 & r_{12} \\ r_{12} & 1 \end{pmatrix} \]

In Medium 2:

The field in medium 2 mentioned above is at z = 0. Now in order to move forward with our calculations, we need to find out the field in medium 2 at z = d (where d is the layer thickness). Once we have this we can then relate the fields at the next interface.

\[ E_2^+(d) = E_2^+(0)e^{-ik_2d} \]

\[ E_2^-(d) = E_2^-(0)e^{+ik_2d} \]

\[ \begin{pmatrix} E_2^+(0) \\ E_2^-(0) \end{pmatrix} = \begin{pmatrix} e^{ik_2d} & 0 \\ 0 & e^{-ik_2d} \end{pmatrix}\begin{pmatrix} E_2^+(d) \\ E_2^-(d) \end{pmatrix} \]

The matrix here is called the propagation matrix.

\[ P_2 = \begin{pmatrix} e^{ik_2d} & 0 \\ 0 & e^{-ik_2d} \end{pmatrix} \]

Now finally at the last interface.

At interface 23:

We can apply the same equations as interface 12. The matrix relationship we will get here is:

\[ \begin{pmatrix} E_2^+(d) \\ E_2^-(d) \end{pmatrix} = I_{23}\begin{pmatrix} E_3^+ \\ E_3^- \end{pmatrix} \]

where the interface matrix is :

\[ I_{23} = \frac{1}{t_{23}}\begin{pmatrix} 1 & r_{23} \\ r_{23} & 1 \end{pmatrix} \]

Now lets combine them all together.

We will start from the first relationshio we had:

\[ \begin{pmatrix} E_1^+ \\ E_1^- \end{pmatrix} = \frac{1}{t_{12}}\begin{pmatrix} 1 & r_{12} \\ r_{12} & 1 \end{pmatrix}\begin{pmatrix} E_2^+ \\ E_2^- \end{pmatrix} \]

Replacing the electric field of medium 2 at z =0 in the above equation with

\[ \begin{pmatrix} E_2^+(0) \\ E_2^-(0) \end{pmatrix} = \begin{pmatrix} e^{ik_2d} & 0 \\ 0 & e^{-ik_2d} \end{pmatrix}\begin{pmatrix} E_2^+(d) \\ E_2^-(d) \end{pmatrix} \]

we get:

\[ \begin{pmatrix} E_1^+ \\ E_1^- \end{pmatrix} = \frac{1}{t_{12}}\begin{pmatrix} 1 & r_{12} \\ r_{12} & 1 \end{pmatrix}\begin{pmatrix} e^{ik_2d} & 0 \\ 0 & e^{-ik_2d} \end{pmatrix} \begin{pmatrix} E_2^+(d) \\ E_2^-(d) \end{pmatrix} \]

And now we replace medium 2 z=d electric fields with the last interface equation, and we get:

\[ \begin{pmatrix} E_1^+ \\ E_1^- \end{pmatrix} = \frac{1}{t_{12}}\begin{pmatrix} 1 & r_{12} \\ r_{12} & 1 \end{pmatrix}\begin{pmatrix} e^{ik_2d} & 0 \\ 0 & e^{-ik_2d} \end{pmatrix} \frac{1}{t_{23}}\begin{pmatrix} 1 & r_{23} \\ r_{23} & 1 \end{pmatrix} \begin{pmatrix} E_3^+ \\ E_3^- \end{pmatrix} \]

This is concisely written as :

\[ \begin{pmatrix} E_1^+ \\ E_1^- \end{pmatrix} = M\begin{pmatrix} E_3^+ \\ E_3^- \end{pmatrix} \]

Here M is called as the tranfer matrix and is:

\[ M = I_{12}P_2I_{23} = \begin{pmatrix} M_{11} & M_{12} \\ M_{21} & M_{22} \end{pmatrix} \]

Now lets assume that the substrate is so thick that there is no field in - direction in the substrate. This means:

\[ E_3^- = 0 \]

Thus,

\[ E_1^+ = M_{11}E_3^+ \]

\[ E_1^- = M_{21}E_3^+ \]

We know that r is the ration of reflected field in medium 1 and incident field of medium 1, thus divide the two equations to get that.

\[ r = \frac{E_1^-}{E_1^+} = \frac{M_{21}}{M_{11}} \]

and t is the ratio of transmitted field in medium 3 and incident field of medium 1,

\[ t = \frac{E_3^+}{E_1^+} = \frac{1}{M_{11}} \]

The Reflectance and Tranmittance is :

\[ R = |r|^2 = \left|\frac{M_{21}}{M_{11}}\right|^2 \]

\[ T = \frac{n_3}{n_1}|t|^2 = \frac{n_3}{n_1}\left|\frac{1}{M_{11}}\right|^2 \]