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\exnumber{2660}
\heading{Scattering matrix - The Mach-Zehnder Interferometer}
\auname{Isaac Weinberg}



\textbf{The problem:}

\Dn

The Mach-Zehnder Interferometer is built from two semi-reflecting
mirrors and two regular mirrors.\\
 A regular mirror has a return amplitude of $1$ so the only effect
on the reflected beam is a phase change of $\pi$.There are a few
passable ways to write the matrix of the semi-reflecting mirror here
are two of them each describes a different mirror , the first symmetric
and the second is not symmetric. (each matrix will gives a different
result but we can build an interferometer from both)\\
\\
 A particles beam is split at the first semi-reflecting mirror to
two beams which go through two different optical courses which cause
a phase difference of $\phi$.\\
 They reunite in the second semi-reflecting mirror and the particles
can exit in one of two channels. 
\begin{enumerate}
\item Write a $4\times4$ matrix which describe a semi-reflecting mirror.
The transmission (T) and reflection (R) are given. Explain why the
semi-reflective mirror can be treated as a two channel junction problem
with $2\times2$ matrix.
\item Write an expression for the two channels.
\item Make sure that the sum of the channels is one. 
\end{enumerate}
\Dn\Dn

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\textbf{The solution:}\\
\emph{original solution by Meny gabbay}
\Dn

(1)

We will use the lower left semi-reflecting mirror as our reference
mirror and describe the 4 channels relative to it.

channel 1 is on the left (the light source)

channel 2 is the upper one.

channel 3 is on the right.

channel 4 is the lower one.

We will define the wave function in each channel as: $\psi_{a}=\frac{{1}}{\sqrt{{v_{a}}}}(A_{a}e^{-ikr}+B_{a}e^{+ikr})$
A is the amplitude of the incoming wave and B of the outgoing wave
(from the mirror)

The S matrix is defined as: $B_{b}=\sum_{a}S_{ba}A_{a}$which is the
matching conditions of the ingoing and outgoing amplitudes;

The properties of our S matrix:
\begin{enumerate}
\item Since the mirror isn\textquoteright{}t absorbing the light the intensity
which goes in equal the intensity which goes out: $\sum{|A_{n}|^{2}}=\sum{|B_{n}|^{2}}$
which mean $SS^{\dagger}=I$
\item Because of the symmetry, a beam coming from channel 1 and reflected
to channel 2 is the same as a beam coming from channel 2 and reflected
to channel 1 which mean: $S_{ij}=S_{ji}$
\item Because the mirror is slanted, a beam cannot be returned to the channel
from which it came: $S_{ii}=0$
\item Because the mirror is slanted, $S_{14}=S_{23}=S_{32}=S_{41}=0$
\end{enumerate}
Due to properties 2,3 and 4 the S matrix has only 8 unknowns ($S_{ij}$
are complex numbers): 
\[
S=\left(\begin{array}{cccc}
0 & S_{12} & S_{13} & 0\\
S_{12} & 0 & 0 & S_{24}\\
S_{13} & 0 & 0 & S_{34}\\
0 & S_{24} & S_{34} & 0
\end{array}\right)
\]


Using property 1 we get the following equations:

\begin{multline*}
|S_{12}|^{2}+|S_{13}|^{2}=1\\
|S_{12}|^{2}+|S_{24}|^{2}=1\\
|S_{13}|^{2}+|S_{34}|^{2}=1\\
|S_{24}|^{2}+|S_{34}|^{2}=1\\
S_{12}S_{24}^{\ast}+S_{13}S_{34}^{\ast}=0\\
S_{12}S_{13}^{\ast}+S_{24}S_{34}^{\ast}=0\\
S_{13}S_{12}^{\ast}+S_{34}S_{24}^{\ast}=0\\
S_{24}S_{12}^{\ast}+S_{13}S_{34}^{\ast}=0\\
\end{multline*}


The scattering matrix determine the matching conditions up to a phase
so we can choose for example that $S_{13}=t$ and is a real number.

We use the rest of the equations above to determine the rest of the
phases of each element of the matrix. after solving the above equation
we get:

\[
S=\left(\begin{array}{cccc}
0 & r & t & 0\\
r & 0 & 0 & t\\
t & 0 & 0 & -r\\
0 & t & -r & 0
\end{array}\right)
\]


\[
R=|r|^{2},T=|t|^{2},R+T=1
\]


The semi-reflecting mirror can be treated as a two channel junction
when the mirror is perpendicular to the beam making reflection to
the same channel possible, and eliminating the channels parallel to
the mirror.

(2)

The amplitude for each channel equals the sum over paths that went
in it:\\


\begin{eqnarray*}
 \text{channel 2: } B_{2} & = & [t^{2}e^{i\pi}e^{i\phi_{1}}-r^{2}e^{i\pi}e^{i\phi_{2}}]A_{0}\\
 \text{channel 1: } B_{1} & = & [rte^{i\pi}e^{i\phi_{1}}+rte^{i\pi}e^{i\phi_{2}}]A_{0}
\end{eqnarray*}


Where $\phi_{1}=k*L_{1}$ and $\phi_{2}=k*L_{2}$ and $A_{0}=1$ is
the amplitude of the beam before it is split.\\
 (3)

From here we get that:\\

\begin{description}
\item [{
\[
|B_{2}|^{2}=r^{2}t^{2}(2+e^{i\phi}+e^{-i\phi})=2r^{2}t^{2}(1+cos(\phi))
\]
}]~
\end{description}
\[
|B_{1}|^{2}=r^{4}+t^{4}-r^{2}t^{2}e^{i\phi}-r^{2}t^{2}e^{-i\phi}=r^{4}+t^{4}-2r^{2}t^{2}cos(\phi)
\]


\[
|B_{1}|^{2}+|B_{2}|^{2}=r^{4}+t^{4}+2r^{2}t^{2}=(r^{2}+t^{2})^{2}=(R+T)^{2}=1
\]
Were $\phi=\phi_{2}-\phi_{1}$.\\
 $ $

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