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\heading{E262: Scattering matrix of I-junction} 

\auname{Alice Shuturov}


{\bf The problem:}

\Dn Three wires are attached in one point. Suppose that the signal
coming from one of the legs is partly transmitted to other legs
and is partly reflected back to the first leg. Define the
scattering matrix as $S$ which is $3\times 3$

\Dn

(1) Write $S$ if given that all the reflection amplitudes are
equal to real number $r$ and all the transmission amplitudes are
equal to positive real number $t$.

(2) Find the numerical values of $r$ and $t$.

(3) Generalize the problem for a case of more than 3 wires.


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}
\Dn

(1) The wave function can propogate in 3 directions. The
scattering matrix is:

\[
\mathcal{S} = \left( \amatrix { r_1 & t_{21} & t_{31} \cr t_{12} &
r_2 & t_{32} \cr t_{13} & t_{23} & r_3}\right)
\]
 We know that the system is symmetric, so we can say that
 $r1=r2=r3$ and also $t_{21}=t_{31}=t_{12}=t_{32}=t_{13}=t_{23}$.
 We obtain the following matrix:
\[
\mathcal{S} = \left( \amatrix { r & t & t \cr t & r & t \cr t & t
& r}\right)
\]
%%%%%%%%%%%%%%%
\Dn

(2)We know that $t$ and $r$ are real and also $t>0$. Requiring
that
\[S^{^{+}}\cdot S=I\]
\[ \left( \amatrix { r & t & t \cr t & r & t \cr t & t & r}\right)
\cdot \left( \amatrix { r & t & t \cr t & r & t \cr t & t &
r}\right)= \left( \amatrix { 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 &
1}\right)
\]
we can obtain:
\[
\left\{\begin{array}{l}r^{2}+2t^{2}=1\\rt+tr+t^{2}=0
\end{array}\right.
\]
Solutions of these equations are:
\[ r=\pm 1/3\]
\[t=\mp 2/3 \]
 Requiring $t>0$ the solutions can be rewritten as:
\[ r=-1/3\]
\[ t=2/3\]

%%%%%%%%%%%%%%
\Dn

(3)If we have more than 3 wires,we can write matrix as:
\[
\mathcal{S} = \left( \amatrix { r_1 & t_{12} & t_{13} & t_{14} &
t_{15} & . & . & . \cr t_{21} & r_2 & t_{23} & t_{24} & t_{25} & .
& . & . \cr t_{31} & t_{32} & r_3 & t_{34} & t_{35} &  . & . & .
\cr . & . & . & . & . & . & . & .}\right)
\]
 We know that the system is symmetric, so we can say that
 $r_1=r_2=r_3=r_4=...=r$ and also $t_{21}=t_{31}=t_{12}=t_{32}=t_{13}=t_{23}=...=t$.
 We obtain the following matrix:
\[
\mathcal{S} = \left( \amatrix { r & t & t & t & t & . & . & . \cr
t & r & t & t & t & . & . & .  \cr t & t & r & t & t & . & . & .
\cr . & . & . & . & . & . & . & .}\right)
\]
We know that $t$ and $r$ are real and also $t>0$. Requiring that
\[S^{^{+}}\cdot S=I\]
\[\left( \amatrix { r & t & t & t & t & . & . & . \cr
t & r & t & t & t & . & . & .  \cr t & t & r & t & t & . & . & .
\cr . & . & . & . & . & . & . & .}\right) \cdot\left( \amatrix { r
& t & t & t & t & . & . & . \cr t & r & t & t & t & . & . & .  \cr
t & t & r & t & t & . & . & . \cr . & . & . & . & . & . & . &
.}\right)= \left( \amatrix { 1 & 0 & 0 & 0 & . & . \cr 0 & 1 & 0 &
0 & . &.  \cr 0 & 0 & 1 & 0 & . & . \cr . & .& .& .& . &.}\right)
\]
we can obtain:
\[
\left\{\begin{array}{l}r^{2}+(n-1)t^{2}=1\\rt+tr+(n-2)t^{2}=0
\end{array}\right.
\]
where {n} is number of the wires. Solutions of these equations
are:
\[ r=\pm (2/n)\mp1\]
\[t=\pm 2/n \]
 Requiring $t>0$ the solutions can be rewritten as:
\[ r=(2/n)-1\]
\[ t=2/n\]



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