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\exnumber{1328}
\heading{Transmission Through a Junction in a One-Dimensional Lattice} 
\auname{Elad Benjamin and Ido Michealovich}


\textbf{The question:}\\

In this problem it is required to find the transmission coefficient $g=|t|^{2}$ of a particle of momentum $k$ through a junction in a one-dimensional lattice.\\
The given parameters are: Lattice constant $a$, ``hopping'' frequency between nearest neighbors $c_{0}$ and ``hopping'' frequency between the $n=\pm1$ sites $c_{J}$.\\
\\(1) In this section we assume an unperturbated lattice, where instead
of a junction we have another site $n=0$. Write down the unperturbated
Hamiltonian $H_{0}$ and the velocity operator $v$ as a function
of the momentum operator $p$.\\
(2) Write down the matrix $H_{n,m}$ representing the perturbated
Hamiltonian. It is sufficient to write the $4x4$ sub-matrix for $n,m=-2,-1,1,2$.\\
(3) Write a system of two equations for the transmission and reflection
amplitudes $(r,t)$ of the junction.\\
(4) Write an expression for the transmission coefficient $g$ as a
function of $k$.\\
(5) For a very large of very small $g\propto|v_{k}|^{2}$, write down
the proportion coefficients.\\

\textbf{The solution:}\\
\\(1)\\
$H_{0}=\left(\begin{array}{ccccc}
\ddots & \ddots & \ddots\\
\ddots & 0 & c_{0} & 0\\
\ddots & c_{0} & 0 & c_{0} & \ddots\\
 & 0 & c_{0} & 0 & \ddots\\
 &  & \ddots & \ddots & \ddots
\end{array}\right)$\\

We notice that:\\
$H_{0}=c_{0}\left(D+D^{\dagger}\right)=2c_{0}\left(\frac{e^{-ipa}+e^{+ipa}}{2}\right)=2c_{0}\cos(ap)$\\

To find v:\\
$v=\frac{dx}{dt}=i[H_{0},x]=-i[x,2c_{0}cos(ap)]=-2c_{0}i(-a\cdot sin(ap))[x,p]=-2c_{0}a\cdot sin(ap)$\\

(2)\\
$H=\left(\begin{array}{cccc}
0 & c_{0} & 0 & 0\\
c_{0} & 0 & c_{J} & 0\\
0 & c_{J} & 0 & c_{0}\\
0 & 0 & c_{0} & 0
\end{array}\right)$\\

\vspace{0.3 cm}

(3)\\
\\We assume as guided:\\
\\$\psi_{n}^{L}=e^{ikan}+re^{-ikan}$\\
$\psi_{n}^{R}=te^{ikan}$\\

From $H\psi=E_{k}\psi$ we get:\\

$\left(\begin{array}{cccc}
0 & c_{0} & 0 & 0\\
c_{0} & 0 & c_{J} & 0\\
0 & c_{J} & 0 & c_{0}\\
0 & 0 & c_{0} & 0
\end{array}\right)\left(\begin{array}{c}
\psi_{-2}\\
\psi_{-1}\\
\psi_{1}\\
\psi_{2}
\end{array}\right)=E_{k}\left(\begin{array}{c}
\psi_{-2}\\
\psi_{-1}\\
\psi_{1}\\
\psi_{2}
\end{array}\right)$\\

We obtain the set of 2 equations:\\

$c_0\psi_{-2}+c_J\psi_{1}=E_k\psi_{-1}$\\
$c_J\psi_{-1}+c_0\psi_{2}=E_k\psi_{1}$\\
\\(Notice that we cannot use the first and last equations since the matrix is in fact infinite and they are incorrect)\\

where $E_k=2c_0 cos(ak)=c_0(e^{ika}+e^{-ika})$\\
\\(Same as in the unperturbated hamiltonian).\\

Inserting the assumed wave functions and energy into the equations, we get:\\

$c_0(e^{-2ika}+re^{2ika})+c_J(te^{ika})=c_0(e^{ika}+e^{-ika})(e^{-ika}+re^{ika})$\\
$c_J(e^{-ika}+re^{ika})+c_0(te^{2ika})=c_0(e^{ika}+e^{-ika})(te^{ika})$\\
$\Downarrow$\\
$\frac{c_J}{c_0}te^{ika}=1+r$\\
$\frac{c_J}{c_0}(e^{-ika}+re^{ika})=t$\\

(4)\\
\\Solving for $t$ from the above system of 2 equations (isolating $r$ in both and equating them) we get:\\

$t=\frac{2ic_0 c_J sin(ka)}{c_J^2 e^{2ika}-c_0^2}$\\
$\Downarrow$\\
$|t|^2=\frac{4c_0^2 c_J^2 sin^2(ka)}{\left|c_J^2 e^{2ika}-c_0^2\right|^2}=\frac{c_J^2 v_k^2/a^2}{\left|c_J^2 e^{2ika}-c_0^2\right|^2}$\\

(5)\\
\\For very large $c_J$:\\
\\$|t|^2=\frac{1}{c_J^2}\frac{1}{a^2}\cdot v_k^2$\\
\\For very small $c_J$:\\
\\$|t|^2=\frac{c_J^2}{c_0^4}\frac{1}{a^2}\cdot v_k^2$\\

\end{document} 
