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\heading{E136: A three site system} 
\auname{Eitan Rothstein}


{\bf The problem:}

\Dn

A three site system is descibed with the standard base $\left|1\right\rangle , \left|0\right\rangle, \left|-1\right\rangle$ 
\Dn 
through the hamiltonian
 
\[\mathcal{H} = \left( \amatrix { 0 & c & 1 \cr c & u & c \cr 1 & c & 0}\right) \] 


\Dn

assume that the site in the center $\left|0\right\rangle$ is disconnected $(c=0)$.

(1) Using the standard base write the eigenstates $\left|S\right\rangle , \left|S_0\right\rangle, \left|A\right\rangle$ and their eigenvalues.

(2) Write the hamiltonian using the new base that you have found.

from now on assume that $u=1$ and $c\neq0$.

(3) Write the eigenvalues $E_1, E_2, E_3$, and the eigenstates $|E_1\rangle , |E_2\rangle, |E_3\rangle$ of the system as a superposition of the states $|S\rangle , |S_0\rangle, |A\rangle$.
 
(4) Write the new eigenvalues using the standard base.

(5) What is the probobilty to find the partice in the state $|0\rangle$ after a time interval $t$ if it is prepered at time $t=0$ in the state $|-1\rangle$.

(6) What is the probobilty to find the partice in the state $|0\rangle$ after a time interval $t$ if it is prepered at time $t=0$ in the state $|A\rangle$.

\Dn\Dn


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

\Dn

(1) 
The eigenstates $\left|S\right\rangle , \left|S_0\right\rangle, \left|A\right\rangle$ are:
\[
\left|S\right\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\0\\1\end{array}\right) 
\]
\[
\left|S_0\right\rangle=\left(\begin{array}{c}0\\1\\0\end{array}\right)
\]
\[
\left|A\right\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\0\\-1\end{array}\right) 
\]

The eigenvalues are:
\[
\mathcal{H}\left|S\right\rangle = |S\right\rangle
\]
\[
\mathcal{H}\left|S_0\right\rangle = u\left|S_0\right\rangle
\]
\[
\mathcal{H}\left|A\right\rangle = -|A\right\rangle
\]



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

(2)
The new hamiltonian is:

\[
\mathcal{\widetilde{H}} = \left( \amatrix { 1 & 0 & 0 \cr 0 & u & 0 \cr 0 & 0 & -1}\right) 
\] 



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

(3)
Let us operate the hamiltonian on the states $\left|S\right\rangle , \left|S_0\right\rangle, \left|A\right\rangle$:
\[
\mathcal{H}\left|S\right\rangle = |S\rangle + \sqrt{2}c|S_0\rangle
\]
\[
\mathcal{H}\left|S_0\right\rangle = \sqrt{2}c|S\rangle + |S_0\rangle
\]
\[
\mathcal{H}\left|A\right\rangle = -|A\rangle
\]

Now we can write the hamiltonian using the base $\left|S\right\rangle , \left|S_0\right\rangle, \left|A\right\rangle$ as a block diagonal matrix:
\[
\mathcal{\widetilde{H}} = \left( \amatrix { 1 & \sqrt{2}c & 0 \cr \sqrt{2}c & 1 & 0 \cr 0 & 0 & -1}\right)
\] 

The eigenstates $\left|E_1\right\rangle , \left|E_2\right\rangle, \left|E_3\right\rangle$ are:

\[
|E_1\rangle = \frac{1}{\sqrt{2}}(|S\rangle + |S_0\rangle)
\]
\[
|E_2\rangle = \frac{1}{\sqrt{2}}(|S\rangle - |S_0\rangle)
\]
\[
|E_3\rangle = |A\rangle
\]

The eigenvalues are:
\[
\mathcal{\widetilde{H}}|E_1\rangle = (1+\sqrt{2}c)|E_1\rangle
\]
\[
\mathcal{\widetilde{H}}|E_2\rangle = (1-\sqrt{2}c)|E_2\rangle
\]
\[
\mathcal{\widetilde{H}}|E_3\rangle = -|E_3\rangle
\]


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

(4)
The eigenstates $\left|E_{1}\right\rangle , \left|E_{2}\right\rangle, \left|E_{3}\right\rangle$ writen in the standard base are:

\[
|E_1\rangle = \frac{1}{2}(|1\rangle + \sqrt{2}|0\rangle + |-1\rangle)
\]

\[
|E_2\rangle = \frac{1}{2}(|1\rangle - \sqrt{2}|0\rangle + |-1\rangle)
\]

\[
|E_3\rangle =\frac{1}{\sqrt{2}} (|1\rangle + |-1\rangle)
\]

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

(5)
Let us write the standard base using $\left|E_{1}\right\rangle , \left|E_{2}\right\rangle, \left|E_{3}\right\rangle$

\[
|1\rangle = \frac{1}{2}(|E_1\rangle + |E_2\rangle + \sqrt{2}|E_3\rangle)
\]
\[
|0\rangle = \frac{1}{\sqrt{2}}(|E_1\rangle - |E_2\rangle)
\]
\[
|-1\rangle = \frac{1}{2}(|E_1\rangle + |E_2\rangle - \sqrt{2}|E_3\rangle)
\]
We know that at $t=0$ the state of the particle is:
\[
\Psi_{(t=0)}=|-1\rangle
\]
Then the state of the particle, after time $t$ will be:
\[
\Psi_{(t)} = \frac{1}{2}(e^{-(iE_1t)}|E_1\rangle +e^{-(iE_2t)}|E_2\rangle - \sqrt{2}e^{-(iE_3t)}|E_3\rangle) =
\frac{1}{2}(e^{-i(1+\sqrt{2}c)t}|E_1\rangle +e^{-i(1-\sqrt{2}c)t}} |E_2\rangle
 - \sqrt{2} e^{it}{2}}|E_3\rangle)
\]
The probobilty of finding the particle in state $|0\rangle$ after a time interval $t$ is:
\[
|\langle x=0|\Psi_{(t)} \rangle | ^ {2} = |\frac{1}{2^{\frac{3}{2}}}( e^{-i(1+\sqrt{2}c)t}}- e^{-i(1-\sqrt{2}c)t}})|^2 = \frac{1}{2}\sin^2(\sqrt{2}ct)
\]


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

(6)
The state $\left|A\right\rangle$ is an eigenstate of the hamiltonian, and therefore it is a stationary stae which we know is a superposition of states $|1\rangle$ and $|-1\rangle$. Therefore the probobilty of finding the particle at state $|0\rangle$ is zero.



\end{eqnarray}



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