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\heading{E140: Two particles in two sites, oscillations} 
\auname{Amir Hirsch}


{\bf The problem:}

\Dn

Two different particles ${a}$ and ${b}$ are given in two different sites.
The system is symetric to transparent and the jump amplitude per time unit is ${\Delta}$. (assume ${\Delta}$ is real without the generality limitation).
The standard basis is: ${\left|x_{a},x_{b}\right\rangle=\left|1,1\right\rangle,\left|1,2\right\rangle,\left|2,1\right\rangle,\left|2,2\right\rangle}$.

\Dn

(1) Write the Hamiltonian ${H}$ in the standard basis.

Note that in an infinite time step only one particle can move between sites.
The Probability for a simultanian movemant is zero.

(2) Write the matrix presentation of the operator ${T}$ which switches the particles.

(3) Define a set of states ${\left|i=L,R,S,A\right\rangle}$ in which ${T_{ij}}$ is diagnol.

From now on assume the particles are identicle with spin zero (bosons!).

(4) Write the Hamiltonian ${H_{ij}}$ in the basis ${\left|i=L,R,S\right\rangle}$.

(5) Write in the same basis the transparent operator ${R_ij}$.

(6) Define a set of states in which ${R}$ is diagnol.

(7) Find the eigensates of the hamiltonian. write them in the standard basis.

(8) What are the eigenenergies ?

The system is prepared at ${t=0}$ to the symetric state ${S}$.

(9) The number of particles in the left site is measured. what are the possible results ?

(10) What is the period of the system oscillations ?

(11) What is the answer to (9) after half a period ?



\Dn\Dn


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{\bf The solution:}

\Dn

(1) 

The Hamiltonian in the standard basis is

\[
\mathcal{H} 
= \left( \amatrix { 0 & \Delta & \Delta & 0 \cr \Delta & 0 & 0 & \Delta \cr \Delta & 0 & 0 & \Delta \cr 0 & \Delta & \Delta & 0}\right) 
\]


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\Dn

(2)

The operator ${T}$ which switches the particles is

\[
\mathcal{T} 
= \left( \amatrix { 1 & 0 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 0 & 1}\right) 
\]


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

(3)

A set of states in which ${T_{ij}}$ is diagnol:

${\left|L\right\rangle=\left|1,1\right\rangle}$

${\left|R\right\rangle=\left|2,2\right\rangle}$

${\left|S\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1,2\right\rangle+\left|2,1\right\rangle\right)}$

${\left|A\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1,2\right\rangle-\left|2,1\right\rangle\right)}$


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

(4)

the Hamiltonian in the basis ${\left|i=L,R,S\right\rangle}$ is

\[
\mathcal{H} 
= \left( \amatrix { 0 & 0 & \Delta \cr 0 & 0 & \Delta \cr \Delta & \Delta & 0 \cr }\right) 
\]


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

(5)

the transparent operator ${R}$ is

\[
\mathcal{R} 
= \left( \amatrix { 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 }\right) 
\]


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

(6)

A set of states in which ${R}$ is diagnol:

${\left|S\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1,2\right\rangle+\left|2,1\right\rangle\right)}$

${\left|+\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1,1\right\rangle+\left|2,2\right\rangle\right)}$

${\left|-\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1,1\right\rangle-\left|2,2\right\rangle\right)}$


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\Dn

(7)

The eigensates of the hamiltonian are


${\left|E_1\right\rangle=\left|-\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1,1\right\rangle-\left|2,2\right\rangle\right)}$


${\left|E_2\right\rangle=\frac{1}{\sqrt{2}}\left(\left|+\right\rangle+\left|S\right\rangle\right)=\frac{1}{\sqrt{2}}\left(\left|1,1\right\rangle+\left|2,2\right\rangle+\left|1,2\right\rangle+\left|2,2\right\rangle\right)}$


${\left|E_3\right\rangle=\frac{1}{\sqrt{2}}\left(\left|+\right\rangle-\left|S\right\rangle\right)=\frac{1}{\sqrt{2}}\left(\left|1,1\right\rangle+\left|2,2\right\rangle-\left|1,2\right\rangle-\left|2,2\right\rangle\right)}$



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\Dn

(8)

The eigenenergies are

${E_1=0}$

${E_2=2\Delta}$

${E_3=-2\Delta}$


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\Dn

(9)

If the system is in the symetric state and we measure how many particles are in the left site the result will be ${1}$.


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\Dn

(10)

The period of the system oscillations is

${T=\frac{\pi}{2\Delta}}$


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\Dn

(11)

If the system is prepared in the state ${\left|S\right\rangle}$ after half a period ${T=\frac{\pi}{4\Delta}}$ it will be in the state ${\left|+\right\rangle}$, and then if we measure the number of particles in the left site the possible results are: ${0,2}$.


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\begin{verbatim}
Remarks:

(4) Missing numerical factor in the Hamiltonian.

\end{verbatim}




\end{document} 
