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\heading{E138: system with three sites} 
\auname{Yogev dahan}


{\bf The problem:}

\Dn

A system with three sites described by the hamiltonian

\[
\mathcal{\hat{H}} 
= \left( \amatrix { 2 & -1 & -1 \cr -1 & 2 & -1 \cr -1 & -1 & 2 }\right) 
\]


 and the position matrix:
\[
\mathcal{\hat{X}} 
= \left( \amatrix { 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 2 }\right) 
\]
.
The purpose of this problem is to find the probability ${P(t)}$ 
to find the particle at site left after time ${t}$ and the current as a function of ${t}$. 

\Dn

(1) Write the Hamiltonian if we add a magnetic flux $ \Phi $ through the ring that connects the right site to the left.

(2) Define the current operator using the formula $ I = -\frac{dH}{d\phi} $.

(3) Write in the standard basis the eigenstates and the energies eigenvalues.note that the hamiltonian is symetrical 

(4) Calculate the probability ${P(t)}$ to find the particle in the left site ufter a while .

(5) Calculate the current $ \langle I \rangle_t $ that flow through the ring.   



\Dn\Dn


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

(1) 
The Hamiltonian is 
\[
\mathcal {\hat{H}} 
= \left( \amatrix { 2 & -1 & -\eexp{ -i\phi } \cr -1 & 2 & -1 \cr -\eexp {i\phi} & -1 & 2 }\right) 
\]

and the phase is $ \phi=e\Phi $


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

(2) The current when $ \Phi=0 $ is:

\[
{I}=-\frac{dH}{d\Phi}
= e \left( \amatrix { 0 & 0 & -i \cr 0 & 0 & 0 \cr i & 0 & 0 }\right) 
\]



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

(3)The momentum eigenvalues are:
$ k_0=0 $ , $ k_+ = \frac{2\pi}{3} $  ,  $ k_-=-\frac{2\pi}{3} $
and the eigenstates are (because the hamiltonian is symmetrical to translations we have $ S $ and $ anti S $ solutions):

\[ 
|k_0 \rangle=\frac {1}{\sqrt{3}}( |0 \rangle + |1 \rangle + |2 \rangle ) \] 



\[ 
|k_{+} \rangle= \frac {1}{\sqrt {3} }( |0 \rangle + \eexp{ i \frac {2 \pi } {3} }|1 \rangle + \eexp { - i \frac {2 \pi}{3} } | 2 \rangle )
\]   

\[
|k_{-} \rangle= \frac {1}{\sqrt {3} }( |0 \rangle + \eexp{ -i \frac {2 \pi} {3}} | 1 \rangle + \eexp {i \frac {2\pi} {3}} | 2 \rangle )
\]  


and the energies are:  
\[
E_0=0
\]
\[
E_+=3
\]
\[
E_-=3 
\]  
%%%%%%%%%%%%%
\Dn

(4)
The probability to find the particle at site left after time ${t}$ is:

\[
\psi (t) = \frac {1}{\sqrt{3}} ( |k_0 \rangle + \eexp{-i\omega t}|k_+ \rangle + \eexp{-i\omega t}|k_- \rangle ) 
\]
where $ \omega = E_+ - E_0 = 3 $.

With some algebra we get :
\[
P(t)= |\langle 0 |\psi(t) \rangle |^2 =| \frac {1}{\sqrt{3}} (\langle 0|k_0 \rangle + \eexp{-i\omega t}\langle 0|k_+ \rangle + \eexp{-i\omega t}\langle 0|k_- \rangle )|^2 =| \frac {1}{\sqrt{3}}(\frac {1}{\sqrt{3}} 
+\eexp{-i\omega t} \frac{1}{\sqrt{3}}+\eexp{-i\omega t} \frac {1}{\sqrt{3}})|^2 
\] 
$ =\frac{1}{9}(5+4\cos(\omega t)) $ 


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

(5) Using $ \psi (t) $ from   $ (4) $ and the definition of $ I $ from $ (2) $ we get the current:
\[
 \psi_0 = \langle 0|\psi(t)\rangle=\frac {1}{3} (1 + \eexp {-i \omega t } + \eexp {-i\omega t} ) 
\]
\[
\psi_2 = \langle 2|\psi(t)\rangle=\frac {1}{3} (1 + \eexp {-i \omega t -i\frac {2\pi}{3} } + \eexp {-i \omega t +i \frac {2\pi}{3}} ) 
\]  
\[ 
\langle I \rangle _t =  \langle \psi (t)| I  | \psi (t) \rangle 
= e ( -i \psi_0 *\psi_2 + i \psi_2* \psi_0 ) = \frac {2}{3} e \sin {\omega t }  
\] 




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