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\heading{E139: A partical in a four site system} 
\auname{Roy Bensimon}


{\bf The problem:}

\Dn
	
A system with four sites described by the Hamiltonian:

\[H= \left( \begin{matrix} {0} & {1 }& {0} & {c} \cr {1} & {0 }& {c} & {0} \cr {0} & {c} & {0} & {1} \cr {c} & {0} & {1} & {0}\end{matrix} \right)\]

 
The electrical charge is $e$. 

\Dn

(1) Write the Hamiltonian if we add a magnetic flux $\Phi$ through the ring that connected site $\left|1\right\rangle$ to site $\left|2\right\rangle$.

(2) Define the current operator using the formula $I=-\frac{dH}{d\Phi}_{\Phi=0}$.
We assume that the flux is zero.

(3) Write in the standart basis the eigenstates and the energies eigenvalues when $c=0$.
Use in ${\left|S_{1}\right\rangle,\left|A_{1}\right\rangle,\left|S_{2}\right\rangle,\left|A_{2}\right\rangle}$.

(4) Write the Hamiltonian in the new basis.
 
(5) Write the energies eigenvalues $\left(E_{1}, E_{2}, E_{3}, E_{4}\right)$ and the eigenstates $\left|E_{1}\right\rangle,\left|E_{2}\right\rangle,\left|E_{3}\right\rangle,\left|E_{4}\right\rangle$ of the system as a superposition of the states $\left|S_{1}\right\rangle,\left|A_{1}\right\rangle,\left|S_{2}\right\rangle,\left|A_{2}\right\rangle$.    

(6) Write the energies eigenstats aforementioned in the standart basis.

\Dn

    The partical is plased in site one  ${\left|\Psi(t=0)\right\rangle=\left|1\right\rangle}$.

\Dn
(7) Calculate the current that flow through the ring as a time function ${\left\langle \Psi(t)\left|I\right|\Psi(t)\right\rangle}$.


\Dn\Dn


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{\bf The solution:}

\Dn

(1) 
Magnetic flux add a phase to the Hamiltonian. 

\[H= \left( \begin{matrix} {0} & {e^{-i\varphi}}& {0} & {c} \cr {e^{i\varphi}} & {0 }& {c} & {0} \cr {0} & {c} & {0} & {1} \cr {c} & {0} & {1} & {0}\end{matrix} \right)\]
\[\varphi=e\cdot\Phi \]

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\Dn

(2)
The current operator :


\[I= \left( \begin{matrix} {0} & {i}& {0} & {0} \cr {-i} & {0 }& {0} & {0} \cr {0} & {0} & {0} & {0} \cr {0} & {0} & {0} & {0}\end{matrix} \right)\]

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\Dn

(3)
For $c=0$. 
The Hamiltonian is :


\[H= \left( \begin{matrix} {0} & {1}& {0} & {0} \cr {1} & {0 }& {0} & {0} \cr {0} & {0} & {0} & {1} \cr {0} & {0} & {1} & {0}\end{matrix} \right)\]


The eigenstates of the Hamiltonian are the states which are symmetrical  or antisymmetrical .


\[ |S_{1} \rangle = \frac{1}{\sqrt{2}} (|1 \rangle + |2 \rangle ) \] 
\[ |A_{1} \rangle = \frac{1}{\sqrt{2}} (|1 \rangle - |2 \rangle ) \]
\[ |S_{2} \rangle = \frac{1}{\sqrt{2}} (|3 \rangle + |4 \rangle ) \]  
\[ |A_{2} \rangle = \frac{1}{\sqrt{2}} (|3 \rangle - |4 \rangle ) \]


The energies eigenvalues :
\[ E_{1}=1 \ ; \[ E_{2}=1 \ ; \[ E_{3}=-1 \ ; \[ E_{4}=-1 \]  


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\Dn

(4)
In order to fined the Hamiltonian in the new basis, we need to find a matrix that takes us from the old basis to the new one.

\Dn
The new basis is :   $\left\{{|S_{1} \rangle,}{|S_{2} \rangle,}{|A_{1} \rangle ,}{|A_{2} \rangle }\right\}$ 
 
\Dn 
The transformation matrix is :

\[ T = \frac{1}{\sqrt{2}}\left( \begin{matrix} {1} & {0}& {1} & {0} \cr {1} & {0}& {-1} & {0} \cr {0} & {1} & {0} & {1} \cr {0} & {1} & {0} & {-1}\end{matrix} \right) \] 
 
\[  H_{new}= T^{-1}H_{old}T = \left( \begin{matrix} {1} & {c}& {0} & {0} \cr {c} & {1}& {0} & {0} \cr {0} & {0} & {-1} & {-c} \cr {0} & {0} & {-c} & {-1}\end{matrix} \right)
 \] 

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\Dn

(5)
The energies eigenstates of the system is :

\[  |E_{1} \rangle = \frac{1}{\sqrt{2}}\left( \left|S_{1}\right\rangle + \left|S_{2}\right\rangle\right)  \] 
\[  |E_{2} \rangle = \frac{1}{\sqrt{2}}\left( \left|S_{1}\right\rangle - \left|S_{2}\right\rangle\right)  \]
\[  |E_{3} \rangle = \frac{1}{\sqrt{2}}\left( \left|A_{1}\right\rangle + \left|A_{2}\right\rangle\right)  \] 
\[  |E_{4} \rangle = \frac{1}{\sqrt{2}}\left( \left|A_{1}\right\rangle - \left|A_{2}\right\rangle\right)  \]

The energies eigenvalues :
\[ E_{1}=1+c \ ; \[ E_{2}=1-c \ ; \[ E_{3}=-1-c \ ; \[ E_{4}=-1+c \]  


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

(6)
The energies eigenstates of the system in the standart basis is :


\[ \left|E_{1}\right\rangle = \frac{1}{2}\left( \left|1\right\rangle + \left|2\right\rangle + \left|3\right\rangle + \left|4\right\rangle \right) \] 
\[ \left|E_{2}\right\rangle = \frac{1}{2}\left( \left|1\right\rangle + \left|2\right\rangle - \left|3\right\rangle - \left|4\right\rangle \right) \]
\[ \left|E_{3}\right\rangle = \frac{1}{2}\left( \left|1\right\rangle - \left|2\right\rangle + \left|3\right\rangle - \left|4\right\rangle \right) \] 
\[ \left|E_{4}\right\rangle = \frac{1}{2}\left( \left|1\right\rangle - \left|2\right\rangle - \left|3\right\rangle + \left|4\right\rangle \right) \]



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\Dn

(7)
\[ \left|\Psi(t=0)\right\rangle = \left|1\right\rangle = \frac{1}{2}\left(\left|E_{1}\right\rangle + \left|E_{2}\right\rangle +\left|E_{3}\right\rangle + \left|E_{4}\right\rangle\right) \]

\[ \left|\Psi(t)\right\rangle = \frac{1}{2}\left(e^{-iE_{1}t}\left|E_{1}\right\rangle + e^{-iE_{2}t}\left|E_{2}\right\rangle + e^{-iE_{3}t}\left|E_{3}\right\rangle + e^{-iE_{4}t}\left|E_{4}\right\rangle \right) \]


\[A=\langle 1|\Psi(t)\rangle=\frac{1}{4}\left(e^{-iE_{1}t}+e^{-iE_{2}t}+e^{-iE_{3}t}+e^{-iE_{4}t}\right)=cos(t)cos(ct)\]
\[B=\langle2|\Psi(t)\rangle=\frac{1}{4}\left(e^{-iE_{1}t}+e^{-iE_{2}t}-e^{-iE_{3}t}-e^{-iE_{4}t}\right)=-isin(t)cos(ct)\]

\[ \left\langle \Psi(t)\left|I\right|\Psi(t)\right\rangle = i(A^*B-B^*A)= 2Im[B^*A] =  2Im[\left\langle 2\left|\Psi(t)\right\rangle^{*}\left\langle 1\left|\Psi(t)\right\rangle] \]



The current that flow through the ring as a time function is :

\[ I_{(t)}= e\cdot sin(2t)cos^{2}(ct)  \] 



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\Dn


 


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