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\heading{E371: Electron(s) in Hall-Effect Geometry, (Moed A, 2004)} 
\auname{Yenon Benhaim}


{\bf The problem:}

\Dn

	Assume a two dimensional rectangle model with length L and cyclic boundary condition on the interval
	${0 < x < L }$, and width W ${\left(-\frac{W}{2}) < y < (\frac{W}{2}\right) }$


    The specimen is located in a magnetic filed of the form :${ \vec{B} = (0,0,B(y)) }$
    The specimen occupies electrons with no spin, mass ${m}$ and charge ${e}$.

\Dn
\begin{enumerate}
	\item write the Hamiltonian of an electron in this geometry.
	
	
	assuming Landu gauge, (${p_x  = \left( {\frac{{2\pi }}{L}} \right)l }$) is constant of motion.

	\item Write the effective potential ${ V_l(y) }$ that determines the functions ${ \varphi_0(y) }$ 
			
	Assume homogenous magntic filed ${ B(y) = B_0 }$ and that: (${\omega _B  = \frac{{eB_0 }}{m} > 0 }$)
	and also assume that only the lowest Landu level is occupied with electrons.
	
	Recall that each Landu level occupies a strip with width ${r_B}$ and the distance between the strips is ${R_B}$

  \item Write explicit expression for ${R_B }$, and, ${r_B }$, and for the condition ${\frac {r_B} {R_B} << 1}$. 
  does the magntic filed need to be strong or weak in order to get a good separation between the levels?
  
  
  Tip: the ground state of the Harmonic oscillator is a Gaussian ${ \varphi _0 (r) \propto \exp ( - \frac{1}{2}m\Omega
   r^2 )}$


\item Assume that the width of the rectangular is   ${ W = 3R_B }$ , so it contains only 3 	
	electrons, write the wave function of each of the electrons, use the following notation: 
	${|0\rangle, |1\rangle, |-1\rangle }$
	 
\item Suppose that the direction of the Magnetic filed changed only in the lower plain 
  ${ ( B(y < 0 ) = -B_0 ) }$ . Draw ${ V_l(y) }$ for  ${ l = 0, +1, -1 }$ Caution!
 
\item Write approximation for the wave function of each of the electrons, use the   ${ |S0\rangle
 ,|S\rangle , |A\rangle }$ notation.
 
\item Guess what would be the energy splitting of the states   ${ |S\rangle , |A\rangle }$.


Hint : find the "hight" of the barrier and the "overlap" factor of the gaussians in the superposition. 

\end{enumerate}
\Dn\Dn



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

\Dn



\begin{enumerate}
	\item  The Hamiltonian is :   ${ H = \frac{1}{2m} ( \vec{p} - \frac{e}{c} \vec{A} )^2 + \hat{V} }$

    using the "spirit" of Landau gage, we can define the vector potential as   ${ \vec{A} =
     (-\int_0^y{B(y')dy'},0,0) }$
     
    (It can be seen that when ${B(y) =B_0}$ we return to the normal Landau gage. )
    We also define the potential to be   ${\hat{V} = \hat{V_{Box}} }$ . and we get:

    \[ \hat{H} = \frac{1}{2m} \left( \hat{p_x} + \frac{e}{c} \vec{A(y)}\right)^2 + 		  
     \frac{1}{2m}\hat{p_y}^2 + \hat{V_{Box}} \]


    it's clear that ${ [\hat{H},\hat{p_x}] = 0 => \hat{p_x} }$ is a constant of motion, so
     we can define an effective potential:
    \[ \hat{V}_{eff} = \frac{1}{2m}(\hat{p_x} + \frac{e}{c} \vec{A_x})^2 + \hat{V_{Box}} \]

 \item   and the Hamiltonian reduce to:
 				 \[ \hat{H} = \frac{1}{2m}(\hat{p_x})^2 + \hat{V}_{eff }\]


\item  If  ${ B(y) = B_0 }$ then we return to the "Normal" Landau Gage  ${ \vec{A}=(-B_o y,0,0) }$ and the Hamiltonian becomes:

      ${\hat{H} = \frac{\hat{p}_y^2}{2M} + \frac{1}{2}M\omega_B^2(y-Y_l)^2 }$           
      
      
      \bigskip
      where :  ${Y_l = \frac{1}{B_0} \frac{2\pi}{L_x}l}$ and ,   ${\omega_B = \frac{eB_0}{Mc} }$
      
 
      we can see that the Hamiltonian is of harmonic oscillation and a free particle, so using the separation of
       variable theorem, the wave function in the ${|l,\nu\rangle }$ base is of the form :

		
    \[ \psi = \frac{1}{\sqrt{L_x}}exp(-iB_0Y_lx) \varphi_0(y-Y_l) \]

    Therefore, we can see that the distance beteen the energy strips are quantumized by units of ${ \frac{1}{B_0} \frac{2\pi}{L_x} }$ , and we can conclude that the distance between each Landau state is exactly that:  
    
    
     ${R_B = \frac{1}{B_0} \frac{2\pi}{L_x} }$


		
    The width of the strip the electrons occupies, is the width of the wave function,
    and in the ground state the wave function of harmonic oscillator is a Gaussian
     with width of :
  
	${r_B = \frac{1}{\sqrt{M\omega_B}} }$   


    we can now calculate the ratio:

    \[ \frac{r_B}{R_B} = \frac{ \frac{1}{\sqrt{M\omega }} }{ \frac{2\pi}{B_0L_x} } 
      = \frac{L}{2\pi} \sqrt{eB_0} \]
     
     
 



    and if we demand that this ratio would be small then it follows that :   ${B_0 << \frac{4\pi^2e}{cL_x^2} }$

\item      When we have only three electrons, we can write these states as follow :


    \[ |0 \rangle = \frac{1}{\sqrt{L_x}} \varphi_0(y)\]

    \[ |+1 \rangle = \frac{1}{\sqrt{L_x}}exp(\frac{i2\pi}{L_x}x) \varphi_0(y+\frac{2\pi}{B_0L_x}) \]

    \[ |-1 \rangle = \frac{1}{\sqrt{L_x}}exp(-\frac{i2\pi}{L_x}x) \varphi_0(y-\frac{2\pi}{B_0L_x}) \]


\item  Now we flip the dirction of ${\vec{B}}$ in the lower plain - this means that the vector potential changed to   					${\vec{A} = (B_0y,0,0) }$ ,in the lower part of the plain , and thus the Hamiltonyan is (in the lower region)
  ${\hat{H}_{Lower} = \frac{\hat{p}_y^2}{2M} + \frac{1}{2}M\omega^2(y+Y_l)^2 }$
  ${\hat{H}_{Uper} = \frac{\hat{p}_y^2}{2M} + \frac{1}{2}M\omega^2(y-Y_l)^2 }$ where   ${Y_l = \frac{1}{B_0} \frac{2\pi}{L_x}l}$ and ,   ${\omega_B = \frac{eB_0}{Mc} }$
  
\bigskip


so we can see tgat : 

\begin{itemize}
	\item when l = 0, then the potential is a parabola centred at the origin. 
	
%Fig.1	

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%	\begin{figure}
%	\includegraphics[scale=0.45]{l=0.eps}
%	\caption{\label{Fig1} l=0, parabola centred at the origin}
%	\includegraphics[scale=0.55]{l=+1.eps}
%	\caption{\label{Fig2} l=+1,parabola elvated a litel }
%	\includegraphics[scale=0.55]{l=-1.eps}
%	\caption{\label{Fig3} l=-1, Two parabolas shfted}
%	\end{figure}
	 
%%%%%%%%%%%%%%%%%%

	\item when l is positive the potential is a parabola which is shifted, at the uper region to the right and in the
	 lower region to the left.
	 and since we demand that the wave function have to be continuous and smooth, then we get back a parabola center at
	  the y = 0 but elevated a little.

 \item When l is negative then also we have 2 parbola shifted, but this time the uper side is shifted left and the lower
  side to the right, and the result is as illustrated below.
  
  
\end{itemize}
\bigskip\bigskip
\item  The energy state of the system:


			for  l= 0, the problem reduce to one dimension potential and the ground state is ${ |S0\rangle = |0\rangle =
			 \frac{1}{\sqrt{L_x}}\phi_0(y)}$
			 
			 for ${l= \pm1}$, the problem reduce to two sites problem, and the solutions are the symmetric and anti-symmetric
			  wave function:
   
    \[ |S\rangle = \frac{1}{\sqrt{2}}\left(|+1\rangle + |-1\rangle\right) = \frac{1}{\sqrt{2L_x}}exp(-i\frac{2\pi }{L_x}x)\left
     (\varphi_0(y-R_B)+\varphi_0(y+R_B)\right) \]

    \[ |A\rangle =\frac{1}{\sqrt{2}}\left(|+1\rangle - |-1\rangle\right) =
     \frac{1}{\sqrt{2L_x}}exp(-i\frac{2\pi }{L_x}x)\left
     (\varphi_0(y-R_B)-\varphi_0(y+R_B)\right) \]
     
     \item The energy splitting reflects the overlap of the two Gaussians
      ${|+1\rangle}$ and ${|-1\rangle}$ that were mixed into ${|S\rangle }$ and
       ${|A\rangle}$

     
      \[ Energy Splitting \propto exp(-(R_B/r_B)^2) \]

     



\end{enumerate}


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