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\heading{E361: Electron in the Hall effect geometry} 
\auname{Maayan Shani}


{\bf The problem:}

\Dn

Assume a two dimensional model with length L and cyclic boundary condition on the interval $0<x<L$.
The electric potential $V(y)=\frac{1}{2}\alpha y^{2}$ is given and a homogeneous magnetic field B at the z direction is applied.
It convenient to express the magnetic field using the dimensionless parameter: $\ b=\frac{B}{\sqrt{m\alpha}}$ . We can see that $b$ is in fact the ratio between two frequency (What are they?)
Inside the model we insert N spinless electrons.

\Dn
(a) Show that as long as $ N<[4(1+b^{2})^{3}m\alpha]^{1/4}L/\pi $ only the first landau band is    
    populated.
    
\Dn
    
(b) Show how you can get the result from (a) using very strong field or very weak field approximations.

\Dn\Dn


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

\Dn

(a) 

\Dn

Let us look at the two frequencies of the system: $ \omega_{B} $ - the frequency of circular motion due to Lorentz force, and $\omega_{osc}$ - the frequency of the oscillations due to the electric potential:

\[
b = \frac{\omega_{B}}{\omega_{osc}} = \frac{(eB)}{(mc)}\sqrt{\frac{m}{\alpha}} = \frac{e}{c}\frac{B}{\sqrt{m\alpha}}
\]


\Dn
The Hamiltonaian that describes the motion of a particle in the Hall bar geometry is :
 
\[
\hat{H}=\frac{1}{2m}(p_{x} + B\hat{y})^2 + \frac{1}{2m}p_{y}^2 + V(y)
\] 

We define the constant of motion:

\[
Y = \hat{x} - \frac{1}{\omega_B}\hat{v}_{y} = - \frac{1}{B}\hat{p}_{x}
\]


and the eigenvalues of $\hat{Y}$ are:

\[
Y_{l} = \frac{1}{B}\frac{2\pi}{L}l
\]

So, our Hamiltonian is:

\[ 
H = \frac{1}{2m}p_{y}^2 + \frac{B^2}{2m}(y-Y_{l})^2 + V(y). 
\]


We define the effective potential:

\[ 
V^{(\ell)}= \frac{1}{2}m\omega_{B}^2(y-Y_{l})^2 + V(y).
\]



\[
V^{(\ell)} = \frac{1}{2}m\omega_{B}^2(y^2 - 2yY_{l} + Y_{l}^2) + \frac{1}{2}\alpha y^2 = \frac{1}{2}m\omega_{B}^2(y^2 - 2yY_{l} + Y_{l}^2 + \frac{\alpha}{m\omega_{B}^2}y^2) = \frac{1}{2}m\omega_{B}^2\left[\left(1 + \frac{1}{b^{2}}\right)y^2 - 2yY_{l} + Y_{l}^2\right].
\]



Now, lets write the equation in quadratic forms
\[ 
\frac{1}{2}m\omega_{B}^2 \left[ \left(\sqrt{1+\frac{1}{b^2}}y -\sqrt{\frac{1}{1+1/b^2}}Y_{l}\right)^2 + Y_{l}^{2}\left(1-\frac{1}{1+1/b^2}\right)\right] = \frac{1}{2}m\omega_{B}^2\left[\left(1+\frac{1}{b^2}\right)\left(y-\frac{1}{1+1/b^2}Y_{l}\right)^2 + Y_{l}^2\left(\frac{1}{b^{2}+1}\right)\right]. 
\]

So, our Hamiltonian is

\[
H = \frac{1}{2m}p_{y}^2 + \frac{1}{2}m\omega_{B}^2\left[\left(1+\frac{1}{b^2}\right)\left(y-\frac{1}{1+1/b^2}Y_{l}\right)^2 + Y_{l}^2\left(\frac{1}{b^{2}+1}\right)\right].
\]


Let us define:

\[
\tilde{y} = y - Y_{l}\frac{1}{1+1/b^2}
\]

and 

\[
\tilde{\omega} = \omega_{B}\sqrt{1+\frac{1}{b^2}},
\]

to get the new Hamiltonian

\[
\tilde{H} = \frac{1}{2m}p_{y}^2 + \frac{1}{2}m\tilde{\omega}^2\tilde{y}^2 + \frac{1}{2}m\omega_{B}^2\frac{1}{b^{2}+1}Y_{l}^2.
\]

The new eigenvalues are:

\[
\tilde{E}_{l,\nu} = \tilde{\omega}(\frac{1}{2}+\nu) + \frac{1}{2}m\omega_{B}^2\frac{1}{b^{2}+1}Y_{l}^2.
\]

The condition for filling the first Landau level is:

\[
E_{l,0} <  E_{0,1}
\]

\[
\frac{1}{2}m\omega_{B}^2\frac{1}{b^{2}+1}Y{_l}^2 < \tilde{\omega}
\]

\[ 
m\omega_{B}^2\frac{1}{b^{2}+1}\left(\frac{2\pi}{BL}\right)^{2}l^{2} < \tilde{\omega} = 2\omega_{B}\sqrt{1+\frac{1}{b^2}}
\]

\[
l^2 < 2\frac{1}{m\omega_{B}}\left(\frac{BL}{2\pi}\right)^{2}(b^{2}+1)\sqrt{1+\frac{1}{b^2}} =   2\frac{1}{m\omega_{B}}\left(\frac{BL}{2\pi}\right)^{2}\frac{(b^{2}+1)^{3/2}}{b}
\] 

\[
l^2 < 2\frac{m}{mB} \left(\frac{BL}{2\pi}\right)^{2} (b^{2}+1)^{3/2}\frac{(m\alpha)^{1/2}}{B} = 2\left(\frac{L}{2\pi}\right)^{2} (b^{2}+1)^{3/2}(m\alpha)^{1/2}
\]

\[
|l| < [4(b^{2}+1)^{3}m\alpha]^{1/4}\frac{L}{2\pi}
\]

\[
N=2|l|
\]

\[
N < [4(b^{2}+1)^{3}m\alpha]^{1/4}\frac{L}{\pi}
\]

\Dn
(b)
\Dn
\[
B<<1:
\]

\Dn
The Hamiltonaian is :
 
\[
\hat{H}=\frac{1}{2m}p_{x}^2 + \frac{1}{2m}p_{y}^2 + V(y)
\] 
The  eigenvalues are:
\[
E_{l,\nu} =  (\frac{1}{2} + \nu)\omega_{osc} + \frac{1}{2m}\left(\frac{2\pi l}{L}\right)^2
\]

The condition for filling the first Landau level is:

\[
\frac{1}{2m}\left(\frac{2\pi l}{L}\right)^2 < \omega_{osc}
\]


\[
l^2 < 2\sqrt{\frac{\alpha}{m}}\left(\frac{L}{2\pi}\right)^{2}m
\]

\[
|l|<(\frac{1}{4}\alpha m)^{1/4}\frac{L}{\pi}
\]

\[
N<(4\alpha m)^{1/4}\frac{L}{\pi}
\]

\[
B>>1 :
\]

For strong magnetic field with constant electric field we use the harmonic approximation then the energes are:

\[
E_{l,\nu} \approx   (\frac{1}{2}+\nu)\omega_{B} + V(Y_{l})  = (\frac{1}{2}+\nu)\omega_{B} + \frac{1}{2}m\omega_{osc}^2 Y_{l}^2
\]

The condition for filling the first Landau level is:

\[
\frac{1}{2}m\omega_{osc}^2 Y_{l}^2 < \omega_{B}
\]

\[
\frac{1}{2}\alpha\left(\frac{2\pi l}{BL}\right)^2 < \frac{B}{m}
\]

\[ 
|l| < \frac{BL}{2\pi}\left(\frac{2B}{m\alpha}\right)^{1/2} = \frac{L}{2\pi}\left(\frac{2B^3}{m\alpha}\right)^{1/2} = \frac{L}{2\pi}\left(\frac{2(\beta\sqrt{m\alpha})^3}{m\alpha}\right)^{1/2}
\]

\[
N < (4b^{6}m\alpha)^{1/4}\frac{L}{\pi}
\]

\end{document} 
