\documentclass[11pt,fleqn]{article}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site 
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Page setup
\topmargin -1.5cm      
\oddsidemargin -0.04cm   
\evensidemargin -0.04cm  
\textwidth 16.59cm
\textheight 24cm 
\setlength{\parindent}{0cm} 
\setlength{\parskip}{0cm} 


% Fonts
\usepackage{latexsym}
\usepackage{amsmath} 
\usepackage{amssymb} 
\usepackage{bm}


% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}

% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}

% Mass symbol
\newcommand{\mass}{\mathsf{m}} 
\newcommand{\Mass}{\mathsf{M}} 

% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}} 
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}} 
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}

% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}

% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}

% heading
\newcommand{\exnumber}[1]{\newcommand{\exnum}{#1}}
\newcommand{\heading}[1]{\begin{center} {\Large {\bf Ex\exnum:} #1} \end{center}}
\newcommand{\auname}[1]{\begin{center} {\bf Submitted by:} #1 \end{center}}


\begin{document} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\exnumber{3640}
\heading{Electrons in a 2 dimensional finite potential well with a magentic field} 
\auname{Guy Alis}


{\bf The problem:}

\Dn

Given a 2 dimensional potential well of size $ L\times L $ and of depth $V _{0} $ :
\[
\begin{array}{@{} r @{} c @{} l @{} }
&V(x,y) &{}=\displaystyle
\begin{cases}
0 &\text{if } x<-\frac{L}{2},y<-\frac{L}{2},\\
-V_{0} &\text{if } -\frac{L}{2}<x<+\frac{L}{2},-\frac{L}{2}<y<+\frac{L}{2},\\
0 &\text{if } x>+\frac{L}{2},y>+\frac{L}{2} .
\end{cases}
\end{array}
\]
Hence,all of the electrons with energy $ E>0 $ are escaping the well.Assume that the electrons which remain in the well have mass $ m $ ,charge $ e $ ,and have no spin.

\Dn

(1) Write the condition $ L_{min}<L $ of which at least one electron will could be in the well\\
For future calculations,assume $ L_{min}<<L $\\
(2) Calculate the number of electrons $ N_{0} $ that can exist in the well\\
A homogeneous magnetic field $ B $ is created perpendicularly to the plane of the well.\\
In the next sections,known results regarding Landau levels can we used (without any proof)\\
(3) Find the maximal magnetic field $ B=B_{2} $ so that two Landau levels will be filled in the well.\\
Denoting the number of particles from section (3) to be $ N_{2} $: \\
(4) What is the ratio of $ \dfrac{N_{2}}{N_{0}} $ ?.

(5) Generalize the result of section (4) for the case of $ n $ full Landau levels.\\
\\
Guidance:\\
The boundary conditions at the edges of the well have negligible effect on the spectrum,\\
so that previous results that are derived from Landau`s solution can be used.


\Dn\Dn


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

\Dn

(1) 
The eigenenergies of a 2D finite well of depth $ V_{0} $ are:


\[ E_{n_{x},n_{y}} \ = \ \dfrac{\pi^{2}}{2mL^{2}}(n^{2}_{x}+n^{2}_{y}) -V_{0} \]
\[ where : n_{x},n_{y} >0\]
In order to stay located in the well,the particle needs to have energy $ E<0 $.The smallest energy level that can occupy one electron is $ E_{1,1} $.substituting these conditions into the expression of the eigenenergies will give:


\[ L > \dfrac{\pi}{\sqrt{mV_{0}}} \]
Therefore:
\[ L_{min} = \dfrac{\pi}{\sqrt{mV_{0}}} \]

%%%%%%%%%%%%%%%
\Dn
(2)
If $ L>>L_{min} $, then $ E<0 $.Applying this condition on $ E_{n_{x},n_{y}} $ will give:
\[ n^{2}_{x}+n^{2}_{y} < \dfrac{2mL^{2}V_{0}}{\pi^{2}} \]
This equation describes a circle with center $ (0,0) $ and radius $ R^{2} = \dfrac{2mL^{2}V_{0}}{\pi^{2}} $.\\
Adding the constraint of $ n_{x},n_{y} > 0 $,$ N_{0} $ will be the quarter of this circle:
\[ N_{0} = \frac{mL^{2}V_{0}}{2\pi} \]


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

(3)
Making use of known results regarding Landau levels,the energy of the $ n^{th} $ Landau level is:
\[ E_{\nu} = (\frac{1}{2}+\nu)\omega_{B} -V_{0}\]
\[ where: \omega_{B}=\frac{eB}{mc},\nu=n-1,\nu=0,1,2,... \]
For 2 Landau levels, $ \nu=1 $ and then , plugging the constraint that $ E<0 $ so that the electrons will stay located in the well,the maximal magnetic field $ B_{2} $ is :
\[ B_{2}=\dfrac{2mcV_{0}}{3e} \]

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

(4)
The number of electrons occupying each Landau level is defined as $ g_{Landau} = \dfrac{L^{2}eB}{2\pi c} $.Due to the fact that the potential floor is "flat" the occupation in each Landau level is the same :\\ $ N_{n} = n\cdot g_{Landau} $ . Therefore:
\[ N_{2} = (2)\cdot g_{Landau} = \dfrac{4}{3}\frac{mL^{2}V_{0}}{2\pi} = \frac{4}{3}N_{0} \]
Now it is immediate that:
\[ \dfrac{N_{2}}{N_{0}} = \frac{4}{3} \]



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

(5)
By a short generalization of the previous sections, the maximal magnetic field $ B_{n} $ so that n Landau levels will be filled in the well:
\[ B_{n}=\frac{mcV_{0}}{e(n-\frac{1}{2})} \]
The number of electrons $ g_{B_{n}Landau} $ that can occupy each Landau level under this condition is :
\[ g_{B_{n}Landau} = \dfrac{L^{2}eB_{n}}{2\pi c} = \frac{1}{n-\frac{1}{2}}N_{0} \]
Yielding the ratio:
\[ \dfrac{N_{n}}{N_{0}} = \dfrac{n}{\dfrac{1}{2}+\nu} = \dfrac{n}{n-\dfrac{1}{2}} \]


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document} 
