\documentclass[onecolumn,fleqn]{revtex4}

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


% Special 
\usepackage{ifthen}
\usepackage{ifpdf}

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

\ifpdf
\usepackage{graphicx}
\usepackage{epstopdf}
\newcommand{\rang}{90}
\else
\usepackage{graphicx}
\usepackage{epsfig}
\newcommand{\rang}{0}
\fi



% Page setup
\setlength{\parindent}{0cm} 
\setlength{\parskip}{0cm} 


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


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



\heading{EX352: Hamiltonian in unform magnetic field} 
\auname{Lev Haim \& Yevgeni Estrin}


{\bf The problem:}

\Dn

(1) A particle is placed in a uniform magnetic field. The field $\vec{B}=(0,0,B) $.\\
Prove that the standart hamiltonian of the particle in the uniform magnetic field $H=\frac{1}{2m} \left( \vec{p}-\frac{e}{c} \vec{A}\right)$ \\
we can write as $H=\frac{p^2}{2m} -\frac{e}{2mc}BL_z + \frac{e^2 B^2}{8mc^2}  \left( x^2+y^2\right)$.\\
(2)Derive using polar coordinate.\\
(3)What is the effective V(r) which is experienced by a particle with $L_z=m'$ .\\
(4)Generalize your answers to the case of non-uniform magnetic field: $B = (0,0, Br^a) $.


\Dn


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

\Dn

(1) Some definitions: $\vec{p}=(p_x,p_y,p_z) $ and  $L_z=\epsilon_{zij}r_ip_j=xp_y-yp_x$  .


The pastulat $ \vec{B}=rot\vec{A} $ is describing three differential equations.\\
By using the gauge transformation $A_z=0$ we simplified the problem to one differential equation :  $\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=B$ ,\\
so we get: $\vec{A}= \left(- \frac{1}{2}By,  \frac{1}{2}Bx,0   \right)$ .




\[
H=\frac{1}{2m} \left( \vec{p}-\frac{e}{c} \vec{A}\right)^2
= \frac{1}{2m} \left\{\left( p_x^2 + \frac{e}{2c}By \right)^2+\left( p_y^2 - \frac{e}{2c}Bx \right)^2 + p_z^2\right\}=
\]


\[
= \frac{1}{2m} \left\{  \left(p_x^2+p_y^2+p_z^2\right)+
\frac{eB}{c}\left( - xp_y+yp_x  \right)+
\left( \frac{eB}{2c} \right)^2   \left(x^2+y^2 \right)  \right\}=
\]

\[=\frac{p^2}{2m} -\frac{e}{2mc}BL_z + \frac{e^2 B^2}{8mc^2}  \left( x^2+y^2\right)
\]


\Dn

(2)To defined the Hamiltonian in cylindrical coordinates we shell use those relationships $x=rcos\varphi$ , $y=r sin\varphi$ \\
while $L_z$ is not change when $(x,y,z)\Rightarrow (r,\varphi ,z)$ and $p^2= p^2_r+\frac{L^2}{r^2}$ so we get:
\[H=\frac{1}{2m} \left(p^2_r+\frac{L^2}{r^2}\right) -\frac{e}{2mc}BL_z + \frac{e^2 B^2}{8mc^2}  r^2
\]


\Dn

(3)The effective V(r) is:
\[V_{eff}= \frac{L^2}{2mr^2}-\frac{e}{2mc}BL_z + \frac{e^2 B^2}{8mc^2}  r^2\]

The effective V(r) which is experienced by a particle with $L_z=m'$ is:

\[V_{eff}|l m'>=\left( \frac{l(l+1)}{2mr^2}  -\frac{eBm'}{2mc} + \frac{e^2 B^2}{8mc^2}  r^2 \right)|lm'>\]

\newpage

(4)By using the gauge transformation we set $A_r=A_z=0$ in cylindrical coordinates\\
from $ \vec{B}=rot\vec{A} $ we get :

\[ \ \ \ \ Br^a=\frac{1}{r}\frac{  \partial \left(rA_{\varphi}\right) }{\partial r}  \ \ \ \ \ \Rightarrow \ \ \ \ \ A_(r)\hat{\varphi}=\frac{B}{a+2}r^{a+1}\]

when  
\[ A_x=-A_{\varphi}sin \varphi =-\frac{yA_{\varphi}}{r} \ \ \ \ \ , \ \ \ \ \
 A_y=A_{\varphi}cos \varphi =\frac{xA_{\varphi}}{r}
\] 

\[H= \frac{1}{2m} \left\{ \left(p_x,p_y,p_z\right) -\frac{e}{c} \left( -\frac{yA_{\varphi}}{r}  , \frac{xA_{\varphi}}{r} ,0   \right)
   \right\}^2=\]

\[=\frac{1}{2m}\left\{ p^2-
\frac{2eB}{(a+2)c} r^a L_z +
\left[\frac{eB}{(a+2)c}\right]^2 r^{2a+2}\right\}\]

It is easy to see that if $a=0$ we have solution for particle in an uniform magnetic field.

And more easy solution can be derived in polar coordinates:
\[H= \frac{1}{2m} \left\{
 \left(p',p_\varphi,p_z\right) -\frac{e}{c} \left(0  , A(r) ,0   \right)   \right\}^2=\]
 
 \[=\frac{1}{2m}  \left\{   p'^2-  \left({p'},p_\varphi,p_z\right) \cdot\frac{e}{c} \left(0  , A(r) ,0   \right)
-\frac{e}{c} \left(0  , A(r) ,0   \right)\cdot \left({p'},p_\varphi,p_z\right) +\left(\frac{eA}{c}\right)^2
\right\}= \]
\[=\frac{1}{2m}\left\{p'^2-\frac{2e}{c}Ap_\varphi +\left(\frac{eA}{c}\right)^2 \right\}=
\]
 \[=\frac{1}{2m}\left\{ p^2_r+\frac{L^2}{r^2}-
\frac{2eB}{(a+2)c} r^a L_z +
\left[\frac{eB}{(a+2)c}\right]^2 r^{2a+2}\right\}\]



\Dn
\Dn
\Dn
\Dn
\begin{verbatim}
Remarks:

(2) is wrong.
(4) is better derived in polar coordinates.


\end{verbatim}




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