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\heading{E3555: Fermions in magnetic field - Landau} 
\auname{Yigal  Tirosh}


{\bf The problem:}

\Dn

     Consider $N$ spinless electrons that have mass $m$ and charge $e$ in a 2$D$ box that has an area $A$ at zero temperature.
     perpendicular magnetic field $B$ is applied. The purpose of this question is to find the magnetization of the system.
\vspace{5mm}

(1) What are the threshold value$B_\nu$  that is required to empty all the $n >\nu$  Landau levels.

(2) Find the energy$ E(B)$ and the magnetization $M(B)$  for strong field $ B > B_o$. Give an optional semicalssical
     derivation to the result assuming that each electron is doing a cyclotron motion with minimal one-particle energy.

(3) Find the energy $E(B)$  and the magnetization $M(B)$ for general $ B_\nu < B < B_{\nu+1}$. Explain why the values $ E(B_\nu)$
     are all equal $E(0)$.

(4) Give a semicalssical derivation to the paramagnetic drops of $M(B)$ at the threshold values $B_\nu$, using the Hall
      formula for the current along the Edge.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{5mm}

{\bf The solution:}

\Dn

(1)
\vspace{1mm}

$g$ - degnerate Landau energy levels.
\begin{equation}     
g = \frac{eB_\nu A}{hc}.
\end{equation}
\begin{equation}     
 g(\nu +1) = N   \rightarrow  B_\nu  = \frac{hc}{eA(\nu +1)}.
\end{equation}

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

(2) 
\begin{equation}     
 B >B_o.
\end{equation}
\begin{equation}     
 B_o = \frac{Nhc}{eA}.
\end{equation}

       All  particles   in  the  lowest  Landau  level.
\begin{equation}     
\frac{ E_o}{N} = \frac{\hbar w_o}{2} =\frac{\hbar eB}{2mc} .
\end{equation}
\begin{equation}     
\frac{ M}{N} = - \frac{e\hbar}{2mc}.
\end{equation}

       semicalssical derivation
\begin{equation}
\frac{ qvB}{c} = \frac{mv^2}{r}  \rightarrow  E = \frac{e^2 B^2 r^2}{ m 2 c^2}
\end{equation}
\begin{equation}
\frac{ M}{N} = \frac{IA}{c}
\end{equation}
\begin{equation}
I =\frac{ev}{2\pi r} = \frac{e^2 B}{2\pi mc}  \rightarrow \frac{ M}{N}  =  \frac{e^2Br^2}{2 m c^2}
\end{equation}

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

(3) 
\begin{equation}
\epsilon_j =2 \mu_o B(j+ \frac{1}{2}) ,\hspace{5mm}    \mu_o = \frac{e\hbar}{2mc}.
\end{equation}
\begin{equation}
B_o = \frac{n h c}{e}, \hspace{5mm}   n = \frac{N}{A}\hspace{5mm},  x = \frac{B}{B_o}.
\end{equation}
\begin{equation}
x < 1  \rightarrow  \hspace{5mm}  \frac{1}{j+2}  < x <  \frac{1}{j +1}.
\end{equation}
j - lowest  landau  level  that  completely  filled.

\vspace{5mm}
$E = Nx\sum_{j =0}^{j}{\epsilon_j} +  [ N - (j +1) Nx ]\epsilon_{j+1} =$
\begin{equation}
= \mu_o NBx [(2j + 3) - (j + 2)(j + 1) x ].
 \end{equation}
Magnetization  per  unit  area.
\begin{equation}
 M = -\frac{N}{A}\frac{\partial E}{\partial B} = \mu_o n[2(j + 1)(j +2)x - (2j +3)].
\end{equation}
 $ E(B_\nu)$  equal  to  the  kinetic  energy  of the  system  that only  influenced  from  magnetic  field  that  dont  do  work  on  the  electron.
In  the  general  case $E = V(y) + \hbar w(\nu +\frac{1}{2}) $
 \vspace{1mm}

$V(y)$  depend  on  the  box  of area A.   

\Dn

%%%%%%%%%%%%%%%
(4)
\vspace{5mm} 

$\mu$
 - chemical  potential.
\begin{equation}
\mu(\nu) = \frac{B_\nu e\hbar}{mc}(\nu +\frac{1}{2})
\end{equation}   
\begin{equation}
I_x = R_{xy} V , \hspace{5mm}  R_{xy} = \frac{e^2}{h}  
\end{equation}   
\begin{equation}
I = \frac{-e}{h }(\mu(\nu) - \mu(\nu-1)) \rightarrow \Delta  M = \frac{IA}{c}  =   \frac{-e^2 A( B_\nu(\nu +\frac{1}{2}) - B_{\nu -1}(\nu - \frac{1}{2}))}{c^2 m}.
\end{equation}   




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