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\begin{document}
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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.
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\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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\end{document}