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\heading{E3550: Magnetic properties of T=0 electrons (Pauli)} 
\auname{Gitit Feingold}


{\bf The problem:}\\
\Dn
$N$ electrons with mass $m$ and spin $\frac{1}{2}$ are placed in a box at zero temperature. A magnetic field B is applied, such that the interaction is $-\gamma B \sigma_{z}$ where $\gamma$ is the gyromagnetic ratio.
Solve the question for:
(I) one-dimensional box with length L. \\
(II) two-dimensional box with area A.\\
(III) three dimensional box with volume V.\\
\begin {itemize}
\item[(a)]
Describe the single particle density of states function, differ spins up from spins down.
\item[(b)]
Determine which graph describes the magnetization $M(B)$ of each case $(I)$,$(II)$,$(III)$ and complete the missing details ($M_s=$?,$B_c=$?,$\chi=$?)
\end {itemize}
Use only $\gamma$,m,N,L,A,V.

\begin{figure}[h]
\begin{center}
\includegraphics[width=4in]{Ex3550}
\end{center}
\end{figure}


\Dn\Dn
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{\bf The solution:}
\begin {itemize}
\item[(a)] One can easily see that the system can occupy more spin up states.
\end {itemize} 
%%%%%%%% 1D %%%%%%%%
\textbf{1D:}\\
For $B<B_c$ there are both spins up and spins down in the system, thus, the total number of spins in an energy level is:
\begin{equation}
N(E)=N_{\downarrow}+N_{\uparrow}
\end{equation}
\begin{equation}
N(E_f)=4\cdot L(2m)^{1/2}E^{-1/2}=4L(2m)^{1/2}((E_f+\frac{\gamma B}{2})^{-1/2}+(E_f-\frac{\gamma B}{2})^{-1/2})
\end{equation}
Magnetization is calculated using the difference between up and down spins: 
\begin{equation}
M(B)=\sigma \gamma(N_{\uparrow}-N_{\downarrow})=4L(2m)^{1/2}\frac{\gamma}{2}[(E_f-\frac{\gamma B}{2})^{-1/2}-(E_f+\frac{\gamma B}{2})^{-1/2}]\cong 2L(2m)^{1/2}\frac{\gamma}{2}E_f^{-1/2}(\frac{\gamma B}{2E_f}+\frac{5}{8}(\frac{\gamma B}{2E_f})^3)
\end{equation}
Calculating the magnetic susceptibility, we use the definition:
\begin{equation}
\chi = \frac{\partial M}{\partial B}=\frac{L}{2}(2m)^{1/2}\gamma ^2[(E_f-\frac{\gamma B}{2})^{-3/2}+(E_f+\frac{\gamma B}{2})^{-3/2}]
\end{equation}
Critical magnetic field $B_c$ is such that all spins are in the up state i.e. low energy level, at $T=0$ it means that $E<E_f$, therefore we can find $B_c$ using the condition:
\begin{equation}
\frac{\gamma B}{2}=E_f \Longrightarrow B_c=\frac{1}{2m}(\frac{\pi N(0)}{8L})^{2}\frac{2}{\gamma}
\end{equation} 
 
%%%%%%%% 2D %%%%%%%%
\textbf{2D:}\\
\begin{equation}
N(E_f)=2\cdot \frac{A\pi}{(2\pi)^2}(2mE)=\frac{2AmE_f}{\pi}
\end{equation}
Using the same method as before the magnetization is:
\begin{equation}
M(B)=\frac{\gamma}{2}\Delta N=\frac{Am}{2\pi}\gamma ^2 B
\end{equation}
Now calculating the magnetic susceptibility:
\begin{equation}
\chi = \frac{\partial M}{\partial B}=\frac{Am}{2\pi}\gamma ^2
\end{equation} 
the critical magnetic field $B_c$ will be:
\begin{equation}
\frac{\gamma B}{2}=E_f \Longrightarrow B_c=\frac{2\pi N(0)}{Am\gamma}
\end{equation} 
%%%%%%%% 3D %%%%%%%%
\Dn
\textbf{3D:}
Again we write an expression for the number of spins as:
\begin{equation}
N(E_f)=2\cdot\frac{4\pi}{3}\frac{V}{(2\pi)^3}\cdot(2mE)^{3/2}=\frac{V}{3\pi^2}(2m)^{3/2}[(E_f+\frac{\gamma B}{2})^{3/2}+(E_f-\frac{\gamma B}{2})^{3/2}]
\end{equation}
The magnetization will be:
\begin{equation}
M(B)=\frac{\gamma}{2}\Delta N=\frac{V}{3\pi^2}(2m)^{3/2}\frac{\gamma}{2}[(E_f+\frac{\gamma B}{2})^{3/2}-(E_f-\frac{\gamma B}{2})^{3/2}]
\end{equation}
\begin{equation}
M(B)=\frac{\gamma}{2}\Delta N\cong \frac{\gamma}{2}\frac{V}{3\pi^2}(2m)^{3/2}\sqrt{E_f}\frac{3}{2}\cdot\gamma B + O(B^3)
\end{equation}
The Magnetization is therefore:
\begin{equation}
M=\frac{V}{4\pi^2}(2m)^{3/2}\sqrt{E_f}\cdot\gamma ^2 B
\end{equation}
where
\begin{equation}
E_f=\frac{1}{2m}(\frac{3\pi ^2\cdot N(0)}{V})^{2/3}
\end{equation}
The susceptibility:
\begin{equation}
\chi =\frac{V^{2/3}\gamma ^2}{4\pi ^2}2m(3\pi ^2 N)^{1/3}=const>0
\end{equation}
The critical magnetic field has the same condition as in the 2D case:
\begin{equation}
\frac{\gamma B}{2}=E_f \Longrightarrow B_c=\frac{1}{m\gamma}(\frac{3\pi ^2 N(0)}{V})^{2/3}
\end{equation} 

\begin {itemize}
\item[(b)]
\end {itemize} 
Using the above calculations we are able to match the right graph to the magnetization functions first graph from the left represent the $1D$ case, the second one refers to the $3D$ case and the third refers to the $2D$ case. 

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