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\begin{document}

\exnumber{3550}
\heading{Fermions in Magnetic Field} 
\auname{Alon Ishakbayev}

\paragraph*{The Problem:}
\begin{flushleft}
$N$ electrons with mass $m$ and spin $\frac{1}{2}$ are placed in
a box at zero temperature $T=0$. A magnetic field B is applied, such
that the interaction is $-\gamma B\sigma_{z}$ where $\gamma$ is
the gyromagnetic ratio. Consider the following cases:
\par\end{flushleft}

\begin{flushleft}
$(a)$ One dimensional box with length $L$
\par\end{flushleft}

\begin{flushleft}
$(b)$ Two dimensional box with area $A$
\par\end{flushleft}

\begin{flushleft}
$(c)$ Three dimensional box with volume $V$
\par\end{flushleft}

\paragraph*{Answer the following questions:}
\begin{flushleft}
Express your results using $\gamma$, $m$, $N$, $L$, $A$, $V$.
\par\end{flushleft}

\begin{flushleft}
$(1)$ What is the single particle density of states. Distinguish
between a spin up and spin down particles.
\par\end{flushleft}

\begin{flushleft}
$(2)$ Which is the graph that describes the magnetization $M(B)$
of each case $(a)$,$(b)$,$(c)$.
\par\end{flushleft}

\begin{flushleft}
Complete the missing details: what are $M_{s}$, $B_{c}$ ,$\chi$.
\par\end{flushleft}

\begin{center}
	\includegraphics[width=0.7\linewidth]{Ex3550Pic}
\end{center}


\paragraph*{The Solution:}
\begin{flushleft}
$(1)$ By definition the density of states is given by: 
\begin{equation}
g(\epsilon)=\frac{\partial N(\epsilon)}{\partial\epsilon}
\end{equation}
\par\end{flushleft}

\begin{flushleft}
In order to find $g(\epsilon)$, we will find $N(\epsilon)$ the number
of states up to energy $\epsilon$ 
\par\end{flushleft}

\begin{flushleft}
The one particle hamiltonian is given by: 
\begin{equation}
H_{1}=\frac{\boldsymbol{p}^{2}}{2m}-\gamma B\sigma_{z}
\end{equation}
\par\end{flushleft}

\begin{flushleft}
Therefore the dispersion relation for spin up $\uparrow$ and down
$\downarrow$ correspondingly : 
\begin{equation}
\epsilon=\frac{\boldsymbol{p}^{2}}{2m}\mp\gamma B
\end{equation}
\par\end{flushleft}

\begin{flushleft}
Assuming periodic boundary conditions, momentum states: $\boldsymbol{p}=\boldsymbol{\frac{2\pi}{L}n}$,
$(n=0,\pm1,\pm2...)$
\par\end{flushleft}

\begin{flushleft}
	Therefore:
\begin{equation}
|n_{\pm}(\epsilon)|=\frac{L}{2\pi}\sqrt{2m(\epsilon\pm\gamma B)}
\end{equation}
\par\end{flushleft}

\begin{flushleft}
The total number of states $N(\epsilon)$ for both spins $\uparrow$
and $\downarrow$ correspondingly:
\par\end{flushleft}

\begin{flushleft}
\begin{equation}
N_{\pm}(\epsilon)=\begin{cases}
2n_{\pm}(\epsilon), & 1D\\
\pi n_{\pm}^{2}(\epsilon), & 2D\\
\frac{4\pi}{3}n_{\pm}^{3}(\epsilon), & 3D
\end{cases}=\begin{cases}
\frac{L}{\pi}[2m(\epsilon\pm\gamma B)]^{\frac{1}{2}}, & 1D\\
\frac{A}{2\pi}m(\epsilon\pm\gamma B), & 2D\\
\frac{V}{6\pi^{2}}[2m(\epsilon\pm\gamma B)]^{\frac{3}{2}}, & 3D
\end{cases}
\end{equation}
\par\end{flushleft}

\begin{flushleft}
Finally, the DOS $g_{\pm}(\epsilon):$ 
\begin{equation}
g_{\pm}(\epsilon)=\frac{\partial N(\epsilon)}{\partial\epsilon}=\begin{cases}
\frac{L}{2\pi}(2m)^{\frac{1}{2}}(\epsilon\pm\gamma B){}^{-\frac{1}{2}}, & 1D\\
\frac{mA}{2\pi}, & 2D\\
\frac{V}{4\pi^{2}}(2m)^{\frac{3}{2}}(\epsilon\pm\gamma B){}^{\frac{1}{2}}, & 3D
\end{cases}
\end{equation}
\par\end{flushleft}

\begin{flushleft}
$(2)$ The magnetization $M(B)$ is given by: 
\begin{equation}
M(B)=<-\frac{\partial H}{\partial B}>
\end{equation}
\par\end{flushleft}

\begin{flushleft}
Total number of electrons at $T=0$ : 
\begin{equation}
N=N_{\uparrow}+N_{\downarrow}=\begin{cases}
\frac{L}{\pi}\sqrt{2m}[(\epsilon_{f}+\gamma B)^{\frac{1}{2}}+(\epsilon_{f}-\gamma B)^{\frac{1}{2}}], & 1D\\
\frac{Am}{\pi}\epsilon_{f}, & 2D\\
\frac{V}{6\pi^{2}}(2m)^{\frac{3}{2}}[(\epsilon_{f}+\gamma B)^{\frac{3}{2}}+(\epsilon_{f}-\gamma B)^{\frac{3}{2}}], & 3D
\end{cases}
\end{equation}
\par\end{flushleft}

\begin{flushleft}
The hamiltonian of $N$ spins:
\begin{equation}
H=\sum_{i=1}^{N}\frac{\boldsymbol{p}_{i}^{2}}{2m}+\gamma B(N_{\uparrow}-N_{\downarrow})
\end{equation}
\par\end{flushleft}

\begin{flushleft}
The total magnetization $M(B)$: 
\begin{equation}
M(B)=\gamma(N_{\uparrow}-N_{\downarrow})=\begin{cases}
\frac{\gamma L}{\pi}\sqrt{2m}[(\epsilon_{f}+\gamma B)^{\frac{1}{2}}-(\epsilon_{f}-\gamma B)^{\frac{1}{2}}], & 1D\\
\frac{\gamma^{2}Am}{\pi}B, & 2D\\
\frac{\gamma V}{6\pi^{2}}(2m)^{\frac{3}{2}}[(\epsilon_{f}+\gamma B)^{\frac{3}{2}}-(\epsilon_{f}-\gamma B)^{\frac{3}{2}}], & 3D
\end{cases}
\end{equation}
\par\end{flushleft}


\begin{flushleft}
\break
\textbf{\underline{Finding the missing details:}} $M_{s}$, $B_{c}$
,$\chi$
\par\end{flushleft}

\begin{flushleft}
From the graphs: above the value $B_{c}$ all the spins are $\uparrow$
, therefore the magnetization $M_{s}$ for all the three cases is
just the number of electrons times the gyromagnetic ratio: 
\begin{equation}
M_{s}=\gamma N
\end{equation}
\par\end{flushleft}

To find the critical magnetic field $B_{c}$ one needs to solve equation
$(10)$.

using the fact that $N_{\downarrow}=0$, means that the fermi energy
is: $\epsilon_{f}=\gamma B_{c}$.

Combining equations $(10+11)$: 
\begin{equation}
M_{s}=\gamma N=\begin{cases}
\frac{\gamma L}{\pi}\sqrt{2m}\cdot(2\gamma B_{c})^{\frac{1}{2}}, & 1D\\
\frac{\gamma^{2}Am}{\pi}\cdot B_{c}, & 2D\\
\frac{\gamma V}{6\pi^{2}}(2m)^{\frac{3}{2}}\cdot(2\gamma B_{c})^{\frac{3}{2}}, & 3D
\end{cases}
\end{equation}

Solving equation $(12)$ for $B_{c}$: 

\begin{equation}
B_{c}=\begin{cases}
\frac{\pi^{2}N^{2}}{4m\gamma L^{2}}, & 1D\\
\frac{\pi N}{mA\gamma}, & 2D\\
\frac{1}{4m\gamma}(\frac{6\pi^{2}N}{V})^{\frac{2}{3}}, & 3D
\end{cases}
\end{equation}

To find $\chi$ we will find the slope of $M(B)$ under the condition
$B_{c}\gg B\simeq0$

The magnetization can be approximated using: $(1+x)^{\frac{1}{2}}\approx1+\frac{1}{2}x$,
$(1+x)^{\frac{3}{2}}\approx1+\frac{3}{2}x$

From $(10)$:
\begin{flushleft}
\begin{equation}
M(B)=\Biggl\{\begin{array}{c}
\frac{\gamma L}{\pi}\sqrt{2m\epsilon_{f}}[(1+\frac{\gamma B}{\epsilon_{f}})^{\frac{1}{2}}-(1-\frac{\gamma B}{\epsilon_{f}})^{\frac{1}{2}}]\\
\frac{\gamma^{2}Am}{\pi}B\\
\frac{\gamma V}{6\pi^{2}}(2m\epsilon_{f})^{\frac{3}{2}}[(1+\frac{\gamma B}{\epsilon_{f}})^{\frac{3}{2}}-(1-\frac{\gamma B}{\epsilon_{f}})^{\frac{3}{2}}]
\end{array}\approx\Biggl\{\begin{array}{c}
\frac{\gamma L}{\pi}\sqrt{2m}\frac{\gamma}{\sqrt{\epsilon_{f}}}\cdot B\\
\frac{\gamma^{2}Am}{\pi}\cdot B\\
\frac{\gamma^{2}V}{2\pi^{2}}(2m)^{\frac{3}{2}}\sqrt{\epsilon_{f}}\cdot B
\end{array}
\end{equation}
\par\end{flushleft}

Finding $\epsilon_{f}$ around $B\simeq0$ using equation $(8)$ of
the total number of electrons $N$:
\begin{flushleft}
\begin{equation}
N=\Biggl\{\begin{array}{c}
\frac{L}{\pi}\sqrt{2m}[(\epsilon_{f}+\gamma B)^{\frac{1}{2}}+(\epsilon_{f}-\gamma B)^{\frac{1}{2}}]\\
\frac{Am}{\pi}\epsilon_{f}\\
\frac{V}{6\pi^{2}}(2m)^{\frac{3}{2}}[(\epsilon_{f}+\gamma B)^{\frac{3}{2}}+(\epsilon_{f}-\gamma B)^{\frac{3}{2}}]
\end{array}\approx\Biggl\{\begin{array}{c}
\frac{L}{\pi}\sqrt{2m}2\cdot\epsilon_{f}^{\frac{1}{2}}\\
\frac{Am}{\pi}\cdot\epsilon_{f}\\
\frac{V}{3\pi^{2}}(2m)^{\frac{3}{2}}\cdot\epsilon_{f}^{\frac{3}{2}}
\end{array}
\end{equation}
\par\end{flushleft}

Therefore: 

\begin{equation}
\epsilon_{f}=\begin{cases}
\frac{\pi^{2}N^{2}}{8mL^{2}} & 1D\\
\frac{\pi N}{Am} & 2D\\
\frac{1}{2m}(\frac{3\pi^{2}N}{V})^{\frac{2}{3}} & 3D
\end{cases}
\end{equation}

Substituting $(16)$ into the magnetization from equation $(14)$:

\begin{equation}
M(B)=\Biggl\{\begin{array}{c}
\frac{4\gamma^{2}L^{2}m}{\pi^{2}N}\cdot B\\
\frac{M_{s}}{B_{c}}\cdot B\\
\frac{3}{2^{\frac{4}{3}}}\frac{M_{s}}{B_{c}}\cdot B
\end{array}=\Biggl\{\begin{array}{c}
\frac{M_{s}}{B_{c}}\cdot B\\
\frac{M_{s}}{B_{c}}\cdot B\\
1.2\frac{M_{s}}{B_{c}}\cdot B
\end{array}
\end{equation}

$\chi$ is the linear slope of $M(B)$ :

\begin{equation}
\chi=\begin{cases}
\frac{M_{s}}{B_{c}} & 1D\\
\frac{M_{s}}{B_{c}} & 2D\\
1.2\frac{M_{s}}{B_{c}} & 3D
\end{cases}
\end{equation}

To fit each graph to its corresponding case we will use equations $(10)+(8)$ to find $B(M)$ (removing dependence of $\epsilon_{f}$):

\begin{equation}
	$B$=\begin{cases}
		\frac{B_{c}}{4}[(1+\frac{M}{M_{S}})^2-(1-\frac{M}{M_{S}}^2)] & 1D\\
		\frac{B_{c}}{2}[(1+\frac{M}{M_{S}})-(1-\frac{M}{M_{S}})] & 2D\\
		\frac{B_{c}}{2^{2/3}}[(1+\frac{M}{M_{S}})^{2/3}-(1-\frac{M}{M_{S}}^{2/3})] & 3D		
	\end{cases}=\begin{cases}
	\frac{B_{c}}{M_{S}}M, & 1D\\
	\frac{B_{c}}{M_{S}}M, & 2D\\
	\frac{B_{c}}{2^{2/3}}[(1+\frac{M}{M_{S}})^{2/3}-(1-\frac{M}{M_{S}}^{2/3})], & 3D
	\end{cases}	
\end{equation}

From equation(19) we can see that the linear middle graph represents $1D$ and $2D$ case whereas the left graph represents the $3D$ case.

\textbf{{\underline{Results Summary}}}:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline 
Case & Density of States & $M_{s}$ & $B_{c}$ & $\chi$ & Graph \tabularnewline
\hline 
\hline 
1D box & $\frac{L}{2\pi}(2m)^{\frac{1}{2}}(\epsilon\pm\gamma B){}^{-\frac{1}{2}}$ & $\gamma N$ & $\frac{\pi^{2}N^{2}}{4m\gamma L^{2}}$ & $\frac{M_{s}}{B_{c}}$ & Middle\tabularnewline
\hline 
2D box & $\frac{mA}{2\pi}$ & $\gamma N$ & $\frac{\pi N}{mA\gamma}$ & $\frac{M_{s}}{B_{c}}$ & Middle\tabularnewline
\hline 
3D box & $\frac{V}{4\pi^{2}}(2m)^{\frac{3}{2}}(\epsilon\pm\gamma B){}^{\frac{1}{2}}$ & $\gamma N$ & $\frac{1}{4m\gamma}(\frac{6\pi^{2}N}{V})^{\frac{2}{3}}$ & $1.2\frac{M_{s}}{B_{c}}$ & Left\tabularnewline
\hline 
\end{tabular}
\par\end{center}
\end{document}
