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\exnumber{3550}
\heading{Fermions in magnetic field - Pauli}
\auname{Yaakov Monsa}
{\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.
Consider the following cases:
\begin {itemize}
\item[(a)] one-dimensional box with length $L$.\Dn
\item[(b)] two-dimensional box with area $A$.\Dn
\item[(c)] three dimensional box with volume $V$.\Dn
\end {itemize}
Answer the following questions.
Express your results using $\gamma$, $m$, $N$, $L$, $A$, $V$.
\begin {itemize}
\item[(1)]
What is the single particle density of states.
Distinguish between a spin up and spin down particles.
\item[(2)]
Which is the graph that describes the magnetization $M(B)$
of each case $(a)$,$(b)$,$(c)$. Complete the missing
details: what are $M_s$, $B_c$ ,$\chi$.
\end {itemize}
\Dn
\begin{center}
\includegraphics[width=0.7\linewidth]{Ex3550FIG}
\end{center}
\Dn\Dn
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{\bf The solution:}
\Dn
(1)
The single particle density of states:
In order to get the density of states $g(\epsilon)=\frac{\partial}{\partial\epsilon}\mathcal{N}(\epsilon)$, we need to find $\mathcal{N}(\epsilon)$, the number of states up to energy $\epsilon$. $\mathcal{N}(\epsilon)$ is given by the volume of a d-dimensional ball with radius $\textbf{n}(\epsilon)$:
\[\mathcal{N}(\epsilon)=\frac{\Omega_{d}}{d}|\mathbf{n}(\epsilon)|^{d},\ \Omega_{d}=\begin{cases}
2, & d=1\\
2\pi, & d=2\\
4\pi, & d=3
\end{cases}\]
where $\textbf{n}(\epsilon)$ is defined from the realtion $\textbf{p} = \frac{2\pi}{L} \textbf{n}(\epsilon)$.
For a free particle the dispersion relation is $
\epsilon=\frac{p^{2}}{2m}$, we have $|\mathbf{n}(\epsilon) |=L\frac{\sqrt{2m}}{2\pi}\epsilon_{}^{\frac{1}{2}}$ so that: \[\mathcal{N}(\epsilon)=\frac{\Omega_{d}}{d}(L\frac{\sqrt{2m}}{2\pi})^{d}\epsilon^{\frac{d}{2}}=\frac{1}{d}\varLambda_{d}\epsilon^{\frac{d}{2}}\]
resulting in:
\[ g\left(\epsilon\right)=\frac{1}{2}\varLambda_{d}\epsilon^{\frac{d}{2}-1}\]
In our case, the one-particle Hamiltonian is $H_{1}=\frac{p^{2}}{2m}-\gamma B\sigma_{z}$, with the dispersion relation $\epsilon=\frac{p^{2}}{2m}\mp\gamma B$. Using $|\mathbf{n}(\epsilon) |=\frac{L}{2\pi}p=L\frac{\sqrt{2m}}{2\pi}\left(\epsilon\pm\gamma B\right)^{\frac{1}{2}}$, we get the number of states up to energy $\epsilon$ for the spin up(+) and down(-) states: $\mathcal{N}_{\pm}\left(\epsilon\right)=\frac{1}{d}\varLambda_{d}\left(\epsilon\pm\gamma B\right)^{\frac{d}{2}}$.\\
Finally, we get for the different dimensions:
\begin {itemize}
\item[(a)] One-dimensional box with length $L$: $g_{\pm}(\epsilon)=L\frac{\sqrt{2m}}{2\pi}(\epsilon\pm\gamma B)^{-\frac{1}{2}}$
\item[(b)] Two-dimensional box with area $A$:
$g_{\pm}(\epsilon)=A\frac{m}{2\pi}$
\item[(c)] Three dimensional box with volume $V$:
$g_{\pm}(\epsilon)=V\frac{(2m)^{\frac{3}{2}}}{4\pi^{2}}(\epsilon\pm\gamma B)^{\frac{1}{2}}$
\end {itemize}
\Dn
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\Dn
(2) We need to use two equations in order to solve this section, the first one relates the total particle and the number of energy states:
At T=0 the particles fills the energy levels up to the fermi energy $\epsilon_f$, so that:
\begin{align}N=\mathcal{N}_{+}+\mathcal{N}_{-}=\frac{1}{d}\varLambda_{d}[(\epsilon_f+\gamma B)^{\frac{d}{2}}+(\epsilon_f-\gamma B)^{\frac{d}{2}}]\end{align}
And the second equation is the total magnetization:
by definition, the one particle magnetization is $M_{1}=\langle -\frac{\partial H_{1}}{\partial B}\rangle =\gamma\langle \sigma_{z}\rangle$. So that, the total magnetizations is:
\begin{align}
M=\sum_{i}M_{i}=\gamma(\mathcal{N}_{+}-\mathcal{N}_{-})=\gamma\frac{1}{d}\varLambda_{d}\epsilon_f^{\frac{d}{2}}[(1+\frac{\gamma B}{\epsilon_f})^{\frac{d}{2}}-(1-\frac{\gamma B}{\epsilon_f})^{\frac{d}{2}}]
\end{align}
\Dn
Finding of $B_{c}$ and $M_s$:
For $B\geq B_{c}$ all the particles have spin up with maximal magnetization $M = M_s$. When $B = B_c$ the particles fills the energy levels up to energy $\epsilon_f = \gamma B_c$ (we mention that because $N$ is given $\epsilon_f$ isn't a constant as function of B, from eq.(1)). From eq.(1) we get
$N=\frac{1}{d}\varLambda_{d}[(2\gamma B_{c})^{\frac{d}{2}}+0]$, so that:
\[\gamma B_{c}=\frac{1}{2}(\frac{dN}{\varLambda_{d}})^{\frac{2}{d}},\ M_{s}=\gamma N\]
\Dn
Finding of $\chi$:
For $B\ll B_{c}$ we can approximate:
\[M\approx\gamma\frac{1}{d}\varLambda_{d}\epsilon_f^{\frac{d}{2}}[(1+\frac{d}{2} \frac{\gamma B}{\epsilon_f})-(1-\frac{d}{2}\frac{\gamma B}{\epsilon_f})]=\gamma\varLambda_{d}\epsilon_f^{\frac{d}{2}-1}\gamma B\]
furthermore, we can postulate that $\epsilon_f\approx\epsilon_{f0}+ O(B)$, ''$\epsilon_{f0}$'' is $\epsilon_f$ for $B=0$. for B=0 we will find $\epsilon_{f0}$ from eq.(1): $ N=\frac{1}{d}\varLambda_{d}[(\epsilon_{f0})^{\frac{d}{2}}+(\epsilon_{f0})^{\frac{d}{2}}]
\Rightarrow\
\epsilon_{f0}=(\frac{dN}{2\varLambda_{d}})^{\frac{2}{d}}=2^{1-\frac{2}{d}}\gamma B_{c}$, for this postulation we get:
\[M\approx\frac{d}{2^{2-\frac{2}{d}}}\frac{\gamma N}{B_c}B\equiv\chi B\]
In order to relate the graphs to the different systems we need to compare the slope of $M=\chi B$ with the slope of the straight line $M=\frac{M_s}{B_c} B$. We can see that in the left graph $\chi>\frac{M_{s}}{B_{c}}$, in the right one $\chi<\frac{M_s}{B_c}$, and in the middle one $\chi=\frac{M_s}{B_c}$. So that, the equation $\chi=\frac{d}{2^{2-\frac{2}{d}}}\frac{M_s}{B_c}$ relates the graphs and the dimensions.
\Dn
Finnaly we have:
for all the three cases $M_{s}=\gamma N$.
\begin {itemize}
\item[(a)] One-dimensional box with length $L$:
$\gamma B_{c}=\frac{\pi^{2}N^{2}}{4mL^{2}},\ \ \chi=\frac{M_{s}}{B_{c}}$ - this fits the middle graph.
\item[(b)] Two-dimensional box with area $A$:
$\gamma B_{c}=\frac{\pi N}{Am},\ \ \chi=\frac{M_{s}}{B_{c}}$ - this fits the middle graph too.
\item[(c)] Three dimensional box with volume $V$:
$\gamma B_{c}=\frac{1}{4m}(\frac{6\pi^{2}N}{V})^{\frac{2}{3}},\ \chi=\frac{3}{2^{2-\frac{2}{3}}}\frac{M_{s}}{B_{c}}$ - this fits the left graph.
\end {itemize}
\Dn
\
\Dn
\textbf{Appendix}: Drawing the graphs.
\Dn
In order to draw the graphs we need to find $M(B)$, but actually it is easier to find $B(M)$. We want to discard $\epsilon_f$. From eq.(1-2) we can get:
\[\epsilon_f+\gamma B=[\frac{dN}{2\varLambda_{d}}(1+\frac{M}{\gamma N})]^{\frac{2}{d}}\]
\[\epsilon_f-\gamma B=[\frac{dN}{2\varLambda_{d}}(1-\frac{M}{\gamma N})]^{\frac{2}{d}}\]
so that the equation that relates $M$ and $B$ is:
\[B=2^{-\frac{2}{d}} B_{c}[(1+\frac{M}{M_{s}})^{\frac{2}{d}}-(1-\frac{M}{M_{s}})^{\frac{2}{d}}]\]
for the approximation of $M\ll M_s$ we can get $\chi$ again:
\[B=2^{-\frac{2}{d}} B_{c}\frac{4}{d}\frac{M}{M_{s}}\ \Rightarrow\ M=\frac{d}{2^{2-\frac{2}{d}}}\frac{M_{s}}{B_{c}}B=\chi B\]
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\end{document}