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\heading{E6040: Electron gas with magnetic field} 
\auname{Yaron de Leeuw}


{\bf The problem:}

\Dn

A box with electrons of mass $m$ is subjected to a magnetic field $B$.
The single particle interaction is described by $-\gamma B \sigma_z$.
The chemical potential of the electrons inside the box is $\mu$.
A hole through one of the walls is drilled.
The electrons that are emitted from the hole with
a velocity in the range ${v < v' < v+dv}$ are filtered,
and subsequently their spin is measured.
The measured current is defined as ${I=I_{\uparrow}+I_{\downarrow}}$.

\Dn\Dn

(a) Find the ratio
${\alpha(B;\mu) =  (I_{\uparrow}-I_{\downarrow}) / I}$.

\Dn\Dn

(b) Find a linear approximation for $\alpha(B;\mu)$
regarded as a function of the magnetic field.

\Dn\Dn

(c) What is the maximal value of $\alpha(B;\mu)/B$,
and what is the range for which the result is valid.

\Dn\Dn


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{\bf The solution:}

\Dn
The flux for $N$ electrons in volume $V$ with velocity $v$ is :
\[ J = \iint_{|\theta|<\frac{\pi}{2}}  \frac{d\Omega}{4\pi}\frac{N}{V}v\cos\alpha = \frac{1}{4}\left(\frac{N}{V}v\right)\]
The number of spin up electrons in the velocity range $v<v'<v+dv$ is given by the Fermi distribution:
\[ N_\uparrow(v<v'<v+dv) = g(\frac{1}{2} m v^2 - \gamma B)f(\frac{1}{2} m v^2 - \gamma B -\mu) \]
So the current of spin up electrons, with velocity $v<v'<v+dv$ is:
\[ I = J\delta A = g(\frac{1}{2} m v^2)f(\frac{1}{2} m v^2 - \gamma B -\mu) \frac{1}{4}v \delta A\]
The spin down current is:
\[ I = J\delta A = g(\frac{1}{2} m v^2)f(\frac{1}{2} m v^2 + \gamma B -\mu) \frac{1}{4}v \delta A\]
We can now find $\alpha(B;\mu)$, the normalized current difference:
\begin{align*}
    \alpha(B;\mu) &= \frac{f(\frac{1}{2} m v^2 - \gamma B -\mu)  -  f(\frac{1}{2} m v^2 + \gamma B -\mu) } {f(\frac{1}{2} m v^2 - \gamma B -\mu) +  f(\frac{1}{2} m v^2 + \gamma B -\mu)} = \frac{\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu -\gamma B)}} -\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu +\gamma B)}}} {\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu -\gamma B)}}+\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu +\gamma B)}}} \\
    &\approx \frac{\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu)}(1 -\beta\gamma B)} -\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu)}{(1 +\beta\gamma B)}}} {\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu)}}+\frac{1}{1+e^{\beta(\frac{1}{2} m v^2 -\mu)}}} \approx \frac{e^{\beta(\frac{1}{2} m v^2 -\mu)}}{e^{\beta(\frac{1}{2} m v^2 -\mu)}+1}\beta\gamma B \\
    &=\frac{1}{1+e^{-\beta(\frac{1}{2} m v^2 -\mu)}}\beta\gamma B+ \mathcal{O}(B^3)
\end{align*}
$\alpha(B;\mu)$ is an odd function of $B$, so there are only odd powers of $B$.

Within the approximation for low $B$ ($\beta\gamma B \ll 1$), the maximal value of $\alpha(B;\mu)/B$ is $\beta\gamma$ and it is valid for $e^{-\beta(\frac{1}{2} m v^2 -\mu)}\ll 1$ meaning we filter velocities at $ v^2\gg \frac{2}{m}(T+\mu)$

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