= Solution
In <Landau gauge for a uniform magnetic field> $\mathbf A=(-By,0,0)$, the <guiding centre> is $Y=\hbar k_x/(eB)$. The <group velocity> is
$$
\boxed{v_x=\frac1\hbar\frac{dE}{dk_x}=\frac{U'(Y)}{eB}.}
$$
Thus the upper edge in the original trajectory diagram carries <electrons> to the right and the lower edge to the left. Reversing the <magnetic field> reverses both arrows. Semiclassically these are skipping trajectories; quantum mechanically the <guiding centres> label propagating <quantum Hall edge states>. The larger orbital-plus-spin <energy> crosses the <Fermi energy> farther toward the interior, because less <confinement> <energy> is needed. Opposite <spins> in one orbital branch are therefore spatially separated.
In a clean bulk with the <Fermi energy> between <Landau levels>, there are no extended bulk states at that <energy>. Occupied bulk cyclotron orbits lie below it; one must not draw them as extra conducting Fermi-level channels. Smooth disorder can instead produce localized bulk contours without adding a direct source-to-drain channel.
\Image[/past-exam-of-the-physics-of-nanoelectronic-systems-course-of-the-university-of-cambridge/2013/paper-1-edge-trajectories.png]
{title=Original guiding-centre trajectories for the four Fermi-level channels on each boundary}
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