= Solution
Use $e>0$, $B>0$ along $+z$, and $U(y)$ for the <electron>'s <confinement> <potential energy>. The spin-resolved <Landau level> branches in the smooth-edge approximation are
$$
\boxed{E_{n,\pm}(y)=\hbar\omega_c(n+\tfrac12)\pm\tfrac12E_Z+U(y),\qquad
\omega_c=eB/m^*.}
$$
The <spin> labels denote upper and lower energies; the sign of a material's effective $g$ factor is not inferred from the stated positive splitting. In units of $\hbar\omega_c$, the four occupied bulk energies are $1/3,2/3,4/3,5/3$. A <chemical potential> in $(5/3,7/3)$ therefore gives <filling factor> four. The branches rise at both confining edges and cross the common <Fermi energy> once at each edge. Each crossing is one spin-resolved <quantum Hall edge state>; there are four on either side. The two <spin> branches within an orbital level are separated by $\hbar\omega_c/3$, while the intervening gap to the next orbital pair is $2\hbar\omega_c/3$.
\Image[/past-exam-of-the-physics-of-nanoelectronic-systems-course-of-the-university-of-cambridge/2013/paper-1-edge-spectrum.png]
{title=Original normalized edge spectrum with four occupied spin-resolved bulk branches}
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