= Solution
Contracting the result of part (c) with the \x[Pauli matrix]{magic}{p} gives
$$
\begin{aligned}
[T^a,T^b]
&=\frac14\sigma^a_{ij}\sigma^b_{kl}
(\delta_{jk}T_{il}-\delta_{il}T_{kj})\\
&=\frac14[\sigma^a,\sigma^b]_{il}T_{il}
=i\varepsilon_{abc}T^c.
\end{aligned}
$$
Thus the $T^a$ form the <Schwinger boson representation> of the $\mathfrak{su}(2)$ commutation relations. The second Pauli matrix gives
$$
T^2=\frac{i}{2}(a_y^\dagger a_x-a_x^\dagger a_y),
$$
and therefore
$$
\boxed{L=2T^2}.
$$
On the energy level $E=N+1$, the total occupation is $N$. The $N+1$ states $|n_x,N-n_x\rangle$ form the spin-$j=N/2$ irreducible representation. Every component of its angular momentum generator has eigenvalues
$$
m=-\frac N2,-\frac N2+1,\ldots,\frac N2.
$$
Since $L=2T^2$, the possible orbital angular momenta are
$$
\boxed{\ell=-N,-N+2,\ldots,N-2,N}.
$$
There are $N+1$ distinct values, exactly matching the degeneracy found in part (a); hence each occurs once. Restoring $\hbar$ multiplies every $\ell$ by $\hbar$.
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