= Solution
{c}
Write $u=\psi_{1,0}=a_1^*\psi$ and $v=\psi_{0,1}=a_2^*\psi$. The <canonical commutation relations> for the <creation and annihilation operators> give
$$
a_1u=\psi,\qquad a_2u=0,\qquad a_1v=0,\qquad a_2v=\psi.
$$
For example, $a_1a_1^*\psi=(I+a_1^*a_1)\psi=\psi$. The expression for $L$ from part (b), with distinct modes reordered, then yields
$$
Lu=i\hbar v,\qquad Lv=-i\hbar u.
$$
Thus $\operatorname{span}\{u,v\}$ is an <invariant subspace>, and the <matrix> of $L$ in the ordered <orthonormal basis> $(u,v)$ is
$$
[L]_{(u,v)}=\hbar\begin{pmatrix}0&-i\\i&0\end{pmatrix}.
$$
Its <characteristic polynomial> is $\ell^2-\hbar^2$, giving the <eigenvalues> and normalized <eigenstates>
$$
\boxed{\ell_+=+\hbar,\qquad \chi_+=\frac{\psi_{1,0}+i\psi_{0,1}}{\sqrt2};\qquad
\ell_-=-\hbar,\qquad \chi_-=\frac{\psi_{1,0}-i\psi_{0,1}}{\sqrt2}.}
$$
Indeed, $L(u\pm iv)=\pm\hbar(u\pm iv)$ and $\|u\pm iv\|^2=2$ by the assumed <orthonormality>. Both <eigenstates> remain in the $N=1$ <eigenspace>.
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