 Solution

Write $u={\psi }_{1,0}=a_1^{\ast }\psi$ and $v={\psi }_{0,1}=a_2^{\ast }\psi$. The canonical commutation relations for the creation and annihilation operators giveFor example, $a_1{a}_1^{\ast }\psi =(I+a_1^{\ast }{a}_1)\psi =\psi$. The expression for $L$ from part (b), with distinct modes reordered, then yieldsThus $\mathrm{span}\{u,v\}$ is an invariant subspace, and the matrix of $L$ in the ordered orthonormal basis $(u,v)$ isIts characteristic polynomial is ${\ell }^2-{\hslash }^2$, giving the eigenvalues and normalized eigenstatesIndeed, $L(u\pm iv)=\pm \hslash (u\pm iv)$ and $\Vert u\pm iv{\Vert }^2=2$ by the assumed orthonormality. Both eigenstates remain in the $N=1$ eigenspace.