 Solution

Choose the first three energy eigenstates consistently as ${\psi }_{0,0}$, ${\psi }_{1,0}=a_{1,+}{\psi }_{0,0}$ and ${\psi }_{0,1}=a_{2,+}{\psi }_{0,0}$. The Fock vacuum is annihilated by both lowering operators, soFor example $a_{1,-}{a}_{1,+}{\psi }_{0,0}={\psi }_{0,0}$, while $a_{2,-}{\psi }_{1,0}=0$, giving the middle formula. Thus the simultaneous eigenstates can be chosen asTheir pairs of eigenvalues $(H,L_3)$ are respectively $(\hslash \omega ,0)$, $(2\hslash \omega ,+\hslash )$ and $(2\hslash \omega ,-\hslash )$. The factor $1/\sqrt{2}$ normalizes the combinations when the original three states are normalized. Mixing only inside the degenerate first-excited energy eigenspace preserves the energy eigenvalue while diagonalizing $L_3$.