= 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, so
$$
\boxed{L_3\psi_{0,0}=0,\qquad
L_3\psi_{1,0}=i\hbar\psi_{0,1},\qquad
L_3\psi_{0,1}=-i\hbar\psi_{1,0}.}
$$
For 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 as
$$
\boxed{\chi_0=\psi_{0,0},\qquad
\chi_+=\frac{\psi_{1,0}+i\psi_{0,1}}{\sqrt2},\qquad
\chi_-=\frac{\psi_{1,0}-i\psi_{0,1}}{\sqrt2}.}
$$
Their pairs of <eigenvalues> $(H,L_3)$ are respectively $(\hbar\omega,0)$, $(2\hbar\omega,+\hbar)$ and $(2\hbar\omega,-\hbar)$. The factor $1/\sqrt2$ 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 $L_3$.
Back to article page
