Solution (source code)

= Solution

With the specified ladder-operator <normalization>, $A_+^\dagger=A_-$ and
$$
A_+A_-=P^2+m^2\omega^2X^2-m\hbar\omega=2mH-m\hbar\omega.
$$
Therefore $\|A_-\psi\|^2=2m(E-\hbar\omega/2)\|\psi\|^2\geq0$. This proves $\boxed{E\geq\hbar\omega/2}$, with equality exactly when $A_-\psi=0$. In position representation this equation is $\hbar\psi'+m\omega x\psi=0$, whose general solution is
$$
\boxed{\psi(x)=C\exp[-m\omega x^2/(2\hbar)].}
$$
For a unit vector $|C|=(m\omega/\pi\hbar)^{1/4}$; its phase is arbitrary. The <commutators> $[H,A_\pm]=\pm\hbar\omega A_\pm$ make $A_+$ raise and $A_-$ lower energies by $\hbar\omega$. Repeated raising of the Gaussian produces all levels $(n+1/2)\hbar\omega$. Conversely lowering any nonzero <energy> <eigenvector> must eventually terminate, since otherwise it would produce an <energy> below the lower bound. A terminating vector is annihilated by $A_-$ and hence has <energy> $\hbar\omega/2$. Thus no other <eigenvalues> occur.

Bilinearity of the two-dimensional <commutators> gives
$$
[P_\theta,X_\varphi]=-i\hbar(\cos\theta\cos\varphi+\sin\theta\sin\varphi)\mathbf1=-i\hbar\cos(\theta-\varphi)\mathbf1,
$$
while all momentum-momentum and position-position <commutators> remain zero. Equal angles give a canonical position-momentum pair. Perpendicular angles give commuting components, which belong to independent coordinate directions.

The orthogonal coordinates $X_+=(X_1+X_2)/\sqrt2$, $X_-=(X_1-X_2)/\sqrt2$ preserve the kinetic sum, and
$$
5(X_1^2+X_2^2)-8X_1X_2=X_+^2+9X_-^2.
$$
Hence $\boxed{\omega_+=\omega,\quad\omega_-=3\omega}$ and $H=H_++H_-$. Every cross-mode position/momentum pair commutes, so $[H_+,H_-]=0$. On the tensor-product oscillator basis,
$$
E=\hbar\omega(n_++\tfrac12)+3\hbar\omega(n_-+\tfrac12)=(n_++3n_-+2)\hbar\omega.
$$
For each $N\geq0$, allowed pairs have $n_-=0,1,\ldots,\lfloor N/3\rfloor$ and $n_+=N-3n_-$. Thus
$$
\boxed{E_N=(N+2)\hbar\omega,\qquad g_N=\lfloor N/3\rfloor+1.}
$$
Each pair contributes one product <eigenvector>; counting them accounts for the <energy-level degeneracy>.