= Solution
The <canonical commutation relations> are $[\widehat x,\widehat p]=i\hbar I$ (in several dimensions $[x_i,p_j]=i\hbar\delta_{ij}I$, with the position-position and momentum-momentum <commutators> zero). For Hermitian $x,p$, the <commutator> is anti-Hermitian, just as $i\hbar I$ is; there is no conflict with their self-adjointness. The <annihilation operator> and its adjoint obey
$$
[a,a^\dagger]=\frac1{2\hbar}[x+ip,x-ip]=1,\qquad
N=a^\dagger a=\frac{x^2+p^2-\hbar}{2\hbar}.
$$
Thus the <quantum harmonic oscillator> <Hamiltonian> is $\boxed{H=\hbar(N+1/2)}$. The <commutators> $[N,a]=-a$, $[N,a^\dagger]=a^\dagger$ and the normalized <ground state> construct
$$
\boxed{|n\rangle=\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle,\quad
E_n=\hbar(n+1/2),\quad
a|n\rangle=\sqrt n|n-1\rangle,\quad
a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle.}
$$
Normalization follows inductively from $aa^\dagger=N+1$. Positivity of $N$ and repeated lowering exclude a noninteger <eigenvalue>: lowering it until it becomes negative would produce a nonzero vector of negative $N$ <eigenvalue>. Every eigenstate therefore lowers to the unique <ground state>, so these normalized ladder states exhaust the oscillator eigenstates.
For two independent oscillators, the <tensor products> $|n_1,n_2\rangle$ have $E=\hbar(n_1+n_2+1)$. The level $n=n_1+n_2$ has \b[degeneracy $n+1$]. In its degenerate subspace the perturbation $\lambda x_1x_2$ reduces to
$$
\frac{\lambda\hbar}{2}(a_1a_2^\dagger+a_1^\dagger a_2),
$$
since terms that raise or lower both numbers leave the subspace. For $n=0$ its matrix is zero. For $n=1$, in $|1,0\rangle,|0,1\rangle$, it is $(\lambda\hbar/2)\begin{pmatrix}0&1\\1&0\end{pmatrix}$. For $n=2$, in $|2,0\rangle,|1,1\rangle,|0,2\rangle$, it is $(\lambda\hbar/2)\begin{pmatrix}0&\sqrt2&0\\\sqrt2&0&\sqrt2\\0&\sqrt2&0\end{pmatrix}$. Their <eigenvalues> give
$$
\boxed{\begin{array}{c|c}
n&\text{energies through first order}\\\hline
0&\hbar+O(\lambda^2)\\
1&2\hbar\pm\lambda\hbar/2+O(\lambda^2)\\
2&3\hbar-\lambda\hbar,\ 3\hbar,\ 3\hbar+\lambda\hbar\quad+O(\lambda^2).
\end{array}}
$$
The $n=1$ states are the symmetric and antisymmetric combinations. At $n=2$, the shifted states are $(|2,0\rangle\pm\sqrt2|1,1\rangle+|0,2\rangle)/2$, and the zero-shift state is $(|2,0\rangle-|0,2\rangle)/\sqrt2$.
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