= Solution
The oscillator operators satisfy $[a,a^\dagger]=1$ and $[a,a]=[a^\dagger,a^\dagger]=0$. Substitution into the kinetic and potential energies gives
$$
\boxed{H=\hbar\omega(a^\dagger a+\tfrac12).}
$$
The <number operator> $N=a^\dagger a$ is nonnegative because $\langle\psi,N\psi\rangle=\|a\psi\|^2$. The commutators $[N,a]=-a$ and $[N,a^\dagger]=a^\dagger$ lower and raise an <eigenvalue> by one. If an <eigenvalue> were a noninteger $\nu\geq0$, repeated lowering would produce a nonzero <eigenvector> with <eigenvalue> between zero and one, and one further lowering would have negative <eigenvalue>; its <norm> is nonzero because $\|a\psi\|^2=\nu\|\psi\|^2$. This contradicts nonnegativity. Thus only nonnegative integers are allowed. The normalized Gaussian solves $a|0\rangle=0$, and raising it supplies every level, with
$$
\boxed{E_n=\hbar\omega(n+\tfrac12),\qquad
|n\rangle=(a^\dagger)^n|0\rangle/\sqrt{n!}.}
$$
For the three-dimensional unperturbed oscillator the product states are $|n_1,n_2,n_3\rangle=|n_1\rangle\otimes|n_2\rangle\otimes|n_3\rangle$, with energy $\hbar\omega(n_1+n_2+n_3+3/2)$.
Write the perturbation as $\lambda W$, where $W=(\hbar\omega/2)\sum_{i<j}(a_i+a_i^\dagger)(a_j+a_j^\dagger)$. Its ground-state expectation is zero. Acting on the <ground state>, it produces $|110\rangle,|101\rangle,|011\rangle$, each with coefficient $\hbar\omega/2$. They lie $2\hbar\omega$ above the <ground state>, so second-order <perturbation theory> gives
$$
\boxed{E_{\rm ground}=\tfrac32\hbar\omega-\tfrac38\lambda^2\hbar\omega+O(\lambda^3).}
$$
In the first excited subspace with basis $|100\rangle,|010\rangle,|001\rangle$, the first-order perturbation <matrix> is $(\lambda\hbar\omega/2)\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$. Its symmetric <eigenvector> has <eigenvalue> $\lambda\hbar\omega$, while the perpendicular two-dimensional space has <eigenvalue> $-\lambda\hbar\omega/2$. Thus the lowest excited triplet splits into
$$
\boxed{\tfrac52\hbar\omega+\lambda\hbar\omega+O(\lambda^2)\quad\text{once},\qquad
\tfrac52\hbar\omega-\tfrac12\lambda\hbar\omega+O(\lambda^2)\quad\text{twice}.}
$$
As an independent check, diagonalizing the quadratic potential gives normal frequencies $\omega\sqrt{1+2\lambda}$ and $\omega\sqrt{1-\lambda}$ twice; their small-$\lambda$ expansions reproduce both results.
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