= Solution
Invert the operator definitions to obtain $a=\sqrt{m\omega/(2\hbar)}\,\hat x+i\hat p/\sqrt{2\hbar m\omega}$ and the adjoint formula for $a^\dagger$. The <canonical commutation relation> yields $[a,a^\dagger]=1$, with $[a,a]=[a^\dagger,a^\dagger]=0$. Substitution gives
$$
\boxed{H=\hbar\omega\left(a^\dagger a+\tfrac12\right)}.
$$
Write $N=a^\dagger a\ge0$. Since $[N,a]=-a$ and $[N,a^\dagger]=a^\dagger$, these are lowering and raising operators. Applying $a$ to a <ground state> would lower its <energy> unless it vanishes, so $a|0\rangle=0$. From the normalized unique <ground state> construct
$$
\boxed{|n\rangle=\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle,
\qquad E_n=\hbar\omega(n+\tfrac12),\quad n=0,1,\ldots}.
$$
The commutator gives the factorial normalization and $a|n\rangle=\sqrt n|n-1\rangle$. For any <energy> eigenstate, repeated lowering cannot pass below $N=0$; positivity of each lowered <norm> forces its $N$ <eigenvalue> to be a nonnegative integer and lowering eventually reaches the <ground state>. Its uniqueness then yields the entire eigenstate ladder. The usual oscillator on the real line has discrete complete spectrum because its potential is confining, so these states span its <Hilbert space>.
For the perturbation $\lambda W$ with $W=\hbar\omega(a^2+a^{\dagger2})$, nondegenerate <time-independent perturbation theory> gives $E_n'=E_n+\lambda W_{nn}+\lambda^2\sum_{m\ne n}|W_{mn}|^2/(E_n-E_m)+O(\lambda^3)$. The first-order term vanishes. Only $m=n-2,n+2$ contribute, with squared <matrix> elements $(\hbar\omega)^2n(n-1)$ and $(\hbar\omega)^2(n+1)(n+2)$ respectively. Thus
$$
\boxed{E_n'=\hbar\omega(n+\tfrac12)-\lambda^2\hbar\omega(2n+1)+O(\lambda^3)}.
$$
The absent downward transition for $n=0,1$ is automatically represented by $n(n-1)=0$. Direct substitution also gives
$$
H'=\frac{1-2\lambda}{2m}\hat p^2+\frac{m\omega^2(1+2\lambda)}2\hat x^2.
$$
For $|\lambda|<1/2$ this is another oscillator with mass $m'=m/(1-2\lambda)$ and frequency $\omega'=\omega\sqrt{1-4\lambda^2}$. Therefore
$$
\boxed{E_n'=\hbar\omega\sqrt{1-4\lambda^2}\,(n+\tfrac12)}.
$$
Its expansion is $E_n[1-2\lambda^2+O(\lambda^4)]$, agreeing with the perturbative result and sharpening its remainder. At the excluded endpoints confinement is lost; beyond them the quadratic <Hamiltonian> is not bounded below.
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