= Solution
The commutator gives $a(a^\dagger)^n=(a^\dagger)^na+n(a^\dagger)^{n-1}$. Hence $a|n\rangle=\sqrt n|n-1\rangle$ and $a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$. Starting with the normalized vacuum, repeated application gives $\langle0|a^n(a^\dagger)^n|0\rangle=n!$, proving unit norm. The <number operator> then has <eigenvalue> $n$, so
$$
\boxed{E_n=\hbar\omega(n+\tfrac12).}
$$
Different levels are orthogonal because the <Hamiltonian> is self-adjoint and their <eigenvalues> are distinct.
For the perturbation, the ground-state diagonal <matrix> element is zero, while $V|0\rangle=\hbar\omega\sqrt{r!}|r\rangle$. The nondegenerate second-order perturbation formula consequently gives
$$
\boxed{\Delta E_0=-\lambda^2\hbar\omega\frac{r!}{r}+O(\lambda^3)=-\lambda^2\hbar\omega(r-1)!+O(\lambda^3).}
$$
For $r=1$, $V|1\rangle=\hbar\omega(|0\rangle+\sqrt2|2\rangle)$. The lower and upper intermediate levels contribute respectively $+\lambda^2\hbar\omega$ and $-2\lambda^2\hbar\omega$, giving \b[$\Delta E_1=-\lambda^2\hbar\omega$ to second order].
For $r=1$ the exact completion of the square is
$$
H_\lambda=\hbar\omega\left[(a^\dagger+\lambda)(a+\lambda)+\tfrac12-\lambda^2\right].
$$
The shifted operators $b=a+\lambda$, $b^\dagger=a^\dagger+\lambda$ have the same commutator and are unitarily related to the original ones by a displacement operator. Their vacuum is the <coherent state> with amplitude $-\lambda$. Therefore \b[every exact level shifts by $-\lambda^2\hbar\omega$].
The <stability of monomial ladder-operator perturbations> gives a necessary qualification to the request for a lowest level for every $r$. For $r>2$ and real nonzero $\lambda$, a <coherent state> of amplitude $Re^{i\theta}$ has <energy> expectation $\hbar\omega[R^2+1/2+2\lambda R^r\cos(r\theta)]$. Choose the phase so the last term is negative. As $R\to\infty$ the expectation tends to minus infinity, so the perturbed <Hamiltonian> has no true ground state. The displayed second-order answer is then the formal perturbative coefficient of the branch from the unperturbed vacuum, not an exact lowest <eigenvalue>. For $r=2$ <stability> requires $|\lambda|<1/2$; for $r=1$ the completed-square <Hamiltonian> is bounded below for every real $\lambda$.
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