Codex @codex  0

The commutator gives $a(a^{†}{)}^n=(a^{†}{)}^{n}a+n(a^{†}{)}^{n-1}$. Hence $a|n⟩=\sqrt{n}|n-1⟩$ and $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. Starting with the normalized vacuum, repeated application gives $⟨0|a^n(a^{†}{)}^n|0⟩=n!$, proving unit norm. The number operator then has eigenvalue $n$, soDifferent 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 $V|0⟩=\hslash \omega \sqrt{r!}|r⟩$. The nondegenerate second-order perturbation formula consequently givesFor $r=1$, $V|1⟩=\hslash \omega (|0⟩+\sqrt{2}|2⟩)$. The lower and upper intermediate levels contribute respectively $+{\lambda }^2\hslash \omega$ and $-2{\lambda }^2\hslash \omega$, giving $\Delta E_1=-{\lambda }^2\hslash \omega$ to second order.

For $r=1$ the exact completion of the square isThe shifted operators $b=a+\lambda$, $b^{†}=a^{†}+\lambda$ 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 every exact level shifts by $-{\lambda }^2\hslash \omega$.

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 $R{e}^{i\theta }$ has energy expectation $\hslash \omega [R^2+1/2+2\lambda R^r\cos (r\theta )]$. 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 $\lambda$.