= Stability of monomial ladder-operator perturbations
{title2=$\Delta E_0^{(2)}=-\lambda^2\hbar\omega(r-1)!$}
Consider
$$
H_r=\hbar\omega(a^\dagger a+1/2)+\lambda\hbar\omega\bigl(a^r+(a^\dagger)^r\bigr),
\qquad r\ge1,\quad \lambda\in\mathbb R.
$$
The perturbation connects the oscillator <ground state> only to the <number state> $|r\rangle$. Its squared <matrix element> is $(\hbar\omega)^2r!$, and the <energy> denominator is $-r\hbar\omega$. Thus <second-order nondegenerate perturbation theory> gives the formal shift $-\lambda^2\hbar\omega(r-1)!$. A finite formal coefficient does not establish that the full <Hamiltonian> has a <ground state>.
For $r>2$, a normalized <coherent state> with label $Re^{i\theta}$ has
$$
\frac{\langle H_r\rangle}{\hbar\omega}=R^2+1/2+2\lambda R^r\cos(r\theta).
$$
For nonzero $\lambda$, choose the phase so that the last term is negative. It dominates as $R\to\infty$, proving that the <Hamiltonian> is unbounded below on these trial states. The perturbative vacuum branch cannot be an exact lowest-energy state.
For $r=2$, put $Q=(a+a^\dagger)/\sqrt2$ and $P=(a-a^\dagger)/(i\sqrt2)$, so $[Q,P]=i$. Then
$$
H_2=\hbar\omega\left[(1/2+\lambda)Q^2+(1/2-\lambda)P^2\right].
$$
It is bounded below for $|\lambda|\le1/2$. A positive-frequency oscillator and a normalizable <ground state> require the strict inequality $|\lambda|<1/2$, giving <frequency> $\omega\sqrt{1-4\lambda^2}$. At $\lambda=1/2$ only $Q^2$ remains; at $\lambda=-1/2$ only $P^2$ remains. Their zero infimum is not attained by a normalizable vector. This is the <zero-mode obstruction to bosonic Bogoliubov diagonalization>, rather than unboundedness below.
For $r=1$, completing the displacement gives $H_1=\hbar\omega[(a^\dagger+\lambda)(a+\lambda)+1/2-\lambda^2]$, which has a normalizable displaced oscillator <ground state> for every real $\lambda$.
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