 Stability of monomial ladder-operator perturbations

The perturbation $\lambda \hslash \omega (a^r+(a^{†}{)}^r)$ couples the oscillator vacuum only to $|r⟩$ at first order. Its squared coupling is $(\hslash \omega {)}^{2}r!$ and the energy denominator is $-r\hslash \omega$, giving the displayed formal second-order coefficient. For $r>2$ and nonzero real $\lambda$, coherent-state expectations contain $R^2+2\lambda R^r\cos r\theta$; choosing its phase negative proves the Hamiltonian is unbounded below. The formal vacuum branch must therefore not be called an exact ground state. For $r=2$ boundedness requires $|\lambda |<1/2$, while the displaced $r=1$ oscillator is stable for every real strength.