 Stability of monomial ladder-operator perturbations

ConsiderThe perturbation connects the oscillator ground state only to the number state $|r⟩$. Its squared matrix element is $(\hslash \omega {)}^{2}r!$, and the energy denominator is $-r\hslash \omega$. Thus second-order nondegenerate perturbation theory gives the formal shift $-{\lambda }^2\hslash \omega (r-1)!$. A finite formal coefficient does not establish that the full Hamiltonian has a ground state.

For $r>2$, a normalized coherent state with label $R{e}^{i\theta }$ hasFor 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 $Q=(a+a^{†})/\sqrt{2}$ and $P=(a-a^{†})/(i\sqrt{2})$, so $[Q,P]=i$. ThenIt is bounded below for $|\lambda |\le 1/2$. 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 $H_1=\hslash \omega [(a^{†}+\lambda )(a+\lambda )+1/2-{\lambda }^2]$, which has a normalizable displaced oscillator ground state for every real $\lambda$.