Zero-mode obstruction to bosonic Bogoliubov diagonalization ($\alpha =|\Delta |$)

For $H=\alpha a^{†}a+\Delta (a^2+a^{†2})/2$, a finite real squeeze requires $\alpha >|\Delta |$. At $\alpha =\Delta >0$, the Hamiltonian instead equals $\alpha Q^2-\alpha /2$, with $Q=(a+a^{†})/\sqrt{2}$. The formal oscillator frequency is zero, but this operator is not constant: $⟨2|H|0⟩=\alpha /\sqrt{2}$. Thus a normalizable oscillator ground state and finite canonical squeeze cannot be obtained by simply substituting zero frequency into the positive-frequency formula.