= Zero-mode obstruction to bosonic Bogoliubov diagonalization
{title2=$\alpha=|\Delta|$}
For $H=\alpha a^\dagger a+\Delta(a^2+a^{\dagger2})/2$, a finite real <Bogoliubov transformation>[squeeze] requires $\alpha>|\Delta|$. At $\alpha=\Delta>0$, the <Hamiltonian> instead equals $\alpha Q^2-\alpha/2$, with $Q=(a+a^\dagger)/\sqrt2$. The formal oscillator frequency is zero, but this <Linear operator>[operator] is not constant: $\langle2|H|0\rangle=\alpha/\sqrt2$. Thus a normalizable oscillator <ground state> and finite canonical <Bogoliubov transformation>[squeeze] cannot be obtained by simply substituting zero frequency into the positive-frequency formula.
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