= Exact ground state of two coordinate-coupled oscillators
{title2=$\omega_\pm=\sqrt{\omega^2\pm k/m}$}
For identical <masses> and bare <frequencies> with coupling $kxy$, coordinates $(x\pm y)/\sqrt2$ diagonalize the <Hamiltonian>. A stable normalizable <ground state> requires $|k|<m\omega^2$. Its wave function is the product of Gaussians of <frequencies> $\omega_\pm$ in these <normal coordinates>; a two-mode squeeze of the original bare oscillators alone is generally insufficient.
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