 Exact ground state of two coordinate-coupled oscillators

For identical masses and bare frequencies with coupling $kxy$, coordinates $(x\pm y)/\sqrt{2}$ 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.