# Schmidt decomposition of coupled oscillators

Consider a system of two coupled oscillators, with Hamiltonian ($\hbar = m = 1$):

\begin{align} \mathcal{H} = \frac{1}{2}(p_1^2 + \omega_0^2 x_1^2) + \frac{1}{2}(p_2^2 + \omega_0^2 x_2^2) +\frac{1}{2}k(x_1 - x_2)^2. \end{align}

Changing to normal coordinates ($y_1 = \frac{1}{\sqrt{2}}(x_1-x_2)$ and $y_2 = \frac{1}{\sqrt{2}} (x_1 + x_2)$), one gets the wave functions to be a product of two decoupled oscillators with eigenfrequencies $\omega_1^2 = \omega_0^2 + 2k$ and $\omega_2^2 = \omega_0^2$. In particular the ground state is \begin{align} \Psi_0 = \left(\frac{\omega_1\omega_2}{\pi^2} \right)^{1/4} \exp(-\frac{1}{2}[\omega_1 y_1^2 + \omega_2 y_2^2] ). \end{align}

It is known that we can perform a Schmidt decomposition on the system, and write \begin{align} \Psi_0 = \sum_{n=0}^\infty \frac{(-\tanh \eta)^n}{\cosh \eta}\Phi_n(x_1) \Phi_n(x_2), \end{align} where $\exp(4\eta) = \omega_1/\omega_2$ and $\Phi_n$ are the oscillator states for a frequency $\tilde{\omega} = \sqrt{\omega_1 \omega_2}$.

I'm not too sure if my question makes sense, but it is: how do we know a priori that the Hilbert space of the system which is a product of two oscillators of frequency $\omega_0$ (is it ??): \begin{align} H = H_{SHO,\omega_0} \otimes H_{SHO,\omega_0} \end{align} can be also written as \begin{align} H = H_{SHO,\omega_1} \otimes H_{SHO, \omega_2} \end{align} as in the decoupled case, and also as \begin{align} H = H_{SHO,\tilde{\omega}} \otimes H_{SHO,\tilde{\omega}} \end{align} as in the last step?

In other words, I wish to decompose the Hilbert space into a tensor product of smaller subspaces, but how do I do that?

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