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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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All harmonic oscillator Hilbert spaces of different frequencies are unitarily equivalent by virtue of the Stone–von Neumann theorem. This theorem implies also that these Hilbert spaces are unitarily equivalent to the free particle Hilbert space; where the transformation is given by means of the Bargmann transform, please , for example, see the definition in the introduction equation (1) of Olafsson-Ørsted article. Thus if the algebra of operators is composed (of a finite number of copies) of the canonical commutation relations, one already knows that the quantization Hilbert space is a tensor product of the same number of oscillator Hilbert spaces

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