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Given two distinguishable modes $a$ and $b$ with $[a,a^\dagger]=1$ and $[b,b^\dagger]=1$ and $[a,b^\dagger]=0$, the two-mode squeezed vacuum state is given by \begin{equation} \exp (\zeta^* a b - \zeta a^\dagger b^\dagger)|0\rangle \end{equation} where $|0\rangle$ is the vacuum state.

My question is, how would the equivalent state for continuous modes $a(\omega)$ and $b(\omega)$ look like with $[a(\omega),a^\dagger(\omega')]=\delta (\omega-\omega')$, $[b(\omega),b^\dagger(\omega')]=\delta (\omega-\omega')$ and $a(\omega),b^\dagger(\omega')]=0$? The state should be normalised, that means we need to introduce some regularizing functions or spectral functions and define normalised modes $a_f^\dagger = \int_0^\infty f(\omega)a^\dagger (\omega)$ and $b_g^\dagger = \int_0^\infty g(\omega)a^\dagger (\omega)$, where $\int_0^\infty |f|^2=\int_0^\infty |g|^2=1$. Would a physically realisable state take the form \begin{equation} \exp (\zeta^* a_f b_g - \zeta a_f^\dagger b_g^\dagger)|0\rangle \end{equation} or would one possibly introduce some entanglement (or correlation) between the frequencies of the two modes? Because in this scenario there is no relation between the frequency of the $a$ mode and the $b$ mode. Are there difference depending on the generation of the squeezed state that one have to additionally take into account?

In a book the only state that came close to what I was looking for was \begin{equation} \exp \left(\int_0^{2\Omega} \zeta (\omega)^* a (\omega)a (2\Omega-\omega)- \zeta(\omega)a^\dagger (\omega)a^\dagger (2\Omega-\omega)\right)|0\rangle \end{equation} This state is not normalized, they only use the $a(\omega)$ mode and there is some correcation between the frequencies.

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