# Theoretical description of a two-mode squeezed state with continuous modes

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 $$$$\exp (\zeta^* a b - \zeta a^\dagger b^\dagger)|0\rangle$$$$ where $$|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 $$$$\exp (\zeta^* a_f b_g - \zeta a_f^\dagger b_g^\dagger)|0\rangle$$$$ or would one possibly introduce some entanglement (or correlation) between the frequencies of the two modes? Because in this scenario there is no relation between the frequency of the $$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 $$$$\exp \left(\int_0^{2\Omega} \zeta (\omega)^* a (\omega)a (2\Omega-\omega)- \zeta(\omega)a^\dagger (\omega)a^\dagger (2\Omega-\omega)\right)|0\rangle$$$$ This state is not normalized, they only use the $$a(\omega)$$ mode and there is some correcation between the frequencies.