Let $\left|0\right>$ be the Bunch-Davies vacuum state of a QFT, for example a free scalar field, in de Sitter space. The creation and annihilation operators w.r.t. this state is a vacuum, i.e. $a^0_n\left|0\right>=0$, are denoted by $a^0_n,(a^0_n)^\dagger$. Now if we perform a Bogoliubov transformation \begin{align} a^\alpha_n&=\cosh{\alpha} a^0_n - \sinh{\alpha}(a^0_n)^\dagger \\ (a^\alpha_n)^\dagger&=\cosh{\alpha} (a^0_n)^\dagger - \sinh{\alpha}a^0_n \end{align} we obtain a different set of creation and annihilation operators $a^\alpha_n, (a^\alpha_n)^\dagger$ for each $\alpha\in \mathbb{R}$, which also fulfill the commutation relations \begin{align} [a^0_n,(a^0_m)^\dagger] = [a^\alpha_n,(a^\alpha_m)^\dagger] = \delta_{nm} \end{align} This is true in curved space but not in flat space, since in flat space the creation and annihilation operators are uniquely determined by Poincare symmetry. Since we now can choose between all these sets of creation and annihilation operators to expand our fields, we get for each choice of $\alpha$ a different vacuum state $\left|\alpha\right>$ by \begin{align} a^\alpha_n\left|\alpha\right> \end{align} It turns out that the different vacuum states for a free scalar field are related by the following transformation \begin{align} \left|\alpha\right> = S\left|0\right>, \qquad S=\exp{\left[\frac{\alpha}{2}\sum_n\left(((a_n^0)^\dagger)^2 -(a_n^0)^2\right)\right]} \end{align} Note that $\left|\alpha=0\right>=\left|0\right>$. My question now is the following:
In QFT the Hilbert space of the theory is the fock space generated by the vacuum. So if we choose $\left|0\right>$ as our vacuum state, meaning we chose the set of creation and annihilation operators $a_n^0, (a_n^0)^\dagger$, we obtain a fock space $H_0$. If instead we chose some $\left|\alpha\right>$ for a fixed $\alpha$, we would obtain some $H_\alpha$ as the Hilbert space. So in particular $\left|\alpha\right>$ and $\left|0\right>$ are not elements of the same Hilbert space, but instead are used to define different Hilbert spaces. Now the transformation $S$ consists of powers of creation and annihilation operators $a_n^0, (a_n^0)^\dagger$. In first order in $\alpha$ for example \begin{align} \left|\alpha\right> = (\mathbf{1}+\frac{\alpha}{2}\sum_n((a^0_n)^\dagger)^2)\left|0\right> \end{align} This state is of course not normalizable, but it looks like a state where in each mode $n$ there are two particles. In particular $S$, consisting of creation and annihilation operators, is a map $H_0\rightarrow H_0$, so how does this map makes sense if $\left|0\right>$ and $\left|\alpha\right>$ do not lie in the same Hilbert space? Isn't $\left|\alpha\right>$ just a multi particle state in $H_0$ from that point of view?
