How I can find the correlation between a vacuum state $\left|0\right>_a$ and a vacuum state $\left|0\right>_b$ in curved space-time?
The field operator in curved space-time can be expanded as
$$\hat{\phi}(x)=\sum_{n}\left(\hat{a}n f_n(x)+\hat{a}^\dagger _n f^ _n (x)\right)$$ where $f(x)$ is the positive frequency mode and $f^(x)$ is the negative frequency mode. We define the vacuum state as $\hat{a}_n \left|0\right>_a=0$ But we can choose a different set of positive and negative frequency modes, and an expansion $$\hat{\phi}(x)=\sum{n}\left(\hat{b}n h_n(x)+\hat{b}^\dagger _n h^ n (x)\right)$$ Now the vacuum would be define as $\hat{b}_n \left|0\right>_b=0$. The mode functions $f_n$ are normalized $(f_m,f_n)=\delta{m n}$, $(f_m,f^_n)=0$ and $(f^_m,f^_n)=-\delta{m n}$. From the two different expansions of $\hat{\phi}$ we have $$\sum_{n}\left(\hat{a}n f_n(x)+\hat{a}^\dagger _n f^ n (x)\right)=\sum{n}\left(\hat{b}_n h_n(x)+\hat{b}^\dagger _n h^ _n (x)\right)$$ Taking the inner product with $f_n$ on each side, we get $$\hat{a}_m=\sum_n (h_n,f_n)\hat{b}_n+\sum_n (h^_n,f^_n)\hat{b}^\dagger_n \equiv \sum_n \alpha{m n}\hat{b}n + \sum_n \beta{m n}\hat{b}^\dagger_n$$ Thus the vacuum $\left|0\right>a$ satisfies $$0=\hat{a}_m \left|0\right>_a=\left(\sum_n \alpha{m n}\hat{b}n + \sum_n \beta{m n}\hat{b}^\dagger_n\right)\left|0\right>_a$$ If we suppose we had just one mode, with a relation $$\left(b+\gamma b^\dagger\right)\left|0\right>_a=0$$ The solution to this equation is of the form: $$\left|0\right>_a=Ce^{\mu \hat{b}^\dagger \hat{b}^\dagger}\left|0\right>_b$$ Where $C$ is a normalization constant and $\mu$ is a number.
My question is how you derive the solution of the equation : $$\left(b+\gamma b^\dagger\right)\left|0\right>_a=0$$ 281