Is the vacuum state different after a Bogoliubov transformation of phase space operators? I am looking at the problem in the context of quantum optics.
Consider this logic:


*

*In the Schrodinger picture, the state evolves in time. 

*The time evolution of a state is given by a unitary that is an exponential of the Hamiltonian as usual. 

*The time evolution of quadratic Hamiltonians can be given by Bogoliubov transformations between the initial creation and annihilation operators to final creation and annihilation operators.

*Squeezing is an example a quadratic Hamiltonian. 

*Squeezing of the vacuum state doesn't give back vacuum but a
superposition over all number states. 

*Squeezing operation can be viewed as the time evolution operator of some quadratic Hamiltonian acting over some time t.

*Time evolution of the vacuum gives back the vacuum. $|0(t=t)\rangle=e^{-i\hat{H}t/\hbar}|0(t=0)\rangle=|0(t=0)\rangle$.

*Squeezing the vacuum state should give back the vacuum.


There is a contradiction between 5 and 8. What am I missing? In my understanding, in order to derive the Bogoliubov transformation for squeezing, the essential step is 4. But clearly squeezing is a counterexample according to the given logic. 
I seem to have some misconception at a fundamental level but don't know what.
 A: Equality in 7 might be wrong. We should distinguish between formal, $|0\rangle$, and physical, $|0\rangle_{phys}$, vacuum vectors. The formal vacuum vector is defined by its main property
$$
\hat{b}^-|0\rangle = 0.
$$
The physical vacuum vector is the least energy stationary state of the system's Hamiltonian. If the Hamiltonian contains "anomalous terms" like $\hat{b}^+\hat{b}^+$ and $\hat{b}^-\hat{b}^-$, then $|0\rangle \neq |0\rangle_{phys}$ and formal vacuum is not invariant under Schrodinger evolution. In this case $\hat{H}|0\rangle \neq 0$ and $e^{-\frac{i}{\hbar}\hat{H}t}|0\rangle \neq |0\rangle$.
Update. I suppose, squeezed states are vacuum vectors for "new" bosonic annihilation operators related to the "old" ones by Bogolyubov transformation. For example, the vector
$$
|0\rangle_{\phi} = A\, e^{\frac12\phi\,\hat{b}^+\hat{b}^+}|0\rangle
$$
is the formal vacuum vector for the operator
$$
\hat{\tilde{b}}\, ^- = \frac1{\sqrt{1-\phi^*\phi}}\left(\hat{b}^- - \phi\, \hat{b}^+\right) = u\, \hat{b}^- - v\, \hat{b}^+
$$
It is straightforward to check that $\hat{\tilde{b}}\, ^- |0\rangle_\phi = 0$.
