# 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:

1. In the Schrodinger picture, the state evolves in time.
2. The time evolution of a state is given by a unitary that is an exponential of the Hamiltonian as usual.
3. 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.
4. Squeezing is an example a quadratic Hamiltonian.
5. Squeezing of the vacuum state doesn't give back vacuum but a superposition over all number states.
6. Squeezing operation can be viewed as the time evolution operator of some quadratic Hamiltonian acting over some time t.
7. Time evolution of the vacuum gives back the vacuum. $$|0(t=t)\rangle=e^{-i\hat{H}t/\hbar}|0(t=0)\rangle=|0(t=0)\rangle$$.
8. 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.

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 $$\hat{\tilde{b}}\, ^- |0\rangle_\phi = 0$$.