Understanding time evolution in Von Neumann's pre-measurement (Ozawa model) I'm studying Quantum Information Theory as a non-physicist and so I'm struggling a bit with simple concepts.
Let's say we have a system $\Psi$ composed of two sub-systems: $\Psi = \Gamma \cup \Xi$ and that this system is prepared into an initial separable state (wlog)
$$
|\Psi\rangle = |\Gamma\rangle \otimes |\Xi_D\rangle 
$$
where $|\Xi_D\rangle$ is a dumb state for the environment and $|\Gamma\rangle$ is the pure state for my main system. I'm considering an ideal PVM whose effects are orthonormal projectors 
$$
\hat{E}_{\gamma} = \{|\gamma \rangle \langle \gamma| \}
$$
associated to an orthonormal basis $\{ |\gamma \rangle \}$.
Let's now consider an Hamiltonian of the form
$$
\hat{H} = \hat{O}_{\Gamma} \otimes \hat{O}_{\Xi} + g \hat{H}_\Xi
$$
where $ \hat{O}_{\Gamma}$ and $\hat{O}_{\Xi}$ are Hermitian operators acting, respectively, on $\mathcal{H}_{\Gamma}$ and $\mathcal{H}_{\Xi}$, and the operator $\hat{O}_{\Gamma}$ is defined as
$$
\hat{O}_{\Gamma} = \sum_{\gamma} \omega_{\gamma} |\gamma \rangle \langle \gamma |
$$
The state $|\Gamma \rangle$ can be written as
$$
|\Gamma \rangle = \sum_{\gamma} c_{\gamma} |\gamma \rangle
$$
Ozawa [1] states that the evolution from time $t_0 = 0$ to time $t$ is described by the unitary operator ($\hbar=1$)
$$
\mathcal{\hat{U}}_{0,t} = e^{-i\hat{H}t} = \exp\{-it[\hat{O}_{\Gamma} \otimes \hat{O}_{\Xi} + g \hat{H}_\Xi]\}
$$
such that
\begin{align}
|\Psi(t)\rangle 
&= \mathcal{\hat{U}}_{0,t} |\Psi \rangle = \exp\{-it[\hat{O}_{\Gamma} \otimes \hat{O}_{\Xi} + g \hat{H}_\Xi]\} \Big ( |\Gamma\rangle \otimes |\Xi_D\rangle  \Big ) \\
&= \exp\{-it[\hat{O}_{\Gamma} \otimes \hat{O}_{\Xi} + g \hat{H}_\Xi]\} \Big ( \sum_{\gamma} c_{\gamma} |\gamma \rangle \otimes |\Xi_D\rangle  \Big ) \\
&\overset{?}{=} \sum_{\gamma} c_{\gamma} |\gamma \rangle \Big ( \exp\{-it[\omega_\gamma \hat{O}_{\Xi} + g \hat{H}_{\Xi} ] \} |\Xi_D\rangle\Big )
\end{align}
I'm totally missing the last passage. Even throwing out the control term $g \hat{H}_{\Xi}$ from the Hamiltonian I'm still finding difficulties in deriving the last equation. In this case, where
$$
\hat{H} = \hat{O}_{\Gamma} \otimes \hat{O}_{\Xi}
$$ 
I guess I should have something like
$$
|\Psi(t) \rangle = \sum_{\gamma} c_{\gamma} |\gamma \rangle \otimes \Big ( e^{-it\omega_\gamma \hat{O}_{\Xi}} |\Xi_D\rangle \Big )
$$
but I'm not able to derive it. Can you help me out? 
Thank you very much guys.
 A: You don't define $\hat H_{\Xi}$, but from the equations you write it seems that it's a Hamiltonian that only acts nontrivially on the second system, that is, $\hat H_\Xi\equiv I\otimes \hat H_\Xi$.
The main step is the following:
$$
  \exp\{-it[\hat{O}_{\Gamma} \otimes \hat{O}_{\Xi} + g \hat{H}_\Xi]\}
  \left( |\gamma \rangle \otimes |\Xi_D\rangle \right)
=
  |\gamma\rangle \otimes \left[
    \exp\left( -it\left[ \omega_\gamma \hat O_\Xi+g\hat H_\Xi \right]\right)
    |\Xi_D\rangle
  \right].
$$
This works because $\hat O_\Gamma|\gamma\rangle=\omega_\gamma |\gamma\rangle$, and therefore
\begin{align}
  (\hat{O}_{\Gamma} \otimes \hat{O}_{\Xi} + g \hat{H}_\Xi)
  \left(|\gamma\rangle\otimes|\Xi_D\rangle\right)
  &=
  \omega_\gamma (|\gamma\rangle\otimes \hat O_\Xi|\Xi_D\rangle) + g |\gamma\rangle\otimes(\hat H_\Xi |\Xi_D\rangle) \\
  &= |\gamma\rangle\otimes (\omega_\gamma \hat O_\Xi+g\hat H_\Xi)|\Xi_D\rangle.
\end{align}
The first equation with the exponentials then follows, as you can see for example by expanding the exponential as a series.
