I have the Hamiltonian evolution operator for a two qubit system :
\begin{equation} \hat{U}= e^{-i\gamma H}\;,\quad H = \frac{1}{2}(I - Z\otimes Z) \end{equation} where the $Z$ gate is applied on both qubits, ($I=I\otimes I$).
I would like to express this evolution operator in terms of CNOT gate and rotations.
I was told that $e^{i \frac{\gamma}{2}Z\otimes Z}$ could be decomposed as follows : \begin{equation} e^{i \frac{\gamma}{2}Z\otimes Z}=\text{CX}\;(I\otimes R_z(\gamma))\;\text{CX}\;,\quad R_z(\gamma) = e^{-i\frac{\gamma}{2}Z} \end{equation} where the CNOT gate operates on the second qubit each time (first qubit being the control one) and the rotation applies to the second one.
I found a related question here but I wonder if this can be simply demonstrated with the CNOT gate decomposition : \begin{equation} \text{CX}= e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)} \end{equation}
The development is the following : \begin{equation} \begin{split} \text{CX}\;(I\otimes e^{-i\frac{\gamma}{2}Z})\;\text{CX} &= e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)}\;(I\otimes e^{-i\frac{\gamma}{2}Z})\; e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)}\\ &=e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)}\;e^{-i\frac{\gamma}{2}I\otimes Z}\; e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)} \end{split} \end{equation}
Then, I am not sure how to proceed, it seems to me not trivial to end up with $e^{i \frac{\gamma}{2}Z\otimes Z}$. This is the question. If the matrices within exponential do not commute, I don't think I can go further.