While trying to simplify a certain 2-qubit quantum circuit, I managed to get it down to this:
But by inspecting the corresponding two-qubit unitary directly, I can come up with the arguably simpler:
where the rotation operator "moved" to the first qubit. I am using the convention $ R_{\theta} := R_y(2\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} $. The rotation $R_{\pi/4}$ is thus $=XH$.
How can I show that the two circuits are equivalent by "elementary" circuit identities, instead of just verifying that they amount to the same unitary? I tried many different simplifications without success.