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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.

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} $.

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.

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.

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Simplification of "rotation swapping" circuit

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} $.

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.