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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.

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Why not make it even smaller?

One two qubit

Initial state:

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Change rotation basis:

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Phase the parity on the top bit instead of the bottom bit:

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Apply XZ = iY to the adjacent CX CZ operations:

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Move the middle CNOT leftward, changing its interaction basis from XZ to XY to ZY to ZY as it moves through the single qubit gates, resulting in it become a CY:

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Cancel the adjacent CYs, then propagate the non-parameterized rotations rightward. This changes the basis of the rotation back to Y, and transforms the NOTC into a CY.

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Cancel the two X rotations and you're done.

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I am going to assume some usual transformations such as the involution of $H$ and CNOT. Rewriting the two gates $R$ with more standard gates, a possible derivation goes like this (details for equalities (1), (2) and (3) are given below):

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I used in particular the identity:

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from which I also derived:

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I also used:

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A very powerful tool for doing this kind of graphical proof is the ZX-Calculus (https://arxiv.org/abs/0906.4725), which I actually used for reasoning on your problem here, before turning the answer in circuit form.

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If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the inverse/dagger of the second one:

This whole thing should be an identity. Note that, for this, we should prove that whole circuit except last $R_y(-\theta) I$ should be equal to $R_y(\theta) I$. By remembering that the SWAP gate consists of 3 CNOT gates and $\text{SWAP} \cdot I R_y(\theta) \cdot\text{SWAP} = R_y(\theta) I$, we will have:

This should be equal to $R_y(\theta) I$. This circuit is equal to $U \cdot R_y(\theta) I \cdot U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $U \cdot R_y(\theta) I \cdot U^\dagger = R_y(\theta) I$. After simplifying the $U$ I have obtained this circuit:

where I took into account that $HXH = Z$ and $I \otimes H \cdot \text{CNOT} \cdot I \otimes H = CZ$ (note that I have omitted the $\otimes$ sign in the previous expressions). If my calculations are right, then $U = \frac{1}{\sqrt{2}}(I \otimes I+iY \otimes X)$. So, by omitting again the $\otimes$ sign:

$$U \cdot R_y(\theta) I \cdot U^\dagger = \frac{1}{2}\left[II+iY X\right]\left[\cos(\theta)I I - i\sin(\theta)Y I\right]\left[I I-iY X\right] = R_y(\theta) I$$

So, the circuits are equivalent.

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