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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 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 $UR_y(\theta) I U^\dagger$$U \cdot R_y(\theta) I \cdot U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $UR_y(\theta) I U^\dagger = R_y(\theta) I$$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:

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

So, the circuits are equivalent.

If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the 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 $UR_y(\theta) I U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $UR_y(\theta) I U^\dagger = R_y(\theta) I$. 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:

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

So, the circuits are equivalent.

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.

deleted 130 characters in body
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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 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 $UR_y(\theta) I U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $UR_y(\theta) I U^\dagger = R_y(\theta) I$. 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:

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

So, the circuits are equivalent.

If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the 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 $UR_y(\theta) I U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $UR_y(\theta) I U^\dagger = R_y(\theta) I$. 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:

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

So, the circuits are equivalent.

If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the 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 $UR_y(\theta) I U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $UR_y(\theta) I U^\dagger = R_y(\theta) I$. 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:

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

So, the circuits are equivalent.

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If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the 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 $UR_y(\theta) I U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $UR_y(\theta) I U^\dagger = R_y(\theta) I$. 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:

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

So, the circuits are equivalent.

If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the 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 $UR_y(\theta) I U^\dagger$, where $U$ is this circuit:

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

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

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

So, the circuits are equivalent.

If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the 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 $UR_y(\theta) I U^\dagger$, where $U$ is this circuit:

Now we should try to prove that $UR_y(\theta) I U^\dagger = R_y(\theta) I$. 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:

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

So, the circuits are equivalent.

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