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This exercise wants me to prove the equivalence of the two circuits using their mathematical representations.

Circuit 1:

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Circuit 2:

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Circuit 1 (q1 CNOT q0) should be represented by $I \otimes P_0 + X \otimes P_1$. Circuit 2 (Hadamard q0 and q1, q0 CNOT q1, Hadamard q0 and q1) should be $(H \otimes H)(P_0 \otimes I + P_1 \otimes X)(H \otimes H)$.

I use the following identities $$P_0 + P_1 = I = P_{+} + P_{-}$$ $$X = P_{+} - P_{-}$$ $$Z = P_0 - P_1$$ $$P_{+} = HP_0H$$ $$P_{-} = HP_1H$$ where $P_0, P_1, P_{+}, P_{-}$ are $|0\rangle\langle 0|$, $|1\rangle \langle 1|$, $|+\rangle \langle +|$, and $| - \rangle \langle - |$ respectively.

I take circuit 1 and get this: $$I \otimes P_0 + X \otimes P_1$$ $$= (P_{+} + P_{-}) \otimes P_0 + (P_{+} - P_{-}) \otimes P_1$$ $$= P_{+} \otimes (P_0 + P_1) + P_{-} \otimes (P_0 - P_1)$$ $$= P_{+} \otimes I + P_{-} \otimes Z$$ $$= HP_0H \otimes I + HP_1H \otimes Z$$

Are my circuit representations correct to begin with? If so, should the Z operator be there? Any help would be appreciated.

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Your derivation is correct and is just missing the final step:

$$ \begin{align} \dots &= HP_0H \otimes I + HP_1H \otimes Z \\ &= HP_0H \otimes HH + HP_1H \otimes HXH \\ &= (H\otimes H) (P_0 \otimes I) (H\otimes H) + (H\otimes H) (P_1 \otimes X) (H \otimes H) \\ &= (H\otimes H) (P_0 \otimes I + P_1 \otimes X) (H \otimes H) \end{align} $$

where we used the identity $HXH=Z$ which is easy to check.

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You can also just convert it to matrix representation and show that the two matrix are the same.

Circuit 1: This have $q_1$ as the controlled qubit and so it has the matrix representation as unitary matrix $U_1$: \begin{align} U_1 = CNOT_{q_1, q_0} &= I \otimes |0\rangle\langle0| + X \otimes |1 \rangle \langle 1| = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix} \end{align} where $X = \begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$ and $|0\rangle\langle 0| = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} $ and $|1\rangle\langle 1| = \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix} $


Circuit 2: Here we have $U_2 = H\otimes H \cdot CNOT_{q_0, q_1} \cdot H \otimes H$

where $H = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1\\ \end{pmatrix}$ and hence $H \otimes H = \dfrac{1}{2} \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\1 & -1 & -1 & 1 \end{pmatrix} $. Therefore, $$U_2 =\begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\1 & -1 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\1 & -1 & -1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix} $$


Thus, $U_1 = U_2$.

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