Show that the two circuits are equivalent mathematically

Question

This exercise wants me to prove the equivalence of the two circuits using their mathematical representations.
Circuit 1:

Circuit 2:

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 $|0ranglelangle 0|$, $|1rangle langle 1|$, $|+rangle langle +|$, 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.

Adam Zalcman · Accepted Answer

Your derivation is correct and is just missing the final step:
$$
begin{align}
dots &= HP_0H otimes I + HP_1H otimes Z 
&= HP_0H otimes HH + HP_1H otimes HXH 
&= (Hotimes H) (P_0 otimes I) (Hotimes H) + (Hotimes H) (P_1 otimes X) (H otimes H) 
&= (Hotimes H) (P_0 otimes I + P_1 otimes X) (H otimes H)
end{align}
$$
where we used the identity $HXH=Z$ which is easy to check.

KAJ226 · Answer

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 |0ranglelangle0| + 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 $X = begin{pmatrix} 0 & 1 1 & 0end{pmatrix}$ and $|0ranglelangle 0| = begin{pmatrix} 1 & 0 0 & 0 end{pmatrix} $ and $|1ranglelangle 1| = begin{pmatrix} 0 & 0 0 & 1 end{pmatrix} $

Circuit 2: Here we have $U_2 = Hotimes H cdot CNOT_{q_0, q_1} cdot H otimes H$
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 & -11 & -1 & -1 & 1 end{pmatrix} $.
Therefore,
$$U_2 =begin{pmatrix}  1 & 1 & 1 & 1 1 & -1 & 1 & -1 1 & 1 & -1 & -11 & -1 & -1 & 1 end{pmatrix} begin{pmatrix}  1 & 0 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 0 & 1 0 & 0 & 1 & 0 end{pmatrix} begin{pmatrix}  1 & 1 & 1 & 1 1 & -1 & 1 & -1 1 & 1 & -1 & -11 & -1 & -1 & 1 end{pmatrix} = begin{pmatrix}  1 & 0 & 0 & 0 0 & 0 & 0 & 1 0 & 0 & 1 & 0 0 & 1 & 0 & 0 end{pmatrix}  $$

Thus, $U_1 = U_2$.

Show that the two circuits are equivalent mathematically

2 Answers

Add your own answers!

Ask a Question