2. CNOT gate and a circuit identity. We define CNOT = CNOT 10 gate as CNOT =CNOT with qubit i as the control and qubit j as the target. (3) For

example, CNOT10|0j) = 10j) and CNOT10|1j) |1j1) or CNOT10la) [b) = (a) la @ b). We use the following diagram to denote the CNOT gate: I H FIG. 1: CNOT=CNOT10 gate. If we want the control and target to be flipped, it would be CNOT01 and we would draw the circuit as Another way to flip the control and target qubits is to apply Hadamard gates to both sides of the CNOT: In other words, the following identity holds valid: (HH) CNOTwo (HH) = CNOTOI (a) Prove this circuit identity by directly/explicitly computing the left-handed and the right-handed sides. (2 points)/nH H FIG. 2: CNOT01 gate. H H = I FIG. 3: One way to flip the control and target qubits is to apply Hadamard gates to both sides of the CNOT. (b) Prove above circuit identity using (4 by 4) matrix representations of H H and CNOT. (1 point) (c) Prove above circuit identity by checking the following examples with IBM Quantum Composer. Screen shot should be fine for this question. (1 point) (H&H) CNOT10 (HH)|00) = CNOT01 100) (HH) CNOT10 (HH)|01) = CNOT01 101) (HH) CNOT10 (H&H)|10) = CNOT01|10) (H&H) CNOT10 (HH)|11) = CNOT01|11) (5) (6) (7) (8)