I have an exercise but not the answer, can somebody tell me if this is correct? Here is the exercise and my answers:
Consider the following unitary operation $$ U = (CNOT_{13} \otimes I_2)(CNOT_{12} \otimes I_3)(H_1 \otimes I_2 \otimes I_3) $$ where the indices $i = 1,2,3$ indicates on which qbit the gates are acting on. $CNOT_{ij}$ means that the qbit $i$ controls the qbit $j$.
(a) What is the Hilbert space of this problem ? What's the dimension ?
solution : $\mathcal{H} = (\mathbb{C}^2)^{\otimes 3}$ and $\dim \mathcal{H} = 2^3 = 8$.
(b) Draw the circuit.
solution :
(c) The initial state of the circuit is $|0\rangle \otimes |0\rangle \otimes |0\rangle$. Compute the final state.
solution : Step by step I find :
1)
$$
(H_1 \otimes I_2 \otimes I_3)(|0\rangle \otimes |0\rangle \otimes |0\rangle) = H|0\rangle \otimes |0\rangle \otimes |0\rangle = |+\rangle \otimes |0\rangle \otimes |0\rangle = |\Psi(t_1)\rangle
$$
2) $$ (CNOT_{12} \otimes I_3)|\Psi(t_1)\rangle = |+\rangle \otimes |+\rangle \otimes |0\rangle = |\Psi(t_2)\rangle $$
3) $$ (CNOT_{13} \otimes I_2)|\Psi(t_2)\rangle = |+\rangle \otimes |+\rangle \otimes |+\rangle = |\Psi(end)\rangle $$
(d) Suppose that the hardware architecture can only make the operation $CNOT_{31}$. Propose a modification of the circuit on the bit $1$ and $3$ such that the new circuit is equivalent to the first one.
solution : I'm not confident about that part but I have that $(CNOT_{31} \otimes I_2)|\Psi(t_2)\rangle = |+\rangle \otimes |+\rangle \otimes |0\rangle$ so in order to retrieve the $|\Psi(end)\rangle$ state we just have to put another Hadamard gate on the third qubit after the CNOT gate. Am I correct ?
Thanks in advance for any help