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 :

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(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 :

$$ (H_1 \otimes I_2 \otimes I_3)(|0\rangle \otimes |0\rangle \otimes |0\rangle) = H|0\rangle \otimes |0\rangle \otimes |0\rangle = |+\rangle \otimes |0\rangle \otimes |0\rangle = |\Psi(t_1)\rangle $$

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


1 Answer 1


your answer to the question (c) is wrong. I won't give you the full correct answer but here's a clue to understand what's wrong :

Applying a $CNOT$ gate on the state $|+\rangle|0\rangle $ does not give the $|+\rangle|+\rangle $. I invite you to check the effect of this two gates on this page : https://en.wikipedia.org/wiki/Bell_state#Creating_Bell_states.

As your final state in answer (c) is wrong, the conclusion made in answer (d) is also not good but here is another clue for this part :
In this question, you can only use the $CNOT_{31}$ but you want to apply a $CNOT_{13}$ transformation. You will therefore have to find a gate or a set of gate to apply before and/or after the $CNOT_{31}$ to make the whole transformation equivalent to the $CNOT_{13}$ gate you want to apply.

  • $\begingroup$ Well yes thank you, it was my mistake. So, now I found the GHZ state for the question c, is that correct ? For the question d, adding a $CNOT_{23}$ gate at the end will do the job right ? $\endgroup$
    – Bozu
    Apr 26, 2022 at 14:46
  • $\begingroup$ Yes, it's better ! Now you can answer the question (d) without any trouble. $\endgroup$ Apr 26, 2022 at 14:56
  • $\begingroup$ I've edited my previous comment, but I think you did not see it, but just adding a $CNOT_{23}$ gate will give us an equivalent circuit right ? $\endgroup$
    – Bozu
    Apr 26, 2022 at 15:01
  • 1
    $\begingroup$ Not exactly : You want to apply a $CNOT_{13}$ gate but you can't. You can apply a $CNOT_{31}$ gate which is the same gate but the qubits were moved, isn't it ? As you can't move the position of qubits in the transformation,one thing you can do is to move the qubits before and after the gate. Do you know a gate that allows you to make this kind of transformation ? $\endgroup$ Apr 26, 2022 at 15:24
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    $\begingroup$ Actually, replacing the $CNOT_{13}$ gate with a $CNOT_{23}$ still gives the GHZ state but, as it is written, I think your teacher wants you to use the given $CNOT_{31}$ gate. $\endgroup$ Apr 26, 2022 at 15:29

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