# (Exercise-verification) Basic exercise on a 3 qubits circuit

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

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.

• 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 ?
– Bozu
Apr 26, 2022 at 14:46
• Yes, it's better ! Now you can answer the question (d) without any trouble. Apr 26, 2022 at 14:56
• 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 ?
– Bozu
Apr 26, 2022 at 15:01
• 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 ? Apr 26, 2022 at 15:24
• 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. Apr 26, 2022 at 15:29