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My question is related to the exercise 4.28 in the book of Nielsen and Chuang (Quantum Computation and Quantum Information). Here is the exercise

For $U=V^2$ with $V$ unitary, construct a $C^5(U)$ gate analogous to that in Figure 4.10, but using no work qubits. You may use controlled-$V$ and controlled-$V^{\dagger}$ gates.

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I think the exercise in the current form is not possible to solved. Since any usage of a controlled-$V$ or $V^{\dagger}$ gate gives even 4 cases (with case I mean a combination of qubits) where $V$ or $V^{\dagger}$ is applied. So we can increase the number of applied $V$ gates with any usage of a controlled-$V$ or decrease the number of applied $V^{\dagger}$ gates with any usage of a controlled-$V^{\dagger}$ by 4. But to obtain the controlled-$U$ gate, we may only have $2$ usages of a $V$ gate. Thus any way we apply the controlled-$V$ or $V^{\dagger}$ gates, in the end we will always have more than 2 cases where a $V$ or $V^{\dagger}$ is applied (with applying I mean that it does not contract with another controlled gate, so it will really be applied at the end).

Note that this problem does not occur for controlled-X gate since an addition can be though as an addition modulo 2, therefore we can decrease the numbers of cases where a qubit is switched.

Since I think that this chapter is heavily influenced by the paper https://arxiv.org/abs/quant-ph/9503016.pdf which constructs a $C^3(U)$ with controlled-$V$ gates where $V^4=U$, I think it should be $V^16=U$ in the exercise. However I am not certain if there really does not exist a solution, maybe one using maybe gates creating a superposition like the Hadamard gate. So my question is, if someone has a solution for this exercise or a proof that it is impossible to do.

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It can be done in the same way as the $$C^3(U)$$ circuit with 4 controls. circuit

Look in arXiv0708.3274 for details

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  • $\begingroup$ This construction is also in the paper I posted in my question (Lemma 7.5), but I thought, in the exercise by "controlled-V" gate a gate with only one control qubit is meant, however the term is vague and could mean your solution and probably does. It is just a bit unsatisfying, because to construct the controlled-U gate you need another controlled-V gate with one less control bit, which is in general also not easy to construct; thus the construction seems to be not very useful. $\endgroup$ – Alex Go Aug 31 '17 at 17:16

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