I’ve been struggling with this exercise for a while now. I realize that one must use the decomposition of a unitary gate $U$ as $e^{i\phi}AXBXC$ where $A,B,C$ are unitaries s.t. $ABC=I$. Nevertheless, I can’t seem to write up a circuit that gives an identity sequence of gates e.g. $ABCCBA=I$ when either of the two control qubits are 0, and $U$ in its decomposition otherwise. Perhaps I am approaching this incorrectly?

The questions asks to prove that a $C^2(U)$ gate can be constructed using at most 8 one-qubit gates and 6 controlled-NOTs.


closed as off-topic by Chris, Jon Custer, sammy gerbil, ACuriousMind Jun 2 '18 at 8:47

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    $\begingroup$ Hi Faris and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Jun 1 '18 at 5:23
  • $\begingroup$ Hi John. I just checked the Meta and it does not appear to say that homework-type questions won’t be answered. I’ll make sure this independent study question fits with the guidelines though. $\endgroup$ – Faris Sbahi Jun 1 '18 at 5:25
  • $\begingroup$ For the identity sequence of gates, couldnt you just chose A=B=C=I ( with I=I_1 tensor I_1). If not, maybe uou.can provide much more needed detail, so.one.could.answer.you Q without consulting the book. When you say 'the question' do yiu mean your question or the priblem.in the book. $\endgroup$ – lalala Jun 1 '18 at 7:23
  • $\begingroup$ No, that does not work. That assumes that $U=I$. The point is that arbitrary $U$ can be decomposed as above and we can design a circuit just keeping in mind this assumption (i.e. no suppositions about $A,B,C$ other than being unitary). Note that $X$ is the Paul $x$ matrix. I'll update my question so it's clearer... Sorry about that $\endgroup$ – Faris Sbahi Jun 1 '18 at 7:27
  • $\begingroup$ A hint: I think I remember that you need to use the square root of $U$ to do this. $\endgroup$ – Peter Shor Jun 1 '18 at 10:28