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Let $U,V$ being any unitary. Is it possible to decompose $\land(UXU^\dagger)$ in one-qubit operations and only a single $\land(X)$?

Something like the following: $\land(UXU^\dagger) \equiv (\mathbb{I}\otimes V)\land(X)(\mathbb{I}\otimes V^\dagger)$

I assumed that $\land(\cdot)$ operation controls over first qubit and operates over second.

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  • $\begingroup$ So $\land(X)$ is the CNOT gate? Is this a common notation? $\endgroup$
    – M. Stern
    Jul 8 at 6:00
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    $\begingroup$ @M. Stern, $\land U$ was a common notation for controlled-$U$. You can find it in old papers like arXiv:quant-ph/9503016 $\endgroup$ Jul 8 at 9:17
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$$V=U^\dagger$$

The trick here is to simply analyse the two cases of what happens if the control qubit is either 0 or 1. If it's 0, then the action you want is $I=V^\dagger V$, which works automatically. If it's 1, the action you want is $UXU^\dagger=V^\dagger XV$. So, you can just read it off.

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  • $\begingroup$ Thank you. Actually also $U=V$ works. I didn't notice how trivial was my question. $\endgroup$ Jul 11 at 14:20
  • $\begingroup$ I don't think $U=V$ does work. It's an issue of being careful about the order of multiplication. Remember that in a circuit going from left to right, you multiply the corresponding matrices going right to left. $\endgroup$
    – DaftWullie
    Jul 12 at 6:54

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