# Is it possible to decompose $\land(UXU^\dagger)$ in one-qubit operations and only a single $\land(X)$?

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.

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

$$V=U^\dagger$$
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.
• Thank you. Actually also $U=V$ works. I didn't notice how trivial was my question. Jul 11 at 14:20
• 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. Jul 12 at 6:54