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I'm going through a paper that uses quantum amplitude estimation, and one of the ingredients to the algorithms is an operator that flips the sign of a state based on the state of a control qubit. Basically,

$S_\psi |x\rangle = |x\rangle$ if least significant bit of $x$ is 0 and $-|x\rangle$ if it's 1..

Now, naively I would have thought that just applying the $Z$ gate to the qubit that represents the relevant bit of $x$ would achieve that, would it not? If my qubit starts out in state $\alpha |0\rangle + \beta |1\rangle$ then after applying $Z$ it'll be in state $\alpha |0\rangle - \beta |1 \rangle$, which is exactly what I need.

The paper, on the other hand, says we need to include an ancilla bit, use the $X$ gate to prepare it in state $1$, then act on this ancilla qubit with a controlled $Z$ gate (controlled by the qubit representing the lsb of $x$) and then apply another $X$ gate to "uncompute" the ancilla.

But if I follow along with that logic, I'm just left with the ancilla unchanged and essentially $Z$ being applied to my original qubit, just as I had in the "naive" implementation.

Am I missing something subtle here or were the authors of the paper overthinking?

Later on, that whole operation that I just describes needs to be in turn controlled via some other bits, but I don't see how my naive implementation would not be able to be turned into a controlled one...

Reference: Section V in the paper "Credit Risk Analysis Using Quantum Computers". arxiv:19007.03044v1 [quant-ph] 5 Jul 2019 https://arxiv.org/abs/1907.03044

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  • $\begingroup$ Please give some references etc.! As you phrase it, a Z looks fine. $\endgroup$ – Norbert Schuch Aug 29 at 21:58
  • $\begingroup$ it's not clear from a quick read of the relevant paragraph, but maybe they want to use the ancilla qubit to determine which state is the target one? Or there might be a reason why they cannot directly use a $Z$ on the relevant qubit? $\endgroup$ – glS Aug 30 at 20:44
  • $\begingroup$ The target qubit is always the same. I don't see why you couldn't use Z directly on the relevant qubit, because it's part of other operations as well. $\endgroup$ – Lagerbaer Sep 3 at 16:03

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