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I've been looking for the OAA paper but I've been unsuccessful in finding it. (so any links will also be helpful)

My question is essentially this, given the state:

$$\vert 0good \rangle + \vert 1bad \rangle$$

If I know the amplitudes of both states, can I amplify the amplitude of $\vert 0good \rangle$ without a reflection by the initial state using the flag qubit only? What conditions do I need to satisfy in order to do this?

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  • $\begingroup$ This should be the paper arxiv.org/abs/1312.1414 $\endgroup$
    – dylan7
    Commented Dec 26, 2021 at 7:06
  • $\begingroup$ You mean you know the amplitudes of both states, or you can construct an oracle (i.e. a circuit that flips some flag if the input state is as desired)? $\endgroup$
    – 3yakuya
    Commented Dec 29, 2021 at 7:19
  • $\begingroup$ @3yakuya I know the amplitudes of both states and have the oracle but it turns out it is not possible to do what I wanted to do either way. $\endgroup$
    – Dani007
    Commented Dec 30, 2021 at 20:05
  • $\begingroup$ If you have the oracle you can use the rotation as in Grover procedure. Do I understand you want to amplify amplitude without the rotation? Not sure if there is a known algorithm for that. $\endgroup$
    – 3yakuya
    Commented Dec 30, 2021 at 23:19
  • $\begingroup$ @3yakuya Yes I was trying to not use the oracle as it was expensive computationally but like you said I don't think it's possible. $\endgroup$
    – Dani007
    Commented Jan 1, 2022 at 5:28

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I do not believe there are known quantum algorithms based on amplitude amplification that do not rely on an oracle to flip the phase of the desired state. Popular algorithms using amplitude amplification, including Grover Search, rely on the same idea proposed by Brassard/Hoyer and Grover, requiring an oracle.

Technically you do not always need an additional oracle qubit per se (see Why is an oracle qubit necessary in Grover's algorithm?) but in practice such qubit simplifies the implementation a lot.

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    $\begingroup$ Don't forget to upvote the question if you think it was worth amswering and was decent. $\endgroup$ Commented Jan 1, 2022 at 22:41

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