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I understand that, according to amplitude amplification, I can amplify states according to a partition over the state space. However, suppose I want to amplify or de-amplify a specific portion of the state after putting it into a superposition, (e.g. I have $\frac{1}{2\sqrt(2)}\sum_{x=000}^{111}\left|x\right>$ and I want to amplify/de-amplify $\left|000\right>$). Is this possible without amplitude amplification?

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    $\begingroup$ I'm not clear why you don't want to use amplitude amplification, which does exactly what you are asking for $\endgroup$ – glS Aug 1 at 11:22
  • $\begingroup$ @glS The reason I don't want to use amplitude amplification is that I want to de-amplify one specific, constant state and I want to avoid the $O\left(\sqrt{N}\right)$ time required to do that completely. $\endgroup$ – Woody1193 Aug 2 at 4:52
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If you know, in advance, that the state you want to deamplify is specifically $|000\rangle$, there are a couple of strategies that you could follow.

For example, introduce an ancilla and perform the multi-controlled not, targeting the ancilla, where it is controlled off every qubit in the original state being in the $|0\rangle$ state. So, you'd be doing $$ \frac{1}{\sqrt{8}}\sum_x|x\rangle|0\rangle\mapsto \frac{1}{\sqrt{8}}\left(|000\rangle|1\rangle+\sum_{x\neq 000}|x\rangle|0\rangle\right). $$

If you want to completely get rid of the $|000\rangle$ term, just measure your ancilla in the standard (0/1) basis. If you get the 0 answer, you've succeeded (here, this happens with probability 7/8). If you get the 1 answer, you've failed. You produce your state again and repeat until success.

If all you want to do is decrease the amplitude of $|000\rangle$ rather than completely remove it, you can correspondingly increase your probability of success. A simple strategy is to apply the POVM/filtering operation $$ \left(\begin{array}{cc} 1 & 0 \\ 0 & \alpha \end{array}\right) $$ for $\alpha<1$. Afterwards, you'd repeat your multi-controlled-not operation to disentangle the ancilla.

These strategies are perhaps conceptually simpler to understand than amplitude amplification, and work well if the amplitude you're trying to de-amplify is small enough. However, if the amplitude you're trying to amplify is too small, you won't do any better than the scaling resulting from amplitude amplification, and you'd be much better off seeing if you can change the circuit that's producing your state so that you don't have so much of the $|000\rangle$ component in the first place. For instance, in your example, instead of applying $H^{\otimes 3}$, you'd find a slightly more complicated circuit such as enter image description here where $U|0\rangle=(\sqrt{3}|0\rangle+2|1\rangle)/\sqrt{7}$ and $V|0\rangle=(|0\rangle+\sqrt{2}|1\rangle)/\sqrt{3}$.

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  • $\begingroup$ I don't see how you can do POVM-filtering. That seems like the best option to me but I don't see how you can build a circuit to do that. Can you please provide more information about this? $\endgroup$ – Woody1193 Aug 6 at 1:46
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    $\begingroup$ @Woody1193 asked and answered onyour behalf: quantumcomputing.stackexchange.com/q/6952/1837 $\endgroup$ – DaftWullie Aug 6 at 11:00
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    $\begingroup$ @Woody1193 Do you mean for the POVM? It's not unitary. It only works with a certain probability of success, but is heralded, meaning that if it works, you know it's worked. $\endgroup$ – DaftWullie Aug 23 at 5:34
  • $\begingroup$ Sorry, I had to look up more information on it to really understand what it was doing. Thanks for the reply $\endgroup$ – Woody1193 Sep 5 at 2:14
  • $\begingroup$ If I had a state $\frac{1}{\sqrt{2}} \left( \left|000\right> + \left|101\right> \right)$ and I used POVM to measure the third qubit with a bias toward $\left|1\right>$, would that make the probability of measuring $\left|101\right>$ increase above 50% or would it not change? Sorry to keep bothering you about this $\endgroup$ – Woody1193 Sep 19 at 9:56
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The probability amplitudes of a quantum system comprising of several qubits can be changed (amplified / de-amplified) with the application of suitable Rotation Gate. However, the sum of probabilities of collapsing to all possible states will remain unity, so if some states are getting amplified then others will get de-amplified. Is there any practical utility of such a gate or is it a hypothetical question?

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  • $\begingroup$ Thanks for the answer. The reason I want to do this is that I have an algorithm which produces a number of states and the state $\left|000\right>$, which is a junk state. However, due to the nature of the algorithm, this state is the most probable. So, I want to de-amplify this state while amplifying the others. $\endgroup$ – Woody1193 Aug 2 at 4:57
  • $\begingroup$ OK, though naively Rotation Gate appears to help in a single qubit case but in multi qubit system, it may not be very helpful and linearity of quantum mechanics may pose a challenge in designing such a gate, i.e. you may not be able to beat performance of Amplitude Amplification technique, though again I have not come across any proofs in this regard...so All the Best.. $\endgroup$ – Ashish Aug 2 at 5:13
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There is a method called inversion about the mean, which is used in Grover's search algorithm (and many more algortihms). In Grover's search algorithm, the preferred outcome (e.g. what you are searching for) is a certain measurement outcome. To increase the possibility of this outcome (compared to other outcomes), the probability of this outcome (e.g. its amplitude) is increased through this 'inversion about the mean' method.

Essentially what it does, is flipping (through phase-inversion) a certain outcome's phase (in your case the $|000\rangle$ state) about the mean of all the states. Then this outcome is further apart from the mean - updating all the states in some way that bring them close to the mean then can increase the amplitude of this outcome.

A nice introduction can be found here, but it doesn't have any graphical explanation. That can be found here instead.

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    $\begingroup$ isn't this essentially the amplitude amplification algorithm? $\endgroup$ – glS Aug 1 at 11:21
  • $\begingroup$ Welp, yes, sorry for that. My bad, I didn't really check the link, this renders this answer obsolete! $\endgroup$ – JSdJ Aug 1 at 13:08

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