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I have a multi qubit state that consists of two basis states and looks like the following for different number of qubits:

  • 2 qubits: $|00⟩+|xx⟩$
  • 3 qubits: $|000⟩+|xxx⟩$
  • 4 qubits: $|0000⟩+|xxxx⟩$

...

(The $x$ simply stands for 0 or 1) That means it's a state that contains two basis states, and I know that $|00⟩$, $|000⟩$, $|0000⟩$ etc., are in the superposition. But I don't know the second basis state denoted with x, but what I do know are the amplitudes of both basis states, which are not equal. Is it possible to amplify the unknown state when I know the amplitudes? Is there something better than Amplitude Amplification for this case?

Edit: It would probably be much better to use a principle that doesn´t use Amplitude Amplification for the case that the state initialization operator would be large, which would not be convenient in State Amplification.

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    $\begingroup$ Is the amplitude on the $|xxx\rangle$ component small? (otherwise, you'd be better off, depending upon application, just measuring different copies until you get an answer that's not 000.) But if you do want to follow the amplitude amplification route, just amplify the $|000\rangle$ term! Using the correct number of iterations, you can de-amplify it too, which is just what you need. $\endgroup$
    – DaftWullie
    Nov 17, 2023 at 15:41
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    $\begingroup$ Yes, the the amplitude of the unknown basis state is small. Can you give me an example how to implement it? And doesn´t deamplifying |000⟩ result in an increase of amplitudes of the wanted basis state and the other ones too? $\endgroup$
    – Manuel
    Nov 17, 2023 at 15:52
  • $\begingroup$ Thankfully amplitude amplification can only change the amplitudes of the bases that are already non-zero (to be more technical it keeps the state in the 2-D subspace defined by the state you want to amplify and the other orthogonal state, which itself can be composed of several basis states) $\endgroup$ Nov 17, 2023 at 16:28
  • $\begingroup$ How can I deamplify a state? $\endgroup$
    – Manuel
    Nov 17, 2023 at 17:13

1 Answer 1

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Let's assume that you have a unitary $U$ the produces the state you want: $$ U|0\rangle=\cos\theta|0\rangle+\sin\theta|x\rangle $$ where $\theta$ is small, and $x$ is some basis state that we don't know.

Now, let me write $$ |\psi(\phi)\rangle=\cos\phi|0\rangle+\sin\phi|x\rangle $$ and define $$ V=I-2|0\rangle\langle 0|. $$ This gives us our usual Grover iterator $$ G=UVU^\dagger V. $$ You can easily show (standard analysis of Grover's algorithm) that $$ G|\psi(\phi)\rangle=|\psi(\phi+2\theta)\rangle. $$ Thus, if we start from $|\psi(\theta)\rangle$ and apply Grover's iterator $R$ times such that $\sin((2R+1)\theta)\approx\pi/2$, we will have created $|x\rangle$ with very high accuracy.

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  • $\begingroup$ Is there a way to do this without using U for the Grover Operator, for the case that U is a large circuit? $\endgroup$
    – Manuel
    Nov 23, 2023 at 14:45
  • $\begingroup$ You need to get your information about $|x\rangle$ from somewhere! $\endgroup$
    – DaftWullie
    Nov 23, 2023 at 14:46

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