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For example in grover search algorithm, assuming that the amplified amplitude is $\alpha$ , is it possible to perform this operation :

$\alpha |0...\rangle + \beta |1...\rangle \longmapsto \beta|0...\rangle + \alpha|1...\rangle$

It seems that the operation may be reversible

Is it possible to apply inverse : $\frac{1}{\alpha},\frac{1}{\beta}$ then how normalize to get the sum of probabilities to 1?

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  • $\begingroup$ Your question doesn't define sufficiently what you want to achieve. For any $x\in\{0,1\}^n$, if you have a particular $\alpha_x|x\rangle$, for what $y\in\{0,1\}^n$ in the output are you wanting the amplitude to be $\alpha_x$? So far, it looks like I could just apply a bit flip on all qubits.... $\endgroup$ – DaftWullie Feb 21 at 14:06
  • $\begingroup$ yes, in fact I did not formulate my question correctly, what I want to do is simply inverse the amplitudes of the states (not only 2 states): $ \frac{1}{\alpha_{x}}, \frac{1}{\alpha_{y}}, \frac{1}{\alpha_{z}}, ....$ and than normalize, for example at the end of the grover algorithm I would like to get the value marked as the least likely to obtain when measuring. $\endgroup$ – rabah Feb 25 at 9:02
  • $\begingroup$ How do you want it to behave if any of the amplitudes are 0. Do you want to ignore them, or should the answer be, for instance, a uniform superposition of the terms with 0 amplitude? $\endgroup$ – DaftWullie Feb 25 at 10:00
  • $\begingroup$ I'm new in this field ... I thought that states with 0 amplitude dosn't exists in the superposition ... In fact I have a quantum registre of 3 qubits containing 4 Integers (when I do mesurement ... not 8), so, I want after searching one of the 4 integer, inverse the amplitudes . $\endgroup$ – rabah Feb 25 at 14:42
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The reversing of the amplitudes is a pretty easy operation by using the Pauli $X$ operator. The result of applying such operation to a qubit $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ is:

$X|\psi\rangle=X(\alpha|0\rangle + \beta|1\rangle)=\alpha X|0\rangle + \beta X|1\rangle=\alpha|1\rangle+\beta|0\rangle=\beta|0\rangle+\alpha|1\rangle$,

and so it performs the amplitude swap you are looking for, and that is why this operation receives the name of bit flip as it flips the bits of the state. If you want to apply an operation like that to a multiple qubit state, that is what I think you are asking for, as it would for example be $|\psi\rangle=\alpha|0010\rangle+\beta|0111\rangle$ then you should apply $X$ gates to the qubits you need to change so that the resultant state is what you are looking for. For the example I gave, it would be

$(I\otimes X\otimes I\otimes X)|\psi\rangle = (I\otimes X\otimes I\otimes X) (\alpha|0010\rangle+\beta|0111\rangle) =\alpha(I|0\rangle X|0\rangle I|1\rangle X|0\rangle)+\beta(I|0\rangle X|1\rangle I|1\rangle X|1\rangle)=\alpha|0111\rangle+\beta|0010\rangle = \beta|0010\rangle+\alpha|0111\rangle$

and the swapped amplitude state that you are looking for is obtained.

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As I understand the question, you're asking if it's possible to produce a transform that looks something like $$ \alpha|000\rangle+\beta|010\rangle+\gamma|011\rangle\mapsto\frac{1}{\alpha}|000\rangle+\frac{1}{\beta}|010\rangle+\frac{1}{\gamma}|011\rangle $$ up to normalisation.

Without further knowledge, this is impossible. To see this, consider the problem where I give you either $$ |0\rangle\qquad\text{or}\qquad \sqrt{1-\epsilon^2}|0\rangle+\epsilon|1\rangle, $$ where $\epsilon$ is small. You don't know which I've given you, but you're tasked with transforming these two into $$ |0\rangle\qquad\text{or}\qquad\sqrt{1-\epsilon^2}|1\rangle+\epsilon|0\rangle $$ respectively. In the case of small $\epsilon$, after the transformation you are almost guaranteed to be able to distinguish between which of the two you were given, which you certainly shouldn't be able to do given the starting position of two practically indistinguishable states. (This violates no-cloning bounds, and similar.)


However, your question mentioned that your state was the output of a Grover Search. In that specific context, you surely can. This is because Grover's Search lets you explore a particular space of the form $$ \cos\theta|\psi_g\rangle+\sin\theta|\psi_b\rangle. $$ So, at a particular point in the algorithm you've produced $M$ terms (the $M$ items you're searching for) with amplitudes $\cos\theta_0/\sqrt{M}$, and another $N-M$ with amplitudes $\sin\theta_0/\sqrt{N-M}$. You just have to find a value of $\theta_1$ such that $$ \cos\theta_1=\frac{\tan\theta_0}{\sqrt{\tan^2\theta_0+(N/M-1)^2}} $$ assuming you know $N,M,\theta_0$. Then you just use your Grover Search to produce that (or, at least, a good approximation. The larger $N$, the better the approximation you can make.). Since the "bad" states now have greater amplitudes, this presumably takes less time than the search itself.

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