# 振幅を2乗した状態または任意のパワーに変換する方法は？

$$（\ alpha | 0 \ rangle + \ beta | 1 \ rangle）（\ alpha | 0 \ rangle + \ beta | 1 \ rangle）| 0 \ rangle =\ alpha ^ 2 | 000 \ rangle + \ alpha \ beta | 010 \ rangle + \ beta \ alpha | 100 \ rangle + \ beta ^ 2 | 110 \ rangle。$$

2つのCNOTゲートが連続している場合、補助キュビットはターゲットキュビットです。

$$\ alpha ^ 2 | 000 \ rangle + \ alpha \ beta | 011 \ rangle + \ beta \ alpha | 101 \ rangle + \ beta ^ 2 | 110 \ rangle。$$

0を測定した場合、これに続いて補助量子ビットの測定が行われ、最初の2つの量子ビットの状態は$$\ frac {\ alpha ^ 2} {\ sqrt {| \ alpha | ^ 4 + | \ beta | ^ 4}} | 000 \ rangle + \ frac {\ beta ^ 2} {\ sqrt {| \ alpha | ^ 4 +| \ beta | ^ 4}} | 110 \ rangle。$$

2番目の量子ビットにCNOTゲートを使用し、最初の量子ビットを制御として使用する

$$\ frac {\ alpha ^ 2} {\ sqrt {| \ alpha | ^ 4 + | \ beta | ^ 4}} | 00 \ rangle + \ frac {\ beta ^ 2} {\ sqrt {| \ alpha | ^ 4 +| \ beta | ^ 4}} | 10 \ rangle =（\ frac {\ alpha ^ 2} {\ sqrt {| \ alpha | ^ 4 + | \ beta | ^ 4}} | 0 \ rangle + \ frac {\ beta ^ 2} {\ sqrt {| \ alpha | ^ 4+ | \ beta | ^ 4}} | 1 \ rangle）| 0 \ rangle$$

$$\ frac {\ alpha ^ 2} {\ sqrt {| \ alpha | ^ 4 + | \ beta | ^ 4}} | 0 \ rangle + \ frac {\ beta ^ 2} {\ sqrt {| \ alpha | ^ 4 +| \ beta | ^ 4}} | 1 \ rangle$$

ただし、補助キビットでの測定は厄介です。任意の数のキュービットを測定せずに、電力供給された振幅状態を取得できますか？

As noted by Mateus in the comments, the transformation you are looking for is non-linear. This cannot be done with any matrix transformation. Thus, you will need more qubits, and your solution shows two (+1 scratch qubit) is sufficient. I guess you might wonder if a two-qubit unitary can do it, though?

The problem is that the transformation you want to implement depends on the input state. You can't do this (unitarily) even with extra qubits. I believe the most general result forbidding such requirements is the No-Programming Theorem.

Also note at, as $$r\to\infty$$, the transformation becomes a projection onto the subspace spanned by the states with highest modulus. You do are doing something like a weak measurement when $$r$$ is finite.

Nearly final observation: you mention you want $$|\psi\rangle$$ to be "unknown". You should be cautious taking your solution (as you generalise requiring more copies of $$|\psi\rangle$$) farther without thinking about no-cloning or more subtle resource counting.

Last thing. A coherent version of something like what you might be looking for is Amplitude Amplification.