4つのキュービットのセットがあるとします。たとえば、 $ q_ {0}、q_ {1}、q_ {2}、q_ {3} $ は、 $ 4 $ ビットの2進数、MSBとして $ q_ {0} $ を使用。これらのキュビットにQFTを適用した後、 $ q_ {0} $ のフェーズでは、2進分数の概念（ $ e ^ {\ frac {a} {2 ^ {n}}} = e ^ {0..an ... a1} $ ）は次のようになります：
$Q(q_0) = \frac{1}{\sqrt{2}}\big(|0\rangle + e^{(0.q_{0}q_{1}q_{2}q_{3})}|1\rangle\big)$
私の質問は： $ Q（q_0）$ のフェーズを変更する量子操作はありますか？math-container "> $ e ^ {（0.q_ {0} q_ {1} q_ {2} q_ {3}）} $ から $ e ^ {（0.q_ {1} q_ {2} q_ {3}）} $ ？
First you need to undo the QFT applied to $|q_3\rangle$, so you would apply a $H$ gate to $Q(|q_3\rangle)$.
$HQ(|q_3\rangle) = |q_3\rangle$
or more generally for an $n$-qbit system
$HQ(|q_n\rangle) = |q_n\rangle$
If you then apply the conditional rotation (on $Q(|q_0\rangle)$ conditioned by $|q_3\rangle$) of angle $R_4^{\dagger}$:
$ \begin{pmatrix} 1 & 0 \\ 0 & e^{-\pi i / 8} \end{pmatrix} $
to $q_0$ this will undo the $q_3$ rotation applied by QFT. You would then want to fix up the $|q_3\rangle$ state to put it back into QFT by applying a $H$ gate again.
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I don't think you need to correct $|q_3\rangle$ for any phase-kickback after applying the conditional rotation to $Q(|q_0\rangle)$ as we're not applying the rotation to an eigenstate, but anyone feel free to correct me!
It seems that such a transformation would not be unitary. Since it drops the information about bit $q_0$ altogether, it would have basis states that differ in that bit transformed into the same state, and that would not be possible to invert.
