# フェーズ$e ^ {（0.q_0 q_1 q_2 q_3）}$を$e ^ {（0.q_1 q_2 q_3）}$に変更する量子演算はありますか？

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：

$$Q(q_0) = \frac{1}{\sqrt{2}}\big(|0\rangle + e^{(0.q_{0}q_{1}q_{2}q_{3})}|1\rangle\big)$$

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.