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Given a set of four qubits, say $q_{0},q_{1},q_{2},q_{3}$ which represent a $4$-bit binary number with $q_{0}$ as the MSB. After applying QFT on these qubits the phase of $q_{0}$using the concept of binary fractions ($e^{\frac{a}{2^{n}}} = e^{0..an...a1}$) becomes:

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

My question is: Is there a quantum operation to change the phase of $Q(q_0)$ from $e^{(0.q_{0}q_{1}q_{2}q_{3})}$ to $e^{(0.q_{1}q_{2}q_{3})}$?

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  • $\begingroup$ If you look at the QFT circuit you can simply apply the principle of reverseable computing and apply the reverse circuit related to the q3 rotation to undo it. This is basically just applying a very selective version of the inverse QFT (which would undo all the rotations). $\endgroup$ – Sam Palmer Jun 24 at 13:37
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    $\begingroup$ Sorry for screwing up those last indices while formatting! $\endgroup$ – JSdJ Jun 24 at 14:39
  • $\begingroup$ @JSdJ No worries. $\endgroup$ – virattara Jun 24 at 16:15
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    $\begingroup$ You could implement your own modified 'QFT' algorithm that for $Q(q_0)$ has all the rotation indexes shift down by 1. $\endgroup$ – Sam Palmer Jun 26 at 19:18
  • $\begingroup$ related on physics.SE by OP: physics.stackexchange.com/q/561544/58382 $\endgroup$ – glS Jun 27 at 15:40
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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.

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  • $\begingroup$ You could write the QFT of $q_0$ to an ancilla to preserve $q_0$? $\endgroup$ – Sam Palmer Jun 27 at 14:58
  • $\begingroup$ Yes, something extra needs to be done, but I'm not sure what off the top of me head. $\endgroup$ – Mariia Mykhailova Jun 27 at 18:44
  • $\begingroup$ Ancilla needs to be preserved as well if it's going to take the value of q0 after measurement. $\endgroup$ – virattara Jun 27 at 19:03
  • $\begingroup$ I think this would require a circular shift-type operation on all the qubits, to preserve information. $\endgroup$ – virattara Jun 27 at 19:41
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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!

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  • $\begingroup$ Thank you for the reply but I think there was a serious mistake in the question. I have corrected it now, the phase transition that I wanted to know is feasible or not is from 𝑒(0.π‘ž0π‘ž1π‘ž2π‘ž3) to 𝑒(0.π‘ž1π‘ž2π‘ž3). Similar to a left shift operator. $\endgroup$ – virattara Jun 24 at 16:15

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