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Is it possible to perform an operation on two qubits with initial states as follows:

$$q_1: 1/\sqrt(2)(|0\rangle + exp(0.a_1a_2a_3)|1\rangle)$$ $$q_2: 1/\sqrt(2)(|0\rangle + |1\rangle)$$

To resultant state:-

$$q_1: 1/\sqrt(2)(|0\rangle + exp(0.a_1a_2)|1\rangle)$$ $$q_2: 1/\sqrt(2)(|0\rangle + exp(0.a_3)|1\rangle)$$

Without knowing the value of $a_3$. Where $a_1,a_2,a_3 ∈ [0, 1].$

The idea is to shift the phase of $q_1$ by $exp(-0.00a_3)$ and $q_2$ by $exp(0.a_3)$ with the unitary operation not being aware of the value of $a_3$.

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  • $\begingroup$ What are the normalization factors on $q_1, q_2$? Are $|0 \rangle, |1\rangle$ of equal probability or...? $\endgroup$
    – C. Kang
    Jul 24 '20 at 22:59
  • $\begingroup$ Yes of equal probability. Have updated the question. $\endgroup$
    – virattara
    Jul 25 '20 at 6:02
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    $\begingroup$ Take two possible different choices to your a coefficients, and assume a unitary exists that implements your desired transformation for those elements. By linearity you now know how it works for all input states. Does this coincide with what you want? (I assume not) $\endgroup$
    – DaftWullie
    Jul 25 '20 at 6:46
  • $\begingroup$ Yes it does, but trying to figure out a unitary that doesn't depend on the value of $a_3$ $\endgroup$
    – virattara
    Jul 25 '20 at 7:31
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No, it's not possible to extract digits of the phase like that. It would violate the Holevo bound. In general there's no way to "amplify" single small phase differences into big phase differences, because of linearity.

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  • $\begingroup$ The Holevo's theorem and Nayak's bound deal with information retrieved from a quantum system after measurement. Whereas in the question asked the system would ultimately require n qubits to retrieve n classical bits of information from one qubit's phase. Any thoughts on this? $\endgroup$
    – virattara
    Jul 28 '20 at 7:45
  • $\begingroup$ @virattara The number of qubits used while extracting the message is irrelevant; what matters is the number of qubits the message was encoded into. $\endgroup$ Jul 28 '20 at 8:14
  • $\begingroup$ So that means if the message encoded isn't in an orthogonal state, there's no way to retrieve information safely at the receiver. This further means a receiver who doesn't has any information of the quantum state received can't convert a non-orthogonal state to orthogonal, even though he increases the Hilbert space. He can only do so if the transmitter provides him with adequate information of how he prepared the state in the first place and that would mean transferring information indirectly thus defeating the purpose of efficient communication. $\endgroup$
    – virattara
    Jul 30 '20 at 8:17

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