Can we use quantum phase estimation (or any other) algorithm to estimate the phase of an arbitrary single-qubit state, without measuring it?

That is: estimate the relative phase ๐œ™ of the qubit ๐‘Ž|0โŸฉ + ๐‘ ๐‘’๐‘–๐œ™|1โŸฉ (a and thus b unknown), maybe transfer ๐œ™ to the state of n ancilla qubits and measure it?

Thank you in advance.

  • $\begingroup$ No, even with measurement. $\endgroup$
    – kludg
    Aug 3, 2022 at 5:32

1 Answer 1


No, it cannot be done because if it could be done, you could distinguish between arbitrarily many linearly dependent states -- which is not possible. Another way to say the same thing is that if you can read off the phase of a qubit, you can encode arbitrarily high number of classical bits in a single qubit -- which cannot be done as proven by Holevo's theorem.

Particularly regarding your suggestion of using quantum phase estimation, quantum phase estimation has nothing to do with the determination of the phase of an arbitrary qubit. It's about determining the eigenvalue $e^{2\pi i\theta}$ associated with a given eigenstate $\vert \psi_\theta\rangle$ of a unitary operator $U$ which is provided to us as an oracle. So, given an arbitrary qubit, you cannot figure out which oracle to construct so that the given state of the qubit will be its eigenstate with the eigenvalue $e^{2\pi i\theta}$ (where $\theta$ is the phase of the given state) which you can determine using QPE -- unless you know the basis in which the given state has been prepared. If you do know the basis in which the state has been prepared then you don't need to do the QPE anyway -- you can just measure the qubit in the appropriate basis and figure out the phase.

  • 1
    $\begingroup$ I like the appeal to Holevoโ€™s theorem $\endgroup$ Aug 3, 2022 at 2:49
  • $\begingroup$ Thank you for your answer @FreeAssange. If I know the basis in which the state has been prepared but I don't want to measure it because I still need it in further calculations, how can I use QPE in this case? $\endgroup$
    – cef
    Aug 3, 2022 at 15:40
  • 1
    $\begingroup$ @cef If you know which basis the state has been prepared in then measuring it with the operator that is diagonalized by the basis won't change the state. For example, if you know that the state has been prepared in either $\vert \pm\rangle$ then you can measure its $X$ spin (thus learning the phase) and you won't disturb the state. $\endgroup$ Aug 3, 2022 at 16:34
  • $\begingroup$ @FreeAssange, nice! $\endgroup$
    – MonteNero
    Aug 4, 2022 at 4:50
  • $\begingroup$ @FreeAssange, thank you for your answer. But I'm not sure why measurement won't disturb the state? Doesn't measurement in a circuit always causes the state to collapse? $\endgroup$
    – cef
    Aug 5, 2022 at 14:26

Your Answer

By clicking โ€œPost Your Answerโ€, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.