Qubits can be represented by a state vector $\psi \in \mathbb{C}^{2^N}$, where $N$ is the number of qubits.

The higher $| \psi_i |$ is, for any index $i$, the more likely it is to be measured. But is there any way to reverse this? The closer the value $| \psi_i |$ is to 0, the more probable the measurement of $i$ is?

In a rephrased way, is there a method to amplify probability amplitudes closer to 0? Such as Grover’s algorithm, which amplifies negative probability amplitudes. But instead for values closer to 0.

  • $\begingroup$ did you mean to write "the close the value $|\psi_i|$ is to 0, the least probable measuring $i$ is"? I mean, you simply have $|\psi_i|=\sqrt{p_i}$ where $p_i$ is the probability of measuring the $i$-th outcome, so yes the two quantities are indeed tightly related in a rather straightforward way $\endgroup$
    – glS
    Oct 19, 2022 at 23:43
  • 1
    $\begingroup$ I'm pretty sure they're asking for a process to amplify low-probability states and minimize high-probability states. Maybe for example some process that takes $|\psi\rangle = \sum_{j} \sqrt{p_j} |j\rangle$ to some state like $|\psi'\rangle \propto \sum_{j} (1-\sqrt{p_j}) |j\rangle$ where you are most likely to measure the originally least likely states. $\endgroup$
    – Chris E
    Oct 19, 2022 at 23:50
  • $\begingroup$ @ChrisE Yes, I am exactly asking that. That is also a process I considered, but I don’t see how it can be imported as a unitary (or even a matrix). $\endgroup$
    – Loic Stoic
    Oct 20, 2022 at 15:30
  • $\begingroup$ @glS I don’t understand how this is a textbook question— I think you maybe misunderstood. I will add Chris E’s rephrased version as well. $\endgroup$
    – Loic Stoic
    Oct 20, 2022 at 15:32
  • $\begingroup$ well, yes, the edit changes quite a bit what you were asking. So by "method" you're asking for a quantum circuit doing that, or something else? Also, are you asking about a coherent operation doing that, or a way to post-process data to do it? What are the circumstances? Surely you can't do it in general: imagine you have $\psi_0=1$ and $\psi_i=0$ for $i>0$. Then with such procedure you'd end up with a highly non-normalised state. $\endgroup$
    – glS
    Oct 20, 2022 at 16:05


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.