I saw on the site Brilliant.org a question witch is:

"Let's say I'm given a quantum wire, I measure it and find it charged "on". If I come back some time later, what state should I measure?"

And the answer was: Charged "on".

And they explain: "If a quantum wire is measured to be "charged", then for any following measurements of the same isolated quantum wire, we would expect to measure the same charge and observe the same state."

  • My question is: doesn't the system evolves between the measurement at t=0 and t>0, then the system would return in a combination states, so we don't expect the same state?? (we can do the same experiment with a qubit)

1 Answer 1


I think this question makes two assumptions:

  1. the system does not evolve intentionally (i.e., no gates or measurements are performed on it), and

  2. there is no noise that would cause the system to evolve in an unintended way.

Of course, if you change the state of the system by applying some gates, or if the system is noisy, this will not be the case.

  • $\begingroup$ Thank you Mariia Mykhailova. Yes, they assume that there is no measurement noise affecting the system. but, the superposition is an intrinsic property of a quantum system, so I think that the system evolves between t=0 and t>0, so he would return in a superposition state, and we couldn't expect the same state. Am I right? $\endgroup$
    – walid
    Jul 10, 2019 at 20:28
  • $\begingroup$ The system doesn't evolve on its own in the absence if nothing is done to it (i.e., no gates are applied and no measurements are done). Think of it as a value of a classical variable - if nothing is done to the register that stores the variable and there is no noise to flip any bits, the value will remain the same. $\endgroup$ Jul 10, 2019 at 21:38

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