I'm playing around with the IBM Q to demonstrate the qunatum Zeno effect.

If we prepare a qubit in the $|0\rangle$ state and apply 5 consecutive $R_y(\pi/5)$ gates, we should end up in state $|1\rangle $ with 100% probability since $R_y(\pi/5)^5=R_y(\pi)$.

enter image description here

This gives output 00100, as expected.

If we now add couplings between each of these gates and measure them (using deferred measurement), then after the first $\pi/4$ rotation we collapse the q[2] qubit into either $|0\rangle $ with probability $\cos^2(\pi/8)=0.905$ or into $|1\rangle $ with probability $\sin^2(\pi/8)=0.095$.

What happens after all 5 have been applied, with measurements after each rotation?

My intuition tels me to create a probability tree such as

enter image description here

which gives us a total probability of 67.4% chance of ending up with a |0> state.

The IBM Q simulator gives a 66.3% chance of measuring $|0\rangle $. Is this just statistical error or is there something wrong with my circuit?

My circuit is available at https://quantumexperience.ng.bluemix.net/share/code/5c3890a72f408b005a0d9f06

enter image description here

  • $\begingroup$ What are your thoughts? $\endgroup$ Jan 14, 2019 at 10:51
  • $\begingroup$ @NorbertSchuch I think it is probably just statistical error, but it just depends whether I've made an error anywhere! $\endgroup$
    – Cameron
    Jan 14, 2019 at 12:29
  • $\begingroup$ Did you run the circuit on a simulator which does not make statistical errors? $\endgroup$ Jan 14, 2019 at 15:11
  • $\begingroup$ @NorbertSchuch as far as I'm aware there doesn't seem to be away to turn off the randomness on the simulator $\endgroup$
    – Cameron
    Jan 19, 2019 at 15:57
  • 1
    $\begingroup$ Then why don't you answer your own question? Might help future users with the same question! (And you'll get my upvote :) ) $\endgroup$ Jan 19, 2019 at 19:27

1 Answer 1


If you want to know whether the deviation of the output of your circuit from the analytically derived value is a statistical error, you can use a circuit simulator which simulates the exact noise-free circuit, such as quirk.


This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .