# Quantum computing explanation for the quantum Zeno effect

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 $$R_y(\pi/5)^5=R_y(\pi)$$.

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

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

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