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Mar 13 '20 at 13:46 comment added Vitality Thank you so much. I accepted your answer :-)
Mar 13 '20 at 13:28 comment added Martin Vesely If I understood it correctly, you put $X$ on qubit $q_1$. If there is no noise, the result should be $10$ always. Accroding to results you posted, it seems that after error correction probability of measuring $10$ increased. Probability of measuring $00$ which is result of spontaneus relaxation from state $10$ decreased significantly. So you are on good way to correct for noise. Of course, you are not able to eliminate noise completely but this is OK as quantum systems are probabilistic.
Mar 13 '20 at 12:23 comment added Vitality Following your answer and the comment by @DavitKhachatryan, I have prepared the code in the question edit. Is the achieved result what I should expect in practice?
Mar 10 '20 at 19:20 history answered Martin Vesely CC BY-SA 4.0