I understand that if we have, for example, the blue path doesn't make any check operator detect any error, but what about the red path? We have the check operator detecting since there is just one qubit afect for each operator.
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$\begingroup$ Where is the figure taken from? The boundaries look incorrect. $\endgroup$– Peter-JanMay 17 at 15:18
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$\begingroup$ I did it. Why are the boundaries incorrect? This forms a torus, doesn't it? $\endgroup$– LittleBlueMay 17 at 15:23
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$\begingroup$ It's not immediately clear which qubits at the left (top) boundary you identify with which qubits at the right (bottom) boundary. $\endgroup$– Peter-JanMay 17 at 21:13
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1 Answer
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If the blue path consists of Pauli X terms, it's undetectable (commutes with all stabilizers). If the red path consists of Pauli Z terms, it's undetectable. So on the blue path you can have logical X and on the red path a logical Z.
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$\begingroup$ And if the red path is X errors or the blue path is Z errors? $\endgroup$ May 17 at 15:04
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$\begingroup$ If the red path is X errors, then it's not a logical operator because it doesn't commute with the stabilizers. Same goes for if the blue path is Z errors. $\endgroup$ May 17 at 15:22
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$\begingroup$ Soooo, it's just some horizontal paths that are actually undetectable? Sorry, i am a little confused. $\endgroup$ May 17 at 15:27
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$\begingroup$ They are not undetectable, they anti-commute with stabilizers, so if you measure the stabilizers you will get non-trivial syndrome. $\endgroup$ May 17 at 21:13
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$\begingroup$ This is because all the articles i have read, they say any horizontal loop is udetectable so i thought i might being missing something... In the end if consider all the horizontal and vertical loops, just half of them are undetectable, right? $\endgroup$ May 17 at 21:16