I am reading Nielsen and Chuang's chapter on error correction. I have a very basic question about this theorem:

Theorem 10.8 (Error-correction conditions for stabilizer codes): Let $S$ be the stabilizer for a stabilizer code $C(S)$. Suppose $\{E_j\}$ is a set of operators in $G_n$ such that $E_j^\dagger E_k \notin N(S)-S$ for all $j$ and $k$. Then $\{E_j\}$ is a correctable set of errors for the code $C(S)$.

I'll illustrate my question with an apparent counterexample. Consider the 3-qubit bit flip code, which has stabilizer $S = <Z_1Z_2,Z_2Z_3> = \{\mathbf{1},Z_1Z_2,Z_2Z_3,Z_1Z_3\}$. Let $\{E_j\} = \{Z_1,X_1\}$ be a set of errors. There are only two products to consider, $Z_1X_1$ and $X_1Z_1$. Both of them anticommute with $Z_1Z_2$ and hence are not in $N(S)$ (and therefore not in $N(S)-S$). Theorem 10.8 then seems to conclude that $\{Z_1,X_1\}$ is a correctable set of errors even though $Z_1$ is not a correctable error by the bit flip code ($Z_1 \in N(S) - S$).

What am I missing?


1 Answer 1


You're forgetting to include the identity as a possible "error". When you do, it shows you also have to consider "pairs" $Z_1$ and $X_1$.

  • $\begingroup$ So it's implicit in this theorem that $\{E_j\}$ always needs to include arbitrary identities on any subsystem for it to hold? Like here the 'valid' set of errors to consider would be $\{\mathbf{1}_1,\mathbf{1}_2,\mathbf{1}_{12},Z_1,X_1\}$? $\endgroup$ Feb 29 at 16:52
  • $\begingroup$ Yes, except that all of those identities are the same thing. The point is that you're trying to say any of the possible states that you could end up in are either the same or have distinct error syndromes. Typically, one of those possible states is the state with no error. (On the other hand, you might want to talk about a situation where you are guaranteed that either $X_1$ or $Z_1$ happens. In that case, your treatment shows that you definitely can correct for that.) $\endgroup$
    – DaftWullie
    Mar 1 at 7:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.