Suppose I have a quantum error correcting code with two stabilizers. Does this mean that I can potentially correct at most 3 distinct errors using this code?

My reasoning is that for each error, we need a syndrome. With only two stabilizers, we have the syndromes $(00, 01, 10, 11)$. The first one is the case of no error so that leaves three more errors.

In general, does the number of stabilizer elements bound the number of correctable errors?


1 Answer 1


Yes, and no.

Yes, in the sense that for a non-degenerate code, it does bound this, and this is essentially the content of the Quantum Gilbert-Varshamov Bound. Non-degenerate codes are exactly those ones for which distinct correctable errors must map to distinct syndromes.

No in the sense that there are non-degenerate codes for which different errors have the same syndrome and the same correction. The Shor code is a good example of this: there are sets of 3 qubits for which a $Z$ error on any of the individual qubits gives the same syndrome. You perform any one of the $Z$s as a correction, and whichever one you do, and whichever the single error was, you've got back to the original state.

  • $\begingroup$ This is because the bit flip error is handled by a repetition code, correct? So you just choose the majority vote to correct the bit flip and that's the source of the non-degeneracy. $\endgroup$
    – JRT
    Oct 2, 2023 at 17:12
  • $\begingroup$ Well, usually for the Shor code the $X$ errors are all non-degenerate, and it's the $Z$ errors that have a degeneracy. This basically comes from blocks of qubits being encoded into states like $|000\rangle+|111\rangle$. Applying $Z$ to any one of the three qubits gives the same output. But that's specific to the Shor code, even though the concept might carry over to other degenerate codes. $\endgroup$
    – DaftWullie
    Oct 3, 2023 at 6:18

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