Is there a simple condition under which X-and-Z-error correctability leads automatically to Y-error correctability?

Question

I had the impression and guess that in a quantum error correction code, once it can correct any single-qubit X and Z errors, it automatically can also correct all single-qubit Y errors. Now after reading these posts post1, post2, I know that this is not true. But I want to ask, is there a simple elegant condition under which it is true? Namely, is there a simple elegant condition, under which once a code can correct any single-qubit X and Z errors, it automatically get the capability to correct all single-qubit Y errors?

Answer 1

As user1936752 said in his comment, "the code being CSS gives you a sufficient condition" (though, not a necessary condition, of course).

This is because CSS codes sort of consist of two classical codes (which may or not be identical) where each code in itself is strong enough in the classical realm to correct at least a single error (if $d \geq 3$, of course). One of them is taken for X-checks, the other is taken for Z-checks. The crucial point is that you construct stabilizers for both codes independently and they thus detect errors independently. By combining them, they can distinguish between X-, Y- and Z-errors.

For example, the Steane code can correct any single X error combined with any single Z error - thus, even a two error combination like this one: $X_2 Z_4$. If X error and Z error happen to be on the same qubit, you effectively have a single Y error.

This procedure does not work for the code of this answer. If you want to look at it through the CSS lense, the Y-checks ($G_0$, $G_1$ and $G_2$) are strong enough (being the classical [7,4,3] Hamming code), but the single X-check is to weak.