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The $[[5,1,3]]$ code is a perfect code basically meaning that the weight-0 and weight-1 error spaces completely fill out the $32$-dimensional Hilbert space.

On the other hand, the $[[7,1,3]]$ Steane code is not perfect. There are $6$ stabilizer generators amounting to $2^6 = 64$ syndromes and yet the weight 0/1 errors only account for $22$ of these, leaving $42$ unused syndromes!

In practice it seems prudent to assign some of the weight-2 errors to these $42$ unused syndromes. However, there are $3^2 \binom{7}{2}=189$ weight-2 errors so we certainly cannot have a syndrome for all of these errors (which makes sense because in that case the code would have distance $d=5$ instead). This means we must make a choice for which weight-2 errors we wish to correct. I have never seen this done but I am sure that someone has done it somewhere (does anyone have a reference)?

Is there some choice that results in a better logical error rate than some other choice? Of course it depends on the error model, for example if a $Z$ error is more likely than an $X$ or $Y$ error then you probably want a syndrome for each of the weight-2 $Z$-type errors and maybe even some weight-3 $Z$-type errors. But in the special case of a depolarizing channel (where $X$, $Y$, and $Z$ occur with the same probability) I would assume that the logical error rate is invariant under this choice. Am I correct?

Edit (2-29-2024): As discussed in the answer by @ChrisD, there seems to be a canonical choice for weight-2 syndromes for the Steane code owing to the fact that it is a CSS code. However, does this choice lead to the best probability of logical error (or more simply, does this choice lead to the best pseudo-threshold) amongst all possible choices of weight-2 syndromes?

Eric Kubischta
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The error correction protocol for any CSS code (such as the Steane code) is described in e.g. Section 10.4.2 of Nielsen & Chuang, although it's much easier to understand in the stabilizer picture. They key point is that X errors and Z errors are corrected separately, so the Steane code can correct any single qubit $X$ error and any single qubit $Z$ error. This includes all weight-1 pauli errors $X_i$, $Y_i$, $Z_i$ and all weight-2 errors $X_i Z_j$ which, when $i \neq j$, give the 42 unused syndromes you are referring to. What it will not correct is weight-2 errors like $X_i X_j$, $Z_i Z_j$ or $X_i Y_j$ and so on.

ChrisD
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