1
$\begingroup$

So I know that, for a stabilizer code, the stabilizer group $S$ has $n-k$ commuting generators.

Is there a general way of knowing what the order of the full group of $S$ is, aside from writing out all the possible combinations of its generators $g_{1}, \dots, g_{n-k}$ which give rise to different/unique stabilizers?

My intuition says no, because $|S|$ is dependent on the stabilizer generators, which are different for each code. For example, for one code $g_{1}g_{2}=g_{3}g_{4}$, and so, we cannot count this stabilizer twice in $S$. However, for another code $g_{1}g_{2} \neq g_{3}g_{4}$, giving two new stabilizers to $S$. However, I would be grateful if somebody could confirm/debunk my suspicions.

$\endgroup$
2
  • 2
    $\begingroup$ If I understand correctly, the cardinality of a full stabilizer group is equal to $2^{n-k}$ $\endgroup$
    – Yunzhe
    Commented Jun 7 at 10:54
  • $\begingroup$ Yes, I think that I was missing the fact that each unique multiplicative combination of stabilizer generators will give rise to unique stabilizer elements, $\endgroup$
    – am567
    Commented Jun 7 at 11:05

1 Answer 1

2
$\begingroup$

For a $[[n, k, d]]$ stabilizer code, the $n - k$ stabilizer generators are not only commuting, but they are also independent.

Taking into account that $g^2 = 1$ for each generator $g$, we have the cardinality $$|S| = 2^{n - k}$$

$\endgroup$

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.