1
$\begingroup$

Let me know if this question is too broad:

I'm learning more about enumerator polynomials for quantum error correcting codes (QECCs). What is the relationship between enumerator polynomials and the distance $ d $ of an $ [[n,k,d]] $ quantum code?

For example is there any condition in terms of enumerator polynomials that is sufficient to imply that $ d \geq 2 $ (error detecting) or $ d \geq 3 $ (error correcting)?

Update: here is my understanding of the method described by @unknown :

To any subgroup $ G $ of the the group $ P_m $ on $ m $ qubit Pauli operators we can assign a polynomial called the weight enumerator which has the form $$ p_G:=\sum_{i=1}^m n_i x^{n_i} $$ where $ n_i $ is the number of weight $ i $ Pauli operators (mod global phase, this restriction is important because, for example, it means the degree 0 coefficient is always 1) in the subgroup $ G $.

Let $ V $ be a code space encoded into $ m $ physical qubits. Then let $ N $ be the subgroup of $ P_m $ that maps $ V $ back into $ V $. Let $ S $ be the subgroup of $ P_m $ that stabilizes every element in $ V $. Then by definition the distance of the code is the minimum weight of an element in $ N\S $. Equivalently that is the minimum degree of the polynomial $$ p_N-p_S $$ note that $ S $ is always the Pauli stabilizer of the code while $ N $ is in general not the normalizer of $ S $ if the code is not a stabilizer code. But this setup seems to work in general, even if the code is not a stabilizer code.

$\endgroup$

1 Answer 1

1
$\begingroup$

For a classical code the enumerator gives the distance directly $p_C = 1 + n_d x^d + \cdots$. So the first nonzero power is the distance; $n_d$ is the number of codewords at that disatance. For a qauntum code there are two polynomials at play : $p_H$ for the stabilizer code and $p_N$ for its normalizer. $p_N-p_H = n_d x^d + \cdots$ gives you the distance. (Assuming linear classical codes and stabilizer codes here)

Here's an example for the $[[5,1,3]]$ code :

The stabilizer is generated by 4 elements; calculating the weights of the corresponding (16 element) group gives $p_H = 1 + 15 x^4$. Note sum of coeffecients is 16.

The normalizer is generated by 6 elements (4 stabilizers + 2 logicals); calculating the weights of the corresponding (64 element group) gives $p_N=1+ 30 x^3 + 15 x^4 + 18 x^5$. Note sum of coefficients is 64.

So $p_N - p_H = 30x^3 +18 x^5$ and the distance is 3.

$\endgroup$
3
  • $\begingroup$ could you say more about the definition of $ p_H $ and $ p_N $? $\endgroup$ Dec 11, 2022 at 21:04
  • $\begingroup$ I@IanGershonTeixeira I added an example that should clear things up I hope $\endgroup$
    – unknown
    Dec 11, 2022 at 21:50
  • $\begingroup$ ok I think I understand your approach it seems like it should work even for non stabilizer codes. I described your method as I understand it and posted the description as an update to my question $\endgroup$ Dec 11, 2022 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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