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.


1 Answer 1


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.

  • $\begingroup$ 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

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