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: Define weight enumerators \begin{align*} A_i &= \frac{1}{4^k} \sum_{E \in \mathcal{E}_i} |Tr(E \Pi)|^2 \\ B_i &= \frac{1}{2^k} \sum_{E \in \mathcal{E}_i} Tr( E \Pi E \Pi) \end{align*} Here $\Pi$ is the code projector and $\mathcal{E}_i$ are the Pauli errors with weight $i$. Then the code has distance $ d $ if and only if $$ A_i=B_i $$ for all $ i \leq d-1 $. In particular note that $ A_0=1=B_0 $ always. So the code has distance $ 2 $ if and only if $$ A_1-B_1=0 $$ And has distance $ 3 $ if and only if $$ A_1-B_1=0=A_2-B_2 $$ As noted in the answer below, the distance $ d $ is the degree of the first nonzero term in the polynomial $ B-A $.


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

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