# enumerators polynomials and distance for a quantum error correcting code

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.

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.

• could you say more about the definition of $p_H$ and $p_N$? Dec 11, 2022 at 21:04
• I@IanGershonTeixeira I added an example that should clear things up I hope Dec 11, 2022 at 21:50
• 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 Dec 11, 2022 at 23:27