Do the coefficients of the weight enumerator polynomial add up to $2^{n-k}$ for any $[\![n,k]\!]$ code?

Question

Let $ C $ be an $ [\![n,k]\!] $ quantum error correcting code encoding $ k $ logical qubits into $ n $ physical qubits. Define the weight enumerator polynomial $ A(x) $ of the code as $$ A(x):=A_0+A_1x+\dots+A_nx^n $$
where $$ A_j:=\frac{1}{(2^k)^2} \sum_{p \in P_n,\,\mathrm{wt}(p)=j} |\mathrm{tr}(p \Pi)|^2. $$ here $ \Pi $ is the projector onto the code space. Is it the case that $$ A(1)=A_0+A_1+\dots +A_n=2^{n-k} $$ for an arbitrary $ [\![n,k]\!] $ code?

Answer 1

Yes, it is the case that $\sum_{i=0}^nA_i=2^{n-k}$ for any $[\![n,k]\!]$ quantum error correcting code. To see this, begin by expanding the code subspace projector $\Pi$ in the Pauli basis and calculate $$ \begin{align} \Pi &= \frac{1}{2^n}\sum_{p\in P_n}\mathrm{tr}(p\Pi)p\tag1\\ \Pi &= \frac{1}{2^n}\sum_{p\in P_n}\mathrm{tr}(p\Pi)p\Pi\tag2\\ \mathrm{tr}(\Pi) &= \frac{1}{2^n}\sum_{p\in P_n}\mathrm{tr}(p\Pi)^2\tag3\\ 2^k&= \frac{1}{2^n}\sum_{i=0}^n\sum_{p\in P_n,\mathrm{wt}(p)=i}\mathrm{tr}(p\Pi)^2\tag4\\ 2^{-k}&= \frac{1}{2^n}\sum_{i=0}^n\frac{1}{(2^k)^2}\sum_{p\in P_n,\mathrm{wt}(p)=i}\mathrm{tr}(p\Pi)^2\tag5\\ 2^{-k}&= \frac{1}{2^n}\sum_{i=0}^nA_i\tag6\\ 2^{n-k}&= \sum_{i=0}^nA_i\tag7\\ \end{align} $$ as expected.