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A non-entangling gate on $ n $ qubits is an element of the group $$ N\Big(\bigotimes_{i=1}^n U(2)\Big)=\bigotimes_{i=1}^n U(2) \rtimes S_n $$ which is generated by $ U(2) $ acting locally on each qubit as well as all $ SWAP $ gates between qubits.

Let $ C_1,C_2 $ be $ [[n,k,d]] $ quantum codes. In other words, $ n_1=n_2, k_1=k_2,d_1=d_2 $. If $ C_1, C_2 $ are equivalent by non-entangling gates then they have the same weight enumerator.** Is the converse true? That is, if $ C_1,C_2 $ have the same weight enumerator must they be related by a non-entangling gate?

Note that

  • the weight enumerator already gives $ d $ (it is the least degree of a nonzero term in $ B(x)-A(x) $, cf. this post
  • $ A(x) $ determines $ B(x) $ by quantum McWilliams identity and gives $ n-k $ (because $ 2^{n-k}=A(1) $, cf. this post) so two different codes on $ n $ qubits with the same weight enumerator must have the same parameters $ [[n,k,d]] $.

**For a reference for this fact see top of page 4 in Quantum Weight Enumerators which in turn refers to the discussion below equation (9) in Quantum MacWilliams Identities

Frederik vom Ende
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Ian Gershon Teixeira
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No. There are 16 inequivalent $[[7,1,3]]$ codes but they correspond to only 10 weight enumerators. For reference see https://arxiv.org/pdf/0709.1780.pdf.

Eric Kubischta
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