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Consider an $ [[n,k,d]] $ code. Take the image of the codespace under some Clifford gate $ C $. Does the new code have the same distance?

If $ C $ is a tensor product of single qubit Cliffords, such as $ H $ and $ P $, the answer is yes. Indeed any tensor product of single qubit unitaries preserves distance.

However what about an entangling gate like $ CNOT $? Does the image of the codespace under a $ CNOT $ always have the same distance?

EDIT:

Reflecting on the answer from squiggles a nice counterexample is the following: Consider the $ [[4,2,2]] $ code with stabilizer generators $ XXXX,ZZZZ $. Applying the Clifford gate $ CNOT $ to this code clearly does not preserve distance since the image under $ CNOT $ gives a $ [[4,2,1]] $ stabilizer code with stabilizer generators $ XIXX,IZZZ $.

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No. For any two $[[n, k]]$ stabilizer codes there is a Clifford mapping between them.

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  • $\begingroup$ Are you sure that is true? That would be saying something like the Clifford group acts transitively on all abelian subgroups of the Pauli group of the same size. Do you have a reference or proof? I guess if you rephrase it as the symplectic group $ Sp(2n,2) $ acts transitively on all dimension $ k $ isotropic subspaces of $ 2^{2n} $ then it sounds kind of obvious. But do you know a more direct/ quantum information style proof? $\endgroup$
    – Ian Gershon Teixeira
    Apr 6 at 19:56
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    $\begingroup$ Sure, you can find a constructive proof here: arxiv.org/abs/quant-ph/9607030. For stabilizer codes $S_1$ and $S_2$ with encoding circuits $C_1$ and $C_2$ (from a canonical input state), you can explicitly convert between the codes by applying $C_2C_1^{-1}$. $\endgroup$
    – squiggles
    Apr 6 at 20:25
  • $\begingroup$ @squiggles but this doesn't show that $C_2 C_1^{-1}$ preserves distance....otherwise all codes are equivalent to the trivial code $\endgroup$
    – unknown
    Apr 7 at 1:56
  • $\begingroup$ It definitely doesn’t preserve distance. $\endgroup$
    – squiggles
    Apr 7 at 6:45

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