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A necessary and sufficient condition for a stabilizer code having transversal $CNOT$ is that the code is a CSS code (see Theorem 11.5 here or the question here).

I know that a sufficient condition for a code having transversal $H$ is that it is a self-dual CCS code (see here)

Is this condition also necessary? If not, is there a necessary condition that captures a family of stabilizer codes having transversal $H$ (like for the $CNOT$ case).

To clarify: I am aware of the necessary conditions that the logical $H$ must be in the normalizer of the stabilizer and that in the code space, it transforms the logical $X$ and $Z$ into each other by conjugation.

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  • $\begingroup$ Great question. Are you also considering stabilizer codes with more than 1 logical qubit? I'm not aware of a k=2 stabilizer code with transversal logical H on each logical qubit. There are trivial examples which are two distinct code blocks, but I'm not aware of an example where the logicals have nontrivial support on the same physical qubits. Finding such a code or showing they do not exist may be a useful step towards answering your question. $\endgroup$
    – Jonas Anderson
    Commented Jun 25 at 12:18
  • $\begingroup$ Yes, I am also considering stabilizer codes with $k\geq 2$ - even if only the blockwise logical Hadamard is transversal. Btw: the [[15,7,3]] Hamming code (errorcorrectionzoo.org/c/stab_15_7_3) is self-dual CSS and hence should have transversal block-wise $H$. $\endgroup$
    – qubitzer
    Commented Jun 25 at 12:40

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