A previous post transversal P (phase) gate shows that codes where all stabilizer elements have weights that are multiple of 4 will have a transversal $P$ gate. "Transversal" seems to have multiple definitions; here a gate $G$ is transversal for the code with stabilizer $S$ if it is in the normalizer of $S$ in the full unitary group : $G \in U_n : G' S G = S$. (note $G$ need not be in the clifford group). This is shown using $P^\dagger X P = \imath XZ, P^\dagger Z P = Z;$. At first guess I thought that codes where all stabilizer elements have weight that's a multiple of 8 would have a transversal $T$ gate. ($T^2=P$); but $T^\dagger X T = w_8 P Z X; T^\dagger Z T=Z$; where $w_8^8=1$. So it's not obvious if a similar argument can be used. Does anyone know how to extend the result for $P$ gate to $T$?

  • $\begingroup$ I think this is impossible due to the Eastin–Knill theorem: en.wikipedia.org/wiki/Eastin%E2%80%93Knill_theorem $\endgroup$ Jun 7, 2022 at 5:47
  • 2
    $\begingroup$ Yes, but it's a part of an active research project of mine that I wouldn't want to give too much away on! In short, look for "triorthogonal codes". $\endgroup$
    – DaftWullie
    Jun 7, 2022 at 6:37
  • 1
    $\begingroup$ @VictoryOmole this doesn't violate Eastin-Knill. It just tells you that you cannot have transversal Hadamard (which in turn says that the code cannot be a self-dual CSS code). $\endgroup$
    – DaftWullie
    Jun 7, 2022 at 6:51
  • $\begingroup$ Ah, I see. Thanks. $\endgroup$ Jun 7, 2022 at 8:41
  • 1
    $\begingroup$ Probably very relevant: arxiv.org/abs/2001.04887 $\endgroup$
    – dabacon
    Jun 9, 2022 at 22:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.