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 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 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 at 6:51
  • $\begingroup$ Ah, I see. Thanks. $\endgroup$ Jun 7 at 8:41
  • 1
    $\begingroup$ Probably very relevant: arxiv.org/abs/2001.04887 $\endgroup$
    – dabacon
    Jun 9 at 22:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.