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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 $T$?

FDGod
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unknown
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  • I think this is impossible due to the Eastin–Knill theorem: https://en.wikipedia.org/wiki/Eastin%E2%80%93Knill_theorem – Victory Omole Jun 07 '22 at 05:47
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    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". – DaftWullie Jun 07 '22 at 06:37
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    @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). – DaftWullie Jun 07 '22 at 06:51
  • Ah, I see. Thanks. – Victory Omole Jun 07 '22 at 08:41
  • @DaftWullie I saw triorthogonal codes mentioned here https://arxiv.org/pdf/1910.09333.pdf where they also mention quantum reed muller codes. I had trouble checking the results of that paper...I'll probably post some (possible) discrepancies I saw in other questions – unknown Jun 07 '22 at 14:27
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    Probably very relevant: https://arxiv.org/abs/2001.04887 – dabacon Jun 09 '22 at 22:34
  • The simplest/ most well known example with transversal $ T $ is the Reed-Muller type $ [[15,1,3]] $ CSS code. As noted above this code is triorthogonal and not self dual. It is specifically mentioned in the link from dabacon. Just wanted to add a nice concrete example. – Ian Gershon Teixeira Jun 29 '22 at 13:56

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The $ 11 $ qubit $ d=3 $ error-correcting quantum code we find here

https://arxiv.org/abs/2310.17652

is probably the smallest code with transversal $ T=\begin{bmatrix} 1 & 0 \\ 0 & e^{i \pi/4}\end{bmatrix} $.

It has codewords $$ |\overline{0} \rangle= \sqrt{\frac{5}{16}} | D^{11}_0 \rangle + \sqrt{\frac{11}{16}} | D^{11}_8 \rangle $$ and $$ |\overline{1} \rangle = \sqrt{\frac{5}{16}} | D^{11}_{11} \rangle + \sqrt{\frac{11}{16}} | D^{11}_3 \rangle $$ where $ | D^{11}_k \rangle $ represents a normalized uniform superposition over all $ 11 $ qubit computational basis kets of weight $ k $. So $ | D^{11}_0 \rangle =|00000000000 \rangle $ while $ | D^{11}_8 \rangle $ is $ 1/\sqrt{\binom{11}{8}} $ times the uniform sum of all $ \binom{11}{8} $ many of the weight $ 8 $ basis kets. Similarly for $ | D^{11}_{11}\rangle $ and $ | D^{11}_3 \rangle $

The logical $ T $ gate for this code is implemented by $ (T^3)^{\otimes 11} $.

Eric Kubischta
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Ian Gershon Teixeira
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  • interesting paper! It was straight forward to verify that $(T^3)^{\otimes 11}$ acts as $[[1,0],[0,E(8)]]$ on $[\bar 0,\bar 1]$. Since this is not a stabilizer code, I don't have any existing code to check the distance so I couldn't verify the $d=3$ part. How did you check the distance? – unknown Mar 29 '24 at 21:59