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The Eastin-Knill Theorem shows that the transversal gates always form a group and that moreover this group is a finite subgroup of the group of all unitaries.

For many codes, for example all self dual CSS codes, the group of transversal gates is exactly the Clifford group.

https://quantumcomputing.stackexchange.com/a/22226/19675

Does anyone know of codes whose group of transversal gates is significantly different from the Clifford group? For example, does anyone know a code on $ n $ qubits whose group of transversal gates is not isomorphic to a subgroup of the Clifford group on $ n $ qubits?

Or failing that, does anyone know any restrictions on which finite groups can occur as the group of transversal gates of a code?

Update: The answer to

Exotic transversal gate group

seems to show that the transversal gate group of an $ [[n,1,d]] $ stabilizer code, for $ d \geq 2 $, must be generated by Clifford gates and/or $ T_k $ gates. So by the classification of finite subgroups of $ PU_2 $ we can conclude that the transversal gate group must be either a dihedral 2 group $ D_{2^k} $ or $ A_4 $ or $ S_4 $ (cyclic transversal gate group is not possible because we have chosen to specialize to stabilizer codes and thus $ X $ and $ Z $ are both transversal and so they generate a noncyclic Klein 4 subgroup and thus the whole transversal gate group must be non cyclic).

Moreover all these groups can indeed be realized of the transversal gate group of some $ [[n,1,d]] $ , $ d \geq 2 $, stabilizer code. Each dihedral 2-group $ D_{2^k} $ arise as the transversal gate group of the corresponding $ [[2^{k+1}-1,1,3]] $ quantum Reed-Muller code. $ A_4 $ is the transversal gate group of the perfect $ [[5,1,3]] $ code see

What are the transversal gates of the [[5,1,3]] code?

And $ S_4 $ (the single qubit Clifford group) is the transversal gate group of the $ [[7,1,3]] $ Steane code see

Transversal logical gate for Stabilizer (or at least Steane code)

1 Answers1

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If you take the Reed-Muller code of 15 qubits, this is a distance 3 CSS code (so has transversal c-NOT, Z and X) but it also has transversal T (and transversal controlled-S and controlled-controlled-Z). What it doesn't have is transversal Hadamard. You'll find this code properly defined in a bunch of places, but, for example, here is the first one that Google threw at me!

DaftWullie
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