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What are the transversal gates of the perfect 5-qubit code ( [[5,1,3]] code) ?

Since this is a stabilizer code we have transversal Pauli gates.

Are there any others?

I am primarily interested in single qubit gates that are transversal for a single block of the code. But I would also be very curious to see any two qubit gates which are transversal for two blocks of the code (and higher for three blocks of the code etc... but I would imagine that is very hard to find).

I know that $ Z $ is transversal because $ |0_L> $ is a superposition of even parity bit strings and $ |1_L> $ is a superposition of odd parity bit strings.

However the phase gate $ S=\begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} $ is not transversal since $ |0_L> $ is not doubly even. See Transversal logical gate for Stabilizer (or at least Steane code)

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  • $\begingroup$ IIRC it's only the Pauli gates. $\endgroup$ Commented Jun 29, 2022 at 14:30

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This paper (page 4) lists Paulis + $HS$ gate.

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  • $\begingroup$ Very interesting! I would be very interested in some sort of conceptual reason why that true. Other than checking directly on logical basis states do you have any idea why that would be true? $\endgroup$ Commented Jun 29, 2022 at 21:59
  • $\begingroup$ I'm afraid not. This paper arxiv.org/pdf/1603.03948.pdf (page 7) puts the HS gate in a family of 8 "octahedral gates"...all are transversal for the $[[5,1,3]]$ code. $\endgroup$ Commented Jun 29, 2022 at 22:13
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    $\begingroup$ Looked into it turns out transversal $ HS $ implements logical $ -HP $ which is interesting. If you want something that implements itself you can have $ \tilde{HP} $ to be the determinant 1 version of $ HS $ and then transversal $ \tilde{HP} $ implements logical $ \tilde{HP} $ . $\endgroup$ Commented Jul 6, 2022 at 22:17
  • $\begingroup$ sorry that I kept switching between $ S $ and $ P $ in my notation I just mean the phase gate $ S=P=diag(1,i) $ $\endgroup$ Commented Jul 7, 2022 at 20:57
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As unknown's answer already says, it's Paulis + all 8 octahedral Clifford gates ($HS$ being one of them).

I will explain why this is the case (leaving out the Pauli part, since all stabilizer codes have transversal Paulis).

Why the octahedral gates

The Paulis map cyclicly into each other (with possible additional $(-1)$ factors) by conjugation via any octahedral gate (and the octahedral gates are the only gates that do so). For example for $HS$

$$(HS) X (HS)^\dagger = - Y$$ $$(HS) Y (HS)^\dagger = - Z$$ $$(HS) Z (HS)^\dagger = X $$

The 8 octahedral gates induce the 8 possible cyclic permutations where $0$ or $2$ minus signs get introduced for the Paulis.

Now, what does this cyclic transformation have to do with the $[[5,1,3]]$ code? To see this, let's not only consider its 4 stabilizer generators, but rather all of its $2^4 = 16$ stabilizers:

$$XZZXI$$ $$YXXYI$$ $$ZYYZI$$

and cyclic permutations in the 5 qubits for each of those 3 stabilizers. This gives 15 stabilizers, +1 for the identity. Notice that transforming $X \to -Y$, $Y \to - Z$, $Z \to X$ transforms any stabilizer into another stabilizer. This works for any octahedral gate induced transformation.

It follows that conjugating a stabilizer with a transversal version of an octahedral gate (like $(HS)^{\otimes 5}$) gives another stabilizer.

This means that the octahedral gates are in the normalizer of the stabilizer.

What remains to show is that the effect of these transversal versions of the octahedral gates on the logical Paulis is the same as for the single qubit case, but that is rather trivial and I will leave it out here.

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