2
$\begingroup$

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)

Improve this question
$\endgroup$
1
  • $\begingroup$ IIRC it's only the Pauli gates. $\endgroup$
    – DaftWullie
    Jun 29, 2022 at 14:30

1 Answer 1

Reset to default
2
$\begingroup$

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

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – Ian Gershon Teixeira
    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. $\endgroup$
    – unknown
    Jun 29, 2022 at 22:13
  • 1
    $\begingroup$ 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} $ . $\endgroup$
    – Ian Gershon Teixeira
    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) $ $\endgroup$
    – Ian Gershon Teixeira
    Jul 7, 2022 at 20:57

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.

Not the answer you're looking for? Browse other questions tagged or ask your own question.