2
$\begingroup$

There are certain codes which have transversal non-Clifford gates. My question is, what are the simplest examples of such codes and how does one check for a non-Clifford gate? Is it a computationally intensive question, compared to checking for Clifford gates?

$\endgroup$
2
  • $\begingroup$ Welcome to QCSE! What do you mean by "checking for a non-Clifford gate" and "checking for a Clifford gate"? $\endgroup$ Jan 3, 2021 at 21:21
  • $\begingroup$ The Most well known example is the $ [[15,1,3]] $ Reed-Muller type CSS code. This code is triorthogonal but not self dual. Its transversal gates are all generated by $ X,T, CNOT $. $\endgroup$ Jun 29 at 14:04

1 Answer 1

2
$\begingroup$

Here's a semi-stupid (partial) answer: the $k$ qubit majority vote codes, which have logical codewords $$ |0_L\rangle=|0\rangle^{\otimes k},\qquad |1_L\rangle=|1\rangle^{\otimes k}. $$ Of course, logical $X$ is just $X^{\otimes k}$. However, for phase gates, you can apply and logical version of $R_z(\theta)$ just by applying $R_z(\theta/k)$ on every qubit. Of course, it's not much use as an error correcting code if you have to protect against phase noise because a single-qubit phase gate is also a logical phase gate.

Part of your question asks how you can check for the existence of a non-Clifford transversal gate (or, at least, that is how I interpret your question). As a general rule, I would expect this to come out by construction - you design your code with respect to certain parameters, and if this is the one you're interested in, you've probably built it in by construction. Anyway, perhaps the best starting point is the Eastin-Knill theorem - if you've got a set of transversal operations that only require one more gate to make them universal, that additional gate cannot be implemented in a transversal manner. This tends to rule out a vast array of logical operations that you might search for. For instance, say that I know how to do logical H and X transversally. I want to know what phase gates I might be able to implement. Certainly T is impossible because that would let you implement a continuous set of gates. As would any phase angle that is not a rational multiple of $\pi$. Most other rational fractions of $\pi$ are also impossible (off the top of my head, I don't know the exact characterisation, but it may be that the only options open are S, Z and S$^\dagger$).

$\endgroup$
1
  • $\begingroup$ About the example with transversal $ X, H $. $ X, H $ in the projective unitary group $ PU_2 $ generate a dihedral group of order 8. So if there are other transversal gates then either the whole group of transversal gates is the single qubit Clifford group in $ PU_2 $ (in this case meaning $ S $ is transversal, exactly the option DaftWullie mentioned) or the group of transversal gates is a dihedral group of order $ 8n $ in $ PU_2 $. Note that $ n $ can be arbitrarily large, it just means you picked some really contrived code that can transversally implement an $ n $th root of the Hadamard. $\endgroup$ Jun 1 at 19:02

Your Answer

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

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