In a paper that I have recently read, a protocol was given that required a CSS code with certain properties for which all logical Clifford gates and standard measurements have a transversal implementation (the 7-qubit would be an example). Later in that same paper a procedure was required that required the controlled $XP^{\dagger}$ ($C-XP^{\dagger}$) gate to be transversal for that specific code.

Now my question is, is that even possible? As far as I know, the set {Clifford group + a single non-Clifford} gate is universal, and that no stabilizer code allows for transversal implementation of a universal set of gates without any further assumptions. The controlled phase gate $C-P$ is a non-Clifford gate and $C-XP^{\dagger} = CNOT*(C-P^{\dagger})$, so that $C-XP^{\dagger}$ also is a non-Clifford gate. Doesn't this make the requirement that the ($C-XP^{\dagger}$) gate be transversal impossible in the first place?

  • $\begingroup$ just to be clear, when you're talking about C-P, you mean the gate which is diag(1,1,1,i)? $\endgroup$ – DaftWullie Oct 16 '20 at 15:52
  • $\begingroup$ Clifford operations are certainly transversal on the [[7,1,3]] CSS code. $\endgroup$ – Condo Oct 16 '20 at 16:03
  • $\begingroup$ @DaftWullie yes, or, equivalently, C-P^{\dagger} with diag(1,1,1,-i). $\endgroup$ – jgerrit Oct 16 '20 at 18:05
  • $\begingroup$ You're right, this seems to be impossible by the Eastin-Knill theorem. $CP$ is in third level of the Clifford hierarchy. Perhaps they use a more general notion of transversality or some other relaxation. Could you link the paper? $\endgroup$ – Markus Heinrich Oct 17 '20 at 13:09
  • $\begingroup$ Could you give us a link to the paper, just so we can check out the context of what they say a bit more carefully? $\endgroup$ – DaftWullie Oct 18 '20 at 15:32

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