For Clifford gates (when performing randomized benchmarking and starting from ground state) the final state is always ground. It is acquired by applying at the end recovery gate, which transfers the state of Clifford gates sequence to the gound state. But how does it work in case we apply non-Clifford gates?

  • $\begingroup$ Perhaps you should make you question more concrete. "Standard" RB always ends with a recovery/inversion gate which, ideally, should give you the initial state back. Are you asking about other RB schemes like interleaved RB? $\endgroup$ Oct 22, 2021 at 11:19
  • $\begingroup$ I don't mind to change the formulation of my question - whatever you like, I just tried to formulate it as simple as possible)) Does the "standard" RB use only Cliffords? I thought, it can be done with any set of gates (not exactly Cliffords). Oh, yes, I'm asking about interleaved RB, but with non-Clifford gates. $\endgroup$
    – Curious
    Oct 22, 2021 at 11:49
  • $\begingroup$ There are simply dozens of different RB schemes -- that's why I wanted you to name the one you're interested in. I would say that "standard" RB means the scheme where the gates are drawn from the Clifford group. However, RB works (in principle) with any group, it might just be more involved to analyse the data. Interleaved RB does not change how the (Clifford) gates are drawn and what "end gate" is applied - it just interleaves every gate with a fixed, usually non-Clifford, unitary. Still, please specify the question. Although the details are involved, papers on interleaved RB can be looked up $\endgroup$ Oct 25, 2021 at 7:27
  • $\begingroup$ @MarkusHeinrich ok, let me clarify few moments before I'll edit the question: first, it is well known, that for any Clifford sequence the resulting inverse matrix always Clifford and applying of which (to the whole sequence) will always transfer the final state to the ground state; second, if I want to apply non-Clifford gates I'd like to ask, how can I estimate the final matrix? what inverse (recovery) gate should I apply to get the probability of success? and what exactly will be the probability of success? $\endgroup$
    – Curious
    Oct 25, 2021 at 16:44
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    $\begingroup$ In the interleaved RB scheme, the end gate is the same as without interleaving. So if you apply random Cliffords, the end gate will still be Clifford. The point is that this is a different scheme, so you're not really interested in returning to the initial state. $\endgroup$ Oct 26, 2021 at 15:36


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