0
$\begingroup$

I am currently trying to understand this paper: https://arxiv.org/abs/2302.07395. I understand that the goal is to move the twist around to perform the Y measurement/phase gate. I see the step by step circuit construction in the paper, it's very helpful. I was wondering if one could get away with performing less gates if we can use neutral atom platforms and just shuttle atoms across? Is there any way of using reconfigurable platforms and modify this scheme to perform the Y measurement or S gate with less time overhead?

Thanks!

$\endgroup$

1 Answer 1

0
$\begingroup$

In the introduction of the paper, it's noted that if you have non-local connectivity then you can perform $R_Y$ and $M_Y$ in constant depth by using a folded surface code. The circuit for doing so is included in the zenodo upload, e.g. r=15,d=15,p=0.003,noise=SI1000,b=Y_folded,rb=0,q=479.stim in circuits.zip.

Here's a simpler version of the circuit (not entirely compiled into CZs) in crumble. It's "just" a (d+1)*d surface code with a layer of Ss along the diagonal and CZs mirrored+shifted across the diagonal at the end of the first round.

enter image description here

$\endgroup$
3
  • $\begingroup$ Thanks for this. Any intuitions behind the staggered/shifted CZs. I recalled in Moussa's constructions with the un-rotated surface code, you perform the CZ across the reflection without the shift. Is this mirrored+shifted approach to performing a Y measurement well known in the literature for the rotated surface code? $\endgroup$
    – user33129
    Commented Nov 19 at 22:11
  • $\begingroup$ @user33129 The shift is for the two body operators to line up correctly, so that they pair up across the fold. $\endgroup$ Commented Nov 20 at 0:02
  • $\begingroup$ Thanks, this is just the first time I have seen this construction. I guess it's not known in the literature previously (for the rotated surface code)? I believe Moussa has to elongate his patch to perform the S gate fold-transversally for the rotated surface code. $\endgroup$
    – user33129
    Commented Nov 20 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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