Doubt about a surface-code paper maybe-typo

Doubt about a surface-code paper maybe-typo Ask Question Asked today Modified today Viewed 30 times 2 To brush up surface codes I'm reading "The Hitchhiker’s Guide to the Surface Code" by Fang Zhang and Jianxin Chen (https://doi.org/10.3390/e28020251) On page 15 and 16 it deals with 3D MWPM taking into account not only qubit errors but also ancilla measurement errors. In the last lines of page 16 and the first of page 17, dealing with short cycles between consecutive layers as difference of equally valid errors interpretations, it states: This also means that we do not care if such a cycle is the difference between the actual error configuration and our correction. For example, suppose that P 1 𝑃 1 and P 2 𝑃 2 are two adjacent stabilizers, and the set of non-zero detectors we observe is { Δ P 1 ,1 ∆ 𝑃 1 , 1 , Δ P 2 ,2 ∆ 𝑃 2 , 2 }, meaning that the measurement results of both P 1 𝑃 1 and P 2 𝑃 2 have flipped once each, but P 2 𝑃 2 flipped one round later than P 1 𝑃 1 . This can be interpreted in two ways: • The qubit error that flips { P 1 𝑃 1 , P 2 𝑃 2 } occurred after round 1, but P2 was measured incorrectly in round 2, thus not properly reflecting the flip until round 3. • The qubit error that flips { P 1 𝑃 1 , P 2 𝑃 2 } occurred after round 2, but P1 was measured incorrectly in round 1, thus seemingly flipping one round earlier. We cannot distinguish between these two cases, but fortunately we do not need to. My doubt is that, in second interpretation, P1 should have been incorrectly measured in round 2, not round 1. Infact: Stabilizer P 𝑃 measurement result at round i 𝑖 is defined as r P,i 𝑟 𝑃 , 𝑖 Stabilizer detector Δ P,i ∆ 𝑃 , 𝑖 is defined as Δ P,i = r P,i ⊕ r P,i+1 ∆ 𝑃 , 𝑖 = 𝑟 𝑃 , 𝑖 ⊕ 𝑟 𝑃 , 𝑖 + 1 and contained in layer between round i 𝑖 and round i+1 𝑖 + 1 Ancilla measurement error at round i 𝑖 flips detectors Δ P,i ∆ 𝑃 , 𝑖 and Δ P,i−1 ∆ 𝑃 , 𝑖 − 1 (a part from boundaries where we have just one detector) A qubit error between round i 𝑖 and round i+1 𝑖 + 1 will flip two stabilizers P 1 𝑃 1 and P 2 𝑃 2 ,and so detectors Δ P 1 ,i ∆ 𝑃 1 , 𝑖 and Δ P 2 ,i ∆ 𝑃 2 , 𝑖 That said: round 1st interpretation PAPER 2nd interpretation MY 2nd interpretation layer-spanning short cycle 3 layer qubit error ⟹ ⟹ Δ P 1 ,2 ∆ 𝑃 1 , 2 , Δ P 2 ,2 ∆ 𝑃 2 , 2 qubit error ⟹ ⟹ Δ P 1 ,2 ∆ 𝑃 1 , 2 , Δ P 2 ,2 ∆ 𝑃 2 , 2 qubit error ⟹ ⟹ Δ P 1 ,2 ∆ 𝑃 1 , 2 , Δ P 2 ,2 ∆ 𝑃 2 , 2 2 P 2 𝑃 2 measure error ⟹ ⟹ Δ P 2 ,2 ∆ 𝑃 2 , 2 , Δ P 2 ,1 ∆ 𝑃 2 , 1 (in layers) P 1 𝑃 1 measure error ⟹ ⟹ Δ P 1 ,2 ∆ 𝑃 1 , 2 , Δ P 1 ,1 ∆ 𝑃 1 , 1 (in layers) P 1 𝑃 1 , P 2 𝑃 2 measure error ⟹ ⟹ Δ P 1 ,2 ∆ 𝑃 1 , 2 , Δ P 1 ,1 ∆ 𝑃 1 , 1 , Δ P 2 ,2 ∆ 𝑃 2 , 2 , Δ P 2 ,1 ∆ 𝑃 2 , 1 (in layers) layer qubit error ⟹ ⟹ Δ P 1 ,1 ∆ 𝑃 1 , 1 , Δ P 2 ,1 ∆ 𝑃 2 , 1 qubit error ⟹ ⟹ Δ P 1 ,1 ∆ 𝑃 1 , 1 , Δ P 2 ,1 ∆ 𝑃 2 , 1 1 P 1 𝑃 1 measure error ⟹ ⟹ Δ P 1 ,1 ∆ 𝑃 1 , 1 (in layer) Please note that, as earlier recapped in point 2, detectors are contained into layers, so detectors flipped by a round measure have to be thought into layers just beneath and just over that round; so, e.g., lower layer in 1st interpretation column will contain Δ P 2 ,1 ∆ 𝑃 2 , 1 two times, and both layers in layer-spanning short cycle column will have every detector twice: it seems to me to be exactly the graphical representation of detectors "mutual cancellation" by means horizontal and vertical edges intersection, given that the layer-spanning short cycle (with all detectors doubled) doesn't produce any visible error trace. So I cannot understand how second column can differs from the first one by means of the last (while imho it works for third column); and even without relying on "graphical thinking", it seems to me that paper 2nd interpretation should result in three non-zero detectors, not only the two stated. Am I wrong? Thank you very much for your help! surface-codeminimum-weight-perfect-matching Share Improve this question Follow asked 5 hours ago baro77 1213 3 bronze badges New contributor Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 0 I’m not the paper’s author, so I can’t say whether this is a typo. That said, your reasoning seems sound, and I’d agree with your interpretation. If you want certainty, contacting the author is probably the best thing you can do. Share Improve this answer Follow answered 1 hour ago Ramos 214 4 bronze badges New contributor Add a comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions surface-codeminimum-weight-perfect-matching See similar questions with these tags. The Overflow Blog Welcome to the “find out” stage of AI Lights, camera, open source! 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