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I'm reading Horsman et al. "Surface code quantum computing by lattice surgery" and I'm wondering about the rotated surface code.

Consider Figure 13:

enter image description here

This is supposed to have distance 5. But in (c), an $X$ error on the top left qubit and the qubit beneath it would be undetectable (brown plaquettes are $X$-stabilizer measurements). If there were a stabilizer measurement on every outside edge, this problem wouldn't occur. What am I misunderstanding?

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3 Answers 3

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an X error on the top left qubit and the qubit beneath it would be undetectable

That error would be detected by the flipping of the four body Z stabilizer adjacent to the lower qubit you operated on:

enter image description here

If you or I got X and Z mixed up, then the error is undetectable, but it corresponds to a topologically trivial cycle from a boundary to itself, so it has no effect on the logical qubit:

enter image description here

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The $X$ error on the top left qubit and the qubit beneath it indeed produce the same measurement syndrome, but this does not mean they are uncorrectable. Lets call $X$ on the top left qubit $X_{\text{topleft}}$ and $X$ beneath it $X_{\text{beneath}}$. Then, $X_{\text{topleft}} X_{\text{beneath}}$ is a stabilizer and acts as the identity gate on the logical qubit.

In the case of a single error $X_{\text{topleft}}$ (or $X_{\text{beneath}}$) we can choose to correct both errors with either $X_{\text{topleft}}$ or $X_{\text{beneath}}$. Both operations $X_{\text{either}}X_{\text{topleft}}$ and $X_{\text{either}}X_{\text{beneath}}$ being logical identities.

Take a look at: How does Surface-17 tell apart Z errors on Db and Dc?

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  • $\begingroup$ How can I see that $X_{topleft}X_{beneath}$ is a stabilizer? $\endgroup$
    – Sam Jaques
    Commented May 6, 2020 at 17:12
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I guess you are confusing Stabilizer types.

Brown plaquettes(X Stabilizer) & Yellow plaquettes(Z Stabilizer)

So if there is Error $Z_{\text{topleft}}$ or Error $Z_{\text{beneath}}$, it gives the same Syndrome. If this is your question,

Imagine if there are only $Z$ Errors occurred on the Rotated Surface Code ( Figure c).

$Z_{\text{topleft}}$ $Z_{\text{beneath}}$ is one of stabilizer. So we can choose $Z_{\text{topleft}}$ or $Z_{\text{beneath}}$ as a correction operator

  1. $Z_{\text{topleft}}$ Error
  • Correction Operator ($Z_{\text{topleft}}$)

    $Z_{\text{topleft}}Z_{\text{topleft}}$ = $I$

  • Correction Operator ($Z_{\text{beneath}}$)

    $Z_{\text{topleft}}Z_{\text{beneath}}$ -> Stabilizer

  1. $Z_{\text{beneath}}$ Error
  • Correction Operator ($Z_{\text{topleft}}$)

    $Z_{\text{beneath}}Z_{\text{topleft}}$ -> Stabilizer

  • Correction Operator ($Z_{\text{beneath}}$)

    $Z_{\text{beneath}}Z_{\text{beneath}}$ = $I$

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