Is there benefit to extra stabilizers in a rotated surface code?

1

1

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?

Sam Jaques

Posted 2020-05-06T13:06:05.653

Reputation: 924

Answers

4

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

Craig Gidney

Posted 2020-05-06T13:06:05.653

Reputation: 11 207

1

The 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 on the top left qubit $_{topleft}$ and beneath it $X_{beneath}$. $X_{topleft} _{beneath}$ is a stabilizer and acts as the identity gate on the logical qubit. So we can choose to correct both errors with either $_{topleft}$ of $X_{beneath}$.

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

Peter-Jan

Posted 2020-05-06T13:06:05.653

Reputation: 189

How can I see that $X_{topleft}X_{beneath}$ is a stabilizer? – Sam Jaques – 2020-05-06T17:12:49.883