1
$\begingroup$

I'm reading this old post surface code and the second answer under error detection and correction it talks about measuring "every stabilizer". It is known that of the $d^2$ $X$ stabilisers only $d^2-1$ are independent (same for $Z$ stabilisers). Do you measure all $d^2$ or just $d^2-1$? If it's the latter then this seems to break the geometric symmetry of the stabilizers : one would need to be skipped (for example the one in the bottom left corner...). What if one of the anyon pairs land on that omitted site?

$\endgroup$

1 Answer 1

1
$\begingroup$

It really doesn't matter. It might be that something in your implementation makes the symmetric implementation easier, or you may just prefer to minimise the total number of operations.

If one anyon from a pair lands on the unmeasured stabilizer, you know it: there must always be an even number of anyons, so if you see an odd number, you know there's one on the region you didn't look at.

$\endgroup$
3
  • $\begingroup$ This sounds right for the surface code. The redundant stabilizer is the product of all the rest so if the error anti-commutes with two of the rest it will commute with the product. I think for general codes things can get more complicated $\endgroup$
    – unknown
    Jun 15, 2022 at 20:09
  • $\begingroup$ No, it isn't any more complicated (at least for stabilizer codes). You're still looking at the case where there's a bunch of stabilizers that product to identity. This means that the product of measuring all of them is +1. If one is missing, its value is the same as the product of all others. That's all there is to it. $\endgroup$
    – DaftWullie
    Jun 16, 2022 at 8:06
  • $\begingroup$ For a general code the product of the stabilizers is not necessarily identity. You could also have more than one redundant stabilizer. At any rate I think I see the general point : you can express the redundant ones in terms of any set of complete independent stabilizers and work out how the error commutes with the full group...thanks for your answer. $\endgroup$
    – unknown
    Jun 16, 2022 at 15:02

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.