It can be shown that clifford gates do not preserve distance.

My question is what if you restrict to local clifford gates, is distance preserved by these?

(by local I mean that they act on each qubit individually; so tensor product of single qubit gates).

(my guess is that they do not but it will be good to see a proof or reference)

  • $\begingroup$ All local unitaries, Clifford or not, on any code, stabilizer or not, preserve distance $\endgroup$ Commented Mar 4 at 14:25

1 Answer 1


Since local Cliffords (LC) do not change the weight of a Pauli operator, the distance of two LC-equivalent stabilizer codes is the same.

As a remark: Any stabilizer code is also LC-equivalent to a graph code. In fact, LC-equivalence of stabilizer codes can be reduced to checking equivalences of graphs by local complementation (Schlingemann, van den Nest et al.).

  • $\begingroup$ nice argument using weights...I'm not too familiar with the graph code form and the (first) paper looks hard to digest but I'll try. $\endgroup$
    – unknown
    Commented Nov 8, 2023 at 15:58
  • $\begingroup$ @unknown I think the second one is a bit simpler to understand. Anyway, it was just a remark in case you want to check LC-equivalence or you like graphs. $\endgroup$ Commented Nov 9, 2023 at 8:05
  • $\begingroup$ I did some more testing and ran across this potential discrepancy : for $$G=[[-1+\imath,-1+\imath],[1+\imath,-1-\imath]]/2$$ it is clifford but maps $X \to -\imath X Z$ and $Z \to X$ so it wouldn't preserve weights. Any idea what's happening? $\endgroup$
    – unknown
    Commented Nov 9, 2023 at 23:15
  • 1
    $\begingroup$ @unknown $-iXZ=-Y$ so it has weight 1 $\endgroup$ Commented Nov 10, 2023 at 8:45
  • $\begingroup$ yes of course...thanks for the clarification $\endgroup$
    – unknown
    Commented Nov 10, 2023 at 15:32

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