This a follow-up question to Conjugating pairs of Paulis to each other with a Clifford

We call a Clifford gate local if it is a tensor product of single qubit Clifford gates.

We call a Clifford gate non entangling if it is generated by single qubit Clifford gates and permutations.

Let $ A,B $ be two Paulis with the same weight $ wt(A)=wt(B) $. Then there always exists some non-entangling Clifford $ C $ such that $$ CAC^\dagger=B $$

Let $ A_1,A_2 $ be two Paulis which anticommute. Let $ B_1,B_2 $ be another pair of Paulis that anticommute. (Or you can assume that both pairs commute the answer shouldn't change.) Moreover suppose that $ wt(A_1)=wt(B_1) $ and $ wt(A_2)=wt(B_2) $.

Does there always exist some non-entangling Clifford $ C $ such that simultaneously $$ CA_1C^\dagger=B_1 $$ $$ CA_2C^\dagger=B_2 $$

  • 1
    $\begingroup$ Are you allowing permutations as well? $ZI$ and $IZ$ aren’t locally Clifford equivalent. $\endgroup$
    – squiggles
    Commented Apr 8, 2023 at 3:42
  • $\begingroup$ @squiggles great point! I should have said equivalent by local Cliffords + permutations (in other words equivalent by non-entangling Cliffords) $\endgroup$ Commented Apr 8, 2023 at 11:18


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.