# Conjugating pairs of Paulis to each other with a non-entangling Clifford

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 $$CA_1C^\dagger=B_1$$ $$CA_2C^\dagger=B_2$$

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