# Are permutations of the Pauli strings unitary operations?

Consider the set of Pauli strings $$P_N=\{\tau \}$$, composed out of tensor products of Pauli matrices $$\sigma_i^\alpha$$ acting on $$N$$ or qubits, e.g. $$\tau=\sigma^x_1 \otimes \mathbb{1}_2 \otimes \sigma^y_3 \otimes \cdots \otimes \mathbb{1}_N$$.

Now consider the set of operations that map a Pauli string to another Pauli string i.e all the operations that can be written as $$f(\tau_i)=\tau_{g(i)}$$ were $$g$$ is a permutation. Because $$g$$ is a permutation, $$f$$ is unitary in the space of the Pauli strings.

But is $$f$$ unitary in the space of the quantum states? i.e for every permutation $$g$$ and every string $$\tau_i$$, is there a $$U_g$$ such that $$U_g\tau_iU_g^\dagger=\tau_{g(i)}$$ ?

• Potentially relevant: (1) The stabilizer formalism & Clifford gates, as those permute Pauli strings. (2) The LU-LC conjecture, see e.g. arxiv.org/abs/0709.1266 Jun 9, 2023 at 9:43

No. This fails because the operation $$U_{g}$$ is not necessarily trace-preserving.

Suppose $$N = 1$$ and $$g(1) = 0$$, i.e. the Permutation that maps $$X$$ to $$\mathbb{I}$$. We thus have $$\mathbb{I} = \tau_{0} = U_{g}\tau_{1}U_{g}^{\dagger}$$.

Then by the cyclic nature of the trace we have:

$$2 = \mathrm{tr}\big[\mathbb{I}\big] = \mathrm{tr}\big[\tau_{0}\big] = \mathrm{tr}\big[U_{g}\tau_{1}U_{g}^{\dagger}\big] = \mathrm{tr}\big[U_{g}^{\dagger}U_{g}\tau_{1}\big].$$

If $$U_{g}$$ where a unitary, we would have $$U_{g}^{\dagger}U_{g} = \mathbb{I}$$, so that we get an inconsistent equation: $$\mathrm{tr}\big[U_{g}^{\dagger}U_{g}\tau_{1}\big] = \mathrm{tr}\big[\tau_{1}\big] = 0$$ but it should also equal two. We can conclude that $$U_{g}^{\dagger}U_{g} \not = \mathbb{I}$$.

If you restrict yourself to the permutations $$g$$ that keeps the all-identity Pauli invariant, then it does become true. We then get a permutation of all traceless Paulis, which is exactly a Clifford operation. The Cliffords are the normalizer of the Pauli group in the unitary group $$U(2^{N})$$ and therefore unitaries themselves.

• Im not sure that the any permutation of traceless Paulis is a Clifford operation. Consider the following: f(1)=1, f(X)=Y, f(Y)=X, f(Z)=Z. Then f(Y)f(Z)=XZ=-iY. But f(YZ)=if(X)=iY. This is not unitary. Jun 13, 2023 at 3:55