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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 $U_g\tau_iU_g^\dagger=\tau_{g(i)}$ ?

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  • $\begingroup$ 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 $\endgroup$ Jun 9, 2023 at 9:43

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Nichola
    Jun 13, 2023 at 3:55

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