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

  • $\begingroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.