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I believe so (caveat: this is not something I've every thought about before). I'm going to rewrite the $p_x$ from your question as $p_{xy}$. So, we have $$p_{xy}=|\langle x|U_1|0^n\rangle|^2\ |\langle y|U_2|0^n\rangle|^2$$ Note that this is two independent probabilities $p_x$ and $p_y$. Now, the probability that $p_{xy}>\alpha'$, which we write as $\... 2 We use the Haar measure, thus for any unitary$U_0$the distributions of$U$and$UU_0$are the same. Hence, the distributions of$|\psi\rangle = U|0^n\rangle$and$|\psi\rangle = UU_0|0^n\rangle$are also the same. Therefore, it's the same distribution$|\psi\rangle = U|\psi_0\rangle$for any state$|\psi_0\rangle$. So, the standard basis plays absolutely ... 2 Note that the quoted relation $$\bar M_i = \sum_\lambda a_\lambda P_\lambda,$$ only holds if the$M_i$also commute with the representation of the symmetric group! Otherwise this can obviously not be true by a simple counting argument: The dimension of the commutant$U\mapsto U^{\otimes n}$is$n!$but there are certainly less partitions of$n\$. Thus, the ...