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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 ...


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