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Consider a quantity \begin{equation} \mathbb{E}\big[\langle z|\rho|z\rangle\big], \end{equation} where $\rho = |\psi \rangle \langle \psi|$ is a Haar-random state $n$-qubit quantum state and $z$ is the label of a fixed $n$-qubit basis vector. Now, consider \begin{equation} \sigma = \underset{\text{diagonal}~U}{\mathbb{E}}\big[U\rho U^{*}\big], \end{equation} where $\rho$ is as defined before, and $U$ is a diagonal unitary matrix such that the diagonal entries are uniformly random complex phases. I am trying to prove that \begin{equation} \mathbb{E}\big[\langle z|\rho|z\rangle\big] = \mathbb{E}\big[\langle z|\sigma|z\rangle\big] \end{equation}

Intuitively, the result is clear as the Haar measure is invariant under left and right multiplication by a unitary. But, the RHS has two expectations - one nested inside the other - and I do not know how to simplify that.

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With the chosen structure of $ U $, i think it's even possible to prove the stronger statement: $$ \langle z| \rho|z \rangle = \langle z| \sigma_\rho|z \rangle, \hspace{0.2em} \text{where} \hspace{0.2em} \sigma_\rho = \mathbb{E}_U \big[U\rho U^\dagger\big] \text{and} \hspace{0.3em} |z\rangle \hspace{0.3em} \text{a computational basis vector.}$$ You may write $ U = \sum_{k \in \{0, 1\}^n} e^{i \phi_k} |k\rangle \langle k| $, for uniform $ \phi_k \in_R [0, 2\pi] $ and calculate \begin{align*} U\rho U^\dagger &= \sum_{k, m} e^{i (\phi_k - \phi_m)} \langle k|\rho|m \rangle \cdot |k\rangle \langle m| \implies \\ \mathbb{E}_U \big[U\rho U^\dagger\big] &= \sum_{k, m} \mathbb{E}_\phi \big[e^{i (\phi_k - \phi_m)}\big] \cdot \langle k|\rho|m \rangle \cdot |k\rangle \langle m| \implies\\ \langle z |\sigma_\rho| z \rangle &= \sum_{k, m} \mathbb{E}_\phi \big[e^{i (\phi_k - \phi_m)}\big] \cdot \langle k|\rho|m \rangle \cdot \langle z|k\rangle \langle m|z \rangle \implies\\ \langle z |\sigma_\rho| z \rangle &= \sum_{k, m} \mathbb{E}_\phi \big[e^{i (\phi_k - \phi_m)}\big] \cdot \langle k|\rho|m \rangle \cdot \delta_{z,k} \delta_{z,m} = \langle z|\rho|z \rangle \end{align*}

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