# Expected value of a Haar random quantum state multiplied by a unitary

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.

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*}