Let $\psi$ and $\phi$ be two uniformly random pure state $\psi, \phi \sim\mathbb{C}^d$. The the following equality holds \begin{align} \mathbb{E}_{\psi, \phi \sim \mathbb{C}^d} {\rm Tr}[\vert \phi \rangle \phi \vert \psi \rangle \langle \psi \vert] = \mathbb{E}_{\psi, \phi \sim \mathbb{C}^d} \vert \langle \phi \vert\psi \rangle\vert^2 = \frac{1}{d} \tag{1}. \end{align} How to prove Eq. (1)?
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3possible duplicate of one of https://quantumcomputing.stackexchange.com/q/28615/55, https://quantumcomputing.stackexchange.com/q/25883/55, or https://quantumcomputing.stackexchange.com/q/28707/55 (observing that by symmetry the question is equivalent to fixing one of the states and only averaging wrt the other one) – glS Oct 09 '23 at 12:43
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Thank you gIS. The Eq. (1) can be obtained using the properties of Haar measure from the above link. – Michael.Andy Oct 10 '23 at 07:22