Consider arbitrary unit vectors $w,x,y,z \in \mathbb{C}^d$. Is there an explicit formula for what this average is?
$$
 \int \mathrm{Tr}( \psi \psi^\* \, \, w x^\* \,\, \psi \psi^\* \,\, y z^\*) d\psi
$$
where the average is over a Haar-random unit vector $\psi \in \mathbb{C}^d$. Here, $\psi \psi^\*$ is the rank-one matrix formed by the outer product of $\psi$ with itself. I'm looking for a formula that relates this average to inner products between $w, x, y, ,z$, for example.


A related question I'm wondering about is what the expected value of the following inner product is: suppose you pick a Haar-random vector $\psi$, followed by a Haar-random vector $\psi^\perp$ that's orthogonal to $\psi$, and then look at $\langle \psi, x \rangle \langle y, \psi^\perp \rangle$. 


Thanks!



Averages of vector inner products over the Haar measure
$\psi\_\alpha$, $\alpha=1,2,\ldots d$, is a column vector of a $d\times d$ unitary matrix $U$; averaging over the Haar measure gives
$$\int d\psi\, \psi\_{\alpha} \psi^\ast\_\beta \psi\_{\alpha'}\psi^\ast\_{\beta'}=\frac{1}{d+d^2}\left(\delta\_{\alpha\beta}\delta\_{\alpha'\beta'}+\delta\_{\alpha\beta'}\delta\_{\alpha'\beta}\right)$$


so your integral equals


$$\int \mathrm{Tr}( \psi \psi^\* \, \, w x^\* \,\, \psi \psi^\* \,\, y z^\*) \,d\psi=
\langle z|\psi\rangle\langle \psi|w\rangle\langle x|\psi\rangle\langle \psi|y\rangle$$
$$\qquad\qquad = \frac{1}{d+d^2}\bigl(\langle z|w\rangle\langle x|y\rangle+ \langle z|y\rangle\langle x|w\rangle\bigr)$$


in answer to your second question: the average of $\langle\psi|x\rangle\langle y|\psi'\rangle$ with $\psi,\psi'$ two different column vectors of $U$ is zero (obviously, because $\psi,\psi'$ and $-\psi,\psi'$ are equally probable)


