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 $\langle \psi, x \rangle \langle y, \psi^\perp \rangle$.



$\psi_\alpha$, $\alpha=1,2,\ldots d$, 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)

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  • $\begingroup$ Great, thanks! Where did you get the first formula, though? Also, as a followup: would you know what kind of measure of concentration is satisfied by these inner products? That is, what is the probability that $\mathrm{Tr}( \psi \psi^* \, \, w x^* \,\, \psi \psi^* \,\, y z^*)$ deviates from the average by more than some amount $\epsilon$? $\endgroup$ – Henry Yuen Mar 30 '16 at 14:22
  • $\begingroup$ for the first formula, see equation B5 of arxiv.org/abs/cond-mat/9612179 ; measure of concentration: for large $d$ the elements of $\psi$ are independent Gaussians (zero mean, variance $1/d$) $\endgroup$ – Carlo Beenakker Mar 30 '16 at 14:28
  • $\begingroup$ Do you have a more precise estimate of the error (as a function of $d$) when you try to treat the elements of $\psi$ as independent Gaussians? $\endgroup$ – Henry Yuen Mar 30 '16 at 14:37

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