What is the expectation value ${\Bbb E}[\langle\psi,O\psi\rangle]$ over the Haar distribution?

Question

What is the average $\mathbb{E}_{\text{Haar}}|\langle\psi|O\psi\rangle|$ of expectation of an arbitrary observable $O$ over the Haar distribution? Let $d$ be the dimension, i.e, the size of $O$. Do we have something similar to $$\mathbb{E}_{\text{Haar}}\langle\psi|O\psi\rangle=\frac{\text{tr}O}{d}?$$

Answer 1

Since a Haar-random $\lvert\psi\rangle=U\lvert0\rangle$ for a Haar-random $U$, your expectation value equals $$ \langle 0 \rvert \Big[\int \mathrm d U\, UOU^\dagger\Big]\lvert0\rangle\ . $$ The integral in the brackets must be proportional to the identity matrix (that's Schur's lemma -- the identity is the only operator which commutes with all unitaries), and the proportionality constant can easily be determined to be $\mathrm{tr}(O)/\mathrm{tr}(\mathrm{Id})=\mathrm{tr}(O)/d$.

Thus, $$ \langle 0 \rvert \Big[\int \mathrm d U\, UOU^\dagger\Big]\lvert0\rangle = \langle 0 \rvert \frac{\mathrm{tr}(O)}{d} \mathrm{Id} \lvert0\rangle = \frac{\mathrm{tr}(O)}{d}\ , $$ as you indeed suspected.