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

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 $$\mathbb{E}_{\text{Haar}}\langle\psi|O\psi\rangle=\frac{\text{tr}O}{d}?$$

• Certainly not the trace, since the lhs has to be positive. Apr 10, 2022 at 18:31
• You’re right. I edited my question. Apr 11, 2022 at 7:29
• Yes, you can check eq.(11) in this paper nature.com/articles/s41467-018-07090-4.pdf . Apr 11, 2022 at 7:36
• That's an entirely different - and much easier - question! Apr 11, 2022 at 8:58

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.
• Basically Schur's Lemma says that $I$ is the only matrix which transforms trivially under $X\mapsto UXU^\dagger$. And everything which transforms non-trivially will averge out to zero. Apr 14, 2022 at 13:37