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

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

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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.

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  • $\begingroup$ I know about Schur's Lemma but I fail to see why should the integral commute with all unitaries, is there some simple argument for that that I'm missing? $\endgroup$ Apr 14 at 7:33
  • $\begingroup$ 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. $\endgroup$ Apr 14 at 13:37

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