Im looking for the moments of Haar random states. Is it true that $\textbf{E}_{\psi\sim \text{Haar}}|\langle x| \psi\rangle|^{2t}\sim \frac{1}{\binom{d}{t}}?$ How does one prove this?


1 Answer 1


The factor in your claim is wrong. It should be $\binom{d+t-1}{t}^{-1}$. The correct claim follows from the identity $$ \int |\psi\rangle\langle\psi|^{\otimes t} d\psi = \binom{d+t-1}{t}^{-1} P_{\mathrm{Sym}^t(\mathbb C^d)} \,, $$ where $P_{\mathrm{Sym}^t(\mathbb C^d)}$ is the projector onto the symmetric subspace in $(\mathbb C^d)^{\otimes t}$. Indeed, setting $D_t:=\binom{d+t-1}{t}$, we find $$ \int |\langle x|\psi\rangle|^{2t} d\psi = D_t^{-1} \langle x|^{\otimes t} P_{\mathrm{Sym}^t(\mathbb C^d)}|x\rangle^{\otimes t} = D_t^{-1} \langle x|x\rangle^{t} = D_t^{-1}. $$

So why does the first formula hold? Observe that the integral $A:= \int |\psi\rangle\langle\psi|^{\otimes t} d\psi$ is symmetric under permutation of tensor factors and hence represents an operator on the symmetric subspace $\mathrm{Sym}^t(\mathbb C^d)$ of dimension $D_t=\binom{d+t-1}{t}$. Moreover, $A$ commutes with any $U^{\otimes t}$ where $U\in U(d)$. Now, since the symmetric subspace is irreducible under the action of $U^{\otimes t}$, Schur's lemma says that the operator $A$ has to proportional to the identity on $\mathrm{Sym}^t(\mathbb C^d)$, i.e. to the projector $P_{\mathrm{Sym}^t(\mathbb C^d)}$. Since $A$ has trace one, we find that the correct proportionality factor is $\mathrm{tr}(P_{\mathrm{Sym}^t(\mathbb C^d)})^{-1} = D_t^{-1}$: $$ \int |\psi\rangle\langle\psi|^{\otimes t} d\psi = D_t^{-1} P_{\mathrm{Sym}^t(\mathbb C^d)} \,. $$

Remark: More general Haar integrals can be computed similarly, c.f. my answer here


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