# Prove that uniformly random states have moments ${\bf E}_\psi|\langle x|\psi\rangle|^{2t}\sim1/\binom d t$

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?

• – glS
Oct 31, 2022 at 1:26

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 $$\mathrm{tr}(P_{\mathrm{Sym}^t(\mathbb C^d)})^{-1} = D_t^{-1}$$: $$\int |\psi\rangle\langle\psi|^{\otimes t} d\psi = D_t^{-1} P_{\mathrm{Sym}^t(\mathbb C^d)} \,.$$