The reason one can explicitly compute such integral is that the measure factors
dμN,K=dμ|ρ|⊗dμΩ (A.5)Trρ=‖ψ‖^2 depends only on the first, radial, coordinate while Trρ
2
(Trρ)^2 depends only
on the second, angular, part
∫
dμN,KTr
(
ρ^2)
=
∫
dμΩTrρ^2
(Trρ)^2∫
dμ|ρ|(Trρ)^2=
∫
dμN,KTrρ^2
(Trρ)^2∫
dμN,K(Trρ)^2 (A.6)( AssumingdμNKnormalized.) The integral in Eq. (A.4) is reduced to (ratio of)
two Gaussian integrals. Wick theorem (for the standard normal distribution) gives
∫
dμN,KTr
(
ρ^2)
=
∑
jkαβ∫
dμψξαjξ ̄βjξ ̄βkξαk=
∑
jkαβ(δαβ+δjk)=NK(K+N) (A.7)
Similarly
∫
dμN,K(Trρ)^2 =
∑
jkαβ∫
dμGξαjξ ̄αjξ ̄βkξβk=
∑
jkαβ(1 +δjkδαβ)=NK(NK+ 1) (A.8)
So finally, ∫
dμN,KTrρ^2
(Tr ρ)^2=
N+K
KN+ 1
(A.9)
This reduces to Eq. (6.9) whenN=K.
The computation of higher moments can be similarly reduced to a (tedious)
combinatoric problem.
B The N dimensional unit cube is almost a ball
The fact thatDNlooks like a ball in most directions is a general fact about convex
bodies in high dimensions. It is instructive to see this happening for the unit cube