Question about Haar random quantum states

In the following, I'll show the evaluation of the probability densities of the transition probabilities: $|langle psi | zrangle^2$ and their pairwise independence. I didn't work out the full mutual independence.

The $n$-qubit pure states span the complex projective space $CP^{N-1}$ with $N=2^n$. Pure $n$-qubit states can be parametrized almost everywhere as: $$|psi(mathbf{zeta}, mathbf{bar{zeta}})rangle = frac{[1, zeta _1,.,.,., zeta _{N-1}]^t}{sqrt{1+mathbf{zeta}^{dagger} mathbf{zeta} }}$$ (The states which cannot be parametrized as above consist of a lower dimensional subspace, thus they correspond to zero probability and they do not contribute to the probabilistic calculations)

The Haar volume element of $CP^{N-1}$ is given by: $$d{mu}_{CP^{N-1}}(mathbf{zeta}, mathbf{bar{zeta}}) = frac{(N-1)!}{pi^{N-1}}frac{prod_{k=1}^{N-1} dzeta_k dbar{zeta}_k}{(1+mathbf{zeta}^{dagger} mathbf{zeta})^N}$$

It is normalized to a unit total volume. $$int_{CP^{N-1}} d{mu}_{CP^{N-1}}(mathbf{zeta}, mathbf{bar{zeta})} = 1$$

In the scalar product $langle z_k|psi(mathbf{zeta}, mathbf{bar{zeta}})rangle $ only one term $zeta_k$ survives. It is exactly at the index $k$ whose binary representation contains ones in the places where the string $z_k$ has ones and zeros where the string $z_k$ has zeros.

Thus, we get the following expression for the transition squared amplitude (for an arbitrary $z$): $$alpha = |langle z_k|psi(mathbf{zeta}, mathbf{bar{zeta}})rangle|^2 = frac{bar{zeta_k} zeta_k }{(1+mathbf{zeta}^{dagger} mathbf{zeta})^N}$$

Thus, the probability density of $alpha$ is given by: $$ f_{alpha}(alpha) = int_{CP^{N-1}} deltaleft(alpha - frac{bar{zeta_k} zeta_k }{(1+mathbf{zeta}^{dagger} mathbf{zeta})}right) , d{mu}_{CP^{N-1}} $$

Where $delta$ is the Dirac delta function. Defining: $$x = sum_{jne k} bar{zeta_j} zeta_j$$ and $$u_k = bar{zeta_k} zeta_k $$ and in addition, expressing the integration elements over $bar{zeta_k}$ and $zeta_k$ in polar coordinates: $$ dzeta_k dbar{zeta}_k = frac{1}{2} du_k dtheta_k$$ We obtain: $$ f_{alpha}(alpha) = frac{(N-1)!}{pi^{N-1}}int_{CP^{N-1}} deltaleft(alpha - frac{u_k }{(1+x)(1+frac{u_k}{(1+x))})}right) , frac{1}{2} du_k d{theta_k} frac{prod_{jne k} dzeta_j dbar{zeta}_j}{(1+x)^N(1+frac{u_k}{(1+x))}))^N}$$ Performing another change of variables: $$v_k = frac{u_k}{1+x}$$ We obtain: $$f_{alpha}(alpha) = frac{(N-1)!}{pi^{N-1}}int_{CP^{N-1}} deltaleft(alpha - frac{v_k }{(1+v_k)}right) , frac{1}{2} dv_k d{theta_k} frac{prod_{jne k} dzeta_j dbar{zeta}_j}{(1+x)^{N-1}(1+v_k)^N}$$ Using the properties of the Dirac delta function: $$deltaleft(alpha - frac{v_k }{(1+v_k)}right) = (1+v_k) deltaleft(v_k- frac{alpha }{(1-alpha)}right) $$ Substituting into the integral (and performing the trivial integral over $theta_k$: $int d{theta_k} = 2pi$, we obtain:

$$f_{alpha}(alpha) = (N-1) (1-alpha)^{N-3} frac{(N-2)!}{pi^{N-2}}int_{CP^{N-2}} frac{prod_{jne k} dzeta_j dbar{zeta}_j}{(1+sum_{jne k} bar{zeta_j} zeta_j)^{N-1}}$$ The integral with its pre-factor is just the normalized volume element of $CP^{N-2}$. i.e., equal to $1$. Thus $$f_{alpha}(alpha) = (N-1) (1-alpha)^{N-3}$$ In the limit $Nrightarrow infty$ $$f_{alpha}(alpha) approx N (1-alpha)^N = N left(1-frac{Nalpha}{N}right)^N approx N e^{-Nalpha} = 2^n e^{-2^nalpha}$$

Pairwise independence

For $lne k$: $$beta = |langle z_l|psi(mathbf{zeta}, mathbf{bar{zeta}})rangle|^2 = frac{bar{zeta_l} zeta_l }{(1+mathbf{zeta}^{dagger} mathbf{zeta})^N}$$ The joint probability density: $$ f_{alpha, beta}(alpha, beta) = int_{CP^{N-1}} deltaleft(alpha - frac{bar{zeta_k} zeta_k }{(1+mathbf{zeta}^{dagger} mathbf{zeta})}right) deltaleft(beta - frac{bar{zeta_l} zeta_l }{(1+mathbf{zeta}^{dagger} mathbf{zeta})}right) , d{mu}_{CP^{N-1}} $$

Pursuing the same method as above, separation of the coordinates $zeta_k$, $zeta_l$ from the other coordinates and defining:

$$x = sum_{jne k,l} bar{zeta_j} zeta_j,$$ then performing the necessary changes of variables and the polar angular trivial integrations, we arrive at:

$$f_{alpha, beta}(alpha, beta) = frac{(N-1)!}{pi^{N-1}}int_{CP^{N-1}} deltaleft(alpha - frac{v_k }{(1+v_k+v_l)}right) deltaleft(beta - frac{v_l }{(1+v_k+v_l)}right) , frac{1}{4} dv_k d{theta_k} dv_l d{theta_l} frac{prod_{jne k} dzeta_j dbar{zeta}_j}{(1+x)^{N-2}(1+v_k+v_l)^N}$$

Again, using the transformation properties of the delta functions:

$$deltaleft(alpha - frac{v_k }{(1+v_k+v_l)}right) deltaleft(beta - frac{v_l }{(1+v_k+v_l)}right)= (1+v_k+v_l)^3deltaleft(v_k- frac{alpha }{(1-alpha - beta)}right) deltaleft(v_l- frac{beta }{(1-alpha - beta)}right)$$ and after the substitution, we have $$dv_k dv_l = frac{dalpha dbeta}{(1-alpha - beta)^3 }$$ Thus, we are left with: $$f_{alpha, beta}(alpha, beta) = (N-1)(N-2) (1-alpha-beta)^{N-6} frac{(N-3)!}{pi^{N-3}}int_{CP^{N-3}} frac{prod_{jne k, l} dzeta_j dbar{zeta}_j}{(1+sum_{jne k,l} bar{zeta_j} zeta_j)^{N-2}}$$ Again, the integral with its pre-factor is just the normalized volume element of $CP^{N-3}$. Thus, we are left with: $$f_{alpha, beta}(alpha, beta) = (N-1)(N-2) (1-alpha-beta)^{N-6}$$ In the limit $Nrightarrow infty$ $$ f_{alpha, beta}(alpha, beta) approx N^2 left(1-alpha- betaright)^N = N^2 left(1-frac{N(alpha+beta)}{N}right)^N approx N^2 e^{-N(alpha+beta)}= 2^n e^{-2^nalpha} 2^n e^{-2^nbeta} approx f_{alpha}(alpha) f_{beta}(beta) $$

Thus, the random variables are pairwise independent.

Without the large $N$ approximation, the joint distribution function is not equal to the product of the individual distributions.