Let $|\psi\rangle$ be a $n$ qubit Haar-random quantum state. I am trying to show that in the limit of large $n$, for each $z_{i} \in \{0, 1\}^{n}$, $$ |\langle 0|\psi\rangle|^{2}, |\langle 1|\psi\rangle|^{2}, \ldots, |\langle 2^{n} - 1|\psi\rangle|^{2} ~\text{are i.i.d random variables and}$$

$$ |\langle z_{i}|\psi\rangle|^{2} \sim \text{PorterThomas}(\alpha),$$ where the probability density function for the Porter Thomas distribution is given by $$ f(\alpha) = 2^{n} e^{-2^{n} \alpha}.$$

For example, look at Fact 10 of this paper. I am specifically interested in why we need a large enough $n$ to have the i.i.d approximation.


1 Answer 1


In the following, I'll show the evaluation of the probability densities of the transition probabilities: $|\langle \psi | z\rangle^2$ 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: $$d{\mu}_{CP^{N-1}}(\mathbf{\zeta}, \mathbf{\bar{\zeta}}) = \frac{(N-1)!}{\pi^{N-1}}\frac{\prod_{k=1}^{N-1} d\zeta_k d\bar{\zeta}_k}{(1+\mathbf{\zeta}^{\dagger} \mathbf{\zeta})^N}$$

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: $$ f_{\alpha}(\alpha) = \int_{CP^{N-1}} \delta\left(\alpha - \frac{\bar{\zeta_k} \zeta_k }{(1+\mathbf{\zeta}^{\dagger} \mathbf{\zeta})}\right) \, d{\mu}_{CP^{N-1}} $$

Where $\delta$ is the Dirac delta function. Defining: $$x = \sum_{j\ne k} \bar{\zeta_j} \zeta_j$$ 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: $$ d\zeta_k d\bar{\zeta}_k = \frac{1}{2} du_k d\theta_k$$ We obtain: $$ f_{\alpha}(\alpha) = \frac{(N-1)!}{\pi^{N-1}}\int_{CP^{N-1}} \delta\left(\alpha - \frac{u_k }{(1+x)(1+\frac{u_k}{(1+x))})}\right) \, \frac{1}{2} du_k d{\theta_k} \frac{\prod_{j\ne k} d\zeta_j d\bar{\zeta}_j}{(1+x)^N(1+\frac{u_k}{(1+x))}))^N}$$ Performing another change of variables: $$v_k = \frac{u_k}{1+x}$$ We obtain: $$f_{\alpha}(\alpha) = \frac{(N-1)!}{\pi^{N-1}}\int_{CP^{N-1}} \delta\left(\alpha - \frac{v_k }{(1+v_k)}\right) \, \frac{1}{2} dv_k d{\theta_k} \frac{\prod_{j\ne k} d\zeta_j d\bar{\zeta}_j}{(1+x)^{N-1}(1+v_k)^N}$$ Using the properties of the Dirac delta function: $$\delta\left(\alpha - \frac{v_k }{(1+v_k)}\right) = (1+v_k) \delta\left(v_k- \frac{\alpha }{(1-\alpha)}\right) $$ Substituting into the integral (and performing the trivial integral over $\theta_k$: $\int d{\theta_k} = 2\pi$, we obtain:

$$f_{\alpha}(\alpha) = (N-1) (1-\alpha)^{N-3} \frac{(N-2)!}{\pi^{N-2}}\int_{CP^{N-2}} \frac{\prod_{j\ne k} d\zeta_j d\bar{\zeta}_j}{(1+\sum_{j\ne k} \bar{\zeta_j} \zeta_j)^{N-1}}$$ 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 $N\rightarrow \infty$ $$f_{\alpha}(\alpha) \approx N (1-\alpha)^N = N \left(1-\frac{N\alpha}{N}\right)^N \approx N e^{-N\alpha} = 2^n e^{-2^n\alpha}$$

Pairwise independence

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

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

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

$$f_{\alpha, \beta}(\alpha, \beta) = \frac{(N-1)!}{\pi^{N-1}}\int_{CP^{N-1}} \delta\left(\alpha - \frac{v_k }{(1+v_k+v_l)}\right) \delta\left(\beta - \frac{v_l }{(1+v_k+v_l)}\right) \, \frac{1}{4} dv_k d{\theta_k} dv_l d{\theta_l} \frac{\prod_{j\ne k} d\zeta_j d\bar{\zeta}_j}{(1+x)^{N-2}(1+v_k+v_l)^N}$$

Again, using the transformation properties of the delta functions:

$$\delta\left(\alpha - \frac{v_k }{(1+v_k+v_l)}\right) \delta\left(\beta - \frac{v_l }{(1+v_k+v_l)}\right)= (1+v_k+v_l)^3\delta\left(v_k- \frac{\alpha }{(1-\alpha - \beta)}\right) \delta\left(v_l- \frac{\beta }{(1-\alpha - \beta)}\right)$$ and after the substitution, we have $$dv_k dv_l = \frac{d\alpha d\beta}{(1-\alpha - \beta)^3 }$$ Thus, we are left with: $$f_{\alpha, \beta}(\alpha, \beta) = (N-1)(N-2) (1-\alpha-\beta)^{N-6} \frac{(N-3)!}{\pi^{N-3}}\int_{CP^{N-3}} \frac{\prod_{j\ne k, l} d\zeta_j d\bar{\zeta}_j}{(1+\sum_{j\ne k,l} \bar{\zeta_j} \zeta_j)^{N-2}}$$ 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 $N\rightarrow \infty$ $$ f_{\alpha, \beta}(\alpha, \beta) \approx N^2 \left(1-\alpha- \beta\right)^N = N^2 \left(1-\frac{N(\alpha+\beta)}{N}\right)^N \approx N^2 e^{-N(\alpha+\beta)}= 2^n e^{-2^n\alpha} 2^n e^{-2^n\beta} \approx f_{\alpha}(\alpha) f_{\beta}(\beta) $$

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.


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