# Question about Haar random quantum states

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.

• please read the excerpt for the tags you use. quantum-information is set to be removed and shouldn't be used
– glS
Nov 22 '20 at 16:55
• They can't be independent if you use the same $|\psi\rangle$. They are obviously independent if you take independent $|\psi_i\rangle$, $i=0,1,...,2^n-1$. Dec 3 '20 at 9:37

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.

• In the last equation, it should be $N^{2} \left(1-\frac{N(\alpha+\beta)}{N}\right)^N$, right? Nov 22 '20 at 16:24
• @BlackHat18 Thank you, I have made the correction. Nov 23 '20 at 8:08