# Independence in state prepared by independently drawn Haar random gates

Consider independently drawn $$2 \times 2$$ Haar random unitaries $$U_1, U_2, \ldots, U_n$$ and

$$V = U_1 \otimes U_2 \otimes \cdots U_n.$$ Consider the state $$\sigma$$ given by

$$\sigma = V \rho V^{*},$$ where $$\rho$$ is an arbitrary $$n$$-qubit quantum state.

Let

$$\sigma = \begin{pmatrix} X_{0,0}&\cdots&X_{0, 2^{n}-1\\}\\\vdots & \ddots&\vdots\\X_{2^n-1,0}&\cdots&X_{2^n-1,2^n-1}\end{pmatrix}.$$

Are the random variables $$X_{i, j}$$ and $$X_{k, l}$$ independent (and/or identically distributed)?

Note that when $$\sigma$$ is an $$n$$-qubit truly Haar random state, this statement holds (the variables are iid complex Gaussians.) I am trying to see how weak we can make the assumption for this to still hold.

This is not the case. In fact, this is not true even in the case of a true Haar-random unitary.

First of all, note that since $$\sigma$$ is a quantum state, it is hermitian. As such, $$X_{a,b}=\overline{X_{b,a}}$$ for any pair $$(a,b)$$, which implies that these two random variables are not independant.

Furthermore, since $$\sigma$$ is a quantum state, its trace is equal to $$1$$. As such, being given $$X_i$$ for $$i\in\left[2^n-1\right]$$, the value of $$X_{2^n-1,2^n-1}$$ is forced, which means that they are not independant. To take another example, if you know that $$X_{0,0}=\frac34$$, then you also know that $$X_{i,i}\leqslant\frac14$$ for $$i\geqslant1$$.

Note that this reasoning still holds no matter what the distribution of $$V$$ is. In particular, this is still true when $$V$$ is a tensor product of Haar-random unitaries.

Also, note that for $$V$$ being Haar-random, it is not true that $$X_{a,b}$$ are drawn from a complex Gaussian. If $$\rho$$ is pure, then $$\sigma$$ is also pure and can be written as $$|\psi\rangle\langle\psi|$$, with: $$|\psi\rangle=\sum_{k=0}^{2^n-1}\psi_k|k\rangle$$ where each $$\psi_k$$ is independently drawn from a complex Gaussian distribution (that is, $$\psi_k=A_k+\mathrm{i}B_k$$ with $$A_k$$ and $$B_k$$ being independently drawn from $$\mathcal{N}(0;1)$$) and then normalized. This means that $$X_{a,b}=\psi_a\psi_b^\dagger$$. This means for instance that each $$X_{i,i}$$ is drawn from a $$\chi^2(2)$$ distribution and then normalised so that their sum equals $$1$$.

However, this means that each $$X_{i,i}$$ are identically distributed, and so are the $$X_{a,b\neq a}$$. If $$\rho$$ is a mixed state, then writing $$\sigma$$ as: $$\sigma=\sum_jp_jV\rho_jV^\dagger=\sum_jp_j\sigma_j$$ shows that this still holds for mixed states.

• Why is the state always selected from the $\chi^2(2)$ distribution and then normalised? Wouldn’t there be dependence on the distribution of $V$? And why is it true when $V$ is a tensor product of Haar random unitaries? Jan 29 at 3:36
• @BlackHat18 My answer wasn't clear about that, but this particular result holds for $V$ being Haar-random. Excluding the normalization, a diagonal coefficient $X_{i,i}$ can be written as $A_{i,i}^2+B_{i,i}^2$ for $A_{i,i}$ and $B_{i,i}$ being drawn from $\mathcal{N}(0;1)$. Thus, $X_{i,i}$ is the sum of two i.i.d. squared standard centered Gaussian variables, which means its law is $\chi^2(2)$. You then normalize them so that the trace is equal to $1$. I've reorganized the answer so that it is less confusing Jan 29 at 11:21