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.


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

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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – BlackHat18
    Commented Jan 29, 2023 at 3:36
  • $\begingroup$ @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 $\endgroup$
    – Tristan Nemoz
    Commented Jan 29, 2023 at 11:21

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