Let $\mathbf{x} \sim \mathcal{CN} (\boldsymbol{\mu}, \boldsymbol{\Sigma})$ and let us consider the general case where $\boldsymbol{\mu} \neq \mathbf{0}$ and $\boldsymbol{\Sigma}$ is a full-rank, non-diagonal matrix.
What can be said about $\mathbf{x}$ if $\mathbb{E}[\mathbf{x} \mathbf{x}^{\mathrm{H}}]$ is a diagonal matrix?
Perhaps the element of $\mathbf{x}$ are somehow independent?
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- Some background information in case it's useful. Let $\mathbf{y} \sim \mathcal{CN} (\bar{\boldsymbol{\mu}}, \bar{\boldsymbol{\Sigma}})$: we have $\mathbb{E}[\mathbf{y} \mathbf{y}^{\mathrm{H}}] = \bar{\boldsymbol{\mu}} \bar{\boldsymbol{\mu}}^{\mathrm{H}} + \bar{\boldsymbol{\Sigma}} = \bar{\mathbf{U}} \bar{\mathbf{D}} \bar{\mathbf{U}}^{\mathrm{H}}$ (after applying the eigenvalue decomposition). Hence, the original vector $\mathbf{x}$ is obtained as $\mathbf{x} = \bar{\mathbf{U}}^{\mathrm{H}} \mathbf{y}$.
Suppose $\mathbf x = \left[ \begin{array}{c} X \\ Y \end{array} \right]$ where $X=\begin{cases} \phantom{-}1 \\ \phantom{-}0 & \text{each with probability } 1/3, \\ -1 \end{cases}$ and $Y=X^2.$
Then $\operatorname E\left[ \mathbf x \mathbf x^H \right] = \left[ \begin{array}{cc} 2/3 & 0 \\ 0 & 2/3 \end{array} \right],$ but the components of $\mathbf x$ are not independent.
The most you will get is that they are uncorrelated.