Consider an arbitrary ensemble $\{p_x\rho_x\}_{x\in X}$ and define the state $$ \rho = \sum_{x\in X} \vert x \rangle\langle x \vert \otimes p_x\rho_x. $$ I am interested in understanding the quantity $$ \inf_{\xi, \hspace{2mm}\xi_x\perp\xi_{x'}} \Vert \rho - \xi \Vert_1, \quad\quad \text{with } \Vert\cdot\Vert_1 = \operatorname{Tr}\vert\cdot\vert, $$ where the infimum is taken over all orthogonal ensembles, that is for the ensemble $\{q_x\xi_x\}_{x\in X}$ we define $\xi = \sum_{x\in X}\vert x \rangle\langle x \vert \otimes q_x\xi_x$ and we assume $\xi_x\perp \xi_{x'}$ for all distinct $x,x'\in X$ in the sense $\operatorname{Tr}\xi_x\xi_{x'} = 0$. In particular, my curiosity is directed at having some intuition as to what a good choice of $\xi$ depending on $\rho$ is when trying to minimize the expression above. For instance, one could let $\{M_x\}_{x\in X}$ denote a set of orthogonal projections and define $\xi_x = \frac{M_x\rho_x M_x}{\operatorname{Tr} M_x\rho_x M_x}$ and $q_x = p_x$. Then \begin{align*} \Vert \rho - \xi\Vert_1 &= \sum_{x\in X} \Vert p_x\rho_x - p_x\xi_x\Vert_1 \\ &= \sum_{x\in X} p_x\left\Vert \left(1 - \frac{1}{\operatorname{Tr} M_x\rho_xM_x}\right)M_x\rho_xM_x - (Id-M_x)\rho_x(Id-M_x)\right\Vert_1 \\ &= \sum_{x\in X} p_x\left(\left(\frac{1}{\operatorname{Tr} M_x\rho_xM_x}-1\right)\operatorname{Tr} M_x\rho_xM_x + \operatorname{Tr} (Id-M_x)\rho_x(Id-M_x)\right) \\ &= 1 - \sum_{x\in X}\operatorname{Tr}(2M_x-Id)p_x\rho_x \\ &= 2\left(1 - \sum_{x\in X}\operatorname{Tr}M_xp_x\rho_x\right) \end{align*} So choosing $\{M_x\}_{x\in X}$ such that $\sum_{x\in X}\operatorname{Tr}M_xp_x\rho_x$ is maximized is a decent choice! However, is this optimal in general? Or are there any other natural constructions of $\xi$, which give a pretty good upper bound on the quantity in question?

Any help is appreciated!



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