Finite ensemble barycenters are density operators

Set $\bar{\rho }={\sum }_{a=1}^n{p}_a{\rho }_a$, the barycenter in [F1]. It is self-adjoint because it is a finite real linear combination of self-adjoint operators. For every $x\in H$, $⟨\bar{\rho }x,x⟩={\sum }_a{p}_a⟨{\rho }_{a}x,x⟩\ge 0$ by [F2]–[F3], so $\bar{\rho }$ is non-negative.

By trace linearity [F4], $\mathrm{tr}(\bar{\rho })={\sum }_a{p}_a\mathrm{tr}({\rho }_a)={\sum }_a{p}_a=1$. Thus $\bar{\rho }$ is a density operator by [F2].

Let $B_1,\dots ,B_m$ partition the outcome indices and put $q_b={\sum }_{a\in B_b}{p}_a$. If $q_b>0$, the weights $p_a\mathrm{/}{q}_b$ within block $B_b$ sum to one, so the argument in steps 1.1–2.1, applied to this finite normalized subensemble, shows that ${\sigma }_b=q_b^{-1}{\sum }_{a\in B_b}{p}_a{\rho }_a$ is a density operator and $q_b{\sigma }_b={\sum }_{a\in B_b}{p}_a{\rho }_a$.

If $q_b=0$, nonnegativity of the weights forces every $p_a=0$ in $B_b$. Choose any density operator ${\sigma }_b$ using [F5]; then $q_b{\sigma }_b=0={\sum }_{a\in B_b}{p}_a{\rho }_a$. Summing the block identities gives ${\sum }_b{q}_b{\sigma }_b={\sum }_a{p}_a{\rho }_a$, so regrouping preserves the barycenter, including zero-weight blocks.