Trace Distance Range And Equality

Let $\mathrm{\Delta }=\rho -\sigma ={\mathrm{\Delta }}_{+}-{\mathrm{\Delta }}_{-}$. Since $\mathrm{tr}\mathrm{\Delta }=0$, [L1] gives $\mathrm{tr}{\mathrm{\Delta }}_{+}=\mathrm{tr}{\mathrm{\Delta }}_{-}=D\ge 0$, proving the lower bound.

If $\mathrm{supp}\rho \perp \mathrm{supp}\sigma$, then $\rho \sigma =0$ and $\mathrm{\Delta }$ acts as $\rho$ on $\mathrm{supp}\rho$, as $-\sigma$ on $\mathrm{supp}\sigma$, and as zero on the orthogonal complement. Hence $\mathrm{\parallel }\mathrm{\Delta }{\mathrm{\parallel }}_1=\mathrm{tr}\rho +\mathrm{tr}\sigma =2$ and $D=1$.

Let $P_{+}$ project onto the strictly positive eigenspaces of $\mathrm{\Delta }$. Then $0\le P_{+}\le I$ and $D=\mathrm{tr}(P_{+}\mathrm{\Delta })=\mathrm{tr}(P_{+}\rho )-\mathrm{tr}(P_{+}\sigma )\le \mathrm{tr}(P_{+}\rho )\le 1$. The first inequality uses $\mathrm{tr}(P_{+}\sigma )\ge 0$ by [L2], and the last follows from $\mathrm{tr}((I-P_{+})\rho )\ge 0$ by [L2] and $\mathrm{tr}\rho =1$.

If $D=0$, [F1] gives $\mathrm{\parallel }\rho -\sigma {\mathrm{\parallel }}_1=0$, so definiteness of the trace norm gives $\rho =\sigma$. Conversely equal states have distance zero by [F1].

Conversely, if $D=1$, then both inequalities in step 2.1 are equalities: $\mathrm{tr}(P_{+}\rho )=1$ and $\mathrm{tr}(P_{+}\sigma )=0$. Thus $\mathrm{tr}((I-P_{+})\rho )=0$ and $\mathrm{tr}(P_{+}\sigma )=0$. Applying [F3] to each positive operator gives $(I-P_{+})\sqrt{\rho }=0$ and $P_{+}\sqrt{\sigma }=0$, so $\mathrm{supp}\rho \subseteq \mathrm{ran}{P}_{+}$ and $\mathrm{supp}\sigma \subseteq \ker P_{+}$. These subspaces are orthogonal.