# Does closeness in trace distance imply close measurement outcomes?

Suppose we have two density matrices $$\rho$$ and $$\rho'$$ such that $$\|\rho - \rho'\|_1 \leq \varepsilon$$. Let $$\{\Lambda, I - \Lambda\}$$ be elements of some POVM. If it holds that

$$Tr(\Lambda\rho) \leq \delta$$

then is it also true that

$$Tr(\Lambda\rho')$$

can be upper bounded in terms of $$\varepsilon$$ and $$\delta$$?

Yes, we can bound $$\mathrm{tr}(\Lambda\rho')$$ in terms of $$\delta$$ and $$\varepsilon$$ as follows \begin{align} \mathrm{tr}(\Lambda\rho')&=\mathrm{tr}(\Lambda\rho)+\mathrm{tr}(\Lambda(\rho'-\rho))\tag1\\ &\le\delta+\mathrm{tr}(\Lambda(\rho'-\rho))\tag2\\ &\le\delta+D(\rho,\rho')\tag3\\ &=\delta+\frac12\|\rho-\rho'\|_1\tag4\\ &\le\delta+\frac{\varepsilon}{2}\tag5 \end{align} where $$D(\rho,\sigma):=\frac12\|\rho-\sigma\|_1$$ is the trace distance between $$\rho$$ and $$\sigma$$. Inequality $$(3)$$ follows from the fact that $$D(\rho,\sigma)=\max_E\mathrm{tr}\left(E(\rho-\sigma)\right)\tag6$$ where the maximization is done over all positive operators $$E\le I$$. For a proof of $$(6)$$ see the text following equation $$(9.22)$$ on page $$404$$ in Nielsen & Chuang. It is an application of Jordan-Hahn decomposition.