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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 $\delta$?

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1 Answer 1

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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.

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  • $\begingroup$ Thank you, that's a neat trick with (6)! $\endgroup$
    – JRT
    Oct 17, 2022 at 1:45

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