The data processing inequality for relative entropy states that

$$D(\rho\|\sigma) \geq D(N(\rho)\|N(\sigma))$$

for some CPTP map $N$ where $\rho$ is a quantum state and $\sigma$ is a positive-semidefinite operator.

What goes wrong if $\rho$ is not normalized (either sub-normalized or has trace larger than 1)?


1 Answer 1


Nothing goes wrong. A data processing inequality for normalized operators implies a data processing inequality for any positive operators. To see this note that if $\rho, \sigma \geq 0$ and $a,b > 0$ then $$ D(a \rho \| b \sigma) = a D(\rho\|\sigma) + a \log \frac{a}{b}. $$ Then define $a = \mathrm{tr}(\rho)$, $b = \mathrm{tr}(\sigma)$, $\hat{\rho} = \rho/a$ and $\hat{\sigma} = \sigma/b$. Then $$ D(\rho\|\sigma) = a D(\hat{\rho} \|\hat{\sigma}) + a \log(a/b). $$ Now consider any CPTP map $N$ and note that $a$ and $b$ are left unchanged by $N$ as $N$ is trace preserving. Finally we get $$ \begin{aligned} D(N(\rho) \| N(\sigma)) &= D(a N(\rho)/a\| b N(\sigma)/b) \\ &= a D( N(\rho)/a\| N(\sigma)/b) + a \log(a/b) \\ &= a D(N(\hat{\rho}) \| N(\hat{\sigma}) ) + a\log(a/b) \\ &\leq a D(\hat{\rho} \| \hat{\sigma}) + a \log(a/b) \\ &= D(\rho \|\sigma) \end{aligned} $$

  • $\begingroup$ I see. In Theorem 11.8.1 of arxiv.org/pdf/1106.1445.pdf there is a restriction to only density operators on the first argument. Strange that the author chose to do so. $\endgroup$
    – JRT
    Apr 8, 2021 at 6:45
  • $\begingroup$ @JRT Well it seems that the author restricted their definition of $D$ to having only states in the first argument, so it's not surprising this restriction is enforced in the subsequent proofs. This restriction is anyway very common as for most applications it is sufficient to define the quantity for states in the first argument. When people define the quantity for general PSD operators then it is also common to define it slightly differently see this work $\endgroup$
    – Rammus
    Apr 8, 2021 at 7:26

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