# Is data processing for relative entropy true when states are not normalized?

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)?

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}
• @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 Apr 8 '21 at 7:26