# Does the relative entropy variance $V(\rho_{AB}\|\rho_A\otimes\sigma_B)$ satisfy an ordering for different $\sigma_B$?

The relative entropy between two quantum states is given by $$D(\rho\|\sigma) = \operatorname{Tr}(\rho\log\rho -\rho\log\sigma)$$. It is known that for any bipartite state $$\rho_{AB}$$ with reduced states $$\rho_A$$ and $$\rho_B$$, it holds that

$$D(\rho_{AB}\|\rho_A\otimes\rho_B)\leq D(\rho_{AB}\|\rho_A\otimes\omega_B)$$

for all choices of $$\omega_B$$. This can be seen by expanding both sides and noting that the relative entropy is nonnegative. Now define the relative entropy variance (see this reference, 2.16)

$$V(\rho \| \sigma):=\operatorname{Tr} [\rho(\log \rho-\log \sigma-D(\rho \| \sigma))^{2}]$$

Does this also satisfy a similar property i.e.

$$V(\rho_{AB}\|\rho_A\otimes\rho_B) \leq V(\rho_{AB}\|\rho_A\otimes\sigma_B)$$

for all $$\sigma_B$$?

EDIT: It seems like the last inequality is not true. But the name variance is suggestive so perhaps there is a non trivial lower bound for $$V(\rho_{AB}\|\rho_A\otimes\sigma_B)$$ using $$V(\rho_{AB}\|\rho_A\otimes\rho_B)$$?

No, such an ordering does not exist. For example, take $$\rho = |\phi\rangle \langle \phi|$$ with $$| \phi \rangle = \cos(\theta) |00 \rangle + \sin(\theta) |11\rangle$$ and $$\theta \in (0,\pi/4)$$. Then take $$\sigma_B = I/2$$, the maximally mixed qubit.
A direct calculation gives $$V(\rho_{AB}\|\rho_A \otimes \rho_B) = 8 \big(\log[\tan(\theta)] \sin(\theta) \cos(\theta)\big)^2$$ and $$V(\rho_{AB}\|\rho_A \otimes \sigma_B) = 4 \big(\log[\tan(\theta)] \sin(\theta) \cos(\theta)\big)^2 \,.$$ So in this case we actually have $$V(\rho_{AB}\|\rho_A \otimes \rho_B) = 2 V(\rho_{AB}\|\rho_A \otimes \sigma_B)$$.
• Thank you. If $\rho_{AB}$ is $d-$dimensional, can $V(\rho_{AB}\|\rho_A\otimes\sigma_B)$ and $V(\rho_{AB}\|\rho_A\otimes\rho_B)$ be arbitrarily far apart or is there a dimension factor that comes into play? In your example, you have a factor of 2 which seems neat so is that somehow optimal? Apr 1, 2021 at 13:45
• I don't know, I've never had to deal much with the relative entropy variance. Maybe it's worth posting as a separate question. I doubt it's true though. Actually, can't you just make the relative entropies arbitrarily different? I'm imagining choosing a $\sigma_B$ such that $D(\rho\|\rho_A \otimes \sigma_B) = \infty$. Apr 1, 2021 at 14:16