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

Question

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

Answer 1

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 $V(\rho_{AB}\|\rho_A \otimes \rho_B) = 2 V(\rho_{AB}\|\rho_A \otimes \sigma_B)$.