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

  • $\begingroup$ 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? $\endgroup$ – user1936752 Apr 1 at 13:45
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    $\begingroup$ 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 $D(\rho\|\rho_A \otimes \sigma_B) = \infty$. $\endgroup$ – Rammus Apr 1 at 14:16
  • $\begingroup$ Thank you - I've asked as another question with your points in mind $\endgroup$ – user1936752 Apr 2 at 5:19

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