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


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

  • $\begingroup$ 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? $\endgroup$ Apr 1, 2021 at 13:45
  • 1
    $\begingroup$ 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, 2021 at 14:16
  • $\begingroup$ Thank you - I've asked as another question with your points in mind $\endgroup$ Apr 2, 2021 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.