Suppose we have a classical quantum state $\sum_x |x\rangle \langle x|\otimes \rho_x$, one can define the smooth-min entropy $H_\min(A|B)_\rho$ as the best probability of guessing outcome $x$ given $\rho_x$. How does this quantity relate to $H(A|B)_\rho$ the standard conditional entropy? If not, how does it relate to the mutual information $I(A:B)_\rho$?

The conditional min-entropy $\text{H}_{\text{min}}(A | B)_{\rho}$ can be defined for an arbitrary state $\rho$ of a pair of registers $(A,B)$ as $$ - \inf_{\sigma} \,\text{D}_{\text{max}}(\rho \| \mathbb{1}\otimes \sigma), $$ where the infimum is over all states $\sigma$ of $B$ and $\text{D}_{\text{max}}$ is the quantum relative max-entropy: $$ \text{D}_{\text{max}}(P\|Q) = \inf\{\lambda\in\mathbb{R}: P\leq 2^{\lambda} Q\}. $$ In contrast, the ordinary conditional entropy $\text{H}(A | B)_{\rho}$ can be expressed as $$ - \inf_{\sigma}\, \text{D}(\rho \| \mathbb{1}\otimes \sigma), $$ where here $\text{D}$ refers to the ordinary quantum relative entropy. (This expression for the conditional entropy simplifies to something more familiar once you know that the infimum is always achieved by $\sigma = \operatorname{Tr}_{A}(\rho)$, which is not necessarily true for the formula for the conditional min-entropy.)

It happens to be the case that for a classical-quantum state $$ \rho = \sum_x p(x) \,|x\rangle \langle x | \otimes \rho_x $$ that the conditional min-entropy $\text{H}_{\text{min}}(A | B)_{\rho}$ is equal to the negative logarithm of the optimal guessing probability.

It is always the case that $$ \text{H}_{\text{min}}(A | B)_{\rho} \leq \text{H}(A | B)_{\rho}, $$ for every state $\rho$ and not just classical-quantum states. This follows from the fact that $$ \text{D}(\rho \| Q) \leq \text{D}_{\text{max}}(\rho \| Q) $$ for every density operator $\rho$ and every positive semidefinite operator $Q$. This inequality follows from the observation that $\rho \leq 2^{\lambda} Q$ implies \begin{align} \text{D}(\rho \| Q) & = \operatorname{Tr}(\rho \log(\rho)) - \operatorname{Tr}(\rho \log(Q))\\ & \leq \operatorname{Tr}(\rho\log(\rho)) - \operatorname{Tr}(\rho\log(2^{-\lambda}\rho))\\ & = \lambda, \end{align} where the inequality makes use of the operator monotonicity of the logarithm function: if $Q \geq 2^{-\lambda}\rho$, then $\log(Q) \geq \log(2^{-\lambda}\rho)$.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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