Does the quantum relative entropy have a direct operational interpretation?

Question

Consider the quantum relative entropy, defined as $$D(\rho\|\sigma) = \operatorname{tr}(\rho\log\rho)-\operatorname{tr}(\rho\log\sigma),$$ for all $\rho,\sigma\ge0$ such that $\operatorname{im}(\rho)\subseteq\operatorname{im}(\sigma)$. It is well-known that $\rho,\sigma$ are states with $[\rho,\sigma]=0$, then $D(\rho\|\sigma)$ equals the classical relative entropy between the probability distribution corresponding to their respective eigenvalues.

The classical relative entropy can be interpreted for example in a communication context as how "inefficient" coding a message using a code optimised for the wrong probability distribution is, as mentioned e.g. in this math.SE post.

Does the quantum relative entropy have any direct operational interpretation? By this, I mean a way to interpret $D(\rho\|\sigma)$ as quantifying a specific property in any kind of communication/coding/informational context. For example, we can understand the von Neumann entropy $S(\rho)$ as the Shannon entropy of the probability distribution obtained measuring $\rho$ in its eigenbasis (Does the von Neumann entropy equal the smallest accessible Shannon entropy?), and we can understand the quantum mutual information $I(X:Y)_\rho$ in terms of entanglement-assisted communication rates (although not directly as just naive "correlations", due to it potentially containing quantum discord).

A possible way to do this is to understand the quantum relative entropy as the classical relative entropy between some pair of classical distributions. Admittedly, given the general expression $$D(\rho\|\sigma) = \sum_{jk} |\langle\lambda_j(\rho)|\lambda_k(\sigma)\rangle|^2 \lambda_j(\rho) \log\left(\frac{\lambda_j(\rho)}{\lambda_k(\sigma)}\right),$$ where $\lambda_j(\rho)$ is the $j$-th eigenvalue of $\rho$, and $|\lambda_j(\rho)\rangle$ the corresponding eigenvector, this doesn't seem to be the case, but I don't know how to rule it out altogether.

Answer 1

I would argue yes, in the context of recoverability. Given a quantum channel $\mathcal{N}: L(A) \to L(B)$ and a state $\sigma$ on $A$, we say that a quantum channel $\mathcal R: L(B) \to L(A)$ is a $(\sigma, \mathcal{N})$-recovery channel if it satisfies $$ \mathcal R \circ \mathcal{N}(\sigma) = \sigma\,. $$ Note that such channels always exist, i.e. $\mathcal R(X) = \mathrm{Tr}[X] \sigma$ is such a channel. However, is it possible to define a recovery channel that recovers multiple states? I.e., can we find an $\mathcal{R}$ that is both a $(\sigma, \mathcal N)$ and $(\rho, \mathcal{N})$ recovery channel. It turns out that such channels exist iff $$ D(\rho\| \sigma) = D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))\,. $$ I refer you to to chapter 12 of Wilde's book for more details.