Is the quantum min-relative entropy $D_{\min}(\rho\|\sigma)=-\log(F(\rho, \sigma)^2)$ or $D_{\min}(\rho\|\sigma)=-\log(tr(\Pi_\rho\sigma))$?

Question

In John Watrous' lectures, he defines the quantum min-relative entropy as

$$D_{\min}(\rho\|\sigma) = -\log(F(\rho, \sigma)^2),$$

where $F(\rho,\sigma) = tr(\sqrt{\rho\sigma})$. Here, I use this question and answer to make the definition simpler although one should note that the linked question uses a different definition of fidelity (squared vs not squared).

On the other hand, one of the early papers introducing this quantity (see Definition 2 of this paper) defines it as

$$D_{\min}(\rho\|\sigma) = -\log(tr(\Pi_\rho\sigma)),$$

where $\Pi_\rho$ is the projector onto the support of $\rho$. It's not clear if these definitions are equivalent since I can change $\rho$ without altering its support.

How are the two definitions related to each other, if at all?

Answer 1

As @rnva points out these are not the same quantities. To give some clarity as to why they are both referred to as $D_{\min}$ it is best to look at the as limiting cases of $\alpha$-R'enyi divergences.

First, we have the sandwiched divergences which for $\alpha \in (0, 1) \cup (1, \infty)$ are defined as $$ \widetilde{D}_{\alpha}(\rho\|\sigma) = \frac{1}{\alpha - 1} \log \mathrm{Tr}\left[ (\sigma^{\frac{1-\alpha}{2\alpha}} \rho \sigma^{\frac{1-\alpha}{2\alpha}} )^\alpha \right]. $$ These divergences are monotonically increasing in $\alpha$ and satisfy the data processing inequality (DPI) for all $\alpha \geq 1/2$. Thus the smallest divergence in this family satisfying the DPI is $$ \widetilde{D}_{\min}(\rho \| \sigma) = \widetilde{D}_{1/2}(\rho \|\sigma) = - \log \mathrm{Tr}[\sqrt{\rho} \sqrt{\sigma}]^2. $$

Another well studied family of divergences are the so-called Petz divergences defined for $\alpha \in (0,1) \cup (1, \infty)$ to be $$ \overline{D}_{\alpha}(\rho \| \sigma) = \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^{\alpha} \sigma^{1-\alpha}]. $$ This family satisfies the DPI for $\alpha \in (0,1) \cup(1,2]$ and they are also monotonically increasing in $\alpha$. Thus, the smallest divergence satisfying the DPI in this family is $$ \overline{D}_{\min}(\rho \| \sigma) = \lim_{\alpha \to 0^+} \overline{D}_{\alpha}(\rho \|\sigma) = -\log \mathrm{Tr}[\Pi_\rho \sigma ]. $$