4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ They are not the same. Later in Watrous' lecture, the second definition is covered and it goes under the name of hypothesis testing relative entropy with $\varepsilon = 1$. The confusion is due to different definitions by different authors. $\endgroup$ – rnva Jul 17 at 17:15
2
$\begingroup$

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 $$ \overline{D}_{\min}(\rho \| \sigma) = \lim_{\alpha \to 0^+} \overline{D}_{\alpha}(\rho \|\sigma) = -\log \mathrm{Tr}[\Pi_\rho \sigma ]. $$

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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