# Does the max-relative entropy satisfy $0 < D_{\max}(\rho \parallel I_A \otimes \sigma_B) < 1$?

The quantum conditional min-entropy is defined as

$$H_{\min}(A|B) = - \inf\limits_{\sigma_B} D_{\max} \left( \rho_{AB} \parallel I_A \otimes \sigma_B \right)$$ where in general $$D_{\max}(\rho \parallel \sigma) = \inf \{ \lambda : \rho \leq 2^\lambda \sigma \}$$

I am trying to understand why there is a negative sign in the definition of $$H_{\min}(A|B)$$. My hunch is that $$D_{\max}(\rho \parallel I_A \otimes \sigma_B) = \inf \{ \lambda : \rho \leq 2^\lambda I_A \otimes \sigma_B \}$$ is a number between $$0$$ and $$1$$, and since $$\log$$ function is involved, we take the negative at the end, because $$0 < \log(x) < 1$$ when $$x<1$$. I would like to come up with a more rigorous argument, and I could not come up with a proof.

Is it true that $$0 < D_{\max}(\rho \parallel I_A \otimes \sigma_B) < 1$$?

• Have you tried coming up with some examples? I'm pretty sure you'll quickly find counterexamples to this. Commented Mar 12, 2023 at 18:11
• I agree that it is highly likely to come up with counterexamples easily. That is why I am a bit confused because I think $H_{\min} (A|B ) \geq 0$ and I am not sure how this inequality is satisfied, if $D_{\max}(\rho \parallel \sigma) > 1$? Basically, I am trying to understand how this definition ensures $H_{\min}(A|B) \geq 0$.
– Josh
Commented Mar 12, 2023 at 18:32
• Why do you think the conditional min entropy is non-negative? I suggest trying to construct your own examples. But spoiler, you'll find that it can indeed be negative. The definition is built to recover the usual min entropy for classical systems. Commented Mar 12, 2023 at 21:25

No. As discussed e.g. in the second lecture of https://cs.uwaterloo.ca/~watrous/QIT-notes/, between pages 16 and 17, if $$\sigma$$ is a state, then $$2^{-D_{\rm max}(\rho\|\sigma)}\in[0,1]$$, or equivalently, $$D_{\rm max}(\rho\|\sigma)\ge0$$.
For example, $$D_{\rm max}(\rho\|\rho)=0$$, and $$D_{\rm max}\left(\frac I2\bigg\| \begin{pmatrix}1-\epsilon&0\\0&\epsilon\end{pmatrix}\right) = -\log(2\epsilon),$$ which diverges for $$\epsilon\to0^+$$.
Similar bounds hold when you compute $$D_{\rm max}(\rho\|I\otimes \sigma)$$ with $$\rho$$ bipartite. For example, $$D_{\rm max}\left(\frac{I\otimes I}4\bigg\| I\otimes\begin{pmatrix}1-\epsilon&0\\0&\epsilon\end{pmatrix}\right) = -\log(4\epsilon).$$ To have $$D_{\rm max}(\rho\|I\otimes \sigma)=0$$, consider the following two-qubit example: $$D_{\rm max}\left(\frac{|00\rangle\!\langle00|+|11\rangle\!\langle11|}{2}\bigg\| I\otimes \frac{I}{2}\right) = 0,$$ I wrote this thinking in terms of the kind of states that give a vanishing conditional entropy.
There's different ways to compute the above examples, but at least in these cases probably the more convenient formulation is $$D_{\rm max}(P\|Q)=\log\|\sqrt{Q^+}P\sqrt{Q^+}\|$$. See again the above linked notes to see where this comes from.