The quantum relative entropy is defined for any quantum states $\rho, \sigma$ as

$$D(\rho\|\sigma) = tr(\rho\log\rho) - tr(\rho\log\sigma)$$

For arbitrary choice of $\rho,\sigma$, the quantum relative entropy can take any nonnegative value. Consider some bipartite state $\rho_{AB}$ and let its marginals be $\rho_A$ and $\rho_B$. If we consider $D(\rho_{AB}\|\rho_A\otimes\rho_B)$, we have the mutual information. Moreover, we have that

$$D(\rho_{AB}\|\rho_A\otimes\rho_B) \leq \min(2\log|A|, 2\log|B|)$$


The one-shot analogue of the relative entropy is the max-relative entropy and is defined as

$$D_{\max}(\rho \| \sigma)=\inf \left\{\lambda \in \mathbb{R}: 2^{\lambda} \sigma \geq \rho\right\},$$

where $A\geq B$ is used to denote that $A-B$ is positive semidefinite. Like the ordinary relative entropy, the max-relative entropy can also take any nonnegative value. If I now consider $D_{\max}(\rho_{AB}\|\rho_A\otimes\rho_B)$, is there an upper bound on the maximum value that it can take?

I believe the answer is yes since the case of $+\infty$ is ruled out due to the support of $\rho_{AB}$ being contained in the support of $\rho_A\otimes\rho_B$ but have not been able to find a bound.

  • $\begingroup$ Just for clarification: How do we interpret the comparison of density matrices in $2^\lambda \sigma \geq \rho$? Diagonalize both in the same basis, and require that the diagonal elements of one all be at least as large as the diagonal elements of the other? Or do we allow them to be in different bases? $\endgroup$ Aug 16, 2020 at 1:10
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    $\begingroup$ @RickyTensor, sorry should have added this. I use the notation $A\geq B$ to denote that $A-B$ is a positive semidefinite matrix. $A$ and $B$ do not have to share a common eigenbasis. $\endgroup$ Aug 16, 2020 at 1:15
  • $\begingroup$ Sorry, but what are A and B in your post (I mean the first inequality you write in the background)? What they have to do with $\rho_{AB}, \rho_{A}, \rho_{B}$? $\endgroup$
    – Agnieszka
    Sep 19, 2020 at 17:08

1 Answer 1


$\renewcommand{ket}[1]{\left| #1 \right\rangle}$ A state that saturates the mutual-information bound is $$\rho_{AB} = \frac{1}{N} \sum_{i = 1}^{N} \ket{a_i}\ket{b_i} $$ where $N = \min(|A|,|B|)$ and $\{\ket{a_i}\}, \{\ket{b_i}\}$ are bases for $A,B$, respectively. Intuitively, this state maximizes the entropy of the marginals while keeping $A$ and $B$ perfectly correlated.

This state gives $I_{\max} = \log_2(N)$. I haven't proven that this is an upper bound, but it seems like a good place to start.


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