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The max-relative entropy between two states is defined as

$$D_{\max }(\rho \| \sigma):=\log \min \{\lambda: \rho \leq \lambda \sigma\},$$

where $\rho\leq \sigma$ should be read as $\sigma - \rho$ is positive semidefinite. In other words, $D_{\max}$ is the logarithm of the smallest positive real number that satisfies $\rho\leq\lambda\sigma$. I would like to understand the following properties of this quantity when the states are bipartite i.e. they live on $H_A\otimes H_B$. In the following, all $\rho$ and $\sigma$ correspond to quantum states (positive semidefinite matrices with unit trace).

A quantity known as the max-information that $B$ has about $A$ is given by

$$I_{\max}(A:B)_\rho = \min\limits_{\sigma_B} D_{\max}(\rho_{AB}||\rho_A\otimes\sigma_B)$$

Let the minimum on the right hand side be achieved by the state $\sigma^\star_B$. My questions are as follows

  1. Can someone provide an example of a state $\rho_{AB}$ for which $\sigma^\star_B \neq \rho_B$?

  2. Is it true that $D_{\max}(\rho_{B}||\sigma^\star_B) \leq D_{\max}(\rho_{B}||\sigma_B)$ for all $\sigma_B$ i.e. is the $D_{\max}$ minimizing state preserved under a partial trace?

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Can someone provide an example of a state $\rho_{AB}$ for which $\sigma^\star_B \neq \rho_B$?

Why not start very easily, with a separable state such as $$ \rho_{AB}=\left(p_0|0\rangle\langle 0|\otimes \tau_0+p_1|1\rangle\langle 1|\otimes \tau_1\right) $$ where $\tau_0$ and $\tau_1$ are different (normalised) single-qubit density matrices. We have that $$ I=\min_{\sigma_B}\log\min_{\lambda}\left\{\lambda:p_0|0\rangle\langle 0|\otimes (\lambda\sigma-\tau_0)+p_1|1\rangle\langle 1|\otimes (\lambda\sigma-\tau_1)\geq 0\right\} $$ Now, by construction, this devolves into two separate questions: pick the common $\lambda$ and $\sigma$ such that both $\lambda\sigma-\tau_i$ are positive semi-definite. However, this is entirely independent of the $p_i$. Hence, the answer must be independent of the $p_i$. By comparison, $\rho_B$ depends on the $p_i$. Thus, unless the answer is highly degenerate (allowing for all possible linear combinations of the $\tau_i$), it cannot be that $\sigma=\rho_B$.

For example, if $\tau_0$ and $\tau_1$ are orthogonal, the best choice must be $\sigma=(\tau_0+\tau_1)/2$ with $\lambda=2$. This is certainly not $\rho_B$ for any $p_0\neq 1/2$.

Is it true that $D_{\max}(\rho_{B}||\sigma^\star_B) \leq D_{\max}(\rho_{B}||\sigma_B)$ for all $\sigma_B$ i.e. is the $D_{\max}$ minimizing state preserved under a partial trace?

This cannot be true, right? $D_\max(\rho_B\|\rho_B)=0$. So in any case where $\sigma^\star\neq \rho_B$, $D_\max(\rho_B\|\sigma^\star)>D_\max(\rho_B\|\rho_B)$. We know, for example, that $\lambda\geq 1$ because $\text{Tr}(\lambda\sigma-\rho)=\lambda-1$, and if $\lambda\sigma-\rho$ is going to be non-negative, then the total of the eigenvalues must be non-negative.

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