Questions on the max-relative entropy $D_{max}(rho||sigma)$

Question

The max-relative entropy between two states is defined as

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

where $rholeq sigma$ 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 $rholeqlambdasigma$. I would like to understand the following properties of this quantity when the states are bipartite i.e. they live on $H_Aotimes H_B$. 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

$$I_{max}(A:B)_rho = minlimits_{sigma_B} D_{max}(rho_{AB}||rho_Aotimessigma_B)$$

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

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

DaftWullie · Answer

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
$$
rho_{AB}=left(p_0|0ranglelangle 0|otimes tau_0+p_1|1ranglelangle 1|otimes tau_1right)
$$
where $tau_0$ and $tau_1$ are different (normalised) single-qubit density matrices. We have that
$$
I=min_{sigma_B}logmin_{lambda}left{lambda:p_0|0ranglelangle 0|otimes (lambdasigma-tau_0)+p_1|1ranglelangle 1|otimes (lambdasigma-tau_1)geq 0right}
$$
Now, by construction, this devolves into two separate questions: pick the common $lambda$ and $sigma$ such that both $lambdasigma-tau_i$ 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 $p_0neq 1/2$.

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 $sigma^starneq rho_B$, $D_max(rho_B|sigma^star)>D_max(rho_B|rho_B)$. We know, for example, that $lambdageq 1$ because $text{Tr}(lambdasigma-rho)=lambda-1$, and if $lambdasigma-rho$ is going to be non-negative, then the total of the eigenvalues must be non-negative.

Questions on the max-relative entropy $D_{max}(rho||sigma)$

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