3

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}{\...


3

Source of the problem The purported contradiction arises due to the use of incorrect assumptions for Klein equality $$ S(\rho||\sigma) \ge 0. $$ The inequality does not require any particular relationship$^1$ between the support of $\rho$ and the support of $\sigma$. However, it does require that $\rho$ and $\sigma$ be states, i.e. unit trace positive ...


2

As you say, $$ \mathrm{Tr}[\rho_{XB} \log \rho_{XB}] = -S(X) + \sum_{x} p(x) \mathrm{Tr}[\rho_{B}^x \log \rho_B^x]. $$ But if you can prove the above statement, then the exact same derivation gives you $$ \mathrm{Tr}[\rho_{XB} \log \sigma_{XB}] = - S(X) + \sum_{x} p(x) \mathrm{Tr}[\rho_B^x \log \sigma_B^x]. $$ If you put both together then you get $$ \begin{...


2

Both limits are dealt with in a fair amount of detail in the work that originally defined the sandwiched entropies: On quantum Renyi entropies: a new generalization and some properties. In particular, you'll find the relevant results in section IV.C.


2

There is a problem in the derivation you presented, since $\rho \leq \lambda \sigma$ is only equivalent to $\sigma^{-1/2} \rho \sigma^{-1/2} \leq \lambda I$ when $\sigma$ is invertible (or at least when $\mathrm{supp}(\rho) \subset \mathrm{supp}(\sigma)$, so that you can restrict the space to $\mathrm{supp}(\sigma)$ instead of the whole Hilbert space). The ...


2

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=\...


2

I think I have an answer. The following should be the CVX code for one of the formulations found in this link. cvx_begin sdp variable X(2, 2) hermitian minimize(trace(id' * X)) % id is eye(2) subject to kron(id, X) >= rho_ab % the tensor product of two density matrices a, b X >= 0 cvx_end The optimal value found in this program is $$\text{optval} = ...


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