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.


1

An easy (but perhaps cheating) answer: The relative entropy $S(A||B)$diverges when $A$ has support over the kernel of $B$ (e.g., wikipedia). Now, the kernel of $B=|0\rangle^{\otimes n}\langle 0|^{\otimes n}$ is everything other than $|0\rangle^{\otimes n}$, i.e., $\mathbb{I}-|0\rangle^{\otimes n}\langle 0|^{\otimes n}$! For the inequality to be meaningful, ...


Only top voted, non community-wiki answers of a minimum length are eligible