# Tag Info

## Hot answers tagged relative-entropy

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, ...

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