I Entangled, mixed state with conditional entropy zero

[greypilgrim]

Hi.
The classical (Shannon) conditional entropy is never negative. It can be written as $H(Y|X)=H(X,Y)-H(X)$ which allows for a quantum generalization using von Neumann entropy. In the case of entangled states, it can become negative.

I guess it should be possible to construct an entangled, mixed (bipartite) state where $H(Y|X)$ is exactly zero (though I don't know how to exactly do that). Does this have a specific meaning?

 

[LittleSchwinger]

Conditional entropy in classical probability is always positive because the entropy of the total system is always greater than or equal to the entropy of its parts. This is no longer true in QM and it can be negative because you can have maximum knowledge of the whole system (it's in a pure state), but less than optimal knowledge of the parts (subsystems are in a mixed state). It vanishing is just the special case where the entropies happen to be equal, it has no additional special meaning.