# Positive conditional quantum entropy for entangled state

The quantum conditional entropy $$S(A|B)\equiv S(AB)-S(A)$$, where $$S(AB)=S(\rho_{\rm AB})$$ and $$S(B)=S(\rho_{\rm B})$$ is known to be non-negative for separable states. For entangled states, it is known that the quantum conditional entropy can attain both negative and positive values. Thus, a negative quantum conditional entropy serves as a sufficient, but not necessary, criterion for the quantum state to be entangled. References where this is stated would be https://arxiv.org/pdf/1703.01059.pdf or https://arxiv.org/pdf/quant-ph/9610005.pdf.

While it is easy to find entangled states which yield negative quantum conditional entropies, such as any pure entangled state, I cannot think of/construct an entangled state which yields a positive quantum conditional entropy. Can someone provide an example?

As you mention pure states will not do. So lets look at a simple example of mixed entangled states, two-qubit Werner states. Let $$\rho_{AB} = q |\Psi^- \rangle \langle \Psi^-| + (1-q) I / 4$$ where $$| \Psi^- \rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$$ and $$q \in [0,1]$$. It is known that this family of states is separable iff $$q \in [0, 1/3]$$.
So let's calculate their conditional entropy. We have $$H(AB) = 2 - \frac34 (1-q) \log(1-q) - \frac14 (1+3q)\log(1+3q)$$ and $$H(B) = 1$$. Putting this together we have $$H(A|B) = g(q)$$ where $$g(q) = 1 - \frac34 (1-q) \log(1-q) - \frac14 (1+3q)\log(1+3q)$$. Now you can check that for a range of values greater than $$1/3$$ the conditional entropy is positive. E.g. $$g(1/2) \approx 0.549$$. Plotting $$g(q)$$ it looks like the entropy becomes positive somewhere just below $$q = 3/4$$.