Positive conditional quantum entropy for entangled state

Question

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?

Answer 1

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 $q = 3/4$.