On Wikipedia, the conditional quantum entropy is defined as follows

$$H(A|B)_{\rho_{AB}} = H(\rho_{AB}) - H(\rho_B)\tag{1},$$

where $\rho_B$ is the partial state after tracing out system $A$ and $H(X)$ is the von Neumann entropy of $X$.

An alternative definition (see eq(3) of this paper) of what the authors call the conditional von Neumann entropy is

$$H(A|B)_{\rho_{AB}} = -D(\rho_{AB}||I\otimes \rho_B)\tag{2},$$

where $D(\rho||\sigma) = \text{Tr}(\rho(\log\rho - \log\sigma))$.

Choose $\rho_{AB} = \sigma_A\otimes\sigma_B,$ a product state. Let $\sigma_A$ be mixed. Then, using the fact that $H(\sigma_A\otimes\sigma_B) = H(\sigma_A) + H(\sigma_B)$ and $\rho_B =\text{Tr}_A(\sigma_A\otimes\sigma_B) = \sigma_B$. we have by (1)

$$H(A|B)_{\rho_{AB}} = H(\sigma_A) > 0.$$

Yet using Definition (2) and the non-negativity of relative entropy, we have

$$H(A|B)_{\rho_{AB}} = -D(\rho_{AB}||I\otimes\sigma_B) \leq 0$$

What is the source of the contradiction? Or are these quantities two different things?


1 Answer 1


The contradiction comes from the statement $$-D(\rho_{AB}||I \otimes \sigma_B) \leq 0$$ which is wrong. Using the general identity $\log(X \otimes Y) = I \otimes \log Y + \log X \otimes I$, we can calculate for an arbitrary density matrix (not necessary of product form) $$-D(\rho_{AB}||I \otimes \rho_{B}) = -\text{Tr}\rho_{AB}(\log \rho_{AB} - \log(I \otimes \rho_B)) \ = H(\rho_{AB}) + \text{Tr}\rho_{AB} \log(I \otimes \rho_B) = H(A|B)_{\rho}$$


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