Von Neumann Entropy of a joint state

[Danny Boy]

Definition 1 The von Neumann entropy of a density matrix is given by $$S(\rho) := - Tr[\rho ln \rho] = H[\lambda (\rho)] $$ where $H[\lambda (\rho)]$ is the Shannon entropy of the set of probabilities $\lambda (\rho)$ (which are eigenvalues of the density operator $\rho$).

Definition 2 If a system is prepared in the ensemble $\{ p_j, \rho_j \}$ then we define the Holevo $\chi$ quantity for the ensemble by $$\chi := S(\rho) - \sum_j p_j S(\rho_j) $$

Short Question : Let $A$ and $B$ be two quantum systems in a state of the form $$\rho^{(AB)} = \sum_k q_i |a_{i}^{(A)} \rangle \langle a_{i}^{(A)} | \otimes \rho_{i}^{(B)} $$
where the states $|a_{i}^{(A)} \rangle$ are orthogonal. What property of the von Neumann entropy implies that the von Neumann entropy of the joint state is $$S(\rho^{(AB)}) = H(\vec{q}) + \sum_i q_i S(\rho_{i}^{(B)})? $$

Thanks for any assistance.

 

[Danny Boy]

Proposal: I'm pretty sure it makes use of the following properties of von Neumann entropy: $$S(\rho_{A} \otimes \rho_{B}) = S(\rho_{A}) + S(\rho_{B})$$ and if $\rho_{A} = \sum_{x} p_x | \phi_x \rangle \langle \phi_x|$ then $$S(\rho_{A} \otimes \rho_{B}) = H(X) + S(\rho_{B})$$ where $X = \{| \phi_x \rangle, p_x \}$. I'm still missing something which leads to the result stated above...

 

[atyy]

Where did you see this?

 

[Danny Boy]

I saw it at the bottom of page 4 of the paper "Limitations on the amount of accessible information in a quantum channel".