# Von Neumann Entropy of a joint state

• A
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.

• atyy and Johny Boy

## Answers and Replies

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
Science Advisor
Where did you see this?