# von Neumann Entropy of a joint state

Definition 1 The von Neumann entropy of a density matrix is given by $$S(\rho) := - \mathrm{Tr}[\rho \ln \rho] = H[\lambda (\rho)]$$ 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 $$\rho^{(AB)} = \sum_i q_i |a_{i}^{(A)} \rangle \langle a_{i}^{(A)} | \otimes \rho_{i}^{(B)}$$ 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 $$S(\rho^{(AB)}) = H(\vec{q}) + \sum_i q_i S(\rho_{i}^{(B)})?$$

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 $X = \{| \phi_x \rangle, p_x \}$.

Thanks for any assistance.

## 1 Answer

$$\rho^{(AB)} = \sum_i q_i |a_i^{(A)}\rangle \langle a_i^{(A)}| \otimes \rho_i^{(B)} \tag 1$$ Note that subsystem $A$ and $B$ are separable, let $$\rho^{(B)}_i = \sum_j \lambda^j_i |{b_i^j}^{(B)}\rangle \langle {b_i^j}^{(B)}|. \tag 2$$ Substitute equation $(2)$ into $(1)$: \eqalignno{ S(\rho^{(AB)}) &= -\sum_{ij} q_i \lambda_i^j \ln(q_i \lambda_i^j) \\ &= -\sum_{ij} q_i \lambda_i^j (\ln q_i + \ln \lambda_i^j) \\ &= -\sum_i q_i \ln q_i - \sum_i q_i \sum_j \lambda_i^j \ln \lambda_i^j \\ &= H(\vec q) + \sum_i q_i S(\rho_i^{(B)}). &(3) }

• Thanks for your response. I just want to confirm the first step after substituting (2) into (1) i.e. $$S(\rho^{(AB)}) = -\sum_{ij} q_i \lambda_i^j \ln(q_i \lambda_i^j).$$ Is the reasoning as follows: After substituting (2) into (1) we obtain \begin{align} \rho^{(AB)} &= \sum_{i}\sum_{j}q_{i} \lambda_{i}^{j} | a_{i}^{(A)} \rangle \langle a_{i}^{(A)}| \otimes \sum_{j}\lambda_{i}^{j}|b_{i}^{j(B)} \rangle \langle b_{i}^{j(B)}| \\ &= \sum_{i}\sum_{j}q_{i}\lambda_{i}^{j}|a_{i}^{(A)} \rangle \langle a_{i}^{(A)}| \otimes |b_{i}^{j(B)} \rangle \langle b_{i}^{j(B)}| \end{align} – user165535 Jan 19 '18 at 12:25
• Hence we use note that since $|a_{i}^{(A)} \rangle$ diagonalizes $\rho^{A}$ and $|b_{i}^{j(B)} \rangle$ diagonalizes $\rho^{B}$ it follows that $|a_{i}^{A} \rangle \otimes |b_{i}^{j(B)} \rangle$ diagonalizes $\rho^{(AB)}$with eigenvalues $q_{i} \lambda_{i}^{j}$.Hence we apply the definition of von Neumann entropy to this? – user165535 Jan 19 '18 at 12:25
• Yeah, the key is that A and B are separable. It's more complicated when systems are correlated. – Mountain Mao Jan 19 '18 at 15:57