# Accessible information of system vs system, apparatus and environment

Suppose we have a quantum system $Q$ with an initial state $\rho^{(Q)}$. The measurement process will involve two additional quantum systems: an apparatus system $A$ and an environment system $E$. We suppose that the system $Q$ is initially prepared in the state $\rho_{k}^{(Q)}$ with a priori probability $p_k$. The state of the apparatus $A$ and environment $E$ is $\rho_{0}^{(AE)}$, independent of the preparation of $Q$. The initial state of the entire system given the $k$th preparation for $Q$ is $$\rho_{k}^{(AEQ)} = \rho_{0}^{(AE)} \otimes \rho_{k}^{(Q)}.$$ Averaging over the possible preparations, we obtain $$\rho^{(AEQ)} = \sum_{k} p_{k} \rho_{k}^{(AEQ)}.$$

In quantum information theory, the accessible information of a quantum system is given by $$\chi := S(\rho) - \sum_{j}P_{j}S(\rho_{j}),$$ where $S$ is the von Neumann entropy of the quantum state. How can we show that if $\rho_{0}^{(AE)}$ is independent of the preparation $k$, that $$\chi^{(AEQ)} = \chi^{(Q)}?$$

Thanks for any assistance.

For density matrices $\rho_A$ and $\rho_B$ having eigenvalues $\lambda^{\left(A\right)}$ and $\lambda^{\left(B\right)}$, \begin{align}S\left(\rho_A\otimes\rho_B\right) &= -\rho_A\otimes\rho_B\ln\left(\rho_A\otimes\rho_B\right)\\ &= -\sum_{j, k}\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\right)\\ &= -\sum_{j, k}\left[\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(A\right)}_j\right) + \lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(B\right)}_k\right)\right]\\ &= -\sum_j\lambda^{\left(A\right)}_j\ln\left(\lambda^{\left(A\right)}_j\right)\sum_k\lambda^{\left(B\right)}_k - \sum_j\lambda^{\left(A\right)}_j\sum_k\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(B\right)}_k\right)\\ &= -\sum_j\lambda^{\left(A\right)}_j\ln\left(\lambda^{\left(A\right)}_j\right) - \sum_k\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(B\right)}_k\right)\\ &= S\left(\rho_A\right) + S\left(\rho_B\right). \end{align}
As $\sum_jp_j = 1$, it follows that $$\chi^{\left(AEQ\right)} = S\left(\rho^{\left(Q\right)}\right)-\sum_jp_jS\left(\rho_j^{\left(Q\right)}\right) = \chi^{\left(Q\right)}$$
• Thanks for the answer. One question, are you thinking in terms of matrices to justify this equation: $$-\rho_A\otimes\rho_B\ln\left(\rho_A\otimes\rho_B\right)\\ = -\sum_{j, k}\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\right)$$ ? – John Doe Apr 28 '18 at 13:29