Information of system vs system, apparatus and environment

[Danny Boy]

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.

 

[thephystudent]

Clasically, entropy is an extensive quantity. If there is no entanglement between the different subsystems, this generalizes towards the von Neumann entropy of quantum systems.

In particular, if
$$\rho^{(AEQ)}=\rho_0^{(AE)}\otimes\rho^{(Q)}$$
then
$$S(\rho^{(AEQ)})= S(\rho_0^{(AE)})+S(\rho^{(Q)}).$$

Does this solve the problem?