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?