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.


1 Answer 1


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}

This gives

\begin{align} \chi^{\left(AEQ\right)} &= S\left(\rho^{\left(AEQ\right)}\right) - \sum_jp_jS\left(\rho_j^{\left(AEQ\right)}\right)\\ &=S\left(\rho^{\left(AE\right)}\right) + S\left(\rho^{\left(Q\right)}\right) - \sum_jp_j\left(S\left(\rho^{\left(AE\right)}\right) + S\left(\rho_j^{\left(Q\right)}\right)\right)\\ &= S\left(\rho^{\left(AE\right)}\right) + S\left(\rho^{\left(Q\right)}\right) - \sum_jp_jS\left(\rho^{\left(AE\right)}\right) -\sum_jp_jS\left(\rho_j^{\left(Q\right)}\right). \end{align}

As $\sum_jp_j = 1$, it follows that $$\chi^{\left(AEQ\right)} = S\left(\rho^{\left(Q\right)}\right)-\sum_jp_jS\left(\rho_j^{\left(Q\right)}\right) = \chi^{\left(Q\right)}$$

  • $\begingroup$ Thanks for the answer. One question, are you thinking in terms of matrices to justify this equation: $$ -\rho_A\otimes\rho_B\ln\left(\rho_A\otimes\rho_B\right)\\ = -\sum_{j, k}\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\right)$$ ? $\endgroup$
    – John Doe
    Commented Apr 28, 2018 at 13:29
  • $\begingroup$ @JohnDoe yes, although there would be more rigorous/detailed ways to show the same. In any case, it still holds true for mixed states (such as your state with different possible preparations, as long as the eigenvalues sum to one, which they will do, unless you're considering subsystems or something) $\endgroup$
    – Mithrandir24601
    Commented Apr 28, 2018 at 14:32

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