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Let $\rho_{AB}$ be the state of a composite quantum system with state space $H_A\otimes H_B$ (two finite dimensional Hilbert spaces). Now assume that $A$ and $B$ are isolated and suffer a unitary evolution given by $U_A$ and $U_B$. If we measure the system $A$ then the probability of observing $x$, one of the eigenvalues of the meassurement operator is:

\begin{equation} P(x)=tr((Pr_x\otimes I)(U_A\otimes U_B)\rho_{AB}(U_A^*\otimes U_B^*)) \end{equation} where $Pr_x$ is the projector on the eigenspace associated with $x$ and $I$ the identity on $H_B$.

I would like to prove that $P(x)$ is independent of the evolution of system $B$ and in particular follows: \begin{equation} P(x)=tr_A(Pr_xU_Atr_B(\rho_{AB})U_A^*) \end{equation} where $tr_A$ and $tr_B$ are the partial traces on systems $A$ and $B$. It is not exactly homework, it is just an unproven statement in a textbook.

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    Just changed it, it is just a common statement.2012-08-06

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$ \begin{align} \def\tr{\operatorname{tr}} P(x) &= \tr((Pr_x\otimes I)(U_A\otimes U_B)\rho_{AB}(U_A^*\otimes U_B^*)) \\ &= \tr_A(\tr_B((Pr_x\otimes I)(U_A\otimes U_B)\rho_{AB}(U_A^*\otimes U_B^*))) \\ &= \tr_A(Pr_x\tr_B((U_A\otimes U_B)\rho_{AB}(U_A^*\otimes U_B^*))) \\ &= \tr_A(Pr_x\tr_B((U_A\otimes I)\rho_{AB}(U_A^*\otimes I))) \\ &= \tr_A(Pr_xU_A\tr_B(\rho_{AB})U_A^*)\;. \end{align} $

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    @joriki I'm sure you are correct, but there's twist worth explaining. $U_B$ can only be cycled that way because it operates only on the subspace being traced over. A general operator would only cycle under the total trace.2013-10-31