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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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Please let us know your definition of composite quantum system. Please don't give orders (as in "Prove..."). –  Rasmus Aug 6 '12 at 8:41
Just changed it, it is just a common statement. –  Euclean Aug 6 '12 at 8:45
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1 Answer

up vote 2 down vote accepted

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

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Thanks for your answer. Why is it in the fourth equality that $tr_B((U_A\otimes U_B)\rho_{AB}(U_A^*\otimes U_B^*))=tr_B((U_A\otimes I)\rho_{AB}(U_A^*\otimes I))$? –  Euclean Aug 6 '12 at 9:13
@Euclean: This is because the trace is invariant with respect to cyclic permutation, and $U_B^*U_B=I$. If you write it out with indices, this is $$\sum_{\alpha\beta ijkl}U_{A,\alpha i}U_{B,\beta j}\rho_{AB,ijkl}U^*_{A,k\alpha}U^*_{B,l\beta}=\sum_{\alpha ik}U_{A,\alpha i}\left(\sum_m\rho_{AB,imkm}\right)U^*_{A,k\alpha}\;,$$ since $\sum_\beta U_{B,\beta j}U^*_{B,l\beta}=\delta_{jl}$. –  joriki Aug 6 '12 at 9:41
I see, thanks for the answer! –  Euclean Aug 6 '12 at 10:01
@Euclean: You're welcome! –  joriki Aug 6 '12 at 10:22
@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. –  Adrian Ratnapala Oct 31 '13 at 20:43
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