Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.