Quantum Computing Asked on December 1, 2020
Let $A, B$ be (finite-dimensional) Hilbert spaces, and $rho$ some mixed state of $A otimes B$. I am trying to show that a measurement performed on the ‘$A$-subsystem’ does not affect $rho^B = text{Tr}_A(rho)$.
I understand a ‘measurement performed on the $A$-subsystem’ as given by some observable $X otimes I$, where $X$ is a self-adjoint operator on $A$ which decomposes as $X = sum m P_m$ (where $P_m$ is the orthogonal projection on the $m$-eigenspace). If this measurement results in outcome $m$, the resulting state should be
$$rho’ = frac{(P_m otimes I) rho (P_m otimes I)}{text{Tr}((P_m otimes I) rho)}$$
and I wish to see that $text{Tr}_A(rho’) = text{Tr}_A(rho)$. Now I can use the cyclicity of trace to see that
$$text{Tr}_A(rho’) = frac{text{Tr}_A((P_m otimes I) rho)}{text{Tr}((P_m otimes I) rho)}$$
but why should this be equal to $text{Tr}_A(rho)$? I did check this is the case if $rho = rho^A otimes rho^B$ decomposes as a product. And I know a general $rho$ will be a linear combination of such cases; but since the equality desired is not linear, it does not seem to follow that it’ll hold (in fact, it seems to indicate it will not hold). Hopefully someone can point to the mistake in my thinking.
What you've denoted as $rho'$ is just an $m$-th possible outcome. We have to write $$ rho'_m = frac{(P_m otimes I) rho (P_m otimes I)}{text{Tr}((P_m otimes I) rho)}. $$ Now, since Bob doesn't know the value of $m$ he has to assume that the new state is a mixture of $rho'_m$ with corresponding probabilities $text{Tr}((P_m otimes I) rho)$. So, the actual $rho'$ is $$ rho' = sum_m rho'_m text{Tr}((P_m otimes I) rho) = sum_m (P_m otimes I) rho (P_m otimes I). $$ Finally, you can see that $$ text{Tr}_A(rho') = sum_m text{Tr}_A((P_m otimes I) rho (P_m otimes I)) = $$ $$ = sum_m text{Tr}_A(rho (P_m otimes I)) = text{Tr}_A(rho sum_m (P_m otimes I)) = text{Tr}_A(rho). $$
Correct answer by Danylo Y on December 1, 2020
Indeed, this equation does not hold. Take for example $rho = |phi^+ranglelangle phi^+|$, where $|phi^+rangle = frac1{sqrt2}(|00rangle+|11rangle)$. If Alice obtains outcome 0, then $rho' = |00ranglelangle 00|$, and $rho^B = |0ranglelangle0|$, and if Alice obtains outcome 1, then $rho' = |11ranglelangle 11|$, and $rho^B = |1ranglelangle1|$. This is the typical example of the nonlocality of wavefunction collapse.
The statement of the no-communication theorem is that when you average over Alice's outcomes, then the result does not depend on which measurement she made, or indeed if she made a measurement at all. Let then $$rho_m = frac{(P_m otimes I) rho (P_m otimes I)}{text{Tr}((P_m otimes I) rho)},$$ the collapsed state after obtaining measurement result $m$, and $$rho^B_m = operatorname{tr}_A (rho_m).$$ The theorem is then that $$sum_m p(m) rho^B_m = rho^B = operatorname{tr}_Arho,$$ for all possible projectors $P_m$, or even POVM elements.
Answered by Mateus Araújo on December 1, 2020
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