Why can't measuring a system $B$ change a correlated system $A$? I am reading Quantum computation & quantum information from Nielsen & Chuang.
On page 187, he talks about the principle of implicit measurement.
He says that if we have two system $A$ and $B$, doing a measurement on $B$ doesn't change the density matrix of $A$.
I agree with it if we do an unread measurement :
I take the projectors : $\Pi_k = |e_k\rangle \langle e_k|$
$$  Tr_B(\sum_k \Pi_k \rho_{AB} \Pi_k)= \sum_{p,k} \langle e_p|e_k \rangle \langle e_k | \rho_{AB} | e_k \rangle \langle e_k|e_p \rangle = \sum_k \langle e_k | \rho_{AB} | e_k \rangle = Tr_B(\rho_{AB})$$
Thus the density matrix of $A$ after we measured an observable on $B$ is the same than the one if we didn't measure anything on $B$.
But the way I understand the paragraph seems to be more general than unread measurement. And for me it is not true that measuring on $B$ won't change the density matrix on $A$, in particular if we have correlations (quantum or classical one).
My question :
Thus I would like to check how to understand precisely what he says. Does he implicitly indeed talks only about unread measurement or it is more general and I didn't get something ?
Also (extra question) :
He says that it physically makes sense that measuring on $B$ won't change the density matrix on $A$.I agree with the math (for unread measurement) but for me it still looks weird because of possible correlations between $A$ and $B$. Why is it what we could naturally expect ? I feel like I understand the math but not the physical meaning behind.
 A: You are correct: the measurement of $B$ does not change the density matrix of $A$ only with respect to an observer who does not know the outcome of the measurement. If the two qubits are initially in a maximally entangled state, e.g., the singlet state $$\frac{1}{\sqrt{2}}(|01\rangle-|10\rangle),$$
then after measurement of qubit $B$, an observer who doesn't know the outcome of the measurement will assign the completely mixed state (density matrix = 1/2 $\times$ identity) to qubit $A$. This just formalizes the fact that this observer knows that the state of $A$ was determined with certainty by the measurement of $B$, but has no information about what that state is.
If instead the observer knows the outcome of the measurement, then they also know the pure state that $A$ now occupies, so in this case the density matrix indeed changes, as you thought it should. I agree that this is not stated particularly clearly in N&C!
A: Answering the last part of the question:

He says that it physically makes sense that measuring on  won't change the density matrix on . I agree with the math (for unread measurement) but for me, it still looks weird because of possible correlations between  and . Why is it what we could naturally expect? I feel like I understand the math but not the physical meaning behind.

The point here is that the sole act of $B$ acquiring information about his part of the system should not affect $A$'s system. Because there are correlations, it means that $B$ will (potentially) have some information about what $A$ will observe, but from the point of view of $A$ nothing changes (unless of course $B$ tells $A$ what he observed).
This is akin to what happens classically if you distribute two marbles, one black and one white, to Alice and Bob. When Bob "measures" his marble's color he finds (say) black, and therefore knows that Alice's marble is white, but this does not mean that the act of Bob's measuring his marble magically changed the state of Alice's one.
In the case of quantum mechanics the situation is a bit murkier, as arguably the act of Bob measuring his part does affect Alice's part of the system. However, it still holds true that without knowing what was observed by Bob, Alice has no way of knowing what Bob did or observed.
