Do sequences of operations (including measurements) applied to different halves of an entangled pair always commute?

Question

Let us say $A$ has one half of an entangled qubit pair, and $B$ has the other half. $A$ may be able to perform any type of operation on their half of the pair, such as unitary operations, entangling the qubit with other qubits, etc... followed by a final measurement of the qubit. $B$ has the same capabilities. However, one of either $A$ or $B$ must finish all their operations and their measurement before the other can begin operating on their qubit. This results in two cases:

Case (1): $A$ performs all of their operations on their half of the qubit pair, then measures their half of the qubit pair, then $B$ performs all of their operations on their half of the qubit pair, then measures their half of the qubit pair.

Case (2): $B$ performs all of their operations on their half of the qubit pair, then measures their half of the qubit pair, then $A$ performs all of their operations on their half of the qubit pair, then measures their half of the qubit pair.

Will the expected outcome of the measurement results in Case (1) and Case (2) always be the same?

EDIT: An example to be more clear about what I mean by 'the measurement results' and how this question does not imply breaking the no-communication theorem:

Case (1) always gives $A$ and $B$ the same measurement results and Case (2) gives $A$ and $B$ different measurement results (with some probability less than or equal to 1 and greater than 0) would classify the 'measurement results' in Case (1) and Case (2) as being 'different' and thus the operations 'not commuting'. Furthermore, this example does not break the no-communication theorem with the following condition: $A$'s likelihood of measuring $1$ or $0$ on their half of the qubit pair is not dependent on whether they are in Case (1) or Case (2), and $B$'s likelihood of measuring $1$ or $0$ on their half of the qubit pair is not dependent on whether they are in Case (1) or Case(2).

Answer 1

No, you can't do this. Any process like this would make quantum mechanics incompatible with special relativity. Different inertial frames would disagree about the ordering, and predict different results. It would let you do experiments to determine an absolute reference frame for the universe.

More mathematically, you can prove that the matrix $A \otimes I$ commutes with the matrix $I \otimes B$. Combined with the deferred measurement principle, this shows the ordering of operations on the two parts of the system cannot matter.

Answer 2

You don't need to make any assumption about the order in which operations are performed by the two parties. Nothing Alice can do will affect the outcomes of whatever Bob does, and vice versa.

More generally, the scenario you are considering (if I'm reading it correctly) is one where Alice and Bob apply some kind of channel on their systems. Call these channels $\Phi_A$ and $\Phi_B$. Note that a quantum channel encodes any kind of operation that can be performed on a state, including unitary evolutions, measurements, post-processing, etc. The overall channel describing the result of the joint operations of Alice and Bob is then $\Phi_A\otimes\Phi_B$. Notably, there is no notion here of whether Alice's operations happened before, after, or simultaneously to Bob's.

An easy way to see why no influence is possible between the two parties, is to realise that if an arbitrary channel $\Phi_A$ is applied by Alice, and the initial shared state is $\rho$, then the resulting overall state is $(\Phi_A\otimes I)(\rho)$. The state on which Bob operates/which Bob observes (assuming no other communication channel between them) is then $$\operatorname{Tr}_A[(\Phi_A\otimes I)(\rho)] = \operatorname{Tr}_A(\rho),$$ which doesn't depend on the channel $\Phi_A$. You can observe this directly by expanding the above expressions and using the property of quantum channels of being trace-preserving.

Answer 3

Yes, otherwise the parties $A$ and $B$ could employ such a protocol to communicate, which is impossible by the No-communication theorem.