2
$\begingroup$

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).

$\endgroup$
3
  • 1
    $\begingroup$ Just to make sure: There is no communication (e.g. of measurement outcomes) between A and B, right? $\endgroup$
    – M. Stern
    Jun 29 at 8:44
  • 1
    $\begingroup$ Yes, as the operations in question occurr on different subsystems, so their actions will commute(provided, like @M.Stern says, there is no communication) $\endgroup$ Jun 29 at 11:24
  • $\begingroup$ correct. no communication during A and B's operations on their qubits. A and B may still decide to share their measurement results after they are both done performing all their operations and measurements on their entangled qubit pair halves. $\endgroup$ Jun 29 at 20:56
3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Ok. The deferred measurement principle is what makes this impossible. Thanks. $\endgroup$ Jun 30 at 16:48
  • $\begingroup$ do you know of a good source showing a proof for this? $\endgroup$ Jun 30 at 16:50
  • $\begingroup$ @QuantumGuy123 I don't know offhand a reference showing that the two tensor products commute. The deferred measurement principle is covered in Mike and Ike. Neither is particularly tricky to prove; you can probably prove them for yourself. $\endgroup$ Jun 30 at 17:33
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ updated my question clarifying why this does not break the no-communication theorem. does what you are saying still apply? $\endgroup$ Jun 29 at 19:48
3
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ updated my question clarifying why this does not break the no-communication theorem. $\endgroup$ Jun 29 at 19:48
  • $\begingroup$ my initial description was a bit vague at the end. $\endgroup$ Jun 29 at 19:49
  • $\begingroup$ I don't really follow your edit, in particular, I don't understand why you claim it doesn't break the no-communication theorem? The statistics can't be different in the scenarios otherwise it would enable A and B to signal each other faster than the speed of light. $\endgroup$
    – Condo
    Jun 30 at 13:16
  • $\begingroup$ This part: "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)." From both $A$ and $B$'s viewpoint, the qubits are the same in Case (1) and Case (2). To figure out which case they are in, they must share their results (which can not be done faster than the speed of light) $\endgroup$ Jun 30 at 16:45
  • 1
    $\begingroup$ Craig answered the question. $\endgroup$ Jun 30 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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