I am CS/Math person I am not quite sure what it means in physics information cannot be transferred faster than light (FTL). I tried to understand the proof of no transfer theorem but I lacked the understanding of quantum mechanics ( actually physics in general ).

So I would be grateful if somebody with physics knowledge answered my questions.

The experiment goes like this:

  1. Alice and Bob is given some pairs of entangled qubits ( A1, A2,... ; B1, B2, ...) Ai are entangled with Bi.

  2. Alice and Bob are both aware of the each other's basis for the measurement beforehand when certain event happens.

  3. (Ai) and (Bi) are separated.

  4. Both agree that they will perform the measurement approximately after certain time T has passed.

  5. After time T if certain event happened in time interval ( 0 , T ) to B , B measures the qubits w.r.t the sequence of basis that A already knows. ex) A knows that if B is injured B will measure the qubits (Bi) with basis ( b1, b2, b3, ... ) , if B is happy and well B will measure with ( c1, c2, c3... ).

In this case is it possible that if something happened to B in (0,T), A and B measures the qubit on some time in ( T, ~ ) ,the information that certain event happened to B transferred FTL to Alice?

  • $\begingroup$ Consider the question which basis A uses, and what happens in either of the cases. $\endgroup$ – Anders Sandberg Jan 10 '19 at 1:46

You are assuming that the result of measurement depends on the choice if basis. It is true for Bob when he measures his qubits but there is no change for Alice. Suppose they share just one entangled qubit for simplicity in the state $$ |{\Phi^+}\rangle = \frac{1}{\sqrt{2}}(|{0_A}\rangle|{0_B}\rangle + |{1_A}\rangle|{1_B}\rangle). $$ Let us say that Bob is injured in the interval $(0,T)$ and they have agreed that Bob will measure in the basis $\big\{(1,0),(0,1)\big\} $. Bob will now measure either a zero or a one with $50\%$ probability. Also because of this measurement Alice's qubits will collapse onto the state Bob measured. When Bob measured $|{1_B}\rangle$ Alice's qubit will become $|{1_A}\rangle$ as well.

After some time $T$ they will both measure the state as agreed. For Bob this is a bit stupid because he already knows what state the qubits are in. Alice doesn't, so she will measure the $|{1}\rangle$ with $100\%$ probability. But this could also have been a $|{0}\rangle$ because the measurement of Bob could also have been a $|{0}\rangle$, it was just chance. So Alice doesn't actually know for certain how Bob felt, just for $50\%$ certainty.

Bob and Alice are actually measuring eigenvalues and eigenvalues don't change under basis transformations so it does not matter if Bob measures in basis $b$ or $c$.

You might be interested in the paper on quantum teleportation if you haven't read it :https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.70.1895. A qubit is actually teleported from Alice to Bob but special relativity is not violated because it always comes at the cost of sending classical information fro A to B which is restricted by the speed of light.


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