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Recently I have been reading about quantum information theory, including how to perform quantum teleportation, logic gate operations, bell experiments and entanglement swapping.

If I understood correctly, a Bell measurement can result in two uncorrelated particles to become entangled by basically projecting the 2 particle state into one of the bell states. This is a crucial element in teleportation and entanglement swapping.

However, experimentally, it seemed Bell measurement rely on the two particles in question to arrive at the two detectors (with setting correspond to a bell basis) at the same time thus allowing a joint correlation to be registered in the computer.

This is so far so good if the particles are photons (since photon speed are invariant across any inertial frames of reference)

Clarification: I am interested in the Bell measurement step in the teleportation scheme. Thus all answers do not need to involve what happens during the teleportation step. That is, for all purpose, we can ignore what happened to that party on the other side.


Question 1

  1. Consider a teleportation scheme using electron spins. At the 1st step of the protocol, electrons 1 and 2 are generated as an entangled pair. Electron 3 is prepared in some state and to be teleported soon.
  2. Now a bell measurement is performed on 2 and 3 to entangle them, and hence breaking the entanglement with 1.
  3. There are two observers A and B. A is stationary wrt the electron beam arriving at the two detectors, while B is moving at some % of the speed of light wrt the detectors
  4. The two electron beams will look simultaneously arriving the detectors as seen from A, thus they should collapse into a bell state
  5. But for B, one electron beam will arrive earlier than the other

(NB To avoid confusion, A and B are in the same lab where the teleportation is prepared. The other party where electron 1 is kept is not the focus of this question)

a. Is simultaneity of arriving at detectors a requirement for a Bell measurement?

b. If no, then does it mean in order for all inertial frames to agree with the physics occurred, all observers will agree a bell state will be produced by the Bell measurement on two uncorrelated particles?


Question 2

  1. Suppose (for curiosity?), the two detectors that does the bell measurement are separated to a distance of 10 light seconds away. Two uncorrelated photon beams were used in this scheme. That is, the two detectors d1 and d2 used to perform a bell measurement are now separated by a distance of 10 light seconds apart.

  2. A Bell measurement is then performed as the two photon beams arrived at the respective detectors simultaneously (This is ensured because the frames where the detectors are situated in are comoving)

But from LOCC, entanglment is not allowed to increase by local measurements in the absence of entanglement swapping. In the above setup, the two detectors are basically spacelike separated (since it took 10 light seconds for any classical information to move between them), thus any detection of the photon on each detector will be in a sense a local operation, thus two uncorrelated particles should not became entangled due to this logic.

But Bell measurements can create entanglement, so how to reconcile between the results of LOCC and bell measurement. Will the two photons became entangled in this setup?

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  • $\begingroup$ just to answer your question (a): simultaneity is not required. After Alice measures her part of the system, Bob can wait as much as he wants to measure his bit. As long of course as his part of the state does not decohere nor interacts with anything else. When eventually he decides to do the measurement, Alice and Bob can compare their result, compute the correlations, and prove Einstein wrong once again. $\endgroup$ – glS Oct 26 '16 at 14:32
  • $\begingroup$ To clarify, A and B are not Alice and Bob. In the edited question, A and B are located in the same lab where electrons 2 and 3 were kept. However I think your answer to (a) "simultaneity is not required" might still hold $\endgroup$ – Secret Oct 26 '16 at 14:45
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Is simultaneity of arriving at detectors a requirement for a Bell measurement?

No, it's not required. For example, here is a circuit that does a Bell measurement one input at a time:

Time-separated bell measurements

However, note that this circuit has to hold two qubits coherent in the time between A and B's interactions. That's hard to do, so it may be experimentally convenient to have the interactions happening at the same time.

all observers will agree a bell state will be produced by the Bell measurement on two uncorrelated particles?

I'm not sure I get what you're asking. Of course a Bell measurement produces a Bell pair, and how fast you are moving won't affect that.

how to reconcile between the results of LOCC and bell measurement. Will the two photons became entangled in this setup?

In order to perform a teleportation, or a non-local / spatially-separated Bell measurement, you need a Bell pair. One bit of entanglement goes in, you use it to do something fancy, and the original entanglement has been burned up.

But actually, because you used it to do a Bell measurement, you also had one bit of entanglement come out. So I wouldn't say you're destroying the bit of entanglement, or creating a new bit of entanglement, I'd say you're remixing the 1 bit of entanglement into a different form. Nothing gained, nothing lost, everything consistent.

(Also note that quantum teleportation and non-local Bell measurements have a classical communication step. So if A and B are galaxies apart, it's going to take a long time to finish remixing that bit of entanglement.)

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  • $\begingroup$ Clarification on "I'm not sure I get what you're asking": I mean for any observers that are moving relative to the Bell measurement set up, hence will see the two electron beams arriving one after the other, or simutaneously, whether they will all agree that the uncorrelated particles became entangled. Based on your answer on a), it seems the answer is yes. So theoretically one can watch the experiment while travelling some % close to speed of light, and as long the particles stay coherent between the interval AB, then all frames will agree the entanglement will be formed? $\endgroup$ – Secret Oct 26 '16 at 17:05
  • $\begingroup$ @Secret The measurement you described is built up at one particular place, i.e. its separation is timelike not spacelike, so all observers will agree on the ordering. However, even if in a setup that wasn't like that (such as the non-local measurement) it wouldn't matter. The two sides of the measurement commute. Doing them in the opposite order or even overlapping will give the same result. $\endgroup$ – Craig Gidney Oct 26 '16 at 17:50
  • $\begingroup$ (hopefully) final question. For the nonlocal /spacelike bell measurement, if I understood correctly, is it that it cannot be done with two unentangled particles, and entanglement is necessary to perform it? $\endgroup$ – Secret Oct 27 '16 at 6:38
  • $\begingroup$ @Secret Right. Otherwise you could trivially create entanglement with classical communication. $\endgroup$ – Craig Gidney Oct 27 '16 at 7:24
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Measurement doesn't entangle the particles, they're already entangled. Measurement does destroy entanglement though. So measurements would need to be simultaneous for the correlation to manifest itself. If measurements are taken in succession, the latter measurement wouldn't be on an entangled system. I'm not sure what your second question is asking.

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  • $\begingroup$ Except possibly for "I'm not sure what your second question is asking", every sentence in this answer is false. To begin with the first sentence, for a mult-particle system, measuring an observable all of whose eigenstates are entangled is of course guaranteed to result in an entangled state, so measurement does entangle the particles. The rest is equally far off. $\endgroup$ – WillO Oct 26 '16 at 15:07
  • $\begingroup$ @WillO So measuring an unentangled system causes entanglement? $\endgroup$ – Yogi DMT Oct 26 '16 at 15:16
  • $\begingroup$ Did you not read my earlier comment? $\endgroup$ – WillO Oct 26 '16 at 15:29
  • $\begingroup$ @WillO Care to elaborate? $\endgroup$ – Yogi DMT Oct 26 '16 at 15:31

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