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A The Spooky Speedometer

  1. Sep 12, 2017 #1
    Hi There,

    I've been spending some of my free time learning about Quantum Mechanics. This was all going somewhat smoothly, until I encountered something known as "Entanglement Swapping" while learning about loophole free bell tests, such as the one conducted by Hensen et al at TU Delft in 2015.

    While I'm no professional physicist and thus it is highly likely that I'm misunderstanding something. It appears that entanglement swapping can be exploited to determine the absolute velocity of object, in defiance of the first postulate of special relativity. To explain why this appears to be the case, I've devised a thought experiment called the Spooky Speedometer.

    Experiment Configuration

    On board a spacecraft, Alice and Bob float into a room dressed in space suits and evacuate all the air out of the room.Alice takes her place on left side of the room, while Bob takes his place on the right. Alice places two tiny diamonds, each containing a nitrogen vacancy, side by side. Bob in turn, sets up an entanglement swapping device at the other end of the room.

    Alice also brought two lasers into the room, which she can use to excite a trapped electron within each of the diamonds, causing each to emit a photon that is entangled with its emitting electron. Once the photons are emitted, Alice can choose at any moment to measure the spin of both her electrons.

    Alice's diamonds are configured so that emitted photons will traverse the vacuum of the room and enter Bob's entanglement swapping device simultaneously. This will result in the electrons, which were not initially entangled with each other, becoming instantaneously entangled.

    Performing the Experiment
    Initially, Alice uses her lasers to cause her electrons to emit entangled photons and then immediately measures her electrons. She repeats this process a number of times, in order to determine with a high degree of certainty, if her electrons are entangled. However, as insufficient time has passed for the photons to reach Bob's entanglement swapping device, they will not be entangled.

    Alice subsequently increases the time elapsed between emitting and measuring, in very small increments,
    performing multiple measurements at each increment, until an elapsed time is reached which produces entangled electrons. This elapsed time is the same time taken for the photons to traverse from Alice to Bob.

    Analysing the result
    If the spacecraft is stationary, the time required for the photons to traverse the room, can be calculated by the below formula:

    tAlice = d/c

    where:
    tAlice: The time it takes for the photons to traverse the room according to Alice's experiment
    d: The distance between Alice and Bob when the spacecraft is at rest.
    c: The speed of light in a vacuum

    However, if the spacecraft has motion along the axis of the photon pathway, the formula is a little more complex for the observer at rest. For instance, the stationary observer will measure the distance between Alice and Bob as reduced due to Lorentz contraction:

    d/γ

    Where:
    γ: The Lorentz factor

    Similarly, Alice's measurement of time is slowed by Lorentz dilation:
    trest = γtAlice

    Where:
    trest: The time it takes for the photons to traverse the room according to a stationary observer

    Additionally, for an observer at rest, the relative speed of the photons in relation to the spacecraft, will vary from speed of light, if the spacecraft is in motion:

    c + v

    Where:
    v: The absolute velocity of the spacecraft along the axis of the photon pathway (motion from Bob to Alice considered positive)

    Thus, for a stationary reference frame, the time taken for the photons to traverse the room is given by the below formula:

    trest = γtAlice = d/(γ(c + v))

    Once Alice has determined the elapsed time via the experiment, she can rearrange the prior formula and determine the absolute velocity of the spacecraft along the axis of photon pathway:

    v = c - ((tAlicec^2)/d)

    [see page three of link for each step: https://drive.google.com/open?id=0B3K5GT7ukRAKM05PYVVmSC1XejA ]

    Alice could have approached measuring the elapsed time in a simpler manner, by asking Bob the time at which the photons reach his entanglement swapping device and deducting from it the time which she emitted them. However, this measurement method would be impacted by time dilation, length contraction, aberration of light and relativity of simultaneity, so that when inputted into the above formula, it would always output the spacecraft as being stationary.

    By utilising entanglement swapping (as I understand it to work), the spooky speedometer experiment can determine the absolute velocity of an object (the spacecraft in this thought experiment), along the axis of the photon pathway.

    Thus, this leads me to conclude that either:
    1. My understanding of entanglement swapping is flawed or,
    2. Special Relativity is flawed.

    Obviously the later would be highly improbable. So, I'm expecting that one of you will be able to articulate why it is the former. Any assistance getting to the bottom of this is much appreciated. If you want me to elaborate, please feel free to ask.
     
  2. jcsd
  3. Sep 12, 2017 #2

    DrClaude

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    How can Alice figure out if the electrons are entangled?
     
  4. Sep 12, 2017 #3
    Alice must preform multiple measurements of the elections (following emission of photons) at each increment. If the electrons do not exhibit correlation across the measuerments, she can be certain the duration is too small.

    If correlation is present, there is chance of a false positive, though this chance is reduced with a greater number of measurements.
     
  5. Sep 12, 2017 #4

    DrClaude

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    There is no way this can be done. There is no measurement Alice cannot perform that will tell her if the particles are entangled. Entanglement can only be figured out after measurements on many identically prepared systems.

    Also, any measurement of an electron entangled with a photon will break the entanglement.
     
  6. Sep 12, 2017 #5
    I agree that no individual measurement can determine if the particles are entangled. It is only once you repeatedly conduct the process (excite electrons with lasers to emit photons) under the same parameters a sufficient number of times, that the data resultant can be analysed to make a conclusion with a reasonable level of confidence.

    Its my fault for not being clearer, but i think that your reading it as "emit photons, measure electrons, measure electrons, measure electrons....." If i was more clear it would be a little more obvious that the process is "emit photons, measure electrons, emit photons, measure electrons...." for each increment. Each of these repetitions is independent from the other; they are as you say " identically prepared systems".
     
  7. Sep 12, 2017 #6

    DrChinese

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    Well, this particular point happens to be incorrect. There is no amount of time T that acts as a barrier to entanglement swapping. You can swap entanglement before, during, or AFTER measurement of Alice's newly entangled electrons. You have not read my statement incorrectly, and it may seem paradoxical that Bob can entangle (via swapping) after Alice has completed her measurements. Elapsed time (or distance) is NOT a factor. See this reference from one of the top teams in this area:

    Experimental Nonlocality Proof of Quantum Teleportation and Entanglement Swapping
    Thomas Jennewein, Gregor Weihs, Jian-Wei Pan, Anton Zeilinger
    https://arxiv.org/abs/quant-ph/0201134

    "...Alice’s [Bob in your example] measurement projects photons 0 and 3 [Alice's measurements in your example] into an entangled state after they have been measured."
     
    Last edited: Sep 12, 2017
  8. Sep 12, 2017 #7

    DrChinese

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    This is correct of course. But I want to explain to the OP why it is correct.

    When you perform entanglement swapping, there are 2 important swapped states we care about. Call one the + state, the other the - state. In one, the resulting entanglement is to a state where both polarizations are the same (+). In the other, the resulting entanglement is to a state where both polarizations are different (-). So to make sense of the results, you need to know which of those 2 states were created. Bob will need to tell Alice that.

    Of course, since Bob and Alice are only a short distance apart, Bob can simply tell Alice classically. So she needs input from Bob to figure out whether there is entanglement or not - he has the essential key for that. Otherwise, Alice is simply looking at a random set of data values with no meaning whatsoever.

    So now, with Bob's help, Alice can determine entanglement. Of course, my post #6 explains that time/distance changes nothing in this experiment. :smile:
     
  9. Sep 12, 2017 #8
    Entanglement isn't an event that happens at a particular time. Therefore, the term "instantaneous" doesn't apply to entanglement. If the two electrons are entangled, then they are entangled at all times, including before they interacted. Whether this is "retrocausality" depends on your interpretation of quantum mechanics, but it can't be used to send signals faster than light or back in time, only to check that signals agree after the fact. QM is well tested by experiments such as the delayed choice quantum eraser, and always shows non-local behavior.
     
  10. Sep 13, 2017 #9
    Thank you for your responses DrChinese,

    You raise some very interesting points. Could we focus on the one made my DrClaude, which you then clarified firstly, the cover off the other major point regarding the instantaneity of entanglement?

    I'm having difficulty understanding how the measurement of Electron A is independent of Electron B if they are in fact entangled with each other. Consider the wiki definition:

    "Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently of the others." ​

    If you can't determine that Electron A and Electron B are entangled from measuring them alone (after multiple repetitions) how can they be entangled?
     
  11. Sep 13, 2017 #10

    Nugatory

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    Do you mean repeated measurements on the same two electrons (measure A repeatedly, measure B repeatedly) or do you mean that we prepare a large number of pairs of electrons in the same way and perform one measurement on each member of each pair (that is, run the experiment from the beginning many times, measuring A once and B once for each repetition)?

    Only in the latter case is it possible to determine whether our experimental setup was producing entangled pairs, and then only after comparing results of the measurements at both sides after many runs of the experiment.
     
  12. Sep 13, 2017 #11

    DrChinese

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    I think in your example, the issue of being able to determine the presence of entanglement from a single example is not what you care about. You are thinking that a repeated series of measurements will indicate entangled correlations. Yes, that is the case. And it doesn't matter for your example that a) this would be determined a bit of time later; or that b) it will take information from Bob to properly filter the measurements into + or - entanglement cases (since Alice can't determine that on her own). In other words: the + entanglement cases are groups to determine correlations, and that would need to be separate from the - cases which will have opposite correlations.

    So let's agree that Alice and Bob working together, after the fact, and with enough trials, they can see that there is entanglement occurring.
     
  13. Sep 13, 2017 #12
    The latter. the Spooky Speedometer is an experiment made of smaller experiments, which I've been referring to as repetitions, so not to confuse it with the larger experiment.
     
  14. Sep 13, 2017 #13
    The efficacy of the Spooky Speedometer is dependent on entanglement swapping being instant.

    Whist this is the key issue, as requirement a) & b) can be satisfied. I'm having trouble understanding the necessity of requirement b). I just can't comprehend how two particles can be considered entangled without consistency between their correlations (always + or -).

    I would like understand this better before moving to the key issue.
     
  15. Sep 14, 2017 #14

    DrChinese

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    There are actually 4 Bell states possible when swapping. Usually only 1 or 2 are used, which for simplicity I refer to as + and -. In the earlier quoted paper (see post #6), they give a fairly detailed explanation which you can read. The important thing to understand is that when the swap occurs, you don't have any control over which of the Bell states result. That is random, but you can see that result and use it. I extracted the following, where the + and - states are called Psi+ and Psi-. The photons labeled 0 and 3 are the ones Alice measures (your example) and those labeled 1 and 2 are the ones your Bob executes the swap on. The below names Alice/Bob/Victor to not correspond to your example, so don't be thrown off by that.

    "Alice subjects photons 1 and 2 to a measurement in a Bell-state analyzer (BSA), and if she finds them in the state |Ψ−>12, then photons 0 and 3 measured by Bob, will be in the entangled state |Ψ−> 03 . If Alice observes any of the other Bell-states for photons 1 and 2, photons 0 and 3 will also be perfectly entangled correspondingly. We stress that photons 0 and 3 will be perfectly entangled for any result of the BSA, an d therefore it is not necessary to apply a unitary operation to the teleported photon 3 as in the standard teleportation protocol. But it is certainly necessary for Alice to communicate to Victor her Bell-state measurement result. This will enable him to sort Bob’s data into four subsets, each one representing the results for one of the four maximally entangled Bell-states."

    So what we have is some subset of all the swapped entangled pairs. Once the subset(s) is(are) properly identified to your Alice, she can see that the appropriate quantum correlations appear.
     
  16. Sep 15, 2017 #15
    I think I'm going to have to do a whole lot more reading (probably this weekend) to try and understand this more.

    I've had a quick second glance at Hensen et al ( https://arxiv.org/pdf/1508.05949.pdf ) as I do not remember Lab C (which is Bob in the spooky speedometer) performing any other measurement than detection of photons. On page two it says:

    "In an event-ready scenario, an additional box C gives a binary output signalling that the boxes A and B were successfully prepared"

    and

    "Setup at location C. The single-mode fibres from locations A and B are connected to the input ports of a fibre-based beam splitter (FBS) after passing a fibre-based polarizer (POL). Photons in the two output ports are detected using single photon counters, and detection events are recorded."

    My understanding is that the fidelity of the fibre optic lines is not perfect and thus a percentage of the photons do not make it to Lab C. Photon detection is required, so that when photons are lost and thus entanglement swapping does not occur, the repetition can be classed as not 'event ready' and data collected from Lab A and B removed from the analysis. Is this correct or am I missing something?
     
  17. Sep 18, 2017 #16

    DrChinese

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    Yes, you will want to read a bit deeper. In the last full paragraph of page 3:

    "We generate entanglement between the two distant spins by entanglement swapping in the Barrett-Kok scheme 26,27
    using a third location C (roughly midway between A and B, see Fig. 1e) ... The two photons are then sent to location C, where they are overlapped on a beam-splitter and subsequently detected. If the photons are indistinguishable in all degrees of freedom, the observation of one early and one late photon in different output ports projects the spins at A and B into the maximally entangled state Psi-..."


    So yes, this is where the event ready determination is made, and it corresponds to a successful entanglement swap.
     
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