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Entanglement Swapping

  1. Feb 23, 2012 #1
    Hi there,

    If we set-up an experiment for entanglement swapping, the two entangled pairs of photons are:
    Group 1: A, B
    Group 2: X, Y

    Take B and Y and perform a Bell-state measurement on them.
    Obviously A and X are now entangled. But for this to occur, must B and Y have no definite polarisation?
     
  2. jcsd
  3. Feb 23, 2012 #2

    DrChinese

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    I think the requirement is that the identities of B and Y are indistinguishable in the sense you don't know which is which.
     
  4. Feb 23, 2012 #3
    Hmm, I wonder how entanglement swapping is dealt with in Fock space, where not only individual particle identity but also particle number is lost.
     
  5. Feb 23, 2012 #4
    Okay:

    Because if we measure particles A and X, the entanglement with B and Y is lost. So when you go to entangle B and Y, because no entanglement exists between A and B, and X and Y, A and X won't become entangled?

    When we talk of teleportation, is it when we measure A, Y takes on the same polarisation?
     
  6. Feb 24, 2012 #5

    DrChinese

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    Just remember that the *ordering* does not matter in a swapping setup. You can measure A and X before you swap the entanglement via B and Y and the results are the same. Strange but true!
     
  7. Feb 24, 2012 #6
    Even if the entanglement is broken between A and B, because you've measured A?
     
  8. Feb 25, 2012 #7
    Okay, I've read this paper: http://www.univie.ac.at/qfp/publications3/pdffiles/2001-06.pdf

    And I am lost!!

    So what I understand: photons 1 and 4 show entanglement correlations; i.e. both are HH or VV, never seeing HV or VH.

    "This is the so-called entanglement swapping , which can also be seen as teleportation either of the state of photon 2 over to photon 4 or of the state of photon 3 over to photon 1."
    This is where I am confused, especially if photons 1 and 4 already have polarisations. Teleporting the state of photon 3 to 1, etc. in this case makes my brain fry.
     
  9. Feb 26, 2012 #8

    DrChinese

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    Keep in mnd that they 1 and 4 photons did not have a well defined polarization when the swapping occurred. Now keep these 2 additional points in your head: forget the ordering, it doesn't matter. And remember that swapping only occurs sometimes, just when a certain results occurs with 2 and 3. I.e. a subset.
     
  10. Feb 27, 2012 #9

    zonde

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    Correlations for photons detected in detectors 1 and 4 show up only when you condition them on detections in detectors 2 and 3. It's very similar to correlations in delayed choice quantum eraser.
    I wouldn't say that teleportation is very good word for entanglement swapping.
     
  11. Feb 27, 2012 #10
    If we implement the same experimental set-up, and measure photons 1 and 4 before measurement occurs on photons 2 and 3 - I assume when the polarising beam splitter combines particles 2 and 3 this is when the entanglement switch occurs. So if we detect photons 1 and 4 in both H (or V polarisations), prior to reaching the filters and detectors photons 2 and 3 take on V (or H polarisations) - the opposite of what the others take?
     
  12. Feb 27, 2012 #11

    zonde

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    Ok, we should detect only HH or VV in detectors 1 and 4 for particular type of correlations observed in that experiment. But we detect HV and VH as well and at the same rate as HH and VV.

    So how they disappear when we perform Bell state measurement? We throw them out because their pair photons reach the same detector (either both reach detector 2 or detector 3) and so we don't have 4-fold coincidence in 1,2,3 and 4.
     
  13. Feb 29, 2012 #12
    I see.

    If we analyse photon 3 in the 45 basis, according to QM it gets entangled with the filter so it both passes and fails at the same time. If we then measure photons 1 and 4, so they show VV or HH, will photon 4 take the opposite polarisation of photons 1 and 4, or will it written in the 45 basis so take on V or H, without dependence on what photon 1 takes?
     
  14. Mar 1, 2012 #13

    zonde

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    I am not sure I understand what measurement settings has polarization analyzer before detector 2. It seems you have mixed up detector numbering somewhere.

    Anyways you get entanglement for photons 1 and 4 only (in 4-fold coincidences) when you measure photons 2 and 3 in +/- (+45/-45) basis after PBS.
     
  15. Mar 8, 2012 #14
    When we detect HV and VH, would that when four-photon entanglement has NOT been created successfully?


    For entanglement swapping: is entanglement created between photons 2 and 3 when they hit the beam splitter (in this experiment scheme: http://www.univie.ac.at/qfp/publications3/pdffiles/2001-06.pdf)?

    Is entanglement swapping the same thing as creating four-photon entanglement?
     
  16. Mar 10, 2012 #15

    zonde

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    When we detect HV in 1 and 4 we have two photons in 3 and no photons in 2.
    When we detect VH in 1 and 4 we have two photons in 2 and no photons in 3.

    And what do you mean by four-photon entanglement?

    No. Besides in this experiment it is explained as creation of entanglement between photons 1 and 4. But basically you can take any two detectors, perform measurements in +/- basis and claim that 4-fold coincidence reveals entanglement between the other two.

    So entanglement swapping happens when you register 4-fold coincidence in coincidence counter i.e. when filtering is performed.
     
  17. Mar 13, 2012 #16
    So in this case, there'd be no entanglement between 1 and 4?

    Four photons entangled.


    Entanglement swapping is NOT a creation of entanglement between photons 1 and 4?

    If I measured photon 2 in 45 basis and 3 in 135 basis, wouldn't that indicate whether the photons were described by the GHZ state? If one passes, the other fails. From the GHZ state, you can't have photon 2 in 45 and photon 3 in 135.
     
  18. Mar 13, 2012 #17

    zonde

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    Right

    Wrong

    If you measure photon 2 in 45 basis and 3 in 135 basis then photons 1 and 4 will have positive correlation for +45 and +135 measurement and negative correlation (minimal rate of four-fold coincidences) for +45 and +45 measurement.
     
  19. Mar 13, 2012 #18

    DrChinese

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    No, they are not the same. You can entangle N particles where N>2. The statistics are different than 2 particle entanglement, because with N particles there is not the "perfect correlations" as in 2 particle states. This does occur sometimes with N=4.

    With entanglement swapping, you have 2 entangled pairs (which is N=4) but the 4 are never entangled all at once.
     
  20. Mar 14, 2012 #19
    So we have photons 1 and 2, and 3 and 4 entangled. We send 2 and 3 to beam splitter and we hope they get combined (I'm assuming the authors do this to create the four photon entanglement). Is this four-photon entanglement? From the title of the paper, I'm sure they're speaking of all four photons, 1,2,3,4 as entangled with each other. So if so, is the entangled state is the |GHZ> given? I fail to see photon 2 taking on 45 and 3 taking on 135 being possible if described by this state, when you 'sum over' (45+135)(45-135)(45-135)(45+135)+(45-135)(45+135)(45+135)(45-135).

    But you can talk about projecting a certain bell-state onto the photons. I want to measure 2 in |45> and 3 in |135> without doing any projecting of bell states. Surely this is possible? I'm interested in what is possible if four photons are entangled, which I assume is described by |GHZ>.

    Of course to ensure you have the |GHZ> state, you need to make sure you detect it. But shouldn't this |GHZ> state exist independent of whether we measure all four photons in the |45> basis and detect 4-fold coincidence?

    "In entanglement swapping 1 and 2 are entangled, and 3 and 4 are entangled. Then 2 and 3 are jointly measured in a particular way. At the end *conditioned on the result of the measurement*, 1 and 4 are entangled. So, indeed, at no time do any of the photons have a definite polarization."

    Conditioned on the result of the measurement? So not all the times 2 and 3 are measured (2 and 3 going to seperate detectors) will 1 and 4 be entangled? I'm thinking I might have that wrong though.... Surely each time you measure 2 and 3 (in whichever basis), it forces them to take on a particular bell-state, and hence 1 and 4 the same bell-state = entanglement each time.
     
    Last edited: Mar 14, 2012
  21. Mar 15, 2012 #20

    zonde

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    You can detect two-fold coincidences in 2 and 3. Only there would be no correlations. What is the use of that?

    Not sure. One rather trivial thing is that two entangled pairs do not have correlated emission. So you have to pick out rather accidental emissions of two pairs.

    I will try to explain phrase "conditioned on the result of the measurement" with idealized experiment:
    Let's assume we have ideal source that emits only pairs of entangled pairs. You get one photon from each entangled pair and I get one photon from each entangled pair.
    You perform measurements on your pair and record results. I perform Bell state measurement with other two photons and record result. My Bell state analyzer is idealized and therefore can distinguish all four possible Bell states. So I have one of four possible outcomes for every pair.
    Before we meet your results show no correlation. Then we meet and based on my result we sort your results in four subsets (we are "conditioning on the result of the measurement") each corresponding to one of the four Bell states [itex]\Psi^{+},\ \Psi^{-},\ \Phi^{+},\ \Phi^{-}[/itex] and within each subset they show expected correlations for that Bell state. And there we are - all of your photons where entangled all along. You just didn't know what pair belonged to what Bell state. :rolleyes:

    And you might try to look at this experiment:
    High-fidelity entanglement swapping with fully independent sources

    In this experiment they have identified two Bell states out of four for D1 and D4. So this setup is a bit closer to that idealized experiment.
     
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