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Why does a BBO crystal create entangled photons?

  1. Oct 14, 2013 #1
    Hi,

    If we pump a beta barium borate (BBO) crystal, we get one circle of vertically polarized photons ##\left|V\right\rangle## and an intersecting circle of horizontally polarized ones ##\left|H\right\rangle##: http://quantum.ustc.edu.cn/old/img/image002.gif
    At the intersection points of the circles, we get entanglement
    $$\left|\psi\right\rangle= \frac{1}{\sqrt{2}}\left(\left|H\right\rangle\left|V\right\rangle+ \left|V\right\rangle\left|H\right\rangle\right)\enspace.$$
    Why do we get an entangled state and not just a mixed state
    $$\rho= \frac{1}{2}\left( \left|H\right\rangle \left|V\right\rangle\left\langle H\right|\left\langle V\right|+ \left|V\right\rangle\left|H \right\rangle\left\langle V\right|\left\langle H\right|\right)\enspace?$$
    I know we can confirm the entanglement in many experiments, but I guess somebody first had to come up with the idea that this actually creates entanglement.

    As a related question, in this case there is no relative phase shift, but I've also read papers where the BBO created
    $$\left|\psi\right\rangle= \frac{1}{\sqrt{2}}\left(\left|H\right\rangle\left|V\right\rangle- \left|V\right\rangle\left|H\right\rangle\right)\enspace,$$
    on what does the relative phase depend?
     
  2. jcsd
  3. Oct 14, 2013 #2

    DrChinese

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    I guess one way to say it is that the sources are indistinguishable. That always means it is a superposition, and therefore an entangled system. Not sure if that answers your question.
     
  4. Oct 14, 2013 #3
    I think it does actually, but leads to some follow-up questions:

    1. When are the photons truly indistinguishable? Say we take the beam from an intersection point and let it pass a huge lens and find out that we missed the intersection point by a tiny distance, so we can see from which circle the photons come from and they are therefore distinguishable. If the intersection point is actually a mathematical point we will always miss it by some distance, so the photons are always distinguishable.

    2. Couldn't we just take photons from two absolutely different sources and make them indistinguishable using a beam splitter and create entanglement? It doesn't work classically since the sources would not be coherent, but if we used single photons?
     
  5. Oct 14, 2013 #4

    DrChinese

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    Great question!

    Actually you CAN do experiments based on this concept. Particles from different laser sources can be entangled using a difficult to achieve technique. If the lasers are pulse locked (synched) the source can be indistinguishable. While you cannot necessarily do anything with that by itself, you can use that source to create entangled pairs and can then swap the entanglement between pairs.

    The final result is an entangled pair known to be from 2 distinct sources (and you know which is the source) that have themselves never interacted. This result can be deduced from my original statement (as you have done) but would not be the kind of thing you would be likely to guess otherwise. See figure 1 in the below, this is some pretty fancy stuff!

    http://arxiv.org/abs/0809.3991
     
  6. Oct 14, 2013 #5

    DrChinese

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    This too is a good question.

    If I take a conic area and route everything in that area someplace, I won't any longer know the exact source point (assuming timing and path length differences being minimized). That is what they do for both intersection points. So the collection area is larger than you might otherwise guess.
     
  7. Oct 14, 2013 #6
    The paper is very interesting, but it again leads to my first question: It may be possible to synchronize the lasers, but how are you going to make the paths 2 and 3 *exactly* the same length? Maybe we cannot distinguish the photons by their arrival time now, but in 100 years we might have much preciser clocks that could tell the source because one path is slightly shorter than the other. Same with the orientation of the BS: You cannot *exactly* get the angle right, so if we make the paths after the beam splitter a few miles longer the beams would actually diverge and we could tell the source.
     
  8. Oct 14, 2013 #7
    I think the photons become distinguishable when they don't hit the PBS (or BS) at exactly the same time (i.e. the paths leading to the BS aren't the same in length).

    I think in most experiments you can get away with the photons not hitting the BS at exactly the same time and still observe interference (thus them becoming entangled), but not sure at what point you can distinguish between the two if they hit at different times.
     
  9. Oct 14, 2013 #8

    DrChinese

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    It doesn't work quite like that (as you really know since otherwise it wouldn't work).

    Getting 2 & 3 to be the same length is a matter of tuning, making a slight adjustment to one path until you see the effect. Like an interferometer setup, which depends on matching as well.

    There is some uncertainty in the system, so you can't expect the source times to be knowable. Ultimately, you end up with some pairs in which 2 & 3 arrive sufficiently contemporaneously to create the desired outcome.
     
    Last edited: Oct 14, 2013
  10. Oct 14, 2013 #9

    Cthugha

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    They do not have to be exactly the same length. It is enough if they overlap with a precision equal to or better than the coherence time of the beam. All photons inside one coherence time are per definition indistinguishable anyway. For femtosecond pulses like those used here the coherence time is the same as the duration of the pulse. This precision can routinely be reached in labs using piezos. Depending on the piezo used, you have accuracy somewhere in the nm range, which corresponds to temporal accuracy somewhere in the attosecond range. More precise clocks will not help. Measurement precision cannot beat the uncertainty principle.
     
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