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3 photons from PDC?

  1. Oct 23, 2005 #1


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    How about this one... I have read of some attempts to create a 3 photon entangled state, but there is a lot of difficulty doing it. Could it be done by progressively splitting photons?

    1. Generate a spin entangled photon pair from a standard parametric down conversion (PDC) setup. This "splits" one pump photon (say at 405 nanometers, a common source wavelength) in a H> & V> mixture passing through a pair of Type I (parallel output) BBO crystals aligned at 90 degrees to each other. The output photons are a pair of H>H> & V>V> at half the frequency (double the wavelength) of the pump photon, so they are now at 810 nm.
    2. Take one of the photons (say the signal) in its superposition of H>H> & V>V>, and run it into another set of BBO crystals, tuned appropriately.
    3. Would that create a new pair (at 1620 nm) in some mixed state like H>H>H> & V>V>V> ? With the original idler now in the same state? Admittedly, there would only be 3 output photons occasionally and therefore the source would seem pretty weak, maybe only 1 set out every 10 seconds.

    This is breaking my brain. Maybe you can't split such a photon yet again. Yet it seems like the only thing you need to create an entangled pair in the first place is an appropriate pump photon. If you did have 3 entangled photons, think of the Bell tests you could do on them! Might be pretty interesting.

    Vanesch, ZapperZ, someone...?
    Last edited: Oct 23, 2005
  2. jcsd
  3. Oct 24, 2005 #2
    From the position of the theoretician an interesting idea but for the experimentalist it looks like lot of work for problematic hard to find gain. Problematic because wouldn’t most claim the second splitting would just collapse the prior wave function?

    Why not ask them to do a simpler search in a more direct way. Currently the “Pair” of photons are created by to crystals one at 0o – 180o the other at 90o – 270o giving two overlapping “cones of light”.
    We could ask they attempt to find crystals that might work together to divide the photon directly into three; from 410 nm into 1230 nm. Crystals aligned at 0o – 180o, 60o – 240o, and 120o – 300o. Might we get three overlapping cones of light?
    Even if the lab guy didn’t believe it could work it would not be near so hard to try.

    What would be the meaning? Asking for some advice from a couple of very smart guys; just remember it me putting words in their mouths, but this is what I’d expect.

    “Einstein” – I should like to borrow this devise once it is made, I shall use one photon to find its speed, the second to find its position, and thus know both with precision for the third. Therefore demonstrating nature is determinant!

    “Niels Bohr” – Ridiculous, QM is non-determinant and so is nature! You shall never be able to create such a device that would imply otherwise. Our theory of QM is as good a theory as can ever be made!

    I’m guessing most experimentalist would say the constraints on their equipment would concur with Niels.
    Last edited: Oct 24, 2005
  4. Oct 24, 2005 #3


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    In the experiment where they they beat the diffraction limit using entangled photons, they did use 3 photons that were "maximally entangled".

    M.W. Mitchell et al., Nature v.429, p.161 (2004).

    I don't quite have the time to refresh my memory yet on how they created this. I'll try to re-read the paper when I have the chance unless you or someone else get to it first.

  5. Oct 24, 2005 #4


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    http://xxx.lanl.gov/PS_cache/quant-ph/pdf/0312/0312186.pdf is the one you posted - I was somewhat familiar with that one. Also there is related:


    which I have not yet read. They are doing some things with their tests that angle the issue a little differently than the direction I was headed.
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  6. Oct 24, 2005 #5
    I’m always a little troubled by experiments of this type –
    Usually looking for which way info or confirming phase matching or whatever.

    But I don’t see where they have run some basic double checks to see if they are getting any interference dark and light patterns in the locations of their photo detectors.
    You have to assume you might get something like that any time you have a double slit or double ‘point source’. Even more so if more than two sources, as are here a laser point source is split into two or more.

    If they are getting patterns in the area of the photo detectors, and they have not carefully accounted for it how can the collected information be useful since the happen stance of placement could then be significant.
    Just seems a little incomplete in the work up to me, I’d assume there is some issue with patterns & detector placement within the pattern would therefore affect observations.

    It’s at least not obvious to me how the possibility can reasonable be ignored.
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  7. Oct 24, 2005 #6


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    In fact, here is a pretty good reference though from 1998:

    http://citebase.eprints.org/cgi-bin/fulltext?format=application/pdf&identifier=oai:arXiv.org:quant-ph/9810035 [Broken]

    They say they got mixtures of H>H>V> & V>V>H> using 4 photons - where one of the photons is used for discriminating the desired entangled states from within trial runs. 4-fold detection indicates the desired state was achieved.


    So, assuming we have that state in hand... If we set polarizers a, b and c at the appropriate angles for the 3 entangled photons, we should see "perfect" correlations. We will call those settings for a, b and c "0 degrees".

    But what will happen if we now set the polarizers at a=0, b=120 and c=240 degrees? According to conventional thinking, the ab, ac and bc correlations should be .25 each (cos^2 law). But this is impossible, since we already know those values violate Bell's Inequality (this is exactly what Bell's Theorem showed, that there are values of a/b/c that cannot work together).

    I.e. With these settings, if ab and ac match exactly 25% of the time as predicted, bc will match 62.5% of the time - but bc should match only 25% just like ab and ac do.

    So what gives? Depending on how you apply the projection postulate, you could get different results. Presumably, measuring at a, b or c causes collapse and now the results at the other 2 are correct.

    But can you control whether measurement a, b or c is what is causing the collapse of the wave function? It didn't matter when there were only 2 entangled photons as the stats work out the same no matter what. But the situation is different with three photons. So I would predict you would see one of the following:

    i) If you could force the "a" measurement to occur first, you would end up with the following stats:

    ab=.25, ac=.25, bc=.625 (from above)


    ii) There is no possibility of choosing whether the collapse occurs at a, b or c, and you end up with the following stats:

    ab=.33, ac=.33, bc=.33 (which are the minimum values that are equal that work out)

    If it is i), then you could use the technique to send FTL signals (maybe :smile: ), so that doesn't seem right. But if it is ii), then you don't violate Bell's Inequality, so that doesn't seem right either.

    Last edited by a moderator: May 2, 2017
  8. Oct 24, 2005 #7
    The key to this one is two pump photons going through together at the same time creating four photons and expecting them to be in superposition with each other. I can’t see any justification for that.

    And if it can be made to look like it is true, it may say more about experimental set ups being able to act more like a magicians trick than we’d like experimental observations to allow.
  9. Oct 25, 2005 #8


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    Here is an updated reference on a test of Bell Inequalities using 4 :surprised PDC photons.

    http://citebase.eprints.org/cgi-bin/fulltext?format=application/pdf&identifier=oai:arXiv.org:quant-ph/0302042 [Broken]

    This is pretty close to what I was speculating about in the OP. They are getting 1 set of 4 every 24 seconds, not far from what I had imagined. They set 3 of the 4 polarizers set at 0 degrees, and then vary the fourth. At the appropriate spot, they get "perfect" correlations.

    So why do they not simply do a three-fold test of polarization at 120 degree angles? I am dying to see a data stream when they are arranged in that pattern, as it seems to be a way to look at the inequalities from the original Bell paper directly. It would seem to me that one could compare the correlation ratio for AB, AC and BC (ignoring D which can be set the same as one of the others, as in:


    The question being, would the correlation ratio be .25 (the QM value) or .33 (the realisitic value). These values are for pairs at different angles.
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  10. Oct 25, 2005 #9
    I guess I’m just missing the point.
    I agree if they are getting 4 way or three way correlations it is very interesting and understanding why it seems that way might be very instructive.
    Because it seems to me that is what needs to be explained why does it “seem” that way.
    Why are they “seeing” entanglement across all four photons in the correlations? When there is no real logical reason to expect to them to be entangled.

    I don’t see where in any of the QM theories where superposition entanglement is used that two photons down converting at the same time should create entanglement across all four resulting photons.
    Is there some idea that two photons traveling side by side should spontaneously become entangled with each other? And then their ‘offspring’ pairs should remain entangled with each other?

    Without explaining that, how they can apply the Bell-theorem rationally?
    I’d see more cause to test for multiple entanglement were it generated more as you had first suggested in the original post.
  11. Oct 25, 2005 #10


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    It has been determined, from trial and error really, that entanglement occurs WHENEVER you lose knowledge of individual particles. That is how the multi-particle states are created. My merging particles together within very short time windows, their individuality is lost. Then the conversation rules become important.
  12. Oct 26, 2005 #11
    I’d call that a assumption rather than a “determination”, it would seem to require a relaxation of the “conservation” rules to get it to happen. And then to turn it back on again for correlations to follow the rules.

    To me this says more about asking the question:
    If tests for entanglement on four particles that can not be all co-entangled can be made to appear as if they are; then might we be able to construct a test that could create the false appearance of entanglement of two particles? I see no justification for ignoring conservation to get the first two pump photons to become entangled, to start with.

    I think the greater value of this type of experiment is to figure out if it is misleading us to just thinking we are seeing entanglement; and most important understanding how the experiment created that ‘misdirection’. Once we understand how, this could be important to double check and verify that the same thing might not have been happening in the single split into two with superposition case, maybe fooling us for decades now.
    Last edited: Oct 26, 2005
  13. Oct 26, 2005 #12


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    Randall, It actually works the other way. They use the "perfect" correlation case to demonstrate entanglement was acheived. It's pretty cool really, because there will definitely NOT be perfect entanglement otherwise. The perfect entanglement case in and of itself it not considered to be a violation of Bell's Inequality.
  14. Nov 2, 2005 #13


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  16. Nov 3, 2005 #15


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  17. Nov 3, 2005 #16


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    What about entanglement of TWELVE (count 'em) photons, DrChinese? :)

    H.S. Eisenberg et al. PRL 93, 193901 (2004).

  18. Nov 3, 2005 #17


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    :surprised :surprised My brain is exploding on this one:

    Quantum entanglement of a large number of photons

    They are talking about states of not only 12, but as many as 100!!!!!
  19. Nov 3, 2005 #18


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    Suppose we are considering an ensemble of N entangled photons (A, B, C, ...), where N>2. Let's suppose that we plan to measure their spin in such a way that we see "perfect" correlation when all polarizers are aligned at an angle we'll call 0 degrees. So we have something like this:

    P(ABC) = V>V>V>... + H>H>H>...

    Now, we already know that it is NOT possible, under QM, to learn anything more about the original state of the entangled particles than what a single measurement of 1 of the N particles gives us. This is what was pointed out in the original EPR paper, and which they assumed could not be correct. Of course, subsequent theory & experiment has supported QM. You cannot beat the HUP.

    So what we end up with instead is this: we measure photon A first and get a +. We can measure photon B, and the correlation with A will be consistent with the relative angle between them, cos^2(ab) where ab is the angle between A and B. That's fine, we always see this relation in normal 2 photon Bell tests. Ditto for correlations between A and C, A and D, etc.

    Now here is the confusion. Say A=0, B=120 and C=240 degrees. The correlation between A and B is .25. The correlation between A and C is also .25. That is because the cos^2 of the angle between is .25. So far, no problem.

    It is clear that in my example, I gave A a privileged spot in the chain, because I specify that I measure it first. Presumably it is physically possible to "appear" to make this happen. If you look at the PDC 4 photon setup, for example, you will see that the distance to the detector apparatus can be arbitrarily large. So make A's distance from the detector small, and the others large in relative terms.

    1. If I try to say AB=AC=BC=.25, which is right at first glance. But it violates Bell's Inequality and even though that shouldn't matter, it will. No set of data points can possibly give this result. So this is wrong.

    2. Now consider the correlation between B and C. This needs to be .25 to be consistent. However, the minimum correlation between B and C when the correlation between AB=AC=.25 is actually .50 (or double what we expected). This can't be right, because then an observer looking at BC alone would be able to detect this ratio. Using a suitable variation on this, one could send a signal faster than the speed of light. So this is wrong too.

    3. So I am not saying that varying the distance to the A detector will or won't make a difference, but... If I then recalculate the correlations assuming that neither A, B or C is measured first (because it is impossible to know which photon measurement caused the collapse of the wave function), I now get AB=AC=BC=.3333 which cannot be right either (because this does not match experiment, which should be .25). If it were correct, then Bell's Inequality is not violated after all. So this is wrong.

    (I have built this by extending one of Mermin's descriptions of the Bell Theorem, his with only 2 entangled photons.)

    If 1. 2. and 3. are wrong, then what really happens? Help!!!

    P.S. I also note that in the actual experiments of Eibl et al, only the data sets in which there is four-fold coincidence are considered. Obviously, you must bring those results together to analyze, and this removes any chance of actual FTL signalling very nicely. If you could see BC=.50 per 2. above, it would be clear evidence of the physical nature of the collapse of the wave function.
    Last edited: Nov 3, 2005
  20. Nov 17, 2005 #19


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    OK, here is a specific 4 photon scenario modeled starting from the Eibl experiment at this link. This post is very similar to my last post above, but I have reformulated it to more closely match the cited experiment.

    The reference includes a diagram (Figure 1 on page 4) which describes the setup in detail. Keep in mind that the 4-fold coincidence rate is very low, on the order of once every 25 seconds. Detections are recorded as either +1 or -1 after passing through polarizing beam splitters.

    They label the 4 measurement subsystems - each which can be set at any angle - as A, A', B and B'. To make my question a little easier to see, I will change B' to C so that we have A, B, C and A' as our four measurement settings. Also, in the actual experiment, the length of the path to the measurement subsystems is not specified. I want to specify that the path to A, B and C is 10 meters and the path to A' is only 1 meter. This is intended to cause the measurement at A' to "precede" the measurements at A, B and C.

    In the actual experiment, maximum coincidence occurs when A=B=C=A'. But we are not concerned with the 4 fold coincidences themselves - in my question, I am considering the subsets of 2 fold coincidences (from the full set of 4 fold coincidences) where measurement settings are A=0, B=120, C=240 and A'=A. My question is this: what are the coincidence rates predicted by QM for AB, AC and most important, BC? Is it one of the following?

    i) I would initially assume that the coincidence rates would be equal: AB=AC=BC and that each of these would be .25 (since the angle between is 120 degrees, and the cos^2(120 degrees)=.25). But that can't be true, because there is no data set for A, B and C in which that relationship holds.

    ii) I might next assume that AB=AC=BC and the each of these would be .33 (which is the lowest value possible where the pairs are equal). This matches the Local Realistic prediction!

    iii) But if I wanted to assume that the measurement at A' specifically caused the collapse of the shared wave state, then we know that A=A' with certainty. If that were true, then AB=AC and both of those will definitely be .25 (since the angle between is 120 degrees, and the cos^2(120 degrees)=.25). That makes a certain amount of sense, but... BC is no longer .25 - it will instead be .50. Is it possible, in an experiment with entangled particles, to force one of the particles to be measured "first"?

    With a pair of entangled particles, there would be no discernible difference in measurement ordering. Might there be a subtle difference if there are more than 2 entangled particles???? (Of course, to actually tell the difference you would need to examine a statistical collection of coincidence events. And that would take hours of data collection with this setup.)
    Last edited: Nov 17, 2005
  21. Nov 17, 2005 #20


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    I think the basic misunderstanding is here.
    You want to do as in the 2-particle case, the "fully correlated" system. This is of course possible, as you write down. However, there is a difference. The state:
    |z+>|z+> + |z->|z-> is ROTATIONALLY INVARIANT.
    This means that the tensor product is mathematically equal to:
    |n+>|n+> + |n->|n-> for n an arbitrary direction.

    All directions are equivalent.
    This is NOT the case anymore with 3 particles:
    |z+>|z+>|z+> + |z->|z->|z-> is NOT rotationally invariant, so it is NOT equal to:
    |n+>|n+>|n+> + |n->|n->|n-> for n an arbitrary direction.
    The explanation is that the 2-photon case is the TOTALLY SYMMETRICAL state, while this is NOT the case for the 3-photon case.
    As the calculations become a bit tedious, I fired up Mathematica.
    I set up the state. Instead of rotating the detectors, I rotated the states: A in the z+/z- direction, B in the 120 degree direction and C in the 240 degree direction, and I DETECTED always in the z-direction for A,B and C. That should be equivalent to having the photons in the z direction, and the detectors rotated ; it is simply easier to write out.
    So I defined a state |n+> = cos(th) |z+> + sin(th)|z->
    and a state |n-> = -sin(th) |z+> + cos(th) |z->
    I set:
    |psi> = |z+>|n120+>|n240+> + |z->|n120->|n240->
    In Mathematica:
    nplus[th_] = {Cos[th], Sin[th]}
    nmin[th_] = {-Sin[th], Cos[th]}
    I work this out in the z basis, which is also my measurement basis, and I find:
    Outer[Times, nplus[0], nplus[2/3 Pi], nplus[-2/3 Pi]] +
    Outer[Times, nmin[0], nmin[2/3 Pi], nmin[-2/3Pi]] // Simplify
    which are the amplitudes to find |+,+,+>,|++->,|+-+>,...|--->
    So, the probability for +++ is 1/16, the probability for ++- is 3/16, the probability for +-+ is 3/16, the probability for +-- = 9/16
    (and of course divided by 2 because I forgot a factor 1/sqrt(2) in front).
    The z-axis is special, because of the non-rotationally symmetrical state (or in this case, measurement). You see this, because the highest probabilities are for +-- and -++ (both 9/32).
    If you want to introduce a rotationally symmetrical (singlet) state, corresponding to the spin-0 state, you will have combinations in the wavefunction which do not have 100% correlation.
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