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Does a beam of entangled photons create interference fringes?

  1. Jan 25, 2006 #1
    Does a single beam of entangled photons create interference?

    Hi,
    This question has come up in an FTL thread going on right now. I've seen two different answers to this question, and I'd be interested if anyone knows the real answer.

    Here's the setip: shine a laser beam on a crystal which performs SPDC (spontaneous parametric down conversion). This produces two beams, A and B. The photons in beam A are entangled with the photons in beam B.

    Here's the question: If you run beam A through a double slit to a detector, will you see an interference pattern?

    I know that SPDC is not perfect, and beams A and B can probably never be perfectly entangled (i.e., some unentangled optical 'junk' gets through), so perhaps this is why there is no definitive experimental evidence. I guess you would need to filter the frequency and do coincidence counting on the two beams to insure that you were only looking at photons that were members of an entangled monochromatic pair.

    Still, this seems like a simple question that should have a clear anwer. I hope someone can provide it.

    Bruce
     
    Last edited: Jan 25, 2006
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  3. Jan 25, 2006 #2

    vanesch

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    Short answer: no.
    The quantum-mechanical answer is of course the entanglement, and the fact that the interference term <psi1 | psi2> includes the in product of the "local" photon and the "remote" photon states, and the remote photon states are orthogonal.
    But the "classical" answer is that the light coming out of a PDC xtal is a "rainbow". So for two different slits, they will receive slightly different components of the rainbow (slightly different colors), which will hence not interfere.
    Now, of course, you can select out a very tiny part of the rainbow, and have it interfere ; but then it will turn out that the entanglement is essentially gone (you've essentially selected out ONE term of the (|a>|b> + |c>|d>) state) and you end up with a separated product state.
     
  4. Jan 25, 2006 #3

    dlgoff

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    What if you only produced two photons at a time from the PDC and let one of the photons go through the slits i.e. one at a time. Do you get interference then?

    Thanks
     
  5. Jan 26, 2006 #4

    DrChinese

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    The result Vanesch mentions is independent of the number of photons going through. I.e. he is already assuming you are considering a pair at a time.

    Anytime you are talking about PDC, you are pretty much guaranteed to see only a lone pair at a time and the data is built from that. You will only see 2 entangled pairs within a specified time window on the order of twice a minute.
     
  6. Jan 26, 2006 #5
    I think the answer yes of course it would.

    Based on the question you asked the SPDC would not be important at all.
    In your set up it’s only a part of creating a beam of photons into an experiment that has nothing to do with entanglement.
    You are throwing away the B side and not using it, right.
    So A is just a beam of photons that will behave as any other in a double slit. When you move the detector and count hits as needed you will accumulate an inference pattern, no reason it shouldn’t.

    But if your intent was to compare some part of that area that the pattern builds up on in some polarization test of correlation with beam B. I sure going though the double slit would collapse & destroy any entanglement.

    (This wasn't a trick question was it?)
     
    Last edited: Jan 26, 2006
  7. Jan 26, 2006 #6
    I have read that if you do coincidence counting between the A and B photon beams (see for example http://grad.physics.sunysb.edu/~amarch/) then the interference pattern is detected. Do you know why the coincidence counting would make a difference in this simple setup (in the delayed choice erasure experiments, the coincidence counting separates out the fringes from the anti-fringes; but in this simple case, I don't think there are any anti-fringes).

    Bruce
     
  8. Jan 26, 2006 #7
    No, it's not a trick. Apparently, the existence of the entanglement changes the behavior of the two photons. I read the book Entanglement : the Greatest Mystery In Physics by Amir D. Aczel, and he says that somehow the two photons are only 'half' photons and that there is no interference unless they are coincidence counted. Unfortunately, Aczel's book is written for the general public (no equations, but lots of anecdotes about Schrödinger's sex life), and I am hoping to get a deeper understanding of the interference phenomenon here.
     
    Last edited: Jan 26, 2006
  9. Jan 26, 2006 #8
    I think yes, there will be an interference pattern. I don't see why there shouldn't be one due to entanglement. Indeed, in Fig. 2 of the Walborn paper it is shown that interference occurs also if you detect the photons in coincidence.

    The interference pattern vanishes in the Walborn experiment because you mark the photons by their polarization and thus you get which-way-information. In Walborn's experiment it's important to have coindicent counts otherwise you can't say which of the photon's which-path-information has been erased.
    It's all about which-path information!


    The reason why anti-fringes and fringes occur can be maybe explained the following way: In Fig. 7 you see a non-interference pattern (bell shaped curve). You will notice that it is somehow the sum of Fig. 4 (fringe) and Fig. 5 (anti-fringe) (I don't know why the counts in Fig. 7 are lower). So actually what you see in Fig. 7 are both the photons from Fig. 5 and 4.
    Or differently explained: Imagine you have the bell-shaped curve, someone tells you that she doesn't see any interference pattern. But if you take the coincidence counts, you will see interference (fringe). Next step, you substract the fringe pattern from the bell-shaped curve, that is Fig. 4 from Fig. 7. Then, in my opinion, you should get the anti-fringe.

    In short: fringe + anti-fringe = bell shaped curve.

    See also here:
    http://www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf
    Look at the last page of the above pdf at the right bottom picture showing the orange fringe and the red anti-fringe pattern under the black bell shaped curve.
     
    Last edited: Jan 26, 2006
  10. Jan 26, 2006 #9
    As near as I can tell, in the very simple setup I described, you will not get interference from a single beam, but you will get an interference pattern if you use coincidence counting.

    So, I'm wondering, 'why is it that coincidence counting between the two entangled beams produces interference, but a single entangled beam alone will not produce interference?'

    Bruce

    (PS - I'm sorry I mentioned the anti-fringes, they don't really apply to the simple setup I described.)
     
  11. Jan 27, 2006 #10

    vanesch

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    Well, the way the question was posted, it was assumed that no preselection was going to take place such that the entanglement was gone ; and that is exactly what you have to do with the light coming out of a SPDC crystal if you want to have a beam of which you can obtain interference
     
  12. Jan 27, 2006 #11
    No fringes (unless you do selection)

    A few threads below, on Quantum Erasure 9903047, this issue has been discussed. Also a search on entanglement will produce similar discussions.

    The answer is that you will NOT observe an interference pattern on pA.

    pA not being the whole object, cannot simply interfere with itself. You will have to include the effect of photon pB in the analysis. Photon pB will act to obscure the interference pattern, so that without the help of pB (through coincidence selection), you will not see the interference pattern for pA on the double slit. Instead what you will see is a gaussian distribution.

    I understand what happens is conceptually as follows: The wave function governing pA also contains equal components from pB. If pA collapses on its own components, you will get a fringe. But there is equal probability that pA will collapse on the wavefunction for pB. In this case pA and pB being asymmetric, such a collapse will generate an anti-fringe.

    You will never know just by looking at pA whether it has collapsed on the fringe or anti-fringe. So you cannot separate out (without the help of pB) the two fringes. The end result is that the anti-fringe will exactly normalize the fringe resulting in a gaussian (normal) distribution, that contains NO information (of the wavelength - momentum).

    However, if you measure pB, transmit the info locally (classically) to pA, you will be able to tell when pA falls on the fringe vs. the anti-fringe, and hence select out the fringe. Note that the frequency you will get is the sum of the frequencies of pA and pB, which shows that pA is acting non-locally.

    In entangled pairs, each element contains no information. It is the sum of the two entangled parts that contain information.

    Yes, the biography in Aczal's book is quite amusing.
     
  13. Jan 27, 2006 #12
    Producing an interference pattern, ala young’s double slit, has nothing to do with “coincidence counting” it uses statistical accumulation of many problematic events to build up an interference pattern. So I have no idea why Bruce things it does.

    The idea that with no “pre-selection”, thus no loss of entanglement; somehow leaves a beam of light, or individual photon in it, abnormal or missing something as compared to another beam able to produce interference when sent through a double slit.

    What exactly do you expect QM says would be shown on a negative given appropriate exposure time with both slits open for a single SPDC created beam?
    If not interference, what is predicted? Seems like very simple experiment to conduct – IF interference is produced would you say QM or some part of QM is falsified?

    I’m positive interference will be seen, but I can see no part of QM such a demonstration would falsify.
     
  14. Jan 27, 2006 #13

    vanesch

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    Imagine an entangled pair:
    |red>|blue> + |blue>|red>

    Note, that in order for the photons to be entangled in the first place, we need them to come in two DIFFERENT flavors (here, red and blue). Indeed, if all the photons in the first beam were in the pure "red" state, we would not have entanglement, because that state could be factored out.

    Ok, this means, if you are only going to look at the first photon, that they will appear to come in a statistical mixture, which is 50% red, and 50% blue. If you send them through a thing to have them interfere, then you will have 50% a "red" interference pattern, and 50% a "blue" interference pattern, which comes down to no pattern at all. That is not surprising, and you don't need to look at the second beam for this, because the first beam simply appears to be "white" (a stochastical mixture of red and blue).

    But you can now put a filter in there. You can filter out only the red photons. But then, you're left (Copenhagen view) only with the |red> |blue> state. Entanglement is gone. Interference is possible.

    So there is nothing "magical" about photons from an entangled pair, which would suddenly make them "avoid to interfere". It is only that, when you view them from one side, they COME IN A STATISTICAL MIXTURE of orthogonal states (here, color). So it is not amazing that the beam mixture doesn't seem to give interference. From the moment you purify the mixture, you select only ONE term of the entanglement (which means that you end up with a product state). Then of course you can now have interference.

    ADDED: that's why the beams coming out of a PDC Xtal come in a *rainbow*.
     
    Last edited: Jan 27, 2006
  15. Jan 27, 2006 #14
    This is interesting, especially since I don't understand SPDC yet. Whether an interference pattern emerges when a beam of photons is sent through a double-slit device depends on the original photon source (eg., the crystals used in SPDC)?

    So, the answer to RandallB's question:What exactly do you expect QM says would be shown on a negative given appropriate exposure time with both slits open for a single SPDC created beam? -- is that no interference pattern will emerge? So, exactly what would you see on the negative after running beam A through the double-slit device -- just a single-slit distribution?

    Sorry to be redundant, but I'm having a difficult time understanding this. These red and blue states of the photons in beam A --- is the beam a combination of photons having different wavelengths?
     
  16. Jan 27, 2006 #15

    vanesch

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    Yes, and/or different polarizations. That's the entire idea! Classically, you can understand the different wavelengths (at different angles) as the different solutions to the pump -> idler + signal conservation of energy and momentum conditions, when the angles are slightly different. That's why the output is a *rainbow*. Let us forget for the moment, polarization, it makes the issue a bit more complicated. Let us assume that we are going to consider the entangled states based upon wavelength (and hence, outgoing angle). In fact, because of the relationship between angle and color, the outgoing photons will indeed be in an entangled state |red>|blue> + |blue> |red>. One could do EPR-like experiments with that (usually it is not the wavelength, but the polarization that is used as discriminating property, but wavelength is simpler to explain).

    This means you have to accept, of course, photons of different wavelength in your experiment! Otherwise, you have already selected out ONE term of the entanglement. But "accepting photons of different wavelength" means, having essentially an incoherent light beam. So if you try your interference experiment on it, you will not see interference. And you won't find that strange, because the beam, to you, looks incoherent. You're not surprised.
    Now, you can then purify your beam, to have a coherent beam. And then of course you will have interference ; but you destroyed the entanglement by purifying the beam! You projected out ONE of the terms, in quantum speak.

    It is of course important that the entangled degree of freedom is the one causing, or not, interference. If the interference is independent of the quantity that is entangled (meaning, if that kind of interference is not going to change wrt the measured property of the entangled partner, and as thus, is not the right experiment to do an EPR demonstration), then of course you can still obtain the interference, with entanglement.

    This is thinkable, for instance, with a pure wavelength entangled pair, the entanglement being the polarization. If we have the state:

    |red+>|blue-> + |red->|blue+> and we do a double-slit interference experiment on the red beam which is INSENSTIVE TO POLARIZATION, then of course we will get an interference pattern. This is because the |red+> state and the |red-> state will give rise to identical interference patterns, and so the statistical mixture of 50% |red+> and 50% |red-> will also yield that interference pattern. But with such an interference experiment, you cannot do any EPR like demonstration, because no matter what you do, and no matter what you subselect on the "blue" side, you'll ALWAYS get the same red interference pattern. That is because the red interference pattern is INDEPENDENT of the entangled degree of freedom (which is, here, polarization) - in fact, it can be factored out, and one could write:

    |red>|blue> (|+>|-> + |->|+>)

    So the "red" and the "blue" degrees of freedom are NOT entangled, only the + and - polarization degrees of freedom are. As such, interference of the "red" degree of freedom is possible, and of the "blue" degree of freedom too.

    But not a polarization-dependent interference pattern. That will appear to be an unpolarized mixture at both sides.
     
  17. Jan 27, 2006 #16
    Just by way of background, when you do a PDC, you shine in a monochromatic laser light and the non-linear crystal produces two photons for each one input. Conservation of energy requires that the frequencies of the new photons sum to the original laser frequency. However, the new photons do not each have 1/2 of the original frequency -- they just need to sum to the original frequency. So, even though you started with a monochromatic laser, your new beams will be incoherent.

    Moving right along, thanks for the nice explanation, Vanesch. I'm still slightly confused by a few things I see in the published experiments, so let me see if I've got this right:

    DetectorA <-- Double-slit <-- FilterA <-- SPDC --> FilterB --> DetectorB

    1. If you have no filters and no coincidence counting, then there will not be any interference at DetectorA.

    2. If you have no filters but you do coincidence counts between DetectorA and DetectorB, there I think there will be an interference pattern revealed by the coincidence counters.

    3. If you have FilterA, and no coincidence counts, then you will see interference (monochromatic light, entanglement lost).

    4. If you have FilterB but not FilterA, and you perform coincidence counts, then you will see interference (since monochormatic B implies monochromatic A).

    The reason I'm not sure about (2) is that in the Walborn et al paper (http://grad.physics.sunysb.edu/~ama...//www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf
     
    Last edited: Jan 27, 2006
  18. Jan 28, 2006 #17
    The entangled photons produced by PDC crystals are entangled in both wavelength and polarization. So, if beam A or B (which are each superpositions of two different wavelengths) is directed through a double-slit device (with no other filtering), then the banding characteristic of a coherent light source directed through a double-slit device won't be seen at screenA or screenB.
    So it isn't entanglement, per se, that disallows an interference pattern sans filtering.
    The PDC created state is:
    |red+>|blue-> + |blue-> |red+>

    ... and I see the difference between the PDC state and:
    |red+>|blue-> + |red->|blue+>

    What sources produce the latter state (substituting whatever colors or wavelengths are characteristic of the emissions)?

    Ok, thanks vanesch. A clear enough explanation for laymen like me (though I did spend about 4 hours yesterday refreshing my memory of the classical optics math involved in interference and diffraction).
     
  19. Jan 28, 2006 #18

    vanesch

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    The PDC Xtal generates a lot of states. The degrees of freedom of a photon are wavevector and polarization ; the wavevector itself consists of direction and wavelength. So you could think of a "photon state" as:
    |direction>|color>|spin>

    Now, a PDC xtal generates a lot of terms of the kind:

    |direction1>|color1>|spin1> |direction2> |color2> |spin2>:

    we have:

    |2-photon state> = weighted sum over
    |direction1_a>|color1_a>|spin1_a> |direction2_a> |color2_a> |spin2_a>
    |direction1_b>|color1_b>|spin1_b> |direction2_b> |color2_b> |spin2_b> +
    |direction1_c>|color1_c>|spin1_c> |direction2_c> |color2_c> |spin2_c>
    +...

    where a,b,c... are different possible cases, which respect the physics of the PDC, that is: conservation of energy, angular momentum and momentum in the |pump> => |c> transition.

    As such, the PDC generates a very complex entangled 2-photon state.

    But the experimental setup selects out usually a single direction for idler and signal beam, and uses a wavelength filter. As such, only ONE set of direction1 and direction2, and color1 and color2 are selected ; what remains are the different possibilities of spin. Which gives us:

    |direction1>|color1>|spin1a>|direction2>|color2>|spin2a>
    + |direction1>|color1>|spin1b>|direction2>|color2>|spin2b>

    which can be rewritten as:

    |direction1>|color1>|direction2>|color2> (|spin1a>|spin2a> +
    |spin1b>|spin2b>)

    As the direction and colors of the two beams are now factored out, this is a non-interesting degree of freedom, and one usually doesn't even mention it.

    cheers,
    Patrick.
     
  20. Jan 28, 2006 #19
    Ridiculous – I’d be absolutely shocked not to get an interference pattern!
    You can make an interference paten with sunlight how much more incoherent do you need. Also, the wavelength diff between the two “different colors” is such a tiny fraction it’s not worthy of being called a “rainbow”.

    “If not interference, what is predicted?”
    Your saying – a single-slit distribution through double slits!

    You quoted my other question but didn’t answer:
    “Seems like very simple experiment to conduct – IF (and IMO when) interference is produced would you say QM or some part of QM is falsified?”

    What part of QM stands to be falsified if your claim fails??

    Before you dig this any deeper or common sense kicks in, PM another mentor to take a look at this.

    This is simple double slit - entanglement has no effect.
     
  21. Jan 28, 2006 #20
    I agree. You would see an interference pattern.

    If you just viewed a screen behind the double slit you would "see" an interference pattern. Of course, the way I understand it you are viewing much more than just entangled photons. PDC (parametric down conversion) only produces a few hundred entangled pairs per second. So the entangled pairs are pretty much washed out by the millions of non-entangled photons (Isn't that correct?)
    But, even if you use a coincidence counter to only record entangled pairs then you will still see an interference pattern, because you have done nothing to determine which-way informaton.
    Now, if you start adding quarter-wave plates and polarizers things get a little crazy. The best explanation I have seen is a link earlier in this thread: http://www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf

    "Anytime you send photons, of any kind or flavor, through a double slit you will see an interference pattern unless you provide a method for which-way information." I believe that statement to be incontrovertible. (I said that, it wasn't in a paper somewhere.)

    Coherence doesn't matter. Talk of colors, frequency, and other stuff is just confusing and not needed. Polarization techniques are the easiest to explain and actually perform.

    Of course, I could be wrong :surprised
     
    Last edited: Jan 28, 2006
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