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Delayed Choice Quantum Eraser

  1. Jan 2, 2015 #1
    I found other threads with the same title on this forum but my question is somehow differ.

    I think this was the most simple DCQE with double slits: http://arxiv.org/pdf/quant-ph/0106078v1.pdf

    For a short summary: the experimental setup uses an entangled pair of photons (p and s). The s-photons goes through a double slit which signs the photons with the which-way information using quarter-wave-plates. On the p-photons we can choose if we erase the ww info by placing a polarizer - this will let go through only half of the photons and erase the polarization info of the p-photons and this way the s-photons too. The s-photons are detected in Ds detector. There is a coincidence counter too, which can be used to find out which p-photon detection is the EPR pair of an s-photon (and vice versa).

    My explanation:
    I found that at the secondary detector (Ds) we can find a Gaussian noise and this fact is never influenced by the polarizer POL1 regardless if placed or not near before the primary detector. The eraser works this way: using the polarizer POL1 with Dp data and and with the coincidence counter data we can find the subset of the s-photons which subset draws an interference pattern. These s-photons are the half of all s-photons. The other half of s-photons draws an other interference pattern and the sum of these two patterns produces the Gaussian noise pattern.

    Question 1: Am I right that all of the Ds data always draws a Gauss distribution?
    Qusetion 2: am I wrong at any point regarding the above explanation?

    Now I try to explain my problem. Let's modify the experiment setup a little bit. Increase the distance between the two slits so the non-interferencing pattern will split into two Gauss distribution. Can we do this? I think we can. With this setup we have two choice:
    1 - not using the eraser (POL1) we got the two Gaussian pattern
    2 - using the eraser we have to got the double slit interference pattern by choosing the appropriate s-photon subset (based on Dp and coincidence counter data)
    The second choice is problematic because a two-slit interference pattern should contain some constructive fringes behind the "wall" which is between the two slits (at x=0 place). But we have two Gaussian distribution with low intensity at x=0. We can not find any subset of s-photons to draw the double slit interference pattern.

    Question 3: Can we increase the distance between the slits to see two Gaussian distribution in Ds (not using POL1)?
    Question 4: If we use POL1 with this setup we see the same distribution in Ds?
    Question 5: What kind of interference pattern subset can be found in this case?

    Thats enough for starting. Thanks in advance for any help.
  2. jcsd
  3. Jan 3, 2015 #2


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    Answer 1: Almost always. If you place the light source really far away from the double slit, you can directly see an interference pattern. However, in this case the subset of photons arriving at the double slit is so small and well defined that this subset cannot be considered entangled with the other subset anymore.
    Answer 2: Yes.

    Answer 3: Yes, that is possible.
    Answer 4: That depends on the exact geometry, but usually you will still see two Gaussians.
    Answer 5: The key to the DCQE is, that in a usual double slit experiment, the likelihood of a photon arriving at some certain position depends on its momentum (or angle relative to the slits). If you place the two slits very far apart, the position where a photon arrives, will not depend on the momentum anymore as there is little to no overlap between the photons arriving from the two slits and there is no such momentum dependence for two independent slits - it is rather more probable that a photon going through slit 1 will be detected close to the slit because the slits are so far apart. If you just have a look at the center, you should still see an interference pattern, but it should be weak as the number of photons arriving there should be incredibly small. So starting with the slits quite close and placing them further apart, you would notice that the visibility of the interference pattern in coincidence counting will gradually become worse.
  4. Jan 4, 2015 #3
    Cthugha, thank you very much for your answer.

    Let me summerize to check if I see the essence.

    In the experiment with a fixed geometry (slits' distance, slits' width, wavelength etc.) we always see the same distribution in Ds regardless of using or not using the POL1 eraser. The interference pattern can be find by seeing only a subset of s-photons. If we increase the distance between the slits then we can see double Gaussian distribution. Now the method to see the interference pattern is the same. Although we see a weird interferenc pattern this is not a contradiction - the reason behind the weird form of the new interference pattern is behind the weird geometry (distant slits). This interference pattern would be drawn in a case of a simple double slit experiment too (with same geometry). And again the sum of the two weird interference pattern gives the all Ds detections distribution. Correct?

    You wrote that we almost always see the Gaussian distribution: "If you place the light source really far away from the double slit, you can directly see an interference pattern." Can you please explain this with a little bit more details? Why should appear the interference? This is very strange for me.

    I don't understand why are the photons differently polarized. There is a degree difference and this is the reason that we can get only the half of them through POL1. Is this difference intended or unavoidable? And is this degree difference the reason of the two "slipped" interference pattern too? I think this is, but don't understand how.

    And another question to make clearer my picture. Let me assume that I don't use the BBO and build only the s-path of the experiment setup. Now the +45 and -45 QWPs will destroy the interference. Is this correct?

    Sorry for the many (and may be wrong) questions and thanks in advance.
  5. Jan 4, 2015 #4


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    Sounds good to me.

    This is nothing that is special to entangled light, but you will see that for any light source. If you have a simple double slit at home, you can try it yourself. Use a rather extended and bright light source and place it directly in front of the slits. You will not see an interference pattern. The reason is, that the light source is very extended and from each point of the surface of the light source you have a different path length difference to the two slits. As this is a crucial parameter for the shape of the final interference pattern, you will see a superposition of all of those patterns which results in no visible pattern at all. If you move the light source far away from the double slit, things change. The spread of angles between the points on the surface of the light source and the slits gets smaller. At some point you will start to see an interference pattern. Basically the finite size of the slits acts as a filter itself.
    This of course also works for initially entangled light. However, if you do further experiments on the initially entangled photons after passing through the double slit, you will find that this subset does not violate Bell inequalities anymore and can thus not be considered entangled. The presence of the slit is similar to a measurement of momentum.

    This is basically a consequence of how the light is created. The sum of the polarization of the two photons is well defined, but in each run it is randomly distributed among the two photons (of course within limits of usual selection rules). So when looking at just one of the photons, you will find that it is randomly polarized. This is somewhat related to the "slipped" interference pattern. The reason for that is simply that the initial path difference between the light source and the two slits matters. If you just put a light source at some position, check the interference pattern and move the light source parallel to the slits, you will notice that the interference pattern changes and moves along. This is basically a function of the phase difference of the light fields at the slit positions. So there will be phase differences which will result in an interference maximum at some point and there will be phase differences resulting in a minimum at the same position. Polarization optics of course also have an effect on the phase of the light field, so you get some effect that way.

    If I get your experiment right, this is correct. Orthogonal polarizations do not interfere.
  6. Jan 4, 2015 #5
    Thank you very much for your helpful answers.

    Your discussion about the distance between the light source and the slits is OK for me. But originally I asked if we always see Gauss distribution while the QWPs at the slits are used. So in this case the interference could not appear regardless of the distance of the source and slits. Am I correct?

    Let me answer a little bit technical question. Given an S monochromathic light source which sends photons to a P plate. On the plate are two slits (A and B) and the photons (p) went through the slits are detected at D.
    Given these data:
    Distance of the source and the plate d(S,P)
    Distance of the plate and the detectors d(P,D)
    Width of the slits on the plate (both are the same): w
    Distance between the center of the slits: d(A,B)
    Wavelength of photons: λ

    1) What is the equation which gives me the distribution the photons at the detector as a function of the place (x)?
    2) I blackout one of the slits. Now what is the distribution equation?

    I tried to find it on the web but no success. It is not trivial for me so thanks if you (or anyone) can help in this question too.
  7. Jan 5, 2015 #6
  8. Jan 6, 2015 #7
    I have got one thing that I can not understand.

    I found that in a single slit expirement we got a Gaussian distribution with the highest intensity "I". I read in the theory that If we open an other slit then the highest intensity of the interference pattern will be 4*I. Now I try to use this fact in the explanation of the Walborn expirement like this:

    When the pathes are signed we got two Gaussian distribution - these two are almost fully overlapped but not fully so the highest intensity value of the pattern in Ds x=0 is a little bit less then 2I.
    Let's use the eraser. We can see an interference pattern of the half of the photons which means the highest intensity value could be now (I/2)*4 which is a little bit higher then it was in the previous case. But the experiment explanation was that the interference pattern is a subset of the Gaussian pattern so the uneqaulity above looks like a contradiction.

    Of course this little quantity can not be measured but the theory is not OK for me. May be the theoretic 4*I is not exact? Can I prepare the Walborn's setup that way to get the 4*(I/2) value when using the eraser? Am I right that while not using the eraser the highest intensity can not be 2*I but same lower value?
  9. Jan 27, 2015 #8
    Dear All,

    I have question about the quantum eraser. I found two quantum eraser experiment with the same effect:
    The effect is: seeing the two interference patterns they are contrary. The sum of the two patterns gives the Gaussian distribution. At the locations of first pattern's local intensity maximum there is a local minimum of the second pattern. So these patterns are somehow slipped away from each other. What is the physical reason of this slipping? The wavelength and other relevant things are the same so I don't understand how can one interpret this slipping effect. What should one change in the experimental setup to change the locations of intensity minimums and maximums?

    Thanks in advance for any help.
  10. Jan 27, 2015 #9


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    It is just the phase. A double slit pattern depends on the relative phase between the light fields at the position of the two slits and all the quantum eraser versions showing fringe and anti-fringe patterns give you some means to vary this parameter. This might be the position of the second detector used for coincidence counting or one might add some half- or quarter-wave plates or other optical elements that shift the phase of the light field.

    If you have zero phase difference, you see the typical symmetric double slit pattern with a peak in the center. Changing the phase difference will introduce an offset to that pattern such that you can have an interference minimum or maximum or anything in between at the center.
  11. Jan 28, 2015 #10
    Thank you very much, Cthugha.

    I will think about your answer... it is not clear yet to me. Here is the experimental setup of Kim:


    I can't see any constraint to be a phase difference between the photons arrived into D1 and D2. This setup seems very simmetrical. Could you point out where or what is the exact reason of phase difference in this setup?
  12. Jan 28, 2015 #11


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    Ok, this is not a double slit based setup, so the phase difference occurs due to differences in reflection and transmission.

    On its way to D1, the blue beam gets reflected twice (Ma and BSc) and transmitted once (BSa), while the red beam gets reflected once and transmitted once.
    On its way to D2, the blue beam gets reflected once and transmitted twice, while the red beam gets reflected twice and transmitted once.

    For 50/50 beam splitters, you get a phase shift of pi for the reflected beam and a phase shift of pi/2 for the transmitted beam, resulting in a difference of +pi/2 between the two beams. On the other path, the role of the two beams is reversed, resulting in a difference of -pi/2. So the total phase difference amounts to pi, which is just the difference between getting a fringe pattern and an anti-fringe pattern.
  13. Jan 28, 2015 #12
    Cthugha, really thanks for the answer, that was the point! I was thinking about this puzzle over weeks. You are really welcome!

    Please let me ask again I think it will be ease for you. What is the point of the similar pi shift in the Walborn experiment.
    There are entangled photons going up (p-path to the Dp detector) and down (s-path to the Ds detector) and the only difference of the two measurement that the POL1 polar filter is turned 90 degrees. The result in Ds draws again a pi phase shift between the two measurement. (There is a coincidence counter too which is missing from the figure and we gather only those Ds detections which have a coincident detection in Dp.) Can you explain how will this POL1 rotation cause a phase shift in the s-path?
    Last edited: Jan 28, 2015
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