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Explaining DCQE - via coherence in layman terms

  1. Jun 7, 2011 #1
    Does the below sound ok?

    1. Interference is caused between coherent waves

    2. When we try to find which-way we break the coherence
    a) When we try to "partially" find which way, we partially break the coherence and hence a "partial" interference pattern

    3. DCQE is explained by the fact that the sub-sample is taken either of coherent or in-coherent bunch of photons (via pairing in the co-incidence counter) depending upon what we did to p-photon ...and thus matches with (what we did to) p-photon.....for the sub-sample of s-photon as well.
    Last edited: Jun 8, 2011
  2. jcsd
  3. Jun 8, 2011 #2


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    I thought the interference patters was simply because of each photon interfering with itself through the double slit. It looked to me like the polarizers at the slits caused a change in the interference patterns on the detector, with each the different polarization interference patterns adding together to look like there was no pattern. Adding another polarizer in the other beam simply took one set of photons out of the results, resulting in the return of the interference pattern, which had always been there but the other set of photons polarized differently added to the results making it look like the interference was gone.
  4. Jun 9, 2011 #3
    ya....but how do we explain going from an interference pattern to a non-interference pattern?....because an interference pattern would have blank spaces between the fringes....how do they get "filled-in" by photon ....when we are getting a sub-set (sub-sample) of the interference pattern.....

    i mean how can a sub-sample of an interference pattern be a non-interference pattern?...how do the blank spaces (between the fringes) get filled-in...

    what am i missing here?

    i.e. in the DCQE we do no-which-way for the s-photons and (a few nano seconds) later do which-way for p-photons...
    Last edited: Jun 9, 2011
  5. Jun 9, 2011 #4


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    When you add the polarizer in the P beam, ONLY the photons that pass through will be recorded. All of these photons will be polarized in the same way. Remember that the coincidence counter only counts when the input is from BOTH detectors. All those photons getting to the S detector but having their entangled P partners blocked by the polarizer at P will NOT be counted. So you have removed whatever pattern would be caused by them by only counting events where both S and P are detected.
  6. Jun 9, 2011 #5


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    No, that is backwards ... at least it is for the Walborn version of the DCQE, which is what I think you are referring to. In that case, the s-photon path is associated with the which-way measurement due to the QWP's that induce right-hand circular polarization on photons passing through the right hand slit and left-hand circular polarization on photons passing through the left hand slit. Thus the two sets of photons are distinguishable, and there is no interference in the single-photon measurements on Ds. If there is no polarizer in the p-photon path, then the photons arriving at the p-detector are also distinguishable, so there is ALSO no interference pattern observed in the coincident two-photon measurements. It is only when the polarizer is in place, and set to the angle of one of the QWP's (i.e. 45 or 135), that the interference fringes are observed in the coincident two-photon measurements.

    I *think* that this is only the choice where the interference is strongest, and that progressively weaker fringes would be observed as the detector is rotated off those values, disappearing entirely for angles of 0 or 90. (My justification for that statment is equation [14] from the http://arxiv.org/abs/quant-ph/0106078" [Broken].)

    As to how the "blanks" in the interference pattern are filled in .. Drakkith had it right when he said that the two-photon count in the absence of the polarizer is a superposition of the interference patterns that are observed at polarizer angles of 45 and 135 degrees. Those patterns are perfectly out of phase, and so the fringes cancel when they are superposed.
    Last edited by a moderator: May 5, 2017
  7. Jun 10, 2011 #6

    Great answers Drakkith and SpectraCat. I think you are right but I don't fully grasp it yet.

    SpectraCat, Yes i am talking about the Walborn experiment and I get it.

    However I still not get it if we do the Walborn experiment "backwards". I am missing something about the photons and the patterns.

    I am still trying to understand both your post (to resolve the confusion, in my mind, I understand your posts one-way) in the meantime I will post what if confusing me.

    There is only one fringe. So I don't understand where the superposition would come from.

    The one fringe is caused by s-photons going through no-which-way. i.e. a fringe caused by a simple bare double slit.

    Now when we do which-way on the p-photon there would be a scatter of p-photons around a band.

    Now when we do the co-incidence counting...there would be no pairing because

    all the s-photons went no-which-way and all the p-photons went which-way?

    So if there is no pairing, would the sheet look blank?
    Last edited by a moderator: May 5, 2017
  8. Jun 10, 2011 #7

    Maybe I am confusing polarization with which-way (or no-which-way) by considering them to be corelated/synonymous.
  9. Jun 11, 2011 #8
    In the "backwards" experiment (of Walborn's), there would not be absence of a polarizer but a presence of a polarizer.

    Let's start with a new simple experiment below:

    we send s-photons through a double slit.....and get an interference pattern

    now we send p-photons through a polarizer to find out which-way (i.e. mark the photon)

    then we do co-incidence count-

    what pattern would s-photons have on the co-incidence counts?
    what pattern would p-photons have on the co-incidence counts?
  10. Jun 11, 2011 #9


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    You realize you have to have both P and S photons being detected at the same time for coincidence counts right?
  11. Jun 11, 2011 #10
    yes......(..not same time though.....but.....correlated to same emission time, i guess that is what you mean to say..).....but see my reply to zonde's post where.....

    you have both interference pattern and which-way.......its the same issue....and ....zonde has expressed what I am talking about….

    its the last post on this link -->

    Last edited: Jun 11, 2011
  12. Jun 11, 2011 #11
    I'm very much a layman but this is the way I see it (apologies if this is wrong, but I'm sure someone will correct me): -
    You can't get which way info simply by looking at the p-photon. You can only get which path information by comparing the polarization of the p-photon with the polarization of the s-photon.
    if p-photon is vertical and s-photon is rotated left then s-photon went through slit 1
    if p-photon is vertical and s-photon is rotated right then s-photon went through slit 2
    if p-photon is horizontal and s-photon is rotated right then s-photon went through slit 1
    if p-photon is horizontal and s-photon is rotated left then s-photon went through slit 2

    So, without QWPs in place I would expect there to be an interference pattern, as we don't have which way information.
  13. Jun 11, 2011 #12

    If the above is correct then you have answered my question. ..:-) thanks Joncon, that's what I was missing


    this ties in with what zonde said...in the link below


    Does this also say/mean that (in a limited sense) for the DCQE:

    a) you can go from which-way to no-which way but you cannot go from

    b) no-which-way to which-way...

    i.e. a) you can get an interferene pattern from a no-interference pattern

    b) but you cannot get a no-interfernce pattern from an interference pattern

    because an interference pattern is sub-set of a non-interference pattern

    and that is why walborn does the experiement in the sequence a) above and not b) above
    Last edited: Jun 11, 2011
  14. Jun 11, 2011 #13
    SpectraCat: my question was about doing the Walborn "backwards".

    Jocon has answered it in post number 11 & 12 above.
    Last edited by a moderator: May 5, 2017
  15. Jun 11, 2011 #14
    Walborn wanted to demonstrate delayed choice eraser as simply as possible, so he avoided the multiple paths of previous experiments by the inspired experimental setup of SPDC entangled pairs, quarter wave plates (to enable which way info) and a (distant) polarizer (to destroy which way)

    The fact that everyone has over-analyzed the experiment makes it appear much more complex than it really is, it's like analysing tests of General Relativity by appealing to statistical issues in the apparatus, yeah they might be valid if the experiment hasn't been devised sufficiently cleanly.

    Subsequent experiments have shown QM to be non-classical in even more stunning ways, so arguments about its implications are kinda silly
  16. Jun 11, 2011 #15
    the DCQE is explainable by sub-samples, SpectraCat/Cthuga are right however their answers (to my questions) were incomplete.
  17. Jun 11, 2011 #16
    The sub-samples of course correlate afterwards, but so what? How are the sub-samples created? Even if you have a polarization beam splitter so that ALL photons are measured, it is still a PROBABILSITISIC law that determines which detector they go to.

    How do the s-photons know which p-photons will go where if there is a delay after the s-photons are measured?

    SpectraCat/Cthuga are WRONG if they think a simple subsampling argument explains this. You HAVE to invoke the nonlocal and/or non-separable nature of QM to explain it.
  18. Jun 11, 2011 #17
    Joncon, what do you think would happen if we placed QWPs (not polarizer) in the path of the p-photon (and none in the path of s-photons) after s-photon has stuck detector Ds (by going through the double slit without any QWPs)?
    Last edited: Jun 11, 2011
  19. Jun 11, 2011 #18
    the s-photons don't know which way p-photons will go. However the probabilities of the p-photons path are frozen at the time s-photons strikes the detector. The entanglement is broken (when s-photon strikes the detector) and the p-photon's behavior/path is "somewhat/probabilistically" determinable.

    i.e. you can tell what is the probability of p-photon taking a particular path, (out of the choices available)

    but this again cannot be used to transmit (volitional) information faster than the speed of light...

    when we do co-incidence count, we are simply picking only those photons (from both s and p) marks/positions that match the pattern. the co-incidence counter acts like a filter.

    the peaks of one interference pattern coincide with the troughs of the other....causing a no-interference pattern.

    and Yoon in his paper also explains it this way.....the sub-sampling way.....
    Last edited: Jun 11, 2011
  20. Jun 11, 2011 #19
    unusualname, how does probability explain the interference patterns in this version of the experiment? http://arxiv.org/abs/quant-ph/9903047

    The thing I don't understand is that the D1 and D2 detectors both show interference fringes and anti-fringes when the sub-samples are examined. What I don't get is that the idler photons encounter a beam splitter before going to either of the detectors. As I understand it, the chance of passing through this BS or reflecting off it is 50/50. So I would expect no interference patterns in these sub-samples.

    To put it another way - how do idler photons, of signal photons which contribute to an interference pattern, always end up at the same detector?
  21. Jun 11, 2011 #20
    I don't see how this can change anything. You still have no information about the s-photon.
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