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Delayed choice experiment

  1. Mar 30, 2005 #1


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    My brain is in great pain from reading Greene's 'Fabric o/t Cosmos'. I'm currently on p.197 (Chapter 7: Time and the Quantum - Shaping the Past).

    He's got the delayed choice setup rigged so that photons with 'which-path' info and no 'which-path' info are hitting the same screen. He says that if you examine *all* the data from all photons, you see no interference pattern - quote" ...not the slightest hint of an interference pattern...", but if you examine a subset of that data, corresponding to the photons that had no which-path data available, you see an interference pattern.

    He goes further to suggest that if you could have the two types of points coloured-coded, a colour-blind observer would see no pattern.

    So, here's my question:

    If half the photons are lining up in an interference pattern, and the sum total have no pattern, that must mean that the remaining photons are impacting preferentially in places where the interference patterns produce dark bands. Basically, they would have to produce a sort of inverse pattern to cancel out the first pattern.

    (See attached diagram)

    (It's something often experienced when spray painting. If you draw some lines (a pattern) on a wall, you can't then expect that a second spraying in a random (no pattern) coverage will cancel it out - you would merely get the sum of the two sprayings. To actually cancel your design out, you would have to preferentially fill in the blank areas until everything was even again.)

    Am I misunderstanding?

    What if you heavily weighted the experiment to produce 99% photons with no 'which-path' data? Would the sum total of the 99% patterned and the 1% unpatterned photons still show no discernible pattern?

    Ow ow.
    Last edited: Nov 28, 2006
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  3. Mar 31, 2005 #2


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    Yes, it is exactly as you say: the total dataset contains the addition of two interference patterns, which are "shifted" one to another, so that the addition shows NO pattern. You need the information on the "which subset" information from the other photon to select one of the two subsets, in which case you will obviously see the interference pattern of the subset.

    It is a bit as if half of the photons are distributed according to sin^2(x) and the other half are distributed according to cos^2(x).
    Together, you get sin^2(x)+cos^2(x) = 1, a flat distribution.
    But if you set up your experiment such that you only look at the first half, using a special trigger using the second photon (giving you "which subset" information), you select out only those with a sin^2(x) distribution.

    The trick is now that in order to obtain the "which subset" information, you need to do a measurement which is incompatible with the "which way" information, so you "erase" it (I really don't like that terminology). And if you know the "which way" information, you cannot also obtain the "which subset" information, and hence you cannot make any subset selection, so that you cannot observe one of the two interference patterns.

    Last edited: Mar 31, 2005
  4. Apr 1, 2005 #3
    And let me put that another way, so it is fully clear why you not only cannot see an interference pattern from the full experiment, but also cannot see an interference pattern from either of the two subsets that each represent an interference pattern.

    First, there are four patterns, not two: two, one from each side, where we know the "which path" information which therefore show no interference; and one from each of the two paths where we "erase" the "which-path" information, which each show interference. But we cannot disentangle either of these pairs of paths without knowing the exact impact time of both the signal photon, which goes to the screen, and the idler photon, which goes to the "which-path" detectors or the "eraser." We therefore cannot tell whether a particular signal photon was an interference photon or a non-interference photon without correlating it with an idler photon.

    Thus, we cannot tell which of the four patterns a signal photon is a member of without correlating its impact with that of an idler photon.

    The second relationship that makes this ironclad is this: even if we replace the splitters that determine whether a photon goes to the "eraser" or to the "which-path" detector with mirrors that send all the idler photons to the "eraser," we still cannot see interference without correlating each signal photon with an idler photon- because there are two interference patterns that exactly cancel one another out and result in the appearance of non-interference!

    If you're interested, I have a link to the paper that is worth seeing, but you have to crawl around some pretty heavy-duty math. You don't need the math to get the point, but it's pretty intimidating.
  5. Apr 5, 2005 #4
    Last edited by a moderator: Apr 21, 2017
  6. Apr 7, 2005 #5
    Yes. And that second is in fact the paper I was referring to.

    Yes. We cannot see those patterns on the screen; instead, we see them superimposed on one another. If you look at them carefully, and note the positions of the peaks and valleys across the bottom (horizontal) axis, you will note that they add together to superimpose a valley from one graph on a peak from the other. With the peak counts and valley counts adding together, you get a peak at every point; and that means that you can't see any peaks or valleys, only the smooth curve of the last figure. Which is (of course) the non-interference case; and you therefore cannot tell the non-interference case from the interference case, until you correlate each erased idler photon with a signal photon. Which is another way of saying the same thing Dr. Chinese did in your other thread.
    Last edited by a moderator: Apr 21, 2017
  7. Jun 13, 2005 #6
    Hello Schneibster,

    Could you try again to answer Edgardo’s posting of 04-05-2005? He asked for a simple explanation of the two interference patterns. I think you begged the question. It seems to me that there must be an essential asymmetry in the two paths that yield "fringes" in R_01 and "anti-fringes" in R_02. Could it have something to do with a difference in polarization or spin of the photons along the two paths?

    I feel a bit awkward here; this is my first attempt to participate in an online forum. I am just a layman who is fascinated by Brian Greene’s description of this experiment.
  8. Jul 21, 2005 #7
    Here's how I see it:

    If you examine how the erasure is actually carried out, you see that at the end, both the left-idler beam and the right-idler beam pass through beam splitter BS from opposite directions. This produces two new beams (let's call them beam 1 and beam 2), which are detected by two separate detectors, which are labeled D1 and D2. Half of the 'erased' idler photons go to D1, and half to D2.

    If you perform a coincidence count between D1 and a positional scan of the signal beam (D0), you will obtain an interference pattern. If you perform a count between D2 and the signal beam, you will obtain an anti-fringe pattern.

    The fringe and anti-fringe counts add up to the big lump you see if you do not perform the erasure (i.e., height of fringe(x) + height of anti-fringe(x) = height of unerased distribution(x)).

    There are two ways to look at the data. One is to say that the positional counts on D0 produce fringes if you coincidence count the signal photons with idlers that end up at D1.

    The other way to look at the data is this: the probability that an idler will be detected at D1 depends on the x-value where the signal was detected. Look at figure 3 in the original paper. When detector D0 was registering photons at x=1.5, 120 photons were detected by detector D1. Figure 4 shows that at the same time, only 40 photons were detected by detector D2. So, for a signal photon with x = 1.5, which is a peak in the fringe pattern, p(idler registers at D1) = .75 and p(idler registers at D2) = .25.

    So, as x changes, p(idler at D1) and p(idler at D2) vary according to the fringe and anti-fringe heights, and so you can interpret the interference in this way.

    What's really going on is that the x-position of the signal photon and the probability that the idler registers at D1 vs. D2 are entangled. The real key is that the erasure produces two new beams, and the interference is manifest through the idler photons' "choice" of which beam they end up in, since the idlers' "choice" is entangled with the signal's x.

    I guess I should point out that this is my own point of view on the subject, and I'm bringing it up here for the first time, so some people might disagree. Still, the data show that when x=1.5, 75% of the idlers go to D1, and when x=1.9, 75% go to D2 instead, so the relationship between x and p(D1) is clearly there.

    Bruce Zweig

    PS -- I should also point out that Greene oversimplified this experiment by leaving out the connections between the detectors and the coincidence counter in his diagram on p. 196. It would be clearer if he showed the fringe pattern at detector D1 and the anti-fringe pattern at detector D2, instead of implying that there was simple interference at D0.
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