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This above diagram is the delayed-choice quantum eraser. My current understanding is as follows:

1) If the experiment is done with $BS_a$ and $BS_b$ both mirrors, then double-clumping is seen on $D_0$.

2) If the experiment is done with $BS_a$ and $BS_b$ both transparent, then an interference pattern is seen on $D_0$.

3) If the experiment is done as it was originally (50% transparency on $BS_a$ and $BS_b$), then clumping in the form of a double-clump, will be seen on $D_0$. However, when you filter the data based on $D_{1-4}$, $D_0$ can be shown as a sum of two interference patterns, and two double-clumps, that individually match what is normally seen in the double-slit experiment.

4) The experiment can be done, with $BS_i$ arbitrarily far away. The results remain the same.

These are my assumptions. They may be wrong (Well, they must be due to the paradox), so please let me know if that's where the issue lies. So 1) and 2) are the normal double-slit result, though the apparatus is more complicated this time. I still understand it. 3) is complicated, but not hard to understand in a way that doesn't break causality. My understanding is that the x-coordinate of a photon on $D_0$ will collapse its wavefunction notably (But not entirely). If the x-coordinate is such that it lies on where the center of the red path clump should be, then its entangled photon will be much more likely to arrive on $D_4$, but it may still land on $D_{1-3}$. Similarly, the x-coordinate of any photon on $D_0$ directly assigns the probability distribution that its entangled partner will have upon hitting the remaining detectors.

However, let's say the apparatus is such that $BS_i$ and the detectors are a light-year away from from G-T prism. What happens if, during transit (Say, a day before photon arrival), we swap out each 50% transparency $BS_i$ with one that is 100% reflective. Now, double clumping must result on $D_0$. But this is a paradox, since we just changed the result of the experiment after the experiment happened.

At the minimum, I notice that (2) must be wrong here. But, why wouldn't an interference pattern develop on $D_0$ when which-path information is lost? This is my first and foremost question.

But, there also appears to be a deeper result. Let's say that $R_i$ is the subset of photons on $D_0$ that match a given $D_i$ (Or to be more precise, say $R_i$ is a probability distribution). Then the above paradox is not possible if $R_1 + R_2 = R_3 + R_4$ (Since the transparency of $BS_i$ would then not affect $D_0$). I believe that my assumptions above are correct, so I believe that my paradox allows me to deduce here that $R_1 + R_2 = R_3 + R_4$ (So that the two interference patterns on $R_3$ and $R_4$ sum to a double-clump). But, why? I guess this also asks the question of, in the original experiment, why do the peaks and troughs of $R_{1-2}$ cancel out in that way (And hence my original question of why (2) is wrong)? But generalizing, why does $R_1 + R_2 = R_3 + R_4$ appear to hold? I ask this from a mathematical perspective, as opposed to the solution that "It's a paradox if they don't".


I haven't read the paper, and there's conflicting info on Wikipedia on this subject. It appears that the original experiment may have been performed so that red and blue paths are both aimed at x-coordinate 0, so that my assumption (1) is incorrect (Single-clumping occurs, not my presumed double clumping). I don't think this matters, since the original experimenters could have had the red and blue paths aimed slightly off from each other (But not so far away that the interference patterns of $R_1$ and $R_2$ are lost). And, the distinction doesn't matter for the $R_1 + R_2 = R_3 + R_4$ theorem.


1 Answer 1


Disclaimer: I am not a physicist, and don't know the details of QM, but have read a lot about the DCQE experiment, and feel I understand how it works intuitively. And I feel that a serious question like this one should not go unanswered.

Your main mistake is assuming that anything changes on the D0 side, depending on what you do on the other side. The D0 detector will continue to see the same (lack of) pattern independent of what you do on the other side. It does not matter what you place on the other side or how far away it is, it does not change what is shown on D0. You can place as many mirrors and detectors and prisms there that you like, it will not change anything at D0. And what is shown on D0 is not an interference pattern, or two bumps, but photons arriving everywhere, with more in the middle, and gradually less when you measure more to the sides.

It is only after you compare the results from the D1 and D2 detectors with results from the D0 detector that an interference pattern becomes apparent. For each time a photon hits detector D1, you look where on D0 its peer landed, and you plot those locations on a graph. Then that graph will show an interference pattern. That is, the subset of the photons on the D0 side whose peers arrive at D1 form an interference pattern.

The photons at D2 form the opposite interference pattern at D0. Where D1 has peaks, D2 has troughs, and the other way around. If you add the interference patterns of D1 and D2 together, then cancel each other out. Added together, they just form the normal (lack of) pattern that you see appearing on D0 for all photons.

Another way to look at it is as follows: Suppose you look at a specific location at D0, which is at a peak of the D1 pattern, and detect a photon landing there. Then the chance of the peer of that photon landing at D1 is greater than it landing at D2.

See this answer for what I think is an excellent explanation. And I encourage you to read the paper itself, it is excellently written, and pretty understandable, except for the mathematical derivations in the middle.

Edit: To answer the question in the comments:

Why, if BSa and BSb are transparent, is D0 still random noise? Shouldn't it be an interference pattern?

The short answer is: because you have not changed anything at the D0 side.

The following is a gross over-simplification, but it catches the idea of what happens. Let's look at the signal (upper) side of the experiment first.

The x position on D0 is really simply a measurement of the phase difference between the two possible paths of the photon. When the photon leaves the crystal, there is a certain phase difference between the paths. Because there is a small difference in length between the two paths from the crystal to D0, it will be in-phase or out-of-phase when arriving at D0. If it is in-phase it can land at that spot, if out-of-phase it will never land there. But the difference in length between the two paths, and therefore where the photon lands on D0, is a measurement of what phase difference it had when leaving the crystal.

In a standard 2-slit experiment, without the crystal, the paths are always in-phase at the 2-slits, and so it will create an interference pattern at D0 - the photons will only land at x coordinates that correspond to paths that are in-phase.

If you put the BBO crystal in for the DCQE experiment, then it splits the photon in two, but it will also impose a random phase difference between the two paths of the photon. That is, the phase difference between the two paths is no longer always in-phase at the start, but random, and therefore it can land anywhere at D0, and no interference pattern is visible at D0. If you don't change anything at the D0 side, this will always be the case. But note that you can still deduce from the x-coordinate where it landed, what the phase difference at the start must have been.

Now have a look at the idler photon (lower) side of the experiment. Luckily, the idler photon starts out with the same phase difference as the signal photon. The idler photon will travel through the equipment at the bottom and reach the BSc mirror. Think of that as a filter for the phase difference. If the phase difference is in-phase, the photon will be send to D1, if the phase difference is out-of-phase, it will be send to D2.

That is, for the photons that arrive at D1, you know that they must have started in-phase. But then its peer signal photon must have started in-phase as well, and therefore can only have landed at certain x-coordinates at D0. So, by detecting the phase difference, and only considering those photons that started in-phase, you have filtered an interference pattern from photons arriving at D0.

  • $\begingroup$ Yes, I noted that landing on $D_0$ causes the wave function to collapse in a way that the particle's chance of landing on $D_1$ or $D_2$ is affected (In the standard interference pattern). My big question is why is (2) wrong. Why, if $BS_a$ and $BS_b$ are transparent, is $D_0$ still random noise? Shouldn't it be an interference pattern? $\endgroup$ Jan 31, 2019 at 15:41
  • $\begingroup$ Forget about cause and effect in QM experiments. As Luboš Motl says in the link in the answer: Correlation is not causation. At the end of the experiment, the wave function is collapsed, but it does not matter what caused it to collapse. See the edit for an answer to your question. $\endgroup$
    – fishinear
    Jan 31, 2019 at 18:00
  • $\begingroup$ Thank you! The information in the edit was perfect for understanding what is happening here. Great answer! A bit of a gap is why the prism imparts a random phase difference between the two paths (Which I'd still like to know), but under that assumption the rest of the explanation was quite helpful. $\endgroup$ Feb 1, 2019 at 0:45
  • $\begingroup$ @NicholasPipitone I guess it's because the BBO crystal actually generates two new photons, each with a fresh phase unrelated to the original phase of the incoming photon. But a physicist might know better. $\endgroup$
    – fishinear
    Feb 2, 2019 at 15:55
  • $\begingroup$ I've been wrestling with this experiment myself these last few hours and came to much the same conclusion as you. Perhaps I'm not crazy after all. $\endgroup$ Aug 8, 2019 at 5:29

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