Preamble: I'm quite aware I've been duped by sci-pop videos and maybe the Wikipedia. I'm asking this question precisely in order to correct the Wikipedia article and to make my own video debunking the retrocausality in DCQE.

I'm desperately trying to get at least a superficial glimpse of quantum mechanics from science popularization YouTube channels (and in some degree from Coursera), and I think there's one important bit of information about the delayed choice quantum eraser (DCQE) experiment missing in all of the descriptions, even on PBS Space Time and FermiLab. Namely, this is difference between the double-slit experiment with electrons (or atoms, or even whole molecules) where detection is performed by hitting an electron (or its absence) with a photon, and DCQE where we have to split a photon into an entangled pair. Even more specifically, I'll be referring to Quantum Eraser Lottery challenge video on PBS Space Time (links [1] and [2] below).

So, the no-communication theorem holds true because without knowing data from ABCD (original DCQE detectors D1 to D4) we cannot see any pattern on the D0 interference screen. But if we compare this to simple double-slit, then we'll see there we didn't need to sort electrons into two piles to observe either interference or clump pattern — we had different total images depending on whether we were bombarding exit of one of the slits with photons or not (I'm intentionally using this wording in order to remind myself that in this case the act of measurement is very materialistic and destructive, and that the so called 'observer effect' is not an explanation of how things actually work in QM).

In DCQE, on the other hand, we have the same total pattern if we add up all I+A- and I+B-detected photon pairs and if we add up all I+C- and I+D- detected pairs — the sum of two clumps exactly matches the sum of two interference patterns and provides just a single gaussian-distributed clump. In the case of electrons we didn't need to distinguish between A-slit electrons and B-slit electrons in order to see a pattern, we were only — as I put it earlier — bombarding or not bombarding the slits with photons.

Wikipedia explains this difference in the following way: BBO crystals used to produce entangled photon pairs have very low efficiency — they only produce an entangled pair once in 10^5 times. So, to detect whether any single photon comes from an entangled pair, we still have to wait for its twin to be detected at A, B, C or D. (The other 99999 times I assume they produce a non-entangled pair, or a squeezed vacuum, since if they left an unsplit high-energy photon, as it is (incorrectly?) stated in Wikipedia, it would be simple to filter out such a photon because it has twice as much energy and frequency.) In other sources I met slightly different numbers, but in all cases it is true that only a statistically small fraction of photons detected on the interference screen can be used in this experiment at all.

This explains to me why we have to bother with the C and D detectors at all: otherwise we could just turn them off, so that 'particle-like' photons are detected at A and B, and 'wave-like' are left undetected at all.

But now the question: is the impossibility of creating a crystal which would provide entangled photon pairs in a statistically significant amounts (say, 75% of all produced pairs) a technical limitation or a fundamental law? As I see it now, creating such a crystal would allow us to break the no-communication theorem in DCQE, and since the no-communication theorem is strictly rooted in the current QM theory, then existence of such a crystal would require us to revisit the QM theory — not some interpretation, but the very equations it is based on, like Schrodinger's equation? If so, then by the means of QM theory we could derive even some theoretical maximum efficiency rate for SPDC process taking place in a BBO crystal or other type of entangle-splitter?

Or is Wikipedia article misleading here and the difference between double-slit and DCQE (or between shooting electrons and photons at the interference screen) here more fundamental, having nothing to do with the efficiency of SPDC process?

Finally, is this in any way connected to the fact that the photon to be split comes to the crystal in a superposition of two states and that superposition needs to be preserved at this stage? Are the two parts of this BBO crystal in any way connected to each other? As I understand, other sources of entangled photon pairs are more efficient than this setup, but for a fair comparison we have to look at the slits + BBO system as producing actually a group of 4 photons, or, better to say, a product of entanglement and superposition, which is a higher order phenomenon than simple entanglement? Or it doesn't play any role in producing entangled photon pairs?

  1. Quantum Eraser Challenge https://www.youtube.com/watch?v=2Uzytrooz44
  2. Quantum Eraser Solution https://www.youtube.com/watch?v=MuvwcsfXIIo&t=240s
  • $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Commented Nov 5, 2021 at 12:57

1 Answer 1


The crystal produces many more pairs than can be detected, the detection apparatus has a small acceptance angle, the crystal produces the pairs in a wide array of angles ... but the apparatus is limited. Yes all the non-converted original photons all filtered out.

Here is part of the wiki article: " The delayed-choice quantum eraser does not communicate information in a retro-causal manner because it takes another signal, one which must arrive by a process that can go no faster than the speed of light, to sort the superimposed data in the signal photons into four streams that reflect the states of the idler photons at their four distinct detection screens.[note 2][note 3]"

  • $\begingroup$ That still doesn't explain the key difference with double slit. When we look at light that passed through 2 slits, we can see the interference pattern. When we look at light that passed 2 slits AND splitter crystal, we don't see it anymore. Is it because after splitting photons move at an angle to previous direction and their phase is out of sync? Why this part doesn't get attention anywhere? $\endgroup$
    – Androniq
    Commented Nov 6, 2021 at 15:05
  • $\begingroup$ physics.stackexchange.com/questions/675002/… $\endgroup$ Commented Nov 6, 2021 at 17:12
  • $\begingroup$ See one of my answers above for a more modern theory of light, it is not so popular because the old classical "interference" theory has been taught for many years. And the old theory also works nicely mathematically as well so it is good for high school and first year university. Feynman spent most of his time on nuclear physics during WWII and afterwards and new light theories have not been that important. $\endgroup$ Commented Nov 6, 2021 at 17:16
  • $\begingroup$ In general photons are already planning their paths even before they have even started. One interpretation is that an excited electron in the atom is already sending "virtual photons" (EM forces) into the EM field, these EM field disturbances can sense the path and eventually the real photon arises and transfers energy to a receiving atom. Whenever the starting atom/electron is able to see multiple paths the pattern will be present. If on the other one of the hand paths is blocked (polarizers or filters) the atom/electron will see one path and there is no pattern seen. $\endgroup$ Commented Nov 6, 2021 at 17:25

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