0
$\begingroup$

I am mostly basing my understanding on this video https://www.youtube.com/watch?v=RQv5CVELG3U (I also have a background in lasers, and a tiny bit of undergraduate QM but nothing too profound)

I can understand why the interference pattern is destroyed when we measure the photons using detectors D1 or D2 in this diagram.

enter image description here

The detection of an entangled photon at D1 or D2 gives conclusive evidence regarding which slit the entangled photon took, so there will be certain that there will be no interference pattern. Similarly, the distributions of the particles associated with the "erasure" detectors D3 and D4 (shown in the video but not the image above) are completely reasonable to me.

My question is, what if there are no detectors at all? What if we just let all the entangled particles going towards the right side diagram just travel off to outer space in a way that no one can ever detect them? Or put a crude black obstruction that would absorb the photons in a way that no one can recover the which way information?

From the sources I read, I get the impression that no matter what you do with the entangled photons going to the right side of the diagram, the total pattern on the screen will be the sum of two non-interacting single slits - in other words no interference.

This leads me to conclude the "crystal" placed after the slits is responsible for destroying interference. The way I see it, the mere existence of that crystal after the slits is profoundly altering the wavefunction of the photon. As soon as the crystal "works its magic", any hope of interference is forever destroyed right then and there. (How accurate is this statement? I am talking about an interference pattern created by all the photons. I am well aware that if we select a subset of photons by an informed matter, for example using detector D3, we could see a pattern.) What is this crystal doing, and why do so many explanations completely omit the seemingly crucial part it plays?

A follow-up question: What if we used a laser that generates many particles, rather than a single particle generator? The right side of the experiment still goes off to infinity where they can never be detected. Do we expect to see an interference pattern in this case? I see no reason the laser case should be different, so no interference. But if we do not observe an interference, will it even be a "laser" beam coming out of the crystal?

I asked many questions, and please feel free to engage with whichever ones you think will be useful for clarifying my misconceptions.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

There IS "interference" when all the detectors (except your screen) are removed. But please know in the "official" DCQE experiment a coincidence circuit was used to observe the interference and there was never a screen, they used a scanning detector for D0. So what you have read is probably that if the detectors are removed there is no way to observe the interference NOT that there is no interference.

For your version .... you would have 2 overlapping "interference" patterns, you have a basic simple double slit experiment but have just doubled the photons going to the screen with the crystal.

Also this was discussed in more detail here: Is This Quantum Eraser Video from Fermilab Incorrect?

$\endgroup$
6
  • $\begingroup$ That was not really the question. I am talking about taking every photon into consideration. Taking every photon into consideration without any grouping, without the "crystal" you get an interference pattern, with the crystal you do not. This is true no matter what you do on the other side of the experiment. So it seems like the mere presence of the "crystal" is disturbing the standard double slit experiment, not merely sampling the photons without disturbing the actual measurement. $\endgroup$ Dec 26, 2021 at 18:57
  • $\begingroup$ You still get an interference pattern with the crystal. But you are correct that there is some deterioration. Consider not a crystal but an optical diffuser instead, the strength of the pattern would depend on the strength of the diffuser, the stronger the diffuser the weaker the pattern. In the actual DCQE there are a few optical components in the path after the slits... all of which have some effect. But the important point is that the double slit has effected the path of many of the photons. $\endgroup$ Dec 26, 2021 at 20:55
  • $\begingroup$ IN the DCQE some photons do not interfere ... the explanation is that they take a direct route thru one of the slits only as we know. These possible paths are possible due to the beam splitter which offers a path directly to D3/D4. Feynman gives us the Feynman Path Integral to calculate the probability of photons travelling certain pathways. $\endgroup$ Dec 26, 2021 at 21:08
  • $\begingroup$ I have seen sources that say something along the lines of "any entanglement destroys interference". The crystal that creates the entangled pair seems to be giving a random phase shift in the moment of creating the entangled pair. Which gives me the impression that if laser light is shined on this crystal. the output would not be laser any longer, it would be incoherent but monochromatic light. Any idea if this is true? $\endgroup$ Dec 27, 2021 at 3:34
  • 1
    $\begingroup$ ..... like when we try and measure which slit the electron went thru .... the electron can't make the pattern as it got entangled with a photon on its way. Please note that the electron still has has wave properties but now its wave properties are brand new having nothing to do with the constraints the double slit and screen put on the electron initially ..... the intended path(s) that were to create the pattern are destroyed, the electron is now free to find a new path unconstrained, i.e. no pattern. $\endgroup$ Dec 27, 2021 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.