Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Delayed choice quantum eraser -- Again...

  1. Sep 15, 2016 #1

    This topic seems to be discussed a lot of times before. But I still have some questions which was not clarified earlier and I will be grateful if someone will give me explanations :)

    First of all, to define a context, original experiment setup:
    (source: https://upload.wikimedia.org/wikipedia/commons/2/21/Kim_EtAl_Quantum_Eraser.svg)

    My understanding of what is happening:
    Laser emits a powerful ray which meets an obstacle with a double slit. At this point, each of photons in that ray propagates through both slit simultaneously (i.e. superposition; or through one exact slit, depending on detectors) and meets BBO crystal, where SPDC process taking place which creates entangled photon pairs. Next, first photon ("signal") travels to the detector D0, and the second photon ("idler") travels to one of the "determined" detectors (D3 or D4), or to the quantum eraser circuitry (randomly chosen detector D1 or D2). Detector D0 all the time sees just a blur. But when we filter all the photons reached D0 by corresponding D1-D4 signals, we will get four images: a "blur" (diffraction pattern) for D3 and D4 (where the idler photon has which-way information), and an interference patterns for D1 and D2 (where the idler photon has no which-way information). Interference patterns for D1 and D2 will be different (in counterphase).

    1. As was noted in few sources, difference between interference patters for D1 and D2 caused by phase shift, which takes place in BSc half-silvered mirror. Is it so? If yes, please explain, how exactly it happens if the system is fully symmetrical?

    2. Consider next image:
    I removed detectors D3-D4 and corresponding half-silvered mirrors BSa-BSb. Detector D0 still sees kinda "blur". But if we filter photons which idler pair hit exactly detector D1 or D2, we will get two interference patterns as in previous case. But this patterns still be different. Am I right?

    3. And finally, most interesting one. Consider this:
    Following the same logic, here we should see an interference patterns on the both screens. Literally "see on the screen" - there is no detectors anymore. There is no which-way information exist. But if so, it suggests violation of causality/faster-than-light interaction! (e.g. if we simply place two polarizers before lens ("mark" photons), interference pattern should disappear)
    So, my question is: what will actually happen and why?

    Thank you very much if you read so far :)
    Best regards.
  2. jcsd
  3. Sep 15, 2016 #2


    User Avatar
    Science Advisor

    It doesn't need to be powerful. It's actually preferable for it to be quite weak, and photons enter one at a time, since you need to figure out the pairing between photons going to two different places.

    Here's a quantum logic circuit that show the same qualitative behavior as the optical experiment:


    I'm going to be answering your questions in terms of that circuit, because I can actually play with it in Quirk to confirm my guesses.

    Here's a beam splitter, with a superposed photon incoming from two directions:

    beam splitter.png

    If the two paths have agreeing phase, the output will be symmetric. Equal chance of both output directions.

    But if the top path is a quarter phase ahead of the left path, and the beam splitter applies a quarter phase when reflecting, then the rightward output will interfere destructively and there will only be a bottomward output. The opposite happens if the top path is a quarter phase behind.

    In effect, you can think of this setup as a way of telling you the relative phase of the inputs. It transitions from "always goes down" to "always goes right" as you adjust the relative phase from quarter-behind to quarter-ahead.

    This is what's happening in the delayed choice experiment: the landing point on the screen determines the relative phase of the idler photon's two paths, then the beam splitter setup is used to tell you their relative phase.

    You get the same ripple pattern when conditioning on D1 or D2. The landing spot determines the relative phase, and D1/D2 tell you about that relative phase. D3 and D4 don't affect that, they were just an alternative measurement that could be done (it told you the relative magnitude of each slit's contribution instead of the relative phase assuming equal contribution).

    I tried this in the circuit by changing the 'choice' from a 50/50-output H to an always-on-output X:

    Screenshot from 2016-09-15 13:55:23.png

    Then you'll just see a lack-of-interference pattern on both screens. If you carefully pair up the photons there will be some correlation in the landing points, same as with any other experiment involving entanglement.
  4. Sep 15, 2016 #3


    User Avatar
    Gold Member

    The system is not fully symmetrical as only one side of the mirror has a coating.

    Interference can be seen directly only for coherent light. Downconverted light is not coherent so there is no way you can see the interference directly.
  5. Sep 15, 2016 #4


    User Avatar
    Gold Member

    Actually laser has to be powerful as very small fraction of laser light is downconverted in BBO crystal.
  6. Sep 15, 2016 #5


    User Avatar
    Science Advisor

    Ah, that would do it.
  7. Sep 17, 2016 #6
    This is not symmetric at all.
    D2 on red line 180+180
    D1 on red line 180+0
  8. Sep 17, 2016 #7
    As for first question.
    Strilanc & dochow, thank you for explanation. Now it looks more clearly. This means D2 detect photons from top slit (red line) with phase shift of π relative to photons from bottom slit (blue line). Detector D1 detect photons from both slits without relative phase shift. Okay. But I still not get - how this fact can affect the result of D0? Difference in interference patterns suggests that one of the waves must have phase shift, but it takes place only on eraser (bottom) side. It seems I'm missing something. :nb)

    Yes, I think so too. Just because of possibility (impossibility, to be precise) of causality violation/faster-than-light interaction. But I can not understand - why? Why we'll get interference with one screen, but will not with two of them? One single photon which passes two slits (as a wave of course) will produce an interference pattern. If we split this photon to two photons with lower energy (and this exactly what happens in BBO crystal during SPDC process), we'll not affect any information of the first one. So, both of new photons (signal and idler) should produce an interference pattern. And they do it in the DCQE experiment, at least one of them while other not being watched (metaphorically). So why adding second screen instead of eraser circuitry should destroy interference?

    Here I must make small remark. As zonde noted,
    saying "screen" I'm not quite right. Lets assume that "screen" will be either screen + very long exposure time camera, or special phosphorescent screen.

    p.s. Strilanc, Quirk seems to be quite nice tool, that's good to know. Thanks.
  9. Nov 13, 2016 #8
    Thank you so much for asking this question! This finally cleared up the matter for me. On the other hand, it is kind of disappointing to understand how trivial the solution is, all the tantalizing mystery is gone. And it makes me wonder why so many highly educated physicists, including some well respected ones, keep making such a fuzz about this.

    So, for future reference, or alternatively, as a chance to be corrected if I have still got it wrong, here is my explanation of the experiment:

    1. A photon is sent through a double slit. After the double slit, the photon is split into an entangled pair of photons. One of the new photons (called the signal photon) is sent on a path to self interfere and to be detected at some position at a "screen". "Unfortunately" though, the the phases of the two parts of the signal photon's wave function corresponding to slits A and B are randomly shifted when the entangled pair is created, so that the interference pattern is different for each consecutive photon that passes through, and cannot be reconstructed simply by accumulating data from multiple photons (in contrast to the original double slit experiment, where the phase of the wave function is not messed with).

    2. In order to recreate the interference pattern, we need to filter the photons by their phase shift. "Fortunately", the phase shift of the signal photon correlates with the phase shift of its entangled partner, so all we need to do is to measure the phase shift of this. For this reason, the experimenters put in a phase shift detector consisting of BSc and D1 and D2. When looking at signal photons corresponding to photons detected at D1 (or D2) only, the expected interference pattern is recreated, since these signal photons have similar phase shift as in the original double slit experiment.

    How disappointingly boring! And worst of all, the delayed choice apparatus is only smoke and mirrors. Literally! The whole thing is no more mysterious than stage magic!

    Please correct me if I am wrong.
  10. Nov 13, 2016 #9
    @nnerik, you're right. Caveat: I haven't studied your description in detail, maybe a nit-picker can find something wrong with it, but yes, it's the right idea. A further point is that when the idler photon hits D3 or D4, we lost the opportunity to determine its phase, therefore can't reconstruct associated interference pattern. The fact that we learn which slit it went through is not directly relevant. What counts is, we didn't learn the phase.
  11. Nov 14, 2016 #10
    @nnerik, Wow! How laconically and accurately. I'm impressed, thank you very much! Your explanation seems to leaves no questions anymore.

    Only one nuance left unclear for me. Please, look at this experiment: http://laser.physics.sunysb.edu/~amarch/eraser/ (paragraphs "Double-slit interference", "Which-Way Marker" and "Quantum Erasure").
    Here is experiment setup:
    In such setup we're able to "see" interference pattern. So, how this fact is consistent with your explanation? I suppose, the key factor is the absence of the double slit before BBO crystal. But I'm not sure how exactly random phase shift happens and correspondingly, the meaning of the double slit before BBO for this process remains unclear for me.
  12. Nov 15, 2016 #11


    User Avatar
    Gold Member

    No, there is no reason to think that you could see interference pattern in S detector alone. If you look at first graph it clearly says it's coincidence counts that appear on the graph.
    I looked at the paper (there is link in the article), for the reason why P detector is phase sensitive. I would say that's because of interference filter, after all light is coherent after interference filter.
  13. Nov 15, 2016 #12


    User Avatar
    Gold Member

  14. Nov 16, 2016 #13
    Thank you, you're right. I missed that.

    Damn, our Universe is so awesome in its own freaky way :smile:
    So many fascinating things happens.

    Thank all of you guys for every answer. I was glad to get a little bit more understanding of what happens in the strange quantum world.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted