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I have come across 'Wheelers Delayed Choice Experiment' which tries to prove that you can work out which Slit a Photon Goes Through in the Double Slit Experiment. But I thought it was impossible to know which Slit a Photon goes through? - So my question is, Why is Wheelers Delayed Choice Experiment Incorrect?

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    $\begingroup$ What's your source? I wouldn't say the DCE tries to "prove" which path a photon goes through or that the DCE is "incorrect". $\endgroup$
    – innisfree
    Dec 3, 2014 at 22:00
  • $\begingroup$ physics.stackexchange.com/questions/15776/… from the related questions is relevant. $\endgroup$
    – hobbs
    Dec 4, 2014 at 6:18

4 Answers 4

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It's not, rather your assumption that it's always impossible to know which slit a photon went through is incorrect. It's only impossible to know which slit it went through in an experiment where you get a double-slit interference pattern--if you wish you can set up an experiment to find out which slit the photon goes through, but the result will then be no interference between the two slits (see this page for a good discussion of both types of observation). This is the idea of 'complementarity' between which-path information and interference, which was formalized by Wootters and Zurek in a 1979 paper as discussed here (their mathematical formulation actually allowed for degrees of which-path information and degrees of interference, with greater which path information corresponding to less interference and vice versa, but in Wheeler's delayed-choice experiment the only two options possible given the experimental setup are either perfect knowledge of the path with zero interference, or zero knowledge of the path with perfect interference).

A version of Wheeler's delayed choice experiment is discussed here. The basic idea is that after the photon has passed through the slit but before it has reached your location, you can make a choice: option A is to place a screen at your location, in which case the probability of detection at different points on the screen will show a wave-like interference pattern and you won't know which slit the photon went through, whereas option B is to remove the screen and instead aim telescopes at each slit, in which case with a sufficiently sensitive detector you could measure the photon coming through one telescope but not the other. In the latter case you'll find out which slit the photon went through, but there won't be any interference pattern to observe, the experiment will give results consistent with the photon having traveled in a particle-like manner. As it says on the site:

Consider the difference in the experimental set up depending on our choice of detection. If we choose to leave the screen in place, we get a particle distribution consistent with the interference pattern that would be produced by two hypothetical symmetrical waves, each emanating from one of the slits. We might say (although we are extremely reluctant to say this) that the photon traveled as a wave from the point of origin, through both slits, and on to the screen.

On the other hand, if we choose to remove the screen, we get a particle distribution consistent with the clumping pattern that would be produced by particle motion from the point of origin through one slit or the other and to the left telescope or to the right telescope. After all, the particle "appeared" (we saw a flash) at one telescope or the other, rather than "appearing" at some other point along the length of the screen.

In summary, we have chosen whether to know which slit the particle went through, by choosing to use the telescopes or not, which are the instruments that would give us the information about which slit the particle went through. We have delayed this choice until a time after the particles "have gone through one slit or the other slit or both slits," so to speak. Yet, it seems paradoxically that our later choice of whether to obtain this information determines whether the particle passed through one slit or the other slit or both slits, so to speak. If you want to think of it this way (I don't recommend it), the particle exhibited after-the-fact wave-like behavior at the slits if you chose the screen; and it exhibited after-the-fact particle-like behavior at the slits if you chose the telescopes. Therefore, our delayed choice of how to measure the particle determines how the particle actually behaved at an earlier time.

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  • $\begingroup$ Note that there is absolutely nothing mysterious about it, provided you believe in quantum mechanics. The mathematics is straightforward. $\endgroup$ Dec 4, 2014 at 6:43
  • $\begingroup$ @Hypnosifl: your last paragraph is MISLADING. You say "we have chosen whether to know which slit the particle went through". The particle is a wave, it goes through both arms of the apparatus. If we put detectors on the arms one of the detectors will make a recording. It DOESN'T MEAN that the particle WAS THERE before the recording. As I said in my answer, the photon is a NON-LOCAL phenomenon. The process of collapse involves both branches of the particle. I recommended you articles, to see experiments done. $\endgroup$
    – Sofia
    Dec 4, 2014 at 11:11
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    $\begingroup$ @Sofia - That last paragraph wasn't my own words, it was a quote from the site I had linked to, as indicated by the light orange box. But I would say that whether the particle was "there" before the recording is really a matter of which interpretation of QM you would choose (I don't think the phrase you quoted is necessarily expressing a definite opinion, 'which slit the particle went through' could just be a shorthand for 'which-path information' as defined in the Wootters/Zurek paper). Note that many-worlds advocates say their interpretation is entirely local, for example. $\endgroup$
    – Hypnosifl
    Dec 4, 2014 at 14:21
  • $\begingroup$ I also don't think the idea of modeling the photon as a wave traveling through both slits itself qualifies as evidence of "non-locality" in the sense of being incompatible with local realism, since classical electromagnetic waves travel through both slits and classical electromagnetism is certainly a local realist theory. $\endgroup$
    – Hypnosifl
    Dec 4, 2014 at 14:23
  • $\begingroup$ Please read the answer I added now. Classical electromagnetic waves, and classical physics in general, are not a way to judge QUANTUM behavior. Won't you do as I said? Write me a letter, and I'll send you articles. NEITHER classical electromagnetism, nor ANY local theory, is able to explain the non-locality we find in quantum mechanics. The most ILLUSTRIOUS MINDS try hard since 1935 until today, without being able to get rid of the non-locality. But, to see this, you have to read proofs. It doesn't go cheaply. $\endgroup$
    – Sofia
    Dec 4, 2014 at 15:07
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I read about Wheelers choice experiment from Wikipedia, not from some article. There is no end to trials as that of Wheelers.

What is wrong with his attempt is that the photon is a WAVE. And as any wave, it takes both paths through the interferometer.

I saw in Wikipedia the statement

"since in the first case the photon is said to "decide" to travel as a particle and in the second case it is said to "decide" to travel as a wave, Wheeler wanted to know whether, experimentally, a time could be determined at which the photon made its "decision." "

All these are nonsenses. The photon makes no decision. It always travels as a wave, though, if we put detectors on the arms of the interferometer, the photon delivers its energy only to ONE of the detectors. WHY so? A tiny particle (called also quantum particle) has NON-LOCAL properties. The phenomenon of delivering all the energy only to one of the detectors, is NON-LOCAL.

This is what we can say. It's not a pleasure, but this is the situation.

To be more explicit, the non-locality consists in that a single particle never produces a recording in both detectors. It is "as if" there is some sort of agreement between the two wave-packets of the photon, which one will answer, and which not. (Asher Peres called once this phenomenon "conspiracy". But, of course, he didn't know exactly how is realized the agreement between the two wave-packets.)

Good luck!

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  • $\begingroup$ What does delivering energy in a localized manner have to do with non-locality? Have you read any physicists explaining this in terms of the concept of non-locality, or is this your own idea? $\endgroup$
    – Hypnosifl
    Dec 3, 2014 at 22:30
  • $\begingroup$ @Hypnosif: it's not my idea, it is the idea adopted by most of the long experienced specialists (it's not the idea of a few people). Are you interested in references? S. M. Tan, D. F. Walls, and M. J. Colett, "Nonlocality of a single photon", Physical Review Letters, Vol. 66, no. 3, page 252, (1991); L. Hardy, "Nonlocality of a single photon revisited", Physical Review Letters, Vol. 73, no. 17, page 2279 (1994). But there are more. $\endgroup$
    – Sofia
    Dec 4, 2014 at 2:15
  • $\begingroup$ (continuation) Peres described the state of a particle that passes through the beam-splitter, as "entanglement with the vacuum", i.e. |$1_{transmitted}$>|$0_{reflected}$> + |$0_{transmitted}$>|$1_{reflected}$>. A beautiful experiment to confirm this idea was performed by F, Sciarrino (see in arXiv quant-ph). $\endgroup$
    – Sofia
    Dec 4, 2014 at 2:23
  • $\begingroup$ But are they saying that the very fact that a photon is found at a single location is itself due to non-locality, or are they saying they've come up with some specific experiment that demonstrates non-locality for a single photon? The abstract sounds more like the latter. $\endgroup$
    – Hypnosifl
    Dec 4, 2014 at 4:00
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    $\begingroup$ Is my understanding correct, that the wave (or wave function) of the photon is non local, e.g. it travels through both slits, interferes with itself and hits both telescopes, BUT it only "collapses" at one location (e.g. one telescope) to give the impression that a particle traveled in a straight path to get there. So if you think of it this way the photon is not retro-actively deciding which slit to take, rather it is a question of what properties of the telescopes (or screen for the interference pattern) is causing the "collapse" and the swapping of them is really irrelevant. $\endgroup$ Dec 4, 2014 at 9:11
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This answer is not about QM but about a classical system that simulates many aspects of QM very well, and might be of some ineterestto you. There are experiments using oil dropplets that bounce up and down a vibrating fluid, the horizontal motion of these droplets is due to bouncing on the waves that move on the fluid's surface, which in turn are influenced by the collision with the droplets. The analogu here is the droplets are the particles and the fluid is the wavefunction. This is a purely classical system, but they were able to replicate the double slit experiment using one droplet at a time. Even if the motion of the droplet has chaotic components and pass through one particular slit at random, the end result using many driplets is an interference pattern similar to that obtained in the double slit experiment in QM.

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    $\begingroup$ This would not reproduce Wheelers Delayed Choice Experiment though would it? If the oil droplet is heading towards a bright spot in the center of the screen but you swapped the screen for the telescopes then the oil droplet would miss the telescopes? $\endgroup$ Dec 4, 2014 at 9:16
  • $\begingroup$ No, you are correct, at least not yet. The one thing they could not reproduce yet is entanglement of two particles, but I am pretty sure they will. Thet already simulated many nonclassical aspects of QM using a classical model. $\endgroup$
    – user65916
    Dec 5, 2014 at 7:34
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I believe it's incorrect but for another reason. I think it's ultimately a question of underestimating the implications of special relativity.

Here is my take on why Wheeler's reasoning is flawed. Feel free to put me in my place for my audacity.

The total distance travelled by a photon to the detector through slit A (Δx₁') from it's frame of reference and due to make an interference fringe at D1 (2 possible paths)

Length contraction: From the point of view of the photon, the distance is 0.

Δx₁' = Δx₁ / γ₁

Δx₁' = Δx₁ / ∞

Δx₁' = 0

(It arrives at the same time it departs from it's frame of reference)

The amount of time (γ₁) that occurs from the frame of reference of photon A:

γ₁ = 1 / (√(1 - β₁²))

γ₁ = 1 / (√(1 - 1²))

γ₁ = 1 / (√(0))

γ₁ = 1 / 0

γ₁ = ∞

or

Δt₁' = Δγ₁t

Δt₁' = ∞t

(i.e. an infinite amount of time occurs from the frame of reference of photon A)

The total distance travelled by photon through slit B to the same destination from it's frame of reference (taking the alternate route to the same detector):

Δx₂' = Δx₂ / γ₂

Δx₂' = Δx₂ / ∞

Δx₂' = 0

Δx₁' = Δx₂' = 0 (regardless of distance)

The amount of time that occurs from the frame of reference of photon B:

γ₂ = 1 / (√(1 - β₂²))

γ₂ = 1 / (√(1 - 1²))

γ₂ = 1 / (√(0))

γ₂ = 1 / 0

γ₂ = ∞

γ₁ = γ₂

or

Δt₂' = Δγ₂t

Δt₂' = ∞t

Δt₁' = Δt₂' = ∞t

It follows:

Δt₁' = Δt₂' = ∞t

Δx₁' = Δx₂' = 0

From the frame of reference of either photon, they both have zero distance to traverse over an infinite amount of time. The order of when they arrive is unintuitive from our point of view (frame of reference). It's 'spooky' to us as Einstein put it. The infinite amount of time is key to my argument. The photons do overlap in time, just not our perception of time - but theirs.

i.e. we perceive the interference to occur at some time t after a photon has already traversed a different path (at what we perceive to be) at a different time. This isn't really what happens from the frame of reference of the photons. It's more intuitive to think about it in terms of time dilation than length contraction arguably.

From the point of view of length contraction, any change that happens to photon must by definition happen at all points in "time" from it's frame of reference because ..they're all happening at the same distance - 0 distance via length contraction - it's all happening instantaneously, the distance is irrelevant, time is irrelevant from it's frame of reference, so it's not violating causality or Bell states as there are no hidden variables to violate, because there is no time from the frame of reference of the photon. From the point of view of the photon, there is no distance, there is no time as such, time is infinite. Of course any change made to either entangled pair will appear to defy causality from our frame of reference because we perceive time linearly as humans. The 2 entangled photons were connected in time at the same frame of reference as soon as their source photon was spawned into existence regardless of whether we perceive this entanglement to have occurred after the source photon was spawned into existence. i.e. since there is no time really for the photons, this link even though appearing to form afterwards propagates right back to the spawning of their origin photon from their point of view - the apparent spookiness of causality being defied.

From it's (photon's) point of view it's nothing more than a circuit branching at the point of entanglement but still linked to the "past" as we perceive it, causality becomes irrelevant. It's not travelling backwards into the past, because it's all instantaneous from it's reference frame.. The past, the present, the future - it means nothing for an entangled pair in the context of such an experiment.

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