Why is Wheelers Delayed Choice Experiment Incorrect?

Question

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?

Answer 1

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.

Answer 2

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!

Answer 3

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.

Answer 4

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.