In the double-slit experiment, quantum mechanics states that if you try to determine which slit the photon goes through, you won't have a resulting wave pattern.

But, knowing the time it took for the photon to go from the source to the observing screen, can you deduct the distance of the photon path and so which slit it passes through (except if the photon impacts the exact middle of the observing screen)?

  • $\begingroup$ How are you going to "know" that time? $\endgroup$ – ACuriousMind Mar 25 '16 at 16:46
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    $\begingroup$ You can measure the time of detection of the photon, but you wouldn't know when it was created as the initial state. As Emilio Pisanty points out correctly, measuring both would destroy the interference. $\endgroup$ – CuriousOne Mar 25 '16 at 17:06
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    $\begingroup$ I guess you can "know" that time by putting a "spontaneous parametric down conversion" between the source and the slits and a detector for the entangled photon. So you can know the "emission time". The observing screen is a kind of detector, so i guess you can have an "impact time". $\endgroup$ – BenLaz Mar 25 '16 at 17:07
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    $\begingroup$ +1 awesome question. hope it gets some more detailed answers, although i feel like emilio pisanty probably hit the core of the problem. but a longer explanation would be nice eg why you need long wavepackets for interference. cant you do it with individual photons? $\endgroup$ – Wolpertinger Mar 25 '16 at 18:09
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    $\begingroup$ @Numrok note that pulse duration and photon number are independent measures. $\endgroup$ – Emilio Pisanty Mar 25 '16 at 20:35

But knowing the time it took for the photon to go from the source to the observing screen, you can deduct the distance of the photon path and so which slit it passes through

... and indeed such information will make it impossible for fringes to appear.

Interference experiments use wavepackets that have a long duration, which makes it impossible to tell from timing information which slit the particle came through, eliminating the problem.

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    $\begingroup$ When you say that "long duration makes it impossible to tell from timing information which slit the particle came through", does it means there is a real impossibility (theorical) or it's just a lack of technology ? $\endgroup$ – BenLaz Mar 25 '16 at 17:18
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    $\begingroup$ No, it's a real impossibility. There needs to be substantial (almost complete) overlap in the probability distribution of arrival times of the two slits. $\endgroup$ – Emilio Pisanty Mar 25 '16 at 20:34
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    $\begingroup$ Interference has been observed with single electrons, neutrons, and bucky balls. Why is timing those objects/particles not an option to get which-path information? $\endgroup$ – Jasper Mar 25 '16 at 22:12
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    $\begingroup$ @Jasper It is an option. And if you use wavepackets short enough to get which-path information, you will destroy the interference. $\endgroup$ – Emilio Pisanty Mar 25 '16 at 23:46

There exist single photon at a time experiments though for a double slit experiment the single electron at a time are easier.

That the interference pattern disappears if one places detectors to check the path is an experimental fact, it is the result of the mathematics of quantum mechanics which up to now really describes the microworld of particles and atoms.

As it is clearer for electrons, I will start with them:

a) There exists a wavefunction a solution of a QM wave equation, which may display wave characteristics of interference given the correct boundary conditions , which pick the solution.

b)The double slit experiment is a boundary value problem for the quantum mechanical regime. It is the boundary problem "electrons of specific momentum scattering off two slits of specific size".

Nature gives the solution as an interference pattern.

If one puts a detector on the path of the electron, the boundary problem changes to " electrons of specific momentum scattering off two slits of specific size and rescattering off detector on the way". Let us suppose that the detector at the screen is an active pixel with good time resolution so only a detector on the way is needed.

What happens is that any detector on the way will scatter the electron, i.e., a different wave function will be describing the setup.The interference pattern is not seen because the secondary scatter spoils the phase of the solution, with respect to the geometry and no interference appears.

That it is the change in the boundary conditions that destroys the phases and thus the interference pattern was made clear in a recent photon experiment, where they show that by trying to locate which way the photon went, the photon reaching the detector loses the phase and leaves the interference pattern.

The same would happen with any detector, either positional or timing detector. The change in the boundary conditions that the detector introduces destroys the original phase correlations that gave the original interference pattern.


Just try to think of a thought experiment by which you will measure the time of travel for the photon:

You will need to start a timer when the photon gets emitted by the source. Thus you'll need to Measure its position once.

Then you'll need to measure it when it reaches the screen to stop the timer.

But here's the catch: photons are indistinguishable. So just by measuring at the start and at end you cannot say it was the same photon. So you'll need to trace it throughout the way. But everytime you measure you are disturbing the system. Thus measuring the photon even with very low energy rays or fields you'll disrupt the process of interference and there will be no fringe pattern.

P.S. Feynmann has a brilliant way of explaining the whole business. Go through Feynman's lectures in Physics, volume 3. It's really entertaining.

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    $\begingroup$ What "it" of Feynmann's are you referring? A link to a webpage or at the very least the name of the book/essay you're referring to would be helpful. $\endgroup$ – R.M. Mar 25 '16 at 17:50
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    $\begingroup$ Maybe feynmanlectures.caltech.edu/III_01.html $\endgroup$ – WorldSEnder Mar 25 '16 at 19:08
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    $\begingroup$ If the source is weak enough there shouldn't be much ambiguity about which photon it is. $\endgroup$ – Owen Mar 25 '16 at 19:08
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    $\begingroup$ This is wrong, surely? We can make sure that the photons are -- with high probability -- emitted one at a time, and each individual photon can be energetic. $\endgroup$ – TonyK Mar 26 '16 at 9:44
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    $\begingroup$ That has nothing to do with photons being indistinguishable. $\endgroup$ – TonyK Mar 26 '16 at 15:23

protected by Qmechanic Mar 25 '16 at 23:42

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