6
$\begingroup$

In a double-slit experiment, interference patterns are shown when light passes through the slits and illuminate the screen. So the question is, if one shoots a single photon, does the screen show interference pattern? Or does the screen show only one location that the single photon particle is at?

$\endgroup$
6
$\begingroup$

The answer is yes to both questions: yes, the screen does show one location for one particle and yes, the accumulated picture after repeating the experiment many, many times does show the interference pattern.

There is a set of beautiful pictures and a video of the double slit experiment in one-particle-per-time mode that can be found here (the experiment is with electron but conceptually there is no difference).

$\endgroup$
  • $\begingroup$ You can also try Chad Orzel's discussion at scienceblogs.com/principles/2012/03/…, which links to a new paper in Nature Nanotechnology that is OpenAccess at the time of writing. Not video, though. $\endgroup$ – Peter Morgan Mar 28 '12 at 14:58
  • 4
    $\begingroup$ Here, you'll find an experimental movie of the gradual interference construction with single photons. Full disclosure : I work in the group who did the experiment, and the authors ar friends of mine. $\endgroup$ – Frédéric Grosshans Mar 28 '12 at 15:00
  • $\begingroup$ @FrédéricGrosshans There is an other point of view. The interaction between the photons and the surface electrons from the slits surface are quantized and the result are fringes on the observations screen. $\endgroup$ – HolgerFiedler Jun 27 '15 at 13:57
7
$\begingroup$

Let me try a slightly different way to answer this (well worn) question.

The photon doesn't have a location, or at least not a well defined location, until you interact with it and cause it to localise.

When the photon hits the photomultiplier, or photographic plate, or whatever you're using as the screen the interaction occurs at a point and that localises the photon. Until then it's somewhat meaningless to talk about the position of the photon. I don't mean the photon has a position but we don't know it, I mean the photon simply doesn't have a position. That's why it doesn't make sense to ask which slit the photon went through. because the photon's position is ill defined it occupies the whole experimental apparatus.

So a single photon does indeed passthrough both slits, but it then interacts with the screen at a point. The point of interaction with the screen is random, with the probability o the position being given by the square of the wavefunction. That's why over time the pattern created by many phtons gives you the interference pattern.

$\endgroup$
  • 1
    $\begingroup$ John, as always thanks for the clear answers you provide. As example, sticking to the case of single photon, e.g. in double slit experiment, people often say the wavefront (the probability wave) is what is being split, or simple put the wavefunction. Although this is not a physical wave, do we still assume the probability wavefront travel at speed of light? $\endgroup$ – user929304 Jul 4 '16 at 14:44
  • $\begingroup$ You would have to define what you mean by probability wavefront. The group velocity of the wavepacket representing the photon travels at the speed of light. The wavefunction is not restricted by the speed of light and can change in time arbitrarily fast. $\endgroup$ – John Rennie Jul 4 '16 at 14:47
  • $\begingroup$ Thanks, makes sense. Can I use an example? I think it will help to clarify what I m looking to understand: in the Michelson experiment, if we have only one photon, after the beamsplitter, its wave-function is split into two components (not the photon itself), thus represented by a state of superposition. Then by analogy to physical waves (EM e.g.) we say the reflection of these components off of the two mirrors interfere, and if the path lengths (leading up to each mirror after the beamsplitter) are different, ... $\endgroup$ – user929304 Jul 4 '16 at 15:01
  • $\begingroup$ the interference will in turn be different. So in a sense it implies the probability waves of the single photon could not have had infinite speed, as else the path lengths would have become irrelevant to the observed interference at the detector. But it is known that the geometry of the setup modifies the interference effect, and I cannot convince myself... :( $\endgroup$ – user929304 Jul 4 '16 at 15:01
  • $\begingroup$ My current take is that: the speed of propagation of a quantum system's wavefunction (which is not a physical wave), depends on the physical system itself, (e.g. if a photon, then bounded by speed of light). $\endgroup$ – user929304 Jul 4 '16 at 15:10
0
$\begingroup$

We don't know whether the light source shoots photons or not. We know that if we turn off power to the light source the interference pattern disappears, and that if we turn down the light intensity enough we eventually start seeing individual events, if we have the right sort of measurement apparatus. Again, if we turn off the power those individual events stop (except for the "dark rate" that is characteristic of the detector), so it's definitely the light source that is causing the individual events, but we don't know what happens in between.

It's possible to account for this simple kind of experiment using a semi-classical model in which there is an electromagnetic field between the source and the detector, and the detector current flips off and on. It's only when we consider more sophisticated experiments, in particular in which we engineer the light sources so that two or more individual events are closely synchronized in time, that we find that neither shoots photons nor there's an electromagnetic field works very well.

Consequently, we might or might not be able to satisfy the premise of "if one shoots a single photon, ...", making it not possible to answer the question with certainty with our current understanding. Nonetheless, I up-voted Slaviks Answer, because that's what is usually said.

$\endgroup$
-1
$\begingroup$

After 2006, no philosophic duality exist, only bad pictures

As @Slaviks says, "the answer is yes to both questions", but I like the modern and experimental (!) Y. Couder interpretation. See by your self (!),

Youtube Couder experiments

The quantum particle HAVE a location (against @Rennie says), there are no "philosofic duality", there are only a limitation in choose a good pictoric model when you is constrained by "wave or particle" picture options: Couder demonstrates that a good picture, of an "intermediary wave/particle object" model, exists!

Imagine a "localizable object" that haven't a well-defined boundary, but have a well-defined distance-limit (lambda) to interact with obstacles (other objects).

There are an article online about the experiment.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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