I can accept that when single photons are used in the double slit experiment that a diffraction pattern results at the target due to their wave property.

What I am puzzled about is exactly what practical device is used to try to measure which slit the photon travels through in a non-intrusive manner. Why is it such a mystery that this observation disturbs the diffraction pattern? Surely any measurement we take will spoil the experiment.

Still are physicists using fields, other photons or what to detect the photon before it travels through the slit?

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    $\begingroup$ One can't "use single photons" in an experiment. They don't exist as independent physical objects. A photon is a number produced by an experiment on a quantum field. It's not a physical entity that somehow magically propagates trough space to leave random footprints on screens. The best way of looking at photons is the same way one would look at a measured quantity like an electron spin, which can be "up" or "down" or in a mixed state. In the same way there are one, two, three etc. photons in a given location at a given time, but neither the spin state nor photon states "propagate". $\endgroup$
    – CuriousOne
    May 17, 2015 at 17:21
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    $\begingroup$ this demonstration contradicts the statement that one cant use single photons sps.ch/en/articles/progresses/… $\endgroup$
    – anna v
    Aug 27, 2018 at 5:28
  • $\begingroup$ @ACuriousOne I don't agree with you but it is an interesting duscussion. Your argument leads to the conclusion that a photon is a quasiparticles. This maybe so but the same argument applies to any fundamental particle or quantized excitation. $\endgroup$
    – my2cts
    Sep 11, 2019 at 12:38

4 Answers 4


It's easier to get at the issue here by using massive things such as neutrons or atoms as the entities that go through the slits and hit the screen. It's not so self-evident that the observation of the location of an atom is guaranteed to disturb the atom's motion sufficiently to wash out the interference pattern. Of course quantum mechanics says that this is what must happen, and thus quantum mechanics is consistent.

One can measure the location of an atom as it flies past by, for example, bouncing light off it and detecting the light. This is called Rayleigh scattering.

If one were to use photons as the entities passing through the slits and showing interference, as in the question posed, then one wants to determine the location of a photon, to sufficient accuracy, without absorbing the photon. In principle this can be done by letting the photon reflect off a very light mirror (prepared in a precise state of motion) and observing the recoil of the mirror. In practice this is too difficult. Instead photons have been detected near one slit or another by a different method. It has been done by using an optical cavity to enhance the photon's interaction with a single atom located in the cavity. The photon flies through and the atom changes state.


In the famous double-slit experiment using photons, you have a couple of configurations:

Configuration A - 2 slits and 1 screen:

After sending 1 photon at a time at the double slit, the photon hits the screen seemingly at random, but over time an interference pattern builds up. But each photon went through on its own, so there are no other photons for it to interfere with. This means the photon must be, in some sense, passing through both slits at the same time and interfering with itself.

Configuration B - 2 slits, 2 photon detectors, 1 behind each slit:

There is no screen in this setup, and if there were, it would be blocked by the detectors. You send 1 photon at a time, but this time, 1 of the detectors registers exactly 1 photon and the other registers 0. This means the photon must only go through one slit at a time. If it went through both, as we expect from experiment A, both detectors would detect something.

The mystery:

Why there is a difference between configurations A and B? The only difference is that in B, we would be able to tell which slit the photon went through. It's like the photon knows it's being watched, which is spooky.

So to address the main concern in your question, we don't measure individual photons non-intrusively. Some advances have been made recently in possible ways to do this (e.g. here and here), but in general it's extremely difficult/barely possible. It's not necessary to detect the photon and let it continue to the screen, we just detect the photon.

  • $\begingroup$ This is not true. The result of A are always single photons hitting the screen, the same result which you get in case B. You will only see fringes after a lot of photons will hit the screen. In case B, if you place instead of two detectors two CCD-chips you get the same result in B as in A $\endgroup$ May 16, 2015 at 22:03
  • $\begingroup$ Yes... I should clarify that the interference pattern builds up over time. Fixed that now. $\endgroup$
    – jamcowl
    May 16, 2015 at 22:14
  • $\begingroup$ So your conclusions are wrong. In both cases you get the same result. The only thing in case B is that you use an instrument with less precision. $\endgroup$ May 16, 2015 at 22:20
  • $\begingroup$ In case A the photons' distribution matches an interference pattern, as if each individual photon was passing through 2 slits and self-interfering. In case B we see that there is only ever 1 photon per slit. That's the difference. $\endgroup$
    – jamcowl
    May 16, 2015 at 22:25
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    $\begingroup$ Why do you think there'd be a diffraction pattern if you knew which slit the photon went through? A single coherent wave source doesn't interfere with itself. You need at least 2 sources, so you need the photon's wavefunction to pass through 2 slits so they can act as 2 sources for interference. So... you can't have an experiment that "reveals a diffraction pattern and Also shows which slit the photon went through". $\endgroup$
    – jamcowl
    May 20, 2015 at 0:47

You wrote,

What I am puzzled about is exactly what practical device is used to try to measure which slit the photon travels through in a non-intrusive manner. Why is it such a mystery that this observation disturbs the diffraction pattern? Surely any measurement we take will spoil the experiment.

You've put your finger right on the crux of the matter: it is not possible to measure which slit the photon travels through, in a non-intrusive matter. All measurement methods that detect a photon either absorb the photon or change it in some way.

Single-photon interferometry experiments are necessarily statistical in nature: the result is obtained by compiling the results of many single-photon detection events.

A good resource for building an understanding of this is the Wikipedia entry on Wheeler's Delayed Choice Experiment. Another is the Wikipedia entry on Delayed Choice Quantum Eraser.


There is no known way to detect photons in flight.

You can only detect a photon when it is absorbed. And not always then.

In theory you might be able to detect a photon when it is emitted, by the effect on the source that emits it. That is usually not possible, usually you tell that photons were emitted by absorbing some of them.

Here is a question -- are photons real, or are they an artifact of detection?

What if light is waves, and it travels through space exactly like waves but when it interacts with molecules (which are quantized, it can't interact with 1.72 molecules), then the result is quantized because the molecules are quantized?

What if molecules always absorb a quantum amount of radiation, leaving behind whatever they don't absorb? What if they radiate a quantum amount of radiation, and no more? (They do.)

Then light would be a wave, and everything we see as photons would happen because of our limited detection methods.

Is that true? I don't know. I can't think of an experiment to test it.

In practice we use whatever theory is most convenient. For geometric optics we assume that light is rays that travel in straight lines. That's most convenient.

For looking at where light actually goes, when diffraction matters, we use wave theory. Because that's most convenient.

When we want look at how light interacts with molecules, we use QED which gives us particles which travel exactly like waves, because that fits easiest into the rest of modern theory. Light as waves or as modern quanta both work perfectly with the data, and we use the version that is most convenient. Light as straight rays does not work as well sometimes. But sometimes it's the simplest and easiest to use.

If two photons are detected at the same time, maybe they are the same wave getting detected twice? So you turn down your emitter until it emits around one photon per second. You put a detector behind each of two slits. And you consider it a double detection if two photons are detected within one nanosecond of each other, which (let's say) is the limit of your ability to time photon detection. Then you can expect to have two photons detected at the same time by accident, because you created two photons around the same time, about once in 10^9 seconds, assuming that photons are produced at random.

On the other hand, suppose that you have your emitter turned down until the wave intensity produces on average one random detection event per second. It could get a detection any time, but the wave is so weak that they are rare. Then the chance that the wave will result in two detections at once is about once in 10^9 seconds.

But imagine you could detect every time your source emits a photon that heads in the right direction. You never get a false positive, when there was no photon emitted but one was detected. You might find that most of the photons are not detected. Maybe you never get two detections when there was only one photon emitted. Or maybe sometimes you do get two detections when there was only one photon emitted.

That would be a great experiment! If you do get two detections, it shows the photon travels through both slits (or that the wave of course did.) If you never get that, then each photon only travels through one slit and the fact that it could have gone through the other one instead bud didn't, somehow affected it. That would be weird and exciting.

  • $\begingroup$ On the topic of quantized absorption causing the perceived "particle" effect, this was arguably the first argument against Einstein, and to my understanding that was what Planck had in mind with his quantization of energy (never mind photons). A combination of rigorous analysis and the discovery of Compton scattering eventually put the photon theory on solid footing. See Non-Einsteinian Interpretations of the Photoelectric Effect by Stuewer: mcps.umn.edu/assets/pdf/5.11_Stuewer.pdf $\endgroup$ Dec 12, 2018 at 20:13
  • $\begingroup$ I don't yet see why this is incompatible with radiation behaving as a wave when it travels, and that molecules emit and absorb quanta from the wave. The historical argument is that wave theory was undeveloped at the time, and quantum theory successfully answered some questions first. I have not seen an argument that wave theory could not be developed to answer those same questions. $\endgroup$
    – J Thomas
    Dec 12, 2018 at 20:25
  • $\begingroup$ What do you mean by "wave theory was undeveloped at the time, and quantum theory successfully answered some questions first?" $\endgroup$ Dec 12, 2018 at 20:39
  • $\begingroup$ The wave theory of the time could not show why there was not an ultraviolet catastrophe, and had no explanation for the Compton Effect. The theory that light came in discrete packets could explain both of those. So it was thought better to consider light in discrete packets, than to find ways to patch up wave theory. Existing wave theory of the time did not explain all the results, therefore all wave theory was considered wrong. That's how it looks to me. $\endgroup$
    – J Thomas
    Dec 12, 2018 at 21:39
  • $\begingroup$ Fortunately that wasnt the case. It took 20 years before photons were accepted, and the physocs community fought tooth and nail to avoid the conclusion. The logic and rationale was always up for discussion, but there remains little doubt of the invalidity of the wave theory. $\endgroup$ Dec 12, 2018 at 22:04

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