I understand the double slit experiment up until the point that we begin "detecting" single photons. What does it mean to detect. You cant place a camera in the slit because that would capture the photon just like the photosensitive plates would.

I've also heard that you can do the double slit experiment with electrons. Perhaps you can use the attribute of charge to detect the electron from a distance.

I have read many times of how the double slit experiment works, but never the specific details of how the experimenters manage to shoot single particles and then detect them moving through the slit. Any links describing the equipment used in these experiments would be appreciated

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    $\begingroup$ Related: physics.stackexchange.com/q/76162. Single photon detection is easy enough with film, phosphorescent screens (with high enough photon energy), PMTs, silicon surface detectors and so on. There is a whole industry in doing this stuff. $\endgroup$ – dmckee --- ex-moderator kitten Mar 6 '14 at 18:46
  • $\begingroup$ Also related: physics.stackexchange.com/q/100829, physics.stackexchange.com/q/10429 In fact all the parts of this question have been asked and answered on the site before. $\endgroup$ – dmckee --- ex-moderator kitten Mar 6 '14 at 18:47
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    $\begingroup$ Slightlyciborg: Do you have elucidated how scientists manage to detect the passage of a single photon through one of the slits without affecting it? For me this question is still unanswered. I suppose that placing whatever detector on one of the slits necessarily alters the conditions of the experiment hence destroying the interference pattern. It is very clear for me, no need to involving conscious observers. It is just the effect caused by the detector. Am I wrong? $\endgroup$ – adone Sep 27 '14 at 11:03
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    $\begingroup$ One way to detect individual photons is a photomultiplier (en.wikipedia.org/wiki/Photomultiplier). But that captures the photon. I too am wondering how you detect a photon at the slit. $\endgroup$ – Moss Sep 22 '15 at 18:04
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    $\begingroup$ Having had the same question, I read, and re-read all the replies. In no way I am closer to the answer then before. It seems, people completely ignore the point and give answers about what they've read somewhere, not about what they been asked. $\endgroup$ – Division by Zero Jan 16 '17 at 20:29

The photons are not detected at the slits. They are detected at a distance from the slits as illustrated here:

double slit

Photons or particles of matter (like an electron) produce a wave pattern when two slits are used

This has been done with single photons, as seen in this video.(after 2' it shows single photon interference), and this publication.. The detector of single photons in the video is a photomultiplier at the screen position.

Generally if a detector is put in one slit, the interference disappears. Here is the build up of single electron interference


Successively longer integration times as electron arrivals (white dots) are recorded.

There have been experiments exploring why when the slit the electron went through is known the interference pattern is destroyed. A recent one used the following method for detecting which slit the electron passed through:

they modified one of the slits by covering it with a filter made of several layers of “low atomic number” material to create a which-way detector for the electrons passing through.

They concluded that the method of detection changes the conditions to the point of destroying interference effects:

Overall, the results suggest that the type of scattering an electron undergoes determines the mark it leaves on the back wall, and that a detector at one of the slits can change the type of scattering. The physicists concluded that, while elastically scattered electrons can cause an interference pattern, the inelastically scattered electrons do not contribute to the interference process.

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    $\begingroup$ This did not answer the question. You start your answer off with "the photons are not detected at the slits but then include a quote that says "and that a detector at one of the slits can change the type of scattering" so are they just guessing that a detector at one of the slits can collapse the wave function? Every time I've read about this experiment there is a detector at the slit. $\endgroup$ – SlightlyCyborg Mar 13 '14 at 15:49
  • $\begingroup$ You misunderstand what you are reading then. The double slit experiment does not have detectors at the slit to start with. There are other experiments that have shown that detectors at the slit destroy the interference pattern, and some new explorations that attribute the spoiling of the interference to the change in the boundary condition of transmission , by interaction with the detectors at the slit. $\endgroup$ – anna v Mar 13 '14 at 17:00
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    $\begingroup$ So I believe what the OP wants to know is the actual mechanics of the detectors. And I am particularly interested to know what methods have been tried for detecting a photon or electron as it passes through the slit. As the OP said, how can you detect the particle and still let it pass by? $\endgroup$ – Moss Sep 22 '15 at 17:56
  • $\begingroup$ I've never really believed that video was an actual experiment. I think it's just a computer simulation. $\endgroup$ – Marty Green May 5 '16 at 2:24
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    $\begingroup$ The test of a theory is whether it can make successful predictions. My theory is that there are no such thing as photons. And so I predicted that your video was not a real experiment. If I am correct, then that is good evidence in support of my theory. $\endgroup$ – Marty Green May 5 '16 at 13:36

As strange as it may be to hear, the double-slit experiment conducted one photon at-a-time with "detectors at the slits" is nothing short of a myth. No such experiment has been reported in a peer-reviewed journal or any reputable book or publication. Talk of this supposed experiment, and even figures illustrating it, can be found throughout the internet, but these refer to no actual experiment being performed.

The origin of the myth is a poor popularization of the quantum eraser experiment, which though "marking" the photons traveling through each of the two slits with which-way information and destroying the interference pattern, involves no detectors or recorders in front of or behind the slits except for the main CCD backplate.

  • $\begingroup$ I've even read about it in books though? $\endgroup$ – Brady Moritz Nov 21 '17 at 18:14
  • $\begingroup$ It would be great to have a reference. $\endgroup$ – vy32 Oct 19 '19 at 1:06
  • $\begingroup$ You will read about it in books. The authors didn't know what they were writing about. How can you detect a photon at the slit without stopping it? $\endgroup$ – vy32 Dec 28 '19 at 21:43

Any measurement you make at the slits is going to destroy the interference pattern. The wavefunction of the photon or electron collapses as soon at it interacts with the environment in a thermodynamically irreversible way. In essence, any type of measurement you could make that would tell you which slit a particle went through would cause the wave function to collapse and the interference pattern would disappear. In order for the interference pattern to exist on the screen, the particles have to remain in a state of superposition all the way to the screen.

The interference pattern exists even when measuring single photons, which is significant because it shows that the effect is quantum in nature and not explained by any classical statistical interference. The wavefunction of a single particle can interfere with itself.

One way of measuring single photons is to use a photomultiplier tube. Photons incident on the tube create an avalanche of electrons in each stage that can be detected as a current pulse. Photomultiplier tubes have sufficiently high gain to detect a single incident photon. However, they do not detect every incident photon since their quantum efficiency is less than 100%. There are probably other ways of detecting single photons but this is just one example.

  • $\begingroup$ Could you give any reference on this comment "as soon at it interacts with the environment in a thermodynamically irreversible way"? Especially interested in examples of thermodynamically reversible interactions. Thank you for any pointers $\endgroup$ – Peter A Feb 21 at 11:41

You can't detect single photons. There is no way to prove that a click in a photodetector represents a single photon versus the response of the system to an electromagnetic wave. There is no way to prove that the appearance of a fleck of silver on a photographic plate represents a single photon versus the response of the silver bromide crystal to an electromagnetic wave. See my blogpost on Quantum Siphoning for an explanation.

And while we're on the topic, you can't create single photons at will either. If you could, you would just shoot photons at a photodetector one by one and count them at will. Never been done. Why not? because there are no pea-shotters for photons (see my blogpost).

  • $\begingroup$ No. You're arguing for the existence of waves over photons, which is besides the point here. And you're missing what the question's really about. It's not bout the photodetector. It's about that newly famous "detection at the slits" to find out supposedly which slit each photon traveled through. Also, the method used isn't about creating single photons at will; it's about creating light that is so low in intensity that literally only one photon will exist between the slits and the plate at any one moment in time. That said, I quite enjoyed your blogpost and paper some time ago. $\endgroup$ – David Reishi May 4 '16 at 16:45
  • $\begingroup$ PSA (to all flaggers): This appears to be an intellegible attempt at answering the question. Please do not flag answers that you think are wrong as "low quality" or "not an answer", but downvote them instead. $\endgroup$ – ACuriousMind May 4 '16 at 20:02
  • $\begingroup$ photomultiplier tubes can detect single photons. $\endgroup$ – vy32 Dec 28 '19 at 21:44
  • $\begingroup$ Yes you can apparently. latimes.com/science/sciencenow/… $\endgroup$ – Russell McMahon May 5 at 8:25

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