Related: Accuracy of various optical instruments

In many books, it's written that knowing which slit a photon passes through (by placing a detector before the slit) in a Young's double-slit setup beforehand destroys the interference (producing two bright fringes). My question is, does this require there to be only one photon in the double slit experiment at a time?

The one-photon interference is explained by there being "half a photon" coming from each slit, and these interfere. If we know which slit it comes through, then it becomes a full photon in one slit and it can't interfere with itself. But can other photons interfere with it? I'm quite sure that light can be either a particle or a wave in such experiments, never both.

Aside from this, how was the photon detected without absorbing it? If I keep a detector behind a slit, yes, it will detect all the photons, but it won't let any past it to the main apparatus. Unless the detector was something else..

UPDATE: A clearer version of this question:

  1. If you know which slit a photon went through, its wavefunction is collapsed. One can say that it ceases to be wave like. This collapsed photon cannot interfere with itself
  2. Can it interfere with others? It is, after all, not really a wave anymore (it is a wave, but collapsed to a point).
  3. In a normal double slit experiment (beam of light, no doodads), are photons interfering with themselves, each other, or both? (See comment: Is Time Significant in the Double Slit Experiment)
  • $\begingroup$ Could whoever downvoted please explain the downvotes in the below two answers? I'm getting doubly confused... $\endgroup$ Commented Feb 16, 2012 at 2:18
  • $\begingroup$ People who have downvoted (not me) are probably too tired of this question to give a written answer. I am pretty new in this forum, so although I may eventually become tired too, I will try. Without the knowledge of the mathematical models, there can be absolutely no healthy meaning for sentences like "the wave function collapses" or "the particle becomes entangled with the detector". Popular books for the layman offer that kind of sentences together with mental images and high school maths, but is all in vain. (I will follow in a new comment) $\endgroup$ Commented Nov 1, 2012 at 12:54
  • $\begingroup$ What happens here is that some scientists in the 1930s invented a set of mathematical rules that predict the same results as nature, when dealing with atomic-scale phenomena. That is called Quantum Mechanics. The model works very good. But neither themselves nor any other person can depict what may be "really" happening at the atomic level with the abilities that evolution has given to our brains in order to survive in the macroscopic world. $\endgroup$ Commented Nov 1, 2012 at 12:57
  • $\begingroup$ @Eduardo: Umm, the answers were downvoted, not the question. There was some controversy over it, that's all. $\endgroup$ Commented Nov 1, 2012 at 13:05
  • $\begingroup$ I do know part of the mathematical formulation of QM, just not enough to understand this issue. I was a bit more comfortable writing it in layman's terms, and I did so. I am currently studying physics, and I'll eventually get there :) Got a course on QM right now, though it's rather elementary. $\endgroup$ Commented Nov 1, 2012 at 13:08

4 Answers 4


After you put a detector, the basis of states becomes a tensor product of the states of the particle (with the two states meaning: came from this or that slit) and the ones of the detector (went through this slit or not).

So the cross terms that appear when you evaluate the probability of having reached a given position at the final screen coming from one or another slit, are now orthogonal (there is no overlapping of the detector states, otherwise the detector would be useless) and don't contribute to the probability, so you recover the classic result.

So basically, you'd get a no-fringe pattern whether or not you let multiple electrons through at a time, as long as you lace a detector

And this gives rise to a very interesting question: because, if you put the detector after the screen (for example a very fine, ideal, bubble chamber that could give you the direction the electron had) the result would be the same, and that is deeply counterintuitive (the classical physicist would scratch his/her head wondering: does the partice see into the future too?)


Firstly, let's hope I understood your questions correctly. To answer them in order -

A) Once a photon's wavefunction has collapsed, you won't see the interference pattern if you send in one photon at a time. So if you have a detector at each slit to see if the photon has passed through that slit, you won't see the interference pattern. This link may illustrate it somewhat.

B) You may be slightly misunderstanding the meaning of a 'wavefunction'. A wavefunction is just a mathematical way of describing a (read carefully) space dependent probability, evolving with time. Meaning to say, the photon (or electron, or whatever) has a probability of being at a certain spot at a certain time. This probability itself changes with time. So a "collapsed" wavefunction just means that the particle has 'picked' which spot it is going to be at. And if we decide to measure the wavefunction, we are basically choosing for it, in a manner of speaking. This is why the interference pattern vanishes. It's not exploring all the possible probabilities.

C) As John Rennie mentioned, photons don't follow number conservation. So ideally, you should be talking about electrons/some other massive particle for a clearer understanding. But in a regular interference experiment, we can think of light as behaving as a wave rather than as quantized particles, since there are just so many damn particles.


You say:

I'm quite sure that light can be either a particle or a wave in such experiments, never both.

but I would take issue with this. Maybe this is unnecessarily philosophical, but it's not the light that is either a wave or a particle, it's the mathematical model we use to describe it that is either a wave or a particle. When the light hits your CCD, photographic plate or whatever your using as a screen it's easiest to describe the light as a particle since the interaction with the screen occurs at a point. When you're calculating the interference pattern it's easiest to think of light as a wave, though you can of course treat is as a particle and do a Feynmann sum of paths and after much labour arrive at the same result as the wave model.

Kartsa is quite correct that you can do a double slit type experiment by using two lasers. See https://journals.aps.org/pr/abstract/10.1103/PhysRev.159.1084 (this site requires a password, but Googling for "pfleegor-mandel experiment" quickly finds a PDF of the paper). In the experiment such low intensities were used that only one photon was present at a time, but that photon can only have come from one laser so you have in effect identified which slit the photon came from. The paper concludes that to understand the interference pattern you need to model the detection process and suggests that:

the localisation of the photon at the detector makes it intrinsically uncertain from which of the two sources it came

which I must say feels a bit unsatisfactory to me, though I admit I only skimmed the paper. Anyhow, this question turned out more interesting than I thought. Thanks :-)

  • $\begingroup$ Wrt the light-wave-particle thing, i was referring to the complimentarity principle. In this case, once you know which slit a photon goes through, it ceases to be wave-like as its wavefunction is collapsed. So it cannot interfere with itself. My question is, can it interfere with other ('pure'/collapsed) photons? $\endgroup$ Commented Feb 16, 2012 at 2:29
  • $\begingroup$ I'm not sure it makes sense to talk about a collapsed photon. Unlike, say, electrons, photon number is not conserved. I'm not sure you can interact with a photon without destroying it (and possibly creating a new photon). $\endgroup$ Commented Feb 16, 2012 at 6:50
  • $\begingroup$ I see how photon number need not be conserved (destructive interference, right?), but isn't that true for electrons as well? What if you entangled the which-slit information in another photon, and detected THAT? (This is done in the quantum erasers, and it does 'collapse the photon', but I'm not sure what happens if you have multiple photons) $\endgroup$ Commented Feb 16, 2012 at 7:22
  • $\begingroup$ Have a look at physics.stackexchange.com/questions/437/… for some background info about photons $\endgroup$ Commented Feb 16, 2012 at 8:01

Very weak or very intense light from ordinary light bulb shines on two slits: interference

Each slit has its own very weak or very intense light bulb light source: no interference

Each slit is illuminated by its own laser beam, laser beams have been carefully made as alike as possible: interference (experimentally confirmed)

Aforementioned two laser beams are very weak: interference (I don't know if this experiment has been done)

Here "very weak light" means that photons are emitted one at a time, and at random times.


Actually no slits are needed when laser beams interfere: two photon interference


If you read Paul Dirac's book "Quantum Mechanics", you will this sentence:

“each photon then interferes only with itself. Interference between different photons never occurs.”

Now the question has been answered: No interference between two photons, ever.

BUT science has evolved and today we know interference of two different photons happens: two photon interference

  • $\begingroup$ I know that, my question has to do with placing detectors behind the slits. More accurately, if I do a YDSE in such a way so that I know the slit each photon goes through, will there be interference? I know that there won't be interference if it is done one photon at a time(the photon cant interfere with itself when it's collapsed), but will there be interfence if many photons are there in between the slits and screen at a time (and I know which slit each photon went through)? $\endgroup$ Commented Feb 15, 2012 at 16:30
  • $\begingroup$ The photon detected near the left slit went through the left slit, we now this already without measurement. There's a photon near the right slit too, it went through the right slit. I guess, if your detectors are identical enough, you can detect, and preserve interference. Because you aren't getting any information. $\endgroup$
    – kartsa
    Commented Feb 15, 2012 at 18:20
  • $\begingroup$ Why aren't we getting any information? $\endgroup$ Commented Feb 15, 2012 at 18:39
  • $\begingroup$ Well I don't see what information is gained. There's no point in detecting "did a photon go through this slit or not", when we know already that the answer is yes. You tell me what information we get. $\endgroup$
    – kartsa
    Commented Feb 15, 2012 at 19:28
  • $\begingroup$ Let's say we point at one photon at a time and ask the detectors: "which slit did this photon go through", and detectors answer us. This does not destroy interference. $\endgroup$
    – kartsa
    Commented Feb 15, 2012 at 20:56

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