There are a lot of questions on this site about the double slit experiment, none of them answer my question specifically. I am not asking about detection or detectors or anything just at the slit. My question is about the photons traveling as partial waves through both slits and the dots on the screen. A lot of the answers claim that:

  1. the photon goes through both slits

That's why it goes through both slits.

Shooting a single photon through a double slit

If a photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

  1. one photon, one dot

In this case , the probability of the single photon to be absorbed by the intensifier completely is 1, by construction. In this case the energy of the single photon will not make a difference as long as it is within the energy levels available on the phosphor screen. One point will appear.

How can we prove that a photon is absorbed only once?

Do photons interact only with single particles or can a single photon be absorbed by 2 or more particles simultaneously?

  1. a photon can only be absorbed once

Yet if you think about it, if the photon travels as a wave, and partial waves travel through both slits, then these partial waves might reach and interact with the screen at different locations.

enter image description here

If you look at the image, you can clearly see a single photon shot at a time, traveling through both slits, as partial waves, and the partial waves reaching the screen at different (shown by two arrows) locations simultaneously. Why does this not leave two dots on the screen?


  1. If a photon truly goes through both slits, then why can't it make two dots (excite two atoms) in the screen?
  • 1
    $\begingroup$ Is your apprehension only for photons? What if the experiment is done using electrons? Do you have the same issue then? $\endgroup$ Sep 4, 2020 at 19:27
  • $\begingroup$ Regardless of how you think it propagates, it is only one photon. $\endgroup$
    – Jon Custer
    Sep 4, 2020 at 19:32
  • 2
    $\begingroup$ What has the photon "going through both slits" to do with the picture of the waves you draw? You can just take a single wavefront and let it hit a screen at more than one point and ask the exact same question about it. Why do you keep asking these endless variations of the question whether or not the photon "really" goes through both slits if you're never going to engage with the answers you've repeatedly gotten that the quantum object "photon" is not a classical point particle and does not have a definite path it takes? $\endgroup$
    – ACuriousMind
    Sep 5, 2020 at 16:28
  • 1
    $\begingroup$ @ACuriousMind why was this question closed? It is about a single photon being absorbed by multiple atoms, which, is if you google it for example, possible. Very little probability, but possible. It is simply about the fact that in this case it is easy to visualize with this double slit diagram. Can you please reopen, the cited duplicate does not answer my question. $\endgroup$ Sep 5, 2020 at 19:03
  • 1
    $\begingroup$ @ÁrpádSzendrei I think some of the downvotes are simply a matter of formatting: your question looks like you breezed through a lot of questions, picked random bits of them without spending too much time trying to comprehend them. Even if it isn't true, the question looks like it. Another reason might be that the question is very broad – it stems from your misunderstanding of QM, so you basically ask that “somebody explains QM to you”. I personally think it's a great pedagogical question, so myself being an upvoter, I can only speculate. Anyway, I also don't think this question should be closed $\endgroup$
    – m93a
    Sep 9, 2020 at 15:44

3 Answers 3


I believe your questions stems from the misunderstanding of quantum superposition and quantum measurement. First of all, you should understand that “photon goes through both slits” is a simplification – it uses words from classical physics to describe a non-classical phenomenon. I'll try to explain the subtlety here, but before that – to make sure we're both on the same page – let us explore the classical case.

Classical analogy

Imagine you have a rubber ball cannon pointing on a wall with two holes. The rubber ball cannon isn't very precise and shoots the rubber balls with various speeds and under various angles and there's a 50:50 chance the rubber ball will pass through each hole. Now imagine you've set up a camera to take a picture of your rubber ball one second after firing. Can you predict, where that ball would be?

Since you don't know the exact position and velocity of the rubber ball, you'd have to track every single point where the ball can be, assign it a probability maybe, and let those points evolve in time. If you understand how phase space works, you could assign a probability amplitude to each point in the rubber ball's phase space and evolve them in time. Then, if you wanted to compute the probability that the ball is in such and such volume, you'd just integrate over it.

Having constructed this analogy, let's forget the ball's actual trajectory (since we don't know it anyway). Instead, let's focus on the probabilistic description and try to answer several questions about it.

  • When we fire a ball, through which slit does it go?

We don't have the information to answer this. Based on our probabilistic description, we can say the ball went through both slits with the same probability. Does that mean that we could see it passing through both slits at the same time? No, of course it doesn't. But it means we could see it passing through either slit, both options are possible.

  • When the ball hits the wall on the other side, can it hit on several places at once?

Our probabilistic description tells us the ball can be in either of many places, and when it hits the wall in the back, it can hit it on one point out of many. But can it hit the wall on two places at once? Of course not, it's just a single ball! If two boys were standing behind the slits, both boys could get hit with the same probability. But they would never get hit at the same time. If we include the boys into our probabilistic model, we can say that either one of them got hit, or that both got possibly hit, but we know they never got hit at the same time. A detail that will become important later: if one boy tells us he got hit, we know with certainty that the other didn't get hit.

The quantum case

All the phenomena I described in the classical analogy carry surprisingly well to the quantum case. Wavefunction is essentially just a probability distribution on the phase space (with several built-in limitations, like the uncertainty principle). There are just two crucial differences. The first difference is that in the classical analogy, there existed one actual trajectory, we just didn't know it. Meanwhile in the quantum case, a single trajectory simply doesn't exist and the probabilistic description is as close as we can get. I'll get to the second difference in a minute.

Now, equipped with a better intuition, let's try answering a few questions about the double-slit experiment. Assume we have a photon gun pointing at a plate with two slits and we measure where the photon lands using a second plate in the back.

  • When we detect that a photon landed, through which slit did it go?

The most natural answer is that it could have gone through either. It has gone through both slits with the same probability – same as the rubber ball.

  • Can the photon hit two places at once?

No, it can't. It's the same as with the ball, one photon can't hit two places at once. However, an interesting thing can happen if we place two perfectly isolated atoms in the space behind the slits, let the photon pass and then try to test which atom got hit.

  • Which atom got hit by the photon?

Same as with the boys – either could get hit. Either one of the atoms is excited by the photon. Both atoms got hit with the same probability, but both never got hit at the same time. And what's really curious: if we measure one of the atoms and find that it was excited by the photon, we know with certainty that the other atom wasn't excited. This is the famous entanglement – the state of one particle depends on the state of a different particle. In fact, a similar procedure called the Rydberg blockade is used to create entangled atoms in a lab.

Hopefully you now understand the analogy with classical physics and it isn't surprising to you that one photon cannot be detected on two places at once.

Now the second difference that I promissed. In the classical model, the probabilities of different outcomes could only add up. If there's a 20 % chance that the balls flies this way and ends here, and a 10 % chance it flies a different way and ends here, the probability of the ball ending here is 30 %. In the quantum case, the probability amplitudes aren't positive real numbers, they are actually complex. That means that probabilities of different outcomes can not only add up, but also cancel each other out. This is the reason why you see an interference pattern in the quantum case, but not in the classical one.

The measurement problem

I know this answer is quite long already, there's just one elephant in this room that needs to be adressed. How do we know when to use the wavefunction description and when to resort to definite description? If somebody placed detectors in front of the two slits, even if they didn't tell us, the probabilistic description would fail and the interference pattern would be lost. It seems that in order to use the probabilistic description, the results must not only be unknown to you as an experimentator, but fundamentally unknowable.

This is called the measurement problem and there is a lot of interpretations and hypotheses about it. My personal favorite interpretation is the relational QM.


It helps if you think in terms of fields instead of particles, at least when it comes to photons. When you scale up the experiment and realize you can get the exact same interference patterns with rather giant multi-atom objects, this gets harder to hold a mental image of, but the fundamental answer of why you will only ever see a single mark on the detection plate corresponding to a single object entering your experiment is thanks to fundamental conservation laws (or symmetries, if you rather).

If you introduce one quanta of energy into the photon field, it's rather meaningless to speak of detecting "half a photon," because one photon is what's required to excite the electron in the atom in the photodetection plate that then changes energy levels and alerts you to its presence. Regardless of what story you want to tell about what the momentum/energy packet you're calling a photon does between emission and absorption, if you only put X amount of energy into the system, that's all you can get back out. This is reflected in a single mark on your detection plate corresponding to each photon emitted.

Put in enough energy for two photons to be created, and you'll see two marks downrange. Put in some more energy and you'll see some positrons and electrons as well. The fact that the counts all add up for Energy in/Energy out is guaranteed by the universal bookkeeping laws.

It's akin to sending a kid to a store with one dollar, in which no items in the store cost less than one dollar. They may wander around the store, pick up a few items to look at, take a few random paths through the aisles and loiter around for a bit... but since they only carry one dollar, they're only going to be able to walk out of the store with one item once they pay.

Once you're comfortable with that, you're still left with the fundamental question of what exactly is happening to quantum objects between the moment of emission and detection? for which you can insert your favored Interpretation of Quantum Mechanics since they all predict the same things.


This aspect of QM somehow remains mysterious to everyone.

There is however one interpretation of QM that helped me think of it in a logical and nautral way. It is a kind of "multi-universe" way of thinking Bohm-DeBroglie theory (it is closer in spirit to Madelung's fluid interpretation, the formalism is the same for both theories). You don't have to subscribe to it, but it may help to feel more at ease with QM logically. What follows is vulgarisation. To get into it, you need to study the Bohm-DeBroglie theory. Note that all these "interpretations" do not differ in terms of prediction from standard QM. They only help to forge an easier mental picture of the theory.

You can see the photon as going through both slits but in different universes. The wave is a cross-universe wave, it is not a wave in your universe. The photon is initially at the same place in all universes, but then it starts going into different directions in each universe. This sounds like a probabilist abstraction (several possibilities but actually only one really happens) but it is not. Because the different universes interact with each-other.

The photon "clones" that have gone into other directions in the other universes, kind of "push" the photon in your universe (officially this is called quantum force or quantum pressure derived from the quantum potential). That's how self-interference happens. And finally you see the photon in a single place because in your universe, it is at a single place.

The photon is like a cross universe fluid, a kind of gas or liquid. Each universe contains a single particle (a photon "clone") of this gas. The particles in the gas kind of push eachother across universes. Even though this is formally a pressure, it does not work like an ordinary pressure. It's a bizarre way of pushing.

Note: what I explained is more accurate for a massive particle (like an electron). Photons are a bit special. But the logics is the same.

Now answers :

If a photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

In each universe, the photon is detected at the slit it went through if you decide to detect it. But if you do so, the interference cannot happen.

How can we prove that a photon is absorbed only once?

This is somehow decoherence theory. After interraction with a macroscopic system, the photon cannot interfere with itself anymore. This can be interpreted as "in each universe, everything will happen like the photon was absorbed by a single atom".


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