I have read this question:

Why does the photon strike at one or another place on the tape?

where PhysicsDave says:

All photons passing thru the slits leave a dot on the screen, this is true for single or multiple photon intensities. Destructive interference is a violation of conservation of energy, destructive interference is best explained by QM, there is low probability of photons arriving in the dark areas and that is why the dark areas are dark.

Now there are two opinions on this site:

  1. When a single photon is shot, it passes through both slits as a wave and the partial waves of the photon interfere with each other, and create constructive (bright area) or destructive (dark area) interference.

  2. Each and every single photon shot leaves a spot on the screen in the bright areas, there is a low probability for a photon to land in the dark area

These are two different meanings. One says that some photons never reach the screen, and never interact with the screen, that is why we see dark areas.

The other one says, that each and every single photon shot will leave a dot on the screen in the bright areas, and it is just that there is low probability for the photons to arrive in the dark areas, that is why we see dark areas.


Which one is right? Does every single photon shot leave a dot on the screen in the bright area?

  • 1
    $\begingroup$ It's always possible for a photon to be reflected $\endgroup$
    – Ruslan
    Oct 7, 2019 at 18:08
  • $\begingroup$ This may help: aapt.scitation.org/doi/full/10.1119/1.4955173 $\endgroup$
    – J. Manuel
    Oct 7, 2019 at 18:19
  • 2
    $\begingroup$ Excpet in the idealized limit of a diffraction grating there isn't a crisp distinction between "light area" and "dark area"; there is smooth transition between light and dark. It is easier to see in a single slit pattern (and it is why we usually map out the location of the dark fringes for single slit patterns). $\endgroup$ Oct 7, 2019 at 18:26
  • $\begingroup$ In the ideal case (no reflection), all photons will interact with the screen as this is required by energy and momentum conservation. The density of interacting points will be proportional to the square of the wave function though. $\endgroup$
    – J. Manuel
    Oct 7, 2019 at 18:29
  • $\begingroup$ The first one doesn't say that some photons never reach the screen. The photon is not a bullet-like (particle-like) object. As it travels, it is the wave, and it reaches the slits and the screen as a wave; but then there's an interaction with the screen that ultimately results in a measurement, and in that process, the photon "manifests" itself as a dot in a random (unpredictable) location. It's just that some locations are more likely then others, and this is a function of the shape of the wave - and that distribution is predictable. $\endgroup$ Oct 8, 2019 at 13:09

2 Answers 2


The photon is a quantum mechanical entity.

Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions .


Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The photons arrive one at a time at the left, the interference appearing slowly with the accumulations on the right, a probability distribution.

The boundary condition problem "photon + two slits of given width a distance apart" defines the wave function of the system,$Ψ$ . The $Ψ^*Ψ$ is the probability distribution for the accumulation of photons.

It will depend on the experiment if every photon that leaves the source and hits the double slits leaves a footprint, the efficiency of the screen. There will be experimental errors. In principle every photon that passes the double slit should end on the screen.

  • 4
    $\begingroup$ Number 1 also dangerously gets close to treating $\Psi$ as something physical $\endgroup$ Oct 7, 2019 at 18:32
  • $\begingroup$ Every photon that passes the double slit should end on the screen. I have a very basic understanding of qm, but couldn't the photon pass through the screen entirely? $\endgroup$
    – Mathemats
    Oct 8, 2019 at 4:59
  • 2
    $\begingroup$ @Mathemats Sure. Or be reflected, and hit the back of the slitted wall. Those are complicating factors we usually don't consider in simplified settings like this idealized double slit experiment. $\endgroup$
    – Arthur
    Oct 8, 2019 at 6:12
  • $\begingroup$ I do have a problem with the terminology of photons arriving. It's the e.m. fields that are propagating and interact with charges in the screen or detectors. With Fermi's golden rules there is a probability a light quantum or "photon" is annihilated. The photon is not traveling. $\endgroup$
    – Jan Bos
    Oct 9, 2019 at 18:02
  • $\begingroup$ @JanBos You may have whatever "picture" you like, but in mainstream physics the photon is an elementary point particle in the table of particles, as much as the electron and the neutrino, and that is the way it is treated in quantum mechanical theories , QED. Not as a traveling electromagnetic wave, although classical electromagnetic waves are shown mathematically to emerge from a confluence of photons. $\endgroup$
    – anna v
    Oct 10, 2019 at 4:10

Which one is right?

I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices.

Without getting into the real physics (i.e., the math):

  • The geometry of the experiment (two slits, and the screen) defines a wave function.
  • Any photon that gets past the slits will make at most one mark on the detector, but the detector probably is less than 100% efficient, and some photons may fail to leave a mark.
  • Considering only those photons that leave a mark, the wave function predicts the spatial distribution of the marks that they leave.

Edit: Oops! I forgot to say, "...and the wavelength of the light." You can't know the wave function if you don't know the wavelength of the photons that you're sending through.


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