I have read this:


It says that elastic scattering creates cylindrical waves, and that creates interference pattern, and the constructive interference will create the bright parts visible on the screen.

It says that inelastic scattering creates spherical waves, and that does not create interference, at least not constructive interference, that would be visible as bright pattern on the screen.

I did not find anything on why the spatial shape of the wave would hinder it from creating an interference.

In this double slit experiment, the cylindrical waves (elastic scattering in air), the ones that go through a slit without any filter, will create visible bright interference pattern.

Now in this experiment, the photons or electrons are shot one at a time, still they create interference, because they travel as waves and the partial waves that go through the slits interfere with each other.

But why does it matter what spatial shape the waves have, shouldn't both type of shape of waves go through the slits and interfere and create an interference pattern? We are talking about single electrons or photons shot at a time. Even a single particle should create an interference pattern, and it should not matter what spatial shape the partial waves have.


  1. Is there and explanation why cylindrical waves create an interference pattern, and why spherical waves do not?

  2. Why does the spatial shape of a wave matter in interference?


4 Answers 4


It is very easy to show that spherical waves do interfere. You can do the experiment at home yourself, using a laser pointer and two closely spaced, very small pinholes in aluminum foil in a dark room. Cylindrical waves interfere too. A spherical wave will interfere with a cylindrical wave as well. All that matters is that the two waves need to be mutually coherent: they must have the same wavelength and a fixed phase relationship in order to form a stationary interference pattern.

  • $\begingroup$ An "interference" pattern implies a cancelling out that produces dark areas in the the pattern. This is shown not to be the case, every photon or electron that passes the slit contributes to the bright spots it the pattern. Also the experiment notes the electrons are fired singly thus 2 waves are not required. $\endgroup$ Aug 20, 2018 at 3:02
  • $\begingroup$ Try the experiment, and you will find that spherical waves do interfere. Two mutually coherent point sources (or two pinholes illuminated from a single monochromatic source) produce a pair of spherical waves, and they produce an interference pattern downstream. A very common example is the pattern that occurs when a spread-out laser beam goes through an uncoated lens and is multiply reflected from the various lens surfaces. Each reflection produces what amounts to a different spherical wave. Downstream you see a bull's eye pattern: the interference between the different spherical waves. $\endgroup$
    – S. McGrew
    Aug 20, 2018 at 3:48
  • $\begingroup$ Another common example is a diverging (spherical) wavefront passing through a flat glass plate, being multiply reflected from the two surfaces. Each of the multiply reflected beams amounts to a spherical wave with an offset center of divergence. Downstream you see a bull's eye pattern. $\endgroup$
    – S. McGrew
    Aug 20, 2018 at 3:52

It appears the spherical wave is not the cause of the of loss of interference. Instead, the inelastic scattering destroys and the interference and produces a spherical wave.

Calling $z$ the dimension along the slit: and elastic scattering event means we don't know where the interaction occurred. In fact, there is amplitude to occur at any $z$, so you add those amplitudes coherently. That means the secondary wave radiated from the entire slit, and the transmitted wave is cylindrical (the Fourier transform [FT] of the slit). Since it remain coherent, an interference pattern is produced. (I suspect there is a phase shift which shifts the peak left/right, as this occurs when you put a piece of glass in front of 1 slit while using laser light. Propagation of light through a material with an index of refraction greater than one is like forward elastic scattering).

Onto inelastic scattering: here the event occurs at a single atom. That localizes the propagation at fixed $z = z_{atom}$. Hence, the secondary wave shape is the FT of a point, which is a spherical wave. As the interaction destroyed coherence: there is no interference pattern.

  • $\begingroup$ What do you mean by FT? $\endgroup$ Aug 19, 2018 at 5:12

You are misinterpreting the statement out of context.

Within the context:

As the physicists explained, an electron undergoing inelastic scattering is localized at the covered slit, and acts like a spherical wave after passing through the slit. In contrast, an electron passing through the unfiltered slit is more likely to undergo elastic scattering, and act like a cylindrical wave after passing through that slit. The spherical wave and cylindrical wave do not have any phase correlation, and so even if an electron passed through both slits, the two different waves that come out cannot create an interference pattern on the wall behind them.

To get a feeling about coherence have a look at this link. It talks about light, but the mathematics are the same for coherent electron beams.

It is evident that the point sources after inelastic scattering for each electron are incoherent between two different electrons, as the phase relation is lost by the inelastic scatter. The electron that only scatters elastically retains a phase relation to the beam and the other electrons coming consecutively, even though one at a time, by construction of the beam.

Interference patterns only appear when there exists a phase between the particles.

(btw, it is only the probability distribution for the electron that can exist for both slits.)

  • $\begingroup$ Thank you, so do I understand correctly, that (in the case of inelastic scattering) there will be bright spots on the screen, but those bright spots will seem to create a random pattern. So this random pattern will not show interference right? $\endgroup$ Aug 19, 2018 at 16:39
  • $\begingroup$ They will not accumulate into a pattern because in the inelastic scattering it is like different incoherent point sources. $\endgroup$
    – anna v
    Aug 19, 2018 at 17:08

For double slit experiments the term interference is employed in the classical/historical explanation for observations. In the modern understanding, whether it's photon or electron, each one has a wave function to describe it's propagation. As a single event a wave/particle is traveling towards the slits and chooses one based on probability, the probability is based on the varying EM field of the wave/particle acting with the varying EM field of the slit (lots of electrons moving in the slit material), the varying causes the randomness of which slit is chosen. The wave propagation function also causes the banding pattern seen, it's a solution of the wave equations with the EM properties and dimensions of the slit. In this experiment by employing a filter some electrons pass normally (interact with the slit) and some electrons slam into the filter and are reradiated starting their own new wavefunction which is their spherical wave ( but its interaction with the slit has been minimal).

I would say the choice of cylindrical vs spherical wave terminology simply means a particle that has interacted normally with the long rectangular slit (cylindrical) verses the reradiated spherical one.

The term "interference" can not explain what happens in a double slit experiment and leads to a lot of debate, it's outdated.


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