# Does a particle interact with the slit in diffraction from a single slit?

The wave explanation of the diffraction pattern from a single slit takes in account the interference of the wavelets of all points between the walls of the slit. In effect it turns so that the wave does not interact with the walls and with nothing at all. There is an effect (change of direction of light) without any cause, because there is no interaction. In Quantum Mechanics the wave is even unphysical and no interaction is evident from the model accepted. How is that acceptable? Because light must interact to change its direction.

Yes the photon/wave/particle interacts with the slit. You will even hear that the light is interacting with itself. Diffraction of light is a truly complex subject, from high school we are taught about "interference" and Huygens principle however these concepts don't explain why single photons fired one after the other still form the pattern (whether its double or single slit diffraction). Also photons "interfering" or canceling is a violation of conservation of energy.

The wave function is a mathematical way of describing the probability of a path, but since there are so many probable paths this function just becomes an average and hence its unphysical nature. But if it were possible to track a single photon it would have a path and this would indeed be a unique wave function but not one that is possible to calculate.

The "interference" word can be misleading since photons don't cancel, Feynman offered that the photon chooses the shortest path that is n multiples of its wavelength (this path would include effects caused by the slit). This explanation therefore still maintains that the observed patterns are due to the wave nature of light but not due to superposition or interference, but more like the harmonic nature of the wave, like a guitar string of a certain length and mass will only play one note.

Single slit diffraction in the quantum mechanical frame, is the quantum mechanical solution of "scattering of a given energy photon, with a slit, of given width'. So there is an interaction. The interaction is the probability of the photon to scatter off the sides of the slit due to the electric and magnetic fields that t atoms on the side of the slit have. So, of course there is an interaction..

The wavefunction describing individual photons comes from the quantum mechanical solutions with the given above boundary conditions, $$Ψ$$, when complex conjugate squared $$Ψ^*Ψ$$ is the probability of the photon to be found at an (x,y,z) after passing the slit. The probability has the diffraction pattern after accumulating many photons.

• It is not obvious from the solution about any interaction with the walls (just the contrary). No matter what a definitions is trying to persuade us. If there is scattering the problem must could to be treated like direct process of redirecting the particle. Or at least interference of paths including scattering off the walls. – Mercury Mar 16 at 22:38
• I deliberately choose a photon because it does not interact with EM fields. So in order to interact it must be localized in the electron of some atom. But then how would the other wall interfere, because the pattern depends heavily on width? – Mercury Mar 16 at 23:10
• You have a confused view of quantum mechanics. Even photons in a scattering experiment have a wavefunction that has the boundary conditions to the solution of scatterig, i.e.slit width slit distance fringe electric and magnetic fields from the atoms of the sides. – anna v Mar 17 at 6:44
• Individual Feynman diagrams would be hard to write, for the photon interacting with the virtual electrons of the orbitals of the atom, but nature does have the solution. – anna v Mar 17 at 6:48
• I wouldn't argue if the photon has a WF. Even if it has, it can not interact by the WF with the atoms. It has been shown that classical scattering off the walls can not bring the observed pattern. – Mercury Mar 17 at 22:22