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Using photons for the double-slit experiment doesn't require a vacuum, therefore the photon's wave function either doesn't collapse upon absorption or it's regained upon emission when interacting with air. I don't know of any way the detector that's in front of the slits could observe the photon unless it is through absorption, and to make it to the other side, it has to be re-emitted.

In either case, if we put a detector in front of the slits, it comes out of the detector as a particle, its wave function having collapsed. So what makes the detector act differently than air?

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When a photon interacts with an atom, three things can happen:

  1. elastic scattering, the photon keeps its energy and phase and changes angle

  2. inelastic scattering, the photon gives part of its energy to the atom, and changes angle

  3. absorption, the photon gives all its energy to the atom and the absorbing electron moves to a higher energy level as per QM

In the case of the detector, it is absorption. The photon seizes to exist. It transforms into the kinetic energy of the electron.

In the case of air, the air's atoms scatter the photons elastically. It is Rayleigh scattering, that is why the sky is blue. In this case, the wavelength of the photons is much bigger then the scattering atoms. This is the only way that the energy and phase of the photons is kept, and you can see images of objects through air without bigger distortions. It is a coherent (specular) way of refraction. In optics, we use the expressions coherent and diffuse (decoherent) for reflection and refraction. In the case of refraction, like through air, this (coherent) means that that not only is the energy and phase of the individual photons kept, but the relative phases of the photons' too. This is the only way that the image we see through air keeps coherent.

Now you edited your question to talk about the detector in front of the slits. That is called the which way experiment. In that case, there is a detector in front of one of the slits. That is inelastic scattering, and that will cause that photon not to create an interference pattern.

So basically when the photon is elastically scattered through air, it will still create an interference pattern, because in elastic scattering, the energy and phase is kept.

When there is a detector in front of the slit, that is inelastic scattering, and in that case that photon does not create an interference pattern, because the photon gives part of its energy to the scattering atom, and changes phase.

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  • $\begingroup$ szendrei Arpaf you are makin git up as you go along. What is specular refraction. Never heard of it. $\endgroup$ – my2cts Aug 18 '18 at 18:33
  • $\begingroup$ @Árpád Szendrei I edited my question to be less ambiguous. I am referring to the detector that's in front of the slits. $\endgroup$ – StackUser20004 Aug 18 '18 at 18:54
  • $\begingroup$ @Árpád Szendrei so does this mean that all photons, after going through inelastic scattering, can never act as waves? $\endgroup$ – StackUser20004 Aug 18 '18 at 22:28
  • $\begingroup$ @StackUser20004 No, after inelastic scattering, the photon still travels as wave, but the energy and phase changes. that is why it can never create constructive interference. It is the constructive interference after the slits, that causes the visible, brighter pattern on the display. Now you might ask, but how does one single photon create interference? Now what happens, is that a single photon travels as wave and the parts of the wave travel through the slits. These partial waves, after the slits, will interfere with each other, and create interference. $\endgroup$ – Árpád Szendrei Aug 18 '18 at 22:54
  • $\begingroup$ @StackUser20004 The constructive interference creates the bright patterns. The destructive interference creates the darker parts on the screen. Now you might ask, but if the photon travels through the slits as partial waves, why does inelastic scattering hinder the photon to interfere with itself? What happens is that inelastic scattering creates spherical waves, and that wave does not create constructive interference. Elastically scattering creates cylindrical waves, and those create constructive interference. $\endgroup$ – Árpád Szendrei Aug 18 '18 at 23:00
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When a photon interacts with an air molecule, it doesn't lose its coherence - that is, while there is some phase shift of the wave function as a result of interaction with the electrons of the molecule, the wave remains a wave. If it actually gets absorbed and re-emitted, the coherence would be lost; in that case you would lose the fringes (the direction of the emitted photon would no longer be the same as the direction of the incident photon).

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  • $\begingroup$ If a photon is detected, it is gone. It cannot continue through the detector. It's a somewhat subtle point that if a photon is absorbed by an air molecule and re-emitted coherently, it is impossible to know which air molecule it interacted with- so, the air molecule doesn't act as a detector. $\endgroup$ – S. McGrew Aug 18 '18 at 17:14
  • $\begingroup$ @S.McGrew I edited my question to be less ambiguous. I am referring to the detector that's in front of the slits. $\endgroup$ – StackUser20004 Aug 18 '18 at 18:55
  • $\begingroup$ This absorption and reemission idea is pushed continuously. No one sofar produced any scientific work to back up this idea. It can safely be ignored. $\endgroup$ – my2cts Aug 18 '18 at 21:08
  • $\begingroup$ @S.McGrew "if a photon is absorbed by an air molecule and re-emitted coherently, it is impossible to know which air molecule it interacted with- so, the air molecule doesn't act as a detector" - This is the first meaningful statement I see on this topic in a long time. Could you please answer my earlier question along these lines here and I'll accept your answer? physics.stackexchange.com/questions/368333/… $\endgroup$ – safesphere Aug 19 '18 at 4:26
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    $\begingroup$ Your answer is unclear. How else a photon can interact with a molecule without being absorbed and reemitted? When people talk about the "elastic scattering" they often overlook the fact that such a "scattering" is described be a Feynman diagram showing absorption and reemission mediated by a virtual electron. Accordingly, the QFT description of this process involves the annihilation and creation operators for the photon. Thus the "scattered" photon is actually a different photon emitted by the molecule after absorbing the original photon. See S. McGrew's comment above for more details. $\endgroup$ – safesphere Aug 19 '18 at 5:00

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