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I read about electron interference that in presence of photons there are no fringes formed but in its absence fringe patterns are formed. Can it be explained like this. An electron travels free as a wave . In absence of photons since they are waves they produce interference pattern. But in presence of photons electron wave transforms to particle so there is no wave pattern observed. Is this Correct? Or if there is some other possible explanation for this phenomena then please explain.

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    $\begingroup$ I think you should give a link on your "I read". There is confusion in what you are asking which cannot be resolved unless one knows what you are quoting. $\endgroup$ – anna v Jan 16 '15 at 17:46
  • $\begingroup$ A quantum object is neither a wave nor a particle, though it carries properties of both. Unless you explain a bit more what you are asking about, this question remains unclear $\endgroup$ – ACuriousMind Jan 16 '15 at 17:59
  • $\begingroup$ @ACuriousMind , I think that it is clear. The user heard something non-clear as she/he asks us what's that. $\endgroup$ – Sofia Jan 16 '15 at 18:24
  • $\begingroup$ Please keep in mind that the wave nature of elementary particles exists in the probability distributions that describe them, which are predicted by the square of the wavefunction that is the solution for the specific boundary conditions. $\endgroup$ – anna v Jan 16 '15 at 19:31
  • $\begingroup$ @anna-v Feynman, for example, says that when electrons are passed through two slits the registration of the interference pattern depends on whether one observes which slit the electron passes through. The observation is done by scattering photons off the electrons after they have passed through the slits. $\endgroup$ – hyportnex Jan 16 '15 at 22:54
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There are a couple of inexact things here:

1) "electrons travel free as a wave".

Yes, for our mathematical treatment sometimes it is convenient to treat the traveling electron as a wave. But if the traveling electron is really a wave in the nature, on this there is a debate of tens of years.

2) "in absence of photons, since they are waves, ..."

If the above description as waves is good for the treatment of the respective experiment, it remains good also if photons yes are present around, however the electron wave is perturbed.

3) "in presence of photons electron wave transforms to particle so there is no wave pattern observed.

The formulation "in presence of photons" isn't good. Photons may be present, but not allowed to influence the electrons. Putting it simply, if we try to "watch" the electrons by means of light or whatever other particles that cross the path of the electrons and interact somehow with them, the pattern may be gone.

As an explanation, for producing an interference pattern we split the electron beam by means, for instance, of a 2slit configuration. Well, if photons, or other particles illuminate at least one of the two beams, the interference pattern will be gone.

One of the causes for which the illumination of the electrons during their travel destroys the pattern, is called "decoherence". This is a drastic disturbation of the phases of the two beams due to interaction of the beams with macroscopic apparatuses or with a huge an uncontrollable number of other particles (e.g. your photons). In this case we won't see anymore fringes. Another cause is simply entanglement with other particles e.g. a few photons. In the latter case a state of a couple of particles appears, one of the particles being the electron, other particle(s) the photon(s). The phases of the two electron beams are "complicated" with the additional particles, s.t. on the photographic plate again we won't get fringes.

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  • $\begingroup$ So you mean to say that electron is not a wave and not a particle at one instant. Always it is a mixture. But there is dominance of one character over the other in the above situation. And this explains the phenomena. $\endgroup$ – user70848 Jan 16 '15 at 18:34
  • $\begingroup$ Also it is the change in phase because of entanglement of electrons with photons that dismisses the pattern $\endgroup$ – user70848 Jan 16 '15 at 18:35
  • $\begingroup$ @user70848 , yeas, I am with you. No, I don't mean that the electron is not a wave. I will explain in continuation. $\endgroup$ – Sofia Jan 16 '15 at 18:47
  • $\begingroup$ @user70848 : we don't know what exactly we have in our apparatuses. There is a debate of tens of years. Sometimes it is more convenient to represent the traveling electron as a wave. Also in the atom we treat it as a wave, but the wave in the atom poses us very difficult problems. Not that a "particle picture" would be better, no, it would be worse. $\endgroup$ – Sofia Jan 16 '15 at 18:52
  • $\begingroup$ The "particle picture" is sometimes convenient. In experiments with high-energy electrons and that pass through ion chambers, the "wave-picture" may be futile. Why? Because the wave-packet of the electron is so small that we can say that we have a particle. But I have to tell you more. $\endgroup$ – Sofia Jan 16 '15 at 18:58

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