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In double slit experiment, if flux of electrons are very low, then..

  1. if one "observes" the electrons before it has reached the slit, it "behaves" as a particle and this is due to the interaction (reflection?) of photons with the electrons.

  2. The screen depicts an interference pattern but the individual spots are still due to particle nature. therefore we can say that whenever waves are interacting it interacts through particle/quanta (in this context).

Here are the questions:

  1. If no one is observing, then the electron is suppose to behave as a wave.

  2. Even when no one is observing, there should be interactions (collisions/reflection) between photons and electrons. If so, then how does it behave as a wave?

Is it something to do with the reference system? i.e. "observers" who observe the reflected photons observe that the electron is a particle and those who don't observe it as a wave?.. if so, why the information of the interaction is carried only by the photons and not retained in the electrons?

Qmechanic
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abhishek bhat
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  • have a look at my answer here http://physics.stackexchange.com/questions/205420/in-double-slit-experiment-how-do-we-observe-electrons-near-slit/205461#205461 – anna v Aug 22 '16 at 03:57
  • Abhiskek Bhat Excellent question. See my question here – HolgerFiedler Aug 22 '16 at 04:38
  • abhishek bhat Thanks to your question I published at the end an elaboration about the history of discovery of electron diffraction and added some conclusions: https://www.academia.edu/27983554/Deflection_of_electron_beams_at_edges – HolgerFiedler Aug 24 '16 at 16:43

1 Answers1

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Electrons are quantum mechanical entities. They are successfully modeled by wavefunctions which are solutions of a wave equation. These wave functions are probabilistic, the complex conjugate square of the function gives the probability of finding the electron at (x,y,z,t) in a dV, where V is the four vector volume element. A probability density function.

If no one is observing, then the electron is suppose to behave as a wave.

Whether an observer exists or not, the probability density is given by the boundary conditions of the experiment at hand. It will have sinusoidal forms which can show interference according to the boundary conditions.

Even when no one is observing, there should be interactions (collisions/reflection) between photons and electrons. If so, then how does it behave as a wave?

The small interactions are very weak with respect to the boundary conditions of "electron impinging on double slit of distance x between slits" and do not destroy the main solution. If one is interested in the solutions of "free electron interacting with ambient photons" it is another story and different boundary conditions apply.

An electron in a bubble chamber leaves an ionization track from the small energy interactions with the hydrogen atoms in the chamber. It is always described by a probability density. Its footprint is that of a particle loosing energy with small interactions with matter. Any sinusoidal behavior is not within the boundary conditions of the small collisions and thus no "wave like" behavior is observed.

anna v
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