It is known that electron behaves as a wave also.
How does electron know that it has to change into a wave?
Are there any factors that influence the behavior of electron changing into wave?

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    $\begingroup$ An electron is neither a wave nor a particle in the classical sense. It is a quantum object that has both particle-like and wave-like properties. There is no conversion from one to the other. An electron is one thing not two, but it is a quantum thing, not classical. What classical analog is more appropriate depends on the situation. In particular a wave description is closer for propagation, and a particle picture is closer for interaction. But there is no conversion from one thing to the other going on. $\endgroup$ – Michael Brown Aug 26 '13 at 14:46
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    $\begingroup$ Why the downvote? Just because a question arises from a misconception, that doesn't mean it's bad to ask the question. $\endgroup$ – Ben Crowell Aug 26 '13 at 15:16
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    $\begingroup$ @BenCrowell Agreed. Just to clarify for the OP: I'm not the downvoter. This is a perfectly legitimate question. $\endgroup$ – Michael Brown Aug 26 '13 at 15:30

The electron is always a wave.

The electron is wave, as experiments of diffraction and interference showed.

Waves come in an infinity of "shapes". Some kinds of shapes have some properties, and others have other properties. Examples of properties are position and momentum. The two shapes of the electron's wave having these properties are

  1. When the wave is concentrated at a point. In this case, it has a definite position. It doesn't have a definite momentum.

  2. When the wave is a plane wave, having a definite frequency (more complex shapes can be obtained by superposing various frequencies). In this case, there is no definite position, since the plane wave extends in all space.

The problem is, when does the electron how to be plane wave, or to be concentrated at a point. The answer is strange. If you measure the frequency (or momentum), you will find that the electron is a plane wave. If you measure the position, you will find that the wave is concentrated at a point.

Yes, you understood well. The electron, and any other particle for that matter, has precisely the kind of shape for which the property you want to measure is defined. Measure another property, which is not compatible with the property previously measured, and you will find it has another shape. Now, this may look strange, but this is how it happens.

Wait, there is more, when more particles of the same kind are present. Then, saying it is a wave is not enough. But this is another story.

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    $\begingroup$ Your last paragraph is the key point. Identifying a particle with a single particle wavefunction is a conceptual mistake that you can only get away with when you only have a single particle system. It leaves you completely powerless to handle entanglement, fermi/bose statistics, etc etc etc. The dimensionality of the wavefunction is exponentially large. It can't possibly represent anything like a collection of classical particles and/or waves. $\endgroup$ – Michael Brown Aug 26 '13 at 15:07
  • $\begingroup$ @Michael Brown. I agree with your comment. I think the question asked by M.Tarun was honest, and I tried to give a simple explanation. I also tried to not leave the impression that this explains everything. On the contrary, much has to be learned about quantum mechanics, to make an idea of what a particle is (or what we know up to this point, because we don't actually know). $\endgroup$ – Cristi Stoica Aug 26 '13 at 15:14
  • $\begingroup$ I think the fact that the wave nature is a probability wave, not a wave on the mass/energy of the particle should be clarified. It is the probability of finding the particle at an (x,y,z,t) or with a fourvector (p_x,p_y,p_z,E) that displays wave properties. The electron itself always appears whole. $\endgroup$ – anna v Aug 26 '13 at 18:21
  • $\begingroup$ @anna v: You are right, before measuring it is a probability wave. But I was talking about what you find when measuring. What you find is an eigenstate of the observable, which is a wave. Why I was talking about what you find? Because this is the part that depends on what you measure, the part when the electron appears to be something that has the measured property. Why did I omit the discussion about probability wave? Because I had to omit many things from the discussion. I don't want to be exhaustive, just to answer the question in the simplest mode I can. Thank you! $\endgroup$ – Cristi Stoica Aug 26 '13 at 18:43
  • $\begingroup$ I have found that for non physicists and for some physicists too the concept of wave raises the vision of the particle being spread out like wave in space time, that is why I always comment on the necessity of stating that each electron appear whole at measurement. $\endgroup$ – anna v Aug 26 '13 at 18:52

You should be aware that you don’t have to accept that the electron is always a wave, or that it’s neither a wave nor a particle. It is true that in the orthodox QM we only have waves (particle-like behavior is a special case). But that is not the only quantum theory that we have.

For example, you can say that the electron is a particle which is guided by wave function (which generates the velocity field on the configuration space). And suddenly, you have the theory (Bohmian mechanics) with particles and trajectories (without measurement axiom, complementarity or other obnoxious things) which is able to reproduce all results of orthodox non – relativistic QM.

If you want to know more about Bohmian mechanics, I would like to recommend you this introductory article: http://math.rutgers.edu/~tumulka/papers/bm.pdf

  • $\begingroup$ By no way is this an answer. You're just saying "Oh, whatever it is; forget about this; Quantum mechanics is wrong, so go Bohmian mechanics!". Bohmian mechanics is an obnoxiously crackpot "theory"; and by no way does it reproduce all the results of Quantum Mechanics! It's deterministic, which is why some people even (irrationally) like it at all! . ? $\endgroup$ – Abhimanyu Pallavi Sudhir Sep 15 '13 at 9:41
  • $\begingroup$ @dmckee, it is not true that Bohmian QM is disfavored by Bell’s inequalities. Can you give me reference for that claim? You can’t. Because Bell was talking about locality in his article and you can say that Bell ruled out “local hidden variable” theories – which is not case with Bohmian mechanics. If you want to clear a misconceptions that you have regarding to Bell’s inequalities, I would like to recommend you an article: J. S. Bell, Bertlmann's socks and the nature of reality, 1980, Speakable and unspeakable in quantum mechanics, Cambridge, 2004, p. 157. $\endgroup$ – postblue Jan 11 '14 at 10:20
  • $\begingroup$ @Dimensio1n0, your claims are completely wrong (any references, please?) because Bohmian mechanics can reproduce all of the results from nonrelativistic QM. If you want to know how, you can read it in this book: springer.com/physics/quantum+physics/book/978-3-540-89343-1 It's very fascinating that people often misunderstand basics of Bell's inequalities and Bohmian mechanics. Perhaps the reason is that they don't actually read the original papers or they don't read at all. Who knows. $\endgroup$ – postblue Jan 11 '14 at 10:25
  • $\begingroup$ It seems that I commented on the basis of a very incomplete understanding. Withdrawn. $\endgroup$ – dmckee Jan 12 '14 at 3:55

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