Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why does electron have wave property

  1. Dec 30, 2007 #1
    I read that a single electron, when passing two slits, would exhibit wave like distribution on target screen. This makes me wonder about the following.

    1) Would this have anything to do with the interference by the slots? I understand that my question may be moot because such interference probably would result in a Gaussian distribution instead of wave-like distributions on the target screen. But still, what's is the role of the slits in this experiment? If we remove the slots, do we still see the wave? If not, why not? If yes, why yes?

    2) I believe there must be some cause-effect behind the fact exhibited by the experiment. Classic physics says that an object would not change its momentum unless a force acts upon it. I'd believe this law should still hold in quantum world. So, what is making electron (or any other particles or even daily-life object) behave like a wave? Or is it because the electron is always subject to some intrinsic field (that accompanies any object) which gives it the wave property? What would this field be? If there is no field whatsoever associated, then what's giving the electron its wave property?

    - Ming
  2. jcsd
  3. Dec 30, 2007 #2
    First of all, you cannot observe the "wave like distribution," or interference, with only one electron. The interference pattern emerges statistically, after many electrons have been detected at the screen.

    As for question 1), the slits cause the wave interference that produces the pattern at the screen. A single slit would produce a different, gaussian-like pattern.

    Question 2) can be answered by considering the uncertainty principle, which describes the non-classical character of momentum, etc.
  4. Dec 30, 2007 #3
    What I was referring to was, an experiment where electron is shot to the screen *one by one*, and observation is made after a large number of electrons have been fired. Thus there is no interference between the electrons.

    This is where my question is. If the "wave-like" property relies on the presence of slits, then electrons *themselves* would not possess a wave property.

    You seem to suggest here that wave property of electron (or any other objects) exist without the slits. While I believe this has been proven in experiments, I wonder what gives matter such a property. Where is the cause and effect? It seems to me the uncertainty principle only provides an incomplete model of the world where some parameters are still missing.

    - Ming
  5. Dec 31, 2007 #4


    User Avatar
    Gold Member

    Hi Ming,
    There are other situations where the wave nature of matter must be used to agree with experiment, so picking on the DS experiment is a mis-targeting.

    Getting correct energy levels for the hydrogen atom is one example where wave mechanics is used, but there are no slits.

    As for 'why' this is so, I wouldn't like to guess. I recommend 'Inward Bound' by A. Pais which covers how the whole thing came about.
  6. Dec 31, 2007 #5
    Hi Mentz,

    Thanks for the suggested reading. Regarding the slit experiment, are you suggesting that it’s an invalid proof or can be misleading?

    Perhaps, electrons (or anything else) vibrate all the time at a given frequency (wave), that is in a discrete range as Quantum mechanics postulates? I am still very puzzled by the claim by some in Quantum mechanics that such a probabilistic wave makes the electron at anywhere at anytime and then “collapses” instantaneously when being acted up.
  7. Jan 1, 2008 #6


    User Avatar
    Gold Member

    Hi Ming,

    This puzzles a lot of people and is the subject of constant debate in this forum and elsewhere. If you search the forum for 'wave function collapse' you'll find many threads discussing foundational and interpretive issues in quantum mechanics.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook