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Uncertainity and the double-slit experiment with electrons

  1. Mar 27, 2006 #1
    Consider the double-slit experiment with electrons.

    It is so, that observing trough wich split the electron goes will destroy the interference pattern.

    This can be seen in a thought experiment in wich one adds a light source near the splits. One can calculate that as soon as the wavelength of the light is such that it can be distinguished trough wich split the electron goes, the uncertainity on the electron's momentum becomes large enough to destroy the interference pattern.

    Now consider this variation on the thought experiment, in wich no light source is used to detect the electrons:

    One might try , instead of bombarding the electron with photons, to make the electron radiate photons by accelerating it (accelerating charges radiate photons), for example by adding a local electrical field. In this way, the momentum of the electron would not be disturbed while it can be known trough wich split the electron went (by adding light detectors on strategical places).

    There must be some error in the above reasoning, and I'd like to know what it is. There must be some mechanism that disturbs the electron's momentum enough to destroy the interference pattern.

    This "solution" is pure speculation on my part. It may be totally wrong. But I'll add it anyway:
    I read that the electrical force (that could be used to accelerate the electron, and, thus, make it radiate photons) is caused by the exchange of particles known as "Virtual photons". (I read that here: http://www.physics.ox.ac.uk/documents/PUS/dis/virtual_photon.htm). Is it so that using an electromagnetic field to accelerate the electrons is not fundamentally different from trying to illuminate them, i.e. both ways involve shooting at the electrons with (virtual) photons? Or is this "solution" totally wrong and is there some other mechanism that is unknown to me?
  2. jcsd
  3. Mar 27, 2006 #2


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    Staff: Mentor

    How can an electron (or anything else!) accelerate without changing its momentum? Remember, momentum is a vector quantity, and the uncertainty principle applies to each component of momentum separately: [itex]\Delta x \Delta p_x < \hbar / 2[/itex], etc.

    If an electron emits a photon (which must have momentum), it must certainly change its momentum as it recoils.
  4. Mar 27, 2006 #3
    You're right, but the change of momentum is known if the electrical field used to accelerate it is known.

    I know.

    Hmmm. That must be it. I should have thought of that. Thank you for this information.
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