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Collapse of Wavefunction, Double-Slit Experiment

  1. Jun 9, 2014 #1
    Now I'm trying to imagine the wavefuction before and after collapse when measured at either slit. Before the particle enters the slits I imagine the wavefunction more as a planar wave with no definite position (single wavenumber and a superposition of positions). If no measurement is made then our particle acts similar to a water wave when it comes to the two slits. It has a very definite wavenumber and hence it can constructively/destructively interfere with itself. If a measurement is made at either slit the wavefunction collapses to a very definite position (either slit A or slit B). This collapsed wavefunction is still a wave, but diffraction or interference would be almost impossible to detect as we now have many spatial frequency components. Do we know why the act of measurement simultaneously collapses the WF to a definitive position, whilst increasing its wavenumber components? Can we think of the particle pre-measurement as "coherent" and post-measurement "de-cohered", I'm confused as how coherence plays a role in the single particle experiment?

    Also..If, after the collapse of the wavefunction, we let the particle propagate would it spread out over time due to the Schrodinger Equation?

    Hope this makes sense!
  2. jcsd
  3. Jun 9, 2014 #2


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    The classical picture will have problems. We can place polarizers in front of each slit to gain information about the photon. When they are parallel, there will be interference and the associated pattern. When crossed, there will be no interference and just the 2 bar pattern building up.

    So there is not the kind of collapse you are envisioning (to a point particle).
  4. Jun 9, 2014 #3


    Staff: Mentor


    You may be confused about a few things.

    There is no collapse in the double slit experiment because the photon etc is destroyed.

    You may want to check out an QM analysis of the experiment:

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