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Super position and mathematical limitations

  1. May 9, 2012 #1
    I've been curious for some time about the concept of super position. my introduction to the idea was through the schrodingers cat paradox, which i assume most know, so i will stick with this example. is it not correct to assume super position is a concept to resolve the perceptual limitations of the observer to know with 100% accuracy an unobserved event? there may be far too much to take into consideration when predicting the behaviors of particles that, mathematically, it cannot be expressed with certainty but the present state of the unobserved event is, none the less, 'locked in'. The cat is either alive or dead but, with our current models, we cannot predict which with 100% accuracy essentially boiling down to "we just wont know until we look" rather than a fundamental characteristic of the universe. could super position just be a 'band-aid theory' to cover up the mathematical limitations of predicting events with a large number of variables to take into account?
     
  2. jcsd
  3. May 9, 2012 #2
    No. There is a distinct difference between a mixture and a superposition.

    In fact this can be illustrated using a simple model of polarised light.
    Let [itex]\left|H\right>[/itex] denote horizontally polarised light and [itex]\left|V\right>[/itex] denote vertically polarised light.
    Consider then, two sources. [itex]S_{1}[/itex] produces photons that are polarised at [itex]45^{o}[/itex] ie the state is [itex]\frac{1}{\sqrt{2}}\left|H\right> + \frac{1}{\sqrt{2}}\left|V\right>[/itex]. [itex]S_{2}[/itex] however produces horizontally polarised light 50% of the time and vertically polarised light 50% of the time. The state is 50% [itex]\left|H\right>[/itex] and 50%[itex]\left|V\right>[/itex].

    If I pass both [itex]S_{1}[/itex] and [itex]S_{2}[/itex] through a vertical polariser or horizontal polariser, in both cases 50% of the time the photons will pass through.
    Now, it would seem that they might be indistinguishable - but wait, if I tilt my polariser at [itex]45^{o}[/itex], [itex]S_{1}[/itex] will pass through 100% while [itex]S_{2}[/itex] will only pass through 50% of the time! So, experimentally, we can detect a difference between a mixture and superposition.

    This analogy aptly carries over to more quantum mechanical domains. Don't fuss too much with Schrodinger's cat. It's just a philosophical gendaken argument, and is not exactly very physical to think about.
     
  4. May 9, 2012 #3
    Are photons from [itex]S_{2}[/itex] able to pass through the [itex]45^{o}[/itex] oriented polariser as a result of the superposition of the the possible orientations it is able to make? Is it then just a compromise between the two orientations, or is [itex]S_{2}[/itex] capable of creating photons that may pass through a polariser with any intermediate orientation between horizontal and vertical(say [itex]15^{o}[/itex] or [itex]30^{o}[/itex])? Also, what would be the reason that photons from [itex]S_{1}[/itex] would pass through a horizontal or vertical polariser? [itex]S_{1}[/itex] creates photons with only a [itex]45^{o}[/itex] orientation so is there another orientation that is being super imposed on the photon from [itex]S_{1}[/itex] to allow it to pass through a strictly horizontal or vertical polariser? I'm not entirely familiar with experimental methods, so I apologize if my question seems a little uninformed.
     
  5. May 10, 2012 #4
    I, too, read "Mr. Thomkins In Paperback" and thought it said more about Goethe's "Indecidability" than Heisenberg's. Yes, it is supposed to illustrate a paradox. Of course, with true paradoxes come the inescapable intellectual limitations. I once had a professor ask me: "Just because Thomkins said there was a cat in the box, but he didn't let you see it before or after, how did you know it was there in the first place?"
     
  6. May 10, 2012 #5
    That has to do with classical electromagnetism.
    In terms of wave theory, the component of the electric field vector parallel to the polarising axis will be allowed through. So, as long as the polarisation of the wave is not perpendicular to the polarising axis, then some "part" of the wave will go through.

    Translating this into photons turns this into a probability distribution.
    For instance, in wave theory, a horizontally polarised light wave passing through a polariser oriented at 45deg will be reduced in intensity by 50% (use malus' law).
    Now, intensity really is just (energy of one photon) X (number of photons) / (area of surface).
    If the energy of the photon is constant (it should be for a source with constant frequency), then intensity is proportional to the number of photons. Thus this means that the photons will have 50% probability of passing through the polariser.
     
  7. May 10, 2012 #6
    Of course. My sunglasses. You know, I thought I knew Q.E.D. (as in Feynmans's book) better than that. Thank you.
     
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