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Is quantum theory a risky house of cards?

  1. Mar 16, 2004 #1
    Caveat Emptor – do not analyze in terms of quantum theory. or, on the other hand, here goes nothing.

    We use Richard Feynman’s ‘Lectures on Physics’ Vol. III, Chapter 5, as our experimental results gospel. The Stern-Gerlach transition results are not disputed and a +S particle exiting an S segment where the two lower channels (‘0’ and ‘-‘), or trajectories, are obstructed, and is then directed through a T segment (rotated around the y-axis direction of travel), the +S state will reform as if the unobstructed T segment were not there, or as, S → S.

    A compass needle always returns to North due to the earth’s magnetic field after being perturbed. In the above transition written, completely as, +S -> T’ -> +S, we see the reformed +S state must necessarily occur at the exit of the T segment and in field free space. There are two events to consider, the polarization when the particle exits the field free volume and enters the field/gradient volume,[/B] and secondly when the particle exits the f/g volume and reenters the field free volume. It appears to me that the T’ state must contain some elements of +S that guarantee the reformation of the +S state. In any event, we may write the S -> T’ transition as S -> T’ = T’(00), where 00 are the elements inferred above and whwich have no physical implications by appearing in the trivial description as written.

    If we say nothing more, it appears that the particle enjoys the status of an inertial platform. Why? The +S state is a polarized orientation of the “spin vector” of the particle and the particle will always take the +S channel of any blocked or unblocked S segment. The particle must take one of the known channels, or trajectories, in the T segment when polarized upon entry, else it wouldn’t make it through the segment.

    Hence, the T’ -> +S transition is necessarily a transfer of “force information” that effectively reorients the “spin vector” to the +S ‘base state’. The quantum theory gods, circa 1926, in their original zealous analysis of the SG experiments, stopped short of completeness. By discarding the ‘rigidly attached, randomly oriented spin vector model’ they arbitrarily excluded the case where the random oriented “spin vector” could take on an ‘unpolarized spherical’ shape, hence automatically aligning with any field directions by virtue of the distortion the field has on the “spherical magnetic spin vector”. Here is analogy to distortion of the charge on an electron in field free space. One needs not to conjure up a ‘quantized space’ model to proceed. The particle will then be guaranteed to move along the z-axis where the direction of this motion is ‘+’, or ‘up’ (or ‘0’ ‘-‘) direction as decided by the particle.

    This isn’t quantum mechanics is it? However, does it not reasonably substitute for qm at least this far?

    Where we include the ‘elements’, certainly unobserved, nonlocal(?) and ψ(+S) = ψ(100f) [ψ( 010) = 0S, ψ(001) = -S] and the ‘00’ are the ‘elements guaranteeing the reformation of the +S state, the polarization, B to depolarization, B-1, goes as,

    Bψ = Bψ(100f) -> ψ(_ 00 _ _ f) -> ψ(1 00 00[T] f)-> B-1ψ(1 00 00[T] f) -> ψ(_ 00 _ _ f) -> Ψ(1 00) = ψ(100f)

    an unambiguous +S state. The 2nd And 3rd terms indicate a step-wise process of polarization; the 5th through 7th terms the inverse B-1.. The 00[T] anticipates the substitution there should the 00 become ‘lost’, as it does when the particle passes through an obstructed T segment. Where, even with the ‘nonlocal aspects’, is the “deep quantum mystery” that Feynman refers to 200 times in ‘Lectures’?
  2. jcsd
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